Lorentz-invariant electric charge?

[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 ?

 

[dextercioby]

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.

sam[/QUOTE]

Here's a question for you:

If the current is conserved, (delta on it zero, where delta is Cartan's codifferential),then the charge is a number, either a real number, or a complex number. It doesn't depend on the point of $M_{4}$ in which one choses to evaluate it. So there's no question on how it should behave under the Lorentz group, since it is not affected by Lorentz transformations, since it is not an $M_{4}$ valued function. In the case of U(1) invariance (either global or gauge), we're talking about a natural number, which, for example in the case of The Dirac's field is the multiple of the absolute value of electron's electric charge.

We're talking about a number. Is this number a frame dependent...? If i rotate the system, or Lorentz boost it, do we get one more electron...?(charge diminishes by 1). I hope not.

The tricky part is when it comes to tensor-type charges, like angular momentum, whose charge is M^{\mu\nu}, a second rank tensor. It is conserved, iff the Noether charge is conserved. It doesn't depend on "x". So why would it be affected by Lorentz transformations...? Are we allowed write

$$M'^{\mu\nu} =\Lambda^{\mu}{}_{\rho} \Lambda^{\nu}{}_{\sigma} M^{\rho\sigma}$$

when $$x^{\mu}\rightarrow x'^{\mu}=\Lambda^{\mu}{}_{\nu} x^{\nu}$$

,when we're talking about real numbers (or even natural ones) which should be in no way affected by Lorentz transforming the inertial frame of reference...?

My point is that charge conservation, which stems from current conservation, makes the discussion over the tensorial/spinorial type of Noether charges simply ridiculous.

Daniel.

 

[dextercioby]

Apparently something is rather unclear to me:

I'm reading page #76 of Paul Roman's "Introduction to Quantum Field Theory" in which he states that, even though $P^{\mu}$ and $M^{\mu\nu}$ are conserved quantities, that is "x" independent, they still change under infinitesimal Lorentz transformations. See formulae on the bottom of page #76 and on top of page #77.

This is weird. That number seems to be frame dependent.

Can someone make it any clrearer to me...?

Daniel.

 

[dextercioby]

OMG, i just realized, I'm such an idiot. Of course those quantities are frame dependent. The momentum, even though it is a number, depends on whether the frame of reference is moving at 5m/s or at 10m/s. Same for orbital angular momentum...

The matter is clear to me now.

Daniel.

 

[reilly]

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

On page 307 of Weinberg Vol I, your proof is given almost exactly. (The only exception I can see is that he uses five equations; you use four. You could argue that W uses only four, as his first eq. effectively defines the the delta, so-to-speak, in terms of a general infitesimal LT.

Your energy-momentum tensor -> Lorentz vector is a very standard homework problem in beginning field theory; I used it several times when I was teaching.

My ulimate point is that it's Nature that determines the various invariance properties: like -- if Maxwell and Lorentz are right, then indeed charge is conserved, and is a global scalar. And Maxwell and Lorentz are right as far as classical E&M goes, are right. Similarly, Nature indicates that Lorentz invariant quantum field theory is the best game in particle physics. Given that, again Noether gives us conserved E&M current and a scalar charge. Nature also gives us a so-called partially conserved axial vector current -- the Goldberger-Trieman relation says that the divergence of the axial current is proportional to the pion field, and, lo and behold, G&T got it right, experiment confirms -- involves the pion-nucleon coupling constant also.

If there is no experimental evidence to support a theory, then it is moot until experiments can be done. We believe in both types of relativity, not because of their superior elegance, but because experiments show them to be empirically valid. Weyl, in Time, Space, Matter discusses a theory of Mie, pre quantum, which tries to extend Lorentz-Maxwell to incude notions of the "interior of an electron", and of a relativistic action, and supposedly suggests that only + and - charges exist. Mie's work is very elegant, very clever, and, in many repsects, quite profound. But. it did not do well with Nature, so we don't hear much about it anymore.

Regards,
Reilly Akinson

 

[samalkhaiat]

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 ?

So the problem now is not the vector nature of the current, rather the problem is the tensorial nature of the fields which make up the current. OK, to resolve this problem, I need to start from the starting point of theotetical physics:
We believe that there exists a correspondence between the physical world and the world of abstract entities (mathematics). That is a morphism from mathematical structures on experimental data. Without such morphism science would not exist.

In the physical world, we observe objects with spin = 0,1 & 1/2 (experimental data). Then we use the representation theory on Minkowski spacetime to identify:
1) the spin-0 object with the Lorentz scalar field(other fields fail to descibe spin zero particle)
2) spin-1 with the Lorentz vector field
3) spin-1/2 with the spinor field.
So we identify Dirac's particle with the spinor field because any other Lorentz field fails to account for the spin-1/2 property of Dirac' particle.
The only "assumption" in this identification is the above-mentioned "inexplicable" correspondence between nature and mathematics.

I wish, I know why nature lend herself for mathematical treatment, but I don't.

By the way, we also do the same thing in mechanics where the "field" is the position 3-vector. Do you ask why the point particle is described by the position vector? and do you regard the proofs based on the vector nature of this position vector as "begging the question"?

If one chooses to apply your line of arguments about any proof in theotetical physics, the conclusion would be: "there are no proofs in theoretical physics".

I understood the thread to be equivalent to the following mathematical exercise:
Given the current conservation
$$\partial_{\mu}J^{\mu}=0$$
Show that
$\int d^{3}x J^0$
is Lorentz invariant.
I gave the answer in post#24. Now, if you to mark the answer, would you marked as wrong. If you do, then show me the right solution to that mathematical problem.


regards

sam

 

[samalkhaiat]

OMG, i just realized, I'm such an idiot. Of course those quantities are frame dependent. The momentum, even though it is a number, depends on whether the frame of reference is moving at 5m/s or at 10m/s. Same for orbital angular momentum...

The matter is clear to me now.

Daniel.

And the time, lenght, the charge density, the volume of dust particle, they all real numbers and frame dependent.

regards

sam

 

[samalkhaiat]

On page 307 of Weinberg Vol I, your proof is given almost exactly. (The only exception I can see is that he uses five equations; you use four. You could argue that W uses only four, as his first eq. effectively defines the the delta, so-to-speak, in terms of a general infitesimal LT.

This must be a very special edition book

On P.307, Weinberg uses arbitrary (not Lorentz) infinitesimal transformatin on the the fields in the Lagrangian and end up showing that the current is conserved, i.e he proved

$$\partial^{\mu} J_{\mu}=0$$

My proof: 1) deal with no Lagrangian. 2) The current conservation is given from the start. i.e I started from Weinberg's end point. 3) I proved that $\int d^3xJ^0$ is Lotrntz invariant.
Since on P.307 Weinberg does not even mention the Lorentz invariance of any quantity,therefore the book you are reading is not Weinberg's book

sam

 

[masudr]

There appears to be some misunderstanding of what the OP wanted. Sam is showing that the charge is a Lorentz scalar -- it takes the same value in all inertial frames. And this is what the title of the thread suggests the OP was looking for. Others are showing the conservation of charge-current [i.e. that the divergence of current density is the time rate of change of current] which is not the same thing as charge being a Lorentz scalar.

 

[samalkhaiat]

There appears to be some misunderstanding of what the OP wanted. Sam is showing that the charge is a Lorentz scalar -- it takes the same value in all inertial frames. And this is what the title of the thread suggests the OP was looking for. Others are showing the conservation of charge-current [i.e. that the divergence of current density is the time rate of change of current] which is not the same thing as charge being a Lorentz scalar.

You see, I could have answered the question in the following way:

Lorentz invariant charge means that the charge of a particle does not change by setting it in motion. So it has to be invariant, Otherwise the neutral nature of matter would be upset by the mere motion of its electrons.

But this answer would not do justice to the OP who clearly asked for proof based not on observation.

Little attention was paid (by manny poster) to the correct mathematical formulation of the question. This was the reason for a lot of mumbo-jumbo-type posts.

regards

sam

 

[vanesch]

By the way, we also do the same thing in mechanics where the "field" is the position 3-vector. Do you ask why the point particle is described by the position vector? and do you regard the proofs based on the vector nature of this position vector as "begging the question"?

Well, if someone were to ask the fundamental theoretical reason why, in classical mechanics, the volume of an object is the same, no matter what orthogonal coordinate system he uses to compute the integral, then I would indeed consider that what is the question really, is why is the space of position vectors an Euclidean space.

While you would quickly show that the Jacobian of an orthogonal transformation of coordinates equals one.

Both answers are correct, but it depends on what the original question was after to pick the best one.

 

[samalkhaiat]

Well, if someone were to ask the fundamental theoretical reason why, in classical mechanics, the volume of an object is the same, no matter what orthogonal coordinate system he uses to compute the integral, then I would indeed consider that what is the question really, is why is the space of position vectors an Euclidean space.

