How can I understand the Einstein summation convention for vector algebra?

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The discussion centers on understanding the Einstein summation convention in vector algebra, specifically in the context of calculating the divergence of the expression Div(rn r) where r = xi + yj + zk. The user, Matt, seeks clarity on how the term xixi simplifies to r, which is derived from the spatial coordinates x2 + y2 + z2. The forum provides resources and explanations, confirming that the summation convention applies when indices are repeated, leading to the correct interpretation of the notation.

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Matt1991
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Hi,

I am just starting to learn vector algebra with Grad, Div, Curl etc and have in passing come across Einstein notation which seems to make things much more concise.

The problem I have is in Finding Div(rn r) where r =xi + yj + zk. The unbold r is the magnitude of r.

I have used some basic Einstein notation to make my working shorter but am stuck understanding a certain part of the notation which must be true to lead to the correct answer.

My Working:

\frac{\partial}{\partial x_{i}}\((r^n x_{i})


product rule:

= \ nr^{n-1} \frac{\partial r}{\partial x_{i}} x_{i}+r^n \frac{\partial x_{i}}{\partial x_{i}}

= \ nr^{n-1}\frac{x_{i}}{r}x_{i}\ <br /> <br /> = \ nr^{n} \r \ + \ 3 r^n


= (n+3)r^n


My problem is in understanding the step where \frac{ x_{i} x_{i}}{ r} becomes r. For this to happen xixi must be evaluated as x2+y2+z2 (in spatial coordinates) which is the part I am having trouble understanding.

An explanation of this or if somebody could point me towards somewhere where I can get a simple explanation of this would be very much appreciated.

Thanks,

Matt

PS Sorry if the laTeX is bad. Its my first time using it.
 
Last edited:
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I have made a correction to my original working that I posted. It is now hopefully correct.

Apologies,

Matt
 
Matt1991 said:
Hi,

I am just starting to learn vector algebra with Grad, Div, Curl etc and have in passing come across Einstein notation which seems to make things much more concise.

The problem I have is in Finding Div(rn r) where r =xi + yj + zk. The unbold r is the magnitude of r.

I have used some basic Einstein notation to make my working shorter but am stuck understanding a certain part of the notation which must be true to lead to the correct answer.

My Working:

\frac{\partial}{\partial x_{i}}\((r^n x_{i})


product rule:

= \ nr^{n-1} \frac{\partial r}{\partial x_{i}} x_{i}+r^n \frac{\partial x_{i}}{\partial x_{i}}

= \ nr^{n-1}\frac{x_{i}}{r}x_{i}\ <br /> <br /> = \ nr^{n} \r \ + \ 3 r^n


= (n+3)r^n


My problem is in understanding the step where \frac{ x_{i} x_{i}}{ r} becomes r. For this to happen xixi must be evaluated as x2+y2+z2 (in spatial coordinates) which is the part I am having trouble understanding.
But you are using coordinates x1, x2, and x3 in place of x, y, and z. By the Einstein summation convention, x_iy_i means x_1y_1+ x_2y_2+ x_3y_3 so that x_iy_i means x_1x_1+ x_2x_2+ x_3x_3= x_1^2+ x_2^2+ x_3^2 which is the same as x^2+ y^2+ z^2.

An explanation of this or if somebody could point me towards somewhere where I can get a simple explanation of this would be very much appreciated.

Thanks,

Matt

PS Sorry if the laTeX is bad. Its my first time using it.
 
Thanks for the response,

Ah right. If that is the convention then I can definitely see why. I am not sure I understand the reasoning behind the convention though.

if xi is simply x1 + x2 + x3 + ...

then I am not sure what the reasoning is behind xi multiplied by itself acting as the einstein notation suggests it does. When I imagine this as its individual spatial coordinates (or as a operations on x1,x2, etc) it seems to me that it should be (x1+x2+...)^2

Obviously my resoning is wrong, I just can't seem to figure out where.

Thanks,

Matt
 
Matt1991 said:
if xi is simply x1 + x2 + x3 + ...

On its own, xi doesn't stand for a sum. It's just one variable: x or y or z. The summation convention only applies when two variables in the same term have the same index:

xixi = x1x1 + x2x2 + x3x3 = x2 + y2 + z2.

aibi = a1b1 + a2b2 + a3b3

(Incidentaly, if you're using this convention and you happen to have two variables with the same index in a term but don't want it to denote a sum, just write "no sum on i" or "no sum over k" or whatever the index is.)
 
You can find out more about the Einstein summation convention here:

1) See http://www.ph.ed.ac.uk/~martin/mp2h/VTF/lecture05.pdf" course.
(Martin Evans, University of Edinburgh, http://www.ph.ed.ac.uk/" )

2) http://www.luc.edu/faculty/dslavsk/courses/phys301/classnotes/einsteinsummationnotation.pdf" by David Slavsky
(Physics 301/Math 355: Mathematical Methods of Physics, Loyola University Chicago)

3) http://www.cs.caltech.edu/~cs20/c/esn-v205.pdf" by Alan H. Barr, California Institute of Technology

4) Via the http://en.wikipedia.org/wiki/Einstein_notation" in which Einstein introduces his notation (page 158 of the document or page 8 of the PDF).

5) John Armstrong explains why Einstein introduced his notation in http://unapologetic.wordpress.com/2008/05/21/the-einstein-summation-convention/" .
 
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