# How can we prove charge invariance under Lorentz Transformation?

We have gravitational force between two massive particles and we have electromagnetic force between two charged particles. When special relativity suggests that mass is not an invariant quantity why we have electric charge as an invariant quantity ?

Let $j=(\rho,\boldsymbol j)$ be the current density of a system. This four numbers are, by hypothesis, a vector. This means that the charge density $\rho$ transforms just like $t$ does, i.e., it gets "dilated" when changing from reference frame to reference frame: $$\rho'\to\gamma \rho \tag{1}$$

Charge is, by definition, the volume integral of the charge density: $$Q\equiv \int\mathrm d\boldsymbol x\ \rho \tag{2}$$

In a different frame of reference the charge is $$Q'=\int\mathrm d\boldsymbol x'\ \rho'=\int\mathrm d\boldsymbol x'\ \gamma \rho \tag{3}$$ where I used $(1)$.

Next, we need to know what $\mathrm d\boldsymbol x'$ is. The trick to evaluate this is to note that the product $\mathrm dt\;\mathrm d\boldsymbol x$ is invariant (in SR). This means that we can write $\mathrm dt'\;\mathrm d\boldsymbol x'=\mathrm dt\;\mathrm d\boldsymbol x$; solving for $\mathrm d\boldsymbol x'$ we get

$$\mathrm d\boldsymbol x'=\frac{\mathrm dt}{\mathrm dt'}\mathrm d\boldsymbol x=\frac{1}{\gamma}\mathrm d\boldsymbol x \tag{4}$$

If we plug this into $(3)$, we find $$Q'=\int\mathrm d\boldsymbol x'\ \gamma \rho=\int\mathrm d\boldsymbol x\ \rho=Q \tag{5}$$ that is, $Q=Q'$.

• would the down-voter care to comment? – AccidentalFourierTransform Apr 11 '16 at 19:02
• I'm not the downvoter, but this argument assumes the conclusion. You've just slightly modified the starting assumption to be "$j^\mu$ is a 4-vector", which is basically the same thing as charge invariance. The better way is to start with a Lagrangian (postulated to be invariant), then apply Noether's theorem. – knzhou Apr 12 '16 at 19:32
• @knzhou I'm not sure what you mean. I didn't make up this proof: it is the standard one, and can be found in many books on SR. My assumption is indeed "$j^\mu$ is a four vector". Using this, together with Lorentz contraction/dilation we easily prove that $Q$ takes the same value in any reference frame. Also, Noether's theorem is an total overkill here, where we have a simple proof that uses physics that any undergrad knows. – AccidentalFourierTransform Apr 12 '16 at 19:40
• But until you learn about $j^\mu$ in special relativity, every proposed 4-vector is carefully proven to actually be a 4-vector; $j^\mu$ is the only one that has to be postulated. Justifying that assumption is the real question, i.e. that's where the physics is. – knzhou Apr 12 '16 at 19:48
• Yes, I agree. I guess I was projecting myself onto the OP; when I learned about special relativity, nobody would tell me about (2) or (3), they just repeated (1) over and over again. At the level of this post (1) is fine. – knzhou Apr 12 '16 at 19:56

The whole electromagnetic theory lies on Maxwell's equations plus the Lorentz force equation. The Lorentz transformation is a geometrical description of how something related to space and time varies as it approaches the speed of light (in the absence of gravity;We are considering only inertial frames here).
The laws of electromagnetism are fundamental to any electromagnetic phenomenon. As one of the postulates of special theory of relativity states that the laws of physics are identical in all inertial frames, then the laws of electromagnetism should also be identical in all inertial frames. This is to (I mean the purpose of relativity) agree what two persons in different frame of reference should see the same physical laws.
So we need that the laws of electromagnetism should be the same for all inertial frames. This can be true only if we treat charge as an invariant quantity only. The only source of electromagnetic theory is the charges. We have the theorem of conservation of charges (we cannot either create or destroy a charge. All we can do is just move it from one point to another). This one law of physics should be valid for all inertial observers. As charge is the fundamental aspect of any electromagnetic phenomenon, two inertial frame should detect the same amount of charge at a definite region of space. Otherwise charge will not be conserved. The conservation of charge came from Maxwell's equations, which means that then all the laws of electromagnetism will appear non-symmetrical (totally different) to different inertial frames.
But you should not that if one person sees electrostatic force in one frame of reference, the other person in some other inertial frame of reference see the same as magnetic force. However both experiences a force. (This thing is one of the important theorem in relativistic electrodynamics : Magnetism is a relativistic phenomenon )

The electrical neutrality of atoms and molecules proves that charge is independent of velocity.A helium atom and a hydrogen molecule are both neutral although the speed of the electrons in the helium atom is almost twice what it is in a hydrogen molecule. So at least experimentally, it is a proven fact that charge is independent of velocity.

• Special relativity is a classical theory, but the electron in an atom doesn't really have a classical velocity, so I'm not convinced by this argument... – Kyle Oman Apr 14 '16 at 10:30
• What I meant by speed is actually the kinetic energy. Kinetic energy is an observable in quantum mechanics. How else then can one account for the fact that atoms are neutral? – Procyon Apr 14 '16 at 11:56

For any conserved quantity, $$\partial_\mu j^\mu =0$$, you can prove that $$Q = \int d^3 x j^0$$ is a scalar.

• Please show your proof. – Derek Seabrooke Mar 24 '20 at 0:59