I Applying Reisenbach Transf. to EM Wave in Microwave Oven

wnvl2
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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?
 
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wnvl2 said:
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.
 
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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.
 
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@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.
 
wnvl2 said:
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.
 
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wnvl2 said:
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.
 
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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.
 
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  • #10
wnvl2 said:
@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.
Wikipedia said:
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

Wikipedia said:
##\mathbf B = \frac{1}{c_0} \hat {\mathbf k} \times \mathbf E##
Source:
https://en.wikipedia.org/wiki/Electromagnetic_radiation#Derivation_from_electromagnetic_theory

 
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  • #11
That is true, it is too long ago that I studied TEM waves.:rolleyes:
 
  • #12
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.
 
  • #13
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.
 
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  • #14
and don't forget that you have necessarily magnetic field components too.
 
  • #15
Ibix said:
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

$$F' = \Lambda F \Lambda^T$$
 
  • #16
Ibix said:
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.
 
  • #17
wnvl2 said:
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.
 
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