Vector calculus — Computing this Divergence

[jorgeluisharo]

Homework Statement

Show that if $\vec{R}=\vec{r}-\vec{r'}$ and $f$ it's a function with good behaviour, then
$$\nabla\cdot\vec{ f(R)}=-\nabla '\cdot\vec{f(R)}$$

Relevant Equations

$$\nabla\cdot\vec{ f(R)}=-\nabla '\cdot\vec{f(R)}$$

I really don't know how to proceed if I'm not using an specific coordinate system, Is there a way of doing this using only indices, in general form?

 

[PeroK]

It's the chain rule. Note that $\nabla$ is defined in Cartesian coodinates:
$$\nabla \cdot \vec f (\vec R) \equiv \frac{\partial}{\partial x}(f_x (\vec R)) + \frac{\partial}{\partial y}(f_y (\vec R)) + \frac{\partial}{\partial z}(f_z (\vec R)) $$

 

[Gaussian97]

Your notation is a little bit confusing. PeroK seems to understand that your $f$ is a vector field and you are asking about divergences.
In my case the first time I read it I understood that $f$ is a scalar field and you want to compute the gradient.
Can you please clarify exactly what do you have to compute?
Anyway, the solution should not be too different in both cases, but there may be minor details.

 

[PeroK]

It's the chain rule. Note that $\nabla$ is defined in Cartesian coodinates:
$$\nabla \cdot \vec f (\vec R) \equiv \frac{\partial}{\partial x}(f_x (\vec R)) + \frac{\partial}{\partial y}(f_y (\vec R)) + \frac{\partial}{\partial z}(f_z (\vec R)) $$

Or:
$$\nabla f (\vec R) \equiv (\frac{\partial }{\partial x}(f (\vec R)), \frac{\partial}{\partial y}(f (\vec R)), \frac{\partial}{\partial z}(f (\vec R)) $$

 

[jorgeluisharo]

It's the chain rule. Note that $\nabla$ is defined in Cartesian coodinates:
$$\nabla \cdot \vec f (\vec R) \equiv \frac{\partial}{\partial x}(f_x (\vec R)) + \frac{\partial}{\partial y}(f_y (\vec R)) + \frac{\partial}{\partial z}(f_z (\vec R)) $$

Thanks, but is there a way of writing and proving this without the use of cartesian coordinates, I mean in a $n$ dimensional space?

 

[jorgeluisharo]

Your notation is a little bit confusing. PeroK seems to understand that your $f$ is a vector field and you are asking about divergences.
In my case the first time I read it I understood that $f$ is a scalar field and you want to compute the gradient.
Can you please clarify exactly what do you have to compute?
Anyway, the solution should not be too different in both cases, but there may be minor details.

Thank you, I want to compute the divergence, f indeed is a vector field.

 

[Gaussian97]

Thanks, but is there a way of writing and proving this without the use of cartesian coordinates, I mean in a $n$ dimensional space?

There are formulas for the divergence in a general coordinate system, but why would you want to complicate your life when you can do the proof in cartesian coordinates? The generalization of cartesian coordinates to $n$ dimension should be rather obvious.
Anyway, I recommend you to prove it first in 3D cartesian coordinates, once you have proven it there, the generalization to $n$ dimension and other coordinates is not complicated.

 

[ergospherical]

It might be easier to understand if you write this as
\begin{align*}
\mathbf{f}(\mathbf{R}) = \mathbf{f}(R_1, R_2, R_3) = \mathbf{f}(x-x', y-y', z-z')
\end{align*}Then\begin{align*}
\dfrac{\partial \mathbf{f}}{\partial x} = \dfrac{\partial \mathbf{f}}{\partial R_1} \dfrac{\partial R_1}{\partial x} = \dfrac{\partial \mathbf{f}}{\partial R_1}
\end{align*}whilst\begin{align*}
\dfrac{\partial \mathbf{f}}{\partial x'} = \dfrac{\partial \mathbf{f}}{\partial R_1} \dfrac{\partial R_1}{\partial x'} = - \dfrac{\partial \mathbf{f}}{\partial R_1}
\end{align*}etc.

 

[PeroK]

Thanks, but is there a way of writing and proving this without the use of cartesian coordinates, I mean in a $n$ dimensional space?

How do you define divergence in the first place? It's defined in Cartesian coordinates. And, Cartesian coordinates extend to $n$ dimensional space. Although, curl is generally not so well defined.

 

[pasmith]

If the question askes you to show x = y, then including x = y as a relevant equation isn't really all that helpful. Relevant equations here would be a definition of the divergence and the chain rule.

 

[Zexuo]

How do you define divergence in the first place? It's defined in Cartesian coordinates. And, Cartesian coordinates extend to $n$ dimensional space. Although, curl is generally not so well defined.

$$\nabla\cdot\mathbf{F} = \lim_{V\rightarrow 0}\frac{1}{V}\oint_S\mathbf{F\cdot n}da$$
$$\nabla\times\mathbf{F} = \lim_{V\rightarrow 0}\frac{1}{V}\oint_S\mathbf{n\times F}da$$

No coordinate system needed... until you want to solve a problem.