# Performing a lorentz boost

B-80

## Homework Statement

i) Show that the wave equation:

[( -1/c^2) d^2/dt^2 + d^2/dx^2 + d^2/dy^2 + d^2/dz^2 ]u(t,x,y,z) = 0

is invariant under a Lorentz boost along the x-direction, i.e. it takes the same form as a partial differential equation in the new coordinates. [Use the chain rule in two variables.]

ii) Show that the wave equation is not invariant under a Galilei transformation.

## Homework Equations

second order partial derivative equation. I can't get Latex to work right now, or I'd write it out, here's a link to it on another thread https://www.physicsforums.com/showthread.php?t=216869

g=gamma, the lorentz factor

x'=g(x-vt)
y'=y
z'=z
t'=g(t-v x/c^2)

## The Attempt at a Solution

So I just haven't ever done this, and I am not super sure where to start.

I tried rearranging the x' and t' eqs to read
x=x' / g + vt
and
t=t'/g +vx/c^2

then using the chain rule which gave me

dt/dt' = 1/g
and
dx/dx' = 1/g

with both the second derivatives being 0

Plugging that into the equations gave me an extra factor of 1/g^2 on the second derivatives with respect to x and t. I am fairly certain this is the wrong approach anyway. I just don't know where to plug what.

I know this is 100% a math exercise, but it's my HW, and I figure this is the best place to put it.

edit: I realized that this is a multivariable problem, and I have been treating it as a single variable problem i.e. u=u(x'(x,t),y',z',t'(x,t))

and therefore

du/dx = (du/dx')(dx'/dx)+(du/dt')(dt'/dx)

but now if I want to take the second derivative of that i.e.
d^2/dx^2 u = d/dx [(du/dx')(dx'/dx)+(du/dt')(dt'/dx)]

how do I evaluate d/dx(du/dx') and d/dx(du/dt') I know to product rule the rest, but i don't quite remember if there's anything fancy about taking the derivative of a function with respect to a variable whose already had it's derivative taken by function depending on that variable.

Last edited:

Homework Helper
Gold Member
then using the chain rule which gave me

dt/dt' = 1/g
and
dx/dx' = 1/g

with both the second derivatives being 0

You need to do more than this. You've probably already seen this type of calculation before when converting differential equations to polar or spherical coordinates. You need to use the chain rule to write the derivatives with respect to the old coordinates in terms of the derivatives with respect to the new coordinates. For example

$$\frac{\partial}{\partial t} = \frac{\partial t'}{\partial t}\frac{\partial}{\partial t'} + \frac{\partial x'}{\partial t}\frac{\partial}{\partial x'} + \frac{\partial y'}{\partial t}\frac{\partial}{\partial y'} +\frac{\partial z'}{\partial t}\frac{\partial}{\partial z'},$$

with similar expressions for the spatial derivatives. Many terms will vanish because of the specific form of the coordinate transformation.