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If $ use $(+,-,-,-)$ sign convention then four position, four velocity become positive but four acceleration becomes negative!

$x_{\mu}x^{\mu}=\tau^2c^2,$

$U_{\mu}U^{\mu}=c^2,$

$a_{\mu}a^{\mu}=-(a\gamma^3)^2,$

in other hands If I use $(-,+,+,+)$ sign convention then four position, four velocity became negative but four acceleration becomes positive!

$x_{\mu}x^{\mu}=-\tau^2c^2,$

$U_{\mu}U^{\mu}=-c^2,$

$a_{\mu}a^{\mu}=(a\gamma^3)^2,$

Isn't it true!?

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why is it this like!? –  Ahmed Mar 12 '13 at 9:45

1 Answer 1

In the $({+}{-}{-}{-})$ signature, positive values of $V^\mu V_\mu$ are referred to as "timelike" and negative ones correspond to "spacelike" 4-vectors. The velocities are timelike (positive) vectors because the world lines of massive objects must have a more timelike angle – that's because massless objects can't exceed the speed of light.

On the contrary, the acceleration vector is negative i.e. spacelike because it's proportional the difference of infinitesimally nearby timelike vectors of the same length. To see why it's spacelike, it's enough to go to the immediate reference frame of the particle. In that frame, $$V^\mu (t=0) = (1,0,0,0),\quad V^\mu(t=dt)\sim (1,0,0,dt\, a)$$ up to terms of higher order than linear in $dt$. Note that the length of both vectors is $1$ within the accuracy. Their difference divided by $dt$ is the acceleration $$a^\mu = (0,0,0,a) $$ which is spacelike, i.e. negative squared norm. I used the acceleration in the $z$-direction but because the norm is rotationally invariant, the same comments on positivity hold for an arbitrary orientation. They also hold in other reference frames because the squared length of a vector is Lorentz-invariant.

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