I can follow the proofs for these identities, but I struggle to intuitively understand why they must be true: $$$$ 1. The curl of a gradient of a twice-differentiable scalar field is zero:

$$\nabla\times\nabla U=\boldsymbol0$$

Conservative forces can be written as gradients of a scalar potential, so this means these forces are irrotational. If $U$ represents a gravitational potential, what's the intuition that the force due to gravity has no 'curliness'?

$$$$ 2. The divergence of a curl of a twice-differentiable vector field is zero: $$\nabla\cdot(\nabla\times \boldsymbol{\mathrm{A}} )=0$$

As I understand it, the curl of a vector field generates another field where each vector is orthogonal to the plane of rotation in the original field, and the divergence is a measure of net flow. Is there an intuitive explanation for why the curl field, $\nabla\times\boldsymbol{\mathrm{A}}$, shouldn't have net flow anywhere?

The idea that fields with high curl should have low divergence makes sense to me (spinning in a circle doesn't generate any outward flow), but this explanation relates the divergence and curl of the same field, not the divergence of the curl field. It seems like this explanation relies on the assumption that curl fields are intrinsically 'curly' - is this valid?

  • $\begingroup$ Several comments removed. Please remember that, if you want to answer a question, you should leave an answer, not a comment. $\endgroup$
    – ACuriousMind
    Apr 22, 2023 at 17:53
  • $\begingroup$ "...but I struggle to intuitively understand..." The problem with this kind of question is that we don't know what exactly you are asking for, or what you will accept as "intuitive." The "intuitive" explanation is presumably going to be some analogy. For example, one answerer provider you a stick-in-the-water analogy, to which you immediately objected because it was two dimensional. How are we supposed to know such an answer is unacceptable? Maybe you are struggling with the difficulty of visualizing things in three dimensions, rather than the difficulty of finding an "intuitive" explanation. $\endgroup$
    – hft
    Apr 24, 2023 at 23:06

5 Answers 5


It is common in vector calculus courses to also consider geometric definitions for the gradient/curl/divergence operators.

Recall the integral theorems: $$\int_\Gamma\boldsymbol{\nabla}\phi\cdot\mathrm d\mathbf r=\phi(\mathbf r_1)-\phi(\mathbf r_0),\qquad(\mathrm{I1})$$where $\Gamma$ is a curve with initial point $\mathbf r_0$ and final point $\mathbf r_1$, $$ \int_\Sigma\boldsymbol{\nabla}\times\mathbf A\cdot\mathrm d\mathbf S=\int_{\partial\Sigma}\mathbf A\cdot\mathrm d\mathbf r,\qquad(\mathrm{I2}) $$where $\Sigma$ is a surface with boundary loop $\partial\Sigma$ and $$ \int_V\boldsymbol{\nabla}\cdot\mathbf B\,\mathrm dV=\int_{\partial V}\mathbf B\cdot\mathrm d\mathbf S,\qquad(\mathrm{I3}), $$where $V$ is a volume with boundary surface $\partial V$.

Basically, the gradient/curl/divergence operations can be defined to be those operations which make the integral theorems $(\mathrm{I1}), (\mathrm{I2})$ and $(\mathrm{I3})$ true infinitesimally.

I give this for the divergence only: $$ (\boldsymbol{\nabla}\cdot\mathbf B)(\mathbf r)=\lim_{V\rightarrow 0}\frac{1}{V}\int_{\partial V}\mathbf B\cdot\mathrm d\mathbf S, $$ where $V$ is some volume centered around the point $\mathbf r$, and the limit $V\rightarrow 0$ is taken such that the volume is contracted onto the point $\mathbf r$.

In short, these operations can be understood through understanding the integral theorems $(\mathrm{I1})$, $(\mathrm{I2})$ and $(\mathrm{I3})$.

First, let's take $\mathbf A:=\boldsymbol{\nabla}\times\boldsymbol{\nabla}\phi$ for some scalar function $\phi$, and let $\Sigma$ be some surface. Then from $(\mathrm{I2})$ and $(\mathrm{I1})$ we get $$ \int_\Sigma\mathbf A\cdot\mathrm d\mathbf S=\int_{\partial\Sigma}\boldsymbol{\nabla}\phi\cdot\mathrm d\mathbf r=\int_{\partial\partial\Sigma}\phi, $$ where $\partial\Sigma$ is the curve bounding $\Sigma$, and $\partial\partial\Sigma$ is the boundary of the curve $\partial\Sigma$, defined formally as "end point $-$ starting point". The last "integral" is interpreted simply as $\int_{\partial\partial\Sigma}\phi=\phi(\mathbf r_1)-\phi(\mathbf r_0)$, where $\mathbf r_1$ and $\mathbf r_0$ are the final and initial points of the curve.

But because $\mathbf A\equiv 0$, this latter "integral" vanishes for all possible choices of scalar field $\phi$, hence we must conclude that $\partial\partial\Sigma=0$. Or stated differently, the boundary of the boundary curve of a surface is zero, i.e. the boundary curve is always closed.

Now let $\psi:=\boldsymbol{\nabla}\cdot\boldsymbol{\nabla}\times\mathbf A$ for some vector field $\mathbf A$, $V$ some volume and apply (I3) and (I2) twice to get $$ \int_V\psi\,\mathrm dV=\int_{\partial V}\boldsymbol{\nabla}\times\mathbf A\cdot\mathrm d\mathbf S=\int_{\partial\partial V}\mathbf A\cdot\mathrm d\mathbf r=0, $$ where this is zero for all choices of $\mathbf A$ due to the identity $\mathrm{div} \circ \mathrm{curl}=0$, and here $\partial\partial V$ is the curve bounding the boundary surface $\partial V$.

Hence we must again conclude that the boundary curve of a boundary surface is zero, i.e. the boundary surface is always closed.

The identities $\boldsymbol{\nabla}\times\boldsymbol{\nabla}\phi=0$ and $\boldsymbol{\nabla}\cdot\boldsymbol{\nabla}\times\mathbf A=0$ are simply infinitesimal expressions of these geometric facts.

It might be illuminating to look at the higher dimensional/more general version of this situation. Let $M$ be an $n$ dimensional "smooth shape" (differentiable manifold). On $M$ we interpret for each $0\le k\le n$ differential $k$-forms whose set is denoted $\Omega^k(M)$. A differential $k$-form is a field on $M$ that can be integrated over $k$-dimensional surfaces. A zero form (an element of $\Omega^0(M)$) is just a smooth function

There is a differential operator $\mathrm d:\Omega^k(M)\rightarrow\Omega^{k+1}(M)$ called the exterior derivative which raises the degree of a differential form by one and it squares to zero: $\mathrm d\mathrm d=0$.

If $M\subseteq\mathbb R^3$ is an open subset of $\mathbb R^3$, then we have the following identifications:

  1. $0$-forms = scalar fields
  2. $1$-forms = vector fields
  3. $2$-forms = vector fields
  4. $3$-forms = scalar fields
  5. $\mathrm d$ mapping $0$-forms to $1$-forms = gradient
  6. $\mathrm d$ mapping $1$-forms to $2$-forms = curl
  7. $\mathrm d$ mapping $2$-forms to $3$-forms = divergence.

Thus $\mathrm d$ generalizes the gradient/curl/divergence into a single operator. We also have the generalized Stokes theorem: $$ \int_\sigma \mathrm d\omega=\int_{\partial\sigma}\omega, $$ where $\omega\in\Omega^{k-1}(M)$ is a $k-1$-form and $\sigma$ is a "k-dimensional surface" (actually something more general called a smooth singular $k$-chain, but I am being a bit handwavy here).

We can also define spaces $\Delta_k(M)$, which consist of formal linear combinations of $k$-dimensional surfaces in $M$ (actually, these are the "smooth singular $k$-chains", but what I say is correct intuitively at least). We have the boundary operator $\partial:\Delta_k(M)\rightarrow\Delta_{k-1}(M)$ which decreases the dimension of the chain by one and squares to zero, i.e. $\partial\partial \sigma=0$.

Let us say that a $k$-cochain in $M$ is a linear function on $k$-chains, i.e. if $\phi\in \Delta^k(M)$ is a $k$-cochain then for each $k$-chain $\sigma\in\Delta_k(M)$, the expression $\phi(\sigma)$ is a real number linear in $\sigma$.

It then follows from linear algebra that the boundary operator $\partial$ has a transpose $\partial^\ast:\Delta^k(M)\rightarrow\Delta^{k+1}(M)$ on cochains which increases the degree by one, squares to zero and is determined by $(\partial^\ast\phi)(\sigma)=\phi(\partial\sigma)$.

If $\omega\in\Omega^k(M)$ is a differential $k$-form, then the map $\sigma\mapsto\int_\sigma\omega$ is a linear function on $k$-chains, hence $\omega$ determines a unique $k$-cochain. Not every cochain is of this form, however. But we see through the generalized Stokes theorem that the coboundary operator $\partial^\ast$ - when acting on cochains determined by differential forms - is just the exterior derivative operator: $\partial^\ast\omega=\mathrm d\omega$.

So we have first the relation $\partial\partial=0$ which is a geometric fact expressing that the boundary of a boundary is zero, then we have $\partial^\ast\partial^\ast=0$ from this by algebraic duality, and then we have $\mathrm d\mathrm d=0$ since $\mathrm d$ is just $\partial^\ast$ when acting on cochains determined by differential forms. Thus $\mathrm d\mathrm d=0$ is just an expression of the geometric relation $\partial\partial=0$.


A simplification of curl is to consider a vector field representing the direction and speed of water flowing. If you throw a stick across the flow and it rotates, then the vector field has curl. This can be because the path of the water turns. It can be because water speed on one side of the flow is larger than on the other side. Either way, the curl vector of the field "passes through" the "curling effect" of the field. In terms of the water surface, just as much curl vector comes in from blow as exits above. Because just as much enters as leaves, the divergence of the curl should be zero.

  • $\begingroup$ I think my issue here is that the water surface is two-dimensional so the curl vector field is essentially a scalar field, which makes the example very intuitive but I struggle to generalise it to three dimensions. $\endgroup$ Apr 22, 2023 at 11:35
  • $\begingroup$ At any 3-D point, you can separate the motion into three perpendicular planes, perhaps considered as projections onto the planes.The curl in the field projection of a plane will correspond to the component of the curl vector perpendicular to, or through, that plane. Curl in the z-direction depends only on vector components in the x-y plane, the plane perpendicular to the z-axis. A similar argument applies to the x and y components of the curl vector. $\endgroup$ Apr 24, 2023 at 0:22

A more physical intuition for the first identity:

One of the important properties of a conservative field is that the total work done by that field on a particle traveling in a closed loop is zero. If gravity were not a conservative force field, then it would be possible to travel in a closed path and gain arbitrary amounts of kinetic energy for free--just keep looping until you have enough speed to get to where you're going in the universe. Since energy is not conserved in this system, this version of gravity cannot be a conservative force field.

Force fields with non-zero curls have such paths. A non-zero curl indicates that there exists closed-loop paths where the force field mostly points in the same direction as the path. More technically, the line integral $\oint \vec{F}\cdot d\vec{r}$ is non-zero.


I like the answer involving the Stokes theorem. Another approach looks at these as symmetric and anti-symmetric tensors cancelling each other out.

Rushed intro to tensors

Assuming that you know what a scalar field and vector field are, a covector field is a linear map from vector fields to scalar fields. These can be thought of in $D$-dimemsional space as a family of regularly spaced parallel $D-1$ dimensional hyperplanes, a vector pushes through 5½ of these planes and thus the inner product is 5½... See Wiki’s linear form article for some illustrations etc.

Given some $D$ basis vectors you can apply a covector to each of these to get $D$ scalars which represent the covector... this eventually leads to the idea of a “dual basis” for the covectors, the basis covector conjugate to basis vector $\hat e_n$ is the plane spanned by all the $\hat e_i,~~i\ne n$, the planes are separated by just enough that $\hat f^i(\hat e_i) = 1,$ or some textbooks use some other constant like 2π etc.

A multi-linear map from $m$ covector fields and $n$ vector fields to a scalar field is an $[m,~n]$-tensor field, depicted with $m$ upper indices and $n$ lower indices, which we use to label its function argument slots, a sort of “wiring diagram” called “abstract index notation.” Since a vector can be used to take covectors to scalars (by applying them to itself) vectors are all [1 0]-tensors $v^\bullet$ with one raised index, similarly covectors are all [0 1]-tensors $u_\bullet$ with one lowered index, and we wire them together (feed the vector to the covector to produce a scalar) by using the same index label for the top and bottom, so $u_a v^a$ is a scalar.

The “dot product” is a [0 2]-tensor $g_{\bullet\bullet}$ called the metric that takes two vectors and gives us their inner product, and it is symmetric in its arguments, meaning that if you give the arguments in reverse you get the same dot product between them, $g(\mathbf u,\mathbf v) = g(\mathbf v,\mathbf u)$ for all $\mathbf u,\mathbf v.$ With abstract indices we can write this without dummy vectors inserted as $g_{ab} = g_{ba}.$ This also corresponds to a “symmetric matrix” via the basis of the vector space as follows: we extract out components $g_{12} = g(\hat e_1,\hat e_2)$ and this gives us a matrix of numbers; that matrix is symmetric along its diagonal. The assumption that the tensor can be perfectly reconstructed from these components can be formally phrased as follows: an $[m, n]$-tensor can always be represented as a finite sum of terms, each of which is an “outer product” of $m$ vectors and $n$ covectors. This then allows you to do tensor contraction, which generalizes the notion of the trace of a matrix in a geometric way. To contract a tensor we repeat an index in the top and bottom, like $u_a = T^b_{\phantom bab}$, this means “express $T$ as a finite sum of outer products, feed all of the vectors that are looking for a covector as the first argument, to all of the covectors that are looking for a vector as the third argument, this creates a sum of scalars times covectors, which sums to a covector.”

The metric tensor can be viewed as a canonical mapping of vectors to covectors, this is a bijection and there is an inverse metric tensor $g^{\bullet\bullet}$ which undoes it. The use of these allows us to raise or lower indices on a tensor ad-hoc.

If all of that mostly made sense to you then we can proceed with the answer! If any of it did not, I have tried to insert a bunch of links where you could learn more about tensor fields.

Differential operators

Taking derivatives of tensor fields is accomplished with a covector operator $\nabla_\bullet$, and the mathematical structure your manifold needs to do it is called a connection. This basically means that there are several good choices for $\nabla$ when you are looking only at what differential operators do, and so you have to impose some extra axioms like “the metric tensor has derivative zero” to get a special $\nabla$ called the “Levi-Civita connection” which always behaves a bit more normally.

What about cross products? In $D$-dimensional space an orientation takes $n$ vectors and produces a scalar, and the sign of that scalar tells you whether that combination of vectors was right handed or left handed. So this is a $[0, D]$-tensor field $\epsilon_{\bullet\bullet\dots\bullet}$.

This latter tensor is anti-symmetric, $\epsilon_{abc} = - \epsilon_{acb} = +\epsilon_{cab},$ swapping any two arguments swaps the sign. And the thing about contracting symmetry with antisymmetry is that it generates a lot of zeros. So let $S^{\bullet\bullet}$ be symmetric in its two indices and $A^{\bullet\bullet}$ be antisymmetric. Then the combination $$ \begin{align} Q&=S^{ab}A_{ab}\\ &=S^{ba} A_{ab}&(\text{symmetry of }S)\\ &=S^{ab} A_{ba}&(\text{relabel dummy indices }a\leftrightarrow b)\\ &=-S^{ab} A_{ab}&(\text{antisymmetry of }A)\\ &=-Q. \end{align} $$ So then $2Q=0, ~~Q=0$ follows immediately after. If there are other tensor indices, it doesn't matter, you get a zero tensor field instead of a zero scalar field.

