# Lorentz-invariant electric charge?

1. Mar 17, 2006

### gulsen

I'm rasining my question in QP on a suggestion made at relativity forum:

Why is charge invariant under Lorentz transformation. Is there a fundamental and theoretical answer to this question? (not the experimental "that's the way nature works") Can QED answer question? Or should I look for an answer in string theories?

2. Mar 18, 2006

### Perturbation

In QED there's a conserved current associated with the current vector of classical EM, the zeroth component of this vector is the charge. This operator is Lorentz invariant and so its eigenvalues are (the charge).

3. Mar 18, 2006

### gulsen

Great, but this's where I've started. Relativistic Maxwell equations also say the same a priori thing.
What I'm asking is not attributes of operator (or what is used for getting charge), it's why is charge (or operator) is Lorentz-invariant.

4. Mar 18, 2006

### masudr

Well, I suppose that answer is that the four-current appears in the Lagrangian the way it does, and also that the 0th component of the four-current is this thing we call charge.

5. Mar 18, 2006

### vanesch

Staff Emeritus
I think that the best answer to my knowledge, is: because spacetime is a 4-dimensional geometrical entity with certain properties ; and as such, all "things" that are defined on it, and are supposed to have a physical existance, must conform to its geometry - which is Lorentz-invariance in the case of flat Minkowski space.
See, it is a bit as if you asked: why is the mass of an object invariant under rotations in space in Newtonian physics ? That is, I take the mass of an object, say, a car, I *rotate my coordinate system*, and I calculate its mass again, and, hey, this comes out the same number. You wouldn't really be surprised, and you wouldn't think that this is a deep and mysterious property of the concept "mass". You'd rather think that this must be evident, because a rotation in space is just *another way of describing the same geometrical object* which is, in this case, space, and the mass density that is defined over it. It's not because you're going to change the coordinates of Euclidean space, that suddenly the mass (integral of the mass density over space) should come out differently, it there is any physical meaning to be attached to mass density in space.

Same with charge, and the geometry of spacetime.

cheers,
Patrick.

6. Mar 19, 2006

### snooper007

Noether's Theorem and conserved current
$$\partial_]\mu j^\mu=0$$+four-dimensional stokes theorem.

7. Mar 19, 2006

### samalkhaiat

Because you can prove it.

Any U(1) Noether charge,

$$Q=\int J^0 d^3x = \int J^\mu d\sigma_{\mu}$$

can be shown to be scalar invariant.

regards

sam

Last edited: Mar 19, 2006
8. Mar 19, 2006

### gulsen

I guess we have a serious communication problem.
I know that all equations are indicating that charge is Lorentz-invariant, and I know that these equations are correct experimentally, so should their assumptions.

vanesch, I see no analogy between "mass variance under rotation" and "charge invariance". Can anyone?
What I know, SR affects all physical phenemona that takes place within space-time, not anything specific such as mass, momentum, energy, etc. And therefore I question: why not charge? Again, I "know" that the answer is "it's invariant", I'm asking "why it's invariant", an answer that does not invole the assumption itself.

samalkhaiat, what good is a proof when it includes the assumption that it's supposed to prove?
What I see on wikipedia, I'm curious that answer is not so simple:

Last edited: Mar 19, 2006
9. Mar 19, 2006

### vanesch

Staff Emeritus
What I tried to outline, is that if you accept that the fabric of space is a certain geometrical structure, that anything that has physical significance must be defined over that structure. My simple example was accepting that space was Euclidean 3-dim space, and that "mass" is a property of a physical function which is mass density ; mass density being a function over Euclidean space. Now, if mass density is a function over Euclidean space, then it can only be function of a *coordinate system* in a special way ; or better, it can only transform under a change of coordinate system in a special way, which means that the integral of the mass density must remain invariant under a rotation (after all, we're integrating the SAME function, over the SAME piece of Euclidean space, but with different coordinates!).

In SR, we accept that the fabric of spacetime is Minkowski space, and if we accept that current is something like a 4-vector over it, then Lorentz transformations are nothing else but changes of coordinates on that same Minkowski space. If we are now going to integrate over a certain lump of Minkowski space, we shouldn't be surprised to find the same result, independent of how we chose our coordinate system over that space.

(and I'm surprised at the Wiki entry...)

10. Mar 20, 2006

### Meir Achuz

This is the right answer. It is proven in advanced EM texts, but the proof is a bit tricky.

11. Mar 20, 2006

### samalkhaiat

12. Mar 20, 2006

### gulsen

First, for the records, I have studied only Classical EMT & QM so far, no QFT or QED.

Let me make an analogy, to clarify what I mean. Suppose, I'm asking a proof for Phythagorean theorem. And one replies:

simple, consider cosine law: $$C^2 = A^2 - 2ABcos(\theta) + B^2$$. Since they are perpendicular, $$\theta=\pi/2$$, therefore $$C^2 = A^2 + B^2$$. Q.E.D.

