Partial derivative w.r.t. another partial derivative

In summary, the conversation discusses how to calculate the partial derivative of a given equation and its implications. The equation in question is not well-defined due to the power operator not being defined for vectors, but it can be rewritten as a dot product. The result is correct except for a factor of 2, and it is mentioned that the A term should be zero.
  • #1
cc94
19
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
  • #2
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.
 
  • #3
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!
 

1. What is a partial derivative with respect to another partial derivative?

A partial derivative with respect to another partial derivative is a mathematical concept that involves calculating the rate of change of a function with respect to one of its independent variables while holding another independent variable constant.

2. How is a partial derivative with respect to another partial derivative calculated?

To calculate a partial derivative with respect to another partial derivative, the chain rule is used. This involves taking the derivative of the function with respect to the first independent variable, and then multiplying it by the partial derivative of the first independent variable with respect to the second independent variable.

3. What is the purpose of calculating a partial derivative with respect to another partial derivative?

The purpose of calculating a partial derivative with respect to another partial derivative is to determine how a function changes when both of its independent variables change simultaneously. This is useful in many fields of science and engineering, such as in optimization problems and in understanding how systems behave.

4. Can a partial derivative with respect to another partial derivative be negative?

Yes, a partial derivative with respect to another partial derivative can be negative. This would indicate that the function is decreasing with respect to one independent variable while the other independent variable is increasing.

5. Are there any practical applications of partial derivatives with respect to another partial derivative?

Yes, there are many practical applications of partial derivatives with respect to another partial derivative. For example, in economics, this concept is used to analyze how changes in multiple variables affect the production and consumption of goods. In physics, it is used to understand how multiple factors contribute to the overall behavior of a system. It is also used in machine learning and data analysis to optimize models and make predictions.

Similar threads

  • Advanced Physics Homework Help
Replies
3
Views
286
  • Advanced Physics Homework Help
Replies
10
Views
458
  • Advanced Physics Homework Help
Replies
9
Views
2K
  • Advanced Physics Homework Help
Replies
9
Views
1K
  • Advanced Physics Homework Help
Replies
10
Views
962
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
822
  • Advanced Physics Homework Help
Replies
1
Views
881
  • Advanced Physics Homework Help
Replies
1
Views
875
  • Advanced Physics Homework Help
Replies
1
Views
760
Back
Top