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The first postulate of relativity

by metalrose
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metalrose
#1
Dec14-12, 07:02 AM
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I just want to make sure that the following assumption is equivaent to the first postulate.

Consider two inertial frames S and S'. By inertial frame, is meant, a frame where all the laws of physics are valid in their simplest form.

Now let an observer in S measure the velocity of S' to be 'v' towards his/her right.
Now we automatically assume, that when an observer in S' measures the velocity of S, he/she will measure it to be 'v' to his/her left.

How can we say that the two observers will agree on the magnitude of the velocities they measure, of the other frame, from their frame?

Is this just an assumption equivalent to the first postulate of relativity which says that :
Given an inertial frame S, a frame S' moving with constant velocity with respect to S, will also be inertial.

According o the above postilate then,

Given S is inertial, and an observer in S measures constant v as the velocity of S', this implies S' is inertial.

Now we know S' is inertial and S is also inertial, so an observer in S' measuring the velocity of S would observe a constant velocity according to the first postulate. It doesn't however garauntee that this velocity's magnitude is going to be v itself.
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phinds
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Dec14-12, 07:24 AM
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That seems to be like saying that if you put a minus sign in front of an integer, you have changed the absolute value of the integer. Just doesn't seem to make sense.
ghwellsjr
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Dec14-12, 10:41 AM
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Quote Quote by metalrose View Post
I just want to make sure that the following assumption is equivaent to the first postulate.

Consider two inertial frames S and S'. By inertial frame, is meant, a frame where all the laws of physics are valid in their simplest form.

Now let an observer in S measure the velocity of S' to be 'v' towards his/her right.
Now we automatically assume, that when an observer in S' measures the velocity of S, he/she will measure it to be 'v' to his/her left.

How can we say that the two observers will agree on the magnitude of the velocities they measure, of the other frame, from their frame?

Is this just an assumption equivalent to the first postulate of relativity which says that :
Given an inertial frame S, a frame S' moving with constant velocity with respect to S, will also be inertial.

According o the above postilate then,

Given S is inertial, and an observer in S measures constant v as the velocity of S', this implies S' is inertial.

Now we know S' is inertial and S is also inertial, so an observer in S' measuring the velocity of S would observe a constant velocity according to the first postulate. It doesn't however garauntee that this velocity's magnitude is going to be v itself.
I think you are just pointing out why the first postulate (the principle of relativity) by itself is not enough to establish Einstein's Theory of Special Relativity which also needs the second postulate (that all light propagates at c in all inertial frames).

Nugatory
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Dec14-12, 11:05 AM
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The first postulate of relativity

Quote Quote by metalrose View Post
How can we say that the two observers will agree on the magnitude of the velocities they measure, of the other frame, from their frame?

Is this just an assumption equivalent to the first postulate of relativity which says that :
Given an inertial frame S, a frame S' moving with constant velocity with respect to S, will also be inertial.
If I understand correctly, you're asking whether SR, like Galilean relativity before it, assumes that A's speed relative to B is equal to B's speed relative to A (and note that I said "speed" not "velocity"); or whether instead that follows from the first principle?

As an assumption, this is so totally unobjectionable (no observed violations, ever) that I doubt many people spend much time considering the question. But if you want to prove it, there are some pretty convincing reductio ad absurdum arguments.
metalrose
#5
Dec14-12, 11:33 AM
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Quote Quote by Nugatory View Post
If I understand correctly, you're asking whether SR, like Galilean relativity before it, assumes that A's speed relative to B is equal to B's speed relative to A (and note that I said "speed" not "velocity"); or whether instead that follows from the first principle?

As an assumption, this is so totally unobjectionable (no observed violations, ever) that I doubt many people spend much time considering the question. But if you want to prove it, there are some pretty convincing reductio ad absurdum arguments.
I'd really like to read upon such arguments. Could you point me to some such arguments??
DaleSpam
#6
Dec14-12, 12:12 PM
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Quote Quote by metalrose View Post
Now let an observer in S measure the velocity of S' to be 'v' towards his/her right.
Now we automatically assume, that when an observer in S' measures the velocity of S, he/she will measure it to be 'v' to his/her left.

How can we say that the two observers will agree on the magnitude of the velocities they measure, of the other frame, from their frame?
Let's assume the opposite. Suppose an observer in S measures the velocity of S' to be v to the right but an observer in S' measures the velocity of S to be u to the left where v≠u. Then, suppose an observer in S constructs a standard gun which fires a standard projectile at a muzzle velocity of v. Then, an observer with an identical gun and projectile could determine if they are in S or S' by measuring the muzzle velocity. That would require that the laws of physics were different in S and S', which violates the principle of relativity. Therefore, the assumption is not compatible with the principle of relativity.
metalrose
#7
Dec15-12, 01:45 AM
P: 126
Quote Quote by DaleSpam View Post
Let's assume the opposite. Suppose an observer in S measures the velocity of S' to be v to the right but an observer in S' measures the velocity of S to be u to the left where v≠u. Then, suppose an observer in S constructs a standard gun which fires a standard projectile at a muzzle velocity of v. Then, an observer with an identical gun and projectile could determine if they are in S or S' by measuring the muzzle velocity. That would require that the laws of physics were different in S and S', which violates the principle of relativity. Therefore, the assumption is not compatible with the principle of relativity.
I didn't get the logic.
By muzzle velocity, i understand the velocity of the bullet with respect to the gun kept at rest in a particular frame, just at the moment the bullet leaves the gun.
This velocity will be fixed for a particular make of the gun.

