# EoM from action, indices confused, (QFT)

1. Dec 25, 2017

### binbagsss

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

Using the E-L equations to get the EoM from the action.

2. Relevant equations

I am using E-L equations in the form:

$\frac{\partial}{\partial_u} \frac{\partial L}{\partial_u \phi}-\frac{\partial L}{\partial \phi}$

where $L$ is the Lagrangian

3. The attempt at a solution

I am a bit confused with the term $\frac{1}{2} \partial_u \phi \partial^u \phi$ of alot of actions.

The correct answer is $\frac{1}{2} \partial_u \partial^u \phi$

As it is for the first term of the EoM from [1] I get:

$\frac{\partial}{\partial_u} \frac{1}{2} \partial^u \phi = \frac{1}{2}\partial_u \partial^u \phi$

However if I write it as $\frac{1}{2}(\partial_u \phi)^2$ , the factor of half cancels, but my indices don't work out :

$\frac{\partial}{\partial_u} \partial_u \phi = \partial_u \partial_u \phi$

Many thanks

2. Dec 25, 2017

### Orodruin

Staff Emeritus
Writing it as $(\partial_u \phi)^2/2$ is a bastard notation often used in physics. At face value you might think it means what you are taking it to mean, but what it really means is $g^{ab}(\partial_a\phi)(\partial_b \phi)/2$.

3. Dec 25, 2017

### binbagsss

i dont understand the differnce, the metric in your expression raises the index to get the same expression, no?

4. Dec 25, 2017

### Orodruin

Staff Emeritus
Yes, the point being that the expression $(\partial_a \phi)^2$ while seemingly containing two covariant indices really contains a contraction between a covariant and a contravariant index and needs to be treated as such.

5. Dec 25, 2017

### binbagsss

oh so as in when i differentiate what i have done is treated them as seperate indices but i need to apply some sort of chain rule instead?

6. Dec 25, 2017

### Orodruin

Staff Emeritus
Right. $\partial^a\phi$ is not indepentent from $\partial_a\phi$.

7. Dec 25, 2017

### Orodruin

Staff Emeritus
You mean when you take the derivative $\partial_u$ of $\partial L/\partial(\partial_u\phi)$? In general yes, but in most cases QFT is done on a Minkowski space background meaning that that term vanishes. In a curved background (or in curvilinear coordinates) you will generally not obtain $\partial_a\partial^a\phi =0$ but $\nabla_a\partial^a\phi =0$, which is fortunate since the first expression is not invariant whereas the second is.

8. Dec 26, 2017

### binbagsss

so looking at the OP, what is the term coming from the chain rule I am missing that you have hinted at.
From the above discussion all I have is:

$\frac{1}{2} \partial_u (g^{ua} \partial_{a} \phi ) =\frac{1}{2} ( g^{ua} \partial_u \partial_{a} \phi + \partial_{a} \phi \partial_u g^{ua} )$

this is equal to what I had previously , sincce the 2nd term is zero...

Please can I have more guidance,

thanks

9. Dec 26, 2017

### Orodruin

Staff Emeritus
Why do you still have the 1/2?

10. Dec 26, 2017

### Orodruin

Staff Emeritus
Also I am not sure I understand what you think the problem is ...

11. Dec 26, 2017

### binbagsss

becuse in genreal the action will include other terms such as $m^2 \phi ^2 + \lambda \frac{\phi}{4!}$ i.e $\frac{\partial L}{\partial \phi } \neq 0$

12. Dec 26, 2017

### Orodruin

Staff Emeritus
Are you dealing with a complex scalar field or a real scalar field? The 1/2 should disappear in the case of a real scalar field because you have two derivative terms. In the case of a complex field you do not have the factor of 1/2 from the beginning. In the case of a real scalar field, the mass term also comes with a 1/2.

13. Dec 26, 2017

### binbagsss

yep the m^2 term comes with a half.
no i mean im stil struggling with this derivative term.

ok so I want $\frac{\partial }{\partial_u} (\partial_u \phi \partial^u \phi) = \frac{\partial }{\partial_u} (\partial_u \phi g^{au} \partial_a \phi)= g^{au}\partial_a \phi (1) + g^{au}\partial_u \phi \frac{\partial}{\partial_u \phi} (\partial_a \phi) + 0$

using the product rule

for the 2nd term I get $\frac{\partial}{\partial_u \phi} (\partial_a \phi) =\delta^{au}$

this must be wrong since this then gives me a factor of $4$ via the trace of the metric and so the index is not raised as required also.

