Applying Reisenbach Transf. to EM Wave in Microwave Oven

[wnvl2]

There is no possible measurement, no matter how clever, that can measure the one way speed of light. It is a synchronization convention. In this topic I would like to apply this idea on a specific case.

I have a microwave oven with width L. In this oven I have a standing wave.

$$E(t,x)=E cos(\omega t) sin( k x) $$

with $k = \frac{\pi}{L}$ and $c = \frac{\omega}{k}$.

We can split this standing wave into a left and a right moving wave.

$$\frac{E}{2} sin (\omega t - kx) + \frac{E}{2} sin (\omega t + kx)$$

Now I want to see what happens if we switch to another synchronization where the speed of the right-moving wave is
$c^{+}$ is $\frac{c}{2\epsilon}$ and the speed of the left moving wave $c^{-}$ is $\frac{c}{2(1-\epsilon)}$.

The Reisenbach synchronisation transformation looks like

$$ \begin{bmatrix}
ct \\
x \\
y \\
z \\
\end{bmatrix} = \Lambda \begin{bmatrix}
ct' \\
x' \\
y' \\
z' \\
\end{bmatrix}$$

with $$ \Lambda = \begin{bmatrix}
1 & \kappa & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix} $$

and $\kappa=2 \epsilon-1$

This means that the electromagnetic field tensor

$$F^{\mu \nu} = \begin{bmatrix}
0 & -\frac{E_x}{c} & -\frac{E_y}{c} & -\frac{E_z}{c}\\
\frac{E_x}{c} & 0 & -B_z & B_y\\
\frac{E_y}{c} & B_z & 0 & -B_x\\
\frac{E_z}{c} & -B_y & B_x & 0\\
\end{bmatrix} $$

will transform according to

$$\Lambda^T F^{\mu \nu} \Lambda $$

This means that $E_x$ transforms into $\kappa E_x$. I assume that in the xyz frame the magnetic field is everywhere 0.

The waves in the new frame x'y'z' transform into

$$\frac{\kappa E}{2} sin (\omega (t'-\frac{\kappa x'}{c}) - kx') + \frac{E}{2} sin (\omega (t'-\frac{\kappa x'}{c}) + kx')$$

$$\frac{\kappa E}{2} sin (\omega t'-(\omega\frac{\kappa}{c} - k)x') + \frac{E}{2} sin (\omega t'- (\omega \frac{\kappa}{c} + k)x')$$

A first conclusion is that I have not anymore a standing wave.

Is this reasoning already correct?

 

[anuttarasammyak]

Now I want to see what happens if we switch to another synchronization where the speed of the right-moving wave is
$c^{+}$ is $\frac{c}{2\epsilon}$ and the speed of the left moving wave $c^{-}$ is $\frac{c}{2(1-\epsilon)}^$.

I have no idea what kind of synchronization it is which let light waves have speed other than c.

 

[wnvl2]

I refer to the reference of Dale in

https://www.physicsforums.com/threads/reichenbach-synchronisation-and-reflection.1013931/

or to this

https://www.physicsforums.com/threa...ed-of-light-by-combined-measurements.1014053/

discussion.

 

[Ibix]

Generally, coordinates in which light speed is anisotropic mean that your clocks are out of sync with respect to Einstein synchronised clocks. Simply applying the coordinate transform $x'=x$ and $t'=t+\kappa x$ I would expect $\sin(\omega t)\sin(kx)$ to transform to $\sin(\omega(t'-\kappa x'))\sin(kx')$. So I suspect you will find you have nodes in the same place, but the wave will be a sine wave with some kind of position dependent phase function. I don't have pen and paper to work out if your result matches that.

 

[DrGreg]

@wnvl2 $$\cos(\omega t) \sin( k x) = \frac{1}{2} \sin (kx - \omega t) + \frac{1}{2} \sin (kx + \omega t)$$which explains what went wrong.

But there's no need for this. Why not just substitute directly into $\cos(\omega t) \sin( k x)$? The $\sin( k x)$ term remains unaltered.

 

[Sagittarius A-Star]

This means that $E_x$ transforms into $\kappa E_x$. I assume that in the xyz frame the magnetic field is everywhere 0.

Electromagnetic waves are transverse waves. Therefore, $E_x = B_x = 0$ in this scenario, with one wave moving in +x direction and the other in -x direction. The $E$ vector and $B$ vector are perpendicular to the x-direction and to each other. The magnetic field is not everywhere 0.

 

[Dale]

A first conclusion is that I have not anymore a standing wave.

It still seems to be a standing wave, although it is more difficult to see that since you split it into two traveling waves.

 

[wnvl2]

@Sagittarius A-Star

Yes, I have $E_x = 0$ and $B_x = 0$.

 

[wnvl2]

I did already some corrections to the opening post. Too many errors

There is no possible measurement, no matter how clever, that can measure the one way speed of light. It is a synchronization convention. In this topic I would like to apply this idea on a specific case.

I have a microwave oven with width L. In this oven I have a standing wave.

$$E(t,x)=E cos(\omega t) sin( k x) $$

with $k = \frac{\pi}{L}$ and $c = \frac{\omega}{k}$.

We can split this standing wave into a left and a right moving wave.

$$-\frac{E}{2} sin (\omega t - kx) + \frac{E}{2} sin (\omega t + kx)$$

Now I want to see what happens if we switch to another synchronization where the speed of the right-moving wave is
$c^{+}$ is $\frac{c}{2\epsilon}$ and the speed of the left moving wave $c^{-}$ is $\frac{c}{2(1-\epsilon)}$.