The last thing you need to know about the orientation is that it is a general analogue for the cross product. So if you choose $\epsilon(\hat e_1,\hat e_2,\hat e_3)=1$ for your orientation in 3D then you get a cross product between two vectors $\mathbf w=\mathbf u\times\mathbf v$ in this notation as $$w^a = g^{ab}~ \epsilon_{bcd} ~u^c~v^d.$$(Note that in this wiring diagram the symmetric $g$ only has one wire going into the antisymmetric $\epsilon,$ not two as we saw with $S$ and $A$. So this is just converting a covector, which is the actual natural form of the cross product, a covector anticipating the scalar triple product, into a vector. No zeros here!)

Then your two expressions are written, $$ \epsilon^{abc}\nabla_b\nabla_c\phi = 0^a $$ The symmetry of second partial derivatives here is something we imposed on the Levi-Civita connection (we insist that it not have “torsion”), so $\nabla_b\nabla_c\phi$ is a symmetric tensor that we have wired into twice from an antisymmetric tensor. Zero! And, $$ \nabla_a~\epsilon^{ab}_{\hphantom{ab}c}~\nabla_b A^c = 0$$ This seems to require a little more, it requires that the orientation tensor has zero derivative when we use the Levi-Civita connection in our manifold, which seems plausible but I don't see an easy proof. (Something maybe like how $\epsilon^{abc}\epsilon_{adf} =\delta^b_d~\delta^c_f -\delta^b_f~\delta^c_d,$ take a derivative of both sides, you know that the derivative of the labeling isomorphism $\delta$ is zero because $\delta^a_c = g^{ab}~g_{bc}$ and the left hand side derivatives don't cancel, can maybe operate on it with another $\epsilon$ and wrestle it onto submission?)

But assumimg that it is because in all of the usual cases it seems to be, then we have the same thing, after the product rule then $\epsilon^{ab}_{\hphantom{ab}c}~\nabla_a~\nabla_b A^c$ and you have wired two symmetric indices to two anti-symmetric ones, you get the $0_c$ tensor contracted with $A^c.$

So, both of these kind of hinge on the idea that your second partial derivatives commute, and you're trying to include a curl which is anti-symmetric but also trying to contract everything together, and the symmetry contracts with antisymmetry to make zero. These relations might not have analogues in 2D where the curl is a scalar or 4D where the curl is a [0 2]-tensor, you might have the wrong number of indices to contract everything away, but in 5D or so you should get something similar?


I have been trying my very best to stay out of this question because @Bence Racskó's answer is already very good, but alas, it nags at me. Now I will be wasting a lot of time just to make a small improvement upon his answer.

Let us work backwards. The generalised Stokes's theorem that encompasses all three as special cases, is $$ \int_\sigma \mathrm d \omega = \int_{\partial\sigma} \omega $$ if $\sigma = \Gamma$ is a 1D curve starting @ $\vec r _1$ and ending @ $\vec r _2$, and $\omega$ is the zero-form, scalar function $\phi$, then we get $$ \int_\Gamma \mathrm d \phi = \int_{\partial\Gamma} \phi\\ \int_\Gamma \vec{\mathrm d r} \cdot \vec{\nabla\phi} = \phi ( \vec r _2 ) - \phi ( \vec r _1 ) \tag{Fundamental Thm of Calculus} $$ if $\sigma = \Sigma$ is a 2D surface with boundary loop $\partial\Sigma$, and $\omega = \vec A \cdot \vec{\mathrm d r}$ is a one-form, $$\tag{Stokes's Theorem} \int_\Sigma \vec{\mathrm d S} \cdot \vec\nabla \times \vec A = \int_{\partial\Sigma} \vec{\mathrm d r} \cdot \vec A $$ and finally, if $\sigma = V$ is a 3D volume with boundary surface $\partial V$, and $\omega = \vec{\mathrm d S} \cdot \vec B$ is a two-form, then $$\tag{Gauß's Law} \int_V \mathrm d V\ \vec\nabla \cdot \vec B = \int_{\partial V} \vec B \cdot \vec{\mathrm d S} $$ Now, if you accept Bence's answer, which boils down to ``There is no boundary of a boundary", which is, in itself, a rather geometric concept written in rigorous maths, then you have a stronger answer than what I am about to present, because I will be assuming nice and smooth stuff for the sake of clarity. I just felt that it is somewhat sad to reduce a beautiful geometric concept to abstract manipulation.

Let us now look at the geometric understanding of the grad. The stereotypical understanding of $\phi$ is the electrostatic potential, which, in turn, was a borrowed concept from gravitational potential. This is a function, everywhere in space, that gives a height. You can visualise smooth hills and valleys of any complexity. The nice part of being able to do this is that you can leverage that you know how hills and valleys are like, and maybe you might know about contour lines. The gradient $\vec{\nabla\phi}$ is a vector field that always points towards the hill-tops and away from valleys, in the direction of steepest ascent, and thus must always be perpendicular to contour lines.

Now let us prove that curl grad vanishes. That means we substitute $\vec A = \vec{\nabla\phi}$ into Stokes's Theorem and try to define things by considering small neighourhoods of every point in space. That is, we pick a very small area patch $\Sigma$, which we must first insist not to be malformed, i.e. should be reasonably circular or squarish, and use the average of that integral to define its value at that point. We will never actually use the surface integral, instead using the boundary loop integral to define the value of the integral. Now, if we happened to have picked a hilltop or valley bottom, then obviously we could have chosen the necessarily existing infinitesimal contour line surrounding the hilltop or valley bottom, and obviously we get a zero because of the perpendicularity of the gradient and the contour lines as mentioned above. If we do not so happen to pick those special spots, then we can break the boundary loop integral into two parts, and that then allows us to apply the Fundamental Theorem of Calculus to the two parts individually. In general, each of the two parts has a higher and lower end points. One part goes on a path from the lower end point to the higher end point, and the other part is a different path from the higher end point to the lower end point. The important thing is that such a sum necessarily cancels because they are the same end points, cancelling in pairs. Thus, the boundary loop integral of the gradient of any scalar function is necessarily zero, which means that the surface integral about any neighbourhood of any point of a curl of a gradient is zero too. This then implies that the curl of a gradient is zero. i.e. $\vec\nabla\times\vec{\nabla\phi}=\vec0$

Continuing on, we should now discard the link between the vector field $\vec A$ from the gradient of any function. Let us study what the curl of a general vector field is. The three components of $\vec\nabla\times\vec A$ at any point are the integral over small circles (boundaries of discs) aligned in the $yz,\ zx\ \&\ xy$ planes respectively of the vector field $\vec A$ centred at the point. Note that if $\vec A$ has any constant vector field parts (think of uniform wind) or any convergence divergence (think of the gradient field hilltop above, which would always be perpendicular to contour lines, and contour lines near smooth hills and valleys are always approximately circular), then those contributions give zero in the loop integral. The only parts that gives a contribution is when the vector field itself has any swirliness, precisely why it is called the curl.

Now we try to understand why div curl is zero. For this purpose, it is preferable to consider a small cube centred at any point to be the definition. A cube has six surface area squares, and each square is made of four lines. The curl at the centre of the surface area squares is defined as the loop integral of its four lines. The divegence of a general vector field is the surface integral of the top minus the bottom, for some convention, right minus the left, and front minus the back, added together. The divergence of the curl is thus some sum and difference of the loop integrals around the edges of the cube. Everything here is linear, so we can superpose solutions. Let us consider just assigning vectors for the top and bottom squares, leaving the vertical lines field-less. If the top and bottom squares are getting integrals that rotate around in the same direction, then the top and bottom integrals cancel, and the sides integral also cancel out, so that is fine. If the top and bottom are rotating in the opposite direction, then although the top and bottom integrals reinforce, the sides integrals will get just enough contributions to cancel those. Hence, any superposition of the "same rotation direction" and "opposing rotation direction", will cancel out. That allows us to have any distribution of vector fields on the top and bottom edges as we like. Since we may also superpose the side faces too, this means that any general distribution of vectors along the edges of the cube will always get a zero integral. This proves that $\vec\nabla \cdot \vec\nabla \times \vec A = 0$

Of course, this complicated argument is far less easy to understand than just seeing that the spherical surface integral of the curl = two hemispherical surface integrals of the curl, and that turns into the criss-crossing equatorial integrals, cancelling out. But this is really just the ``there is no boundary of a boundary" argument, which I had initially rejected as being too abstract.

After all that work, I wanted to point out that we have a notation nightmare. Of course, notation nightmares are really impossible to avoid because physicists had just been inventing new maths before mathematicians could have had the time to properly study what those new tools should be unified as.

If you know some tensors and differential geometry (especially forms), then you will know from tensors that we have such a thing as contravariance and covariance. We have been very severely abusing the notation, treating non-vectorial quantities as vectors, causing a lot of confusion. So here I will quickly cover a few facts.

Scalar functions are zero-forms. The gradient is most naturally a one-form. When we say that it "points in the direction of steepest ascent", it really cannot point anywhere because it is still a one-form. That is, in $\mathrm d \phi = \mathrm d x^a \nabla_a \phi$, the $\nabla_a \phi$ is a covariant quantity that is really more like layer cake, and cannot be pointing anywhere. Instead, we must use the metric to raise the index into a contravariant quantity, and then we can identify it with vectors, for it to point anywhere. i.e. $g^{ab} \nabla_b \phi$ is finally a vector that can point towards the direction of steepest ascent.

This extends to momentum $p$ too. Because in quantum theory we have $p_a x^a$, where position coördinates $x^a$ are contravariant by definition, the momenta must be covariant. Or better, the Noether's theorem conserves the covariant components. If you want to see where the momentum points to, then, again, you use the metric to convert it to a vector, but the contravariant version of the momentum have components that are NOT conserved. To make life incredibly sickening, the universally accepted convention is such that contravariant momentum is all positive signs, which means that covariant momentum inherits a negative sign, even though it is the one that is natural. Same with the vector potential.

The electrodynamics vector potential is actually naturally a one-form, which goes well with the momentum in both classical Hamiltonian mechanics and quantum theory. Thus, the Faraday tensor is $$F = \mathrm d A = \mathrm d x^a \wedge \mathrm d x^b F_{ab} = \mathrm d x^a \wedge \mathrm d x^b \left ( \partial_a A_b - \partial_b A_a \right )$$ Or maybe its Hodge dual. Since $\mathrm d \mathrm d = 0$, this means $\mathrm d F = 0$ and that is the internal constraint pair of the Maxwell's equations. The other pair is in $\mathrm d \star \mathrm d A = J$, where $J$ is the current three-form, just ripe for integration. Note that it has the correct units---charge density is just ripe for volume integration. If you want to find the arrow direction of this current density, you have to take the Hodge dual, $\star J$, and then raise the indices to contravariant.

That is, if you want to insist to map 1-forms and vector fields, yes, you have a one-to-one mapping. But then the E and B fields of Maxwell's electrodynamics, is really 2-forms in Minkowski spacetime, and they really are not vectors at all. Considering the curl, which, as seen in the small circles earlier, are really $yz,\ zx\ \&\ xy$ bivectors, if they are even to be sent back as contravariant quantities to look at. This is the source of why we have that silly polar v.s. axial vector nonsense. If you leave them as they are naturally, as 2-forms or as bivectors, then you will not have this problem. Note that forces, like momentum, is naturally a 1-form because for a charged particle moving with velocity $v^b$, the Lorentz force law reads as $F_a = \frac{\mathrm d\ }{\mathrm d t} p_a \propto F_{ab} v^b = F_{ab} g^{bc} p_c / E$

For completeness, the basis vectors of contravariant stuff is $\partial_a$, and for zero-forms, $\partial_a = \nabla_a$. For example, velocity vector acting on potential $\phi$ is $v^a \nabla_a \phi$.

These are the correct stuff, but I have neither the time nor space to explain why. I hope they can still help.


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