Special relativity - please can someone clear this up?

[John_M]

I don't think this is an original point - but it seems an obvious problem with the idea of time dilation for which I haven't heard a satisfactory answer. I don't have any knowledge of physics beyond GCSE (low level high school) so keep it simple please!

The 'twin paradox'. I'm sure everyone's familiar with it. One twin stays on earth, the other flies off somewhere at high speed and returns. Moving clocks run slow and therefore the twin in the spaceship has not aged as much as the twin who's stayed on Earth.

Isn't it the case that this theory simply assumes the Earth is an 'absolute' frame of reference, directly contradicting the relativity principle? Why can't you say that the twin on Earth is moving, the twin in the spaceship is still, and that the Earth twin's clock should therefore 'run slow'? How can you make the general statement that 'moving clocks run slow' without violating the relativity principle - whether something is moving or not depends on your frame of reference?

 

[Planetary Nebula]

It isn't based merely on saying that "moving clocks run slow", but on working out the whole roundtrip problem, considering the times from the standpoints of both reference frames. In the end, both twins agree about the difference in total time of the trip and which twin has logged the most time. The relativity principle isn't violated either. But the case needs to be closely argued.

 

[jcsd]

I don't think this is an original point - but it seems an obvious problem with the idea of time dilation for which I haven't heard a satisfactory answer. I don't have any knowledge of physics beyond GCSE (low level high school) so keep it simple please!

The 'twin paradox'. I'm sure everyone's familiar with it. One twin stays on earth, the other flies off somewhere at high speed and returns. Moving clocks run slow and therefore the twin in the spaceship has not aged as much as the twin who's stayed on Earth.

Isn't it the case that this theory simply assumes the Earth is an 'absolute' frame of reference, directly contradicting the relativity principle? Why can't you say that the twin on Earth is moving, the twin in the spaceship is still, and that the Earth twin's clock should therefore 'run slow'? How can you make the general statement that 'moving clocks run slow' without violating the relativity principle - whether something is moving or not depends on your frame of reference?

No, it doesn't consider the Earth as an absolute frame of refernce, BUT there's an importnat difference between the two refrence frames: the refrence frame on the Earth is an inertial refernce frame, but the refernce frame of the spaceship is non-inertial (i.e. it accelerates). The principle of relativity only applies to inertial reference frames.

You shouldn't make the statement 'moving clocks run slow', as it may not be true: i.e. moving relatuive to what? in which frame of rest frame doe sit run slow?

 

[zoobyshoe]

John,

Here are two quotes from another thread in which the twin paradox was finally explained to me in a way that made sense. The key to my understanding was the concept of the space-time interval. The space-time interval for one twin is not symetrical to that for the other.

A path in spacetime is called an interval.

The length of an interval is called its proper time. If you have a clock follow some path, the clock will measure that much time having been elapsed when moved along that path.

This is not really true -- a distant observer will measure a clock as running slowly when it is moving at high relative velocity to the observer. According to the clock, however, everything is just fine.

Imagine Picard is flying along in the Enterprise at 0.9c with respect to the Earth. An observer on the Earth will measure Picard's clock as running slow compared to an identical Earth-bound clock. Picard, however, will see everything on the bridge of the Enterprise as running completely normally, but will measure the Earth-bound clock as running slowly.

If you think about it, it has to be that way... if it weren't, then some cosmic ray particle moving at 0.9c with respect to you somewhere in the depths of space would somehow affect YOUR clock!

- Warren

The longer path ages more, if you measure "length" using the spacetime interval.

Consider the case where the Earth twin stays at home for 10 years according to his own clock. The other twin travels at 80% of light speed, traveling 4 lightyears in 5 years according to the Earth twin, then returns the same way (after an instantaneous deceleration and acceleration back home).

The Earth twin's worldline is a line from (t,x) = (0,0) to (10,0). The traveling twin's worldline consists of two line segments, one from (0,0) to (5,4), the other from (5,4) to (10,0).

The proper time measured by the Earth twin is the spacetime length of his worldline,

<br /> \tau = \sqrt{{\Delta t}^2-{\Delta x}^2} = \sqrt{(10-0)^2-(0-0)^2} = 10<br />

The proper time measured by the traveling twin is the spacetime length of his worldline,

<br /> \begin{equation*}<br /> \begin{split}<br /> \tau &amp;= \tau_1+\tau_2 = \sqrt{{\Delta t_1}^2-{\Delta x_1}^2}+<br /> \sqrt{{\Delta t_2}^2-{\Delta x_2}^2} \\<br /> &amp;= \sqrt{(5-0)^2-(4-0)^2}+\sqrt{(10-5)^2-(0-4)^2}\\<br /> &amp;= 3+3 = 6<br /> \end{split}<br /> \end{equation*}<br />

The traveling twin ages 6 years to the Earth twin's 10 years.

variable speed of light - Physics Help and Math Help - Physics Forums
Address:https://www.physicsforums.com/showthread.php?t=8769&page=1&pp=20[/QUOTE]

 

[John_M]

"No, it doesn't consider the Earth as an absolute frame of refernce, BUT there's an importnat difference between the two refrence frames: the refrence frame on the Earth is an inertial refernce frame, but the refernce frame of the spaceship is non-inertial (i.e. it accelerates). The principle of relativity only applies to inertial reference frames."

I don't see this. Why can't you say that the spaceship is inertial reference frame, relative to which the Earth is accelerating? Doesn't your answer just restate the same problem?

I agree with your point about 'moving clocks run slow' - but that was exactly my question - I'm trying to get to grips with SR and a good deal of the internet is covered in the phrase! :)

 

[jcsd]

I don't see this. Why can't you say that the spaceship is inertial reference frame, relative to which the Earth is accelerating? Doesn't your answer just restate the same problem?

You can't say this in Newtonian physics and you can't say this in special relativity, as I said before the principle of relativity specifically only applies to inertial frames of refernbce not accelerating frames of refernce and there is a definite difference between the two.

The keypoint is in special relativity the laws of physics look different in accelerated frames of refreence than they do in inertial refrence frames, hence the need for general relativity.

edited for typing errors and to clarify:

what I mean by "hence the need for general relativity", is NOT that you need general relativity to resolve the paradox as that can be done using special relativity alone, BUT you need general relativity in order to express the laws of physics in a frame invariant manner.

 

[John_M]

jcsd...

OK, I remember something about this. Coffee sloshing all over you when you're accelerating etc. But...two questions.

1. The principle of relativity still applies to inertial reference frames. So - assuming you believe that the space-twin ages slower - are you saying that any process of 'slower ageing' taking place on the space-twin's clock occurs only when he is accelerating?

2. I don't really understand zoob's post (Too much maths - I only have a basic knowledge of maths and physics!) But it seems to offer a different explanation to the problem than yours does. It specifically avoids the problem of acceleration/ deceleration:

"The other twin travels at 80% of light speed, traveling 4 lightyears in 5 years according to the Earth twin, then returns the same way (after an instantaneous deceleration and acceleration back home)."

Do you agree with zoob's post?

 

[jcsd]

1. No I'm not saying that 'slow aging' only take place during acceleration. Imagine a spaceship traveling past the Earth at a constant velocity; a clock on the spaceship will appeared to be running slow to someone on the Earth, but it's important to note that there is still symmetry as a clock on the Earth will appear to be running slow to somoen on the Earth.

2) I agree with zoob's post, notice the spaceship is chaning it's inertial frame while the Earth does not.

 

[John_M]

1. No I'm not saying that 'slow aging' only take place during acceleration. Imagine a spaceship traveling past the Earth at a constant velocity; a clock on the spaceship will appeared to be running slow to someone on the Earth, but it's important to note that there is still symmetry as a clock on the Earth will appear to be running slow to somoen on the Earth.

2) I agree with zoob's post, notice the spaceship is changing it's inertial frame while the Earth does not.

I'm still lost...

With reference to point (1) - A clock on the spaceship will appear to be running slow to someone on the Earth but a clock on the Earth will appear to be running slow to someone on the (I assume you mean 'spaceship' here!)

This is exactly my problem. If there are no absolute frames of reference, what possible reason could there be for saying that, when the frames of reference are synchronised, one particular twin will have aged faster than the other?

 

[jcsd]

I'm still lost...

With reference to point (1) - A clock on the spaceship will appear to be running slow to someone on the Earth but a clock on the Earth will appear to be running slow to someone on the (I assume you mean 'spaceship' here!)

correct.

This is exactly my problem. If there are no absolute frames of reference, what possible reason could there be for saying that, when the frames of reference are synchronised, one particular twin will have aged faster than the other?

Because frames one of the twins has changed his frame of refrence while the other one has not, the symmetry is broken.

 

[zoobyshoe]

2. I don't really understand zoob's post (Too much maths - I only have a basic knowledge of maths and physics!)

It is much less hairy than it looks. I, too, know hardly any math, but those equations actually involve only very basic algebra, and basic algebra is just slightly fancy basic math. You just need to know what the variables stand for, and to be able to handle the square/square root button on your calculator. There's no calculus or anything like that here.

The stay at home twin has only one world line because he doesn't go anywhere. The rocket twin has two worldlines: one for the trip out, one for the trip back.

Ambitwistor's main point is that the stay at home twin ends up experiencing 10 years to the traveling twin's 6 years.

Stuff like the triangle symbol shouldn't scare you off. It just stands for "the change in" or "the difference between".

 

[zoobyshoe]

I'm still lost...

Actually, you should be. It shows you're thinking logically through this, and that I forgot to quote one very important thing by Chroot.

This is exactly my problem. If there are no absolute frames of reference, what possible reason could there be for saying that, when the frames of reference are synchronised, one particular twin will have aged faster than the other?

