View Full Version : "The wave under consideration is therefore no less a spherical wave"

chinglu1998

2011-Feb-26, 01:42 AM

Einstein made the following claim.

We now have to prove that any ray of light, measured in the moving system, is propagated with the velocity c, if, as we have assumed, this is the case in the stationary system; for we have not as yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity.At the time t=τ=0, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then

x² + y² + z² = c² t²

Transforming this equation with the aid of our equations of transformation we obtain after a simple calculation

ξ² + η² + ς² = c² τ²

The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system. This shows that our two fundamental principles are compatible.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

Yet, as anyone can see, cτ (Tau) is not a constant and thus not a light sphere. Therefore, Einstein's proof is false.

macaw

2011-Feb-26, 02:09 AM

Einstein made the following claim.

We now have to prove that any ray of light, measured in the moving system, is propagated with the velocity c, if, as we have assumed, this is the case in the stationary system; for we have not as yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity.At the time t=τ=0, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then

x² + y² + z² = c² t²

Transforming this equation with the aid of our equations of transformation we obtain after a simple calculation

ξ² + η² + ς² = c² τ²

The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system. This shows that our two fundamental principles are compatible.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

Yet, as anyone can see, cτ (Tau) is not a constant and thus not a light sphere. Therefore, Einstein's proof is false.

You made the same false claim in this (http://www.bautforum.com/showthread.php/110006-if-photon-not-valid-frame-in-SR-why-SR-make-thoughts-about-light?p=1825657#post1825657) locked thread. At that time , your error was explained to you. Once again, in his paper, Einstein explains how

x² + y² + z² = c² t²

transforms into:

ξ² + η² + ς² = c² τ²

through the Lorentz transforms.

chinglu1998

2011-Feb-26, 02:19 AM

You made the same false claim in this (http://www.bautforum.com/showthread.php/110006-if-photon-not-valid-frame-in-SR-why-SR-make-thoughts-about-light?p=1825657#post1825657) locked thread. At that time , your error was explained to you. Once again, in his paper, Einstein explains how

x² + y² + z² = c² t²

transforms into:

ξ² + η² + ς² = c² τ²

through the Lorentz transforms.

Yes it tansforms into ξ² + η² + ς² = c² τ² but Einstein said,

"The wave under consideration is therefore no less a spherical wave"

Now, we all know what a spherical wave is.

Here, I will let Einstein explain it,

"when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom,"

So, we find the wave in the moving frame does not satisfy the properties of a light sphere or as Einstein put it, "a spherical wave".

macaw

2011-Feb-26, 02:24 AM

A. Yes it tansforms into ξ² + η² + ς² = c² τ²

B. So, we find the wave in the moving frame does not satisfy the properties of a light sphere or as Einstein put it, "a spherical wave".

Q1: Do you realize that you are contradicting yourself in a very short paragraph.

Q2: Do you realize that you are repeating the same false claim you made here (http://www.bautforum.com/showthread.php/110006-if-photon-not-valid-frame-in-SR-why-SR-make-thoughts-about-light?p=1825657#post1825657)

chinglu1998

2011-Feb-26, 02:27 AM

Q1: Do you realize that you are contradicting yourself in a very short paragraph.

Q2: Do you realize that you are repeating the same false claim you made here (http://www.bautforum.com/showthread.php/110006-if-photon-not-valid-frame-in-SR-why-SR-make-thoughts-about-light?p=1825657#post1825657)

I know that a spherical wave requires ct to be a constant.

I find that ct' is not a constant using Einstein's proof. This you cannot refute.

macaw

2011-Feb-26, 02:31 AM

I know that a spherical wave requires ct to be a constant.

False, the wavefront PROPAGATES (with speed equal to c)

I find that ct' is not a constant using Einstein's proof. This you cannot refute.

I just did.

chinglu1998

2011-Feb-26, 02:37 AM

False, the wavefront PROPAGATES (with speed equal to c)

Yes, you are correct.

But, you are not on the subject.

Let's think about the spherical wave in the rest frame.

We all know that ct is a constant or we are crackpots, correct?

Now, you and Einstein claimed based on this spherical wave in the rest frame, the wave is no less spherical in the moving frame.

However, we find ct', based on LT, is not a constant and thus is not a spherical wave, a contradiction.

macaw

2011-Feb-26, 02:45 AM

Yes, you are correct.

But, you are not on the subject.

Sure I am, I have explained the same exact fallacy in the locked thread.

Let's think about the spherical wave in the rest frame.

We all know that ct is a constant or we are crackpots, correct?

The wavefront PROPAGATES. "ct" VARIES with time. So, contrary to your repeated claims, ct is NOT constant.

Now, you and Einstein claimed based on this spherical wave in the rest frame, the wave is no less spherical in the moving frame.

And so do millions of people that understand basic physics. You should study the thread on fallacies.

However, we find ct', based on LT, is not a constant and thus is not a spherical wave, a contradiction.

According to the very basic application of the Lorentz transforms, the spherical wavefront in frame F transforms into a spherical wavefront in frame F'. This is what Einstein explained and millions of people understood without any trouble.

chinglu1998

2011-Feb-26, 02:51 AM

Sure I am, I have explained the same exact fallacy in the locked thread.

The wavefront PROPAGATES. "ct" VARIES with time. So, contrary to your repeated claims, ct is NOT constant.

And so do millions of people that understand basic physics. You should study the thread on fallacies.

According to the very basic application of the Lorentz transforms, the spherical wavefront in frame F transforms into a spherical wavefront in frame F'. This is what Einstein explained and millions of people understood without any trouble.

OK, we all know a spherical wave means ct is a constant. Yet we find using LT we cannot convert this spherical wave into the spherical wave in the moving frame. Einstein said this was needed to prove SR.

He failed. Einstein said, "The wave under consideration is therefore no less a spherical wave".

macaw

2011-Feb-26, 02:54 AM

OK, we all know a spherical wave means ct is a constant.

False, it doesn't. (Repeated) fallacies don't make up a theory, not even an ATM.

Yet we find using LT we cannot convert this spherical wave into the spherical wave in the moving frame. Einstein said this was needed to prove SR.

Most people that know physics understood the above without any trouble.

chinglu1998

2011-Feb-26, 02:57 AM

False, it doesn't. (Repeated) fallacies don't make up a theory, not even an ATM.

Most people that know physics understood the above without any trouble.

OK are you claiming ct does not have to be a constant and can define a light sphere?

Now, given ct is a constant, ct' is not a constant. So, Einstein's claim is wrong.

chinglu1998

2011-Feb-26, 03:09 AM

False, it doesn't. (Repeated) fallacies don't make up a theory, not even an ATM.

I looked at the logical fallacies, actually I know all of them, and found the one Einstein committed.

Dicto simpliciter

Here is how this fallacy works.

You have a bag of coins. You pull out a penny and claim only pennies are in the bag.

Einstein pulled out one light beam calculated with LT and claimed all light beams measure ct' as a constant, a spherical wave in the moving frame.

macaw

2011-Feb-26, 03:21 AM

OK are you claiming ct does not have to be a constant and can define a light sphere?

Yes, of course. Why do you have so much trouble with this simple concept?

chinglu1998

2011-Feb-26, 03:25 AM

OK are you claiming ct does not have to be a constant and can define a light sphere?

Yes, of course. Why do you have so much trouble with this simple concept?

It my adherence to rules of geometry.

Please define a light sphere with ct not being a constant as you said.

I will wait.

macaw

2011-Feb-26, 03:27 AM

It my adherence to rules of geometry.

Please define a light sphere with ct not being a constant as you said.

I will wait.

I really don't understand why you have so much trouble with basic concepts:

x^2+y^2+z^2=(ct)^2

chinglu1998

2011-Feb-26, 03:31 AM

I really don't understand why you have so much trouble with basic concepts:

x^2+y^2+z^2=(ct)^2

Do you understand a sphere must have ct as a constant?

Here, I took some time for your class studies.

http://education.yahoo.com/reference/dictionary/entry/sphere

Mathematics A three-dimensional surface, all points of which are equidistant from a fixed point.

macaw

2011-Feb-26, 03:35 AM

Do you understand a sphere must have ct as a constant?

No, it doesn't. Especially when the subject is PROPAGATING spherical wavefronts.

Attempting to learn math and physics via google is not going to help you.

pzkpfw

2011-Mar-06, 04:37 AM

Politeness works both ways. Everyone needs to be a bit more careful how things are written, even if they are getting frustrated. Thread open...

Van Rijn

2011-Mar-06, 07:03 AM

Do you understand a sphere must have ct as a constant?

Here, I took some time for your class studies.

http://education.yahoo.com/reference/dictionary/entry/sphere

Mathematics A three-dimensional surface, all points of which are equidistant from a fixed point.

I, for one, don't understand your argument. Are you claiming that "t" (time) is a constant? That is, the "t" part of "ct"? Or are you just arguing about the wording of the text, and not making any claim about the math itself? Would you accept the statement, "An expanding sphere does not require ct to be constant"?

Geo Kaplan

2011-Mar-06, 07:24 AM

Do you understand a sphere must have ct as a constant?

Here, I took some time for your class studies.

http://education.yahoo.com/reference/dictionary/entry/sphere

Mathematics A three-dimensional surface, all points of which are equidistant from a fixed point.

You are erroneously equating a sphere with a spherical wave. These are not the same thing. A spherical wave propagates (with a radius that expands with time). The "spherical" part is a description of its symmetry. It does not imply that the radius is fixed.

chinglu1998

2011-Mar-06, 05:36 PM

No, it doesn't. Especially when the subject is PROPAGATING spherical wavefronts.

Attempting to learn math and physics via google is not going to help you.

But, this has nothing to do with what Einstein wrote.

He wrote:

At the time t=t'=0, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then......

He then wrote:

The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

I think everyone is clear on the meaning of a spherical wave in the context of Einstein's statement, it is a perfect sphere.

Einstein then used then same exact terminology to describe the LT mapped object transformed from the spherical wave in the stationary system. But, the spherical wave in the rest frame does not LT translate into a spherical wave in the moving system.

chinglu1998

2011-Mar-06, 05:40 PM

You are erroneously equating a sphere with a spherical wave. These are not the same thing. A spherical wave propagates (with a radius that expands with time). The "spherical" part is a description of its symmetry. It does not imply that the radius is fixed.

I agree with your statement. But, if that is true, then Einstein's proof would need to use the derivative in his proof to be consistent with your statement. He did not. You do not use a static proof analysis to prove or model a dynamic process such as light sphere expansion. Therefore, it is clear, he was referring to a spherical object that is a sphere.

chinglu1998

2011-Mar-06, 05:45 PM

I, for one, don't understand your argument. Are you claiming that "t" (time) is a constant? That is, the "t" part of "ct"? Or are you just arguing about the wording of the text, and not making any claim about the math itself? Would you accept the statement, "An expanding sphere does not require ct to be constant"?

I am not claiming t is a constant. I agree with the statement that an expanding sphere is not consistent with a constant time.

And, actually, to be more specific, other than proving the light like space time interval is invariant, I can't find any other value to the proof.

In other words, all he did was prove a light beam measures c if and on if the LT mapped light beam measures c.

He did not prove the wave "....is no less a spherical wave when viewed in the moving system."

http://www.fourmilab.ch/etexts/einstein/specrel/www/

pzkpfw

2011-Mar-06, 06:22 PM

... Einstein then used then same exact terminology to describe the LT mapped object transformed from the spherical wave in the stationary system. But, the spherical wave in the rest frame does not LT translate into a spherical wave in the moving system.

chinglu1998, the problem you have with your ATM threads is you never really explain the important bits. The above snip is an example.

You write things like A and B and C, therefore D. Now, A and B might be "common knowledge", but if C isn't, you can't simply carry on and say you've shown D.

Pretty much the point of the paper you've been quoting, is the opposite of what you are claiming. You need to put much much more effort into showing:

... He did not prove the wave "....is no less a spherical wave when viewed in the moving system." ...

Strange

2011-Mar-06, 06:23 PM

Are you saying that one of the wavefronts will not be spherical? (I don't see how that can be, as SR is based on the postulates that the speed of light is constant in all frames of reference.)

macaw

2011-Mar-06, 06:28 PM

But, this has nothing to do with what Einstein wrote.

He wrote:

At the time t=t'=0, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then......

He then wrote:

The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

I think everyone is clear on the meaning of a spherical wave in the context of Einstein's statement, it is a perfect sphere.

Err, no. It is a spherical WAVEFRONT. Do you understand the difference?

Q1: What is the difference between a sphere and a SPHERICAL WAVEFRONT?

Q2: Show (with math) how a sphere transforms under LT. Show (with math) how a spherical wavefront transforms under LT

But, the spherical wave in the rest frame does not LT translate into a spherical wave in the moving system.

Your claim is false. Repeating it multiple times doesn't make it correct.

chinglu1998

2011-Mar-06, 10:30 PM

chinglu1998, the problem you have with your ATM threads is you never really explain the important bits. The above snip is an example.

You write things like A and B and C, therefore D. Now, A and B might be "common knowledge", but if C isn't, you can't simply carry on and say you've shown D.

Pretty much the point of the paper you've been quoting, is the opposite of what you are claiming. You need to put much much more effort into showing:

I can see your comment, but I thought the relativity of simultaneity was common knowledge.

I provide more analysis.

Let's assume there is a spherical wave in the stationary frame.

Then, we find if (r,0,0) is just acquired by that spherical wave, then (-r,0,0) is also just acquired by the spherical wave.

In general, in the view of the stationary from, the spherical wave is a distance r in all directions from the origin. That is what is meant bu the wave is spherical in the view of a frame.

We now apply LT to (r,0,0) and (-r,0,0).

(r,0,0) -> r ( 1 - v/c )γ

(-r,0,0) -> r ( 1 + v/c )γ

Hence, the LT mapped spherical wave is not spherical in the view of the moving system since r ( 1 - v/c )γ ≠ r ( 1 + v/c )γ as claimed.

This is also consistent with the relativity of simultaneity.

chinglu1998

2011-Mar-06, 10:35 PM

Are you saying that one of the wavefronts will not be spherical? (I don't see how that can be, as SR is based on the postulates that the speed of light is constant in all frames of reference.)

I am trying to show the prove by Einstein only showed a light beam measures c if and only if the LT mapped light beam measures c and nothing more.

And I believe SR is based on more than above the space time interval is invariant and this cannot ever be refuted.

But, that is not sufficient to prove SR preserves the spherical wave.

chinglu1998

2011-Mar-06, 10:43 PM

Err, no. It is a spherical WAVEFRONT. Do you understand the difference?

Q1: What is the difference between a sphere and a SPHERICAL WAVEFRONT?

Q2: Show (with math) how a sphere transforms under LT. Show (with math) how a spherical wavefront transforms under LT

Your claim is false. Repeating it multiple times doesn't make it correct.

Q1: What is the difference between a sphere and a SPHERICAL WAVEFRONT?

If Einstein is trying to prove the spherical wavefront is preserved across frames, where is the proof? I only see a proof that selects one point from THE LIGHT SPHERE and maps it with LT and claims this proves it is a light sphere in the view of the moving frame. Did you find something in the proof beyond this?

Q2: Show (with math) how a sphere transforms under LT. Show (with math) how a spherical wavefront transforms under LT

I already showed with math how the sphere transforms. It is not spherical by selecting only two points.

Einsteins' proof did used one point. You cannot use one point and prove anything about a a spherical wavefront.

This thread is about Einstein's proof and what it says.

Geo Kaplan

2011-Mar-06, 11:10 PM

Q1: What is the difference between a sphere and a SPHERICAL WAVEFRONT?

If Einstein is trying to prove the spherical wavefront is preserved across frames, where is the proof? I only see a proof that selects one point from THE LIGHT SPHERE and maps it with LT and claims this proves it is a light sphere in the view of the moving frame. Did you find something in the proof beyond this?

Q2: Show (with math) how a sphere transforms under LT. Show (with math) how a spherical wavefront transforms under LT

I already showed with math how the sphere transforms. It is not spherical by selecting only two points.

Einsteins' proof did used one point. You cannot use one point and prove anything about a a spherical wavefront.

This thread is about Einstein's proof and what it says.

You are stubbornly insisting that you, and you alone, understand what Einstein is saying. But as you point out, if Einstein were saying what you claim, it would be foolish. If, instead, you interpret the term "spherical wave" as different from "sphere," then everything makes sense.

So, you have set up a straw man, and continue to insist on bashing it. Entertaining to you, perhaps, but a colossal waste of time.

chinglu1998

2011-Mar-06, 11:18 PM

You are stubbornly insisting that you, and you alone, understand what Einstein is saying. But as you point out, if Einstein were saying what you claim, it would be foolish. If, instead, you interpret the term "spherical wave" as different from "sphere," then everything makes sense.

So, you have set up a straw man, and continue to insist on bashing it. Entertaining to you, perhaps, but a colossal waste of time.

Maybe you are true.

But, this implies you can show me any math proof in which someone took one and only one point and proved an entire theory.

That is what Einstein did.

macaw

2011-Mar-07, 12:01 AM

Q1: What is the difference between a sphere and a SPHERICAL WAVEFRONT?

If Einstein is trying to prove the spherical wavefront is preserved across frames, where is the proof? I only see a proof that selects one point from THE LIGHT SPHERE and maps it with LT and claims this proves it is a light sphere in the view of the moving frame. Did you find something in the proof beyond this?

Yes, his proof is complete. Now, please answer the question.

Q2: Show (with math) how a sphere transforms under LT. Show (with math) how a spherical wavefront transforms under LT

I already showed with math how the sphere transforms.

This is not what Einstein talks about in his paper. He doesn't say anything about sphere transformations.

It is not spherical by selecting only two points.

Incorrect, you do not understand his proof. Your persistent misrepresentation of his proof does not count as errors in his proof.

Einsteins' proof did used one point.

Incorrect, Eibnstein shows how spherical WAVEFRONTS transform. Now, please answer the question.

Geo Kaplan

2011-Mar-07, 12:07 AM

Maybe you are true.

But, this implies you can show me any math proof in which someone took one and only one point and proved an entire theory.

That is what Einstein did.

No.

macaw

2011-Mar-07, 12:08 AM

I can see your comment, but I thought the relativity of simultaneity was common knowledge.

I provide more analysis.

Let's assume there is a spherical wave in the stationary frame.

Then, we find if (r,0,0) is just acquired by that spherical wave, then (-r,0,0) is also just acquired by the spherical wave.

This is not a spherical wave, this is just two points on the x-axis. Study Einstein's proof in his paper and try answering the question correctly.

chinglu1998

2011-Mar-07, 12:09 AM

Yes, his proof is complete. Now, please answer the question.

No, his proof is not complete.

Here is my post

Let's assume there is a spherical wave in the stationary frame.

Then, we find if (r,0,0) is just acquired by that spherical wave, then (-r,0,0) is also just acquired by the spherical wave.

In general, in the view of the stationary from, the spherical wave is a distance r in all directions from the origin. That is what is meant bu the wave is spherical in the view of a frame.

We now apply LT to (r,0,0) and (-r,0,0).

(r,0,0) -> r ( 1 - v/c )γ

(-r,0,0) -> r ( 1 + v/c )γ

Hence, the LT mapped spherical wave is not spherical in the view of the moving system since r ( 1 - v/c )γ ≠ r ( 1 + v/c )γ as claimed.

This is also consistent with the relativity of simultaneity.

This is not what Einstein talks about in his paper. He doesn't say anything about sphere transformations.

Maybe you are true. What does he mean by this then?

If (x, y, z) be a point just attained by this wave

The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

This way, I can understand your question.

chinglu1998

2011-Mar-07, 12:13 AM

This is not a spherical wave, this is just two points on the x-axis. Study Einstein's proof in his paper and try answering the question correctly.

I studied his proof. He used one point. If this is false, prove it.

Now, let's see how that works. Assume I handed you a bag of marbles.

You pulled one point (sorry one marble) out of the bag and it was red.

Can you conclude all marbles in the bag are red? This is what Einstein's proof demands.

One point is sufficient to prove all truths in the universe. That is Einstein's claim.

chinglu1998

2011-Mar-07, 12:16 AM

No.

OK, show me where he used more than one point.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

macaw

2011-Mar-07, 12:18 AM

I studied his proof. He used one point. If this is false, prove it.

No, (x,y,z,ct) is the SET of POINTS on the wavefront. The fact that you continue to misrepresent a well known proof means that :

-you don't understand it

-your contention is false

-Einstein's proof is correct.

macaw

2011-Mar-07, 12:21 AM

Maybe you are true. What does he mean by this then?

If (x, y, z) be a point just attained by this wave

The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

This way, I can understand your question.

It means that, contrary to your misunderstandings and misrepresentations, Einstein showed how a spherical wavefront transforms into a spherical wavefront under LT.

