Partial derivative w.r.t. another partial derivative

cc94
Messages
19
Reaction score
2

Homework Statement


Given
$$L = \left(\nabla\phi + \dot{\textbf{A}}\right)^2 ,$$

how do you calculate $$\frac{\partial}{\partial x}\left(\frac{\partial L}{\partial(\partial\phi / \partial x)}\right)?$$

Homework Equations


By summing over the x, y, and z derivatives, the answer is supposed to show that ##\nabla^2 \phi = 0##.

The Attempt at a Solution


My vector calculus isn't too good, so here's my guess ( ##\partial\phi / \partial x \equiv \phi_x##):

$$\begin{align}
\frac{\partial L}{\partial(\phi_x)} &= 2(\nabla\phi + \dot{\textbf{A}})\cdot\left[\frac{\partial}{\partial\phi_x} (\nabla\phi + \dot{\textbf{A}})\right] \\
&= 2\left(\langle\phi_x + A_x,\phi_y + A_y,\phi_z + A_z\rangle\right)\cdot\left[\frac{\partial}{\partial(\phi_x)}\langle\phi_x + A_x, \phi_y + A_y, \phi_z + A_z\rangle\right] \\
&= 2\left(\langle\phi_x + A_x,\phi_y + A_y,\phi_z + A_z\rangle\right)\cdot\langle 1,0,0\rangle\\
&= 2(\phi_x + \dot{A}_x)
\end{align}
$$

and so

$$\frac{\partial}{\partial x}\frac{\partial L}{\partial(\phi_x)} = 2(\phi_{xx} + \dot{A}_{xx})$$

and summing gives ##\nabla^2\phi + \nabla^2 \dot{A}##.

So did I calculate ##\frac{\partial L}{\partial(\phi_x)}## correctly? If so, why is there still an A term at the end?
 
Physics news on Phys.org
cc94 said:
$$L = \left(\nabla\phi + \dot{\textbf{A}}\right)^2 ,$$
This is a meaningless equation. The power operator is not defined for vectors. ##\vec x \vec x = ## ?
If it were written as ##L =|\nabla\phi + \dot{\textbf{A}}|^2##, it would be equivalent to ##L =\left(\nabla\phi + \dot{\textbf{A}}\right) \cdot \left(\nabla\phi + \dot{\textbf{A}}\right)##. Then your result would be correct except for a factor of 2.
 
tnich said:
This is a meaningless equation. The power operator is not defined for vectors. ##\vec x \vec x = ## ?
If it were written as ##L =|\nabla\phi + \dot{\textbf{A}}|^2##, it would be equivalent to ##L =\left(\nabla\phi + \dot{\textbf{A}}\right) \cdot \left(\nabla\phi + \dot{\textbf{A}}\right)##. Then your result would be correct except for a factor of 2.

Er, yeah the power is supposed to be the dot product like you wrote. Ok, so I just have to figure out why the A term is zero. Thanks!
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top