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 existence, 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
$$\partial_]\mu j^\mu=0$$+four-dimensional stokes theorem.

 

[samalkhaiat]

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?

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

 

[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:

The origin of charge invariance (indeed, all relativistic invariants) is under speculation presently. There may be some hints proposed by string/M-theory.

 

[vanesch]

vanesch, I see no analogy between "mass variance under rotation" and "charge invariance".

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]

Because you can prove it.

Any U(1) Noether charge,

$$Q=\int J^0 d^3x$$

can be shown to be scalar invariant.

regards

sam

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

 

[samalkhaiat]

samalkhaiat, what good is a proof when it includes the assumption that it's supposed to prove?

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

What I see on wikipedia, I'm curious that answer is not so simple

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: $$C^2 = A^2 - 2ABcos(\theta) + B^2$$. 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.

 

[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 density.

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.

 

[vanesch]

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?"

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]

current as Lorentz-vector ==> charge as Lorentz-scalar.
This should answer your question. If not, then there is a problem
sam

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]

A proof

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

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.

 

[Meir Achuz]

Yes, both charge and its Lorentz invariance can in fact already be
derived from the classical wave equation:
[tex]
Regards, Hans.

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.

 

[Hans de Vries]

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.

?

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-Pointcaré
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
as "read a textbook".


Regards, Hans.

 

[Meir Achuz]

?

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-Pointcaré
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
as "read a textbook".
Regards, Hans.

The que3stion was about charge.
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.

 

[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.

 

[Hans de Vries]

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.

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.

 

[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.

 

[samalkhaiat]

[

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

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, don't 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

 

[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.

Here is my proof which is simpler and shorter than any textbook's proof:
Under infinitesimal Lorentz transformation;

$$\bar{x}^{\mu}=x^{\mu}+ \delta{x}^{\mu}$$

the conserved vector $J^{\mu}$ changes according to:

$$\delta J^{\mu}= \bar{J}^{\mu}(x)-J^{\mu}(x) = \partial_{\nu}(J^{\nu} \delta x^{\mu} - J^{\mu} \delta x^{\nu})$$

Put $\mu =0$, integrate over 3-volume, use Gauss's theorem and get,

$$\int d^3x\delta J^0 = \int d^3x\partial_{i}( J^i \delta x^0 - J^0 \delta x^i ) = 0$$

Thus

$$\delta Q = 0$$

Quod Erat Demonstrandum.

Since you talk about derivations, here is an exercise for you to do in a rainy afternoon. It is another special case of the general theorem stated in my post #11.
Let the conserved Noether current be a second rank tensor $T^{\mu\nu}$ (the energy-momentum tensor). Show that the time-independent noether charge$\int d^3x T^{0\nu}$ is a Lorentz vector(the energy-momentum 4-vector).

regards

sam

 

[samalkhaiat]

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.

You mean 3-volume integral of its 0'th component. But the question was why does charge transform as scalar under Lorentz group? The man did not ask us to introduce electrodynamics, did he?
I have proved in post#24 that the charge is a Lorentz-scalar.

regards

sam

 

[reilly]

There's no proof at all. The plain fact is that the assumption that the electric charge is a relativistic scalar, along with the whole apparatus of E&M works. It's a great fit to the data, just like the idea that the electric field is a three-vector in non-rel E&M, or part of an antisymmetric tensor in 4-space. We venerate Maxwell because his theory delivers the goods. Always there are assumptions. (If we had a theory that explained charge, we might be in a different situation.)

If e were not a scalar, what else could it be? And, could a non-scalar e theory account for all E&M?

Regards,
Reilly Atkinson

 

[samalkhaiat]

There's no proof at all.

Sweet and simple proof can be found in post #24.

Always there are assumptions.

When one goes about the business of proving Lorentz covariance of Maxwell equations, one needs to regard the charge density as a time-component of 4-vector. However, this "assumption" does not mean that the volume integral of $J^0$ (the charge) is a scalar. Why should it be? The charge is time-independent and scalar only because of

$\partial_{\mu}J^{\mu}=0$.

