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Question

Asked 4th May, 2016

# What is the physical meaning of curl of gradient of a scalar field equals zero?

A vector field that has a curl cannot diverge and a vector field having divergence cannot curl.

But the curl (rotation about a point) of maximum space rate of increase in a scalar field equals zero? What does this mean?

## Most recent answer

A scalar field is single valued. That means that if you go round in a circle, or any loop, large or small, you end up at the same value that you started at. The curl of the gradient is the integral of the gradient round an infinitesimal loop which is the difference in value between the beginning of the path and the end of the path. In a scalar field there can be no difference, so the curl of the gradient is zero.

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## All Answers (21)

University of Santiago de Compostela

Dear Suhas,

There are no physical meaning behind so mathematical identity, which is in fact a very special application of the Poincare's lemma: the inner product of a derivative by its co-derivative is always zero if you are working in simple connected differential manifold.

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Visvesvaraya National Institute of Technology

Dear Suhas and Daniel,

The physical interpretation involves the understanding of rotation and convergence.

**At any point**some physical quantity that exhibits the rotation, at that point it can never converge or diverge.For a scalar field which varies in space, its variation is direction dependent as it may vary differently in different directions of the space. The peak variation (or maximum rate change) is a vector represented by the gradient. Curl of gradient is zero-> means the rotation of the maximum variation of scalar field at any point in space is zero.

Tezpur University

*"Curl of gradient is zero-> means the rotation of the maximum variation of scalar field at any point in space is zero. "*

This means the gradient of a scalar field does have a net divergence/convergence, but no rotation, right?

FH Münster

Hello Suhas, in case you like to think pictorially: Imagine a twodimensional scalar field. The associated gradient field consist of vectors pointing to the direction of maximum rise. If you start at a point where the scalar field has a low value and you follow the vectors, you will necessarily end up at a (local) maximum of the scalar field. Now, if the gradient field had a nonzero curl, you could follow *closed* paths which are always ascending (like a spiral staircase). But this is impossible since there is only *one* vector assigned to each point of the plane.

But generally, a vector field can have both non-zero divergence *and* non-zero curl. If you add a vector field with divergence but zero curl and a second vector field with curl but zero divergence the result is again a vector field having the divergence of the first and the curl of the second field. Besides, (one form of) the Helmholtz theorem states that every (threedimensional) vector field is completely defined by its divergence and its curl (plus a homogeneous component which in physics is assumed to be zero).

FH Münster

Correction to the last sentence of the first paragraph above: "... there is only *one* scalar assigned to each point ..."

Tezpur University

Dear Daniel sir,

*"the inner product of a derivative by its co-derivative is always zero if you are working in simple connected differential manifold."*

Can you please give me an example. I don't understand.

University of Santiago de Compostela

Dear Sushas,

If you calculate on the components of

curl grad =0

div curl =0

just us operators. It is possible to generalize this property the concept of boundary and co-boundary. In one simple connected manifold, what that says is that, that a close form is also exact. And so on. This is a famous lemma given by Henry Poincare.

A physical example where these identities are used if for defining the vector potential A, because

div B=0

in Maxwell's equations, then you can define

B = curl A

But Dirac notice that not necessarily is true GLOBALLY if you want to define a magnetic monopole. But in any case this is pure mathematics if we refer to the properties of the equality which makes the product of the derivative by co-derivative (or vice verse) zero.

University of Jyväskylä

There is no

**the**physical meaning but instead one may find many concretisations of (the abstract property) "curl grad is identically zero" into physics.One of them is easily found from electrostatics: Take a (point) charge and move it slowly around a "loop" in an electrostatic field. The work done in moving the charge is the integral of F = qE around the loop. (As is well known, E is here a gradient field.) Since there is no electric perpetual motion machine, this work has to vanish for all loops. (If there was work involved, by reverting the direction you would get work out of nothing.)

Next, take the limit of such integrals of E around the boundaries of "patches", where the patches converge towards a point and you get point wise curl E = 0. In this case one ends up with

**a**physical interpretation that follows from the conservation of energy.University of Santiago de Compostela

Dear Lauri,

The Coulomb electric field is a central field (conservative). This means that the work made by one electric charge through infinite trajectories which have the same initial and final points is the same. This is quite crazy because if you imagine the Earth as a charge Q and airplain as another q, then the airplain can be given all the tours that it wants but if it come back to the same initial airport, it doesn't do any work at all. This is only a property of the Coulomb law!

But it is not related with the pure mathematical property of the differential operators which do div.curl=0 and curl.grad=0 locally.

University of Santiago de Compostela

Dear Ariadne,

You need to take care because you have forgotten the integral sign in many exercices, for instead in ones devoted to the curl. I like to mention the Stokes theorem for given the idea of generality instead of solving integrals without too much meaning.

Tezpur University

del X del V = 0

del X (-E) = 0 -> This is true for electrostatic fields.

But for time varying fields,

del X E = - dee B/ dee t

del X (-del V) = - dee B/ dee t not equal to 0.

So, is the curl of gradient of a scalar field equals zero only

**applicable**for**static fields**?University of Santiago de Compostela

Dear Suhas,

If you have dependence of time, then the electric field depends also the time derivative of the vector potential A. The equality curl grad follows to be true but not fully applicable to this electric field because you have a new term,- curl (time derivative of time A).

Summarizing, this equality is applicable always and to every field, static or not!

University of Jyväskylä

Daniel, sure, fully agree with you. My example was a particular interpretation, since asking for what is

**the**interpretation ofcurl grad = 0

in physics is not a well established question. Compare, it would not make much sense to ask, what is

**the**interpretation ofy=az, where a is a constant

in physics. (For, there are plenty of such interpretations, but none of them is canonical.)

For an interpretation, I think it is a better idea to look into geometry. Operators curl and grad are metric counterparts of the exterior derivative d, and according to the (generalised) Stokes law

dd is identical to null

is dual to saying "the boundary of a boundary always vanishes".

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University of Santiago de Compostela

Dear Lauri,

I have written a book on the exterior algebra of differential forms, as you can see in my contributions, where the Poincaré's lemma dd=0, or co-d.co-d=0, are the co-homological basic outcoms associated to the boundary of the co-boundary also zero in the homology part. This is a very fruitful form to generalize the usual vectorial calculus usually used in Electrodynamics because it takes into account topological solutions as instantons, monopoles, etc.

I just tried to respond within the context of vectorial calculus, but clearly these mathematical identities are the iceberg of a deeper mathematics as perhaps the question has behind it.

Tezpur University

*"The equality curl grad follows to be true but not fully applicable to this electric field because you have a new term,- curl (time derivative of time A)."*

And this (dee (del X A) / dee t = dee B / dee t) cannot be equal to zero as magnetic flux is time varying.

*"Summarizing, this equality is applicable always and to every field, static or not!"*

Then how can it be applicable for time varying fields?

University of Santiago de Compostela

Dear Suhas,

Curl

**E**=-(time derivative)(curl**A**) using curl grad V=0, as it always ought to be. Thus in such a case the electric field is not conservative as the Coulombic one and therefore the energy of a particle moving within it depeds of the trajectory (not only of the initial and final points).

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University of Jyväskylä

Daniel, I'm well aware of your remarkable contributions. Found your papers already at the time when worked with my PhD.

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Creo Medical Ltd

A scalar field is single valued. That means that if you go round in a circle, or any loop, large or small, you end up at the same value that you started at. The curl of the gradient is the integral of the gradient round an infinitesimal loop which is the difference in value between the beginning of the path and the end of the path. In a scalar field there can be no difference, so the curl of the gradient is zero.

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