Do Total Current and Total Charge form a Lorentz Covariant Vector.

[Phrak]

And if so, How?

From the post 15 and 16 of the thread https://www.physicsforums.com/showthread.php?t=474719"

yes. Charge is the timelike component of the four-current. So it is relative the same way that the components of any four-vector is.

Wait, you mean the charge *density*, right? Charge is a Lorentz scalar; this is verified to extremely high precision because the hydrogen atom is electrically neutral.

But total charge and total current, Q and I, do form a 4-vector, don't they? There seem to be two ways to solve this, but I can't figure out which one is right.

Properly speaking, charge density in 3 space is a pseudo scalar and current density is a pseudo vector.

\rho = \rho_{ijk}\ dx^i dx^j dx^k \

j = j_{ij}\ dx^i dx^j \ *

To each of these, there is a corresponding dual.

\hat{\rho_n}= \epsilon^{ijk}\rho_{ijk}

\hat{j}_i = {\epsilon_i}^{jk} j_{jk}

Where j can be though of as current flux density, \hat{j} might be called the "current strength" or "current intensity". It would be nice to call \rho a current flux, but is doesn't sound very good in 3 dimensions of space unlike the case in spacetime.

Raising the index on j and combining to a 4-vector,

(\hat{\rho}, \hat{j}^i )

could form a proper Lorentz invariant 4-vector. Integrating over a 4-volume could then yield (Q, I), give or take a negative sign.

The second way would be the inverse sequence of operations above. Integrate rho and j over a 4-volume then raise the index on the spatial components.

* To be precise, these should be a directed volume and a directed area,
\rho = \rho_{ijk} \ dx^i \wedge dx^j \wedge dx^k
j = j_{ij} \ dx^i \wedge dx^j \ .

 

[bcrowell]

I think there's a pretty straightforward proof that (Q,I) can't be a four-vector. We know that Q is a Lorentz scalar, and therefore under a Lorentz transformation it has to stay the same. So (Q,I) would be a four-vector whose timelike component was invariant under Lorentz transformations. But that can't be right, because it's always possible to change the timelike component of a given four-vector by performing some Lorentz transformation.

 

[Dale]

Charge is a scalar, so it cannot be a component of any four-vector.

I don't think that current is a component of any four-vector since it is equal to a scalar times a three-vector. But I am not certain about that.

 

[Phrak]

I think there's a pretty straightforward proof that (Q,I) can't be a four-vector. We know that Q is a Lorentz scalar, and therefore under a Lorentz transformation it has to stay the same. So (Q,I) would be a four-vector whose timelike component was invariant under Lorentz transformations. But that can't be right, because it's always possible to change the timelike component of a given four-vector by performing some Lorentz transformation.

Charge is a scalar, so it cannot be a component of any four-vector.

I don't think that current is a component of any four-vector since it is equal to a scalar times a three-vector. But I am not certain about that.

Well, velocity itself can be coerced into a 4-vector.

If not charge, then what quantity, X, satisfies (X,I) is a Lorentzian 4-vector?

I'm very shocked myself to find what looks like charge to be sitting in a 4-vector. So I still have some doubts. Charge is conserved according to the charge continuity equation directly derived from Maxwell, but this this is not the same as saying charge is a Lorentz scalar.

Charge Q satisfies

Q = \int_\Omega dx^4 \rho \ .

Charge density \rho is not a Lorentz scalar, so charge cannot be a Lorentz scalar, right?

 

[atyy]

Charge Q satisfies

Q = \int_\Omega dx^4 \rho \ .

Charge density \rho is not a Lorentz scalar, so charge cannot be a Lorentz scalar, right?

I'd do it over d³x, which should transform such that Q is the same (I haven't checked, but in principle ...).

 

[bcrowell]

Charge density \rho is not a Lorentz scalar, so charge cannot be a Lorentz scalar, right?

Charge is a Lorentz scalar, as demonstrated to extremely high precision by the fact that hydrogen atoms are neutral. Charge density is not a Lorentz scalar. Charge density is the timelike component of a four-vector made out of the charge density and the current density. You don't need to speculate about all this. You can find it in any standard relativity text.

