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Equations of Motion

  1. Aug 25, 2006 #1
    If for the "geodesic" equation of motion we have the compact form:

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

    My question is if we can put the Equation of motion [tex] \R _\mu \nu =0 [/tex] as the "LIe derivative" or " Covariant derivative" or another Tensor, vector or similar involving the "momentum density" [tex] \pi _a b [/tex] and the metric elements [tex] g_ ab [/tex]
  2. jcsd
  3. Aug 25, 2006 #2


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    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.

    [tex]\nabla_u u = 0 \Rightarrow \nabla_u (m u) = 0 \Rightarrow \nabla_u P = 0[/tex]

    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 [itex]\nabla_a[/itex] 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 [itex]\nabla_t x = t^a \nabla_a x[/itex], 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
  4. Aug 26, 2006 #3
    -Thank you..by the way if we have the Lagrangian [tex] L= \sqrt (-g) R [/tex]

    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:

    [tex] \sum_{i=1,2,3,0} \partial _ {x_i }( \frac{ \partial L}{ \partial g_ab} - \frac{ \partial L}{ \partial g_ab} [/tex] or something similar.
    Last edited: Aug 26, 2006
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