On the Speed of Light Again!

1. Aug 16, 2010

Anamitra

Let us examine the speed of light in relation to the Equivalence Principle . Let me first state the Equivalence Principle:

"At every space-time point in an arbitrary gravitational field it is possible to choose a "locally inertial coordinate system" such that,within a sufficiently small region of the point in question, the laws of nature take the same form as in unaccelerated coordinate systems in the absence of gravitation."

We consider a point in curved space-time and a "local inertial frame" associated with it.
Now by some suitable transformation we move to some other reference frame at that that point.This frame in general could be a non-inertial one.

We write the metrics for the two cases:
1)For the local inertial frame:
ds^2=g(00)dt^2-g(1,1)dx1^2 - g(1,1)dx2^2 - g(1,1)dx3^2

2) In the transformed frame:
ds'^2=g'(00)dt'^2-g'(1,1)dx1'^2 - g'(1,1)dx2'^2 - g'(1,1)dx3'^2

Now,

ds^2=ds'^2
For a light ray ds^2=0=ds'^2

Speed of light in the local inertial frame
=[g(1,1)dx1^2 +g(1,1)dx2^2 + g(1,1)dx3^2]/g(00)dt^2=1 [c=1 in the natural units]

Speed of light in the non inertial frame:
=[g'(1,1)dx1'^2 +g'(1,1)dx2'^2 + g'(1,1)dx3'^2]/g'(00)dt'^2=1 [c=1 in the natural units]

So light has the same speed in inertial or non-inertial frames if you consider them locally!

[It is to be noted that we have to choose the transformation in such a manner that that product terms like dx(i)dx(j) are absent in the transformed metric.]

2. Aug 16, 2010

JesseM

But isn't this true of Rindler coordinates? According to this the Rindler line element is ds2 = -x2dt2 + dx2 + dy2 + dz2. And yet p. 150 of Wald's General Relativity says that in Rindler coordinates, a null geodesic is given by t = ±ln(x) + constant, so clearly dx/dt for a light beam is not a constant in Rindler coordinates.

3. Aug 18, 2010

Anamitra

Let us write the metric for the Rindler coordinates:

ds^2=-x^2 dt^2+dx^2+dy^2+dz^2
It is important to note that dt is not the physical time element

The physical time element is given by dT=xdt.
The spatial element is given by dL^2=dx^2+dy^2+dz^2

For the null geodesic,
ds^2=0 and we have,

0=-x^2dt^2+dx^2+dy^2+dz^2
=>
0=-dT^2+dx^2+dy^2+dz^2=>
0=-dT^2+dL^2
=>dL/dt=1 [c=1 in the natural units]

It would be a serious mistake to interpret dx/dt as physical speed of light in the Rindler coordinates. It represents "coordinate speed". This can be different from "c" in any system of coordinates.

4. Aug 19, 2010

Passionflower

Some comments on the definition you use:

"At every space-time point in an arbitrary gravitational field it is possible to choose a "locally inertial coordinate system" such that,within a sufficiently small region of the point in question, the laws of nature take the same form as in unaccelerated coordinate systems in the absence of gravitation."

What is sufficiently small?
The term sufficient is not proper to use in a scientific definition because it can become a moving goalpost: "It is always true that a corrupt official will grant you amnesty provided you give him sufficient compensation". How can you prove that something like that is true or false? If you find a case without amnesty you can always claim the compensation was not sufficient.

How about a region of with a diameter of 1.6 x 10-34 m and a time frame of 10-42 seconds? (about 10 times the Planck time and distance)
Looks very small so we can 'fuzz' it and say it is pretty darn flat right? Well how about sufficiently (I use the same term here) close to a singularity?

In the GTR (except for a flat spacetime) there are no inertial frames, but in the neighborhood of a spacetime point they are approximately inertial.

Last edited: Aug 19, 2010
5. Aug 19, 2010

Anamitra

It is important to note that the concept of the Rindler Coordinates has been used incorrectly wherever and whenever dx/dt [as we find in the Rindler Coordinates] has been interpreted as the physical speed of light[or the physical speed of light and the coordinate speed of light have been used in an interchangeable way] This is serious mistake which can only serve the purpose of propagating errors with carefree abandon.

