Vector calculus index notation

[Physgeek64]

Homework Statement

prove grad(a.grad(r^-1))= -curl(a cross grad (r^-1))

Homework Equations

curl(a x b)= (b dot grad)a - (a dot grad)b +a(div b) - b(div a )

The Attempt at a Solution

Im trying to use index notation and get
di (aj (grad(r^-1))j)
=grad(r^-1) di(aj) +aj(di grad(r^-1))j

which is obviously not right. I've tried attacking the problem from the reverse direction and haven't had much luck there either.

Thank you :)

 

[jedishrfu]

Can you provide more details? Is A a constant vector not dependent on r, theta, phi?

Have you looked at the vector identities in spherical coordinates for curl, div and grad?

Try to solve it without going to index notation.

 

[jedishrfu]

Try starting with ##\nabla( A \cdot B ) = (B \cdot \nabla ) A + (A \cdot \nabla) B + B \times (\nabla \times A) + A \times (\nabla \times B)##

 

[Physgeek64]

Try starting with ##\nabla( A \cdot B ) = (B \cdot \nabla ) A + (A \cdot \nabla) B + B \times (\nabla \times A) + A \times (\nabla \times B)##

Hi thank you for replying!
Sorry, a is a constant vector and r is (X,y,z)

I have not come across this identity. Any hints on how to prove it?

 

[jedishrfu]

Here's a more complete list of identities:

https://en.wikipedia.org/wiki/Vector_calculus_identities

Off hand I don't know how to prove this particular identity but often when presented with a problem such as yours you can use the identities transform the left had side to the right hand side or vice versa.to "prove" it.

 

[PeroK]

Homework Statement

prove grad(a.grad(r^-1))= -curl(a cross grad (r^-1))

Homework Equations

curl(a x b)= (b dot grad)a - (a dot grad)b +a(div b) - b(div a )

The Attempt at a Solution

Im trying to use index notation and get
di (aj (grad(r^-1))j)
=grad(r^-1) di(aj) +aj(di grad(r^-1))j

which is obviously not right. I've tried attacking the problem from the reverse direction and haven't had much luck there either.

Thank you :)

Time to learn some latex, I suggest!

 

[jedishrfu]

Some latex codes:

http://web.ift.uib.no/Teori/KURS/WRK/TeX/symALL.html

in particular look at vector operators \nabla for the del operator,\cdot for the inner-product and \times for the cross-product. This site uses mathjax for display. Mathjax looks for expressions bracketted by double # characters.

 

[Physgeek64]

Here's a more complete list of identities:

https://en.wikipedia.org/wiki/Vector_calculus_identities

Off hand I don't know how to prove this particular identity but often when presented with a problem such as yours you can use the identities transform the left had side to the right hand side or vice versa.to "prove" it.

I think proving the identity was the purpose of the question? I could be wrong but it's to do with index notation

 

[jedishrfu]

You problem has a specific function ##\nabla(1/r)## in it so its seemed to me that its not an identity but an example where you apply the identity to get the result you need to prove.

Its true I could be wrong here. Perhaps @Mark44 or @PeroK could comment more on this.

 

[PeroK]

You problem has a specific function ##\nabla(1/r)## in it so its seemed to me that its not an identity but an example where you apply the identity to get the result you need to prove.

Its true I could be wrong here. Perhaps @Mark44 or @PeroK could comment more on this.

I assumed it did depend on the second vector being ##\nabla(1/r)##. It must. I evaluated each side for the x-term (and then used symmetry). It's not too bad if you're careful with your differentiation.

 

[PeroK]

I assumed it did depend on the second vector being ##\nabla(1/r)##. It must. I evaluated each side for the x-term (and then used symmetry). It's not too bad if you're careful with your differentiation.

I checked with the second vector being ##\nabla f## and it reduces to Laplace's equation ##\nabla ^2 f = 0##

 

[Physgeek64]

I checked with the second vector being ##\nabla f## and it reduces to Laplace's equation ##\nabla ^2 f = 0##

Its okay, I've managed to do it with index notation. Thank you guys for helping though- I really appreciate it :)