Curved Space-time and Relative Velocity

[Anamitra]

The end points of the stick measured by the observer [simultaneously] at each and every moment [individually] is \frac{d}{\gamma}. He sees the ray passing over the stick.

We get this directly from the Lorentz transformations:

Let the coordinates of the end points of the stick,A and B ,as seen by the moving observer be {x_{1}}^{'} and{x_{2}}^{'} at the same instant of time,say,t^{'}
The corresponding values wrt the ground frame are:
[x_{1}{,}t_{1}] and [x_{2}{,}t_{2}]

We have,
x_{1}{=}{\gamma}{(}{{x_{1}}^{'}}{+}{v}{t_{1}}{'}{)}

x_{2}{=}{\gamma}{(}{{x_{2}}^{'}}{+}{v}{t_{1}}{'}{)}

Subtracting the second equation from the first we have,
{x_{2}{-}x_{1}}{=}{\gamma}{(}{x_{2}}^{'}{-}{x_{1}}^{'}{)}

{d}{=}{\gamma}{(}{x_{2}}^{'}{-}{x_{1}}^{'}{)}

{x_{2}}^{'}{-}{x_{1}}^{'}{=}{(}{{1}{/}{\gamma}}{)}{d}

"d" is the uncontracted length as observed by the person in the ground frame.

The above relation is true for any instant t_{'} observed in the moving frame.

 

[Dale]

Yes, it is true, but it is not relevant. You are not interested in the distance between the two stations, you are interested in the distance that the light travels. They are not the same. Draw the diagram and you will see it immediately.

Here is a relevant derivation:
In the unprimed frame a pulse of light satisfies
\Delta t=\frac{\Delta x}{c}

By the Lorentz transform
\Delta x'=\gamma (\Delta x - v \Delta t)

By substitution
\Delta x' = \gamma \left(1-\frac{v}{c}\right) \Delta x \ne \frac{\Delta x}{\gamma}

 

[Anamitra]

Exactly correct, the endpoints must be simultaneous. Here we are interested in the distance traveled by the light. Are the endpoints of the distance traveled by the light simultaneous? Clearly they are not. In fact, there is no frame where the endpoints of the distance traveled by the light are simultaneous. Therefore, the length contraction formula is never valid for determining the distances traveled by light in two different frames. The proof follows from the Lorentz transform in just a few lines.

We consider a marble rolling over a stick. The stick itself is in uniform motion wrt my frame[inertial]. I want to calculate the time required for the marble to reach from one end to the other.

What we need to know:
1)The speed of the stick wrt my frame.
2) The speed of the marble wrt to my frame

What I do:
1) I do calculate the speed of the marble relative to the ruler[Using the Special Relativity Rule for relative speed].

2)Divide the contracted length{d}{/}{\gamma} of the ruler by the relative speed.

What is not required:The length of the ruler with respect to the moving marble.
[In the case of a light ray the relative speed is "c" wrt to the observer or the ruler.]

I do not need to know what the light ray feels about the length of the ruler.

 

[Dale]

Anamitra, please draw the spacetime diagram as I have suggested 4 times now and examine my proof above. You can even check my derivation by plugging the formula into the problem we were working on earlier. Then see if you can identify where your mistake is. I will check in again later this afternoon.

 

[Anamitra]

[Incidentally I noticed your last thread after posting my last thread]

I will definitely draw the diagram as suggested by you. But I will get nothing different from what I got in thread-->https://www.physicsforums.com/showpost.php?p=2899896&postcount=201
[I mean, the first set of calculations,using the Lorentz transformations.]

Here is a relevant derivation:
In the unprimed frame a pulse of light satisfies
{\Delta} {t}{=}\frac{\Delta x}{c}

By the Lorentz transform
\Delta x'=\gamma (\Delta x - v \Delta t)

By substitution
\Delta x' = \gamma \left(1-\frac{v}{c}\right) \Delta x \ne \frac{\Delta x}{\gamma}

{\Delta}{x}^{'}{\neq}\frac{{\Delta}{x}}{\gamma}

If \frac{{\Delta}{x}}{\gamma} represents the distance from A to B as observed from the moving frame[the contracted stick] then,it appears to me, that light has not reached its destination[or overshot it] from the point of view of the moving observer[by traveling the distance {\Delta}{x}^{'}.

