Find the range of ##k## for which ##\omega## is imaginary

[JD_PM]

Homework Statement

Given the relation (treat $k, \omega$ as variables and $v_0, \omega_p$ as constants).


$$y^2=1+2\alpha^2 \pm(1+8\alpha^2)^{1/2}, \ \text{where} \ \alpha:= kv_0/\omega_p, \ y:= \sqrt{2}\omega/\omega_p$$



a) Find the range of wave numbers $k$ for which the oscillation frequency $\omega$ is imaginary



b) Find $k_{max}$ and $\text{Im}(\omega_{max})$. Hint: sketch the solution for $\text{Im}(\omega)$

Relevant Equations

N/A

Out of the given information I see no way of solving neither of the two sections.

So I did some reading and it turns out that the given relation is a solution of the following function (I won't prove it here)

$$\epsilon (k, \omega) = 1 - \frac 1 2 \left[ \frac{\omega_p^2}{(\omega-kv_0)^2} + \frac{\omega_p^2}{(\omega+kv_0)^2}\right] = 0$$

So I think that finding the range of wave numbers $k$ for which the oscillation frequency $\omega$ is imaginary on the $\epsilon$ function should be equivalent. I did some manipulations

$$\frac{\omega_p^2}{(\omega-kv_0)^2} + \frac{\omega_p^2}{(\omega+kv_0)^2} = 2 \Rightarrow \omega_p^2\left( \frac{\omega^2+2kv_0+k^2v_0^2+\omega^2-2kv_0+k^2v_0^2}{(\omega^2-k^2v_0^2)^2} \right) = 2$$

$$\frac{(\omega^2+k^2v_0^2)}{(\omega^2-k^2v_0^2)^2} = \frac{1}{\omega_p^2}$$

But got stuck in the above step...

I strongly think we better avoid using the function and focus on the given relation to answer the questions.

Might you please guide me through?

Thanks.

 

[Mark44]

If $y^2 < 0$, then y will be imaginary, and since $y = \sqrt 2 \frac \omega {\omega_p}$, its seems likely that $\omega$ will also be imaginary, but without knowing whether $\omega_p$ is real, I can't say for sure.

Assuming for the moment that y being imaginary implies that $\omega$ will be imaginary, I'll continue on. Assuming that $\alpha$ is real, $y^2 < 0$ when $1 + 2\alpha^2 - \sqrt{1 + 8\alpha^2} < 0$; i.e., when $1 + 2\alpha^2 < \sqrt{1 + 8\alpha^2}$ . Solving this inequality gives you a range of values for $\alpha$, and presumably you can find a range of values for k.

 

[JD_PM]

Let's assume $\omega_p, k, v_0$ to be real.

$$(1 + 2\alpha^2)^2 < 1 + 8\alpha^2 \Rightarrow \alpha^2 + \alpha^4 < 2\alpha^2 \Rightarrow \alpha^2(\alpha^2 - 1) < 0$$

Hence $\alpha^2 < 0$ or $\alpha^2 < 1$. We assumed $\omega_p, k, v_0$ to be real so, by definition, $\alpha$ is real. Then, we stick to $\alpha^2 < 0$.

Assuming $v_0 \neq 0, \omega_p \neq 0$

$$k^4 < 0$$

Mmm but I expected to get a shorter range, I think I made a mistake somewhere.

 

[Mark44]

Hence $\alpha^2 < 0$ or $\alpha^2 < 1$. We assumed $\omega_p, k, v_0$ to be real so, by definition, $\alpha$ is real. Then, we stick to $\alpha^2 < 0$.

If $\alpha$ is real, then its square can never be negative. On the other hand, $\alpha^2 < 1 \Rightarrow -1 < \alpha < 1$.

 

[JD_PM]

If $\alpha$ is real, then its square can never be negative.

That was my naïve mistake indeed! Thanks.

Hence the $k$ range is given by

$$-\left| \frac{\omega_p}{v_0}\right| < k < \left| \frac{\omega_p}{v_0}\right|$$

If you agree, let's tackle the next section

b) Find $k_{max}$ and $\text{Im}(\omega_{max})$. Hint: sketch the solution for $\text{Im}(\omega)$

About the hint: do they want us to sketch the imaginary solution for the relation

$$y^2=1+2\alpha^2 \pm(1+8\alpha^2)^{1/2}, \ \text{where} \ \alpha:= kv_0/\omega_p, \ y:= \sqrt{2}\omega/\omega_p$$

?

That is

$$0 > y^2=1+2\alpha^2 - (1+8\alpha^2)^{1/2}$$

With $\omega$ in the vertical axis and $k$ in the horizontal axis.

 

[Mark44]

Hence the k range is given by $$-\left| \frac{\omega_p}{v_0}\right| < k < \left| \frac{\omega_p}{v_0}\right|$$

Looks fine. Let me think about your other question...

 

[Mark44]

Given that $y = \frac{\sqrt 2}{\omega_p}\omega$, then
$\left(\frac{\sqrt 2}{\omega_p}\omega \right)^2 = 1 + 2\alpha^2 - \sqrt{1 + 8\alpha^2}$
Solve for $\omega$ and then plug in some well-chosen values for $\alpha$. That's what I would do. Note that due to the range of values for k, the graph should lie between the left- and right bounds for k shown above.

 

[JD_PM]

Given that $y = \frac{\sqrt 2}{\omega_p}\omega$, then
$\left(\frac{\sqrt 2}{\omega_p}\omega \right)^2 = 1 + 2\alpha^2 - \sqrt{1 + 8\alpha^2}$
Solve for $\omega$ and then plug in some well-chosen values for $\alpha$.

We have

$$\omega = \pm \sqrt{\frac{\omega_p^2}{2} + \omega_p^2 \alpha^2 - \frac{\omega_p^2}{2} \sqrt{1 + 8\alpha^2} }$$

We know that $-1 < \alpha < 1$, so I plotted the $\omega$ solution (y axis $\omega$, x-axis $\alpha$; I set $\omega_p:= 1$ for simplicity)

Note that due to the range of values for k, the graph should lie between the left- and right bounds for k shown above.

Mmm I am doing something wrong, because within the left- and right bounds for $k$ I get nothing...

I thought of plugging special values for $\alpha$, such as $\alpha = 1$, but I get the trivial solution $\omega = 0$ ofc...

 

[Mark44]

$\left(\frac{\sqrt 2}{\omega_p}\omega \right)^2 = 1 + 2\alpha^2 - \sqrt{1 + 8\alpha^2}$

It's probably better to solve for $\omega^2$, and plot $\omega^2$ (vertical axis) vs. $alpha$ (horizontal axis). Remember my assumption that $\omega$ would be imaginary.

Here are some values I got in Excel, with $\alpha$ in the first column, and $\omega^2$ in the second column. I don't know the value for $\omega_p$, so I just used 1, which seems to be what you did, as well.

-0.9​	-0.05748​
-0.8​	-0.09693​
-0.7​	-0.11905​
-0.6​	-0.12489​
-0.5​	-0.11603​
-0.4​	-0.09498​
-0.3​	-0.06574​
-0.2​	-0.03446​
-0.1​	-0.00962​
-1.4E-16​	0​
0.1​	-0.00962​
0.2​	-0.03446​
0.3​	-0.06574​

 

[JD_PM]

It's probably better to solve for $\omega^2$, and plot $\omega^2$ (vertical axis) vs. $alpha$ (horizontal axis). Remember my assumption that $\omega$ would be imaginary.

.This is what I get for the $\omega^2$ (vertical axis) vs. $alpha$ (horizontal axis) plot (with $\omega_p:= 1$)

However, the hyperbolic behavior of the plot does not match your excel data...

Given that we are interested in plotting $\text{Im}(\omega)$ to eventually find $k_{max}$ and $\text{Im}(\omega_{max})$, wouldn't be best to plot $\omega$ instead of $\omega^2$? I would say that if we were to plot the former then $\text{Im}(\omega)$ would simply correspond to the $-1 < \alpha < 1$ range. Then, to find $\text{Im}(\omega_{max})$ we would just have to look at the maximum value of $\omega$ in that range and $k_{max}$ would be equivalent to find $\alpha_{max}$ (by definition of $\alpha$).

 

[Mark44]

Hence $\alpha^2 < 0$ or $\alpha^2 < 1$.

Which I explained by saying that the first inequality couldn't occur for any real $\alpha$, and the second one meant that $-1 < \alpha < 1$.

However, the hyperbolic behavior of the plot does not match your excel data..

Sure it does. In the spreadsheet I was interested only in values of $\alpha$ in the interval (-1, 1). For these values, the corresponding values of $\omega^2$ are negative, hence the values of $\omega$ are imaginary. Think back to the title you chose for this thread.

 

[JD_PM]

Do you see the following plot OK? (with the axis as mentioned)

.This is what I get for the $\omega^2$ (vertical axis) vs. $alpha$ (horizontal axis) plot (with $\omega_p:= 1$)

View attachment 295082

Sure it does. In the spreadsheet I was interested only in values of $\alpha$ in the interval (-1, 1). For these values, the corresponding values of $\omega^2$ are negative, hence the values of $\omega$ are imaginary. Think back to the title you chose for this thread.

