# Deriving Gravitoelectromagnetism Lorentz Force Equation

I'm attempting a problem from Wald (Chapter 4, problem 3a) and having quite a bit of trouble.

## Problem

The text states, "If we assume that the time derivatives of $$\bar{\gamma}_{ab}$$ are negligible, then the space-space components of $$\bar{\gamma}_{ab}$$ vanish, and we find that to linear order in the velocity of the test body, the geodesic equation now yields $$\mathbf{a} = -\mathbf{E} - 4\mathbf{v} \space \times \space \mathbf{B}$$

where $$\mathbf{E}$$ and $$\mathbf{B}$$ are defined in terms of $$A_a$$ by the same formulas as in electromagnetism."

## Assumptions and Definitions

We're assuming the spacetime metric is $$g_{ab} = \eta_{ab} + \gamma_{ab}$$

where $$\eta_{ab}$$ is the metric for flat space time and $$\gamma_{ab}$$ is a small deviation, and we're only concerned about terms that are linear in $$\gamma_{ab}$$.

The Christoffel Symbols are given by $$\Gamma^c_{ab} = \frac{1}{2}\eta^{cd}(\partial_a\gamma_{bd} + \partial_b\gamma_{ad} - \partial_d\gamma_{ab})$$

We also know that the vector potential $$A_a$$ is defined to be $$A_a := -\frac{1}{4}\bar{\gamma}_{ab}t^b$$

with $$\bar{\gamma}_{ab} = \gamma_{ab} - \frac{1}{2}\eta_{ab}\gamma$$

## My Attempts

I've tried starting with the geodesic equation $$\frac{d^2x^\mu}{dt^2} + \Gamma^\mu_{\sigma\nu} \frac{dx^\sigma}{dt} \frac{dx^\nu}{dt} = 0$$

and plugging in for the Christoffel Symbols, and then substituting the partials of $$\gamma_{ab}$$ in terms of $$A_a$$ but am a bit lost.

Any guidance?

• Apr 16, 2022 at 18:46

I ended up figuring it out and just wanted to post the answer for anyone who may be interested.

# Goal

We want to show that linearized gravity predicts that the motion of masses produces gravitational effects similar to those of electromagnetism. Namely, to linear order in the velocity of a test body, the geodesic equation yields an analog of the Lorentz Force equation:

$$\mathbf{a} = - \mathbf{E} - 4\mathbf{v} \times \mathbf{B}.$$

# Assumptions

These assumptions are given on page 74 of Wald, but I will summarize them here.

## Metric

We are approximating gravity as "weak", so we assume that our metric is given by

$$g_{ab} = \eta_{ab} + \gamma_{ab}, \;\;\;\;\;\;(1)$$

where $$\eta_{ab}$$ is our flat metric and $$\gamma_{ab}$$ is a small perturbation of this metric.

We will also be raising and lowering indices with our flat metric $$\eta_{ab}$$ and $$\eta^{ab}$$ instead of $$g_{ab}$$ and $$g^{ab}$$.

## "Linearized" Gravity

For our computations, we will only keep terms that are of linear order in $$\gamma_{ab}$$, assuming that higher order terms will be sufficiently small.

## Covariant Derivative $$\nabla_a$$

In our first order approximations, we can write $$\partial_a$$ in place of $$\nabla_a$$ if applied to $$\gamma_{ab}$$ or an expression of $$\gamma_{ab}$$. This is because the Christoffel symbol terms will be not be first order in $$\gamma_{ab}$$.

## Christoffel Symbols

Because $$\partial_a \eta_{bc} = 0$$, when we write the Christoffel symbols in terms of our flat metric plus our perturbation $$\gamma_{ab}$$, we get

$$\Gamma^c_{ab} = \frac{1}{2}\eta^{cd}(\partial_a \gamma_{bd} + \partial_b \gamma_{ad} - \partial_d \gamma_{ab}). \;\;\;\;\;\;(2)$$

## Gamma Bar $$\overline{\gamma}_{ab}$$

To simplify computation, we define

$$\overline{\gamma}_{ab} = \gamma_{ab} - \frac{1}{2} \eta_{ab} \gamma, \;\;\;\;\;\;(3)$$

where $$\gamma$$ is defined as $$\gamma_a^a$$ as usual.

