Timeline for What is $\epsilon(\infty)$ in $\epsilon(\omega) = \epsilon(\infty)[1-\bar\omega_p^2/\omega^2]$, in Kittel's solid state book?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 11, 2016 at 12:57 | comment | added | Robin Ekman | @EmilioPisanty I agree completely. | |
| Oct 11, 2016 at 12:49 | comment | added | valerio | I agree with @EmilioPisanty. Until we ask Kittel himself, this is just notation and we should not be too concerned about it. | |
| Oct 11, 2016 at 12:40 | comment | added | Emilio Pisanty | @RobinEkman Perhaps I wasn't clear earlier: The reasons for the notation do not matter. Only the physics does, and the notation is not the physics. | |
| Oct 11, 2016 at 12:38 | comment | added | Robin Ekman | I think it's $\epsilon(\infty)$ because it could be seen as assuming infinitely massive ions, compared to electrons. Cf. the notation for the Rydberg constant. | |
| Oct 11, 2016 at 10:18 | comment | added | Emilio Pisanty | @Sad_lab_rat This is all terminology - there's nothing to see there, really. Your argument works, but it's only notation (and, in any case $\infty$ is actual infinity unless you preface it with "except you shouldn't go too high", which Kittel doesn't do). The physics is the only thing that actually matters. | |
| Oct 11, 2016 at 10:16 | comment | added | Sad_lab_rat | Kittel does use the phrase "$\epsilon(\infty)$ is essentially constant upto frequencies well above $\omega_p$". Doesn't that mean that he's using the "constant" nature of background from low frequency and going up to high frequencies? The high frequencies needn't be so high but they're sufficiently higher than $\omega_p$ to be considered $\infty$ wrt $\omega_p$ | |
| Oct 11, 2016 at 10:10 | comment | added | Emilio Pisanty | @valerio92 That's of limited use, though - the assumption that the background permittivity is flat only works for low frequencies, so getting $A$ from $\omega\to\infty$ only works in the vaguest, most symbolic of senses. | |
| Oct 11, 2016 at 10:05 | comment | added | valerio | I think he uses that symbol just because once you assume $\epsilon(\omega)=A(1-\omega_p^2/\omega^2)$ where $A$ is some constant you get $\epsilon(\omega \to \infty)=A$ | |
| Oct 11, 2016 at 10:03 | vote | accept | Sad_lab_rat | ||
| Oct 11, 2016 at 9:56 | history | answered | Emilio Pisanty | CC BY-SA 3.0 |