Relative Velocity II: Defining Relative Velocity in GR

In summary: But this is really all that is required to define relative velocity in GR.If we assume (for the sake of argument) that the observers are not at the same event in space-time, then we must somehow compare their 4-velocity vectors. The easiest way to do this is to "parallel transport" the 4-velocity vector along the geodesic connecting the two observers. This means that we take the 4-velocity vector of O and send it directly to the 4-velocity vector of O' by transporting it at the speed of light. Doing this will result in a change in the direction of the 4-velocity vector (
  • #1
Oxymoron
870
0
The term "Relative Velocity" has a somewhat vague definition in General Relativity - perhaps it doesn't really have a definition at all! What is the relative velocity of two observers with respect to one another when the spacetime that they are moving in is curved? In SR one compares the velocity vector of one observer with the 4-velocity vector of the other by parallel transporting it. In flat space there is exactly one way to do this - parallel transport the 4-velocity vector along the straight line between the two. As a result there is a definitive notion of relative velocity between two observers in SR. In GR, the spacetime is necessarily curved and hence there is perhaps a great number of ways to parallel transport the 4-velocity vector, hence a great number of ways to define what one means by relative velocity.

I have not read anywhere an attempt to give a proper definition to relative velocity in GR. I am certainly not saying that this is possible or that relative velocity should even be given a definition in GR - it may be rediculous! Never-the-less I will attempt the task in this thread, hopefully with the help of others.

My main source of information is the book "General Relativity" by Robert Wald which Pervect suggested I read, from the thread in this forum titled "Relative Velocity". This thread becomes the second part in my quest to define relative velocity in GR.

Wald does spend a bit of time discussing redshift, which I like - particularly chapter 6.3. However, he does not explicitly discuss relative velocity - but I feel that from what he has written I can make the following proposal:

relative velocity between two observers (In GR) is a physical quantity derived from the frequency shift of a light ray connecting the two observers.

That is, I want to define relative velocity in GR as being directly related to the redshift. I understand that this will not be quite universal, that is, I don't think I will be able to define a frequency shift explicitly in every GR setting - BUT - in the cases where I am able to define a frequency shift I WILL be able to define relative velocity. This is the definition I want to give to the term "Relative Velocity" of two observers in GR.

Any thoughts?
 
Physics news on Phys.org
  • #2
Oxymoron said:
I feel that from what he has written I can make the following proposal:

relative velocity between two observers (In GR) is a physical quantity derived from the frequency shift of a light ray connecting the two observers.
Any thoughts?
How do you differentiate between 'relative velocity' Doppler shift and gravitational red shift?

Garth
 
  • #3
Posted by Garth:

How do you differentiate between 'relative velocity' Doppler shift and gravitational red shift?

Some rhetorical questions first:

Q: Given two observers O and O', what is the relative velocity of O' as observed by O?

This is the big question, right? How does one define the relative velocity of one observer with respect to another? I suppose there should be no answer at all as I have not given any reasonable clue as to what these "observers" are. Let me make the following hypotheses:

1. The "observers" exist in a 4-dimensional space-time manifold with Lorentzian metric and the Levi-Civita connection.

2. There exist a unique geodesic between every pair of points p and q.

An observer in this space-time is a timelike worldline which I will denote simply as O. Every point along this worldline has a 4-velocity (a future-pointing timelike unit vector field U defined on O and tangent to O). That is, the vector field U assigns a 4-velocity vector [itex]u \in T_p\mathcal{M}[/itex] to each point along the worldline O.

Now, given two observers O and O' at the same event in space-time with 4-velocities [itex]u_p[/itex] and [itex]u'_p[/itex], then there is a unique vector [itex]v \in u^{\perp}[/itex] (that is, there exists a unique vector v in the orthogonal space of u) and a unique positive real number [itex]\gamma[/itex] (simply the gamma factor associated with the "length" of v) such that:

[tex]u'_p = \gamma(u_p + v)[/tex]

This is simply a matter of geometry and "connecting" up the vectors into a quadrilateral. And this is very simple to do and requires nothing but vector geometry IF (and this is a big "If") and only if the two observers happen to be at the exact same event in space-time because no parallel transport is required to compare the two 4-velocity vectors.
 
