Curved Space-time and Relative Velocity

[Dale]

That quantity can certainly exceed c, no problem, even in flat spacetime in a standard inertial frame. There is no particular "physical significance".

Just because some particular quantity has units of Length/Time and happens to be greater than c does not imply that the speed of light has been exceeded in the sense of superluminal travel or superluminal information transfer.

 

[Passionflower]

We consider the metric:

$${ds}^{2}{=}{g_{00}}{dt}^{2}{-}{g_{11}}{{dx}_{1}}^{2}{-}{g_{22}}{{dx}_{2}}^{2}{-}{g_{33}}{{dx}_{3}}^{2}$$

We may define ds as the propertime interval[noting, c=1]on the infinitesimal scale.
Locally it is nothing different from what we find in Special Relativity.Interestingly ds is an invariant and it should not be different in any other frame.

$${s}{=}{\int{ds}}$$
[We are considering stationary fields]

Integration extends between the points concerned

Proper speed=$${\frac{dL}{ds}}$$

$${dL}^{2}{=}{g_{11}}{{dx}_{1}}^{2}{+}{g_{22}}{{dx}_{2}}^{2}{+}{g_{33}}{{dx}_{3}}^{2}$$

Well since you are limiting this to a stationary metric, then I think the interpretation, which was your question, it is the same as in SR namely the total ruler distance over proper time and this value is obviously unlimited.

 

[Anamitra]

On Proper Time and Proper Speed

We consider a train running from New York(A) to Boston(B).It accelerates from A to P over a small interval of time ,$${\nabla}{t}$$.Then it continues from P to Q with uniform speed and finally it retards over a the same small time interval,$${\nabla}{t}$$, from Q to B. We assume, PQ>>AP and PQ>>QB. During the uniform motion the traveler experiences shortened time intervals. Logically this shortening should have occurred over the distance AP. The shortened interval at P continues over the distance PQ. Over the interval QB the intervals again become longer. But the shortened intervals predominate over the elongated intervals. On arriving at Boston the traveler ‘s clock and the clock at Boston station should disagree. The observer now should find it natural to consider the uncontracted length from New York to Boston. His recorded time [$${\tau}$$] is approximately the proper time due to the small period of elongation of the time intervals at the end of the journey. The natural idea of the speed estimate would be to divide the uncontracted distance from New York to Boston by the value
[$${\tau}$$].This could exceed the value of “c” by a large amount and it is meaningful from the point of view of utility.
An alternative way to consider the acceleration or the retardation would be to replace them by a gravitational field over the small distances AP and QB. Over AP the field should act from A to P and over QB it should act from B to Q.From P to Q there is no gravitational field Over the distance AP we are moving from a higher to a lower potential. The time intervals should decrease. Over PQ the contracted time intervals continue. From Q to B we are moving from lower to a higher potential. The time intervals should become larger. On getting off at Boston the clocks of the traveler and the Boston station should disagree.

 

[Passionflower]

That sounds fine Anamitra but I fail to see the point you try to make?

 

[Anamitra]

We are not violating any law of physics but in an effective way we are traveling faster than light!

 

[Dale]

His recorded time [$${\tau}$$] is approximately the proper time

The only mistake is this. The recorded time is exactly the proper time as long as the time is recorded on a good clock. The rest is fine.

We are not violating any law of physics but in an effective way we are traveling faster than light!

No, any light arrives first. At no point are you traveling faster than light nor are you traveling faster than light overall. As I said before, just because some particular quantity has units of Length/Time and happens to be greater than c does not imply that the speed of light has been exceeded in the sense of superluminal travel or superluminal information transfer.

Btw, this has nothing to do with curved spacetime.

 

[pervect]

Proper velocity , also sometimes called celerity, is a logical means of measuring velocity, and it has the advantage of requiring only one clock - avoiding a mostly pointless extended discussion of how to synchronize clocks. The disadvantage is that said clock must be on the traveling object, which isn't always possible.

