Recently I've been trying to learn General and Special Relativity by myself. There is an specific thing I do not understand perfectly, proper time in the metric of the space-time.

Take the case of an empty space-time:

$$-c^2 \mathrm d \tau^2 = -c^2 \mathrm d t^2 + \mathrm d x^2 + \mathrm d y^2 + \mathrm d z^2$$ where $\tau$ is the proper time of an object.

I don't understand when $c^2 \mathrm d \tau^2$ is used as $ds^2$. Why is this possible? It is related to the worldline traced by an object? Could you calculate the line integral in order to find the length of the path the object? This line integral would give the path length of a geodesic?

Could you get the speed of an object by doing this (correct me if I'm wrong): $$-c^2 \mathrm d \tau^2 = -c^2 \mathrm d t^2 + \mathrm d x^2 + \mathrm d y^2 + \mathrm d z^2 \rightarrow \\ -c^2 \left( \frac{\mathrm d \tau}{\mathrm d t} \right)^2 = -c^2 + \left( \frac{\mathrm d x}{\mathrm d t} \right)^2 + \left( \frac{\mathrm d y}{\mathrm d t} \right)^2 + \left( \frac{\mathrm d z}{\mathrm d t} \right)^2 $$ If you assume that: $\dot{x}^2 + \dot{y}^2 + \dot{z}^2 = ||v||^2$, then by rearranging a little bit the equation, you could get the velocity of an object.


3 Answers 3


In relativity, notions like "space" and "time" depend on what coordinate system you use. However, there are some physical quantities which all observers will agree on. Say there is a clock which travels along some trajectory, from one point $x_i$ in spacetime to another point $x_f$. (Points in spacetime have an invariant meaning, but the way we label them will depend on the coordinate system we use, i.e. the spacetime point $x$ could be written in some coordinates as $(x^0, \vec{x})$ or in some other coordinates as $({x^0}', \vec{x}')$.)

All observers will agree on what time the clock reads when it starts its journey and what time it reads when it ends its journey. The time elapsed will be given by the proper time integral from the start point to the end point $$ \Delta \tau = \int_{x_i}^{x_f} d \tau. $$ This is what the spacetime metric gives you: an invariant measure of how much proper time passes when a clock travels along a path.


$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$

If the result is positive, the equation corresponds to a spacelike worldline segment. It is always possible to find a frame, so that $dt = 0$ and two events linked by the segment are simultaneous for this frame, happening at different spatial locations.

If the result is negative, it is a timelike segment. It is always possible to find a frame, so that $dx^2 + dy^2 + dz^2 = 0$ and two events linked by the segment happen in the same place for this frame but at different times.

In this last situation, the quantity $dt^2 = d\tau^2$, is the proper time. The time of a clock at rest in the frame.


To your question in the last paragraph: yes. You lost a factor of -1 somewhere, though. Once you put it back in, try solving for $d\tau$ and simplifying. The right hand side will be $\gamma ^{-1}dt$, the Lorentz transformation for time.


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