Problem with dimensionless quantities in an equation requiring M^2 (Super Hamiltonian Formulation for Geodesic Motion)

[TerryW]

Homework Statement

Super Hamiltonian Formulation for Geodesic Motion (MTW Ex 25.2)

Relevant Equations

What do the choices $\mathcal H = -½, 0, -½\mu^2, +½$ mean for the geodesic and the choice of parameterisation?

I've worked my way through this exercise but I am a bit puzzled by the last line "What do the choices $\mathcal H = -½, 0, -½\mu^2, +½$ mean for the geodesic and the choice of parameterisation?"

I've worked to produce:

$\big (\frac {dr}{d\lambda} \big ) ^2 = E^2 - \big (1- \frac {2M}{r^2} \big ) \big ( \frac { L^2}{r^2} - 2\mathcal H \big )$ or

$\big (\frac {dr}{d\tau} \big ) ^2 = \tilde E^2 - \big (1- \frac {2M}{r^2} \big ) \big ( \frac { \tilde L^2}{r^2} - 2\frac{\mathcal H}{\mu^2} \big )$


Box 25.6 is helpful for clarifying what this means for $\mathcal H = -½ \mu^2$ and by following a similar logic, I think I am OK with $\mathcal H = 0$ (Hyperbolic path if $\tilde E^2 < \big (1- \frac {2M}{r^2} \big ) \big ( \frac {\tilde L^2}{r^2} \big )$ or falling into the black hole if $\tilde E^2 > \big (1- \frac {2M}{r^2} \big ) \big ( \frac {\tilde L^2}{r^2} \big )$

But I can't figure out what is going on when $\mathcal H = +½ or -½$.

When $\mathcal H = -2\mu^2$, it has the dimensions of [Mass]^2 . +½ and -½ are dimensionless.

How does this work with these equations?

Can anyone shed any light on this?


Cheers

 

[JimWhoKnew]

To understand the role of the signs, take a look at Eq. 25.10. For example, if
$$\mathcal H \equiv \frac 1 2 g^{\mu \nu}p_\mu p_\nu=0$$ what does it tell us about the vector $p^\mu$ ?

The reason for the ellimination of $\mu^2$ by reparametrization, is described in section 25.3, above and below Eqs. 25.15.

BTW, the exercise expects you to derive the affinely parametrized geodesic equations via Hamilton equations. I assume you've done that.

 

[TerryW]

Hi JimWhoKnew,

I've worked out why the values of $\mathcal H$ given are not all dimensionless!

Basically, $\big (\frac {dr}{d\tau} \big ) ^2 = \tilde E^2 - \big (1- \frac {2M}{r^2} \big ) \big ( \frac { \tilde L^2}{r^2} - 2\frac{\mathcal H}{\mu^2} \big )$ produces $\big (\frac {dr}{d\lambda} \big ) ^2 = E^2 - \big (1- \frac {2M}{r^2} \big ) \big ( \frac { L^2}{r^2} - 2\mathcal H \big )$ if you multiply throughout by $\mu ^2$, so I don't have an issue with dimensions any more.


Regards


TerryW

 

[JimWhoKnew]

Hi JimWhoKnew,

I've worked out why the values of $\mathcal H$ given are not all dimensionless!

Basically, $\big (\frac {dr}{d\tau} \big ) ^2 = \tilde E^2 - \big (1- \frac {2M}{r^2} \big ) \big ( \frac { \tilde L^2}{r^2} - 2\frac{\mathcal H}{\mu^2} \big )$ produces $\big (\frac {dr}{d\lambda} \big ) ^2 = E^2 - \big (1- \frac {2M}{r^2} \big ) \big ( \frac { L^2}{r^2} - 2\mathcal H \big )$ if you multiply throughout by $\mu ^2$, so I don't have an issue with dimensions any more.


Regards


TerryW

Yes. This is the difference between choosing $\mathcal H = -1/2$ and $\mathcal H=-\mu^2/2$ . Note that by rescaling ($\lambda'=a \lambda+b$) you can set $\mathcal H$ to be any arbitrary value you wish (with same sign), but these 2 options are more useful than others.

I suspect you don't approach exercise 5.2 properly. The exercise is about a general metric, not necessarily the Schwarzschild solution. Box 13.3 shows how to derive the geodesic equation by applying the Euler-Lagrange equations to a Lagrangian. In exercise 5.2 you are expected to do something similar: derive the geodesic equation by applying the Hamilton equations to the given super-Hamiltonian $\mathcal H$. Then you are asked to show that $\mathcal H$ is a constant of motion. Timelike, spacelike and null geodesics are not the same. How does this difference appear in $\mathcal H$?

 

[TerryW]

You are absolutely right. I haven't gone about this the right way. I'll have a go at doing what you suggest and see how it goes. I'll get back to you either to tell you that I've sorted it or maybe ask for a few more hints if things don't become clear. I may not be troubling you for a little while as it is now school hols and quite a bit of my time will be taken up entertaining my grandson (who unfortunately is not yet old enough to be helping me out with this).

Regards


TerryW

 

[JimWhoKnew]