# How does a change of reference frame affect a 1D (non-relativistic) electrodynamic analysis?

I am a bit confused as to how to handle the effect of a change of reference frame on a 1D electrodynamic analysis that I am looking at, in relation to a plasma.

Say I have an infinite tube containing some sort of 'ionic plasma' - equal quantities of positive and negative ions (quasi-neutral), with equivalent drift rates. I am trying to consider this as a 1D case (and assuming such a plasma can exist).

If I apply an electric field in the positive $$x$$ direction (along the tube axis), then the positive ions will drift in that direction and the negative ions in the $$-x$$ direction. So, a net current will flow in the $$+x$$ direction.

Now, say I change to a frame of reference that is moving with the negative ions - how does that affect how the electric field should be considered in the analysis? In the new frame, the negative ions are stationary, so the apparent field acting on them is zero. However, now the apparent speed of the positive ions is doubled, so the apparent E field acting on them seems to be double. So, do I now need to have two different E fields in the new frame, for the positive/negative ions? Or, is it the apparent charges of the particles that have changed?

Also, just to clarify, this is not relativistic - I am interested in 'classical' velocities. There is probably a simple answer to this based on electrodynamics, but would greatly appreciate any pointers on the correct way to consider this.

I haven't worked this out directly before, should be fun....

The scalar and vector potentials in relativity combine into a four vector that gets Lorentz-boosted. In the nonrelativistic limit, when you shift reference frames by speed $$\mathbf v$$, this will shift your scalar and vector potentials like (SI units) $$\phi'=\phi-\mathbf v\cdot \mathbf A,\\ \mathbf A' =\mathbf A-\mathbf v~ \phi/c^2.$$ From this it can be seen that starting from a frame with no magnetic field where $$\mathbf A=0,$$ the scalar potential stays intact to first order, and the fields should only change by $$\mathbf E'=\mathbf E+\frac{\mathbf v}{c^2}\frac{\mathrm d\phi}{\mathrm dt},\\ \mathbf B' = \frac{1}{c^2} \mathbf E\times\mathbf v.$$ If I'm understanding you right, in your case you further have a constant field over time and a reference frame change in the direction of that field, and this means that you don't see the fields change at all. If you were to introduce a field perpendicular to the motion, you might see a magnetic field pop up... but the electric field to first order would remain approximately constant.

Now, you do appear to have a further confusion. For some reason you think that the electrostatic force is velocity dependent. It is not.

Hope that helps!

• Hi, thanks for your answer. This makes sense and I can see that the force due to the electric field on the charged particles would not change with the change of frame. However, in this case I am actually considering a 'dense' plasma, where the charged particles would reach a constant drift velocity $v = \mu E$ due to particle collisions (I probably should clarify that in the question). Feb 17, 2022 at 18:24
• Ok, I think I'm just being silly then. There's no need to make any change to the E field in the new ref. frame - $v' = \mu E - v_f$, where $v_f$ is the frame velocity. Feb 17, 2022 at 18:30
• Ah. Then you need to consider the drift velocity relative to the particles they are colliding with, $v=v_0+\mu E,$ and your reference frame transform is modifying $v_0$ too. Feb 17, 2022 at 18:31

Under a proper Lorentz transformation, the Lorentz force 3-vector and 4-vector is not frame independent. Suppose we move to a new reference frame with velocity $$\mathbf{V}_{o}$$, the electric and magnetic field (in cgs units now) 3-vectors transform as: \begin{align} \mathbf{E}' & = \gamma \left( \mathbf{E} + \frac{ \mathbf{V}_{o} }{ c } \times \mathbf{B} \right) - \frac{ \gamma^{2} }{ \gamma + 1 } \frac{ \mathbf{V}_{o} }{ c } \left( \frac{ \mathbf{V}_{o} }{ c } \cdot \mathbf{E} \right) \tag{0a} \\ \mathbf{B}' & = \gamma \left( \mathbf{B} - \frac{ \mathbf{V}_{o} }{ c } \times \mathbf{E} \right) - \frac{ \gamma^{2} }{ \gamma + 1 } \frac{ \mathbf{V}_{o} }{ c } \left( \frac{ \mathbf{V}_{o} }{ c } \cdot \mathbf{B} \right) \tag{0b} \end{align} while the 3-vector velocity transforms according to the addition of velocities and $$\gamma$$ is the Lorentz factor. One can see that Equations 2a and 2b reduce to 0a and 0b in the limit $$V_{o} \ll c$$ because $$\gamma \rightarrow 1$$ and the 2nd term in Equations 2a and 2b are 2nd order in $$\tfrac{ V_{o} }{ c }$$. The $$\tfrac{ \mathbf{V}_{o} }{ c } \times \mathbf{E}$$ drops out when you convert back to SI units because $$B \rightarrow c \ B$$ thus there is a factor of $$\tfrac{ V_{o} }{ c^{2} }$$ in that term.