I'd say: No! That's because, in order to derive law of cosines, you have used Phythagorean theorem, and it's assumptions (Euclidian assumptions), therefore albeit it's mathematically valid, it cannot be a real proof.

Similarly, one can reply my question with laws of classical relativistic electrodynamics. One can show that according to Maxwell laws, the charge is Lorentz-invariant. But that's a priori statement, and it's actually based on a priori assumption made by Coloumb. As we go to the lowest level, we see that classical relativistic electrodynamics assumed that charge is Lorentz-invariant -because that's what is observed in nature-. Therefore, just like in the previous case, laws of classical relativistic electrodynamics cannot be used to prove that charge is Lorentz invariant.

Since I haven't studied QED, I don't know how things are derieved (out of air?). I guess, QED takes a major part of QFT and classical laws of electrodynamics for granted. If it is, since classical electrodynamics cannot be used to prove that charge is Lorentz invariant, same applies to QED.
However, it is if what I guessed was correct. But I don't know if it's true or not, and that's why I'm asking to those who have studied QED.

13. Mar 20, 2006

### masudr

This is what I said earlier; I didn't say Noether's theorem etc. because it looks like we all know about that already. The crucial point is the way the four-current appears in the Lagrangian for the EM field theory.

It's important to ask questions like "why?" but it's partially meaningless in this case, because you could equally ask "Why does energy/momentum transform the way it does?" The answer someone would give is that these quantities are components of a 4-vector etc. etc. but that doesn't really explain why.

The fundamental reason appears to be that nature HAS CHOSEN to represent energy-momentum by this geometric object we call a tensor. Similarly, nature HAS CHOSEN to use the Lagrangian formalism to work out what happens (in the context of classical EM, at least), and has chosen the Lagrangian density to take the form it does. From that we get Noether theorem, and can show (as has been done) that charge is constant.

QED takes classical field theory and quantizes it. In particular, it assumes the Lagrangian takes the form it does, so looking for a proof from QED is probably (note: QED is not my area of expertise so do not take my words as gospel) not going to be very fruitful. You need to explain why the four-current appears the way it does (which is it multiplies the field, i.e. $j^\mu A_\mu$), and this IS put in by hand at this stage. You could say gauge theory provides us with the interaction term, but then you've just shifted the explanation onto WHY nature likes to go about this gauge business in the first place.

If you can derive EM from a more fundamental assumption, then you'll be famous.

Last edited: Mar 20, 2006
14. Mar 21, 2006

### vanesch

Staff Emeritus
Of course a chain in "why?" will always end in a "meaningless" question, but I tried to address this issue partly by saying that, if you accept that physical objects are defined over a 4-manifold (in this case, Minkowski space), then it FOLLOWS automatically that it is represented by a mathematical object which is a representation of the symmetry group of that 4-manifold (in other words, that it has some Lorentz invariance to it)...

So the deeper reason to why our Lagrangian needs to be a Lorentz scalar, defined by tensor operations of objects which are representations of the Lorentz group, is that these objects must be defined over Minkowski space.

15. Mar 21, 2006

### Meir Achuz

16. Mar 21, 2006

### Hans de Vries

A proof

Yes, both charge and its Lorentz invariance can in fact already be
derived from the classical wave equation:

$$\mbox{Classical Wave equation:} \qquad \frac{\partial^2 \psi}{\partial t^2}\ =\ v^2 \frac{\partial^2 \psi}{\partial x^2} \qquad \qquad (1)$$

The derivatives in time and space are proportional by a constant
which stems from the characteristic speed of the medium. This simply
means that the equation is satisfied by any arbitrary function which
shifts along with a speed v (or -v). We can expand the equation to
three dimensions, for instance for the electric potential field V:

$$\mbox{Electric Potential:} \qquad \frac{\partial^2 V}{\partial t^2}\ =\ c^2 \frac{\partial^2 V}{\partial x^2} + c^2 \frac{\partial^2 V}{\partial y^2} + c^2 \frac{\partial^2 V}{\partial z^2}\qquad \qquad (2)$$

Where c is the speed of light. The same expression holds for the three
components of the magnetic vector potential. Again these equations
are satisfied by any arbitrary function which shifts along with the
characteristic speed c: The electro magnetic waves.

