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I Constructing a vector in another basis

  1. Mar 11, 2017 #1
    Suppose we have defined a vector ##V## at a point ##x##, so it has components ##V^\mu(x)## at ##x##. Let ##y## be another point, such that ##y^\mu = x^\mu + \epsilon \zeta^\mu(x)##, ##\epsilon## a scalar. Now, since ##x## and ##y## are coordinate points, the vector ##V## should not depend on them. So we have the familiar transformation of a vector from one point to another (a prime denotes quantities in ##y##):
    $$ V'^\mu (y) = V^\nu (x) \frac{\partial y^\mu (x)}{\partial x^\nu}$$

    We would end up with ##V'^\mu (y) = V^\mu (x) + \epsilon V^\nu (x)\partial_{\nu}\zeta^{\mu}(x)##. My question is if it's correct to assume that relation between ##x## and ##y##.
     
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  3. Mar 11, 2017 #2

    Orodruin

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    Your function ##\zeta## is not a vector. It is a collection of functions that describe the relation between two coordinate systems.
     
  4. Mar 11, 2017 #3
    Thanks. I have edited my post, because this has become not relevant to me
    Is it ok to take ##\zeta^\mu(x)## such that ##\zeta^\mu(x) = 0## but its first derivatives non-zero at ##x##?
     
  5. Mar 11, 2017 #4

    PeterDonis

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    Why not? You've only defined ##V## at the point ##x##; you haven't defined it at the point ##y##. Vectors are "attached" to particular points.

    Now you're talking as though ##y## denotes the same point as ##x##, just in a different coordinate chart. This is not the same as ##y## being a different point from ##x##, expressed in the same chart. Which is it?
     
  6. Mar 11, 2017 #5
    Oh yes. I'm thinking of a point ##P## on a manifold, where the vector ##V## depends on ##P##. That is why I assumed the vector to be the same. ##x## and ##y## are different maps to ##P##.
     
  7. Mar 11, 2017 #6

    stevendaryl

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    I think you're mixing up two different concepts:
    1. The components of a vector in two different bases
    2. The value of a vector field at two different points.
    If you use a coordinate basis, and you change coordinate systems from [itex]x^\mu[/itex] to [itex]y^\mu[/itex], then you would use the transformation equation:

    [itex]V'^\mu = \frac{\partial y^\mu}{\partial x^\nu} V^\nu[/itex]

    On the other hand, if you are talking about two different, nearby points, [itex]\mathcal{P}[/itex] with coordinates [itex]x^\mu[/itex] and [itex]\mathcal{P'}[/itex], with coordinates [itex]x^\mu + \epsilon \zeta^\mu[/itex] (you're keeping the same coordinate system), then you could use:

    [itex]V^\mu(\mathcal{P'}) \approx V^\mu(\mathcal{P}) + \epsilon \zeta^\nu \frac{\partial V^\mu}{\partial x^\nu}[/itex]

    That only makes sense for a vector field (a function assigning a vector to each point in space), not a vector.

    There is yet a third concept that you might be thinking of: parallel transport of a vector. If you have a vector [itex]V^\mu[/itex] at one point [itex]\mathcal{P}[/itex] with coordinates [itex]x^\mu[/itex]), then what is the result of transporting that vector to a nearby point [itex]\mathcal{P'}[/itex] with coordinates [itex]x^\mu + \epsilon \zeta^\mu[/itex]?
     
  8. Mar 11, 2017 #7
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