[It is to be noted that A and B are always at the two ends of the stick. This is true for both the frames]

The distance between the two stations wrt the moving observer seems to have different values from two different interpretations.

 

[Anamitra]

The Problem:

A moving observer[in the primed frame] may calculate the spatial distance between a pair of points,fixed in the unprimed frame by two procedures:

1)He undertakes a simultaneous consideration of the two points from his frame.[ A standard procedure]

2)A light signal is sent between the two points. The coordinates are noted in the unprimed frame. We transform them to the primed frame and take the difference between the spatial coordinates.

We get different results.This fact is expressed in the following threads:

https://www.physicsforums.com/showpost.php?p=2899896&postcount=201
https://www.physicsforums.com/showpost.php?p=2901615&postcount=211

 

[JesseM]

Distance of B from A,wrt ground frame=d
Speed of train from the ground frame=0.8c
{\gamma}=1/0.6
Light is flashed from A at time,t=0
Position coordinate of A=0
Event 1:(0,0) ---------------------[light is flashed from A]
Event2: (d,t)=(d,d/c) -----------------[light is received at B]

Transformed coordinates from the train
Event1:(0,0)
Event2:(1/3 d,1/3 t)

According to the Lorentz transformation, with x=d and t=d/c, the x' and t' coordinates would be:

x' = gamma*(x - vt) = (5/3)*(d - 0.8c*d/c) = (5/3)*(1/5)*d = (1/3)*d
t' = gamma*(t - vx/c^2) = (5/3)*(d/c - 0.8c*d/c^2) = (5/3)*(1/5)*d/c = (1/3)*t

So, this looks right.

Alternatively,
Distance from A to B wrt to the moving observer: \frac{d}{\gamma}
Time taken[by light ray/pulse]from the passenger's point of view={\frac{{d}{/}{\gamma}}{c}{=}{0.6t}{\neq}{1/3}{t}

No, you forget that both A and B are moving in the passenger's frame, at a speed of 0.8c in the -x' direction. So if the light is emitted at t'=0 from x'=0, and at this moment B is at position d/gamma = 0.6*d, then the position as a function of time for the light ray is x'(t') = c*t', while the position as a function of time for B is x'(t') = 0.6*d - 0.8c*t' (so at t'=0, B is at position x'=0.6*d, but at later times B is at different positions because it is moving at 0.8c in this frame). So, to find when the light reaches B in the passenger's frame we must find out at what t' both equations give the same value for x', which we can find by setting them equal:

c*t' = 0.6*d - 0.8c*t'
1.8c*t' = 0.6*d
t' = (1/3)*d/c

Which agrees with what you found when you calculated things from the perspective of the ground frame and used the Lorentz transformation to find the coordinates in the passenger frame.

 

[JesseM]

He observes the distance between the stations like a moving stick of fixed length,{\frac{d}{\gamma}}. The stick moves in the backward direction along with the ground frame[speed =-v=-.08c].The light ray moves with a speed c in the forward direction from A to B.The relative speed of light wrt the "stick" is c.

What do you mean by "relative speed of light wrt the stick"? Do you mean:

1) the speed of light in the rest frame of the stick
2) the speed at which the distance between the light ray and the end of the stick is changing in the frame where the stick is moving at 0.8c

The two are quite different in relativity! If you want to calculate the speed at which the distance between two objects is changing in a frame where neither is at rest, that's a quantity sometimes called the "closing speed", and it's not in general equal to the speed of one object in the other object's rest frame. In the frame of the passenger who sees the stick moving at 0.8c, the "closing speed" between the light ray and the end of the stick is actually 1.8c, meaning every second the distance between them decreases by 1.8 light-seconds (because every second the light ray moves 1 light-second in the +x' direction while the end of the stick moves 0.8 light seconds in the -x' direction...for example, the equations giving x'(t') for the light ray and B in my previous post show that at t'=0 seconds the light ray is at x'=0 while B is at x'=0.6*d, so the distance between them is 0.6*d - 0 = 0.6*d, but at t'=1 second the light ray is at x'=1 while B is at x'=0.6*d - 0.8 so the distance between them is (0.6*d - 0.8) - 1 = 0.6*d - 1.8)

