# Wave Equation / Damping / Phase Velocity

• piano.lisa
I think you are missing something in the right hand side. The correct expression is:\nu^2 = (u + i\omega)(u + i\omega) = u^2 + 2i(u\omega) - \omega^2 -1 = u^2 + 2i(u\omega) - \omega^2 -iand that should allow you to solve for alpha and beta in terms of u and \nu^2.Thank you for pointing that out. I have now solved for \alpha and \beta. Is the phase velocity

## 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.

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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.

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

piano.lisa said:
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.

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 $$v_{phase} = v$$?

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piano.lisa said:
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.

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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 $$\alpha$$.

piano.lisa said:
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

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

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.

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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:
$$\alpha = \frac{\omega (u + i\omega)}{u + 2i\omega}$$

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piano.lisa said:
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.

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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.

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piano.lisa said:
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

piano.lisa said:
$$k^2 = \omega^2 / v^2 = \frac{(\alpha + i\beta)^2}{(u + i\omega)^2}$$
$$= \frac{(\alpha^2 - \beta^2 + 2i\alpha \beta)(u^2 - \omega^2 - 2iu\omega)}{(u^2 - \omega^2 + 2iu\omega)(u^2 - \omega^2 - 2iu\omega)}$$
$$k^2 = \frac{(u^2 - \omega^2)(\alpha^2 - \beta^2) + 4\alpha\beta u\omega}{(u^2 + \omega^2)^2} + i\frac{(u^2 - \omega^2)(2\alpha\beta) - (2u\omega)(\alpha^2 - \beta^2)}{(u^2 + \omega^2)^2}$$

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.

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

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

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

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

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

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

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

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

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

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

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

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

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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.

Here is what I have for the square norms calculation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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.

## 1. What is the wave equation?

The wave equation is a mathematical formula that describes the behavior of waves. It is used to calculate the properties of different types of waves, such as sound waves, light waves, and water waves.

## 2. What is damping in relation to waves?

Damping is the gradual decrease in the amplitude of a wave over time. It is caused by the dissipation of energy due to factors such as friction and resistance. Damping can affect the intensity and speed of a wave's movement.

## 3. How is phase velocity related to wave speed?

Phase velocity is the speed at which a specific phase of a wave, such as the crest or trough, moves through a medium. It is related to wave speed, which is the speed at which the entire wave moves, by the formula phase velocity = wave speed / wavelength.

## 4. What factors affect the phase velocity of a wave?

The phase velocity of a wave is affected by the properties of the medium through which it is traveling, such as density and elasticity. It is also influenced by the frequency and wavelength of the wave.

## 5. How is the wave equation used in real-world applications?

The wave equation is used in a variety of fields, including physics, engineering, and telecommunications. It is used to study and predict the behavior of waves in different mediums, and is essential for understanding and developing technologies such as fiber optics and seismic imaging.