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Vectors and covectors

  1. May 22, 2012 #1
    Hello! I have a question regarding the tangentvector belonging to a parameterized curve, at a certain point p.

    Lee has argued in his book 'Introduction to Smooth Manifolds' that the gradient earlier encountered as:

    [itex](\partial_\mu)f ,[/itex]

    is not a tangentvector since it is not always coordinate independent in the tangentvector basis when we cannot define a metric. Instead these components är coordinateindependent in the dual vector basis [itex]dx^\mu[/itex].

    When trying to describe the velocity of a particle moving along a parameterized curve these components are the ones to use, so...
    Does this mean that for manifolds without any metric, the velocity of, lets say a particle, always should be described as a covector (the differential df) or that the velocity in manifolds without any defineable metric is coordinate independent.. :/ Am I talking about two different things here maybe?

    A second question is regarding the notation of vectors i a basis by:

    [itex]\vec{x} = x^\mu \hat{e}_\mu,[/itex]

    maybe this is a stupid question, but I just want to be sure. Our components x^\mu is not a contravariant vector here, right? The index is instead just put in the upper position for the Einstein summation convention?

    Thanks for your support in my studies! :)
  2. jcsd
  3. May 22, 2012 #2
    Hi kontilera,

    df is well defined on any differential manifold. Keep this in mind, the differential df is just the best linear approximation of f(x)-f(p) and is calculated from the same formula no matter what coordinates your are using. The gradient is another story. Its definition relies on a metric: <grad f, v> = df(v). You need a metric for the inner product on the left side to make sense. Think of it another way. grad f is a vector perpendicular to the level surfaces of f. Its length is the rate of change of f in that direction. "Perpendicular" and "length" are words that only make sense with a metric.

    Curve velocities are vectors not covectors and are well defined on any differential manifold. Vector components are indeed contravariant under coordinate changes. Suppose you have coordinates x^1, x^2,...,x^n. Then the ith basis vector is the differential operator $\partial_{x^i}$. This notation together with the chain rule shows how to transform vector components.

    One last thing: There are confusing "conventions" regarding the gradient. I just looked up covariant/contravariant on Wikipedia and it lists the gradient as a covariant vector field. But really the thing they are talking about is not the gradient at all, it is the differential. Lets just clarify the difference with an easy standard example. If f is a function on the Euclidean plane, then in polar coordinates:

    grad f = (f_r)e_r + (1/r^2) f_(theta)e_(theta).
    df = f_r dr + f_theta dtheta

    In the formula above, e_theta is not the unit vector, but the vector that operates on functions f by e_theta(f) = f_theta (differentiation with respect to theta). Its length is r.
  4. May 22, 2012 #3
    Thank you Vargo! I guess my confusion is due to the fact that Lee says that 'grad f' needs a metric to be coordinate independent, but the curve velocity seems to have the same components and is a vector. Therefore it should be coordinatedependent for some situations without metric..

    Maybe they don't have the same components though, a second look shows:

    [itex]V^\mu = \frac{dx^\mu}{d\lambda}[/itex], while
    [itex]grad f = \frac{df}{dx^\mu}.[/itex]

    Looks like I was confused..right?. :)

    Thanks again!
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