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Electrocusionist
2011-May-23, 09:32 AM
I've been reading wikis and very simple papers on introductions to Special Relativity.

On the 9th episode of Astronomy Cast, titled 'Einstein's Theory of Special Relativity', Dr. Gay gave an example with a common-to-SR moving train theme: A is inside the moving train with two mirrors placed perpendicular to A's body, one next to the crown of her head, one next to her feet. B is outside the train, on a field, with a similar mirror setup.

SR continues with respect time dilation with 'a moving clock ticks more slowly than when it is at rest with respect to the observer'.

Now, my dilemma. I wanted to convince myself with a little math, so I used one of the few equations I can handle comfortably: v = d/t

Since speed of light is the same for all observers, I gave v a meager 10 m/s for the sake of my calculating prowess. Let's say for B light travels 20 m in 2 s. So,

B: 10 m/s = 20 m / 2 s

Near A, light has to travel the extra distance because the train is moving so I gave it 25 m (correct me if this 'assumption', or all assumptions so far such as using 'sub-standard' formulas, is baseless).

Since v (= 10 m/s) shall remain the same, it's a matter of finding out the time light took to travel 25 m, which in this case is: t = 25m / 10 m/s = 2.5 s

The final time measurements read, A = 2.5 s, B = 2 s.

For the life of me, I can't understand how the moving clock, at A's position, is slower in this case.

I know I'm missing something and I'd be grateful if someone pointed out the error of my ways. Also, I'm terrible at keeping things concise, so apologies there.

Thank you, once more, for your time!

Strange
2011-May-23, 09:56 AM
The final time measurements read, A = 2.5 s, B = 2 s.

OK. Now think of this from A's point of view: if she uses this number and the distance travelled (which will still appear to be 20m to her, only B sees A's light travel 25m) then she would get a different value for the speed of light. Because the starting point is that the speed of light must be constant, A's clock must actually read 2s.

So B see the events happen in A's frame of reference and take 2.5s but also sees that A's clock only reads 2s and is therefore running slow.

Does that make sense (sorry if it doesn't - my brain feels a bit woolly today!)

Electrocusionist
2011-May-23, 10:58 AM
Okay, so let me see if I got this right.

A clock IN the train will read 2 seconds because it perceives light travels 20m and not 25m, and a clock outside the train observing light travel the diagonal length of 25m will measure 2.5 s.

That does make sense! I guess I just rephrased what you said, woolly brain prevails, thank you!

I also ran into a bit of trouble in the same podcast episode about the conservation of momentum and its implications in travelling at the speed of light, if I can't resolve it by myself I hope you won't mind if I bother you.

caveman1917
2011-May-23, 10:59 AM
The conceptual problem you have is that you take a "bird's view" of the situation and decide that A is "really" moving while B is "really" stationary. The main point in relativity is that there is no such "bird's view" and motion is relative. We can say A is moving relative to B, or just as well that B is moving relative to A. But we can't say this one is moving and that one isn't. You always need to be strict about who's perspective you're using.

With that in mind, let's do the example again.

First let's take the perspective of A.
He sees a mirror above and below him, and gets 10m for 1s. The light goes up and down, so his clock reads 2s afterwards. To him B is the moving one, and light has to travel further there. He sees a stationary train but the earth is moving backwards below it. So he says B's light takes (to A's clock) 2.5s.

Then let's take the perspective of B.
It is equivalent. His clock reads 2s for his own light, but 2.5s for A's light.

So A will say that B's clock is running slow, it "should have" ticked 2.5s but only did 2s. And conversely B will say that A's clock is running slow. Each one will say that the stationary clock (ie his own) is ticking right, but the moving clock (ie the other's) is ticking slower.

ETA: to visualise this, suppose that instead of wristwatches they have those huge neon light clocks. A looks at his own and it reads 2.5s and looks at B's and it reads 2s. He concludes B's clock (the moving one according to him) is slower. Conversely B looks at his own clock and sees 2.5s and looks at A's to see 2s. He concludes A's clock (the moving one according to him) is running slower.

Strange
2011-May-23, 11:10 AM
Much better. I wish I had said that :)

caveman1917
2011-May-23, 11:25 AM
Much better. I wish I had said that :)

Thanks :)

I hope it helps the OP, that first conceptual step is always the hardest.

Electrocusionist
2011-May-23, 12:07 PM
it did indeed caveman, that was a wonderful answer.

the 'bird's view' was my fundamental mistake and your further explanation cleared up a lot of unmentioned doubts I had. Not trying to overuse the phrase, haha, but thank you, again.

I think my previous post is still being moderated and so this, along with that, should take a while to get published but I read both your answers and got me doubts cleared much earlier.

caveman1917
2011-May-23, 01:06 PM
Yes, that previous post wasn't there when i posted my answer.

About conservation of momentum (and other conservation laws), note that this is always for some single observer. For example A will have some value for total momentum at the start and he will have the same value at the end. B will also have that. The values for A and B can be different, this is no problem, the point is that it remains the same value for the same observer.

macaw
2011-May-23, 10:57 PM
I've been reading wikis and very simple papers on introductions to Special Relativity.

On the 9th episode of Astronomy Cast, titled 'Einstein's Theory of Special Relativity', Dr. Gay gave an example with a common-to-SR moving train theme: A is inside the moving train with two mirrors placed perpendicular to A's body, one next to the crown of her head, one next to her feet. B is outside the train, on a field, with a similar mirror setup.

SR continues with respect time dilation with 'a moving clock ticks more slowly than when it is at rest with respect to the observer'.

Now, my dilemma. I wanted to convince myself with a little math, so I used one of the few equations I can handle comfortably: v = d/t

Since speed of light is the same for all observers, I gave v a meager 10 m/s for the sake of my calculating prowess. Let's say for B light travels 20 m in 2 s. So,

B: 10 m/s = 20 m / 2 s

Near A, light has to travel the extra distance because the train is moving so I gave it 25 m (correct me if this 'assumption', or all assumptions so far such as using 'sub-standard' formulas, is baseless).

Since v (= 10 m/s) shall remain the same, it's a matter of finding out the time light took to travel 25 m, which in this case is: t = 25m / 10 m/s = 2.5 s

The final time measurements read, A = 2.5 s, B = 2 s.

For the life of me, I can't understand how the moving clock, at A's position, is slower in this case.

I know I'm missing something and I'd be grateful if someone pointed out the error of my ways. Also, I'm terrible at keeping things concise, so apologies there.

Thank you, once more, for your time!

The transformation of time from the train frame to the "outside the train frame" is:

t'=\gamma(t+vx/c^2)

The transformation of time from the "outside the train frame" to the train frame is:

t=\gamma(t'-vx'/c^2)

Differentiating, you get:

dt'=\gamma(dt+vdx/c^2)
dt=\gamma(dt'-vdx'/c^2)

If you want to find out how the time intervals transform from one frame to another, consider a fixed clock in the train frame (so dx=0) and a fixed clock in the "outside" frame (so dx'=0). Then, the above transformations simplify to:

dt'=\gamma(dt)
dt=\gamma(dt')

The first transformation tells you that the observer on the "outside" measures a longer interval of time on his clock than the elapsed time on the clock moving on the train. The reason is that \gamma>1
The second transformation tells you that the observer riding on the train measures a longer interval of time on his clock than the elapsed time on the clock in the "outside the train" frame.

csmyth3025
2011-May-24, 12:57 AM
I've been reading wikis and very simple papers on introductions to Special Relativity....

...Now, my dilemma. I wanted to convince myself with a little math, so I used one of the few equations I can handle comfortably: v = d/t...
Thanks Macaw for your wonderfully mathematical explanation:

The transformation of time from the train frame to the "outside the train frame" is:

http://www.codecogs.com/eq.latex?t'=\gamma(t+vx/c^2)

The transformation of time from the "outside the train frame" to the train frame is:

http://www.codecogs.com/eq.latex?t=\gamma(t'-vx'/c^2)

Differentiating, you get:

http://www.codecogs.com/eq.latex?dt'=\gamma(dt+vdx/c^2)
http://www.codecogs.com/eq.latex?dt=\gamma(dt'-vdx'/c^2)
...
I think Electrocutionist and I are still trying to work our way up from "v=d/t" so it might take us a while to appreciate the equations you've provided.

Chris

macaw
2011-May-24, 01:33 AM
Thanks Macaw for your wonderfully mathematical explanation:

I think Electrocutionist and I are still trying to work our way up from "v=d/t" so it might take us a while to appreciate the equations you've provided.

Chris

You are welcome. Studying physics requires that one is proficient in math. Once you get the math proficiency, the rest comes easy. Math knowledge facilitates elegant explanations.

Ken G
2011-May-24, 02:39 PM
Personally, I think the train scenario is actually an unnecessarily awkward way to understand SR, even though Einstein used it himself. It is inevitable that you will fall into a certain sequence of pitfalls that we have all seen many times, and each time you fall into one, you will again be confused. So if you just want to understand the "moving clocks run slow" element, I feel it is better to adopt Feynman's "light clock" approach.

Here, you imagine that the clock you are using to say how fast "time is progressing" is a light clock, which is two mirrors with a rigid rod between them a length L apart, and the "tick" is a light pulse that bounces back and forth between the mirrors in time 2L/c (using only your v=d/t formula). Each observer has such a clock, which is at rest with respect to themself, but the two observers have a relative speed v. Picking either frame, you can use equations with the spirit of v=d/t to analyze the tick rate of each clock from each observer's point of view, subject only to the requirement that all observers perceive light as traveling at a fixed speed c.

First of all, by symmetry, it should be clear that if observer A concludes the B clock is ticking slow, then observer B must conclude that the A clock is slow, because nothing distinguishes the observers-- they both see another clock moving at v. This is the first of many seeming paradoxes-- surely if A thinks B's clock is slow, B must think A's clock is fast, right? No, only if time is absolute, but it's not-- time is owned by a clock, and different clocks are measuring different time, which is called "proper time". (More precisely, time is an attibute of a path between two events, and the clocks are following different paths and experiencing different events.) So before you even try to get time dilation, it is important to recognize that time is an attribute of a path between two events and measured by a clock following that path, it is not a property of the two events themselves and it certainly isn't some global property that interconnects all events the way we used to imagine it was.

Now it gets a little tricky, because v has a direction, and the two mirrors of the light clock can be aligned along that direction, or perpendicular to it. So give each observer one clock of each orientation, and say c is always fixed. You're done, it's now just the algebra of v=d/t applied to the two moving clocks, subject to the constraint that light moves at c and the physical requirement that if the moving observer sees their two perpendicular clocks ticking at the same rate, then the stationary observer must be able to analyze the situation and also conclude that the moving observer will see both perpendicular clocks ticking at the same rate. Note the path of the light in one of the moving clocks is forward a larger distance then back a smaller distance, whereas the path of the light in the other (perpendicular) moving clock is diagonal up then diagonal back, creating a zig zag pattern. But it's still all v=d/t applied to these complicated paths, and you can figure out how long the moving clock takes to tick, in the stationary frame, using c and v.

The next surprise you get is that there's no way to make it work unless you do two things-- not only does the moving clock have to tick more slowly, but the rigid rod that separates the mirrors must seem shorter to the stationary observer (ETA: in the case that the rod is along v, not in the case that the rod is perpendicular to v). So you have to invoke both time dilation and length contraction, they come together in the same package, to make it all work out.

The last surprise appears in more complicated situations like the train example, where you also find the relativity of simultaneity-- you must assert that clocks that seem synchronized for one observer will not seem synchronized to an observer in relative motion, and the degree of asynchronization will be proportional to the separation between the clocks. The light clock situation is set up to avoid encountering that additional oddity, because neither observer is analyzing time in any frame but their own.

I'm using the standard language here that the "moving clock is ticking slow" because you see that a lot. Personally, I don't like that language at all, because it belies the very point of how time is treated in relativity-- time has meaning along a path. A moving clock reliably measures time along the path taken by that clock, the idea that it is ticking slow is actually a kind of mistake made by the "stationary" observer who hasn't really learned relativity well enough to recognize that their own concept of time applies only to where their own clock is and has no business being extrapolated to where the other clock ends up. That kind of extrapolation is called "choosing a coordinate system", and has no more physical importance than deciding the north pole is a special place because it has no longitude.

StupendousMan
2011-May-24, 02:59 PM
If you'd like to see some diagrams to accompany Ken's words (and some additional words, too), you could try looking at some of the pages from my "Introduction to Special Relativity" class:

http://spiff.rit.edu/classes/phys200/lectures/dilation/dilation.html
http://spiff.rit.edu/classes/phys200/lectures/contraction/contraction.html

uncommonsense
2011-May-25, 06:12 PM
I've been out of the loop for awhile. just shaking out the cob webs. Can someone remind me how Ken G's explanation and then StupendousMan's visuals don't use c + additional velocity as an explanation to SR? My mind is way out of shape :-)

speedfreek
2011-May-25, 07:46 PM
I've been out of the loop for awhile. just shaking out the cob webs. Can someone remind me how Ken G's explanation and then StupendousMan's visuals don't use c + additional velocity as an explanation to SR? My mind is way out of shape :-)

Hi!

Picking either frame, you can use equations with the spirit of v=d/t to analyze the tick rate of each clock from each observer's point of view, subject only to the requirement that all observers perceive light as traveling at a fixed speed c.

Followed by:

The next surprise you get is that there's no way to make it work unless you do two things-- not only does the moving clock have to tick more slowly, but the rigid rod that separates the mirrors must seem shorter to the stationary observer (ETA: in the case that the rod is along v, not in the case that the rod is perpendicular to v). So you have to invoke both time dilation and length contraction, they come together in the same package, to make it all work out.

