# From field equation to equation of motion

1. Dec 20, 2012

Dear fellow relativiters,

I never fully got how to get from the field equations of Einstein's
$R_{ \mu \nu} - \frac{1}{2} g_{ \mu \nu} R= -\frac{8 \pi G}{c^4} T_{ \mu \nu}$

to a special metric, let's say the FRW metric

$ds^2 = c^2 dt^2 - a(t)^2 \cdot (\frac{dx^2}{1-kx^2} + x^2 d\Omega^2)$
and from there the equation of motion for a particle facing gravitational force.

In the above I left out $\Lambda g_{\mu\nu}$ the term that represents that the metric itself is a solution and that gravity is repulsive on scales of vacuum energy.
So in Newtonian approximation I should get something like the inverse square law + $\Lambda \cdot \vec{r}$ ?

Thanks

2. Dec 20, 2012

### Nabeshin

To get from the EFE to a specific metric, you just have to solve the EFE! In practice this means specifying a stress energy tensor. In the case of FRW, it's a perfect fluid. Then you impose homogeneity and isotropy, and literally compute the metric from the differential equations.

Since gravitational force don't exist in GR, you kind of have to do a Newtonian approximation to get such a thing to pop out.

3. Dec 20, 2012

### elfmotat

The field equations are highly non-linear second-order differential equations. Only a few exact solutions are known, so you're going to be hard-pressed to try to solve them for a specific stress-energy.

As for your question regarding a Newtonian Cosmological Constant, the form of the field equations with zero cosmological constant is:

$$G_{\mu \nu}=\kappa T_{\mu \nu}$$

Compare this with the analogous Poisson equation for Newtonian gravity:

$$\nabla^2 \phi =4\pi G\rho$$

The field equations with nonzero CC are:

$$G_{\mu \nu}=\kappa T_{\mu \nu}-g_{\mu \nu}\Lambda$$

Because the metric is analogous to the Newtonian potential, this suggests a modified Poisson equation of the form:

$$\nabla^2 \phi =4\pi G\rho-\phi \Lambda$$

However, if we assume the gravitational field is weak (which is the only time when Newtonian gravity is accurate anyway) and we linearize the field equations with $g_{\mu \nu }=\eta_{\mu \nu }+h_{\mu \nu }$ and $|h_{\mu \nu}|\ll 1$, then we get the following field equations:

$$G_{\mu \nu}=\kappa T_{\mu \nu}-\eta_{\mu \nu}\Lambda$$

where $G_{\mu \nu}=-\frac{1}{2} \partial^{\alpha} \partial_{\alpha}(h_{\mu \nu}-\frac{1}{2}\eta_{\mu \nu} h^{\sigma}_{~\sigma })$, though the form of the Einstein Tensor isn't really important for our purposes. These field equations imply (because the 00 component of the Minkowski metric is of magnitude 1) a Newtonian Poisson equation of the following form:

$$\nabla^2 \phi =4\pi G\rho-\Lambda$$

This is clearly much simpler than the previous form we considered. If we switch to regular units and apply Gauss' Law to a spherically symmetric mass then we obtain a modified version of Newton's Law:

$$\nabla^2 \phi=\nabla \cdot (\nabla\phi )=\nabla \cdot(-\mathbf{g})$$
$$\int \nabla\cdot \mathbf{g}dV=\int \mathbf{g}\cdot d\mathbf{A}=\int (-4\pi G \rho +\Lambda c^2)dV$$
$$g\int dA=-4\pi G\int \rho dV + \Lambda c^2 \int dV$$
$$4\pi r^2g=-4\pi GM + \frac{4}{3}\pi r^3 \Lambda c^2$$
$$g=-\frac{GM}{r^2} + \frac{\Lambda c^2 r}{3}$$

$$\ddot{\mathbf{r}}=\left ( \frac{\Lambda c^2}{3}- \frac{GM}{r^3}\right )\mathbf{r}$$

4. Dec 20, 2012

### atyy

The equation of motion for matter in a gravitational field generally requires separate postulates. However, specifying the form of the stress energy tensor in terms of the matter fields does place some constraints on the equations of motion for the matter fields, since the stress-energy tensor is divergence free according to the EFE. In some cases, the constraint is strong enough to specify the equations of motion for the matter fields completely. There is a discussion starting on p3 of http://arxiv.org/abs/1006.3903 .

Last edited: Dec 20, 2012
5. Dec 20, 2012

### elfmotat

Well, it's relatively straightforward to derive the equations of motion straight from the metric:

$$\left (\frac{\partial}{\partial x^\mu } -\frac{d}{d\lambda} \left [\frac{\partial}{\partial(dx^\mu/d\lambda )} \right ] \right ) g_{\alpha \beta} \frac{dx^\alpha}{d\lambda} \frac{dx^\beta}{d\lambda }=0$$

6. Dec 20, 2012

### PAllen

Ignoring EM, there is no need for a separate postulate - first approximately circa 1938, Einstein, Infeld and Hoffman showed the geodesic postulate was unnecessary. It is has been proved a number of rigorous ways since then.

Even with EM, the way EM figures in the stress energy tensor leads to the result that solving the EFE for a body with charge and mass reproduces motion consistent with Lorentz force law.

For example, an early reference for the charged case is: http://prola.aps.org/abstract/PR/v95/i1/p243_1

Last edited: Dec 20, 2012
7. Dec 20, 2012

### atyy

Yes, I forgot about that. I changed my post while you were replying to my original version.

8. Dec 21, 2012

Hi, thanks that helped me but could you please explain what makes the metric an analogy to a potential. I mean the metric just implies the distance (dot prodict) of two line elements.

9. Dec 21, 2012

Is $$\lambda$$ the parameter of the particle's path?
How to derive this formlular?
thanks

Last edited: Dec 21, 2012
10. Dec 21, 2012

### andrien

11. Dec 21, 2012

### elfmotat

The field equations with zero CC simply say:

(second derivatives of the metric) = (constant)*(energy-momentum distribution)

The solution to this differential equation gives you a particular metric. Now, think about what Poisson's equation for Newtonian gravity says:

(second derivatives of the potential) = (constant)*(mass distribution)

The solution to this differential equation gives you a particular potential. The jump from Newtonian to Relativistic gravity is essentially one where the gravitational potential becomes a rank-2 tensor field (the metric) instead of just a scalar field.

λ is a general affine parameter. You derive that equation from the principle of extremal proper time. Proper time is given by a path integral which we can parametrize with an affine parameter λ:

$$\tau = \int \sqrt{-g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}}d\lambda$$

So we see that the square root can be thought of as a Lagrangian (which we'll call L) that we can just plug into the Euler-Lagrange equation. It's easy to show that an equivalent Lagrangian is given by -L2, which helps simplify things a bit.

12. Dec 21, 2012

### samalkhaiat

13. Dec 27, 2012