(see for instance http://arxiv.org/abs/physics/0608040 for defintions of various different post-relativistic concepts of ways to describe velocity and their names.)

It's not the usual definition of velocity, however, so one must be sure to call it by the right name to avoid confusion.

The concept of celerity also makes the concept of isotropy easier to "pin down" exactly. An isotropic clock synchronization is where the ratio between celerity and velocity, which will be in general some number greater than one, does not depend on the direction of travel. The ratio can and does depend on the velocity, but not on direction.

 

[Anamitra]

May I refer to the considerations in thread #183.Let us assume that a light ray has been flashed from New York towards Boston just at the moment the traveler is about to board the train.The light ray is expected to arrive at Boston at time t=T[Boston clock].If the proper speed of the traveler exceeds the value of "c", and as we know that his clock registers a time less than that of the Boston station clock when he has come out to the Boston platform, he has indeed reached Boston before the arrival of the light ray.Incidentally the two clocks were synchronized at New York.
[We have assumed a straight line travel between the two stations.]

 

[pervect]

Let's make things simpler. If you are at Boston, and you board a train, and at the instant you do (I guess we'll assume you jump into the moving train, a bit like a mail package used to be picked up), you send a light signal to New York - which will arrive first? You, or the light signal? The answer is hopefully obvious - if we disagree, there's some much deeper issue. The light will arrive first, the train is always slower than light, if light and the train always run the same course.

So - it should be clear that the train really is slower than light - and you've not measured the velocity accurately. So, you've set up an unfair comparison, a bad way of comparing speeds. There are lots of ways of doing this.

IT appears to me that you are hoping, somehow, to save the notion of "universal time". The following situation, if you take it seriously, should be able to disabuse you of that notion, if that is in fact what you are doing. Its the usual "twin paradox". If you synch your watch with the one at Boston, and get on the train to New york, then take the first train back to Boston, when you compare your watch with the clock at Boston, you'll find it reads differently. This should be good enough to show you that your watch and the clock at Boston are not measuring some sort of "universal time". If you keep on trying to assume that it must, you're just going to confuse yourself :-(.

If we assume that you spend zero time at New York - and hop right back on a train to Boston - then the reason for the distinction between velocity and celerity should be clear. Your wristwatch isn't measuring the same time as the one on the station. Thus, it becomes vital to know which one we are using - the station clocks (velocity), or your wristwatch (celerity). Also, the prescription for synchronizing the clocks in order to properly measure velocity in the "station frame" should become clear. There will be only one way to synch clocks to make the ratio between celerity and proper time independent of your direction of travel. You can do this without regard to external signals, even, just by traveling back and forth on the train - if you've got an accurate enough and shock-resistant enough watch.

 

[Anamitra]

There is clear reversal in the temporal order of events.

The ground observer[the one at New York] sees/calculates the light pulse reaching Boston first and then the traveler coming off the train.. But the traveler is expected to receive the light pulse after reaching Boston.

Event A:Traveler reaches Boston
Event B: Light pulse reaches Boston

Are these events causally connected?

Even if you could connect them by some procedure we must accept the fact that superluminal motion allows the breakdown of the principle of causality!

 

[pervect]

Two events that occur at different locations "at the same" time in some specific inertial frame are not casually connected at all.

Causal connection in relativity is defined in terms of the light cones. The idea that there is some universal ordering of events according to some universal time doesn't work in relativity, because there isn't any universal time to order them by.

Two spatially separated events that occur "at the same time" in one frame (the Earth frame) can and do occur at different times in the train frame.

If you factor this into your scenario properly, you'll see that this is what you failed to take into account - you didn't include the effects due to the relativity of simultaneity.

 

[Dale]

If the proper speed of the traveler exceeds the value of "c", and as we know that his clock registers a time less than that of the Boston station clock when he has come out to the Boston platform, he has indeed reached Boston before the arrival of the light ray.

