Affine parametrization for null geodesic?

  • #51
WannabeNewton
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Okay, I saw a definition on the internet, that two metrics [itex]g_{\mu \nu}[/itex] and [itex]\tilde{g}_{\mu \nu}[/itex] are "conformally equivalent" if you can get from one to the other by a scaling factor (in general, position-dependent). In other words:

[itex]\tilde{g}_{\mu \nu} = e^\phi g_{\mu \nu}[/itex]

But what's the significance of that? Are the mass/energy distributions that give rise to those two metrics related in some simple way?
You can also express it as [itex]\tilde{g}_{\mu \nu} = \Omega^2 g_{\mu \nu}[/itex] for some smooth scalar field ##\Omega##, which is how you will usually see it. It's the same thing in the end anyways.

Geometrically, conformally equivalent metrics preserve angles but not lengths. I have already proven one significance of them, which is that conformally equivalent metrics agree on null geodesics (post #45). The converse is true as well: two metrics which agree on null geodesics are conformally equivalent. Also, Maxwell's equations in curved space-time (as well as certain other physical equations in curved space-time) remain invariant under conformal transformations.

The most important application of conformal transformations, in the context of general relativity, is probably in the definition of asymptotic flatness which involves a conformal isometry of a certain space-time onto an open subset of another space-time satisfying certain conditions (for example a conformal isometry of Minkowski space-time onto an open subset of the Einstein static universe).
 
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  • #52
WannabeNewton
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A light ray in the Schwarzschild geometry expressed in isotropic coordinates satisfies ##0 = -(\frac{1 - M/2\tilde{r}}{1 + M/2\tilde{r}})^{2}dt^{2} + (1 + M/2\tilde{r})^{4}(d\tilde{r}^{2} + d\tilde{\Omega}^{2})##. When put into the form (4.75) in page 159 of Padmanabhan's text we have ##0 = dt^{2} + \frac{(1 + M/2\tilde{r})^{6}}{-(1 - M/2\tilde{r})^2}(d\tilde{r}^{2} + d\tilde{\Omega}^{2})##. Hence the light ray satisfies the geodesic equation with an affine parameter ##t## (the coordinate time) in a three-dimensional space with metric ##\frac{(1 + M/2\tilde{r})^{6}}{(1 - M/2\tilde{r})^2} = \Omega^{2}(1 + M/2\tilde{r})^{4}## which is obviously not the spatial metric on the ##t = \text{const.}## slices in isotropic coordinates even though it is conformally equivalent to it; this is simply a specific example of (4.78) and the passage directly below it in page 159 of said text. I don't know how much clearer I can make it.
 
  • #53
WannabeNewton
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What? Bill was saying the exact same thing I've been saying over and over. It's trivial to extrapolate from what I said and connect it with what Bill said. We take the metric ##g_{\mu\nu} = \text{diag}(-(\frac{1 - M/2\tilde{r}}{1 + M/2\tilde{r}})^{2},(1 + M/2\tilde{r})^{4})## and go to the conformally equivalent space-time with metric ##\tilde{g}_{\mu\nu} = (\frac{1 + M/2\tilde{r}}{1 - M/2\tilde{r}})^{2}g_{\mu\nu} = \text{diag}(-1,\frac{(1 + M/2\tilde{r})^{6}}{(1 - M/2\tilde{r})^{2}})##. Then we evaluate the line element for the specific case of a light ray and end up at (4.78) along with the conclusion following it. It's literally a matter of elementary algebra.
 
  • #54
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What? Bill was saying the exact same thing I've been saying over and over. It's trivial to extrapolate from what I said and connect it with what Bill said. We take the metric ##g_{\mu\nu} = \text{diag}(-(\frac{1 - M/2\tilde{r}}{1 + M/2\tilde{r}})^{2},(1 + M/2\tilde{r})^{4})## and go to the conformally equivalent space-time with metric ##\tilde{g}_{\mu\nu} = (\frac{1 + M/2\tilde{r}}{1 - M/2\tilde{r}})^{2}g_{\mu\nu} = \text{diag}(-1,\frac{(1 + M/2\tilde{r})^{6}}{(1 - M/2\tilde{r})^{2}})##. Then we evaluate the line element for the specific case of a light ray and end up at (4.78) along with the conclusion following it. It's literally a matter of elementary algebra.
Sorry I just saw my silly mistake, deleted my rants too.
 
  • #55
pervect
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Hmm. But what is the significance of the parameter s, since it's not proper time? I guess in flat spacetime it could very well be coordinate time.
If you imagine light as a plane wave following a null geodesic, it may be useful to think of the parameter s as being proportional to the optical path length along said geodesic.

(You can find cases where light doesn't follow a null geodesic exactly, or if it does, it doesn't do it as a plane wave. Then you wouldn't want to use the visualization I suggest).

Visualize a plane electromagnetic wave propagating along the null geodesic. Then the planes comprising the plane wave mark out "equal intervals" along the lightlike geodesic. I.e. from the wiki on plane waves, http://en.wikipedia.org/wiki/Plane_wave

http://upload.wikimedia.org/wikipedia/commons/2/20/Plane_wave_wavefronts_3D.svg

The frequency of the wave sets the spacing between the planes. Differently moving observers will assign different spacings between the plane waves (and a different frequency to the monochromatic plane wave), but they'll all agree that the planes are uniformly spaced.

The "uniform spacing" has physical signficance in that all observers agree on the uniformity (with the previously mentioned caveats). Physically, observers can detect non-uniformity in the spacing by physical and mathematical means (for the later, a failure to satisfy the geodesic equations - physically, one might use an interferometer, though the space-time geoemtry would have to be nearly static for that approach to work).
 
  • #56
Hello. I found a tetrad formalism of relativistic point particles, which might be useful for your question. Partial information about this formalism can be found in the textbook by Polchinski, "String Theory."
This formalism is useful in that it is valid also for massless particles, and you may find the answer in eq.(1.2.7) in Volume 1 of the textbook. There, the tetrad is given by squared world velocity divided by squared mass, both of which are zero in massless case.
Maybe the answer is: the affine parameter for massless particles is the massless limit of the proper time of massive point particles divided by the mass.

I am sorry if my interpretation is incorrect. I could not find a reference in which your question is explicitly stated. Yeah, there are tremendously many questions you naturally have but rarely find the answer in references. It is very frustrating. I appreciate that you posted this question in this open forum. Thank you.
 
  • #57
WannabeNewton
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Maybe the answer is: the affine parameter for massless particles is the massless limit of the proper time of massive point particles divided by the mass.
But the proper time of any massive particle depends only on the space-time geometry and the 4-velocity of the particle,
thus for a freely falling particle the proper time is determined entirely by the space-time geometry alone since the space-time geometry also determines the time-like geodesics. How then would the ratio of the proper time to the mass of the particle converge to a finite value in the limit as the mass goes to zero?

Rather, an affine parameter ##\lambda## for a null geodesic is simply defined as one such that ##\frac{dx^{\mu}}{d\lambda} = p^{\mu}## holds.
 
  • #58
Hello, Mr/Ms WannabeNewton. Thank you for your mention.

As you said, it is completely right to define an affine parameter as one such that the geodesic equation in that form holds.

However, it seems to me that Mr/Ms stevendaryl, who is the original questioner, wants to get physical characteristics of null affine parameter. In massive case, an affine parameter can be defined similarly so that geodesic equation in that form holds, but at the same time it has physical characteristics that it is the proper time of the particle. So, it seems natural that one seeks such characteristics also in massless case. This is why I emphasized the connection between massive and massless cases.

Next, I want to explain the term "massless limit" I used. As you said, the geodesic equation does not depend on the mass; it is the property of the spacetime itself. However, practically there are many non-gravitational forces in the world, and to such forces the response of the particle depends on its mass; the smaller the mass is, the bigger the acceleration is. As a result, the speed of the particle becomes near the light velocity. I used the term "massless limit" in this practical sense. This is the phenomenological explanation, and the following is more formal explanation. Yeah, I should have clarified what is kept fixed when we take the massless limit. Your way of thinking is to take the limit with the world line fixed. I took another way of limitation with "the proper time of massive point particle divided by the mass" kept fixed, not with the world line fixed. This corresponds to changing the initial speed of the particle bigger as we change the mass smaller. Although the geodesic equation is independent of the mass, the initial value of the particle can depend on the mass if we like to do.

At the same time, I agree that it is sometimes dangerous or even meaningless to seek analogue between massive and massless cases. The formulation and the interpretation to it for massive particles is valid only for massive particles themselves. Physically, it is of great difference whether or not the rest frame of the particle exists. After all, what is certain is that a massless particle draws some world line. We simply parametrise it so that the geodesic equation of motion holds in a definite form (the form of eom can be determined essentially by its transformation property), and call the parameter an affine parameter.

I am still learning the subject, and my understanding gets improved through this conversation. Thanks.
 
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