The Reisenbach synchronisation transformation looks like

$$ \begin{bmatrix}
ct \\
x \\
y \\
z \\
\end{bmatrix} = \Lambda \begin{bmatrix}
ct' \\
x' \\
y' \\
z' \\
\end{bmatrix}$$

with $$ \Lambda = \begin{bmatrix}
1 & \kappa & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix} $$

and $\kappa=2 \epsilon-1$

This means that the electromagnetic field tensor

$$F^{\mu \nu} = \begin{bmatrix}
0 & 0 & -\frac{E_y}{c} & 0\\
0 & 0 & -B_z & 0\\
\frac{E_y}{c} & B_z & 0 & 0\\
0 & 0 & 0 & 0\\
\end{bmatrix} $$

will transform according to

$$\Lambda^T F^{\mu \nu} \Lambda $$

This means that $E_y$ remains $E_y$

The waves in the new frame x'y'z' transform into

$$-\frac{\kappa E}{2} sin (\omega (t'-\frac{\kappa x'}{c}) - kx') + \frac{E}{2} sin (\omega (t'-\frac{\kappa x'}{c}) + kx')$$

$$-\frac{\kappa E}{2} sin (\omega t'-(\omega\frac{\kappa}{c} - k)x') + \frac{E}{2} sin (\omega t'- (\omega \frac{\kappa}{c} + k)x')$$

A first conclusion is that I still have a standing wave.

 

[Sagittarius A-Star]

@Sagittarius A-Star

Yes, I have $E_x = 0$ and $B_x = 0$. If I have E_y, does that correspond to $B_z = \frac {E_y}{c}$ or $B_z = \frac {-E_y}{c}$?

• For a normal wave it depends on which direction you call "y" and which "z". In physics, the usual default for the orientation of the x-, y-, z- axes is as cited below. In this case $B_z = \frac {E_y}{c}$ is correct.

The orientation is usually chosen so that the 90 degree angle from the first axis to the second axis looks counter-clockwise when seen from the point (0, 0, 1); a convention that is commonly called the right hand rule.

Source:
https://en.wikipedia.org/wiki/Cartesian_coordinate_system#Three_dimensions

$\mathbf B = \frac{1}{c_0} \hat {\mathbf k} \times \mathbf E$

Source:
https://en.wikipedia.org/wiki/Electromagnetic_radiation#Derivation_from_electromagnetic_theory

• For a standing wave, both is wrong. The "electric standing wave" has, compared to the "magnetic standing wave", a phase shift of T/4 in time and of $\lambda / 4$ in space. The nodes of the "magnetic standing wave" are in the middle between the nodes of the "electric standing wave".

 

[wnvl2]

That is true, it is too long ago that I studied TEM waves.

 

[wnvl2]

When I calculate $\Lambda^T F^{\mu \nu} \Lambda$, I get

$$
F'^{\mu \nu} = \begin{bmatrix}
0 & 0 & -\frac{E_y}{c} & 0\\
0 & 0 & -\frac{\kappa E_y}{c}-B_z& 0\\
\frac{E_y}{c} & \frac{\kappa E_y}{c}+B_z & 0 & 0\\
0 & 0 & 0 & 0\\
\end{bmatrix}
$$

This means that the value of the electric field does not change, but the value of the magnetic field changes after the synchronisation transformation.

 

[Ibix]

You don't need the $\mu\nu$ if you are using matrix notation. Or, rather, you should put indices on your $\Lambda$s too.

I think you'll find that because $E_y=E_y(x,t)$ and you want $E'_y(x',t')$ then your transformed electric field is slightly different.

 

[vanhees71]

and don't forget that you have necessarily magnetic field components too.

 

[wnvl2]

You don't need the $\mu\nu$ if you are using matrix notation. Or, rather, you should put indices on your $\Lambda$s too.

Unbelievable how many mistakes I make

$$F^{'\alpha\beta} = \Lambda^{\alpha}_{\mu} \Lambda^{\beta}_{\nu} F^{\mu \nu} $$

or

$$F' = \Lambda F \Lambda^T$$

 

[wnvl2]

I think you'll find that because $E_y=E_y(x,t)$ and you want $E'_y(x',t')$ then your transformed electric field is slightly different.

I get

$$F' = \begin{bmatrix}

0 & 0 & -\frac{E_y}{c} & 0\\

0 & 0 & -\frac{\kappa E_y}{c}-B_z& 0\\

\frac{E_y}{c} & \frac{\kappa E_y}{c}+B_z & 0 & 0\\

0 & 0 & 0 & 0\\

\end{bmatrix}$$

I don't see that change in electric field, it is still $E_y$. For the magnetic field in the z direction I see a change.

 

[Ibix]

I get

$$F' = \begin{bmatrix}

0 & 0 & -\frac{E_y}{c} & 0\\

0 & 0 & -\frac{\kappa E_y}{c}-B_z& 0\\

\frac{E_y}{c} & \frac{\kappa E_y}{c}+B_z & 0 & 0\\

0 & 0 & 0 & 0\\

\end{bmatrix}$$

I don't see that change in electric field, it is still $E_y$. For the magnetic field in the z direction I see a change.

Yes, but $E_y=E\cos(kx)\sin(\omega t)=E\cos(kx')\sin(\omega(t'-\kappa x'))$. The change in the functional form of $E$ is (in this case) solely from the change in the functional form of the coordinates. For $B$ it is both the transformation of $F$ and of the coordinates.