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When the del operator is involved, does the meaning of dot and cross product of vectors change? That is, $\nabla \cdot \vec h$ is defined as below:

$$\nabla \cdot \vec h = \frac{\partial h_x}{\partial x} + \frac{\partial h_y}{\partial y} + \frac{\partial h_z}{\partial z}.$$

How can this be interpreted in the form form $ \vec{a}\cdot \vec{b}=|a| |b| \cos(\theta)$?

Second part of the question is, where else is the dot product used in physics other than the definition of work and in Maxwell's equations? Is it possible to get an exhaustive list?

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    $\begingroup$ I don't think it can. The $\nabla$ is an operator, which has no meaning on its own without it being operated with another vector, so you could figure why it having a magnitude wouldn't make any sense. So, no, I don't think an interpretation in terms of $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos \theta$. $\endgroup$ – vs_292 Oct 1 '17 at 17:05
  • $\begingroup$ The $\nabla$ operator is used whenever you require a gradient of some function, or the divergence, or the circulation of a vector (curl). It forms the basis of vector calculus, and it appears in every continuity equation. (you can figure out why by working out the math.) $\endgroup$ – vs_292 Oct 1 '17 at 17:09
  • $\begingroup$ Are you interested in the general geometric definition of $\nabla_\alpha E^\beta$ in the context of differential geometry and its trace $\nabla_\alpha E^\alpha?$ It is a story which does not make very complicated use of this $\|\vec a\|~\|\vec b\|~\cos\theta$ definition so it will probably be unenlightening, but it is the geometric picture for what $\nabla$ truly does and it has the surprising drama that $\nabla_\alpha$ is only defined up to a $[1,2]$-valence tensor field. $\endgroup$ – CR Drost Oct 1 '17 at 22:37
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Yes it is the usual dot product, but no it is not $ \vec{a}\cdot \vec{b}=|a| |b| \cos(θ)$.

The thing is that $ \vec{a}\cdot \vec{b}=|a| |b| \cos(θ)$ is not a definition of the inner product, but a relationship you can derive from a definition when the vectors involved are members of $\mathbb{R}^n$ or some other field with the same algebraic properties. But $\nabla$ is a member of a very different space.

So the lesson is that it s an error to think of $ \vec{a}\cdot \vec{b}=|a| |b| \cos(θ)$ as a defining property of the dot product.

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  • $\begingroup$ I would not write $\vec{\nabla}$ since the whole point of your argument is that $\nabla$ is an operator and not part of $\mathbb R^n$, so putting an arrow on top kind of contradicts that. $\endgroup$ – JamalS Oct 1 '17 at 17:30
  • $\begingroup$ @JamalS it is much better to write $\vec\nabla$ as it is a "vector" in the sense that, if you feed it a scalar, the outcome is a vector, and you can "take the inner product" of $\vec\nabla$ with a vector to get a scalar. Of course, the symmetry property $\vec a\cdot \vec b=\vec b\cdot\vec a$ does not work so it's "not really" a vector. $\endgroup$ – ZeroTheHero Oct 1 '17 at 17:31
  • $\begingroup$ @ZeroTheHero It's not consistent notation actually, as we don't write the Laplacian $\nabla \cdot \nabla$ as $|\vec \nabla|^2$ :) $\endgroup$ – JamalS Oct 1 '17 at 17:32
  • $\begingroup$ @JamalS well... $\vec r\cdot \vec r=r^2$ is pretty common, as is $\vec\nabla\cdot\vec\nabla=\nabla^2$. I agree one must be careful but many identities of vector calculus are very analogous to identities of vector algebra. To use with care but careful usage can be quite insightful. $\endgroup$ – ZeroTheHero Oct 1 '17 at 17:35
  • $\begingroup$ I think @JamalS's is fundamentally correct, but when I was a student my instructors used that as a kind of syntactic sugar to remind the students that 'this symbol partakes of the algebra of vectors', and it seem to help my students as well. So I am in the habit of writing it so on a casual basis. $\endgroup$ – dmckee Oct 1 '17 at 17:35
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There is no interpretation of $\nabla \cdot \vec E$ as $\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta$ due to the fact that $\nabla$ is an operator and it is not part of a vector space in the way that $\vec E \in \mathbb R^3$.

The $\cos\theta$ relation is based on a geometric view of the inner product on $\mathbb R^3$ and this is not applicable as $\nabla$ as aforementioned cannot be characterised as some geometrical entity in the vector space.

We can think of the divergence as a map $\mathbb R^n \to \mathbb R$ as it produces a scalar field from a single vector field. In other contexts - for example - $\nabla^2$ can be seen as belonging to a Weyl algebra.

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Here is a complimentary point of view. In some circumstances $\nabla\cdot \vec E$ does have an interpretation (almost) as a standard dot product. Say $\vec E(x,t)$ is a plane wave $$ \vec E(x,t) = \vec E(k\cdot x - c t) $$ where $k$ is wavenumber and $c$ is wave velocity. Then, the divergence of $\vec E$ takes the form $$ \nabla\cdot\vec E = k\cdot \vec E' $$ where $\vec E'$ is the derivative of $\vec E$ with respect to the phase $k\cdot x - ct$. Then nothing holds you back from saying that $$ \nabla\cdot\vec E = \lvert k\rvert \lvert\vec E'\rvert \cos \theta $$ where $\theta$ is the angle between $\vec E'$ and $k$. Note however that $k$ and $\vec E'$ have different units and belong to two "very different spaces" as was stressed in the answer of dmckee.

If you take $\vec E$ to be a harmonic wave, things get even simpler (Think of how derivatives change under a Fourier transform).

More generally, the $\nabla$ operator, whether denoting a divergence, a curl or a Laplacian, will behave much like vectors do for plane waves. For other kinds of waves/fields, its action is more intricate but not fundamentally all that different I would say.

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  • $\begingroup$ As you say a feature of the way exponential function behave under differentiation, but a very common case so worth knowing. $\endgroup$ – dmckee Oct 1 '17 at 18:45
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@JamalS has provided the more formal reason as to why you don't want to think of $\nabla$ as a vector, but here are some reasons (convenient but not formal) as to why it can be useful to do so.

$\vec\nabla\cdot\vec h$ does have some of the meaning of a scalar product in the sense that $$ \vec a\cdot \vec b= a_xb_x+a_yb_y+a_zb_z $$ gets copied to $$ \vec\nabla \cdot \vec h=\frac{\partial}{\partial x}f_x+ \frac{\partial}{\partial y}f_y+\frac{\partial}{\partial z}f_z $$ if you think of $\vec \nabla=\hat x\frac{\partial}{\partial x}+ \hat y\frac{\partial}{\partial y}+\hat z\frac{\partial}{\partial z}$, i.e. if you think of $\vec\nabla$ as a vector with components $\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$.

Moreover, there are several vector identities that have an vector calculus interpretation if you think of $\vec \nabla$ as a vector. For instance, since $\vec a\times\vec a=0$, you can "export" this to get insight into $$ \vec\nabla\times \vec \nabla f=0\, , $$ for any scalar function $f$. You can also "export" $\vec a\cdot (\vec a\times \vec b)=0$ to intuitively understand $$ \vec\nabla\cdot \left(\vec\nabla\times \vec A\right)=0\, . $$ Also a vector multiplied by a scalar yields a vector, much like $\vec \nabla f$ is a vector.

However, not all properties of the usual scalar product can be "exported" to $\vec\nabla$. For instance, $\vec a\cdot \vec b=\vec b\cdot \vec a$ but $$ (\vec\nabla \cdot \vec h)f\ne (\vec h\cdot\vec\nabla)f\, . $$ So, while convenient to think of $\vec \nabla\cdot\vec h$ as a scalar product, and convenient to think of $\vec \nabla$ as a vector, one should do so with care else blind manipulations may get you in trouble.

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protected by Qmechanic Oct 1 '17 at 18:15

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