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

 

[TerryW]

HI Jim,

I've had a look at the derivation of the Hamiltonian equations and, following 21.109, I now have the second equation as:
$\frac {dp_\mu}{d \lambda} + \frac {\partial \mathcal H}{\partial x^{\mu}} = 0$.

The problem that I am now wresting with is that with the change of sign, my workings to get the equation of the geodesic is now going wrong and I lose the all important $\frac {d^2 x^{\alpha}}{d \lambda ^2}$ term!

I'll continue working on this to find the mistake.

If, in the meantime, you have a couple more comments on my workings, pass them on.

Regards


TerryW

 

[JimWhoKnew]

OK. Let's summarize what we've got so far:
$$ \frac{dx^\mu}{d \lambda} = \frac{\partial \mathcal H}{\partial p_{\mu}} $$ $$ \frac{dp_\mu}{d \lambda} = -\frac{\partial \mathcal H}{\partial x^{\mu}} $$ These are Hamilton's equations of motion. Together with the given $$\mathcal H \equiv \frac 1 2 g^{\mu \nu}p_\mu p_\nu$$ they are all you need for this exercise (don't go back to Euler-Lagrange!).

We said one step at the time, so now you are asked to show that $\mathcal H$ is a constant of motion.

Spoiler: Hint

For a physical quantity $A$ which is defined on a curve $\gamma(\lambda)=\left\{x^\mu(\lambda)\right\}$ : $$\dot{A}\equiv \frac{d A}{d\lambda} = \frac{\partial A}{\partial x^\mu} \dot{x} ^\mu + \frac{\partial A}{\partial p_\mu} \dot{p} _\mu + \frac{\partial A}{\partial \lambda} $$

For the step after (i.e. derivation of the geodesic equations), note that $\{x^\mu , p_\nu\}$ are the independent variables, and that the $p^\sigma$ are functions of them. You've already found $\frac{d x^\mu}{d\lambda}$ before. So what is $\frac{d^2 x^\mu}{d\lambda^2}$ ?

Spoiler: Another hint

Terms containing $g^{\mu\nu}_{~~~~ ,\,\rho}$ can be converted by using $\left( g^{\mu\nu}g_{\nu\sigma}\right)_{,\,\rho}=0$

 

[TerryW]

Hi Jim,

I've just tried to see what happens if I use $\mathcal L = ½ p^{\mu}p_{\mu}$. Looks like I get the geodesic equation! I'll run a check. I haven't looked at your spoilers yet. Regards TerryW

Reference: https://www.physicsforums.com/threa...nian-formulation-for-geodesic-motion.1063880/

 

[JimWhoKnew]

Why $\mathcal L$ ?

 

[TerryW]

Well it does give the result I want but I realise I am just repeating Box 13.3.

I've had a go at using the Hamilton's equations and applying them to $\mathcal H$ and I now have every term I want, except that one of my terms is $-g_{\mu \nu , \alpha }p^{\mu}p^{ \nu}$ when it needs to be $+g_{\mu \nu , \alpha }p^{\mu}p^{ \nu}$. (To give me the $\Gamma$ term.)

I'll continue to pore over my workings to see if I can spot my error. Would your "Another Hint' help with this?

Cheers

 

[JimWhoKnew]

I wrote the "Another hint" because I thought it could be useful. It's up to you whether to look at it.
I can't tell why you got the wrong sign without seeing your work, but my guess is that you were not careful enough with the covariant and contravariant indices. Note, for example, that in the equation for $\dot p_\mu$ , $\mathcal H = \frac 1 2 g^{\mu \nu}p_\mu p_\nu$ is "processed" differently than $\mathcal H = \frac 1 2 g_{\mu \nu}p^\mu p^\nu$ , although they are the same. Can you tell why? Using the former is simpler. If you get stuck, look at the "Another hint".

 

[TerryW]

Hi Jim,

I think I've cracked the second part, proving that H is a constant of motion for the general case rather than the specific case.


