1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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