# Homework Help: Wave Equation / Damping / Phase Velocity

1. Dec 2, 2006

### piano.lisa

1. The problem statement, all variables and given/known data
Consider the simplified wave function: $$\psi (x,t) = Ae^{i(\omega t - kx)}$$
Assume that $$\omega$$ and $$\nu$$ are complex quantities and that k is real:
$$\omega = \alpha + i\beta$$
$$\nu = u + i\omega$$
Show that the wave is damped in time. Use the fact that $$k^2 = \frac{\omega^2}{\nu^2}$$ to obtain expressions for $$\alpha$$ and $$\beta$$ in terms of $$u$$ and $$\omega$$. Find the phase velocity for this case.

2. Relevant equations
i $$\psi (x,t) = Ae^{i(\omega t - kx)}$$
ii $$\omega = \alpha + i\beta$$
iii $$\nu = u + i\omega$$
iv $$k^2 = \frac{\omega^2}{\nu^2}$$

3. The attempt at a solution
I substitued ii into i to obtain the expression: $$\psi (x,t) = Ae^{-\beta t}e^{i(\alpha t - kx)}$$. Therefore, the factor of $$e^{-\beta t}$$ represents the damping of the wave in time. There is no damping in the position of the wave.
I cannot seem to find expressions for $$\alpha$$ and $$\beta$$ in terms of $$u$$ and $$\omega$$. I have tried rearranging the given equations in many such ways, but have not come up with any conclusive result.

Any suggestions are greatly appreciated. Thank you.

Last edited: Dec 2, 2006
2. Dec 2, 2006

### OlderDan

If $$k^2 = \frac{\omega^2}{\nu^2}$$ means the ratio of the actual squares of the complex numbers, and not the squares of their norms, then the condition that k is real imposes a restriction that allows you to solve for $\alpha$ and $\beta$ in terms of the u and the norm of $\omega$. It's a bit of tediuous algebra to square out the complex numbers and rationalize the denominator and set the imaginary part of $$k^2 = \frac{\omega^2}{\nu^2}$$ to zero.

3. Dec 2, 2006

### piano.lisa

Why is it tedious to square the complex numbers?
Do I have to do, for example, (v)(v*) ?

4. Dec 2, 2006

### OlderDan

(v)(v*) is not bad. My question is, does v² mean (v)(v*) or does it mean (v)(v)? If it means (v)(v*), I have not yet found a way of solving for $\alpha$ and $\beta$. However, if it means (v)(v), then demanding that k² is real leads to a solution.

5. Dec 2, 2006

### piano.lisa

Using (v)(v) to solve,
I obtained:
$$\beta = \frac{\alpha}{2u\omega}[(\omega^2 - u^2) \pm (u^2 + \omega^2)]$$
However, $$\beta$$ is still in terms of $$\alpha$$, so I'm not sure what I'm doing wrong.

Therefore, I obtained
$$\beta = \frac{\alpha \omega}{u}$$ OR $$\beta = -\frac{\alpha u}{\omega}$$
Likewise, $$\alpha = \frac{\beta u}{\omega}$$ OR $$\alpha = -\frac{\beta \omega}{u}$$.

Assuming those are right, how can I find the phase velocity from this point? If k is real, does that imply that $$v_{phase} = v$$?

Last edited: Dec 2, 2006
6. Dec 2, 2006

### OlderDan

I'm not sure what you did, but I should have been more explicit about being consistent in interpreting the square. If v² = vv rather than vv*, then ω² = ωω rather than ωω*.

If k² = ωω/vv then the numertor and denominator are both complex. If you expand them and rationalize the denominator you get k² = Re(k²) + iIm(k²). If you then demand Im(k²) = 0 you get an equation that can be solved for beta in terms of u and ωω*. But you also know from the complex expression for ω that the sum of the squares of alpha and beta is also ωω*, so you can use that to solve for alpha in terms of u and ωω*.

The assumption that v² = vv and ω² = ωω is equivalent to simply saying that k = ω/v, and one can get the same equations for alpha and beta by taking k = ω/v = ωv*/vv* and demanding that the imaginary part vanish. I don't know why they would give you k² = ω²/v² if they really meant k = ω/v. On the other hand, if k = ω/v is valid, then v = ω/k and you have the usual expression for phase velocity. Of course in this case, v is complex.

I'm not at all confident that the assumption is valid, so it would be good if you had a way of determining for sure what they mean by k² = ω²/v² . I am also bothered by an apparent dimensional inconsistency in the definition of v. Could it be that your equation $\nu = u + i\omega$ is missing something?

I will take another look at it assuming they mean the squares of the norms and see if I can find a solution.

Last edited: Dec 2, 2006
7. Dec 2, 2006

### piano.lisa

I think that's what I did...
Here, I'll show you:
$$k^2 = \omega^2 / v^2 = \frac{(\alpha + i\beta)^2}{(u + i\omega)^2}$$
$$= \frac{(\alpha^2 - \beta^2 + 2i\alpha \beta)(u^2 - \omega^2 - 2iu\omega)}{(u^2 - \omega^2 + 2iu\omega)(u^2 - \omega^2 - 2iu\omega)}$$
$$k^2 = \frac{(u^2 - \omega^2)(\alpha^2 - \beta^2) + 4\alpha\beta u\omega}{(u^2 + \omega^2)^2} + i\frac{(u^2 - \omega^2)(2\alpha\beta) - (2u\omega)(\alpha^2 - \beta^2)}{(u^2 + \omega^2)^2}$$

From there, I set $$Im(k^2) = 0$$ and solved for $$\beta$$ with the quadratic formula. However, $$\beta$$ was still in terms of $$\alpha$$. I could not make $$\beta$$ in terms of $$\omega\omega *$$ without it also being in terms of $$\alpha$$.

8. Dec 2, 2006

### OlderDan

When you write
$$\nu = u + i\omega$$
you have to remember that
$$\omega = \alpha + i\beta$$
is complex. When you substitute you get
$$\nu = u + i(\alpha + i\beta) = (u - \beta) + i\alpha$$
The real part of this is
$$Re(\nu) = (u - \beta)$$
and the imaginary part is
$$Im(\nu) = \alpha$$

If you do the calculation again your rationalization should give you an imaginary part that will lead you to

$$\beta = \frac{{\omega \omega^*}}{u}$$

and then

$$\alpha = \sqrt {\omega \omega^*\left( {1 - \frac{{\omega \omega^*}}{{u^2 }}} \right)}$$

I have to go, but I'll check back in tomorrow morning.

Last edited: Dec 2, 2006
9. Dec 2, 2006

### piano.lisa

Taking into account what you said,
I simplified further using the fact that w is complex, and I obtained:
$$\beta = \frac{u \pm \sqrt{u^2 - 4\alpha^2}}{2}$$
Substituting $$\alpha = \omega - i\beta$$ into the above,
I obtained:
$$\beta = \frac{\omega^2}{u + 2i\omega}$$
Which is slightly the same as yours, if I take $$Im(\beta) = 0$$.

However, I also obtain:
$$\alpha = \frac{\omega (u + i\omega)}{u + 2i\omega}$$

Last edited: Dec 2, 2006
10. Dec 3, 2006

### OlderDan

I don't think you can salvage your calculation by putting complex ω in at the end. If you have complex ω in your denominator, then when you rationalize it you have to replace i with -i and you also have to replace ω with ω*. I suggest you write v in the usual Re(v) + iIm(v) form and work through it again. There is really no need to use k² = ω²/v² if "actual" squares are bing assumed, since that is equivalent to assuming k = ω/v. If you just rationalize ω/v and demand k real, you will get the result.

If you will make one more attempt at that, I will post everything I have done, since I don't have a good feeling about this "actual squares" assumption. I have a solution assuming square norms instead of actual squares, but I cannot eliminate k from the solution. Maybe they intended for the solution to include k. It is a bit ugly. As I said before, I don't like the apparent dimensional inconsistency in v, so I did the square norm calculation again assuming v = u + iω/k. The result still depends on k, but it is a simpler looking result than one gets without the k.

Last edited: Dec 3, 2006
11. Dec 3, 2006

### piano.lisa

This was my method:
$$k^2 = \frac{\omega^2}{v^2} = \frac{(\alpha + i\beta)^2}{(u - \beta +i\alpha)^2}$$

Then, I expand the brackets, and rationalize the denominator.
$$k^2 = \frac{(\alpha^2 - \beta^2 + 2i\alpha\beta)[(u^2 + \beta^2 - \alpha^2 - 2u\beta) - i(2u\alpha - 2\beta\alpha)]}{[(u^2 + \beta^2 - \alpha^2 - 2u\beta) + i(2u\alpha - 2\beta\alpha)][(u^2 + \beta^2 - \alpha^2 - 2u\beta) - i(2u\alpha - 2\beta\alpha)]}$$

$$k^2 = \frac{(\alpha^2 - \beta^2 + 2i\alpha\beta)(u^2 + \beta^2 - \alpha^2 - 2u\beta) - i(\alpha^2 - \beta^2 + 2i\alpha\beta)(2u\alpha - 2\beta\alpha)}{(u^2 + \beta^2 - \alpha^2 - 2u\beta)^2 + (2u\alpha + 2\beta\alpha)^2}$$

From there, I have
$$k^2 = \frac{REAL + i(2\alpha\beta u^2 - 2\alpha\beta^2 u - 2\alpha^3 u)}{denominator}$$
I then set $$Im(k^2) = 0$$ and obtained the solutions I stated previously for $$\alpha$$ and $$\beta$$

$$2\alpha\beta u^2 - 2u\alpha\beta^2 - 2\alpha^3 u = 0$$

$$\alpha^2 - \beta u + \beta^2 = 0$$

$$\beta = \frac{u \pm \sqrt{u^2 - 4\alpha^2}}{2}$$

Then, I substitute $$\alpha = \omega - i\beta$$ to obtain $$\beta(-4u - 8i\omega) + 4\omega^2 = 0$$
This gives:
$$\beta = \frac{\omega^2}{u + 2i\omega}$$ and $$\alpha = \frac{\omega (u + i\omega)}{u + 2i\omega}$$

You're right though... my answer would be the same as doing k=w/v... So what is the problem with my method?
If I do $$k^2 = \frac{\omega\omega *}{vv*}$$, I end up with a real value for k, obviously... But I cannot solve for $$\alpha$$ and $$\beta$$ in terms of $$u$$ and $$\omega$$.

Thank you.

Last edited: Dec 3, 2006
12. Dec 3, 2006

### OlderDan

Perhaps I am misunderstanding, but this

looks very different to me than what you get from expanding

$$k^2 = \frac{\omega^2}{v^2} = \frac{(\alpha + i\beta)^2}{(u - \beta +i\alpha)^2}$$

When I did it, every term that involved an αβ product canceled out. I hope I did not make an algebra mistake. Here is what I have. Check it over.

$$\omega = \alpha + i\beta$$

$$v = u + i\omega = u + i\left( {\alpha + i\beta } \right) = u - \beta + i\alpha$$

$$k^2 = \frac{{\omega ^2 }}{{v^2 }} = \frac{{\alpha ^2 - \beta ^2 + 2i\alpha \beta }}{{u^2 - 2u\beta + \beta ^2 + 2i\alpha \left( {u - \beta } \right) - \alpha ^2 }}$$

$$k^2 = \frac{{\left( {\alpha ^2 - \beta ^2 + 2i\alpha \beta } \right)\left( {u^2 - 2u\beta + \beta ^2 - \alpha ^2 - 2i\alpha \left( {u - \beta } \right)} \right)}}{{\left| {u^2 - 2u\beta + \beta ^2 - \alpha ^2 + 2i\alpha \left( {u - \beta } \right)} \right|^2 }}$$

$$Im \left( k^2 \right) = 0 = \frac{{2i\alpha \beta \left( {u^2 - 2u\beta + \beta ^2 - \alpha ^2 } \right) - 2i\alpha \left( {u - \beta } \right)\left( {\alpha ^2 - \beta ^2 } \right)}}{{\left| {u^2 + \beta ^2 - 2u\beta - \alpha ^2 + 2i\alpha \left( {u - \beta } \right)} \right|^2 }}$$

$$0 = \beta \left( {u^2 - 2u\beta + \beta ^2 - \alpha ^2 } \right) - \left( {u - \beta } \right)\left( {\alpha ^2 - \beta ^2 } \right)$$

$$0 = \beta u^2 - 2u\beta ^2 + \beta ^3 - \beta \alpha ^2 - u\alpha ^2 + u\beta ^2 + \beta \alpha ^2 - \beta ^3$$

$$0 = \beta u^2 - u\beta ^2 - u\alpha ^2$$

$$\beta u = \alpha ^2 + \beta ^2 = \omega \omega ^*$$

$$\beta = \frac{{\omega \omega ^* }}{u}$$

$$\alpha ^2 = \omega \omega ^* - \beta ^2 = \omega \omega * - \left( {\frac{{\omega \omega ^* }}{u}} \right)^2 = \omega \omega *\left( {1 - \frac{{\omega \omega ^* }}{{u^2 }}} \right)$$

$$\alpha = \sqrt {\omega \omega ^* \left( {1 - \frac{{\omega \omega ^* }}{{u^2 }}} \right)}$$

Last edited: Dec 3, 2006
13. Dec 3, 2006

### piano.lisa

Thank you so much.

We do have the same expansion of $$k^2$$.
Except, at the point where $$0 = \beta u - \beta^2 - \alpha^2$$, I solved for $$\beta$$ using the quadratic formula, and you used a more efficient method.

14. Dec 3, 2006

### OlderDan

Here is what I have for the square norms calculation

$$\omega = \alpha + i\beta$$

$$v = u + i\omega = u + i\left( {\alpha + i\beta } \right) = u - \beta + i\alpha$$

$$k^2 = \frac{{\omega ^2 }}{{v^2 }}$$

$$\omega ^2 = \alpha ^2 + \beta ^2$$

$$v^2 = u^2 - 2u\beta + \beta ^2 + \alpha ^2 = u^2 - 2u\beta + \omega ^2$$

$$\frac{{\omega ^2 }}{{k^2 }} = u^2 - 2u\beta + \omega ^2$$

$$\omega ^2 = k^2 \left( {u^2 - 2u\beta + \omega ^2 } \right)$$

$$\left( {1 - k^2 } \right)\omega ^2 = k^2 u\left( {u - 2\beta } \right)$$

$$u - 2\beta = \left( {\frac{{1 - k^2 }}{{k^2 }}} \right)\frac{{\omega ^2 }}{u}$$

$$2\beta = u - \left( {\frac{{1 - k^2 }}{{k^2 }}} \right)\frac{{\omega ^2 }}{u}$$

$$\beta = \frac{u}{2} - \left( {\frac{{1 - k^2 }}{{k^2 }}} \right)\frac{{\omega ^2 }}{{2u}}$$

$$\alpha ^2 = \omega ^2 - \beta ^2 = \omega ^2 - \left[ {\frac{u}{2}^2 - \left( {\frac{{1 - k^2 }}{{2k^2 }}} \right)\omega ^2 + \left( {\frac{{1 - k^2 }}{{2k^2 }}} \right)^2 \left( {\frac{{\omega ^2 }}{u}} \right)^2 } \right]$$

$$\alpha ^2 = \omega ^2 - \omega ^2 \left[ {\frac{{u^2 }}{{2\omega ^2 }} - \left( {\frac{{1 - k^2 }}{{2k^2 }}} \right) + \left( {\frac{{1 - k^2 }}{{2k^2 }}} \right)^2 \left( {\frac{\omega }{u}} \right)^2 } \right]$$

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

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

15. Dec 3, 2006

### OlderDan

And here is what happens if you throw a 1/k into the imaginary part of v to fix the apparent dimensional inconsistency (pure speculation on my part). This is a square norm calculation.

$$\omega = \alpha + i\beta$$

$$v = u + i\frac{\omega }{k} = u + i\left( {\frac{\alpha }{k} + i\frac{\beta }{k}} \right) = u - \frac{\beta }{k} + i\frac{\alpha }{k}$$

$$k^2 = \frac{{\omega ^2 }}{{v^2 }}$$

$$\omega ^2 = \alpha ^2 + \beta ^2$$

$$v^2 = u^2 + \left( {\frac{\beta }{k}} \right)^2 - 2u\frac{\beta }{k} + \left( {\frac{\alpha }{k}} \right)^2 = u^2 - 2u\frac{\beta }{k} + \frac{{\alpha ^2 + \beta ^2 }}{{k^2 }}$$

$$v^2 = u^2 - 2u\frac{\beta }{k} + \frac{{\omega ^2 }}{{k^2 }} = u^2 - 2u\frac{\beta }{k} + v^2$$

$$0 = u^2 - 2u\frac{\beta }{k}$$

$$\frac{{2\beta }}{k} = u$$

$$\beta = \frac{{ku}}{2}$$

$$\alpha ^2 = \omega ^2 - \beta ^2 = \omega ^2 - \left( {\frac{{ku}}{2}} \right)^2$$

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

16. Dec 3, 2006

### piano.lisa

Thank you for that part as well.
Another student also recognized that dimensional inconsistency, and has asked the professor, but is still waiting for a response.

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