Affine parametrization for null geodesic?

[TrickyDicky]

You can parametrize any curve, null or not, geodesic or not, with any parameter you choose. But for geodesics you only get the simple geodesic equation as it is usually written if you use the proper time/proper distance/affine parameter. Use of any other parametrization will produce an extra term.

Exactly.
Null geodesics have the particularity that the equation with the extra term will produce also a geodesic, as all null paths are geodesics. This is not the case for timelike or spacelike geodesics, they need the nice simple equation.

Affine parameters remain unchanged under a coordinate transformation. However under a conformal transformation, although the null geodesics themselves are preserved, the affine parameters are not.

That's what this book by Padmanabhan is doing. He's replacing Schwarzschild by a conformally related spacetime,

ds² = (1 - 2m/r) ds'²

ds'² = dt² - (1 - 2m/r)^-2 dr² - r²/(1 - 2m/r) d²Ω

In this spacetime since g₀₀ = 1, the energy integral is simply E = dt/ds, and t is indeed an affine parameter. But that's in a different spacetime, not in Schwarzschild.

Not exactly.
It's not a different spacetime. It is the spatial hypersurface(wich is conformally flat) of the exterior Schwarzschild spacetime in conformally flat explicit form thru the coordinate change I gave and then making use of the fact that for the Schwarzschild null geodesic $(1-2m/r)dt^2=dr^2/(1-2m/r)-r^2dΩ^2$

 

[WannabeNewton]

Exactly.
Null geodesics have the particularity that the equation with the extra term will produce also a geodesic, as all null paths are geodesics. This is not the case for timelike or spacelike geodesics, they need the nice simple equation.

That's not what Bill is saying. A geodesic is a curve that satisfies the equation $\xi^a \nabla_a \xi^b = \alpha \xi^b$ for some scalar field $\alpha$. This is true for any geodesic whatsoever. However one can show easily that any such curve can be reparametrized so as to satisfy $\xi^a \nabla_a \xi^b = 0$. This is called an affine parametrization for geodesics.

Back to the original question: as I have tried to state, any search for a general physical meaning, in arbitrary curved space-times, for arbitrary affine parameter will be futile, and for obvious reasons from elementary differential geometry. Arbitrary curve parameters are not meant to be physical and in general have no physical meaning whatsoever. They are bookkeeping parameters that is all. Some special cases allow for physical meanings, such as the arc-length parametrization (which is where proper time comes from) but for null geodesics such a parametrization does not exist.

 

[Bill_K]

Sorry, TrickyDicky, I stand by my remarks.

 

[WannabeNewton]

It's not a different spacetime. It is the spatial hypersurface(wich is conformally flat) of the exterior Schwarzschild spacetime in conformally flat explicit form thru the coordinate change I gave and then making use of the fact that for the Schwarzschild null geodesic $(1-2m/r)dt^2=dr^2/(1-2m/r)-r^2dΩ^2$

No it's a conformally related space-time.

 

[TrickyDicky]

That's not what Bill is saying.

I know that's what I'm saying.

A geodesic is a curve that satisfies the equation $\xi^a \nabla_a \xi^b = \alpha \xi^b$ for some scalar field $\alpha$. This is true for any geodesic whatsoever. However one can show easily that any such curve can be reparametrized so as to satisfy $\xi^a \nabla_a \xi^b = 0$. This is called an affine parametrization for geodesics.

This is understood, I was adding to this my reason to stress the case of null geodesics.

 

[TrickyDicky]

Sorry, TrickyDicky, I stand by my remarks.

No it's a conformally related space-time.

ugh! You guys are missing my point.
The spacetime is a conformally related spacetime, right. But for the problem at hand(the possibility of using coordinate time as affine parameter for null geodesics in a specific spacetime) we can concentrate on the space hypersurface for physical and empirically confirmed results(Cassini probe)-this is physicsforums right?, we only need the spatial part of the spacetime and the spatial part is the same in both spacetimes(in the region where it is possible to do such foliation).
WBN I believe you have Padmanabhan text, check exercise 4.8
Anyway it is silly to argue about this beyond this point, so I won't insist.

