Special relativity - please can someone clear this up?

[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' = 0} = \int \gamma dct'
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' + \gamma \beta x'
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' + \int \gamma \beta dct'
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...

 

[robphy]

In my opinion, one of the best articles on the Twin Paradox is:

"The Clock Paradox in Relativity Theory"
Alfred Schild
American Mathematical Monthly, Vol. 66, No. 1 (Jan., 1959) , pp. 1-18

If your institution has access to JSTOR, it is available here
http://links.jstor.org/sici?sici=0002-9890%28195901%2966%3A1%3C1%3ATCPIRT%3E2.0.CO%3B2-L

 

[John_M]

I'm a law student in the UK...my institution certainly doesn't have access.

I suppose you couldn't send me a copy? (hope I don't get banned from the forum for asking that...)

 

[zoobyshoe]

I think your best bet, John, is to get a grasp on those worldlines that ambitwistor explained to me. They are pretty easy, they explain everything, and they cut through the fact that everyone seems to have a different take on the acceleration explanation.

 

[pervect]

I'm a law student in the UK...my institution certainly doesn't have access.

I suppose you couldn't send me a copy? (hope I don't get banned from the forum for asking that...)

It may not be the best thing ever written on the twin paradox, but you could start by finding and reading the sci.physics.faq entry on the twin paradox.

If you pick up any text on relativity, it will probably have some discussion on this as well.

Basically, it should be easy to find a lot of material on the subject.

It's really not that hard to understand. As the sci.physics.faq points out, perhaps the biggest obstacle is that there are so many diifferent ways of explaining the paradox. Personally I favor pointing out that the "paradox" is closely related to the triangle identity. In Euclidiean geometry, the shortest idistance between two points is a straight line, and as a consequence of this, the sum of two sides of a triangle is always longer than the direct route.

In relativity, the shortest distance between two points has the longest proper time, because the sign of the time term in the metric is inverted.

So if two different clocks take different routes to the same definition, the one that takes the straight-line path reads the longest, all other clocks read a lower amount of time.

 

[clj4]

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...

John

Here is one of the best explanations complete with a beautiful solution:

http://sheol.org/throopw/sr-ticks-n-bricks.html

All free. have fun!