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From Nature Of Photon:

Electromagnetic field

The set of Maxwell equations [2] for vacuum is:
$$\begin{align} \mathrm{rot} \mathbf{E} &= -∂\mathbf{B}/c∂t, \tag{1} \\ \mathrm{rot} \mathbf{B} &= ∂\mathbf{E}/c∂t, \tag{2} \\ \mathrm{div} \mathbf{E} &= 0, \tag{3} \\ \mathrm{div} \mathbf{B} &= 0 \tag{4} \end{align}$$
where:
$\mathbf{E}$ – vector of electric field,
$\mathbf{B}$ – vector of magnetic field,
$t$ – time,
$c$ – speed of light.

In the case of a monochromatic wave the expression for electric field $\mathbf{E}$ is:
$$\mathbf{E} (x, t) = \mathbf{E}_0 \sin (ωt), \tag{5}$$ where: $\mathbf{E}_0$ – amplitude of electric field.

A physically correct solution can be obtained if in the equation (2) the expression of electric field $\mathbf{E}$ is used from (5), i.e.,
$$\mathrm{rot} \mathbf{E} = ∂(\mathbf{E}_0 \sin (ωt)) /c∂t.$$

The result is:
$$\mathbf{B} (x, t) = \mathbf{B}_0 \cos (ωt). \tag{6}$$

Vector $\mathbf{E}$ is shifted according to vector $\mathbf{B}$ by 90 degrees (Fig. 1.).
enter image description here
Fig. 1. Electric $\mathbf{E}$ and magnetic $\mathbf{B}$ field of the photon.

Where's that wrong?

Sebastiano
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HolgerFiedler
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  • What do you think is wrong? – garyp Jul 16 '19 at 20:41
  • @garyp IF that is a possible solution I would prefer to see in the Pointing vector a statistical or summarising value and to accept that the near field image of the EM radiation and the propagation of photons are shown in the sketch above correct. – HolgerFiedler Jul 16 '19 at 20:47
  • Your answer is incomplete, but not yet wrong. What do you think is wrong with it? – garyp Jul 17 '19 at 10:25

1 Answers1

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The proposed solution for $\vec{E}$ needs to include the spatial variation if you want to be able to solve for $\vec{B}$. Otherwise you cannot evaluate curls. For a travelling plane wave, it would be of the form $$\vec{E} = \hat{x} E_0 \sin(\omega t - kz)$$ assuming $\vec{E}$ is along $\hat{x}$ and propagation is along $\hat{z}$.

The solution to $$\vec{\nabla}\times \vec{B} = \frac{1}{c} \frac{\partial\vec{E}}{\partial t}$$ is $$\vec{B} = \hat{y} \frac{\omega}{c k} E_0 \sin(\omega t - kz)$$ From the other curl equation you can find $k=\pm\omega/c$ for self-consistency.

If the spatial variation of the fields is different, we need to how it is different before we can comment. When there is a superposition of forward and backward traveling waves with the same amplitude, there will be a $90^\circ$ phase difference between $\vec{E}$ and $\vec{B}$ as your figure suggests.

Puk
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  • Puk, it’s not a figure of mine, I’ve only cited it ;-) – HolgerFiedler Jul 17 '19 at 04:55
  • Your last sentence is mistaken: a standing wave also won't have any phase difference. – Ruslan Jul 17 '19 at 13:57
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    Ruslan: I believe you are mistaken. When propagation is along a single axis and the amplitudes of forward and backward traveling waves are equal, the wave impedance will always be imaginary, meaning a $\pm 90^\circ$ phase difference between $\vec{E}$ and $\vec{H}$. Another way to see this is that in this case the time-averaged Poynting vector will be zero, which for non-zero $\vec{E}$ and $\vec{H}$ can only happen if their phase difference is $\pm 90^\circ$. – Puk Jul 17 '19 at 18:37
  • Puk, wouldn't you have to add to your answer that k=0 is also a solution? – HolgerFiedler Jul 18 '19 at 03:52
  • No. For any non-zero $\omega$, $k = 0$ is not allowed. This can be seen from the fact that it makes $\vec{B}$ blow up. It is only allowed if $\omega = 0$, in which case $k = \pm \omega/c$ already gives zero. – Puk Jul 18 '19 at 05:44
  • @Puk How the time derivative of E with sin(wt) (without kz) blows up B? – HolgerFiedler Jul 18 '19 at 08:16
  • See the expression for $\vec{B}$ in my answer. The denominator has $k$ in it, which cannot be zero. The two curl equations cannot hold simultaneously if neither $\vec{E}$ nor $\vec{B}$ have a spatial variation (non-zero curl), unless $\vec{E}$ and $\vec{B}$ are both static (i.e. have no time variation). – Puk Jul 18 '19 at 08:27
  • Puk, a derivation of a summ of something + zero is definitely not a denominator with zero. – HolgerFiedler Jul 18 '19 at 11:18
  • I don't know what that means, and I don't know what else I can say to convince you other than "do the math yourself." $\vec{E}$ and $\vec{B}$ as you give them in your question are not solutions to Maxwell's equations because they have time variation but no spatial variation, which in vacuum cannot happen. Unless you indicate to me that you are legitimately looking for help and not just resisting what I'm trying to tell you for no apparent reason, I'm done here. – Puk Jul 18 '19 at 17:43