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

1. Mar 20, 2011

### Phrak

And if so, How?

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

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,
$$\rho = \rho_{ijk} \ dx^i \wedge dx^j \wedge dx^k$$
$$j = j_{ij} \ dx^i \wedge dx^j \ .$$

Last edited by a moderator: Apr 25, 2017
2. Mar 20, 2011

### bcrowell

Staff Emeritus
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.

3. Mar 20, 2011

### Staff: Mentor

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.

4. Mar 20, 2011

### Phrak

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?

5. Mar 20, 2011

### atyy

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

6. Mar 20, 2011

### bcrowell

Staff Emeritus
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.

7. Mar 21, 2011

### Phrak

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?

8. Mar 21, 2011

### Phrak

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

9. Mar 21, 2011

### Staff: Mentor

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.

10. Mar 21, 2011

### homology

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 :(

11. Mar 21, 2011

### Physics Monkey

If you are interested in using a very geometric language, then a nice way to write Maxwell's equations is
$$d*F = J$$
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.

12. Mar 22, 2011

### Phrak

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 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.

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

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?

13. Mar 22, 2011

### Physics Monkey

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.

14. Mar 23, 2011

### 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?

15. Mar 23, 2011

### Phrak

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 spacial parts
$$Q = \int_\Sigma \rho_{xyz}dxdydy$$
seems to be what we measure, and call charge.

Last edited: Mar 23, 2011
16. Mar 24, 2011

### Physics Monkey

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.

17. Mar 25, 2011

### Phrak

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 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.

18. Mar 26, 2011

### Phrak

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 + cjzdz + cjydy + cjxdx

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 + cIzdz + cIxdx + cIydy

(-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?

Last edited: Mar 26, 2011
19. Mar 26, 2011

### bcrowell

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

20. Mar 26, 2011

### Phrak

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 = (QR, -cI)​

This obtains Q2 = QR2 - c2I2

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

This is exactly the same derivation to obtain m2c4 = mR2c4 - c2p2 from energy rather than charge, where we would never claim that relativistic mass were Lorentz invariant.

Last edited: Mar 26, 2011