Your flawed attempt at proving the contrary uses two POINTS, not a spherical wavefront. You need to learn the basics before you challenge SR.

chinglu1998

2011-Mar-07, 12:22 AM

No, (x,y,z,ct) is the SET of POINTS on the wavefront. The fact that you continue to misrepresent a well known proof means that :

-you don't understand it

-your contention is false

-Einstein's proof is correct.

No, he only used one point.

And if you consider the set, I already proved with a post and the relativity of simultaneity the wave is not spherical in the view of the moving frame. Einstein said it was.

chinglu1998

2011-Mar-07, 12:24 AM

It means that, contrary to your misunderstandings and misrepresentations, Einstein showed how a spherical wavefront transforms into a spherical wavefront under LT.

Your flawed attempt at proving the contrary uses two POINTS, not a spherical wavefront. You need to learn the basics before you challenge SR.

Yes, I used two points and Einstein used one in his failed proof.

I proved the wave is not spherical in the view of the moving frame.

Are you claiming this is false?

macaw

2011-Mar-07, 12:25 AM

No, he only used one point.

Instead of repeating the same mistake over and over, try understanding the corrections to your mistakes.

And if you consider the set, I already proved with a post and the relativity of simultaneity the wave is not spherical in the view of the moving frame. Einstein said it was.

No, you proved absolutely nothing. To show you that:

Q3: What is the equation of a spherical wavefront?

Q4: How does a spherical wavefront transform under LT?

Please answer the questions.

chinglu1998

2011-Mar-07, 12:42 AM

Instead of repeating the same mistake over and over, try understanding the corrections to your mistakes.

I did not make any mistakes. If I did, you would list them.

No, you proved absolutely nothing. To show you that:

Q3: What is the equation of a spherical wavefront?

Q4: How does a spherical wavefront transform under LT?

Please answer the questions.[/QUOTE]

Q3: What is the equation of a spherical wavefront?

x² + y² + z² = c² t²*

Q4: How does a spherical wavefront transform under LT?

x'² + y² + z² = c² t'²*

( x - vt )²γ² + y² + z² = c² ( t - vx/c² )²γ²*

So, what does this have to do with the fact that Einstein selected one point in the universe and claimed to prove all truths regarding intertial motion? Why do you refuse to address this?

macaw

2011-Mar-07, 12:49 AM

I did not make any mistakes. If I did, you would list them.

No, you proved absolutely nothing. To show you that:

Q3: What is the equation of a spherical wavefront?

Q4: How does a spherical wavefront transform under LT?

Please answer the questions.

Q3: What is the equation of a spherical wavefront?

x² + y² + z² = c² t²*[/quote]

This is correct.

Q4: How does a spherical wavefront transform under LT?

x'² + y² + z² = c² t'²*

Almost correct. The correct equation is:

x'² + y'² + z'² = c² t'²

This means, as Einstein showed more than 100 years ago, that a spherical wavefront transforms into a spherical wavefront.

( x - vr/c )²γ² + y² + z² = c² ( t - vx/c² )²γ²*

Incorrect. Try again.

chinglu1998

2011-Mar-07, 12:51 AM

Q3: What is the equation of a spherical wavefront?

x² + y² + z² = c² t²*

This is correct.

Almost correct. The correct equation is:

x'² + y'² + z'² = c² t'²

Incorrect. Try again.[/QUOTE]

Almost correct. The correct equation is:

x'² + y'² + z'² = c² t'²

We are only considering the standard configuration.

I said that, so I am correct.

x'² + y² + z² = c² t'²

chinglu1998

2011-Mar-07, 12:54 AM

[B]

You still refuse to explain why Einstein picked one point in the universe and proved all facts about intertial frames.

I explained this is impossible and you claim this is correct.

This is where you and I are.

macaw

2011-Mar-07, 12:56 AM

You still refuse to explain why Einstein picked one point in the universe and proved all facts about intertial frames.

He didn't. He used the derivation that I am trying to teach you.

chinglu1998

2011-Mar-07, 12:59 AM

He didn't. He used the derivation that I am trying to teach you.

That is the first thought I had when I woke this morning.

But, then I realized Einstein picked one and only one point.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

Perhaps you can show me where he picked more than one point in the rest frame. That would mean you are correct.

macaw

2011-Mar-07, 01:03 AM

That is the first thought I had when I woke this morning.

But, then I realized Einstein picked one and only one point.

Nope, he used the same exact equation that I've been trying to teach you. You still don't understand what he did.

Let's try again:

Q6: If you start with x^2+y^2+z^2=(ct)^2 (as Einstein did in his paper)

how do you end up with:

x'^2+y'^2+z'^2=(ct')^2 (as Einstein did)?

Please do the appropriate calculations. You will learn a little SR this way.

chinglu1998

2011-Mar-07, 01:09 AM

Nope, he used the same exact equation that I've been trying to teach you. You still don't understand what he did.

Let's try again:

Q6: If you start with x^2+y^2+z^2=(ct)^2 (as Einstein did in his paper)

how do you end up with:

x'^2+y'^2+z'^2=(ct')^2 (as Einstein did)?

Please do the appropriate calculations. You will learn a little SR this way.

Yea, this is correct. I agree.

Wait, how does this prove the wave is no less spherical when viewed in the moving frame?

The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

macaw

2011-Mar-07, 01:11 AM

Yea, this is correct. I agree.

Wait, how does this prove the wave is no less spherical when viewed in the moving frame?

Q7: What does the equation

x'^2+y'^2+z'^2=(ct')^2

tell you?

chinglu1998

2011-Mar-07, 01:15 AM

Q7: What does the equation

x'^2+y'^2+z'^2=(ct')^2

tell you?

Standard configuration.

x'^2+y^2+z^2=(ct')^2 = k

It tells me as one point satisfies this equation.

Should I now conclude for all points on the rest light sphere

x'^2+y^2+z^2= k

That would mean the moving frame views a spherical wave as claimed by Einstein.

I already refuted this.

macaw

2011-Mar-07, 01:19 AM

Standard configuration.

x'^2+y^2+z^2=(ct')^2 = k

It tells me as one point satisfies this equation.

Wrong. Think some more, don't rush to answer.

Should I now conclude for all points on the rest light sphere

x'^2+y^2+z^2= k

Incorrect.

Q8: If x^2+y^2+z^2=(ct)^2 is the spherical wavefront in frame F, what does x'^2+y'^2+z'^2=(ct')^2 mean in frame F'?

chinglu1998

2011-Mar-07, 01:22 AM

Wrong. Think some more, don't rush to answer.

No, keep thinking.

my math is available for discretion.

Einstein used one point and claimed to prove all knowledge of inertial frames.

This is crackpottery.

Geo Kaplan

2011-Mar-07, 01:55 AM

my math is available for discretion.

Einstein used one point and claimed to prove all knowledge of inertial frames.

No.

This is crackpottery.

I fully agree, if your "this" refers to your continued mischaracterizations of what Einstein is saying.

Swift

2011-Mar-07, 03:20 AM

Wrong. Think some more, don't rush to answer.

This is crackpottery.

Both of you knock off the rude and condescending comments or we'll infract both of you.

Van Rijn

2011-Mar-07, 04:19 AM

I am not claiming t is a constant. I agree with the statement that an expanding sphere is not consistent with a constant time.

So then, you also agree that ct is not a constant? Are you retracting the claim that a "spherical wave means ct is a constant"?

Van Rijn

2011-Mar-07, 04:24 AM

You still refuse to explain why Einstein picked one point in the universe and proved all facts about intertial frames.

Could you provide more detail on your claim that "Einstein picked one point in the universe"? It isn't clear to me where he did so.

grav

2011-Mar-07, 11:03 PM

x'² + y² + z² = c² t'²

( x - vt )²γ² + y² + z² = c² ( t - vx/c² )²γ²Finish that. The Einstein simultaneity convention is defined as the simultaneity that allows all inertial frames to measure the speed of light isotropically at c, so in effect light emitted simultaneously and propagating in all directions from a point will necessarily appear spherical to every inertial frame, as is evident in the Lorentz transformations and Minkowski metric. So we get

x'^2 + y'^2 + z'^2 = c^2 t'^2

γ^2 (x - v t)^2 + y^2 + z^2 = c^2 γ^2 (t - v x / c^2)^2

y^2 + z^2 = γ^2 [c^2 (t - v x / c^2)^2 - (x - v t)^2]

y^2 + z^2 = γ^2 [c^2 t^2 - 2 t v x + v^2 x^2 / c^2 - x^2 + 2 x v t - v^2 t^2]

y^2 + z^2 = γ^2 [(c^2 - v^2) t^2 - x^2 (1 - (v/c)^2)]

y^2 + z^2 = γ^2 [(1 - (v/c)^2) c^2 t^2 - x^2 (1 - (v/c)^2)]

y^2 + z^2 = γ^2 (1 - (v/c)^2) [c^2 t^2 - x^2], where γ = 1 / sqrt(1 - (v/c)^2), so

y^2 + z^2 = c^2 t^2 - x^2

x^2 + y^2 + z^2 = c^2 t^2

chinglu1998

2011-Mar-07, 11:09 PM

So then, you also agree that ct is not a constant? Are you retracting the claim that a "spherical wave means ct is a constant"?

No, I do not retract my statement. And your connection of "spherical wave" meaning a collection of light spheres, ie the light cone, as making some difference to argument does not.

If you can't even prove the moving frame views a light sphere, how can you possibly prove the moving frame is viewing a collection of light spheres?

Maybe I am wrong, but it seems you have already agreed the moving frame is not viewing a light sphere.

If the moving frame does not view one light sphere, how does one pass onto the concept that the moving frame is viewing a collection of light spheres?

chinglu1998

2011-Mar-07, 11:13 PM

Could you provide more detail on your claim that "Einstein picked one point in the universe"? It isn't clear to me where he did so.

At the time t=t'=0, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then.....

chinglu1998

2011-Mar-07, 11:21 PM

Finish that. The Einstein simultaneity convention is defined as the simultaneity that allows all inertial frames to measure the speed of light isotropically at c, so in effect light propagating simultaneously in all directions from a point will necessarily appear spherical to every inertial frame, as is evident in the Lorentz transformations and Minkowski metric. So we get

x'^2 + y'^2 + z'^2 = c^2 t'^2

γ^2 (x - v t)^2 + y^2 + z^2 = c^2 γ^2 (t - v x / c^2)^2

y^2 + z^2 = γ^2 [c^2 (t - v x / c^2)^2 - (x - v t)^2]

y^2 + z^2 = γ^2 [c^2 t^2 - 2 t v x + v^2 x^2 / c^2 - x^2 + 2 x v t - v^2 t^2]

y^2 + z^2 = γ^2 [(c^2 - v^2) t^2 - x^2 (1 - (v/c)^2)]

y^2 + z^2 = γ^2 [(1 - (v/c)^2) c^2 t^2 - x^2 (1 - (v/c)^2)]

y^2 + z^2 = y^2 (1 - (v/c)^2) [c^2 t^2 - x^2], where γ = 1 / sqrt(1 - (v/c)^2), so

y^2 + z^2 = c^2 t^2 - x^2

x^2 + y^2 + z^2 = c^2 t^2

Looks OK, This is a proof of the invariance of the light like space time interval.

All I see is that LT mapped light beams measure c. This is not prove a spherical wave.

Since you have claimed this proof proves the moving frame is viewing a spherical wave, I would like to see the light sphere in the moving frame based on LT.

See, when I am in the rest frame, I view at any time t, light equidistance in all directions ie a light sphere.

So, if the moving frame is "viewing a spherical wave" llike you claimed you proved, I would like to see that by using only LT just like I see one in the rest frame, which anyone can understand what that means in the rest frame.

Thanks.

grav

2011-Mar-07, 11:34 PM

Looks OK, This is a proof of the invariance of the light like space time interval.

All I see is that LT mapped light beams measure c. This is not prove a spherical wave.

Since you have claimed this proof proves the moving frame is viewing a spherical wave, I would like to see the light sphere in the moving frame based on LT.

See, when I am in the rest frame, I view at any time t, light equidistance in all directions ie a light sphere.

So, if the moving frame is "viewing a spherical wave" llike you claimed you proved, I would like to see that by using only LT just like I see one in the rest frame, which anyone can understand what that means in the rest frame.

Thanks.SR is derived with all inertial frames measuring an isotropic speed of light c as one of its postulates. That means that as viewed from any inertial frame, light emitted simultaneously in all directions from a point will be measured to travel at c in every direction, forming a spherical wave. The Lorentz transformations and Minkowski metric when applying the Einstein simultaneity convention are natural consequences that follow from there. As far as showing that x^2 + y^2 + z^2 = c^2 t^2 defines a spherical wave, we know that x^2 + y^2 + z^2 = r^2 defines a sphere of radius r, so with time t the sphere expands and the radius increases such that r(t) = c t.

chinglu1998

2011-Mar-07, 11:44 PM

SR is derived with all inertial frames measuring an isotropic speed of light c as one of its postulates. That means that as viewed from any inertial frame, light emitted simultaneously in all directions from a point will be measured to travel at c in every direction, forming a spherical wave. The Lorentz transformations and Minkowski metric when applying the Einstein simultaneity convention are natural consequences that follow from there. As far as showing that x^2 + y^2 + z^2 = c^2 t^2 defines a spherical wave, we know that x^2 + y^2 + z^2 = r^2 defines a sphere of radius r, so with time t the sphere expands and the radius increases such that r(t) = c t.

OK, since you know it is a spherical wave, let's see it. Use LT and show it to me.

I want to see a light sphere of radius r' in the moving frame. This way I know the light wave is no less spherical when viewed in the moving frame.

grav

2011-Mar-08, 12:04 AM

OK, since you know it is a spherical wave, let's see it. Use LT and show it to me.

I want to see a light sphere of radius r' in the moving frame. This way I know the light wave is no less spherical when viewed in the moving frame.In the rest frame we have

x(t)^2 + y(t)^2 +z(t)^2 = c^2 t^2 = r(t)^2

where x(t), y(t), and z(t) are points on a sphere of radius r(t) at a time t according to the rest frame's clocks.

For a frame that is moving in respect to the rest frame we have

x'(t')^2 + y'(t')^2 + z'(t')^2 = c^2 t'^2 = r'(t')^2

where x'(t'), y'(t'), and z'(t') are points on a sphere of radius r'(t') at a time t' according to the moving frame's clocks.

Light will always form a spherical wave in any inertial frame since that is the way SR is defined according to its postulates and set up using the Einstein simultaneity convention. Light or anything else that is emitted simultaneously and propagating from a point in every direction at the same speed as measured by a frame will form an expanding sphere according to that frame. The Lorentz transformations and Minkowski metric were also used earlier to show this, but they are derived after the fact since we start with the postulate that light travels at c according to every inertial frame and go from there, thereby light must always travel as a spherical wave as a given in SR, which is supported by aberration experiments and MMX.

At the time t=t'=0, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then.....

You do realize that the (x,y,z) convention means any and all possible values for x, y, and z? That "single point" you claim is actually an infinite number of points.

You limit this artificially by using (r,0,0) and (-r,0,0), thereby locking down two of the axes' values. Einstein imposed no such limit.

chinglu1998

2011-Mar-08, 12:14 AM

In the rest frame we have

x(t)^2 + y(t)^2 +z(t)^2 = c^2 t^2 = r(t)^2

where x(t), y(t), and z(t) are points on a sphere of radius r(t) at a time t according to the rest frame's clocks.

For a frame that is moving in respect to the rest frame we have

x'(t')^2 + y'(t')^2 + z'(t')^2 = c^2 t'^2 = r'(t')^2

where x'(t'), y'(t'), and z'(t') are points on a sphere of radius r'(t') at a time t' according to the moving frame's clocks.

Light will always form a spherical wave in any inertial frame since that is the way SR is defined according to its postulates and set up using the Einstein simultaneity convention. Light or anything else that is emitted simultaneously and propagating from a point in every direction at the same speed as measured by a frame will form an expanding sphere according to that frame. The Lorentz transformations and Minkowski metric were also used earlier to show this, but they are derived after the fact since we start with the postulate that light travels at c according to every inertial frame and go from there, thereby light must always travel as a spherical wave as a given in SR, which is supported by aberration experiments and MMX.

No, I already provided a counter example where we took points from the light sphere in the rest frame and they were not equidistant to the origin of the moving frame.

So, this does not work.

Since you made the claim, can you provide proof that LT can produce a light sphere of r' in the view of the moving frame? That means all the points on the light sphere not just one.

chinglu1998

2011-Mar-08, 12:19 AM

You do realize that the (x,y,z) convention means any and all possible values for x, y, and z? That "single point" you claim is actually an infinite number of points.

You limit this artificially by using (r,0,0) and (-r,0,0), thereby locking down two of the axes' values. Einstein imposed no such limit.

Sure, I know this means any (x,y,z). I said he proved the invariance of the light like space time interval. That means for all selected points translated light beam measure c. But, that is all it shows.

You limit this artificially by using (r,0,0) and (-r,0,0), thereby locking down two of the axes' values. Einstein imposed no such limit.

I only used this to show a light sphere in the rest frame does not translate to a light sphere in the moving frame, nothing more.

At this point, we have no way to prove the moving frame is viewing a spherical wave.

LT needs to show somehow that the moving frame views a light sphere of radius r' for any r'.

grav

2011-Mar-08, 01:25 AM

Oops, sorry. I tried to delete my post quickly so I could work out the example before you posted. Here it is again. I will work it out using the two points (r(t), 0, 0, t) and (r'(t'), 0, 0, t').

The points on the sphere are equidistant according to both the rest frame and the moving frame. Taking y=0 and z=0, then in the rest frame we have the two points (-r(t), 0, 0, t) and (r(t), 0 , 0, t), and in the moving frame we have the two points (-r'(t'), 0, 0, t') and (r'(t'), 0, 0, t'). These relate in each frame by the Lorentz transformations and Minkowski metric as demonstrated earlier in post #59.

chinglu1998

2011-Mar-08, 01:27 AM

Oops, sorry. I tried to delete my post quickly so I could work out the example before you posted. Here it is again. I will work it out using the two points (r(t), 0, 0, t) and (r'(t'), 0, 0, t').

No problem, I deleted my response.

Take your time.

chinglu1998

2011-Mar-08, 01:36 AM

Oops, sorry. I tried to delete my post quickly so I could work out the example before you posted. Here it is again. I will work it out using the two points (r(t), 0, 0, t) and (r'(t'), 0, 0, t').

Actually, what we are trying to achieve is to somehow view a spherical wave in the moving frame.

Einstein claimed his proof supplied it. Yet no human can see it mentally.

We need to see r' in the view of the moving system. Then we are viewing a spherical wave in the moving frame as Einstein claimed.

As we all know, scientific humans do not accept assertions on blind faith.

grav

2011-Mar-08, 01:39 AM

Okay, so we have the two points (r, 0, 0, t) and (-r, 0, 0, t) at time t (it is understood this applies to r(t) and -r(t) at time t for an expanding sphere). We will use the Lorentz transformation to find the coordinates in the moving frame.

x^2 + y^2 + z^2 = c^2 t^2 = r^2 for the rest frame, where y=0 and z=0, so x^2 = r^2

x'^2 + y'^2 + z'^2 = c^2 t'^2 = r'^2 for the moving frame, where y'=0 and z'=0, so x'^2 = r'^2, which gives us

x'^2 = c^2 t'^2

γ^2 (x - v t)^2 = c^2 γ^2 (t - v x / c^2)^2

(x - v t)^2 = c^2 (t - v x / c^2)^2

x^2 - 2 x v t + v^2 t^2 = c^2 t^2 - 2 t v x + v^2 x^2 / c^2

x^2 (1 - (v/c)^2) = c^2 t^2 (1 - (v/c)^2)

x^2 = c^2 t^2

whereas x^2 = c^2 t^2 = r^2, so for +/- r in the rest frame, we also have +/- r' in the moving frame, equidistant at a given time t in the rest frame and at a given time t' in the moving frame.

grav

2011-Mar-08, 01:51 AM

Actually, what we are trying to achieve is to somehow view a spherical wave in the moving frame.

Einstein claimed his proof supplied it. Yet no human can see it mentally.

We need to see r' in the view of the moving system. Then we are viewing a spherical wave in the moving frame as Einstein claimed.

