Lorentz-invariant electric charge?

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?

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).

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.

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.

vanesch

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.

snooper007

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

samalkhaiat

Because you can prove it.

Any U(1) Noether charge,

[tex]Q=\int J^0 d^3x = \int J^\mu d\sigma_{\mu}[/tex]

can be shown to be scalar invariant.

regards

sam

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:

vanesch

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.

That's what I meant. I don't know if it answers, or doesn't answer, your question.

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

Meir Achuz

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

samalkhaiat

[QUOTE]Look, your question is a simple field theory exercise. In general, if the conserved Noether current is a Lorentz tensor of rank n, then the assiciated charge is time-independent Lorentz tensor of rank (n-1).
Now, the tensorial character of the current is determined by, and only by, the transformatiom properties of the dynamical variables (the fields) with respect to Lorentz group. In other words, if you know the transformation law for the fields, then you can workout the transformation law for the current. This, in turn, tells you what kind of quantity the charge is.
In your case;
Lorentz transformation of fields ==> current as Lorentz-vector ==> charge as Lorentz-scalar.
This shoud answer your question. If not, then there is a problem

Trust me, the wikipedia entry is a total garbage.

regards

sam

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: [tex]C^2 = A^2 - 2ABcos(\theta) + B^2[/tex]. Since they are perpendicular, [tex]\theta=\pi/2[/tex], therefore [tex]C^2 = A^2 + B^2[/tex]. 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.

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. [itex]j^\mu A_\mu[/itex]), 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.

vanesch

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.

Meir Achuz

[QUOTE=samalkhaiat]This step requires that the four div=0, and a bit of derivation.
Why don't you break the rules and read a print textbook?
No QED is needed for this.

Hans de Vries

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

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

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:

[tex]
\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)
[/tex]

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:

[tex]
\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
[/tex]


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

[tex]
\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}
[/tex]

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.

Meir Achuz

Your derivation does not prove
"Why is charge invariant under Lorentz transformation. Is there a fundamental and theoretical answer to this question?"
Look in a textbook.