
#1
Jun2513, 08:16 AM

P: 20

Hello,
I am a 15 year old high school student, We were being taught about current, My teacher said its a scalar quantity, I had a doubt on that that since wires bound the charges, we shouldn't say that since wire's orientation doesn't change the magnitude of current, its Scalar quantity. For clarifications I searched some books, that given me a much bigger confusion. Halliday Resnick Walker Fundamentals of Physics, like many other high school physics books, says this too that current is a scalar quantity because it can be added algebraically on page no 683 Chapter 26 section 262. On the other hand, David J. Griffiths in his book, Introduction to Electrodynamics Third Edition on page no. 208 , Chapter 5, section 5.1.3 say that current is a vector!. I am quoting its lines: " Current is actually a Vector: I = λv (where λ is charge density and v is the velocity of the particles.) since the path of the flow is dictated by the shape of wire, most people don't bother to display the vectorial character of I explicitly....". Now what to believe, what not to? both the mentioned books are a great deal so they can't be Wrong!. Then what's the truth. Please Clarify me on this. THANKS in Advance 



#2
Jun2513, 08:34 AM

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hello aayushgsa!
for example, in the amperemaxwell law (one of the four maxwell's equations), curl of the magnetic field equals the current plus the rate of change of the electric field, and they are all vectors




#3
Jun2513, 08:47 AM

P: 20

Thanks tiny tim for a quick reply but why there is such a paradox in all the books. The water's momentum analogy is a special case, seeing this case one cannot say the momentum is scalar, we have taught from starting that momentum is vector then why current is said scalar in halliday resnick for just the special case of wire?




#4
Jun2513, 09:02 AM

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Current vector or scalar.. different in different bookswater current is exactly like electric current in this respect it's totally unhelpful!there's no point in writing momentum or current equations as vector equations unless we're willing to add in all the other vectors acting on the flow … since that just introduces a lot of unknowns, which we ultimately solve to produce a scalar equation anyway, we might as well have started with scalars! 



#5
Jun2513, 10:03 AM

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Strictly speaking one must be a bit more careful in how to define "current". The quote from Griffiths is a bit sloppy, but it's the usual slang of physicists one must excuse him. One should, however once explain it correctly:
You should strictly distinguish currents (scalar quantities) and current densities (vector fields). Also I should stress that I talk about tensor properties in the sense of Euclidean threedimensional quantities (not manifestly covariant fourtensors in Minkowski space). In my opinion the simpler idea is the current density, [tex]\vec{j}(t,\vec{x})=\rho(t,\vec{x}) \vec{v}(t,\vec{x}).[/tex] The physical meaning is the following: Take a infinitesimally small area element [itex]\mathrm{d} \vec{F}.[/itex] You may have in mind a tiny rectangle of area [itex]\mathrm{d} \vec{F}[/itex] with the direction of [itex]\mathrm{d} \vec{F}[/itex] perpendicular to it (the choice of the direction or orientation of the area element is arbitrary, either one of the two is fine). Then the amount of charge per unit time running through this area is given by [tex]\mathrm{d} i=\rho(t,\vec{x}) \vec{v}(t,\vec{x}) \cdot \mathrm{d} \vec{F}.[/tex] Here [itex]\mathrm{d} i[/itex] is a infinitesimal part of the current through the surface element. The total current through a finite surface is then given as the sum over many infinitesimal surfaces making up the big surface or, in the limit of truely infinitely small and infinitetly many area elements, the surface integral [tex]i(t)=\int_{F} \mathrm{d} \vec{F} \cdot \vec{j}(t,\vec{x}).[/tex] The total current is positive (negative) if there is a positve net charge per unit time is running in the direction (against the direction) of the surface vectors. The physical meaning is of course, independent of the orientation of the surface vectors, because if you flip the orientation the sign of the current, [itex]i(t)[/itex], flips, but that's how it should be, because the meaning of net positive charge per unit time running in one specific direction stays the same, no matter how you choose the orientation, and flipping the orientation thus must flip the sign of [itex]i(t)[/itex]. 



