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Vectors, one-forms and gradients

  1. Jul 22, 2005 #1
    I'm currently working through Schutz's "A first course in general relativity" as a preparation for a graduate course in General Relativity based on Carroll's notes. I'm a little confused about vectors, one-forms and gradients.

    Schutz says the gradient is not a vector but a one-form, because it maps vectors into the reals in a linear way and explains that you should represent a gradient (and a one-form in general) by a series of surfaces. When you contract a one-form with a vector, the number you get is the number of surfaces the vector crosses.

    If this was all I knew about the subject, it would be quite clear, however, my advanced calculus says the gradient is a vector pointing in the direction of fastest increase and it's a vector that's, when evaluated at a point p, is perpendicular to the tangent plane of the surface at point p. Schutz says there's nothing really wrong with this definition, because in normal Euclidean space, vectors and one-forms are the same, but how can something be both a vector (an arrow) and a one-form (a series of surfaces)?

    Later on, Schutz says that a vector can also be seen as a linear map from one-forms into the reals (Carroll also says the dual of a dual vector space is the original vector space), so this way it seems whether to call something a vector or a one-form is totally arbitrary (as long as you do so in a consistent way), whether you're dealing with Euclidean space or not... :grumpy:

    When I turn to other texts I get confused only more, for example, in some text on Differential Geometry a vector is defined as a linear operator on a function space that produces a real number...? I find this a strange definition, wouldn't this imply that a function is actually a one-form (and a vector, since it's totally up to you what to call the original vector space and what it's dual).

    I think it's all equivalent in some peculiar way, but I really don't see how...
     
    Last edited: Jul 22, 2005
  2. jcsd
  3. Jul 22, 2005 #2
    Think about the size of the surfaces. By Schutz's definition the number you get when you contract is the number of surfaces that the vector passes. This would imply that you would only get rational numbers, unless the surfaces were infintesmal. Now how can you define a surface in Euclidian space? By a vector peripendicular to it! Then you take the product of the two vectors and get a number.

    Yes. Duals of tensors are those that contract the tensors into the original numbers.

    Read on, when you get to tensors everything will be more clear.
     
    Last edited: Jul 22, 2005
  4. Jul 22, 2005 #3
    the gradient, [tex] \vec{\nabla f}[/tex] is a vector. it has vector components [tex]\nabla f ^i = g^{ij} \frac{\partial f}{\partial x^i}[/tex]

    whereas [tex]df[/tex] is a one-form with covector components [tex]\frac {\partial f}{\partial x^i}[/tex]

    df is a one-form and so it resides in the dual space with the basis [itex]dx^i[/itex]:

    [tex]df = \frac{\partial f}{\partial x^i} dx^i[/tex]

    the gradient is a vector in the tangent space with the basis [itex]\frac{\partial}{\partial x^i}[/itex], defined as:

    [tex]df(\vec{v}) = <{\nabla f}, \vec{v}>[/tex]

    so the gradient vector is the vector whose inner product with [itex]\vec{v}[/itex] maps to the same real value as acting the one-form [itex]df[/itex] on the vector [itex]\vec{v}[/itex]

    it is an important distinction because they live in entirely different spaces, being related by the metric tensor.
     
  5. Jul 22, 2005 #4
    think of it this way.

    [tex]\alpha: V \rightarrow \mathbb{R}[/tex]

    so the one-form (a dual-vector) alpha takes a vector and maps it to a real. it is a linear functional.

    [tex]\alpha(\vec{v}) = a[/tex]

    so what would be my vector acting on the one-form be? simple, let's define it this way:

    [tex]\vec{v}(\alpha) = \alpha(\vec{v}) = a[/tex]
     
  6. Jul 22, 2005 #5
    these are very good questions that you are asking.

    yes, the vector can be viewed as a differential operator. what operator? how 'bout the directional derivative:

    [tex]\vec{v}(f) = v^i \frac{\partial f}{\partial x^i}[/tex]

    which is a real number. this is different than how you may normally think of the directional derivative. before, you probably thought about the function being fixed, and varying the vector to get the derivative. here we are saying that the tangent vector is fixed and that we can vary the function.

    if we can vary the function, then we can define our vector as:

    [tex]\vec{v} = v^i \frac{\partial}{\partial x^i}[/tex]

    in order to see this, consider constructing a map from a tangent vector on one manifold to another, by way of a function that maps points from one manifold to the other. you will see that neither the coordinate system nor the function f matter in constructing the tangent vector, and that the two tangent vectors will be mapped by the Jacobian matrix, in essence a linear transformation between tangent vectors.
     
