Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Galilean transformation

  1. Mar 8, 2007 #1
    1. The problem statement, all variables and given/known data
    1) Show that the electromagnetic wave equation,
    d^2(phi)/dx^2 + d^2(phi)/dy^2 + d^2(phi)/dz^2 –(1/c^2)( d^2(phi)/dt^2) = 0
    is not invariant under Galilean transformation.
    Note: here d is a partial differential operator.

    2. Relevant equations

    3. The attempt at a solution

    I have the solution but I couldn’t understand one particular step. The solution is as follows:
    The equation will be invariant if it retains the same form when expressed in terms of the new variables x’,y’,z’,t’. From Galilean transformation we have,
    dx’/dx=1, dx’/dt=-v, dt’/dt=dy’/dy=dz’/dz=1, dx’/dy= dx’/dz= dy’/dx= dt’/dx=0
    From chain rule and using the above results we have,
    d(phi)/dx= [d(phi)/dx’][dx’/dx] + [d(phi)/dy’][dy’/dx] + [d(phi)/dz’][dz’/dx] + d(phi)/dt’][dt’/dx] = d(phi)/dx’
    d^2(phi)/dx^2= d^2(phi)/dx’^2
    d^2(phi)/dy^2= d^2(phi)/dy’^2 &
    d^2(phi)/dz^2= d^2(phi)/dz’^2

    d(phi)/dt= -v[d(phi)/dx’] + d(phi)/dt’
    Differentiating the above equation with respect to t ,
    d^2(phi)/dt^2 = d^2(phi)/dt’^2 -2v[d^2(phi)/dx’dt’] + v^2[d^2(phi)/dx’^2]
    This is where I have a doubt. I differentiated in the following way:

    d^2(phi)/dt^2= -v[d^2(phi)/dx’dt] - [d(phi)/dx’][dv/dt] + [d^2(phi)/dt’^2]
    = -v[(d^2(phi)/dx’^2)(dx’/dt)] –[d(phi)/dx’][dv/dt] + [d^2(phi)/dt’^2]
    = (v^2)[ d^2(phi)/dx’^2] –[d(phi)/dx’][dv/dt] + [d^2(phi)/dt’^2]
    I am able derive 2 of the terms but how to derive the third term -2v[d^2(phi)/dx’dt’] from –[d(phi)/dx’][dv/dt]. Could somebody please help me with this derivation?
  2. jcsd
  3. Mar 8, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's incorrect. Recall that, as you already used for the first derivative,
    [tex] \frac{d}{dt} (anything) = \frac{d}{dt'} (anything) -v \frac{d}{dx'}(anything) [/tex]
    In addition, you may use that the derivative of "v" is zero (v is a constant in a Galilean transformation).

    Apply what I just wrote above to the time deriavtive of your two terms appearing in d(phi)/dt= -v[d(phi)/dx’] + d(phi)/dt’ and you will get the answer.

  4. Mar 9, 2007 #3
  5. Aug 17, 2008 #4
    bump... have same question
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook