arestes
Jan21-11, 08:37 AM
Hi guys!
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
ds^2 = -c^2dt^2 + dx^2
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
ds^2 = -\theta_0^2dt^2 + t^2d\theta^2
where t stands for a "radial" coordinate and \theta is the angle in Minkowski spacetime and \theta_0 is a real constant that is there for dimensionality reasons . This way the interval is
\Delta s = \int_{p_1}^{p_2}d\theta \sqrt{-\theta_0^2 t'^2 + t^2}
where t' is the derivative of t wrt \theta. Now, I will use the Euler-Lagrange equations to get the stationary path
\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}
Operating I get (I've checked this many times but please re-check it)
-\frac{t''}{t} + 2\frac{t'^2}{t^2} = \frac{1}{\theta_0^2}
which, by changing variables t \rightarrow 1/y
y'' - \frac{1}{\theta_0^2} y = 0
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
t = t_0 sec[\frac{1}{\theta_0} \theta + \phi]
which means that the differential equation for y should have a PLUS sign in front of 1/\theta_0^2.
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.
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
ds^2 = -c^2dt^2 + dx^2
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
ds^2 = -\theta_0^2dt^2 + t^2d\theta^2
where t stands for a "radial" coordinate and \theta is the angle in Minkowski spacetime and \theta_0 is a real constant that is there for dimensionality reasons . This way the interval is
\Delta s = \int_{p_1}^{p_2}d\theta \sqrt{-\theta_0^2 t'^2 + t^2}
where t' is the derivative of t wrt \theta. Now, I will use the Euler-Lagrange equations to get the stationary path
\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}
Operating I get (I've checked this many times but please re-check it)
-\frac{t''}{t} + 2\frac{t'^2}{t^2} = \frac{1}{\theta_0^2}
which, by changing variables t \rightarrow 1/y
y'' - \frac{1}{\theta_0^2} y = 0
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
t = t_0 sec[\frac{1}{\theta_0} \theta + \phi]
which means that the differential equation for y should have a PLUS sign in front of 1/\theta_0^2.
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.