As we all know, scientific humans do not accept assertions on blind faith.Einstein's proof supplies it because it is a given, an assumption upon which the proof is based, basically that light is measured to travel as expanding spherical waves according to any inertial frame. He started by applying that assumption as a postulate and then found what must be true once accepted, which leads directly to the Lorentz transformations and Minkowski metric. In other words, SR does not prove that light travels isotropically at c according to all inertial frames, but SR is built upon that postulate and that the laws of physics are the same in every inertial frame as givens and proceeds from there. So if SR is true, then naturally light must expand as a spherical wave because that is what SR assumes in the first place. And of course we know that SR is supported by experiment, which demonstrate so far that light can be measured isotropically by any inertial frame as SR states.

chinglu1998

2011-Mar-08, 01:53 AM

Okay, so we have the two points (r, 0, 0, t) and (-r, 0, 0, t) at time t (it is understood this applies to r(t) and -r(t) at time t for an expanding sphere). We will use the Lorentz transformation to find the coordinates in the moving frame.

x^2 + y^2 + z^2 = c^2 t^2 = r^2 for the rest frame, where y=0 and z=0, so x^2 = r^2

x'^2 + y'^2 + z'^2 = c^2 t'^2 = r'^2 for the moving frame, where y'=0 and z'=0, so x'^2 = r'^2, which gives us

x'^2 = c^2 t'^2

γ^2 (x - v t)^2 = c^2 γ^2 (t - v x / c^2)^2

(x - v t)^2 = c^2 (t - v x / c^2)^2

x^2 - 2 x v t + v^2 t^2 = c^2 t^2 - 2 t v x + v^2 x^2 / c^2

x^2 (1 - (v/c)^2) = c^2 t^2 (1 - (v/c)^2)

x^2 = c^2 t^2

whereas x^2 = c^2 t^2 = r^2, so for +/- r in the rest frame, we also have +/- r' in the moving frame, equidistant at a given time t in the rest frame and at a given time t' in the moving frame.

So let me see if I understand your proof.

What is simultaneous in the rest frame is also simultaneous in the moving frame. You would have proved the relativity of simultaneity is false.

Is that what you proved?

The answer is no.

You proved the light like space time interval is invariant.

chinglu1998

2011-Mar-08, 01:55 AM

Einstein's proof supplies it because it is a given, an assumption upon which the proof is based. He started by applying that assumption as a postulate and then found what must be true once accepted, which leads directly to the Lorentz transformations and Minkowski metric as well as spherically expanding waves of light. In other words, SR does not prove that light travels isotropically at c according to all inertial frames, but SR is built upon that postulate and that the laws of physics are the same in every inertial frame as givens and proceeds from there. So if SR is true, then naturally light must expand as a spherical wave because that is what SR assumes in the first place. And of course we know that SR is supported by experiment, which demonstrate so far that light can be measured isotropically by any inertial frame as SR states.

Good, then you will have no problem showing me a light sphere of radius r' in the moving frame based on LT.

That way, I am viewing a spherical wave in the moving frame.

Support your claims.

grav

2011-Mar-08, 02:00 AM

So let me see if I understand your proof.

What is simultaneous in the rest frame is also simultaneous in the moving frame. You would have proved the relativity of simultaneity is false.

Is that what you proved?

The answer is no.

You proved the light like space time interval is invariant.Well no, the rest frame measures the radius of the sphere as r at a time t while the moving frame measures the radius of the sphere as r' at a time t'. They do not measure the same radius in the same time. r and r', t and t' are related by the Lorentz transformations between the two frames in the usual way.

chinglu1998

2011-Mar-08, 02:05 AM

Well no, the rest frame measures the radius of the sphere as r at a time t while the moving frame measures the radius of the sphere as r' at a time t'. They do not measure the same radius in the same time. r and r', t and t' are related by the Lorentz transformations between the two frames in the usual way.

That is nice. But that is not what your math shows.

The whole point of Einstein's proof is that the moving system is viewing a spherical wave.

That means at least the moving frame can view a light sphere of radius r' based on LT. I mean why not.

Can you do this? I want to see this spherical wave.

grav

2011-Mar-08, 02:06 AM

Good, then you will have no problem showing me a light sphere of radius r' in the moving frame based on LT.

That way, I am viewing a spherical wave in the moving frame.

Support your claims.

That is nice. But that is not what your math shows.

The whole point of Einstein's proof is that the moving system is viewing a spherical wave.

That means at least the moving frame can view a light sphere of radius r' based on LT. I mean why not.

Can you do this? I want to see this spherical wave.That has already been done. The coordinates of photons emitted simultaneously in all directions from a point are (x, y, z, t) where x^2 + y^2 + z^2 = c^2 t^2 = r^2 according to the rest frame and (x', y', z', t') where x'^2 + y'^2 + z'^2 = c^2 t'^2 = r'^2 according to the moving frame. These are the definitions of spheres according to either frame and will remain spheres for any time t and t'.

chinglu1998

2011-Mar-08, 02:09 AM

That has already been done. The coordinates of photons emitted simultaneously in all directions from a point are (x, y, z) where x^2 + y^2 + z^2 = c^2 t^2 = r^2 according to the rest frame and x'^2 + y'^2 + z'^2 = c^2 t'^2 = r'^2 according to the moving frame. These are the definitions of spheres according to either frame and so will remain spheres for any time t and t'.

OK, I am simple minded.

Show me a spherical wave of r' in the moving frame based on LT, r' a constant.

I keep asking you keep claiming, but you do not deliver.

I wonder if I know the answer.

grav

2011-Mar-08, 02:13 AM

OK, I am simple minded.

Show me a spherical wave of r' in the moving frame based on LT, r' a constant.

I keep asking you keep claiming, but you do not deliver.

I wonder if I know the answer.It is a wave, so r' is not constant. The wave is travelling outward in every direction from the point of origin and so the sphere is expanding over time t' according to the equation c^2 t'^2 = r'^2. As t' becomes greater, so does r' in proportion. Only the measured speed c of the wave remains constant.

chinglu1998

2011-Mar-08, 02:19 AM

It is a wave, so r' is not constant. The wave is travelling outward in every direction from the point of origin and so the sphere is expanding over time t' according to the equation c^2 t'^2 = r'^2. As t' becomes greater, so does r' in proportion. Only the measured speed c of the wave remains constant.

well, in the rest frame, I can set an r and everything works. I see the rest frame light sphere is equidistant to the origin and all works with r.

Are you saying the moving frame does not see this? If so, then SR is false.

chinglu1998

2011-Mar-08, 02:26 AM

It is a wave, so r' is not constant. The wave is travelling outward in every direction from the point of origin and so the sphere is expanding over time t' according to the equation c^2 t'^2 = r'^2. As t' becomes greater, so does r' in proportion. Only the measured speed c of the wave remains constant.

I want to see a wave of r' in the moving frame.

Are you claiming this is impossible?

grav

2011-Mar-08, 02:36 AM

well, in the rest frame, I can set an r and everything works. I see the rest frame light sphere is equidistant to the origin and all works with r.

Are you saying the moving frame does not see this? If so, then SR is false.You can only set r at a particular time t in the rest frame. The light travels outward from the point of origin at a speed c, so the sphere expands with time. The sphere's surface is equidistant from the origin in the rest frame and it is also equidistant in the moving frame.

chinglu1998

2011-Mar-08, 02:40 AM

You can only set r at a particular time t in the rest frame. The light travels outward from the point of origin at a speed c, so the sphere expands with time. The sphere's surface is equidistant from the origin in the rest frame and it is also equidistant in the moving frame.

Good, I want to see r' for the moving frame as a light sphere.

Why do you continue to avoid this?

Simply show this and then we can see you know what you are talking about.

We want to see a light sphere of radius constant r' in the moving frame. That way we know SR is true.

grav

2011-Mar-08, 02:41 AM

I want to see a wave of r' in the moving frame.

Are you claiming this is impossible?The moving frame measures c^2 t'^2 = r'^2. The rest frame measures c^2 t^2 = r^2. This is a given in SR. Regardless of what frame is moving relative to any other frame, if we are in some frame and we measure the speed of something to travel in all directions at the same speed from a point of origin, then it will form an expanding sphere. One can easily mentally picture that, and that is really all there is to it.

Geo Kaplan

2011-Mar-08, 02:44 AM

Good, I want to see r' for the moving frame as a light sphere.

Why do you continue to avoid this?

Simply show this and then we can see you know what you are talking about.

We want to see a light sphere of radius constant r' in the moving frame. That way we know SR is true.

By default in ATM, mainstream physics is assumed to be valid. The burden of proof lies with the ATM proponent (here, that would be you), so it is you who must persuade, not the other way around.

Swift

2011-Mar-08, 03:36 AM

Simply show this and then we can see you know what you are talking about.

As Geo Kaplan said, but I'll make it official. No one is obliged to show you anything about mainstream physics, though they may if they wish. You certainly can not request it. It is up to you to prove your idea, not the other way around.

macaw

2011-Mar-08, 04:45 AM

OK, since you know it is a spherical wave, let's see it. Use LT and show it to me.

I want to see a light sphere of radius r' in the moving frame. This way I know the light wave is no less spherical when viewed in the moving frame.

Please answer Q7 and Q8 and you will be able to prove it to yourself. As it has been explained to you, you are not in position to demand any explanations from us. So, please answer Q7,Q8. Thank you

Strange

2011-Mar-08, 09:44 AM

Good, I want to see r' for the moving frame as a light sphere.

Why do you continue to avoid this?

Simply show this and then we can see you know what you are talking about.

We want to see a light sphere of radius constant r' in the moving frame. That way we know SR is true.

r' is constant (at any given time) by definition. If you don't accept that then I don't know what sort proof anyone can give you that you will accept, because you don't accept the premises the proof is based on.

Strange

2011-Mar-08, 09:58 AM

What do you mean by r' being "constant"? It is obviously varying with time, so I assume you mean "same in all directions"? Well, c is defined to the same in all directions and so r' is "constant". Now, what was the question?

chinglu1998

2011-Mar-08, 10:41 PM

The moving frame measures c^2 t'^2 = r'^2. The rest frame measures c^2 t^2 = r^2. This is a given in SR. Regardless of what frame is moving relative to any other frame, if we are in some frame and we measure the speed of something to travel in all directions at the same speed from a point of origin, then it will form an expanding sphere. One can easily mentally picture that, and that is really all there is to it.

Right, but this thread is about whether LT provides a light sphere for the moving frame.

We now apply LT to (r,0,0) and (-r,0,0).

(r,0,0) LT-> r ( 1 - v/c )γ

(-r,0,0) LT-> r ( 1 + v/c )γ

This is not a light sphere.

chinglu1998

2011-Mar-08, 10:44 PM

Please answer Q7 and Q8 and you will be able to prove it to yourself. As it has been explained to you, you are not in position to demand any explanations from us. So, please answer Q7,Q8. Thank you

I doubt that.

Q7

http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1859876#post1859876

Q8 Q8: If x^2+y^2+z^2=(ct)^2 is the spherical wavefront in frame F, what does x'^2+y'^2+z'^2=(ct')^2 mean in frame F'?

If we map all points on the light sphere x^2+y^2+z^2=(ct)^2 with LT, we get an x-axis elongated ellipsoid in the moving frame, F'.

chinglu1998

2011-Mar-08, 10:45 PM

r' is constant (at any given time) by definition. If you don't accept that then I don't know what sort proof anyone can give you that you will accept, because you don't accept the premises the proof is based on.

Sure, r' is a constant at some given time. Except, Einstein's proof never used this as a basis of anything.

Right, but this thread is about whether LT provides a light sphere for the moving frame.

We now apply LT to (r,0,0) and (-r,0,0).

(r,0,0) LT-> r ( 1 - v/c )γ

(-r,0,0) LT-> r ( 1 + v/c )γ

This is not a light sphere.

Exactly, because you have constrained your transformations to two points, not all values for x, y, and z.

I doubt that.

Q7

http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1859876#post1859876

Q8 Q8: If x^2+y^2+z^2=(ct)^2 is the spherical wavefront in frame F, what does x'^2+y'^2+z'^2=(ct')^2 mean in frame F'?

If we map all points on the light sphere x^2+y^2+z^2=(ct)^2 with LT, we get an x-axis elongated ellipsoid in the moving frame, F'.

No, wrong. This has been shown to you several times.

chinglu1998

2011-Mar-08, 10:57 PM

What do you mean by r' being "constant"? It is obviously varying with time, so I assume you mean "same in all directions"? Well, c is defined to the same in all directions and so r' is "constant". Now, what was the question?

I simply meant, in the moving frame, there exists a light sphere at say r', then r' + ∆r' and so on.

This is what it means to be a spherical wave in the view of the moving frame.

There is no paper nor any proof of LT showing an equation such that the moving frame is viewing this.

If there di exists such a proof, that would mean the moving frame is viewing a spherical wave and everyone could inspect that LT concluded r' for any r'.

Do they have some form of tex here?

Logically, it is impossible for LT to provide a spherical wave in the view of the moving frame. LT calculates and claims the origin of F' is moving. The view of the F' frame is that the origin is not moving.

There concepts are not compatible.

chinglu1998

2011-Mar-08, 11:02 PM

Exactly, because you have constrained your transformations to two points, not all values for x, y, and z.

Well, let's now use y = r and y = -r.

We have in the rest frame.

(r,0,0)

(-r,0,0)

(0,r,0)

(0,-r,0)

In the moving frame, LT maps these to

( r ( 1 + v/c )γ,0,0)

( r ( 1 - v/c )γ,0,0)

(0,r,0)

(0,-r,0)

I have graphed this also and it is an ellipsoid. Though these four points do prove an ellipsoid, they disprove a light sphere.

I will see if I can make a graph and use the picture.

Be back later.

Strange

2011-Mar-08, 11:10 PM

I simply meant, in the moving frame, there exists a light sphere at say r', then r' + ∆r' and so on.

This is what it means to be a spherical wave in the view of the moving frame.

There is no paper nor any proof of LT showing an equation such that the moving frame is viewing this.

If there di exists such a proof, that would mean the moving frame is viewing a spherical wave and everyone could inspect that LT concluded r' for any r'.

Do they have some form of tex here?

Logically, it is impossible for LT to provide a spherical wave in the view of the moving frame. LT calculates and claims the origin of F' is moving. The view of the F' frame is that the origin is not moving.

There concepts are not compatible.

but the Lorentz transform is irrelevant because we are dealing with light; c is defined to be constant in all directions to all observers so, by definition, all observers will see a spherical wavefront.

If you were talking about a sound wave, then you could apply the LT and produce a non-spherical result.

grav

2011-Mar-08, 11:20 PM

Right, but this thread is about whether LT provides a light sphere for the moving frame.

We now apply LT to (r,0,0) and (-r,0,0).

(r,0,0) LT-> r ( 1 - v/c )γ

(-r,0,0) LT-> r ( 1 + v/c )γ

This is not a light sphere.Okay, I think I see what you are trying to do here. In the rest frame we have a light sphere expanding at c in all directions, so r = c t. So at time t, we have two photons that have travelled to -r and r while the moving observer has travelled to v t along the positive x direction from the origin. So according to the rest frame, one photon at -r is a distance of r + v t from the moving observer while the other photon at r is a distance of r - v t, right? So since r = c t in the rest frame, those can be rewritten as

r + v t = r + v c t / c = r + v r / c = r (1 + v / c) and

r - v t = r - v c t / c = r - v r / c = r (1 - v / c)

Then we just apply the Lorentz contraction between the frames to get what the moving frame measures with

r (1 - v / c) γ and r (1 + v / c) γ

Is that right? Is that what you have done to find these?

grav

2011-Mar-09, 12:25 AM

Assuming what I wrote in my last post is how you are working it out, it is correct as far as the distance the moving observer measures for each of the photons, but not at the same time in the moving frame, because it does not include relativity of simultaneity. That is, while the rest observer measures -r and r at time t for the photons, the moving observer measures different times for each of the photons to travel to the distances you have indicated. That is because according to the rest observer, the moving observer's clocks read greater times in the negative x direction than the positive x direction, so different times at those locations for the photons than the rest frame's clocks read. Also taking that into account, however, while we will still have x'^2 + y'^2 + z'^2 = r'^2 = c^2 t'^2 for what the moving observer measures, I mistakenly wrote x'^2 = r'^2 in post #72 with x' = +/- r' for both of the photons that the rest frame measures at +/- r at t, so I will have to rework that using the Lorentz tranformations for -r and r individually, x1 = -r at t and x2 = r at t for the rest frame and x1' at t1' and x2' at t2' for the moving frame, so here it is.

For x1', we have x1'^2 = c^2 t1'^2

γ^2 (x1 - v t1)^2 = c^2 γ^2 (t1 - v x1 / c^2)^2, where x1 = r, so

(r - v t1)^2 = c^2 (t1 - v r / c^2)^2

r^2 - 2 r v t1 + v^2 t1^2 = c^2 t1^2 - 2 t1 v r + v^2 r^2 / c^2

r^2 (1 - (v/c)^2) = c^2 t1^2 (1 - (v/c)^2)

r^2 = c^2 t1^2

For x2', we have x2'^2 = c^2 t2'^2

γ^2 (x2 - v t2)^2 = c^2 γ^2 (t2 - v x2 / c^2)^2, where x2 = - r, so

(-r - v t2)^2 = c^2 (t2 + v r / c^2)^2

r^2 + 2 r v t2 + v^2 t2^2 = c^2 t2^2 + 2 t2 v r + v^2 r^2 / c^2

r^2 (1 - (v/c)^2) = c^2 t2^2 (1 - (v/c)^2)

r^2 = c^2 t2^2

Now for the rest frame, we have t1 = t2 = t, but the times aren't the same for what the moving frame measures, x1' and x2' at t1' and t2', although it still works out to x'^2 + y'^2 + z'^2 = c^2 t'^2 = r'^2 for any given time t' in the moving frame. It's a little confusing, I know, but that's how it works out and it is consistent. I will work it out the way you were doing it next, but include the relativity of simultaneity.

chinglu1998

2011-Mar-09, 12:29 AM

No, wrong. This has been shown to you several times.

This place will not use equal units for x and y. But, if it did, the LT mapped object is an ellipsoid.

Example used.

r = 1

v = .6c

γ = 5/4

z = 0

All distance units are in light seconds.

( x + 3/4 )^ 2 * 16/25 + y^2 = 1

http://www.wolframalpha.com/input/?i=%28+x+%2B+3%2F4+%29%5E+2+*+16%2F25+%2B+y%5E2+%3 D+1

chinglu1998

2011-Mar-09, 12:34 AM

but the Lorentz transform is irrelevant because we are dealing with light; c is defined to be constant in all directions to all observers so, by definition, all observers will see a spherical wavefront.

If you were talking about a sound wave, then you could apply the LT and produce a non-spherical result.

The light postulate in each frame says the wave will be spherical.

But, that is not what LT calculates based on the light sphere in the rest frame. I provided a plot.

grav

2011-Mar-09, 12:35 AM

This place will not use equal units for x and y. But, if it did, the LT mapped object is an ellipsoid.

Example used.

r = 1

v = .6c

γ = 5/4

z = 0

All distance units are in light seconds.

( x + 3/4 )^ 2 * 16/25 + y^2 = 1

http://www.wolframalpha.com/input/?i=%28+x+%2B+3%2F4+%29%5E+2+*+16%2F25+%2B+y%5E2+%3 D+1 Right, it is an ellipsoid before relativity of simultaneity is accounted for, mapped out at different times in the moving frame, but when the points are mapped simultaneously in the moving frame according to the moving frame's own clocks, it forms a sphere.

chinglu1998

2011-Mar-09, 12:39 AM

Okay, I think I see what you are trying to do here. In the rest frame we have a light sphere expanding at c in all directions, so r = c t. So at time t, we have two photons that have travelled to -r and r while the moving observer has travelled to v t along the positive x direction from the origin. So according to the rest frame, one photon at -r is a distance of r + v t from the moving observer while the other photon at r is a distance of r - v t, right? So since r = c t in the rest frame, those can be rewritten as

r + v t = r + v c t / c = r + v r / c = r (1 + v / c) and

r - v t = r - v c t / c = r - v r / c = r (1 - v / c)

Then we just apply the Lorentz contraction between the frames to get what the moving frame measures with

r (1 - v / c) γ and r (1 + v / c) γ

Is that right? Is that what you have done to find these?

Close.

x' = ( x - vt )γ

Set x = r, t = r/c

x' = ( r - vr/c )γ = r ( 1 - v/c )γ

Now set x = -r, t = r/c

x' = ( -r - vr/c )γ = -r ( 1 + v/c )γ

I can see I made a mistake before and accidentally omitted the - sign for the above.

chinglu1998

2011-Mar-09, 12:51 AM

Assuming what I wrote in my last post is how you are working it out, it is correct as far as the distance the moving observer measures for each of the photons, but not at the same time in the moving frame, because it does not include relativity of simultaneity. That is, while the rest observer measures -r and r at time t for the photons, the moving observer measures different times for each of the photons to travel to the distances you have indicated. That is because according to the rest observer, the moving observer's clocks read greater times in the negative x direction than the positive x direction, so different times at those locations for the photons than the rest frame's clocks read. Also taking that into account, however, while we will still have x'^2 + y'^2 + z'^2 = r'^2 = c^2 t'^2 for what the moving observer measures, I mistakenly wrote x'^2 = r'^2 in post #72 with x' = +/- r' for both of the photons that the rest frame measures at +/- r at t, so I will have to rework that using the Lorentz tranformations for -r and r individually, x1 = -r at t and x2 = r at t for the rest frame and x1' at t1' and x2' at t2' for the moving frame, so here it is.

