Vector calculus and order of operations

[PhysicalProof]

I understand how to do (a). However, I'm having trouble with the rest, here. I've never properly learned the order of operation for vectors like this.

For example, for (b) does one dot (r - r') with itself BEFORE using the del operator?

Thanks guys, help appreciated. Even if you point me to a good source for understanding this stuff I will be grateful.

 

[Dick]

Yes, (r-r').(r-r') then find the gradient. That's really the only way to interpret it.

 

[HallsofIvy]

I understand how to do (a). However, I'm having trouble with the rest, here. I've never properly learned the order of operation for vectors like this.

For example, for (b) does one dot (r - r') with itself BEFORE using the del operator?

The order of operations for vectors is the same as for scalars. In particular, r- r' is in parentheses. Of course, you do that first. That is what parentheses mean. Finally, [math]\nabla f[/math], written in that way, is a scalar operator. You are essentially "multiplying" the scalar f by the vector \nabla. Since it is to be applied to a scalar, not a vector you must take the dot product first to get a scalar.

For (c), note the "dot" between \nabla and the vector. That is a dot product which has to be done between vectors. Now you are taking the dot product of the two "vectors" [math]\nabla[/math] and r- r'.

You may not have learned it yet but you should shortly learn \nabla\times \vec{f}(x,y,z), the "cross product" of \nabla and the vector function \vec{f}(x,y,z).

Thanks guys, help appreciated. Even if you point me to a good source for understanding this stuff I will be grateful.

 

[yungman]

I understand how to do (a). However, I'm having trouble with the rest, here. I've never properly learned the order of operation for vectors like this.

For example, for (b) does one dot (r-r') with itself BEFORE using the del operator?

Thanks guys, help appreciated. Even if you point me to a good source for understanding this stuff I will be grateful.

I am no expert, I'll try!

I don't know b) the way you wrote it. Do you mean \nabla |\vec{r}-\vec {r}'|^2? If so then it is \nabla [ (\vec{r}-\vec {r}') \cdot (\vec{r}-\vec {r'}) ].

c) Is just (\frac{\partial}{\partial x}\widehat{x} + \frac{\partial}{\partial y}\widehat{y}) \cdot (\vec{r}-\vec {r}')

d) Remember \vec{a} \cdot \vec {b} =\vec{b} \cdot \vec {a}. so it is just (\nabla \cdot \widehat{r} ) \widehat{r}

 

[PhysicalProof]

Thanks for your assistance everyone! I'm trying to work through them now. Are you sure about C youngman? Aren't there three components in each of those r vectors that we need to use?

 

[yungman]

Thanks for your assistance everyone! I'm trying to work through them now. Are you sure about C youngman? Aren't there three components in each of those r vectors that we need to use?

I am sorry, I just use R2. Yes R3(3 dimemsion in space ) is exactly the same. Just add the z component or in your case use x1, x2 and x3.

\nabla \cdot (\vec{r}-\vec{r}') = (\frac{\partial}{\partial x_1} \hat{x}_1 + \frac{\partial}{\partial x_2} \hat{x}_2 + \frac{\partial}{\partial x_3} \hat{x}_3) \cdot (\vec{r}-\vec{r}')

I am just assuming we are using rectangular coordinates. It would be a little different with cylindrical and spherical coordinates. I thought your question concentrate on the \nabla operator so I really did not get into the coordinates.