Prove this identity? Am I allow to do it like this?

[flyingpig]

Homework Statement

[PLAIN]http://img560.imageshack.us/img560/5384/unledkb.jpg

The Attempt at a Solution

For 21. I simply did

$$div(\mathbf{F} + \mathbf{G}) = \vec{\nabla} \cdot (\mathbf{F} + \mathbf{G}) = \vec{\nabla} \cdot \mathbf{F} + \vec{\nabla} \cdot \mathbf{G} = div(\mathbf{F}) + div(\mathbf{G})$$

My book proves it using partial derivatives. But I don't think I am wrong.

For 23. I don't understand what my book did with partial derivatives

[PLAIN]http://img10.imageshack.us/img10/3527/unlednx.jpg

I tried doing

$$div(f\mathbf{F})= \vec{\nabla} \cdot f\mathbf{F} = f(\vec{\nabla} \cdot \mathbf{F}) = fdiv(\mathbf{F})$$

I couldn't get the other dot product.

 

[Sethric]

You aren't actually proving the relations using your method, just either restating them, or using the relation to prove the relation. Your book is using partials because, at this stage, it's what you will have to do to prove the relation.

What your book did for the second one was actually show the calculations by using the components of the vector field and just multiplying out the steps. Once the derivations were done, it regrouped terms appropriately. You will need to follow suit.

 

[I like Serena]

div F is a synonym for ∇·F and both are shorthand notations for a commonly used set of partial differentials.
And although ∇·F looks like a regular inner product, it is not!
It is a derivative operator, defined by the partial differentials as shown in your given solution.

This is also the reason that you cannot simply say that ∇·(fF) = f∇·F.
That is because f is a function dependent on x, y, and z, and the derivative needs to be taken (product rule).

To proof the equations, the derivative needs to be written out in the x, y, and z coordinates, and afterward it can be turned back into the shorthand notation.

 

[flyingpig]

I don't understand, aren't I just trying to prove the vector properties?

 

[SammyS]

Yes, the important word being prove. But, you're assuming the vector operators behave like vectors .

 

[flyingpig]

Yes, the important word being prove. But, you're assuming the vector operators behave like vectors .

Aren't they?

 

[SammyS]

Judging by the example you showed, you should not be assuming that the vector operators behave like vectors. Use the definitions of divergence, curl, and gradient as defined by partial derivatives and vector components.

The fact that the vector operators behave like vectors, is one result you can get after doing the proofs in this exercise.

 

[flyingpig]

Oh then was it wrong that I added the arrow on top of the gradient operator?

 

[I like Serena]

Oh then was it wrong that I added the arrow on top of the gradient operator?

No, you were right to add the arrow on top of the gradient operator.
That is, in al lot of respects the gradient operator looks and behaves like a vector does.
So it is good practice to put an arrow on top to stress this fact.

However, there are a couple of cases where the gradient operator behaves differently and you need to know when that is exactly, and how it behaves then.

 

[SammyS]

...
My book proves it using partial derivatives. But I don't think I am wrong.

For 23. Isn't this #25 ? I don't understand what my book did with partial derivatives

[PLAIN]http://img10.imageshack.us/img10/3527/unlednx.jpg
...

What did the book do with partial derivatives that you don't understand?