For x1', we have x1'^2 = c^2 t1'^2

γ^2 (x1 - v t1)^2 = c^2 γ^2 (t1 - v x1 / c^2)^2, where x1 = r, so

(r - v t1)^2 = c^2 (t1 - v r / c^2)^2

r^2 - 2 r v t1 + v^2 t1^2 = c^2 t1^2 - 2 t1 v r + v^2 r^2 / c^2

r^2 (1 - (v/c)^2) = c^2 t1^2 (1 - (v/c)^2)

r^2 = c^2 t1^2

For x2', we have x2'^2 = c^2 t2'^2

γ^2 (x2 - v t2)^2 = c^2 γ^2 (t2 - v x2 / c^2)^2, where x2 = - r, so

(-r - v t2)^2 = c^2 (t2 + v r / c^2)^2

r^2 + 2 r v t2 + v^2 t2^2 = c^2 t2^2 + 2 t2 v r + v^2 r^2 / c^2

r^2 (1 - (v/c)^2) = c^2 t2^2 (1 - (v/c)^2)

r^2 = c^2 t2^2

Now for the rest frame, we have t1 = t2 = t, but the times aren't the same for what the moving frame measures, x1' and x2' at t1' and t2', although it still works out to x'^2 + y'^2 + z'^2 = c^2 t'^2 = r'^2 for any given time t' in the moving frame. It's a little confusing, I know, but that's how it works out and it is consistent. I will work it out the way you were doing it next, but include the relativity of simultaneity.

I already include the relativity of simultaneity in that LT is used.

Anyway, just because x'^2 + y'^2 + z'^2 = c^2 t'^2 is true based on LT output, that does not imply the LT mapped object is a light sphere. In fact, it is impossible. Given LT asumes the F' origin is moving, and the relativity of simultaneity is true, then it is impossible that x'^2 + y'^2 + z'^2 = c^2 t'^2 os a sphere nased on LT because LT believes the subject origin is moving. It is impossible for LT to conclude the moving frame is viewing a spherical wavefront unless LT claims the light sphere is moving and centered at the origin of the moving frame as in Ritz's theory.

But, to achieve a measure at c for each light beam in the moving frame, and the relativity of simultaneity, it must be the case that the simultaneity of the rest frame is mapped into an interval in the moving frame.

So, this is how it works.

Moving frame using rest frame information and applying LT.

The light beam measure c

-r'..........O'

Now, O' is moving in the calculations of LT, so for the front beam to measure c, it strikes later.

-r'..........|.....O'...........r'

chinglu1998

2011-Mar-09, 12:53 AM

Right, it is an ellipsoid before relativity of simultaneity is accounted for, mapped out at different times in the moving frame, but when the points are mapped simultaneously in the moving frame according to the moving frame's own clocks, it forms a sphere.

No, relativity of simultaneity is already included.

pzkpfw

2011-Mar-09, 01:07 AM

... claims the light sphere is moving and centered at the origin of the moving frame ...

Why not? A person sitting in a chair may see a car drive past, but the driver of the car is entitled to see themselves as stationary and the chair moving the other way. ("Relativity".)

If a flash bulb goes off in the (according to the chair person) "moving" car, the driver is entitled to see the origin of that sphere of light as being stationary, with them.

The driver and the chair-sitter won't be seeing the same sphere at the same time (assuming there was some way for them to figure "same time"), but they'll both see spheres.

Edit: if you don't accept this, presumably you'd be claiming that when the flash bulb goes off, the origin of that flash is somehow fixed to the chair-sitters frame. You are essentially claiming the chair sitter (and only the chair sitter) is in a "stationary" frame. To which I'd point out the Earth is spinning, orbiting, etc. What you'd need to do is somehow find the "one true" stationary frame - which would be the only place where (according to your claims) a wave of light from some source would be truely spherical. In doing so, you'd be defining a special "one of a kind" frame... which breaks another postulate of relativity.

chinglu1998

2011-Mar-09, 01:22 AM

Why not? A person sitting in a chair may see a car drive past, but the driver of the car is entitled to see themselves as stationary and the chair moving the other way. ("Relativity".)

If a flash bulb goes off in the (according to the chair person) "moving" car, the driver is entitled to see the origin of that sphere of light as being stationary, with them.

The driver and the chair-sitter won't be seeing the same sphere at the same time (assuming there was some way for them to figure "same time"), but they'll both see spheres.

This I all agree. You are listing the individual truths in each frame.

But, LT combines all these truths. LT provides the conjunction of these truths.

This is the "cost" of the conjunction.

Moving frame using rest frame information and applying LT.

The light beam measure c

-r'..........O'

Now, O' is moving in the calculations of LT, so for the front beam to measure c, it strikes later.

-r'..........|.....O'...........r'

In the view of the rest frame, simultaneity in the moving frame is not simultaneous to it as we all know.

But, as we can see in the condensed picture, the price of simultaneity in the moving frame according to LT is the the moving origin must move in order for simultaneity in the moving frame to occur from the LT calculation. This is all that is needed to prove LT does not believe the moving frame is viewing a light sphere.

And, that is the whole issue here. It is not about what each individual frame believes, it is what LT believes since it represents the truth of all of it whereas the individual frames can only speak for themselves.

Now, this thread does not claim to falsify SR based on this. It is just to point out that Einstein made a sweeping claim that LT cannot mathematically support. However, Einstein absolutely proved LT preserves the constant speed of light for each mapped light beam.

At this point, there is no evidence that LT must preserve the light sphere across frames since that would imply the relativity of simultaneity is false. In other words, what is simultaneous in the rest frame and then mapped by LT is simultaneous in the moving frame. We all know that is false.

So, it is perfectly consistent that LT does not preserve the light sphere.

grav

2011-Mar-09, 01:25 AM

Close.

x' = ( x - vt )γ

Set x = r, t = r/c

x' = ( r - vr/c )γ = r ( 1 - v/c )γ

Now set x = -r, t = r/c

x' = ( -r - vr/c )γ = -r ( 1 + v/c )γ

I can see I made a mistake before and accidentally omitted the - sign for the above.Yes, okay. We would be taking the absolute values for the distances the moving frame measured before, but it is better to include the sign. Okay, so the rest frame measures x1 = r and x2 = -r at time t. The moving frame measures x1' = r (1 - v/c) γ and x2' = - r (1 + v/c) γ at t1' and t2'. The moving frame does not measure the photons at these distances simultaneously, otherwise it would be an ellipsoid as you said. So let's find out at what times the moving frame will measure these using the relativity of simultaneity.

When the photons are at -r and r at time t according to the rest frame, the rest frame says the moving observer's clock reads t / γ and that the moving observer is at the location v t. A second observer is in the same frame as the moving observer but at the location r at time t. Using the relativity of simultaneity, the rest frame says the second observer's clock reads

t1' = t / γ - x1' v / c^2

= t / γ - [r (1 - v/c) γ] v / c^2

= t / γ - [c t (1 - v/c) γ] v / c^2

= (t / γ) [1 - γ^2 (1 - v/c) (v/c)]

= (t / γ) [1 - (v/c) / (1 + v/c)]

= (t / γ) [(1 + v/c) - (v/c)] / (1 + v/c)

= (t / γ) / (1 + v/c)

A third observer in the moving frame is at the position -r at time t according to the rest frame, so the third observer's clock reads

t2' = t / γ + x2' v / c^2

= t / γ + [r (1 + v/c) γ] v / c^2

= t / γ + [c t (1 + v/c) γ] v / c^2

= (t / γ) [1 + γ^2 (1 + v/c) (v/c)]

= (t / γ) [1 + (v/c) / (1 - v/c)]

= (t / γ) [(1 - v/c) + (v/c)] / (1 - v/c)

= (t / γ) / (1 - v/c)

Okay, so the times on the second and third observers' clocks that coincide with the two photons indicate the times the photons reach those distances according to the moving frame. So the moving frame says one photon is at a distance x1' = r (1 - v / c) γ at a time t1' = (t / γ) / (1 + v/c) and the other photon is at a distance x2' = r (1 + v/c) γ at a time t2' = (t / γ) / (1 - v/c), so we have

x1' / t1' = [r (1 - v/c) γ] / [(t / γ) / (1 + v/c)]

= r (1 - v/c) (1 + v/c) γ^2 / t

= r / t

= c

x2' / t2' = [r (1 + v/c) γ] / [(t / γ) / (1 - v/c)]

= r (1 + v/c) (1 - v/c) γ^2 / t

= r / t

= c

So the distances the moving frame measures for the distances combines with the relativity of simultaneity such that the moving frame still measures the speed of light at c the same as the rest frame does, but just measures them at different times than the rest frame. The moving frame measures the photons to be travelling the same speed in both directions, so with the same distance travelled in the same time, whereby the photons will be equidistant simultaneously in the moving frame at any time t' according to the moving frame's own clocks.

grav

2011-Mar-09, 01:35 AM

I already include the relativity of simultaneity in that LT is used.Right, you did for x1 = r and x2 = -r, giving x1' = r (1 - v/c) γ and x2' = - r (1 + v/c) γ, but you must include the times t1' and t2' as well. The moving frame does not measure those two photons simultaneously at those positions as the rest frame does. I found for the time dilation and relativity of simultaneity separately in my last post, but go ahead and find for the times t1' and t2' using the transforms, then find for x1' / t1' and x2' / t2'.

chinglu1998

2011-Mar-09, 01:37 AM

Yes, okay. We would be taking the absolute values for the distances the moving frame measured before, but it is better to include the sign. Okay, so the rest frame measures x1 = r and x2 = -r at time t. The moving frame measures x1' = r (1 - v/c) γ and x2' = - r (1 + v/c) γ at t1' and t2'. The moving frame does not measure the photons at these distances simultaneously, otherwise it would be an ellipsoid as you said. So let's find out at what times the moving frame will measure these using the relativity of simultaneity.

When the photons are at -r and r at time t according to the rest frame, the rest frame says the moving observer's clock reads t / γ and that the moving observer is at the location v t. A second observer is in the same frame as the moving observer but at the location r at time t. Using the relativity of simultaneity, the rest frame says the second observer's clock reads

t1' = t / γ - x1' v / c^2

= t / γ - [r (1 - v/c) γ] v / c^2

= t / γ - [c t (1 - v/c) γ] v / c^2

= (t / γ) [1 - γ^2 (1 - v/c) (v/c)]

= (t / γ) [1 - (v/c) / (1 + v/c)]

= (t / γ) [(1 + v/c) - (v/c)] / (1 + v/c)

= (t / γ) / (1 + v/c)

A third observer in the moving frame is at the position -r at time t according to the rest frame, so the third observer's clock reads

t2' = t / γ + x2' v / c^2

= t / γ + [r (1 + v/c) γ] v / c^2

= t / γ + [c t (1 + v/c) γ] v / c^2

= (t / γ) [1 + γ^2 (1 + v/c) (v/c)]

= (t / γ) [1 + (v/c) / (1 - v/c)]

= (t / γ) [(1 - v/c) + (v/c)] / (1 - v/c)

= (t / γ) / (1 - v/c)

Okay, so the times on the second and third observers' clocks that coincide with the two photons indicate the times the photons reach those distances according to the moving frame. So the moving frame says one photon is at a distance x1' = r (1 - v / c) γ at a time t1' = (t / γ) / (1 + v/c) and the other photon is at a distance x2' = r (1 + v/c) γ at a time t2' = (t / γ) / (1 - v/c), so we have

x1' / t1' = [r (1 - v/c) γ] / [(t / γ) / (1 + v/c)]

= r (1 - v/c) (1 + v/c) γ^2 / t

= r / t

= c

x2' / t2' = [r (1 + v/c) γ] / [(t / γ) / (1 - v/c)]

= r (1 + v/c) (1 - v/c) γ^2 / t

= r / t

= c

So the distances the moving frame measures for the distances combines with the relativity of simultaneity such that the moving frame still measures the speed of light at c the same as the rest frame does, but just measures them at different times than the rest frame. The moving frame measures the photons to be travelling the same speed in both directions, so with the same distance travelled in the same time, whereby the photons will be equidistant simultaneously in the moving frame at any time t' according to the moving frame's own clocks.

All this math is not needed. I showed you the calculations using LT and to come back to the other direction, the LT matrix is invertible.

Tobin Dax

2011-Mar-09, 01:46 AM

This place will not use equal units for x and y. But, if it did, the LT mapped object is an ellipsoid.

Example used.

r = 1

v = .6c

γ = 5/4

z = 0

All distance units are in light seconds.

( x + 3/4 )^ 2 * 16/25 + y^2 = 1

http://www.wolframalpha.com/input/?i=%28+x+%2B+3%2F4+%29%5E+2+*+16%2F25+%2B+y%5E2+%3 D+1

This is in the moving frame? Then it is incorrect. y'≠y on this spherical wavefront, since t and t' are different. Your equation should be (x+3/4)2*16/25 + y'2 = 1.

(This is a direct question.) How long does it take the light wave to travel a distance of x' in the moving frame? How far does the wavefront move in the y' direction in that time? in the z' direction? Please show calculations for all three answers.

chinglu1998

2011-Mar-09, 01:52 AM

This is in the moving frame? Then it is incorrect. y'≠y on this spherical wavefront, since t and t' are different. Your equation should be (x+3/4)2*16/25 + y'2 = 1.

(This is a direct question.) How long does it take the light wave to travel a distance of x' in the moving frame? How far does the wavefront move in the y' direction in that time? in the z' direction? Please show calculations for all three answers.

No, you are wrong.

This is the standard configuration, so y' = y.

http://en.wikipedia.org/wiki/Lorentz_transform#Lorentz_transformation_for_frame s_in_standard_configuration

Your direct question lists variables. This is also an error.

I cannot possible give a direct answer with a direct question which does not include exact values.

chinglu1998

2011-Mar-09, 01:58 AM

Right, you did for x1 = r and x2 = -r, giving x1' = r (1 - v/c) γ and x2' = - r (1 + v/c) γ, but you must include the times t1' and t2' as well. The moving frame does not measure those two photons simultaneously at those positions as the rest frame does. I found for the time dilation and relativity of simultaneity separately in my last post, but go ahead and find for the times t1' and t2' using the transforms, then find for x1' / t1' and x2' / t2'.

Go ahead and calculate the t primes. Feel free.

Given y = 0 = z, then it is natural,

x1' / t1' = c and x2' / t2' = c

what are you trying to claim?

korjik

2011-Mar-09, 02:12 AM

No, you are wrong.

This is the standard configuration, so y' = y.

http://en.wikipedia.org/wiki/Lorentz_transform#Lorentz_transformation_for_frame s_in_standard_configuration

Your direct question lists variables. This is also an error.

I cannot possible give a direct answer with a direct question which does not include exact values.

There is more than enough information to rather trivially answer the question. Here is a start for you:

It takes light x'/c seconds to travel a distance of x' in the moving frame.

grav

2011-Mar-09, 02:12 AM

Go ahead and calculate the t primes. Feel free.

Given y = 0 = z, then it is natural,

x1' / t1' = c and x2' / t2' = c

what are you trying to claim?Okay well, we'll go from there then. So x1' / t1' = c gives the distance x1' travelled by one of the photons according to the moving frame in the time t1' according to the moving frame also, and with x2' / t2' = c, x2' is the distance travelled by the other photon in the time t2' according to the moving frame as well, right? Each distance divided by the time to reach that distance gives the same speed that the moving frame measures for the photons in each direction along the x axis, correct? So if the moving frame measures the photons to be travelling at the same speed in either direction along the x axis, then at any given time in the moving frame, the distance each of the photons travels is the same as measured simultaneously by the moving frame, so the distances travelled are always equidistant from the origin according to the moving frame, right?

chinglu1998

2011-Mar-09, 02:19 AM

Okay well, we'll go from there then. So x1' / t1' = c gives the distance x1' travelled by one of the photons according to the moving frame in the time t1' according to the moving frame also, and with x2' / t2' = c, x2' is the distance travelled by the other photon in the time t2' according to the moving frame as well, right? Each distance divided by the time to reach that distance gives the same speed that the moving frame measures for the photons in each direction along the x axis, correct? So if the moving frame measures the photons to be travelling at the same speed in either direction along the x axis, then at any given time in the moving frame, the distance each of the photons travels is the same as measured simultaneously by the moving frame, so the distances travelled are always equidistant from the origin according to the moving frame, right?

Yes, I am going to go with this. But more, "so the distances travelled are always equidistant from the origin according to the moving frame" and LT agrees with this from the rest frame for the moving frame.

grav

2011-Mar-09, 02:21 AM

Yes, I am going to go with this. But more, "so the distances travelled are always equidistant from the origin according to the moving frame" and LT agrees with this from the rest frame for the moving frame.:)

chinglu1998

2011-Mar-09, 02:23 AM

There is more than enough information to rather trivially answer the question. Here is a start for you:

It takes light x'/c seconds to travel a distance of x' in the moving frame.

Yes, but in the view of the rest frame, the moving frame origin is moving. We are talking about what LT thinks.

So you need to apply this to all waves with x'/c with x' calculated from LT. What do you get when you use x' calculate from the rest frame?

macaw

2011-Mar-09, 02:38 AM

I doubt that.

Q7

http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1859876#post1859876

Q8 Q8: If x^2+y^2+z^2=(ct)^2 is the spherical wavefront in frame F, what does x'^2+y'^2+z'^2=(ct')^2 mean in frame F'?

If we map all points on the light sphere x^2+y^2+z^2=(ct)^2 with LT, we get an x-axis elongated ellipsoid in the moving frame, F'.

Q9: Prove it. Do the calculations. This is the third time I have challenged you to do the calculations.

korjik

2011-Mar-09, 03:04 AM

Yes, but in the view of the rest frame, the moving frame origin is moving. We are talking about what LT thinks.

So you need to apply this to all waves with x'/c with x' calculated from LT. What do you get when you use x' calculate from the rest frame?

Are you being evasive or just misunderstanding? The question dosent ask about the stationary frame, only the moving frame.

Not only that, but when you calculate x', you will pretty much always come up with x' for your answer. So, how long does it take light to travel the distance x', and how far does the light travel in y' and z' in that time?

pzkpfw

2011-Mar-09, 03:46 AM

chinglu1998, your original claim (post #1) was "Yet, as anyone can see, cτ (Tau) is not a constant and thus not a light sphere. Therefore, Einstein's proof is false.", which you seem to have moved away from.

Your sequence was something like [ct not constant], [only one point was used] and now [LT does not give the result I expect].

This has the appearance of evasion, as every time your point is refuted you simply move on to some other point.

You are also "answering" some clear questions with questions, when actual answers should not be hard to give.

Bear in mind, this thread will be locked at 30 days; and you will be struggling to find another way to present your case. If you really wish to convince anyone, you need to be much clearer - and get to the point.

tusenfem

2011-Mar-09, 08:41 AM

To add to pzkpfw's comments, any more questioning mainstream to try and support your "misty" ATM will lead to an infraction.

Strange

2011-Mar-09, 08:53 AM

The light postulate in each frame says the wave will be spherical.

This is wrong - or at least, it is incomplete. The speed of light is constant as seen from any frame (regardless of which frame the light was emitted in). Therefore the wavefront, by definition will be spherical to any observer. If you haven't fully understood the postulates that SR is built on, it is not surprising that you keep coming with these basic misunderstandings.

Tobin Dax

2011-Mar-09, 08:27 PM

No, you are wrong.

This is the standard configuration, so y' = y.

http://en.wikipedia.org/wiki/Lorentz_transform#Lorentz_transformation_for_frame s_in_standard_configuration

No, it's not the standard configuration since the wavefront is not a solid object. The positions of the points (0,y',0) and (0,0,z') depend on the time of travel of the light wave, t'.

Your direct question lists variables. This is also an error.

I cannot possible give a direct answer with a direct question which does not include exact values.

Nope, no error. In my post, I used a specific example equation that you used. That example has specific values for all variables involved in my question. Feel free to use those values. But please answer my questions. I'll restate them for you. I am also going to add a question. Words alone are not sufficient to answer these questions. Please use equations as well for each answer.

TD1) What is the length of time (in the moving frame), t', that it takes for the light wave to propagate from (0,0,0) to (x',0,0)?

TD2) How far does the wavefront move in the y direction (to point (0,y',0)) in that time? in the z' direction (to point (0,0,z'))?

TD3) What is the length of time in the rest frame, t, that is takes the light wave to propagate to (x,0,0)?

TD4) How does time t compare to the dilated time t'? How does y' compare to y? How does z' compare to z?

chinglu1998

2011-Mar-09, 11:19 PM

Q9: Prove it. Do the calculations. This is the third time I have challenged you to do the calculations.

I provided a graph and an example. So, your statements are not accurate.

Here is the equation. We assume the light postulate is true in the rest frame and has acquired a radius r.

So, we will use as the domain the set of points that satisfy x² + y² + z² = r² and assume the standard configuration. Hence, x = ±√(r² - y² - z²).

