The difference between a metric and the metric tensor
I have started to teach myself General Relativity and I have been pointed to a book by Robert M. Wald called General Relativity. I really like it actually, I like how it doesn't skip the math behind the theory. It makes it appear more beautiful to me. However I think the book is quite vague about the difference between a metric and the metric (as defined on page 22 if anyone has the book, it shouldn't be neccesary however). I understand that a tensor field is a quite general concept. For each point p in a manifold it simply specifies a tensor over the tangent space. I understand that a metric is a nondegenerate, symmetric tensor field of type (0,2) (that is: it takes 2 tangent vectors and turns it into a real number). Thus it defines a pseudo (not neccesarily positive definite) inner product at each tangent space. It is not neccessary to specify this tensor field to be linear, because due to the linearity of tensors, all tensor fields are linear at a given point. How should i interpret the metric however? Is the tensor assigned to all points allowed to change, that is: Given two different tangent spaces, the metric gives two different tensors? Is it constant? What excactly is the metric tensor, is it just the same as the metric?
Thank you 
Re: The difference between a metric and the metric tensor
Quote:
The metric and the metric tensor are essentially the same, but you can think of the metric as the general idea of a thing that tells you the distances and angles on a manifold, while the metric tensor is the precise mathematical implementation of this idea. 
Re: The difference between a metric and the metric tensor
Thank you for answer. I think there might be some insight hidden in some mathematical precision, so let me see if I get this straight:
Let T(k,l) be the set of all tensors of type (k,l). Let M be a manifold. A tensor field of type (k,l) is then a map, [tex]F : M \rightarrow T(k,l)[/tex] (where the output tensor is over the tangent space). A metric, g, is a tensor field of type (0,2) such that for each [tex]p \in M[/tex] the assigned tensor is symmetric and non degenerate. The tensor assigned to the point p is called the metric tensor. Is this definition retarded or flawed? 
Re: The difference between a metric and the metric tensor
It's good. I just have a couple of minor nitpicks:
That map F is usually defined to take a point p to the pair (p,g_{p}) where g_{p} is a (0,2) tensor on the tangent space at p. I think the only point of that is to make sure that the tensor field can be described as a section of a fiber bundle. (Not sure why that's desirable, and physics books often use your definition). If you meant to define "metric" as a tensor field, and "metric tensor" as a tensor on a specific tangent space, I disagree slightly with that terminology. I'd say that when people use the word "metric" in a GR context, they always mean "metric tensor", and when they use the word "tensor" they often mean "tensor field". So "metric" is often just a sloppy way to say "metric tensor field". I don't mind this little abuse of terminology, since it doesn't really cause any confusion (except for students who assume that the book they're reading never abuses the terminology :smile:). 
Re: The difference between a metric and the metric tensor
Thank you, your post really helped a lot. It seems to me that Wald isn't completely consistent with his own definitions of a metric and a metric tensor since he use the 2 words as synonyms.
I am not quite sure what you mean when you say that the tensor field F takes a point to the pair [tex](p,g_p)[/tex] because we want the tensor to be a section of a fiber bundle. I am not familiar with the term fiber bundle. 
Re: The difference between a metric and the metric tensor
Quote:
You can think of a fiber bundle E as some sort of space B (called the base space) with another space (the fiber) attached to each point. For example, a circle with a line segment attached to each point can be a cylinder, but if you twist or stretch the fibers as you go around the base space, you can get something else, e.g. a Möbius strip. E can always be described as a bunch of pieces of the form U×F glued together, where U is an open subset of B, but E is usually not equal to B×F. If it is, it's said to be trivial. A cylinder is a trivial bundle, and a Möbius strip is one of the simplest examples of a nontrivial bundle. I'm not going to go through all the technical stuff in the definition (which takes a lot of time to understand), but I'll mention that the main ingredient is a surjection [itex]\pi:E\rightarrow B[/itex] called the projection. Such a function ensures that E can be expressed as the union of all sets of the form [itex]\pi^{1}(b)[/itex], and that those sets are all disjoint. If all the conditions mentioned in the technical definition of a fiber bundle are met, then those sets (which may have additional structure (they can e.g. be topological spaces or vector spaces)) are all isomorphic to F (and therefore to each other). A section of a fiber bundle [itex](E,B,F,\pi:E\rightarrow B)[/itex] is a function [tex]X:U\rightarrow E[/tex] (where U is an open subset of B) such that [tex]\pi(X(b))=b[/tex] for all b in U. An example of a fiber bundle (to be more precise, a vector bundle, since the fibers have vector space structure) is the tangent bundle TM of a manifold M. It consists of ordered pairs (p,v) such that [itex]p\in M[/itex] and [itex]v\in T_pM[/itex]. (Note that it's the same p in both cases). It's a tangent bundle with E=TM, B=M, F=[itex]\mathbb R^n[/itex] (because each T_{p}M is isomorphic to [itex]\mathbb R^n[/itex]) and [itex]\pi[/itex] defined by [itex]\pi(p,v)=p[/itex] for all p. A vector field is defined as a section of the tangent bundle. 
Re: The difference between a metric and the metric tensor
Great, thank you for taking time to write such a detailed answer.
I have a book by John M. Lee on Smooth Manifolds that I have studied a bit. I remember seeing a chapter about the tangent bundle, though I haven't read it. Maybe I should spend some time studying this book some in detail. It is secondary to my goal however as I'm writing my bachelor thesis about black holes and not the math behind it all (though I find it very interesting in it's own right). 
Re: The difference between a metric and the metric tensor
Yes, it's nice to take some time later on to understand fibre bundles; GR is naturally described by the tangent bundle as soon as you start working with vielbeins and the spin connection. Here the fibres depend on the base manifold (space time).
Also the standardmodel is naturally described by fibre bundles. However, the fibres here are completely independent of the base manifold and are Lie algebra's. 
Re: The difference between a metric and the metric tensor
As a recommendation, if you are teaching yourself this book is really good. Gravity: An Introduction to Einstein's General Relativity by James B. Hartle.
P.S. I am teaching myself the GR also. Thanks Matt 
Re: The difference between a metric and the metric tensor
Ah yes the Hartle book is also very good, I had the book in an introductory course to GR. Thank you for your recommendation though. While I find it very well written it has a major flaw: It doesn't teach you how to solve Einsteins equation. It postulates the solutions. This is the price you have to pay for the simplified math I guess. It is a very good book to get familiar with the whole theory though. I find Wald's book satisfying due to the fact that it doesn't simplifies all the math in order to make the theory more intuitive. I have been looking at S. Chandresakhars book: Mathematical theory of black holes, and it seems in line with Wald's. I have been using it as cross reference for the parts where I found Wald's book unclear (mainly about the difference between tensors and tensor fields so far).

