A particle with mass $m$ and velocity $\vec{V}$ has a kinetic energy $T=\frac{1}{2}mV^2$.

In an Euclidean space with Cartesian coordinates using Pythagoras theorem I can write the following $$T=\frac{1}{2}mV_{x}^{2}+\frac{1}{2}mV_{y}^{2}+\frac{1}{2}mV_{z}^{2}=T_x+T_y+T_z$$ My understanding/interpretation for this expression is that although kinetic energy is not a vector quantity, based on the magnitude of velocity in different directions the kinetic energy is kind of "partitioned" between orthogonal directions of space and total kinetic energy is sum of all.

Now my question is, what would happen in a non-Euclidean (or in a non-orthogonal or a curved) space where Pythagoras theorem (as we know it) cannot be written? What happens, in particular, if there is a "space" where a Pythagoras like theorem, say, $$a^{2}=b^{2}+c^{2}-d^{2}$$ exists?

  • 1
    $\begingroup$ What is your precise definition of non-Euclidean space? $\endgroup$ – Qmechanic Dec 15 '17 at 12:06
  • $\begingroup$ where Euclid's fifth postulate does not hold or maybe where Pythagoras theorem does not hold $\endgroup$ – physicopath Dec 15 '17 at 12:09
  • $\begingroup$ A Riemannian manifold $(M,g)$ has positive (non-relativistic) kinetic energy $T=\frac{1}{2}g_{ij}\dot{x}^i\dot{x}^j$ if we identify metric and mass tensor. $\endgroup$ – Qmechanic Dec 15 '17 at 12:15
  • $\begingroup$ That is awesome. But I am more interested in what is the correct interpretation of $T=T_x+T_y+T_z$ and whether is it possible to have a "manifold" where such a Pythagoras like theorem exists that one can write say, $T=T_x+T_y-T_z$ $\endgroup$ – physicopath Dec 15 '17 at 12:22
  • $\begingroup$ Then look up pseudo-Riemannian manifolds. $\endgroup$ – Qmechanic Dec 15 '17 at 12:33

Pythagoras' equation in special relativity has exactly the form you describe:

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

This equation is called the metric (in this case the Minkowski metric) and the length $ds$ being calculated is called the proper length. The catch is of course that the dimension with the negative sign is the time dimension not one of the spatial dimensions.

And this is the problem with your idea. Any dimension that appears in the metric with a negative sign is going to be a timelike dimension not a spatial one, so you are never going to be able to have a spatial three vector with a length squared less than zero.

  • $\begingroup$ Thanks for the answer +1. So what you are saying is that because there is no negative length, there cannot be a metric with a $-d^2$ like term. Am I right? And also do you want to say something about how to interpret $T=T_x+T_y+T_z$ equation? $\endgroup$ – physicopath Dec 15 '17 at 12:44

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.