6. Aug 19, 2010

JesseM

What does "physical speed" mean? Surely it is just the coordinate speed in a locally inertial frame? Would you say that a slower-than-light object has a "physical speed" too, or does only light have a physical speed?

7. Aug 19, 2010

Anamitra

Are all local frames inertial?The answer is no.
Of course we can find frames that are "locally inertial". Also by suitable transformations we may find frames that are "locally non-inertial".
[In fact the Rindler coordinates relate to uniformly accelerating frames in flat space-time
http://en.wikipedia.org/wiki/Rindler_coordinates] [Broken]

Physical and Coordinate speeds
Let us consider a pair of points a and b lying on the x1-axis. The physical distance[this is the distance as we know in the physical world] between a and b along the x1 axis is given by:

physical distance= integral [from a to b] g(1,1)dx1

If a particle travels from a to b along the x1 axis the physical time is given by:

physical time= integral [from t1 to t2 ] g(0,0)dt

physical speed =physical distance/physical time

Now to the coordinate speed.The coordinates are simply like labels.The quantity integral[from a to b] dx1 along the x1 axis in general not the distance between a and b as we mean in the physical world.
Thus

coordinate distance=integral[from a to b] dx1, [and we are moving along the x1 axis]
temporal separation[from the coordinate point of view]= integral[from t1 to t2]dt
[dx1 and dt are coordinate separations throughout this posting]
Coordinate speed=coordinate separation/temporal separation(in the coordinate sense)
Important to note, that for a pair of events the coordinate times are the same for all observers[does not matter where they are standing]. But the physical times are different.

dt(A)=dt(B) --------------- for coordinate times[for the points A and B]

=>dT(A)/g(0,0) at A=dT(B)/g(0,0) at B----------------- [dT(A) and dT(B) are the physical times at A and B]

By integrating the above two sides and using a suitable initial condition we may try to synchronize the clocks at any two clocks in our frame.

When we say that clocks run at different rates at two distinct points we always refer to the physical time and never to the coordinate time.The difference in the rates is due to the different values of the metric coefficients at the two points.

[In the local case one may just leave out the integration signs keeping everything else in tact]

Last edited by a moderator: May 4, 2017
8. Aug 19, 2010

Anamitra

On the Local Inertial Frames

We consider a complicated metric:

ds^2=g(0,0)dt^2-g(1,1)dx1^2-g(2,2)dx2^2-g(3,3)ds^2

The above metric is not a flat space-time metric.Can we convert such a metric to a flat space-time metric by some suitable transformation?The answer is "yes". But there is a restriction.We can achieve our goal[quite accurately] but to a limited region of space-time.And we get a local inertial frame.Does this frame have a true physical existence? Our answer is again "yes". We may give a complicated motion some frame locally and achieve the effect.

A frame at rest in a gravitational field is in effect equivalent to an accelerating frame. Our new frame ,if it is an accelerating one ,cancels the existing "in effect" acceleration of the rest frame. The net effect is that we get a true inertial frame.

The fact that gravity is equivalent to an acceleration is an interesting fact. May I refer to the papers in the Physical Review and the Scientific American that contain the experimental details substantiating the above facts. These papers have been cited by Robert Resnick in "Introduction to Special Relativity", Supplementary Topic B, Section B5---->"An Experimental Test"

May I also refer to Steven Weinberg,Gravitation and Cosmology",Chapter 3,The Equivalence Principle.Section 1-->Statement of the Principle

The Papers cited by Robert Resnick:
1)J. Bronowski, "The Clock Paradox", Scientific American,February,1963
2)Walter Kundig, Phys Rev. ,129,2371(1963)
3)J.J. Hay,J.P. Schiffer,T.E. Cranshaw and P.A. Egelstaff,Phys Rev Letters,4,165(1960) discussed in reference (1)

Last edited: Aug 19, 2010
9. Aug 19, 2010

Passionflower

It would be helpful if you could take the trouble to use Latex.

