Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gradient equal to multiplying by vector?

  1. Oct 10, 2015 #1
    Hi guys!
    So I have been researching the electric field, and I have come upon some interesting equations that confused me a little (all from wikipedia): fe40e6e5030963130cd36c93ca29f634.png , 400073dd9ef3633c03d01c35c9f6281e.png , and 8757428f7aeb787795811caa40897baa.png with V being the same as psi. Doing the algebra, I would get (Q/4πε0)*(r-hat/|r|2)=-∇(Q/4πε0r)-∂A/∂t. Now in the case that A does not change with time, we would get that taking the negative gradient is equal to multiplying by r-hat/|r|. Is this correct?
     
  2. jcsd
  3. Oct 10, 2015 #2
    Yes, this is true for the gradient of ##\frac{1}{|r|}##. $$-\nabla \frac{1}{|r|} = \frac{\hat{\mathbf{r}}}{|r|^2}, \;\;\;\;\;\;\;\;r\neq0.$$
     
  4. Oct 11, 2015 #3
    Would the negative gradient of 2|r|\frac{2}{|r|} be 2r-hat over |r|2?
     
  5. Oct 11, 2015 #4
    Sorry, I don't know what happened with the code. I was asking bout 2/|r|
     
  6. Oct 11, 2015 #5
    you can always take a moltiplicative constant out of the gradient.

    --
    lightarrow
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook