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Confused with directional derivative definition of vectors

  1. Jul 31, 2012 #1
    Hi, been reading about GR and am quite confused about the new definition of vectors.
    My main problem with this is that the text uses partial derivatives as the vector basis, I understand this is related to directional derivatives but cannot see the direct mathematical relation. Secondly, how are physical quantities defined in GR if we cannot use the line-segment definition of vectors?
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  3. Jul 31, 2012 #2


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    When I was younger I worried a great deal about how vectors were defined on the surface of a sphere! Did they "curve" around the sphere or did they go through the sphere? It finally dawned on me that vector "live" in the tangent plane. And that means that vectors are always tangent vectors for some curve. And that, in turn, means that vectors are derivatives. In particular, if a curve on a given surface is given by [itex]f(t)\vec{i}+ g(t)\vec{k}+ h(t)\vec{k}[/itex], with parameter t, then the tangent vector to that curve is [itex]f'(t)\vec{i}+ g'(t)\vec{j}+ h'(t)\vec{k}[/itex].

    If, on the other hand, you are given a function of three variables, f(x,y,z), as, perhaps, the potential energy at a point, then [tex]\nabla f= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}[/tex] is the force vector.
    Last edited by a moderator: Aug 1, 2012
  4. Jul 31, 2012 #3
    Could a more mathematically precise explanation be given?
  5. Aug 1, 2012 #4


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    Wald's "General Relativity" has a rather precise explanation of what mathematical properties a derivative operator must have. But I'm not sure how listing them would help you. Your questions don't sound like the sort that need mathematical precision but rather physical interpretation.
  6. Aug 1, 2012 #5


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    Let M be a differentiable manifold, p a point on that manifold. Let F be the set of all functions from R to M such that f(t0)= p for some t. (Equivalently, let F be the set of all differentiable functions from R to M such that f(0)= p. Since if f(t0)= p, we can define F(t)= f(t+ t0) and have F(0)= f(t0)= p, those are the same set.) We say that two such functions are equivalent if and only if their derivatives at t0 (equivalently, at 0) are the same. That can be shown to be an equivalence relation and so defines a collection of equivalence classes. We define a "vector" (more precisely, a tangent vector at p) to be one of those equivalence classes.

  7. Aug 1, 2012 #6


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    I don't think HallsOfIvy grasped the question you were asking. He has used the definition of vector as a derivative of coordinates. Given a differentiable curve parametrised as [itex]x^\alpha(s)[/itex] you can obtain a vector[tex]

    \frac{d\textbf{x}}{ds} = \frac{dx^\alpha}{ds} \textbf{e}_\alpha

    [/tex] where e0 is the vector (1,0,0,0) etc.

    If f is a differentiable scalar field defined in a region enclosing the curve then, by the chain rule, we have the equation[tex]

    \frac{df}{ds} = \frac{dx^\alpha}{ds} \frac{\partial f}{\partial x^\alpha}

    [/tex]along the curve. Or, to put it another way we have the operator [tex]

    \frac{d}{ds} = \frac{dx^\alpha}{ds} \frac{\partial}{\partial x^\alpha}

    [/tex]acting on the set of differentiable scalar fields in the region. Comparing this equation with the first one, we make a isomorphism between dx/ds and d/ds, and under that isomorphism each of the four basis vectors ea vectors is identified with the corresponding ∂/∂xa operator, and we identify the vector[tex]

    \textbf{V} = V^\alpha \textbf{e}_\alpha

    [/tex]with the (directional derivative) operator[tex]

    V^\alpha \frac{\partial}{\partial x^\alpha}

    [/tex]Under this way of thinking a "vector" then becomes an operator in the 4D space spanned by the four ∂/∂xa operators, and you can show this definition is independent of the choice of coordinate system.
  8. Aug 1, 2012 #7


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    I don't really understand what you're confused about in the first part of the question, I hope some of the other responses has been helpful.

    As far as the second part of your question, it may or may not be helpful for you to know that you can define a map from the tangent space (where the vector basis are defined) to the manifold. This is called the "exponential map".