Any metric that is not Riemann flat cannot be converted into a Riemann flat metric.

It is not equivalent, it can be similar but it is not equivalent.

10. Aug 19, 2010

Anamitra

An infinitesimally small curve may be represented by a straight line segment though the two elements can never have the same equation no matter how small they are.But in the final stage of a calculus problem [say an integration]we perform a limiting process which succeeds in removing all errors incurred in replacing an infinitesimally small curve by a straight line.You may apply this logic to my statement in thread #8-->"We can achieve our goal[quite accurately] but to a limited region of space-time."

ds^2=g(0,0)dt^2-g(1,1)dx1^2-g(2,2)dx2^2-g(3,3)dx3^2

There was a mistake in the last term of the metric in my last thread
[The latex toolbar is simply not working]

11. Aug 19, 2010

JesseM

For a slower-than-light particle this is coordinate dependent, no? You seem to be defining it by picking a line of constant t,x2,x3 (and varying x1) between the positions of two events on the particle's worldline, and then a line of constant x1,x2,x3 (and varying t) between the times of two events on the particle's worldline, but of course this depends on how your x1,x2,x3,t axes are defined, it isn't coordinate-invariant like if you integrated the metric's line element along the actual worldline of the particle (where multiple coordinates may be varying). For example, two events on a timelike worldline which have a nonzero distance in one frame may happen at the same position coordinate, and thus have zero distance, in another.

Also, I don't see why the above definitions would imply the "physical speed" of a light beam is equal to 1. For the Rindler metric we have g(0,0)=-x and g(1,1)=1, right? And one geodesic for a light beam in Rindler coordinates might be t=-ln(x), so two points on the path of this light beam could be x=2.71828, t=-1 and x=1, t=0. Can you calculate the "physical time" and "physical distance" between these points according to your above definitions?

12. Aug 19, 2010

Anamitra

Coordinates at the initial point=(-1,e)
Coordinates at the final point=(0,1)

Temporal separation(physical)=integral[-1 to 0]xdt
Now ,
0=-x^2dt^2 + dx^2
dx=(+or-)xdt
We take the plus sign first:
dt=dx/x

Physical separation of time=integral[from -1 to 0] x dx/x
= integral[from e to 1] dx
=1-e
Physical separation of distance = 1-e
Therefore,

Speed of light =1 [natural units]

If we take the negative sign:

dt=-dx/x
Physical separation of time=- integral[from -1 to 0] x dx/x
= -integral[from e to 1] dx
=-(1-e)
Physical separation of distance = 1-e

|Speed of light|=1
Again we get a consistent result.

13. Aug 19, 2010

JesseM

I don't see how it's valid to use the equation which relates dt and dx along the worldline of a light beam, 0=-x^2dt^2 + dx^2, to use as a substitution in the integral of the line element along a timelike worldline of constant x between two times t=-1 and t=0, which I thought is how you were calculating the "physical time". For example, if we have a line of constant x=e with endpoints x=e,t=-1 and x=e,t=0, then along this line we have ds^2 = -e*dt^2, so ds=i*sqrt(e)*dt, so if we integrate ds along this line we have integral[from -1 to 0] i*sqrt(e) dt = 0*i*sqrt(e) - (-1)*i*sqrt(e) = i*sqrt(e). And ds is always imaginary for a timelike separation, so this means sqrt(e) is the proper time along this worldline, I thought this would correspond to what you meant by the "physical time" between a pair of events with times t=-1 and t=0.

Similarly, if you have a line of constant t=0 with endpoints x=e,t=0 and x=1,t=0, then along this line we have ds^2 = dx^2, so ds=dx, so if we integrate ds along this line we have integral[from e to 1] dx = 1-e, which is what you got for the "physical distance".

