# Need Good book on Vector Calculus

1. Apr 22, 2005

### newbie101

Hi All,

I need some suggestion on a good book for vector calculus/advanced vector calculus.
current book im reading just give equations like

del x ( A x B ) = A del.B - Bdel.A + (B.del)A - (A.del)B

A x ( B x C ) = B(del.A) - C(A.B)

del x (f A) = f del x A + del f x A

etc

however they dont show the proof
Is there any book or maybe a website which gives the proof step by step

thanks
newbie101

* if there is a free book i could download .. it would be fantastic

2. Apr 22, 2005

### dextercioby

They can be proven on components.All three of them are vector identities,so it suffices to prove only for one scalar component.The second one is really easy if u use cartesian tensors...The same goes for the 3-rd.

Daniel.

3. Apr 23, 2005

### newbie101

dextercioby,

please show me how they are proven....

thanks
newbie

4. Apr 23, 2005

### dextercioby

The first u've written there is incorrect...

Come up with the correct form.

1

$$\vec{A}\times\left(\vec{B}\times\vec{C}\right)=...?$$

$$\vec{B}\times\vec{C}=\epsilon_{ijk}B_{i}C_{j}\vec{e}_{k}$$

$$\vec{A}\times\left(\vec{B}\times\vec{C}\right)=\epsilon_{lkn}A_{l}\left(\vec{B}\times\vec{C}\right)_{k}\vec{e}_{n}=\epsilon_{lkn}\epsilon_{ijk}A_{l}B_{i}C_{j}\vec{e}_{n}$$

$$\epsilon_{lkn}\epsilon_{ijk}=-\epsilon_{lnk}\epsilon_{ijk}=-\left(\delta_{li}\delta_{nj}-\delta_{ni}\delta_{lj}\right)=\delta_{ni}\delta_{lj}-\delta_{li}\delta_{nj}$$

Therefore,making the summations with the delta Kronecker

$$\vec{A}\times\left(\vec{B}\times\vec{C}\right)=B_{i}A_{l}C_{l}\vec{e}_{i}-A_{l}B_{l}C_{j}\vec{e}_{j}=\left(\vec{A}\cdot\vec{C}\right)\vec{B}-\left(\vec{A}\cdot\vec{B}\right)\vec{C}$$

Q.e.d.

Daniel.

5. Apr 23, 2005

### dextercioby

2

$$\nabla\times\left(A\vec{B}\right)=\epsilon_{ijk}\partial_{i}\left(AB_{j}\right)\vec{e}_{k}=\epsilon_{ijk}\left(\partial_{i}A\right)B_{j}\vec{e}_{k}+\epsilon_{ijk}A\left(\partial_{i}B_{j}\right)\vec{e}_{k}=\left(\nabla A\right)\times\vec{B}+A\left(\nabla\times\vec{B}\right)$$

Q.e.d.

Daniel.

6. Apr 23, 2005

### dextercioby

Okay.I'll make reference to post #4 in which the simply contracted tensor product of Levi-Civita tensor appears.

3

$$\nabla\times\left(\vec{A}\times\vec{B}\right)=\epsilon_{ijk}\partial_{i}\left(\epsilon_{lmj}A_{l}B_{m}\right)\vec{e}_{k}=\epsilon_{ijk}\epsilon_{lmj}\left[\left(\partial_{i}A_{l}\right)B_{m}+A_{l}\left(\partial_{i}B_{m}\right)\right]\vec{e}_{k}$$

$$=-\epsilon_{ikj}\epsilon_{lmj}\left[\left(\partial_{i}A_{l}\right)B_{m}+A_{l}\left(\partial_{i}B_{m}\right)\right]\vec{e}_{k} =\left(\delta_{im}\delta_{kl}-\delta_{il}\delta_{km}\right)\left[\left(\partial_{i}A_{l}\right)B_{m}+A_{l}\left(\partial_{i}B_{m}\right)\right]\vec{e}_{k}$$

$$=\left(\partial_{m}A_{k}\right)B_{m}\vec{e}_{k}+A_{k}\left(\partial_{m}B_{m}\right)\vec{e}_{k}-\left(\partial_{l}A_{l}\right)B_{k}\vec{e}_{k}-A_{i}\left(\partial_{i}B_{k}\right)\vec{e}_{k}$$

$$=\left(\vec{B}\cdot\nabla\right)\vec{A}+\vec{A}\left(\nabla\cdot\vec{B}\right)-\vec{B}\left(\nabla\cdot\vec{A}\right)-\left(\vec{A}\cdot\nabla\right)\vec{B}$$

