Gradient Question: A,B Vectors & e-xr-2 \widehat{r}

• likephysics
In summary, the conversation revolved around a vector identity and the calculation of its gradient. The original problem was to derive the vector identity and the conversation discussed the various steps and techniques involved in solving it. The final solution involved the use of the product and chain rules to calculate the gradient of a vector.
likephysics

Homework Statement

This is not a homework problem, just a question
$$\nabla$$(A.B) = (B.$$\nabla$$) A +(A.$$\nabla$$)B+Bx($$\nabla$$xA)+Ax($$\nabla$$xB)

A,B are vectors

The Attempt at a Solution

I can't make sense of the first 2 terms on the right hand side - is (B.$$\nabla$$)
just div of B?

Also, how do I solve, $$\nabla$$(e-xr-2 $$\widehat{r}$$)
Can I treat the unit vector $$\widehat{r}$$ as constant?

likephysics said:
I can't make sense of the first 2 terms on the right hand side - is (B.$$\nabla$$)
just div of B?

No, it's more complicated than that. In Cartesian Coordinates, the Del operator is

$$\mathbf{\nabla}=\hat{\mathbf{i}}\frac{\partial}{\partial x}+\hat{\mathbf{j}}\frac{\partial}{\partial y}+\hat{\mathbf{k}}\frac{\partial}{\partial z}$$

So,

$$\textbf{B}\cdot\mathbf{\nabla}=B_x\frac{\partial}{\partial x}+B_y\frac{\partial}{\partial y}+B_x\frac{\partial}{\partial z}$$

And so,

$$(\textbf{B}\cdot\mathbf{\nabla})\textbf{A}=B_x\frac{\partial \textbf{A}}{\partial x}+B_y\frac{\partial \textbf{A}}{\partial y}+B_x\frac{\partial \textbf{A}}{\partial z}$$

Also, how do I solve, $$\nabla$$(e-xr-2 $$\widehat{r}$$)
Can I treat the unit vector $$\widehat{r}$$ as constant?

The gradient of a vector is a second rank tensor. Is this really what you are trying to calculate? What is the original problem?

gabbagabbahey said:
The gradient of a vector is a second rank tensor. Is this really what you are trying to calculate? What is the original problem?

The original problem is :

Derive the Vector identity:

(Ji.$$\nabla'$$)$$\nabla'$$(e-$$\gamma$$r/r) =

[$$\gamma$$2((Ji.$$\hat{r}$$)$$\hat{r}$$+3/r($$\gamma$$+1/r)(Ji.$$\hat{r}$$)$$\hat{r}$$ - Ji/r($$\gamma$$+1/r)]e-$$\gamma$$r/r

I solved
$$\nabla'$$ (e-$$\gamma$$r/r) and got

($$\gamma$$+1/r) (e-$$\gamma$$r/r2) $$\hat{r}$$

Last edited:
likephysics said:
Derive the Vector identity:

(Ji.$$\nabla'$$)$$\nabla'$$(e-$$\gamma$$r/r) =

[$$\gamma$$2((Ji.$$\hat{r}$$)$$\hat{r}$$+3/r($$\gamma$$+1/r)(Ji.$$\hat{r}$$)$$\hat{r}$$ - Ji/r($$\gamma$$+1/r)]e-$$\gamma$$r/r

$$(\textbf{J}\cdot\mathbf{\nabla})\left(\mathbf{\nabla}\frac{e^{-\gamma r}}{r}\right)=\left[\gamma^2(\textbf{J}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}+\frac{3}{r}\left(\gamma+\frac{1}{r}\right)(\textbf{J}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\frac{1}{r}\left(\gamma+\frac{1}{r}\right)\textbf{J}\right]\frac{e^{-\gamma r}}{r}$$

^^^ Is this what you meant? I assume that $\textbf{r}$ is just the usual position vector, $r$ is its magnitude and $$\hat{\mathbf{r}}[/itex] is its direction? If so, why does your expression have primes next to the nablas? I solved [tex]\nabla'$$ (e-$$\gamma$$r/r) and got

($$\gamma$$+1/r) (e-$$\gamma$$r/r2) $$\hat{r}$$

Again, I can only assume that you mean

$$\mathbf{\nabla}\left(\frac{e^{-\gamma r}}{r}\right)=-\left(\gamma+\frac{1}{r}\right)\frac{e^{-\gamma r}}{r^2}\hat{\mathbf{r}}$$

If so, then yes, that's correct.

