Calculating Varying an Action: \delta I

[latentcorpse]

So we have an action
$I[x,p;e]=\int d \lambda \left( \dot{x}^\mu p_\mu - e (p^2+m^2) \right)$
where e is the einbein and $p^2=g^{\mu \nu} p_\mu p_\nu$

We're asked to find $\delta I$ given the variations
$\delta x^\mu=\epsilon(\lambda) \xi^\mu(x)$ and $\delta p_\mu = - \epsilon(\lambda) \partial_\mu \epsilon^\nu p_\nu$

So I find that $\delta \dot{x}^\mu = \dot{\epsilon}(\lambda) \xi^\mu(x)$

and we have that

$\delta I = \int d \lambda \left( \delta ( \dot{x}^\mu p_\mu ) - \delta ( ep^2+em^2) \right)$
$=\int d \lambda \left( \delta \dot{x}^\mu p_\mu + \dot{x}^\mu \delta p_\mu -2e g^{\mu \nu} p_\mu \delta p_\nu \right)$ since $\delta m=0$

Then, this gives by substitution

$= \int d \lambda \left( \dot{\epsilon} \xi^\mu p_\mu - \epsilon \dot{x}^\mu \partial_\mu \xi^\nu p_\nu + 2eg^{\mu \nu} p_\mu \epsilon \partial_\nu \xi^\rho p_\rho \right)$

But this appears to be going "off course" so I reckon I've messed up but I can't see anything wrong with it!

 

[fzero]

So I find that $\delta \dot{x}^\mu = \dot{\epsilon}(\lambda) \xi^\mu(x)$

Isn't

$\delta \dot{x}^\mu = \dot{\epsilon}(\lambda) \xi^\mu(x) + \epsilon(\lambda) \partial_\nu\xi^\mu \dot{x}^\nu?$

 

[latentcorpse]

Isn't

$\delta \dot{x}^\mu = \dot{\epsilon}(\lambda) \xi^\mu(x) + \epsilon(\lambda) \partial_\nu\xi^\mu \dot{x}^\nu?$

oh yeah.

since $\delta \dot{x}^\mu = \frac{d}{d \lambda} \delta x^\mu = \dot{\epsilon}(\lambda) \xi^\mu(x) + \epsilon \frac{d}{d \lambda} \xi^\mu(x) = \dot{\epsilon}(\lambda) \xi^\mu(x) + \epsilon(\lambda) \partial_\nu\xi^\mu \dot{x}^\nu$ by chain rule on 2nd term, right?

However, that does leave us with

$\delta I = \int d \lambda \left( \dot{\epsilon} \xi^\mu p_\mu + 2 \epsilon e p^\nu ( \partial_\nu \xi^\rho) p_\rho \right)= \int d \lambda \left( \dot{\epsilon} \xi^\mu p_\mu + 2 \epsilon e p^\nu p^\rho( \partial_\nu \xi_\rho) \right)$
so we need to fix that 2nd term

I should mention that we are trying to show the result

$\delta I = \int d \lambda \left( \epsilon e p^\mu p^\nu D_\mu \xi_\nu + \dot{\epsilon} \xi^\mu p_\mu \right)$

 

[dextercioby]

Who's $D_{\mu}$ ?

 

[fzero]

$\delta I = \int d \lambda \left( \dot{\epsilon} \xi^\mu p_\mu + 2 \epsilon e p^\nu ( \partial_\nu \xi^\rho) p_\rho \right)= \int d \lambda \left( \dot{\epsilon} \xi^\mu p_\mu + 2 \epsilon e p^\nu p^\rho( \partial_\nu \xi_\rho) \right)$
so we need to fix that 2nd term

But

$$p^\nu ( \partial_\nu \xi^\rho) p_\rho \neq p^\nu p^\rho( \partial_\nu \xi_\rho) .$$

You can't just raise and lower indices under a derivative. In fact, carefully treating the action of the derivative on the metric should give you the expression with the covariant derivative.

 

[dextercioby]

But

$$p^\nu ( \partial_\nu \xi^\rho) p_\rho \neq p^\nu p^\rho( \partial_\nu \xi_\rho) .$$

You can't just raise and lower indices under a derivative. In fact, carefully treating the action of the derivative on the metric should give you the expression with the covariant derivative.

Ah, ok, so it's the covariant derivative probably wrt the Levi-Civita connection. Then he must first write all the terms correctly and then try to shift indices so that he gets eventually derivatives acting on the metric tensor he could then use for the covariant derivative.

 

[latentcorpse]

But

$$p^\nu ( \partial_\nu \xi^\rho) p_\rho \neq p^\nu p^\rho( \partial_\nu \xi_\rho) .$$

You can't just raise and lower indices under a derivative. In fact, carefully treating the action of the derivative on the metric should give you the expression with the covariant derivative.

ok. so if we look at the 2nd term

$-2eg^{\mu \nu}p_\mu \delta p_\nu$
$=2eg^{\mu \nu}p_\mu \epsilon \partial_\nu (\xi^\rho) p_\rho$
$=2 \epsilon e g^{\mu \nu} p_\mu p_\rho \partial_\nu \xi^\rho$
$=2 e \epsilon p^\nu p_\rho \partial_\nu ( \xi_\lambda g^{\rho \lambda})$
$=2 e \epsilon p^\nu p_\rho ( \partial_\nu \xi_\lambda ) g^{\rho \lambda} + 2 e \epsilon p^\nu p_\rho \xi_\lambda \partial_\nu g^{\rho \lambda}$
$=2 e \epsilon p^\nu p^\lambda ( \partial_\nu \xi_\lambda ) + 2 e \epsilon p^\nu p_\rho \xi_\lambda \partial_\nu g^{\rho \lambda}$
So if I can justify getting rid of the 2nd term then I think I am done. Why would it vanish though? We don't know that $\partial_\nu g^{\rho \lambda}=0$, do we?

 

[fzero]

No, $$\partial_\nu g^{\rho \lambda}\neq 0$$. There's another term in the variation of the action that you're missing. This comes from varying the metric in

$$\delta(p^2) = p^\mu p^\nu \delta g_{\mu\nu} + 2 p^\mu \delta p_\mu .$$

If you add this term, you should get an expression that you can relate to the covariant derivative of the Killing vector.

As an aside, I can't reconcile the result we have for $$\delta \dot{x}^\mu$$ with $$\delta p_\mu$$. They're supposed to be related by

$$p_\mu = \frac{1}{2e} g_{\mu\nu} \dot{x}^\nu.~~(*)$$

However, I don't exactly see this posing a problem in deriving the result here, but you might keep this in mind if you're still having trouble. You could also derive the answer by using (*) to rewrite the action in terms of $$\dot{x}^\mu$$ only.

 

[latentcorpse]

No, $$\partial_\nu g^{\rho \lambda}\neq 0$$. There's another term in the variation of the action that you're missing. This comes from varying the metric in

$$\delta(p^2) = p^\mu p^\nu \delta g_{\mu\nu} + 2 p^\mu \delta p_\mu .$$

If you add this term, you should get an expression that you can relate to the covariant derivative of the Killing vector.

As an aside, I can't reconcile the result we have for $$\delta \dot{x}^\mu$$ with $$\delta p_\mu$$. They're supposed to be related by

$$p_\mu = \frac{1}{2e} g_{\mu\nu} \dot{x}^\nu.~~(*)$$

However, I don't exactly see this posing a problem in deriving the result here, but you might keep this in mind if you're still having trouble. You could also derive the answer by using (*) to rewrite the action in terms of $$\dot{x}^\mu$$ only.

well the metric is a function of the coordinates $g_{\mu \nu}(x)$

so $\delta g_{\mu \nu} = \frac{\delta g_{\mu \nu}}{\delta x^\lambda} \delta x^\lambda$ right?

but how do we compute $\frac{\delta g_{\mu \nu}}{\delta x^\lambda}$?

 

[fzero]

well the metric is a function of the coordinates $g_{\mu \nu}(x)$

so $\delta g_{\mu \nu} = \frac{\delta g_{\mu \nu}}{\delta x^\lambda} \delta x^\lambda$ right?

but how do we compute $\frac{\delta g_{\mu \nu}}{\delta x^\lambda}$?

That's normal calculus.

 

[latentcorpse]

That's normal calculus.

But we don't know what $g_{\mu \nu}$ is so how can we take its derivative?
I'm not sure if $\frac{\partial g_{\mu \nu}}{\partial x^\lambda} = \frac{\delta g_{\mu \nu}}{\delta x^\lambda}$ and even if it does, just leaving my answer as $g_{\mu \nu , \lambda}$ isn't going to help.
You were saying I should be able to write it in terms of covariant derivatives somehow?

