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I Kind of basis

  1. Dec 11, 2016 #1
    Is it correct, at least in the context of general relativity, to say that in a coordinate basis, the inner product between space-like basis vectors will be 1, and in a non-coordinate basis the inner product will be defined by the corresponding component of the metric? Can I take this conditions as a definition of coordinate and non-coordinate basis?
     
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  3. Dec 11, 2016 #2

    PeterDonis

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    No. For a simple counterexample, consider the inner product of, say, ##\partial / \partial r## and ##\partial / \partial \theta## in Schwarzschild coordinates. [Edit: deleted incorrect statement.]

    You have it backwards. The non-coordinate basis is the one in which we carefully pick the basis vectors to be unit vectors. Coordinate basis vectors are the ones whose inner products are defined by the appropriate metric coefficients and which are not, in general, unit vectors.
     
    Last edited: Dec 11, 2016
  4. Dec 11, 2016 #3
    Can you give me a example of a line element in a non-coordinate basis?
     
  5. Dec 11, 2016 #4

    PeterDonis

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    No, because line elements are written in terms of metric coefficients as functions of the coordinates, so the concept of a line element in a non-coordinate basis doesn't even make sense.

    It might help if you would give more details about why you are interested in the difference between a coordinate and non-coordinate basis.
     
  6. Dec 11, 2016 #5

    pervect

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    A line element? Not really. But I can give some examples in polar coordinates that might help. Our polar coordinates will be ##r, \theta## and the line element will be ##dr^2 + r \, d\theta^2##.

    Now let's discuss the notation for a bunch of interesting quantities. First, let's consider a unit vector in the r direction. The notation for this would be ##\hat{r}## in pre-tensor notation. In tensor notation, we represent vectors by partial derivative operators - this is probably covered in your text, but you might want to review it. I believe it's discussed in Sean Caroll's online lecture notes (or the book he wrote based on them) as well. If you want a link to Caroll's online notes, let me know.

    For reasons that hopefully will become more clear later, we are also interested in the dual basis. A dual vector space is a map from a vector space to a scalar, something else that should be mentioned in your text. We usually write the dual of ##\hat{r}## as dr. Texts differ on the notation here, it'd be good to look up the notation your text uses.

    There's a tricky point here. dr could be either a number, or a dual vector. Recall that dual vectors are maps from vectors to a number. So the value of dr is always a number in either case, but if we interpret dr as something that operates on a vector to give a number, then dr is a one-form. When it doesn't operate on anything, then dr is just a number.

    We might also write ##\omega_r## instead of dr, as it's fairly standard to use ##\omega## to denote a one-forms.

    WIth these preliminares given a cursory exposition, let's get on to the confusing part. What about ##\theta##?

    Well, a unit vector ##\hat{\theta}## is ##(1/r) \, \partial / \partial \theta##. This is probably not familiar looking. A unit one-form would be ##r \, d \theta##. Which hopefully is a bit more familiar looking.

    A set of basis vectors simply means that at every point, we have vectors - in this example, unit vectors - ##\hat{r}## and ##\hat{\theta}## - at every point.

    We could also write a dual basis, a basis of one forms at every point.

    Let's go back to discussing the difference between coordinate and non-coordinate basis vectors for a minute - since that was closely related to the original question. The coordinate basis at every point in our 2d manifold associates a pair vectors to every point. These vectors are ##\partial / \partial r## and ##\partial / \partial \theta##. These are not unit vectors. In our pretensor notation, they'd be ##\hat{r}## and r ##\hat{\theta}##.

    To have unit length basis vectors at every point, we need to define a non-coordinate basis. A lot of physics texts will define a non-coordinate basis of unit one forms, this is a very handy technique to pick up. "An orthonormal basis of one-forms" might sound scary, but it really isn't.
     
  7. Dec 11, 2016 #6

    George Jones

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    Just to be clear here, do you mean the inner product between a spacelike coordinate basis vector and itself? The inner product between two spacelike basis vectors (coordinate or non-coordinate) can be positive, zero, or negative. The inner product between any spacelike (basis) vector and itself (coordinate or non-coordinate) is always positive (with appropriate sign convention).
     
