View Full Version : A question about relativity

Doe, John

2004-Jul-16, 09:14 AM

I'm not sure this is the right forum for this question, but astronomy and physics are two closely related disciplines and the question is kinda from left field, so here goes.

If two space ships accelerating with reciprocal vectors (not sure I have the terminology right) each have a velocity of .51c then to an observer on one of said ships would not the other ship be exceeding the speed of light?

If this is too speculative for discussion I apologize but this question has been bugging me ever since I thought of it. :-s

obtw this bb rocks.

AstroSmurf

2004-Jul-16, 10:08 AM

Quick version: No, because under relativity you don't add velocities by simple addition.

Long version: We have three guys, A and B in spaceships, and C who is the one measuring their velocity.

According to C, A has velocity u = 0.51c and B has velocity w = -0.51c. Now, according to relativity, C also sees time for A and B slowed down, and further, their measurements of distances are different.

In the end, you arrive at the formula for relativistic velocity addition:

v = (w - u)/(1 - wu/c2)

With u and w as stated above, this gives us the velocity of A as B sees it: v = 0.81c. For A, switching places for w and u gives us B's velocity v = -0.81c, as we would expect.

cable

2004-Jul-16, 01:21 PM

I'm not sure this is the right forum for this question, but astronomy and physics are two closely related disciplines and the question is kinda from left field, so here goes.

Welcome to BABB.

You are at the right forum to discuss Relativity.

Doe, John

2004-Jul-16, 02:54 PM

Quick version: No, because under relativity you don't add velocities by simple addition.

Long version: We have three guys, A and B in spaceships, and C who is the one measuring their velocity.

According to C, A has velocity u = 0.51c and B has velocity w = -0.51c. Now, according to relativity, C also sees time for A and B slowed down, and further, their measurements of distances are different.

In the end, you arrive at the formula for relativistic velocity addition:

v = (w - u)/(1 - wu/c2)

With u and w as stated above, this gives us the velocity of A as B sees it: v = 0.81c. For A, switching places for w and u gives us B's velocity v = -0.81c, as we would expect.

Ok, I got the math, but doesn't this show their relative speeds according to the rest of the universe? Since according to C, A and B are both experiencing time at different rates wouldn't their perceptions be different than the universe at rest? Man thinking about this makes my brain hurt.

I guess what I don't get is why you have to bring C into the picture.

Doe, John

2004-Jul-16, 02:58 PM

Thanks cable

SeanF

2004-Jul-16, 03:13 PM

I guess what I don't get is why you have to bring C into the picture.

Actually, you brought C into the picture, in your original post:

If two space ships accelerating with reciprocal vectors (not sure I have the terminology right) each have a velocity of .51c . . .

They "each have a velocity of .51c" relative to what? :)

Doe, John

2004-Jul-16, 04:40 PM

I guess what I don't get is why you have to bring C into the picture.

Actually, you brought C into the picture, in your original post:

If two space ships accelerating with reciprocal vectors (not sure I have the terminology right) each have a velocity of .51c . . .

They "each have a velocity of .51c" relative to what? :)

<light slowly dawning

So it's BECAUSE they both experience time at different rates that their velocities can't just be added together?? :-k

Normandy6644

2004-Jul-16, 04:56 PM

I guess what I don't get is why you have to bring C into the picture.

Actually, you brought C into the picture, in your original post:

If two space ships accelerating with reciprocal vectors (not sure I have the terminology right) each have a velocity of .51c . . .

They "each have a velocity of .51c" relative to what? :)

<light slowly dawning

So it's BECAUSE they both experience time at different rates that their velocities can't just be added together?? :-k

It's more a consequence of the "cosmic speed limit" imposed by light. Since you have to speak of velocities (at least at speeds close to c) as relative to something, you can't add them quite so directly. That's what the equation says to, it defines the velocities to be added as relative to either the object or to an observer. For example, let's say you were on a ship travelling at .9c and you threw a baseball at .5c (you have a good arm), how fast does the baseball appear to travel according to an observer watching the ship fly past. See how the question is phrased? That makes all the difference. Classical reasoning (Galilean) would have you think that the answer would be 1.4c, but we know that's not possible! Think you can take a stab at the answer?

Doe, John

2004-Jul-16, 05:40 PM

I guess what I don't get is why you have to bring C into the picture.

Actually, you brought C into the picture, in your original post:

If two space ships accelerating with reciprocal vectors (not sure I have the terminology right) each have a velocity of .51c . . .

They "each have a velocity of .51c" relative to what? :)

<light slowly dawning

So it's BECAUSE they both experience time at different rates that their velocities can't just be added together?? :-k

It's more a consequence of the "cosmic speed limit" imposed by light. Since you have to speak of velocities (at least at speeds close to c) as relative to something, you can't add them quite so directly. That's what the equation says to, it defines the velocities to be added as relative to either the object or to an observer. For example, let's say you were on a ship travelling at .9c and you threw a baseball at .5c (you have a good arm), how fast does the baseball appear to travel according to an observer watching the ship fly past. See how the question is phrased? That makes all the difference. Classical reasoning (Galilean) would have you think that the answer would be 1.4c, but we know that's not possible! Think you can take a stab at the answer?

Well, AstroSmurf provided a formula for relativistic velocity addition; v = (w-u)/(1 - wu/c^2). So I imagine I could just plug in values of .9c for w and .5c for u to arrive at a value of v= .73c? :-s

No, that can't be right, how about .5c for w and .9c for u?, that would give a result of v= -.73c :x

I hit the cranial capacity limit switch, system reset #-o

Ricimer

2004-Jul-16, 05:44 PM

yeah, you brought the C into it.

If I head towards you at 10 miles an hour, I could say you're heading towards me at 10 miles per hour, I'm stationary.

Now, if you say I'm heading towards you at 10mph, AND you're heading towards me at 10mph, we've implied a common reference frame (usually the surface of the earth in our case).

Now, I, being very egocentric, think you're heading towards me at 20mph, and the earth is moving at 10mph. Same can go for you.

The reason we can't just add them aritmetically is because the speed of light speed limit. Now, the reasons that's a speed limit is because space and time change, and can only change so much (go to no distance, and no passage of time).

But yeah, you're on the right track, essentially.

The important thing to remember is that all velocities are measured in reference to something that is assumed to be stationary. Nothing is inherently motionless, though, so you have to choose a reference object to be "stationary". When you say that Ship A has a velocity of .51c, and Ship B has a velocity of -.51c, then you autmoatically assume there to be an observer C in such a frame as to view the two ships, A and B, moving at the given velocities relative to his/her/it-self. Otherwise, how do you know how fast the ships are going?

cable

2004-Jul-16, 06:00 PM

yeah, you brought the C into it.

The reason we can't just add them aritmetically is because the speed of light speed limit. Now, the reasons that's a speed limit is because space and time change, and can only change so much (go to no distance, and no passage of time).

I disagree.

C, the speed of light in vacuum as a limit is a postulate in Relativity.

It's value bears uncertainty due to errors in measurment.

What we can say is, no experiment shows a speed exceeding C

Ricimer

2004-Jul-16, 06:07 PM

We have shown that C doesn't change depending on our relative motion (MM experiments and others).

We have shown that the rate of time slows as you go faster (cosmic ray decay rates, particle accelerator results).

Length contraction...I don't think we've proved yet.

Anyway, if you accept that the speed of light is constant, in all reference frames (implied by maxwell's laws and the relativity principle) then the time dilation and length contraction logically follow (as the way it remains constant).

The logical consequence of time dilation and length contraction is that you cannot exceed the speed of light.

If you accept Maxwell's equation as valid, and the relativity principle as valid (that everybody uses the same physics everywhere, under all conditions) the rest fall out from that.

Doe, John

2004-Jul-16, 06:26 PM

I guess what I don't get is why you have to bring C into the picture.

Actually, you brought C into the picture, in your original post:

If two space ships accelerating with reciprocal vectors (not sure I have the terminology right) each have a velocity of .51c . . .

They "each have a velocity of .51c" relative to what? :)

<light slowly dawning

So it's BECAUSE they both experience time at different rates that their velocities can't just be added together?? :-k

It's more a consequence of the "cosmic speed limit" imposed by light. Since you have to speak of velocities (at least at speeds close to c) as relative to something, you can't add them quite so directly. That's what the equation says to, it defines the velocities to be added as relative to either the object or to an observer. For example, let's say you were on a ship travelling at .9c and you threw a baseball at .5c (you have a good arm), how fast does the baseball appear to travel according to an observer watching the ship fly past. See how the question is phrased? That makes all the difference. Classical reasoning (Galilean) would have you think that the answer would be 1.4c, but we know that's not possible! Think you can take a stab at the answer?

Well, AstroSmurf provided a formula for relativistic velocity addition; v = (w-u)/(1 - wu/c^2). So I imagine I could just plug in values of .9c for w and .5c for u to arrive at a value of v= .73c? :-s

No, that can't be right, how about .5c for w and .9c for u?, that would give a result of v= -.73c :x

I hit the cranial capacity limit switch, system reset #-o

I think I got it. If I change the equation so that the velocities are added together rather than subtracted, v = (w+u)/(1-uw/c^2)which would be the case in your example as contrasted to mine then I come up with value for v of .97c which imo is a much more believable number :)

Normandy6644

2004-Jul-16, 06:32 PM

I guess what I don't get is why you have to bring C into the picture.

Actually, you brought C into the picture, in your original post:

If two space ships accelerating with reciprocal vectors (not sure I have the terminology right) each have a velocity of .51c . . .

They "each have a velocity of .51c" relative to what? :)

<light slowly dawning

So it's BECAUSE they both experience time at different rates that their velocities can't just be added together?? :-k

It's more a consequence of the "cosmic speed limit" imposed by light. Since you have to speak of velocities (at least at speeds close to c) as relative to something, you can't add them quite so directly. That's what the equation says to, it defines the velocities to be added as relative to either the object or to an observer. For example, let's say you were on a ship travelling at .9c and you threw a baseball at .5c (you have a good arm), how fast does the baseball appear to travel according to an observer watching the ship fly past. See how the question is phrased? That makes all the difference. Classical reasoning (Galilean) would have you think that the answer would be 1.4c, but we know that's not possible! Think you can take a stab at the answer?

Well, AstroSmurf provided a formula for relativistic velocity addition; v = (w-u)/(1 - wu/c^2). So I imagine I could just plug in values of .9c for w and .5c for u to arrive at a value of v= .73c? :-s

No, that can't be right, how about .5c for w and .9c for u?, that would give a result of v= -.73c :x

I hit the cranial capacity limit switch, system reset #-o

I think I got it. If I change the equation so that the velocities are added together rather than subtracted, v = (w+u)/(1-uw/c^2)which would be the case in your example as contrasted to mine then I come up with value for v of .97c which imo is a much more believable number :)

The only thing is I thought it was a plus in the denominator. I'll have to look it up to be sure.

cable

2004-Jul-16, 06:46 PM

Length contraction...I don't think we've proved yet.

If you accept Maxwell's equation as valid, and the relativity principle as valid (that everybody uses the same physics everywhere, under all conditions) the rest fall out from that.

Time delatation was proven by atomic clocks.

As for lenght contraction, I think you're right.

The only thing we can say, is our laws of physics are valid within our solar system, or at it's vacinity.

We are making a huge extrapolation on the whole universe.

Gravity may not be the same everywhere, especially at the edge of galaxy.

Ricimer

2004-Jul-16, 09:15 PM

in order for science to work at all, i.e. to predict anything, it has long been accepted that we must make one of several very basic assumptions (and it is just that, cause we can't know, as you said) and that's the law of universiality or somesuch.

Basically what holds here on earth holds everywhere.

If you deny that, then you have removed all predictive powers of science, and no longer support cause and effect relationships. As such, why bother? So it becomes a basic underpinning to science, as fundamental to physics as 1+1=2 is to math. Heck, probably moreso.

Doe, John

2004-Jul-16, 11:40 PM

Oops, it is a plus in the denominator and that is the way I did the calculation. My bad

Still the discussion has definately helped my understanding of the concept. Thanks all

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