Galilean transformation

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.

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’
And,
d^2(phi)/dx^2= d^2(phi)/dx’^2
Similarly,
d^2(phi)/dy^2= d^2(phi)/dy’^2 &
d^2(phi)/dz^2= d^2(phi)/dz’^2

Moreover,
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^2(phi)/dt’^2] ------> {dv/dt=0}
= (v^2)[ d^2(phi)/dx’^2] + [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’] . Could somebody please help me with this derivation?

This post has been edited by plasma on Mar 10 2007, 06:44 PM

I think you tried to shortcut the process too much toward the end. When you went through the full process earlier, like

that was the right way to go.

If you apply the same process to v[d(phi)/dx'] + d(phi)/dt', the first term gives d(v[d(phi)/dx'])/dt =
[d(v[d(phi)/dx'])/dx'][dx'/dt] + [d(v[d(phi)/dx'])/dy'][dy'/dt] + [d(v[d(phi)/dx'])/dz'][dz'/dt] + d(v[d(phi)/dx'])/dt'][dt'/dt] =
v[d^2(phi)/dx'^2]v + v[d^2(phi)/dx'dt']1.

The second term gives something similar, and both terms turn out to contribute one copy of v[d^2(phi)/dx'dt'], which gives you what you want.

For problems like this, there is generally an easier way to do the calculation. Just factor out the original function and look at what the derivative operator does. d/dx'(phi) = d/dx(phi), so d/dx' = d/dx. Similarly, d/dy' = d/dy and d/dz' = d/dz. However, d/dt(phi) = -v[d(phi)/dx'] + d(phi)/dt'. Therefore, d/dt = v*d/dx' + d/dt'.

Now usually you have to be very careful combining differential operators, because the operator on the left will differentiate any functions that appear in the operator on the right. But in this case, the only function in the d/dt operator is v, which is a constant.

Therefore, you can do it like this:
d^2/dx^2 =
(d/dx)(d/dx) = (d/dx')(d/dx') = d^2/dx'^2,
and likewise,
d^2/dy^2 = d^2/dy'^2
and
d^2/dz^2 = d^2/dz'^2.
But
d^2/dt^2 =
(d/dt)(d/dt) =
(v*d/dx' + d/dt')(v*d/dx' + d/dt') =
v^2*(d/dx'^2) + 2v*(d/dx')(d/dt') + (d/dt')^2 =
v^2*d^2/dx'^2 + 2v*d^2/dx'dt' + d^2/dt'^2.

Now you just put the (phi) back in everywhere.

Hope this helps!

--Stuart Anderson

I tried your first method. But I got 4 extra terms other than d^2(phi)/dt’^2 -2v[d^2(phi)/dx’dt’] + v^2[d^2(phi)/dx’^2]
For the first term,
-d(v[d(phi)/dx'])/dt = v^2[d^2(phi)/dx’^2] – v[d^2(phi)/dx’dy’][dy’/dt] – v[d^2(phi)dx’dz’][dz’/dt] – v[d^2(phi)/dx’dt’]

For the second term,
(d/dt)[d(phi)/dt'] = -v[d^2(phi)/dx’dt’] + [d^2(phi)/dy’dt’][dy’/dt] + [d^2(phi)/dz’dt’][dz’/dt] + d^2(phi)/dt’^2
How to eliminate the terms containing dy’/dt & dz’/dt ?

From the derivatives that you gave to define the Galilean transformation, it looks like you are boosting in the x-direction. You forgot to list the time derivatives of y' and z'. They should be part of the Galilean transformation, and they should vanish if you are boosting in the x-direction.

But 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 defined for 3 dimensions, isn't it? So can we take the velocity in the y and z direction as zero?

This post has been edited by plasma on Mar 20 2007, 09:40 PM

If by "the velocity" you mean "the boost velocity", then you are free to choose any direction you like. Since your wave equation is isotropic, you only need to choose one direction, and the isotropy guaruntees the same result for all directions (ah, the beauty of symmetry). And, this idea can be carried a step further in that, you are free to choose the x-direction as your boost direction. So, in short, you may set dy/dt and dz/dt to zero in your Galilean transformation.