# Equations of Motion

1. Aug 25, 2006

### lokofer

If for the "geodesic" equation of motion we have the compact form:

$$\nabla _ u u =0$$ usign the "Covariant derivative"... as a generalization of Newton equation with F=0 (no force or potential) $$\frac{du}{ds}=0$$ where "u" is the 4-dimensional momentum...

My question is if we can put the Equation of motion $$\R _\mu \nu =0$$ as the "LIe derivative" or " Covariant derivative" or another Tensor, vector or similar involving the "momentum density" $$\pi _a b$$ and the metric elements $$g_ ab$$

2. Aug 25, 2006

### pervect

Staff Emeritus
If the mass (invariant mass) is a constant, just multiply thorugh by m, and you have the covariant derivative of the energy-momentum 4 vector being 0.

I.e
$$\nabla_u u = 0 \Rightarrow \nabla_u (m u) = 0 \Rightarrow \nabla_u P = 0$$

In component notation
$$\nabla_u P^i = u^j \nabla_j P^i = 0$$

which makes it clear this gives 4 equations, one for each component of the energy-momentum 4-vector.

The notation for the covariant derivative $\nabla_a$ seems funny at first, but it's extremely useful. Lets say we have a scalar quantity x, and we want to find how fast it changes with time. With an ordinary derivative operator, we write dx/dt. With the covariant derivative operator, one writes $\nabla_t x = t^a \nabla_a x$, where t is a unit vector in the time direction. The covariant derivative operator actually generates a vector from the scalar. One must take the dot product of this vector with the direction one desires the derivative in order to get a scalar quantity. The reason for not doing a contraction automatically is that it can be ambiguous. Sometimes one sees the contracted covariant derivative written with a capital D to make the notation more simlar, but the math is the same.

Last edited: Aug 25, 2006
3. Aug 26, 2006

### lokofer

-Thank you..by the way if we have the Lagrangian $$L= \sqrt (-g) R$$

can you split the Lagrangian into a "Kinetic" and a "potential" part?..

- And What would be the "Euler-Lagrange" equation for Einstein Lagrangian?..i think you have:

$$\sum_{i=1,2,3,0} \partial _ {x_i }( \frac{ \partial L}{ \partial g_ab} - \frac{ \partial L}{ \partial g_ab}$$ or something similar.

Last edited: Aug 26, 2006