Euler Lagrange equation as Einstein Field Equation

[Black Integra]

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) :
$T_{\mu\nu}=\begin{pmatrix} \nabla E& \dot{E}\\ \nabla p & \dot{p}\end{pmatrix}=\begin{pmatrix} \nabla K& \dot{K}\\ \nabla p & \dot{p}\end{pmatrix}+\begin{pmatrix} \nabla V& \dot{V}\\ 0 & 0\end{pmatrix}=K_{\mu\nu}+V_{\mu\nu}$
for kinetic energy K and potential energy V

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

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

for $\kappa=8\pi G$ and set $c=1$

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

apply this to [a], we have

$\epsilon^{\mu\nu}G_{\mu\nu}=2\kappa\nabla V$

This is very beautiful equation but I'm not sure that I'm doing it right. So, am I doing it right?

 

[clamtrox]

You can find the Einstein field equations by minimizing the action
$$S = \int \sqrt{-g} R d^d x + S_M$$
where S_M is the action for matter fields, and geodesic equations by minimizing the proper time,
$$\tau = \int \sqrt{-g_{\mu \nu}\frac{dx^{\mu}}{d \lambda}\frac{dx^{\nu}}{d\lambda}} d \lambda$$

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?

 

[Black Integra]

Oh, I understand the second equation, Thank you :) but I doubt about the first one

hmm How does $\sqrt{-g}R$ 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.

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

- because $\frac{\partial}{\partial x}L - \frac{\partial}{\partial t}(\frac{\partial L}{\partial \dot{x}}) =\nabla L - \dot{p}=L_{00}-L_{11}$......(I use the fact that$\frac{\partial L}{\partial \dot{x}} = p$)

 

[clamtrox]

Oh, I understand the second equation, Thank you :) but I doubt about the first one

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

$\sqrt{-g}$is the determinant of the metric, and together with $d^d x$ they make up the differential volume element. $R$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.

Answer your questions,

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

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

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

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?

 

[Black Integra]

Then why do you say that $κT^{μν}=G^{μν}$? That seems weird, as this doesn't seem to reproduce Newtonian gravity in the appropriate limit.

I don't understand. $κT^{μν}=G^{μν}$ is the Einstein field equation, isn't it?
Or you are trying to say that the definition $E = K+V$ can't be used in General relativity?
I'm new to GR.

kinetic energy is certainly a coordinate-dependent quantity

I forget this point. Thanks for an enlightenment.

 

[clamtrox]

I don't understand. $κT^{μν}=G^{μν}$ is the Einstein field equation, isn't it?
Or you are trying to say that the definition $E = K+V$ can't be used in General relativity?
I'm new to GR.

I mean that you seem to be defining the energy-momentum tensor as $T^{\mu \nu} = \partial^{\mu} p^{\nu}$, 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 $T^{\mu \nu} = (p + \rho) u^{\mu} u^{\nu} + p g^{\mu \nu}$ where p is pressure, ρ is energy density and $u^{\mu}$ is the four-velocity field of the fluid. This is clearly quite different from your definition

 

[Black Integra]

Thanks I get them all now :)