Path that extremizes interval computed with polar coordinates metric in flat space

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  • #1
arestes
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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

[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.
 

Answers and Replies

  • #2
Mentz114
5,432
292


I don't understand your metric.

Going from Cartesian to polar coords, you could work in the (t,r) plane instead of (t,x), in which case the metric is
[tex]
ds^2=-c^2dt^2+dr^2
[/tex]

The surface of a sphere of radius R is

[tex]
ds^2=-c^2dt^2+R^2d\theta^2+R^2\sin(\theta)^2d\phi^2
[/tex]

and you need two spatial dimensions. The 'straightest' path is a segment of a great circle connecting two points.
 
Last edited:
  • #3
K^2
Science Advisor
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When people say polar coordinates in flat space-time, they usually mean the following metric:

[tex]ds^2 = c^2dt^2 - dr^2 - r^2d\theta^2 - r^2sin^2(\theta)d\phi^2[/tex]

I've never seen anyone try and include time in polar coordinates. There is a hyper-spherical metric where there is a boost hyper-angle corresponding to Lorentz boost, but it still looks very different from what you're defining.
 
  • #4
Dale
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Insights Author
2021 Award
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10,229


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

[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 .
What is your coordinate transformation to your "polar coordinates"? Using the standard polar coordinate transformation:
[tex]x=r \; cos(\theta)[/tex]
[tex]t=r \; sin(\theta)[/tex]
I am getting a metric which is much more complicated than the one you posted, even setting c=1. It involves cross terms that do not cancel out:
[tex]d\theta^2 r^2 \cos (2 \theta )+dr^2 (-\cos (2 \theta ))+2
d\theta dr r \sin (2 \theta )[/tex]
 
Last edited:
  • #5
arestes
80
3


Hi guys! I just figured out my mistake... It was in the very metric you all said was wrong. To make it clear, I didn't want to work on the t-r plane. Also, I shouldn't have used "t" as the radial coordinate because it is not time as in cartesian coordinates. I didn't mean to work with time in polar coordinates, it gets mixed up with the spatial coordinate.

In any case, I get the same metric as DaleSpam with a minus sign. Using
[tex]
x=r \; cos(\theta)
[/tex]

[tex]
ct=r \; sin(\theta)
[/tex]
I get
[tex]
ds^2 = -d\theta^2 r^2 \cos (2 \theta )+dr^2 (\cos (2 \theta ))- 2d\theta dr r \sin (2 \theta )
[/tex]
and from here I get the desired result, namely-

[tex]
r = r_0 sec\left[\theta + \phi_0\right]
[/tex]
which describes a straight line in Minkowski spacetime.
I just did this for fun, since I knew that changing coordinates should give me the same result as using the standard Minkowski metric.

thanks!
 

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