Are you still having difficulty carrying out the derivative $$(\textbf{J}\cdot\mathbf{\nabla})$$ on this expression?

The expressions do have primes next to nabla. The nabla operates on primed coordinates. Source is at the primed coordinate and effect is at the unprimed coordinates.

Yes, the difficulty i was having was with the J.nabla
Also, say if you try to take the gradient of
$$-\left(\gamma+\frac{1}{r}\right)\frac{e^{-\gamma r}}{r^2}\hat{\mathbf{r}}$$
how would you deal with direction vector r^

likephysics said:
The expressions do have primes next to nabla. The nabla operates on primed coordinates. Source is at the primed coordinate and effect is at the unprimed coordinates.

Surely this means that $\textbf{r}$ isn't the position vector, but rather the separation vector between the position vectors of the field and source points;

$$\textbf{r}=(x-x')\hat{\mathbf{x}}+(y-y')\hat{\mathbf{y}}+(z-z')\hat{\mathbf{z}}[/itex] ...right? Yes, the difficulty i was having was with the J.nabla Also, say if you try to take the gradient of [tex]-\left(\gamma+\frac{1}{r}\right)\frac{e^{-\gamma r}}{r^2}\hat{\mathbf{r}}$$
how would you deal with direction vector r^

First, the negative sign shouldn't be there if my above assumption is correct.

Second, you aren't taking the gradient of that (if you did, you would end up with a second rank tensor, not a vector). To take the partial derivative of something like $$f(r)\hat{\mathbf{r}}$$, you will have to use the product and chain rules. For example,

\begin{aligned}\frac{\partial}{\partial x'} \left[f(r)\hat{\mathbf{r}}\right] &= \frac{\partial f}{\partial x'}\hat{\mathbf{r}}+f(r)\frac{\partial \hat{\mathbf{r}}}{\partial x'} \\ &= \left(\frac{\partial f}{\partial r}\right)\left(\frac{\partial r}{\partial x'}\right)\hat{\mathbf{r}}+f(r)\frac{\partial}{\partial x'}\left[\frac{(x-x')\hat{\mathbf{x}}+(y-y')\hat{\mathbf{y}}+(z-z')\hat{\mathbf{z}}}{r}\right] \\ &= f'(r)\left(\frac{\partial r}{\partial x'}\right)\hat{\mathbf{r}}+f(r)\left[\frac{-1}{r}\left(\frac{\partial r}{\partial x'}\right)\hat{\mathbf{r}}-\frac{1}{r}\hat{\mathbf{x}}\right]\end{aligned}

Yes, you are right
$$\textbf{r}=(x-xsingle-quote)\hat{\mathbf{x}}+(y-ysingle-quote)\hat{\mathbf{y}}+(z-zsingle-quote)\hat{\mathbf{z}}$$

$$\left(\gamma+\frac{1}{r}\right)\frac{e^{-\gamma r}}{r^2}\hat{\mathbf{r}}$$

Thanks for explaining the second rank tensor expression. I thought of the product rule, but then didn't pursue it coz I couldn't figure out how to evaluate d(r hat)/dt

A gradient is a mathematical vector that represents the rate of change of a function at a specific point. It is represented by a vector symbol (∇) and is also known as a nabla or del operator.

2. How is the gradient calculated?

The gradient is calculated by taking the partial derivatives of a multivariable function with respect to each of its variables. These partial derivatives are then combined into a vector, which represents the direction and magnitude of the function's steepest ascent at that point.

3. What are A and B vectors in gradient equation?

A and B vectors represent the components of the gradient vector. In the equation ∇f = A∙i + B∙j, A and B represent the partial derivatives of the function f with respect to the x and y variables, respectively.

4. What is e-xr-2 in the gradient equation?

e-xr-2 is a term that represents the exponential function of the distance (r) from the point of interest. This term is often used in gradient equations to describe the rate of change of a function in relation to distance from a certain point.

5. What does the symbol ^r mean in the gradient equation?

The symbol ^r, also known as a hat symbol, indicates that the vector is a unit vector in the direction of the vector r. In gradient equations, ^r is often used to represent the direction in which the function is changing at a given point.

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