 

[fzero]

But we don't know what $g_{\mu \nu}$ is so how can we take its derivative?
I'm not sure if $\frac{\partial g_{\mu \nu}}{\partial x^\lambda} = \frac{\delta g_{\mu \nu}}{\delta x^\lambda}$ and even if it does, just leaving my answer as $g_{\mu \nu , \lambda}$ isn't going to help.

Yes it will.

You were saying I should be able to write it in terms of covariant derivatives somehow?

Actually you were the one that said that back in post #3.

 

[latentcorpse]

Yes it will.

Actually you were the one that said that back in post #3.

well plugging all this is i now get

$\delta I = \int d \lambda \dot{\epsilon} \xi^\mu p_\mu + 2 e \epsilon p^\mu \partial_\mu \xi^\nu p_\nu - e p^\mu p^\nu g_{\mu \nu , \lambda} \delta x^\lambda$

$= \int d \lambda \dot{\epsilon} \xi^\mu p_\mu + 2 e \epsilon p^\mu \partial_\mu \xi^\nu p_\nu - e \epsilon p^\mu p^\nu g_{\mu \nu , \lambda} \xi^\lambda$

Now in that second term, is the derivative acting only on the $\xi^\nu$ or is it acting on $\xi^\nu p_\nu$? How can we tell? If it was just acting on $\xi^\nu$ then I would suggest we take the $p_\nu$ to the front of that term?

I can't see how to connect this to covariant derivatives? The derivative of the metric suggests I might be able to get a Christoffel symbol somehow but I can't see it...and there's also that factor of 2...

 

[fzero]

well plugging all this is i now get

$\delta I = \int d \lambda \dot{\epsilon} \xi^\mu p_\mu + 2 e \epsilon p^\mu \partial_\mu \xi^\nu p_\nu - e p^\mu p^\nu g_{\mu \nu , \lambda} \delta x^\lambda$

$= \int d \lambda \dot{\epsilon} \xi^\mu p_\mu + 2 e \epsilon p^\mu \partial_\mu \xi^\nu p_\nu - e \epsilon p^\mu p^\nu g_{\mu \nu , \lambda} \xi^\lambda$

Now in that second term, is the derivative acting only on the $\xi^\nu$ or is it acting on $\xi^\nu p_\nu$? How can we tell? If it was just acting on $\xi^\nu$ then I would suggest we take the $p_\nu$ to the front of that term?

You should go over your calculation and put in parentheses wherever you have a derivative so that you don't get confused. You've had this type of problem before.

I can't see how to connect this to covariant derivatives? The derivative of the metric suggests I might be able to get a Christoffel symbol somehow but I can't see it...

There's a number of ways that you can introduce a covariant derivative. You could write

$$\partial_\mu \xi^\nu = g^{\nu \sigma} D_\mu \xi_\sigma + \cdots$$

and try to simplify the resulting expressions.

 

[latentcorpse]

You should go over your calculation and put in parentheses wherever you have a derivative so that you don't get confused. You've had this type of problem before.



There's a number of ways that you can introduce a covariant derivative. You could write

$$\partial_\mu \xi^\nu = g^{\nu \sigma} D_\mu \xi_\sigma + \cdots$$

and try to simplify the resulting expressions.

The problem kind of stemmed from the fact that i was given the definition

$\delta p_\mu = - \epsilon \partial_\mu \xi^\nu p_\nu$
which wasn't clear about where the derivative acted on.

However, I think I'll assum it's just hitting the $\xi^\nu$ since otherwise he would probably have written $\partial_\mu ( \xi^\nu p_\nu)$

So that sorts that out (hopefully!)