Here Chroot explains the "absolute" that you're looking for. (I really should have included this in the post above):

The interval is a very important quantity in relativistic physics because it is invariant. No matter what coordinate system you use, or which observers you consider to be at rest, the interval they will measure for some path \Gamma is always the same. The interval is independent of observers and is a fixed quantity for any particular path through spacetime.

- Warren

 

[pmb_phy]

Isn't it the case that this theory simply assumes the Earth is an 'absolute' frame of reference, directly contradicting the relativity principle?

No. Why would you think it implies that? Choose a particular inertial frame of reference. Call that frame S. Now let there be a spacestation and a spaceship each at rest in frame S, each of which are at location A. Suppose now there are two twins. One twin is in the ship and other other in the station. The ship now starts to acclerate. The people in the ship feel the ship accelerating and know that they are not at rest in any inertial frame of reference. However the people in the spacestation are always in an inertial frame. There is a broken symnmetry here which identifies one observer as the traveling twin. When the ship returns the twin inside the ship is younger than the person in the spacestation. This does not single out one inertial frame over the other since there is only one inertial frame in what I've just described, and that's the frame in which the spacestation is a rest.

A path in spacetime is called an interval.

A path is spacetime is called a worldline. An interval is the \Delta s (or to some the (\Delta s)^2) in the expression

(\Delta s)^2 = c^2 (\Delta x)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2

Why can't you say that the spaceship is inertial reference frame, relative to which the Earth is accelerating? Doesn't your answer just restate the same problem?

Because if the ship leaves point A and comes back to point A then it must have turned around. During that time the ship was in a non-inertial frame.

Pete

 

[zoobyshoe]

Hi Pete,

Terminology confusion, I guess. Would the two word term spacetime interval be synonymous with the term worldline?

-Zooby

 

[John_M]

correct.

Because frames one of the twins has changed his frame of refrence while the other one has not, the symmetry is broken.

I'm sorry but it still isn't clear to me! So this is what happens in the twin experiment:

1. Twins A and B are in the same frame of reference.
2. Twin B changes his frame of reference (flying off)
3. The twins are now in a different inertial frame of reference. From A's perspective B's clock is running slower and therefore B ages slower. From B's perspective A's clock is running slower and therefore A ages slower.
4. Twin B reverts to Twin A's frame of reference.
5. Twin B has aged slower than Twin A.

Where, in this sequence of events, has the process of B ageing slower than A taken place? Surely not during point (3) - otherwise you're violating the relativity principle!

Are you suggesting something along the lines of 'because B has changed his frame of reference whereas A has not, B's frame of reference is the 'true' frame of reference?' Or something like that??

zoobyshoe...I'll have another look at those equations...I still find the triangles rather scary though!

Thanks for your help guys!

 

[pmb_phy]

Hi Pete,

Terminology confusion, I guess. Would the two word term spacetime interval be synonymous with the term worldline?

-Zooby

No. "Interval" is a general expession. "Spacetime interval" refers to an interval in spacetime.

Pete

 

[selfAdjoint]

The interval is the spacetime "distance" between two points, quotes because it includes time as well as space distance. Interval is a geometric, not a coordinate fact, and thus will have the same value in every inertial frame. The name for this is Lorentz invariant, or since it's a number, Lorentz scalar. The square of the interval is (in one signiture) -c^2(t_1-t_2)^2 + (x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2

Along a world line, each differential element is a differential interval:

ds^2 = -cdt^2 + dx^2 + dy^2 = dz^2.

 

[John_M]

pmb_phy

"There is a broken symnmetry here which identifies one observer as the traveling twin."

This is similar to what jcsd was saying. Do you mean the following by it:

"Since Twin B has accelerated whereas Twin A has not, it is possible to state in absolute terms that B is moving whereas A is still."

Another point. You say the following:

"This does not single out one inertial frame over the other since there is only one inertial frame in what I've just described, and that's the frame in which the spacestation is a rest."

Right. So in your twin experiment, is it correct to state that B's spaceship, on its voyage, never moves at constant speed - and does not therefore constitute an inertial frame of reference at any point?

 

[pmb_phy]

Interval is a geometric, not a coordinate fact, and thus will have the same value in every inertial frame.

The interval has the same value in all spacetime coordinate systems, not just those which correspond to inertial ones.

Pete

 

[zoobyshoe]

No. "Interval" is a general expession. "Spacetime interval" refers to an interval in spacetime.

OK.

In Ambitwistors post the symbol \tau is "tau", correct? And stands for "worldline"?? Or?

 

[Janus]

I'm sorry but it still isn't clear to me! So this is what happens in the twin experiment:

1. Twins A and B are in the same frame of reference.
2. Twin B changes his frame of reference (flying off)
3. The twins are now in a different inertial frame of reference. From A's perspective B's clock is running slower and therefore B ages slower. From B's perspective A's clock is running slower and therefore A ages slower.
4. Twin B reverts to Twin A's frame of reference.
5. Twin B has aged slower than Twin A.

Where, in this sequence of events, has the process of B ageing slower than A taken place? Surely not during point (3) - otherwise you're violating the relativity principle!

Assuming that Twin B returns to Earth, this happens, according to Twin B, during that phase of the trip when he stops and turns around to return to the Earth. During this turn around period he will see twin A aging very fast. Enought so that it more than compensates for the fact that he sees Twin A age slower during the rest of the trip. How much faster B sees A age depends both on how quickly he does the turn around and how far he is from Twin A when he does it.

Are you suggesting something along the lines of 'because B has changed his frame of reference whereas A has not, B's frame of reference is the 'true' frame of reference?' Or something like that??

No, there is no 'true' frame of reference. Each Twin measures a certain sequence of events, but they both agree on the end result. Twin A always sees Twin B as aging slower, While Twin B sees Twin A age slow during part of the time and fast for part of the time. Both sets of observations are equally valid and you cannot say that one or the other is the "correct" one.

zoobyshoe...I'll have another look at those equations...I still find the triangles rather scary though!


Thanks for your help guys!

The triangle are the greek letter "delta", it stands for "change in".

Thus if a plane is flying at an altitude of A= 2000 ft and climbs to an altitude of A= 3000 ft, then

\Delta A=1000ft

 

[pmb_phy]

Do you mean the following by it:

"Since Twin B has accelerated whereas Twin A has not, it is possible to state in absolute terms that B is moving whereas A is still."

No, because there may be more than one inertial observers which are not in the same inertial frame.

Right. So in your twin experiment, is it correct to state that B's spaceship, on its voyage, never moves at constant speed - and does not therefore constitute an inertial frame of reference at any point?

No. Its possible that the acceleratio of turnaround occurs in a very short time and the ship travels at constant speed otherwise.

Pete

 

[Nenad]

the way time knows who is moving and who isn't is simple. It is because of the acceleration. The twin in the spaceship has to accelerate in order to fly off into space. Both twions look at one another and both twins see that the other twins time is running slower than theirs. But once the twin in the spaceship comes back, he sees that he is wrong and he has aged less than his brother. So youre right, it is during step 3) that the twin on the spaceship has his time slow down and age less.

 

[jcsd]

I'm sorry but it still isn't clear to me! So this is what happens in the twin experiment:

1. Twins A and B are in the same frame of reference.
2. Twin B changes his frame of reference (flying off)
3. The twins are now in a different inertial frame of reference. From A's perspective B's clock is running slower and therefore B ages slower. From B's perspective A's clock is running slower and therefore A ages slower.
4. Twin B reverts to Twin A's frame of reference.
5. Twin B has aged slower than Twin A.

Where, in this sequence of events, has the process of B ageing slower than A taken place? Surely not during point (3) - otherwise you're violating the relativity principle!

Are you suggesting something along the lines of 'because B has changed his frame of reference whereas A has not, B's frame of reference is the 'true' frame of reference?' Or something like that??

zoobyshoe...I'll have another look at those equations...I still find the triangles rather scary though!

Thanks for your help guys!

4) is where the change takes place but it's worth noting that instantaneous acceleration is unphhysical, tho' it helps to illustarte what is happening.

 

[John_M]

"the way time knows who is moving and who isn't is simple. It is because of the acceleration. The twin in the spaceship has to accelerate in order to fly off into space. Both twions look at one another and both twins see that the other twins time is running slower than theirs. But once the twin in the spaceship comes back, he sees that he is wrong and he has aged less than his brother. So youre right, it is during step 3) that the twin on the spaceship has his time slow down and age less."

I'm sure your post contradicts what some of the other people on here have been saying...

 

[John_M]

Let me update my list a bit.

1. Twins A and B are in the same frame of reference.
2. Twin B changes his frame of reference (flying off)
3. The twins are now in a different inertial frame of reference. From A's perspective B's clock is running slower and therefore B ages slower. From B's perspective A's clock is running slower and therefore A ages slower.
4. Twin B turns round and heads back towards Earth.
5. The same as (3) but Twin B is moving in the opposite direction relative to Twin A.
6. Twin B reverts to Twin A's frame of reference.
7. Twin B has aged slower than Twin A.

Next question. If the key elements producing the end result (asymmetrical ageing) are not points (3) or (5), then surely this is not a question of special relativity at all, since acceleration is outside the scope of that theory?

(If they are points (3) and (5) as Nenad seems to be suggesting - surely you have big problems with the relativity principle?)

Quote from pmb_phy:

"This does not single out one inertial frame over the other since there is only one inertial frame in what I've just described, and that's the frame in which the spacestation is a rest."

Doesn't this statement imply that the twin paradox is outside the scope of special relativity?

 

[selfAdjoint]

Let me update my list a bit.

1. Twins A and B are in the same frame of reference.
2. Twin B changes his frame of reference (flying off)
3. The twins are now in a different inertial frame of reference. From A's perspective B's clock is running slower and therefore B ages slower. From B's perspective A's clock is running slower and therefore A ages slower.
4. Twin B turns round and heads back towards Earth.
5. The same as (3) but Twin B is moving in the opposite direction relative to Twin A.
6. Twin B reverts to Twin A's frame of reference.
7. Twin B has aged slower than Twin A.

Next question. If the key elements producing the end result (asymmetrical ageing) are not points (3) or (5), then surely this is not a question of special relativity at all, since acceleration is outside the scope of that theory?

(If they are points (3) and (5) as Nenad seems to be suggesting - surely you have big problems with the relativity principle?)

Quote from pmb_phy:

"This does not single out one inertial frame over the other since there is only one inertial frame in what I've just described, and that's the frame in which the spacestation is a rest."

Doesn't this statement imply that the twin paradox is outside the scope of special relativity?

It is false that acceleration is outside of relativity. The fact that a curved timelike worldline has lower proper time than a straight one arises exactly from the relativistic definition of the interval, so that the "Pythagoras Theorem" of relativity is the difference of two squares instead of the sum.

 

[quantumdude]

OK.

In Ambitwistors post the symbol \tau is "tau", correct? And stands for "worldline"?? Or?

It stands for proper time.

 

[Janus]

Let me update my list a bit.

1. Twins A and B are in the same frame of reference.

A and B measure each other aging as the same rate as themselves

2. Twin B changes his frame of reference (flying off)

A and B each see each others aging gradually begin to slow down. B however sees A age a little bit slower than A sees B age.(addition slowing seen be B is caused by the fact that he is accelerating away from A)

3. The twins are now in a different inertial frame of reference. From A's perspective B's clock is running slower and therefore B ages slower. From B's perspective A's clock is running slower and therefore A ages slower.

Okay.

4. Twin B turns round and heads back towards Earth.

A sees B's aging gradually speed up until it matches his again as B slows to a stop before he starts his return journey. He then sees B's aging gradually slow down again as B accelerates back up to speed on the return trip.
B sees A's aging speed up to much faster than his own during the whole of the turn around manuver.

5. The same as (3) but Twin B is moving in the opposite direction relative to Twin A.

okay

6. Twin B reverts to Twin A's frame of reference.

A and B each see each other's aging gradually speed up until they both age at the same rate, in a reversal of stage 1.

7. Twin B has aged slower than Twin A.

Yes, since A always either sees B age more slowly than or the same as himself and B during stage 4 sees A age greatly (more than enough to make up for stages 3 and 5)

Next question. If the key elements producing the end result (asymmetrical ageing) are not points (3) or (5), then surely this is not a question of special relativity at all, since acceleration is outside the scope of that theory?

acceleration is not outside the scope of SR.

 

[zoobyshoe]

It stands for proper time.

Wow. Interesting. So the "worldline" is, in effect, what, something like the "proper spacetime."?

 

[John_M]

acceleration is not outside the scope of SR.

there are a few misleading websites suggesting it is...oh well.

 

[jcsd]

there are a few misleading websites suggesting it is...oh well.

It's a common fallacy, but thionk of it this way: speical relativity wouldn't be much of a kinematic theory if it couldn't deal with acceleration!

 

[pmb_phy]

It's a common fallacy, but thionk of it this way: speical relativity wouldn't be much of a kinematic theory if it couldn't deal with acceleration!

He's right. SR can handle accelerating particles. It just doesn't handle accelerating frames.

For the motion of a uniformly accelerating particle see

http://www.geocities.com/physics_world/sr/uniform_accel.htm

Pete

 

[jcsd]

I can't see why you would say it can't handle acclerated frames, for example we can use SR to find the proper time of an accelerated observer (which you have in on your own website). Just because the principle of relativity does not hold true in accelerated frames and the laws of physics that hold true for inertial frames of reference do not hold true in accelrated frames of refernce and must be modified does not mean that SR cannot handle them.

I suppose the difference between inertial and non-inertial frames of reference is the same difference as viewing Minkowskian spacetime in rectangular coordinates and viewing it in some other orthogonal curvilinear coordinate system.

 

[pmb_phy]

I can't see why you would say it can't handle acclerated frames, for example we can use SR to find the proper time of an accelerated observer (which you have in on your own website). Just because the principle of relativity does not hold true in accelerated frames and the laws of physics that hold true for inertial frames of reference do not hold true in accelrated frames of refernce and must be modified does not mean that SR cannot handle them.

I suppose the difference between inertial and non-inertial frames of reference is the same difference as viewing Minkowskian spacetime in rectangular coordinates and viewing it in some other orthogonal curvilinear coordinate system.

The principle or relativity states that the laws of physics are the same in all inertial coordinate systems. Since that is alll you know then you can't go beyond that and into non-inertial frames. Going from one inertial frame to another requires transformations in the Lorentz group. I guess it took Einstein a while to figure out how to transform to non-inertial frames or that he didn't know how/if the laws of physics could take on tensor form and thus be valid in all frames/coordinate systems. Then when he figured it out he called it the principle of general covariance and called the theory which included this principle the general theory of relativity. It was at that time when he postulated that the laws of physics must take on tensor form and thus hold in all coordinate systems.

