Can the geodesic equation be derived from the EFE in a certain limit?

[vanhees71]

This doesn't seem right, because Dixon is taking moments of $T^{\alpha \beta}$ and saying they correspond to things like the total energy/momentum, total angular momentum, etc., not just quantities corresponding to the fields. If anything, the equation $\nabla_{\alpha} \tilde{T}^{\alpha \beta}=-F^{\alpha \beta} J_{\beta}$ suggests to me something like the Lorentz force equation, which would mean $T^{\alpha \beta}$ on the LHS would be the stress-energy of the particles, not the fields.

Argh, ok. I got wrong by the sign (maybe because Dixon uses the east-coast convention). So it seems to be the matter-field EM tensor with the em.-field taken out. I've to look at the papers.

[EDIT:] Indeed, in the 1st of the quoted papers in the very beginning Dixon exactly says this. In (1.1) $T^{\mu \nu}$ is the EM tensor of the charged matter without the em. field EM tensor. Though he doesn't state it explicitly he then must use the east-coast convention of the metric.

 

[TSny]

I just noticed that Dixon's first paper has an appendix where notation and sign conventions are given. His metric has signature -2.

 

[vanhees71]

But then I'm really puzzled. Let's check the energy-momentum balance equation for charged particles in SR and in standard pseudo-Cartesian coordinates.

The symmetric EM tensor of the em. field reads
$$T^{\mu \nu}={F^{\mu}}_{\rho} F^{\rho \nu}+\frac{1}{4} F_{\rho \sigma} F^{\rho \sigma} \eta^{\mu \nu},$$
using the west-coast convention $(\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)$, which means signature 1=3=-2.

Now to derive Poynting's theorem one simple takes the four-divergence (which in flat space and pseudo-Cartesian coordinates is just calculated with the partial derivative). This gives
$$\partial^{\mu} T_{\mu \nu}=F^{\rho \nu} \partial_{\mu} {F^{\mu}}_{\rho} + {F^{\mu}}_{\rho} \partial_{\mu} F^{\rho \nu} + \frac{1}{2} F_{\rho \sigma} \partial^{\nu} F^{\rho \sigma}. \qquad (1)$$
The Maxwell equations tell us (using natural Heavsiside-Lorentz units with $c=1$)
$$\partial_{\mu} F^{\mu \nu}=j^{\nu}$$
and
$$\partial_{\mu} \epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma}=0.$$
The latter equation can be rewritten in the "cyclic form"
$$\partial_{\mu} F_{\rho \sigma}+\partial_{\rho} F_{\sigma \mu} + \partial_{\sigma} F_{\mu \rho}=0.$$
From this we get
$$F^{\rho \sigma} \partial_{\mu} F^{\rho \sigma}+F^{\rho \sigma} (\partial_{\rho} F_{\sigma \mu}-\partial_{\sigma} F_{\rho \mu})=F^{\rho \sigma} \partial_{\mu} F^{\rho \sigma} + 2 F^{\rho \sigma} \partial_{\rho} F_{\sigma \mu}=0$$
or
$$F^{\rho \sigma} \partial_{\mu} F^{\rho \sigma}=-2F^{\rho \sigma} \partial_{\rho} F_{\sigma \mu}.$$
So the last two terms on the right-hand side of (1) cancel and the remaining part gives
$$\partial_{\mu} T^{\mu \nu}=F^{\rho \nu} j_{\rho}=-F^{\nu \rho} j_{\rho}.$$
With $\Theta^{\mu \nu}$ the energy-momentum tensor of the charged particles the conservation of energy and momentum tells us
$$\partial_{\mu} \Theta^{\mu \nu}=-\partial_{\mu} T^{\mu \nu}=+F^{\nu \rho} j_{\rho}=f^{\nu}.$$
This is the opposite sign than given in Dixon's 1st paper in Eq. (1.1).

Let's check the force density at the final result. For the spatial components we have
$$f^a=F^{a \rho} j_{\rho}=\rho F^{a 0}+F^{ab} j_b.$$
Now
$$F^{a 0}=\partial^a A^0-\partial^0 A^a=-\partial_a A^0-\partial^0 A^a=E^a$$
and
$$F^{ab}=\partial^a A^b-\partial^b A^a=-\partial_a A^b + \partial_b A^a = -\epsilon^{abc} B^c$$
and thus
$$f^a=\rho E^a -\epsilon^{abc} B^c j_b=\rho E^a + \epsilon^{abc} j^b B^c=\rho E^a + (\vec{j} \times \vec{B})^a,$$
which is the correct Lorentz force.

So where is the sign mistake?

 

[TSny]

It looks like Dixon defines the components $F^{\mu \nu}$ to have opposite signs to your definition.

You have

$$F^{a 0}=\partial^a A^0-\partial^0 A^a=-\partial_a A^0-\partial^0 A^a=E^a$$
and
$$F^{ab}=\partial^a A^b-\partial^b A^a=-\partial_a A^b + \partial_b A^a = -\epsilon^{abc} B^c$$

whereas Dixon defines the components in his appendix as

You write $\partial_{\mu} F^{\mu \nu} = j^{\nu}$ and $f^a=F^{a \rho} j_{\rho}$.

I guess Dixon would write $\partial_{\nu} F^{\mu \nu} = j^{\mu}$ and $f^a=-F^{a \rho} j_{\rho}$

 

[vanhees71]

Ok, that makes it really very difficult to read these papers. I thought at least the definition of the em. field tensor is the same in all the literature, but of course, you can define something differently than any other, someone will come and do it ;-)).

One example is von Laue. I love his two books on relativity. I consider them as some of the best books on the subject. What makes me crazy, however, is that he defines the orientation of the surface-normal vectors in the usual 3D Gauß integral theorem in the opposite way than any other textbook or paper I know, i.e., he let's the surface-normal vectors point into the volume. The sign confusion is perfect :-(.

 

[A. Neumaier]

Isn't the classical point-particle picture an idealization?

The paper

• Damour, Soffel, and Xu, General-relativistic celestial mechanics. I. Method and definition of reference systems, Physical Review D, 43 (1991), 3273.

discusses ''the general-relativistic celestial mechanics of systems of N arbitrarily composed and shaped, weakly self-gravitating, rotating, deformable bodies'' and ''a new (and, in our opinion, improved) derivation of the Lorentz-Droste-Einstein-Infeld-Hoffmann equations of motion''. (quoted from the abstract).