Was I mistaken about your procedure for calculating "physical time"? Is the "physical time" not equal to the integral of ds along a timelike worldline of constant x whose endpoints occur at the same time-coordinates as the two events on the worldline of the particle that you want to calculate the physical time between? Are you integrating ds along a worldline where neither x or t is held constant? (and if so why do you only include g(0,0) in the integral but not g(1,1)? I thought the justification for not including g(1,1) in the integral was that dx=0 along the worldline you're integrating along, i.e. the x-coordinate is constant along that worldline, so the general line element ds^2 = g(0,0)dt^2 + g(1,1)dx^2 could be reduced to ds^2 = g(0,0)dt^2 along this worldline)

14. Aug 19, 2010

Anamitra

Interestingly (-1,e) and (0,1) are points on a null geodesic[the time coordinate appearing first]:

t=-lnx

For the Rindler coordinates the null geodesics are given by:

t=(+or-)lnx + constant [Wald---page 150]

In your case you have chosen the negative sign and made the constant zero.Where from do you get a timelike separation?

"Also, I don't see why the above definitions would imply the "physical speed" of a light beam is equal to 1. For the Rindler metric we have g(0,0)=-x and g(1,1)=1, right? And one geodesic for a light beam in Rindler coordinates might be t=-ln(x), so two points on the path of this light beam could be x=2.71828, t=-1 and x=1, t=0. Can you calculate the "physical time" and "physical distance" between these points according to your above definitions?"

15. Aug 19, 2010

starthaus

t does not represent time in Rindler coordinates, T does. t is simply a dimensionless quantity, look at the transformation that produced it :

$$t=arctanh(T/X)$$

As to "c=1", that is by definition (by the way the original metric was constructed), look at the starting point:

$$ds^2=-dT^2+dX^2+dY^2+dZ^2$$

For $$ds=0$$ you get immediately: $c^2=\frac{dX^2+dY^2+dZ^2}{dT^2}=1$

so Anamitra's result is a direct consequence of the definition of the original metric. The reparametrization of the metric via $t=arctanh(T/X)$, etc. will not change the value for which $ds=0$.

Last edited: Aug 19, 2010
16. Aug 19, 2010

JesseM

Of course, that was why I picked them! Remember, I said "And one geodesic for a light beam in Rindler coordinates might be t=-ln(x), so two points on the path of this light beam could be x=2.71828, t=-1 and x=1, t=0"
My understanding was that to get the "physical time" between two events on the worldline of a light beam E1 and E2, you would pick a different pair of events which don't both lie on the light beam's worldline, but which instead lie on a line of constant position coordinate whose endpoints have the same time-coordinates as E1 and E2. For example, if E1 is (x=e, t=-1) and E2 is (x=1, t=0), then we could pick two events A and B with coordinates (x=e, t=-1) and (x=e, t=0), and find the integral of ds along the path of constant x=e between them (which will just be equal to i times the proper time between A and B along this path). You would agree the path between A and B is timelike, yes?

If this isn't your procedure for calculating the "physical time" between E1 and E2, then what, physically, are you calculating? You can't be integrating ds between E1 and E2 on the worldline of the light beam itself, since by definition the integral of ds along the wordline of a light beam is always 0. Are you determining "physical time" by integrating ds along some non-lightlike worldline, yes or no? If you're not integrating ds along some worldline, why then why does your integral include g(0,0) which appears in the expression for ds^2? And if you are integrating ds along some worldline, which worldline is it, if it's not the worldline of the light beam itself and it's not the type of timelike worldline I suggested where the position coordinate is held constant and only the time coordinate varies?
Yes, but your calculation doesn't make any sense to me, which is why I am asking the questions above about the physical meaning of what you are doing. And you didn't answer my other question from that same post which was about the "physical speed" of a slower-than-light particle:

"For a slower-than-light particle this is coordinate dependent, no?"

