# Is electric current relative?

Motion as we know is relative. According to this current which is the flow of charges should be also relative . That means that if someone is moving with the same velocity with repsect to the velocity of electrons (current) would he see no current at all ?

• – The Photon Dec 16 '18 at 17:35
• One problem is that the speed of the electrons in a conductor isn't the speed of the current. The current propagates at an appreciable fraction of light speed, while the electrons move quite slowly - on the order of a meter per hour, or much slower than walking speed. And that's for a direct current: with alternating current, the electrons just move back and forth a very short distance: en.wikipedia.org/wiki/Drift_velocity – jamesqf Dec 16 '18 at 20:44
• @jamesqf, To say that "the current propagates..." I think is misleading. What moves down the wire at near light speed is signals. That is, changes in current and voltage. – Solomon Slow Dec 17 '18 at 0:44
• @Solomon Slow: I suppose that's inherent in trying to describe things in English. One can think of it as signal propagation, or (if you've worked in electric power transmission) as the propagation of power. You might think of an analogy with a garden hose: if the hose is empty, it takes a few seconds for water to come out the end after you turn on the faucet; if it's full, water comes out practically instantaneously, but it's not the water that just came out of the faucet. – jamesqf Dec 17 '18 at 3:40
• @Solomon Slow: But my point was to show the problems with the question. We DO regularly move faster than the electrons in a DC circuit (while in an AC circuit they don't really move at all, just wiggle back & forth), yet we don't see any change in their function. – jamesqf Dec 17 '18 at 3:44

Charge density $$\rho$$ and current density $$\vec{J}$$ form a Lorentz four-vector $$(c\rho, \vec{J})$$ that transforms under a Lorentz transformation just like $$(ct, \vec{r})$$ does.

For example, if two frames are in relative motion in the $$z$$-direction, with the primed frame having velocity $$v\hat{\mathbf{z}}$$ relative to the unprimed frame, then the transformation is

\begin{align} c\rho^\prime &= \gamma\left(c\rho - \frac{v}{c}J_z\right)\\ J_x^\prime &= J_x \\ J_y^\prime &= J_y \\ J_z^\prime &= \gamma\left(J_z - \frac{v}{c}(c\rho)\right) \end{align}

where $$\gamma=1/\sqrt{1-v^2/c^2}$$.

If you had a beam of electrons in space, and you observed it from a reference frame moving with the same velocity as the beam, there would be no current.

Since charge density and current density are frame-dependent, so are electric and magnetic fields. (For example, in the frame moving along with the beam, there is no magnetic field.) However, the transformation rules for the fields are a bit more complicated; they transform as components of a two-index four-tensor rather than a one-index four-vector.

• But why would there be no current? Or is this answer in the formulas you provided? – undefined Dec 17 '18 at 10:43
• @undefined It's in the formulas but the intuition is pretty obvious. If you are going with the current you don't feel it (if the current speed is not constant you could feel G-force). In water it would be the same as standing in still water. In electricity you would see stationary charges. It's classic relativity of motion. – luk32 Dec 17 '18 at 12:00
• @luk32 thank you for your comment. Is this phenomena independent of volt and ampere? – undefined Dec 17 '18 at 12:31
• @undefined The water analogues still hold here, Volt unit are used to describe velocity of the current, and Ampere the amount - as in number of particles in the flow. So I'd say it's not independent, especially relation to Volts seems to be quite direct. – luk32 Dec 17 '18 at 12:36

Typically, we observe current in a solid, such as copper wire. So if you move with the same velocity as the average velocity of electrons in a current-carrying copper wire you will observe current of nuclei and bound electrons.

• This doesn't answer the question, unless the implication is supposed to be that current is frame-independent, which would be incorrect. – Ben Crowell Dec 16 '18 at 22:28
• @BenCrowell : The OP asked: "if someone is moving with the same velocity with repsect to the velocity of electrons (current) would he see no current at all ?" My answer means "no", so it is a direct answer to the OP's question, although a partial one, as it does not answer the question in the title. An answer does not have to be complete. And I did not say that the current is frame-independent. – akhmeteli Dec 16 '18 at 22:37

$$\let\g=\gamma \let\lam=\lambda$$ I'd like to expand G. Smith's answer by applying it to a metal wire.

Assume a straight wire along $$z$$-axis. If it carries a current it will contain stationary positive charges and moving electrons. The wire is neutral, i.e. charge densities of positive and of negative charges balance each other: $$\lam_+ + \lam_- = 0.\tag1$$ There will be a non-null current carried by electrons $$I_-\ne0$$, whereas $$I_+=0\$$ ($$I = I_+ + I_-$$). Note that by $$\lam$$ here I mean linear charge density, by $$I$$ the usual electric current.

It's useful to bear in mind signs. Of course $$\lam_+>0$$, $$\lam_-<0$$. As to currents we may choose. If $$I>0$$ is taken, i.e. $$I_->0$$, then $$v<0$$ (in order to have a positive current electrons have to move in the negative direction).

As far as Lorentz transformations along $$z$$ are concerned, trasformation laws of $$\lam$$, $$I$$ are the same as for $$\rho$$, $$J_z$$: \eqalign{ \lam'_\pm &= \g \left(\!\lam_\pm - {v \over c^2}\,I_\pm\!\right) \cr I'_\pm &= \g\,(I_\pm - v\,\lam_\pm).\cr} \tag2

If we transform to average electrons rest frame, where $$I'_-=0$$, we have $$I_- = v\,\lam_-$$ and substituting into the first of (2) $$\lam'_- = \g \left(\!\lam_- - {v^2 \over c^2}\,\lam_-\!\right) = {\lam_- \over \g}.$$ This is nothing but Lorentz contraction. Charge is invariant, so that charge density transforms inversely to length. As far as electrons are concerned, length in the lab frame is contracted wrt length in their rest frame and charge density increases passing from the latter to the former: $$\lam_-=\g\,\lam'_-$$. The opposite happens to $$\lam_+$$: $$\lam'_+ = \g\,\lam_+$$ since $$I_+=0$$ and positive charges are at rest in the lab frame.

Summarizing, in electrons rest frame we have $$\lam' = \lam'_+ + \lam_- = \g\,\lam_+ + {\lam_- \over \g} = \left(\!\g - {1 \over \g}\!\right) \lam_+ = \g\,{v^2 \over c^2}\,\lam_+ > 0$$ (I've used (1)). In that frame the wire is positively charged. As to current $$I' = I'_+ + I'_- = -v\,\g\,\lam_+ = v\,\g\,\lam_- = \g\,I_- = \g\,I.$$ Of course this is the second of (2) written for $$\lam$$, $$I$$ with $$\lam=0$$. We see that in electrons rest frame an increased current is observed (btw this also implies an increased magnetic field).