Wave Equation & Wave Displacement Invariance: Modern Physics

[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.

 

[robphy]

F is a scalar field.

 

[greswd]

F is a scalar field.

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

 

[greswd]

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

 

[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.

 

[greswd]

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.

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:

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

 

[Orodruin]

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.

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.

 

[greswd]

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.

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?

 

[Orodruin]

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?

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.

 

[greswd]

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.

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

 

[Orodruin]

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

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

 

[greswd]

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.

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?

 

[greswd]

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

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

 

[Orodruin]

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.

 

[greswd]

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.

Thanks for the clear answer.

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)?

 

[Orodruin]

Thanks for the clear answer.

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)?

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

 

[greswd]

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

?
Which is 'they'?