# Help with proof

1. Jan 20, 2004

### jlmac2001

How would I solve E1jk without the summation? I know how to solve it using the summation symbol but don't know howto do it without it.

Also, I need help proving that |torque|^2 = |r x F|^2= r^2F^2sin@(thetarF ). r dot F = rF cos (thetarF . Would I have to use (r x F) dot (r x F)?

2. Jan 20, 2004

### himanshu121

The Qs are not clear to me
Does E represents Electric field

And What do u want to prove for Torque

3. Jan 21, 2004

### jlmac2001

here is an attachment with the questions

I need help with the third one in number 1 and numbers 2(proof) and 3. For number 3, how would I expand?

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• ###### hw3.pdf
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4. Jan 21, 2004

### himanshu121

I still dont know what does epsilon delta represents may u have some notations

But as far as Q3 goes
The angle between $$\vec r x \vec F$$ is zero
Hence $$(\vec r x \vec F).(\vec r x \vec F) = |(\vec r x \vec F)^2|$$

Last edited: Jan 21, 2004
5. Jan 21, 2004

### HallsofIvy

Staff Emeritus
As himashu121 pointed out, in order to find &epsilon;1jk, you have to know what &epsilon;ijk means! I suspect I do know what it means since it is just a matter of looking up a definition, it would be much better for you to do that.

Right out the formula for &epsilon;ijk, and substitute i= 1- in fact, write out all the components and then just copy down those that have i=1.

6. Jan 21, 2004

### himanshu121

I still want to know epsilon and delta i believe these are vector components

Though for Part2:

Write a=wx(wxr) = (w.r)w-(w.w)r where all are vectors and x is a cross product.

If r is perpendicular than w.r=0

7. Jan 22, 2004

### HallsofIvy

Staff Emeritus
Since jlmac2001 hasn't responded: In tensor analysis, &delta;ij is the tensor represented by the unit matrix: 1 if i=j, 0 otherwise.

&epsilon;ijk is the "alternating" tensor. It is defined to be: 1 if ijk is an even permutation of 123, -1 if ijk is an odd permutation of 123, 0 otherwise (i.e. if any one of the indices is repeated).

&epsilon;sub]1jk[/sub] is therefore:
&epsilon;111= 0
&epsilon;112= 0
&epsilon;113= 0
&epsilon;121= 0
&epsilon;122= 0
&epsilon;123= 1
&epsilon;131= 0
&epsilon;132= -1
&epsilon;133= 0

Written as a matrix this would be:
[0 0 0]
[0 0 1]
[0 -1 0]