  8. Dec 11, 2016 #7
    pervect, I really appreciated your very detailed answer.
    In this case, the non-coordinate basis vectors would also be mutually orthogonal? That is, a set of non-coordinate basis vectors would also be a set of orthonormal basis vectors?

    No, I was considering the inner product between any two out of the three vectors.
    I though the time basis vector were the only with a negative component, isn't it?
     
  9. Dec 11, 2016 #8
    Just to know the difference between them, not a particular case.
     
  10. Dec 11, 2016 #9

    pervect

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    In this case, yes, ##\hat{r}## is orthogonal to ##\hat{\theta}##, and they're both unit length so we say it's an orthonormal basis. A general basis wouldn't necessarily be orthogonal, or normal, or a coordinate basis. But while orthonormal bases and coordinate bases are special cases, they're useful special cases.
     
  11. Dec 12, 2016 #10

    Ibix

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    A coordinate system lets you define a map between points in the manifold and a regular grid of points on a standard Euclidean plane. For example, mapping where the latitude and longitude lines on the globe cross onto a square grid on a piece of paper is the standard Mercator projection.

    A coordinate basis is the standard Euclidean basis on the piece of paper, inverse mapped into the manifold. A non-coordinate basis is pretty much anything else. The vielbeins ("four-legs" - one for each direction in spacetime) that you may be aware of are the 4d equivalent of just dropping a set-square on the ground and using that to define directions (so three rods and a clock).

    If you think about how the Mercator coordinate basis works, you'll see that the easterly vector gets longer as you move away from the equator. It's an orthonormal system at the equator, but only orthogonal elsewhere. So this is a good way to draw (nearly) everything, but poor for local navigation.

    Conversely, you can always drop a set square on the ground and choose two directions. But if you try to extend that system you will quickly run into trouble because different routes from A to B lead to different notions of which way the set square should be pointing at B. This is useful for local work, but a poor fit for global pictures.

    In flat Minkowski spacetime this is a distinction without a difference. But in any curved spacetime you have to choose an approach that works for what you are doing, basically.
     
  12. Dec 12, 2016 #11
    I see. Thank you!
     
  13. Dec 13, 2016 #12
    I tried to apply these ideas of ortoghonal vectors in a 3d software program (3ds max). I noticed that in all ways I curved the surface, the grid lines, wich I think, can represent the coordinate lines, remained at 90° from each other. See below an Image of what I did.

    wb8wKNb.jpg

    So if the basis vectors are tangent to the coordinate lines at each point, they would be orthogonal to each other, despite the curvature of the manifold itself. What makes they not to be orthogonal in general? It's because they live in a four-dimensional manifold?
     
    Last edited: Dec 13, 2016
  14. Dec 13, 2016 #13

    Ibix

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    Note that that is a collection of flat spaces with discontinuities in the differentials at the joints. So any problems will occur at corners.

    Take a look at the top corner of the depression where it meets the flat plane. Measure the angle between the basis vectors at that corner. The three quadrants outside the depression are 90°, but the corner that soans the depression is not. Remember that you need to measure in the surface - so the angle in that quadrant is the sum of the angles in the corners of the sloped polygons - which are not 45°.

    Another thing to note is that the distance from one corner to the next is not the same as the distance across the middle of the depression. This is due to the extra distance from going up and down the sloped polygons. An obvious coordinate basis for this is to pick vectors in what the software calls the x and y directions (you can state that more formally if you like). Everywhere except the sloped polygons this corresponds exactly to the "vielbein" basis. But in the sloped polygons, one unit of distance in the x-direction is more than one unit in the plane. That means that the coordinate basis is orthogonal but not orthonormal there.

    Note also that you've embedded your manifold in a higher dimensional space, which isn't a concept that is available in GR. So beware of relying on it...
     
  15. Dec 13, 2016 #14
    Thanks Ibix for clarifying these things to me.

    The corner is not at 90° because I used the smooth option to make it more beautiful. :smile:
     
  16. Dec 13, 2016 #15

    George Jones

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    Define new coordinates for the x-y plane in Minkowski spacetime by u = x and v = x + y. Then, the coordinate vectors associated with u and v are both spacelike, and the inner product between the coordinate vectors associated with u and v is negative.
     
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