Question

Divergence of a constant vector.

I tried calculating the divergence of a constant vector but I get it equal to zero. The sample solution in my book says the following:

Answer 1

what does your book says?

Answer 2

I'm typing it now, it will take a bit to type the formulas.

Answer 3

\[( \nabla \cdot B_0 ) r = B_0 \cdot \nabla r\]

Answer 4

Where B0 is a constant vector.

Answer 5

what is r?

Answer 6

Another vector, I suppose the radius vector.

Answer 7

The divergence of a constant vector will be 0.

Answer 8

So then is the textbook wrong?

Answer 9

This would make more sense:\[( \nabla \cdot B_0 r) = B_0 \cdot \nabla r\]

Answer 10

Yes, indeed.

Answer 11

Hmmm, maybe it's constant with respect to something else?

Answer 12

I think the book is mistaken, or we are talking about something far more complicated

Answer 13

Let me copy the whole problem. Maybe I'm missing something else.

Answer 14

The connection between the magnetic field B and vector potential A is \[ B = \nabla \times A\] What is the magnetic field if \[A = \frac{ 1 }{ 2 }B_0 \times r\] where B0 is a constant vector?

Answer 15

So, the solution says:\[B = \nabla \times ( \frac{ 1 }{ 2 }B_0 \times r) = \frac{ 1 }{ 2 }B_0 ( \nabla \cdot r) - \frac{ 1 }{ 2 } ( \nabla \cdot B_0 ) r \]

Answer 16

And then they claim that \[( \nabla \cdot B_0 ) r = B_0 \cdot \nabla r\]

Answer 17

Does it make any sense now, @wio ?

Answer 18

Maybe \(B_0\) isn't constant then? To be honest I find it a bit strange.

Answer 19

Like, maybe it's constant for time, but not for position

Answer 20

So \(B_0 = f(x,y,z)\) but \(d/dt B_0 = 0\)

Answer 21

Right, I see. Well, I guess I should ask the teacher, then. Thank you for taking the time :)

Answer 22

Okay, one way to think about this is: \[
(B_0)_xr_1+(B_0)_yr_2 = (B_0)_1r_x + (B_0)_2r_y
\]

Answer 23

I'm not sure I understand what you did. Why are the components of B0 and r different - once expressed through x and y, and in the other case - through 1 and 2?