$\frac {d \mathcal H}{d \lambda} = \frac {\partial {\mathcal H}}{\partial {x^{\mu}}} \frac {dx^{\mu}}{d \lambda} + \frac {\partial \mathcal H}{\partial{p_\mu}} \frac {dp_{\mu}}{d \lambda} + \frac {\partial \mathcal H}{{d \lambda}}$

Using the Hamilton equations:

$\frac {d \mathcal H}{d \lambda} = \frac {\partial {\mathcal H}}{\partial {x^{\mu}}} \frac {\partial \mathcal H }{\partial p_{\mu}} + \frac {\partial \mathcal H}{\partial{p_\mu}} \big ( - \frac {\partial \mathcal H}{\partial x^{\mu}} \big ) + \frac {\partial \mathcal H}{{d \lambda}}$

So

$\frac {d \mathcal H}{d \lambda} = \frac {\partial \mathcal H}{{d \lambda}}$ and $\frac {\partial {\mathcal H}}{\partial {x^{\mu}}} \frac {dx^{\mu}}{d \lambda} + \frac {\partial \mathcal H}{\partial{p_\mu}} \frac {dp_{\mu}}{d \lambda} = 0$

But $\frac {\partial {\mathcal H}}{\partial {x^{\mu}}} \frac {dx^{\mu}}{d \lambda} + \frac {\partial \mathcal H}{\partial{p_\mu}} \frac {dp_{\mu}}{d \lambda} = \frac {\partial \mathcal H}{{d \lambda}}$

Therefore $\frac {\partial \mathcal H}{{d \lambda}} = 0$ so $\frac {d \mathcal H}{d \lambda} = 0$

If $\frac {d \mathcal H}{d \lambda} = 0$ then $\mathcal H =$constant.


Is this all OK?


In the meantime, I'll press on with finding the geodesic equations.


Regards


Terry

PS your patience is much appreciated!

 

[JimWhoKnew]

proving that H is a constant of motion for the general case rather than the specific case.

In general, $\mathcal H$ is not necessarily a constant of motion. See below.

$\frac {d \mathcal H}{d \lambda} = \frac {\partial {\mathcal H}}{\partial {x^{\mu}}} \frac {dx^{\mu}}{d \lambda} + \frac {\partial \mathcal H}{\partial{p_\mu}} \frac {dp_{\mu}}{d \lambda} + \frac {\partial \mathcal H}{{d \lambda}}$

Using the Hamilton equations:

$\frac {d \mathcal H}{d \lambda} = \frac {\partial {\mathcal H}}{\partial {x^{\mu}}} \frac {\partial \mathcal H }{\partial p_{\mu}} + \frac {\partial \mathcal H}{\partial{p_\mu}} \big ( - \frac {\partial \mathcal H}{\partial x^{\mu}} \big ) + \frac {\partial \mathcal H}{{d \lambda}}$

So

$\frac {d \mathcal H}{d \lambda} = \frac {\partial \mathcal H}{{d \lambda}}$ and $\frac {\partial {\mathcal H}}{\partial {x^{\mu}}} \frac {dx^{\mu}}{d \lambda} + \frac {\partial \mathcal H}{\partial{p_\mu}} \frac {dp_{\mu}}{d \lambda} = 0$

That's right so far (except for the typos, it should be $\frac {d \mathcal H}{d \lambda} = \frac {\partial \mathcal H}{\partial \lambda}$ ).

Conclusion: $\mathcal H$ is a constant of motion if and only if $\frac {\partial \mathcal H}{\partial \lambda}=0$ .

But $\frac {\partial {\mathcal H}}{\partial {x^{\mu}}} \frac {dx^{\mu}}{d \lambda} + \frac {\partial \mathcal H}{\partial{p_\mu}} \frac {dp_{\mu}}{d \lambda} = \frac {\partial \mathcal H}{{d \lambda}}$

If this was true in the general case, we could have written
$\frac {d \mathcal H}{d \lambda} = 2\left( \frac {\partial {\mathcal H}}{\partial {x^{\mu}}} \frac {dx^{\mu}}{d \lambda} + \frac {\partial \mathcal H}{\partial{p_\mu}} \frac {dp_{\mu}}{d \lambda}\right)$
from the start.
(It is trivially true, however, if $\frac {\partial \mathcal H}{\partial \lambda}=0$ ).

