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I Function describing a traveling wavefield

  1. Oct 29, 2016 #1
    Hello Forum,

    A traveling wave is something that travels and transfers energy from one location to another. There are many waves and many wave equations. There are also standing wave that are waves trapped within some cavity of boundaries (also called standing waves). A standing wave can be made by superposing two traveling waves moving in opposite directions. Two standing waves can form a traveling wave, etc.

    That said, for one-dimensional traveling waves, introductory books show that the mathematical function describing a traveling wave must have an argument like ##(x-vt)## where the spatial and the time variables are combined into one argument. For example, ##sin(x- \omega t)## is a plane wave traveling along the x-direction. If the two variables , spatial ##x## and time ##t##, are separated, the wave is not traveling. Is that generally true?
    For example, a function ##f(x,t)## that is separable,i.e. ##f(x,t) = g(x)*p(t)## cannot be a traveling wave but can be a standing wave. However, I then think of something like ##e^{i(x-\omega t)}= e^{ix}e^{-i\omega t}##, which also a traveling wave and is also separable...

    What general observations can we make about the function that is supposed to represent a traveling wave? What about the spatial and time variables x and t?

    Thanks!
     
  2. jcsd
  3. Oct 29, 2016 #2

    Simon Bridge

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    If you have some function in space ##y=f(x)## and you want that relationship to travel through space with speed ##v## (in the +x direction) then the position of each bit of ##f(x)## gets updated as time goes by. For instance, if ##f(x)## has a peak at ##x_0## when ##t=0##, then then at time ##t>0##, that peak will have moved to ##x_0-vt##. The height of the peak is the same as it was at t=0, it is just in a different place. (Here I am considering that the shape of the function does not change in time, only it's position).

    It follows that the overall time dependence is given by replacing x in the original expression by x-vt ... so ##y=f(x-vt)##.

    If f(x) is a sine wave, then ##f(x)=A\sin kx## ... the A is needed to make the amplitude something other than 1, and the k is needed to convert units of x into radians.
    Following above, a travelling sine wave would be ##y(x,t)=f(x-vt) = A\sin k(x-vt) = A\sin(kx-kvt) = A\sin(kx-\omega t)##
    ... that last step is the definition of ##\omega##.

    If you have ##f(x)=Ae^{ikx}## then ##y(x,t)=Ae^{ik(x-vt)} = Ae^{ikx}e^{-i\omega t}= \psi(x)\chi(t)## following the reasoning above.
    ... this is a plane-wave solution to the Schrodinger equation and is used a lot in quantum mechanics.

    Notice that ##e^{ikx} = \cos kx + i\sin kx## ... this describes a phasor whose rotation angle depends on it's position in space.

    This help?
     
  4. Oct 30, 2016 #3
    Hi Simon Bridge,

    Thank you for your helpful comments. I agree with you. What I am not sure about is a traveling wave described by a mathematical expression where the space variable x and the time variable t are not combined in the argument (x-vt). As you mentioned, as the variable t grows, the x variable has to grow to to maintain the argument (x-vt) constant. That implies that the wavefield is moving.

    I would tend to judge any function that is separable, f(x,t) =g (t)*p(x) as not describing a traveling wavefield if we cannot recast the variables x and t together in the argument (x-vt). Is that incorrect?
     
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