Talk:Four-current

message: I'm sorry, but identifying charge/current density with a 4-vector is just wrong. One can quote Schwart, Jackson, or Griffith and be wrong with authority, but still wrong.
Four current density, J = J_uvp dx^u dx^v dx^p, wedds current *density* and charge *density*, an antisymmetric tensor with lower indices. It is a directed quantity as it should be, obeying CPT symmetry.
To see how this works, it's useful to examine the units of the 3-current and the charge density in a space-like slice of spacetime. Current density has units J = J[Q/(D^2T)], and charge density rho=rho[Q/D^3]. Each has units of charge in the numerator and 3 units spacetime in the denominator. These quantities are assigned to elements of a tensor object. To obtain the invariant tensor, a basis is needed. With basis having units of spatial displacement, dx^i, J_ij = J_ijt[Q/TD^2] dx^i dx^j dt[TD^2]. The tensor object has units of [Q]. For charge density, rho=rho_xyz[Q/D^3]dx dy dz[D^3]. This tensor object also has units of charge, [Q].
In combination as a 4-tensor (density), this object also has units of charge, [Q].
On the other hand, forcing current density to be a three-vector requires an unnatural assignment of units to the vector components having bases partial/partial x^i [1/D]. This force units of [Q/D] in order for the current density of the tensor object to have units of Q. It's even crazier if you want the elements of your vector to have units of [Q/TD^2].
There is, however a pseudo-vector density that does qualify, as long as we know it's an not actually a vector, but an axial vector, and not a tensor object but a density symbol.
Where J is the tensor density three form, take the Hodge dual density to form it's dual, *J. This is a 1-form density. Multiply by the metric and the we have a 4-pseudo-vector density that loosely qualifies as a 4-vector, but is not really an honest vector under CPT symmetry or general covariance.
166.250.64.235 (talk) 07:11, 19 September 2011 (UTC)
replies: text: The gist of this makes sense: what we identify as charge and current density appears to be a 3-form, must transform accordingly, and I guess we should denote its components Jαβγ=J[αβγ] in tensor form.  As this ties in very closely with the physical interpretation of a density (which can be integrated over a 3-D space), it would be nice if a reference could be found that presents it from this perspective. The article unfortunately is very limited. This problem might occur with other quantities; perhaps they were defined before forms were really understood. Quondumtalkcontr 05:48, 15 December 2011 (UTC)
thread_title:  Overhaul 

message: Is this four-vector defined as contravariant?
I think J^a = \left(c \rho, \mathbf{j} \right) is the same as J^a = \rho \frac{dx^a}{dt}, but as far as dt is not invariant, it should be J^a = \rho \frac{dx^a}{dt_0}, yielding J^a = \gamma\left(c \rho, \mathbf{j} \right).
Am I right? -Daniel bg (talk) 18:10, 10 May 2011 (UTC)
thread_title:  Contravariant? 

message: I think there is not much more to write about four-current in special relativity, but maybe an expert could write more on the general relativity approach. -Daniel bg (talk) 13:23, 14 May 2011 (UTC)
thread_title:  General Relativity 

message: Problems and intended fixes: 
#indices should be Greek, convention for space and time, while Latin for space only, this will confuse readers... 
#unnecessary inclusion of definitions (like the 4-gradient and D'Alembert operator, just link them, their definitions are subjects of other articles, else why have those articles?)
#half the links are no use and charge (physics) will not be much help (good for a "see also" link, but if this is principally about electric charge link to electric charge. A separate section on 4-currents for anything can easily be done).
#article genuinely padded and a mess...
#Physical interpretation should be nearer the beginning, all readers can read text, not all will even care about the formalism which follows from the definition, and as it stands is not clear enough:
*"This four-vector unifies charge density (related to electricity) and current density (related to electricity magnetism) in one electromagnetic entity." <-- fine.
*"As studied in electrostatics, charges or distributions of charge moving only through time, thus having no velocity, have only charge density, while if they are moving through space too, thus having some velocity v, they have current density too." <-- Not clear: what does "moving through time and "thus" have no velocity" mean to a typical reader? "If they are moving through space "too", they "thus" have velocity and current density"... only very experts (accomplished or up-and-coming) will know this really means:
:"charges (free or as a distribution) at rest will appear to remain at the same spatial position for some interval of time (as long as they're stationary). When they do move, this corresponds to changes in position, therefore the charges have velocity, and the motion of charge constitutes an electric current."
*"This means that charge density is related to time, while current density is related to space." At least the beginning and end is nice.
I can't say much about general relativity though... maybe someone else will come back to it soon...
Anyway it'll be rewritten and the "attention/confusing" templates will be removed. If anyone disagrees please explain. F = q(E+v×B) ⇄ ∑ici 19:50, 10 April 2012 (UTC)
thread_title:  problems with the article...