Last edited:
  • #4
In my previous post I showed that it is possible to derive a formula for the relative velocity of one observer with respect to another if the worldlines of the observers intersect at a single point in space-time. Once there, we effectively freeze time and compare their 4-velocity vectors. One will obviously coincide with the other if and only if the two observers have the same direction and velocity at the same event (ie. if there are in fact one in the same)

However, if they are indeed distinct observers then u' at p will obviously differ from u at p by a matter of vector geometry. One will equal the other plus, what I call "the relative velocity vector", v, scaled by the gamma factor due to the relativistic effects of motion. Here is the formula again:

[tex]u'_p = \gamma(u_p + v)[/tex]

where

[tex]\gamma = -g(u',u) = \frac{1}{\sqrt{1-|v|^2}}[/tex]
 
  • #5
The question now becomes:

Q: What is the relative velocity of observer O' at event q with respect to observer O at event p in space-time?

A: No solution.

Well, I believe there is a solution given certain conditions.

One of these conditions is that there must exist a null geodesic, or light ray, between p and q. This means that the space-time manifold must be convex.

In such a space-time, for any two observers, at any two distinct events, the relative velocity of one observer with respect to the other at each event will only make sense if there is a light ray joining the two events.
 
  • #6
Keep in mind, my main goal is to come up with some meaningful definition of relative velocity in a hope to possibly combine Doppler and gravitational redshift.

A light ray, by definition (see wikipedia for example), is simply a null geodesic [itex]\lambda[/itex] with a future-pointing null vector field F defined on [itex]\lambda[/itex], tangent to [itex]\lambda[/itex] and parallelly transported along [itex]\lambda[/itex]. That is,

[tex]\nabla_F F = 0[/tex]

by definition.Imagine, then that an observer O sends out one of these light rays at a point p. At that point, there exists a 4-velocity vector u_p (since the worldline comes equipped with a vector field which assigns a 4-velocity vector to each point) which defines the velocity of O at that particular instant in time and space.

Suppose also, that the light ray intersects a second observer O' at q different from p. By the same token, u'_q represents O' 's velocity at q. The question is: What is the velocity of O' as observed by O? In the Newtonian sense, the answer is simply u'_q! But Newton did not have any idea of curved space. The Special Relativist will say: "Why not parallelly transport u_p along the straight line to q?". But who is to say that the shortest path exists? Is even distinct? (Gravitational Lensing) Or if it is even the correct representation of relative velocity?

I am here to say that parallelly transporting the velocity vector along some sort of path to q is correct. But which path? The null geodesic! That is, the relative velocity of O' at q with respect to O at p will depend on how the 4-velocity vector is parallelly transported along the joining null geodesic. It may change, it may not. It would be reasonable to say that it won't change if the manifold is flat between p and q. But what if it is curved? What if there is a star between p and q?

Physically speaking, I am to propose that the relative velocity of O' w.r.t O will depend on not only how the velocity vector changes under parallel transport between the two events but more so on how the frequency of the actual light ray itself is changed upon traveling between the two events.

What I'd like to say is this: If observer O sends out a light signal [itex]\lambda[/itex] with frequency [itex]\omega[/itex] at event p and intersects a second observer O' at q and O' measures the frequency of the intercepted ray to be [itex]\omega'[/itex] then the relative velocity of O' w.r.t O depends solely on the ratio of [itex]\omega[/itex] with [itex]\omega'[/itex].

Note: I am not saying that this is the right way to do it, and I am certainly not saying that it is possible all the time. I merely want to prescribe a consistent method of relative velocity in GR. And why not base the concept of relative velocity upon the frequency shift of the light ray joining the two events and observers!?
 
Last edited:
  • #7
Oxymoron said:
Some rhetorical questions first:

Q: Given two observers O and O', what is the relative velocity of O' as observed by O?

This is the big question, right? How does one define the relative velocity of one observer with respect to another? I suppose there should be no answer at all as I have not given any reasonable clue as to what these "observers" are. Let me make the following hypotheses:

1. The "observers" exist in a 4-dimensional space-time manifold with Lorentzian metric and the Levi-Civita connection.

2. There exist a unique geodesic between every pair of points p and q.

This is a pretty big assumption. It won't work, for example, in the solar system. Any possible orbit will essentially be a geodesic. It's easiest to consider the weak field, where we can use our intuition of Newtonian orbits. If we consider orbits in space, any time we have two orbits with the same transit time between two points, your assumption will be violated.

