Divergence is represented by dot product. How is the divergence related to dot product? And curl is represented by cross product. How is the curl related to cross product?

# [Physics] Dot product and divergence

homework-and-exercisessoft-questionVector Fieldsvectors

#### Related Solutions

They are completely different things. One returns a scalar and the other returns vector; one is commutative, the other anticommutative. You shouldn't even compare them (the question why one and not the other is similar to *why apples and not pianos*).

A derivation of a physical law doesn't *guess* what to use but actually uses a certain reasoning along the way. You need to look into theoretic linear algebra, not just physics. Each of them has a geometric meaning. The dot product involves projection, extraction of components of vectors, involves things that need to be parallel... the quantity it produces is a scalar and thus orientation-independent. It doesn't have a direction. Work is a scalar quantity. It concerns how much force is used to move an object in the direction of the force (hence, the projection). Dot product is actually very ubiquitous, most of the relationships between quantities you know involve some kind of work with vector's components. In fact, this connection to projection runs very deeply: it works in any number in dimensions, and even in abstract spaces, such as spaces of functions and spectra... the concept of projection is needed everywhere.

On the other hand, the cross product is antisymmetric - it is sensitive to orinetation of the space. It involves chiral phenomena, circulation & rotation, inherently three-dimensional phenomena that use up all three spatial dimensions. It's about *perpendicular* things instead of parallel. Its result is a vector - something orientation dependent. Torque involves rotation: it distinguishes clockwise and counterclockwise. It has an axis, perpendicular to the plane of motion, it has a direction... the force and the distance to the axis must be non-parallel if you want a torque at all. Similar examples here are everything involving magnetic field (here, the consequences are very deep and fundamental, all the way back to Maxwell's equations).

I deliberately went with the essay approach. You've got equations and stuff covered by other posters.

There is a sense where you can start with just a scalar product.

If you assume that $v^2=vv=v\cdot v=|v|^2$ holds for any vector then that is a scalar product for a vector with itself.

From that you can get the scalar product of any two vectors $v$ and $w$, $v \cdot w= \frac{1}{2}\left((v+w)^2-v^2-w^2\right)=\frac{1}{2}\left(vw+wv\right)$ where for the last equality we assume the same distributive laws as for matrices.

So the simplest assumption of having a scalar product of any vector with itself can easily give us a scalar product of any two vectors (we can use the scalar products of $v+w, v$ and $w$ with themselves).

But we can also have a general product and allow it to have the regular rules of matrix algebras (associative, distributive, multiplication, addition, scaling by scalars, etcetera). Then we note that for vectors $v$ and $w$ we have $vw=\frac{1}{2}\left(vw+wv\right)+\frac{1}{2}\left(vw-wv\right)$ or $vw=v\cdot w+\frac{1}{2}\left(vw-wv\right)$ and we can notice that the last term $\frac{1}{2}\left(vw-wv\right)$ is different than a scalar or a vector, specifically its square is negative. For vectors $v$ and $w$ we can denote $\frac{1}{2}\left(vw-wv\right)$ by $v \wedge w$.

These new objects, like $v\wedge w$ naturally represent the plane spanned by the vectors $v$ and $w$. They are useful in physics, though for historical reasons we often use the complete accident that in 3d there are vectors orthogonal to planes to represent $v \wedge w$ by the vector orthogonal to it and then call it the cross product.

When you have both products $v \cdot w$ and $v\wedge w$ you have the full information to get the product $vw=v \cdot w+v\wedge w$, and from that you can solve for $v$ or $w$ if you have the other one. For instance $v=vw\frac{w}{w^2}=\left(v \cdot w+v\wedge w\right)\frac{w}{w^2}$. So therefore if you know the scalar product and the wedge product and you know $w$ then you can find out what $v$ was. Neither of the products by themselves allow you that. Many vectors can give the same inner product, and many vectors can give the same wedge product, but knowing both can allow you know the full relative relationship between the two.

It all follows from just the scalar product, but you get the whole package if you want it.

tl;dr

The dot product (symmetric part of the one product) tells you how much they have in common. The other product (antisymmetric part of the one product) tells you how much they orthogonal, specially how much you have to rotate one line to get align it with the other, and if you don't live in just a plane it also tells you the plane in which you need to rotate to send one into the other. That directionality is something you don't get from just a scalar.

But you don't need two products. One product is enough as long as you do it the invertible way. And that way naturally creates numbers, lines, planes and even higher dimensional objects as they come up. Further details above.

## Best Answer

It is pretty much simply a short way to notate both vector field operations by looking at $\nabla$ as a vector operator by writing \begin{equation} \nabla=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right) \end{equation} in $\mathbb{R}^3$, or equivalently \begin{equation} \nabla=\frac{\partial}{\partial x}\hat{\imath}+\frac{\partial}{\partial y}\hat{\jmath}+\frac{\partial}{\partial z}\hat{k}. \end{equation} Performing this vector operator on a scalar field gives you the expression for that field's gradient, whereas applying it to a vector field via a dot product gives you the vector field's divergence (analogoulsy for the cross product, which gives you the field's curl instead).

It is important to note, however, that unlike with regular three-vectors, this expression for divergence is not commutative, as the $\nabla$ operator is not a vector in $\mathbb{R}^3$.