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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?

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closed as unclear what you're asking by garyp, G. Smith, John Rennie, Jon Custer, Emilio Pisanty Jul 18 at 12:46

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  • $\begingroup$ What do you think is wrong? $\endgroup$ – garyp Jul 16 at 20:41
  • $\begingroup$ @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. $\endgroup$ – HolgerFiedler Jul 16 at 20:47
  • $\begingroup$ Your answer is incomplete, but not yet wrong. What do you think is wrong with it? $\endgroup$ – garyp Jul 17 at 10:25
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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.

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  • $\begingroup$ Puk, it’s not a figure of mine, I’ve only cited it ;-) $\endgroup$ – HolgerFiedler Jul 17 at 4:55
  • $\begingroup$ Your last sentence is mistaken: a standing wave also won't have any phase difference. $\endgroup$ – Ruslan Jul 17 at 13:57
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    $\begingroup$ 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 $\pm 90^\circ$. $\endgroup$ – Puk Jul 17 at 18:37
  • $\begingroup$ Puk, wouldn't you have to add to your answer that k=0 is also a solution? $\endgroup$ – HolgerFiedler Jul 18 at 3:52
  • $\begingroup$ 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. $\endgroup$ – Puk Jul 18 at 5:44

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