  7. Jul 22, 2005 #6
    Schutz's book is basic text on General Relativity. It doesn't cover differential forms in detail. His text was the first I studied on the subject and I found it helpful to read it parallel with a book on differential geometry.
     
  8. Jul 22, 2005 #7

    matt grime

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    there are many things in maths that are the same but which we wish to keep separate, and this is just one of them. one of the reasons is that the isomorphism from the tangent space at a point on a manifold to its cotangent space (tangent vectors and 1 forms resp) is that it is contravariant: duality reverses the direction of the arrows. these directions are important! so a map from manifolds M to N induces a map on the tangent vectors at each point from M to N but induces a map from the one forms of N to the one forms on M (hope i got that the right way round). if maps pushfoward on tangents and pull back on cotangents (one forms). years since i did this, apologies for any gross inaccuracies.

    or another way of thinking is that whilst they may have the same underlying structure they do different things.
     
    Last edited: Jul 22, 2005
  9. May 4, 2006 #8
    The problem is that the components of the gradient transform like a covector, and till now i have not seen a satisfy solution for this fact, to interprete the gradient as a vector or why he could exist at all as a vector. :yuck:
     
  10. May 8, 2006 #9

    nrqed

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    I am grateful for this post....

    I usually hear saying that the gradient is a one-form!!!:yuck: Which is very confusing and it disagreement with what quetzalcoatl9 wrote here.

    If I understand correctly, the components of df (in spherical coordiantes, say) would be [itex] { \partial f \over \partial r}, {\partial f \over \partial \theta}, {\partial f \over \partial \phi} [/itex].

    However, I would never have called these the components of the gradient in spherical coordinates, of course! These are [itex] {\partial f\over \partial r}, {1 \over r } {\partial f \over \partial \theta}, {1 \over r \sin(\theta} {\partial f \over \partial \phi} [/itex]!

    So the gradient (in the sense of what is being taught in introductory physics classes) *is* a vector.

    Now I am sure someone will come along to tell me I have it completely wrong:surprised

    But that's fine, my goal is to get it right!
     
  11. May 8, 2006 #10

    mathwonk

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    this is a problem of duality, of having two ways to picture something.

    imagine a heat source like a radiator. do you see it as a concentric family of spheres of constant temperature with the radiator at the center? or do you see it as stream of heated particles flowing radially away from the radiator?

    the two points of view are dual and equivalent to each other. either you prefer to represent ypour phenomenon by the family of surfaces of constant temperature, or by the flow lines of greatest increase (or decrease) of temperature.
    trictly speaking the word "gradient" seems to correctly refer to the directions of greatest increase, but for some reason this name has been taken over also by some people to refer to the level surfaces of the temperature function.

    i.e. some people including myself, carelessly refer to the one form df as the "gradient of f", when really the gradient is the object dual to df. once one becomes familiar with the stuff, it becomes more tempting to be careless in ones use of language.
     
  12. Jun 29, 2006 #11

    nrqed

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    I know this is all old stuff and knowledgeable people here find this type of questions tedious and incredibly boring, but I really want to understand diff geometry (I will go completely insane if I don't sort out all the notation and terminology!!!!) and I will be posting tons of *really* stupid questions, mostly about notation. I apologize ahead of time for that and if that kind of questions make you cranky, just ignore me:shy:


    Firstset of stupid questions: In the book by Baez and Muniain, they write (p.41)

    [tex] df(v) = vf [/tex]
    (where vf is there notation for the directional derivative of the scalar field f along the vector v)

    and then they write below Think of this as a slick way to write [itex] \nabla f \cdot v = vf[/itex]).