Re: The difference between a metric and the metric tensor
I'm having some trouble understanding Wald's definition of a derivative operator (also called the covariant derivative). When Wald attaches a lower index to the derivative operator what does that mean? How should I interpret his requirement 4 and 5: [tex]t(f)=t^a\nabla_a f[/tex] and [tex]\nabla_a \nabla_b f=\nabla_b\nabla_a f[/tex]? Is there any reason why he attaches the subscript a to the derivative operator in requirement 4? Couldn't it just as well be [tex]t(f)=t\nabla f[/tex]? I don't quite get requirement 5 either: Does it mean that [tex]\nabla_a[/tex] and [tex]\nabla_b[/tex] are 2 different derivative operators (that is they both satisfy requirement 14 but are not equal)?

Re: The difference between a metric and the metric tensor
Other texts define the connection as a function [itex]\nabla[/itex] that takes two vector fields to one vector field (and satisfies certain properties).
[tex](X,Y)\mapsto\nabla_XY[/tex] We can use this to define a map [itex]\nabla_X[/itex] for each X. [itex]\nabla_X[/itex] is defined as the map [tex]Y\mapsto\nabla_XY[/tex] Wald's [itex]\nabla_a[/itex] is [itex]\nabla_X[/itex] with X equal to the vector field that Wald writes as [itex]\partial_a[/itex], i.e. one of the partial derivative operators [tex]\frac{\partial}{\partial x^\mu}[/tex] corresponding to a coordinate system x. Note that Wald uses "the abstract index notation" which means e.g. that he would write X^{a} instead of X. The indices are there to show us what kind of tensor we're dealing with, by telling us if how many arguments it takes and if they are tangent vectors or cotangent vectors. For example [itex]T^a{}_b{}^c[/itex] would be a tensor that takes three arguments. The first and third are cotangent vectors, and the second is a tangent vector. 
Re: The difference between a metric and the metric tensor
I just realized my post before was a mess, so I have edited it to make it more clear what I mean.
I understand the abstract index notation, though I still have some difficulty in interpreting the subscript on the covariant derivative. The subscript mirrors the "extra" covector associated with the covariant derivative? Since it is a map of tensor fields of type (k,l) to (k,l+1) the subscript indicates what this extra covector is? 
Re: The difference between a metric and the metric tensor
I to am teaching myself GR. I am looking to get into more of the math behind it and am under the impression that GR is applied differential geometry? I need to find 1 more math class to get a minor and that is offered. Good pick? I seem to remember seeing a lot of this stuff in a book I borrowed from our library called "A First Course in Geometric topology and differential geometry" by Ethan D Bloch. The book is meant to be a starting point for those with a dif eq/liner algebra background.

Re: The difference between a metric and the metric tensor
This is my first time in this forums, I'm a student of Theoretical Physics. My references when I studied General Relativity were:
1. I don't like this one too much, but I have to accept it is necessary because it goes through the math necessary and give you quite a good notion of the physics that the math works. 2. My personal Favourite: SpaceTime and Geometry: An Introduction to General Relativity by Sean Carrol. I think this book is brilliant, Once you use d'Inverno to cover the Math and some notions of the Physics behind the math, This book enclosures your mind to the Physics all. I think that with this book I'm strongly in love with Sean Carrol. LOL. 3. My third reference was: Tensors Analysis on Manifold by Richard L. Bishop. This was to me a very useful book as a Mathematical Reference. Hope this could help anyone looking for references in GR. 
Re: The difference between a metric and the metric tensor
One of the conditions on [itex]\nabla[/itex] mentioned in the definition of a connection is that
[tex]\nabla_X(fY)=(Xf)Y+f\nabla_XY[/tex] Note that Xf and fX are not the same. Xf is the function [itex]p\mapsto X_pf[/itex] and fX is the vector field [itex]p\mapsto(p,f(p)X_p)[/itex]. The effect of [itex]\nabla_X[/itex] on a function is defined as [tex]\nabla_Xf=Xf[/tex] Use this definition with [itex]X=\partial_\mu[/itex], and you get [tex]\nabla_{\partial_\mu}f=\partial_\mu f[/tex] so [tex]t(f)=t^\mu\partial_\mu f=t^\mu\nabla_{\partial_\mu}f[/tex] Note also that the definition allows us to rewrite the first equation as [tex]\nabla_X(fY)=(\nabla_Xf)Y+f\nabla_XY[/tex] The other two requirements in the definition of a connection are [tex]\nabla_X(Y+Z)=\nabla_XY+\nabla_XZ[/tex] [tex]\nabla_{fX+gY}Z=f\nabla_XZ+g\nabla_YZ[/tex] 
Re: The difference between a metric and the metric tensor
Quote:

Re: The difference between a metric and the metric tensor
Ups... Sorry. The First Book is: Introducing Einstein's Relativity by Ray D'Inverno.

All times are GMT 5. The time now is 12:54 PM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums