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From field equation to equation of motion 
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#1
Dec2012, 02:27 PM

P: 22

Dear fellow relativiters,
I never fully got how to get from the field equations of Einstein's [itex] R_{ \mu \nu}  \frac{1}{2} g_{ \mu \nu} R= \frac{8 \pi G}{c^4} T_{ \mu \nu} [/itex] to a special metric, let's say the FRW metric [itex] ds^2 = c^2 dt^2  a(t)^2 \cdot (\frac{dx^2}{1kx^2} + x^2 d\Omega^2) [/itex] and from there the equation of motion for a particle facing gravitational force. In the above I left out [itex] \Lambda g_{\mu\nu} [/itex] 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 + [itex] \Lambda \cdot \vec{r} [/itex] ? Thanks 


#2
Dec2012, 04:18 PM

Sci Advisor
P: 2,193

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
Dec2012, 04:46 PM

P: 260

The field equations are highly nonlinear secondorder differential equations. Only a few exact solutions are known, so you're going to be hardpressed to try to solve them for a specific stressenergy.
As for your question regarding a Newtonian Cosmological Constant, the form of the field equations with zero cosmological constant is: [tex]G_{\mu \nu}=\kappa T_{\mu \nu}[/tex] Compare this with the analogous Poisson equation for Newtonian gravity: [tex]\nabla^2 \phi =4\pi G\rho[/tex] The field equations with nonzero CC are: [tex]G_{\mu \nu}=\kappa T_{\mu \nu}g_{\mu \nu}\Lambda[/tex] Because the metric is analogous to the Newtonian potential, this suggests a modified Poisson equation of the form: [tex]\nabla^2 \phi =4\pi G\rho\phi \Lambda[/tex] 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 [itex]g_{\mu \nu }=\eta_{\mu \nu }+h_{\mu \nu }[/itex] and [itex]h_{\mu \nu}\ll 1[/itex], then we get the following field equations: [tex]G_{\mu \nu}=\kappa T_{\mu \nu}\eta_{\mu \nu}\Lambda[/tex] where [itex]G_{\mu \nu}=\frac{1}{2} \partial^{\alpha} \partial_{\alpha}(h_{\mu \nu}\frac{1}{2}\eta_{\mu \nu} h^{\sigma}_{~\sigma })[/itex], 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: [tex]\nabla^2 \phi =4\pi G\rho\Lambda[/tex] 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: [tex]\nabla^2 \phi=\nabla \cdot (\nabla\phi )=\nabla \cdot(\mathbf{g})[/tex] [tex]\int \nabla\cdot \mathbf{g}dV=\int \mathbf{g}\cdot d\mathbf{A}=\int (4\pi G \rho +\Lambda c^2)dV[/tex] [tex]g\int dA=4\pi G\int \rho dV + \Lambda c^2 \int dV[/tex] [tex]4\pi r^2g=4\pi GM + \frac{4}{3}\pi r^3 \Lambda c^2[/tex] [tex]g=\frac{GM}{r^2} + \frac{\Lambda c^2 r}{3}[/tex] [tex]\ddot{\mathbf{r}}=\left ( \frac{\Lambda c^2}{3} \frac{GM}{r^3}\right )\mathbf{r}[/tex] 


#4
Dec2012, 05:58 PM

Sci Advisor
P: 8,660

From field equation to equation of motion



#5
Dec2012, 06:17 PM

P: 260

[tex]\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[/tex] 


#6
Dec2012, 06:22 PM

Sci Advisor
PF Gold
P: 5,060

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 


#7
Dec2012, 06:34 PM

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P: 8,660




#8
Dec2112, 05:00 AM

P: 22




#9
Dec2112, 05:02 AM

P: 22

How to derive this formlular? thanks 


#10
Dec2112, 06:52 AM

P: 1,020



#11
Dec2112, 11:59 AM

P: 260

(second derivatives of the metric) = (constant)*(energymomentum 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 rank2 tensor field (the metric) instead of just a scalar field. [tex]\tau = \int \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}}d\lambda [/tex] 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 EulerLagrange equation. It's easy to show that an equivalent Lagrangian is given by L^{2}, which helps simplify things a bit. 


#12
Dec2112, 04:20 PM

Sci Advisor
P: 910

http://www.physicsforums.com/showthr...65#post3616065 


#13
Dec2712, 05:48 AM

P: 22




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