    Ok. Here are my stupid questions.

    What is df, explicitly? Let's work in spherical coordinates (so that the metric is not trivial!!)

    I thought that df was simply [tex] {\partial f \over \partial r } dr + {\partial f \over \partial \theta} d \theta + {\partial f \over \partial \phi} d \phi [/tex]. Is that correct or not?

    Let's say v is [tex] v_r {\partial \over \partial r} + v_\theta {\partial \over \partial \theta} + v_\phi {\partial \over \partial \phi} [/tex].

    Then I would think that feeding v to df would simply give [tex] v_r {\partial f \over \partial r }+ v_\theta {\partial f \over \partial \theta} + v_\phi {\partial f \over \partial \phi} [/tex]

    However, this is (at first sight) not at all what we learned in introductory maths the directional derivative of a scalar field was, viz. [tex]v_r {\partial f \over \partial r }+ v_\theta {1 \over r} {\partial f \over \partial \theta} + v_\phi {1 \over r sin \theta} {\partial f \over \partial \phi} [/tex]

    So I am missing something surely obvious :mad: .

    One thing that bothers me is that the components of a vector field using the "diff geometry language" uses for basis the partial derivatives. But that means that these components cannot be the same as the components of a vector field using the }. "intro maths approach" with basis [itex] {\hat e_r}, {\hat e_{\theta} }, {\hat e_{\phi}} [/itex], just by looking at the dimensions ([itex] {\hat e_r} [/itex] and [itex] {\partial \over \partial r}[/itex] don'thave the same dimensions).
    So what am I missing?

    Another thing that bothers me is that the expression df(v), thought as feeding a vector to a differential form is supposed to be completely metric independent (right?). But looking at the standard (i.e. "intro maths") expression for the directional derivative of a scalar field, it does look (at least at first sight) as if it does depend on the metric, through the presence of the 1/r and 1/(r sin theta) terms. What am I missing?


    I know this is all extremely basic and just a question of notation, but it is impossible to focus on the maths and the physics if the notation is not clear.

    Thanks to anyone who will be willing to help!!!

    Patrick
     
    Last edited: Jun 29, 2006
  13. Jun 29, 2006 #12

    George Jones

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    It doesn't have to.

    "Diff geometry language" is not restricted to coordinate bases, so it also can use this basis. If [itex]\left\{v_{r}, v_{\theta}, v_{phi}\right\}[/itex] are the components of [itex]v[/itex] with respect to an orthonormal basis, and [itex]\left\{v'_{r}, v'_{\theta}, v'_{phi}\right\}[/itex] with respect to a coordinate basis, then everything should work out OK.
     
  14. Jun 29, 2006 #13

    nrqed

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    Thanks.

    Then I have to understand "coordinate basis" vs "orthonormal basis". I have always seen the partial derivatives used as basis for vectors in differential geometry, I thought that it was the only thing people worked with, at a fundamental level (i.e. if a metric is not introduced).

    I am confused because it seems to me that in order to define an orthonormal basis, one needs to introduce a scalar product, which means a metric must be introduced.

    And then what is the explicit relation between the components of the vectors in the different basis?

    Thanks!
     
  15. Jun 29, 2006 #14

    nrqed

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    Oh...now it is slowly starting to sink in. Now I *think* I understand a bit better. My mistake is that I was assuming that the components of a vector field in the basis [itex] {\partial \over \partial_i} [/itex] were the same as the components in the basis of the usual unit vectors.
    Of course it makes sense that they will not be (:blushing: ). Now, it must be that the relation between the two involves the metric (or its inverse). How does one establish the relation between the components in the two basis?

    Patrick
     
  16. Jun 29, 2006 #15

    George Jones

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    Yes, a metric is needed. Differential geometry sometimes deals with manifolds that don't have metrics, but it is not restricted to such manifolds.