 

[Phrak]

Charge is a Lorentz scalar, as demonstrated to extremely high precision by the fact that hydrogen atoms are neutral. Charge density is not a Lorentz scalar. Charge density is the timelike component of a four-vector made out of the charge density and the current density. You don't need to speculate about all this. You can find it in any standard relativity text.

This is exactly how this issue came up in the first place. Charge density combined with current density is not a 4-vector, it is a 3-form assignable to each point of a pseudo Riemann manifold. There is a complementary one-form obtained from acting the Levi-Civita tensor on the 3-form. If this one-form cannot be composed of Q and I, then what?

 

[Phrak]

I'd do it over d³x, which should transform such that Q is the same (I haven't checked, but in principle ...).

Yes, you're right. The correction doesn't seem to change expectations, but maybe...

 

[Dale]

Well, velocity itself can be coerced into a 4-vector.

AFAIK the three velocity is not the space like part of any four vector. It is limited to c, so it doesn't transform right. Gamma times v is, but not v.

 

[homology]

This is exactly how this issue came up in the first place. Charge density combined with current density is not a 4-vector, it is a 3-form assignable to each point of a pseudo Riemann manifold. There is a complementary one-form obtained from acting the Levi-Civita tensor on the 3-form. If this one-form cannot be composed of Q and I, then what?

I understand what you're saying, however bcrowell is correct, its pretty standard to define J^{\alpha}=(c\rho, \vec{J}). But now you've got me curious so I took out an old friend: "The Geometry of Physics" He agrees with everyone. He defines his current 4-vector but also ties it to an associated 3-form by contracting the 4-volume with the current vector. If you have a copy its section 7.2b on page 199 in my edition.

I'd say more but while I'm studying EM and relativity the course I'm in doesn't have this lovely geometric side :(

 

[Physics Monkey]

If you are interested in using a very geometric language, then a nice way to write Maxwell's equations is
<br /> d*F = J<br />
where F = dA is the field strength (a 2-form) and J is a current 3-form. Local current conservation is written dJ = 0.

Total charge is the Lorentz scalar obtained from the natural pairing between J and a spatial slice \Sigma as Q = \int_{\Sigma} J. I think this point of view is useful because it clarifies why there is no natural way to make an invariant current i.e. you're already using all of J just to make Q.

Hope this helps.

 

[Phrak]

AFAIK the three velocity is not the space like part of any four vector. It is limited to c, so it doesn't transform right. Gamma times v is, but not v.

Yes, good point. Raising velocity from a 3-vector to a 4-vector results in an unusual looking 4-vector. I'll have to see why that is. The energy momentum 4 vector should also be interesting to understand; it should have an equivalent energy density + momentum flux density counterpart.

I understand what you're saying, however bcrowell is correct, its pretty standard to define J^{\alpha}=(c\rho, \vec{J}). But now you've got me curious so I took out an old friend: "The Geometry of Physics" He agrees with everyone. He defines his current 4-vector but also ties it to an associated 3-form by contracting the 4-volume with the current vector. If you have a copy its section 7.2b on page 199 in my edition.

I'd say more but while I'm studying EM and relativity the course I'm in doesn't have this lovely geometric side :(

I don't have a copy, though it sounds like a good text to have. However, other than sign convention, I'm not sure there's any room left to define the 4-current once the 4-current density is defined, as Physics Monkey presents in the post directly following yours; that is J=dG. I expect to find out, one way or the other, though.

If you are interested in using a very geometric language, then a nice way to write Maxwell's equations is
<br /> d*F = J<br />
where F = dA is the field strength (a 2-form) and J is a current 3-form. Local current conservation is written dJ = 0.

I've spent tedious hours converting electromagnetism back and forth between differential forms and vector calculus. So, I'm OK with charge continuity.

Total charge is the Lorentz scalar obtained from the natural pairing between J

and a spatial slice \Sigma as Q = \int_{\Sigma} J. I think this point of view is useful because it clarifies why there is no natural way to make an invariant current i.e. you're already using all of J just to make Q.

Hope this helps.

Yes, thank you. I don't have a good grasp of how k-forms change between subspaces, let alone spacelike slices, but I'm getting there. Changing integrals to and from subspaces is a challenge.

Rephrased, my question is as follows: Given J=dF, define K=*J. Raise the index in K and separate into spacelike and timelike parts, or one scalar and a 3-vector. What are these two elements physically identified with?