Now, say I change to a frame of reference that is moving with the negative ions - how does that affect how the electric field should be considered in the analysis?

Electric currents and electromagnetic fields are not Lorentz invariants, so it should not be surprising that the current density (a 3-vector in nonrelativistic cases and a 4-vector in relativistic) changes in relativistic cases.

In the nonrelativistic case, you can think of electric currents as being the relative charge density flux between different species. That is, the nonrelativistic current density can be defined as: $$\mathbf{j} = \sum_{s} q_{s} \ n_{s} \ \mathbf{v}_{s} \tag{1}$$ where $$q_{s}$$ is the charge [e.g., C] of species $$sS (including sign),$$n_{s}$$is the number density [e.g.,$$cm^{-3}$$] of species$$sS, and $$\mathbf{v}_{s}$$ is the 3-vector velocity [e.g., km/s] of species \$sS in the reference frame of interest.

If you transform to a frame where $$\mathbf{v}_{-i} = 0$$, that is perfectly okay, but note that $$\mathbf{v}_{+i}$$ will change to account for the loss of the contribution due to $$\mathbf{v}_{-i}$$. In the absence of nuclear/quantum effects (e.g., pair production), the number density and charges are all conserved so a Galilean transformation only changes $$\mathbf{v}_{s}$$. Since it's a linear operation (i.e., subtract all $$\mathbf{v}_{s}$$ by $$\mathbf{v}_{-i}$$), the change is only in the bulk flow velocities of each species.

Now does it remain constant? Let's try a simple thought experiment. Suppose you have two, oppositely signed charge carriers drifting along $$\hat{\mathbf{x}}$$ relative to each other. Further suppose they have equal but opposite charges and equal number densities. Does there exist a reference frame where the net current is zero? If both species drift in the same direction then $$\mathbf{j} \rightarrow 0$$ when the magnitudes of their drift velocities match, i.e., there is no relative drift between the two species. If the two species have different charge states (e.g., alpha-particles and electrons), then they need not drift together to have a zero net current so long as they drift the same direction at the correct ratio of speeds.

So, do I now need to have two different E fields in the new frame, for the positive/negative ions?

No, as I said above the Lorentz force is not a Lorentz invariant. That is, the forces acting on each charge species will change in different reference frames.

Or, is it the apparent charges of the particles that have changed?

No, total charge is a conserved scalar quantity under a Lorentz transformation. Charge density, however, can change under a Lorentz transformation. Though the charge density cannot change arbitrarily, as it is part of the 4-vector with the 3-vector current density to make up what's sometimes called the 4-current. Under Galilean transformations, the charge density is conserved.

So in the limit as $$\tfrac{ V_{o} }{ c } \rightarrow$$ very small (i.e., Galilean transformations), velocities, and the electric and magnetic fields transform as: \begin{align} \mathbf{E}' & = \mathbf{E} + \mathbf{V}_{o} \times \mathbf{B} \tag{2a} \\ \mathbf{B}' & = \mathbf{B} \tag{2b} \\ \mathbf{v}' & = \mathbf{v} - \mathbf{V}_{o} \tag{2c} \end{align}

Note that Equation 2a shows that there can exist reference frames with no electric fields, if the only electric fields present are quasi-static.