Register to reply

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

by Phrak
Tags: charge, covariant, current, form, lorentz, vector
Share this thread:
Phrak
#1
Mar20-11, 07:22 PM
P: 4,513
And if so, How?

From the post 15 and 16 of the thread http://www.physicsforums.com/showthread.php?t=474719
Quote Quote by DaleSpam View Post
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.
Quote Quote by bcrowell View Post
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.

[tex]\rho = \rho_{ijk}\ dx^i dx^j dx^k \ [/tex]

[tex]j = j_{ij}\ dx^i dx^j \ [/tex] *

To each of these, there is a corresponding dual.

[tex]\hat{\rho_n}= \epsilon^{ijk}\rho_{ijk}[/tex]

[tex]\hat{j}_i = {\epsilon_i}^{jk} j_{jk}[/tex]

Where j can be though of as current flux density, [itex]\hat{j}[/itex] might be called the "current strength" or "current intensity". It would be nice to call [itex]\rho[/itex] 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,

[tex](\hat{\rho}, \hat{j}^i )[/tex]

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,
[tex]\rho = \rho_{ijk} \ dx^i \wedge dx^j \wedge dx^k[/tex]
[tex]j = j_{ij} \ dx^i \wedge dx^j \ .[/tex]
Phys.Org News Partner Science news on Phys.org
Climate change increases risk of crop slowdown in next 20 years
Researcher part of team studying ways to better predict intensity of hurricanes
New molecule puts scientists a step closer to understanding hydrogen storage
bcrowell
#2
Mar20-11, 07:32 PM
Emeritus
Sci Advisor
PF Gold
bcrowell's Avatar
P: 5,583
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.
DaleSpam
#3
Mar20-11, 07:49 PM
Mentor
P: 16,952
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
#4
Mar20-11, 09:56 PM
P: 4,513
Do Total Current and Total Charge form a Lorentz Covariant Vector.

Quote Quote by bcrowell View Post
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.
Quote Quote by DaleSpam View Post
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

[tex]Q = \int_\Omega dx^4 \rho \ .[/tex]

Charge density [itex]\rho[/itex] is not a Lorentz scalar, so charge cannot be a Lorentz scalar, right?
atyy
#5
Mar20-11, 10:17 PM
Sci Advisor
P: 8,343
Quote Quote by Phrak View Post
Charge Q satisfies

[tex]Q = \int_\Omega dx^4 \rho \ .[/tex]

Charge density [itex]\rho[/itex] 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
#6
Mar20-11, 10:41 PM
Emeritus
Sci Advisor
PF Gold
bcrowell's Avatar
P: 5,583
Quote Quote by Phrak View Post
Charge density [itex]\rho[/itex] 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
#7
Mar21-11, 12:00 AM
P: 4,513
Quote Quote by bcrowell View Post
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
#8
Mar21-11, 12:08 AM
P: 4,513
Quote Quote by atyy View Post
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...
DaleSpam
#9
Mar21-11, 04:56 AM
Mentor
P: 16,952
Quote Quote by Phrak View Post
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
#10
Mar21-11, 06:25 PM
homology's Avatar
P: 307
Quote Quote by Phrak View Post
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 [tex]J^{\alpha}=(c\rho, \vec{J})[/tex]. 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
#11
Mar21-11, 07:00 PM
Sci Advisor
HW Helper
Physics Monkey's Avatar
P: 1,332
If you are interested in using a very geometric language, then a nice way to write Maxwell's equations is
[tex]
d*F = J
[/tex]
where [tex] F = dA [/tex] is the field strength (a 2-form) and [tex] J [/tex] is a current 3-form. Local current conservation is written [tex] dJ = 0 [/tex].

Total charge is the Lorentz scalar obtained from the natural pairing between [tex] J [/tex] and a spatial slice [tex] \Sigma [/tex] as [tex] Q = \int_{\Sigma} J [/tex]. 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 [tex] J [/tex] just to make [tex] Q [/tex].

Hope this helps.
Phrak
#12
Mar22-11, 09:51 AM
P: 4,513
Quote Quote by DaleSpam View Post
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.


Quote Quote by homology View Post
I understand what you're saying, however bcrowell is correct, its pretty standard to define [tex]J^{\alpha}=(c\rho, \vec{J})[/tex]. 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.


