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Preservation of the infinitesimal element of length 
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#1
Nov1112, 10:39 AM

P: 450

My two questions are related to the title. The problematic is: "How can we connect the coordinates ... x^{α}... for α = 0, 1, 2 and 3 to the coordinates of the same object in another frame, say ...y^{λ} for λ = 0, 1, 2, 3 in preserving the quantity η(x)_{αβ}. dx^{α}.dx^{β} = η(y)_{λμ}. dy^{λ}. dy^{μ}?
Can someone explain clearly: 1°) why we have supposed that the coordinates transformations should absolutely be linear (and not, e.g. a Taylor's developpment including terms of highter degree)? > Poincaré group; 2°) why didn't we prefered to work with the relations proposed in 1869 by E.B. Christoffel for such problematic? Thanks 


#2
Nov1212, 07:42 AM

P: 1,020

It is supposed that the infinitesimals will satisfy the linear transformation i.e. the connection should be affine in character and that is where coefficient of affine connection comes into picture (Christoffel symbols).which is the key to parallel displacement without embedding suface into higher dimensional euclidean spaces.



#3
Nov1212, 01:20 PM

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#4
Nov1212, 11:59 PM

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PF Gold
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Preservation of the infinitesimal element of length



#5
Nov1312, 06:21 AM

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#6
Nov1312, 11:55 AM

P: 450

@ PeterDonis
@andrien: thanks for reinsisting on the link "LeviCivita connection  parallelism". I shall reread again some historical documents with the hope to understand that link better (the question of the embedding). Otherwise, I think (do you think it is correct?) the Christoffel's work can a priori be reused for nonsymmetric differential expressions 


#7
Nov1312, 01:35 PM

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PF Gold
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As far as the question of why we use linear transformations, I believe it's because those are the only ones that preserve geometric invariants. (The line element is only one of these, btw.) There's another thread currently active that touches on this topic: http://physicsforums.com/showthread.php?t=651640 It's only talking about flat spacetime, though. As far as the question of why the metric tensor is symmetric, that follows from the use of the LeviCivita connection. I believe there have been proposals to construct a theory like General Relativity but using a different connection, which would allow the metric tensor to be nonsymmetric (since it's a rank2 tensor, that basically means there would be a symmetric part and an antisymmetric part, the "torsion"the LeviCivita connection is "torsionfree"). My understanding is that all such proposals to date make exactly the same physical predictions as standard GR does, so I don't really see the point. But I am not up to date in this area. You might try Googling "gravity with torsion" or something similar and see what comes up. 


#8
Nov1312, 04:37 PM

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I'm not terribly sure I understand BlackForest's notation, but I think he's starting out asking for a group of isometries, i.e. diffeomorphisms that preserve the metric, assuming that's what he means by the [itex]\eta_{\alpha\beta}[/itex]. And going on from there.
I think the existence of such diffeomorphisms depends on the nature of the space itself, in particular it's Killing vectors. If you've got at least one Killing vector, you can write down at least one such diffeomoprphism. But if you don't have any Killing vectors at all, the answer to the "How can you" question becomes "You can't" rather than "how". So it might be good to start out with asking in general "can you" before asking "how", unless you're thinking specifically of flat spacetimes. As I recall from Wald (I didn't look it up) every one parameter group of diffeomorphisms is associated with a vector field, but only if the group of diffeomrphisms is an isometry is the vector field a Killing vector field. And you can think of the Killing Vector as the generator of the group, an infinitesimal translation (or rotation  or whatever) that generates the whole group. I haven't read Christoffel's original work, I just use the symbols, so I don't know how it applies to this problem. My intuition is saying to me that the isometries might be linear only in flat, homogeneous spaces, but I'm not sure it's right at this point. 


#9
Nov1312, 05:18 PM

P: 3,001

The OP title talks about preservation of infinitesimal length, and that suggests he is referring to infinitesimal isometries, so I would say the transformations would be linear regardless of the curvature of the manifold.



#10
Nov1312, 06:57 PM

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#11
Nov1312, 10:16 PM

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#12
Nov1312, 10:34 PM

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[tex]v = g_{ab} v^a v^b[/tex] Write this out for a simple 2dimensional space: [tex]v = g_{00} v^0 v^0 + g_{01} v^0 v^1 + g_{10} v^1 v^0 + g_{11} v^1 v^1 = g_{00} ( v^0 )^2 + ( g_{01} + g_{10} ) v^0 v^1 + g_{11} ( v^1 )^2[/tex] This will be welldefined and unique for any vector [itex]v[/itex] whether or not [itex]g_{ab}[/itex] is symmetric, because the "cross terms" are effectively symmetrized anyway. In other words, the inner product only "sees" the symmetric part of the metric, so the presence of a nonzero antisymmetric part doesn't affect it at all. At least, that's how it looks to me. 


#13
Nov1312, 10:42 PM

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#14
Nov1412, 12:47 AM

P: 450

So, the Christoffel's work is centered on the preservation of quantities like p_{αβ}. x^{α}. x^{β} where when you read the details of the work one must have p_{αβ} = p_{βα}. Now, except if I did misinterpret this work (of Christoffel), it is possible to rewrite it in supposing that p is not automatically symmetric; i. e. p could be antisymmetric, implying p_{αβ} + p_{βα} = 0. This eventuality is reducing the differential expression to a sum on β of p_{ββ}. (x^{β})^{2}... It is probable that Christoffel's work concern real numbers but it could be meaning full to relook it for complex numbers and why not for quaternions... this would invalid the argument y^{α}. y^{β} = y^{β}. y^{α}... and make everything quite more complicated, offering a natural link with a noncommutative world... So I must work  Come later again. 


#15
Nov1612, 04:24 PM

P: 450

In fact, I think this discussion should be continued here: http://physicsforums.com/showthread.php?t=651640&page=2
"Showing that Lorentz transformations are the only ones possible" As I read it in the other discussion, the answer to my first question is not trivial... 


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