It's easy to see that such geodesic congurences should exist

a----------M-----------b

consider an elliptical orbit from a to b around M. By rotating the plane of the ellipse, we can have an orbit with equal travel time from a to b, hence we have two different geodesics connecting a and b.

Other such situations will give equal-time goedesics. We can have, for instance, a body that goes from a to b to a to b, making 1.5 round trips, and a different orbit that goes from a to b with a longer semi-major axis so that it arrives at the same time.

I think I mentioned my solution last go-around. It involves the notion of static observers, which requires a static space-time, so it's not general. I doubt a general defintion even exists.

Anyway, the idea is that a static observer follows a time-like Killing vector of the static metric. This idea fell afoul of the fact that time-like Killing vectors are not unique. But when we add in the reqiurement that the time-like Killing vectors have an orthogonal space-like hypersurfaces (an extra condition that one imposes with static metrics that stationary metric don't have) I think that non-uniqueness goes away.

Considering the example of a rotating frame of reference. Last time we fell afoul of the fact that a body in a rotating frame, at constant 'r' but with time-varying theta, would also be following a time-like Killing vector.

However, the fact that we can't synchronize our clocks via the Einstein convention in a rotating frame of reference tells us its rotating. So requring that we have space-like hypersurfaces where we can synchronize our clocks via the Einstein convention should rule out these rotating frame of references. So I think the "hypersurface orthogonality" solves the problem, making the time-like Killing vectors unique.

Wald mentions that the notion of "holding station" makes sense in a static space-time, and he also mentions that the static observers follow an orbit of the Killing field, but he didn't prove that the Killing field was unique. If the Killing field is uniquely specified when one adds in the condition of hypersurface orthogonality, then everything works. Otherwise it all falls apart again.
 
Last edited:
  • #8
Posted by Pervect

This is a pretty big assumption. It won't work, for example, in the solar system. Any possible orbit will essentially be a geodesic. It's easiest to consider the weak field, where we can use our intuition of Newtonian orbits. If we consider orbits in space, any time we have two orbits with the same transit time between two points, your assumption will be violated.

Do you think posts #3 through #6 still make sense if I remove the second hypothesis?

Posted by Pervect:

I think I mentioned my solution last go-around. It involves the notion of static observers, which requires a static space-time, so it's not general. I doubt a general defintion even exists.

I agree. I extremely doubt a general definition exits. The best I can hope for is a mostly general definition. One which works in a static spacetime would be great!
 
Last edited:
  • #9
Ok, let me recap.

Wald proposes that given two static observers O_1 and O_2 with 4-velocity vectors u_1 and u_2 we can define these in terms of the static Killing field:

[tex]u^a_1 = \frac{X^a}{\sqrt{-X^bX_b}} \mbox{ at } p_1[/tex]

[tex]u^a_2 = \frac{X^a}{\sqrt{-X^bX_b}} \mbox{ at } p_2[/tex]

...assuming, of course, that one may write the frequency of emission as

[tex]\omega_1 = k_a u^a_1[/tex]

Im guessing that Wald's expression of u_1 and u_2 in terms of the Killing field is some sort of normalization? This is what I don't get. I mean, how can one write the 4-velocity vector in terms of the Killing vector field!?

Perhaps it is because u_1 points in the same direction as X^a. But u_1 is unit, so you have to divide by the square root of the scalar product of X^a. Yes this must be it!
 
Last edited:
  • #10
Oxymoron said:
In my previous post I showed that it is possible to derive a formula for the relative velocity of one observer with respect to another if the worldlines of the observers intersect at a single point in space-time. Once there, we effectively freeze time and compare their 4-velocity vectors. One will obviously coincide with the other if and only if the two observers have the same direction and velocity at the same event (ie. if there are in fact one in the same)

However, if they are indeed distinct observers then u' at p will obviously differ from u at p by a matter of vector geometry. One will equal the other plus, what I call "the relative velocity vector", v, scaled by the gamma factor due to the relativistic effects of motion. Here is the formula again:

[tex]u'_p = \gamma(u_p + v)[/tex]

where

[tex]\gamma = -g(u',u) = \frac{1}{\sqrt{1-|v|^2}}[/tex]


Your notation is confusing me. 4-velocites should be superscripted, i.e. ua. If we replace your subscripts with superscripts, ,there's still the question of v, which you call a vector, not having any indices.

Also I don't know the correct formula offhand, I'll have to work it out.
 
  • #11
If we have two four-velocities represented by 4-vectors u and v, I think that the relative velocity, a scalar, can be most easily computed by the relationship

[tex]\vec{u} \cdot \vec{v} = |u| |v| \cosh \theta = \cosh \theta[/tex]

as the magnitude of the 4-velocites is |u|=|v|=1. The relative velocity [itex]\beta[/itex] can be recovered from the angle theta by the formula

[tex]\theta = \tanh^{-1} \beta[/tex]

This means that

[tex]\vec{u} \cdot \vec{v} = \cosh( \tanh^{-1}(\beta) ) = 1/\sqrt{1-\beta^2}[/tex]

or

[tex] \beta = \sqrt{1 - 1/(\vec{u} \cdot \vec{v})^2}[/tex]

I'm not sure yet how this compares with your result. the dot product is the standard dot product, in component notation

[tex]\vec{u} \cdot \vec{v} = g_{ij} u^i u^j [/tex]
 
Last edited:
  • #12
pervect said:
I'm not sure yet how this compares with your result.

Oxymoron's result for the relative velocity between two coincident observers is correct.

He uses the 4-velocity [itex]u_p[/itex] to split spacetime into space and time. Then 4-velocity [itex]u'_p[/itex] has a time part [itex]\gamma u_p[/itex] parallel to [itex]u_p[/itex], and a space part [itex]\gamma v[/itex] g-orthogonal to [itex]u_p[/itex].

Then,

[tex]g \left(u'_p , u_p \right) = g \left(\gamma \left( u_p + v \right) , u_p \right) = g \left(\gamma u_p , u_p \left) + g \right( \gamma v , u_p \right) = \gamma.[/tex]

This reults, however, is not of much use when the observers are not coincident.

His supscript p refers to the event p of coindence. It is not a covariant index.
 
Last edited:
  • #13
If I'm not mistaken, I think Pervect's calculation gives the correct magnitude of Oxymoron's spatial-velocity vector. It might be worth pointing out that the relative velocity vector of observer A according to observer B differs from the corresponding vector of B according to A, although their magnitudes are equal.
 
  • #14
If I'm not mistaken, I think Pervect's calculation gives the correct magnitude of Oxymoron's spatial-velocity vector.

Yes, Oxymoron and pervect calculate, using different notations, the same thing.

Oxymoron, the book Relativity on Curved Manifolds, by De Felice and Clarke, make a thorough attempt (perhaps overwhelmingly thorough) at separating the parts of a wavelength shift that are due to relative velocity and curvature.

Chapter 9 is titled Physicsal Measurements in Spacetime, with 9.6 Measurement of Relative Velocities and 9.7 The Velocity Composition Law.
 
Last edited:
  • #15
Posted by Pervect:

POST #11

I believe this is correct. Same thing as what I did, but in a different notation.
Posted by George Jones:

Oxymoron's result for the relative velocity between two coincident observers is correct.

He uses the 4-velocity u_p to split spacetime into space and time. Then 4-velocity u'_p...~...This result, however, is not of much use when the observers are not coincident.

His supscript p refers to the event p of coindence. It is not a covariant index.

This is exactly correct.

The 4-vector v in this interpretation is what I call the relative velocity of observer O' w.r.t. observer O when their two worldlines intersect at a single event in space-time. George's explanation of how I did this is much more concise than mine. Remember, all that I have proposed is a definition of relative velocity at the intersection of two worldlines.

Posted by Robphy:

If I'm not mistaken, I think Pervect's calculation gives the correct magnitude of Oxymoron's spatial-velocity vector. It might be worth pointing out that the relative velocity vector of observer A according to observer B differs from the corresponding vector of B according to A, although their magnitudes are equal.

I think this is also correct and a good point to make.
 
Last edited:
  • #16
Now I will attempt to explain (in a similar fashion) my definition of relative velocity of one observer w.r.t. another when they do not intersect.

Do you agree that a light ray is is a lightlike geodesic [itex]\lambda[/itex] with a future-pointing lightlike vector field, let's call it F (for frequency as we will see) defined on [itex]\lambda[/itex]. F is obviously tangent to [itex]\lambda[/itex] and [itex]\nabla_F F = 0[/itex].