    If wiki's explanation isn't terribly clear and you find the idea relevant to your question, feel free to ask more.

    Physical quantites are defined in GR via different means. Many physical quantities are defined by the value of some tensor or tensor density at some point in space-time. I have the possibly incorrect feeling that you may be concerned with the issue of how distances are defined in GR, though.

    Distances in GR are usually defined as the least upper bound of the length of a curve (informally, the length of the shortest curve) lying in a surface of constant time connecting two points. It's surprisingly hard to find a good formal discussion of this, for instance Wald only talks about how to find the lengths of curve, he never defines how to go from that to determning which curve you find the length of to measure a distance.

    The definition of the surface of constant time depends on one's coordinate choice.
  9. Aug 1, 2012 #8
    I'm not too good at LATEX, so see attached Microsoft Word document.

    Attached Files:

  10. Aug 1, 2012 #9


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    Chestermiller, see this post for an intro to using LaTeX at Physics Forums.

    GarageDweller, you might want to check out the book "Modern differential geometry for physicists" by Chris Isham. I think it has a pretty nice introduction to tangent vectors. It defines two different vector spaces associated with a point p in a manifold, and shows that the two spaces are isomorphic. One of these spaces consists of equivalence classes of curves through p, and the other consists of derivative functionals. The latter is easier to work with, but it's hard to see why we would want to call it "the tangent space at p". The former, on the other hand, can obviously be thought of as the tangent space at p. So the best way to understand why derivative functionals can be thought of as tangent vectors is to understand the isomorphism between these spaces. This post has some information about the "equivalence classes of curves" approach and the isomorphism between the two spaces.

    This post and the ones linked to in it might be useful too.
    Last edited: Aug 1, 2012
  11. Aug 4, 2012 #10
    This shed a lot of light on the subject for me, but i still have a question.
    Now since vectors are (1,0) tensors, they are functions of one forms and produce scalars.
    Here, I see you have defined vectors as instead functions of scalars that produce scalars.
    Is there something I'm missing here?
  12. Aug 4, 2012 #11


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    The way I think about it, the fundamental objects are (1) real scalar fields, which are functions [\itex]\Phi(P)[/itex] that take a point in space (or spacetime) and return a real number, and (2) parametrized paths, which are functions [itex]P(s)[/itex] which go in the opposite direction; they take a real number and return a point in spacetime. An example of a real scalar field on the surface of the Earth is the altitude at each point. An example of a parametrized path on the surface of the Earth is a highway, which associates a point on the Earth as a function of the distance down the highway (from some designated starting point).

    We can combine these two kinds of objects to get an ordinary function from reals to reals: If [itex]\Phi(P)[/itex] is a real scalar field, and [itex]P(s)[/itex] is a parametrized path, then the combination [itex]\Phi(P(s))[/itex] is a function from reals to reals. For the examples given, [itex]s[/itex] would be distance down the highway, and [itex]\Phi(P(s))[/itex] would give the altitude as a function of [itex]s[/itex].

    Now, we can define [itex]\dfrac{dP}{ds}[/itex] (the "velocity vector" or "tangent vector" for the parametrized path [itex]P(s)[/itex]) to be the operator [itex]V[/itex] defined by:

    For any real scalar field [itex]\Phi[/itex], [itex]V(\Phi) = \dfrac{d}{ds} \Phi(P(s))[/itex]

    Then, we can define [itex]\nabla \Phi[/itex] (the "gradient" or "one-form" associated with the scalar field [itex]\Phi[/itex]) to be that operator [itex]\omega[/itex] defined by:

    For any tangent vector [itex]V[/itex], [itex]\omega(V) = V(\Phi)[/itex].

    With this way of defining things, a tangent vector is an operator on real scalar fields, while a one-form is an operator on tangent vectors.
    Last edited: Aug 4, 2012
  13. Aug 4, 2012 #12
    Er i get it now, but it's sad when you have to leave an old friend like classical vectors
  14. Aug 4, 2012 #13
    Who said you have to "leave" classical vectors? These are the same vectors you're used to dealing with. The difference is the basis their components are related to. Tangent vectors use basis vectors that are tangent to their respective coordinate axes. Cotangent vectors ("one-forms") use a differnt basis--the cotangent basis vector [itex]e^x[/itex] is perpendicular to the hypersurface formed by the t, y, and z axes.

