Vector calculus identity

  • Thread starter plasmoid
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  • #1
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Can someone help me prove the identity

[tex]\ u \times (\nabla \times u) = \nabla(u^2 /2) - (u.\nabla)u[/tex]


without having to write it out in components?
 

Answers and Replies

  • #2
chiro
Science Advisor
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  • #4
chiro
Science Advisor
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Yes I have, but that doesn't give me the 1/2 in the first term .....

So the first term is given by del (u . u) Where . is the dot product.

Now u . u = |u|^2.

Also you have to realize that u^2 doesn't make sense if u is a vector (at least in the traditional sense). Using the result above which is valid for any inner product, you will get

del (u . u) = |u|^2 (del) where |u| is the norm of u.

Maybe you should check the syntax again, based on the arguments I've presented above.
 
  • #5
160
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You can't really use the vector triple product since del isn't actually a vector. Your identity is related to it though, and with care will give you the right answer, since the method of proof for your identity is essentially identical to the proof of the triple product.

Do you know about the Levi-Civita or completely antisymmetric symbol [itex]\epsilon_{ijk}[/itex]? Using this is the easiest way to prove this sort of identity.
 
  • #6
15
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So the first term is given by del (u . u) Where . is the dot product.

Now u . u = |u|^2.

Also you have to realize that u^2 doesn't make sense if u is a vector (at least in the traditional sense). Using the result above which is valid for any inner product, you will get

del (u . u) = |u|^2 (del) where |u| is the norm of u.

Maybe you should check the syntax again, based on the arguments I've presented above.

How is del (u . u) = |u|^2 (del) ?
 
  • #7
15
0
You can't really use the vector triple product since del isn't actually a vector. Your identity is related to it though, and with care will give you the right answer, since the method of proof for your identity is essentially identical to the proof of the triple product.

Do you know about the Levi-Civita or completely antisymmetric symbol [itex]\epsilon_{ijk}[/itex]? Using this is the easiest way to prove this sort of identity.

Yes, but using the Levi-Civita symbol will essentially mean writing out the components, and I was trying to prove it without doing that...
 
  • #8
160
2
Yes, but using the Levi-Civita symbol will essentially mean writing out the components, and I was trying to prove it without doing that...

Yes, fair point. I took your aim to be avoiding having to laboriously write out every single term in full.

Certainly a nice goal to try to prove this without ever introducing coordinates, but I suspect this will be quite hard. To start with, you'll need a coordinate free definition of curl. This is likely to either be very nasty (limits of integrals or that sort of thing) or require a lot of machinery, drawing from the language of differential forms. Even then it's a bit of a mess. This is partly because the curl and the vector cross product are very special objects particular to 3D, with no natural extension to other dimensions, and partly because the objects in the identity use a lot of the structure of the space (it uses the metric and the orientation, used for Hodge dual, covariant derivative, and musical isomorphism if you like some jargon).

I'd be very interested if anyone has an idea for a cunning strategy!
 

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