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]

 

[TerryW]

Hi Jim,

Thanks to your prod, I think I have now cracked Ex 25.2 - I've attached a PDF with my solution.

 

[JimWhoKnew]

Now you are on the right track, but not entirely there yet. I have too many comments on your work, so I suggest we'll address one step at the time. Blink twice if you agree.

In the first step (deriving the Hamilton equations from the action), you took a shortcut and got the wrong sign in one equation. See Eq. 21.109, or consult an introductory textbook such as section 8.5 in Goldstein's (3rd ed.), to do it right. You can also google "modified Hamilton's principle".

Note that in the super-Hamiltonian formalism $x^0$ is a dynamical variable, and $\lambda$ takes the role of the curve's parameter (replaces the non-relativistic $t$).

 

[TerryW]

(is that blinking twice?) - I'll follow up what you've said