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Geometric interpretation of metric tensor 
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#1
Nov312, 10:59 PM

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Hello,
can anyone suggest a geometric interpretation of the metric tensor? I am also interested to know how we could "derive" the metric tensor (i.e. the matrix <a_{i},a_{j}>) from some geometric considerations that we impose. 


#2
Nov412, 03:25 PM

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I'm not sure what you mean by "geometric considerations". The metric tensor is, of course, from the 'metric' or distance between two given points. For example, if points p and q are on the surface of a sphere of radius R, we can set up parametric equations using "spherical coordinates" with [itex]\rho= R[/itex]. That is,
[itex]x= Rcos(\theta)sin(\phi)[/itex] [itex]y= Rsin(\theta)sin(\phi)[/itex] [itex]z= Rcos(\phi)[/itex] Then [itex]dx= Rsin(\theta)sin(\phi)d\theta+ Rcos(\theta)cos(\phi)d\phi[/itex] [itex]dy= Rcos(\theta)sin(\phi)d\theta+ Rsin(\theta)cos(\phi)\phi[/itex] [itex]dz= Rsin(\phi)d\phi[/itex] and so [tex]dx^2= R^2sin^2(\theta)sin^2(\phi)d\theta^2 2R^2sin(\theta)cos(\theta)sin(\phi)cos(\phi)d\theta d\phi+ R^2cos^2(\theta)cos^2(\phi)d\phi^2[/tex] [tex]dy^2= R^2cos^2(\theta)sin^2(\phi)d\theta^2+ 2R^2sin(\theta)cos(\theta)sin(\phi)cos(\phi)d\theta d\phi+ R^2sin^2(\theta)cos^2(\phi)d\phi^2[/tex] so that [tex]dx^2+ dy^2= R^2sin^2(\phi)d\theta^2+ R^2cos^2(\phi)d\phi^2[/tex] and since [tex]dz^2= R^2sin^2(\phi)d\phi^2[/tex] [tex]ds^2= dx^2+ dy^2+ dz^2= R^2sin^2(\phi)d\theta^2+ R^2 d\phi^2[/tex] Which tells us that the metric tensor for the surface of a sphere of radius R is [tex]\begin{bmatrix}R^2sin^2(\phi) & 0 \\ 0 & R^2\end{bmatrix}[/tex] 


#3
Nov612, 09:13 AM

P: 280

Or a more compact way to calculate the metric tensor would be the transpose of the Jacobian of your parametrization times the Jacobian:
[tex]g_{ij}=J^TJ[/tex] The metric tensor is a function that defines how you measure the distance of two points in a given manifold. For instance, in the earth's surface you measure distance in geodesic arcs, since it's curved. If you are in a city however and you want to calculate your car's travel path, you would have to measure distance in terms of 'city blocks', since you can't simply travel along any path you please in a city. Intuitively you can see that even though your manifold is the same (the Earth), there are more than one ways to calculate the distance between two points. This is the role of the metric tensor. 


#4
Nov812, 12:53 PM

P: 626

Geometric interpretation of metric tensor
Hello,
thanks for both answers. It seems that the notion of metric tensor naturally arises when one tries to calculate the differential element of arc in a manifold (as Hallsofivy showed). I find my question a bit difficult to formulate but basically I am interested in knowing if there is another geometrical concept (besides the differential element of arc length) from which the metric tensor naturally arises. For example: in the previous reasoning we have done the following: (1) let's try to calculate the differential element of arc in a manifold (2) notice that the result depends on a quantity [itex]g_{ij}[/itex] that we call "metric tensor". I want to replace "differential element of arc " in (1) with something else and see if I obtain again the metric tensor. Meldraft came very close to answering my question because we know that the columns (or rows?) of the Jacobian matrix represent the edges of a "differential (oriented) parallelepiped" which is a sort of measure of how stretched is the manifold in a specific point w.r.t. to any direction. Note also that the determinant of the Jacobian would be the (hyper)volume of the parallelepiped. However I don't exactly know what would be (in terms of parallelepipeds) the geometric interpretation of [itex]J^T J[/itex], which happens to be precisely the metric tensor! 


#5
Nov812, 01:43 PM

P: 350

I am not sure if this satisfies your curiosity but this explains the formula (J^T)J.
If v is a vector in your coordinates, then Jv is it image on the manifold in R^n. Given another vector w, if you want to find the dot product of the two images Jv, Jw, you get [itex] \langle Jv, Jw\rangle = \langle J^TJv, w\rangle.[/itex] If you want to involve the parallelepiped in the explanation, you can understand it as a geometrical representation of the linear map J. 


#6
Nov912, 08:44 AM

P: 280

The Jacobian of a function describes the orientation of a tangent plane to the function at a given point. The assignment of a metric tensor on a manifold introduces a scalar product [itex]\langle X,Y\rangle[/itex] of covariant vectors [itex]X,Y[/itex] on the tangent space at that point, defined as a bilinear function [itex]g_p(X,Y)[/itex], where [itex]g_p[/itex] is the value of the field g at the point p, so:
[tex]\langle X,Y\rangle=g_{ij}(p)X^iY^j, X={X^i}, Y={Y^j}[/tex] Intuitively you can see that defining a metric at a manifold gives you a way to 'navigate' through each point of the manifold, and get to another. The Jacobian enters the scene since it helps define the vectors at the tangent space. Basically, the manifold is there, but unless you assign a metric to it you can't really do anything with it. 


#7
Nov912, 10:41 AM

Sci Advisor
P: 1,727

To me the idea of metric tensor arises from the idea that even though the geometry of a space may not be Euclidean overall, in small regions it is well approximated by Euclidean geometry. The flat Euclidean space obeys the law of Similar Triangles which in turn implies the existence of an inner product. This suggests that an infinitesimal inner product should exist at each point of the noneuclidean space.
Remarkably, this infinitesimal inner produxt suffices to recover the geometry of spaces that are well approximated by Eulidean geometry in small regions. The same idea applies in General Relativity where in a free fall frame of reference, geometry is well approximated by Euclidean geometry/Minkowski geometry for small time intervals and small regions of space. I would guess that this line of thinking led to the use of inner products on the tangent space as a tool for describing the laws of gravity. 


#8
Nov1012, 12:01 PM

P: 1,309

The metric is like a dot product that you get at each point by flattening out the manifold because when you zoom in, it's like Euclidean space. As such, the proper geometric interpretation really ought to be in terms of projection from one vector to a unit vector, rather than in terms of a parallelepiped.
Another good way to think of a metric is that it measures (infinitesimal) distances and angles. If you know infinitesimal distances, than you can measure any distance by integration. 


#9
Nov1112, 02:43 PM

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#10
Nov1112, 03:05 PM

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#11
Nov1112, 09:27 PM

P: 280

I would not expect it to be true for pseudoRiemannian manifolds though (could it?).



#12
Nov1112, 11:50 PM

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#13
Nov1212, 09:17 AM

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