# Derivative of first term in Lagrangian density for real K-G theory

1. Nov 11, 2014

Hey guys,

This is really confusing me cos its allowing me to create factors of 2 from nowhere!

Basically, the first term in the Lagrangian for a real Klein-Gordon theory is

$\frac{1}{2}(\partial_{\mu}\phi)(\partial^{\mu}\phi)$.

Now let's say I wana differentiate this by applying the $\partial_{\mu}$ operator. Using the chain rule, I get:

$\frac{1}{2} \left[ (\partial_{\mu}\partial_{\mu}\phi)(\partial^{\mu}\phi)+(\partial_{\mu}\phi)(\partial_{\mu}\partial^{\mu}\phi)\right]$

Which must be wrong because this cancels the factor of $\frac{1}{2}$ outside the square brackets!

My conclusion is that one term must be 0, or I'm doing something horribly wrong (or both :( ) can someone please correct me?

thank you!

2. Nov 12, 2014

### ShayanJ

Does it also confuse you that $\frac{d}{dx}( \frac 1 2 x^2)=x$?
Because its in fact the same thing. The product derivatives term in the KG Lagrangian is in fact $|\partial_\mu \phi|^2=(\partial_\mu \phi)^\dagger (\partial^\mu \phi)=(\partial_\mu \phi)(\partial^\mu \phi)^\dagger$.

3. Nov 12, 2014

### Orodruin

Staff Emeritus
In addition, you really should use a different index for your derivative or your expression is not very well defined within the Einstein summation convention (i.e., your $\mu$s are not the same).

@Shyan He has a real scalar field and thus the correct kinetic term.

4. Nov 12, 2014

Okay I sort of see but I still dont know how to evaluate that derivative, even if I change the index?

5. Nov 12, 2014

### Orodruin

Staff Emeritus
The question is, as Shyan said, why you think that it is wrong that the factor of 1/2 in front is cancelled - because it should be. Changing the summation index, you would have
$$(\partial_\nu\phi)(\partial^\nu\partial_\mu \phi).$$
I do not see that you can do very much else with this without putting it back into the context where you encountered it.

6. Nov 12, 2014

Yea I think I'll post the problem I'm trying to solve. I've come quite far into the solution I think but these indices are catching me out.

Thanks guys!

7. Nov 12, 2014

### ShayanJ

No difference. You should just remove the $\dagger$s in what I wrote because for a real field, $\phi^\dagger=\phi$.
The important point is that the kinetic term is in fact quadratic in derivatives so its not strange that a 2 shows up in the differentiation.

8. Nov 12, 2014

### Orodruin

Staff Emeritus
Classically, having a factor of 2 or not does not really matter. You end up with the same EoM for the free fields. Upon quantization, things are easier if you use canonical normalization for your kinetic term, which for a real scalar field includes a factor 1/2 in front.