 

[PAllen]

as all null paths are geodesics. This is not the case for timelike or spacelike geodesics, they need the nice simple equation.

Just as a side note, this is not true. Consider the null path (but not geodesic) in Minkowski space:

x = r sin( (c/r) t)
y = r cos((c/r) t)

Also, you can find any geodesic using either the simple equation or the one with an extra term. There is nothing different about null versus timelike or spacelike. Solving the simple form of the equation gives you a definition of the curve in terms of an affine parameter. Any description in terms of an arbitrary parameter will be a solution of the more general equations with an extra term.

 

[TrickyDicky]

Just as a side note, this is not true. Consider the null path (but not geodesic) in Minkowski space:

x = r sin( (c/r) t)
y = r cos((c/r) t)

Lightlike paths are always geodesic.

Also, you can find any geodesic using either the simple equation or the one with an extra term. There is nothing different about null versus timelike or spacelike. Solving the simple form of the equation gives you a definition of the curve in terms of an affine parameter. Any description in terms of an arbitrary parameter will be a solution of the more general equations with an extra term.

If you had followed the thread you'd notice that we were dealing with the case when the equation is modified by a conformal transformation, it is in this case when only the null geodesics conserve the geodesic condition(but with a non-affine parameter).

 

[WannabeNewton]

WBN I believe you have Padmanabhan text, check exercise 4.8

I'll try to illustrate his point more clearly and make it more lucid for you. Consider a static space-time $(M,g)$. We know that the existence of a time-like killing field $\xi$ such that $\xi^{\flat} \wedge d\xi^{\flat} = 0$ allows us to write the space-time manifold (locally) as $M = \mathbb{R}\times \Sigma$ and the metric as $g = -\gamma^2 dt^2 + h$ where $\Sigma$ is a space-like foliation and $h$ is the spatial metric on $\Sigma$; note in these coordinates that $\xi = \partial_t$.

Now for light rays in the geometrical optics approximation with tangent $k$ we have $g(k,k) = 0 = -\gamma^2 (k^t)^2 + h_{ij}k^{i}k^{j}$, which if you use that $k^{\mu} = \frac{dx^{\mu}}{d\lambda}$, is the same as $0 = dt^2 + \frac{h_{ij}}{-\gamma^2}dx^i dx^j$ which is what Padmanabhan has. Anyways, if I now let $\tilde{h}_{ij} = \frac{h_{ij}}{\gamma^2} = \Omega^2 h_{ij}$ then following his calculation on page 159 we end up with $\tilde{h}_{ij}\frac{\mathrm{d} ^2 x^{j}}{\mathrm{d} t^2} + \frac{1}{2}(\partial_{k}\tilde{h}_{ij} + \partial_{j}\tilde{h}_{ik} - \partial_{i}\tilde{h}_{jk})\frac{\mathrm{d} x^{j}}{\mathrm{d} t}\frac{\mathrm{d} x^{k}}{\mathrm{d} t} = 0$.

This means that light rays in this geometry, which travel on null geodesics of $(M,g)$, travel on curves of extremal coordinate time $t$ in $(\Sigma,\tilde{h})$ hence $t$ is an affine parameter for said curves in the conformally related space $(\Sigma,\tilde{h})$, not the space $(\Sigma,h)$. This is what Padmanabhan is saying; he is definitely not saying that $t$ is an affine parameter for the null geodesics of $(M,g)$, which is of course false in general.

 

[PAllen]

Lightlike paths are always geodesic.

False, and I gave you an example of a non-geodesic light like path.

[Edit: and if you can't follow my simple example, here is a whole paper on features of null curves that are not geodesics:

http://arxiv.org/abs/gr-qc/0005096

]

 

[stevendaryl]

Lightlike paths are always geodesic.

No, they're not. A lightlike geodesic is the path taken by a pulse of light that has no interactions other than gravity. If the light pulse bounces off a mirror, the resulting path is still lightlike, but isn't a geodesic.

What's different about lightlike paths is that the definition of a geodesic as an extremum of proper time breaks down (since any lightlike path has zero proper time, but only some of them are geodesics.)

