Wave Equation / Damping / Phase Velocity

[piano.lisa]

Homework Statement

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.

Homework 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}

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.

 

[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.

 

[piano.lisa]

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

 

[OlderDan]

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

(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.

 

[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?

 

[OlderDan]

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?

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.

 

[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.

 

[OlderDan]

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.

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

<br /> \beta = \frac{{\omega \omega^*}}{u}<br />

and then

<br /> \alpha = \sqrt {\omega \omega^*\left( {1 - \frac{{\omega \omega^*}}{{u^2 }}} \right)} <br />

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

 

[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}

 

[OlderDan]

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}

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.

 

[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.

 

[OlderDan]

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.

Perhaps I am misunderstanding, but this

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}

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 }}

I am \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)}

 

[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.

 

[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 }

 

[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 }

 

[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.