$\mathcal H$ depends on $\lambda$ because it is a function of $x^\mu$ and $p_\mu$ which are functions of $\lambda$ . But it may also contain an additional explicit dependence. $\frac {\partial \mathcal H}{{\partial \lambda}}$ is the differentiation with respect to the explicit dependence. For example, if
$\mathcal H'=p_\mu x^\mu \sin \lambda$ ,
then
$\frac {\partial \mathcal H'}{{\partial \lambda}}=p_\mu x^\mu \cos \lambda$ .

Spoiler: hint for the derivation of geodesic equation

1. Use $(g^{\mu\nu}g_{\nu\sigma})_{~,\rho}=0~$ to express $g^{\mu\nu}_{~~~~,\rho}$ in terms of $g_{\mu\nu,\rho}$ and $g^{\mu\nu}$ .
2. Write the Hamilton equation $\dot p_\mu=-\frac{\partial\mathcal H}{\partial x^\mu}~$ as $~\frac d {d\lambda}\left( g_{\mu\nu}p^\nu \right) = -\frac \partial {\partial x^\mu}\left( \frac 1 2 g^{\rho\sigma}p_\rho p_\sigma \right)$ .
3. Work it out using the result of (1) above, and $\frac{\partial p_\mu}{\partial x^\nu}=0$ .

 

[TerryW]

Yes, I've managed to get to the required solution using your hint. I went a slightly different route which ended up in a bit of a mess (most of the bits required but not quite the answer! There must be a small mistake in my working somewhere but I haven't been able to spot it so far.

One thing that I have been left trying to get my head around, without success so far, is the difference between $\frac {dA}{d \lambda}$ and $\frac {\partial A}{\partial \lambda}$.​

​

Although I have been able to find refernces to $\frac {dA}{d \lambda} = \frac {\partial A}{\partial x^{\mu} } \frac {d x^{\mu}}{d \lambda} + \frac {\partial A}{\partial p_{\mu} }\frac {dp_{\mu}}{d \lambda} + \frac {\partial A}{\partial \lambda}$ on the internet haven't been able to find anything to clarify why this is true. I can't find an example of this in MTW but on page 262, there is a definition of $\frac {d}{d \lambda}$ which is $\frac {d}{d \lambda} = \frac {d x^{\mu}}{d \lambda} \frac {\partial }{\partial x^{\mu} }$.​

​

I'm also puzzled as to how you deal with $\frac {\partial A}{\partial \lambda}$.​

​

Do you have any links to pages that might clarify this for me?​

​

Once again, thanks for all your input on this,​

​

​

Regards​

​

​

Terryw​

​

​

 

[JimWhoKnew]

Yes, I've managed to get to the required solution using your hint.

​

There must be a small mistake in my working somewhere but I haven't been able to spot it so far

If you'll post your calculation, I'll try to find what's missing. I prefer LaTeX, but if you attach a handwritten pdf, please make it readable and don't leave me guessing whether a certain Greek index is covariant or contravariant.

One thing that I have been left trying to get my head around, without success so far, is the difference between $\frac {dA}{d \lambda}$ and $\frac {\partial A}{\partial \lambda}$.

I thought it was obvious, but maybe it should be said explicitly: In the context of exercise 25.2 (super-Hamiltonian formalism), $\lambda$ parameterizes a curve in the 8-dimensional "generalized phase-space" (9D, if we count $\lambda$ too). The math used is ordinary calculus. As you saw, in the derivation of the geodesic equation, the covariant differentiation emerges through the $g_{\mu\nu,\rho}~$.
Moreover, we use the first Hamilton equation to get $~p^\mu=\frac{dx^\mu}{d\lambda}~$. These act as 4 constraints, so we end up with the geodesic equation where $\lambda$ now parameterizes a curve in the 4D spacetime. Also, the $~p_{\mu,\nu}~$ are not necessarily zero now.

I'll extend my example from post #17:
Suppose $A\left(x^\mu(\lambda),p_\mu(\lambda),\lambda\right)~$ is a physical scalar quantity in generalized phase-space, defined on the curve parameterized by $\lambda$, and given by:
$A\equiv p_\mu x^\mu \sin \lambda \quad$.
Then:
$\frac {dA}{d \lambda} = \frac {\partial A}{\partial x^{\mu} } \frac {d x^{\mu}}{d \lambda} + \frac {\partial A}{\partial p_{\mu} }\frac {dp_{\mu}}{d \lambda} + \frac {\partial A}{\partial \lambda} = p_\mu\frac{dx^\mu}{d\lambda}\sin\lambda+x^\mu\frac{dp_\mu}{d\lambda}\sin\lambda+p_\mu x^\mu \cos \lambda\quad$.
Can you see now why $\frac {d A}{d \lambda}~$ is not the same, in general, as $\frac {\partial A}{\partial \lambda}$ ?

Do you have any links to pages that might clarify this for me?

I expect any introductory textbook or course (or youtube lectures) on multivariable calculus, to explain these ideas.

 

[TerryW]

Can you see now why dAdλ is not the same, in general, as ∂A∂λ ?

Yes, I can see that clearly now and am kicking myself for not working this out earlier! I used to have a little book on partial differentiation from my Uni days but I can't find it at the moment - possibly taken by my son during his Uni days and never returned!

I'll have one more look at my failed derivation and put up the bat signal if I can't find the error.


Regards

TerryW

 

[JimWhoKnew]

and am kicking myself for not working this out earlier!

No need for violence

I'll have one more look at my failed derivation

Remember that $~\frac{\partial p^\mu}{\partial x^\nu}=\frac{\partial \left(g^{\mu\rho}p_\rho\right)}{\partial x^\nu}=g^{\mu\rho}_{~~~~,\nu}p_\rho~$ .

____________
With the geodesic equation at hand, we now have an additional way of showing that $\mathcal H$ is constant on the geodesic in spacetime. It is closely related to exercise 13.5, which is crowned by MTW as "the easiest exercise in this book!"

 

[TerryW]

I've located my error in the derivation of the geodesic.

If you start from the equation $\frac {d}{d \lambda}(g_{\mu \nu}p^{\nu}) = - \frac {\partial}{\partial x^{\mu}}(\frac {1}{2} g^{\rho \sigma}p_{\rho} p_{\sigma})$, it all comes out nicely.

Instead I chose $\frac {d}{d \lambda}(g_{\mu \nu}p^{\nu}) = - \frac {\partial}{\partial x^{\mu}}(\frac {1}{2} g_{\rho \sigma}p^{\rho} p^{\sigma})$ which will of course get to the same result if you don't make a mistake working out the terms which involve $\frac {\partial p^{\rho}}{\partial x^{\mu}}$


Thanks for all your assistance with this problem.

With the geodesic equation at hand, we now have an additional way of showing that H is constant on the geodesic in spacetime.

I'll have a think about this.

It is closely related to exercise 13.5, which is crowned by MTW as "the easiest exercise in this book!"

I've had a look at my original answer to Ex 13.5. It looked OK at the time but maybe needs a rethink in the light of this exercise.


Regards


TerryW

 

[JimWhoKnew]

I've located my error in the derivation of the geodesic.

If you start from the equation $\frac {d}{d \lambda}(g_{\mu \nu}p^{\nu}) = - \frac {\partial}{\partial x^{\mu}}(\frac {1}{2} g^{\rho \sigma}p_{\rho} p_{\sigma})$, it all comes out nicely.

Instead I chose $\frac {d}{d \lambda}(g_{\mu \nu}p^{\nu}) = - \frac {\partial}{\partial x^{\mu}}(\frac {1}{2} g_{\rho \sigma}p^{\rho} p^{\sigma})$ which will of course get to the same result if you don't make a mistake working out the terms which involve $\frac {\partial p^{\rho}}{\partial x^{\mu}}$

I had a feeling this was the problem, and was trying to draw your attention to it since post #11. Try re-reading post #15.

Thanks for all your assistance with this problem.

You're welcome

I've had a look at my original answer to Ex 13.5. It looked OK at the time but maybe needs a rethink in the light of this exercise.

... and maybe it is really OK. Why not?

I find it recommended to rethink "old" problems in light of new understandings. It may lead to deeper insight.

Spoiler: Hint for showing that $\mathcal H$ is constant on geodesics

1. The geodesic equation is $~\mathbf{\nabla _p p} =0~$ in a coordinate-free description. Convince yourself that using coordinates, it becomes $~p^\nu p^\mu _{~~;\nu}=0~$ .

