# I Transformation of Wave Equation and Wave Displacement Value?

1. Oct 31, 2016

### greswd

This question concerns a section from the book Modern Physics by James Rohlf.

http://srv3.imgonline.com.ua/result_img/imgonline-com-ua-twotoone-Bs4zgy7pruqG.png

He shows that the form of the Wave equation for light remains invariant under a Lorentz boost (4.42):

$\frac{∂^2F}{∂x'^2}+\frac{∂^2F}{∂y'^2}+\frac{∂^2F}{∂z'^2}=\frac{1}{c^2}\frac{∂^2F}{∂t'^2}$

What I am confused about is the lack of F' in these derivations. There is no F', only F.

I always thought that if one wants to show that the form of the Wave equation remains invariant under a Lorentz boost, shouldn't the final equation be:

$\frac{∂^2F'}{∂x'^2}+\frac{∂^2F'}{∂y'^2}+\frac{∂^2F'}{∂z'^2}=\frac{1}{c^2}\frac{∂^2F'}{∂t'^2}$ ?

Why is it F instead of F'?

Also, I have to apologize for all the poor concepts and erroneous physics that I post on these forums.

Last edited: Oct 31, 2016
2. Oct 31, 2016

### robphy

F is a scalar field.

3. Oct 31, 2016

### greswd

But can't a scalar field also transform between frames?

4. Oct 31, 2016

### greswd

Like this:
\begin{align} & E'_x = E_x & \qquad & B'_x = B_x \\ & E'_y = \gamma \left( E_y - v B_z \right) & & B'_y = \gamma \left( B_y + \frac{v}{c^2} E_z \right) \\ & E'_z = \gamma \left( E_z + v B_y \right) & & B'_z = \gamma \left( B_z - \frac{v}{c^2} E_y \right). \\ \end{align}

5. Oct 31, 2016

### James R

Here, $F$ is any function that obeys the wave equation. In one coordinate system it is a function of $(x,y,z,t)$, while in the other it is a function of $(x',y',z',t')$, but in both systems it is the same function.

6. Oct 31, 2016

### greswd

That appears to be what James Rohlf (James R? Same as you! ) is saying.

I've always thought that all the quantities in the primed frame should be primed as well. Especially when we're talking about an EM wave.

In his seminal paper on relativity, On the Electrodynamics of Moving Bodies, Einstein makes everything in the primed frame prime when utilizing Maxwell's equations:

The E and B components (or F in Rohlf's example) are transformed in the manner I brought up in #4:

7. Oct 31, 2016

### Orodruin

Staff Emeritus
When you talk about an EM field you are essentially talking about a rank two tensor field. Generally a tensor's components depends on the basis you use. When you talk about $F$ above, it is a scalar field. By definition, scalars (rank zero tensors) are invariant under coordinate transformations.

8. Oct 31, 2016

### greswd

Is it acceptable to plug any of the 6 E and B components into the d'Alembertian? Wouldn't that make F just a stand in for any component of the E or B field?

9. Oct 31, 2016

### Orodruin

Staff Emeritus
This adds further complication. You would also have to take into account that the components transform between frames. You can say that the E' components following the wave equation in S' will imply that the E' components will follow the wave equation in S (keeping talking about components of the E-field relative to the primed basis). It is then a matter of collecting the E' and B' components into the E and B components in order to conclude that the E and B components (relative to the unprimed system) also satisfy the wave equation.

10. Oct 31, 2016

### greswd

So we can mix primes and un-primes within the same d'Alembertian? Cool..

11. Oct 31, 2016

### Orodruin

Staff Emeritus
The d'Alembertian is a linear operator. This should not come as a surprise.

12. Oct 31, 2016

### greswd

Rohlf has shown that it $F$ can satisfy the d'Alembertian in both primed and unprimed frames.
We can replace $F$ with $E_x$, since $E_x'=E_x$, and get a similar result.

Am I right to say that we need the transformations in #4 to satisfy Maxwell's equations, but we don't need them to satisfy the d'Alembertian?

13. Oct 31, 2016

### greswd

Not really a surprise, but an interesting result. Especially to an ignoramus like myself, who finds it difficult to obtain answers.

14. Oct 31, 2016

### Orodruin

Staff Emeritus
The d'Alembertian is a differential operator. It is not something you can "satisfy".

It is involved in the wave equation, which is something a quantity can satisfy.

In general, you know that E and B are linear combinations of E' and B'. Thus, if E' and B' satisfy the wave equations, then so do E and B by the linear property of the d'Alembertian.

15. Oct 31, 2016

### greswd

To reiterate my question in #12, am I right to say that we need the transformations in #4 to satisfy Maxwell's equations, but we don't need them to satisfy the wave equation (for E and B fields)?

16. Oct 31, 2016

### Orodruin

Staff Emeritus
They will automatically satisfy the wave equation of the original fields do.

17. Oct 31, 2016

### greswd

?
Which is 'they'?