Quote Quote by Physics Monkey View Post
If you are interested in using a very geometric language, then a nice way to write Maxwell's equations is
[tex]
d*F = J
[/tex]
where [tex] F = dA [/tex] is the field strength (a 2-form) and [tex] J [/tex] is a current 3-form. Local current conservation is written [tex] dJ = 0 [/tex].
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 [tex] J [/tex]

and a spatial slice [tex] \Sigma [/tex] as [tex] Q = \int_{\Sigma} J [/tex]. 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 [tex] J [/tex] just to make [tex] Q [/tex].

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
#13
Mar22-11, 10:16 PM
Sci Advisor
HW Helper
Physics Monkey's Avatar
P: 1,332
Quote Quote by Phrak View Post
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
[tex] J = \rho \, dx dy dz + j_x \, dt dy dz + j_y \, dt dx dz + j_z \, dt dx dy [/tex] and the dual is obtained by replacing the 3 d's with the fourth missing d as appropriate. [tex] \rho [/tex] is the charge density and [tex] j_i [/tex] is the current density in a particular frame. In principle, for a given hypersurface you need all four pieces of information to compute [tex] Q [/tex].

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

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

More physically, integrating [tex] J [/tex] over [tex] \Sigma_2 [/tex] contains several terms: integrals over [tex] \rho [/tex] at different times in different regions of space and integrals over the the "sides" of the box involving [tex] j_i [/tex]. You'll notice that I need to integrate over dt dx dy, for example, to pick up a contribution from [tex] j_z [/tex]. 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 [tex] j_z [/tex]. To summarize, I get the same answer for [tex] \Sigma_2 [/tex] 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 [tex] j_i [/tex]. 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
#14
Mar23-11, 06:08 AM
P: 603
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
#15
Mar23-11, 10:43 PM
P: 4,513
Quote Quote by Physics Monkey View Post
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
[tex] J = \rho \, dx dy dz + j_x \, dt dy dz + j_y \, dt dx dz + j_z \, dt dx dy [/tex] and the dual is obtained by replacing the 3 d's with the fourth missing d as appropriate. [tex] \rho [/tex] is the charge density and [tex] j_i [/tex] is the current density in a particular frame. In principle, for a given hypersurface you need all four pieces of information to compute [tex] Q [/tex].

Here is a nice example. Let [tex] \Sigma_1 [/tex] be the equal time surface in x,y,z,t coordinates above. Let [tex] \Sigma_2 [/tex] be the hypersurface obtained from [tex] \Sigma_1 [/tex] by "bumping" a little bit of [tex] \Sigma_1 [/tex] 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 spacial parts
[tex]Q = \int_\Sigma \rho_{xyz}dxdydy[/tex]
seems to be what we measure, and call charge.
Physics Monkey
#16
Mar24-11, 10:34 AM
Sci Advisor
HW Helper
Physics Monkey's Avatar
P: 1,332
Quote Quote by Phrak View Post
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
[tex]Q = \int_\Sigma \rho_{xyz}dxdydy[/tex]
seems to be what we measure, and call charge.
I agree, that is what we call the charge and is precisely what hypersurface [tex] \Sigma_1 [/tex] 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 [tex] \Sigma_1 [/tex] 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 [tex] \Sigma_2 [/tex] 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
#17
Mar25-11, 09:37 PM
P: 4,513
Quote Quote by bcrowell View Post
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
#18
Mar26-11, 03:15 PM
P: 4,513
Quote Quote by Physics Monkey View Post
I agree, that is what we call the charge and is precisely what hypersurface [tex] \Sigma_1 [/tex] 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 [tex] \Sigma_1 [/tex] 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 [tex] \Sigma_2 [/tex] 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 [itex]\Sigma_1[/itex] and [itex]\Sigma_2[/itex] but not the other 3-surface boundaries. Thinking of our bump function as a charged particle, the charge is the same on [itex]\Sigma_1[/itex] and [itex]\Sigma_2[/itex] 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 QD-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?


Register to reply

Related Discussions
Battery (amp hours, total charge, total current) Introductory Physics Homework 6
Why is charge/current density 4 vector a twisted differential 3-form? Special & General Relativity 2
Total charge through point on wire due to changing current on other wire Advanced Physics Homework 1
Total Flux Density VS Total Field Intensity VS Current Density Classical Physics 0
Covariant form of the Lagrangian for Lorentz force. Classical Physics 0