 

[Anamitra]

Regarding Point 2 in thread #216:

One may connect the two points in the unprimed frame by a "slower than light" signal.By transforming the coordinates to the primed frame he gets a value of spatial separation[in the primed frame], different from what he got when he used a light signal.
Relative speed between the frames:0.8c
Let the speed of the signal be c/2 in the unprimed frame.
Unprimed frame
Event 1:(0,0)
Event 2:(d,2d/c)

Primed Frame

Event 1:(0,0)
Event 2:(-d,4/3 * d/c)

Spatial separation in the primed frame:=d [I have taken the modulus]

If a light signal is used the spatial separation in the primed frame=d/3

 

[JesseM]

Regarding Point 2 in thread #216:

One may connect the two points in the unprimed frame by a "slower than light" signal.By transforming the coordinates to the primed frame he gets a value of spatial separation[in the primed frame], different from what he got when he used a light signal.
Relative speed between the frames:0.8c
Let the speed of the signal be c/2 in the unprimed frame.
Unprimed frame
Event 1:(0,0)
Event 2:(d,2d/c)

Primed Frame

Event 1:(0,0)
Event 2:(-d,4/3 * d/c)

Spatial separation in the primed frame:=d [I have taken the modulus]

That's the spatial separation between the two events (each occurring at opposite ends the stick) in the primed frame--but since the two events happened at different times and the stick is moving in the primed frame, this is not equal to the length of the stick in the primed frame!

 

[Anamitra]

There has been a mistake in Point 2 in thread #216. One has to make simultaneous measurements

 

[Anamitra]

Yes, it is true, but it is not relevant. You are not interested in the distance between the two stations, you are interested in the distance that the light travels. They are not the same. Draw the diagram and you will see it immediately.

The moving observer simply sees the light ray traveling from one station to the other.He knows the distance between the stations wrt to his frame. This should be the distance traveled by the light ray. The speed of light is the same in all frames.
Accordingly he should calculate the time taken.

 

[Anamitra]

I believe DaleSpam was correct. I had calculated the time of travel by light wrt the stick. It is different from the time if the actual distance traveled by light is considered,since the stick is a moving one.

 

[Dale]

I am glad that you were able to work it out! Was drawing the diagram helpful or did you figure it out algebraically?

 

[Anamitra]

It was very simple to figure it out algebraically.I took the time wrt to the stick[the moving stick]since the relative velocity formula considers the time of the frame on the first particle[wrt which the relative speed is being calculated,the stick in this case].But the distance ,I mean the contracted distance,was wrt the moving observer.The value I got by division was quite misleading. This was due to inadvertence.

Of course the time for the passage of light as calculated by the moving observer is 1/3 year.When he was in New York he got 1 year. When he alights from the train,what should he feel now that his clock time has changed[though he has got back the ticking rate]?[If he is allowed to calculate the same travel of the light ray. ]
[He remembers the calculations he performed at New York and he is back in the same frame but at a different place with a changed time on his own clock]
At the present moment he should/may think that the corresponding clock at New york has also changed time[tick rate remaining the same]. The light ray started off earlier and he gets by calculation that the light pulse passed off before his arrival at Boston.Actually in this calculation he replaces all clocks in the ground frame by his new time, keeping the tick rate the same.

Calculations:
Speed of train: 0.8c
Time taken by the train to travel between the two stations: 1.25 yr
passenger arrives at Boston:0.75 yr[ by his clock]
Time of travel by light as estimated by passenger while in motion:1/3 yr
On alighting passenger arranges to adjust all clocks on the ground frame.
New york time becomes 0.75 yr instead of 1.25 yr
Both passanger and light ray started off at t=-0.5yr instead of t=0
Light ray reached Boston at t=+0.5 yr ie 0.25 yr before the passenger's arrival.

Possibly there is some mistake in my calculations.[ It denies the fact that the time of travel according the passenger's clock is .75 yr] I am requesting DaleSpam to check the matter.