    GR is often the first real exposure to these concepts of tangent vs. cotangent vectors, one that is often tacitly ignored in electrodynamics because flat space makes the consequences fairly minimal, but the differences shouldn't be overstated. I understand Steven's idea about tangent and cotangent vectors being "operators," but I don't think that's a necessary viewpoint. They're just different kinds of vectors, and you can combine them with dot products like you've always done.

    Basically, if you say you have a position vector [itex]x = x^0 e_0 + x^1 e_1 + x^2 e_2 + x^3 e_3[/itex], then, [itex]\partial x/\partial x^0 = e_0[/itex] picks out the 0th basis vector. It's just a way to find the basis related to your coordinates without knowing [itex]e_0[/itex] beforehand, and same for the others.
  15. Aug 4, 2012 #14


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    I don't necessarily think that what you are saying is wrong, but I think it gives the misleading impression that vectors and 1-forms are the same thing, expressed in different bases. They're not, even though, if you have a metric tensor, you can convert one into the other.

    To illustrate how different they are, let me take an example of a manifold that has no associated metric. Suppose we are doing thermodynamics, and the state of the system is defined by two variables, [itex]P[/itex], the pressure, and [itex]V[/itex], the volume. As time goes on, the state S changes as a function of time. We can describe the "velocity vector" for the system using the coordinate basis [itex]e_P[/itex] and [itex]e_V[/itex]:

    [itex]\dfrac{dS}{dt} = \dfrac{dP}{dt} e_P + \dfrac{dV}{dt} e_V[/itex]

    Now, let's consider a scalar field [itex]T(P,V)[/itex] which gives the temperature as a function of [itex]P[/itex] and [itex]V[/itex]. Then we can use the 1-form basis [itex]\omega^P[/itex] and [itex]\omega^V[/itex]. We can express the gradient of [itex]T[/itex] using this basis:

    [itex]\nabla T = \dfrac{\partial T}{\partial P} \omega^P + \dfrac{\partial T}{\partial V} \omega^V[/itex]

    Without a metric tensor, there is no way to relate the 1-form basis [itex]\omega^P, \omega^V[/itex] to the tangent vector basis [itex]e_P, e_V[/itex]. Gradients can only be expressed in the first basis, and tangent vectors can only be expressed in the second basis.

    In particular, if [itex]A = A^i e_i[/itex] is a vector, and [itex]B = B_i \omega^i[/itex] is a one-form, we can make sense of [itex] A . B [/itex], but there is no meaning to [itex]A . A[/itex] or [itex]B . B[/itex].
  16. Aug 4, 2012 #15
    Right, I didn't mean to imply that even in non-metric spaces they're the same objects expressed in different bases per se. I'm just trying to point out that vectors in the cotangent space (cotangent vectors, one-forms, co-vectors) are not fundamentally different from vectors in the tangent space (tangent vectors, ordinary "vectors"). They still obey all the usual vector space axioms and such, so for me personally, it's more sensible to refer to both as vectors, just ones living in different, incompatible spaces that can't be related to each other without a metric. That may be just me, though.
  17. Aug 4, 2012 #16


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    Do you understand that this is only true in the sense that there's an isomorphism from the tangent space to the vector space of (1,0) tensors? The latter isn't the tangent space, it's the dual space of the dual space of the tangent space, as explained here:
    I think the post I linked to at the end of post #9 would be good for you. It's a decent place to get an overview of the basics (if you read the posts I linked to in there as well; the quote above is from one of those posts).
  18. Aug 4, 2012 #17
    Dydxforsn's Law of Vector/Tensor Analysis in Differential Geometry: Invariably, every conversation about vectors and tensor (in the context of general curvilinear coordinates) devolves into a discussion about the differences and similarities (usually semantics) between "vectors" and "one-forms".
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