While you would quickly show that the Jacobian of an orthogonal transformation of coordinates equals one.

Both answers are correct, but it depends on what the original question was after to pick the best one.

If I understood you correctly, You still seem to question the unquestionable fact that our world can be understood mathematically!

Why space-time can be modeled by differentiable manifold? Who knows?
Physics can not answer this question. C.N.Yang said:
"There appears to be no reason why the world should be comprehensible mathematically. The fact remains that, to a large degree, it is."
Philosophers may have something to say about that fact, though B. Russel saw no "philosophical significance" in it.
I believe the secrete lies in the abtract nature of mathematics, As B. Russel put it:
"only mathematics and mathematical logic can say as little as the physicist means to say."

regards

sam

 

[JabberWalkie]

There is no proof people. To the people that keep using the 4-current to show charge invariance, you need to assume that rho*c transforms as a time component to get charge invariance and a working definition of 4-current. This is from the angle where you start with the classical form of Maxwell's equations and move to the covariant forms. If you start with the covariant forms, you implicitly assume charge invariance.

So, as far as classical E&M goes you MUST take charge invariance from experimental observations. See Jackson pg 554,555 3rd ed.

Regards,

Jab

 

[PhySong]

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)


"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

Hi Sam

I am a little bit confuse about how you get
\delta J^{\mu}= \partial_{\nu}(J^{\nu} \delta x^{\mu} - J^{\mu} \delta x^{\nu})
Could you please be more specific ?

And the 4D gauss' theorem to show "time-independent charge" is clear for me, but I am not quite understand your way.


Best regards
Song

 

[samalkhaiat]

Hi Sam

I am a little bit confuse about how you get
\delta J^{\mu}= \partial_{\nu}(J^{\nu} \delta x^{\mu} - J^{\mu} \delta x^{\nu})
Could you please be more specific ?

And the 4D gauss' theorem to show "time-independent charge" is clear for me, but I am not quite understand your way.Best regards
Song

Hi Song, welcome to PF...

Start with expanding the transformation law with $\bar{x} = x + \delta x$:
$$\bar{J}^{\mu}(x + \delta x) = \frac{\partial}{\partial x^{\nu}}\left( x^{\mu} + \delta x^{\mu} \right) \ J^{\nu}(x).$$
To first order in $\delta x$, we get
$$\bar{J}^{\mu}(x) + \delta x^{\nu}\partial_{\nu}J^{\mu}(x) = \delta^{\mu}_{\nu}J^{\nu}(x) + J^{\nu}\partial_{\nu}(\delta x^{\mu}).$$
Using $\partial_{\rho}J^{\rho}= 0$, you can write the above as
$$\bar{J}^{\mu}(x) - J^{\mu}(x) = \partial_{\nu}\left( J^{\nu} \delta x^{\mu} \right) - \delta x^{\nu}\partial_{\nu}J^{\mu}. \ \ (1)$$
Now, for Lorentz transformation, we have
$$\delta x^{\nu} = \omega^{\nu}{}_{\rho}x^{\rho}.$$
So, because of the anti-symmetry of $\omega_{\mu \nu}$, we find
$$\partial_{\nu}( \delta x^{\nu} ) = \omega^{\nu}{}_{\rho}\delta^{\rho}_{\nu} = \omega^{\nu}{}_{\nu} = 0$$
With this, we can rewrite eq(1) in the wanted form:
$$\delta J^{\mu}(x) = \partial_{\nu}\left( \delta x^{\mu}J^{\nu} - \delta x^{\nu} J^{\mu} \right).$$

Also, have a look at post#7 in

www.physicsforums.com/showthread.php?t=606174Regards
Sam

 

[PhySong]

Hi Sam

Thanks very much indeed.
It is very helpful for me.

Regards
Song

 

[vanhees71]

I'm not so sure about what you wrote about the transformation properties of the em. current. It's a four-vector field. So its transformation property is
$$\bar{J}^{\mu}(\bar{x})={\Lambda^{\mu}}_{\nu} J^{\nu}(x),$$
where
$$\bar{x}=\Lambda x$$
and $\Lambda$ is the Lorentz-transformation matrix.