When you write;

$$q= \int d^3x \bar{\psi} {\gamma}^0 \psi$$

Do you just assume that this integral is scalar, guess its behaviour under Lorentz group, or prove it to be Lorentz invariant?
An assumption that can be proven need not be made.

(If we had a theory that explained charge, we might be in a different situation.)

This does not mean that we giveup investigating the mathematical properties of charge.

If e were not a scalar, what else could it be? And, could a non-scalar e theory account for all E&M?

Mathematics should and can answer such questions.

In the Lagrangian field theory, knowing the tensorial character of all Noether (spacetime & internal) charges is paramount.



regards

sam

 

[vanesch]

In the Lagrangian field theory, knowing the tensorial character of all Noether (spacetime & internal) charges is paramount.

My impression was that the OP wanted to know WHY these fields (like the Noether current) had these tensorial characters and transformed the way they did under a change of reference frame. I think he understood that once one ACCEPTS these quantities to be tensorial quantities under Lorentz transformations, that the conservation of charge follows.
I don't think there is any mathematical proof for that: it is a fundamental assumption, isn't it ?

 

[Meir Achuz]

Here is my proof which is simpler and shorter than any textbook's proof:
Under infinitesimal Lorentz transformation;

$$\bar{x}^{\mu}=x^{\mu}+ \delta{x}^{\mu}$$

the conserved vector $J^{\mu}$ changes according to:

$$\delta J^{\mu}= \bar{J}^{\mu}(x)-J^{\mu}(x) = \partial_{\nu}(J^{\nu} \delta x^{\mu} - J^{\mu} \delta x^{\nu})$$

Put $\mu =0$, integrate over 3-volume, use Gauss's theorem and get,

$$\int d^3x\delta J^0 = \int d^3x\partial_{i}( J^i \delta x^0 - J^0 \delta x^i ) = 0$$

Thus

$$\delta Q = 0$$

Quod Erat Demonstrandum.

Since you talk about derivations, here is an exercise for you to do in a rainy afternoon. It is another special case of the general theorem stated in my post #11.
Let the conserved Noether current be a second rank tensor $T^{\mu\nu}$ (the energy-momentum tensor). Show that the time-independent noether charge$\int d^3x T^{0\nu}$ is a Lorentz vector(the energy-momentum 4-vector).

regards

sam

Sam,
I think we (of all the posters) are almost in agreement.
Your proof is essentially the textbook derivation. It is only shorter because it leaves out a crucial step and diagram. Because
$$\partial_\mu j_\mu=0$$, a surface integral over a 4-D hypersurface will vanish. But to prove Lorentz invariance of charge, you have to pick the closed Gaussian hypersurface appropriately (to "use Gaus's theorem"). The derivation will prove either charge conservation or LI of charge (They are different concepts.) depending on how you choose the hypersurface. This is explained most clearly in the Franklin text.

The corresponding proof for any rank tensor is just the same.
As with charge, you do not have to assume time independence. It can be proven by choosing the appropriate hypersurface.

 

[reilly]

It is well known that there is a conserved "charge" connected with a conserved current -- GellMan's Current Algebras and SU3, etc. You have simply demonstrated the connection between global and local symmetries, that is you have proved something that's been known for over a century.

Actually, I was thinking more about the electron's charge as a a scale seting constant parameter. If our E&M theory is to work, then this parameter must be a scalar. And then, there's that simple matter that the validity of Maxwell's equations is a matter of empirical observations,thus the proof that Maxwell is classically right is not a matter of logic. The various invariance properties are after the fact. Hence, again, my claim that there is not mathematical proof. But there is extraordinary empirical support for the Lorentz transformation properties of E&M fields and currents.

About "what else could it be?" See Weinberg's Vol I of Quantum Theory of Fields for an account of Noether's work, and he gives your proof on page 307 and on. Also, in 10.6, Weinberg gives an excellent discussion of "what else could it be?" in connection with nucleon form factors and the Rosenbluth nucleon-electron scattering crossection -- the idea is, classically or "quantumly" that interaction terms must have well defined transformation properties under translations, rotations and under lorentz boosts, etc. For example there are only so many four vectors associated with a one particle state --for Dirac this means gamma mu, p mu, and sigma mu nu p nu -- that can couple to the electromagnetic four vector potential. Weinberg gives the standard arguments to get the most general electromagnetic current for a nucleon -- it's all about symmetry.

