During a discussion with my professor, he asked me about relativistic resistance transformation. Firstly I started with the formula $R=\frac{\rho L}{s}$, where $\rho$ is electric resistivity. So if some resistor moves with relativistic speed its length contracts. And this means that its resistance changes like its length. Its resistivity doesn't change because of Ohm's law $E=\rho I$. Current changes like an electric field and that means that $\rho=\text{const}$. But then the professor said that to find this transformation I can't use the first formula, and I should find it from some laws that have resistance in it. So I used $$P=I^2R \tag{2}$$ and $$R=\frac{U}{I} \tag{3}.$$ From formula (3) we find that resistance doesn't change because difference of potentials and current change the same way. But from formula (2) we find that resistance changes like $\frac{R}{\gamma^3} $ (where $\gamma$ is Lorentz factor). So basically I have three variants, but the professor said that I need to find one result. Any ideas?


2 Answers 2


Repeating what @Dale said in a slightly different way, if you want relativistic laws, it is better to work with relativisitic quantities.

The response of matter to applied electromagnetic field is generally expressed through charge density ($\rho$) and current density ($\mathbf{J}$). The 4D equivalent is the four-current:

$J^\mu=\left(c\rho, \mathbf{J}\right)^\mu$

Next, the electric and magnetic fields are not covariant quantities in SR, so use the electromagnetic tensor: $F^{\mu\nu}$ (https://en.wikipedia.org/wiki/Electromagnetic_tensor)

Now, conductivity, which in general is a tensor, arises as a result of postulating a linear relationship between the applied electric field and the current density.

So we could say:


Now this general definition hoovers up pretty much all the simple types of electromagnetic response. But that's the point, what looks like conductivity response to one observer will look very different to another one.

You could postulate that in your rest frame ($\bar{S}$) the response is solely that of the isotropic conductivity ($\sigma$), then the only non-zero components are ($c$ is the speed of light):




Then consider the lab-frame relative to which the rest-frame is moving along x with speed $v$. You then transform your tensor $\sigma^\mu_{\eta\nu}=\frac{\partial x^\mu}{\partial \bar{x}^\alpha}\frac{\partial \bar{x}^\beta}{\partial x^\eta}\frac{\partial \bar{x}^\kappa}{\partial x^\nu}\bar{\sigma}^\alpha_{\beta\kappa}$. The only components to change are the ones involving x:

I get



So you could say that in lab-frame the conductivity in x direction will change by $\gamma=1/\sqrt{1-\left(v/c\right)^2}$. This is the second equation, but that's not the end of the story. There is the first equation, which in non-relativistic terms would lead to:

$\rho=\frac{v\sigma}{c^2}\gamma\cdot E_x$

Where $E_x$ is the x-component of the electric field. So you would see a build-up of charge density due to applied electric field.

So, as you now see, the picture is more complex than just finding an equation of resistance $R$

  • $\begingroup$ Clearly if you wanted resitance you would also have to integrate the whole thing in space, which would bring into play the length contraction. $\endgroup$
    – Cryo
    Commented May 29, 2019 at 23:53
  • $\begingroup$ I would add a factor $\frac{1}{2}$ in front of your relativistic equation (the one involving the conductivity tensor). This is usual when we contract two antisymetric tensors. This way, there's no factors of $1/2$ in the transformed conductivity. Also, please, consider adding a tag to your equations, using the LaTeX command \tag{#}. $\endgroup$
    – Cham
    Commented May 30, 2019 at 2:13
  • 3
    $\begingroup$ In the case of an isotropic material, the conductivity tensor could be written explicitely using the material's four-velocity and the Minkowski metric: $$\sigma_{ab}^c = \sigma \, (\eta_a^c \, u_b - \eta_b^c \, u_a).$$ Then: $$J^c = \frac{1}{2} \, \sigma_{ab}^c \, F^{ab} \equiv \sigma \, F^{cb} \, u_b,$$ which is Ohm's law in relativistic form (for an isotropic material of proper conductivity $\sigma$). For a general four-velocity, this tensor gives $\sigma_{10}^0 = \sigma \gamma \, v$ and $\sigma_{10}^1 = \sigma \gamma$ (using $c \equiv 1$), as your formulae. No need for coords. transf. $\endgroup$
    – Cham
    Commented May 30, 2019 at 2:25
  • $\begingroup$ Very nice. I didn't have the patience to find the good form. Thank you $\endgroup$
    – Cryo
    Commented May 30, 2019 at 8:37
  • $\begingroup$ This is it. Expect things to become tensorial. When you see something acting weird like that, giving funny, inconsistent results in a context like this - sniff around - could be a tensor lurking. $\endgroup$ Commented May 30, 2019 at 17:00

So basically I have three variants, but professor that I need to find one result. Any ideas?

Circuit theory is inherently nonrelativistic and so there is no relativistic circuit theory. This should actually not be surprising since circuit theory does not use the concept of space at all and relativity is a theory of spacetime.

The issue is in the assumptions that circuit theory is based on. The three assumptions are: (1) there is no net charge on any circuit element, (2) there is no magnetic flux outside of any circuit element, (3) the circuit is small so that electromagnetic influences can be assumed to propagate instantaneously.

Assumption (3) is expressly prohibited by relativity, and since a current density in one frame is both a current and a charge density in another then (1) will generally not be satisfied in all frames. Therefore, it is fatally flawed to attempt to find a relativistic version of Ohms law based on the equations of circuit theory since those laws are inherently non relativistic.

However, you could take the form of Ohm’s law that is based directly on Maxwell’s equations: $\mathbf J = \sigma \mathbf E$. This law is based on Maxwell’s equations, which are completely relativistic. Since we know how $\mathbf J$ and $\mathbf E$ both transform then we can calculate how $\sigma$ transforms.

  • $\begingroup$ J and E change the same way. So sigma doesn't change, it means R change like length? $\endgroup$ Commented May 29, 2019 at 20:09
  • 7
    $\begingroup$ I would not recommend going from 𝜎 to 𝑅. Circuit theory is not relativistic so it doesn’t make sense to find transformation laws for quantities in circuit theory. Just leave it as 𝜎. By the way, J does not transform like E. $\endgroup$
    – Dale
    Commented May 29, 2019 at 20:37

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