The metric tensor

Discussion in 'Physics & Math' started by QuarkHead, Apr 20, 2016.

  1. QuarkHead Remedial Math Student Valued Senior Member

    There has recently been a lot of talk about the metric tensor field in the General Theory of Relativity- you can blame me for that (though I am unapologetic!). Since a tensor field is (simply put) the assignment of a tensor to every point in some manifold, it may be useful to know what a metric tensor actually is.

    I will first give a Mickey Mouse version (for motivation - do not take it too seriously!)

    But even firster, a refresher in elementary linear algebra

    Suppose first \(V\) an "ordinary" real vector space - that is, it is defined over \(\mathbb{R}\), the Real umbers. (of course there is no such thing as an "ordinary" vector space, but I merely mean it may not be tangent to some non-trivial manifold).

    I define the mapping \(G:V \times V \to \mathbb{R}\) by \(G(v,w) \in \mathbb{R}\). I will assume this map is bilnear - linear in each argument taken separately.

    Now if and only if it is the case that

    1. \( G(v,w) = G(w,v)\) (symmetry) - note this is literally true only for Real vector spaces, and

    2. For all \(v \in V \ne 0\) there exists some \(w \in V\) such that \(G(v,w) \ne 0 \in \mathbb{R}\) (non-degeneracy), and

    3. \(G(v,w) \ge 0 \in \mathbb{R}\)

    then I will call this an inner product on \(V\).

    Note that \(G(v,v) =0 \Rightarrow v =0\).

    Under this circumstance only, I define the "length" of a vector here as the positive square root \((G(v,v))^{\frac{1}{2}} = ||v||\). In a more general context this is called the vector norm

    I remind you that in Euclidean space the length of the vector \(v=\sum\nolimits_j \alpha^je_j\) (the set \(\{e_i\}\) are called basis vectors - they span the space) is

    \(||v|| = (\sum\nolimits_j\alpha^j)^2)^{\frac{1}{2}}\) - since length is just a real number, just like the scalar components of any vector (here the \(\alpha^i\)), we not concerned with the set of basis vectors \(\{e_j\}\).

    From which we may infer that \(||v||^2 =(\sum\nolimits_j \alpha_j)^2\) which we call a "quadratic form" - it is a polynomial of degree 2 in each term.

    Then from the ordinary cosine rule from trigonometry, if we know the relative lengths of any 2 vectors - or rather the projection of one along the other - we know the angle between them

    Thus we have a complete metric on our vector space \(V\).

    Question is: will this carry over to the case for the tangent vector space \(T_pM\) where \(M\) is a non-trivial manifold and \(p\in M\) is a point. In particular, will this give us a metric on the entire manifold, rather than just on its tangent spaces? (You may want me to explain what I mean by a "non-trivial manifold" - if so, just ask)

    There is more - much more - to come if you want. In the meantime if I have been unclear (or wrong) just say
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  3. QuarkHead Remedial Math Student Valued Senior Member

    I confess I am surprised there has been so little interest in this thread, the more so since the thread title has been much discussed on this forum. Maybe what I have posted so far is so familiar to you all, you have nothing particular to say?

    So I will try once more by introducing a sort of vector space that may be a little less familiar to some.

    First recall that, merely for simplicity, I am talking about Real vector spaces - those defined over the field of Real Numbers. It is a fact that many of the vector spaces used in applications are not Real but Complex. This presents no real problems, so we can continue in the "Real world".

    So, for every Real vector space \(V\) there exists another vector space \(V^*\), which is the space of all linear mappings \(V^*:V \to \mathbb{R}\). These mappings are called "linear functionals" and the vector space \(V^*\) is called the dual space.

    So let's return to our inner product \(G:V \times V \to \mathbb{R},\,\, G(v,w) \in \mathbb{R}\). Let us now fix some \(v \in V\) and write \(G(v,\,\cdot)\) and call this as \(G_1(w) \in \mathbb{R}\) for any \(w \in V\). Thus, \(G_1 \in V^*\).

    Now for some fixed \(w \in V\) define \(G_2\) by \(G_2= G(\cdot, w)=G_2(v) \in \mathbb{R}\) Thus also \(G_2 \in V^*\).

    Putting this together we will have that \(G_1(w)G_2(v) \in \mathbb{R}\) which implies that \(G=G_1 \otimes G_2:V \times V \to \mathbb{R}\).

    Since the symbol \(\otimes\) defines the tensor product, we can call this our metric tensor.

    Now suppose we set \(G_1= \sum\nolimits_j g_j\epsilon^j\) and \(G_2= \sum\nolimits_k g_k \epsilon^k\), where the \(\{\epsilon^i\}\) are a basis for \(V^*\) and the \(g_i\) are scalars. This is just \(G=G_1 \otimes G_2 = \sum\nolimits_{j,k}g_{jk} \epsilon^j \otimes \epsilon^k\)

    It is customary, especially in applications, to drop all reference to the bases, and to refer to the metric tensor as \(g_{jk}\), even though in reality it refers to the scalar components of this tensor.