As he often does, vanesch got it right(Post 28).

Regards,
Reilly Atkinson

 

[samalkhaiat]

My impression was that the OP wanted to know WHY these fields (like the Noether current) had these tensorial characters and transformed the way they did under a change of reference frame. I think he understood that once one ACCEPTS these quantities to be tensorial quantities under Lorentz transformations, that the conservation of charge follows.
I don't think there is any mathematical proof for that: it is a fundamental assumption, isn't it ?

I am afraid, this post (unlike your previous ones) contain inaccurate and confused statements:

1) The representation theory of Lorentz group determines the tensorial characters (the transformation law) of the fields. It puts them in scalar, vector, spinor, 2nd rank tensor, ...(no assumptions)
2) We use the above fields and their transformation laws to help us construct Lorentz-scalar actions (Lagrangians).
3) If the, now constructed, Lagrangians possesses further symmetry, say internal U(1) symmetry, then Noether theorem leads to a current satisfying $\partial_{\mu}J^{\mu}=0$.
4) The transformation law (tensorial nature) of this current is determined entirely by the tensorial characters of the fields in (1).
For example, in QED, you can show that the U(1) noether current
$$J^\mu = \int d^3x \bar{\psi} {\gamma}^{\mu} \psi$$
is a 4-vector.This is because $(\psi , \bar{\psi})$ are the bispinor representation of Lorentz group. So the tesorial character of the current is not an assumption but derived from the transformation laws of Dirac's fields.
5) As for "conservation of charge" Well, it follows from, and only from,$$\partial_{\mu}J^{\mu}=0$$
It does not depend on the tensorial nature of the fields or the current;
$$\frac{dQ^a}{dt}= \int d^3x \partial_{0} J^{0a} = - \int d^3x \partial_{j} J^{ja} = 0$$
Notice that (a) can be any index, multi-index or no index. So the constancy of charge in time (charge conservation) has nothing to do with how the fields (thus the current) transform under Lorentz group.
6) The issue of this thead is the tensorial (scalar) nature of the charge NOT its conservation. This tensorial character depends only on the (derived) tensorial character of the current according to the theorem:
Noether charge is Lorentz tensor of rank (n-1), where n is the rank of Noether current.
In all of the above there is no (fundamental or otherwise) assumptions.

regareds

sam

 

[samalkhaiat]

Sam,
I think we (of all the posters) are almost in agreement.

Do tou think so? Well, at least one person think that a proof does not exist.

Your proof is essentially the textbook derivation. It is only shorter because it leaves out a crucial step and diagram.

No, My version of the proof is entirely different because:
1) No diagram is needed.
2) No use of the 4D Gauss' theorem is made.
3) The only "trivial" (not essential) step, that is missing, is arriving at the infinitesimal transformation of the current from the finite one:
$$\bar{J}^{\mu}(\bar{x}) = \partial_{\nu} \bar{x^{\mu}}J^{\nu}(x)$$
(I assumed, you know how to Tylor expand the above to the 1st order)

you do not have to assume time independence.