For an infinitesimal Lorentz transformation you have
$${\Lambda^{\mu}}_{\nu} =\delta_{\nu}^{\mu}+{\delta \omega^{\mu}}_{\nu}$$ with $\delta \omega_{\mu \nu}=g_{\mu \rho} {\delta \omega^{\rho}}_{\nu}$ antisymmetric.

This gives
$$\bar{J}^{\mu}(\bar{x})=J^{\mu}(x)+{\delta \omega^{\mu}}_{\nu} J^{\nu}(x) = J^{\mu}(x')-{\delta \omega^{\rho}}_{\sigma} \bar{x}^{\sigma} \bar{\partial}_{\rho} J^{\mu}(\bar{x})+{\delta \omega^{\mu}}_{\nu} J^{\nu}(x).$$

The current-conservation law (continuity equation), $\partial_{\mu} J^{\mu}=0$, ensures that the total charge
$$Q=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} J^0(x)$$
is a scalar (i.e., Lorentz invariant) quantity.

 

[samalkhaiat]

I'm not so sure about what you wrote about the transformation properties of the em. current.

Why is that? I started with the transformation law of a covariant vector field with respect to the group of general coordinate transformations. That is
$$\bar{J}^{\mu}( \bar{x}) = \frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} \ J^{\nu}(x). \ \ (1)$$
For general infinitesimal coordinates transformation,

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

I wrote eq(1) as

$$\bar{J}^{\mu}( x + \delta x ) = J^{\nu}(x) \frac{\partial}{\partial x^{\nu}}\left( x^{\mu} + \delta x^{\mu}\right). \ \ (2)$$
Since,
$$\frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} = \frac{\partial}{\partial x^{\nu}}\left( x^{\mu} + \delta x^{\mu}\right) = \delta^{\mu}_{\nu} + \partial_{\nu}\left( \delta x^{\mu}\right),$$
I can rewrite eq(2) as
$$\bar{J}^{\mu}(x + \delta x ) = J^{\mu}(x) + J^{\nu}(x) \partial_{\nu}\left( \delta x^{\mu}\right).$$
Since, $\partial_{\nu}J^{\nu} = 0$, I can write the above equation in the form,
$$\bar{J}^{\mu}(x + \delta x ) = J^{\mu}(x) + \partial_{\nu}\left( \delta x^{\mu}J^{\nu}(x) \right). \ \ (3)$$
Expanding the LEFT HAND SIDE of eq(3), to first order in $\delta x$, leads to
$$\bar{J}^{\mu}( x + \delta x ) \approx \bar{J}^{\mu}(x) + \delta x^{\rho}\partial_{\rho}\bar{J}^{\mu}(x). \ \ (4)$$
Now, I claim that,
$$\delta x^{\rho}\partial_{\rho}\bar{J}^{\mu}(x) \approx \delta x^{\rho} \partial_{\rho}J^{\mu}(x). \ \ (5)$$
Why? Well, the calculus of infinitesimals tells you so. If the two functions, $\bar{J}(x)$ and $J(x)$, differ by an infinitesimal $\omega$, i.e., if
$$\bar{J}(x) \approx J(x) + \mathcal{O}(\omega ),$$
then,
$$\omega \ \bar{J}(x) \approx \omega \ J(x) + \mathcal{O}( \omega^{2}).$$
Thus, to first order in $\omega$, we write
$$\omega \ \bar{J}(x) \approx \omega \ J(x),$$
which proves the claim in eq(5). The same is true for $J(\bar{x})$ and $J(x)$, i.e. we can, to the first order, write
$$\omega \ J(\bar{x}) \approx \omega \ J(x). \ \ \ (R)$$

Ok, now putting eq(4) and eq(5) into eq(3) leads to
$$\bar{J}^{\mu}(x) - J^{\mu}(x) = \partial_{\nu}\left( \delta x^{\mu}J^{\nu}(x) \right) - \delta x^{\rho}\partial_{\rho}J^{\mu}(x). \ \ (6)$$
To specialized eq(6) to infinitesimal Lorentz transformation,
$$\delta x^{\mu} = \omega^{\mu}{}_{\nu}x^{\nu},$$
I use the fact (which follows from $\omega_{\mu\nu}= -\omega_{\nu\mu}$) that
$$\partial_{\nu}\left( \delta x^{\nu}\right) = 0. \ \ (7)$$
Using eq(7) in eq(6), we arrive at the infinitesimal LORENTZ transformation of the CONSERVED vector current
$$\delta J^{\mu}(x) = \partial_{\nu}\left( \delta x^{\mu}J^{\nu}(x) - \delta x^{\nu}J^{\mu}(x) \right). \ \ (8)$$