Has it clicked yet? If not, perhaps you could help and explain why you think the explanations might use c + additional velocity? :)

uncommonsense
2011-May-25, 08:02 PM
In the case where the mirrors are lined up along v,

As the Red Men continue to watch and make their measurements, the laser beam shoots back towards the left wall. But since the left wall is coming towards the beam, it doesn't take quite so long for the beam to reach the wall this time"

The laser beam is moving at c, and the left wall is moving toward the laser beam at v. These velocities are added to conclude "it doesn't take quite so long for the beam to reach the wall this time"

What am I missing. This sounds like adding c plus v (violation od SR)

Strange
2011-May-25, 08:22 PM
The laser beam is moving at c, and the left wall is moving toward the laser beam at v. These velocities are added to conclude "it doesn't take quite so long for the beam to reach the wall this time"

What am I missing. This sounds like adding c plus v (violation od SR)

This is closing speed rather than a different speed of light. It takes less time because the wall is moving towards the light while the light is moving towards the wall. This is from the frame of reference of the Red Men (who see the light moving at c) NOT the wall; from the frame of reference of the wall; the light is approaching it at c, not c+v. Does that help?

speedfreek
2011-May-25, 08:25 PM
For some reason I am having trouble connecting to StupendousMan's site at the moment, which doesn't help!

The laser beam is moving towards the left wall at c, whilst the left wall is moving towards the laser beam at v.

Obviously, the laser beam will take less time to reach the wall in this scenario, than it would if the left wall were not moving towards the laser beam. So to an observer who is stationary in relation to the wall (in the same reference frame), the light takes longer to reach the wall than it does for an observer in a frame of reference where the wall is moving.

I cannot comment on the addition of velocities as I cannot read the webpage to see what it says - can you say what it is about Ken's post that you infer to mean the same thing, instead?

uncommonsense
2011-May-25, 08:37 PM
I thought the postulate meant that the speed of light is constant for all observers regardless of their motion relative to the light source. The "closing" speed you mention is the sum of c + velocity of wall. This sum is used to fomulate time of travel of laser beam over a distance.

uncommonsense
2011-May-25, 08:54 PM
I cannot comment on the addition of velocities as I cannot read the webpage to see what it says - can you say what it is about Ken's post that you infer to mean the same thing, instead?

I reread Ken's post and conclude StupendousMan's visuals are different in that:

Ken G, for the most part, removes "time" from the analysis and explains what's really going on, distance/length contraction.

My bad, StupendousMan said his links explained what Ken G said. They are actually different.

speedfreek
2011-May-25, 09:11 PM
Okay, I see what you mean now. (Is it just me having trouble connecting to http://spiff.rit.edu/classes/phys200/lectures/dilation/dilation.html ?)

The speed of light is constant for all observers regardless of their motion relative to the light source.

To an observer sitting on the train, the walls at each end of the carriage are stationary, so they only have to deal with the time it takes the light to reach the wall. The distance to the wall remains fixed, whilst the light propagates towards it. That distance, divided by c, will be the time it took the light to reach the wall.

However, to an observer standing on an embankment as the train passes, the light is moving at c, whilst the train is moving at v. Here, the observer has to deal with the left wall moving towards the light, so the closing speed will be faster than c.

Nothing is moving faster than c.

uncommonsense
2011-May-25, 09:23 PM
However, to an observer standing on an embankment as the train passes, the light is moving at c, whilst the train is moving at v. Here, the observer has to deal with the left wall moving towards the light, so the closing speed will be faster than c.

Nothing is moving faster than c.

But ummm..........how is "closing speed" being faster than c acceptable? What is "closing speed" It sounds to me like it is the speed of light from the wall's perspective......? i.e., faster than c.

I think any example must center on length contraction (after all, "time" is simply the ratio of motion and distance)

Strange
2011-May-25, 09:35 PM
But ummm..........how is "closing speed" being faster than c acceptable? What is "closing speed" It sounds to me like it is the speed of light from the wall's perspective......? i.e., faster than c.

If you see two trains rushing towards each other at 90% of c, you will consider them to be closing at 180% of c. From the point of view of either train, the other one will be approaching it at something less (sorry, its too late for arithmetic!) than c.

Similarly the wall (or someone on the wall) would see the light approaching at c. But as someone watching both the light and the wall moving, we can see their relative closing speed is greater than c.

I think any example must center on length contraction (after all, "time" is simply the ratio of motion and distance)

No. You will need both time dilation and length contraction for everything to work out.

uncommonsense
2011-May-25, 09:42 PM
Yes.......I think I need to go back and refamiliarize myself with the postulate re constancy of speed of light.

uncommonsense
2011-May-25, 09:47 PM
Yes...yes....thankyou. closing speed.............yes. got it. thankyou.(getting old)

speedfreek
2011-May-25, 09:58 PM
Yup! It is easily done, accidentally mixing frames without realising it - take it from someone who knows from experience! ;)

Strange
2011-May-25, 10:02 PM
And some people never get it; this seems to be the basis of an awful lot of "Einstein was wrong" claims (I think there might be one in ATM at the moment ...)

StupendousMan
2011-May-26, 01:30 AM
My university is testing the electrical systems this week, so my web site will be down for a few days. Sorry about that. It will return, have no fear.

CaptainToonces
2011-May-27, 12:31 AM
Differentiating, you get:

dt'=\gamma(dt+vdx/c^2)
dt=\gamma(dt'-vdx'/c^2)

If you want to find out how the time intervals transform from one frame to another, consider a fixed clock in the train frame (so dx=0) and a fixed clock in the "outside" frame (so dx'=0). Then, the above transformations simplify to:

dt'=\gamma(dt)
dt=\gamma(dt')

What is "differentiating it" and why is that a necessary step here?

You are multiplying every term by some constant d?

macaw
2011-May-27, 02:52 AM
What is "differentiating it" and why is that a necessary step here?

You are multiplying every term by some constant d?

No, "d" stands for the differentiation operator. You would need an introduction to calculus in order to understand what I am doing. I can suggest a couple of websites, are you interested?

grapes
2011-May-27, 03:55 AM
What is "differentiating it" and why is that a necessary step here?

You are multiplying every term by some constant d?Differentiation is typically introduced early in the study of calculus. The differentiation of a function basically results in a new function that gives the rate of change for the original function. The derivative of x3 is 3x2, which means that the slope of x3 at x=4 is 3(42), or 48.

CaptainToonces
2011-May-27, 06:14 PM
No, "d" stands for the differentiation operator. You would need an introduction to calculus in order to understand what I am doing. I can suggest a couple of websites, are you interested?

yes, that's why i asked. What does the differentiation operator perform, is it multiply by exponent and decrement exponent by 1? Why is this necessary in understanding the lorenz transform?

Strange
2011-May-27, 06:21 PM
yes, that's why i asked. What does the differentiation operator perform

In general terms, it gives the rate of change of a function (or the gradient of a graph) at some point.

is it multiply by exponent and decrement exponent by 1

That is one of the shortcut rules.

Why is this necessary in understanding the lorenz transform?

In this case, it is going from a measure of the time (t) to the rate of "flow" of time (dt) - or how fast your clock goes.