 

[Physics Monkey]

Yes, thank you. I don't have a good grasp of how k-forms change between subspaces, let alone spacelike slices, but I'm getting there. Changing integrals to and from subspaces is a challenge.

Rephrased, my question is as follows: Given J=dF, define K=*J. Raise the index in K and separate into spacelike and timelike parts, or one scalar and a 3-vector. What are these two elements physically identified with?

Great to hear. Here are some additional details. Let me ignore signs, etc and just sketch the basic structure. The 3-form J looks like this
J = \rho \, dx dy dz + j_x \, dt dy dz + j_y \, dt dx dz + j_z \, dt dx dy and the dual is obtained by replacing the 3 d's with the fourth missing d as appropriate. \rho is the charge density and j_i is the current density in a particular frame. In principle, for a given hypersurface you need all four pieces of information to compute Q.

Here is a nice example. Let \Sigma_1 be the equal time surface in x,y,z,t coordinates above. Let \Sigma_2 be the hypersurface obtained from \Sigma_1 by "bumping" a little bit of \Sigma_1 forward in time. It helps to draw a picture.

Since \Sigma_2 is a continuous deformation of \Sigma_1, the charges Q_1 and Q_2 are the same. Q_1 = \int_{\Sigma_1} J = \int_{\Sigma_2} J + \int_{M_{1 \cup 2}} dJ = Q_2 where M_{1 \cup 2} is a four dimensional subspace with boundary given by the differentce between \Sigma_1 and \Sigma_2 i.e. a little box.

More physically, integrating J over \Sigma_2 contains several terms: integrals over \rho at different times in different regions of space and integrals over the the "sides" of the box involving j_i. You'll notice that I need to integrate over dt dx dy, for example, to pick up a contribution from j_z. This is simply telling me that I need to figure out the total charge (integral over dt) that went through a given plane (integral over dx dy) in terms of the appropriate current density j_z. To summarize, I get the same answer for \Sigma_2 because even though I look at the charge density in one region (the bump) at a later time, I subtract off all the charge that entered that region via the terms involving j_i. However, it also shows that I really need all the components of J to make a lorentz invariant quantity.

I hope that wasn't too elementary.

 

[PhilDSP]

I'm not saying I agree with Pauli, but in his "Theory of Relativity" he states, or rather restates from a paper by Minkowski that charge is not invariant between coordinate systems. At what point did that become clarified or corrected in the literature?

 

[Phrak]

Great to hear. Here are some additional details. Let me ignore signs, etc and just sketch the basic structure. The 3-form J looks like this
J = \rho \, dx dy dz + j_x \, dt dy dz + j_y \, dt dx dz + j_z \, dt dx dy and the dual is obtained by replacing the 3 d's with the fourth missing d as appropriate. \rho is the charge density and j_i is the current density in a particular frame. In principle, for a given hypersurface you need all four pieces of information to compute Q.

Here is a nice example. Let \Sigma_1 be the equal time surface in x,y,z,t coordinates above. Let \Sigma_2 be the hypersurface obtained from \Sigma_1 by "bumping" a little bit of \Sigma_1 forward in time. It helps to draw a picture.

That helps a great deal. I need all the visual tools I can get, and simplifying the indexing to x,y,z,t helps put things in more manageable perspective

I'm not sure I agree with your answer, though. Integrating over spatial parts
Q = \int_\Sigma \rho_{xyz}dxdydy
seems to be what we measure, and call charge.

 

[Physics Monkey]

That helps a great deal. I need all the visual tools I can get, and simplifying the indexing to x,y,z,t helps put things in more manageable perspective

I'm not sure I agree with your answer, though. Integrating over spatial parts
Q = \int_\Sigma \rho_{xyz}dxdydy
seems to be what we measure, and call charge.

I agree, that is what we call the charge and is precisely what hypersurface \Sigma_1 directly computes. Following the bump construction I mentioned above, define B to be the bumped spatial region and A to the be the rest of space. Integrating over \Sigma_1 tells me that the total charge Q is (charge in A at time t0 + charge in B at time t0). This is how we normally evaluate the total charge. What happens with \Sigma_2 is simply that we write the total charge Q as (charge in A at time t0 + charge in B at time t1 > t0 - charge that flowed into B between t1 and t0). Either way we get the same total charge as a consequence of charge conservation.

