Friday, March 13, 2015

Particle energy in general relativity?


Assume that a particle is moving with four-velocity $u^\mu$ through a spacetime with metric $g_{\mu\nu}$ ($-+++$ signature).


Let us also assume that there exists a time-like vector field $t^\mu=(1,0,0,0)$.


Is it generally true that the energy $E$ of a particle is given by the following expression:


\begin{eqnarray*} E &=& -mt_\mu u^\mu\\ &=& -mg_{\mu\nu} t^\nu u^\mu\\ &=& -mg_{00}u^0\\ &=& -mg_{00}\frac{dt}{d\tau}? \end{eqnarray*}


In the special case that $t^\mu$ is a Killing vector and $u^\mu$ is tangent to a geodesic then $E$ is a constant of motion of the particle. But the energy of a particle need not remain constant in the general case - is that true?



If $E=-mt_\mu u^\mu$ is indeed the particle mass/energy then does it act as a source of spacetime curvature given that it depends on the $00$-component of the spacetime metric ifself? I suppose that is not a problem.




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