 

[TrickyDicky]

False, and I gave you an example of a non-geodesic light like path.

[Edit: and if you can't follow my simple example, here is a whole paper on features of null curves that are not geodesics:

http://arxiv.org/abs/gr-qc/0005096

]

We are talking about physics here, so you know very well what my sentence meant by the context, yes there are null curves that are not geodesic, and lightlike velocity curves that are not geodesic but when I said null paths I was clearly referring to the paths of light rays, unperturbed photons, which are geodesics. Now you may continue nitpicking out of context.
Good catch by the way.

 

[TrickyDicky]

I'll try to illustrate his point more clearly and make it more lucid for you. Consider a static space-time $(M,g)$. We know that the existence of a time-like killing field $\xi$ such that $\xi^{\flat} \wedge d\xi^{\flat} = 0$ allows us to write the space-time manifold (locally) as $M = \mathbb{R}\times \Sigma$ and the metric as $g = -\gamma^2 dt^2 + h$ where $\Sigma$ is a space-like foliation and $h$ is the spatial metric on $\Sigma$; note in these coordinates that $\xi = \partial_t$.

Now for light rays in the geometrical optics approximation with tangent $k$ we have $g(k,k) = 0 = -\gamma^2 (k^t)^2 + h_{ij}k^{i}k^{j}$, which if you use that $k^{\mu} = \frac{dx^{\mu}}{d\lambda}$, is the same as $0 = dt^2 + \frac{h_{ij}}{-\gamma^2}dx^i dx^j$ which is what Padmanabhan has. Anyways, if I now let $\tilde{h}_{ij} = \frac{h_{ij}}{\gamma^2} = \Omega^2 h_{ij}$ then following his calculation on page 159 we end up with $\tilde{h}_{ij}\frac{\mathrm{d} ^2 x^{j}}{\mathrm{d} t^2} + \frac{1}{2}(\partial_{k}\tilde{h}_{ij} + \partial_{j}\tilde{h}_{ik} - \partial_{i}\tilde{h}_{jk})\frac{\mathrm{d} x^{j}}{\mathrm{d} t}\frac{\mathrm{d} x^{k}}{\mathrm{d} t} = 0$.

This means that light rays in this geometry, which travel on null geodesics of $(M,g)$, travel on curves of extremal coordinate time $t$ in $(\Sigma,\tilde{h})$ hence $t$ is an affine parameter for said curves in the conformally related space $(\Sigma,\tilde{h})$, not the space $(\Sigma,h)$. This is what Padmanabhan is saying; he is definitely not saying that $t$ is an affine parameter for the null geodesics of $(M,g)$, which is of course false in general.

Just one question,bro
$(\Sigma,\tilde{h})$ and $(\Sigma,h)$ are according to you different metrics or a metric expressed in different form? In other words do you agree that the spatial hypersurface of the static region of the Schwarzschild spacetime is conformally flat?

 

[stevendaryl]

We are talking about physics here, so you know very well what my sentence meant by the context, yes there are null curves that are not geodesic, and lightlike velocity curves that are not geodesic but when I said null paths I was clearly referring to the paths of light rays, unperturbed photons, which are geodesics. Now you may continue nitpicking out of context.
Good catch by the way.

So when you said "Lightlike paths are always geodesic" you meant "Lightlike geodesics are always geodesics".

 

[WannabeNewton]

$(\Sigma,\tilde{h})$ and $(\Sigma,h)$ are according to you different metrics or a metric expressed in different form?

They are different but conformally equivalent metrics.

In other words do you agree that the spatial hypersurface of the static region of the Schwarzschild spacetime is conformally flat?

This is actually not a well-posed question. See: http://relativity.livingreviews.org/open?pubNo=lrr-2000-5&page=articlesu6.html

Also it might be worth noting that on the space-time level (i.e. for Lorentzian 4-manifolds), if two space-time metrics are conformally equivalent then they agree on null geodesics (the converse of this is true as well): let $\tilde{g}_{\mu\nu} = \Omega^2 g_{\mu\nu}$ and let $\gamma$ be a null curve with respect to $g_{\mu\nu}$ (in which case it is also null with respect to $\tilde{g}_{\mu\nu}$) with tangent vector field $k^{\mu}$. Finally, denote by $\tilde{\nabla}_{\mu}$ the derivative operator associated with $\tilde{g}_{\mu\nu}$. Then there exists a smooth tensor field $C^{\gamma}_{\mu\nu}$ such that $k^{\mu}\tilde{\nabla}_{\mu}k^{\gamma} = k^{\mu}\nabla_{\mu}k^{\gamma} - C^{\gamma}_{\mu\nu}k^{\mu}k^{\nu}$.

It is easy to show that for conformally related metrics, $C^{\gamma}_{\mu\nu} = -\Omega^{-2}[\delta^{\gamma}{}{}_{(\mu}\nabla_{\nu)}\Omega^{2} - \frac{1}{2}g_{\mu\nu}g^{\gamma \beta}\nabla_{\beta}\Omega^{2}]$. Upon plugging this back in, we find that $k^{\mu}\tilde{\nabla}_{\mu}k^{\gamma} = k^{\mu}\nabla_{\mu}k^{\gamma} + \Omega^{-2}k^{\gamma}k^{\mu}\nabla_{\mu}\Omega^{2}$. Therefore $\gamma$ is a null geodesic with respect to $\nabla_{\mu}$ if and only if it is a null geodesic with respect to $\tilde{\nabla}_{\mu}$.

 

[TrickyDicky]

They are different but conformally equivalent metrics. This is actually not a well-posed question. See: http://relativity.livingreviews.org/open?pubNo=lrr-2000-5&page=articlesu6.html

Right.
Thanks for the very good livingreviews reference, the thing is that I don't know if it is an unconscious giveaway or you are simply pulling my leg.

Both (59) and (60) are presented as intervals of Schwarzschild spacetime, agree? Both are the same spacetime their diference is a coordinate change.
(60) is the one used in exercise 4.8 b of the Padmanabhan text as the metric that has a refractive index, it is the one where coordinate time is an affine parameter along the light path: dt =[f/√|g00|]dl and δ ∫dt =0

 

[TrickyDicky]

he is definitely not saying that $t$ is an affine parameter for the null geodesics of $(M,g)$, which is of course false in general.

Just a correction, my claim(and Padmanabhan) is not that it is true in general for any $(M,g)$, it is only true for spherically symmetric and static $(M,g)$ that is a solution of the EFE without cosmological constant when using the adequate coordinates (like the radial coordinate in isotropic coordinates form of Schwarzschild metric shown in the aforementioned exercise 4.8) and Schwarzschild spacetime $(M,g)$ is the obvious example.
Let's try not to misconstrue each other's claims.

 

[stevendaryl]

They are different but conformally equivalent metrics.

Pardon my ignorance, but what does "conformally equivalent" mean?

 

[stevendaryl]

Pardon my ignorance, but what does "conformally equivalent" mean?

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

\tilde{g}_{\mu \nu} = e^\phi g_{\mu \nu}

But what's the significance of that? Are the mass/energy distributions that give rise to those two metrics related in some simple way?

 

[WannabeNewton]

Just a correction, my claim(and Padmanabhan) is not that it is true in general for any $(M,g)$, it is only true for spherically symmetric and static $(M,g)$

And I'm saying that your claim is incorrect as others have already stated. It is not hard to see why it is incorrect. You are misunderstanding Padmanabhan's statements on page 159 and I have already explained to you why. Instead of jumping the gun just take a half an hour or an hour to actually go through the calculations and the exercise and compare with what I said in post #39.

 

[WannabeNewton]

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

\tilde{g}_{\mu \nu} = e^\phi g_{\mu \nu}

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 \tilde{g}_{\mu \nu} = \Omega^2 g_{\mu \nu} 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).

 

[WannabeNewton]

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.

 

[WannabeNewton]

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.

 

[TrickyDicky]

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.

 

[pervect]

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).

 

[Toru Kikuchi]

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.

 

[WannabeNewton]

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.

 

[Toru Kikuchi]

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.