You can't get very far in physics without at least basic calculus, I'm afraid...

Ken G
2011-May-28, 01:17 AM
macaw is hitting a nail with a sledge hammer, basically, but for those who are used to using calculus it doesn't seem like any big deal. There is no need to use differentiation in special relativity when one restricts to motion at constant speed, which is almost always the case in elementary SR and is certainly the case in this thread. If you don't follow the meaning of the differentiation, don't think that you need to-- just ignore it, you don't need calculus to do this level of SR, the latter is purely an exercise in linear algebra (matrix algebra and so on, and you can get the guts of it with simple 2-by-2 systems). The expressions without the "d" in them are all you need to understand if you want to "get" time dilation, length contraction, and the relativity of simultaneity, in SR for constant relative motion between observers.

macaw
2011-May-28, 02:13 AM
macaw is hitting a nail with a sledge hammer, basically, but for those who are used to using calculus it doesn't seem like any big deal.

Yes, what is the point in using elegant solutions when you can get away with hacky, ugly ones?
What is the point in using any math at all when you can just talk about things? After all, physics doesn't need any solid mathematical basis.

Ken G
2011-May-28, 04:58 AM
Yes, what is the point in using elegant solutions when you can get away with hacky, ugly ones?I fear you miss the point. There is simply no need for any calculus in special relativity for observers in constant relative motion. There just isn't, it has nothing at all to do with "elegance." Indeed, what you did was take a derivative of a global linear transformation (in the context of how that transformation is being used in this thread, constant relative motion). What is actually elegant is understanding how completely superfluous the derivative of a global linear transformation actually is. Now, for those who understand calculus, that superfluousness is no particular issue, but to those who don't, it's a completely unnecessary stumbling block. Hence my comment.

For those reading this who do not know a shred of calculus, fear not, there is no need for you to know any at all to understand the special relativity of observers in constant relative motion. That's just a fact.

macaw
2011-May-28, 05:12 AM
I fear you miss the point. There is simply no need for any calculus in special relativity for observers in constant relative motion.

Really? Would you care to show how the electromagnetic wave equation transforms between frames? Please use only the Lorentz transforms. You are not allowed to use partial derivatives because they are "calculus" and you just claimed that "There is simply no need for any calculus in special relativity for observers in constant relative motion".

How about you derived the law of velocity composition without using derivatives?

There just isn't, it has nothing at all to do with "elegance."

This is demonstrably false, see the little exercises I asked you to do above. I don't see what entitles you to comment on the method, especially in the context of you never posting any calculations. It is very easy to read the results derived by others and to post the answers as if you knew the subject.

What is actually elegant is understanding how completely superfluous the derivative of a global linear transformation actually is.

No, it isn't. It is the standard approach in deriving length contraction/ time dilation. This is how it's done in the better books.

Now, for those who understand calculus, that superfluousness is no particular issue, but to those who don't, it's a completely unnecessary stumbling block. Hence my comment.

According to your logic, we shouldn't use covariant derivatives in GR. We should use tensors either. They are also "completely unnecessary".

For those reading this who do not know a shred of calculus, fear not, there is no need for you to know any at all to understand the special relativity of observers in constant relative motion. That's just a fact.

Fear not, why waste your time learning math when you can talk about physics without ever deriving any quantitative results?

Ken G
2011-May-28, 05:27 AM
Really? Would you care to show how the electromagnetic wave equation transforms between frames? Please use only the Lorentz transforms.Goodness, the Lorentz transformation between any two frames in constant relative motion is simply a linear transformation that depends only on the relative constant velocity between the frames. No calculus there, none. I'm sure you know what the Lorentz transformation is, so I'm at a loss how you could imagine anything else.

No, it isn't. It is the standard approach in deriving length contraction/ time dilation. This is how it's done in the better books.Wrong, it is not at all the standard approach for constant relative motion. The "better books" that you refer to are built to generalize to relative motion that is not constant, and include many other mathematical sledgehammers that are equally irrelevant to the question asked by the OPer.

According to your logic, we shouldn't use covariant derivatives in GR.Um, no, that is nothing like my logic. Rather, my logic is, if someone asks to have SR explained at the level of v=d/t, as in this very thread, it is quite straightforward to do so. I sketched how to do it above, and StupendousMan gave a link to the details. Maybe you should take a look.

macaw
2011-May-28, 05:30 AM
Goodness, the Lorentz transformation between any two frames in constant relative motion is simply a linear transformation that depends only on the relative constant velocity between the frames. No calculus there, none. I'm sure you know what the Lorentz transformation is, so I'm at a loss how you could imagine anything else.

You are not paying attention to the questions (or you don't even understand them). Let's see how you solve the two exercises above without using calculus. Please.

Ken G
2011-May-28, 05:37 AM
You are not paying attention to the questions (or you don't even understand them). Let's see how you solve the two exercises above without using calculus. Please.
I understand the questions completely. Let me say it more clearly. If we have a given electromagnetic field in one frame, and we want to know it another frame in constant relative motion, we can perform a linear transformation on the known fields, without a single scrap of calculus. Yes or no?

You won't bother to answer of course, because the answer is obviously "yes", as I claim. What you are actually talking about is something totally different-- you are talking about the covariant form of the constituitive equations that allow us to solve for the electromagnetic fields in any frame. That is completely irrelevant to anything the OP has asked, you may as well claim that Galilean relativity requires calculus because the Lagrange equations require calculus and respect Galilean relativity. That will of course come as a surprise to all the young physics students who learn Galilean relativity before they ever even hear of calculus.

To others reading this, I can help clarify the issue-- we have physics in a single frame, and no need for any relativity, except that the form of the laws must respect it. But if you are already given the form of the laws, and just want to do physics in one frame, you need no relativity. Then if you want to understand SR for constant relative motion, all you need is the linear transformation (matrix algebra) that lets you transform everything you solved in one frame into another. So clearly, it matters not that the solution in the first frame may have required calculus (or may not), when all you want to understand is the SR of how it transforms to the frame of the other observer. Everything the OP asked is of that nature, and so there is much to understand about the meaning of an SR transformation to another reference frame in constant relative motion, without knowing any calculus at all. This is well understood by anyone involved in teaching SR to students who don't know calculus. Obviously.

macaw
2011-May-28, 05:43 AM
I understand the questions completely. Let me say it more clearly. If we have a given electromagnetic field in one frame, and we want to know it another frame in constant relative motion, we can perform a linear transformation on the known fields, without a single scrap of calculus. Yes or no?

I asked you to transform the wave equation. It is a partial differential equation , last I checked (http://en.wikipedia.org/wiki/Electromagnetic_wave_equation). Yes or no?

How about the simpler problem. the derivation of the velocity composition?

macaw
2011-May-28, 05:45 AM
To others reading this, I can help clarify the issue-- we have physics in a single frame, and no need for any relativity, except that the form of the laws must respect it. But if you are already given the form of the laws, and just want to do physics in one frame, you need no relativity. Then if you want to understand SR for constant relative motion, all you need is the linear transformation (matrix algebra) that lets you transform everything you solved in one frame into another. So clearly, it matters not that the solution in the first frame may have required calculus (or may not), when all you want to understand is the SR of how it transforms to the frame of the other observer. Everything the OP asked is of that nature, and so there is much to understand about the meaning of an SR transformation to another reference frame in constant relative motion, without knowing any calculus at all. This is well understood by anyone involved in teaching SR to students who don't know calculus. Obviously.

This is, of course, incorrect.

Ken G
2011-May-28, 06:33 AM
I guess you'll just have to have your own wrong opinion then. Fortunately, I have taught SR to students who know no calculus many times, and they had many of the same questions as the OPer, and I wager that StupendousMan has too, judging from the calculus-free character of his website that he referenced and you ignored. So we can just leave it up to the OPer to decide which explanation is of more value to them, the simple constructions they can do to fully understand time dilation and length contraction with light clocks, knowing naught but the formula v=d/t that they themselves offered, or the calculus "explanation" you gave. I will simply repeat that taking derivatives of global linear transformations is a completely pointless thing to do, as anyone who understands calculus knows, and as anyone who doesn't understand calculus doesn't need to know.

MicVR
2011-May-28, 07:39 AM
Ken G is of course correct.
In an ordinary university setting I would (and do) teach it to my students using calculus.
But if the premise was to not use calculus and if the specific circumstance of the OP was given it could certainly be done without using it.

Ken G made the parameters of his/her analysis clear and considering those he/she is correct.

Ken G
2011-May-28, 12:53 PM
Thank you MicVR. This might be a good point to summarize the key issues from the above digression so the thread can return to its main thrust:
1) If you don't know calculus already, take heart, you do not need to learn it in order to appreciate the central lessons of special relativity.
2) If you want to learn calculus, what you gain in your understanding of relativity is twofold: you can apply relativity to more complicated motions than constant relative motion between observers (and ultimately to gravity), and you can appreciate the full restrictions that relativity places on the actual laws of physics themselves (which generally require calculus to formulate).
However, even in regard to the laws of physics themselves, you can still get some insights into what the restrictions of relativity are without any calculus-- just as you can apply famous algebraic equations like F = ma without calculus. You can, for example, understand why F = ma cannot be a law of physics in every reference frame, and you won't need calculus to see why. So it's not just a an idle dispute between macaw and myself, it is quite important for understanding what aspects of relativity you should be able to appreciate without having to learn calculus first, even though it is certainly true that if you learn calculus, your insights will apply to a broader class of problems.

caveman1917
2011-May-28, 01:58 PM
You're right that we can do the basic stuff with matrix algebra, but let's not completely downplay the importance of calculus either. During one's way through science (not just SR) it's one of the most important tools to have.

macaw
2011-May-28, 02:03 PM
I guess you'll just have to have your own wrong opinion then. Fortunately, I have taught SR to students who know no calculus many times, and they had many of the same questions as the OPer, and I wager that StupendousMan has too, judging from the calculus-free character of his website that he referenced and you ignored. So we can just leave it up to the OPer to decide which explanation is of more value to them, the simple constructions they can do to fully understand time dilation and length contraction with light clocks, knowing naught but the formula v=d/t that they themselves offered, or the calculus "explanation" you gave. I will simply repeat that taking derivatives of global linear transformations is a completely pointless thing to do, as anyone who understands calculus knows, and as anyone who doesn't understand calculus doesn't need to know.

So, you cannot do either of the two simple exercises I challenged you to do.

macaw
2011-May-28, 02:08 PM
Ken G is of course correct.
In an ordinary university setting I would (and do) teach it to my students using calculus.
But if the premise was to not use calculus and if the specific circumstance of the OP was given it could certainly be done without using it.

Ken G made the parameters of his/her analysis clear and considering those he/she is correct.

KenG claimed that one never needs calculus in the case of SR applied to uniform motion. I gave him a couple of exercises that prove him wrong.

Ken G
2011-May-28, 02:12 PM
You're right that we can do the basic stuff with matrix algebra, but let's not completely downplay the importance of calculus either. During one's way through science (not just SR) it's one of the most important tools to have.I'm certainly not saying a physicist doesn't need calculus. I'm saying that someone who wants to understand the basic lessons of relativity doesn't need calculus, which they don't. This is important because there are a whole lot of people who don't know calculus and never will, but there's no reason that they shouldn't all understand the mysteries of the relativity of two observers in constant relative motion. No reason at all, it's exactly the same reason that we teach Newton's laws to people who will never know calculus.

Ken G
2011-May-28, 02:18 PM
So, you cannot do either of the two simple exercises I challenged you to do.Since you cannot hear a single thing I'm saying, there's little point in continuing. For those who can, let me repeat what I'm saying once more: the transformation of all physical variables between two observers in constant relative motion is a global linear transformation called the Lorentz transformation. Understanding its action is a simple exercise in matrix algebra, generally only 2-by-2 in complexity. That does not require calculus, in fact calculus has absolutely nothing to add to the understanding of global linear transformations. So anyone can easily understand the guts of transforming between inertial observers in constant relative motion (which is what the OP, and this whole thread, is all about), in particular time dilation, length contraction, and the relativity of simultaneity, knowing zero calculus. In fact, people do it all the time, as StupendousMan knows, as MicVR knows, and as you apparently do not know. But yes, if you confuse the act of transforming between frames with the act of doing complicated physics in either frame, you wouldn't see that, now would you, but of course the OPer didn't ask about that and likely has no interest in it at all (as they said their understanding of dynamics is at the v=d/t level, which is actually just fine for understanding the answer to all their questions).

macaw
2011-May-28, 02:21 PM
Since you cannot hear a single thing I'm saying, there's little point in continuing. For those who can, let me repeat what I'm saying once more: the transformation of all physical variables between two observers in constant relative motion is a global linear transformation called the Lorentz transformation.

But that was not the question. The questions were a direct challenge to your repeated false claims (http://www.bautforum.com/showthread.php/115854-An-attempt-to-understand-Special-Relativity?p=1894804#post1894804) that "SR applied to uniform motion does not require calculus". The truth is that it does. You should not use Q&A in order to promote your personal ATMs.

Understanding its action is a simple exercise in matrix algebra, generally only 2-by-2 in complexity. That does not require calculus,

Actually, the two simple challenges that I gave you, do require calculus. There is no way to solve them without using calculus.
The fact that you have spent several hours avoiding answering the direct challenges shows that you don't know how. If you did, you would have answered by now. If you answered (using math rather than your prose) you would have proven yourself wrong.

Ken G
2011-May-28, 03:08 PM
But that was not the question.Correction, that was not your question. I've summarized above quite clearly what calculus is and is not needed for, post #45 leaves little for me to add now.

macaw
2011-May-28, 03:37 PM
Correction, that was not your question. I've summarized above quite clearly what calculus is and is not needed for, post #45 leaves little for me to add now.

"There is no need to use differentiation in special relativity when one restricts to motion at constant speed"

The false statement was repeated in post 36:

"There is simply no need for any calculus in special relativity for observers in constant relative motion."

I have have already given you two counterexamples to your false claims. Here is another one:

Demonstrate the invariance of the Minkowski metric ds^2=(cdt)^2-(dx^2+dy^2+dz^2). According to you, you can do that without using calculus. Please show the math or withdraw your much repeated ATM claim.

MicVR
2011-May-28, 04:03 PM
To me, this seems to be a "battle" more on an emotional than a factual level.

As a neutral observer I can state that Ken G has made his points very clear. I found his posts very carefully worded, leaving little to no room for interpretation.
He carefully and clearly described the parameters for the range of problems that can be solved or described without the use of calculus.

The message that he obviously wanted to bring across he did bring across (for me anyways) and it is correct.

macaw
2011-May-28, 04:10 PM
To me, this seems to be a "battle" more on an emotional than a factual level.

As a neutral observer I can state that Ken G has made his points very clear. I found his posts very carefully worded, leaving little to no room for interpretation.
He carefully and clearly described the parameters for the range of problems that can be solved or described without the use of calculus.

The message that he obviously wanted to bring across he did bring across (for me anyways) and it is correct.

KenG made sweeping, sloppy and misleading statements, as in "There is simply no need for any calculus in special relativity for observers in constant relative motion". I gave three counterexamples that prove his statements to be false. There are many more. If you think otherwise, I would challenge you to solve the three challenges without using calculus. I'll even get you started:

In frame F, the velocity of a particle is v. What is the velocity v' of the same particle in frame F' moving with velocity V with respect to F? You are not allowed to use calculus in solving the challenge.

Ken G
2011-May-28, 05:08 PM
That is precisely the kind of problem that I am talking about, which requires no calculus. Simply look at, for example, http://en.wikipedia.org/wiki/Velocity-addition_formula, or http://www.desy.de/pub/www/projects/Physics/Relativity/SR/velocity.html, not an iota of calculus in either one. I wouldn't dream of using calculus in deriving the velocity addition formula, even if I were teaching it to physics graduate students.

I guess the challenge for macaw is now to find why those two websites are both wrong. On ATM, please.

Ken G
2011-May-28, 05:31 PM
By the way, interested readers may wish to check out http://www.physorg.com/news111075100.html, an article on PhysOrg which states:

Now Joel Gannett, a Senior Scientist in the Applied Research Area of Telcordia Technologies in Red Bank, New Jersey, has found that Einstein didn’t have to do the work the hard way. A researcher in optical networking technologies, Gannett has shown that the Lorentz transformations and velocity addition law can be derived without assuming the constancy of the speed of light, without thought experiments, and without calculus.
The stress there is on different axioms, the lack of a need of calculus is kind of a trivial add-on because it was perfectly obvious anyway.

The AAPT also weighs in (that's the organization for physics teachers) at http://tpt.aapt.org/resource/1/phteah/v47/i4/p231_s1?isAuthorized=no, where a paper in The Physics Teacher spells out:

The purpose of this paper is to show that the new philosophy can be carried out in a noncalculus physics course, by demonstrating that all of the principal results of special relativity theory can be obtained by simple algebra.

Can we get back to the real thread now? If he has any more questions, macaw can start an ATM thread.

macaw
2011-May-28, 06:17 PM
That is precisely the kind of problem that I am talking about, which requires no calculus.

Well, prove it. Derive the formula. The links you gave use the formula, they don't derive it. I've asked you to derive it (without using calculus, as you keep claiming). To be precise, you are being challenged to derive this formula (http://en.wikipedia.org/wiki/Velocity-addition_formula#General_case) without using calculus. Since there is no derivation in the wiki entry, you will need to do the work. The funny thing is that wiki cites ref (http://books.google.com/books?id=D_zYVBu0KAIC) which in turns copies the derivation from page 53 in C.Moller's "The Theory of Relativity". The derivation, as expected, relies heavily on..... differentiation !

I see that you have chosen the easier challenge. How about the one on the invariance of the electromagnetic wave equation? Care to find a non-calculus answer to that challenge as well?

How about the invariance of the Minkowski metric? Still searching the web for an answer?

grapes
2011-May-28, 06:44 PM
Here's the usual translation of Einstein's original paper that first described special relativity: On the Electrodynamics of Moving Bodies (http://www.fourmilab.ch/etexts/einstein/specrel/www/). Not much calculus there, before the electromagnetic stuff.

Ken G
2011-May-28, 06:45 PM
Well, prove it. Derive the formula. The links you gave use the formula, they don't derive it.I guess you better read them a little more carefully. I suggest you look at the sections marked "derivation." Or, consider the words of MicVR, the website of StupendousMan, the two links I gave above, the PhysOrg or The Physics Teacher articles I cited, or just understand what calculus is and isn't. You might also want to think a little about decomposing the general vector situation into components parallel and perpendicular to the direction of the relative velocity vector, since if you do that, you'll quickly see that the general formula involves no more calculus than does the parallel formula. It's all algebra, just muliplying and dividing, and nothing beyond the spirit of v=d/t.
I've asked you to derive it (without using calculus, as you keep claiming). It is perfectly trivial, and is indeed done in those links. Just read the place in the Wiki where it says "To translate the formula of the previous section to three-vector notation, replace u1 with the component of U parallel to V" for example. I think you are a little confused about the difference between calculus and vector algebra.

To be precise, you are being challenged to derive this formula (http://en.wikipedia.org/wiki/Velocity-addition_formula#General_case) without using calculus. Since there is no derivation in the wiki entry, you will need to do the work. You might want to use Sommerfeld's reference (http://en.wikisource.org/wiki/On_the_Composition_of_Velocities_in_the_Theory_of_ Relativity). Dang, he used calculus. Yes, that proves a lot-- if I can use calculus to derive something, then that something requires calculus. Nice logic there. So you believe the AAPT and PhysOrg articles are both incorrect? This doesn't bother you? You're just digging yourself deeper.

I see that you have chosen the easier challenge. How about the one on the invariance of the electromagnetic wave equation? Care to find a non-calculus answer to that challenge as well?

How about the invariance of the Minkowski metric? Still searching the web?No, I have spent no time on either of those irrelevant issues, as they were all perfectly well covered in my post #45, as I already pointed out.

macaw
2011-May-28, 06:47 PM
Just read the place in the Wiki where it says "To translate the formula of the previous section to three-vector notation, replace u1

You have been challenged to solve the general case. Not to translate the particular case in vector notation. The equation you are challenged to derive is the one midway in the paragraph.

with the component of U parallel to V" for example. I think you are a little confused about the difference between calculus and vector algebra.

Would you please stop evading and do the derivation for the general case?

macaw
2011-May-28, 06:53 PM
Here's the usual translation of Einstein's original paper that first described special relativity: On the Electrodynamics of Moving Bodies (http://www.fourmilab.ch/etexts/einstein/specrel/www/). Not much calculus there, before the electromagnetic stuff.

Correct. Yet, when it comes to the equations of electromagnetism, one cannot avoid calculus, no matter how hard KenG tries. Since KenG has evaded answering the em challenge, I gave him the two easier challenges.

Ken G
2011-May-28, 07:05 PM
To clarify. The laws of electromagnetics are themselves expressed using calculus. This was true before relativity, I reference Maxwell. So why anyone would think it is the least bit relevant to relativity to try and redescribe the laws of electromagnetism without calculus is completely beyond me. The OP was not about the laws of electromagnetism, which are used to find electromagnetic fields, it was about transforming between frames, and the resulting issues like time dilation. This can also be done for electromagnetic fields without any calculus. Read that as many times as it takes. Your request for me to prove something silly is perfectly irrelevant, was covered in post #45, and I plan to continue to ignore such a pointless issue.

What is of more relevance to this thread is whether or not the velocity addition formula, which does indeed relate to time dilation and other issues of transforming physical quantities between observers, requires calculus to derive. By my count, I have now told you how to do it, given two links that do it, and cited two papers that say exactly that you can do it. Despite all this, you maintain the contrary, and all apparently on the grounds that Maxwell's equations involve calculus. Fascinating.

macaw
2011-May-28, 07:13 PM
To clarify. The laws of electromagnetics are themselves expressed using calculus.

So, how do you justify the ATM claim that you made repeatedly: "There is simply no need for any calculus in special relativity for observers in constant relative motion"?

it was about transforming between frames, and the resulting issues like time dilation. This can also be done for electromagnetic fields without any calculus.

Prove it. This is all I challenged you. Take Maxwell's equation of em waves and show that it is Lorentz-invariant. Do that without using calculus.

macaw
2011-May-28, 07:19 PM
What is of more relevance to this thread is whether or not the velocity addition formula, which does indeed relate to time dilation and other issues of transforming physical quantities between observers, requires calculus to derive. By my count, I have now told you how to do it, given two links that do it, and cited two papers that say exactly that you can do it.

This is false as well. The general case requires calculus. The most amusing fact is that the wiki link references a book (ref) that uses....differentiation in deriving the formula :-)
The other reference (http://www.desy.de/pub/www/projects/Physics/Relativity/SR/velocity.html) doesn't derive anything, it uses the formula.

so, the challenge stands, how would you support your claim and how would you derive the formula without resorting to calculus. Repeating the same ATM assertions and pointing at links that contradict your claims doesn't work.

Ken G
2011-May-28, 07:40 PM
The general case requires calculus. Here you are referring to the velocity addition formula, which is a very simple and clear-cut case so let's review the evidence for your making this false claim. You have one source that used calculus to derive it, which made no claim at all it was required (and I'm sure the author doesn't think it was required). I have two sources that didn't use calculus, a mathematical explanation of why you would never need calculus (the Lorentz transformation is globally linear for inertial observers in constant relative motion in the absence of gravity), and two articles that make the specific point that you don't need it. What does that score add up to for you?
The most amusing fact is that the wiki link references a book (ref) that uses....differentiation in deriving the formula. The reference in the Wiki goes well beyond the derivation of the formula, and is designed for further reading by those who wish to go into the mathematical proofs that generalize into other areas that really do need calculus. But the derivation itself is right there in black and white, requires nothing but multiplication and division, and you really should go through it some time if you want to claim to know what you are talking about. I'll sketch it for you. Follow the algebra of the parallel velocity case, then recognize that this gives you the way to transform the spatial displacement along the direction of the relative velocity between the observers. The transformation will involve both the relativity of simultaneity and length contraction, as per the parallel Lorentz transformation. The perpendicular displacements need no transforming, as you should know. Then say the transformed velocity has coordinates of the ratio of the spatial displacement components, divided by the transformed time interval, which itself involves time dilation and the relativity of simultaneity along the parallel direction. No formula other than the linear Lorentz transformation is ever needed, and the statement that a coordinate velocity component is a ratio, the ratio of the spatial component in that direction, divided by the coordinate time interval. Multiplication and addition, nothing else, just as plain as day. Clever pre-calculus high school students are quite capable of it, which is precisely the point of The Physics Teacher article I cited and you have not understood.

Everyone else has of course no interest in the pointlessness of this dispute, but the takeaway message for them is that they are all perfectly capable of deriving the general velocity addition formula if they know how to add, multiply, and divide. I think they should not be deceived on that matter, because if they are not in a position to learn calculus, it might cause them to give up on understanding one of the most profound of all physical truths that humanity has ever discovered.

Ken G
2011-May-28, 07:50 PM
The message that he obviously wanted to bring across he did bring across (for me anyways) and it is correct.I am indebted to you MicVR for making me feel like I have not been wasting my time, thank you.

macaw
2011-May-28, 09:24 PM
Here you are referring to the velocity addition formula, which is a very simple and clear-cut case so let's review the evidence for your making this false claim.

No more prose , please, use math.

I'll sketch it for you. Follow the algebra of the parallel velocity case, then recognize that this gives you the way to transform the spatial displacement along the direction of the relative velocity between the observers.

No "sketching" , please, show that you can derive the formula without using calculus. You made the claim, now prove it by doing the derivation.

You are either unable to do it, meaning that your claim is baseless or you have seen it derived in various books (see Rindler, Moller, Landau and Lif****z, Tolman, Pauli, Feynman) where the derivation is invariably based on calculus (differentiation), meaning that your claim is false.

While you are at it, solve the other two challenges as well.

Ken G
2011-May-28, 11:09 PM
No more prose , please, use math.
We've had that debate before, prose is math, when it describes exactly what to do in a mathematical calculation. Which I did. And which the Wiki did, etc. I won't humor you any more, you are totally wrong and not getting any righter, and frankly I don't see anyone around who thinks you are.
You are either unable to do it, meaning that your claim is baseless or you have seen it derived in various books (see Rindler, Moller, Landau and Lif****z, Tolman, Pauli, Feynman) where the derivation is invariably based on calculus (differentiation), meaning that your claim is false. I've already destructed the poor logic you are using here. Those authors did not have the goal of deriving the formula without using calculus, they had the goal of applying calculus, because they knew perfectly well they were going to use it for more advanced calculations shortly, and they alreadly know calculus. That they used calculus in no way proves that had to, nor they they even thought they had to, which I'm sure they did not. You are capable of better logic than that, your argument is a litany of holes and fallacies.

Let me summarize the truth here. The Lorentz transformation between observers in constant relative motion in the absence of gravity is a global linear transformation from Minkowski space to Minkowski space. True or false? (True.)

The derivative of a global linear transformation is the same transformation, it doesn't do anything to take a derivative of a global linear transformation. True of false? (True.)

Ergo, taking a derivative of a global Lorentz transformation is about as insightful as doing a Taylor expansion of a polynomial. True or false? (True.)

When doing velocity transformations between frames, the resulting coordinate velocity is a ratio of a spatial coordinate displacement to a time interval. Ratios involve division only. Coordinate displacements come from the Lorentz transformation from the original frame, which are linear operations. Linear operations are algebraic, involving only multiplication and addition. True or false? (True.)

Ergo, any coordinate velocity one derives via Lorentz transformations, regardless of how the formula works out, represents an algorithm that involves only multiplication, addition, and division. True or false? (True.)

Ergo, and Q.E.D., any formula for velocity transformations between frames in SR are not going to require calculus. Calculus comes in if the velocity transformations are only instantaneous, so only apply for an infinitesmal time "t", so that transformations over a finite time must account for changes in velocity using calculus. This is indeed what calculus is for in transforming between frames. Calculus is also used in relativity in one other way-- relativity is a philosophical constraint on the laws of physics themselves, just as it was for Galilean relativity (just a different constraint, involving the constancy of c). Since the laws of physics themselves require calculus to state in their most general forms, understanding how relativity constrains them involves calculus. None of that has anything directly to do with the OP or time dilation, as it would be equally true in a universe without time dilation, just with a different relativity. Also, much of introductory physics is taught without calculus quite routinely, and the constraints that relativity imposes on that flavor of introductory physics is equally devoid of the need for calculus.

I just can't say it any clearer. Yes, in prose, and it's pure math.

macaw
2011-May-29, 01:01 AM
We've had that debate before, prose is math.

Not for mainstream physics.
So you are unable to prove your ATM claims.

Those authors did not have the goal of deriving the formula without using calculus, they had the goal of applying calculus,

Not really, any mainstream physicist will tell you that the challenges I gave you cannot be solved without using calculus (differentiation, to be more precise). You claimed the opposite, yet you are unable to perform any of such derivations.

Look,

I'll even get you started on another one of them, the Lorentz invariance of the wave equation:

(\nabla-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}) \vec{E}=0

You will need to start by showing that:

\frac{\partial}{\partial x}=\gamma \frac{\partial}{\partial x'}-\frac{v}{c^2}\frac{\partial}{\partial t'}

and

\frac{\partial}{\partial t}=\gamma \frac{\partial}{\partial t'}-v\frac{\partial}{\partial x'}

You cannot do the proof without calculus. If you think otherwise, show your math.

Ken G
2011-May-29, 01:26 AM
Back onto Lorentz invariance again I see, so you're accepting that you could find no flaw in my proof above that the general velocity addition formula doesn't require calculus? Yes or no, it either does or it doesn't. That's the beauty of mathematics.

macaw
2011-May-29, 01:31 AM
Back onto Lorentz invariance again I see,

Yes, I am pointing out the errors in your statements.

so you're accepting that you could find no flaw in my proof above that the general velocity addition formula doesn't require calculus?

Of course it requires calculus, you have produced no "proof". Prose doesn't constitute proof. Mainstream literature contradicts your fringe claim. Derive the velocity composition formula without employing calculus. Let's see it.

Yes or no, it either does or it doesn't. That's the beauty of mathematics.

You never produced one iota of mathematics. Not one.

Ken G
2011-May-29, 01:51 AM
I'm afraid you can't even tell a mathematical proof when it's right in front of you. The post I gave was a logical sequence of statements, each of which must be either true or false, such that if they are all true, the conclusion must also be true. That is mathematical logic. All I see of importance here is that you have failed to point to any false statements in that chain, nor any flaw in its logical flow, so you must accept its conclusion. This is called logic, which when coupled with the mathematical underpinnings of linear transformations and the algorithms involved in multiplication, addition, and division, as it clearly was, we have something we have a term for: mathematics. What I can see is that you are confused about the differences between the following concepts: calculus, algebra, mathematics, and arithmetic. I doubt I can help you there, frankly. No one else seems to think that the general velocity addition formula requires calculus, so I can just leave you in your own private Idaho on that score, since you don't seem to care if you know the truth or not. As we are just going in circles and starting to waste bandwidth, this will be my final post in the thread.

macaw
2011-May-29, 01:54 AM
I'm afraid you can't even tell a mathematical proof when it's right in front of you.

Mathematical proof requires math. You don't have any in any of your 16,000+ posts.
It is very simple, derive the velocity addition formula, derive the Lorentz invariance of the em wave equation. Let's see you do it. If you are unable to do it, admit it and retract your false claim from post 36.

CaptainToonces
2011-May-31, 12:41 AM
Macaw, can you please answer the questions I posed to you in post 32? Or are Strange's answers in post 33 sufficient to express your views?

macaw
2011-May-31, 01:07 AM
yes, that's why i asked. What does the differentiation operator perform, is it multiply by exponent and decrement exponent by 1?

No, it is not. Did you have calculus (http://en.wikipedia.org/wiki/Derivative) in school?

Why is this necessary in understanding the lorenz transform?

It is not necessary in understanding the Lorentz transform. It is instrumental in learning physics.

macaw
2011-May-31, 01:15 AM
In general terms, it gives the rate of change of a function (or the gradient of a graph) at some point.

That is one of the shortcut rules.

In this case, it is going from a measure of the time (t) to the rate of "flow" of time (dt) - or how fast your clock goes.

You can't get very far in physics without at least basic calculus, I'm afraid...

Yep. I agree with all your points.

CaptainToonces
2011-May-31, 08:13 AM
No, it is not. Did you have calculus (http://en.wikipedia.org/wiki/Derivative) in school?

Didn't go to school. What operation does this differentiation operator perform then, if not "multiply by exponent and decrement exponent by 1"?

Strange
2011-May-31, 08:57 AM
Didn't go to school. What operation does this differentiation operator perform then, if not "multiply by exponent and decrement exponent by 1"?

In the case of a simple expression like xn you can use that rule. Other functions have different derivatives (and then there are rules for combining them). More here: http://en.wikipedia.org/wiki/Derivative#Computing_the_derivative

Shaula
2011-May-31, 11:02 AM
Basically:
Imagine a function as a graph of x vs y
You can define a gradient (or slope) between two points on the line (call the points (x0,y0) and (x1, y1)) so:
m = (y1-y0) / (x1-x0)
If you take all the points on the line and get this gradient you could draw another graph of x vs m
The function that fits this defines the instantaneous gradient at any point on the line. This is known as the derivative.

It is a little more complex than this but that is a sort of basic explanation. Differentiation is a mathematical way to derive this function - so the result of dy/dx is a function that, it you plug in any x, gives you the gradient at that point.

Mercier
2011-May-31, 11:39 AM
In the case of a simple expression like xn you can use that rule.

That is completely correct. Just to avoid one possible point of confusion, this rule only works when the variable is not in the exponent, but down below. The rule for 2x is different.

Strange
2011-Jun-14, 05:02 PM
XKCD: Newton and Leibniz (http://xkcd.com/626/)

macaw
2011-Jul-06, 02:09 AM
Ken G is of course correct.
In an ordinary university setting I would (and do) teach it to my students using calculus.
But if the premise was to not use calculus and if the specific circumstance of the OP was given it could certainly be done without using it.

Ken G made the parameters of his/her analysis clear and considering those he/she is correct.

Do you teach electromagnetism? If yes, how do you get around using calculus?
If not, what qualifications do you have to pass judgement?