# Is the Wave Equation Invariant Under a Lorentz Boost?

• B-80
Once you have everything in terms of the derivatives with respect to the new coordinates, you can plug the expressions into the wave equation and see what you get. In summary, using the chain rule and the given coordinate transformations, we can show that the wave equation is invariant under a Lorentz boost along the x-direction, but not under a Galilei transformation. This involves writing the derivatives in terms of the new coordinates and plugging them into the equation.
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:
B-80 said:
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.

## What is a Lorentz boost?

A Lorentz boost is a transformation used in special relativity to convert measurements of space and time between two different inertial frames of reference. It is used to account for the effects of relative motion on measurements of time and space.

## Why is a Lorentz boost necessary?

A Lorentz boost is necessary because according to the theory of special relativity, the laws of physics are the same in all inertial frames of reference. This means that the measurements of time and space can differ between different frames of reference, and a Lorentz boost is needed to convert between them.

## How is a Lorentz boost performed?

A Lorentz boost is performed by applying a set of equations that relate the measurements of time and space in one inertial frame to those in another inertial frame. These equations take into account the relative velocity between the two frames.

## What are the effects of a Lorentz boost?

The effects of a Lorentz boost can include changes in the measurements of time and space between two frames of reference. These changes can include time dilation, length contraction, and changes in the order of events.

## What are some real-world applications of a Lorentz boost?

A Lorentz boost has many real-world applications, including in particle accelerators, GPS systems, and space travel. It is also used in the design of high-speed trains and planes to account for the effects of relative motion on time and space measurements.

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