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Mass term in wave equation

  1. Aug 23, 2007 #1
    I know how to write down solutions of wave equation

    \partial^2_t u(t,x) = \partial^2_x u(t,x)

    for given initial [itex]u(0,x)[/itex] and [itex]\partial_t u(0,x)[/itex] like this

    u(t,x) = \frac{1}{2}\Big( u(0,x+t) + u(0,x-t) + \int\limits^{x+t}_{x-t} \partial_t u(0,y) dy\Big),

    but what about

    \partial^2_t u(t,x) = \partial^2_x u(t,x) - mu(t,x)

    where m is some constant? Is there similar formula for this?
    Last edited: Aug 23, 2007
  2. jcsd
  3. Aug 23, 2007 #2
    The equation is the Klein-Gordon equation. It has (relativistic) plane wave solutions, and is linear.
  4. Aug 23, 2007 #3
    I'm aware of this, and that is where the motivation comes. But for example for m=0 the plane wave solutions don't tell the same things as the expression I wrote in the OP. The fact that u(t,x) depends only on what is in the interval [x-t, x+t] at time zero is quite manifest now, but plane waves don't tell that directly.
    Last edited: Aug 23, 2007
  5. Aug 23, 2007 #4
    In the m=0 case, the plane wave solutions may be used to build up the solution, to any (reasonable) initial condition.
  6. Aug 23, 2007 #5
    My point was that the expression for solution in my OP tells things that the plane wave solutions don't tell.
  7. Aug 23, 2007 #6
    Such as?
  8. Aug 23, 2007 #7
    [itex]u(t,x)[/itex] depends on [itex]u(0,x)[/itex] and [itex]\partial_t u(0,x)[/itex] in the interval [itex][x-t,x+t][/itex]. Initial configuration outside this interval doesn't matter.
  9. Aug 23, 2007 #8

    George Jones

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    The equivalence of the general (m = 0) plane wave solution (i.e., solution using Fourier transforms) to the solution in your original post is posed as an exercise (with hints) in my favourite Fourier analysis book.
    Last edited: Aug 23, 2007
  10. Aug 25, 2007 #9

    Chris Hillman

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    Recommend two books

    Just thought I'd add that the solution of the one-dimensional wave equation, given in terms of an integral over initial data, which Jostpuur mentioned in his post, represents the tip of a rather large iceberg. As many here probably already know, one of the most illuminating approaches to either ODEs or PDEs involves Dirac deltas, Green's functions, integral transforms, and so on. Jostpuur's example illustrates one reason why this approach is so important: comparing analogous solutions for higher dimensional wave equations yields valuable insight into "accidents of low dimensions". It turns out that the behavior of solutions of the wave equation in dimensions 1,2,3 have completely different characters! (Hint: consider the light cone.)

    I have yet to see a book covering Dirac deltas for PDEs which I really like, but a superb book from the applied perspective is Dean G. Duffy, Transform Methods for Solving Partial Differential Equations, CRC Press, 1994. Among textbooks providing readable introductions to the equations of mathematical physics generally, I like Guenther and Lee, Partial Differential Equations of Mathematical Physics and Integral Equations, Dover, 1996 (reprint of 1988 original), which is notable for a wise selection of the most useful topics from this vast subject. Both of these books should help readers gain insight into the effects of adding additional terms to say the wave equation.
    Last edited: Aug 25, 2007
  11. Aug 28, 2007 #10
    I wonder if you have being using these equations to grapple with the physics of our moving through time?

    [itex] \tau [/itex] = proper time
    [itex] \tau_c [/itex] = proper time at centre of a mass distribution
    [itex] m = 1/ \tau_0 [/itex] = mass

    A: [tex]u(t,x) e^{ims} = v(s,t,x) [/tex] is a more general idea so that:
    0 = (\partial^2_t - \partial^2_x + m^2) v = (\partial^2_t - \partial^2_x - \partial^2_s) v
    where [itex] s = \tau [/itex]. So mass m lives throughout time s.
    The time of space-time is sitting out there and mass may have a large temporal extent (width).

    B: [tex] u(t,x) e^{-(ms)^2/2} = w(s,t,x) [/tex]
    0 = (\partial^2_t - \partial^2_x - \partial^2_s) w = (\partial^2_t - \partial^2_x + m^2[1 - (ms)^2]) w
    where [itex] s = \tau - \tau_c; \ and \ |s| < \tau_0 [/itex]. So mass m only lives within the tiny temporal extent of |s| trapped by a matter constraining SHM potential like [1 - (ms)^2]. So we move through a temporal surface of space-time with a tiny temporal extent.