So given any event [itex]p \in \lambda[/itex] along the light ray and any observer O with a 4-velocity [itex]u_p[/itex] at the same event, then there exists a unique vector [itex]w \in u_p^{\perp}[/itex] and a unique positive real number [itex]\nu[/itex] such that

[tex]F_p = \omega(u_p + w)[/tex]

in exactly the same fashion as I did before. Have a guess what I call w? This is the relative velocity of the light ray as observed by the observer O at the point p. And just as we did before, the nu factor is simply the frequency of the light ray as observed by O at the point p.

The only difference between this scenario and the previous one is that now the other observer is a light ray traveling along a null geodesic. But I have shown that now we have an extra property. We can still measure the relative velocities of one w.r.t. the other (because they are coincident at a single event) but now, we have what I call the frequency of the light ray as observed by O. And it equals

[tex]\omega = -g(F_p,u)[/tex]

This is not much different to how Wald defines his frequency as the negative of the rate of change of the phase of the wave. I simply define it as the negative of the metric.
 
  • #17
Oxymoron said:
Have a guess what I call w?

The spatial direction of the light ray according to O.

This is the relative velocity of the light ray as observed by the observer O at the point p.

Right. Note that the speed of the light ray is fixed, so (spatial) velocity is just a direction.

Now, at p, introduce your second observer O' with 4-velocity u'_p. From this, the Doppler effect is easily derived.
 
  • #18
Posted by George Jones:

Right. Note that the speed of the light ray is fixed, so (spatial) velocity is just a direction.

Now, at p, introduce your second observer O' with 4-velocity u'_p. From this, the Doppler effect is easily derived.

Yes! Given a second observer O' at the same event p and using

[tex]F_p = \omega(u_p + w)[/tex]

we can have

[tex]\omega(u_p + w) = F_p = \omega'(u'_p+w')[/tex]

where [itex]\omega[/itex] and [itex]\omega'[/itex] are the frequencies of the light ray [itex]\lambda[/itex] as measured by O and O' respectively, and w and w' are the relative velocities of the light ray as measured by O and O' respectively. Now since [itex]u'_p = \gamma(u_p + v)[/itex] we have the Doppler Effect Equation:

[tex]\omega' = \left(\frac{u+w}{u'+w'}\right)\omega[/tex] EDIT: Typo fixed

Hmm, I don't like w's and omega's in the same equation...
 
Last edited:
  • #19
Oxymoron said:
Now since [itex]u'_p = \gamma(u_p + v)[/itex] we have the Doppler Effect Equation:

[tex]\omega' = \left(\frac{u+w}{w'+w'}\right)\omega[/tex]

Is that a ratio of vectors?
 
  • #20
Now it is possible for me to explain what the relative velocity of an observer is, with respect to another observer, at two different events in space-time so long as they are on the same light ray.

The fact that they must be connected by a light ray is crucial to this definition. As I said at the start, my definition will be for special cases. I thought, since there is no definition for relative velocity at all in GR, why not define one - at least for a special case. I think having one is better than having none at all (as long as when we use the term "relative velocity" we know what we are talking about).

Suppose that [itex]\lambda[/itex] is a light ray between two events p and q in space-time. Also, let there be two observers O and O' who happen to intersect these two points. At each point, each observer has a 4-velocity vector u_p and u'_q respectively. Then we can talk about the relative velocity of O' as observed by O.

How? Simply parallel transport the 4-velocity vector of O' along the light ray to p. That is, parallel transport u'_q along [itex]\lambda[/itex] to p and compare with u_p.

This means that the relative velocity of O' as observed by O is a unique vector [itex]v \in u^{\perp}_p[/itex] such that the parallel transport of the 4-velocity vector [itex]u'_q[/itex] is...well we know what it is...

[tex]\mbox{Parallel transport of }u'_q = \gamma( u_p + v)[/tex]

where [itex]\gamma[/itex] is the gamma factor related to the velocity v and v is the relative velocity of O' as measured by O.
 
  • #21
I have to echo robphy here. How is it possible to divide by lightlike 4-vectors?

