Is this the only form of the Minkowski metric?

[PeterDonis]

the additional assumption that should be made is that light travels with the same speed in all directions. This is enough to kill off the non diagonal terms in the metric.

It's certainly sufficient, but it's stronger than necessary. Try considering my previous post first. (Hint: the geodesic equation applies to null geodesics, that light travels on, as well.)

 

[kent davidge]

for all possible combinations of indices $\beta$ and $\gamma$

but what if say, $dx^1 / d\tau = 0$? Then $\Gamma^\alpha{}_{1 \delta}$ could be different from zero.

 

[PeterDonis]

what if say, $dx^1 / d\tau = 0$? Then $\Gamma^\alpha{}_{1 \delta}$ could be different from zero

Free particles can travel in all directions, and you must meet the conditions for all possible 4-velocities for a free particle. That means you can't pick out any particular component of $dx / d\tau$ and say it's zero. You must allow for the possibility of all of them being nonzero (because there are free particle 4-velocities for which all of them are nonzero).

 

[kent davidge]

Free particles can travel in all directions, and you must meet the conditions for all possible 4-velocities for a free particle. That means you can't pick out any particular component of $dx / d\tau$ and say it's zero. You must allow for the possibility of all of them being nonzero (because there are free particle 4-velocities for which all of them are nonzero).

Ah, I got it. But since the $\Gamma$ are a sum of derivatives of the metric, wouldn't $\Gamma = 0$ only imply that the sum of these terms is zero? (And not necessarily the terms themselves, that is, the derivatives).

 

[PeterDonis]

since the $\Gamma$ are a sum of derivatives of the metric, wouldn't $\Gamma = 0$ only imply that the sum of these terms is zero?

For a single $\Gamma$, yes, that's all you could conclude. But we have that all of the $\Gamma$ are zero.