- #1

arestes

- 80

- 3

I was reviewing some basic stuff in Special Relativity, specifically the part where it can be proven that a straight line connecting two events is the path that maximizes the interval between these two events. The proof is easy using the metric with cartesian coordinates

[tex] ds^2 = -c^2dt^2 + dx^2 [/tex]

but I should get the same using polar coordinates (I'm working with two dimensions for simplicity). In that case I should work with the interval element

[tex] ds^2 = -\theta_0^2dt^2 + t^2d\theta^2 [/tex]

where t stands for a "radial" coordinate and [tex]\theta[/tex] is the angle in Minkowski spacetime and [tex] \theta_0[/tex] is a real constant that is there for dimensionality reasons . This way the interval is

[tex] \Delta s = \int_{p_1}^{p_2}d\theta \sqrt{-\theta_0^2 t'^2 + t^2} [/tex]

where [tex]t'[/tex] is the derivative of t wrt [tex]\theta[/tex]. Now, I will use the Euler-Lagrange equations to get the stationary path

[tex]\frac{d}{d\theta} \frac{\partial \sqrt{-\theta_0^2 t'^2 + t^2}}{\partial t'} = \frac{\partial \sqrt{-\theta_0^2 t'^2 + t^2}}{\partial t} [/tex]

Operating I get (I've checked this many times but please re-check it)

[tex] -\frac{t''}{t} + 2\frac{t'^2}{t^2} = \frac{1}{\theta_0^2} [/tex]

which, by changing variables [tex] t \rightarrow 1/y [/tex]

[tex] y'' - \frac{1}{\theta_0^2} y = 0 [/tex]

which has as solutions hyperbolic sines and cosines or, equivalently, a multiple of a hyperbolic cosine with a phase. This should be wrong because what I expect is to get a straight line in Minkowski spacetime as solution. A straight line in polar coordinates is something of the form

[tex] t = t_0 sec[\frac{1}{\theta_0} \theta + \phi] [/tex]

which means that the differential equation for y should have a PLUS sign in front of [tex]1/\theta_0^2[/tex].

What is wrong here?

Also, I would like to know as to what the best way to see whether the stationary function obtained by the E-L equations is a maximum or a minimum (or just a "saddle point") is.