Q.e.d.

Daniel.

7. Apr 23, 2005

### dextercioby

4

$$\vec{A}\times\left(\nabla\times\vec{B}\right)+\vec{B}\times\left(\nabla\times\vec{A}\right)+\left(\vec{B}\cdot\nabla\right)\vec{A}+\left(\vec{A}\cdot\nabla\right)\vec{B}$$

$$=\epsilon_{ijk}A_{i}\left(\nabla\times\vec{B}\right)_{j}\vec{e}_{k}+\epsilon_{ijk}B_{i}\left(\nabla\times\vec{A}\right)_{j}\vec{e}_{k}+B_{m}\left(\partial_{m}A_{l}\right)\vec{e}_{l}+A_{m}\left(\partial_{m}B_{l}\right)\vec{e}_{l}$$

$$=\epsilon_{ijk}A_{i}\left(\epsilon_{lmj}\partial_{l}B_{m}\right)\vec{e}_{k}+ \epsilon_{ijk}B_{i}\left(\epsilon_{lmj}\partial_{l}A_{m}\right)\vec{e}_{k}+B_{m}\left(\partial_{m}A_{l}\right)\vec{e}_{l}+A_{m}\left(\partial_{m}B_{l}\right)\vec{e}_{l}$$

$$=\left(\delta_{im}\delta_{kl}-\delta_{il}\delta_{km}\right)A_{i}\left(\partial_{l}B_{m}\right)\vec{e}_{k}+\left(\delta_{im}\delta_{kl}-\delta_{il}\delta_{km}\right)B_{i}\left(\partial_{l}A_{m}\right)\vec{e}_{k}+B_{m}\left(\partial_{m}A_{l}\right)\vec{e}_{l}+A_{m}\left(\partial_{m}B_{l}\right)\vec{e}_{l}$$

$$=A_{m}\left(\partial_{l}B_{m}\right)\vec{e}_{l}-A_{l}\left(\partial_{l}B_{m}\right)\vec{e}_{m}+B_{m}\left(\partial_{l}A_{m}\right)\vec{e}_{l}-B_{l}\left(\partial_{l}A_{m}\right)\vec{e}_{m}+B_{m}\left(\partial_{m}A_{l}\right)\vec{e}_{l}+A_{m}\left(\partial_{m}B_{l}\right)\vec{e}_{l}$$

$$=B_{m}\left(\partial_{l}A_{m}\right)\vec{e}_{l}+A_{m}\left(\partial_{l}B_{m}\right)\vec{e}_{l}=\partial_{l}\left(\vec{A}\cdot\vec{B}\right)\vec{e}_{l}$$

$$=\nabla\left(\vec{A}\cdot\vec{B}\right)$$

Q.e.d.

Daniel.

Last edited: Apr 23, 2005
8. Apr 23, 2005

### newbie101

Thanks dextercioby

It will take a while for me to go through this... but you've been a great help!!!

9. Apr 23, 2005

### dextercioby

After that "while",if u become at ease with euclidean tensor calculus & its application to proving nasty vector identities,then u can deal with this one

$$\left(\vec{A}\times\vec{B}\right)\times\left(\vec{C}\times\vec{D}\right)=\left(\vec{A},\vec{C},\vec{D}\right)\vec{B}-\left(\vec{B},\vec{C},\vec{D}\right)\vec{A}$$

,where the (...,...,...) stands for mixed vector product.

Daniel.