Pete

 

[jcsd]

That's different though, because the principle of relativity doesn't prohibit us from out what the laws of special relativity should be in non-inetrial frames and again the fact that the laws of physics are not tensors when transforming into accelarted frames of refrence doens't prohibit us from finding out what they are in these frames. If we could not go beyond that into non-inertial frames we could not expect to get a meaningful answer to the proper time as experinced by an accelerated observer.

 

[pmb_phy]

That's different though, because the principle of relativity doesn't prohibit us from out what the laws of special relativity should be...

Can you re-write that for me please. What does "...prohibit us from out what ..." mean?

in non-inetrial frames and again the fact that the laws of physics are not tensors when transforming into accelarted frames of refrence doens't prohibit us from finding out what they are in these frames. If we could not go beyond that into non-inertial frames we could not expect to get a meaningful answer to the proper time as experinced by an accelerated observer.

Regardless of that, SR refers only to inertial frames. If you want to find laws in some other non-inertial frame or motion in non-inertial frame then give it another name. Call it jcsd's principle of relativity if you will.

Please show me what the electric and magnetic fields are in an accelerating frame without using tensors. As soon as you use tensors then you're using the principle of general covariance and that is general relativity.

Pete

 

[jcsd]

Can you re-write that for me please. What does "...prohibit us from out what ..." mean?

from "finding out..."

Regardless of that, SR refers only to inertial frames. If you want to find laws in some other non-inertial frame or motion in non-inertial frame then give it another name. Call it jcsd's principle of relativity if you will.

No it doesn't, if this were the case clearly the twin paradox could not be resolved in SR.

Please show me what the electric and magnetic fields are in an accelerating frame without using tensors. As soon as you use tensors then you're using the principle of general covariance and that is general relativity.

Pete

this is from John Baez's website (infact he even confirms what i guessed about curvilinear coordinate systems in an earlier post).
http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

 

[pmb_phy]

No it doesn't, if this were the case clearly the twin paradox could not be resolved in SR.

Sure it can. Nonone needs to transform to an accelerating frame of reference. SR can handle accelerating particles and clocks, it just is not within the domain of SR. To see how this is treated in GR see

http://assets.cambridge.org/0521422701/sample/0521422701WS.pdf

Baez tries to encorporate accelerating frames into SR by changing the definition of SR.

Einstein defined SR as "relativity in inertial frames".

Pete

 

[jcsd]

By finding the proper time of an accelrating clock you are specifically looking into a non-inertial frame, there's no getting around it.

You can treat accelrated frmaes without having to make any new postulates, yes you may start to use concepts such as curvilinear coordinates and metric tensors, which starts to make the problems look more like general relativity, BUT the key point is the postulates of SR not of GR are what you're still using.

 

[DW]

As silly as someone not knowing the difference between tank tops and leotards is, jcsd is correct. Constructing physics from the perspective of an accelerated frame can be done completely within the context of special relativity merely by asserting the physics from known from SR for an inertial frame and then translating that into what one would observe from the perspective of a curvalinear coordinate frame. This is why such a curvalinear frame is constructed for the accelerated observer in chapter 6 of MTW for the case of constant proper acceleration where they clearly are stating that special relativity can handle acceleration. Just to generally and mathematically prove my point I'll describe step by step how one may do special relativity for arbitrarily accelerated frames. First start out with an inertial frame according to which the equations for the laws of physics are known. They all can be reduced to equations in the form f(ct,x,y,z) = 0. What one can to do to describe the physics according to an accelerated is simply transform the time and space coordinates into coordinates that are appropriate for an accelerated frame observer and see what the equation results in. This will be the physics as reckoned according to the accelerated frame observer. The only thing left to do is to come up with a choice of curvalinear coordinate frame that is appropriate for an accelerated observer. In other words you want a frame according to which time locally corresponds to what is read on his wrist watch and lengths localy correspond to what he would measure by a ruler which has the globally property of reducing to Lorentz transformations for zero acceleration. First I will consider motion along the x, x' axis, but then generalise the final results. To find how his the wrist watch time is expressed according to an inertial frame observer simply use special relativistic time dilation:
ct|_{x&#039; = 0} = \int \gamma dct&#039;
From this, the above discussed conditions can be satisfied if we make for the global choice of accelerated frame time coordinate the relation:
ct = \int \gamma dct&#039; + \gamma \beta x&#039;
where \gamma and \beta are expressed as functions of ct'. Along with the global choice of accelerated frame x' coordinate made as
x = \gamma x&#039; + \int \gamma \beta dct&#039;
These last two equations are then the coordinate transformations for arbitrarily proper time dependent accelerations along the x, x' axis. One then generalises the coordinate transformations for accelerations in arbitrary directions as well according to equations 5.4.4 at
http://www.geocities.com/zcphysicsms/chap5.htm#BM65
Now that you have the coordinate transformation to an arbitrarily accelerated frame (with the exception of spin which is easy to add in) all you need to do is apply the transformation to the coordinates in the inertial frame equation of the law of physics f(ct,x,y,z) = 0 to find out the law of physics according to the accelerated frame. This proved that special relativity is sufficient to encorporate accelerated frames.
QED
Recap:
With the above transformations genralised to 5.4.4 that reduce to MTW equation 6.17 in the case of constant proper acceleration and to Lorentz transformations in the case of zero acceleration and with the known equations for the laws of physics according to the inertial frame (unprimed coordinates) one can apply the transformation to find the law of physics according to a frame accelerated in arbitrary directions with arbitrary time dependence.

 

[jcsd]

DW, hmm I could of sworn that you were wearing a leotard on you're martial arts site, maybe I'm getting you mixed up with someone else

 

[Al68]

I don't think this is an original point - but it seems an obvious problem with the idea of time dilation for which I haven't heard a satisfactory answer. I don't have any knowledge of physics beyond GCSE (low level high school) so keep it simple please!

The 'twin paradox'. I'm sure everyone's familiar with it. One twin stays on earth, the other flies off somewhere at high speed and returns. Moving clocks run slow and therefore the twin in the spaceship has not aged as much as the twin who's stayed on Earth.

Isn't it the case that this theory simply assumes the Earth is an 'absolute' frame of reference, directly contradicting the relativity principle? Why can't you say that the twin on Earth is moving, the twin in the spaceship is still, and that the Earth twin's clock should therefore 'run slow'? How can you make the general statement that 'moving clocks run slow' without violating the relativity principle - whether something is moving or not depends on your frame of reference?

Hi John,

I've seen a lot of answers to your question that talk about acceleration of the twin on the ship and his change of reference frames. I understand why this doesn't make sense. After all, you can eliminate acceleration and the change of frames by simply rewording the problem. For example, make it a one-way trip, start the clocks when the ship passes earth(or the Earth passes the ship) already in relative motion, and have an observer at the turnaround point, and have the ship signal it's clock reading when it reaches the turnaround point. Here, there is no acceleration or change of frames of either the ship or the earth, and the twin on the ship will still age less. The reason is that the situation is still not symmetrical. If the distance between the Earth and the turnaround point is measured in the Earth's frame, then that fact makes the Earth's frame "prefered". Because the distance between the Earth and the turnaround point will be shorter as measured in the frame of the ship, due to length contraction. So in the so-called "twin's paradox", the Earth's rest frame is prefered, not because it is at absolute rest, but because it is at rest relative to the frame in which the distance to the turnaround point is decided. Note that the twin that travels farther has to age more as long as both twins agree on their relative speed, since v=d/t, where d=distance traveled and t=time.

Also note that if you were to measure the distance to the turnaround point in the frame of the ship (maybe by towing a buoy with a rope), then the ship's frame would be prefered, the twin on Earth would measure the distance traveled to be less than that measured by the ship's twin, and the Earth's twin would age less.

I hope this helps, and I hope I didn't make any mistakes. If I did, I'm sure someone will point them out :).

Alan

 

[DW]

DW, hmm I could of sworn that you were wearing a leotard on you're martial arts site, maybe I'm getting you mixed up with someone else

Its a tank top shirt in the site you are probably talking about.

 

[pervect]

I'm sorry but it still isn't clear to me! So this is what happens in the twin experiment:

1. Twins A and B are in the same frame of reference.
2. Twin B changes his frame of reference (flying off)
3. The twins are now in a different inertial frame of reference. From A's perspective B's clock is running slower and therefore B ages slower. From B's perspective A's clock is running slower and therefore A ages slower.
4. Twin B reverts to Twin A's frame of reference.
5. Twin B has aged slower than Twin A.

Where, in this sequence of events, has the process of B ageing slower than A taken place?

There's no universal way to compare A and B's age while they are a long distance apart. Since B has flown off, and then stopped, B is a long distance away from A. You should really try and get A & B back together again, so they can compare clocks directly, and avoid this issue.

So let's consider the following _two_ scenarios

#1

A and B are having coffee together. B jumps into a spaceship, travles at a high velocity for a long time, turns around, and comes back.

result: B will be younger than A when they meet - the case that's most often discussed. But now let's consider

#2
A and B are having coffee together. B jumps into a spaceship, travels at a high velocity for a long time. A sees that B forgot his watch, and jumps into an even faster spaceship to catch up with B.

result: When A hands B his watch, A will be the younger.

So you see, when A and B are separated, it's not clear who is going to be younger when they next meet, unless you specify how it is that they do meet.

The moral of the story is this. In space, the shortest distance between two points is a straight line. So if you go from A to B in a straight line, you measure the shorest distance possible. If you go from A to C to B, you travel a longer distance. This is not shocking at all, it's sometimes called the triangle inequality.

In space-time, the longest time interval between two points is a straight line. So if you go from A to B in a straight line, your clock will have the most elapsed time. If you go from A to C, then from C to B, your clock will read less time than A's.

There is no such thing as "universal time", so you can't really ask "when" did the slowdown occur. The best you can do is compare the two clocks (whcih you can only do unambiguously when they are close), and note that they will disagree, and predict that the person who traveled the furthest will have the least elapsed time on his clock (be the youngest).

This is necessary and sufficient to resolve the "paradox". Trying to ask "when" the clock slowed down is going to lead to confusion. From the clocks point of view, it never slows down, it just ticks along. Because time and space are not invariant to all observers, different people will observe different things in their own coordinate systems when they tell the tale of the clocks - but when you get them back together, everyone will agree about the result.

 

[Al68]

It seems to me that the easiest, simplest way to calculate elapsed time for any observer, is to use t=d/v, where d=distance traveled as measured by the same observer, v=relative velocity, and t=time elapsed by the same observer. This clears up the "twins paradox" nicely, without considering acceleration or change of frames. And this method makes it more clear who will age less and why, without making the problem more complicated than it has to be by talking about acceleration, or change of reference frames. Because it will always be the case that the observer who traveled farther, as measured from his own frame, will have more elapsed time, regardless of who is considered at rest and who is moving, since of course neither is at absolute rest.
I know this sounds too simple, but it seems to me it really is a lot simpler than a lot of people make it out to be. At least the part about who will really age more.

Alan

 

[jcsd]

It seems to me that the easiest, simplest way to calculate elapsed time for any observer, is to use t=d/v, where d=distance traveled as measured by the same observer, v=relative velocity, and t=time elapsed by the same observer. This clears up the "twins paradox" nicely, without considering acceleration or change of frames. And this method makes it more clear who will age less and why, without making the problem more complicated than it has to be by talking about acceleration, or change of reference frames. Because it will always be the case that the observer who traveled farther, as measured from his own frame, will have more elapsed time, regardless of who is considered at rest and who is moving, since of course neither is at absolute rest.
I know this sounds too simple, but it seems to me it really is a lot simpler than a lot of people make it out to be. At least the part about who will really age more.

Alan

Your confusing me, How can you travel at all in your own frame? Do you actually mean your final inertial frame?

also as accelration is involved in the twin paradox how can you say t = d/v (unless you have v as the average relative velcotiy, but this is still not clear as often in the twin paradox if you take d and v as vectors they are the same for both observers!).

I think I see what your saying though, you are saying the person who has traveled most according to their final inertial frames will of experinced the least time, but this is still not avoiding the issue of acceleration, as by admitting you have traveled from the point of view of your final inertial frame you are admitting that you have accelerated.

 

[donnie]

This might look strange, but, I have the view that, in analysing physics, and as in all other cases of assessment, it's human nature to focus on the PROCESS getting to a desired outcome and not the ELEMENTS/DYNAMICS AFFECTING the OUTCOME?

If I was a child trying to understand these concepts, would it not make sense to say:
I'm sitting on a neighbourhood curb one day, watching an old lady leave her house, to walk to the post box a block away to her left, to deliver a letter. Her neigbour, also leaving to deliver a letter, get's in his car to drive to the post box, a block to his right (opposite direction of the first post box). How does he get their sooner than her? Why? Who would get BACK sooner? Why?

Surely it comes down to:
1. the MODE of travel and its dynamics of speed in RELATION to EACH OTHER?
2. And then DIRECTION IN RELATION TO EACH OTHER.
3. And ENVIRONMENT based on the direction.

In my mind the equation is not necessary/cannot exist without the INFLUENCING factors and not the OUTCOMES of how the old lady returns and how her neighbour views her when driving away (smaller and slower) and coming back (larger and closer to normal).

What a mind job, I know!

 

[Al68]

Your confusing me, How can you travel at all in your own frame? Do you actually mean your final inertial frame?

also as accelration is involved in the twin paradox how can you say t = d/v (unless you have v as the average relative velcotiy, but this is still not clear as often in the twin paradox if you take d and v as vectors they are the same for both observers!).

I think I see what your saying though, you are saying the person who has traveled most according to their final inertial frames will of experinced the least time, but this is still not avoiding the issue of acceleration, as by admitting you have traveled from the point of view of your final inertial frame you are admitting that you have accelerated.

By distance traveled, I mean relative to the frame to which you will compare your elapsed time to. And of course, I know you cannot ignore acceleration, but in the case of the "twins paradox", The twin on the ship would age less even if there were no acceleration involved, for example if it were a one way trip and the ship were already in (relative) motion when the clocks started. And the reason d will not be the same for both observers is because, as commonly presented, the twins paradox defines d as measured from Earth's rest frame. So the distance neccessarily has to be less as measured from the ship's frame due to length contraction. And you could include acceleration in the calculations, but it would make the math more difficult without helping anyone understand why there was less elapsed time for the ship. And the change of frames for the ship is very relavent, because the location and time of this change is different for each observer, because it occurs at a different distance from the starting point, from the point of view of each observer.

Alan

 

[John_M]

Is there a book you can buy on the twin paradox which outlines all the main arguments - and who's made them?

One that includes how to deal with SR in accelerating frames of reference...