Talk:Einstein notation/Archive 1

Archive 1

A Table Comparing Different Notations

It might be helpful to have a table with columns showing objects written in Einstein/indicial notation, traditional notation, and a computer code such as MATLAB/OCTAVE. I don't know the math syntax for Wikipedia, but I can type it out in Word and make an image file. -bmunden

??

1. The text explaining the meaning of notations like "aμνbα" is rather confusing.

2. The examples given seem to illustrate Einstein notation, whereas the formulae in the discussion above them do not (I see no superscripts in them at all).

3. Also, please write for English clear and proper grammar and complete sentences.

4. For the section on "=== Multiplication of a vector by a matrix ===" that part about the swapping of the indicies to compute the transpose of A multiplied by vector v only makes sense if V is a square matrix.

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The text I copied from tensor said: both raised and lowered on the same side. When I did Lagrangian mechanics, it was twice on the same side, which allow dot and cross products to be written in this way -- Tarquin 20:13 Mar 13, 2003 (UTC)

In general, either are used but in several fields (i.e. spacetime geometry) the up/down is required and correspond to a 'tensor contraction'. -- looxix 20:36 Mar 13, 2003 (UTC)

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Can we add this funny Einstein's comment stolen from Wolfram:

The convention was introduced by Einstein (1916, sec. 5), who later jested to a friend, "I have made a great discovery in mathematics; I have suppressed the summation sign every time that the summation must be made over an index which occurs twice..." (Kollros 1956; Pais 1982, p. 216).

-- looxix 00:14 Mar 14, 2003 (UTC)

Don't see why not -- Tarquin 11:40 Mar 14, 2003 (UTC)

http://mathworld.wolfram.com/EinsteinSummation.html

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I will write the next section, but not until tomorrow at least. I may not be up to the last section, however. Both of these are briefly described in Appendix 3 of Wald's 1994 QFT in CS & BHT book. So you can look there if you want to write them! ^_^ -- Toby 07:16 Mar 18, 2003 (UTC)

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IE6 does not show ⊗ (& otimes;) - Patrick 11:00 Mar 18, 2003 (UTC)

I try to explain in the surrounding context what the missing symbol would be when I do something like this. But there are alternatives short of going to full-blown texvc. Can you give some advice for your browser:

1. V ⊗ W (relying on context to know what the box represents);
2. V (x) W (long used by mathematicians in ASCII contexts like Usenet);
3. V x W (relying on context to know what kind of multiplication is involved);
4. V $\otimes$ W (a less distorted picture than that produced by texvc).

-- Toby 00:05 Mar 20, 2003 (UTC)

The box is not clear, the same one appears for every symbol that can not be represented. #3 and #4 both are fine. - Patrick 01:06 Mar 20, 2003 (UTC)

some details

The basic rule is:

   v = vi ei.

In this expression, it is assumed that the term on the right side is to be summed as i goes from 1 to n, because the index i appears on both sides. In that case, the equation is indeed true.

I guess what is meant is "because the index i appears twice", or maybe "because the index i doesn't appear on both sides".

Here, the Levi-Civita symbol e (or ε) satisfies eijk is 1 if (i,j,k) is a positive permutation of (1,2,3), -1 if it's a negative permutation, and 0 if it's not a permutation of (1,2,3) at all.

Shouldn't it be odd and even permutations, rahter than positive and negative? (I don't know if the terms 'positive' and 'negative' are common use, but if it is the case then they should appear in the "permutations" page...)

I think this question from a year or two ago has been answered -- the main page uses a different phrasing when presenting the Levi-Civita symbol. However, just in cast: the 1 and -1 are not describing the permutation, they're describing the value of $\epsilon$ based on the indices of $\epsilon$. Permutations are a relevant concept, but you don't need to think about permutations to use the definition. In other words, $\epsilon$ has values from the set {1, 0, -1} -- the system would not make sense if the 1 and -1 were replaced with odd and even. For example, how do you multiply by "odd"? RaulMiller 17:04, 3 October 2005 (UTC)

Make more simple?

I think this article could be made simpler. For example, it would be greatly enhanced with the Riemman summation symbol at least once in the article (preferably in the definition), and a strategic use of the word "implicit". (preferably in the definition) Kevin Baas | talk 22:11, 2004 Jul 31 (UTC)

I'm removing the entire "examples" section following the Formal Definitions section. Overall, the presentation was haphazard, and confused. For example, discussion of the Levi-Civita symbol $\ \epsilon$ does not belong in a section titled Elementary vector algebra and matrix algebra, as it is neither a vector nor a matrix. (Column vectors and row vectors are a mechanism to cast vector operations as matrix operations and simplify the scope of the discussion, but if you're going to talk about a rank 3 tensor this is a complication not a simplification, if it could be said to be valid at all.) If we're going to have examples, they should be real, concrete examples, not formally broken attempts at formal presentation. As it is, the only people capable of following these "examples" would have more than enough information from the preceeding sections. However, numerical examples and/or functional examples might be appropriate, if I have time, perhaps I'll add some of those. RaulMiller 14:58, 3 October 2005 (UTC)

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I've removed the following two paragraphs from the main page:

We have also used a superscript for the dual basis, which fits in with a convention requiring summed indices to appear once as a subscript and once as a superscript. In this case, if L is an element in V*, then:

L = L_i eⁱ.

