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Why is the proper distance written like this? - General Relativity

Physics Asked on August 1, 2021

I came across the following equation for proper distance:

$$ds^{2} = -c^{2}dt^{2} + [dx – v(t)f(r(t))dt]^{2} + dy^{2} + dz^{2}$$

where the function $v(t)$ tells you the velocity of the warp spaceship, $r$ is the radius function from the ship to the warp bubble shell, and $f$ is the shape of the warp bubble. Similar: https://www.sfu.ca/~adebened/funstuff/warpdrive.html (please scroll down to the first equation since it is similar to the one above).

I want to understand why the 2nd term is written in this form: $[dx – dt]^{2}$ mathematically speaking and how it relates to General Relativity.

Is it because there is a nonzero value in the metric that is not a part of the main diagonal or the trace of the metric? For example:

$$g_{00}, g_{11}, g_{22}, g_{33}, g_{31}$$

are nonzero while all other entries are zero.

If this is the case, can you please provide some examples of how to write the proper distance when there are nonzero values in the metric that are not in the trace?

Please excuse me if I am asking a basic/simple question, I am only a beginner at General Relativity.

One Answer

The line element equation (the equation for $ds^2$ in terms of coordinate changes) and the metric are just two ways of talking about the same thing. For the line element is $$ ds^2 = g_{mu nu} dx^mu dx^nu $$ so to write $ds^2$ in terms of $dx^a$ IS to write the metric; the coefficients are the elements of the metric. In the example you furnished (after two edits) the metric under consideration is $$ [g_{ab}] = left( begin{array}{cccc} v^2 f^2 - c^2 & -v f ;& 0; & 0 -v f & 1 & 0 & 0 0 & 0 & 1 & 0 0 & 0 & 0 & 1 end{array} right) $$ If this metric is connected with a discussion of warp drives etc. then there is a strong possibility that it cannot occur in nature, but I have not checked that. You have to beware, in this area, that sometimes people announce results that are based on the assumption that something could have negative mass, or something like that. The assumption may be hidden however, and one way to hide it is to propose a metric. But a metric implies a configuration of matter, and if the configuration of matter has negative mass then basically you can forget it. An assumption of negative mass is questionable even as a tentative hypothesis, since it leads to other considerations that the proposer might not have thought about, such as that spacetime then becomes unstable. Unless consideration is given to that, the discussion fails to have any useful contribution to make to understanding the actual universe---the one that science is concerned with.

You might want to note also that when the metric is not diagonal it requires careful interpretation. Often the coordinate labelled '$t$' does not represent time in any straightforward sense.

Correct answer by Andrew Steane on August 1, 2021

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