In addition, x' must satisfy, *x' = ( x - vr/c )γ.

x' = ( ±√(r² - y² - z²) - vr/c )γ

x'/γ + vr/c = ±√(r² - y² - z²)

(x'/γ + vr/c)² = (r² - y² - z²)

(x'/γ + vr/c)² + y² + z² = r²

(x'/γ - (-vr/c))²/r² + y²/r² + z²/r² = 1

(x' - (-vrγ/c))² / (r²γ²) + y²/r² + z²/r² = 1

Like I said, the LT translated object is an x-axis elongate ellipsoid.

*

chinglu1998

2011-Mar-09, 11:26 PM

Are you being evasive or just misunderstanding? The question dosent ask about the stationary frame, only the moving frame.

Not only that, but when you calculate x', you will pretty much always come up with x' for your answer. So, how long does it take light to travel the distance x', and how far does the light travel in y' and z' in that time?

I do not need to be evasive.

It may be that I do not understand your question or intent.

I cannot tell you how long to travel to x' unless you are claiming y=0 and z=0.

If so, SR claims t=x'/c=y/c=z/c.

chinglu1998

2011-Mar-09, 11:29 PM

This is wrong - or at least, it is incomplete. The speed of light is constant as seen from any frame (regardless of which frame the light was emitted in). Therefore the wavefront, by definition will be spherical to any observer. If you haven't fully understood the postulates that SR is built on, it is not surprising that you keep coming with these basic misunderstandings.

I cannot see how this is any different from what I wrote.

I wrote, "The light postulate in each frame says the wave will be spherical."

Perhaps you can explain your statement that this is a basic misunderstanding of SR.

chinglu1998

2011-Mar-09, 11:36 PM

No, it's not the standard configuration since the wavefront is not a solid object. The positions of the points (0,y',0) and (0,0,z') depend on the time of travel of the light wave, t'.

What?

Nope, no error. In my post, I used a specific example equation that you used. That example has specific values for all variables involved in my question. Feel free to use those values. But please answer my questions. I'll restate them for you. I am also going to add a question. Words alone are not sufficient to answer these questions. Please use equations as well for each answer.

TD1) What is the length of time (in the moving frame), t', that it takes for the light wave to propagate from (0,0,0) to (x',0,0)?

x'/c

TD2) How far does the wavefront move in the y direction (to point (0,y',0)) in that time? in the z' direction (to point (0,0,z'))?

y'=y, z'=z.

Did you read the standard configuration?

http://en.wikipedia.org/wiki/Lorentz_transformation#Lorentz_transformation_for_ frames_in_standard_configuration

z'=z, y'=y

TD3) What is the length of time in the rest frame, t, that is takes the light wave to propagate to (x,0,0)?

x/c

TD4) How does time t compare to the dilated time t'? How does y' compare to y? How does z' compare to z?

Insufficient information to answer.

I need x,y,z,t in the rest frame and then we simply apply LT.

macaw

2011-Mar-10, 12:04 AM

I provided a graph and an example. So, your statements are not accurate.

Here is the equation. We assume the light postulate is true in the rest frame and has acquired a radius r.

So, we will use as the domain the set of points that satisfy x² + y² + z² = r²

Incorrect. Must satisfy x² + y² + z² = (ct)²

and assume the standard configuration. Hence, x = ±√(r² - y² - z²).

In addition, x' must satisfy, *x' = ( x - vr/c )γ.

x' = ( ±√(r² - y² - z²) - vr/c )γ

x'/γ + vr/c = ±√(r² - y² - z²)

(x'/γ + vr/c)² = (r² - y² - z²)

(x'/γ + vr/c)² + y² + z² = r²

(x'/γ - (-vr/c))²/r² + y²/r² + z²/r² = 1

(x' - (-vrγ/c))² / (r²γ²) + y²/r² + z²/r² = 1

Like I said, the LT translated object is an x-axis elongate ellipsoid.

*

Incorrect.

chinglu1998

2011-Mar-10, 12:09 AM

Incorrect. Must satisfy x² + y² + z² = (ct)²

We are LT mapping the light sphere in this thread. Therefore, x² + y² + z² = r² is correct for some fixed r.

Incorrect.

The math is right there, feel free. Math is either correct and incorrect. In this proof, it is correct.

Strange

2011-Mar-10, 01:05 AM

I wrote, "The light postulate in each frame says the wave will be spherical."

OK. So you agree that the moving observer will see the wavefront as spherical and the stationary observer will see the wavefront as spherical. Is there anything else to say?

Your question seems to be: "why do I get the wrong answer when I try and use the Lorentz transform?" There is another thread in Q&A currently that answers this very question.

chinglu1998

2011-Mar-10, 01:11 AM

OK. So you agree that the moving observer will see the wavefront as spherical and the stationary observer will see the wavefront as spherical. Is there anything else to say?

Your question seems to be: "why do I get the wrong answer when I try and use the Lorentz transform?" There is another thread in Q&A currently that answers this very question.

There is a lot to say.

I provided a proof that LT translates the light sphere into an ellipsoid.

Now, if I have the wrong answer as you claim, you will have no problem supporting your accusations and refuting my proof.

In any event, as we both say, each frame is viewing a light sphere, Einstein claimed by selecting one point the moving frame is viewing a light sphere, I proved the LT mapped light sphere from the rest frame is not a light sphere in the view of the moving frame. It is an x-axis elongated ellipsoid.

So, that is where we are.

macaw

2011-Mar-10, 01:48 AM

We are LT mapping the light sphere in this thread.

You are NOT "mapping a light sphere. You are repeating the same mistake over and over.

The math is right there, feel free. Math is either correct and incorrect. In this proof, it is correct.

You math, like your understanding of physics is incorrect. Incorrect understandings don't make valid theories.

chinglu1998

2011-Mar-10, 02:30 AM

You are NOT "mapping a light sphere. You are repeating the same mistake over and over.

Yes, we are mapping the light sphere. That is the point of this thread. I said, Einstein picked one point and make an unsupported conclusion that the moving frame was viewing a light sphere. I proved that false.

You math, like your understanding of physics is incorrect. Incorrect understandings don't make valid theories.

Since I am from math, you can prove this true or false. Where is your refutation?

Here is the equation. We assume the light postulate is true in the rest frame and has acquired a radius r.

So, we will use as the domain the set of points that satisfy x² + y² + z² = r² and assume the standard configuration. Hence, x = ±√(r² - y² - z²).

In addition, x' must satisfy, x' = ( x - vr/c )γ.

x' = ( ±√(r² - y² - z²) - vr/c )γ

x'/γ + vr/c = ±√(r² - y² - z²)

(x'/γ + vr/c)² = (r² - y² - z²)

(x'/γ + vr/c)² + y² + z² = r²

(x'/γ - (-vr/c))²/r² + y²/r² + z²/r² = 1

(x' - (-vrγ/c))² / (r²γ²) + y²/r² + z²/r² = 1

Like I said, the LT translated object is an x-axis elongate ellipsoid.

macaw

2011-Mar-10, 02:35 AM

Yes, we are mapping the light sphere.

But SR and Einstein mapped a spherical WAVEFRONT. You don't know the difference, this is why you are repeating the same mistake over and over.

chinglu1998

2011-Mar-10, 02:39 AM

But SR and Einstein mapped a spherical WAVEFRONT. You don't know the difference, this is why you are repeating the same mistake over and over.

How do you prove a wavefront with one point as selected by Einstein.

Let's look again.

At the time t=t'=0, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then

x2+y2+z2=c2t2.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

macaw

2011-Mar-10, 02:49 AM

How do you prove a wavefront with one point as selected by Einstein.

Einstein did not use a point, he used the equation of a WAVEFRONT.

Q10: what is the equation of a spherical wavefront?

chinglu1998

2011-Mar-10, 02:55 AM

Einstein did not use a point, he used the equation of a WAVEFRONT.

Q10: what is the equation of a spherical wavefront?

Nope If (x, y, z) be a point that is what Einstein did.

If I am wrong,feel free to show everyone and they will laugh at me.

Now, I will teach you, any proof of a spherical wavefront requires the derivative.

chinglu1998

2011-Mar-10, 02:57 AM

Einstein did not use a point, he used the equation of a WAVEFRONT.

Q10: what is the equation of a spherical wavefront?

Can you show me in his paper when he used this?

Then I can understand it.

Thanks.

macaw

2011-Mar-10, 02:58 AM

Nope If (x, y, z) be a point that is what Einstein did.

If I am wrong,feel free to show everyone and they will laugh at me.

Please answer the question you have been asked (Q10).

Now, I will teach you, any proof of a spherical wavefront requires the derivative.

Q11: What "derivative"?

macaw

2011-Mar-10, 03:00 AM

Can you show me in his paper when he used this?

Then I can understand it.

Thanks.

Paragraph 3. You even cited it multiple times but you are unable to understand it.

pzkpfw

2011-Mar-10, 03:16 AM

Nope If (x, y, z) be a point that is what Einstein did. ...

We've been over this, see posts #66 and #68 of this thread. Where x, y, z are used that way, they refer to all points on the sphere. You are twisiting Einstein to match your meaning.

(e.g. If we say Y = 2 * X, we may use X = 4 as an example (and show we get Y = 8), but X could be any value. We have not limited our formula to just one value of X.)

And now mod words:

... Now, I will teach you ...

You've been infracted before for that kind of talk. Knock it off.

Tensor

2011-Mar-10, 03:19 AM

We are LT mapping the light sphere in this thread. Therefore, x² + y² + z² = r² is correct for some fixed r.

You keep claiming this. Can you please show that for some fixed r, x² + y² + z² = r² will transform as a four vector as x2+y2+z2=ct2 does?

The math is right there, feel free. Math is either correct and incorrect. In this proof, it is correct.

The math is correct, the assumption the math is based on is incorrrect, which invalidates your argument. Using a fixed r does not allow x² + y² + z² = r² to transform as a space-time four vector, which is why you get an ellipsiod, instead of a sphere.

Your asking us to disprove your equation with r in it is tantamount to your assuming r is equal to p and thus, 3r = 3p. When in reality, r is equal to s, invalidating your whole equation.

Try looking here (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html) for some basic four vector information. Here (http://mathworld.wolfram.com/LorentzTransformation.html) is some information on Lorentz transformations using four vectors. Feel free to show where the mathematics of either of those two sites are wrong. And before you say anything about that's not what Einstein said, remember that Einstein's algebraic equations can be and were transfered into four vector notation by Minkowski. Those four vectors are required to obey the Lorentz symmetry, which Quantum Field Theory also depends on for it's precision. Don't you think it's possible that you're misunderstanding something, rather than over 100 years of physicist's missing something?

Geo Kaplan

2011-Mar-10, 03:20 AM

How do you prove a wavefront with one point as selected by Einstein.

Let's look again.

At the time t=t'=0, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then

x2+y2+z2=c2t2.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

You read, but do not understand, Yes, (x,y,z) is a point, but the equation clearly shows that the relationship holds for all (x,y,z). That's what the equals sign means.

This is all very elementary and conventional usage. Open up any math book, for example, and they'll say something like "let (x,y) be a point on a plane. Then the distance to the origin is the root-sum-square of x and y." The relationship holds for any and all (x,y).

You are getting hung up by a simple convention, and instead of questioning your own understanding (or lack thereof), you make a grand and wholly unsupported leap to "Therefore Einstein is wrong."

Dial down your confidence a bit, and consider this: Do you really believe that, after a century of scrutiny, no one else in the history of science ever found such an elementary mistake of Einstein? If you think you've identified such an obvious error in Einstein's work, you are almost certainly wrong.

Think about it.

Tobin Dax

2011-Mar-10, 03:35 AM

What?

TD1) What is the length of time (in the moving frame), t', that it takes for the light wave to propagate from (0,0,0) to (x',0,0)?

x'/c

TD2) How far does the wavefront move in the y direction (to point (0,y',0)) in that time? in the z' direction (to point (0,0,z'))?

y'=y, z'=z.

Did you read the standard configuration?

http://en.wikipedia.org/wiki/Lorentz_transformation#Lorentz_transformation_for_ frames_in_standard_configuration

z'=z, y'=y

TD3) What is the length of time in the rest frame, t, that is takes the light wave to propagate to (x,0,0)?

x/c

TD4) How does time t compare to the dilated time t'? How does y' compare to y? How does z' compare to z?

Insufficient information to answer.

I need x,y,z,t in the rest frame and then we simply apply LT.

You have enough information to answer question 4. We can use an equation which you have used previously: x' = x( 1 - v/c )γ. Since ( 1 - v/c )γ < 1, x' < x. As you correctly answered, t' = x'/c and t = x/c. That means that t' < t since x' < x. The wavefront has traveled for a shorter time in the moving frame than in the rest frame.

The speed of this wavefront is the speed of light, which is the same in all directions in all inertial frames of reference. That means that the wavefront expands at the same rate (c) in both frames. Since the radius of the wavefront then depends only on the travel time, and the travel time in the moving frame is less than the time in the rest frame, all points on the entire spherical wavefront are closer to the origin in the moving frame than in the rest frame.

Your Lorentz transform doesn't show this because it only transforms the spatial coordinates of the points along the direction of motion of the frame. The Lorentz transform does not account for the effect of time dilation on the photons in the directions perpendicular to the direction of motion of the moving frame. Time dilation is the key here, not length contraction, and that's what gives us the spherical wavefront in all frames.

The object considered here is not a solid sphere, but the position of a sphere-shaped wavefront expanding in time.

tusenfem

2011-Mar-10, 07:56 AM

Oh this is pointless, chinglu read up on the difference between a sphere (x,y,z,r) and a lightsphere (x,y,z,ct) and what the difference in transformation is (let me give you a hint, one has r, one has ct, one of the two will Lorentz transform).

Then study hard what everybody has been trying to explain to you. I get the feeling that you are willfully not understanding what is perfectly clear to anyone else.

After that, you can publish your findings on another board, because this is the same thing going round and round, for several threads.

Discussion closed.

No more.

tusenfem

2011-Mar-10, 08:05 AM

A discussion in moderator space came to the conclusion that you can have your full 30 days, but that's it.

So start working on a real proof.

Thread reopened.

chinglu1998

2011-Mar-10, 01:05 PM

Einstein did not use a point, he used the equation of a WAVEFRONT.

He may have used the EQUATION for a WAVEFRONT, but he selected one point in TIME. I then used all those points in that TIME and on the light sphere for that one point in TIME and those poiints satisfy x² + y² + z² = r².

Q10: what is the equation of a spherical wavefront?

This I have clearly answered. It is in this proof.

So, we will use as the domain the set of points that satisfy x² + y² + z² = r² and assume the standard configuration. Hence, x = ±√(r² - y² - z²).

In addition, x' must satisfy, x' = ( x - vr/c )γ.

x' = ( ±√(r² - y² - z²) - vr/c )γ

x'/γ + vr/c = ±√(r² - y² - z²)

(x'/γ + vr/c)² = (r² - y² - z²)

(x'/γ + vr/c)² + y² + z² = r²

(x'/γ - (-vr/c))²/r² + y²/r² + z²/r² = 1

(x' - (-vrγ/c))² / (r²γ²) + y²/r² + z²/r² = 1

Like I said, the LT translated object is an x-axis elongate ellipsoid.

chinglu1998

2011-Mar-10, 01:12 PM

Please answer the question you have been asked (Q10).

Q11: What "derivative"?

∂x/∂t = ± c²t /( c²t² - y² - z² )

∂y/∂t = ± c²t /( c²t² - x² - z² )

∂z/∂t = ± c²t /( c²t² - x² - y² )

∂t/∂x = x / (c √(x² + y² + z²) )

∂t/∂y = y / (c √(x² + y² + z²) )

∂t/∂z = z / (c √(x² + y² + z²) )

Strange

2011-Mar-10, 01:14 PM

He may have used the EQUATION for a WAVEFRONT, but he selected one point in TIME.

Really? What "one point in time" was that then?

tusenfem

2011-Mar-10, 01:18 PM

He may have used the EQUATION for a WAVEFRONT, but he selected one point in TIME. I then used all those points in that TIME and on the light sphere for that one point in TIME and those poiints satisfy x² + y² + z² = r².

This I have clearly answered. It is in this proof.

So, we will use as the domain the set of points that satisfy x² + y² + z² = r² and assume the standard configuration. Hence, x = ±√(r² - y² - z²).

In addition, x' must satisfy, x' = ( x - vr/c )γ.

x' = ( ±√(r² - y² - z²) - vr/c )γ

x'/γ + vr/c = ±√(r² - y² - z²)

(x'/γ + vr/c)² = (r² - y² - z²)

(x'/γ + vr/c)² + y² + z² = r²

(x'/γ - (-vr/c))²/r² + y²/r² + z²/r² = 1

(x' - (-vrγ/c))² / (r²γ²) + y²/r² + z²/r² = 1

Like I said, the LT translated object is an x-axis elongate ellipsoid.

you keep on putting ct = r and thus looking at a sphere and not a spherical wave. This has been explained to you numerous times already! The Lorentz transformation works on the four vector (x,y,z,t) [and use whatever metric you want] thus you also MUST transform t to t', otherwise you are just doing a botched up Galilean transformation.

chinglu1998

2011-Mar-10, 01:18 PM

We've been over this, see posts #66 and #68 of this thread. Where x, y, z are used that way, they refer to all points on the sphere. You are twisiting Einstein to match your meaning.

(e.g. If we say Y = 2 * X, we may use X = 4 as an example (and show we get Y = 8), but X could be any value. We have not limited our formula to just one value of X.)

I thought SR was about space-TIME. Einstein picked one point in TIME and then one arbitrary point on that light sphere.

Given the relativity of simultaneity, you cannot decide anything about the moving frame given one point in time in the rest frame.

Simultaneity in the moving frame is not simultaneous in the rest frame. So how can you possibly claim the moving system is viewing a light sphere or spherical wave given one point in time in the rest frame. What is simultaneous in the rest frame is not simultaneous in the moving frame frame.

Therefore, it takes an interval of time in the rest system to describe simultaneity in the moving system.

tusenfem

2011-Mar-10, 01:20 PM

Really? What "one point in time" was that then?

I guess the "point" was (x,y,z)

and the "time" was t

chinglu1998

2011-Mar-10, 01:56 PM

You keep claiming this. Can you please show that for some fixed r, x² + y² + z² = r² will transform as a four vector as x2+y2+z2=ct2 does?

If you like we can rewrite the equation as

x' - (-vtγ))² / ((ct)²γ²) + y²/(ct)² + z²/(ct)² = 1

For any given t, this is what that point in time in the rest frame translates to in the moving system in terms of the light sphere.

Then, to calculate t' for each point, simply use

t' = √(x'² + y² + z²)/c

Your asking us to disprove your equation with r in it is tantamount to your assuming r is equal to p and thus, 3r = 3p. When in reality, r is equal to s, invalidating your whole equation.

r = ct. I am simply stating the obvious fact that there is a point in time in the rest frame. For example, Einstein picked a point "just attained by the spherical light wave". See, this is on point in time unless that point is attained by the light save more than one.

So, it is a valid assumtion and valid math to pick one point in time in one of the frames.

Try looking here (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html) for some basic four vector information. Here (http://mathworld.wolfram.com/LorentzTransformation.html) is some information on Lorentz transformations using four vectors. Feel free to show where the mathematics of either of those two sites are wrong. And before you say anything about that's not what Einstein said, remember that Einstein's algebraic equations can be and were transfered into four vector notation by Minkowski. Those four vectors are required to obey the Lorentz symmetry, which Quantum Field Theory also depends on for it's precision. Don't you think it's possible that you're misunderstanding something, rather than over 100 years of physicist's missing something?

Hopefully, you agree with a 4 vector, I can look at some point in time say t1 and select all the points (t1,x,y,z) based on the light sphere. That is all I did and this is valid math and consistent with SR.

You will note, by derivation, my equation is consistent with the invariance of the light like space time interval and hence obeys the rules of the Minkowski space-time.

chinglu1998

2011-Mar-10, 02:01 PM

You read, but do not understand, Yes, (x,y,z) is a point, but the equation clearly shows that the relationship holds for all (x,y,z). That's what the equals sign means.

This is all very elementary and conventional usage. Open up any math book, for example, and they'll say something like "let (x,y) be a point on a plane. Then the distance to the origin is the root-sum-square of x and y." The relationship holds for any and all (x,y).

You are getting hung up by a simple convention, and instead of questioning your own understanding (or lack thereof), you make a grand and wholly unsupported leap to "Therefore Einstein is wrong."

Dial down your confidence a bit, and consider this: Do you really believe that, after a century of scrutiny, no one else in the history of science ever found such an elementary mistake of Einstein? If you think you've identified such an obvious error in Einstein's work, you are almost certainly wrong.

Think about it.

Excellent, then you have no choice but to agree that it is consistent to select all (x,y,z) on the light sphere at some point in time in the rest frame. That is all I did.