    So what has all this do with inner products (bilinear forms) and norms - "lengths" - which I called quadratic forms? Specifically, how does this give a sense of "distance" on a manifold which may very well not be flat. The mathematics gets a little challenging, but I am willing to attempt it.......
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  5. Schmelzer Valued Senior Member

    I think there is not much interest simply because a forum is not a good place to teach mathematics.
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  7. origin In a democracy you deserve the leaders you elect. Valued Senior Member

    I would say the reason for so little interest in the thread is due to what I like to call the mathematician problem.
    Mathematics is essentially a language and if you do not know the language you will not understand the math. Mathematicians often assume that the audience understands the language and often are dissmisive of those that do no know the language.
    So I think there are 2 groups; those that are familiar with math and those that do not understand what you are talking about.

    Your post starts with this:
    Since a tensor field is (simply put) the assignment of a tensor to every point in some manifold
    What is a tensor field, what is a tensor, what is a manifold? - why start off with 3 undefined terms.

    Then you follow with this:
    I will first give a Mickey Mouse version...
    Translation - any idiot can understand this. Not very conducive to soliciting questions

    You then dive into the math using words and symbols that have a very specific meaning in mathematics but you do not define them.

    I would hazard a guess that half the members on this site have never even worked with any math utilizing vectors, so they read about 3 lines of your post, their eyes glazed over and the moved on. The people that have learned about tensors, shrug their shoulders and say yeah, yeah.

    What you are trying to do is admirable, but to be honest I would have been surprised if anyone had asked a question about your first post.
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  8. arfa brane call me arf Valued Senior Member

    In my lecture notes from 2nd year calculus:
    Example 1.1

    Let \( \mathbb P \) denote the set of polynomials over \( \mathbb R \), then

    \( \mathbb P = \{a_0x^n + ... +\, a_{n-1}x + a_n\, |\, a_i \in \mathbb R \} \)
    A basis for \( \mathbb P \) is \( \{1,\,x,\,x^2,\,...\} \)."

    Is there a standard way to define the inner product, and is there a notion of distance?
    And that's polynomials, what about functions in general?
    Example 1.2

    Let \( C[a,\,b] \) be the set of real-valued continuous functions on the closed interval \( [a,\,b] \). Clearly if \( f,g \in C[a,\,b] \) then \( f + g \in C[a,\,b] \) and \( \alpha f \in C[a,\,b] \) for all \( \alpha \in \mathbb R\).
    It is not immediately clear what would constitute a basis for \( C[a,\,b] \).

    Since ... vector spaces can consist of functions, we can define inner product spaces on functions."

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  9. QuarkHead Remedial Math Student Valued Senior Member

    Sure. Let \( P_1,\,\,P_2 \in \mathbb{P}\) with \(P_1 = \sum\nolimits_{j=0}^n a_jx^j\) and \(P_2=\sum\nolimits_{k=0}^nb_k x^k\).

    Then the inner product (a Real number, recall) is \(a_0b_0+a_1b_1+a_2b_2+.......+a_nb_n\) So the norm is as I gave it - replace my \(v\) with \(P_i\)
    No, you are correct, this is a deep question. However
    We can, by \(\int_a^b f(x)g(x)\,dx\) for any \(x \in [a,\,b]\)

    Last edited: Apr 23, 2016
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  10. Schneibster Registered Member

    I'll be by shortly, QuarkHead, but I been very busy at work and haven't had time to look this over; thanks for posting this, I know I asked for it, and I'll put in some time to make your effort worthwhile before long.
  11. arfa brane call me arf Valued Senior Member

    Ok. According to the notes I have, you can also introduce a weight function w(x), as

    \( (f,g) = \int_a^b w(x)f(x)g(x)\,dx\)

    In fact you already did, if w(x) = 1.
    Last edited: Apr 23, 2016
  12. QuarkHead Remedial Math Student Valued Senior Member

    While this is true, I don't believe I introduced undefined terms - if I did, say which, and I will explain.

    But look - this is the "Physics & Mathematics" forum. Physicists use a lot of mathematics, some it it very, very hard. Obviously mathematicians do too. So why is a mathematical thread out of place here, since it seems reasonable to assume that our visitors have an interest in one or other of these subjects?

    Surely it cannot be that you, or anyone else here for that matter, believes that physics can be done by pure intuition, or analogy, or pictures or what-all-have-you without any recourse at all to mathematics?
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  13. origin In a democracy you deserve the leaders you elect. Valued Senior Member

    I do not disagree at all.
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  14. Lucas Registered Senior Member

    i have only some elementary knowledge of linear algebra, but I'd be interested on learning about the metric tensor.
    Would it be correct to say that the metric tensor is the tensor product of two linear functionals? If it is so, is this the same kind of metric tensor as the metric tensor used in General Relativity?
  15. arfa brane call me arf Valued Senior Member

    I could follow everything in your OP up until you introduce the tangent vector space. About the only kind of tangent space I can visualize is over a sphere--each point has a set of directions which is "tangent" to the sphere, or a circle of directions is "over" each point.

