Shortest distance between two points is a straight line in non-Euclidean metric

[naele]

Homework Statement

Consider the metric in polar coordinates
ds=\frac{2}{1-r^2}\sqrt{dr^2+r^2d\phi^2}
Show that the shortest path from the origin to any other point is a straight line.

Homework Equations

Euler-Lagrange equations
\frac{\partial F}{\partial y} - \frac{d}{dx}\frac{\partial F}{\partial y'}=0
in polar coordinates
\frac{\partial F}{\partial \phi} -\frac{d}{dr}\frac{\partial F}{\partial \phi'}=0

The Attempt at a Solution

By inspection, there's no dependence on \phi, just r,\phi'. Re-writing ds a little bit gives
ds=\frac{2}{1-r^2}\sqrt{1+r^2\phi'^2}dr
The euler-lagrange equations reduce to a first integral where \partial F/\partial \phi'=C is
\frac{2}{1-r^2}\frac{r^2\phi'}{\sqrt{1+r^2\phi'^2}}=C

I think this is the correct way to go about doing it, but I feel like I'm missing something. To show that the path is straight, I need to show that \phi'=0. But the resulting equations I get whether I integrate wrt r first, or differentiate wrt r gives me very messy equations and solving for \phi' is too difficult.

 

[turin]

To show that the path is straight, I need to show that \phi'=0.

This is not generally true. This is the condition for a straight line that passes through the origin. For a general straight line that does not pass through the origin, each point will have a different φ (i.e. φ will generally change if r changes).

EDIT: Sorry. I reread your problem statement and realized that is not the issue. I'll get back with you shortly.

As far as I know, you must integrate dφ/dr for an arbitrary value of C, and then show that Δφ=0 implies C=0 (which in turn implies that φ'=0). By setting r²=x, I think that you can get an integrand of the form

\frac{1-x}{x\sqrt{|1-bx+x^2|}}

which can be split into two pieces and looked up in a table of integrals. I did not do this because I don't have time right now. Anyway, I expect that you will get an expression for Δφ as a monotonic function of r which can only vanish if C=0 (or r=0).

Some points to be careful about: r=1 may be a problem point. Also, the absolute value in the denominator of the integrand may require you to choose one of two different antiderivatives from the table, and this may actually change from the beginning to the end of the integration. You may need to replace r=0 with r=ε for your lower limit, and then take the limit as ε→0.

 

[gabbagabbahey]

The euler-lagrange equations reduce to a first integral where \partial F/\partial \phi'=C is
\frac{2}{1-r^2}\frac{r^2\phi'}{\sqrt{1+r^2\phi'^2}}=C

I think this is the correct way to go about doing it, but I feel like I'm missing something. To show that the path is straight, I need to show that \phi'=0. But the resulting equations I get whether I integrate wrt r first, or differentiate wrt r gives me very messy equations and solving for \phi' is too difficult.

I'm not sure you need to solve this differential equation. According to the problem statement, you are only concerned with solutions that pass through the origin (r=0). There is only one value of C for which those solutions are possible.

 

[naele]

This is not generally true. This is the condition for a straight line that passes through the origin. For a general straight line that does not pass through the origin, each point will have a different φ (i.e. φ will generally change if r changes).

EDIT: Sorry. I reread your problem statement and realized that is not the issue. I'll get back with you shortly.

As far as I know, you must integrate dφ/dr for an arbitrary value of C, and then show that Δφ=0 implies C=0 (which in turn implies that φ'=0). By setting r²=x, I think that you can get an integrand of the form

\frac{1-x}{x\sqrt{|1-bx+x^2|}}

which can be split into two pieces and looked up in a table of integrals. I did not do this because I don't have time right now. Anyway, I expect that you will get an expression for Δφ as a monotonic function of r which can only vanish if C=0 (or r=0).

Some points to be careful about: r=1 may be a problem point. Also, the absolute value in the denominator of the integrand may require you to choose one of two different antiderivatives from the table, and this may actually change from the beginning to the end of the integration. You may need to replace r=0 with r=ε for your lower limit, and then take the limit as ε→0.

Thanks for the reply, I took your suggestion to and letting b=2 then I get an expression that's similar but a little different
\frac{(x-1)c}{\sqrt{b^2x^2-c^2x^3+2c^2x^2-c^2x}}

With a plus and minus version. Unless my algebra is off which is quite possible.

 

[naele]

I'm not sure you need to solve this differential equation. According to the problem statement, you are only concerned with solutions that pass through the origin (r=0). There is only one value of C for which those solutions are possible.

Well if we're only considering the case when r=0 then C must be zero as well. So if that's the case, then \phi'=0 as well. If that's the case then we have a straight line for the shortest path from the origin to any other point.

 

[gabbagabbahey]

Well if we're only considering the case when r=0 then C must be zero as well.

Careful with your wording here...We're not only considering the point r=0, but rather, we're only considering solutions \phi(r) which pass through r=0. At r=0, your DE becomes 0=C, and hence C must be zero for all solutions that pass through the origin.

 

[naele]

Careful with your wording here...We're not only considering the point r=0, but rather, we're only considering solutions \phi(r) which pass through r=0. At r=0, your DE becomes 0=C, and hence C must be zero for all solutions that pass through the origin.

Ah yes, I think I see the distinction. Thanks for the help!

 

[turin]

When the path passes through the origin, isn't there a discontinuity in φ, and therefore φ'→±∞, and therefore the product of r²φ' remains undetermined (because it is "zero times infinity")? Well, to be more rigorous, I guess that I should say that φ' is simply undefined at that point.

 

[gabbagabbahey]

When the path passes through the origin, isn't there a discontinuity in φ, and therefore φ'→±∞, and therefore the product of r²φ' remains undetermined (because it is "zero times infinity")? Well, to be more rigorous, I guess that I should say that φ' is simply undefined at that point.

First, I should point out that I was rather loose with my wording; the curve doesn't "pass through" the origin, rather the origin is one of its endpoints (as per the problem statement). However, this doesn't alleviate the problem. Due to the definition of \phi\equiv\tan^{-1}\left(\frac{y}{x}\right), not only is \phi'(r=0) undefined (as you would expect from an endpoint), but \phi itself is also undefined at the origin.

I suspect that the questioner intends the student to fudge over this unfortunate problem (due entirely to a poor choice of coordinate systems), and simply pretend that

\phi(r=0)=\lim_{r\to 0^{+}}\phi(r)

and that \phi'(r) (or at least r^2\phi'(r)) goes to zero at r=0.

I'm sure that there is a mathematically rigorous solution to the problem if the student is concerned about this, but without having put a whole lot of thought into this yet, I can only guess it would have to involve switching to Cartesian coordinates. (I'm not sure that any other method will solve the problem of \phi(r=0) being undefined, as IIRC \phi(r) being defined on a closed interval including the endpoints is a necessary condition for the Euler-Lagrange method to apply)