Freeman Posted September 2, 2005 Posted September 2, 2005 I am sorry to say that I do not understand the metric tensor one bit. Isn't it the Kronecker Delta in Euclidean space? And wouldn't that be a collection of row vectors?
DQW Posted September 13, 2005 Posted September 13, 2005 I am sorry to say that I do not understand the metric tensor one bit.Do you understand what a metric is ? Very roughly speaking, a metric is a function that describes the "distance" between points in some set. As a function g(x,y) a metric must satisfy certain properties, such as non-negativity, symmetry and something that states that the distance of any point from itself must be zero (and the converse). Now, this function may also be viewed as a tensor, in which case this tensor must also obey a bunch of corresponding properties (for instance, it must be symmetric and positive definite). In any space, the components of the metric tensor tell you how to calculate general infinitesimal displacements in that space. Isn't it the Kronecker Delta in Euclidean space? Yes, the components of the metric tensor for R^n are Kronecker Deltas. This simply follows from the generalization of Pythgoras. [math](dl)^2 = \sum_{i,j} g_{ij}~dx_i~ dx_j[/math] For R^n, you simply have [math] (dl)^2 = \sum_{i=1}^n (dx_i)^2 [/math] So, from above, this gives [math]g_{ij} = 1,~if~i=j~and~~ g_{ij}=0~otherwise [/math] This is nothing but the Kronecker Delta [imath]\delta _{ij} [/imath] And wouldn't that be a collection of row vectors?I don't understand this question. To me it looks like the metric tensor for Eucliedean space will simply be the identity matrix [imath]\mathbf{1}_n [/imath] (but I might be mistaken). But so far, this is only something to give you an intuitive picture based on more familiar stuff. To really understand the metric tensor, you must know how it is rigorously defined. This is not something I'm entirely comfortable with (haven't looked into it in ages), so I'd leave it to the likes of matt, if you have more questions.
Xerxes Posted September 15, 2005 Posted September 15, 2005 Yes, DQW gave a nice exposition. I might add that the [math] g_{ij} [/math] assigns a coefficient to the derivatives of the generalised Pythagorean. 1 or zero is usual in Euclidean 3-space (hence the identity with the K. delta). As I recall, Lovelock & Rund have a nice and full definition. See if I can dig it out. Hmm..why didn't my LaTeX work? EDIT Cheers DQW!
DQW Posted September 15, 2005 Posted September 15, 2005 Cheers ! I don't believe there is a way to delete a post entirely.
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