Anyway if we focus on the two terms that aren't "right" yet we get

$2e \epsilon p^\mu p_\nu \partial_\mu \xi^\nu - e \epsilon p^\mu p^\nu g_{\mu \nu , \lambda} \xi^\lambda$

and we have $\partial_\mu \xi^\sigma = g^{\nu \sigma} D_\mu \xi_\nu - \Gamma^\sigma{}_{\mu \nu} \xi^\nu$

which can be used to give

$=2e \epsilon p^\mu p_\nu g^{\nu \sigma} D_\mu \xi_\sigma - 2 e \epsilon p^\mu p_\nu \Gamma^\nu{}_{\mu \sigma} \xi^\sigma - e \epsilon p^\mu p^\nu g_{\mu \nu , \lambda} \xi^\lambda$
$=2e \epsilon p^\mu p_\nu g^{\nu \sigma} D_\mu \xi_\sigma - 2 e \epsilon p^\mu p^\lambda g_{\nu \lambda} \frac{1}{2} g^{\nu \kappa} ( g_{\mu \kappa , \sigma} + g_{\sigma \kappa , \mu} - g_{\mu \sigma , \kappa} ) \xi^\sigma - e \epsilon p^\mu p^\nu g_{\mu \nu , \lambda } \xi^\lambda$
$=2e \epsilon p^\mu p_\nu g^{\nu \sigma} D_\mu \xi_\sigma - e \epsilon p^\mu p^\lambda \delta_\lambda{}^\kappa ( g_{\mu \kappa , \sigma} + g_{\sigma \kappa , \mu} - g_{\mu \sigma , \kappa} ) \xi^\sigma - e \epsilon p^\mu p^\nu g_{\mu \nu , \lambda } \xi^\lambda$
$=2e \epsilon p^\mu p_\nu g^{\nu \sigma} D_\mu \xi_\sigma - e \epsilon p^\mu p^\kappa ( g_{\mu \kappa , \sigma} + g_{\sigma \kappa , \mu} - g_{\mu \sigma , \kappa} ) \xi^\sigma - e \epsilon p^\mu p^\kappa g_{\mu \kappa , \lambda } \xi^\lambda$
$=2e \epsilon p^\mu p_\nu g^{\nu \sigma} D_\mu \xi_\sigma - e \epsilon p^\mu p^\kappa ( 2g_{\mu \kappa , \sigma} + g_{\sigma \kappa , \mu} - g_{\mu \sigma , \kappa} ) \xi^\sigma$

Then I'm stuck again!

 

[fzero]

Anyway if we focus on the two terms that aren't "right" yet we get

$2e \epsilon p^\mu p_\nu \partial_\mu \xi^\nu - e \epsilon p^\mu p^\nu g_{\mu \nu , \lambda} \xi^\lambda$

The sign of the second term is wrong. I think you have to vary the metric in the $$\dot{x}^\mu p_\mu$$ term, which should contribute a similar term with a $$+2$$ factor. It's weird that you'd have to do this, but as I said before, it's also strange to vary both $$\dot{x}^\mu$$ and $$p_\mu$$ in an action.

 

[latentcorpse]

The sign of the second term is wrong. I think you have to vary the metric in the $$\dot{x}^\mu p_\mu$$ term, which should contribute a similar term with a $$+2$$ factor. It's weird that you'd have to do this, but as I said before, it's also strange to vary both $$\dot{x}^\mu$$ and $$p_\mu$$ in an action.

I don't see why there's a sign wrong?

$D_\mu \xi^\nu = \partial_\mu \xi^\nu + \Gamma^\nu{}_{\mu \rho} \xi^\rho$
$\Rightarrow \partial_\mu \xi^\nu=D_\mu \xi^\nu -\Gamma^\nu{}_{\mu \rho} \xi^\rho$
$\Rightarrow \partial_\mu \xi^\nu=g^{\nu \lambda} D_\mu \xi_\lambda -\Gamma^\nu{}_{\mu \rho} \xi^\rho$

so i should be ok with a minus right?

 

[fzero]

The sign doesn't come from that part of the calculation. As I said back in post 8, the variation rule for $$p_\mu$$ is strange when compared to $$\delta\dot{x}^\mu$$. So I did the calculation from the equivalent action

$$I = \int d\lambda\left( \frac{1}{4e} (\dot{x})^2 - e m^2 \right)$$

and found that

$$\delta I = \int d\lambda \frac{1}{4e} \left[ 2 \dot{\epsilon} \dot{x}^\mu \xi_\mu + \epsilon \dot{x}^\mu \dot{x}^\nu \left( g_{\mu\nu,\sigma} \xi^\sigma + 2 g_{\mu\sigma} {\xi^\sigma}_{,\nu} \right) \right].$$

This disagrees with your result (which I confirmed modulo the other variation of the metric that I mentioned). It does however give the expected answer.