Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 0 down vote favorite

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.

share|improve this question

1 Answer

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.

share|improve this answer
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 at 16:50
1  
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 at 17:02
Indeed, that was a typo. Thanks. – Frederic Brünner Jan 9 at 17:03
Thank you, Frederic and @LubošMotl. – Greta Jan 9 at 19:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.