Is it allowed to have the zeroth-component of a four-velocity be negative? I presume the answer is yes, but I just want to make sure. Many thanks.
This is referring to $V^0$ for a curved space metric with signature $-+++$.
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1$\begingroup$ If you use the -+++ metric sign convention then the sign of the zeroth component of a four vector depends on whether you mean $V^0$ or $V_0$ so you need to specify your question a little more clearly. $\endgroup$– MichaelCommented Jan 21, 2013 at 1:58
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$\begingroup$ @MichaelBrown: Thank you for commenting, Michael! I am referring to $V^0$ for a curved space metric with signature $-+++$. $\endgroup$– GretaCommented Jan 21, 2013 at 1:59
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The four velocity is $$ U^\mu = \frac{\mathrm{d}x^\mu}{\mathrm{d}\tau} $$ where $\tau$ is the proper time. $U^0 < 0$ implies that $\frac{\mathrm{d}t}{\mathrm{d}\tau}<0$, i.e. the particle is moving backwards in time. There is an out of date interpretation, popularised by Feynman, that a particle moving backward in time can be thought of as an antiparticle moving forward in time, but this has fallen out of favour because it causes a great deal of unnecessary confusion. Normally the situation you describe is considered unphysical since the energy of the particle would be negative and unbounded from below.