 

[Anamitra]

Whenever we are in motion our clock rate changes. On coming back to rest in the original frame we get back the clock rate. But we don't get back the time.In such a situation observers at rest in the same frame should have different times[depending on the history of their motion], though the clock rate is the same.How should this influence physics?

 

[Passionflower]

Whenever we are in motion our clock rate changes.

Motion by itself does not change clock rates a change in motion does.

Whenever we change our speed wrt an inertial observer our clock rate changes wrt that observer's clock.

If the change is constant then the relation between the two clocks for two dimensional movement is:

<br /> d t = {1 \over \alpha} \sinh(\alpha d\tau) = {dw \over \alpha}<br />

Where w is the proper velocity or celerity

 

[Anamitra]

Actually I meant the changes in clock rate due to acceleration/retardation at the beginning/end of the uniform motion.

In fact if we take a particle from one point to another in a gravitational field and bring it back to to the original point by some different path a similar effect could take place

 

[Dale]

What you are describing in your last few posts is a non-inertial reference frame. Unfortunately, there is no standard way to assign coordinates in a non-inertial reference frame. So you simply have to adopt some convention and arbitrarily declare that it is what you mean when you are talking about the perspective of your non-inertial observer. My favorite approach is that taken by Dolby and Gull:
http://arxiv.org/abs/gr-qc/0104077

However, although I like their approach and consider it to be the most natural extension of the Einstein synchronization convention for a non-inertial observer, it is important to emphasize that it is merely a convention and any other convention may be chosen just as well. My preference is a personal preference and not a physical preference.

Once you have defined your coordinate system then the only remaining thing to do is to determine the metric in the coordinate system. Once you have determined the metric then all of the normal processes can be used.

 

[JDoolin]

What you are describing in your last few posts is a non-inertial reference frame. Unfortunately, there is no standard way to assign coordinates in a non-inertial reference frame. So you simply have to adopt some convention and arbitrarily declare that it is what you mean when you are talking about the perspective of your non-inertial observer. My favorite approach is that taken by Dolby and Gull:
http://arxiv.org/abs/gr-qc/0104077

However, although I like their approach and consider it to be the most natural extension of the Einstein synchronization convention for a non-inertial observer, it is important to emphasize that it is merely a convention and any other convention may be chosen just as well. My preference is a personal preference and not a physical preference.

Once you have defined your coordinate system then the only remaining thing to do is to determine the metric in the coordinate system. Once you have determined the metric then all of the normal processes can be used.

That is fascinatingly arbitrary, and could this go under the "not even wrong" category?
I think the problem lies with the question--determining distant simultaneity for a non-inertial frame--any answer will be arbitrary.

I think, though, if you are trying to determine synchronization, there is one convention that stands out above all others, and could almost be called non-arbitrary.

In particular, given any moment in time of the traveling twin, take the reference frame where the traveling twin is momentarily at rest, and use this reference frame to determine synchronization.

Of course, as is pointed out in the paper, "if Barbara’s hypersurfaces of simultaneity at a certain time depend so sensitively on her instantaneous velocity as these diagrams suggest, then she would be forced to conclude that the distant planets swept backwards and forwards in time whenever she went dancing!" (I see no problem with that.)

Some may find this to be an uncomfortable idea, but you should also consider that every time Barbara goes dancing, distant objects would seem to swing back and forth over large distances as she turned back and forth as well.

Another possibility to make this even less arbitrary is to attach a 360^o camera to Barbara's hat. Then you would consider only events which were just now becoming visible in Barbara's field-of-vision. Ask what Barbara sees instead of just what events she deems as simultaneous. We need not ask Barbara to invoke some arbitrary definition of non-inertial simultaneity,

...just as we do not need her to invoke some arbitrary definition of "rotational forward"

 

[Dale]

I think, though, if you are trying to determine synchronization, there is one convention that stands out above all others, and could almost be called non-arbitrary.

In particular, given any moment in time of the traveling twin, take the reference frame where the traveling twin is momentarily at rest, and use this reference frame to determine synchronization.

No, that convention is also completely arbitrary. It is certainly the convention that most people think of when talking about simultaneity in a non-inertial reference frame, but there is nothing which makes it physically preferable to any other convention. And there are some features that make it mathematically a very bad convention.

Of course, as is pointed out in the paper, "if Barbara’s hypersurfaces of simultaneity at a certain time depend so sensitively on her instantaneous velocity as these diagrams suggest, then she would be forced to conclude that the distant planets swept backwards and forwards in time whenever she went dancing!" (I see no problem with that.)

The bigger problem than time going backwards is the problem of multiple time coordinates being assigned to the same event. This violates some of the essential mathematical features of a coordinate chart, specifically that it be a 1-to-1 map of events on the manifold to coordinates. This means that the chart is non-invertible, or in other words, you can go from a coordinate to an event on the manifold but not from an event on the manifold to a coordinate.

 

[JDoolin]

Of course, as is pointed out in the paper, "if Barbara’s hypersurfaces of simultaneity at a certain time depend so sensitively on her instantaneous velocity as these diagrams suggest, then she would be forced to conclude that the distant planets swept backwards and forwards in time whenever she went dancing!" (I see no problem with that.)

No, that convention is also completely arbitrary. It is certainly the convention that most people think of when talking about simultaneity in a non-inertial reference frame, but there is nothing which makes it physically preferable to any other convention. And there are some features that make it mathematically a very bad convention.

The bigger problem than time going backwards is the problem of multiple time coordinates being assigned to the same event. This violates some of the essential mathematical features of a coordinate chart, specifically that it be a 1-to-1 map of events on the manifold to coordinates. This means that the chart is non-invertible, or in other words, you can go from a coordinate to an event on the manifold but not from an event on the manifold to a coordinate.

Let me point out the difference in strengths of these statements: "forced to conclude" vs. "essential mathematical features of a coordinate chart." The former forces you into making logical conclusions, whereas the other forces you to make up arbitrary coordinate systems which will be mathematically convenient.


Don't get me wrong. Dolby and Gull's mapping of events is very pretty. It has some aesthetic satisfactoriness. And it meets your criteria of giving Barbara a way of defining each event in her history with unique coordinates. Yay. But...

First of all, this mapping arbitrarily uses the reference frame of Barbara's trip at the middle of her acceleration process. So instead of using all of the reference frames that Barbara goes through, Dolby and Gull just picked one of them, arbitrarily. There was no physical significance to making this choice (except perhaps for symmetry).

But it was a choice of an infinite number of physically significant reference frames. The reference frames that Barbara goes through during her acceleration.

 

[JDoolin]

The bigger problem than time going backwards is the problem of multiple time coordinates being assigned to the same event. This violates some of the essential mathematical features of a coordinate chart, specifically that it be a 1-to-1 map of events on the manifold to coordinates. This means that the chart is non-invertible, or in other words, you can go from a coordinate to an event on the manifold but not from an event on the manifold to a coordinate.

It is okay for multiple x, y, and z coordinates being assigned to the same event. When one person is facing north, and one person is facing west, they will have different coordinates. It is also okay for multiple t coordinates being assigned to the same event.

 

[JesseM]

Let me point out the difference in strengths of these statements: "forced to conclude" vs. "essential mathematical features of a coordinate chart." The former forces you into making logical conclusions, whereas the other forces you to make up arbitrary coordinate systems which will be mathematically convenient.Don't get me wrong. Dolby and Gull's mapping of events is very pretty. It has some aesthetic satisfactoriness. And it meets your criteria of giving Barbara a way of defining each event in her history with unique coordinates. Yay. But...

First of all, this mapping arbitrarily uses the reference frame of Barbara's trip at the middle of her acceleration process. So instead of using all of the reference frames that Barbara goes through, Dolby and Gull just picked one of them, arbitrarily. There was no physical significance to making this choice (except perhaps for symmetry).

But it was a choice of an infinite number of physically significant reference frames. The reference frames that Barbara goes through during her acceleration.