Take the equation

Oxymoron said:
[tex]\omega(u_p + w) = F_p = \omega'(u'_p+w')[/tex]

and "dot" both sides (i.e, "g" both sides) with [itex]u'_p[/itex]. On the right, leave [itex]u'_p[/itex] as [itex]u'_p[/itex]. On the left, expand [itex]u'_p[/itex] as

[itex]u'_p = \gamma(u_p + v)[/itex]
 
  • #22
Posted by Robphy and George Jones:

Is that a ratio of vectors?

No. The equation I wrote is the classic Doppler Effect equation. The equation I get looks something like:

[tex]\omega' = \gamma(1 - g)\omega[/tex]

Posted by George Jones:

Take the equation ... and "dot" both sides (i.e, "g" both sides) with...

Do you mean like this:

[tex]\omega(u_p + w)\cdot u'_p = \omega'(u'_p + w')\cdot u'_p \quad \Rightarrow[/tex]
[tex]\omega(u_p + w)\cdot \gamma(u_p + v) = \omega'(u'_p + w') \cdot u'_p \quad \Rightarrow[/tex]
[tex]\omega \gamma (u_p + w)\cdot (u_p + v) = \omega'(u'_p + w') \cdot u'_p \quad \Rightarrow[/tex]
[tex]\omega \gamma [(u_p + w)\cdot v + (u_p + w)\cdot u_p] = \omega'(u'_p + w') \cdot u'_p \quad \Rightarrow[/tex]
[tex]\omega \gamma (u_p \cdot v + v \cdot w) + \omega\gamma(u_p + w)\cdot u_p = \omega'(u'_p + w')\cdot u'_p \quad\Rightarrow[/tex]
[tex]\omega \gamma (1 - g(v,w)) + \omega\gamma(u_p + w)\cdot u_p = \omega'(u'_p + w')\cdot u'_p[/tex]
 
Last edited:
  • #23
Oxymoron said:
No. The equation I wrote is the classic Doppler Effect equation. The equation I get looks something like:

[tex]\omega' = \gamma(1 - g)\omega[/tex]

Above, and at the end of post#18, I don't understand your notation.

[tex]
\begin{align}
g\left( \omega \left( u_{p}+w\right) ,u_{p}^{\prime }\right) &= g\left( \omega ^{\prime }\left( u_{p}^{\prime }+w^{\prime }\right) ,u_{p}^{\prime }\right)\\
g\left( \omega \left( u_{p}+w\right) ,\gamma \left( u_{p}+v\right) \right) &= g\left( \omega ^{\prime }\left( u_{p}^{\prime }+w^{\prime }\right) ,u_{p}^{\prime }\right)\\
\omega \gamma \left( g\left( u_{p},u_{p}\right) +g\left( w,u_{p}\right) +g\left( u_{p},v\right) +g\left( w,v\right) \right) &= \omega ^{\prime }\left( g\left( u_{p}^{\prime },u_{p}^{\prime }\right) +g\left( w^{\prime },u_{p}^{\prime }\right) \right)\\
\omega \gamma \left( -1+g\left( w,v\right) \right) &=-\omega ^{\prime }\\
\omega ^{\prime } &=\omega \gamma \left( 1-w\cdot v\right)
\end{align}
[/tex]
Three special cases are:

1) the spatial vectors [itex]v[/itex] and [itex]w[/itex] are parallel, so [itex]w\cdot v= \left| v \right|[/itex];

2) the spatial vectors [itex]v[/itex] and [itex]w[/itex] are anti-parallel, so [itex]w\cdot v= \left| -v \right|[/itex];

3) the spatial vectors [itex]v[/itex] and [itex]w[/itex] are perpendicular to each other, so [itex]w\cdot v=0[/itex].

The last case is the transverse Doppler effect, a purely relativistic effect due to time dilation.
 
Last edited:
  • #24
Ah, I understand what you are saying now. I wasn't sure what values you intended to put in the metric [itex]g(\,,\,)[/itex], so I tried dot product...but I didnt get far with that obviously. What you wrote is correct and matches my equation which you quoted above. Except I forgot to write in the vectors.

[tex]\omega' = \gamma(1-g(v,w))\omega[/tex]
 
  • #25
Recap

Ok, so far we have defined the relative velocity of one observer with respect to a second observer at two different events of the same light ray. That is, if two observers' worldlines intersect a light ray then we may use the method of parallel transport to "move" the 4-velocity vector to the other and compare. As a result we obtain another vector which we now call the relative velocity.

