- #1

- 459

- 5

## Main Question or Discussion Point

What does the subscript [tex](\theta, \phi)[/tex] mean on the laplace operator? i.e.

[tex]{\nabla}^2 V_{(\theta, \phi)}[/tex]

[tex]{\nabla}^2 V_{(\theta, \phi)}[/tex]

- Thread starter Swapnil
- Start date

- #1

- 459

- 5

What does the subscript [tex](\theta, \phi)[/tex] mean on the laplace operator? i.e.

[tex]{\nabla}^2 V_{(\theta, \phi)}[/tex]

[tex]{\nabla}^2 V_{(\theta, \phi)}[/tex]

- #2

- 5,490

- 720

Can you provide the context of this notation?

- #3

- 441

- 0

- #4

- 932

- 0

[tex]\nabla^2_{r'} V(r-r')[/tex]

where the subscript is to remind us that, as CPL.Luke says, that we are differentiating with respect to the dashed variables (or undashed, as it is in your example)

- #5

- 459

- 5

Ooops.. I meant to put the subscript on the operator not on the function. Sorry about that.

[tex]\nabla^2_{r'} V(r-r')[/tex]

where the subscript is to remind us that, as CPL.Luke says, that we are differentiating with respect to the dashed variables (or undashed, as it is in your example)

- #6

- 459

- 5

- #7

quasar987

Science Advisor

Homework Helper

Gold Member

- 4,773

- 8

[tex]\nabla^2_{xy}=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}[/tex]

- #8

- 162

- 5

This is what i would say. i believe the theta and phi in the subscript in the original post imply spherical coordinates.

[tex]\nabla^2_{xy}=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}[/tex]

- #9

Meir Achuz

Science Advisor

Homework Helper

Gold Member

- 2,169

- 63

without the radial derivatives.

- Last Post

- Replies
- 18

- Views
- 3K

- Replies
- 5

- Views
- 269

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 6

- Views
- 2K

- Replies
- 43

- Views
- 7K

- Last Post

- Replies
- 8

- Views
- 7K

- Last Post

- Replies
- 8

- Views
- 2K

- Replies
- 24

- Views
- 11K

- Replies
- 20

- Views
- 5K

- Last Post

- Replies
- 0

- Views
- 1K