No he has not, what would make you think that?

Anamitra, if you are confused with this kind of very basic flat spacetime SR situation then you should start a new thread and resolve those confusions before proceeding on in curved spacetimes. There is no hope of understandin GR if you are making these kinds of SR mistakes.

 

[Anamitra]

The observer at the New York Station[say,the station master] never "sees" the light ray reaching Boston. He can only calculate and make predictions about its time of arrival at Boston. The station master at Boston observers the traveler coming off the train first and then he notices/records the arrival of the light pulse.

The traveler may reflect the light back to the New York observer.This person[I mean, the New York observer] feels surprised because he did not take into account the effects of superluminal motion.

 

[Dale]

The station master at Boston observers the traveler coming off the train first and then he notices/records the arrival of the light pulse.

No, the light arrives first. What would make you think otherwise?

You should learn SR before GR.

 

[Passionflower]

The observer at the New York Station[say,the station master] never "sees" the light ray reaching Boston. He can only calculate and make predictions about its time of arrival at Boston. The station master at Boston observers the traveler coming off the train first and then he notices/records the arrival of the light pulse.

The traveler may reflect the light back to the New York observer.This person[I mean, the New York observer] feels surprised because he did not take into account the effects of superluminal motion.

Dalespam is absolutely correct, here is an example that shows you are wrong:

On planet A one light year removed from B, a light pulse is sent to B, at the same time a traveler goes with a velocity of 0.6c towards B.

So we have:

Time for light to go from A to B: 1 year.

The travelers proper velocity: 0.75
Coordinate Time for the traveler to go from A to B: 1 2/3 years
Proper Time for the traveler to go from A to B: 1 1/3 years

Thus the observers on planet B will see that the traveler arrives 2/3rd of a year after the light pulse was observed.

Increasing the velocity will not help as you can see below:

Instead of 0.6c the traveler travels at 0.8c

Thus we have:
The travelers proper velocity: 1 1/3
Coordinate Time for the traveler to go from A to B: 1.25 years
Proper Time for the traveler to go from A to B: 0.75 years

Thus the observers on planet B will see that the traveler arrives 0.25 years after the light pulse was observed.

 

[Anamitra]

The traveler on alighting from the train regains the clock rate at Boston/New York.But his actual time as referenced from the New York clock or the Boston clock is different. In the same frame we have people with different times[of course the rate of ticking is the same now].This is not related to any type of universal time.

 

[Anamitra]

Thus the observers on planet B will see that the traveler arrives 0.25 years after the light pulse was observed.

This is seriously incorrect. The observer at planet B[the observer being the traveler himself] will consider his arrival 0.25 years before the arrival of the light pulse,as referenced from his clock.

 

[Passionflower]

This is seriously incorrect. The observer at planet B will see the traveler 0.25 years before the arrival of the light pulse.

I am afraid I cannot convince you of the contrary. I noticed your postings and it appears you are very sure of yourself, that, in my opinion, obstructs the ability to learn. Even the smartest people make mistakes sometimes, I make a lot of them myself and the worst thing is that I am not even smart.

 

[Dale]

This is seriously incorrect. The observer at planet B[the observer being the traveler himself] will consider his arrival 0.25 years before the arrival of the light pulse,as referenced from his clock.

No Anamitra, you are wrong. Passionflower is correct. Also, as referenced by the traveller's clock the light pulse arrives at the planet after .333 years, much earlier than the .75 year mark when he arrives. In fact, the traveler will get the reflection from the event where the original light pulse reaches the destination when his clock reads .667, a full month before he arrives according to his time.

This is very basic stuff here.

 

[pervect]

I have the feeling that it may be time to abandon this thread soon, but let me say one simple thing first.

You've got your stations, and your train, and your clock. And you take a round trip, and you notice that your clock is always slow by some amount when you return. Let's say 200 units, it doesn't really matter what the units are.