We can also derive an expression for the Doppler effect:

[tex]\omega' = \gamma(1 - g(v,w))\omega[/tex]

where [itex]\omega[/itex] and [itex]\omega'[/itex] are the frequency 4-vectors of the light ray as observed by their respective observers O and O'. And v is the relative velocity of O' w.r.t. O and w is the relative velocity of O w.r.t. O'. And finally, [itex]\gamma[/itex] is the gamma factor corresponding to the velocity [itex]|v|[/itex].

The question is can we interweave this definition of relative velocity and Doppler with the idea of Gravitational Redshift!?...
 
  • #26
Gravitational Redshift Intro

This is where I will be incorporating some of the stuff Pervect and I were talking about in the thread entitled "Relative Velocity" some weeks ago. We came to conclusion that the simplest setting in which we could define such as thing as grav redshift would be for Static Observers. However, as Pervect and I discussed, Stationary Observers may work in certain symmetrical space-times. But for now, as Wald does, we will work with Static observers.

Now I can make some observations at this stage without having to introduce a metric just yet.

Given two observers O and O' and a light ray joining them, call it [itex]\lambda[/itex] then the frequency...

remember I define the frequency of the light ray as the 4-vector, [itex]\omega[/itex], defined on [itex]\lambda[/itex] by a future-pointing lightlike vector field F defined on [itex]\lambda[/itex], tangent to [itex]\lambda[/itex], and [itex]\nabla_F F = 0[/itex]

...of the joining light ray could be different for one observer as it is for the other. If it is then this is due to the Gravitational Redshift. Remember, the observers are stationary, so they are in effect, at rest w.r.t to one another. There is no Doppler effect, but if the frequency were to change, then this is due to Grav. Redshift.
 
  • #27
Static Schwarzschild Spacetime and Gravitational Redshift

Given two Static Observers O and O' then we can easily define a formula for the change between emitted and received frequencies of the light ray between the two observers - that is, as Wald says, a formula for the Gravitational Redshift.

Wald explains: Consider two static observers O and O' (i.e. observers whose 4-velocity is tangent to the static Killing field [itex]X^a[/itex]) whose 4-velocity vectors are [itex]u^a[/itex] and [itex]u'{}^a[/itex]. Then suppose that O emits a light ray at event p which is received by O' at event q.

Wald then says that this light ray travels along a null geodesic, which is fine, but then he assumes that the frequency of emission (that is, the frequency of the light ray as observed by O, which we call [itex]\omega[/itex], as so too does Wald) is given by

[tex]\omega = -(k_au^a)|_p[/tex]

where [itex]k_a[/itex] is the tangent 4-vector of the light ray [itex]\lambda[/itex] at p.

If we assume that the frequency of emission is this (which I can't quite grasp how he defines this just yet, maybe someone could help with that - Wald says that this is simply the negative rate of change of phase of the wave [itex]\Psi_a = C_a \exp(iS)[/itex], i.e. [itex]\omega = - u^a \partial_a S = -k_au^a[/itex] as given. Yet I don't understand this fully) then it is a simple matter of algebra and some facts to show that

[tex]\frac{\omega}{\omega'} = \frac{\sqrt{-X^bX_b}|_q}{\sqrt{-X^bX_b}|_p} = \frac{\sqrt{1-2M/r_2}}{\sqrt{1-2M/r_1}}[/tex]

This derivation is done in Wald Appendix C.3.1 but it is fairly straightforward, assuming [itex]X^bX_b = g_{tt} = -(1-2M/r)[/itex] for the Schwarzschild space-time.

Outline of how one would obtain the Gravitational Reshift Formula:

Given a Lorentz manifold with the Schwarzschild metric we have a Killing vector field (since time translations are isometries here). Then

[tex]u^a = \frac{X^a}{\sqrt{1-2M/r}}[/tex]

that is, a static observer. Obviously the Schwarzschild coordinates are adapted to this observer with 4-velocity vector u and they becomes singular at r =3m where the photon sphere occurs. So u is the 4-velocity of an observer who stands in a higher gravitational potential (a greater r). Consider a light signal that goes from the bottom of say, a long tower on the Earth, to the top where the worldline of our observer intersects the light ray traveling upward. The observer at the top sees a redshift (given that he knows what the frequency was when it was emitted). Since we have a Killing vector field [itex]X^a[/itex], we have effectively

