How does the Einstein tensor contribute to the equations of motion in 4d?

[alejandrito29]

Hello, in a paper http://www-library.desy.de/preparch/conf/theo-ws/workshop2004/data/Chatillon.pdf .

says that.

1) In 4d, it is a total derivative, then does not contribute to the equations of motion

A total derivative respect to time does not contribute to equation of motion?, ¿or other parameter??

2) Only R^2 order combination giving equations of motion
with no derivatives of higher order than two and divergence free, like the Einstein tensor.

¿why einstein tensor gives equation of motion of second derivative of time?...I think that varying the action \int \sqrt{g_{uv}dx^u dx^v} i get the equation of motion, but i don't understand but the einstein tensor too...

 

[Bill_K]

alejandrito29, One way to derive the Einstein equations is to take as the action S = ∫√-g R, and vary this with respect to g_μν. It turns out that δS/δg_μν = √-g G^μν. It's generally true that for any action of the form S = ∫√-g L, where L is a scalar, that δS/δg_μν will be a divergence-free tensor.

The paper considers adding to the Einstein action a term α(R² − 4R_μνR^μν +R_μνρσR^μνρσ)

In four dimensions the variation of this additional term vanishes identically, so it makes no contribution to the equations of motion.

 

[alejandrito29]

alejandrito29, One way to derive the Einstein equations is to take as the action S = ∫√-g R, and vary this with respect to g_μν. It turns out that δS/δg_μν = √-g G^μν. It's generally true that for any action of the form S = ∫√-g L, where L is a scalar, that δS/δg_μν will be a divergence-free tensor.

The paper considers adding to the Einstein action a term α(R² − 4R_μνR^μν +R_μνρσR^μνρσ)

In four dimensions the variation of this additional term vanishes identically, so it makes no contribution to the equations of motion.

very thank , but,
1. ¿a divergence-free tensor does not contribute to the equations of motion?

2. ¿the einstein tensor then does not contribute to the equations of motion?. I understand by Einstein equation G_{uv}= R_{uv}-\frac{1}{2}g_{uv}R=k T_{uv} but by equation of motion \ddot{x} + \Gamma^v_{su} \dot{x}^s \dot{x}^u=0

 

[Bill_K]

¿a divergence-free tensor does not contribute to the equations of motion?

¡I didn't say that! A term T in the Lagrangian which can be written as a total divergence, T = V^μ_;μ, does not contribute to the equations of motion. Lagrangians are not unique. Two Lagrangians that differ by a total divergence will yield exactly the same equations of motion.

¿the einstein tensor then does not contribute to the equations of motion?

You're confusing two things that have nothing to do with each other. S = ∫√-g R d⁴x is the action for the field, and its variation δS/δg_μν yields the Einstein equations. ∫√g_μνdx^μdx^ν is a Lagrangian that can be used to describe geodesics, and its variation yields the equations of motion for a test particle.

 

[alejandrito29]

You're confusing two things that have nothing to do with each other. S = ∫√-g R d⁴x is the action for the field, and its variation δS/δg_μν yields the Einstein equations. ∫√g_μνdx^μdx^ν is a Lagrangian that can be used to describe geodesics, and its variation yields the equations of motion for a test particle.

ok, very very thanks, but
what is the equation of motion in general relativity? G_{uv}= k T_{uv} or \ddot{x}+ \Gamma^u_{s v} \dot{x}^s \dot{x}^v ?,

or other??

 

[member 11137]

or other

The first proposition you give is the Einstein's field equation.
The second proposition, if vanishing, is the equation describing a parallel transported speed vector.
Solutions of the field equation is of the following type:
g_ab. v^a. v^b = constant where v is the speed of the particle and where a, b = 0, 1, 2 and 3.
If you prefer a formulation like in classical mechanics, with forces, then you have to write the equation Dv = forces to recover the solutions.

 

[WannabeNewton]

If your lagrangian is of the form L = \frac{1}{2}g_{\mu \nu }\dot{x^{\mu }}\dot{x^{\nu }} then you can use the euler lagrange equations \frac{\partial L}{\partial x^{\mu }} - \frac{\mathrm{d} }{\mathrm{d} \lambda }(\frac{\partial L}{\partial \dot{x^{\mu }}}) = 0 (where lambda is an affine parameter) to arrive at the geodesic equation \ddot{x}^{\mu } + \Gamma ^{\mu }_{\alpha \beta }\dot{x^{\alpha }}\dot{x^{\beta }} = 0 If you use the Einstein lagrangian L = (-g)^{1/2}R then variation of the lagrangian will yield the vacuum field equations G^{\mu \nu } = 0 and the related bianchi identity \triangledown _{\mu }G^{\mu \nu } = 0.