2. In spacetime: $~\frac{d\mathcal H}{d\lambda}=\frac{d x^\mu}{d\lambda}\mathcal H _{,\mu}=p^\mu \mathcal H _{;\mu}~$ .

3. Use (1) to find $~\frac{d\mathcal H}{d\lambda}~$ on the geodesics.

Next step: since $~\mathcal H ~$ is a constant of motion, it is equal to some constant $C$ on a geodesic. If we change the parameterization of this geodesic by $~\lambda ' = a\lambda+b~$ , what will be the new constant value of $\mathcal H'$ ?

 

[TerryW]

My 'easy' answer was that in the timelike region of spacetime, $u^2 < 1$ and even with constant
acceleration, a particle with mass will never reach $u^2 = 1$. For the spacelike region, $u^2 > 1$
and even with constant deceleration, a particle with mass will never reach $u^2 = 1$. Particles on null
geodesics have zero mass and cannot therefore be accelerated.

I haven't been able to find a way of using the geodesic equation to demonstrate this.



" Use (1) to find $d\mathcal H$/dλ on the geodesics."

$\mathcal H_{,\nu} = - \frac {dp_{\nu}}{d \lambda} = -p^{\alpha} \frac{\partial p_\nu}{\partial x^\alpha} = 0$

$\frac {d{\mathcal H}}{d \lambda} = p^{\nu} \mathcal H_{,\nu} = 0$

Next step: since $~\mathcal H ~$ is a constant of motion, it is equal to some constant $C$ on a geodesic. If we change the parameterization of this geodesic by $~\lambda ' = a\lambda+b~$ , what will be the new constant value of $\mathcal H'$ ?

If $\mathcal H$ is a constant, it is independent of $\lambda$ so $\mathcal H'$ = $\mathcal H$

Is all this OK now and are there any more steps?

Regards


TerryW

 

[JimWhoKnew]

My 'easy' answer was that in the timelike region of spacetime, $u^2 < 1$ and even with constant
acceleration, a particle with mass will never reach $u^2 = 1$. For the spacelike region, $u^2 > 1$
and even with constant deceleration, a particle with mass will never reach $u^2 = 1$. Particles on null
geodesics have zero mass and cannot therefore be accelerated.

Seems like you've forgotten the basics of SR. I think it will be fruitful for you to refresh your memory before going on.
A smooth timelike worldline, accelerated or not, always has $~\mathbf u ^2=-1~$ . Similarly, for spacelike curves $~\mathbf u ^2=1~$ (and zero for smooth null curves).

" Use (1) to find $d\mathcal H$/dλ on the geodesics."

$\mathcal H_{,\nu} = - \frac {dp_{\nu}}{d \lambda} = -p^{\alpha} \frac{\partial p_\nu}{\partial x^\alpha} = 0$

$\frac {d{\mathcal H}}{d \lambda} = p^{\nu} \mathcal H_{,\nu} = 0$

Wrong. We are no longer in "generalized phase space". We are reasoning in spacetime now. See post #19.

Spoiler: Another hint that $\mathcal H$ is constant on geodesics

$\frac{d\mathcal H}{d\lambda}=p^\mu \mathcal H _{;\mu}=p^\mu \left( \frac 1 2 g_{\rho\sigma}p^\rho p^\sigma\right) _{;\mu}=\frac 1 2 p^\mu \left(g_{\rho\sigma ; \mu}\right)p^\rho p^\sigma+\left(p^\mu p^\rho _{~~;\mu}\right) g_{\rho\sigma} p^\sigma ~~.$
The terms in parentheses on the RHS are zero. Why?

If $\mathcal H$ is a constant, it is independent of $\lambda$ so $\mathcal H'$ = $\mathcal H$

As a constant of motion, $\mathcal H$ is really independent of $\lambda$ on a given geodesic, but it is not invariant under reparametrizations.
$\mathcal H'=\frac 1 2 g_{\mu\nu}\frac{dx^\mu}{d\lambda'}\frac{dx^\nu}{d\lambda'}=\left(\frac{d\lambda}{d\lambda'}\right)^2\left(\frac 1 2 g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}\right)=\frac{\mathcal H}{a^2}~~.$
So you can set $\mathcal H$ to equal any constant you wish by reparametrization, as long as the sign is conserved (timelike geodesics remain timelike, etc.).

Is all this OK now and are there any more steps?

