Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Euler Lagrange equation as Einstein Field Equation

  1. Apr 23, 2012 #1
    I want to prove that Euler Lagrange equation and Einstein Field equation (and Geodesic equation) are the same thing so I made this calculation.
    First, I modified Energy-momentum Tensor (talking about 2 dimension; space+time) :
    [itex]T_{\mu\nu}=\begin{pmatrix} \nabla E& \dot{E}\\ \nabla p & \dot{p}\end{pmatrix}=\begin{pmatrix} \nabla K& \dot{K}\\ \nabla p & \dot{p}\end{pmatrix}+\begin{pmatrix} \nabla V& \dot{V}\\ 0 & 0\end{pmatrix}=K_{\mu\nu}+V_{\mu\nu}[/itex]
    for kinetic energy K and potential energy V

    Then, I defined new tensor that I call Lagrangian-momentum Tensor where
    [itex]L_{\mu\nu}=\begin{pmatrix} \nabla L& \dot{L}\\ \nabla p & \dot{p}\end{pmatrix}=K_{\mu\nu}-V_{\mu\nu}=T_{\mu\nu}-2V_{\mu\nu}[/itex]

    Substitute this for [itex]T_{\mu\nu}[/itex] in Einstein Field Equation, we have
    [a]..... [itex]\frac{1}{\kappa}G_{\mu\nu}-2V_{\mu v}=L_{\mu\nu}[/itex]

    for [itex]\kappa=8\pi G[/itex] and set [itex]c=1[/itex]

    Now, consider Euler Lagrange Equation
    [itex]\frac{\partial}{\partial x}L - \frac{\partial}{\partial t}\frac{\partial}{\partial \dot{x}}L = 0[/itex]
    Or written in Lagrangian Tensor form :
    [itex]L_{00} - L_{11} = 0 \rightarrow \epsilon^{\mu\nu}L_{\mu\nu}=0; \epsilon^{\mu\nu} = \begin{pmatrix} 1& 0\\ 0 & -1\end{pmatrix}[/itex]

    apply this to [a], we have

    [itex]\epsilon^{\mu\nu}G_{\mu\nu}=2\kappa\nabla V[/itex]

    This is very beautiful equation but I'm not sure that I'm doing it right. So, am I doing it right?
  2. jcsd
  3. Apr 23, 2012 #2
    You can find the Einstein field equations by minimizing the action
    [tex] S = \int \sqrt{-g} R d^d x + S_M [/tex]
    where SM is the action for matter fields, and geodesic equations by minimizing the proper time,
    [tex] \tau = \int \sqrt{-g_{\mu \nu}\frac{dx^{\mu}}{d \lambda}\frac{dx^{\nu}}{d\lambda}} d \lambda [/tex]

    In your calculation,

    -what are E and p?
    -how do you split E into K and V?
    -are you sure L is a tensor?
    -how does E-L equation imply L00-L11=0?
  4. Apr 23, 2012 #3
    Oh, I understand the second equation, Thank you :) but I doubt about the first one

    hmm How does [itex]\sqrt{-g}R[/itex] come up in the equation? what is the physical meaning of this expression?

    Answer your questions,

    - E is a total energy (Hamiltonian) , so I can split it into K+V. And p is momentum.

    - [itex]L_{\mu\nu}[/itex] has similar structure with [itex]T_{\mu\nu}[/itex], thus it should be a tensor ([itex]L_{\mu\nu}[/itex] and [itex]L[/itex] are not the same object, anyway)

    - because [itex]\frac{\partial}{\partial x}L - \frac{\partial}{\partial t}(\frac{\partial L}{\partial \dot{x}}) =\nabla L - \dot{p}=L_{00}-L_{11}[/itex]........................(I use the fact that[itex]\frac{\partial L}{\partial \dot{x}} = p[/itex])
  5. Apr 24, 2012 #4
    [itex] \sqrt{-g} [/itex]is the determinant of the metric, and together with [itex] d^d x [/itex] they make up the differential volume element. [itex]R[/itex]is the Ricci scalar. One can justify using this action by saying that it's the simplest action which fulfils a set of requirements (for example, it only depends on second derivatives of the metric) but at the end of the day, it's just a guess.

    So what you can do then is to minimize the action. You can either assume that connection is Levi-Civita and minimize wrt the metric, or you can take the connection to be an independent degree of freedom, and minimize wrt both connection and metric. The former option gives you standard general relativity, and latter gives so called Palatini formulation.

    Then why do you say that [itex] \kappa T^{\mu \nu}= G^{\mu \nu} [/itex]? That seems weird, as this doesn't seem to reproduce Newtonian gravity in the appropriate limit.

    Defined in the usual way, kinetic energy is certainly a coordinate-dependent quantity, so it cannot possibly be a tensor. How do you define it then?
  6. Apr 24, 2012 #5
    I don't understand. [itex]κT^{μν}=G^{μν}[/itex] is the Einstein field equation, isn't it?
    Or you are trying to say that the definition [itex]E = K+V[/itex] can't be used in General relativity?
    I'm new to GR.

    I forget this point. Thanks for an enlightenment.
    Last edited: Apr 24, 2012
  7. Apr 24, 2012 #6
    I mean that you seem to be defining the energy-momentum tensor as [itex]T^{\mu \nu} = \partial^{\mu} p^{\nu} [/itex], and this definition again does not seem to be a tensor. You'd need to replace the partial derivative with covariant derivative, and then you'd get all sorts of connection terms.

    For example, the energy-momentum tensor for an ideal fluid is [itex] T^{\mu \nu} = (p + \rho) u^{\mu} u^{\nu} + p g^{\mu \nu} [/itex] where p is pressure, ρ is energy density and [itex]u^{\mu}[/itex] is the four-velocity field of the fluid. This is clearly quite different from your definition
  8. Apr 25, 2012 #7
    Thanks I get them all now :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook