Differentiate w/ Respect to Affine Parameter

[Haorong Wu]

TL;DR Summary

Differentiation with respect to an affine parameter with an orthonormal tetrad.

Hi, there. I am doing differentiation with respect to an affine parameter $s$, I am not sure whether my idea is right or wrong.

Let $C$ be a geodesic for light and the path length $s$ on it be the affine parameter. Now I need to calculate $\frac {\partial f}{\partial s}$, with $f$ being some function. Meanwhile, I have constructed an orthonormal tetrad $\{\mathbf e_\mu \}$ along the geodesic $C$ where $\mathbf e_3$ is along the direction of the wave vector $\mathbf k$.

I think since at every point, I have a tetrad and $\mathbf e_3$ is tangent to the geodesic, I could just calculate $\frac {\partial f}{\partial s}=\frac {\partial f}{\partial x^3}$. Is this correct?

Thanks ahead.

 

[PeterDonis]

Let $C$ be a geodesic for light and the path length $s$ on it be the affine parameter.

You can't use path length along a null geodesic as an affine parameter; the path length is zero at all events on the geodesic.

I have constructed an orthonormal tetrad $\{\mathbf e_\mu \}$ along the geodesic $C$ where $\mathbf e_3$ is along the direction of the wave vector $\mathbf k$.

I think since at every point, I have a tetrad and $\mathbf e_3$ is tangent to the geodesic

This is impossible. A vector tangent to a null geodesic has length zero. There is no such thing as an orthonormal tetrad with one member tangent to a null geodesic.

 

[Haorong Wu]

Thanks, @PeterDonis . I used a wrong concept. I should say let $C$ be the integral curve of $\mathbf k$. It is the curve along which the light propagates.

 

[PeterDonis]

I should say let $C$ be the integral curve of $\mathbf k$. It is the curve along which the light propagates.

You already said this in the OP, just in different words. That is not the problem. You are not addressing the actual issue I raised.

 

[PeterDonis]

I think since at every point, I have a tetrad and $\mathbf e_3$ is tangent to the geodesic, I could just calculate $\frac {\partial f}{\partial s}=\frac {\partial f}{\partial x^3}$. Is this correct?

No. Even leaving aside the issue I already raised, having an orthonormal tetrad (more precisely a tetrad field) is not the same as having a coordinate chart. In general you cannot even be sure of finding a correspondence between an orthonormal tetrad field and a coordinate chart.

 

[Haorong Wu]

Hi, @PeterDonis. I am reading this article, Phase evolution of the photon in Kerr spacetime .

In the first paragraph of Section IV, the author wrote "Under the geometrical optics approximation, the wave function of a photon depends only on its propagating path length. Let $C$ be the integral curve of $\mathbf k$ and let the path length $s$ parametrize this curve."

Also, in the paragraph before Eq. (3.6), they wrote "Let the unit vectors $\{\mathbf e_i \}$, $i =1,2, 3$ form an orthogonal tetrad with $\mathbf e_3$ along the direction of the wave vector $\mathbf k$."

My idea mainly comes from these two paragraphs. Could you point out where I understand wrong when you are available? Thanks!

 

[PeterDonis]

I am reading this article, Phase evolution of the photon in Kerr spacetime .

Unfortunately it's behind a paywall. If you can find the arxiv preprint, that would be helpful.

Note that Phys Rev D is notorious for publishing lower quality papers that don't get the same level of peer review, so you can't always trust them to correctly take into account well known properties of what they are studying.

In the first paragraph of Section IV, the author wrote "Under the geometrical optics approximation, the wave function of a photon depends only on its propagating path length. Let $C$ be the integral curve of $\mathbf k$ and let the path length $s$ parametrize this curve."

I would need to see the context to know what they mean by "path length". Unless this is one of those papers that Phys Rev D didn't do enough checking on (see above), I would expect them to define a "path length", or equivalently an affine parameter, that does not involve arc length along the curve itself, but is defined some other way.

Also, in the paragraph before Eq. (3.6), they wrote "Let the unit vectors $\{\mathbf e_i \}$, $i =1,2, 3$ form an orthogonal tetrad with $\mathbf e_3$ along the direction of the wave vector $\mathbf k$."

By $\mathbf{k}$ do they mean a spatial 3-vector in some particular frame, or a 4-vector?

If they mean a 4-vector, this is obviously wrong since the wave 4-vector of light is null, as I've already said, so it can't possibly be a vector in an orthonormal tetrad.

However, it looks like they are only defining 3 vectors in the "orthogonal tetrad", not 4. So they might mean a spatial 3-vector in some particular frame.

 

[Haorong Wu]

@PeterDonis

I am sorry I could not find the paper on arxiv, and I think it is inappropriate to post the pdf file here.

The "path length" is not defined before this paragraph. But later, the author wrote that "the unit wave vector $\mathbf n_k=\nabla s$." It appears that the direction of $s$ is parallel to the direction of $\mathbf k$.

Also, the paper mainly works in three-dimensional space so $\mathbf k$ is indeed a 3-vector. I think the authors use the tetrad to represent a set of 3-dimensional orthonormal basis, which is different from what I learned earlier.

Oh, PRD is so bad? No wonder it publishes my two papers.

 

[vanhees71]

You can't use path length along a null geodesic as an affine parameter; the path length is zero at all events on the geodesic.This is impossible. A vector tangent to a null geodesic has length zero. There is no such thing as an orthonormal tetrad with one member tangent to a null geodesic.

Indeed, for the description of the motion of "massless particles" under the sole influence of gravity you just may use the "square form" of the variational principle, i.e.,
$$L=\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu},$$
where the dot refers to the derivative wrt. an arbitrary parameter of the worldline and which has the advantage that the "Hamiltonian"
$$H=p_{\mu} \dot{x}^{\mu} -L=const,$$
because $L$ doesn't explicitly depend on time.

Thus automatically $\lambda$ is an affine parameter, and you get the geodesic equation in the usual form using an affine parameter as world-line parameter.

For massive particles you can choose $L=c^2/2$. Then $\lambda=\tau$ is the proper time of the particle along its trajectory.

For massless particles you have $L=0$, and the overall scaling factor of $\lambda$ is to be determined by the initial conditions. Usually one uses the massless-particle description as a kind of "photon" to describe the propagation of electromagnetic waves in the gravitational field (or curved spacetime). Then the equation can be interpreted as describing electromagnetic waves in the leading-order eikonal approximation of the Maxwell equations, and then the overall scale is determined by the identification of $\dot{x}^{\mu}=k^{\mu}$, i.e., with the four-wave vector of the em. wave.

For more about the more careful interpretation of the "pseudo-photons" used to describe light propgation in a gravitational field within GR, see

https://itp.uni-frankfurt.de/~hees/pf-faq/gr-edyn.pdf

 

[Haorong Wu]

Thanks, @vanhees71

From the identification of $\dot x^i=k^i$ (suppose I have performed the 3+1 decomposition of spacetime and I focus only on the space part), I have $$\frac {d}{d\lambda}=\frac {dx^i}{d\lambda}\frac{d}{dx^i}=k \frac{d}{dx^3}, $$ where $\mathbf k$ is assumed to be along the direction of $x^3$.

 

[vanhees71]

You have
$$D_{\lambda} k^{\mu}=\dot{k}^{\mu} +\Gamma_{\alpha \beta}^{\mu} k^{\alpha} k^{\beta}=0.$$
For details, see

https://itp.uni-frankfurt.de/~hees/pf-faq/gr-edyn.pdf