Solving E1jk & Proving |Torque|^2 w/o Summation

[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)?

 

[himanshu121]

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

And What do u want to prove for Torque

 

[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?

 

[himanshu121]

I still don't 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|

 

[HallsofIvy]

As himashu121 pointed out, in order to find ε_1jk, you have to know what ε_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 ε_ijk, and substitute i= 1- in fact, write out all the components and then just copy down those that have i=1.

 

[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

 

[HallsofIvy]

Since jlmac2001 hasn't responded: In tensor analysis, δ_ij is the tensor represented by the unit matrix: 1 if i=j, 0 otherwise.

ε_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 anyone of the indices is repeated).

εsub]1jk[/sub] is therefore:
ε₁₁₁= 0
ε₁₁₂= 0
ε₁₁₃= 0
ε₁₂₁= 0
ε₁₂₂= 0
ε₁₂₃= 1
ε₁₃₁= 0
ε₁₃₂= -1
ε₁₃₃= 0

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