 

[Phrak]

Charge is a Lorentz scalar, as demonstrated to extremely high precision by the fact that hydrogen atoms are neutral. Charge density is not a Lorentz scalar. Charge density is the timelike component of a four-vector made out of the charge density and the current density. You don't need to speculate about all this. You can find it in any standard relativity text.

This all gets better and better. I don't have a standard text on special relativity, but I do have Special Relativity by Albert Shadowitz:

The first question to arise is--do A and B [in relative motion] each measure the same total charge Q? [...] It turns of that for charge, however, there is excellent experimental evidence that the charge is, indeed, invariant: both A and B will measure the same value of Q regardless of the value of v. One such piece of evidence is provided by the electrical neutrality of both He and H. Each of these contain two electrons and two protons but the motions of the particles are quite different n the helium atom than in the hydrogen molecule.

The evidence is misapplied. Shadowitz is wrong. It might be simplest to consider a single atom within a region V. As long as the entire system of charges is within the region, the total charge stays constant. If any charge leaves the region, there is an accompanying current leaving the box.

For a system of identically charged particles, the charge covaries with the number count. The number density and the number count are not Lorentz invariant. The number flux and number count together, both form a Lorentz invariant vector. This fact should be present in a sufficiently enlighten book on relativity, I would think. I don't have such a text, so I'm just guessing.

 

[Phrak]

I agree, that is what we call the charge and is precisely what hypersurface \Sigma_1 directly computes. Following the bump construction I mentioned above, define B to be the bumped spatial region and A to the be the rest of space. Integrating over \Sigma_1 tells me that the total charge Q is (charge in A at time t0 + charge in B at time t0). This is how we normally evaluate the total charge. What happens with \Sigma_2 is simply that we write the total charge Q as (charge in A at time t0 + charge in B at time t1 > t0 - charge that flowed into B between t1 and t0). Either way we get the same total charge as a consequence of charge conservation.

OK. This is good. Thanks for providing me some motivation in resolving this. The experimental setup is a bump function within a spacetime 4-volume. The bump function intersects the bounding hyersurfaces \Sigma_1 and \Sigma_2 but not the other 3-surface boundaries. Thinking of our bump function as a charged particle, the charge is the same on \Sigma_1 and \Sigma_2 as long as we don't let the particle out of the 3-volume we have it in. Otherwise there is a current flux out of the box. So we can see that this particular bump function is a special case, right?

In make sense of this, I may have found a remarkably easy way to simplify manipulation of k-forms, so I'll attempt to use it. It also has the very useful property of making the signs and the permutation count evident.

Again, using simplified notation and orthonormal coordinates in dx, dy, dz, and d(ct), partition the n-form J, where n=4 dimensions into spatial and temporal parts. (It makes sense to define J as an n-form, as you will see.)

J = J d(ct)^dx^dy^dz. The components have units of Q·D^-4 and the bases have units of D⁴. So, overall, the tensor has units of charge. I’ve dropped the subscripts for clarity.

Grouping cJ dtdxdydz, the way we are used to, for charge density and current density:

J = 6(cJdt)dxdydz + 6c(Jdtdxdy)dz + 6c(Jdtdydz)dx + 6c(Jdtdzdx)dy

Note that in regrouping, the terms in parenthesis constitute the tensor components. The terms standing to the right of the parenthesis are the basis covectors.

(1/6)J = ρ dxdydz + cj_zdz + cj_ydy + cj_xdx

Grouping, instead, for total charge and total current:

J = -6(cJdxdydz)dt + 6c(Jdtdxdy)dz + 6c(Jdtdydz)dx + 6c(Jdtdzdx)dy

(1/6)J = -Qdt + cI_zdz + cI_xdx + cI_ydy

(-Q, cI) is a covector.
(Q, cI) is a vector.
Q is the total charge.
I is the total current.
c(Jdxdydzdt) is the Lorentz invariant charge pseudo scalar of charge and current.

A little work and we should have a properly formulated Lorentz invariant Kirchoff current law.

Notice in replacing units of charge with units of energy we would end up with a covector of (-E, cp), and should eventually discover that -J² = m²c⁴= E²-c²p². What do you think?