Re-read exercise 25.2 and see for yourself.

 

[TerryW]

I think my SR is OK. I specifically used $u^2$ rather than $\mathbf u ^2$ to indicate the three space components of $~\mathbf u$ as per the MTW typographic conventions so my statement is correct. But I grant you that your statement is more succinct.

Wrong. We are no longer in "generalized phase space". We are reasoning in spacetime now. See post #19.

I tried for a while to find a way to use $p^\nu p^\mu _{~~;\nu}=0$ to prove $\frac {d{\mathcal H}}{d \lambda} = 0$ for a general $\mathcal H$ but didn't get anywhere. I understand why the solution I finally came up with isn't correct. I can see that if you use the specific $\mathcal H = \frac {1}{2}g^{\mu \nu}p_\mu p\nu$ then $\mathcal H = 0$

So you can set H to equal any constant you wish by reparametrization, as long as the sign is conserved

Yes, I see that now.

I can't see anything more that needs to be resolved with this particular exercise. Thank you again for your patience and guidance on this problem, which has been very valuable.

Regards


TerryW

 

[JimWhoKnew]

I specifically used $u^2$ rather than $\mathbf u ^2$ to indicate the three space components of $~\mathbf u$ as per the MTW typographic conventions so my statement is correct.

So please help me understand your arguments in the beginning of post #24. For the following 4 geodesics (straight lines) in flat spacetime (Cartesian coordinates, metric $~\eta_{\mu\nu}~$), what are the values you get for $~u^2~$?
1) Timelike: $~x^\mu(\tau)=(\tau,0,0,0)\quad,\quad u^\mu=(1,0,0,0)$
2) Timelike: $~x^\mu(\tau)=(2\tau,\sqrt 3 ~\tau,0,0)\quad,\quad u^\mu=(2,\sqrt 3,0,0)$
3) Spacelike: $~x^\mu(\lambda)=(0,\lambda,0,0)\quad,\quad u^\mu=(0,1,0,0)$
4) Null: $~x^\mu(\lambda)=(\lambda/\sqrt 2,\lambda/\sqrt 2,0,0)\quad,\quad u^\mu=(1/\sqrt 2,1/\sqrt 2,0,0)$

____
Note that from the reparametrization freedom we see that $~\mathcal H~$ is proportional to $~\mathbf u ^2=u_\mu u^\mu~.$ That's why showing here that $~\frac{d\mathcal H}{d\lambda}=0~$ on a geodesic in spacetime, also solves exercise 13.5 (AKA "the easiest").

 

[TerryW]

Oh dear. $v^2 < 1$ but $u^2 = v^2/(1-v^2)$, so $u$ gets very large as v nears 1.

Sometimes I knock up a little spreadsheet to see what happens when actual numbers are applied but I never bothered with these relatively simple formulae, so I failed to spot this.



Maybe you should change your name to JimWhoKnows!


Regards


TerryW

 

[JimWhoKnew]

Oh dear. $v^2 < 1$ but $u^2 = v^2/(1-v^2)$, so $u$ gets very large as v nears 1.

After all the discussion in this thread, you can now re-visit the OP and have one more attempt at its question: "What do the choices $\mathcal H = -½, 0, -½\mu^2, +½$ mean for the geodesic and the choice of parametrization?"

Spoiler: Answer

From Hamilton equations $~p^\mu=\frac{dx^\mu}{d\lambda}~$ so $~\mathcal H=\frac 1 2 \mathbf{p\cdot p}=\frac 1 2 p_\mu p^\mu$ .

1. $~\mathcal H=-\frac 1 2 ~$ : timelike, $~\lambda=\tau~$ , $~\mathbf p=\mathbf u~$ , free fall geodesic does not depend on the mass of the falling object.

2. $~\mathcal H=-\frac 1 2 \mu^2~$ : timelike, $~\lambda=\frac \tau \mu~$ , $~\mathbf p=\mu \mathbf u~$ is the mechanical 4-momentum for which $~-\mathbf{p\cdot p}=E^2-\vec p^2=\mu^2~$ is satisfied.

3. $~\mathcal H=\frac 1 2 ~$ : spacelike, $~\mathbf p=\mathbf u~$ . Loosely, $~\lambda~$ measures proper length.