"time-independent charge" was not an assumption. Simply dQ/dt=0 is irrelevant for the exercise (if you follow my version). Also, I assumed that you can show:
$$\partial_{\mu}J^{\mu}=0 \Rightarrow dQ/dt=0$$
there are two methods for showing this:
1) Easy, as given in post#31
2)Not so easy, it uses the 4D gauss' theorem and goes like
$$0 = \int_{\Omega} d^4x \partial_{\mu}J^{\mu}= \int_{\partial{\Omega}} J^{\mu}d {\sigma}_{\mu}$$
Now, the current is supposed to decrease outside the world-tube sufficiently rapidly, so that integrals over regoins outside the world-tube will vanish. Also, one should consider those coordinate systems in which spaces of constant time, $x^0=const.$, intersect the world-tube in simply-connected regions.Now, choose as region of integration two non-intersecting hyperplanes $x^0=t$ and $x^0= \bar{t}$ connected by suffaces outside the world-tube:
$$\partial{\Omega}=\partial{\Omega}(t) U \partial{\Omega}(\bar{t}) U \partial{\Omega}(\infty)$$
this leads to
$$\int_t d^3x J^0 = \int_{\bar{t}}d^3xJ^0$$
which means that the charge has the same value for both hyperplanes. That is, it is independent of time $\dot{Q}=0$.
Now, we are in a position to prove that Q is Lorentz invarian.By the way, this is going to be the textbooks version of the proof that you mentioned few times.It was first proved by F. Klein(1917), modified by H.Weyl(1920).
Consider a second Lorentz system K' inside the world-tube. K' has only to satisfy the condition that surfaces $\bar{x^0} = const.$ intersect the world-tube in simply-connected regions.
Now take as region of integration the non-intersecting hyperplanes
$$x^0 = const., and \bar{x^0} = const.$$
In this case, the above 4D Gauss' theorem gives you
$$\int_{K} d^3x J^0 = \int_{\bar{K}} d^3x \bar{J^0}$$
i.e
$$\delta{Q} = 0$$
As you clearly see, this "Klein's proof" is entirely different from my version.
The question now, which version is simpler.

regards

sam

 

[samalkhaiat]

You have simply demonstrated the connection between global and local symmetries,

Sir, I have domonstrated no such thing.I have proved that the charge is Lorentz invariant. Such proof does not depend on the global/local nature of the symmetry group. Both local and global U(1) symmetry lead to the same charge:

$$Q = \int J^0 d^3x$$

that is you have proved something that's been known for over a century.

To be accurate, The Lorentz invariance of Q was first proved in 1917.

The above integral (the charge) has four properties:

(1) It is conserved(constant in time):

$$\frac{dQ}{dt}=[P^0 , Q]=0$$

(2) It is Lorentz invariant (it has the same value in all Lorentz frames i.e scalar):

$$\delta Q = [M^{\mu\nu} , Q] = 0$$

(3) It is the generator of the (internal) symmetry transformation:

$$\delta\phi(x) = [Q , \phi(x)]$$

(4) It is invariant under 3D translations:

$$\nabla Q = [\vec{P} , Q]=0$$

All textbooks on field theory prove (1),(3) and (4).

{ (1)+(2)+(4) = invariance under Poincare' group}

Unfortunately, only few textbooks prove (2). I say unfortunately because (2) is the issue of this thread. The proof of (2) given in some books is due to H. Weyl(1920). However, similar proof was first given by F. Klein in 1917. In 1921, W. Pauli showed that Q is invariant under both Lorentz and general coordinate transformations.
In 1995, I made my version of the proof(post#24).
Klein, Weyl and Pauli all said that one can prove Q to be Lorentz invariant. Yet, you say "there's no proof at all". Forget about me, who is saying the right thing you or those great physicists?

Actually, I was thinking more about the electron's charge as a a scale seting constant parameter.

As I said somewhere on this thread, if one has problem with the integral represenation of Q, one should think of the charge as (1/137). In this case the question of invariance is meaningless.

See Weinberg's Vol I of Quantum Theory of Fields for an account of Noether's work, and he gives your proof on page 307

You shocked me, where is my version or any version of proving (2)? I can not see it. On page 307, Weinberg proves then states Noether's theorem:
At the end of the page he writes "symmetries imply conservation laws"

Noether's theorem is a standard subject in all field theory books.

Look, Proving Noether theorem is one thing (Wienberg did this), and proving the scalar nature of Noether charge is another thing (I did this).
So please sir donn't say that my proof is on pade 307 or any page of THAT Wienberg's book.
In the same book, footnote on page 253, he states "without proof" the general theorem that I mentioned at the end of post#31.

Now check this one: In his book "Gravitation and Cosmology" Weinberg used Weyl method (not mine) to prove that Q is Lorentz invariant.

"what else could it be?"

