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I Vector notion in GR

  1. May 12, 2017 #1
    Hello! I read this definition of vectors in my GR book: "To each point p in spacetime we associate the set of all possible vectors located at that point; this set is known as the tangent space at p, or ##T_p##". This means that each point in space time is viewed as the origin for the 4 dimensional space and all vectors going from p to any other point form this ##T_p##?
     
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  3. May 12, 2017 #2

    PeroK

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    The vectors in the tangent space don't join up points in spacetime. The tangent space is a separate vector space with only the one point from your spacetime, point p.

    It's a generalisation of the tangent line on a simple 1D curve, which just touches the curve at a single point.
     
  4. May 12, 2017 #3
    So what is a vector, inside the tangent plane?
     
  5. May 12, 2017 #4

    PeroK

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    That is a vector! It doesn't join points in curved spacetime. It's defined in the tangent space.
     
  6. May 12, 2017 #5
    Wait, sorry I am confused. From what I understand, the tangent space is a vector space, which contains more than one vector (I guess an infinite number of vectors usually). So, if we take for simplicity a 2D manifold and choose a point, this tangent space is a plane, tangent at that point. And we consider all the vectors inside that plane (is this correct?). So to each point on the 2D manifold, we associate an infinite number of vectors. I am not sure I understand what these vectors mean. Like is the point to which the plane is tangent the origin of these vectors? And what a vector in this plane (a,b) means (picking a basis, what does it point to?). Thank you!
     
  7. May 12, 2017 #6

    PeroK

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    If you imagine a particle constrained to move in 2D along a curve. At each point it has a velocity in the direction of the curve. But that direction doesn't point along the curve for any finite length - unless the curve is a straight line. So, you could ask "where is that velocity vector pointing?"

    Tangent spaces are a 4D space-time generalisation of this. A four velocity vector cannot point through curved spacetime for any finite interval. It's direction only exists infinitesimally in spacetime.

    One approach, therefore, is to consider the tangent space in which to define and manipulate these vectors. You need the tangent space to allow you to do more mathematics with the vectors, which you can't do with just intuitive hand-waving about vectors having a magnitude and direction.
     
  8. May 12, 2017 #7

    Orodruin

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    You can think of the vectors as directions (with magnitude) at the given point, eg, velocities. The velocity points in the direction of motion. The vector does not point "to" another point (this is a lie to children also in the Euclidean case because it works there), it points in a direction along the manifold.

    Of course, there are more formal definitions of vectors, but you seem to be looking more for a physical intuition.
     
  9. May 12, 2017 #8

    Orodruin

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    I must interject that there is nothing particular about 4D or space-time. Tangent spaces are a general concept applicable to smooth manifolds.
     
  10. May 12, 2017 #9

    DrGreg

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    To expand on what PeroK and Oroduin have said, displacements on a curved manifold can't be treated as vectors. For example, on the Earth's surface, if you travel 300 km north and then 400 km east, you end up at a different point than if you had travelled 400 km east and then 300 km north. On the other hand it is valid to add together the 2D horizontal momentum vectors of particles colliding at a single point, so horizontal momentum is a 2D vector, residing in the tangent space at that point.

    And so it is in GR, where the 4-momentum (a.k.a. energy-momentum) of a particle at an event is a 4-vector lying in the tangent space at that event.
     
    Last edited: May 12, 2017
  11. May 12, 2017 #10

    pervect

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    The usual example might help. To make the problem less abstract, lets talk about vectors not in a curved 4-d space=time, but on the curved 2-d surface of a 3-d sphere. The sphere is 3 dimensional, it's surface is 2 dimensional.

    We can imagine a 2-d plane tangent to the sphere at some point. P, as in the image below which I borrowed from google images:

    image022.jpg

    We can imagine and manipulate vectors in the tangent plane as we would on any other plane. The abstract property of these vectors is that they can be multiplied by scalars, and that they add, and that both processes are commutative. We'll focus on the addition property in particular - if ##\vec{A}## and ##\vec{B}## are vectors, ##\vec{A} + \vec{B} = \vec{B} + \vec{A}##.

    This is true for the vectors in the tangent plane, but it's not exactly true for displacments on the surface of the sphere. I'm not sure whether to go into more detail or not - my judgment call is that it's best to go into a little more detail with an example, but not a lot of detail at this point.

    If our displacements are "small" compared to the diameter of the sphere, the vectors "almost commute". But this sort of fuziness is not suitable for a rigorous mathematical treatment. To define vectors that always commute, no matter how large, we need to introduce the tangent space.

    For a very crude example, consider a displacement so large that it wraps halfway around the surface of the sphere, to the antipodal point. If the displacement were a vector, we would say ##\vec{V}## + ##\vec{V}## = ##\vec{0}##, where ##\vec{0}## is the identity element of a vector space. If displacments were vectors, simple algebra would allow us to say that ##\vec{V} = 0##. But the antipodal displacement operator is not zero, it's not an identity element, so we have mathematical inconsistencies.

    So, the bottom line is that we regard vectors on curved manifolds as being defined in the tangent space. Consiering the example above hopefully demonstrates why this is necessary.
     
  12. May 12, 2017 #11

    Orodruin

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    The sphere is two-dimensional. You are thinking of a ball.
     
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