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Is the metric tensor constant in polar coordinates?

  1. Feb 15, 2014 #1
    I've been watching the Stanford lectures on GR by Leonard Susskind and according to him the metric tensor is not constant in polar coordinates. To me this seems wrong as I thought the metric tensor would be given by:
    [tex]
    g^{\mu \nu} =
    \begin{pmatrix}
    1 & 0\\
    0 & 0\\
    \end{pmatrix}
    [/tex]
    Since [itex]g_{\mu \nu}x^{\nu}x^{\mu}[/itex] is supposed to give the squared length of the vector [itex]x^{\mu}[/itex] and indeed it does:
    [tex]
    g_{\mu \nu}x^{\nu}x^{\mu} =
    \begin{pmatrix}
    r & \theta \\
    \end{pmatrix}
    \begin{pmatrix}
    1 & 0\\
    0 & 0\\
    \end{pmatrix}
    \begin{pmatrix}
    r\\
    \theta\\
    \end{pmatrix}
    =
    \begin{pmatrix}
    r & \theta \\
    \end{pmatrix}
    \begin{pmatrix}
    r\\
    0\\
    \end{pmatrix}
    = r^2
    [/tex]
    I'm not sure if he simply made a mistake in the lecture or if I've misunderstood something. I'd appreciate any feedback :)
     
  2. jcsd
  3. Feb 15, 2014 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    It's not ##g_{\mu \nu} x^{\mu} x^{\nu}##, it's ##g_{\mu \nu} dx^{mu} dx^{\nu}##, where the dx's are coordinate *differentials*. That is, if you have an infinitesimal line segment, the metric tensor tells you how long that line segment is in terms of the amount of change in each of the coordinates. In a flat manifold, we can get away with considering straight line segments of finite length as "vectors", but that's a bad habit to get into because it doesn't work in curved manifolds.

    The metric tensor in polar coordinates on a plane is [Edit: fixed indexes to subscripts instead of superscripts]

    $$
    g_{\mu \nu} =
    \begin{pmatrix}
    1 & 0\\
    0 & r^2\\
    \end{pmatrix}
    $$

    A purely radial vector, such as the one you are considering, has a nonzero ##dr## but zero ##d\theta## (i.e., the radial coordinate changes along the line segment but the angular coordinate does not). So its squared length will be

    $$
    g_{\mu \nu} dx^{\mu} dx^{\nu} =
    \begin{pmatrix}
    dr & 0 \\
    \end{pmatrix}
    \begin{pmatrix}
    1 & 0\\
    0 & r^2\\
    \end{pmatrix}
    \begin{pmatrix}
    dr\\
    0\\
    \end{pmatrix}
    =
    \begin{pmatrix}
    dr & 0 \\
    \end{pmatrix}
    \begin{pmatrix}
    dr\\
    0\\
    \end{pmatrix}
    = dr^2
    $$

    The angular part doesn't come into play because the ##d\theta## component of the line segment is zero, not because the angular part of the metric tensor is zero.
     
    Last edited: Feb 15, 2014
  4. Feb 15, 2014 #3
    Ah okay thank you that clears things up.
     
  5. Feb 15, 2014 #4
    Hi Peter. I'm not the expert that you are, so I could easily be wrong, but didn't you mean to have subscripts on the g in the above equation. If they are truly superscripts, shouldn't the θ-θ component of be 1/r2.

    Chet
     
  6. Feb 15, 2014 #5

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Oops, yes, you're right, what I wrote was the matrix for ##g_{\mu \nu}##, with subscripts. With superscripts it would be like you said, yes.

    [Edit: My previous post was still within the edit window so I went back and fixed it to have subscripts.]
     
