- #1

- 98

- 13

I think it is equal to ##\frac{1}{2}(\delta(\omega_p+\omega_v-\omega)+\delta(\omega_p-\omega_v-\omega)+\delta(\omega_v-\omega_p-\omega)+\delta(-\omega_p-\omega_v-\omega))##, is it right?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Salmone
- Start date

- #1

- 98

- 13

I think it is equal to ##\frac{1}{2}(\delta(\omega_p+\omega_v-\omega)+\delta(\omega_p-\omega_v-\omega)+\delta(\omega_v-\omega_p-\omega)+\delta(-\omega_p-\omega_v-\omega))##, is it right?

- #2

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

- 15,860

- 2,515

Do you have some reason to think your answer is not correct?

I think it is equal to ##\frac{1}{2}(\delta(\omega_p+\omega_v-\omega)+\delta(\omega_p-\omega_v-\omega)+\delta(\omega_v-\omega_p-\omega)+\delta(-\omega_p-\omega_v-\omega))##, is it right?

- #3

- 98

- 13

No, I just want to be sure.Do you have some reason to think your answer is not correct?

- #4

Science Advisor

- 3,203

- 1,569

What does ##\Omega\simeq\omega## mean?

- #5

- 98

- 13

- #6

- 5,695

- 2,475

- #7

- 98

- 13

I can't edit the post no longer, but let's say ##\Omega=\Omega_1## and ##\omega=\omega_p-\omega_v=\Omega_2##.

- #8

- 5,695

- 2,475

ok fine if you put ##\Omega_2## instead of ##\omega## then I guess you are free to use ##\omega## as the variable of the Fourier transformI can't edit the post no longer, but let's say ##\Omega=\Omega_1## and ##\omega=\omega_p-\omega_v=\Omega_2##.

Share:

- Replies
- 0

- Views
- 608

- Replies
- 7

- Views
- 1K

- Replies
- 4

- Views
- 1K

- Replies
- 4

- Views
- 1K

- Replies
- 3

- Views
- 804

- Replies
- 8

- Views
- 3K

- Replies
- 1

- Views
- 1K

- Replies
- 4

- Views
- 3K

- Replies
- 6

- Views
- 9K

- Replies
- 2

- Views
- 2K