Note that taking the trace on both sides of (3) results in:

$$\overline{\gamma}_a^{\,\,a} = \gamma_a^{\,\,a} - \frac{1}{2}\eta_a^{\,\,a}\gamma$$

$$\Rightarrow \overline{\gamma} = \gamma - \frac{1}{2} (4) \gamma = -\gamma.$$

Plugging back into (3) gives

$$\gamma_{ab} = \overline{\gamma}_{ab} - \frac{1}{2} \eta_{ab} \overline{\gamma}. \;\;\;\;\;\;(4)$$

## Einstein's Equation in the Analog of the Lorentz gauge

Because linearized gravity has a gauge freedom (read page 75 of Wald for more details), we can make a gauge transformation to simplify Einstein's Equation to

$$\partial^c \partial_c \overline{\gamma}_{ab} = -16\pi T_{ab}, \;\;\;\;\;\;(5)$$

for stress-energy tensor $$T_{ab}$$.

## Stress-Energy Tensor

We assume that only lower order effects of motion are taken into account and neglect stresses. Thus, our stress-energy tensor $$T_{ab}$$ can be approximated to linear order in velocity $$t^a$$ by

$$T_{ab} = 2t_{(a}J_{b)} - \rho t_a t_b, \;\;\;\;\;\;(6)$$

where $$J_b = -T_{ab}t^a$$ is the mass current density 4-vector.

Assuming coordinates, for small velocities we can approximate $$t^{\mu} = (1,0,0,0)$$, so in "matrix" form, our stress-energy tensor can be thought of as

$$T_{\mu \nu} = \begin{bmatrix} \rho & p_x & p_y & p_z \\ p_x & 0 & 0 & 0 \\ p_y & 0 & 0 & 0 \\ p_z & 0 & 0 & 0 \end{bmatrix},$$

where $$\mathbf{p} = (p_x, p_y, p_z)$$ is momentum.

## Analog of Maxwell's Equations

Looking at just the time-time component and space-time components of Einstein's equation, we now have

$$\partial^a \partial_a \overline{\gamma}_{0 \mu} = 16\pi J_{\mu}. \;\;\;\;\;\;(7)$$

Comparing to Maxwell's first equation of E&M in flat 4d-spacetime

$$\partial^a \partial_a A_b = -4\pi J_b,$$

we define

$$A_a \equiv -\frac{1}{4} \overline{\gamma}_{ab} t^b, \;\;\;\;\;\;(8)$$

or for small velocities where $$t^{\mu} \approx (1,0,0,0)$$,

$$A_{\mu} = -\frac{1}{4} \overline{\gamma}_{0\mu}, \;\;\;\;\;\;(9)$$

We can now define an analog of $$\mathbf{E}$$ and $$\mathbf{B}$$ in terms of $$A_a = (-\phi, \mathbf{A})$$:

$$\mathbf{E} = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t} \;\;\;\;\;\;(10)$$ $$\mathbf{B} = \nabla \times \mathbf{A}. \;\;\;\;\;\;(11)$$

# Derivation

## Geodesic Equation

The motion of our test body is given by the geodesic equation,

$$\frac{d^2 x^{\mu}}{d\tau^2} + \Gamma^{\mu}_{\rho \sigma} \frac{d x^{\rho}}{d\tau} \frac{d x^{\sigma}}{d\tau} = 0, \;\;\;\;\;\;(12)$$

where $$x^{\mu}(\tau)$$ is the world line of the particle in global inertial coordinates. Assuming the motion of our test body is much slower than the speed of light, we approximate proper time $$\tau$$ by the coordinate time $$t$$. So, we may write

$$\frac{d^2 x^{\mu}}{dt^2} + \Gamma^{\mu}_{\rho \sigma} \frac{d x^{\rho}}{dt} \frac{d x^{\sigma}}{dt} = 0. \;\;\;\;\;\;(13)$$

Denoting $${d x^{\alpha}}/{dt}$$ as $$v^{\alpha}$$ and moving the second term of (13) to the right side, we get

$$\frac{d^2 x^{\mu}}{dt^2} = -\Gamma^{\mu}_{\rho \sigma} v^{\rho} v^{\sigma}. \;\;\;\;\;\;(14)$$

Lastly, changing $$\mu$$ to $$i$$, where $$i = 1, 2, 3$$, and denoting $$a^i = {d^2 x^i}/{dt^2}$$, we have

$$a^i = -\Gamma^{i}_{\rho \sigma} v^{\rho} v^{\sigma}, \;\;\;\;\;\;(15)$$

where it is important to note that while $$i$$ only runs along the spatial indices, $$\rho$$ and $$\sigma$$ still run from 0 to 3.