#6
Jun2613, 05:15 AM

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#7
Jun2613, 05:20 AM

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Current is a scalar; it cannot be a vector for it does not obey the basic properties of a vector. Current density on the other hand is a vector. Don't worry about tensors for now.




#8
Jun2613, 05:50 AM

P: 20

So the thing tinytim said was wrong? It sounded quite convincing...




#9
Jun2613, 07:15 AM

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The problem is that most physicists are pretty sloppy when talking about these quantities. Sometimes they say "current" when in fact they mean "current density". E.g., it's common slang (at least in the highenergyphysics theory community) to talk about "conserved currents". Dependent on the context, what's meant is that there is a density and the associated current density which fulfill the continuity equation,
[tex]\partial_t \rho+\vec{\nabla} \cdot \vec{j}=0.[/tex] Neither the density nor the current density are of course conserved. Nevertheless the continuity equation is the local form of a conservation law, but what's conserved is in facto the charge associated with its density and current density. This you see by integrating the continuity equation all over space and using Gauß's integral theorem. Since the whole space has no boundary, you immediately get [tex]\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \partial_t \rho(t,\vec{x})= \frac{\mathrm{d}}{\mathrm{d} t} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \rho(t,\vec{x})=\frac{\mathrm{d} Q}{\mathrm{d} t}=0.[/tex] This shows that indeed the total charge is conserved. For the beginner in field theory, I think, it's however pretty important to get a good grip both on the formal and the intuitive side of these matters. One should also be careful to distinguish scalars (vectors) and scalar (vector) fields. In the context of the threedimensional Euclidean space a scalar doesn't change under rotations while a vector transforms as the position vector (i.e., it's rotated). For fields, you have to take care of both the quantity itself as for the position argument. A scalar field (by definition!) transforms under rotations [itex]\vec{x} \rightarrow \vec{x}'=\hat{R} \vec{x}[/itex] according to the law [tex]s'(\vec{x}')=s(\vec{x})=s(\hat{R}^{1} \vec{x}).[/tex] Here [itex]\hat{R}[/itex] is a rotation matrix (i.e., a special orthogonal matrix, obeying [itex]\hat{R}^{1}=\hat{R}^T[/itex], [itex]\mathrm{det} \hat{R}=+1.[/itex] For a vector field you have [tex]\vec{V}'(\vec{x}') = \hat{R} \vec{V}(\vec{x})=\hat{R} \vec{V}(\hat{R}^{1} \vec{x}).[/tex] 



#10
Jun2613, 07:32 AM

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technically, griffiths is correct …
of course, looking at the individual electrons is ridiculously unhelpful, so we treat electricity as a continuous fluid, and only the current density matters as vanhees71 points out, maxwell's equations use the current density, J, and we integrate it across a surface to get the dot product I = ∫ J.dF, which is a scalar, the current across the surface there isn't really any practical use for vector current … but i think it exists just as much as i do! 



#11
Jun2613, 08:00 AM

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I never said there is no practical use for the current density. To the contrary, I prefer the local form of Maxwell's Laws compared to the more complicated integral forms. After all the local form is the most fundamental form!
Finally of course also a single electron has a current density, but it's pretty singular: [tex]\vec{j}=e \vec{v} \delta^{(3)}[\vec{x}\vec{y}(t)].[/tex] Here [itex]e<0[/itex] is the charge of an electron. 



#12
Jun2613, 09:12 AM

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See Ohm's law in local or microscopic form: J=σE where J is current density, E is the electric field and σ is the conductivity. In anisotropic materials σ is a tensor and this "law" represents the relationship between the J and E vectors. To study this in terms of currents will be a bit awkward and probably of little use. Of course, "practically" we measure currents more often than current density. But density (kg/m^3) is not measured directly either. But it still have its practical uses. 