In our world however we also see things which are stationary or move at
other speeds than the speed of light. If we go to three (or more) space
dimensions then such solutions become possible. A stable solution which
shifts along with an arbitrary speed v in the x direction will satisfy both
(1) with a speed of v and (2). We can use this to eliminate the time
dependency by substitution:

$$\left(1-\frac{v^2}{c^2}\right)c^2 \frac{\partial^2 V}{\partial x^2}\ +\ c^2 \frac{\partial^2 V}{\partial y^2}\ +\ c^2 \frac{\partial^2 V}{\partial z^2}\ =\ 0$$

This shows that the solutions are Lorentz contracted in the direction
of v by a factor $\gamma$, The first order derivatives are higher by a factor
$\gamma$ and the second order by a factor $\gamma^2$. Velocities higher then c are
not possible. The solution for v=0 is:

$$\frac{\partial^2 V}{\partial x^2}\ +\ \frac{\partial^2 V}{\partial y^2}\ +\ \frac{\partial^2 V}{\partial z^2}\ =\ 0, \qquad \Rightarrow \qquad V\ =\ \frac{1}{r}$$

Which is the electro static potential. The equation is satisfied at
all points except for r=0 where we have a singularity. This
singularity is now associated with the classical (point)charge.
Without it there would be no solutions at sub-luminal speeds.

The charge is defined by what we measure, the fields. Since charge
is conserved (does not change in time) and the fields are real
scalars it is sufficient to use the Lorentz contraction as a prove
for Lorentz invariance.

The total solution is an arbitrary superposition of 1/r functions.
This includes the Quantum Mechanical fields where charge is spread
out over the wavefunction.

Regards, Hans.

Last edited: Mar 21, 2006
17. Mar 21, 2006

### Meir Achuz

"Why is charge invariant under Lorentz transformation. Is there a fundamental and theoretical answer to this question?"
Look in a textbook.

18. Mar 21, 2006

### Hans de Vries

?

It does show that the potential fields transform as they should do according
to Special Relativity, and why they do so.

If you want to refer the readers to the good old work of Lorentz-PointcarÃ©
on the Maxwell equations or the work of Lienard-Wiechert on the potential
fields of an arbitrary moving charge, then you might be a tad more specific

Regards, Hans.

19. Mar 21, 2006

### Meir Achuz

Try: Jackson "Classical Electrodynamics", Sec.11.9 (in the 2nd Ed.)
or Franklin "Classical Electromagnetism", Sec. 14.10.2,
or Panofsky & Phillips "Classical E & M", Sec. 17-2.
I'm off for a week, so you're on your own.

20. Mar 21, 2006

### masudr

Hans, you didn't really show why the electromagnetic potentials satisfy the wave equation. To do so, you have to start from Maxwell's equations, which are essentially derived from the Lagrangian.

21. Mar 21, 2006

### Hans de Vries

That's true. I presume this as given. Just like you presume the Maxwell
equations as given when you start from there. I prefer to start with the
potentials since they are the more fundamental ones, shown by Lienard/
Wiechert back in 1900 and confirmed in Quantum Mechanics via the
Aharonov Bohm effect.

Regards, Hans.

22. Mar 23, 2006

### dextercioby

Well, i got an idea: let's say we have a bunch of massive electrically charged particles that have either spin 0 (and thus described by a complex scalar field), spin 1/2 (thus described by Dirac fields) and maybe 3/2 spin (described by a massive Rarita-Schwinger field). We know by experience that they have certain value (in terms of "e") of their electric charge. Since electric charge is intimately linked with the electromagnetic interaction, we have to conclude the following

a. Any possible electromagnetic interaction between such particles is mediated by a U(1) gauge theory massless spin 1 particle: the photon.

b. If the matter theory also has a global (rigid) U(1) invariance, then two things follow

1. By Noether's theorem we get a conserved current which is a genuine vector both under $$\tilde{\mathcal{L}}_{+}^{\uparrow}$$ and also $$\tilde{\mathcal{L}}$$. Its zero'th component is a genuine scalar under both symmetry groups.

2. The coupling is typically $A_{\mu}j^{\mu}$ (except for the seagull term in SED) which delivers the equations of motion for the gauge field

$$\partial^{\mu}F_{\mu\nu} = j_{\nu}$$

and, by analogy with classical electrodynamics, we have to perceive $j$ as the electric charge 4-vector, that is a source for the em-field...

c. If the matter theory doesn't have a global (rigid) U(1) invariance, then the existence of possible couplings to a massles spin U(1) gauge field has to be studied more carefully, perhaps using the elegant BRST formalism of generating interactions...

Daniel.

23. Mar 23, 2006

### samalkhaiat

[

All my statements were in the contex of classical field theory.
Sir, the answer to your question does not need even physics. All that needed is tensor calculus. That is, if you "agree" that the charge is given by the integral:
$$\int d^3xJ^0(x)$$
and that,$\partial_{\mu}J^{\mu}=0$.
However, if you, for some personal reason, donot agree with the above integral representation, then think about the charge as some dimensionless number(coupling constant). Such number does not change under any coordinate transformation, does it?

sam

24. Mar 23, 2006

### samalkhaiat

Last edited: Mar 23, 2006
25. Mar 23, 2006

### samalkhaiat

Last edited: Mar 23, 2006