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Far as I can tell, the force generated by electromagnetic field can be seen as an electrical force or as a magnetic force depending on the observer, i.e., these forces are actually the same phenomenon experienced differently, according to special relativity.

Do any of the other fundamental forces exhibit such relationship?

If not, what's so special about the photon field that it would exhibit such a behavior when nothing else does?

(I realize that at higher levels of energy, the weak interaction also unifies with emf. Is this phenomenon perhaps related?)

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    $\begingroup$ An electromagnetic field is always an electromagnetic field, independent of the observer. A magnetic field does not turn into an electric field and vice versa by changing the coordinate system. $\endgroup$ Commented Mar 28, 2023 at 9:12
  • $\begingroup$ That's not what I meant. The field is the field, yes. But the forces observed as experienced by a charged particle depend on the relative velocity between the observer and the particle (I think...) $\endgroup$ Commented Mar 28, 2023 at 10:04

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This answer shows the relativistic formulation of electromagnetism and is heavily based on this document by David Tong. In the final paragraph I somewhat give an answer to your question and hope others can correct/expand in case it's wrong.

In order to construct any relativistic theory you need to satisfy the postulates of special relativity, one of which requires the laws of physics to be the same in all inertial (not undergoing acceleration) frames of reference, or in other words your equations need to be Lorentz covariant. You achieve that by constructing your equations out of 4-vectors (e.g. the position vector $x^{\mu}=(ct,x,y,z)^T$) and tensors of higher rank.

As you can see, a 4-vector has 4 components, while the electric and magnetic fields $\vec{E}$ and $\vec{B}$ contain 3 components each - you need to do something extra in order to construct a 4-vector. Luckily, you know that electric and magnetic fields must obey Maxwell's equations, and you can define a scalar ($\phi$) and a vector ($\vec{A}$) potentials such as: \begin{equation} \vec{E} = -\nabla\phi-\partial{\vec A}/ \partial{t} \end{equation} \begin{equation} \vec{B} = -\nabla\times\vec{A}. \end{equation} You can play around with these, e.g. shift $\vec{A}$ and make a corresponding shift in $\phi$ without changing the values of $\vec{E}$ and $\vec{B}$ in the process. I am writing all this because you can now define a new 4-vector using these potentials, $A^\mu= (\phi/c,\vec{A})^T$, and cast your equations in a Lorentz covariant form using it. More specifically, you can define the electromagnetic tensor as: \begin{equation} F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu = \begin{pmatrix} 0 & E_x/c & E_y/c & Ez/c \\ -E_x/c & 0 & -B_z & B_y \\ -E_y/c & B_z & 0 & -B_x \\ -E_z/c & -B_y & B_x & 0\\ \end{pmatrix}. \end{equation}

If you apply a Lorentz boost to this object in order to go into a different inertial frame of reference, e.g. along the $x$-direction with velocity $v$, you will find the results (5.17) and (5.18) of Tong's document - no change in the $E_x$ and $B_x$, and mixing in the $y$ and $z$ components, e.g. $E_y'=\gamma(E_y-vB_z)$ - what appears to be the electric field in the primed coordinate along the $y'$-axis is a mixture of the electric field along the $y$ axis of the other frame in addition to some part of the magnetic field component along $z$. You can also define the 4-current using the tensor - $\partial_\nu F^{\nu\mu} = J^\mu$, where $J^\mu = (c\rho,\vec{j})$, $\rho$ - charge density, $\vec{j}$ - current.

About the question for other forces - there are equivalent to $F_{\mu\nu}$ tensors for the other forces as well (with some complications due to the weak and strong interactions being non-abelian), with potentials equivalent to $A_\mu$ and currents obtainable from the equations of motion. However, I am not aware of definitions of classical fields as components of the tensor (probably because definitions of classical fields makes less sense due to the short ranges of the weak/strong interactions).

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  • $\begingroup$ "definitions of classical fields makes less sense due to the short ranges of the weak/strong interactions" That is quite interesting insight. I had to look up non-abelian (it's been a while... :) ), but this is very solid info. +1 for now. (I'll wait a day or two before ticking.) $\endgroup$ Commented Mar 28, 2023 at 13:27
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Photons an electromagnetic wave does exhibit wave-particle duality depending when it strikes the object it exhibit as a particle. Through vacuum it travels as a resultant electromagnetic wave form. Photon exhibit this is because it is electromagnetic wave.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – ZaellixA
    Commented Mar 28, 2023 at 9:51
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    $\begingroup$ This is not what I'm asking at all. :) I (think I) understand how electromagnetism works. I'm asking if there are other forces that exhibit similar behaviors. $\endgroup$ Commented Mar 28, 2023 at 10:05

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