If instead every index is required to be a subscript, then a different letter must be used for the dual basis, say d_i = eⁱ.

The real purpose of the Einstein notation is for formulas and equations that make no mention of the chosen basis. For example, if L and v are as above, then

L(v) = L_i vⁱ,

and this is true for every basis.

This seems unclear. In particular, L is defined as both a vector and a scalar -- e is a rank 2 tensor, and v is a rank 1 tensor. RaulMiller 17:54, 3 October 2005 (UTC)

References

The article needs some solid references. I've added one already, namely Einstein's 1916 Annalen der Physik paper. I found this post http://www.lepp.cornell.edu/spr/2003-05/msg0051086.html which contains some useful info. about quotes from Abrahams Pais' biography of Einstein. Perhaps some of this material can be used. MP (talk) 21:20, 12 September 2006 (UTC)

I don't agree. There are no controversial claims in this article. A reference at the bottom to one or two handbooks should suffice. Classical geographer 14:27, 31 January 2007 (UTC)

Agree with Classical geographer --Berland 09:30, 2 April 2007 (UTC)

Vector identity proofs

Might it be a good idea to have some brief proofs of vector identities (calculus ones too) using this notation? After all, the Einstein summation convention is an excellent way to prove CAB-BAC rule and so on.. (Aihadley 19:06, 2 April 2007 (UTC))

Let's show:

$\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )=\mathbf {b} \cdot (\mathbf {c} \times \mathbf {a} )$

using the einstien summation convention, where $\mathbf {a} ,\mathbf {b}$ and $\mathbf {c}$ are vectors in 3 dimensional Cartesian space.

The $i^{th}$ component of the cross product $\mathbf {b} \times \mathbf {c}$ is given by $(\mathbf {b} \times \mathbf {c} )_{i}=\epsilon _{ijk}b_{j}c_{k}$ and

the dot product of $\mathbf {a}$ with another vector $\mathbf {d}$ is given by $\mathbf {a} \cdot \mathbf {d} =a_{i}d_{i}$, so that:

$\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )=a_{i}\epsilon _{ijk}b_{j}c_{k}=b_{j}\epsilon _{ijk}c_{k}a_{i}=b_{j}\epsilon _{jki}c_{k}a_{i}=\mathbf {b} \cdot (\mathbf {c} \times \mathbf {a} )$

where we have used the fact that the Levi-Cevita symbol is invariant under an even permutation of the indices i, j and k.

To derive some of the more useful identities from vector calculus requires formulas for the contraction of two Levi-Cevita symbols. Again we assume 3-dimensional Cartesian coordinates.

We start with the tensor product of two Levi-Cevita Symbols, which we expand in Kronecker deltas:

$\epsilon _{ijk}\epsilon _{lmn}=\delta _{il}\delta _{jm}\delta _{kn}+\delta _{im}\delta _{jn}\delta _{kl}+\delta _{in}\delta _{jl}\delta _{km}-\delta _{il}\delta _{jn}\delta _{km}-\delta _{im}\delta _{jl}\delta _{kn}-\delta _{in}\delta _{jm}\delta _{kl}$

Now we set a pair of indices in each Levi-Cevita symbol equal, so we have:

$\epsilon _{ijk}\epsilon _{imn}=\delta _{ii}\delta _{jm}\delta _{kn}+\delta _{im}\delta _{jn}\delta _{ki}+\delta _{in}\delta _{ji}\delta _{km}-\delta _{ii}\delta _{jn}\delta _{km}-\delta _{im}\delta _{ji}\delta _{kn}-\delta _{in}\delta _{jm}\delta _{ki}$

But,$\delta _{ii}=3$ and $\delta _{jn}\delta _{jm}=\delta _{nm}$, so that:

$\epsilon _{ijk}\epsilon _{imn}=3\delta _{jm}\delta _{kn}+\delta _{km}\delta _{jn}+\delta _{jn}\delta _{km}-3\delta _{jn}\delta _{km}-\delta _{jm}\delta _{kn}-\delta _{kn}\delta _{jm}$

Combining terms we have:

$\epsilon _{ijk}\epsilon _{imn}=\delta _{jm}\delta _{kn}-\delta _{jn}\delta _{km}$

Armed with this result we can derive the triple cross product identity from vector calculus. Consider a vector field $\mathbf {g}$ in 3-dimensional Cartesian coordinates:

$(\nabla \times \nabla \times \mathbf {g} )_{i}=\epsilon _{ijk}\partial _{j}\epsilon _{klm}\partial _{l}g_{m}$

$(\nabla \times \nabla \times \mathbf {g} )_{i}=\epsilon _{kij}\epsilon _{klm}\partial _{j}\partial _{l}g_{m}$

$(\nabla \times \nabla \times \mathbf {g} )_{i}=(\delta _{il}\delta _{jm}-\delta _{im}\delta _{jl})\partial _{j}\partial _{l}g_{m}$

$(\nabla \times \nabla \times \mathbf {g} )_{i}=\partial _{m}\partial _{i}g_{m}-\partial _{j}\partial _{j}g_{i}$

$(\nabla \times \nabla \times \mathbf {g} )_{i}=\nabla _{i}(\mathbf {\nabla } \cdot \mathbf {g} )-\nabla ^{2}g_{i}$

Converting from components to vectors we have:

$\nabla \times \nabla \times \mathbf {g} =\mathbf {\nabla } (\mathbf {\nabla } \cdot \mathbf {g} )-\nabla ^{2}\mathbf {g}$

--Sfitzsi (talk) 02:24, 9 July 2008 (UTC)

I think this is a very good idea. On top of that, we should stress the point that the summation convention was motivated by the fact that you can only sum together a covariant with a contravariant element, that means that $a_{i}b_{i}$ does not make any sense in this context. Hence, this distinction gave rise to the summation convention in the first place, so it should be noted. When calculating in Euclidean space, one often disregards the fact, because $g_{ij}=\delta _{ij}$ or wherever you like to put the indices. The article could be more specific and give an example for that.Tprosser 19:21, 14 May 2007 (UTC)

Scope

Although this article is associated to physics, but in fact it is talking about a notation which applicable to any linear space. So, for any multidimension linear space, this notation is useful. Any objection? Jackzhp 13:54, 2 June 2007 (UTC)

three items

if V=a_i*b_i*c_i sum for all i, then how to write it in einstein notation? Jackzhp 13:54, 2 June 2007 (UTC)

Template

There are a number of pages that have a boilerplate note saying they use this convention. There should be a Template:Einstein notation. 155.212.242.34 21:40, 2 October 2007 (UTC)

Created: Template:Einstein_summation_convention

Examples

Suggest:

• derivatives —DIV (128.250.80.15 (talk) 10:47, 7 May 2008 (UTC))

Verification of claim/weasel word problem

The article states in the vector representations section: "This point is frequently confused." Are there any citations about this and who confuses this point so often? --Hydraton31 (talk) 20:58, 7 May 2008 (UTC)

raised vs. lowered index confusion

This seems like total nonsense to me (in the introduction):

$y=c_{i}x^{i}\,$

actually means

$y=\sum _{i=1}^{3}c_{i}x_{i}.$

In cases where you distinguish raised and lowered indices, $x^{i}\neq x_{i}$ and the equation above is simply wrong.

(This has since been fixed. — Steven G. Johnson (talk) 15:46, 29 September 2009 (UTC))

No new ideas?

In the lead it says It is important to keep in mind that no new physical laws or ideas result from using Einstein notation; but it enforces covariance doesn't it? Is that not a physical idea? --catslash (talk) 10:23, 16 April 2009 (UTC)

It expresses covariance compactly; it is not the source of the covariance idea but is rather the product (historically, it seems pretty clear that the basic covariance idea preceded Einstein notation).

However, I'm also skeptical of the value of that sentence. On the one hand, why is that caveat in this article specifically, as opposed to every other article on any mathematical notation? In a pedantic sense, notation expresses ideas but is not the source of them. Notation is certainly never the source of "physical laws," which after all result from nature itself. On the other hand, the history of mathematics and physics shows that the invention of compact, convenient, and suggestive notations was extremely conducive to the development of new ideas and the discovery of unknown physical principles. — Steven G. Johnson (talk) 17:11, 6 September 2009 (UTC)

Would it be correct (and relevant) to say: An expression written using the Einstein notation is necessarily covariant - it remains the same under a change of coordinates. Hence, if the expression represent a supposed physical law, then the law automatically respects the principle of isotropy if it is in three dimensional Euclidean space, or the general principle of relativity if it is in four dimensional space-time.?

No, I do not think so. It is only a convention. Therefore I do not think that it should be elevated to having covariance properties in general. Unfortunately I do not have a good counterexample. TomyDuby (talk) 05:38, 27 September 2009 (UTC)

Covariance follows automatically for contractions if the notation is being applied correctly to tensors. If you make a "four vector" of charge, mass, temperature, and time, contractions involving it won't be covariant just because you use index notation. If you mess up which indices are raised and which are lowered then you also won't get covariant expressions; notation isn't a panacea. — Steven G. Johnson (talk) 15:41, 29 September 2009 (UTC)

OK, and it explains the importance of the covariant nature of the operation in the introduction. I may delete the sentence in the lead at some point. --catslash (talk) 16:22, 29 September 2009 (UTC)

Would be much clearer to do the Euclidean case first (no raised/lowered distinction)

In the common case (Euclidean manifolds) where there is no distinction between contra- and co-variant vectors or raised/lowered indices, then having both of them is just pointlessly confusing. In that common case, it is much simpler and clearer to exclusively use subscripts and simply say that repeated subscripts are summed. This is not my notation; it's a common simplification of the Einstein notation as far as I can tell; see e.g. Classical Mechanics by Herbert Goldstein (Addison-Wesley, 1950) for one authoritative text using this convention.

I wish the article would simply start with the common case, use all subscripts, and the simplified version of the notation where repeated indices are summed.

Then do the "advanced" case, e.g. in relativity or differential geometry on curved manifolds, where you have to distinguish between raised/lowered indices to contrast co/contra-variance. In this case there is a reason for the rule, in that only contracting (summing) a raised/lowered index combination leads to a coordinate-system-independent result. As it stands, raised/lowered indices seem to be presented (at least initially) as a pointless typographical convention.

—Steven G. Johnson (talk) 22:10, 16 February 2009 (UTC)

Agreed. Perhaps you would like to rewrite it? --catslash (talk) 16:24, 29 September 2009 (UTC)

I've never seen Einstein summation used with superscripts. I was taught you just put all the indices as subscripts, and indeed that's what MathWorld, and most of the google hits for 'einstein notation' say. I think it would be better to change the article to use only sub-scripts, and then mention that sometimes superscripts are used. TimmmmCam (talk) 12:00, 26 October 2009 (UTC)

Distinguishing raised and lowered indices is necessary to get coordinate-system-independent ("covariant") expressions in non-Euclidean geometries. But I agree that in the common Euclidean case it is merely confusing, and common to simplify to a subscripts-only rule; see my comments above. — Steven G. Johnson (talk) 16:08, 26 October 2009 (UTC)

Why Einstein?

The naming of this article is to my understanding completely wrong. Tensor calculus was developed by Tullio Levi-Civita and Gregorio Ricci-Curbastro independent of Einstein and before it was applied to relativity theory. The idea that the tensor methods were useful for relativity theory was originally also not from Einstein but certainly adopted by him!

Stamcose (talk) 21:21, 23 December 2010 (UTC)

It's merely the practice of suppressing the summation sign that is being attributed to Einstein - it the article does not make this sufficiently clear, then it needs fixing. --catslash (talk) 22:32, 23 December 2010 (UTC)

Rhetorical questions.

If the matrix A is represented as $\mathbf {A} _{n}^{m}$, How is the transpose of A represented? If the matrix product "A times B" is represented as
$\mathbf {A} \mathbf {B} =A_{j}^{i}\,B_{k}^{j}$
Then how is "A transpose times B" represented?
$\mathbf {A} ^{t}\mathbf {B} =A_{?}^{?}\,B_{?}^{?}$

Some sources use the following notation
$\mathbf {A} =A_{\ j}^{i},\ \ \mathbf {A} ^{t}=A_{j}^{\ i},\ \ \mathbf {A} \mathbf {B} =A_{\ j}^{i}\,B_{\ k}^{j},\ \ \mathbf {A} ^{t}\mathbf {B} =A_{j}^{\ i}\,B_{\ k}^{i}$

I thought that placing the indices directly over each other was only allowable for Hermitian matrices. — Preceding unsigned comment added by NOrbeck (talk • contribs) 11:21, 18 April 2011 (UTC)

Fine tunning

$A^{ij},{A^{i}}_{j},{A_{i}}^{j},A_{ij}$ are the four types indexing rank 2 tensors which correspond to its bilineal maps the cases:

$A:V^{*}\times V^{*}\to \mathbb {R}$, constructed as $A=A^{kl}e_{k}\otimes e_{l}$

$V^{*}\times V\to \mathbb {R}$, similarly $A={A^{k}}_{l}e_{k}\otimes e^{l}$

$V\times V^{*}\to \mathbb {R}$, as homework

$V\times V\to \mathbb {R}$, also.

Well these are for real vector spaces--kmath (talk) 17:09, 8 October 2011 (UTC)