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Can anyone see a reason for $$\left(1+{U_\rho U^\rho\over c^2}\right)\left(U_\nu{d^2 U^\nu\over d\tau^2}\right)=0$$?

Here $U^\rho$ is the 4-velocity for a particle and $\tau$ the proper time. The context is for a particle moving in an electromagnetic field.

I believe it may be useful to introduce the antisymmetric tensor $F_{\mu\nu}$ -- the electromagnetic field tensor.

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up vote 4 down vote accepted

The left parentheses are equal to zero due to $U_{\rho}U^{\rho}=-c^2$. This is true for timelike vectors in the (-1,1,1,1) signature.

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Thanks for answering, Fredric. However, I don't understand the answer. Firstly, for the LHS bracket to be $0$, surely we need $U_{\rho}U^{\rho}=-c^2$. Secondly, I don't understand how time-like vectors would guarantee that. Many thanks. – Greta Jan 9 '13 at 16:50
Right, Frederic meant or should have meant your value of $U_\rho U^\rho$. It describes the length (proper time) of a time-like 4-vector normalized to de facto unit length. More precisely, the length is $c$. Only the direction (of the velocity, the direction of the world line) is variable; the normalization is fixed by convention. – Luboš Motl Jan 9 '13 at 17:02
Indeed, that was a typo. Thanks. – Frederic Brünner Jan 9 '13 at 17:03
Thank you, Frederic and @LubošMotl. – Greta Jan 9 '13 at 19:42

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