Galilean transformation

  • Thread starter Amith2006
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Homework Statement


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.



Homework Equations





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’
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(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?
 

Answers and Replies

  • #2
nrqed
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Homework Statement


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.



Homework Equations





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’
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]
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.

Patrick
 
  • #3
427
2
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]
I didn't get your point. Could u please explain it in detail?
 
  • #4
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0
bump... have same question
 

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