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- jango_IN_DTOWN

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- jango_IN_DTOWN

##### 1 Attachment

- jango_IN_DTOWN

@ganeshie8

- Astrophysics

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## More answers

- FakeReality

.

- Yttrium

This is an 0-8-4. If you know what I mean. Hehehe

- FakeReality

Oh you.. XD

- jango_IN_DTOWN

what is it by the way? @Yttrium?

- ganeshie8

let me tag @Kainui

- ganeshie8

I'm not sure what it means to take gradient of a vector function...
is this from tensor analysis ?

- jango_IN_DTOWN

no vector calculus.

- jango_IN_DTOWN

\[*=\times \]

- jango_IN_DTOWN

cross product

- Astrophysics

and the . is dot product?

- Kainui

@jango_IN_DTOWN Before we start with anything, are you familiar with what all the symbols in the equation mean?

- ganeshie8

how do you define gradient of a vector ?
I only dealt with taking gradients of scalar functions... never worked with gradients of vector functions..

- Kainui

We're specifically looking at: https://en.wikipedia.org/wiki/Vector_calculus_identities#Vector_dot_product
Where they say "As a special case, when A = B,"

- jango_IN_DTOWN

hi @Kainui which one?

- Kainui

https://upload.wikimedia.org/math/7/1/6/7167818e448d0cc60dfee192dc31697a.png

- jango_IN_DTOWN

How will we prove it

- Kainui

Do you know what this means? \[(A \cdot \nabla) A\]

- jango_IN_DTOWN

Let A=ai+bj+ck
(A. ▽)A=

- jango_IN_DTOWN

|dw:1447402149677:dw|

- jango_IN_DTOWN

@Kainui

- Kainui

Yeah, continue through with writing the steps out and you will be able to show that for any vector A they both give you the same result.

- Kainui

In your problem they are taking the gradient of \(f^2\) but really you should write this as the gradient of \(\vec f \cdot \vec f\) to help simplify your calculations so everything is in terms of the same symbols.

- jango_IN_DTOWN

ok now to prove the identity, whose special case is the one you gave?

- Kainui

Yeah, what I mean is just take the vector f, write it in component form, and calculate the left side of the equation and calculate the right side of the equation and see that you will get the same answer both ways, and that proves it.
Since we're using the cross product, all your vectors must be 3D, so make sure you don't do this with a different number of components!
Also there might be a prettier way of proving this but nothing comes to mind.

- jango_IN_DTOWN

##### 1 Attachment

- jango_IN_DTOWN

this identity?

- Kainui

No, I am just talking about your question, that's a more general case of what you're proving, it's not really important to you

- jango_IN_DTOWN

##### 1 Attachment

- jango_IN_DTOWN

using the identity its a 3 step problem.. but how to prove the identity

- Kainui

First calculate \((\vec f \cdot \vec \nabla ) \vec f\) using \(\vec f = a \hat i + b \hat j + c \hat k\) .
then calculate \(\tfrac{1}{2} \vec \nabla (\vec f \cdot \vec f) - \vec f \times (\vec \nabla \times \vec f)\) using \(\vec f = a \hat i + b \hat j + c \hat k\) .
Are they equal? If yes, then you've proven it.
The only way I know how to prove this outside of this uses the Levi-Civitta tensor, so I'm afraid this is the best you can do unless there's some way I'm not seeing.

- jango_IN_DTOWN

yeah both came equal but the problem is huge. @Kainui
.. however using the identity the problem is too small..

- Kainui

Using the identity is circular reasoning, it's not a proof at all, just rewriting it.

- jango_IN_DTOWN

oh I see...:)... then to prove the identity we have to assume A=a1i+a2j+a3k and B=b1i+b2j+b3k and then again prove lhs=rhs?

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