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Use of chain rule in showing invariance.

  1. Feb 7, 2014 #1
    Hi, I’m a bit confused.
    I am familiar with the chain rule: if y=f(g(t,x),h(t,x)) then dy/dt=dy/dg*dg/dt+dy/dh*dh/dt
    To show that an equation is invariant under a galiliean transform, it’s partially necessary to show that the equation takes the same form both for x and for x’=x-v(T). So if you have a wave equation for E which applies for x, and t, you want to show that the wave equation, with all of its first and second derivatives also applies for x’ and t’.

    For example if you look at question 16 b in the following link, they ask to show that the wave equation is not invariant under Galilean transforms. What I don’t understand is in this question why are they taking the derivative of E with respect to x and t, rather than with respect to x’ and t’. We already know the wave equation takes the correct form for x and t. We want to show that it doesn’t take the correct form for x’ and t’, so then why start off taking the derivative with respect to x and t, and muck about using the chain rule rather than taking the derivative with respect to x’, and t’ (which is what you’re really interested in).

    http://stuff.mit.edu/afs/athena/course/8/8.20/www/sols/sol1.pdf

    I just don’t get why they take the derivative with respect to x and t, when that’s not what you’re really interested in, you’re interested in the form of the derivatives with respect to x’ and t’. Where does the idea to use the chain rule in this case comes from at all? All of the solutions for these types of problems seem to use the chain rule, I don’t get where the natural impetus to take the derivative in the wave equation with respect to variables you already know will work in the wave equations, rather than the ones which have undergone the Galilean transform.

    It’s been a very long time since I’ve done this type of math/physics. Thanks.
     
  2. jcsd
  3. Feb 7, 2014 #2

    strangerep

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    The transformation is ##x' = x - vt,~~~~ t'=t##. This is easily inverted to give ##x=x'+vt',~~~~t=t'##.

    You have a wave equation involving things like ##\partial_x^2 E##, so it's easier to simply find what that looks like in the new coordinates, and substitute that back into the original equation.

    [Actually, that sect 16(b) is not completely explicit. For that, you'd have to go one step further and show that the transformed quantities are not a simple (common) multiple of the originals. I'm guessing the author probably considers that to be obvious.]
     
  4. Feb 9, 2014 #3

    strangerep

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    You transcribed that incorrectly. The 2nd order derivative terms act on the ##E##, but you've moved ##E## to the front. (??)

    The 1st order derivative can be written as the operator equation:
    $$
    \partial_t ~=~ \partial_{t'} - v \partial_{x'} ~,
    $$where ##v## is a constant. So you just have to evaluate
    $$
    \partial^2_t ~=~ \Big( \partial_{t'} - v \partial_{x'} \Big)^2 ~.
    $$Of course, to understand this properly, you've got to make both sides act on something.

    (BTW, make an effort to learn how to use basic Latex for math on PF. Nobody wants to read that ugly ascii math. There's a faq under the site info menu that will help you get started.)
     
  5. Feb 11, 2014 #4
    What you're doing is comparing the movement of the object you're interested in with the movement of the wave crest. Or rather what the wave looks like to your object if the object is moving. The Galilean transformation which strangerep stated can be used to represent the motion of the object. The wave equation in x and t can be translated to a wave equation in x' and t'. That wave equation in x' and t' is the one which applies from the object's perspective. (Or in coordinates which move with the object).

    You find the wave equation in x' and t' by applying the chain for derivatives indicated by the Galilean transformation. If the two wave equations have the same form where x' can replace x and t' can replace t, then the wave equation is invariant.

    Notice that you probably don't want the equation to be invariant for most waves such as sound waves. If that were the case then Doppler Shift would not occur. Doppler Shift can be determined directly from the transformed wave equation using the Galilean transformation.
     
  6. Feb 11, 2014 #5
    What you're trying to check is whether:

    [tex]\frac{\partial ^2 E}{\partial {x'}^2}-\frac{1}{c^2}\frac{\partial ^2 E}{\partial {t'}^2}=\frac{\partial ^2 E}{\partial {x}^2}-\frac{1}{c^2}\frac{\partial ^2 E}{\partial {t}^2}[/tex]

    You can do that by that by applying a coordinate transformation to either side of the equation to try to obtain the corresponding expression in terms of the independent variables on the other side of the equation. What they did in the reference you cited is the standard method of doing such a coordinate transformation.
     
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