    A & B are at extreme ends of visualising our temporal extent as we travel through time.

    I think A is the way I was brought up as undergraduate student just learning SR and time-independent QM & statistical mechanics. The spatial extent of gas molecules in a box is the box itself (similarly for electrons in a metal) - so by analogy the temporal extent is like a 4th dimension of the ‘box’.

    Of course B is a bit of a hotch-potch. The middle way to visualise the extent of the mass is by considering A and the inter-related gaussian distributions of (E, p & m) in (t, x, s). So mass does not drift forever back into the past and future but peaks at the ‘now’.

    [tex] m = m = m = m = m = m = m = m = m = m = m = m = m = m = m = m = m [/tex]
    Last edited: Aug 28, 2007
  12. Aug 30, 2007 #11
    MikeL#, your post was confusing and interesting. At quick glance, to me it seems that solutions of

    (\partial_t^2 - \partial_x^2 + m^2)u(t,x) = 0

    could be written as fourier transforms of some solutions of

    (\partial_t^2 - \partial_x^2 - \partial_m^2)u(t,x,m)=0

    As if n-dimensional Klein-Gordon equation could be reduced to (n+1)-dimensional wave equation withouth the mass term. I haven't got into detail of this yet, but will probably do it at some point.
  13. Aug 31, 2007 #12

    Chris Hillman

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    Not here, please!

    I for one would appreciate it, MikeL, if you would confine this kind of speculation to this subforum:


    Jostpuur, are you still interested in the mathematics of wave equations? One topic which hasn't been mentioned is factoring the wave operator by passing to a larger algebra.
  14. Oct 16, 2007 #13
    I know how to arrive to the solution of the wave equation with a factorization [itex]\partial_t^2 - \partial_x^2=(\partial_t - \partial_x)(\partial_t + \partial_x)[/itex], but I don't what to do with the mass term. How do you factorize sum of three squares?

    A^2-B^2+C^2 = (A+\sqrt{B^2-C^2})(A-\sqrt{B^2-C^2}) = (A+B+C)(A-B+C) - 2AC

    both seem quite useless here at least.
    Last edited: Oct 17, 2007
  15. Oct 17, 2007 #14
    hmhm... I was paying attention to the "factoring" part, and missed the "larger algebra". Did you actually mean the Dirac's equation, or some kind of lower dimensional analogue to it?
  16. Oct 18, 2007 #15

    \partial_t^2-\partial_x^2+m^2 & 0 \\
    0 & \partial_t^2-\partial_x^2+m^2 \\
    m & \partial_t-\partial_x \\
    -\partial_t-\partial_x & m \\
    m & -\partial_t+\partial_x \\
    \partial_t + \partial_x & m \\


    hm :confused:

    Well this looks like it is going to lead somewhere, but it's probably not something that I could see in few seconds. That is a modified transport equation, after all. Let's see how this goes...
    Last edited: Oct 18, 2007
  17. Oct 22, 2007 #16
    mass term in transport equation

    For given initial configuration u(0,x) the solution of the transport equation

    \partial_t u(t,x) - a\partial_x u(t,x) = 0

    is given trivially by


    From a book I could learn that the non-homogenous transport equation

    \partial_t u(t,x) - a\partial_x u(t,x) = f(t,x)

    for fixed f(t,x) is given by

    u(t,x) = u(0,at+x) + \int\limits_0^t f(s,x+(t-s)a)ds

    Now I want to find solution to the transport equation with a mass term,

    \partial_t u(t,x) - a\partial_x u(t,x) = mu(t,x)

    I could substitute [itex]f(t,x)=mu(t,x)[/itex], but that doesn't give the solution to the initial value problem.

    Any clues what to do with this?
  18. Oct 22, 2007 #17
    It seems that an attempt


    solves it.
  19. Oct 28, 2007 #18
    I think I found a way to write down the solution to my original problem using some infinite series of some strange integral expressions. It could be, that it should be called a perturbation series with respect to m. But the stuff got quite laborious, and I haven't started fighting with the details yet.
  20. Nov 20, 2007 #19
  21. Nov 20, 2007 #20
    Differential equations

    Hi,if i could try to solve for the wavefunction in schrodingers equation in 3 dimension,would it take me time?
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