Choice of action to geodesic equations

The second choice is equivalent to the first if you reparameterise your curve such that the curve parameter is an affine parameter (i.e., essentially you are taking the curve length as the parameter). This is of course a choice you can always make and it will not change the curve that results.

 

The two approaches are not exactly the same. If you minimize the first action integral, you obtain the equations of motion:

[itex]\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau} + \frac{dx^\mu}{d\tau} \frac{d}{d\tau} (\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}}) = 0[/itex]

where [itex]\Gamma^\mu_{\nu \lambda}[/itex] is a combination of derivatives of [itex]g_{\mu \nu}[/itex].

The second action integral leads instead to:

[itex]\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau} = 0[/itex]

They are only the same if
[itex]\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}} = 0[/itex]

which means

[itex]\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}} = K[/itex]

for some constant [itex]K[/itex], which in turn implies:

[itex]d\tau = \frac{1}{K} \sqrt{g_{\nu \lambda} dx^\nu dx^\lambda}[/itex]

which means that [itex]\tau[/itex] is linearly related to proper time.

The first integral works for any parameter [itex]\tau[/itex], while the second only works when [itex]\tau[/itex] is proper time.

 

Just to nitpick, I suspect you intended to write:
$$
\frac d{d\tau}\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}} = 0
$$

 

Yes.

 

The relation between proper time and the space-time interval is the same as the relation between curve length and the space interval in Riemann geometry. In Riemann geometry, you could always pick the length along the curve as your curve parameter. In the same way, you can here pick the proper time along the world line as your curve parameter.

 

It's the first integral where you have to make the assumption that [itex]d\tau = dS[/itex] in order to get rid of the term

[itex]\frac{d x^\mu}{d\tau} \frac{d}{d\tau} (\frac{1}{\sqrt{g_{\nu \lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau}}})[/itex]

Using the Lagrangian equations of motion, with [itex]L = \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}[/itex], you find

[itex]\frac{\partial L}{\partial (\frac{dx^\mu}{d\tau})} = \frac{g_{\mu \nu} \frac{dx^\nu}{d\tau}}{\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}}[/itex]

Now, take [itex]\frac{d}{d\tau}[/itex] and you get two terms:
[itex]\frac{d}{d\tau} \frac{\partial L}{\partial (\frac{dx^\mu}{d\tau})} = \frac{\frac{d}{d\tau} (g_{\mu \nu} \frac{dx^\nu}{d\tau})}{\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}} + g_{\mu \nu} \frac{dx^\nu}{d\tau} \frac{d}{d\tau} (\frac{1}{\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}})[/itex]
The first term leads to the usual geodesic equation (once you multply through by [itex]g^{\mu \mu'} \sqrt{g_{\mu'' \nu}\frac{dx^{\mu''}}{d\tau} \frac{dx^\nu}{d\tau}}[/itex]). The second term is an additional term not found in the geodesic equation that you derive from the second action integral.

In the first integral, [itex]\tau[/itex] is not necessarily proper time. It is any parameter whatsoever that increases continuously along the path. The first integral is invariant under parameter changes.

 

Except that in the first integral, it isn't necessary to pick [itex]\tau[/itex] to be proper time.

 

Yes, and that allows you to throw away any derivatives of that expression (since they'll be zero).

 

Which is why I used "can" and not "must". :)
It is the same in Riemann geometry, you can, but must not, use the curve length as the curve parameter. For example, I could parameterise the line ##x = y## as ##x = t^3## and ##y = t^3##.

 

Just getting rid of the square root is wrong, as other posters have mentioned. However, there is a cute trick, known as an einbein, that allows one to accomplish the same end legitimately.

See for instance http://arxiv.org/pdf/hep-ph/9708319v1.pdf

Maybe more later if I have time.

 

##d\tau## in the first integral is a priori only a curve parameter and does not have to be an affine parameter for the proper time. It is only after setting ##d\tau = dS## that you can interpret it as proper time.

 

Yes. You can always make such a reparameterisation. It is a matter of solving the differential equation ##dS/d\tau = 1##.

 

I'd like to point out though, that usually the notation ##d\tau## is for a proper time. For a non-proper time parametrization one uses usually ##dt##. People will be confused if you use ##d\tau## to mean an arbitrary curve parametrization.