# Is current a vector or scalar?

1. Feb 27, 2008

### manjuvenamma

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. Feb 27, 2008

### pam

Current density $${\vec j}$$ is a vector.
The current through a surface is given by
$$\int{\vec j}\cdot\vec{dA}$$, which is a scalar.
But a differential current element in a wire in the Biot-Savart law
is $$I\vec{dL}$$, which is a vector.

3. Feb 28, 2008

### Andy Resnick

This is a good one, I hope we get some good discussion because I'm a little unsure.

"Current" is a 3-form. 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. Feb 28, 2008

### Defennder

I thought current was just defined as the charged particle flux through a cross-sectional 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. Feb 28, 2008

### pam

Flux is a scalar.

6. Feb 28, 2008

### Andy Resnick

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 1-form, B a 2-form, 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 3-form.

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. Mar 1, 2008

### Phrak

This stuff has been bothering me too. In spacetime, E and B are elements of a 2-form called the electromagnetic field tensor on occasion.

If current density is a 3-form, it's dual in 3 dimensional space is a scalar (0-form).

In spacetime, the current density in combination with charge density would be 3-form. But in 4 dimenional spacetime its dual is a 1-form.

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 per-length-squared, its dual will have units of length.

I really must sort all of this stuff out, but ... so many questions, so little brain.

Last edited: Mar 1, 2008
8. Mar 1, 2008

### tiny-tim

Galilean tensors

Yes.

In 4-dimensional space-time, E and B are the six components of a 2-form, $$(E_x\,,\,E_y\,,\,E_z\,,\,B_x\,,\,B_y\,,\,B_z)$$.

Loosely speaking, a 2-form is a tensor (a 4 x 4 square) with all the diagonal terms zero, and the above-diagonal terms equal to minus the below-diagonal 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 $$(E_x\,,\,E_y\,,\,E_z)$$ and 3 of which are $$(B_x\,,\,B_y\,,\,B_z)$$.

E and B separately are 3-dimensional 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 3-dimensionally they are vectors, but 4-dimensionally they aren't, though together they are the parts of a tensor (or 2-form).

Oh, and this all works in Galilean 4-dimensional space-time as well as Einsteinian - physicists knew about it long before Einstein came along!

[size=-2](Current isn't a 3-form. A 1-form is a vector. In 4 dimensions, a 3-form is the dual of a 1-form: so the dual of a vector is an "axial vector". I think angular momentum is a 3-form.)[/size]

9. Mar 1, 2008

### Phrak

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. Mar 1, 2008

### pmb_phy

Current is a Cartesian scalar but it is not a Lorentz scalar. The difference depends on what transformations you're considering which differs according to whether you're considering this question from a nonrelativistic mechanics or from a relativistic mechanics point of view. In relativity current is neither a scalar nor a vector nor a tensor of any type.

Pete

11. Mar 1, 2008

### tiny-tim

… only surface knowledge of surfaces …

http://en.wikipedia.org/wiki/Current_density:
Let's see …

mmm … beats me … I only what they are, and a limited usage of them in special relativity.

12. Mar 1, 2008

### Phrak

Thanks, Tiny.

I guess I'm going it alone.

-deCraig (so many questions, so little brain)

13. Mar 2, 2008

### manjuvenamma

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

14. Mar 2, 2008

### tiny-tim

Ok: one number can be a scalar.

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 four-dimensional 2-form (also called exterior product or wedge product or antisymmetric tensor), and it can be split into two three-dimensional vectors.

And forget about 3-forms (in four dimension, they are the "duals" of vectors) - you won't need them.

15. Mar 2, 2008

### Phrak

I've really enjoyed all the talk about differential forms. It's helped me clarify some issues but... it seems the problem in question is about axial vectors vs. polar vectors.

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.

$$\bar{J}=\nabla \times \bar{B} - \partial\bar{E}/\partial t$$

For torque,

$$\bar{M}=\bar{F}\times\bar{r}$$ .

Both M and the B part of J are axial vectors; also called psuedo-vectors.

Now, angular momentum is also a psuedo-vector.

I vaguely recall that angual momentums add in peculiar ways. Maybe I was dreaming...

Last edited: Mar 2, 2008
16. Mar 2, 2008

### tiny-tim

axial vectors vs. polar vectors

Hi Phrak!

Yes, you're right: technically, angular momentum and torque are axial vectors (or pseudo-vectors), which are 2-forms in three-dimensional space.

And current-plus-charge-density is a 3-form in four-dimensional space, because it is the curl (the four-dimensional cross-product derivative, "∆ x") of the electro-magnetic field (which is a 2-form in four-dimensional space).

No. In three-dimensional space, the geometry of axial vectors (2-forms) is exactly the same as of ordinary vectors (1-forms), only "inside-out" (mathspeak: "opposite chirality"). So "vector addition" still works.

And in four-dimensional space, the geometry of 3-forms is the same as of ordinary vectors (1-forms), only "inside-out".

Which is why my recommendation is to forget about the difference!

17. Mar 2, 2008

### Phrak

http://en.wikipedia.org/wiki/Pseudovector

Here's an extract from the above link.

"To the extent that physical laws are the same for right-handed and left-handed 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.

18. Mar 2, 2008

### Hurkyl

Staff Emeritus
IMHO, it's better to understand how things are the same and how they are different, rather than pretending everything is the same.

Yes, you should recognize that the set of (tangent) vectors and the set of 3-forms are both four-dimensional spaces.

But you should also recognize that vectors correspond to directions, and 3-forms to volume elements. e.g. in an orthonormal coordinate system, the standard basis for 3-forms are the volume elements $dt \, dx \, dy, dx \, dy \, dz, dy \, dz \, dt,$ and $dz \, dt \, dx$.

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.

Last edited: Mar 2, 2008
19. Mar 2, 2008

### tiny-tim

Too much torque …

Current density is a vector: http://en.wikipedia.org/wiki/Current_density
And current itself is a scalar (as Pam said): http://en.wikipedia.org/wiki/Current_density
I don't think so, not even for a current density. One talks for example of the divergence of a current density, and I don't think you can define that for a vector plus a psudeovector.

(Though there may be situations where there are two separate current densities, one a vector and t'other a pseudovector.)

This is a bit Misner-on-Gravitation-ish.

Technically, of course, it's correct - but you might as well say that torque (a three-dimenionsal 2-form) is an area element!

For practical physics, torque is a vector, not an area element, and you'd only notice the difference if for some reason you wanted to measure torque in a pair of mirror-image experiments!

It is in that sense that I still recommend that anyone not doing high-energy physics will understand torque perfectly adequately as a vector!

The OP himself asked for explanation with simplification:
Sometimes it helps to point out the differences, sometimes it helps to point out the similarities.

Too much information (as in some other threads) can be detrimental to understanding.

20. Mar 2, 2008

### Hurkyl

Staff Emeritus
Yes he did -- for the purpose of helping him understand why some things are neither scalars nor (tangent) vectors.