Equation for acceleration in MTW

[jfy4]

Hi,

In MTW, there is a box (3.1) where certain equations for acceleration are used as a model for determining the components of the curvature tensor. the equations are

\frac{d^2x^a}{d\tau^2}=\frac{e}{m}F^{a}_{b}u^b

and

\frac{d^2\xi^a}{d\tau^2}=-R^{a}_{bcd}u^b\xi^cu^d

However, I was under the impression that the latter equation was the equation for geodesic deviation, and that the first equation was the equation of motion for E&M.

What's the relationship MTW is trying to make between these two?

 

[tiny-tim]

Hi Judah!

I don't have a copy of MTW, but the equation I've seen elsewhere for geodesic deviation in an electromagnetic field is

\frac{D^2\,\delta x^{\alpha}}{D\tau^2}\ =\ \frac{q}{m}\,F^{\alpha}_{\ \mu\,;\,\beta}\,V^{\mu}\,\delta x^{\beta}

(where δx^α is the gap between two world-lines).

 

[Bill_K]

MTW is trying to draw a parallel (a rather vague parallel!) between the effect of a gravitational field vs an electromagnetic field, and at the same time give an operational significance to the Riemann tensor. While the electromagnetic field produces an acceleration on a charged particle, the gravitational field produces instead a differential acceleration on a cloud of neighboring particles. In more detail, the structure of the second equation shows that in the rest frame of the particles it is the 'electric' components of the Riemann tensor, R_a0c0, that do this, and that the effect is a quadrupole shear.

 

[jfy4]

MTW is trying to draw a parallel (a rather vague parallel!) between the effect of a gravitational field vs an electromagnetic field, and at the same time give an operational significance to the Riemann tensor. While the electromagnetic field produces an acceleration on a charged particle, the gravitational field produces instead a differential acceleration on a cloud of neighboring particles. In more detail, the structure of the second equation shows that in the rest frame of the particles it is the 'electric' components of the Riemann tensor, R_a0c0, that do this, and that the effect is a quadrupole shear.

Thanks,

To try and get a handle on what it means to call those components the "electric" parts, let me try something here. I have here from one of my reference books Exact Solutions to Einstein's Field Equations by Stephani et al... a relationship between the Weyl Conformal tensor (which is a component of the Riemann curvature tensor) and it's "electric" and "magnetic" parts.

C^{\ast}_{abcd}u^bu^d=E_{ac}+iB_{ac}

where

u_bu^b=-1

and

C^{\ast}_{abcd}=C_{abcd}+iC^{\sim}_{abcd}

where

C^{\sim}_{abcd}=\frac{1}{2}\varepsilon_{cdef}C_{ab}{}^{ef}

Is it ok to draw a parallel between these two (the Riemann and Weyl) tensors and their "electric" and "magnetic" parts?

EDIT:
That is, can I identify these corresponding parts (analogously from the Weyl tensor)?

R_{abcd}u^bu^d=E_{ac}

and

\frac{1}{2}R_{ab}{}^{ef}\varepsilon_{cdef}u^bu^d=B_{ac}

 

[Mentz114]

Try writing C_abcd in terms of R_abcd and plugging that into the expressions for the Weyl E and B.

Somewhere you'll find a formula like

R_abcd = C_abcd - X_abcd/2 - R/6 Y_abcd

where X is a function of the Ricci tensor and Y of the metric.

The answer should be easiest to get in the case where Rab=0.

 

[jfy4]

Try writing C_abcd in terms of R_abcd and plugging that into the expressions for the Weyl E and B.

Somewhere you'll find a formula like

R_abcd = C_abcd - X_abcd/2 - R/6 Y_abcd

where X is a function of the Ricci tensor and Y of the metric.

The answer should be easiest to get in the case where Rab=0.

I have

C_{abcd}=R_{abcd}-N_{abcd}-G_{abcd}

where R_{abcd} is the Riemann curvature tensor,

N_{abcd}=\frac{1}{2}(g_{ac}S_{bd}+g_{bd}S_{ac}-g_{ad}S_{bc}-g_{bc}S_{ad})

and

G_{abcd}=\frac{1}{12}\mathcal{R}(g_{ac}g_{bd}-g_{ad}g_{bc})=\frac{1}{12}\mathcal{R}\,g_{abcd}

where

S_{ab}=R_{ab}-\frac{1}{4}\mathcal{R}\, g_{ab}

where \mathcal{R} is the Curvature scalar and R_{ab} is the Ricci tensor.

Then when I insert this into the electric and magnetic relation I have three tensors responsible for the electric and magnetic parts, not just the Riemann tensor, which is what I want to know. Currently I have

(R_{abcd}-N_{abcd}-G_{abcd})u^bu^d=E_{ac}

and

\frac{1}{2}\epsilon_{cdef}(R_{ab}^{\;\;\;ef}-N_{ab}^{\;\;\;ef}-G_{ab}^{\;\;\;ef})u^bu^d=B_{ac}

If there is a way to get the electric and magnetic components solely in terms of the Riemann tensor, I need a little push forward.

 

[Mentz114]

I might not be understanding the question, but what you've now got is the E and B parts of the Weyl tensor in terms of the Riemann tensor and its contractions.

I thought that was what you wanted.

 

[jfy4]

I might not be understanding the question, but what you've now got is the E and B parts of the Weyl tensor in terms of the Riemann tensor and its contractions.

I thought that was what you wanted.

Almost! as you said, I have the E and B parts of the Weyl tensor, I want the E and B parts of the Riemann tensor. I was asking earlier if in fact it was what I had posted

R_{\alpha\beta\gamma\delta}u^\beta u^\delta=E_{\alpha\gamma}

and

<br /> \frac{1}{2}R_{\alpha\beta}{}^{\epsilon\zeta}\varepsilon_{\gamma\delta\epsilon\zeta}u^\beta u^\delta=B _{\alpha\gamma}<br />

where u^{\alpha} is a time-like vector.