Homework Help: Quantum theory, show variation of S zero, integrate by parts

1. Jan 11, 2017

binbagsss

1. The problem statement, all variables and given/known data

Hi,

I am trying to show the second equality , expressing all as a total derivative (I can then show that $\delta S =$)

2. Relevant equations

See above

3. The attempt at a solution

So the $m$ term is pretty obvious, simply using the chain rule.

It is the first term I am stuck on. So looking by the sign, it looks like we have done integration by parts twice.

My working so far is to go by parts initially as:

$w=\partial^{u}\phi$
$\partial w = \partial_{v}\partial^{u} \phi$
$\partial z = \partial_{u}\partial_{v} \phi$
$z= \partial_{u} \phi$

to get, since we are allowed to assume vanishing of the field $\phi$ at inifnity:

$- \int \partial_{u} \phi ( \partial_{v} \partial^{u} \phi )$

I am now stuck of what to do, I can't see a move that will get the desired expression for a choice of integration by parts, which makes me question whether this was the correct first move to make.

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2. Jan 12, 2017

Orodruin

Staff Emeritus
Is $a$ constant? It is just the same type of relation as the mass term. Just switch the order of the derivatives.

3. Jan 12, 2017

binbagsss

The $\partial_{u}$ is not hitting the $\partial^{u} \phi$ though?

I have the first line as $a^{v}\partial_{u}\partial_{v}\phi \partial^{u} \phi = a^{v}(\partial^2 _{uv} \phi)(\partial^{u} \phi )$,
whereas I have the bottom line as $(a^{v}\partial^{2}_{uv}\phi)(\partial_{v}\partial^{u} \phi)$..

thanks

4. Jan 12, 2017

Orodruin

Staff Emeritus
No this is not the bottom line - it does not contain four derivatives ... Just use the same argument as for the mass term, or more generally $f\, df/dx = 0.5 df^2/dx$.

5. Jan 12, 2017

binbagsss

erm, so I should have used the product rule on the bottom line?
Which would give arise to two terms, whereas the top line has one term..?

6. Jan 12, 2017

Orodruin

Staff Emeritus
I don't get your meaning. Both the bottom and top lines have two terms. One from the mass and one from the kinetic term.

7. Jan 12, 2017

binbagsss

apologies, corresponding to the kinetic term, I am only referring to here.

8. Jan 12, 2017

Orodruin

Staff Emeritus
Is your problem that you do not see that
$$(\partial_a \partial_b \phi) (\partial^a \phi) = \frac{1}{2} \partial_b [(\partial_a \phi)(\partial^a\phi)] ?$$

9. Jan 12, 2017

Orodruin

Staff Emeritus
The kinetic term has one term only in both lines in your OP.

10. Jan 12, 2017

binbagsss

oh right got it !
by two terms I was reffering to the product rule, but then you sea-saw giving the factor of $1/2$, thank you !

11. Jan 12, 2017

binbagsss

[QUOTE="binbagsss, post: 5663557, member: 252335"

My working so far is to go by parts initially as:

$w=\partial^{u}\phi$
$\partial w = \partial_{v}\partial^{u} \phi$
$\partial z = \partial_{u}\partial_{v} \phi$
$z= \partial_{u} \phi$

[/QUOTE]

May I ask, in general how should you approach integration by parts when four -derivaitves are involved.
Should I choose a different index like $\partial_a$ than u and v which are already used in the problem
Am I okay to use $\partial_v$, but then I must be consistent throughout? i.e. not change to $\partial_u$
I'm a bit confused..
Many thanks !

12. Jan 12, 2017

Orodruin

Staff Emeritus
There is nothing different from the usual integration by parts, you just do it in the direction of the given index. In other words, $\partial_\mu$ is really a lot of different terms. You can do partial integration with respect to each direction separately. In reality, it really is just the divergence theorem.