So, if it is logical to select any arbitrary single (x,y,z) just attained by the light sphere, then it is logical to select all (x,y,z) just attained the light sphere and map all those points instead of just one.

chinglu1998

2011-Mar-10, 02:09 PM

You have enough information to answer question 4. We can use an equation which you have used previously: x' = x( 1 - v/c )γ. Since ( 1 - v/c )γ < 1, x' < x. As you correctly answered, t' = x'/c and t = x/c. That means that t' < t since x' < x. The wavefront has traveled for a shorter time in the moving frame than in the rest frame.

The speed of this wavefront is the speed of light, which is the same in all directions in all inertial frames of reference. That means that the wavefront expands at the same rate (c) in both frames. Since the radius of the wavefront then depends only on the travel time, and the travel time in the moving frame is less than the time in the rest frame, all points on the entire spherical wavefront are closer to the origin in the moving frame than in the rest frame.

Your Lorentz transform doesn't show this because it only transforms the spatial coordinates of the points along the direction of motion of the frame. The Lorentz transform does not account for the effect of time dilation on the photons in the directions perpendicular to the direction of motion of the moving frame. Time dilation is the key here, not length contraction, and that's what gives us the spherical wavefront in all frames.

The object considered here is not a solid sphere, but the position of a sphere-shaped wavefront expanding in time.

Your basic premise above is there is no specific point in time in the rest frame. That is false.

You also made the false claim that t' < t for all cases.

Let t be some time in the rest frame say t1.

Then, (-ct1,0,0 ) is on the light sphere.

Translate to t'.

t' = ( t - vx/c² )γ

t' = ( t1 + vt1/c )γ

t' = t1( 1 + v/c )γ

Since ( 1 + v/c )γ > 1, then t' > t1.

The Lorentz transform does not account for the effect of time dilation on the photons in the directions perpendicular to the direction of motion of the moving frame

I cannot translate this into anything that makes sense to me.

chinglu1998

2011-Mar-10, 02:12 PM

Really? What "one point in time" was that then?

That is easy. That point is just attained by the light sphere.

That would be t = √(x² + y² + z²)/c

chinglu1998

2011-Mar-10, 02:17 PM

you keep on putting ct = r and thus looking at a sphere and not a spherical wave. This has been explained to you numerous times already! The Lorentz transformation works on the four vector (x,y,z,t) [and use whatever metric you want] thus you also MUST transform t to t', otherwise you are just doing a botched up Galilean transformation.

A wave is a a continuous interval of r.

That is nothing mathematically wrong with looking at the set of points in the 4 vectors as all (tf,x,y,z) where tf is some fixed time subject to the constraint x² + y² + z² = c²tf²

Then, to view of entire wavefront, r=ctf increases.

But, for each of these fixed tf or r =ctf, this translate to an x-axis elongated ellipsoid.

This is all valid math.

macaw

2011-Mar-10, 02:23 PM

∂x/∂t = ± c²t /( c²t² - y² - z² )

∂y/∂t = ± c²t /( c²t² - x² - z² )

∂z/∂t = ± c²t /( c²t² - x² - y² )

∂t/∂x = x / (c √(x² + y² + z²) )

∂t/∂y = y / (c √(x² + y² + z²) )

∂t/∂z = z / (c √(x² + y² + z²) )

None of the above is correct. Moreover, none of them is used in the Einstein paper (for good reason, since they are incorrect).

Q12: Where in Einstein paper did you see these "derivatives"?

macaw

2011-Mar-10, 02:28 PM

He may have used the EQUATION for a WAVEFRONT, but he selected one point in TIME.

Q13: Where exactly in the paper did "he" "select a point"?

I then used all those points in that TIME and on the light sphere for that one point in TIME and those poiints satisfy x² + y² + z² = r².

Q14: Where do you see any "r" in the Einstein paper?

This I have clearly answered. It is in this proof.

So, we will use as the domain the set of points that satisfy x² + y² + z² = r²

x² + y² + z² = r² is not the equation of a wavefront. This has been explained to you multiple times. This is why your attempt to rubbish the Einstein paper falls flat on its face.

chinglu1998

2011-Mar-10, 02:49 PM

None of the above is correct. Moreover, none of them is used in the Einstein paper (for good reason, since they are incorrect).

Q12: Where in Einstein paper did you see these "derivatives"?

The derivatives are all correct.

And, I do not see derivatives in the paper. I should. To prove a spherical wavefront in the primed frame, all derivatives for the primed frame based on the spherical wavefront in the rest frame would have to be constant in all directions.

macaw

2011-Mar-10, 02:55 PM

The derivatives are all correct.

None of them is correct.

And, I do not see derivatives in the paper.

In an earlier post you claimed you did, this is why I asked you.

I should. To prove a spherical wavefront in the primed frame, all derivatives for the primed frame based on the spherical wavefront in the rest frame would have to be constant in all directions.

Q15: Prove it. Before you do, you may want to get your "derivatives" right.

chinglu1998

2011-Mar-10, 02:56 PM

Q13: Where exactly in the paper did "he" "select a point"?

....At the time , when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then....

http://www.fourmilab.ch/etexts/einstein/specrel/www/

Q14: Where do you see any "r" in the Einstein paper?

I do not.

x² + y² + z² = r² is not the equation of a wavefront. This has been explained to you multiple times. This is why your attempt to rubbish the Einstein paper falls flat on its face.

This is the same strategy as the line cone. The various t's define the expanding light sphere. I only introduced t to make it easier to understand. I do not need it.

Translating the light sphere from the rest frame.

(x' - (-vtγ))² / ((ct)²γ²) + y²/(ct)² + z²/(ct)² = 1

y'=y

z'=z

t' = √(x'² + y² + z²)/c

Then, the light sphere expands in the rest frame as normal. Let t be any point in time, then the equations above translate the entire light sphere for that t point in time. And, t is arbitrary and increasing.

chinglu1998

2011-Mar-10, 03:05 PM

In an earlier post you claimed you did, this is why I asked you.

That would be a translation problem possible on my part. I intended to mean I should have seen the derivative.

And the derivatives are correct. Here is an example, ∂x/∂t. I had to substitute x for t because that is the way the system works at least as far as I can tell.

http://www.wolframalpha.com/input/?i=sqrt%28+c%5E2x%5E2+-+y%5E2+-+z%5E2+%29

Then here is ∂t/∂x

http://www.wolframalpha.com/input/?i=sqrt%28+x%5E2+%2B+y%5E2+%2B+z%5E2+%29%2Fc

Q15: Prove it. Before you do, you may want to get your "derivatives" right.

Once you understand the derivatives are correct, I will prove it.

grav

2011-Mar-10, 03:42 PM

Like I said, the LT translated object is an x-axis elongate ellipsoid.That is true when only mapping out the distances for what the rest frame measures simultaneously at t, transforming the sphere that the rest frame views into an ellipsoid in the moving frame. But you understand that due to the relativity of simultaneity, the readings of the moving frame's clocks that coincide with those points vary with distance along the x axis as well, so we must also transform the times in the moving frame for each of the ellipsoid points with the moving frame's clocks reading greater times in the negative x direction according to the rest frame, producing a sphere when the moving frame maps out the paths of the photons simultaneously in its own frame, right?

Tensor

2011-Mar-10, 04:09 PM

If you like we can rewrite the equation as

x' - (-vtγ))² / ((ct)²γ²) + y²/(ct)² + z²/(ct)² = 1

For any given t, this is what that point in time in the rest frame translates to in the moving system in terms of the light sphere.

Then, to calculate t' for each point, simply use

t' = √(x'² + y² + z²)/c

How exactly did you get to the t' equation above? Are you claiming ct'2 = x'2 + y2 + z2? If it is, this is exactly where your assumption, and thus your proof, is wrong. While you have the r' correct (x' = ( x - vr/c )γ in post 149 to macaw), ct' also has to satisfy the following condition: ct' = (ct-vx/c)γ (in your example, r' = (r-vx/c)γ).

r = ct. I am simply stating the obvious fact that there is a point in time in the rest frame.

You can make any substitution you want. But your substitution does not produce a four vector as your interval is not invariant (as demonstrated by the difference in the two shapes). You have to transform ct, and you don't, it's that simple.

For example, Einstein picked a point "just attained by the spherical light wave". See, this is on point in time unless that point is attained by the light save more than one. So, it is a valid assumtion and valid math to pick one point in time in one of the frames.

You can pick any point in time you want, but you still have to transform it as ct would transform to produce a valid equation, and you don't. As a result, although the math of the equation you provided is correct, the actual equation is wrong, because the necessary transformation has not been done.

Hopefully, you agree with a 4 vector, I can look at some point in time say t1 and select all the points (t1,x,y,z) based on the light sphere. That is all I did and this is valid math and consistent with SR.

No, it's not. You didn't transform ct. All you did was set r = to ct. No where in your equations in post 149 do you have the transformations the provide you with the needed equations for ct'.

You will note, by derivation, my equation is consistent with the invariance of the light like space time interval and hence obeys the rules of the Minkowski space-time.

Again, not its not consistent. See here (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html#c2) specifically the space-time transformation. Note the requirements for ct to transform into ct'. Something you didn't do and is not in your equations.

macaw

2011-Mar-10, 05:17 PM

....At the time , when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then....

http://www.fourmilab.ch/etexts/einstein/specrel/www/

Do you understand that he means an ARBITRARY point? Generations of people over the past 106 years understood that.

I do not.

Q16: Then why did you claim earlier that Einstein did?

This is the same strategy as the line cone.

Q17: What do your incorrect "derivatives" have to do with light cones?

Translating the light sphere from the rest frame.

(x' - (-vtγ))² / ((ct)²γ²) + y²/(ct)² + z²/(ct)² = 1

y'=y

z'=z

t' = √(x'² + y² + z²)/c

This is another incorrect attempt.

Geo Kaplan

2011-Mar-10, 06:17 PM

Excellent, then you have no choice but to agree that it is consistent to select all (x,y,z) on the light sphere at some point in time in the rest frame. That is all I did.

So, if it is logical to select any arbitrary single (x,y,z) just attained by the light sphere, then it is logical to select all (x,y,z) just attained the light sphere and map all those points instead of just one.

Excellent. So we are making progress. You have now withdrawn your erroneous claim that the original equation in your first post does not describe a spherical wavefront. That is good.

In the meantime, you have changed your argument quite a bit. Shifting targets is very bad form, but we are where we are. You have moved from declaring that Einstein's equation does not describe a spherical wavefront to now declaring that SR fails to preserve the spherical symmetry. I point out that you have utterly failed to show anything of the kind, and that numerous people -- most recently Tensor -- have painstakingly identified your error. In short, if you correctly apply SR, you get -- surprise! -- the right answer.

Fortis

2011-Mar-10, 07:16 PM

chinglu1998, I'm coming a little late to this thread but,

Do you agree that the Lorentz Transform preserves the interval ds2 = dx2+dy2+dz2-c2.dt2 = dx'2+dy'2+dz'2-c2.dt'2 ?

chinglu1998

2011-Mar-10, 11:59 PM

Proof there exists spherical light waves in the rest frame that are not spherical light waves in the moving frame.

As with Einstein, I chose to pick a point just attained by the light sphere. I choose this point to be (-xp,0,0) where xp > 0.

I am going to prove, on the rest frame time interval [ xp/c, (xp(c+v))/(c(c-v)) ) not one single light sphere is built by LT for the moving frame.

In the view of the rest frame, clearly, we have a spherical wave because xp(c+v))/(c(c-v)) > xp/c. Hence, in the view of the rest frame, the spherical wave translates to a light cone or a concentric set of light spheres.

Next tp' is calculated with LT.

tp' = ( t - vx/c² )γ = (xp/c + vxp/c²)γ

A light sphere of radius ctp' then in the moving frame includes all the points such that

x'² + y² + z² = c² tp'²

It is now proven, one point is missing from the moving frame light sphere of radius ctp' based on the rest frame time interval [ xp/c, (xp(c+v))/(c(c-v)) ).

So, we have t < (xp(c+v))/(c(c-v)) and assume y=z=0, hence x = ct.

When z=y=0,

t' = ( t - vx/c² )γ = ( t - v(ct)/c² )γ = ( t - vt/c )γ = t ( 1 - v/c )γ

By assumption

t < (xp(c+v))/(c(c-v))

Multiply both sides by ( 1 - v/c )γ

t' = t ( 1 - v/c )γ < (xp(c+v))/(c(c-v))( 1 - v/c )γ

t' < (xp(c+v))/(c(c-v))( (c - v)/c )γ

t' < ( xp(c+v))/c² )γ = ( xp/c + vxp/c² )γ = tp'

Hence, t' < tp' .

Hence, a point was found that is not yet on the light sphere of radius ctp' in the moving frame even though the rest frame elapsed time on the interval [ xp/c, (xp(c+v))/(c(c-v)) ).

So, obviously, the rest frame time interval proves a spherical light wave.

However, as proven above, not one light sphere of radius ctp' was completed by LT. Obviously, a larger radius light sphere in the moving frame could not have been built. Hence, on the rest frame time interval, [ xp/c, (xp(c+v))/(c(c-v)) ) the moving frame did not view even one complete light sphere. Hence, the moving could not view a spherical wave front.

Hence, the rest frame viewed a spherical wave front on the time interval [ xp/c, (xp(c+v))/(c(c-v)) ) but the moving frame did not.

Therefore, this proves a case in which the rest frame viewed a spherical wave front but the moving frame did not contradicting Einstein's conclusion that the moving frame always views a spherical wave front based on the spherical light wave in the rest frame and calculations of LT.

chinglu1998

2011-Mar-11, 12:00 AM

chinglu1998, I'm coming a little late to this thread but,

Do you agree that the Lorentz Transform preserves the interval ds2 = dx2+dy2+dz2-c2.dt2 = dx'2+dy'2+dz'2-c2.dt'2 ?

Absolutely, I depend on this.

chinglu1998

2011-Mar-11, 12:02 AM

That is true when only mapping out the distances for what the rest frame measures simultaneously at t, transforming the sphere that the rest frame views into an ellipsoid in the moving frame. But you understand that due to the relativity of simultaneity, the readings of the moving frame's clocks that coincide with those points vary with distance along the x axis as well, so we must also transform the times in the moving frame for each of the ellipsoid points with the moving frame's clocks reading greater times in the negative x direction according to the rest frame, producing a sphere when the moving frame maps out the paths of the photons simultaneously in its own frame, right?

I completely agree with this.

And, what is simultaneous (the light sphere) to the rest frame will not be simultaneous in the moving frame. Yes, there will be different primed frame clocks times at thoses positions.

chinglu1998

2011-Mar-11, 12:18 AM

How exactly did you get to the t' equation above? Are you claiming ct'2 = x'2 + y2 + z2? If it is, this is exactly where your assumption, and thus your proof, is wrong. While you have the r' correct (x' = ( x - vr/c )γ in post 149 to macaw), ct' also has to satisfy the following condition: ct' = (ct-vx/c)γ (in your example, r' = (r-vx/c)γ).

You can make any substitution you want. But your substitution does not produce a four vector as your interval is not invariant (as demonstrated by the difference in the two shapes). You have to transform ct, and you don't, it's that simple.

You can pick any point in time you want, but you still have to transform it as ct would transform to produce a valid equation, and you don't. As a result, although the math of the equation you provided is correct, the actual equation is wrong, because the necessary transformation has not been done.

No, it's not. You didn't transform ct. All you did was set r = to ct. No where in your equations in post 149 do you have the transformations the provide you with the needed equations for ct'.

Again, not its not consistent. See here (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html#c2) specifically the space-time transformation. Note the requirements for ct to transform into ct'. Something you didn't do and is not in your equations.

this whole post is wrong and you have no valid math objections to the math proof. If you neeed to see ct, fine, it makes no mathematical difference. Here it is with ct.

So, we will use as the domain the set of points that satisfy x² + y² + z² = (ct)² and assume the standard configuration. Hence, x = ±√((ct)² - y² - z²).

In addition, x' must satisfy, x' = ( x - vt )γ.

x' = ( ±√((ct)² - y² - z²) - vt )γ

x'/γ + vt = ±√((ct)² - y² - z²)

(x'/γ + vt)² = ((ct)² - y² - z²)

(x'/γ + vt)² + y² + z² = (ct)²

(x'/γ - (-vt))²/(ct)² + y²/(ct)² + z²/(ct)² = 1

(x' - (-vtγ))² / ((ct)²γ²) + y²/(ct)² + z²/(ct)² = 1

Now, the objective is to translate one light sphere to see what it looks like. Hence, ct is a constant and thus, the figure is am ellipsoid.

The invariance of the light like space time interval is preserved since LT was used.

korjik

2011-Mar-11, 12:20 AM

I completely agree with this.

And, what is simultaneous (the light sphere) to the rest frame will not be simultaneous in the moving frame. Yes, there will be different primed frame clocks times at thoses positions.

You do realize you just disagreed with everything you said in this entire thread? That to measure in the primed frame requres you to use the primed time? So the fact that the ellipsoid in primed position is irrelevant because its surface isnt at a constant primed time?

chinglu1998

2011-Mar-11, 12:23 AM

Do you understand that he means an ARBITRARY point? Generations of people over the past 106 years understood that.

I do. And all that was proven by the arbitrary point is that the light like space time interval is invariant and nothing more.

That means a light beam in the rest frame measures c if and only if the light beam measures c in the moving frame.

Q16: Then why did you claim earlier that Einstein did?

I said already, that was not my intention. If I made a mistake as I said before, then fine.

Q17: What do your incorrect "derivatives" have to do with light cones?

I showed you links the derivatives are correct. Do you think the online mathematica is wrong?

I provided a proof with a different way of looking at what I was saying. We can discuss that.

chinglu1998

2011-Mar-11, 12:26 AM

Excellent. So we are making progress. You have now withdrawn your erroneous claim that the original equation in your first post does not describe a spherical wavefront. That is good.

In the meantime, you have changed your argument quite a bit. Shifting targets is very bad form, but we are where we are. You have moved from declaring that Einstein's equation does not describe a spherical wavefront to now declaring that SR fails to preserve the spherical symmetry. I point out that you have utterly failed to show anything of the kind, and that numerous people -- most recently Tensor -- have painstakingly identified your error. In short, if you correctly apply SR, you get -- surprise! -- the right answer.

I am not shifting anything.

And there are no math errors in my proof. Tensor is mistaken.

If you have software that graphs objects, then you can prove for yourself a rest frame light sphere maps to an x-axis elongated ellipsoid in the moving frame.

Have you ever graphed it?

chinglu1998

2011-Mar-11, 12:33 AM

You do realize you just disagreed with everything you said in this entire thread? That to measure in the primed frame requres you to use the primed time? So the fact that the ellipsoid in primed position is irrelevant because its surface isnt at a constant primed time?

No I did not. I never said one time in the rest frame is one time in the moving frame. So you are wrong.

And if you look at my proof, I specifically wrote,

t' = √(x'² + y² + z²)/c

Because I am well aware that t' will vary all over the place to "view" the rest frame light sphere.

And the fact remains, a light sphere of one constant radius in the rest frame maps to an x-axis elongated ellipisoid in the "view" of the moving frame. It simply does not happen at one time in the moving frame time.

pzkpfw

2011-Mar-11, 12:33 AM

... If you have software that graphs objects, then you can prove for yourself a rest frame light sphere maps to an x-axis elongated ellipsoid in the moving frame. ...

But will an observer in the moving frame see a light sphere?

(Not the "same" light sphere, just a light sphere?)

chinglu1998

2011-Mar-11, 12:43 AM

But will an observer in the moving frame see a light sphere?

(Not the "same" light sphere, just a light sphere?)

Yes. An observer in the moving frame based on the light postulate views a spherical wavefront exactly like the rest frame with no differences and it is the same light sphere.

However, I am questioning LT and its mapping abilities in providing the correct view of the moving frame.

pzkpfw

2011-Mar-11, 01:10 AM

Yes. An observer in the moving frame based on the light postulate views a spherical wavefront exactly like the rest frame with no differences and it is the same light sphere.

However, I am questioning LT and its mapping abilities in providing the correct view of the moving frame.

How does that map to the claim you made when you began this thread?

Einstein made the following claim.

We now have to prove that any ray of light, measured in the moving system, is propagated with the velocity c, if, as we have assumed, this is the case in the stationary system; for we have not as yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity.At the time t=τ=0, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then

x² + y² + z² = c² t²

Transforming this equation with the aid of our equations of transformation we obtain after a simple calculation

ξ² + η² + ς² = c² τ²

The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system. This shows that our two fundamental principles are compatible.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

Yet, as anyone can see, cτ (Tau) is not a constant and thus not a light sphere. Therefore, Einstein's proof is false.

(My bolds.)

* Are you saying his conclusion was correct, but his proof of it was false?

chinglu1998

2011-Mar-11, 01:37 AM

How does that map to the claim you made when you began this thread?

