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Do Total Current and Total Charge form a Lorentz Covariant Vector. 
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#1
Mar2011, 07:22 PM

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And if so, How?
From the post 15 and 16 of the thread http://www.physicsforums.com/showthread.php?t=474719 But total charge and total current, Q and I, do form a 4vector, 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 4vector, [tex](\hat{\rho}, \hat{j}^i )[/tex] could form a proper Lorentz invariant 4vector. Integrating over a 4volume 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 4volume 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] 


#2
Mar2011, 07:32 PM

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PF Gold
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I think there's a pretty straightforward proof that (Q,I) can't be a fourvector. 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 fourvector 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 fourvector by performing some Lorentz transformation.



#3
Mar2011, 07:49 PM

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Charge is a scalar, so it cannot be a component of any fourvector.
I don't think that current is a component of any fourvector since it is equal to a scalar times a threevector. But I am not certain about that. 


#4
Mar2011, 09:56 PM

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Do Total Current and Total Charge form a Lorentz Covariant Vector.
If not charge, then what quantity, X, satisfies (X,I) is a Lorentzian 4vector? I'm very shocked myself to find what looks like charge to be sitting in a 4vector. 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? 


#5
Mar2011, 10:17 PM

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#6
Mar2011, 10:41 PM

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#7
Mar2111, 12:00 AM

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#8
Mar2111, 12:08 AM

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#9
Mar2111, 04:56 AM

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#10
Mar2111, 06:25 PM

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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
Mar2111, 07:00 PM

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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 2form) and [tex] J [/tex] is a current 3form. 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. 


#12
Mar2211, 09:51 AM

P: 4,512

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 3vector. What are these two elements physically identified with? 


#13
Mar2211, 10:16 PM

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[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. 


#14
Mar2311, 06:08 AM

P: 612

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
Mar2311, 10:43 PM

P: 4,512

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. 


#16
Mar2411, 10:34 AM

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#17
Mar2511, 09:37 PM

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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
Mar2611, 03:15 PM

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In make sense of this, I may have found a remarkably easy way to simplify manipulation of kforms, 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 nform J, where n=4 dimensions into spatial and temporal parts. (It makes sense to define J as an nform, 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 D^{4}. 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_{z}dz + cj_{y}dy + cj_{x}dx 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_{z}dz + cI_{x}dx + cI_{y}dy (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 J^{2} = m^{2}c^{4}= E^{2}c^{2}p^{2}. What do you think? 


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