#13
Jun2613, 10:47 AM

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#14
Jun2613, 11:49 AM

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Griffiths uses the definition of current as a vector ##\vec I = \lambda \vec v## in the context of discussing the magnetic force on a straight wire: ##\vec F = \vec I l \times \vec B##.
For this, Halliday & Resnick use ##\vec F = I \vec l \times \vec B##, that is, they associate the "vectorness" with the length of the wire rather than with the current. Griffiths notes (soon after his first definition) that one can also write it this way. Griffiths introduces the current density ##\vec J## and the general definition of current as a scalar $$I = \int {\vec J \cdot d \vec a}$$ somewhat later. In this general case, it makes no sense to talk about current as a vector because both ##\vec J## and ##d \vec a## can be in different directions at different points on the surface of integration. Neither of them has a unique direction. With a wire, on the other hand, we usually assume the current density is always perpendicular to the crosssection of the wire, so the direction of the overall current is welldefined, and parallel to the orientation of the wire. So the current can be thought of has having a vectorlike character in this special case. However, as noted, this vectorlike character can be associated with the length of the wire instead. We usually do it this way when we integrate the magnetic force along a curved wire, and move the constant current outside the integral: $$\vec F = I \int {d \vec l \times \vec B}$$ 



#15
Jun2613, 02:26 PM

P: 4

Current is BOTH a scalar (since it has size) and vector (since it has to be moving in a certain direction). In some problems all that is asked for is the size since the direction is obvious from the circuit it is down the wire. In other problems we need to know the direction for example when Kirchhoff's laws are used.




#16
Jun2613, 06:53 PM

P: 284

Some years ago, when I was at school, the terminology changed. The term phasor (not the startrek thing), was introduced. The diagrams that showed the relationship between resistive and reactive components looked and behaved exactly like vector diagrams but were now called phasor diagrams. This terminology was introduced precisely for the reasons given above.




#17
Jun2713, 03:00 AM

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jtbell's example is of course only a special case, which by the way, is not as simple mathematically as it seems, because here we need a singular current density. What he describes is the model of an infinitely thin wire (at rest). We can describe it with help of a parameter as a curve
[tex]C:\vec{y}=\vec{y}(\lambda).[/tex] Locally, the current is running in direction of the tangent vectors and only along the infinitely thin wire. Note that here an arbitrary direction already has been chosen. The sign of the current is of course relative to this choice of direction. The currentdensity (vector!) is thus given in terms of the current [itex]i(t)[/itex] via [tex]\vec{j}(t,\vec{x})=i(t) \int_{\mathbb{R}} \mathrm{d} \lambda \frac{\mathrm{d} \vec{y}}{\mathrm{d} \lambda} \delta^{(3)}[\vec{x}\vec{y}(\lambda)].[/tex] One should note that this is a definition of the current density that is independent of the choice of the parametrization. Again, the current is a scalar quantity und the current density is a vector field, as it should be. Using HeavisideLorentz units, the force density on a chargecurrent distribution is given by the Lorentzforce law [tex]\vec{f}(t,\vec{x})=\rho(t,\vec{x}) \vec{E}(t,\vec{x})+\frac{\vec{j}(t,\vec{x})}{c} \times \vec{B}.[/tex] The total force on the wire in a magnetic field thus is given by [tex]\vec{F}(t) =\frac{i(t)}{c} \int_{\mathbb{R}} \mathrm{d} \lambda \frac{\mathrm{d} \vec{y}}{\mathrm{d} \lambda} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \delta^3[\vec{x}\vec{y}(\lambda)] \times \vec{B}(t,\vec{x})[/tex] [tex] = \frac{i(t)}{c} \int_{\mathbb{R}} \mathrm{d} \lambda \frac{\mathrm{d} \vec{y}(\lambda)}{\mathrm{d} \lambda} \times \vec{B}[t,\vec{y}(\lambda)][/tex] [tex]=\frac{i(t)}{c} \int_{C} \mathrm{d} \vec{r} \times \vec{B}(t,\vec{r}).[/tex] This is precisely the integral jtbell has given. 


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