How to calculate this dispersion relation

Click For Summary
The discussion focuses on the calculation of a dispersion relation, highlighting confusion regarding its derivation and the significance of the plus and minus signs in the formula. It notes that non-trivial solutions require the determinant of a 2x2 matrix to equal zero, but there are errors in the matrix elements that affect dimensional correctness. An extra factor of omega is mentioned, and the need to handle two cases separately based on the relationship between omega and omega_c is emphasized. The first case yields a minus sign in the denominator, while the second case is more complex, likely resulting in a plus sign. Overall, the conversation reveals challenges in deriving dispersion relations and the necessity for careful consideration of specific cases.
Karl86
Messages
40
Reaction score
3
Homework Statement
Let ##\mathbf{E}## be an electric field that behaves like ##e^{i\mathbf{k}\cdot\mathbf{r} -\omega t}## and consider the equation relating ##\mathbf{E}## to the displacement field ##\mathbf{D}##: $$\mathbf{k}^2 \mathbf{E}=\frac{\omega^2}{c^2}\mathbf{D}$$ ##\mathbf{D}## is related to ##\mathbf{E}## by a permittivity matrix ##\epsilon_{ij}## such that ##D_i=\sum_j\epsilon_{ij} E_j##, but this should not be relevant. If I know that the solutions ##\mathbf{E}## to the equation that are transverse to ##\mathbf{k}##, are given by the ##(E_1,E_2)## such that
$$ \begin{pmatrix} \omega^2 - c^2 k^2 -\frac{\omega_p^2 \omega^2}{\omega^2 - \omega_c^2} & i\frac{\omega_p^2 \omega_c \omega^2}{\omega^2-\omega_c^2} \\ -i\frac{\omega_p^2 \omega_c \omega^2}{\omega^2-\omega_c^2} & \omega^2 - c^2 k^2 -\frac{\omega_p^2 \omega^2}{\omega^2 - \omega_c^2} \end{pmatrix} \begin{pmatrix} E_1 \\ E_2 \end{pmatrix} = 0 $$ It is claimed that the dispersion relation of such a wave solution is $$c^2 k^2 = \omega^2 \left(1 - \frac{\omega_p^2}{\omega(\omega \pm \omega_c)}\right) $$ I don't understand this.
Relevant Equations
##\omega^2=\Omega^2(k)## ??
I have no idea how this dispersion relation was deduced, and also what's the meaning of including plus and minus in the formula.
 
Physics news on Phys.org
The non-trivial solutions have the determinant of the 2x2 matrix on the left equal to zero. Perhaps that helps. Otherwise, for a complete treatment, see Ichimaru's Basic Principles of Plasma Physics A Statistical Approach Section 3.1.
 
  • Like
Likes Karl86
Somewhere, there seems to be an error or two in that 2x2 matrix. The (1,2) and (2,1) elements are not dimensionally correct.
 
Charles Link said:
Somewhere, there seems to be an error or two in that 2x2 matrix. The (1,2) and (2,1) elements are not dimensionally correct.
There is one extra ##\omega## factor in both but I can no longer edit the post. It all worked out anyway, except that I thought there was a general strategy to obtain dispersion relations for generic waves, instead it looks like there are ad hoc methods, so to speak.
 
Yes, I see it now also. It should be first power of ## \omega ## in the numerator of the (2,1) and (1,2) terms.
 
And for this one, you need to do two cases separately: ## \omega > \omega_c ##, and ## \omega < \omega_c ##. I worked the first case and got the minus sign in the denominator. The second one is trickier, but it no doubt gives the plus in the denominator.
 
  • Like
Likes Karl86
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K