# MTW Exercise 25.5 b) - killing vectors

## Main Question or Discussion Point

hi there,
In this Ex ( see attached snapshot ), point b), the poisson bracket equation is not so straightforward to obtain.
Please correct my Poisson Bracket expansion here :
The first one which is provided is simpler :
[ε,η] = εμδμηρ - ημδμερ = ζη

and the monster one :

[pε,pη] = [pμεμ,pβηβ] = (pμεμ)∂μ(pβηβ) -
(pμημ)∂μ(pβεβ)

Thanks,

#### Attachments

• 60.7 KB Views: 551

## Answers and Replies

Related Special and General Relativity News on Phys.org
WannabeNewton
What you have written down doesn't make sense. Thankfully for you it's still quite simple :)

$[p_{\mu}\xi^{\mu},p_{\mu}\eta^{\mu}] = \frac{\partial }{\partial x^{\lambda}}(p_{\mu}\xi^{\mu})\frac{\partial }{\partial p_{\lambda}}(p_{\nu}\eta^{\nu})-\frac{\partial }{\partial p_{\lambda}}(p_{\mu}\xi^{\mu})\frac{\partial }{\partial x^{\lambda}}(p_{\nu}\eta^{\nu})$ and keep in mind that $\xi^{\mu}=\xi^{\mu}(x^{\nu})$, $\eta^{\mu} = \eta^{\mu}(x^{\nu})$ as well as $\frac{\partial p_{\mu}}{\partial p_{\nu}} = \delta^{\nu}_{\mu}$. The rest is a straightforward calculation.

EDIT: I must admit that this entire exercise is a bit weird. Showing that the commutator of two killing fields is also a killing field is way more straightforward than this.

Last edited:
hey Mate,
Thanks for your reply. Well it seems I got confused here. I had based my second commutation on the first one. In my very first attempt I had indeed written your expression but had forgotten that the other variable was x not either of the killing vectors ε or η....
Damn that got me wrong all the way !
At the beginning I had started with the Jacobi identity but did not check out ...
Thanks very much !

What you have written down doesn't make sense. Thankfully for you it's still quite simple :)

By the way, why it does not make sense. It does if it were a commutation relation and not a Poisson bracket. What do you think ?

WannabeNewton
Yeah what you wrote down is fine for a Lie bracket of vector fields on a smooth manifold but what you want is a Poisson bracket of scalar fields that are functions of phase space coordinates (cotangent bundle to the smooth manifold).

Did you get the final result by the way? Feel free to ask if you get stuck elsewhere.

Ok let us expand out your expression above :

(pμξμ)$\frac{∂(p_{v}η^{v})}{∂p_{λ}}$−(pvηv)$\frac{∂(p_{v}ε^{v})}{∂p_{λ}}$ =
( pμεμ + pμεμ ) ( ηλ + pv$\frac{∂(η^{v})}{∂p_{λ}}$ ) -
( pvηv + pvηv ) ( ελ + pv$\frac{∂(ε^{v})}{∂p_{λ}}$ )
Now we have : $\frac{∂(ε^{v})}{∂p_{λ}}$ = 0 = $\frac{∂(η^{v})}{∂p_{λ}}$
we are left with :
( pμεμ + pμεμ ) ( ηλ ) -
( pvηv + pvηv ) ( ελ )
This gives :
pμεμηλ + pμεμ ηλ - pvηvελ - pvηvελ

Now the second and the last term would yield the result (i.e. -pζ based on the commutation of ε and η) which means that the first and the third terms would add up to 0. How do you think ?

WannabeNewton
Remember that we are working in phase space coordinates $\{x^{\mu},p_{\mu}\}$ so $\frac{\partial p_{\mu}}{\partial x^{\nu}} = 0$. This makes the first and third terms vanish.

EDIT: I must admit that this entire exercise is a bit weird. Showing that the commutator of two killing fields is also a killing field is way more straightforward than this.
Yes there is a much shorter version using Lie derivatives indeed...

Remember that we are working in phase space coordinates $\{x^{\mu},p_{\mu}\}$ so $\frac{\partial p_{\mu}}{\partial x^{\nu}} = 0$. This makes the first and third terms vanish.
Well then it is done !

See you in another MTW Ex hopfully !
As a gift to you :http://www.webofstories.com/play/john.wheeler/76
Enjoy wheeler's stories by mighty Wheeler,
Thanks Sir.

WannabeNewton