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Contravariant and covariant vectors transformations

  1. Jun 25, 2012 #1
    Hi all,

    I am new to General Relativity and I started with General Relativity Course on Youtube posted by Stanford (Leonard Susskind's lectures on GR).

    So first thing to understand is transformation of covariant and contravariant vectors. Before I can understand a transformation, I would like to know how do they look like. I mean what is the difference between covariant and contravariant vectors? And what is their role in understanding GR?

    In addition, I would like to know what is the application of Einstein's Field Equations? I mean, can I get one problem where they are actually used?

    Thanks!

    Joe W.
     
  2. jcsd
  3. Jun 25, 2012 #2
    Contravariant or "tangent" vectors lie parallel (tangent) to the axis in question. The [itex]e_x[/itex] vector lies parallel to the x-axis, for example.

    Covariant or "cotangent" vectors lie perpendicular to all the other axes. In 3D, take the plane defined by the y and z axes, and the vector [itex]e^x[/itex] is perpendicular to that plane.

    In cartesian coordinates, covariant and contravariant vectors aren't different, but in more general, curvilinear coordinates or non-orthogonal frames and curved spaces, the two types of vectors become different and have to be kept track of.

    One of the main concepts that is introduced in GR is the invariance of physical laws under arbitrary coordinate transformations. This is not actually unique to GR, but it commonly comes into play here because the concept of spacetime allows for even time derivatives to be treated under this theory. The idea is as follows:

    Often, the position vector is just called [itex]x =x^0 e_0 + x^1 e_1 + x^2 e_2 + x^3 e_3[/itex]. We don't say what kind of coordinate system this is (whether it's cartesian, spherical, or something else). But in general, if we want to change coordinates, we introduce a new position vector [itex]x' = f(x)[/itex], where [itex]f[/itex] is some function. For instance, converting between cartesian and polar coordinates, you might do something like

    [tex]x' = \sqrt{(x^1)^2 + (x^2)^2} e_1 + \arctan \frac{x^2}{x^1} e_2 = r e_1 + \theta e_2[/tex]

    This is a simple example of a 2D transformation. At any rate, an arbitrary transformation has a Jacobian, which relates the partial derivatives of the new coordinates to the old coordinates: the Jacobian [itex]\underline f[/itex] is given by [itex]\underline f(a) = a \cdot \nabla f(x)[/itex], where [itex]a[/itex] is some vector to be transformed. The vector [itex]\underline f(a) = a'[/itex] gives the transformation of a tangent vector. This is usually written in index notation as

    [tex]a'^i = a^j \frac{\partial x'^i}{\partial x^j}[/tex]

    The prototypical example of a tangent vector is the four-velocity [itex]u[/itex], from which we get the transformation law [itex]u' = \underline f(u)[/itex].

    On the other hand, there are cotangent space objects, one of which is the derivative [itex]\nabla[/itex]. One can prove that [itex]\overline f(\nabla') = \nabla[/itex], where the overbar denotes the transpose, and this in general gives the transformation law for objects in the cotangent space.

    So you can see that under arbitrary coordinate transformation laws, we have these two different kinds of objects--covariant (cotangent) vectors and contravariant (tangent) vectors. In general, one converts between the two using the metric of the space. There is an operator [itex]\underline g(a)[/itex] that takes a tangent vector [itex]a[/itex] and converts it into a cotangent vector, and the inverse does the opposite. This gives the flexibility to work with vectors in the most convenient space as appropriate.

    As for the Einstein equations, usually they're used to get a solution for the metric and from there various results are derived. The different time dilation on the Earth compared to satellites in orbit, for instance, is calculated knowing that the Earth yields a certain solution to the Einstein equations in its immediate neighborhood and how that affects the perception of time at various heights.
     
  4. Jun 25, 2012 #3

    Mentz114

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    I assume you know that a vector belongs in a vector space, and that a vector space can be characterised by a set of basis vectors. It is an axiom that every point on curve on a manifold has two vector spaces, one has a basis formed by the tangent vectors, and the other has a basis of 1-forms ( also called covectors). The vector spaces are connected by a metric ( if there is one), so any vector can be written in either basis. Contravariant vectors are written in the tangent space basis and covariant vectors in the 1-form basis. The transformation is linear, vμ = gμα vα.

    This is a machinery to ensure that the contractions formed from tensors are coordinate independent. Like the norm of a 4-vector ( which is a rank-1 tensor) n = vμvμ = gμα vμvα.

    I recommend Sean Carroll's book or lecture notes on this subject.

    [Posted simultaneously with Muphrid]
     
  5. Jun 25, 2012 #4
    Thank you all very much for such comprehensive replies!

    So as I can conclude from your posts the covariant and contravariant vectors look like this:

    http://img841.imageshack.us/img841/9872/contavcovarcompl.png [Broken]

    In this case, we may think of components as vectors themselves. In green is covariant while in blue is contravariant.

    The Tensor itself is a product of two Vectors, isn't it? So the output of the Cross Product of two vectors is a Tensor? If so, what are its components?

    Also, why do we need a metric in a spherical coordinates when we can use Great Circle distance-formula to calculate distance between any two points on a sphere?
     
