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Homework Help: Wave Equation / Damping / Phase Velocity

  1. Dec 2, 2006 #1
    1. The problem statement, all variables and given/known data
    Consider the simplified wave function: [tex]\psi (x,t) = Ae^{i(\omega t - kx)}[/tex]
    Assume that [tex]\omega[/tex] and [tex]\nu[/tex] are complex quantities and that k is real:
    [tex]\omega = \alpha + i\beta[/tex]
    [tex]\nu = u + i\omega[/tex]
    Show that the wave is damped in time. Use the fact that [tex] k^2 = \frac{\omega^2}{\nu^2}[/tex] to obtain expressions for [tex]\alpha[/tex] and [tex]\beta[/tex] in terms of [tex]u[/tex] and [tex]\omega[/tex]. Find the phase velocity for this case.


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

    3. The attempt at a solution
    I substitued ii into i to obtain the expression: [tex]\psi (x,t) = Ae^{-\beta t}e^{i(\alpha t - kx)}[/tex]. Therefore, the factor of [tex]e^{-\beta t}[/tex] 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 [tex]\alpha[/tex] and [tex]\beta[/tex] in terms of [tex]u[/tex] and [tex]\omega[/tex]. 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
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  3. Dec 2, 2006 #2

    OlderDan

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    If [tex] k^2 = \frac{\omega^2}{\nu^2}[/tex] 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 [itex]\alpha[/itex] and [itex]\beta[/itex] in terms of the u and the norm of [itex]\omega[/itex]. It's a bit of tediuous algebra to square out the complex numbers and rationalize the denominator and set the imaginary part of [tex] k^2 = \frac{\omega^2}{\nu^2}[/tex] to zero.
     
  4. Dec 2, 2006 #3
    Why is it tedious to square the complex numbers?
    Do I have to do, for example, (v)(v*) ?
     
  5. Dec 2, 2006 #4

    OlderDan

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    (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 [itex]\alpha[/itex] and [itex]\beta[/itex]. However, if it means (v)(v), then demanding that k² is real leads to a solution.
     
  6. Dec 2, 2006 #5
    Using (v)(v) to solve,
    I obtained:
    [tex]\beta = \frac{\alpha}{2u\omega}[(\omega^2 - u^2) \pm (u^2 + \omega^2)][/tex]
    However, [tex]\beta[/tex] is still in terms of [tex]\alpha[/tex], so I'm not sure what I'm doing wrong.

    Therefore, I obtained
    [tex]\beta = \frac{\alpha \omega}{u}[/tex] OR [tex]\beta = -\frac{\alpha u}{\omega}[/tex]
    Likewise, [tex]\alpha = \frac{\beta u}{\omega}[/tex] OR [tex]\alpha = -\frac{\beta \omega}{u}[/tex].

    Assuming those are right, how can I find the phase velocity from this point? If k is real, does that imply that [tex]v_{phase} = v[/tex]?
     
    Last edited: Dec 2, 2006
  7. Dec 2, 2006 #6

    OlderDan

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    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 [itex]\nu = u + i\omega[/itex] 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
  8. Dec 2, 2006 #7
    I think that's what I did...
    Here, I'll show you:
    [tex] k^2 = \omega^2 / v^2 = \frac{(\alpha + i\beta)^2}{(u + i\omega)^2}[/tex]
    [tex] = \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)}[/tex]
    [tex] 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}[/tex]

    From there, I set [tex]Im(k^2) = 0[/tex] and solved for [tex]\beta[/tex] with the quadratic formula. However, [tex]\beta[/tex] was still in terms of [tex]\alpha[/tex]. I could not make [tex]\beta[/tex] in terms of [tex]\omega\omega *[/tex] without it also being in terms of [tex]\alpha[/tex].
     
  9. Dec 2, 2006 #8

    OlderDan

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

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

    [tex]
    \beta = \frac{{\omega \omega^*}}{u}
    [/tex]

    and then

    [tex]
    \alpha = \sqrt {\omega \omega^*\left( {1 - \frac{{\omega \omega^*}}{{u^2 }}} \right)}
    [/tex]

    I have to go, but I'll check back in tomorrow morning.
     
    Last edited: Dec 2, 2006
  10. Dec 2, 2006 #9
    Taking into account what you said,
    I simplified further using the fact that w is complex, and I obtained:
    [tex] \beta = \frac{u \pm \sqrt{u^2 - 4\alpha^2}}{2}[/tex]
    Substituting [tex]\alpha = \omega - i\beta[/tex] into the above,
    I obtained:
    [tex] \beta = \frac{\omega^2}{u + 2i\omega}[/tex]
    Which is slightly the same as yours, if I take [tex]Im(\beta) = 0[/tex].

    However, I also obtain:
    [tex] \alpha = \frac{\omega (u + i\omega)}{u + 2i\omega}[/tex]
     
    Last edited: Dec 2, 2006
  11. Dec 3, 2006 #10

    OlderDan

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    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
  12. Dec 3, 2006 #11
    This was my method:
    [tex] k^2 = \frac{\omega^2}{v^2} = \frac{(\alpha + i\beta)^2}{(u - \beta +i\alpha)^2}[/tex]

    Then, I expand the brackets, and rationalize the denominator.
    [tex] 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)]}[/tex]

    [tex] 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}[/tex]

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

    [tex]2\alpha\beta u^2 - 2u\alpha\beta^2 - 2\alpha^3 u = 0[/tex]

    [tex]\alpha^2 - \beta u + \beta^2 = 0[/tex]

    [tex]\beta = \frac{u \pm \sqrt{u^2 - 4\alpha^2}}{2}[/tex]

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


    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 [tex]k^2 = \frac{\omega\omega *}{vv*}[/tex], I end up with a real value for k, obviously... But I cannot solve for [tex]\alpha[/tex] and [tex]\beta[/tex] in terms of [tex]u[/tex] and [tex]\omega[/tex].

