Affine parametrization for null geodesic?

  • #26
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Actually, that shows that coordinate time is NOT the affine parameter. Changing from the affine parameter to coordinate time as an independent (not affine) parameter allows the treatment of static gravity as an optical medium. It doesn't change that the affine parameter (for all types of geodesics) is unique up to linear function (affine parameter means the simple - parallel transport - form of geodesic equation must be satisfied using the parameter). Thus, in SC coordinates r is an affine parameter and t is not an affine parameter.
Clearly in SC coordinates r is an affine parameter, not t. I never made a restriction to specific coordinates. Obviously a change of coordinates must be introduced for the spatial part of the metric that puts it in conformally flat form(spatial part that in the case of the statiic Schwarzschild metric is made equal to the time part in the case of null geodesics), wich is perfectly valid, now you are claiming that particular coordinates are fundamental?
In general such conformal tranformations change the form of the geodesic equation and one doesn't obtain a geodesic, but the exception are null geodesics, but then the parameter is no longer affine, again static spacetimes are an exception to that and one can recover the affine parameter, in this case the coordinate t, after the transformation to a conformally flat spatial part.
Quoting once more from Padmanabhan page 160:
"....This is the same as a geodesic equation with an affine parameter t in a three-dimensional space with metric Hαβ. It follows that the null geodesics in a static spacetime can be obtained from the extremum principle for coordinate time δ ∫dt =0 , (4.79) "
 
  • #27
stevendaryl
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Clearly in SC coordinates r is an affine parameter, not t. I never made a restriction to specific coordinates. Obviously a change of coordinates must be introduced for the spatial part of the metric that puts it in conformally flat form(spatial part that in the case of the statiic Schwarzschild metric is made equal to the time part in the case of null geodesics), wich is perfectly valid, now you are claiming that particular coordinates are fundamental
Nobody said that particular coordinates are fundamental. The claim is that for a specific geodesic, there are particular parametrizations that are fundamental, namely the affine ones. And an affine parametrization for an outgoing/incoming lightlike geodesic in Schwarzschild spacetime happens to be the coordinate [itex]r[/itex] of the Schwarzschild coordinates.

The fact that you're using [itex]r[/itex] as the parameter is unrelated to whether you are using Schwarzschild coordinates. It's an affine parameter for any coordinates whatsoever.
 
  • #28
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Nobody said that particular coordinates are fundamental.
I know but you and PAllen seem to be implying it wrt the Schwarzschild coordinates. See below.

The claim is that for a specific geodesic, there are particular parametrizations that are fundamental, namely the affine ones.
But I thought you understood that this is not so for null geodesics that are the subject of your OP.
Affine parametrization is not fundamental at all for null geodesics, you can parametrize them with any parameter, affine or not affine. That is a key difference wrt timelike and spacelike geodesics.
It is a particular consequence of the fact that Schwarzschild spacetime is static that one can obtain coordinate time as an affine parameter for the null geodesics. Otherwise it would not be affine.

And an affine parametrization for an outgoing/incoming lightlike geodesic in Schwarzschild spacetime happens to be the coordinate [itex]r[/itex] of the Schwarzschild coordinates.
I'm not disputing this at all, did you read my phrase "Clearly in SC coordinates r is an affine parameter, not t" in the previous post? The reason as explained by Bill K is that in these coordinates to express r as a function of t can't be done in closed form, you get an ugly infinite expression.
The fact that you're using [itex]r[/itex] as the parameter is unrelated to whether you are using Schwarzschild coordinates. It's an affine parameter for any coordinates whatsoever.
In general that would be true, but note that we are using here a parameter that hapens to be a coordinate, so for instance in coordinates that didn't use the r coordinate explicitly or in wich r was not an independent variable as it is the special case with null geodesics of Schwarzschild spacetime you wouldn't be able to use the radial coordinate as affine parameter in those particular coordinates or in the case of timelike geodesics of that space.
 
  • #29
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This is the coordinate change that allows to use coordinate time as affine parameter for null geodesics: r=ρ(1+2m/2ρ)^2

The caveat as pointed out by Bill is that this change is not valid for r<2m, only r is a valid affine parameter in that case.

But then again there the spacetime is not static so it is kind of obvious.
 
  • #30
Bill_K
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Affine parametrization is not fundamental at all for null geodesics, you can parametrize them with any parameter, affine or not affine. That is a key difference wrt timelike and spacelike geodesics.
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.

It is a particular consequence of the fact that Schwarzschild spacetime is static that one can obtain coordinate time as an affine parameter for the null geodesics. Otherwise it would not be affine.
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.
 
  • #31
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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. :biggrin:
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. :frown:
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##
 
  • #32
WannabeNewton
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Exactly. :biggrin:
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.
 
  • #33
Bill_K
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Sorry, TrickyDicky, I stand by my remarks.
 
  • #34
WannabeNewton
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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.
 
  • #35
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That's not what Bill is saying.
I know that's what I'm saying.:tongue:

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.
 
  • #36
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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.
 
  • #37
PAllen
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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.
 
  • #38
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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).
 
  • #39
WannabeNewton
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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.
 
  • #40
PAllen
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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

]
 
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  • #41
stevendaryl
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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.)
 
  • #42
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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.
 
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  • #43
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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:wink:
##(\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?
 
  • #44
stevendaryl
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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".
 
  • #45
WannabeNewton
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##(\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&amp;page=articlesu6.html [Broken]

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}##.
 
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  • #46
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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&amp;page=articlesu6.html [Broken]
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.:rolleyes:

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
 
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  • #47
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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.
 
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  • #48
stevendaryl
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They are different but conformally equivalent metrics.
Pardon my ignorance, but what does "conformally equivalent" mean?
 
  • #49
stevendaryl
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Pardon my ignorance, but what does "conformally equivalent" mean?
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?
 
  • #50
WannabeNewton
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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.
 

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