# How is dot or cross product possible using the del operator?

Yesterday in class my teacher told me that the del operator has a direction but no value of its own (as its an operator). So it can't be called exactly a vector. But in vector calculus we see that div $\phi$ is the dot product of the del operator and $\phi$. Also in Calculating curl we use the cross product involving the del operator.

If the operator isn't a vector, then how can we have a dot or cross product of it? What is the direction of the operator?

• The $\nabla$ is not a value vector, but a vector of operators. As in $$\nabla = \pmatrix{ \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} }$$ May 16, 2018 at 19:47

The divergence of a vector field is not a genuine dot product, and the curl of a vector field is not a genuine cross product.

$\nabla \cdot \vec A$ is just a suggestive notation which is designed to help you remember how to calculate the divergence of the vector field $\vec A$. The notation is nice, because it looks like a dot product, but as you say $\nabla$ is not actually a vector.

If it helps, you can use the alternate notation $$\operatorname{div}(\vec A) = \partial_x A_x + \partial_y A_y + \partial_z A_z$$

which makes it easier to see that $\operatorname{div}(\bullet)$ is just an operator which eats a vector field and spits out a scalar field. Curl can be defined similarly, though it's a pain to write out in its entirety.

In Cartesian coordinates, the dot and cross products look like this: $$\vec A \cdot \vec B = \sum_i A_i B_i$$ $$\left[\vec A \times \vec B\right]_i = \sum_{j,k}\epsilon_{ijk}A_jB_k$$

while the divergence and curl operations look like this: $$\operatorname{div}(\vec B) = \sum_i \partial_i B_i$$ $$\big[\operatorname{curl}(\vec B)\big]_i = \sum_{j,k}\epsilon_{ijk} \partial_j B_k$$

The striking similarity leads one to define the vector operator $\nabla$, whose components are just the partial derivatives ($\nabla_i = \partial_i$). However, as pointed out in the comments, this similarity does not generally hold up if you switch to a new coordinate system.

For example, in cylindrical coordinates $(\rho,\phi,z)$, the dot product of two vectors becomes $$\vec A \cdot \vec B = A_\rho B_\rho + A_\phi B_\phi + A_z B_z$$ just like before, but the divergence looks like this:

$$\operatorname{div}(\vec B) = \frac{1}{\rho}\partial_\rho(\rho B_\rho) + \frac{1}{\rho}\partial_\phi B_\phi + \partial_z B_z$$

• Perhaps using index notation might be useful to note the similarities between $A\cdot B$ and $\nabla\cdot A$ (and for the cross product)? May 16, 2018 at 12:13
• Not being a genuine cross product can be seen for example with curl in spherical coordinates. May 16, 2018 at 13:11
• J. Murray Can you tell me the direction of the operator? May 16, 2018 at 14:12
• @AsifIqubal An operator does not have a direction. May 16, 2018 at 14:35
• I don't think the dot product in cylindrical coordinates is $\vec A \cdot \vec B = A_\rho B_\rho + A_\phi B_\phi + A_z B_z$ because you can't add things of different units. Some are length^2 and some are angle^2. I think the dot product needs to account for the metric where $A \cdot B = A^\top \mathrm{ M} B$ where $$\mathrm{M} = \left| \matrix{1 & & \\ & {r^2} & \\ & & 1} \right|$$ May 16, 2018 at 19:36

I love this question and was very curious about it, so I built on a 3blue1brown video to answer it with the video below. I’d argue one doesn’t fully understand/appreciate divergence, curl or Maxwell’s equations unless they get this. Here’s the nutshell version:

This connection is difficult to conceptualize because like you said, the del operator, $$\nabla ,$$ is not a typical vector. Like a parasite or virus, it is meaningless alone, and needs a ‘host’ to ‘operate’ on. While you can’t think of it as a real vector, you can treat it like one and use the standard processes for dot and cross products. This yields the equations for the divergence and curl as shown below, which simply boil down to finding 4 components: $$\frac{\partial P}{\partial x}, \frac{\partial Q}{\partial y}, \frac{\partial Q}{\partial x},$$ and $$\frac{\partial P}{\partial y}$$.

If your $$\frac{\partial P}{\partial x}$$and $$\frac{\partial Q}{\partial y}$$ are positive, this means the $$x$$ and $$y$$ components of your vectors are getting larger when you move in the $$x$$ and $$y$$ direction, respectively, which corresponds to positive divergence. If your $$\frac{\partial Q}{\partial x}$$and $$\frac{\partial P}{\partial y}$$ are positive and negative, respectively, this means the $$y$$ and $$-x$$ components of your vectors are getting larger when you move in the $$x$$ and $$y$$ direction, respectively, which corresponds to counterclockwise rotation, or positive curl.

The vector field illustrated below has a larger positive $$\frac{\Delta P}{\Delta x}$$ value than it’s negative $$\frac{\Delta Q}{\Delta y}$$value, corresponding to a slightly positive divergence. It also has a positive $$\frac{\Delta Q}{\Delta x},$$and negative $$\frac{\Delta P}{\Delta y}$$, corresponding to a positive (counterclockwise) curl.

I attempt to explain this more clearly in the video if you’d like to check it out, and let me know if you’d like to discuss further! https://youtu.be/k7WyPNWerN0

• I think its important to point out that its not that $\nabla \cdot A$ is merely suggestive, it is completely correct. Its the definition of $\nabla$ that is untrue. I mean of course it is correct in Cartesian coordinates as you say, just not for other coordinate systems. But that is only because you do not use the full definition of $\nabla$ which is $\nabla_\alpha = \partial_\alpha + \Gamma^\beta_{\alpha\gamma}$. Just because it is merely an operator, doesn't make it any less a vector. Apr 30, 2020 at 16:54
• Thanks for the clarification. I iunderstand the definition doesn't transfer to other coordinate systems. What's the $\Gamma_{\alpha\gamma}^{\beta}$ symbol refer to? I'd say it's not a vector in the same way that the $\partial$ symbol is not a variable. I mean, 2D vector fields have vectors associated with each $(x,y)$ combination. $\nabla$ obviously doesn't have that. So I would certainly say it is 'less' of a vector, where the standard definition of a vector is a "quantity having direction as well as magnitude", obviously, $\nabla$ has neither. May 4, 2020 at 19:29
• To go along with the virus analogy, the definition of 'life' includes the ability to reproduce. A virus cannot reproduce unless it first attaches itself to a 'host'. In the same way, the $\nabla$ cannot 'act' like a vector unless it first attaches itself to one. May 4, 2020 at 19:32
• The definition does transfer to other coordinate systems, if the correct definition is used, which is the covariant derivative $\nabla$. It allows you to find the derivatives of vectors in curved space. The $\Gamma$ is called a Christoffel symbol, which sort of gives a connection to finding tangent (parallel) vectors in curved space, i.e. the differential on the curve. These symbols disappear in flat space because you can always get a parallel vector. These symbols are defined by the geometry of the space, so it allows a complete, consistent definition of $\nabla$ regardless of curvature. May 4, 2020 at 19:38
• I would also argue that, the virus does indeed need a host to reproduce, it needs something to act on, but that doesn't stop it being a virus had it not found a host. Similarly, just as $\nabla$ needs something to operate on, but that doesn't make it any less a vector. $\nabla$ has direction, it has dimension, I'd say it was a vector in its own right. May 4, 2020 at 19:44