It's a four-vector field. So its transformation property is
$$\bar{J}^{\mu}(\bar{x})={\Lambda^{\mu}}_{\nu} J^{\nu}(x),$$
where

This Lorentz transformation follows from the general case of eq(1), when you specify
$$\frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} = \Lambda^{\mu}{}_{\nu}.$$

$$\bar{x}=\Lambda x$$
and $\Lambda$ is the Lorentz-transformation matrix.

which means that YOU can write
$$\Lambda = \frac{\partial \bar{x}}{\partial x}.$$

For an infinitesimal Lorentz transformation you have
$${\Lambda^{\mu}}_{\nu} =\delta_{\nu}^{\mu}+{\delta \omega^{\mu}}_{\nu}$$ with $\delta \omega_{\mu \nu}=g_{\mu \rho} {\delta \omega^{\rho}}_{\nu}$ antisymmetric.

YOU can write this as,
$$\frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} = \frac{\partial x^{\mu}}{\partial x^{\nu}} + \omega^{\mu}{}_{\nu},$$
or,
$$\omega^{\mu}{}_{\nu} = \frac{\partial}{\partial x^{\nu}}\left( \bar{x}^{\mu} - x^{\mu}\right) = \frac{\partial}{\partial x^{\nu}}\left( \omega^{\mu}{}_{\rho}x^{\rho}\right).$$
So, YOU can write YOUR infinitesimal Lorentz transformation in the form
$$\bar{x}^{\mu} = x^{\mu} + \delta x^{\mu},$$
with
$$\delta x^{\mu} = \omega^{\mu}{}_{\nu}x^{\nu},$$
so that YOUR Lorentz condition, $\omega_{\mu\nu}= - \omega_{\nu\mu}$, become equivalent to
$$\partial_{\mu}\left( \delta x^{\mu}\right) = 0.$$

This gives
$$\bar{J}^{\mu}(\bar{x})=J^{\mu}(x)+{\delta \omega^{\mu}}_{\nu} J^{\nu}(x) = J^{\mu}(x')-{\delta \omega^{\rho}}_{\sigma} \bar{x}^{\sigma} \bar{\partial}_{\rho} J^{\mu}(\bar{x})+{\delta \omega^{\mu}}_{\nu} J^{\nu}(x).$$

This equation is correct but UGLY. This is why it doesn’t look like eq(8). OK, let us turn it into something BEAUTIFUL. First instead of YOUR $\delta \omega$, I will use $\omega$; it is the infinitesimal parameter, so there is no need to stick $\delta$ in front of it.
As explained above, we write
$$\omega^{\mu}{}_{\nu} = \bar{\partial}_{\nu}\left( \omega^{\mu}{}_{\rho}\bar{x}^{\rho}\right) = \bar{\partial}_{\nu}\left( \delta \bar{x}^{\mu}\right),$$
and eq(R) (remember the infinitesimal stuff),
$$\omega^{\mu}{}_{\nu} \ J^{\nu}(x) \approx \omega^{\mu}{}_{\nu} \ J^{\nu}(\bar{x}).$$
This transforms YOUR equation into
$$\bar{J}^{\mu}(\bar{x}) – J^{\mu}(\bar{x}) = - \delta \bar{x}^{\nu}\bar{\partial}_{\nu}J^{\mu}(\bar{x}) + J^{\nu}(\bar{x}) \bar{\partial}_{\nu}\left( \delta \bar{x}^{\mu}\right). \ \ \ (Y)$$
Next, we use current conservation,
$$\bar{\partial}_{\nu}J^{\nu}(\bar{x}) = 0,$$
and the Lorentz condition
$$\bar{\partial}_{\nu} \left( \delta \bar{x}^{\nu}\right) = 0.$$
With these, eq(Y) becomes
$$\delta J^{\mu}(\bar{x}) = \bar{\partial}_{\nu}\left( \delta \bar{x}^{\mu}J^{\nu}(\bar{x}) - \delta \bar{x}^{\nu}J^{\mu}(\bar{x}) \right).$$
Isn’t this eq(8)?

The current-conservation law (continuity equation), $\partial_{\mu} J^{\mu}=0$, ensures that the total charge
$$Q=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} J^0(x)$$
is a scalar (i.e., Lorentz invariant) quantity.

Well Sir, the whole thread was about proving that very statement.

Regards
Sam