4. $~\mathcal H=0 ~$ : null, $~\mathcal H ~$ is invariant under (linear) reparametrization.

Maybe you should change your name to JimWhoKnows!

but then it wouldn't sound like $~g_{\mu\nu}~$ .

 

[TerryW]

It has taken me a little while to come to terms with the idea that $\frac {dx^\mu}{d\lambda}$ can be $p^\mu$ rather than $u^\mu$.

Now that I'm comfortable with that, I've had another go at the OP as follows:

$\mathcal H = -\frac {1}{2} = \frac {1}{2}p_\mu p^\mu$

$p_\mu p^\mu = -1 = m^2 \textbf u^2$

But $\textbf u^2 = -1$

So $p_\mu p^\mu = -1 = -m^2$

So $m^2 = 1, \quad \textbf p = \textbf u, \quad \frac {d}{d\lambda} = \frac {d}{d\tau} , \quad \lambda = \tau$

$\mathcal H = -\frac {1}{2}\mu^2$

$\textbf p \cdot \textbf p = -\mu^2$

$\textbf p \cdot \textbf p = \mu^2 \textbf u \cdot \textbf u$

$\textbf p = \mu \textbf u$

$\frac {d}{d\lambda} = \mu \frac {d}{d\tau}$

$\mathcal H = 0 ,$

$\frac{1}{2} p_\mu p^\mu = m^2 \textbf u ^2 = - m^2 = 0$

$m = 0$

Only photons have zero mass. Photons follow null geodesics.


$\mathcal H = +\frac{1}{2}$

Following the same process as used above:

$\frac{1}{2} p_\mu p^\mu = \frac{1}{2}$

$p_\mu p^\mu = m^2 \textbf u ^2 = -m^2$

$m^2 = -1$

So \textbf p = \textbf u

I'm not sure where this leaves me regarding $\lambda$!

but then it wouldn't sound like gμν .

Too subtle for me


TerryW

 

[JimWhoKnew]

$\mathcal H = -\frac {1}{2} = \frac {1}{2}p_\mu p^\mu$

$p_\mu p^\mu = -1 = m^2 \textbf u^2$

But $\textbf u^2 = -1$

So $p_\mu p^\mu = -1 = -m^2$

So $m^2 = 1, \quad \textbf p = \textbf u, \quad \frac {d}{d\lambda} = \frac {d}{d\tau} , \quad \lambda = \tau$
...

What does the $~m~$ in $~m^2~$ stand for?

 

[TerryW]

m is the rest mass. Should have stuck to $\mu$.

 

[JimWhoKnew]

m is the rest mass. Should have stuck to $\mu$.

So how can you get $~m^2=-1~$ as in post #30 ?

Take for example the trivial case $~\lambda=\tau~$ in Minkowski SR spacetime, where the geodesic is the straight line $~x^\mu(\tau)=(\tau,0,0,0)~$ . You get $~p^\mu=\frac{dx^\mu}{d\lambda}=u^\mu=\delta^\mu_0~$ and $~\mathcal H=-\frac 1 2~$ . But how can you infer a mass from it?

Edit: Addition
$p^\mu=\frac{dx^\mu}{d\lambda}~$ is the vector field on the curve of tangent vectors to it, whose components (in a given chart) are scaled by the parametrization $~\lambda~$. It is proportional to $~u^\mu~$ , but the proportionality factor is not $~\mu~$ (except for the case where $~\mathcal H=-\frac 1 2 \mu^2~$ ), so in general it is not equal to the "mechanical 4-momentum" $~\mu \mathbf u~$ . Moreover, what sense does it make to talk of masses in the case of spacelike geodesics?

 

[TerryW]

I know that $m^2 = -1$ is problematical! It's just that the maths lead to that result.

I looked on the internet for some kind of discussion of what is happening in the spacelike realm of spacetime and didn't find anything useful. I have hazy memories of people postulating tachyons which accelerate when a retarding force is applied and decelerate when a positive force is applied (negative mass but not imaginary mass?).

 

[JimWhoKnew]

Hi TerryW

After 3 months on this thread, I give up.
I apologize for accepting the challenge (which turned out to be beyond my capabilities), since by doing so I possibly deprived you from getting a better help.

Thanks for your pleasant company

Regards,

JimWhoKnew