# 4-velocity of an observed object in GRT

1. Nov 15, 2014

### birulami

In general relativity we have $c^2ds^2 = c^2dt^2 - dx^2 -dy^2-dz^2$. From this we can derive the not commonly used equation:
$$(c\frac{ds}{dt})^2 + (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2 =c^2 \qquad\qquad\qquad (1)$$ I think this is the "velocity" squared of an object relative to an observer who's own proper time is $t$. I find this equation slightly remarkable, because of the constancy of this relative velocity which forces a tradeoff between speed through space and speed through time $ds/dt$: the slower you move the faster you get older ;).

Now I wonder what the same would be for general relativity. The respective equation for $ds$ is $ds^2 = g_{\mu\nu} dx^\mu dx^\nu$. With $i,j\in\{1,2,3\}$ and $t:=x^0$ I can expand the sum on the right a bit more explicit to get:
$$ds^2= g_{ij} dx^i dx^j + 2 g_{0j} dt dx^j + g_{00}dt^2$$ There I have the $dt^2$ again and could just divide by it, where it not for the mixed term. Is it possible to "simplify" this equation more into the direction of (1) such that all the $dt$ are in the denominator? Under which circumstances would the $g_{0j}$ be zero such the cross terms with $dt$ disappear gracefully?

2. Nov 15, 2014

### DrGreg

You can write it as$$\left( \frac{ds}{dt} \right)^2= g_{ij} \frac{dx^i}{dt} \frac{dx^j}{dt} + 2 g_{0j} \frac{dx^j}{dt} + g_{00}$$You have a quadratic form instead of a sum of squares.
When the 0th coordinate is orthogonal to all the other coordinates. This means that for any given direction the one-way coordinate speed of light equals the two-way coordinate speed of light.

3. Nov 15, 2014

### Staff: Mentor

But this is frame-dependent, because the time $t$ is coordinate time; you can change the tradeoff between your "speed through time" and your speed through space just by picking different coordinates. This will create problems if you try to read too much physical meaning into this "speed through time".