Equations of Motion with Covariant Derivatives and Momentum Density

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SUMMARY

The discussion focuses on the application of covariant derivatives in the context of the equations of motion, specifically the geodesic equation represented as \nabla_u u = 0. It establishes that if the invariant mass is constant, the covariant derivative of the energy-momentum 4-vector is also zero, leading to four equations corresponding to each component of the energy-momentum vector. The conversation also touches on the utility of the covariant derivative operator in generating vectors from scalars and discusses the potential separation of the Einstein Lagrangian into kinetic and potential components, as well as the formulation of the Euler-Lagrange equation for the Einstein Lagrangian.

PREREQUISITES
  • Understanding of covariant derivatives and their notation, specifically \nabla_a.
  • Familiarity with the concept of energy-momentum 4-vectors in general relativity.
  • Knowledge of the Einstein Lagrangian and its components.
  • Basic grasp of the Euler-Lagrange equation in the context of field theory.
NEXT STEPS
  • Explore the mathematical properties of covariant derivatives in differential geometry.
  • Study the derivation and implications of the geodesic equation in general relativity.
  • Investigate the separation of the Einstein-Hilbert action into kinetic and potential terms.
  • Learn about the application of the Euler-Lagrange equation in the context of general relativity.
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Researchers, physicists, and students in theoretical physics, particularly those focusing on general relativity, differential geometry, and field theory applications.

lokofer
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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]
 
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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.

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

In component notation
[tex] \nabla_u P^i = u^j \nabla_j P^i = 0[/tex]

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. Let's 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.
 
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-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.
 
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