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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 do we have electric charge as an invariant quantity?

Qmechanic
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aargiee
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5 Answers5

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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, $Q=Q'$.

AccidentalFourierTransform
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  • would the down-voter care to comment? – AccidentalFourierTransform Apr 11 '16 at 19:02
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    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
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    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
  • @knzhou well, we have to postulate something. We can postulate that 1) $j^\mu$ is a four vector, 2) the Lagrangian of a point particle is $L=Q u_\mu A^\mu$, 3) that $\psi\to\mathrm e^{i\theta Q}\psi$ is a symmetry of the QED Lagrangian, 4) etc. But we need some input. I believe that option 1) is the most natural one (at the level of OP's knowledge), but of course other people might disagree. You like 3). I wonder what would have OP thought if I had said "let $L=\bar\psi(\not\partial-m)\psi+\cdots$" – AccidentalFourierTransform Apr 12 '16 at 19:54
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    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
  • @AccidentalFourierTransform I'm curious why didn't you use the relation $\partial_{\mu}J^{\mu}=0$ since here the discussion was on Lorentz transformation we could have the idea of Lorentz contraction and dilation but applying the above train of thoughts seems hard in GR. – aitfel Jun 29 '20 at 19:07
  • The integrals defining $Q$ and $Q'$ take place in different spatial hypersurfaces on spacetime (i.e., different observers have different notions of constant time). However, this answer seems to assume that they are taken on the same spatial hypersurface. – Níckolas Alves Jun 28 '22 at 04:04
  • @NíckolasAlves It makes no difference since the current is conserved. $Q(\Sigma)=\int_\Sigma j$ is invariant under deformations of $\Sigma$ thanks to $\mathrm dj=0$. – AccidentalFourierTransform Jul 07 '22 at 16:48
  • But then, is the space integral of energy-density also a Lorentz scalar? This same argument would work with the energy density four-vector too. Doesn't that make the Hamiltonian from quantum field theory a Lorentz scalar? – Ryder Rude Jul 09 '22 at 10:22
  • @RyderRude No, the Hamiltonian is the integral of a rank-2 tensor $T^{\mu\nu}$ (the energy-momentum tensor), but the electric charge is the integral of a rank-1 tensor $j^\mu$ (the charge current). As such, the Hamiltonian is the zeroth component of a vector, and the electric charge is a scalar. (The integration reduced the rank by 1, so $T$ becomes a vector and $j$ a scalar). – AccidentalFourierTransform Jul 22 '22 at 17:15
  • @AccidentalFourierTransform But, for energy alone, we are only interested in the four-vector $T^{\mu 0}$. The other $T^{\mu i}$ relate to momentum currents. So, the space integral of the 0th component of the four-vector $T^{\mu 0}$ should be the total energy, right? That makes it identical to total charge which is also l the space integral of the 0th component of a four vector – Ryder Rude Jul 22 '22 at 18:03
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    @RyderRude The components $T^{\mu0}$ are not a four-vector. Mathematics do not care about what you are interested in. You may only care about $T^{\mu0}$, but that doesn't change the fact that $T^{\mu\nu}$ is a rank-2 tensor. A subset of components of a certain tensor does not transform like a lower-rank tensor. You may choose to forget that those components come from a larger object, but that doesn't change the transformation laws of the same. Under a Lorentz transformation, a vector transforms as $j\to \Lambda j$, and $T^{\mu0}$ transforms as $T\to \Lambda T\Lambda$. – AccidentalFourierTransform Jul 24 '22 at 15:17
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    @AccidentalFourierTransform Sorry I thought Noether's theorem always returned four-vectors. Much thanks. – Ryder Rude Jul 26 '22 at 14:06
  • The fact the current J is a four-vector is not by hypothesis but rather from imposing Lorentz covariance on the continuity equation. For $\partial_\mu J^\mu=0$ to be Lorentz covariant $J^\mu$ must be a four-vector. In Quantum Field Theory one can show that the commutator of the charge operator Q commutes with the generators of the Proper Homogenous Lorentz transformations. The proof, of course, also uses the fact that $J^\mu$ is a four-vector, that the vector component is a polar vector, and that the vector J dies off at infinity quickly enough. – Roberto Vega Mar 05 '23 at 22:24
  • In classical E&M it is obvious that $\rho(\vec{x},t),d^3x$ is Lorentz invariant as, $\rho(\vec{x},t)=\Sigma_n q_n\delta^3(\vec{x}_n-\vec{x}(t))$ for an n-particle system. This assumes that $q_n$, the charge of each individual particle, is independent of position and time. If this were not true then translational invariance would be broken. – Roberto Vega Mar 05 '23 at 22:34
  • That $J^\mu$ is a four-vector also follows from the fact that it is the Noether current associated with the gauge invariance associated with a four-vector field $A^\mu$ – Roberto Vega Mar 05 '23 at 22:41
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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.

Procyon
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  • 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
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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 )

UKH
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Total electric charge is a quantity that labels representations of the gauge group $\mathrm{U}(1)$ of electromagnetism. If Lorentz transformations acted in any way on this electric charge, then this would imply that Lorentz transformations and the global gauge transformations of electromagnetism did not commute.

But, very generally, the Coleman-Mandula theorem states that relativistic physical theories cannot have symmetry groups of any form other than the direct product of the Poincaré group with the internal symmetry groups (like the gauge group of electromagnetism). In particular, the different factors of products of groups commute, so charge transforming under Lorentz transformations in a non-trivial way would imply a violation of this theorem.

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

Proof:

$Q$ is position independent as it is a volume integral. $Q$ is also independent of time, as $$\partial_t Q = \int d^3 x ~ \partial_t j^0 = -\int d^3 x ~ {\vec \nabla} \cdot {\vec j}\,.$$ This is zero because $$\int dx dy dz ~\partial_x j^x\,,$$ $$\int dx dy dz ~\partial_y j^y$$ and $$\int dx dy dz ~\partial_z j^z$$ are zero.

my2cts
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