(My bolds.)

* Are you saying his conclusion was correct, but his proof of it was false?

1) How does this reflect up against the OP.

In this proof

http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1861622#post1861622

Given c(Tau) is not a constant, this requires an interval of time in the rest frame to complete the light sphere of radius c(Tau). Hence, as I already knew, I could create intervals of time in which no light sphere of some radius could have been viewed in the moving frame.

Further, to show this problem, I posted the ellipsoid proof in which a light sphere of radius ct constant mapped to an ellipsoid further raising doubt on the Einstein proof.

2) I am saying his proof and conclusion are false. The moving frame based on LT can be configured to not even view a single light sphere of any radius over a time interval in the rest frame. Again, this is all based on LT calculations.

a) His proof is false because to consider a light wave, you cannot use a single arbitrary point, you must use a time interval since a light wave can only be created on an interval. So, his methodology is wrong.

b) Since the proof does not include an interval of time, the conclusions of the proof cannot be concluded. They are open questions until the proof I posted in

http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1861622#post1861622

then the conclusions are false.

Tobin Dax

2011-Mar-11, 02:51 AM

chinglu1998, have you read the part of the section of Einstein's paper preceding the part that you quoted in the OP? Einstein shows there that η≠y and ς≠z, rather η and ς depend on cτ. The distances to the wavefront in the y- and z-direction depend on the time passed in the relevant frame. Einstein sets this up first to support his conclusion that the wavefront in both frames is spherical.

macaw

2011-Mar-11, 02:57 AM

I do.

Good, because it means that all your previous claims are false.

chinglu1998

2011-Mar-11, 03:05 AM

Good, because it means that all your previous claims are false.

What?

Prove your case,I do.

macaw

2011-Mar-11, 03:08 AM

That would be a translation problem possible on my part. I intended to mean I should have seen the derivative.

Q18: You keep changing your tune. Why would you think that there should be any of your "derivatives" in the paper?

And the derivatives are correct. Here is an example, ∂x/∂t. I had to substitute x for t because that is the way the system works at least as far as I can tell.

http://www.wolframalpha.com/input/?i=sqrt%28+c%5E2x%5E2+-+y%5E2+-+z%5E2+%29

The Mathematica page disagrees with what you posted here (http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1861382#post1861382) in post 150.

Besides, you never managed to connect ANY of the derivatives with the OP.

chinglu1998

2011-Mar-11, 03:09 AM

chinglu1998, have you read the part of the section of Einstein's paper preceding the part that you quoted in the OP? Einstein shows there that η≠y and ς≠z, rather η and ς depend on cτ. The distances to the wavefront in the y- and z-direction depend on the time passed in the relevant frame. Einstein sets this up first to support his conclusion that the wavefront in both frames is spherical.

Yes, I am aware of these. However, he provided a more general case and also did not.

When the x-axis are the same for the frames, then y'=y and z'=z. This is called the standard configuration.

His example was of the standard configuration but yet the equations were of the more general setting leading some to think it is not the case y'=y and z'=z for his example.

chinglu1998

2011-Mar-11, 03:12 AM

The Mathematica page disagrees with what you posted here (http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1861382#post1861382) in post 150.

Besides, you never managed to connect ANY of the derivatives with the OP.

Show me where I am wrong with the derivative and not just say it. Many might think when folks say things they are automatically true.

Also, I provided an alternative and proof in

http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1861622#post1861622

.

This should be sufficient for you. I notice you have not addressed it. Is there a reason?

If a math proof is not addressed in the math world that means it is accepted.

macaw

2011-Mar-11, 03:22 AM

Show me where I am wrong with the derivative and not just say it.

I put the Mathematica page side by side with your post. Your error is obvious to anybody who knows calculus.

Also, I provided an alternative and proof in

http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1861622#post1861622

.

This should be sufficient for you. I notice you have not addressed it. Is there a reason?

If a math proof is not addressed in the math world that means it is accepted.

No, you did not provide any proof, so please do so.

macaw

2011-Mar-11, 03:25 AM

Proof there exists spherical light waves in the rest frame that are not spherical light waves in the moving frame.

As with Einstein, I chose to pick a point just attained by the light sphere. I choose this point to be (-xp,0,0) where xp > 0.

I am going to prove, on the rest frame time interval [ xp/c, (xp(c+v))/(c(c-v)) ) not one single light sphere is built by LT for the moving frame.

In the view of the rest frame, clearly, we have a spherical wave because xp(c+v))/(c(c-v)) > xp/c. Hence, in the view of the rest frame, the spherical wave translates to a light cone or a concentric set of light spheres.

Next tp' is calculated with LT.

tp' = ( t - vx/c² )γ = (xp/c + vxp/c²)γ

A light sphere of radius ctp' then in the moving frame includes all the points such that

x'² + y² + z² = c² tp'²

It is now proven, one point is missing from the moving frame light sphere of radius ctp' based on the rest frame time interval [ xp/c, (xp(c+v))/(c(c-v)) ).

So, we have t < (xp(c+v))/(c(c-v)) and assume y=z=0, hence x = ct.

When z=y=0,

t' = ( t - vx/c² )γ = ( t - v(ct)/c² )γ = ( t - vt/c )γ = t ( 1 - v/c )γ

By assumption

t < (xp(c+v))/(c(c-v))

Multiply both sides by ( 1 - v/c )γ

t' = t ( 1 - v/c )γ < (xp(c+v))/(c(c-v))( 1 - v/c )γ

t' < (xp(c+v))/(c(c-v))( (c - v)/c )γ

t' < ( xp(c+v))/c² )γ = ( xp/c + vxp/c² )γ = tp'

Hence, t' < tp' .

Hence, a point was found that is not yet on the light sphere of radius ctp' in the moving frame even though the rest frame elapsed time on the interval [ xp/c, (xp(c+v))/(c(c-v)) ).

So, obviously, the rest frame time interval proves a spherical light wave.

However, as proven above, not one light sphere of radius ctp' was completed by LT. Obviously, a larger radius light sphere in the moving frame could not have been built. Hence, on the rest frame time interval, [ xp/c, (xp(c+v))/(c(c-v)) ) the moving frame did not view even one complete light sphere. Hence, the moving could not view a spherical wave front.

Hence, the rest frame viewed a spherical wave front on the time interval [ xp/c, (xp(c+v))/(c(c-v)) ) but the moving frame did not.

Therefore, this proves a case in which the rest frame viewed a spherical wave front but the moving frame did not contradicting Einstein's conclusion that the moving frame always views a spherical wave front based on the spherical light wave in the rest frame and calculations of LT.

Aside from the fact that the above is incorrect you have no "derivatives" in it. So, your claim is false.

chinglu1998

2011-Mar-11, 03:31 AM

Aside from the fact that the above is incorrect you have no "derivatives" in it. So, your claim is false.

So, let's see you counter.

Otherwise, you accept the proof.

chinglu1998

2011-Mar-11, 03:33 AM

I put the Mathematica page side by side with your post. Your error is obvious to anybody who knows calculus.

Yea, I think I know calculus. Why not show everyone with proof rather than false accusations.

No, you did not provide any proof, so please do so.

I did provide the proof and link.

Swift

2011-Mar-11, 03:34 AM

This thread is locked while there is some discussion among the moderation team

Swift

2011-Mar-13, 03:09 AM

What?

Prove your case,I do.

chinglu1998,

After a lot of moderator discussion, we have decided to give you one more chance with this thread. I will tell you it was a difficult decision, and there was a lot of consideration to either permanently closing this thread, infracting you severely, or both.

If this discussion is to continue you must very carefully follow our rules. That means you must be polite, and that you may not demand that other people prove mainstream theories. It is solely up to you to prove your idea. If you fail to follow our rules, further actions will be taken.

Fortis

2011-Mar-13, 06:15 PM

Absolutely, I depend on this.

OK. So

ds2 = dx2+dy2+dz2-c2.dt2 = dx'2+dy'2+dz'2-c2.dt'2

is an invariant under the LT.

If we have a pulse of light emitted from (x0,y0,z0,t0) in the frame O, then the equation of the light-cone, i.e. the hypersurface carved out by the pulse of light in spacetime, is given by

dx2+dy2+dz2 = c2.dt2

where dx=x-x0, dy=y-y0, dz=z-z0, dt=t-t0.

Do you agree?

chinglu1998

2011-Mar-13, 09:27 PM

OK. So

ds2 = dx2+dy2+dz2-c2.dt2 = dx'2+dy'2+dz'2-c2.dt'2

is an invariant under the LT.

If we have a pulse of light emitted from (x0,y0,z0,t0) in the frame O, then the equation of the light-cone, i.e. the hypersurface carved out by the pulse of light in spacetime, is given by

dx2+dy2+dz2 = c2.dt2

where dx=x-x0, dy=y-y0, dz=z-z0, dt=t-t0.

Do you agree?

So, far so good.

Except, I am using the standard configuration, so (x0,y0,z0,t0) is (0,0,0,0).

chinglu1998

2011-Mar-16, 11:24 PM

I thought I would post this again because some may have missed it since it is a severe refutation of SR.

Proof there exists spherical light waves in the rest frame that are not spherical light waves in the moving frame.

As with Einstein, I chose to pick a point just attained by the light sphere. I choose this point to be (-xp,0,0) where xp > 0.

I am going to prove, on the rest frame time interval [ xp/c, (xp(c+v))/(c(c-v)) ) not one single light sphere is built by LT for the moving frame.

In the view of the rest frame, clearly, we have a spherical wave because xp(c+v))/(c(c-v)) > xp/c. Hence, in the view of the rest frame, the spherical wave translates to a light cone or a concentric set of light spheres.

Next tp' is calculated with LT.

tp' = ( t - vx/c² )γ = (xp/c + vxp/c²)γ

A light sphere of radius ctp' then in the moving frame includes all the points such that

x'² + y² + z² = c² tp'²

It is now proven, one point is missing from the moving frame light sphere of radius ctp' based on the rest frame time interval [ xp/c, (xp(c+v))/(c(c-v)) ).

So, we have t < (xp(c+v))/(c(c-v)) and assume y=z=0, hence x = ct.

When z=y=0,

t' = ( t - vx/c² )γ = ( t - v(ct)/c² )γ = ( t - vt/c )γ = t ( 1 - v/c )γ

By assumption

t < (xp(c+v))/(c(c-v))

Multiply both sides by ( 1 - v/c )γ

t' = t ( 1 - v/c )γ < (xp(c+v))/(c(c-v))( 1 - v/c )γ

t' < (xp(c+v))/(c(c-v))( (c - v)/c )γ

t' < ( xp(c+v))/c² )γ = ( xp/c + vxp/c² )γ = tp'

Hence, t' < tp' .

Hence, a point was found that is not yet on the light sphere of radius ctp' in the moving frame even though the rest frame elapsed time on the interval [ xp/c, (xp(c+v))/(c(c-v)) ).

So, obviously, the rest frame time interval proves a spherical light wave.

However, as proven above, not one light sphere of radius ctp' was completed by LT. Obviously, a larger radius light sphere in the moving frame could not have been built. Hence, on the rest frame time interval, [ xp/c, (xp(c+v))/(c(c-v)) ) the moving frame did not view even one complete light sphere. Hence, the moving could not view a spherical wave front.

Hence, the rest frame viewed a spherical wave front on the time interval [ xp/c, (xp(c+v))/(c(c-v)) ) but the moving frame did not.

Therefore, this proves a case in which the rest frame viewed a spherical wave front but the moving frame did not contradicting Einstein's conclusion that the moving frame always views a spherical wave front based on the spherical light wave in the rest frame and calculations of LT.

macaw

2011-Mar-16, 11:43 PM

I thought I would post this again because some may have missed it since it is a severe refutation of SR.

Proof there exists spherical light waves in the rest frame that are not spherical light waves in the moving frame.

As with Einstein, I chose to pick a point just attained by the light sphere. I choose this point to be (-xp,0,0) where xp > 0.

I am going to prove, on the rest frame time interval [ xp/c, (xp(c+v))/(c(c-v)) ) not one single light sphere is built by LT for the moving frame.

In the view of the rest frame, clearly, we have a spherical wave because xp(c+v))/(c(c-v)) > xp/c. Hence, in the view of the rest frame, the spherical wave translates to a light cone or a concentric set of light spheres.

Next tp' is calculated with LT.

tp' = ( t - vx/c² )γ = (xp/c + vxp/c²)γ

A light sphere of radius ctp' then in the moving frame includes all the points such that

x'² + y² + z² = c² tp'²

It is now proven, one point is missing from the moving frame light sphere of radius ctp' based on the rest frame time interval [ xp/c, (xp(c+v))/(c(c-v)) ).

No, all you have showed is that xp/c<t<xp/c(c+v)/(c-v) transforms into xp'/c<t'<xp'/c(c+v)/(c-v) .

You have been shown several correct demonstrations that a spherical wavefront transforms into a spherical wavefront.

When do you plan to answer my questions?

chinglu1998

2011-Mar-16, 11:54 PM

No, all you have showed is that xp/c<t<xp/c(c+v)/(c-v) transforms into xp'/c<t'<xp'/c(c+v)/(c-v) .

You have been shown several correct demonstrations that a spherical wavefront transforms into a spherical wavefront.

When do you plan to answer my questions?

You made this assertion,

No, all you have showed is that xp/c<t<xp/c(c+v)/(c-v) transforms into xp'/c<t'<xp'/c(c+v)/(c-v) .

Can you explain this? I do not get these results universally as you have claimed. This implies both frames will always agree on the order of events. That means the relativity of simultaneity is false. That would be something if you could prove this.

Also, you wrote,

You have been shown several correct demonstrations that a spherical wavefront transforms into a spherical wavefront.

Can you show me this proof?

I feel pretty good I can refute this.

macaw

2011-Mar-17, 12:03 AM

You made this assertion,

No, all you have showed is that xp/c<t<xp/c(c+v)/(c-v) transforms into xp'/c<t'<xp'/c(c+v)/(c-v) .

Can you explain this?

I simply used your math.

This implies both frames will always agree on the order of events.

Yes, SR preserves causality.

That means the relativity of simultaneity is false.

Nope, relativity of simultaneity is a different issue and does not contradict causality.

Also, you wrote,

You have been shown several correct demonstrations that a spherical wavefront transforms into a spherical wavefront.

Can you show me this proof?

You can look them up in this thread. Or, you can look up Einstein's correct proof from 106 years ago.

I feel pretty good I can refute this.

You've tried , yet you failed repeatedly.

chinglu1998

2011-Mar-17, 12:11 AM

I simply used your math.

Not, my math does not do this.

Don't run from your own statements and try to attach their absurdities to me.

In short, retract your false statement.

Yes, SR preserves causality.

good.

Nope, relativity of simultaneity is a different issue and does not contradict causality.

What does this have to do with this proof?

You can look them up in this thread. Or, you can look up Einstein's correct proof from 106 years ago.

Yes, well the subject now in this thread is the proof above.

So, do you plan to mathematically agree or disagree with the proof.

macaw

2011-Mar-17, 12:19 AM

Not, my math does not do this.

This means that you don't understand your own formulas?

What does this have to do with this proof?

Nothing, it corrects your expressed misconception that by retaining order of events somehow it disproves relativity of simultaneity.

Yes, well the subject now in this thread is the proof above.

You asked for proofs disproving your claims, I simply gave them to you.

So, do you plan to mathematically agree or disagree with the proof.

The math is fine, it is your conclusions that are incorrect.

chinglu1998

2011-Mar-17, 12:36 AM

The math is fine, it is your conclusions that are incorrect.

I understand.

With math, conclusions are true or false.

So, do you plan to support your position mathematically that the conclusions are false?

Or, do you plan to retract your false statement?

korjik

2011-Mar-17, 12:39 AM

Not, my math does not do this.

Don't run from your own statements and try to attach their absurdities to me.

In short, retract your false statement.

good.

What does this have to do with this proof?

Yes, well the subject now in this thread is the proof above.

So, do you plan to mathematically agree or disagree with the proof.

Your math is wrong, and has been since the OP. When you transform the rest frame sphere into the moving frame, you get an ellipsoid. The different points on the ellipsoid are at different t' in the moving frame's time. Transforming the time also gives you back a sphere, but it is a different sphere than the original.

chinglu1998

2011-Mar-17, 12:45 AM

Your math is wrong, and has been since the OP. When you transform the rest frame sphere into the moving frame, you get an ellipsoid. The different points on the ellipsoid are at different t' in the moving frame's time. Transforming the time also gives you back a sphere, but it is a different sphere than the original.

I am not sure I understand.

You transform a light sphere to an ellipsoid in the moving frame.

Then you transform this back to the rest frame? You will get the same light sphere in the rest frame.

"Transforming the time also gives you back a sphere, but it is a different sphere than the original."

Are you asserting there are multiple light spheres? Wow!

Anyway, to your original logic, yes it is the case the ellipsoid for the primed frame has different t'.

I have said this over and over.

It is still not a sphere.

macaw

2011-Mar-17, 02:39 AM

I understand.

With math, conclusions are true or false.

So, do you plan to support your position mathematically that the conclusions are false?

I have already done so by correcting every mistake that you have made in this thread and by giving you the correct solution.

chinglu,

I see that you want to find an error in relativity, you have started several threads and all of them boiled down to your misunderstandings about SR. Let me give you a piece of advice, the only way you can falsify a theory is through an experiment that contradicts the theory. So, making up ATMs will never manage to falsify SR (or GR).

Tensor

2011-Mar-17, 03:29 AM

I am not sure I understand.

You transform a light sphere to an ellipsoid in the moving frame.

Then you transform this back to the rest frame? You will get the same light sphere in the rest frame.

No, not the rest frame, the moving frame. You have still not transformed the cdt' coordinate yet, have you? You know how I know? You don't have gamma anywhere, where you transform cdt'. You have to transform both the x' and the cdt' when you redraw your moving sphere. Something you still haven't done.

"Transforming the time also gives you back a sphere, but it is a different sphere than the original."

Are you asserting there are multiple light spheres? Wow!

Yes, there is a light sphere in the x,y,z,t frame and a different sphere in the x',y',z',t' frame. What is so hard about that? Other than the fact that you have yet to transform cdt' when you transform x'. Transform cdt' when you transform x', BEFORE you redraw your sphere, then tell us what you get.

Anyway, to your original logic, yes it is the case the ellipsoid for the primed frame has different t'.

I have said this over and over.

It is still not a sphere.

And you have been wrong, over and over, because you haven't done a proper transformation of t'. You have been correct in stating that x' has to satisfy the condition x' = ( x - vr/c )γ. But, no where in your proofs have you stated that the cdt' has to statisfy the condition cdt' = (cdt-vx/c)γ (or in the example you keep using r' = (r-vx/c)γ) ) Can you point out anywhere in your proofs where the condition cdt' = (cdt-vx/c)γ (or r' = (r-vx/c)γ) as you like to say) is a stated condition for cdt' or shows up in your proofs?

korjik

2011-Mar-17, 03:54 AM

I am not sure I understand.

You transform a light sphere to an ellipsoid in the moving frame.

Then you transform this back to the rest frame? You will get the same light sphere in the rest frame.

"Transforming the time also gives you back a sphere, but it is a different sphere than the original."

Are you asserting there are multiple light spheres? Wow!

Anyway, to your original logic, yes it is the case the ellipsoid for the primed frame has different t'.

I have said this over and over.

It is still not a sphere.

You still dont get that the sphere from the primed frame is an ellipsoid where different parts of the ellipsoid are at different t's in the t' frame. When you force the time to dilate also, then the sphere properly transforms into the sphere in the primed frame.

Strange

2011-Mar-17, 07:59 AM

I am not sure I understand.

We are sure you don't :)

Are you asserting there are multiple light spheres? Wow!

No. The spherical wavefront seen in the moving frame is not the same spherical wavefront seen in the stationary frame. But they are both spherical. Obviously.

tusenfem

2011-Mar-17, 08:22 AM

chinglu, please read carefully the comments that are given to you about the Lorentz transformation (LT) of the light sphere that you are looking at and try to understand them.

Transforming a sphere from frame R to moving frame M will result in an ellipsoid, this is just galilean transformation.

However, the light sphere has also time component to it (see comments by Tensor and Strange).

Do not forget that the Lorentz transformation works on a four vector and that the light sphere is impilictly dependent on time.

Fortis

2011-Mar-17, 06:40 PM

So, far so good.

Except, I am using the standard configuration, so (x0,y0,z0,t0) is (0,0,0,0).

Fair enough. So, do you agree that the hypersurface defined by

dx2+dy2+dz2 = c2.dt2

describes a spherical surface expanding from the point with (x0,dy,dz) at time t0 at the speed of light?

chinglu1998

2011-Mar-17, 07:27 PM

OK, I think I am following everyone's point here and it looks like everyone is making the same point.

If this excludes any of your questions, please let me know.

When you use 4-D space-time, you will get a distance from the origin in that space and a clock value at that coordinate.

When you take that distance / the clock time at that position (t'), you get c everytime.

Since, every light beam measures c in all directions, that would mean the mapped light wave is spherical.

Does this sum up all of your views correctly?

Please feel free to add to this in any way you like. I want to make sure I am understanding all of your positions.

Thanks.

chinglu1998

2011-Mar-17, 07:30 PM

Fair enough. So, do you agree that the hypersurface defined by

dx2+dy2+dz2 = c2.dt2

describes a spherical surface expanding from the point with (x0,dy,dz) at time t0 at the speed of light?

Many apologies.

What does this mean? (x0,dy,dz)

Is x0 some constant and dx and dz variables?

Fortis

2011-Mar-17, 07:49 PM

Many apologies.

What does this mean? (x0,dy,dz)

Is x0 some constant and dx and dz variables?

I'm afraid it was a typo. Should have said (x0,y0,z0). The meaning is as per my earlier post.

grav

2011-Mar-17, 09:07 PM

OK, I think I am following everyone's point here and it looks like everyone is making the same point.

If this excludes any of your questions, please let me know.

When you use 4-D space-time, you will get a distance from the origin in that space and a clock value at that coordinate.

When you take that distance / the clock time at that position (t'), you get c everytime.

Since, every light beam measures c in all directions, that would mean the mapped light wave is spherical.

Does this sum up all of your views correctly?

Please feel free to add to this in any way you like. I want to make sure I am understanding all of your positions.

Thanks.Yes, yes. :) That's it exactly.

korjik

2011-Mar-17, 09:23 PM

OK, I think I am following everyone's point here and it looks like everyone is making the same point.

If this excludes any of your questions, please let me know.

When you use 4-D space-time, you will get a distance from the origin in that space and a clock value at that coordinate.

When you take that distance / the clock time at that position (t'), you get c everytime.

Since, every light beam measures c in all directions, that would mean the mapped light wave is spherical.

Does this sum up all of your views correctly?

Please feel free to add to this in any way you like. I want to make sure I am understanding all of your positions.

Thanks.

Not really. If you have a spherical wave emitted at the speed of light, and after a time t measure the sphere, then transform (both time and position) into a moving frame, you end up with a smaller sphere.

What you keep missing is that in the primed frame, the the unprimed sphere is position trandformed into an ellipsoid, but that surface if not all at the same propagation time t' in the primed frame. When you also transform the time, so that you are actually looking at all parts of the spherical wave at the same t', the wave propagating in the primed frame transforms back into a sphere.

chinglu1998

2011-Mar-17, 10:12 PM

I'm afraid it was a typo. Should have said (x0,y0,z0). The meaning is as per my earlier post.

Yes, I agree then.

chinglu1998

2011-Mar-17, 10:18 PM

Not really. If you have a spherical wave emitted at the speed of light, and after a time t measure the sphere, then transform (both time and position) into a moving frame, you end up with a smaller sphere.

What you keep missing is that in the primed frame, the the unprimed sphere is position trandformed into an ellipsoid, but that surface if not all at the same propagation time t' in the primed frame. When you also transform the time, so that you are actually looking at all parts of the spherical wave at the same t', the wave propagating in the primed frame transforms back into a sphere.

I thought I said this. But, this is what I meant.

Except, to conclude you have a spherical wave, you take the LT primed distance / LT t' = c.

You are basing your conclusion that you have a spherical wave on the fact each beam measures c and not on any distance measurements because these alone construct an ellpsoid.

When you factor the associated t', you get a constant c.

So, to summarize, everyone is basing their conclusion that the moving frame is viewing a spherical wavefront solely because each light beam transformed by LT measures c in any direction. Since light is c in every direction, it must be a spherical wavefront.

Is this what you are thinking?

Fortis

2011-Mar-17, 10:34 PM

Yes, I agree then.

So, in the un-primed frame, for an expanding spherical shell expanding at c we find that it is described by

dx2+dy2+dz2 = c2.dt2

and, a hypersurface described by

dx2+dy2+dz2 = c2.dt2

corresponds to a spherical shell expanding at c.

Correct?

Now ds2 = dx2+dy2+dz2 - c2.dt2

is an invariant under the LT, so if ds2 = 0 (which corresponds to a spherical shell expanding at c) then ds'2 = 0 in the primed frame, and hence in the primed frame the world-sheet is given by

dx'2+dy'2+dz'2 = c2.dt'2

Thus, as we have agreed, it corresponds to a spherical surface expanding at c in the primed frame.

So it is spherical in all inertial frames.

Agreed?

chinglu1998

2011-Mar-17, 10:58 PM

So, in the un-primed frame, for an expanding spherical shell expanding at c we find that it is described by

dx2+dy2+dz2 = c2.dt2

and, a hypersurface described by

dx2+dy2+dz2 = c2.dt2

corresponds to a spherical shell expanding at c.

Correct?

Now ds2 = dx2+dy2+dz2 - c2.dt2

is an invariant under the LT, so if ds2 = 0 (which corresponds to a spherical shell expanding at c) then ds'2 = 0 in the primed frame, and hence in the primed frame the world-sheet is given by

dx'2+dy'2+dz'2 = c2.dt'2

Thus, as we have agreed, it corresponds to a spherical surface expanding at c in the primed frame.

So it is spherical in all inertial frames.

Agreed?

Sorry no, I cannot agree with your conclusions.

The invariance of the light like space time interval only proves each transformed light beam measures c. To conclude any further would implement the logical fallacy of proving one case and assuming all.

To prove the moving frame is viewing a spherical light wave, you would use a constructive process of mapping a set of light beams and proving a spherical wavefront as one method.

That was done here.

http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1864365#post1864365

However, this proof demonstrated there exists time intervals in the time of the rest frame in which not one single light sphere is constructed in the moving frame by LT.

Therefore, just because each translated light beam measures c, this alone is not enough to prove the moving frame is viewing a spherical wave front based on LT.

Now, when we take the light postulate in both frames, each frame claims they are viewing a spherical wavefront on any time interval.

However, for some rest frame x coordinate, xp on the time interval [ xp/c, (xp(c+v))/(c(c-v)) ), LT does not construct a single constant radius light sphere for the moving frame.

So, LT is not matching the conclusions of the light postulate in the moving frame.

More specifically, if not one single light constant radius sphere is constructed by LT on the rest frame time interval [ xp/c, (xp(c+v))/(c(c-v)) ), it is impossible for LT to claim it supports a spherical wavefront in the moving frame.

macaw

2011-Mar-17, 11:59 PM

Sorry no, I cannot agree with your conclusions.

The invariance of the light like space time interval only proves each transformed light beam measures c. To conclude any further would implement the logical fallacy of proving one case and assuming all.

To prove the moving frame is viewing a spherical light wave, you would use a constructive process of mapping a set of light beams and proving a spherical wavefront as one method.

That was done here.

http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1864365#post1864365

No, it wasn't. I explained your fallacies in the very next post.

However, this proof demonstrated there exists time intervals in the time of the rest frame in which not one single light sphere is constructed in the moving frame by LT.

You are repeating the same fallacies even after having been proven wrong.

Therefore, just because each translated light beam measures c, this alone is not enough to prove the moving frame is viewing a spherical wave front based on LT.

If you construct incorrect "proofs" don't be surprised if you get contradicting conclusions. You have been shown the correct proof, through several different methods, the fact you you cannot understand it doesn't validate your claims, quite the opposite.

Now, when we take the light postulate in both frames, each frame claims they are viewing a spherical wavefront on any time interval.

However, for some rest frame x coordinate, xp on the time interval [ xp/c, (xp(c+v))/(c(c-v)) ), LT does not construct a single constant radius light sphere for the moving frame.

Your error has been explained to you in the very next post.

More specifically, if not one single light constant radius sphere is constructed by LT on the rest frame time interval [ xp/c, (xp(c+v))/(c(c-v)) ), it is impossible for LT to claim it supports a spherical wavefront in the moving frame.

You already have the explanation of your fallacy.

chinglu1998

2011-Mar-18, 12:02 AM

No, it wasn't. I explained your fallacies in the very next post.

You are repeating the same fallacies even after having been proven wrong.

If you construct incorrect "proofs" don't be surprised if you get contradicting conclusions. You have been shown the correct proof, through several different methods, the fact you you cannot understand it doesn't validate your claims, quite the opposite.

Your error has been explained to you in the very next post.

You already have the explanation of your fallacy.

Can you point me to a post of yours where you proved LT constructs at least one constant radius light sphere on the rest frame time interval [ xp/c, (xp(c+v))/(c(c-v)) )?

That way I can evaluate your assertions. I do not recall such a post.

macaw

2011-Mar-18, 12:07 AM

Can you point me to a post of yours where you proved LT constructs at least one constant radius light sphere on the rest frame time interval [ xp/c, (xp(c+v))/(c(c-v)) )?

That way I can evaluate your assertions. I do not recall such a post.

Follows immediately after the post where you made the fallacious claims. Look up your post and look at the next post where I corrected your misconceptions.

Strange

2011-Mar-18, 12:08 AM

chinglu1998, did you read post #207 (http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1864511#post1864511) ?

chinglu1998

2011-Mar-18, 12:12 AM

Follows immediately after the post where you made the fallacious claims. Look up your post and look at the next post where I corrected your misconceptions.

I found this. I am not sure if this is the correct one. I have to do a guessing game.

No, all you have showed is that xp/c<t<xp/c(c+v)/(c-v) transforms into xp'/c<t'<xp'/c(c+v)/(c-v) .

You have been shown several correct demonstrations that a spherical wavefront transforms into a spherical wavefront.

When do you plan to answer my questions?

I cannot find here where you showed a constant radius light sphere was constructed for the primed frame on the specified interval.

That is what it would take to refute my proof.

chinglu1998

2011-Mar-18, 12:18 AM

chinglu1998, did you read post #207 (http://www.bautforum.com/showthread.php/113056-quot-The-wave-under-consideration-is-therefore-no-less-a-spherical-wave-quot?p=1864511#post1864511) ?

Yes.

It does not disprove my proof.

My proof shows conclusively not one single constant radius light sphere was constructed by LT in the moving frame on the specified rest frame interval.

Regardless of any comments, there exists a time interval in the rest frame in which no constant radius light sphere exists in the moving frame.

It is impossible to claim the moving frame is viewing a spherical wavefront when not one single light sphere exists on the rest frame interval.

I am being extremely specific.

If there exists at least one constructed constant radius light sphere in the moving frame on the specified rest frame time interval, then I am wrong.

Otherwise all of you are wrong.

korjik

2011-Mar-18, 12:28 AM

I thought I said this. But, this is what I meant.

No, this whole thread is you saying this isnt true

Except, to conclude you have a spherical wave, you take the LT primed distance / LT t' = c.

You really ought to learn the proper terminology. If you mean (x'/gamma)*t'=c then No, you are very very wrong.

You are basing your conclusion that you have a spherical wave on the fact each beam measures c and not on any distance measurements because these alone construct an ellpsoid.

There is no measuring here, and you miss the point. The ellipsoid in the primed frame is not all at the same t' in the primed frame. Last I checked, you have to use the same time for all points to find out where something is at one point in time.

When you factor the associated t', you get a constant c.

No. Constant c is a postulate. It is a fixed quantity in SR, and if it changes, you cant use SR

So, to summarize, everyone is basing their conclusion that the moving frame is viewing a spherical wavefront solely because each light beam transformed by LT measures c in any direction. Since light is c in every direction, it must be a spherical wavefront.

Is this what you are thinking?

No, you still have it wrong. The moving frame sees a spherical wave front cause the wave front is spherical. It is never not spherical, you simply wont do the full transform from normal to moving coordinates.

chinglu1998

2011-Mar-18, 12:31 AM

No, this whole thread is you saying this isnt true

You really ought to learn the proper terminology. If you mean (x'/gamma)*t'=c then No, you are very very wrong.

There is no measuring here, and you miss the point. The ellipsoid in the primed frame is not all at the same t' in the primed frame. Last I checked, you have to use the same time for all points to find out where something is at one point in time.

No. Constant c is a postulate. It is a fixed quantity in SR, and if it changes, you cant use SR

No, you still have it wrong. The moving frame sees a spherical wave front cause the wave front is spherical. It is never not spherical, you simply wont do the full transform from normal to moving coordinates.

Care to change the bold above before I respond?

korjik

2011-Mar-18, 12:33 AM

Yes.

It does not disprove my proof.

My proof shows conclusively not one single constant radius light sphere was constructed by LT in the moving frame on the specified rest frame interval.

Regardless of any comments, there exists a time interval in the rest frame in which no constant radius light sphere exists in the moving frame.

It is impossible to claim the moving frame is viewing a spherical wavefront when not one single light sphere exists on the rest frame interval.

I am being extremely specific.

If there exists at least one constructed constant radius light sphere in the moving frame on the specified rest frame time interval, then I am wrong.

Otherwise all of you are wrong.

You are being specifically wrong. Very specifically wrong. You cannot mix primed and unprimed coordinates and get real results. The fact that there is no moving sphere at a specific non-moving time is completely irrelevant. It has no point or basis in reality. The point in the paper in the OP is that when you transform (CORRECTLY) you still get a sphere in the moving frame.

korjik

2011-Mar-18, 12:35 AM

Care to change the bold above before I respond?

No I dont. The word 'if' in this context means that there is some confusion with your kludgy, generally incorrect terminology.

chinglu1998

2011-Mar-18, 12:45 AM

You are being specifically wrong. Very specifically wrong. You cannot mix primed and unprimed coordinates and get real results. The fact that there is no moving sphere at a specific non-moving time is completely irrelevant. It has no point or basis in reality. The point in the paper in the OP is that when you transform (CORRECTLY) you still get a sphere in the moving frame.

All I did was use the invariance of the light like space time interval over an interval of time in the rest frame.

That is not mixing.

And, in the OP I showed give one instant of time in the rest frame. I proved this translated to an ellipsoid in the moving frame over a time interval.

Many claimed one takes the distance calculated by LT / t' calculate by LT and c results and therefore this proves a spherical wave front in the moving frame.

I then wrote a proof that there exists an interval of time in the rest frame in which not one constant radius light sphere is constructed in the moving frame.

Hence, the moving frame is not viewing a spherical wave front in the moving frame.

With this proof, I disconnected space-time and thus the light travel distances hold in the moving frame since space-time has intervals from the rest frame that are not spherical wave fronts in the moving frame.

Hence, logically, the light sphere in the rest frame transforms to an ellipsoid in the moving frame.

To prove this logic false, al you need do is prove on the specified time interval in the rest frame, there exists at least one LT constructed constant radius light sphere ion the moving frame.

chinglu1998

2011-Mar-18, 12:46 AM

No I dont. The word 'if' in this context means that there is some confusion with your kludgy, generally incorrect terminology.

OK, your call.

chinglu1998

2011-Mar-18, 12:51 AM

No, this whole thread is you saying this isnt true

You really ought to learn the proper terminology. If you mean (x'/gamma)*t'=c then No, you are very very wrong.

There is no measuring here, and you miss the point. The ellipsoid in the primed frame is not all at the same t' in the primed frame. Last I checked, you have to use the same time for all points to find out where something is at one point in time.

No. Constant c is a postulate. It is a fixed quantity in SR, and if it changes, you cant use SR

No, you still have it wrong. The moving frame sees a spherical wave front cause the wave front is spherical. It is never not spherical, you simply wont do the full transform from normal to moving coordinates.

(x'/gamma)*t'=c

First off, LT already includes time dilation and so your added \gamma converts the equation to Galilean.

I will explain.

x'/\gamma = ( x - vt)*\gamma/\gamma = ( x - vt).

That is your first error.

Even if you got this correct, your second error is fundamental.

x'*t'=c

How can you possibly ever calculate x'*t' = c?

macaw

2011-Mar-18, 12:54 AM

I cannot find here where you showed a constant radius light sphere was constructed for the primed frame on the specified interval.

You won't because this is (one of) your (many) persistent misconception(s). Spherical wavefronts do not have constant radius, they have variable radius =ct. This is what defines them as wavefronts, chinglu. This misconception of yours has been corrected several times in this thread alone.

chinglu1998

2011-Mar-18, 01:05 AM

You won't because this is (one of) your (many) persistent misconception(s). Spherical wavefronts do not have constant radius, they have variable radius =ct. This is what defines them as wavefronts, chinglu. This misconception of yours has been corrected several times in this thread alone.

Spherical wavefronts do not have constant radius.

Let's see if I understand you correctly.

If I am in the rest frame viewing the wavefront and time t elapses on my clock, the wave front will not be a constant radius ct in all directions?

This is a cpt violation. You just disproved (ATM) relativity.

The spherical wave front is a collection of constant radius light spheres or there exists a cpt violation.

macaw

2011-Mar-18, 01:10 AM

Spherical wavefronts do not have constant radius.

Let's see if I understand you correctly.

Based on what you write next, you clearly don't.

If I am in the rest frame viewing the wavefront and time t elapses on my clock, the wave front will not be a constant radius ct in all directions?

See that "t" in "ct"? Do you understand the notion of "variable dependency on t"?

This is a cpt violation.

It isn't. Please stop using buzzwords whose meaning you don't understand.

Tobin Dax

2011-Mar-18, 01:11 AM

Spherical wavefronts do not have constant radius.

Let's see if I understand you correctly.

If I am in the rest frame viewing the wavefront and time t elapses on my clock, the wave front will not be a constant radius ct in all directions?

If "t elapses," does that mean that it changes or that it stays constant?

chinglu1998

2011-Mar-18, 01:13 AM

If "t elapses," does that mean that it changes or that it stays constant?

I would guess it changes.

macaw

2011-Mar-18, 01:15 AM

I would guess it changes.

So, "changes" means "constant" or "variable"?

Tobin Dax

2011-Mar-18, 01:16 AM

I would guess it changes.

So then is ct constant?

chinglu1998

2011-Mar-18, 01:16 AM

Based on what you write next, you clearly don't.

See that "t" in "ct"? Do you understand the notion of "variable dependency on t"?

It isn't. Please stop using buzzwords whose meaning you don't understand.

Sure, I understand the dependency on t.

Let's grant such a dependency on t'.

Yet, on the specified rest frame time interval, not one light sphere is constructed in the moving frame.

Specifically, how are you defining a spherical wave front. A spherical wave front does not contain one light sphere? How do you make your "definition" happen?

chinglu1998

2011-Mar-18, 01:17 AM

So, "changes" means "constant" or "variable"?

What is your point here? I cannot do a guessing game.

chinglu1998

2011-Mar-18, 01:18 AM

So then is ct constant?

It can be.

Tobin Dax

2011-Mar-18, 01:22 AM

It can be.

No, ct cannot be constant in time. As time elapses, ct increases.

chinglu1998

2011-Mar-18, 01:25 AM

No, ct cannot be constant in time. As time elapses, ct increases.

So, are you claiming there is not such thing as a light sphere?

I certainly can take all the points for some time t such that x² + y² + z² = c² t².

Are you claiming this is not permitted?

Tobin Dax

2011-Mar-18, 01:34 AM

So, are you claiming there is not such thing as a light sphere?

I certainly can take all the points for some time t such that x² + y² + z² = c² t².

Are you claiming this is not permitted?

That depends on how you define "light sphere."

What did you mean when you said that ct could be constant? Under what conditions do you think ct would be constant?

chinglu1998

2011-Mar-18, 01:50 AM

That depends on how you define "light sphere."

What did you mean when you said that ct could be constant? Under what conditions do you think ct would be constant?

This is absurd.

macaw

2011-Mar-18, 01:52 AM

Sure, I understand the dependency on t.

Then, contrary to your repeated incorrect claims it isn't constant, it is variable. This is the point that you are failing to grasp.

Let's grant such a dependency on t'.

Yet, on the specified rest frame time interval, not one light sphere is constructed in the moving frame.

Over 200 posts and you are still unable to grasp the concept of wavefront.

Specifically, how are you defining a spherical wave front. A spherical wave front does not contain one light sphere? How do you make your "definition" happen?

I and several others have explained the notion wavefront to you many times in this thread. You need to make an effort and learn the concept. You can't refute Einstein if you don't know first thing he's talking about.

macaw

2011-Mar-18, 01:55 AM

It can be.

"t" is a variable so, contrary to your false claim, it can't be.

chinglu1998

2011-Mar-18, 01:59 AM

Then, contrary to your repeated incorrect claims it isn't constant, it is variable. This is the point that you are failing to grasp.

Over 200 posts and you are still unable to grasp the concept of wavefront.

I and several others have explained the notion wavefront to you many times in this thread. You need to make an effort and learn the concept. You can't refute Einstein if you don't know first thing he's talking about.

I am OK to some degree to with the wave front being a variable.

However, for the specified time interval, regardless of the billions of posts otherwise, not one single constant radius light sphere exists in the wave front of the moving frame.

So, this proves the moving frame is not viewing a spherical wave front.

If there is no sphere in spherical, then it is not spherical.

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