[tex]\frac{\omega}{\omega'} = \frac{\sqrt{-X^bX_b}|_q}{\sqrt{-X^bX_b}|_p} = \frac{\sqrt{1-2M/r_2}}{\sqrt{1-2M/r_1}}[/tex]
 
Last edited:
  • #28
However, in order to derive an explicit expression for Gravitational Redshift in Schwarzschild Space-time we had to have begun with no ordinary space-time but one with a Killing vector field. Luckily, if one possesses a Stationary Space-time then one possesses a timelike Killing vector field. Furthermore, if that space-time is Static (as we have hypothesized) then not only does our space-time come equipped with its own timelike Killing vector field, it also possesses a hypersurface, call it [itex]\Sigma[/itex], orthogonal to the orbits of the isometries. This is what we need, this orthogonal hypersurface is the key, this is why we needed static-ness.

Of course, everything we have explained so far still works in a static space-time: [1] The idea of the relative velocity of two observers coincident at a single event in space-time, [2] the idea of relative velocity of two observers connected by a light ray, [3] the Doppler effect still works, very well!, and of course [4] the formula for Gravitational Redshift still works too.

However, what we also get, which is the final piece in the puzzle, is a formula for the relative velocity of two galaxies! Yes, that's right, two galaxies. Well, at least I think it does, for a R-W universe at least. But more on this later.
 
Last edited:
  • #29
The big question is this: Since we have essentially defined the formula for Gravitational Redshift using the Doppler effect, that is, to get

[tex]\omega = \frac{A}{B}\omega'[/tex]

in the Schwarzschild space-time, then is the Gravitational Redshift a particular case of the generalized Doppler effect?
 
  • #30
This may be useful

American Journal of Physics -- November 2006 -- Volume 74, Issue 11, pp. 1017-1024
Redshifts and Killing vectors
Alex Harvey, Engelbert Schucking, and Eugene J. Surowitz
http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000074000011001017000001&idtype=cvips&gifs=Yes [Broken]
"Current approaches to physics stress the importance of conservation laws due to spacetime and internal symmetries. In special and general relativity the generators of these symmetries are known as Killing vectors. We use them for the rigorous determination of gravitational and cosmological redshifts."
 
Last edited by a moderator:
  • #31
Thanks Robphy I'll have a look at that one next time I'm at the uni library.
 

1. What is the concept of relative velocity in general relativity?

Relative velocity in general relativity refers to the motion of an object relative to another object in a curved spacetime. It takes into account the effects of gravity and the curvature of spacetime on the velocity of an object.

2. How is relative velocity defined in general relativity?

In general relativity, relative velocity is defined as the rate of change of the spatial distance between two objects in a curved spacetime. This is different from the classical definition of relative velocity in Newtonian physics, which only takes into account the objects' velocities in a flat spacetime.

3. Can relative velocity be measured in general relativity?

Yes, relative velocity can be measured in general relativity using various techniques such as radar ranging, laser ranging, and Doppler shift measurements. However, the measurements may be affected by the curvature of spacetime and the gravitational fields of the objects involved.

4. How does the concept of relative velocity in general relativity differ from special relativity?

In special relativity, relative velocity is defined as the difference in velocities of two objects measured in an inertial frame of reference. In general relativity, however, the concept of an inertial frame of reference doesn't exist, and relative velocity is defined in terms of the curvature of spacetime.

5. Can relative velocity in general relativity be greater than the speed of light?

No, according to the principles of general relativity, the speed of light is the maximum speed at which any object can travel in spacetime. Therefore, relative velocity in general relativity cannot exceed the speed of light.

Similar threads

  • Special and General Relativity
Replies
17
Views
438
  • Special and General Relativity
Replies
3
Views
821
  • Special and General Relativity
Replies
11
Views
938
  • Special and General Relativity
3
Replies
76
Views
4K
  • Special and General Relativity
Replies
8
Views
1K
  • Special and General Relativity
Replies
20
Views
746
  • Special and General Relativity
Replies
13
Views
1K
  • Special and General Relativity
Replies
14
Views
1K
  • Special and General Relativity
2
Replies
46
Views
2K
  • Special and General Relativity
Replies
5
Views
771
Back
Top