Are Derivatives of the Metric Different in Flat Spacetime?

[etotheipi]

The general metric is a function of the coordinates in the spacetime, i.e. $g = g(x^0, x^1,\dots,x^{n-1})$. That means that in the most general case we can't simplify an expression like $\partial g_{\mu \nu} / \partial x^{\sigma}$. But, what about the special case of the flat spacetime metric$$\frac{\partial \eta_{\mu \nu}}{\partial x^{\sigma}} = \dots \,?$$can we simplify that? I thought it might be zero [since it is coordinate independent], but it is perhaps not the case. Also, I was under the impression that$$\frac{\partial x^{\rho}}{\partial x^{\sigma}} = \frac{\partial x_{\sigma}}{\partial x_{\rho}} = \delta^{\rho}_{\sigma}$$is this correct? Furthermore, in order to find what $$\frac{\partial x^{\rho}}{\partial x_{\sigma}} \quad \text{and} \quad \frac{\partial x_{\rho}}{\partial x^{\sigma}}$$are, it will be necessary to understand the answer to $\partial {\eta}_{\mu \nu} / \partial x^{\sigma}$. Thanks!

 

[PeterDonis]

I thought it might be zero, but it is perhaps not the case.

It is zero in the coordinates you chose (the ones in which the metric tensor is $\eta_{ab}$).

Not all of the derivatives would be zero in other coordinates (e.g., spherical).

As far as figuring out what those derivatives are in general, you seem to be making it a lot harder than it needs to be. Each of the ten independent metric components is a function of the coordinates. So you just take the derivatives of those ten functions with respect to each of the coordinates. For the special case of $\eta_{ab}$, all ten components happen to be constants, i.e., constant functions of all four coordinates. (Four of the constants are $\pm 1$ and the other six are $0$.)

 

[etotheipi]

Thanks! The coordinates are also taken to be functions of $\tau$ only, i.e. $x^{\mu} = x^{\mu}(\tau)$. Then, since $u^{\mu} = u^{\mu}(\tau)$, the 4-velocity should have no explicit functional dependence on the coordinates, and we should have for instance$$\partial^a u_b = 0$$I'm struggling to understand why this isn't the case [e.g. for example, in the Lagrangian formulation the position and velocity are taken to be independent variables]

 

[PeterDonis]

The coordinates are also taken to be functions of $\tau$ only

This only applies to a parameterized curve. If you are just looking at the metric tensor in general as a function of the coordinates, there is no $\tau$.

the 4-velocity should have no explicit functional dependence on the coordinates

If you are only looking at one particular timelike worldline, yes. But that's not sufficient for evaluating the derivatives of the metric with respect to the coordinates; for that you need an open neighborhood.

I'm struggling to understand why this isn't the case

Because $\partial^a u_b$ is not an equation that is even meaningful on a single timelike worldline; you need an open neighborhood. See above.

 

[PeterDonis]

that's not sufficient for evaluating the derivatives of the metric with respect to the coordinates; for that you need an open neighborhood.

I suggest reading the first section of the following Insights article:

https://www.physicsforums.com/insights/precession-in-special-and-general-relativity/

Note the distinction it draws between "fields" and "particle properties"; the former are defined "everywhere" (by which the article really means "in an open neighborhood on which we have defined some coordinate chart"), the latter are defined on a particular curve (the worldline of the particle whose properties, such as 4-velocity, we are investigating). The partial derivatives of the metric (and the 4-velocity) with respect to the coordinates are things of the first type, not the second. (Note in particular what the second paragraph of the article says about them.)

 

[etotheipi]

Ahh, okay, I think I see what you're getting at! Cool, I'll have a read of the article. Thanks for being patient, I'll get there eventually (maybe )

 

[PeterDonis]

Moderator's note: Thread level changed to "I".