What isotropy means is that you can synchronize the clocks at New York and Boston so that you loose 100 units on the trip either way. It doesn't matter which way you go, you loose 100 time units when you take the trip.

Also when you do this, you'll also find that this method of synchronizing clocks is compatible with the Einstein method of synchronizing clocks with light or telegraph signals sent from the midpoint.

So - this way of synchronizing clocks is encouraged. IT's interesting you can do it with a robust clock, and that this "robust clock" matches with the Einstein convention.

You're never going to get around the time loss problem no matter how you fiddle with synchronizing the clocks. The best you can do is even it out, so that the time loss is independent of your direction of travel. The round-trip example proves that there's something going on in realtivity that's not going to be explainable by less radical means.

 

[Anamitra]

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)

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=$${\frac{{d}{/}{\gamma}}{c}{=}{0.6t}{\neq}{1/3}{t}$$
In either case the moving observer expects the light ray to pass before his arrival at B[0.75t] but the expected times are different from the two calculations above.
[It is quite possible that I have some mistake somewhere and I would like to be helped]

 

[Passionflower]

What do you mean by the expected times are different. What do you think is different?

 

[Dale]

Alternatively, Distance from A to B wrt to the moving observer: $$\frac{d}{\gamma}$$

Under what conditions is this formula valid? Are those conditions met here?

FYI: I never recommend using the simplified length contraction and time dilation formulas for exactly this reason. It is too easy to accidentally use them in a situation where they do not apply. Instead, I recommend always using the Lorentz transform. It will automatically simplify to the length contraction and time dilation formulas whenever appropriate, and you will avoid falling into this easy trap. There is no benefit to using the abbreviated formulas.

 

[Anamitra]

Let the moving observer[v=0.8c] make simultaneous measurements of the spatial coordinates of A and B[Please refer to thread #201] at some particular instant of time.He gets the distance $${\frac{d}{\gamma}}$$ between A and B. "d" is the distance between A and B as measured from the ground frame.The value,$${\frac{d}{\gamma}}$$ is fixed for all instants wrt the moving observer.
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. Time taken is $$\frac{{d}{/}{\gamma}}{c}{=}{0.6t}$$.Here t=d/c

The moving observer should record this time,ie,0.6t

 

[Dale]

The relative speed of light wrt the "stick" is c.

What exactly do you mean by this?

Btw, you didn't answer the previous question. When is the length contraction formula valid?

 

[Anamitra]

Light traveling from A to B is light traveling over the stick ,viewed from either frame.In one frame the stick is stationary while in the other it is a moving one

Our "stick" is the distance from A to B. In the ground frame its length is "d". With respect to the observer in the moving frame its length is $$\frac{d}{\gamma}$$ at any particular instant of motion. This value is obviously the same for all instants individually.

[Length contraction formula is valid for simultaneous measurements of the spatial coordinates of A and B by the moving observer]

 

[Dale]

[Length contraction formula is valid for simultaneous measurements of the spatial coordinates of A and B by the moving observer]

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.

Our "stick" is the distance from A to B. In the ground frame its length is "d". With respect to the observer in the moving frame its length is $$\frac{d}{\gamma}$$ at any particular instant of motion.

Yes, the length of the stick is indeed contracted, but that is a red herring since the length of the stick is not the distance that the light travels in the moving frame. To convince yourself of this, simply draw the spacetime diagram.

 

[Anamitra]

If one end of the stick is A and the other end B,it appears, that the distance between A and B is represented by the length of the stick. Of course the speed of light relative to the stick does not change.It is c.

 

[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.

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.

Of course,it is true that we can never fix a frame on a light ray to ascertain what it feels about the world. That is true for all experiments.

Is it possible for a light ray to "see" any distance ,regardless of simultaneity?Does it "understand" the distance "d" between the stations A and B wrt to the ground frame?

 

[Dale]

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.

It sounds like you didn't draw the spacetime diagram.