I am thinking about this.

 

[Mark44]

Do you see the following plot OK? (with the axis as mentioned)

Yes, I saw the plot in that post. The part between the two parts of the hyperbola is precisely the part where $\omega$ is imaginary, which is what you asked about at the start of this thread.

 

[SammyS]

Do you see the following plot OK? (with the axis as mentioned)

I am thinking about this.

The spreadsheet values that @Mark44 posted are for $\omega^2$ versus $\alpha$, whereas it appears that your graphing package takes the square root of the values it calculates for $\omega^2$ and plots that versus $\alpha$.

 

[JD_PM]

Hey guys, sorry for the late reply. After some thoughts and discussion with a smart colleague we got an answer!

The plots are not required to solve the exercise, are just suggested. Let's try to avoid them for now.

b) Find $k_{max}$ and $\text{Im}(\omega_{max})$. Hint: sketch the solution for $\text{Im}(\omega)$

To find $k_{max}$ we can simply take $\frac{d y}{d \alpha} = 0$. This leads to $\alpha_{max} \pm \frac{2 \alpha_{max}}{\sqrt{1 + 8 \alpha_{max}^2}} = 0 \Rightarrow \alpha_{max} = \frac{\sqrt{6}}{4}$. We have defined $\alpha:= kv_0/\omega_p$, so it follows that $k_{max} = \frac{\sqrt{6}}{4} \frac{\omega_p}{v_0}$

Let's find $\text{Im}(\omega_{max})$. We already know that $\alpha_{max} = \frac{\sqrt{6}}{4}$. We have defined $y:= \sqrt{2}\omega/\omega_p$. Hence plugging $\alpha_{max}$ into $y$

$$\begin{align*}
\text{Im}(\omega_{max}) &= \pm \frac{\omega_p}{\sqrt{2}}\sqrt{1 + 2\alpha_{max}^2 \pm \sqrt{1 + 8\alpha_{max}^2}} \\
&= \pm \frac{\omega_p}{\sqrt{2}} \sqrt{1 + \frac 3 4 \pm 2} \\
&= \pm \frac{\omega_p}{2} \sqrt{\frac{15}{2}}
\end{align*}$$

Where I have used the fact that the imaginary part of a complex number is real.

Let me know your thoughts on the solution if you wish

 

[SammyS]

Hey guys, sorry for the late reply. After some thoughts and discussion with a smart colleague we got an answer!

The plots are not required to solve the exercise, are just suggested. Let's try to avoid them for now.
To find $k_{max}$ we can simply take $\frac{d y}{d \alpha} = 0$. This leads to $\alpha_{max} \pm \frac{2 \alpha_{max}}{\sqrt{1 + 8 \alpha_{max}^2}} = 0 \Rightarrow \alpha_{max} = \frac{\sqrt{6}}{4}$. We have defined $\alpha:= kv_0/\omega_p$, so it follows that $k_{max} = \frac{\sqrt{6}}{4} \frac{\omega_p}{v_0}$

Let's find $\text{Im}(\omega_{max})$. We already know that $\alpha_{max} = \frac{\sqrt{6}}{4}$. We have defined $y:= \sqrt{2}\omega/\omega_p$. Hence plugging $\alpha_{max}$ into $y$

$$\begin{align*}
\text{Im}(\omega_{max}) &= \pm \frac{\omega_p}{\sqrt{2}}\sqrt{1 + 2\alpha_{max}^2 \pm \sqrt{1 + 8\alpha_{max}^2}} \\
&= \pm \frac{\omega_p}{\sqrt{2}} \sqrt{1 + \frac 3 4 \pm 2} \\
&= \pm \frac{\omega_p}{2} \sqrt{\frac{15}{2}}
\end{align*}$$

Where I have used the fact that the imaginary part of a complex number is real.

Let me know your thoughts on the solution if you wish

Not quite right.

Remember, $y$, and thus $\omega$, is imaginary only when

$1 + 2\alpha_{max}^2 \pm \sqrt{1 + 8\alpha_{max}^2} < 0$

That happens only for the minus, not the plus, of the "$\pm$" and if $0<\alpha^2<1$.

Added in Edit 40 minutes later:

I think you will also find that $\dfrac{d y}{d \alpha} = 0$ only for minus, not for the plus.
... except, of course, at $\alpha =0$.

 

[SammyS]

In particular, notice that $\sqrt{1 + \dfrac 3 4 + 2} = \sqrt{\dfrac{15}{2}}$ is a real number,

whereas $\sqrt{1 + \dfrac 3 4 - 2} = \sqrt{\dfrac{-1}{4}} = \dfrac{1}{2} i \$ is imaginary.