  7. Feb 18, 2014 #6
    Talking about vectors first. In euclidean space the unit vectors are always pointing in the same direction. In for example spherical coordinates the unit vectors move and their directions change as you go around the sphere. The best I can say is that is just the way it is. So the unit vectors move in non-euclidean space. The metric tensor does not define curvature though. The Riemann tensor does that, which you can contract and get the Ricci tensor, the first term in Einsteins field equations. If you take the co-variant derivative you have to add the Christoffel symbol(it is not a tensor) because tensors do not change in any coordinate system. In euclidean space the Christoffel symbol which depends on the metric tensor is zero. On a sphere the Christoffel symbol does not vanish, because again you have to take into account that the tensors move in that coordinate system. If you mean by constant that the metric tensor in polar coordinates is not all ones, then you are right. The co-variant derivative of the metric tensor is always zero, no matter the coordinate system, that is the definition of a tensor. In euclidean coordinates the metric tensor does change when you move around. The Minkowski metric does not change because it is the trace(-1,1,1,1) of a matrices. I hope I helped but it is confusing to students taking physics also. Peter above has the correct definition of the metric tensor for polar coordinates. If you can just remember how vectors act then you can remember the rest. This is just an example a vector and a tensor are different animals. I am not using proper terminology because I am trying to explain it the best I can without going into great detail. So someone will criticize me, but if I can only get the general idea to you that is Ok.
     
    Last edited: Feb 18, 2014
  8. Feb 18, 2014 #7
    There is a great book on vectors and tensors called "A students Guide to Vectors and Tensors" by Daniel Fleisch. It is a cheap book and is really easy to understand and he gives all the answers to all the problems so you can make sure you did the problems right. He even has a video that explains the difference between vectors and tensors. Good Luck.
     
  9. Feb 18, 2014 #8
    I think what you are saying about Euclidean space is not quite what you mean. Polar coordinates on a flat plane features unit vectors that change with spatial position, and Christoffel symbols that are not zero. Yet this is a Euclidean space. However, the Riemann tensor is zero for a flat plane and in any flat space (including Euclidean space and Minkowski space).

    Chet
     
  10. Feb 19, 2014 #9
    I am so sorry but I was only trying to get across and idea. The more detail I go into the more I would lose his point. I could have gone into the fact that in polar coordinates it has extrinsic curvature and zero intrinsic curvature or intrinsically flat. But why? I could have told him that we are only interested in the intrinsic curvature of the universe, but why? I could have showed him how to solve Einsteins Field equations using Cartan's Structure Equations but there is not enough time. I could have gone into AdS/CFT correspondence. I was only trying to get across the point that only in flat space do the basis vectors stay the same. In my experience most students do not really understand that in spherical coordinates the basis vectors move. I even admitted that I was not totally technically accurate. Also the manifold for polar coordinated can be mapped to R^2. Sorry but I was only trying to make him understand this fact and at the same time giving the definition of a tensor. I also gave him a book that is one of the best for students like him to understand vectors and tensors, I have used it on students. I could have gone into Holonomic coordinate bases and Nonholonomic bases and why the forms approach is easier to use. A lot of PhD's do not know about this, especially today when computers can compute the metric, Ricci and Riemann tensors, I know some other PhD's in physics that do not know this. This is why I do not answer questions on here. I predicted that someone would find fault in my post and I was right. You are the expert on here so I will bow out. In polar coordinates there are two Christoffel symbols, -r and 1/R, and when plugged into the geodesic equation gives two differential equations who's solution is a line. So I said it right, best I can do off the top of my head. At his level why does he need to know all this. Your correction was correct so I showed why you are correct. Yes you were correct, but I now showed how to prove you are correct. I am not going to take the time to write out the equations, not used to the editor you use. I use different editors and also the one in Mathematica. The terminology I used earlier was not totally correct and I said it at the time. His question really was only a basic one and knowing about the mapping of manifolds would not help him. SO YOU WERE RIGHT AND I WAS WRONG. But I know how to prove you were right.
     
    Last edited: Feb 19, 2014
  11. Feb 19, 2014 #10
    Please accept my apology. No offense was intended. Since the OP seems to be new to this material, I was trying to avoid any confusion on his part. You obviously have much more experience than I do in this area, and I acknowledge that.

    Chet
     
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