    Thank you.
     
    Last edited: Dec 3, 2006
  13. Dec 3, 2006 #12

    OlderDan

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    Perhaps I am misunderstanding, but this

    looks very different to me than what you get from expanding

    [tex] k^2 = \frac{\omega^2}{v^2} = \frac{(\alpha + i\beta)^2}{(u - \beta +i\alpha)^2}[/tex]

    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.

    [tex] \omega = \alpha + i\beta [/tex]

    [tex] v = u + i\omega = u + i\left( {\alpha + i\beta } \right) = u - \beta + i\alpha [/tex]

    [tex] 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 }} [/tex]

    [tex] 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 }} [/tex]

    [tex] 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 }} [/tex]

    [tex] 0 = \beta \left( {u^2 - 2u\beta + \beta ^2 - \alpha ^2 } \right) - \left( {u - \beta } \right)\left( {\alpha ^2 - \beta ^2 } \right) [/tex]

    [tex] 0 = \beta u^2 - 2u\beta ^2 + \beta ^3 - \beta \alpha ^2 - u\alpha ^2 + u\beta ^2 + \beta \alpha ^2 - \beta ^3 [/tex]

    [tex] 0 = \beta u^2 - u\beta ^2 - u\alpha ^2 [/tex]

    [tex] \beta u = \alpha ^2 + \beta ^2 = \omega \omega ^* [/tex]

    [tex] \beta = \frac{{\omega \omega ^* }}{u} [/tex]

    [tex] \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) [/tex]

    [tex] \alpha = \sqrt {\omega \omega ^* \left( {1 - \frac{{\omega \omega ^* }}{{u^2 }}} \right)} [/tex]
     
    Last edited: Dec 3, 2006
  14. Dec 3, 2006 #13
    Thank you so much.

    We do have the same expansion of [tex]k^2[/tex].
    Except, at the point where [tex] 0 = \beta u - \beta^2 - \alpha^2[/tex], I solved for [tex]\beta[/tex] using the quadratic formula, and you used a more efficient method.
     
  15. Dec 3, 2006 #14

    OlderDan

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    Here is what I have for the square norms calculation

    [tex] \omega = \alpha + i\beta [/tex]

    [tex] v = u + i\omega = u + i\left( {\alpha + i\beta } \right) = u - \beta + i\alpha [/tex]

    [tex] k^2 = \frac{{\omega ^2 }}{{v^2 }} [/tex]

    [tex] \omega ^2 = \alpha ^2 + \beta ^2 [/tex]

    [tex] v^2 = u^2 - 2u\beta + \beta ^2 + \alpha ^2 = u^2 - 2u\beta + \omega ^2 [/tex]

    [tex] \frac{{\omega ^2 }}{{k^2 }} = u^2 - 2u\beta + \omega ^2 [/tex]

    [tex] \omega ^2 = k^2 \left( {u^2 - 2u\beta + \omega ^2 } \right) [/tex]

    [tex] \left( {1 - k^2 } \right)\omega ^2 = k^2 u\left( {u - 2\beta } \right) [/tex]

    [tex] u - 2\beta = \left( {\frac{{1 - k^2 }}{{k^2 }}} \right)\frac{{\omega ^2 }}{u} [/tex]

    [tex] 2\beta = u - \left( {\frac{{1 - k^2 }}{{k^2 }}} \right)\frac{{\omega ^2 }}{u} [/tex]

    [tex] \beta = \frac{u}{2} - \left( {\frac{{1 - k^2 }}{{k^2 }}} \right)\frac{{\omega ^2 }}{{2u}} [/tex]

    [tex] \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] [/tex]

    [tex] \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] [/tex]

    [tex] \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 } [/tex]

    [tex] \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 } [/tex]
     
  16. Dec 3, 2006 #15

    OlderDan

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    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.

    [tex] \omega = \alpha + i\beta [/tex]

    [tex] 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} [/tex]

    [tex] k^2 = \frac{{\omega ^2 }}{{v^2 }} [/tex]

    [tex] \omega ^2 = \alpha ^2 + \beta ^2 [/tex]

    [tex] 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 }} [/tex]

    [tex] v^2 = u^2 - 2u\frac{\beta }{k} + \frac{{\omega ^2 }}{{k^2 }} = u^2 - 2u\frac{\beta }{k} + v^2 [/tex]

    [tex] 0 = u^2 - 2u\frac{\beta }{k} [/tex]

    [tex] \frac{{2\beta }}{k} = u [/tex]

    [tex] \beta = \frac{{ku}}{2} [/tex]

    [tex] \alpha ^2 = \omega ^2 - \beta ^2 = \omega ^2 - \left( {\frac{{ku}}{2}} \right)^2 [/tex]

    [tex] \alpha = \omega \sqrt {1 - \left( {\frac{{ku}}{{2\omega }}} \right)^2 } [/tex]
     
  17. Dec 3, 2006 #16
    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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