All choices about what coordinate system an observer should use are a matter of human convention--even if I am moving in a purely inertial way in SR, nothing physical forces me to use the inertial coordinate system where I am at rest to do calculations, this is just a convention physicists tend to use for convenience. Saying that Barbara "goes through" different inertial frames makes it sounds like some sort of objective physical fact (like going through a tunnel), but of course no one is "in" any particular coordinate system in any non-conventional sense, a coordinate system is labeling system, not a physical entity. Do you disagree with any of the above?

 

[Dale]

It is okay for multiple x, y, and z coordinates being assigned to the same event.

No, it is not. A coordinate chart needs to be 1-to-1.

 

[JDoolin]

All choices about what coordinate system an observer should use are a matter of human convention--even if I am moving in a purely inertial way in SR, nothing physical forces me to use the inertial coordinate system where I am at rest to do calculations, this is just a convention physicists tend to use for convenience. Saying that Barbara "goes through" different inertial frames makes it sounds like some sort of objective physical fact (like going through a tunnel), but of course no one is "in" any particular coordinate system in any non-conventional sense, a coordinate system is labeling system, not a physical entity. Do you disagree with any of the above?

Yes I disagree with one thing you said. 'no one is "in" any particular coordinate system in any non-conventional sense.'

To the contrary, at all times, you ARE in a particular reference frame.

You are facing forward. You are stopped (at rest in your own frame of reference). You can turn right, but after you turn right, you'll still be facing forward. You can accelerate, but once you accelerate, your velocity is zero.

It is your current facing and your current velocity which determine what events you see and where you see them.

Now, we can easily deal with a change in rotation; we know what that looks like--if I hand you a map turned in the wrong direction, you can just turn it around until it's facing the same way you are. You also need to locate yourself on the map ("You Are Here") so you can tell what is in front of you, and what is behind you.

With relativity, if I gave you a space-time map, oriented in the wrong rapidity, you still need to "turn" it until it's "facing" the same way you are, (that is, until your world-lines are perpendicular to your lines of simultaneity) but the turning has to be a Lorentz Transformation instead of a rotation. You also need to locate your current event on the space-time map, so you can see what is in your future and your past.

Thanks, by the way, for seeking clarification.

 

[Dale]

To the contrary, at all times, you ARE in a particular reference frame..

Reference frames are not boxes that you can be "in" or "out" of. Reference frames are coordinate systems. You may be "at rest in" or "moving in" a given coordinate system, but there is no sense in which you are "outside" of a coordinate system where you are not at rest.

At all times you can construct an inertial reference frame in which you are momentarily at rest. This is called your momentarily co-moving inertial frame. Generally we just refer to that frame as "your frame", but it is simply shorthand and there is no sense in which you are "in" or "out" of it.

 

[JDoolin]

No, it is not. A coordinate chart needs to be 1-to-1.

It is still preposterous to try to create a coordinate chart where you are simultaneously facing North, East, and South. It is just as preposterous to try to create a coordinate chart where you are simultaneously traveling at three different rapidities.

 

[JDoolin]

Reference frames are not boxes that you can be "in" or "out" of. Reference frames are coordinate systems. You may be "at rest in" or "moving in" a given coordinate system, but there is no sense in which you are "outside" of a coordinate system where you are not at rest.

At all times you can construct an inertial reference frame in which you are momentarily at rest. This is called your momentarily co-moving inertial frame. Generally we just refer to that frame as "your frame", but it is simply shorthand and there is no sense in which you are "in" or "out" of it.

I must directly contradict you. It is not just a shorthand. In every sense, you are "in" that frame in which you are momentarily at rest.

If you want to calculate what light you are currently seeing, and in particular, where the events are that produced that light, you must use the coordinates of events as calculated "in" that momentarily comoving frame.

 

[TrickyDicky]

It is still preposterous to try to create a coordinate chart where you are simultaneously facing North, East, and South. It is just as preposterous to try to create a coordinate chart where you are simultaneously traveling at three different rapidities.

But nobody is trying to do that, you can't be facing all directions at the same time in any particular coordinates, nor traveling at different speeds in a determinate coordinate chart, by definition a coordinate chart makes you be in a particular situation for that particular chart.
But of course depending on how you define the coordinates you can be facing north with one of them and south with the other without turning around.
This seems pretty trivial to understand, doesn't it?