# What does the Gradient “with respect to a position vector” mean?

I was studying John R. Taylor's book on Classical Mechanics and he introduced a confusing concept in page 140 (2005 edition):

$$\nabla_1 = (\frac{\partial}{\partial x_1}+\frac{\partial}{\partial y_1}+\frac{\partial}{\partial z_1})$$

Where

$$\mathbf r_1 = (x_1+y_1+z_1)$$

He calls this "the gradient with respect to the coordinates of $\mathbf r_1$". In general, what is the gradient with respect to a position vector? Isn't the gradient (in physics) just dependent on where we sit our x-axis and our y-axis? In other words, I've always known the gradient as:

$$\nabla = (\frac{\partial}{\partial x}+\frac{\partial}{\partial y}+\frac{\partial}{\partial z})$$

(If context gives a clue, he was talking about the gradient of a potential $U$ in an isolated, two body system where the positions of the two particles are $\mathbf r_1$ and $\mathbf r_2$, and the claim is that the force on particle 1 due to particle 2 $\mathbf F_1 = -\nabla_1U(\mathbf r_1-\mathbf r_2)$. The same would apply for particle 2 interchanging 1 with 2 in the last formula.)

• Is it a directional derivative? I'm asking this since I just found that it has the same notation but the book does not mention that word a single time. – Francisco Russo Apr 20 '18 at 0:39

Your formula for gradient works for a function that depends on position $(x,y,z)$. But, position of what? In this situation, there are two particles, each with its own $(x,y,z)$, so your formula doesn't make sense.

The potential $U$ depends on the positions of both particles. $\nabla_1$ just means he's keeping the second particle fixed, to take the derivative with respect to the first particle's position. It's a form of "partial derivative". (I haven't read the book, though.)

A directional derivative is a scalar, but this gradient is a vector (as any force must be).

The force equation is saying that the first particle accelerates in the direction that would decrease the potential energy of the system.

• Precisely, what do you mean by taking the derivative with respect to a position? That's where I'm having trouble. Is this just a "trick" (such as labeling the direction of the motion of an object the x-axis) to simplify the problem by translating r2 to a desired place? – Francisco Russo Apr 20 '18 at 1:30
• @FranciscoRusso -- The potential is a function $U = f(x_1,y_1,z_1,x_2,y_2,z_2)$. You can take the partial derivatives of a function. I don't quite understand your question. – mr_e_man Apr 20 '18 at 1:40
• Okay the first paragraph really cleared up everything. – Francisco Russo Apr 20 '18 at 1:44 This picture illustrates the meaning.

According to [Page 8, 2], "The gradient of a vector $\vec {b}$ results in a tensor $\textbf T$:

$$grad \ \vec {b} = \nabla \otimes \vec {b} =$$

\begin{bmatrix}\frac{\partial b_{x}}{\partial x}&\frac{\partial b_{y}}{\partial x}&\frac{\partial b_{z}}{\partial x} \\ \frac{\partial b_{x}}{\partial y}&\frac{\partial b_{y}}{\partial y}&\frac{\partial b_{z}}{\partial y} \\ \frac{\partial b_{x}}{\partial z}&\frac{\partial b_{y}}{\partial z}& \frac{\partial b_{z}}{\partial z} \end{bmatrix}

You would just have to multiply the potential if it is a scalar.

On a side note:

To be honest, I am am a newbie and not quite sure about your situation, but it reminds me of the divergence of a vector[1-3]. The divergence of a vector gives a scalar[Chapter 2, 2].

$$div \ \vec {r} =\nabla \bullet \vec {r} = \sum_{i=1}^{3} \frac{\partial}{\partial x_{i}} r_{i} = \frac{\partial r_{1}}{\partial x_{1}} + \frac{\partial r_{2}}{\partial {x_{2}}}+\frac{\partial r_{3}}{\partial x_{3}} = \frac{\partial r_{1}}{\partial x} + \frac {\partial r_{2}}{\partial y}+ \frac{\partial r_{3}}{\partial z}$$

Since I am talking about divergence, I will provide a definition from Anton.

div $\vec {F}(x,y,z) =$ flux density of $\vec {F}$ at (x,y,z). The flux density at (x,y,z) is a measure of the rate at which fluid is approaching or departing from point (x,y,z).

I hope this helps a little.

Edit: Also, I believe your potential with the vector might be expanded by the following product rule if U is a scalar.

$$\nabla \bullet (\vec {r}U) = \vec {r} \bullet \nabla U + U \nabla \bullet \vec{r} \ [Page \ 9,2]$$

References:

 Tobias Holzmann. Mathematics, Numerics, Derivations, and OpenFOAM. Retreived (2017, Nov 17). Available from: holzmann-cfd[de]. Available from: https://holzmann-cfd.de/publications/mathematics-numerics-derivations-and-openfoam ; Available from: https://www.scribd.com/document/371630911/Mathematics-Numeric-s-Derivations-and-Open-Foam

 Anton, Howard. (1992). Calculus with analytic geometry-4th Edition. Anton Textbooks, Inc.

• Thanks for the answer; but it is not quite what I was looking for; my problem is about a gradient (not divergence). I really appreciate your answer though. – Francisco Russo Apr 20 '18 at 1:36

I always think this as the following; (especially when we talk about internal forces that are derivable from a potential)

Consider a mass at infinity from the origin of our chosen reference frame.

Q1: What is the magnitude of the force field at the origin by the gravity of the mass at infinity ? In other words, the the force that unit mass at origin would feel.

It is zero.

Q2: What about when the mass at infinity comes to position $$x=1$$ ?

It is not zero.

Hence it is clear that, we cannot express $$V$$ as $$V=V(x,y,z)$$ independent of the position of the particles in our system.

To be clear, there might be cases where we can express $$V$$ without the position of the particles, but in that case, $$V$$ also needs to depend on the times, and the dynamics of the system, which will be dependent on the initial conditions etc., so it is always much easier to consider the $$V=V(\vec r_1, \vec r_2, ..., \vec r_N)$$.

That is why, we need the partial derivatives of $$V$$ wrt $$x_1, x_2, x_3,$$ while keeping the the other position of the particles constant, so that we can calculate the work done on a single particle by total force acting on the particle.

This might be outdated. But I just happened to see it so I try to show my opinion and welcome to correct me if something is not going right. I will try to use most common notation, for instance position column vector $$\mathbf r=[x,y,z]^T$$, so the gradient of the position vector $$\begin{bmatrix}\frac{\partial {x}}{\partial x}&\frac{\partial {y}}{\partial x}&\frac{\partial {z}}{\partial x} \\ \frac{\partial {x}}{\partial y}&\frac{\partial {y}}{\partial y}&\frac{\partial {z}}{\partial y} \\ \frac{\partial {x}}{\partial z}&\frac{\partial {y}}{\partial z}& \frac{\partial {z}}{\partial z} \end{bmatrix}=[\mathbb I]=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}.$$

This can be used to prove the identity which is seemingly impossible:

$$(\mathbf v\cdot\nabla)(\mathbf r\times\mathbf c)=\mathbf v\times \mathbf c.$$ where $$\mathbf c$$ is a constant vector.

The proof can be $$(\mathbf v\cdot\nabla)(\mathbf r\times\mathbf c)=v_\ell\partial_\ell\varepsilon_{ijk}r_jc_k=\varepsilon_{ijk}v_\ell c_k\partial_\ell r_j=\varepsilon_{ijk}v_jc_k=\mathbf v\times \mathbf c$$.