    I think you should dumb it down a little; you're trying to explain how the notion of distance between points generalises. We looked at \(\mathbb R^n \) and the inner product and norm (in a LA course at 2nd year I took recently), but not tensors. I have a naive idea that tensors are somehow connected to rowspaces and columnspaces. Finally, in linear systems of equations, is the set of solutions a manifold?
  16. QuarkHead Remedial Math Student Valued Senior Member

    Yes, it would.
    Yes it is, but, although the explanation I gave is perfectly adequate for "flat" spaces, things get a little more tricky for the non-flat case
  17. Farsight

    Quarkhead, I think the lack of interest is because you're defining abstract things in terms of other abstract things, and you aren't referring to real space and gravitational fields enough.
  18. QuarkHead Remedial Math Student Valued Senior Member

    I don't imagine you are the only person who finds it hard to visualize anything other than a 2-manifold.

    The essential point, though, is that the naive picture of a tangent has it (the tangent) "sticking out" into some surrounding space - the so-called embedding space. This is in general undesirable (what is the embedding space for spacetime?). The trick is to use as a basis for the tangent space the differential operators \(\frac{\partial}{\partial x},,\,\, \frac{\partial}{\partial y}\) etc. We can very very loosely think of this as the "tendency" of some point to "move" in the direction of the chosen coordinates. The "magnitude" of this "tendency" is given by scalar multiplication.

    For example, a vector in the vector space tangent to some point \(p \in M\) - we write \(T_pM\) for this space - is written \(v = \sum\nolimits_j\alpha^j \frac{\partial}{\partial x^j}\), where the \(\{x^j\}\) is the set of coordinates for the chosen point.

    As a contraction some people write \(\frac{\partial}{\partial x^j}\) as \( \partial_j\).

    As a further contraction, some also suppress the summation signal and simply write \(v=\alpha^j\partial_j\) Yet others also drop all reference to to the bases and say that \(\alpha^j\) IS the vector.

    And also all the above to tensors of higher rank (the rank of a tensor is the number of vectors that enter into the tensor product that defines it)
  19. QuarkHead Remedial Math Student Valued Senior Member

    Mathematics are under no obligation to relate their craft to what you yourself might call the "real world".

    I would go so far as to say that to insist on it leads to logical problems, as the following simple cases may illustrate.

    We know that 1 + 1 = 2 , an abstraction. Suppose we insist this only always true in the "real world. But what is 1 Einstein + 1 Einstein? 2 Einsteins? Is there, could there ever have been, such a thing?

    How about 1 automobile + 1 dream about automobiles? What sort of object is this?

    And yet most of us consider 1 + 1 = 2 a universal truth.
  20. arfa brane call me arf Valued Senior Member

    Nonetheless, 2-manifolds are used as examples. I know an hyperboloid of one sheet is a set of straight lines,in fact every point has two straight lines through it, which can't be possible on a sphere. Does that mean it has a metric which is easier to define than a sphere?
  21. QuarkHead Remedial Math Student Valued Senior Member

    Ya know what, arfa? I am starting to like you, as you are thinking, and asking intelligent questions.

    The answer is no - that is because the metric is defined in terms of tangent vector spaces and the norm of its members

    Consider this. Choose a point on your hyperboloid and define a tangent vector space there. Allow one of your basis vectors at this point to be tangent to one of the so-called straight lines that passes through it. Then if you insist, as you usually do, that the second basis vector is orthogonal to the first (your hyperboloid is of course a 2-surface), then this second basis vector cannot possibly be tangent to any line that can be called straight in the same sense.

    In other words, we are dealing with curvilinear coordinates, just as in the case of the 2-sphere.
  22. arfa brane call me arf Valued Senior Member

    Ok, so the metric has to be locally defined, or, you want something that doesn't depend on the overall geometry (sphere vs hyperboloid); tangent spaces are tangent locally. That is, you want a linear map so you take small areas of the surface "locally diffeomorphic" to Euclidean 2-space . . .?

    A sphere has no boundary, an hyperboloid of one sheet has two boundaries, does something special need doing at the boundaries, do you need to take some kind of limit? Do you avoid the boundaries altogether?
  23. arfa brane call me arf Valued Senior Member

    The kind of vector space I got coziest with is the one used in analog circuit analysis.

    Now, it should be obvious to anyone who did a bit of geometry and trigonometry, that sin(x) and cos(x) are orthogonal functions, or loosely they are perpendicular for any value of x. Indeed, Wikipedia (that bastion of foreknowledge) tells us that Fourier series have ("are") an orthonormal basis:

    But, where is the manifold? Is it just the (Maxwellian) electromagnetic field? What does distance correspond to in an electronic circuit (say we start with a network of purely resistive elements)--is there a "distance between voltages"?
    Last edited: Apr 25, 2016

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