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Einstien Summation Convention

  • Thread starter bon
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  • #1
bon
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Homework Statement



Ok so I'm meant to answer:

To what scalar or vector quantities do the following expressions in suffix notation correspond? (expand and sum where possible).

1) aibjci
2) aibjcjdi
3) dijaiaj
4) dijdij
5) eijkaibk
6)eijkdij


Homework Equations





The Attempt at a Solution



Okay so I'm pretty stuck to say the least..any help would be great..Here are some thoughts:

1)doesnt this just mean we sum that components of a and c and b can take any value? I'm confused..does this mean the result is a vector.?
2)so we sum a and d and b and c..but how do i simplify this?
3)ok so this one i think i know....isn't it just ajaj so sum the components of a? a.a?
4)doesnt this just equal dij?
5) Not sure
6) Again,not sure :(

thanks
 

Answers and Replies

  • #2
Cyosis
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Does the d_ij represent the Kronecker delta? Are a, b, c and d numbers or can they also be operators. If they are just numbers remember that order doesn't matter. How would you write down the dot product in index notation, what about the cross product?

As for 4). Your answer is not right. Keep in mind that you're summing over both i and j so when i != j you get 0, which doesn't contribute to the sum. Therefore the only entries that contribute to the sum are the entries where i=j. Try to write it out by letting i and j run from 0 to 1 and note that your answer can't be right.
 
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  • #3
bon
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yes d_ij is k delta.

i think abc are just numbers..

dot product would be a_ib_i
cross would be e_ijk a_i b_j e_k?

where the fist e is an epsillon..

but i dont see how to do the problems..:S
 
  • #4
bon
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so is it d_ii?
 
  • #5
Cyosis
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Yes d_ii is correct. However note that d_ii has repeated indices so we are summing all it's diagonal entries together. Can you express d_ii as a number?

Your dot and cross products are correct. So for 1) we get [itex]a_i b_j c_i=a_ic_ib_j=(\mathbf{a} \cdot \mathbf{c})b_j[/itex]. So what kind of object do we get in the end?
 
  • #6
bon
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Aha..so is d_ii = 3?

And so for 1) do we get a vector in the end? so should i write (a.c) (b1, b2, b3)?

Thanks a lot for your help..starting to make more sense..

So for 2) is it just a number, namely (a.d)(b.c)?

3)a.a?
4)3
5)still not sure?
6) ditto..

thanks again
 
  • #7
bon
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Also in another question, im asked to find the div and curl of this expression, where r = (x,y,z) and a and b are fixed vectors..

(a.r)b

so i took the div, which i found i could rearrange to be (a.r)(del.b) but can i simplify further? what does it mean when it says a and b are fixed vectors? does it mean that all components are constants, in which case, del.b = 0? thanks :)
 
  • #8
cristo
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cross would be e_ijk a_i b_j e_k?
Your dot and cross products are correct.
This cross product is incorrect. Remember that the cross product of two vectors is itself a vector, and so much have a free index. So, if [itex]{\bm a}\times{\bm b}={\bm c}[/itex], then [itex]c_i=\epsilon_{ijk}a_jb_k[/itex]
 
  • #9
Fredrik
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I think he meant that e_k is a basis vector, and in that case it's correct.

Bon, I'm not sure what you find confusing about 1. It's really trivial if you think about what the notation means. For example, what does bj mean? Perhaps it would help if you write aibjci with a summation sigma and some parentheses?
 
  • #10
cristo
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I think he meant that e_k is a basis vector, and in that case it's correct.
Oh, well then that's incredibly confusing to use the same symbols to mean different things!
 
  • #11
Cyosis
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Cristo what you have written is the i-th component of the cross product. What bon has written down is the full vector with e being the unit vector.

Example:
[tex]
\epsilon_{ijk}a_ib_j\mathbf{e_k}=a_2b_3\mathbf{e_1}-a_3b_2\mathbf{e_1}+a_3b_1\mathbf{e_2}-a_1b_3\mathbf{e_2}+a_1b_2\mathbf{e_3}-a_2b_1\mathbf{e_3}=(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-b_2a_1)
[/tex]
 
  • #12
cristo
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Cristo what you have written is the i-th component of the cross product. What bon has written down is the full vector with e being the unit vector.

Example:
[tex]
\epsilon_{ijk}a_ib_j\mathbf{e_k}=a_2b_3\mathbf{e_1}-a_3b_2\mathbf{e_1}+a_3b_1\mathbf{e_2}-a_1b_3\mathbf{e_2}+a_1b_2\mathbf{e_3}-a_2b_1\mathbf{e_3}=(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-b_2a_1)
[/tex]
Yes, if you right things out properly then it is obvious. What was written down above was ambiguous.
 
  • #13
bon
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Many thanks all..
so is anyone able to verify the comments i made in post 6? Also any thoughts on post 7?

Thank you
 
  • #14
Cyosis
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The first four look good now. As for 5 compare it to your definition of the cross product. For 6 do something similar to 4) and think what it means for the Levi-Civita symbol if two of the indices are the same.
 
  • #15
bon
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aha ok get them now..

5) = jth component of a x b

and 6) = 0

finally ( i promise) any thoughts on my question in post 7? ty!
 
  • #16
bon
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dont worry if you arent sure about that.. i have another q in the meantime which is:

im meant to find div and curl of (a x r) where a is a fixed vector and r = (x,y,z)

after simplifying the expression for div i get:

e_ijk d/dx_i x_j a_k

but isnt this (del x r)times a which is a vector times a vector with no dot/cross = illegal?
 
  • #17
Cyosis
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6) is correct. 5) is not entirely right remember that the Levi-Civita symbol is antisymmetric.

Yes that is illegal, but it is not what they are asking.

We have [itex]\nabla \times (a \times b)[/itex] first compute axb, which yields a vector and then compute the curl of that vector which again yields a vector. For the divergence we have [itex]\nabla \cdot (axb)[/itex]. First compute (a \times b) again which yields a vector then take the divergence which yields a scalar.
 
  • #18
Cyosis
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Also in another question, im asked to find the div and curl of this expression, where r = (x,y,z) and a and b are fixed vectors..

(a.r)b

so i took the div, which i found i could rearrange to be (a.r)(del.b) but can i simplify further? what does it mean when it says a and b are fixed vectors? does it mean that all components are constants, in which case, del.b = 0? thanks :)
Yes it means that all components are constant, but r is not a constant vector so you can't just put it in front of the del operator.
 
  • #19
bon
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hmm so is it -(axb)_j i.e. -1 times the jth component?
 
  • #20
Cyosis
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Yes and -(axb) is?
 
  • #21
bon
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ok so for div (a.r) b my working is:

d/dx_i e_i . (a_j x_j b_k e_k)

why cant i just rewrite this as a_i x_i d/dx_j bj?

how do i deal with it then?

thanks
 
  • #23
Cyosis
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Yes that is correct.

why cant i just rewrite this as a_i x_i d/dx_j bj?
For the same reason you can't write [itex]\frac{d}{dx}f(x)[/itex] as [itex]f(x)\frac{d}{dx}[/itex]. You are differentiating when taking the divergence.
 
  • #24
bon
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Yes that is correct.



For the same reason you can't write [itex]\frac{d}{dx}f(x)[/itex] as [itex]f(x)\frac{d}{dx}[/itex]. You are differentiating when taking the divergence.
ah ok i see..well im not sure how to deal with it in that case..so far when doing divergence ive just dotted the e_i in the divergence term with the later e_j and simplified to d_ij etc..

what can i do now? thanks
 
  • #25
Cyosis
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The handling of the unit vectors are fine, they do not depend on x y or z. Could you compute the divergence of r for me with r=(x,y,z) the 'normal' way (with that I mean don't use index notation)?
 

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