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Are $\epsilon_{0}$ (permittivity) and $\mu_{0}$ (permeability) invariant between different inertial frames? If yes, what hard evidence does it support that? If no, why would $c$ (velocity of light) be invariant between different inertial frames and what is the need for Lorentz transformation?

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To add to C.R. Drost's Answer: Given the experimental fact of the invariance of Maxwell's equations, the invariance of $\mu_0$ arises by definition; it defines our unit of current. An intuitive and covariant form of Maxwell's equations is:

$$\Box\,A^\alpha = \mu_0\,J^\alpha$$

where $A^\alpha$ is the (contravariant) four potential and $J^\alpha$ the four current.

You can see $\mu_0$ defines the relationship between our units for current and units for potential / EM field by telling us how much field we get for how much current. We define our units so that the ampere is the current that would, if present in two straight parallel conductors of infinite length, of negligible circular cross section, and 1 meter apart in vacuum, give rise to a force between these conductors of $2\times10^{-7}{\rm N}$ per meter of length of conductor. This definition of the ampere fixes $\mu_0$ to its SI value of $4\pi\times10^{-7}{\rm H\,m^{-1}}$. Implicit in this definition is that this scaling is defined in a frame at rest relative to the conductors. Given the empirical fact of the covariance of Maxwell's equations, we are free to choose a frame that is convenient for measurement, just as we define the invariant mass of a particle through its total energy in a frame at rest to the particle.

The invariance of $\epsilon_0$ then of course follows from that of $c$ and that of $\mu_0$.

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  • $\begingroup$ More generally it's $\nabla_\beta F^{\beta\alpha}=\mu_0 J^\alpha$, but in a flat spacetime that reduces to your given equation provided we work in the gauge $\nabla_\alpha A^\alpha=0$. $\endgroup$ – J.G. Jan 24 '17 at 7:30
  • $\begingroup$ @J.G. Absolutely true: The question was about inertial frames and I simply wanted a form that most evoked a "one way" (in the sense of RHS + boundary conditions fully define the LHS) current to implied solution relationship. One strictly gets that with the more general form, but the quoted form is a little more obvious IMO. $\endgroup$ – WetSavannaAnimal Jan 24 '17 at 9:56
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The magnetic constant is fixed by our choice of SI units as $4\pi\cdot10^{-7} \text N/\text A^2.$

The electric constant could vary if $c$ varies and the hard evidence to suggest it doesn't is precisely the hard evidence that suggests a fixed $c$.

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