# I Is this the only form of the Minkowski metric?

#### kent davidge

But "represent well" is an additional assumption. Are you making it or aren't you?
Ah, yes, I am
And if so, what do you mean by it?
By representing well, I have in mind that this coordinate system should give us constant velocity for a free particle.

#### PeterDonis

Mentor
By representing well, I have in mind that this coordinate system should give us constant velocity for a free particle.
Ok. Have you considered what mathematical conditions this implies for the metric coefficients? Hint: consider what the geodesic equation looks like.

#### kent davidge

Have you considered what mathematical conditions this implies for the metric coefficients? Hint: consider what the geodesic equation looks like.
That was one issue I was facing when I started the thread. The equations will be $$\frac{d^2 x^\alpha}{d \tau^2} = 0$$ but this does not imply that the metric is constant or does not have off diagonal components. On the other hand, if we start out assuming that the metric components are constant, then we are lead to the above equation, and the particle is seen with constant velocity. This leaves open the possibility for off diagonal terms, though.

#### kent davidge

Hei, taking a look back at @DrGreg post #2, it seems that 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.

#### PeterDonis

Mentor
The equations will be
$$\frac{d^2 x^\alpha}{d \tau^2} = 0$$

but this does not imply that the metric is constant or does not have off diagonal components.
They do if you put back the crucial terms you left out:

$$\frac{d^2 x^\alpha}{d \tau^2} + \Gamma^\alpha{}_{\beta \delta} \frac{d x^\beta}{d \tau} \frac{d x^\delta}{d \tau} = 0$$

But, you say, those terms in the Christoffel symbols vanish in Minkowski coordinates! Yes, they do, but once more, if all you were concerned with were Minkowski coordinates, there would be no point in this thread! The whole point is that you want to investigate other possibilities for the coordinates, which means you have to include those other terms.

Once you include those other terms, and then impose your "represents well" requirement, which is that $dx^\alpha / d \tau$ must be constant for a geodesic, so that $d^2 x^\alpha / d \tau^2 = 0$, then you can see that we must have

$$\Gamma^\alpha{}_{\beta \delta} \frac{d x^\beta}{d \tau} \frac{d x^\delta}{d \tau} = 0$$

for all possible combinations of indices $\beta$ and $\gamma$. What does that imply about the metric? (Remember that the Christoffel symbols $\Gamma^\alpha{}_{\beta \delta}$ are constructed from derivatives of the metric.)

#### PeterDonis

Mentor
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

Mentor
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

Mentor
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.

"Is this the only form of the Minkowski metric?"

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