    Last edited by a moderator: May 6, 2017
  6. Jun 25, 2012 #5
    I wouldn't think of components as vectors themselves; it's just that you can express a single vector either in terms of the tangent basis vectors and components or in terms of the cotangent basis and those components. You don't need the metric to convert between the two--you can do it the hard way if you already know both the tangent basis and the cotangent basis--but you get the same answer as if you'd just used the metric. This is part of why the metric is convenient.

    Any distance formula implicitly uses the metric--without a metric, the concept of distance doesn't exist.

    Beyond 3D, there isn't a product of vectors that yields another vector. The "cross product" is abandoned, and we deal in wedge products instead. If you have two vectors [itex]a[/itex] and [itex]b[/itex], the wedge product [itex]a \wedge b[/itex] describes the plane that the two vectors span.

    Typically, this is where one resorts to index notation. For [itex]C = a \wedge b[/itex], the components of [itex]C[/itex] would be [itex]C^{ij} = a^i b^j - a^j b^i[/itex] for [itex]i \neq j[/itex]. While you can write a vector in terms of its components and basis vectors, for this object [itex]C[/itex] you would need a concept beyond basis vectors--you'd need something that covers unit planes and such. While these concepts do exist, they're not typically considered in usual formulations of GR. People just tend to deal in components.
     
    Last edited: Jun 25, 2012
  7. Jun 25, 2012 #6

    Mentz114

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    The components of the tensor product are the product of the components of the tensors being multiplied.
    [tex]
    (a,b) \otimes (c,d) = \left[ \begin{array}{cc}
    ac & ad \\\
    bc & bd
    \end{array} \right]
    [/tex]

    The diagram you linked shows the parrallel projected components and the perpendicular projected components and is a way to understand them. The important thing is that the the inner product of those vectors is a geometric invariant.
     
  8. Jun 25, 2012 #7
    So the output of the Cross Product of two vectors is a Tensor, or not?
     
  9. Jun 25, 2012 #8

    Mentz114

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    Yes ! The tensor product of tensors is a tensor. Remember that 4-vectors are rank-1 tensors.

    Tensors can be made by multiplying tensors or by differentiating tensors or contracting tensors.
     
  10. Jun 25, 2012 #9
    Just one confusion that I would like to clear:

    Susskind told that curvature is an obstruction to flatten out the coordinates. So in this sense a sphere would have a curvature, right? Then why does Riemann tensor for sphere is zero?
     
  11. Jun 25, 2012 #10
    What do you mean when you're talking about a sphere? If you're talking about the 2D surface of a ball, it has curvature, and the Riemann tensor is nonzero. If you're talking about the volume of a ball, that space has no curvature because it's the same as flat space.
     
  12. Jun 25, 2012 #11

    WannabeNewton

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    The Riemann curvature tensor doesn't identically vanish for the 2 - sphere.
     
  13. Jun 25, 2012 #12
    I was talking about the volume of a ball, a 3-D sphere. I read that Gammas (Christoff. symb.) were like 1/r etc. And at the end, the components of the Riemann tensor ended up being zero.
     
  14. Jun 25, 2012 #13
    You've run into a bit of geometric pedantry. To distinguish between the surface of a spherical object and the volume that that object contains, "sphere" and "ball" can sometimes no longer mean the same thing. WannabeNewton just did that in his last post: "2-sphere" means "the 2D surface of a 3D spherical object". The object that contains the 2-sphere and the volume within it is called the 3-ball.

    In short, "sphere" = boundary of a "ball".

    At any rate, in flat space but using spherical coordinates, some of the Christoffel symbols are nonzero, but the curvature is still zero. This is just an artifact of the coordinates being used. There's really nothing special about the 3D ball. It may as well be infinite, unbounded space. Curvature here means something more fundamental--that the space and time are stretched or distorted in ways that coordinate system changes don't capture.
     
  15. Jun 25, 2012 #14
    Ok here is a metric for polar coordinates:

    ds[itex]^{2}[/itex]=dx[itex]^{2}[/itex]+dy[itex]^{2}[/itex]

    x=r*cos(θ)

    y=r*sin(θ)

    dx=cos(θ)dr-r*sin(θ)dθ

    dy=sin(θ)dr+r*cos(θ)dθ


    ds[itex]^{2}[/itex]=dr[itex]^{2}[/itex]+r[itex]^{2}[/itex]dθ[itex]^{2}[/itex]

    So this is a metric in polar coordinates. Now if I want to actually use it I will need to get rid of differential s to normal s by integrating the metric, right? So that would give me an arc length in polar coordinates. So

    ∫ds=s=∫sqr(dr[itex]^{2}[/itex]+r[itex]^{2}[/itex]dθ[itex]^{2}[/itex])

    So what is this equal to?

    Also, is it true that dr[itex]^{2}[/itex] term vanishes if I am calculating the arc length on a circle?

    Thank you all!
     
  16. Jun 25, 2012 #15
    In practice, I believe you generally consider the path as having some parametric form, so [itex]r = r(\tau)[/itex] and [itex]\theta = \theta(\tau)[/itex]. Using [itex]\tau[/itex], the proper time, is most common, but you don't need to know it to parameterize the coordinates--indeed, doing this calculation is one way you find proper time.
     
  17. Jun 25, 2012 #16
    ∫ds=s=∫sqr(dr^2+r^2dθ^2)


    How to solve this?
     
  18. Jun 25, 2012 #17
    Got it. Never mind.
     
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