I did not ask this question, you did and I answered you by saying that mathematical physics should and can answer such questions.

in connection with nucleon form factors and the Rosenbluth nucleon-electron scattering crossection -- the idea is, classically or "quantumly" that interaction terms must have well defined transformation properties under translations, rotations and under lorentz boosts, etc. it's all about symmetry.

This is exactly what I meant by getting answers from mathematical physics. and it is exactly why the tensorial nature of Noether charges is of fundamental importance in the Lagrangian field theories.



regards

sam

 

[dextercioby]

You mean 3-volume integral of its 0'th component.

Yes, of course.

But the question was why does charge transform as scalar under Lorentz group?

Just as long as you describe what "electric charge" means in classical field theory (built with fields transforming under proper Lorentz transformations of their arguments after finite dimensional reps. of the universal covering group of the proper Lorentz group), the answer is trivial.

Electric charge is this animal

$$Q=\int d^{3}x j^{0}\left( \{Q^{a}(x) \} ,\{\partial_{\mu}Q^{a}(x) \} \right)$$

Since j^{0}\left( \{Q^{a}(x) \} ,\{\partial_{\mu}Q^{a}(x) \} \right) is the "0"-th component of a genuine 4-vector for any classical field theory, then it's a Lorentz scalar. This is a group theory proof.


Daniel.

 

[vanesch]

I am afraid, this post (unlike your previous ones) contain inaccurate and confused statements:

This could very well be the case, in which case I'm learning and stand corrected. But I'm affraid I don't see where...

1) The representation theory of Lorentz group determines the tensorial characters (the transformation law) of the fields. It puts them in scalar, vector, spinor, 2nd rank tensor, ...(no assumptions)

You don't call this an assumption ?? It is the assumption that the fields are entities defined over a 4-dim geometrical entity that makes that they must be tensorial in character. As the question was, why there was a lorentz invariance of charge, and you use the lorentz invariance of other quantities to prove it, that's begging the question, no ?

The proof you outlined was, how to go from these these tensorial quantities, using the definition of charge (inspired by the Noether theorem - as you point out correctly - for ANOTHER use of the charge, namely the quantity that remains conserved under time evolution) to the conservation of charge under lorentz transformation.

2) We use the above fields and their transformation laws to help us construct Lorentz-scalar actions (Lagrangians).

And the structure of this Lagrangian, on top of the use of fields which are tensorial in nature, is also an assumption, of course, no ?

3) If the, now constructed, Lagrangians possesses further symmetry, say internal U(1) symmetry, then Noether theorem leads to a current satisfying $\partial_{\mu}J^{\mu}=0$.

Right, this follows from the two assumptions: the tensorial character of the fields, and the U(1) gauge symmetry.

4) The transformation law (tensorial nature) of this current is determined entirely by the tensorial characters of the fields in (1).
For example, in QED, you can show that the U(1) noether current
$$J^\mu = \int d^3x \bar{\psi} {\gamma}^{\mu} \psi$$
is a 4-vector.This is because $(\psi , \bar{\psi})$ are the bispinor representation of Lorentz group. So the tesorial character of the current is not an assumption but derived from the transformation laws of Dirac's fields.

What you write is correct of course, but begs the question! If I ask, why is j^mu a 4-vector, do I want to see the derivation from its definition, and of the transformation laws of the Dirac field - or do I want to know why, in the first place, the Dirac field has to be a representation of the Lorentz group ?

5) As for "conservation of charge" Well, it follows from, and only from,$$\partial_{\mu}J^{\mu}=0$$
It does not depend on the tensorial nature of the fields or the current;
$$\frac{dQ^a}{dt}= \int d^3x \partial_{0} J^{0a} = - \int d^3x \partial_{j} J^{ja} = 0$$

Ok, what I wanted to say by "conservation of charge" was not "during time evolution" but "under Lorentz transformation". You are correct, the right word is not conservation, but invariance.

In all of the above there is no (fundamental or otherwise) assumptions.

I reiterate: isn't the requirement for fields to be representations of the Lorentz group, a fundamental assumption ?