# Derivative of lagrangian density

1. Nov 25, 2015

### kawillzocken

i have a mathematical question which is quite similar to one asked before, still a bit different
https://www.physicsforums.com/threa...agrangian-density-for-real-k-g-theory.781472/

the first term of KG-Lagrangian is: $\frac{1}{2}(\partial^{\mu} \phi)(\partial_{\mu} \phi)$
when i try do find $\frac{\partial L}{\partial(\partial^{\mu}\phi)}$, i have two different options, where of course only one can be right.

1) $\partial^{\mu} \phi$ and $\partial_{\mu} \phi$ are different things, so one gets: $\frac{\partial L}{\partial(\partial^{\mu}\phi)} = \frac{1}{2}\partial_{\mu} \phi$

2) $\partial_{\mu} \phi = \eta_{\mu \nu} \partial^{\nu} \phi \rightarrow \frac{\partial L}{\partial(\partial^{\mu}\phi)} = \frac{\partial}{\partial(\partial^{\mu}\phi)}(\frac{1}{2}\eta_{\mu \nu}(\partial^{\nu} \phi)(\partial^{\mu} \phi)) = \partial_{\mu} \phi$

i am confused. several sources tell me that 2) is right, but my understanding of partial derivative tells me to do 1).

as i think of it, one may not simply put equations into others when doing the partial derivative, because it then changes. like z(x, y(x)) = 2x + y(x). partial derivative with respect to x gives 2, but it changes when i put in y(x).

thank you

2. Nov 25, 2015

### Orodruin

Staff Emeritus
Option 2 is correct. The quantities $\partial_\mu\phi$ and $\partial^\mu\phi$ may not be equal, but they do not correspond to different degrees of freedom (they are related by a sign). Compare this to differentiating $-x$ with respect to $x$, $-x$ is not equal to $x$, but it is directly dependent.

3. Nov 25, 2015

### stevendaryl

Staff Emeritus
No, they are definitely not independent. If you write it with the metric, then this becomes more obvious:

$\mathcal{L} = \frac{1}{2}g^{\mu \nu} (\partial_{\mu} \phi\ \partial_{\nu} \phi)$

So $\frac{\partial \mathcal{L}}{\partial (\partial^{\sigma} \phi)} = \frac{1}{2}g^{\mu \nu} (\delta^\sigma_\mu \partial_{\nu} \phi + \delta^\sigma_\nu \partial_{\mu} \phi) = \frac{1}{2} (g^{\sigma \nu} \partial_{\nu} \phi + g^{\mu \sigma} \partial_{\mu} \phi) = \frac{1}{2} (\partial^{\sigma} \phi + \partial^{\sigma} \phi) = \partial^{\sigma} \phi$

4. Nov 25, 2015

### kawillzocken

i do not say they are independent. y(x) and x are not indepentent either, still in a partial derivative of z with respect to x, one does not consider the y(x).

so orodruin, i see that they are related by the metric. y(x) is also related to x by its defining function. so i do still not get the point, where my z-function analogy is failing.
do y and x correspond to different degrees of freedom in z?

regards, kawillzocken

5. Nov 25, 2015

### stevendaryl

Staff Emeritus
Let's make it a lot simpler, by considering only 1 space dimension and 1 time dimension. Then the lagrangian is:

$\mathcal{L} = \frac{1}{2}(\frac{\partial \phi}{\partial t} \frac{\partial \phi}{\partial t} - \frac{\partial \phi}{\partial x} \frac{\partial \phi}{\partial x})$

I think it's pretty obviousthat $\frac{\partial \mathcal{L}}{\partial (\frac{\partial \phi}{\partial t})} = \frac{\partial \phi}{\partial t}$. Right?

6. Nov 25, 2015

### stevendaryl

Staff Emeritus
In a second-order differential equation for $\phi$ (which is what the lagrangian equations of motion produce), to know $\phi$ everywhere, it is good enough to specify $\phi, \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}, \frac{\partial \phi}{\partial t}$ at a single point $(x_0, y_0, z_0, t_0)$. Those are 5 completely independent choices.