Now an observer in S could fire the gun, and measure the muzzle velocity to be v_m, say.
Now, an observer in S' , with an identical gun, may obsetve the same consequences.
But this has no bearing on the velocities of the others' frame that each of them observes ...
DaleSpam
#8
Dec15-12, 11:08 AM
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Quote Quote by metalrose View Post
.This velocity will be fixed for a particular make of the gun.
Not if the principle of relativity doesn't hold. Then it will be fixed for a particular make and reference frame.
metalrose
#9
Dec15-12, 01:30 PM
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Quote Quote by DaleSpam View Post
Not if the principle of relativity doesn't hold. Then it will be fixed for a particular make and reference frame.
Ok, i get that.
To say that the principle of relativity doesn't hold is the same as saying that the laws of physic are differnt in the two frames.
That would lead to the gun behaving differently in the two frames.
Which means the two observers will measure different muzzle velocities.

But this still doesn't lead to the my initial problem of v≠u.
Fredrik
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Dec15-12, 01:57 PM
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Quote Quote by metalrose View Post
Now let an observer in S measure the velocity of S' to be 'v' towards his/her right.
Now we automatically assume, that when an observer in S' measures the velocity of S, he/she will measure it to be 'v' to his/her left.
The first postulate is the principle of relativity, which says roughly that all inertial frames are equivalent. Since that's a rather loosely stated idea, you can't prove rigorously that it's equivalent to (or implies) some other idea. But I would still consider the idea that you're describing to be an aspect of the principle of relativity, since this is something we would expect to hold, if the principle of relativity holds.

Consider e.g. two identical objects A and B attached to a compressed spring that's prevented from expanding e.g. by a string. If you cut the string, the two objects will be shot away from each other. These objects are being subjected to essentially the same thing, so it would be pretty weird if the speed of A relative to B is not the same as the speed of B relative to A.

If those speeds are different, then what would be the reason for that difference? Since the objects are identical, something else would have to differ. For example, there could be something fundamentally different about the two positions in space where the objects started, or the two objects could have started out with a non-zero velocity relative to some kind of invisible "stuff" that fills up all of space, and the speed of one of the objects relative to the "stuff" increased while the speed of the other relative to the "stuff" decreased. These are the sort of things that the principle of relativity is used to rule out.
metalrose
#11
Dec15-12, 02:17 PM
P: 126
Quote Quote by Fredrik View Post
The first postulate is the principle of relativity, which says roughly that all inertial frames are equivalent. Since that's a rather loosely stated idea, you can't prove rigorously that it's equivalent to (or implies) some other idea. But I would still consider the idea that you're describing to be an aspect of the principle of relativity, since this is something we would expect to hold, if the principle of relativity holds.

Consider e.g. two identical objects A and B attached to a compressed spring that's prevented from expanding e.g. by a string. If you cut the string, the two objects will be shot away from each other. These objects are being subjected to essentially the same thing, so it would be pretty weird if the speed of A relative to B is not the same as the speed of B relative to A.

If those speeds are different, then what would be the reason for that difference? Since the objects are identical, something else would have to differ. For example, there could be something fundamentally different about the two positions in space where the objects started, or the two objects could have started out with a non-zero velocity relative to some kind of invisible "stuff" that fills up all of space, and the speed of one of the objects relative to the "stuff" increased while the speed of the other relative to the "stuff" decreased. These are the sort of things that the principle of relativity is used to rule out.
Lets attach frame S to the object A and frame S' to the object B.
Now your argument definitely makes sense because theres a spring thats doing the same thing to both A and B, and so, unless theres some weird stuff going on related to initial positions and stuff, by symmetry, the relative velocities of A wrt B and vice versa should be the same.

But this entire argument depends upon the common spring, doing the same thing to the two frames.

What if there isn't a spring? Just frame S and S' , having constant relative velocities w.r.t. Each other? How do i now prove that the velocity of S w.r.t. S' is the same as the velocity of S' w.r.t. S ??
Fredrik
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Dec15-12, 04:58 PM
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Quote Quote by metalrose View Post
Lets attach frame S to the object A and frame S' to the object B.
Now your argument definitely makes sense because theres a spring thats doing the same thing to both A and B, and so, unless theres some weird stuff going on related to initial positions and stuff, by symmetry, the relative velocities of A wrt B and vice versa should be the same.