## Christoffel Symbols

Approximating time derivatives to be 0, we now calculate our nonzero Christoffel Symbols.

From (2),

$$\Gamma^i_{00} = \frac{1}{2}\eta^{ij}(-\partial_j \gamma_{00})$$ $$=-\frac{1}{2}\partial^i \gamma_{00}.$$

From (4) and (9) and recalling that $$A_0 = -\phi$$, we find

$$\gamma_{00} = \overline{\gamma}_{00} - \frac{1}{2}\eta_{00 }\overline{\gamma}$$ $$=-4A_0 + \frac{1}{2}(4A_0)$$ $$=4\phi - 2\phi$$ $$=2\phi.$$

So,

$$\Gamma^i_{00} = -\partial^i \phi. \;\;\;\;\;\;(16)$$

Using (2) again and (9), we also have

$$\Gamma^i_{0j} = \Gamma^i_{j0} = \frac{1}{2}\eta^{ik}(\partial_j \gamma_{0k} - \partial_k \gamma_{0j})$$ $$= \frac{1}{2}(\partial_j \gamma_0^i - \partial^i \gamma_{0j})$$ $$= \frac{1}{2}(-4)(\partial_j A^i - \partial^i A_j)$$ $$= 2(\partial^i A_j - \partial_j A^i).$$

Thus, using the fact that $$F_{ab} = \partial_a A_b - \partial_b A_a$$, we get

$$\Gamma^i_{0j} = \Gamma^i_{j0} = 2F^i_{\,\,j}. \;\;\;\;\;\;(17)$$

## Finishing the Computation

Recall that $$v^{\alpha} = dx^{\alpha}/{dt}$$, so $$v^0 = 1$$. Therefore, plugging our results from (16) and (17) into (15) gives

$$a^i = -\Gamma^{i}_{\rho \sigma} v^{\rho} v^{\sigma}$$ $$= -\Gamma^i_{00} - \Gamma^i_{0j}v^j - \Gamma^i_{j0}v^j$$ $$\Rightarrow a^i = \partial^i \phi - 4F^i_{\,\,j}v^j. \;\;\;\;\;\;(18)$$

Writing (10) and (11) in index notation and approximating the time derivative of $$\mathbf{A}$$ to be 0, we get

$$E^i = -\partial^i \phi \;\;\;\;\;\;(19)$$ $$B^i = \epsilon^{ijk}\partial_j A_k, \;\;\;\;\;\;(20)$$

where $$\epsilon^{ijk}$$ is the fully antisymmetric Levi Civita symbol (detailed on pages 432 and 433 of Wald).

Lastly, we write $$\mathbf{v} \times \mathbf{B}$$ in index notation and find

$$(\mathbf{v} \times \mathbf{B})_i = \epsilon_{ijk} v^j B^k$$ $$= \epsilon_{ijk} v^j \epsilon^{klm}\partial_l A_m$$ $$= (\epsilon^{klm} \epsilon_{kij}) v^j \partial_l A_m$$ $$= 2\delta^{[l}_i \delta^{m]}_j v^j \partial_l A_m$$ $$= 2\frac{1}{2}(\delta^{l}_i \delta^{m}_j v^j - \delta^{m}_i \delta^{l}_j v^j) \partial_l A_m$$ $$=v^m \partial_i A_m - v^l \partial_l A_i.$$

So,

$$(\mathbf{v} \times \mathbf{B})_i = F_{ij}v^j. \;\;\;\;\;\;(21)$$

Finally, plugging in our results from (19) and (21) into (18) gives

$$\mathbf{a} = - \mathbf{E} - 4\mathbf{v} \times \mathbf{B}. \;\;\;\;\;\;(22)$$

And we're done.

• Nice answer, and welcome to the site! Apr 16, 2022 at 18:42
• @knzhou Thanks! Apr 16, 2022 at 19:20