 

[bcrowell]

The evidence is misapplied. Shadowitz is wrong. It might be simplest to consider a single atom within a region V. As long as the entire system of charges is within the region, the total charge stays constant. If any charge leaves the region, there is an accompanying current leaving the box.

For a system of identically charged particles, the charge covaries with the number count. The number density and the number count are not Lorentz invariant. The number flux and number count together, both form a Lorentz invariant vector. This fact should be present in a sufficiently enlighten book on relativity, I would think. I don't have such a text, so I'm just guessing.

As far as I know there is no controversy on this point in the literature. Shadowitz is right, and charge is a Lorentz invariant.

 

[Phrak]

As far as I know there is no controversy on this point in the literature. Shadowitz is right, and charge is a Lorentz invariant.

Shadowitz jumped to a conclusion without a supporting argument. Do you have a valid argument to supply?

My argument is simple enough. If you have a charge in any state of motion in a bound region of space, the charge doesn't change upon its state of motion. This is the same evidence supplied by the equivalence of charge of He and H₂. The conclusion made is that charge, so defined as charge density times volume, is Lorentz invariant. But upon what grounds? If there is anything in the literature that can validly support this claim better than hand waving arguments, I'd be wildly surprised.

If you can find any valid errors in my argument I would like to hear it.

Now, you should be interested to know that there is a Lorentz invariant charge; a vector quantity. Using more intuitive units than I did in my previous post,

Q = (Q_R, -cI)​

This obtains Q² = Q_R² - c²I²

Q is the invariant charge. Q_R is the relativistic charge. In the claim made by Shadowitz, it is Q_R that is invariant.

This is exactly the same derivation to obtain m²c⁴ = m_R²c⁴ - c²p² from energy rather than charge, where we would never claim that relativistic mass were Lorentz invariant.

 

[bcrowell]

Shadowitz jumped to a conclusion without a supporting argument. Do you have a valid argument to supply?

Yes, see #6.

My argument is simple enough. If you have a charge in any state of motion in a bound region of space, the charge doesn't change upon its state of motion. This is the same evidence supplied by the equivalence of charge of He and H₂. The conclusion made is that charge, so defined as charge density times volume, is Lorentz invariant.

Yes, except that there is no reason to define charge as charge density times volume. If a hydrogen atom accelerates in an electric field, it's charged. If it doesn't, it's not.

If you can find any valid errors in my argument I would like to hear it.

The argument you gave in #17 doesn't make sense.

For a system of identically charged particles, the charge covaries with the number count. The number density and the number count are not Lorentz invariant.

It's true that the number density is frame-dependent, but that doesn't connect in any logical way to anything else you've said.

Now, you should be interested to know that there is a Lorentz invariant charge; a vector quantity.

There is no such thing as a Lorentz-invariant vector. A Lorentz-invariant quantity is a rank-0 tensor. A vector is a rank-1 tensor. Rank-1 tensors transform according to the tensor transformation law, so they can't be invariant.

For an example of a standard textbook presentation of the fact that charge is Lorentz invariant, see Jackson, Classical Electrodynamics, 3rd ed., section 11.9, "Invariance of Electric Charge; Covariance of Electrodynamics." Jackson is the standard graduate text on E&M. Of course it's possible that Jackson has made a mistake of truly epic proportions, that he repeated it in all three editions of his book, and that nobody ever corrected it -- but if you want to convince us of that, it's going to take an extraordinary effort. Another standard textbook discussion of this is given in Purcell, Electricity and Magnetism.

 

[Phrak]

Yes, see #6.


Yes, except that there is no reason to define charge as charge density times volume. If a hydrogen atom accelerates in an electric field, it's charged. If it doesn't, it's not.


The argument you gave in #17 doesn't make sense.


It's true that the number density is frame-dependent, but that doesn't connect in any logical way to anything else you've said.


There is no such thing as a Lorentz-invariant vector. A Lorentz-invariant quantity is a rank-0 tensor. A vector is a rank-1 tensor. Rank-1 tensors transform according to the tensor transformation law, so they can't be invariant.

For an example of a standard textbook presentation of the fact that charge is Lorentz invariant, see Jackson, Classical Electrodynamics, 3rd ed., section 11.9, "Invariance of Electric Charge; Covariance of Electrodynamics." Jackson is the standard graduate text on E&M. Of course it's possible that Jackson has made a mistake of truly epic proportions, that he repeated it in all three editions of his book, and that nobody ever corrected it -- but if you want to convince us of that, it's going to take an extraordinary effort. Another standard textbook discussion of this is given in Purcell, Electricity and Magnetism.

Good. Can you post it in any form?

 

[atyy]

Notice in replacing units of charge with units of energy we would end up with a covector of (-E, cp), and should eventually discover that -J² = m²c⁴= E²-c²p². What do you think?

To run the same argument for mass, you should start from the energy density, just like you started from the charge density. The energy density is some component of the stress-energy tensor.

 

[bcrowell]

Good. Can you post it in any form?

No. It's copyrighted. This is what libraries are for.

 

[Phrak]

You would do better as an authority to do more than appeal to authority.

I challenge you to obtain the relativistic charge-current equation from the relativistic Maxwell equations.

 

[Phrak]

To run the same argument for mass, you should start from the energy density, just like you started from the charge density. The energy density is some component of the stress-energy tensor.

Yes, it is, thank you. It is also the component of a generally covariant type(0,3) tensor.

 

[bcrowell]

You would do better as an authority to do more than appeal to authority.

I did. I pointed out the error in your argument, and I gave an argument to the contrary.

Referring you to a book is not an appeal to authority. If you get out of your chair, walk to the library, and pull the book off the shelf, you can read it for yourself and judge for yourself whether your believe the argument.

Telling you that this is a totally standard statement in every textbook is not an appeal to authority. It's just a warning that you're being foolish. Similarly, if you try to tell me that the Nile river is in South America, I will tell you that a vast number of sources of information contradict you. That's not an appeal to authority, it's an appeal to evidence. If you want to make an argument that the Nile is in South America, feel free to go ahead and knock yourself out, but it's going to take an extraordinarily strong argument. Don't expect anyone else to take you seriously during your process of constructing such an argument, especially if you make other elementary errors along the way.

One of the problems here is that you seem to have a poor grasp of extremely basic issues, such as Lorentz scalars and vectors and the tensor transformation laws, but rather than fixing your confusion about those issues you keep insisting on trying to talk about things at an even higher level of mathematical sophistication. If you don't understand that a vector can't be invariant, then you aren't going to understand all the more sophisticated formalism that you're trying to write down.

 

[Antiphon]

As long as we're hand-waiving (Shadowitz that is) I suggest that the neutrality of the H atom means nothing in this context. Consider the fact that the proton and electron have wildly different motion yet neither radiates. Should we conclude that radiation is independent of motion?

This has to resolved rigorously and this thread is on a good track. My intuition (about people, not physics) tells me that indeed if charge were not a relativistic invariant someone would have rushed to invalidate the textbooks and make a name for themselves. Now let's get to the bottom of it by derivation.

 

[atyy]

Yes, it is, thank you. It is also the component of a generally covariant type(0,3) tensor.

Do you get the rest mass to be an invariant if you run your argument for charge density (and current density) on the energy density (and energy flux)? (Sorry, for being so elliptic, I'm not familiar with your notation, just vaguely remembered having some trouble myself in old-fashioned notation, and thought this might be similar.)

 

[Phrak]

Do you get the rest mass to be an invariant if you run your argument for charge density (and current density) on the energy density (and energy flux)? (Sorry, for being so elliptic, I'm not familiar with your notation, just vaguely remembered having some trouble myself in old-fashioned notation, and thought this might be similar.)

Yes, certainly. Differential forms and the exterior algebra are a very powerful tools. The old way results in some very suspicious combination of terms such as combining a volumetric density and a flux density into a single 4-vector. For instance, current density is a type(0,3) tensor in a spacelike hypersurface and current density is a type(0,2) tensor. These are combined into a type(1,0) tensor in spacetime. What's that all about?

In differential forms we get a far less peculiar development. Promoting these objects to four dimensions, they combine as a single type(0,3) tensor. From here, a dual tensor of type (0,1) is had on applying the Hodge duality operator--or through regrouping of terms, as I did above. Multiply by the metric tensor and the type(1,0) is had.