But this entire argument depends upon the common spring, doing the same thing to the two frames.

What if there isn't a spring? Just frame S and S' , having constant relative velocities w.r.t. Each other? How do i now prove that the velocity of S w.r.t. S' is the same as the velocity of S' w.r.t. S ??
"Prove" is the wrong word in my opinion, since we're not dealing with mathematical statements. This is just a matter of whether a mathematical statement should be thought of as making an aspect of a loosely stated idea mathematically precise. Because of that, I don't think we even need an argument. I would say that if the velocity of S' in S is v, and the velocity of S in S' isn't -v, then that is by itself a good enough reason to not consider S and S' equivalent as far as physics is concerned. If two observers describe each other more differently than this, why would we call them equivalent?
Studiot
#13
Dec15-12, 05:06 PM
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Consider two inertial frames S and S'. By inertial frame, is meant, a frame where all the laws of physics are valid in their simplest form.
That's a strange definition, don't you find it a bit woolly?

I have seen

An inertial frame is one which is not accelerating.

and

An intertial frame is one that is either at rest or moving with constant velocity.
metalrose
#14
Dec16-12, 01:18 AM
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Quote Quote by Fredrik View Post
"Prove" is the wrong word in my opinion, since we're not dealing with mathematical statements. This is just a matter of whether a mathematical statement should be thought of as making an aspect of a loosely stated idea mathematically precise. Because of that, I don't think we even need an argument. I would say that if the velocity of S' in S is v, and the velocity of S in S' isn't -v, then that is by itself a good enough reason to not consider S and S' equivalent as far as physics is concerned. If two observers describe each other more differently than this, why would we call them equivalent?
That makes sense i guess. Thanks a lot.
metalrose
#15
Dec16-12, 01:20 AM
P: 126
Quote Quote by Studiot View Post
That's a strange definition, don't you find it a bit woolly?

I have seen

An inertial frame is one which is not accelerating.

and

An intertial frame is one that is either at rest or moving with constant velocity.
Accelerating with respect to what ? Rest or constant velocity with respect to what ?

The definition i used is a textbook definition.
An inertial frame is one where all laws of physics are valid.
Now, if some other frame i moving with respect to this inertial frame at constant velocity, then even that becomes an inertial frame.
strangerep
#16
Dec16-12, 02:21 AM
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Quote Quote by metalrose View Post
Accelerating with respect to what ?
Acceleration is detectable locally.
The definition i used is a textbook definition.
An inertial frame is one where all laws of physics are valid.
No, an inertial frame is one in which the observer feels no acceleration.
See below...
Quote Quote by Wolfgang Rindler
A frame of reference is a conventional standard of rest, relative to which measurements can be made and experiments described. By "at rest" we understand that an observer at the origin of a reference frame feels no acceleration.
Similarly, an (Einsteinian) inertial frame is a reference frame in which spatial relations, as determined by rigid scales at rest in the frame, are Euclidean and in which there exists a universal time in terms of which free particles remain at rest or continue to move with constant speed along straight lines (i.e., in terms of which free particles obey Newton's first law).

With these definitions, we can formulate (still following Rindler):

Postulate 1 (Principle of Relativity):
The laws of physics are identical in all inertial frames, or, equivalently, the outcome of any physical experiment is the same when performed with identical initial conditions relative to any inertial frame.
strangerep
#17
Dec16-12, 02:29 AM
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Quote Quote by Fredrik View Post
[...] if the velocity of S' in S is v, and the velocity of S in S' isn't -v, then that is by itself a good enough reason to not consider S and S' equivalent as far as physics is concerned.
FYI, I just generalized some of my calculations and verified that from the requirements of inertial motion, spatial isotropy, and the assumption that boosts along a given axis form a 1-parameter Lie group, one can derive the result that a boost ##\Lambda(v)## has inverse given by ##\Lambda(-v)##.

I.e., one does not need to assume that the inverse corresponds to the parameter ##-v##, (as is done in some group theoretic derivations).

Afaict, this establishes the frame reciprocity mentioned above.
metalrose
#18
Dec16-12, 02:57 AM
P: 126
Quote Quote by strangerep View Post
FYI, I just generalized some of my calculations and verified that from the requirements of inertial motion, spatial isotropy, and the assumption that boosts along a given axis form a 1-parameter Lie group, one can derive the result that a boost ##\Lambda(v)## has inverse given by ##\Lambda(-v)##.

I.e., one does not need to assume that the inverse corresponds to the parameter ##-v##, (as is done in some group theoretic derivations).

Afaict, this establishes the frame reciprocity mentioned above.
None of the above made any sense to me, thats because im still a junior in undergrad.
Any resources i could use at this point to understand the above ??


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