# Weak field Geodesic equation

1. Nov 15, 2013

### darida

Geodesic equation:

$m_{0}\frac{du^{\alpha}}{d\tau}+\Gamma^{\alpha}_{\mu\nu}u^{\mu}u^{\nu}= qF^{\alpha\beta}u_{\beta}$

Weak-field:

$ds^{2}= - (1+2\phi)dt^{2}+(1-2\phi)(dx^{2}+dy^{2}+dz^{2})$

Magnetic field, $B$ is set to be zero.

I want to find electric field, $E$, but don't know where to start, so could someone give me a procedure to find it?

Thank you

Last edited: Nov 16, 2013
2. Nov 15, 2013

### WannabeNewton

$F_{\mu\nu}$ is a covariant object but $E^{\mu}$ and $B^{\mu}$ are not i.e. they are frame dependent quantities. They only have meaning if you have a background observer with 4-velocity $u^{\mu}$ who can decompose $F_{\mu\nu}$ into $E^{\mu}$ and $B^{\mu}$ relative to $u^{\mu}$.

Once you have such a $u^{\mu}$, then $E^{\mu} = F^{\mu}{}{}_{\nu}u^{\nu}$ and $B^{\mu} = -\frac{1}{2}\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}u_{\nu}$.

3. Nov 16, 2013

### vanhees71

That's not entirely true. You can build a formalism of classical electromagnetics with socalled bivectors by representing the electromagnetic field with $\mathbb{C}^3$ vector fields,
$$\vec{\mathfrak{E}}=\vec{E} + \mathrm{i} \vec{B}.$$
This makes use of the group-isomorphism $\mathrm{SO}(1,3)^{\uparrow} \simeq \mathrm{SO}(3,\mathbb{C}).$

Thus the product
$$\vec{\mathfrak{E}}^2=\vec{E}^2-\vec{B}^2 + 2\mathrm{i} \vec{E} \cdot \vec{B}$$
is Lorentz invariant wrt. to proper orthochronous Lorentz transformations, and indeed also the four-vector formalism shows that the corresponding invariants given by the real and imaginary part of the bivector scalar product are proportional to the invariants $F_{\mu \nu} F^{\mu \nu}$ and $\epsilon_{\mu \nu \rho \sigma} F^{\mu \nu} F^{\rho \sigma}=F^{\mu \nu} F^{\dagger}_{\mu \nu}$.

A nice review on this funny bivector formalism can be found here:

http://arxiv.org/abs/1211.1218