14. Dec 28, 2017

### binbagsss

please can somebody help with my working here, is the delta term wrong? thanks

15. Dec 28, 2017

### haushofer

You're using very confusing notation, which I think is the reason people are not responding. I fix your notation and get

$\frac{\partial }{\partial_c} (\partial_b \phi \partial^b \phi) = \partial_c \Bigl(\eta^{ab} \partial_a \phi \partial_b \phi \Bigr) = \eta^{ab} \Bigl( (\partial_c \partial_a) \phi \partial_b \phi + \partial_a \phi ( \partial_c \partial _b \phi ) \Bigr) = \Bigl( (\partial_c \partial_a) \phi \partial^a \phi + \partial^a \phi ( \partial_c \partial_a \phi ) \Bigr)$

This equals

$2 \Bigl( (\partial_c \partial_a) \phi \partial^a \phi \Bigr)$

16. Dec 28, 2017

### stevendaryl

Staff Emeritus
Why do you want that? The Euler-Lagrange equations of motion are:

$\partial_u \frac{\partial \mathcal{L}}{\partial (\partial_u \phi)} = \frac{\partial \mathcal{L}}{\partial \phi}$

But $\frac{\partial \mathcal{L}}{\partial (\partial_u \phi)} = \partial^u \phi$ (assuming $\mathcal{L} = \frac{1}{2} \partial_u \phi \partial^u \phi$ + terms involving powers of $\phi$)

So unless you have a strange Lagrangian, you shouldn't have to compute $\partial_u (\partial_v \phi \partial^v \phi)$

17. Dec 29, 2017

### binbagsss

okay well I was trying to incrorporate Oroduin's hints since, as said in the OP, i can't get rid of the factor of 1/2 in the derivative term.

To give a specfic example if I consider: $-1/2 \partial_u \phi \partial^u \phi - 1/2 m^2 \phi^2$
then, my previous method, before making this thread was:
$\frac{\partial L}{\partial \phi}= m^2 \phi$
$\frac{\partial L}{\partial_u \phi}= \partial^u \phi (-1/2)$

therefore yielding $\partial_u \frac{\partial L}{\partial_u \phi} - \frac{\partial L}{\partial \phi} = -1/2 \partial^u\partial_u \phi + m^2 \phi$
whereas the EoM is : $\partial^u \partial_u \phi -m^2 \phi^2$

But now, trying to use Oroduin's hint I rewrite as $L=-1/2 \partial_u \phi g^{au} \partial_a \phi +m^2term...$
Now looking at the derivative term I get $\frac{\partial L}{\partial_u \phi} = =-1/2g^{au} ( \partial_a \phi \frac{\partial}{\partial_u \phi}(\partial_u \phi) + \partial_u \phi \frac{\partial}{\partial_u \phi}(\partial_a \phi))=-1/2 g^{au} ( \partial_a \phi (1) + \partial_u \phi \delta_{au})$
where I have used the product rule, and taken out the metric since it is the minkoski...

$= -1/2 (\partial^u \phi + g^{uu}\partial_u \phi)$...

Last edited: Dec 29, 2017
18. Dec 29, 2017

### stevendaryl

Staff Emeritus
You have to be careful about indices. If you look at just the term $\frac{1}{2} g^{au} (\partial_a \phi) (\partial_u \phi)$ and you take a derivative with respect to $\partial_u \phi$, you have to realize that in $\frac{1}{2} g^{au} (\partial_a \phi) (\partial_u \phi)$, $u$ is a dummy index. You can replace it by $v$, say. So you have:

$\frac{\partial}{\partial(\partial_u \phi)} \frac{1}{2} g^{av} (\partial_a \phi) (\partial_v \phi)$

Now, $\frac{\partial}{\partial(\partial_u \phi)} (\partial_a \phi) = 0$ unless $a=u$. So we can summarize it as:

$\frac{\partial}{\partial(\partial_u \phi)} (\partial_a \phi) = \delta^u_a$

Similarly,

$\frac{\partial}{\partial(\partial_u \phi)} (\partial_v \phi) = \delta^u_v$

So using the product rule gives you:

$\frac{\partial}{\partial(\partial_u \phi)} \frac{1}{2} g^{av} (\partial_a \phi) (\partial_v \phi) = \frac{1}{2} g^{av} (\delta^u_a \partial_v \phi + \delta^u_v \partial_a \phi)$

Then we can use $g^{av} \delta^u_a = g^{uv} \delta^u_a$. Summing over $a$ gives $g^{uv}$. Similarly, $g^{av} \delta^u_v = g^{au} \delta^u_v$. Summing over $v$ gives $g^{au}$. So this simplifies to:

$\frac{1}{2} (g^{uv} \partial_v \phi + g^{au} \partial_a \phi)$

Those two terms are the same, since $a$ and $v$ are dummies and $g^{au} = g^{ua}$. So we have:
$g^{uv} \partial_v \phi$

The EoM are just:

$\partial_u (g^{uv} \partial_v \phi) = \pm m^2 \phi$

(I don't remember whether the right side is a + or a -)

19. Dec 29, 2017

### stevendaryl

Staff Emeritus
In flat spacetime, you can take the $g^{uv}$ outside the derivative.