# Deriving the metric

1. Sep 27, 2014

### PhyAmateur

Hello,

I have a simple question about deriving $$g_{ij} \frac{\partial x^i}{\partial t}\frac{\partial x^j}{\partial t}$$
with respect to time t.

I have noticed that the first term after derivation turns out to be$$\frac{\partial g_{ij}}{\partial x^k} \frac{\partial x^k}{\partial t}\frac{\partial x^i}{\partial t}\frac{\partial x^j}{\partial t}$$

Does $$g_{ij}$$ depend on time? If so, why can't we just write $$\frac{\partial g_{ij}}{\partial t}$$ Why did we plug in the $$x^k$$?
Thanks!

Last edited: Sep 27, 2014
2. Sep 27, 2014

### Orodruin

Staff Emeritus
Generally $g_{ij}$ is a function of all space-time coordinates. Along a world line, the $t$ coordinate is changing just as well as the spatial coordinates. It thus makes a difference if you are differentiating $g$ along a world line or just with respect to the time coordinate. The existence of the coordinate derivatives seem to indicate that you do want to differentiate with respect to a world line in which case the total derivative follows the chain rule:
$$\frac{d}{dt} = \frac{dx^i}{dt}\frac{\partial}{\partial x^i} = \frac{\partial}{\partial t} + \vec v \cdot \vec\nabla.$$

3. Sep 27, 2014

### PhyAmateur

Thank you very much for your reply. You mean that what was supposed to be written is $$\tau$$ instead of t? And thus in this case, I should use the chain rule?

4. Sep 27, 2014

### Orodruin

Staff Emeritus
Well, you can also parameterise the world line with the coordinate $t$ (assuming that there is only one point on the world line for every $t$). It may be that that is simply what is assumed about the parameterisation.

5. Sep 27, 2014

### stevendaryl

Staff Emeritus
There are several different types of derivatives used in relativity, and I think your question might be mixing up two different types. If you have a field, which is a quantity (real number, or a vector or a tensor) defined at every point in spacetime, you use partial derivatives or covariant derivatives. If you have a spacetime path $\mathcal{P}(s)$ and you have some quantity defined at each point along the path (where s is the path parameter), then you don't use partial derivatives, since the quantity only depends on one parameter, $s$. You use a total derivative, $\frac{d}{ds}$. In nonrelativistic physics, the path variable is typically the time $t$, but in relativistic physics, it is often proper time $\tau$ instead.

You write $\frac{\partial x^i}{\partial t}$. But what is $x^i$? If you mean that it is a coordinate, then what does it mean to take a partial derivative with respect to $t$? Partial derivatives only make sense for fields--quantities defined at each point in spacetime.

What I'm guessing that you mean is that you have a path parametrized by $t$ (Do you really mean the time $t$, or do you mean the proper time $\tau$?). So $x^i(t)$ is the coordinate describing the location along the path at time $t$. In that case, you don't actually want a partial derivative, you want an ordinary derivative, $\frac{dx^i}{dt}$.

The other question about what you wrote is what the index $i$ or $j$ means in $x^i$ and $g_{ij}$. In the mathematics of relativity, people usually combine spatial coordinates and the time coordinate $t$ into $x^\mu$, where, $\mu$ can be 0, 1, 2, or 3, and where $(x^0, x^1, x^2, x^3) = (t, x, y, z)$.

So a more typical quantity that you might encounter in relativity is

$g_{\mu \nu} \frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}$

This expression involves both fields ($g_{\mu \nu}$ is a tensor field) and path quantities ($x^\mu(\tau)$ is a function of a path variable, the proper time $\tau$). It doesn't actually make sense to take a derivative of this quantity unless you make sense of it as a path quantity, or as a field. The way to make sense of it as a path quantity is to understand that $g_{\mu \nu}$ is supposed evaluated at the point $x^\mu(\tau)$. So you really have:

$g_{\mu \nu}(x^\lambda(\tau)) \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}$

Now, everything is a function of $\tau$ and we can take a derivative with respect to $\tau$. To take a derivative of $g_{\mu\nu}(x^\lambda(\tau))$ with respect to $\tau$, you have to use the chain rule:

$\frac{d}{d\tau} g_{\mu\nu}(x^\lambda(\tau)) = (\frac{\partial}{\partial x^\lambda} g_{\mu \nu}) \frac{d x^\lambda}{d\tau}$

Last edited: Sep 27, 2014
6. Sep 27, 2014

### PhyAmateur

Thank you very much for making this clearer!