# The meaning and use of $dx^\mu$ in the metric of General Relativity

Inspired by this answer, I start toying with the general equation of the metric tensor $$ds^2 = g_{\mu\nu}dx^\mu dx^\nu .$$

Let $$g$$ be diagonal, i.e. $$g_{\mu\nu}=0$$ for $$\mu\neq\nu$$ and let $$x^0=ct$$. For a light ray, where $$ds=0$$, we get after dividing by $$dt^2$$ $$\begin{equation} 0 = c^2\cdot g_{00} + \sum_{i=1}^{3} g_{ii}(\frac{dx^i}{dt})^2 .\tag{1} \end{equation}$$

According to the cited answer, if the light ray goes along $$x^1$$ alone we get the velocity of light as $$\begin{array} . \frac{dx^1}{dt} &= c\sqrt{-\frac{g_{00}}{g_{11}}}, \\ \frac{dx^2}{dt} &= 0\\ \frac{dx^3}{dt} &= 0 \end{array}$$

If we now let the light ray go along $$x^2$$, the values of the $$\frac{dx^i}{dt}$$ change for all $$i$$.

This is what I don't understand: are the $$dx^\mu$$, which are rather $$dx^\mu(x^0,x^1,x^2,x^3)$$, not some property of the spacetime point $$(x^0,x^1,x^2,x^3)$$? How can $$dx^\mu/dx^\nu$$ then depend on the direction of a light ray passing through that point. I guess that there is some step missing in the derivation that makes the direction of the light ray explicit. How is this done? Should the direction somehow be included in equation (1)?

• I'm not exactly sure what you are asking. Partly because $dx^\mu$ can mean different things to different authors, and event to the same authors depending on the context (and the mood they are in). But for introductory purposes, think of them as the typical infinitesimal displacements in spacetime. I like to think of $\vec{c}$ as a 3-vector whose magnitude is is the speed of light, and whose direction is that of the light propagation. In that case $\vec{c}=\frac{dx^i}{dt}e_i$. BTW, do you understand the distinction between relative time and proper time? Dec 23, 2019 at 17:09
• This doesn't seem like a question about relativity. It seems like a question about notation and multivariable calculus.
– user4552
Dec 23, 2019 at 18:36

The $$dx^\mu$$ are not, in fact, properties of the spacetime point $$x^\mu$$. The "$$d$$" in "$$dx$$" means "small change in." There is no change, large or small, in a single point. The equation $$ds^2 = g_{\mu\nu}x^\mu x^\nu$$ is not about one point in spacetime. It is about the interval between two (differentially) different points.
Which two points? In your example, two points that are connected by the light ray. If the light ray is traveling in the $$x^1$$ direction, then the two points will be different in the $$x^1$$ coordinate and therefore $$dx^1\ne 0$$. On the other hand, these same two points are sure to have equal values of $$x^2$$ and $$x^3$$, and therefore $$dx^2=dx^3=0$$.