Euler Lagrange equation as Einstein Field Equation

In summary, the speaker is trying to prove that the Euler Lagrange equation and Einstein Field equation (and Geodesic equation) are equivalent. To do this, they define a modified Energy-momentum Tensor and a new Lagrangian-momentum Tensor, which they substitute into the Einstein Field equation. By applying the Euler Lagrange equation and simplifying, they arrive at a beautiful equation that they are not sure is correct and ask for clarification. The other person explains the physical meaning of the expressions and points out some potential issues with the definitions of energy and momentum. The speaker thanks them for their help and understanding.
  • #1
Black Integra
56
0
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?
 
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  • #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?
 
  • #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])
 
  • #4
Black Integra said:
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?

[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.


Black Integra said:
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 [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.

Black Integra said:
- [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)

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?
 
  • #5
clamtrox said:
Then why do you say that [itex]κT^{μν}=G^{μν}[/itex]? That seems weird, as this doesn't seem to reproduce Newtonian gravity in the appropriate limit.

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.
clamtrox said:
kinetic energy is certainly a coordinate-dependent quantity

I forget this point. Thanks for an enlightenment.
 
Last edited:
  • #6
Black Integra said:
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 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
 
  • #7
Thanks I get them all now :)
 

1. What is the Euler-Lagrange equation?

The Euler-Lagrange equation is a mathematical expression that describes the relationship between the positions, velocities, and accelerations of a system. It is commonly used in classical mechanics and field theory to determine the equations of motion for a system.

2. How is the Euler-Lagrange equation related to the Einstein Field Equation?

The Einstein Field Equation is a set of equations in general relativity that describe the relationship between the curvature of space-time and the distribution of matter and energy. The Euler-Lagrange equation can be used to derive the Einstein Field Equation by taking the variation of the Lagrangian with respect to the metric tensor.

3. What is the significance of the Euler-Lagrange equation in physics?

The Euler-Lagrange equation is significant because it allows us to determine the equations of motion for a system by considering the total energy of the system. This is useful in a variety of fields, including classical mechanics, quantum mechanics, and field theory.

4. How is the Euler-Lagrange equation used in practice?

The Euler-Lagrange equation is used in practice by setting up a Lagrangian for the system in question, which is a function that describes the energy of the system in terms of its positions, velocities, and accelerations. By taking the variation of the Lagrangian with respect to these variables, the Euler-Lagrange equation can be solved to determine the equations of motion.

5. Are there any other applications of the Euler-Lagrange equation?

Yes, the Euler-Lagrange equation has many other applications in addition to classical mechanics and general relativity. It is also used in quantum field theory, where it helps to determine the equations of motion for particles and fields, and in optimal control theory, where it is used to find the optimal path for a system to follow. It is a fundamental tool in theoretical physics and has numerous applications in various fields.

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