0
$\begingroup$

In my Electrodynamics and Electromagnetism course, the professor is deriving Maxwell equations and the electromagnetic field tensor from differential geometry and wants to show how special relativity is built in Maxwell equations. To be honest the mathematical steps aren't that hard to follow but rather the physical meaning behind the terms. for example in three dimensional Minkowski space the line element is:

$$ds^2 = -(cdt)^2 + (dx^1)^2 + (dx^2)^2 + (dx^3)^2$$

you can either use transformation or simply use the chain rule to change coordinates, let's say for a arbitrary moving coordinates, we get:

$ds^2 = g_{\mu \nu}\check{dx}^\mu\check{dx}^\nu $

and lets say there is another stationary coordinates

$ds^2 = -(cdt)^2$

if we take 2 events, let's say the stationary coordinates moved through time for a period, then the distance in space-time (which is a vector) would be

$ds^2 = -(cdt)^2$

which is equal to its proper time (I would also love a an easy definition of proper time).

but then you could say that the moving coordinates agrees that the events happened and the distance between them is

$g_{\mu \nu}\check{dx}^\mu\check{dx}^\nu $

but then what is meaning of the last term? is it the measurement of the space-time distance the stationary coordinates moved with respect to this coordinates? if so then why would it be equal to the proper time of the stationary coordinates?

I don't think I formulated the question properly but this is kind of a new language to me. I appreciate the help, Thanks!

$\endgroup$
0
$\begingroup$

In the full framework of general relativity, you have to disengangle some concepts from Newtonian mechanics.

In particular, the space itself and its tangent space are two different things. In particular, tangent vectors are attached to a particular point, and each point in the spacetime has its own tangent plane. So the displacement between two spacetime points doesn't have a direct meaning, it is only interpretable relative to some curve connecting them. The line element lets you measure the distance along this curve.

In flat space(time), the line element is just the diagonal matrix, and the spacetime has the same structure as the tangent space, so you can cheat on a lot of this, but in a general curved spacetime, this is not the case, and you lose this difference between vector displacement and scalar distance. it's all scalar distances relative to curves. The only curveball you have here is that you have this idea of a "geodesic" which is an extremal distance between the two points, either the minimum or maximum distance you can travel.

$\endgroup$
0
$\begingroup$

Neither; the line element is a scalar.

$\endgroup$
2
  • $\begingroup$ ok then what is the physics behind it? and if its not too much to ask what's the geometry of it in the space time? $\endgroup$ May 19 at 19:05
  • $\begingroup$ It allows you to calculate physical quantities in spacetime, via the metric tensor. But please refer to a book, I am no teacher! $\endgroup$
    – m4r35n357
    May 19 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.