17. Aug 19, 2010

JesseM

If you're referring to the conventions used in the wikipedia article, T represents the time coordinate in Minkowski coordinates (you can tell because it appears in the Minkowski metric $$ds^2 = -dT^2 + dX^2 + dY^2 + dZ^2$$), while t is the time coordinate in Rindler coordinates.
No, the wikipedia article simply sets c=1 and g=1, where g is the acceleration of the Rindler observer who starts at position X=c2/g at T=0 in Minkowski coordinates, and who is the unique observer whose proper time matches up with Rindler coordinate time (see below). The full equation would be $$t=(c/g)arctanh(cT/X)$$. See this book, which has a section on "coordinates for the accelerated frame" starting on page 240, and on page 246 they write:
Then on page 247 the author points out that this family of accelerating observers will share common planes of simultaneity, so the coordinate time of an event on any observer's worldline can be defined as "our" proper time (where 'we' are the observer with acceleration g) at a time simultaneous with the distant event on one of these shared planes of simultaneity:
Finally on page 250, equations (7.26) show the transformation between the coordinate time $$\bar{t}$$ and position $$\bar{x}$$ in the accelerated frame back to time t and x in the inertial frame:
The relevant bit is that first equation from 7.26, $$\bar{t} = \frac{1}{g} th^{-1} \frac{t}{x}$$. You can see this is the same as the equation in the wikipedia article but with 1/g out in front, and they mentioned earlier that they were using units where c=1 so it would really be c/g. c/g has units of time, so $$\bar{t}$$ does as well.
But here you are using the X,Y,Z,T coordinates of the inertial frame--it's obviously true that any particle moving on a null geodesic has velocity c=1 in an inertial frame, but Anamitra was claiming it was still in some sense true that it has c=1 in the coordinates of the accelerated frame.

Last edited: Aug 19, 2010
18. Aug 19, 2010

DrGreg

I haven't come across Anamitra's technique before, but I think this is what is happening.

Given a metric of the form

$$ds^2 = g_{00}\,dt^2 + g_{ij}\,dx^i\,dx^j$$​

(where i, j take values 1,2,3 only) define two new metrics:

$$dT^2 = -g_{00}\,dt^2$$
$$dL^2 = g_{ij}\,dx^i\,dx^j$$​

T is being called "physical time" and L is being called "physical length". Both are evaluated by integrating along the same spacetime worldline that you would integrate ds along. And both are dependent on your choice of coordinate system.

It should be clear that if you were to evaluate T along a worldline of constant x1,x2,x3 it would equal proper time. If you were to evaluate L along a curve of constant t it would equal proper length. But for an arbitrary worldline you evaluate both along the same worldline.

If you calculate |dL/dT| along a null worldline you get 1.

Last edited: Aug 19, 2010
19. Aug 19, 2010

JesseM

OK, that makes sense. So in that case both g(0,0) and g(1,1) will be varying along a light beam's worldline--for example, if we choose a beam given by t=-ln(x) and x=e^-t in Rindler coordinates, and we want to find the "physical time" between events (x=e, t=-1) and (x=1, t=0), then since g(0,0) = -x = -e^-t, the square of the physical time (the integral of dT^2 along the light beam's worldline) would be equal to integral(from -1 to 0) -e^-t dt = e^-(0) - e^-(-1) = 1 - e (which means the physical time is actually $$\sqrt{1 - e}$$, but that's a minor quibble). And since g(1,1) = 1, the square of the physical distance is equal to integral(from e to 1) 1 dx = 1 - e, so the physical distance is $$\sqrt{1 - e}$$ as well, and "physical distance"/"physical time" is 1.

I'm not sure what physical meaning could be assigned to these "physical" measures calculated using these altered metrics, which are different from the normal spacetime metric. Maybe if you divide an arbitrary worldline into a lot of short segments, and then for each segment you consider a short timelike worldline of constant position coordinate which goes through the midpoint of the segment and whose endpoints have the same time coordinates as the endpoint of the segment, then if you add up the proper time along all the little timelike worldlines (calculated using the normal metric), in the limit as the size of each segment approaches zero (so the number of segments approaches infinity) the sum of proper times will approach the "physical time" calculated with the altered metric? So it's sort of like approximating the smooth worldline by a "pixellated" line and then adding the vertical height of all the pixels, and considering the limit as the number of pixels goes to infinity.

20. Aug 19, 2010

DrGreg

Yes, that's exactly the picture I have.