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Is current a vector or scalar? 
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#1
Feb2708, 02:13 AM

P: 93

I see in several physics papers and articles current, current density represented by vectors. But in one book it was mentioned clearly that current is not a vector because it does not obey vector law of addition. Can some one clarify this point with an example and show that current does not obey vector law of addition? Thanks.



#2
Feb2708, 06:22 AM

P: 455

Current density [tex]{\vec j}[/tex] is a vector.
The current through a surface is given by [tex]\int{\vec j}\cdot\vec{dA}[/tex], which is a scalar. But a differential current element in a wire in the BiotSavart law is [tex]I\vec{dL}[/tex], which is a vector. 


#3
Feb2808, 07:52 AM

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This is a good one, I hope we get some good discussion because I'm a little unsure.
"Current" is a 3form. I'll be honest I have no idea what that means, other than it's sort of a dual to a vector. In practical use, the current is a vector because behind the scenes the metric tensor was used to convert it. 


#4
Feb2808, 08:03 AM

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P: 2,616

Is current a vector or scalar?
I thought current was just defined as the charged particle flux through a crosssectional area. That appears to be the case if you define I to be surface integral of current density over a given surface. Now is flux a vector or scalar? I think it is neither.



#5
Feb2808, 09:52 AM

P: 455

Flux is a scalar.



#6
Feb2808, 11:50 AM

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My only references on this are E.J. Post's "Formal Structure of Electromagnetics" and vague memories of MTW's "Gravitation". In Post's book, E is a 1form, B a 2form, D a vector, and H a bivector (dyadic). Biot Savart's law relates the gradient of the magnetic field B to the current, and so current is a 3form.
Or maybe I am confusing the issue in elementary (i.e. undergrad) treatments, the current (like velocity) is a vector, while flux (like quantity) is a scalar. 


#7
Mar108, 12:14 AM

P: 4,512

If current density is a 3form, it's dual in 3 dimensional space is a scalar (0form). In spacetime, the current density in combination with charge density would be 3form. But in 4 dimenional spacetime its dual is a 1form. I'm not sure where it makes sense to talk about electromagnetism in the context of forms that live in three dimensions. But the units in question do attach themselves to the basis of the form, so that if you take the dual of something in three dimenisons having with perlengthsquared, its dual will have units of length. I really must sort all of this stuff out, but ... so many questions, so little brain. 


#8
Mar108, 01:51 AM

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In 4dimensional spacetime, E and B are the six components of a 2form, [tex](E_x\,,\,E_y\,,\,E_z\,,\,B_x\,,\,B_y\,,\,B_z)[/tex]. Loosely speaking, a 2form is a tensor (a 4 x 4 square) with all the diagonal terms zero, and the abovediagonal terms equal to minus the belowdiagonal terms. So instead of 16 independent terms (= 4 x 4), 4 are zero, and 6 of the remaining 12 are minus the other 6, leaving only 6 independent terms, 3 of which are [tex](E_x\,,\,E_y\,,\,E_z)[/tex] and 3 of which are [tex](B_x\,,\,B_y\,,\,B_z)[/tex]. E and B separately are 3dimensional vectors  they do obey the vector law of addition. But if the observer changes to a different velocity, then E and B get mixed up with each other  so 3dimensionally they are vectors, but 4dimensionally they aren't, though together they are the parts of a tensor (or 2form). Oh, and this all works in Galilean 4dimensional spacetime as well as Einsteinian  physicists knew about it long before Einstein came along! (Current isn't a 3form. A 1form is a vector. In 4 dimensions, a 3form is the dual of a 1form: so the dual of a vector is an "axial vector". I think angular momentum is a 3form.) 


#9
Mar108, 08:55 AM

P: 4,512

Tiny! Where have you been?
Could you possibly check out the thread, "Exterior Calculus and Differential Forms?" under Tensor Analysis & Differential Geometry? The (*) are the Hodge dual operator. 


#10
Mar108, 09:39 AM

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Pete 


#11
Mar108, 12:27 PM

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http://en.wikipedia.org/wiki/Current_density:
Have you tried the poster of thread "Hodge operator and adjoints"?: 


#12
Mar108, 02:39 PM

P: 4,512

Thanks, Tiny.
I guess I'm going it alone. deCraig (so many questions, so little brain) 


#13
Mar208, 04:18 AM

P: 93

I thought a quantity can either be scalar or vector. some one said it is neither. And I do not know what is 2form and 3form etc. If some one can simplify this, that would help a great deal!



#14
Mar208, 04:48 AM

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Three numbers (or one number and a direction, because a direction counts as two numbers  think of latitude and longitude) can be a vector, but it would have to obey the vector law of addition when combined with other sets of three numbers. (And in relativity, four numbers can be a vector.) Six numbers can be a fourdimensional 2form (also called exterior product or wedge product or antisymmetric tensor), and it can be split into two threedimensional vectors. And forget about 3forms (in four dimension, they are the "duals" of vectors)  you won't need them. 


#15
Mar208, 12:02 PM

P: 4,512

A vector cross product is a polar vector. It's not a 'real' vector. It's sign depends on the chirility (or parity, or handedness) of the coordinate system. (And, by the way, this all makes more sense of in the calculus of differential forms.) Within Maxwell's equations the current is obtained from the cross product of the magnetic field. [tex]\bar{J}=\nabla \times \bar{B}  \partial\bar{E}/\partial t[/tex] For torque, [tex]\bar{M}=\bar{F}\times\bar{r}[/tex] . Both M and the B part of J are axial vectors; also called psuedovectors. Now, angular momentum is also a psuedovector. I vaguely recall that angual momentums add in peculiar ways. Maybe I was dreaming... 


#16
Mar208, 12:34 PM

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Yes, you're right: technically, angular momentum and torque are axial vectors (or pseudovectors), which are 2forms in threedimensional space. And currentpluschargedensity is a 3form in fourdimensional space, because it is the curl (the fourdimensional crossproduct derivative, "∆ x") of the electromagnetic field (which is a 2form in fourdimensional space). And in fourdimensional space, the geometry of 3forms is the same as of ordinary vectors (1forms), only "insideout". Which is why my recommendation is to forget about the difference! 


#17
Mar208, 01:50 PM

P: 4,512

http://en.wikipedia.org/wiki/Pseudovector
Here's an extract from the above link. "To the extent that physical laws are the same for righthanded and lefthanded coordinate systems (i.e. invariant under parity), the sum of a vector and a pseudovector is not meaningful." Current is the sum of a vector and pseudovector. manjuvenamma, maybe you could quote what your source had to say about addition of currents. 


#18
Mar208, 02:05 PM

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PF Gold
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Yes, you should recognize that the set of (tangent) vectors and the set of 3forms are both fourdimensional spaces. But you should also recognize that vectors correspond to directions, and 3forms to volume elements. e.g. in an orthonormal coordinate system, the standard basis for 3forms are the volume elements [itex]dt \, dx \, dy, dx \, dy \, dz, dy \, dz \, dt,[/itex] and [itex]dz \, dt \, dx[/itex]. I think it is a bad idea to advise people to mentally ignore the differences; the "proper" method is no more difficult to use than the "sloppy" method... you just have to be willing to do it. 


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