    Let's continue on with this example, i.e., the [itex]\mathbb{R}^3[/itex] with metric for which [itex]\right\{\partial / \partial x, \partial / \partial y, \partial / \partial x \right\}[/itex] is an orthonormal basis. Then, e.g.,

    [tex]\frac{\partial}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial}{\partial y} + \frac{\partial z}{\partial r} \frac{\partial}{\partial z}[/tex].

    Dot this [itex]\partial / \partial r[/itex] with itself, and use the orthonormality of the coordinate basis to find the (squared) length of [itex]\partial / \partial r[/itex]. Divide [itex]\partial / \partial r[/itex] by its length to find [itex]\hat{e}_r[/itex]. Do the same for the other two coordiante.

    In this case we're a bit lucky in that the coodinate vectors are othogonal, but it doesn't always have to be this way.

    Now,

    [tex]
    v = v_{r} \hat{e}_{r} + v_{\theta} \hat{e}_{\theta} + v_{\phi} \hat{e}_{\phi} = v'_{r} \frac{\partial}{\partial r} + v'_{\theta} \frac{\partial}{\partial \theta} + v'_{\phi} \frac{\partial}{\partial \phi}.[/tex]

    Writing either the hatted basis vectors or the unhatted basis vectors in terms of the other set gives the relationship between the compoments.
     
    Last edited: Jun 29, 2006
  17. Jun 29, 2006 #16

    nrqed

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    Thanks by the way for being patient withme:redface: I appreciate.

    I understand this.
    I guess that I was saying this because it seemed to me (from what I was reading in Baez and Muniain) that one important lesson they wanted to make was that the directional derivative was defined before introducing any metric (which they discuss for the first time 32 pages later than the above quote). So I wanted to see how get to the usual expression of the directional derivative without invoking the metric.

    I am guessing that it will cancel out when I see the derivation in full (in the sense that there will be the metric contracted with its inverse).


    Ok, but my dilemma is the following. I don't know how to take the scalar product between a pair of [itex] {\partial \over \partial_i} [/itex]. I can guess what the result should be (<partial_x,partial_x> =1 and so on) but I don't want to make any guess.

    so that becomes my question: how is the scalar product between partial derivatives (I guess these are what you call coordinate bases?) is defined?

    That's my stumbling block for now.

    Thanks a lot!!
     
  18. Jun 29, 2006 #17

    nrqed

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    Just to add to my previous post.

    First let me emphasize again that I am not trying to be difficult. I am just trying to *really* understand (as opposed to doing calculations and getting sometimes the correct results without feeling that I am really understanding what I am doing).

    Just to add a comment: I can understand how to get from the [itex] \partial_r, \partial_\theta, \partial_\phi [/itex] to the [itex] \partial_x, \partial_y,\partial_z [/itex] of course.

    What bothers me is that once I have the [itex] \partial_x, \partial_y,\partial_z [/itex], I have to switch back to thinking of them as the usual i,j,k and to do scalar products and finding magnitudes. It bothers me for two reasons.

    First, I know that we are in R^3 and that there *is* a metric around, but I thought that the point of working with the "partial derivative basis" was that they did not require a metric to be define. So I feel strange about calculating scalar products directly in terms of these.

    Second, if I am allowed to simply make the correspondence [itex] \partial_x \rightarrow {\hat i} [/itex] and so on , why was it possible to simply make the correspondence [itex] \partial_r \rightarrow {\hat r} [/itex] and so on?

    The answer is surely that there is a non trivial step in going from the partial derivative as a basis to unit vectors....But that step becomes probably so trivial in the case of the x,y,z basis that it's not even mentioned. But to really understand what is going on I would need to see that step clearly explained, I am afraid.

    I hope this helps clarify my confusion!

    Thanks!

    Patrick
     
  19. Jun 29, 2006 #18

    George Jones

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    A bit of care has to taken when using the term "scalar product." Modern differential geometry emphasizes that 1-forms exist without metrics, and that a 1-form acts on a vector to give a scalar. All of this can be done interms of coordinate bases. If a metric does happen to exist, then there is a natural correspondence between getting a scalar by letting a 1-form act on a vector, and by letting the metric act on two vectors.

    It is possible!

    In order to have explicitly a manifold with metric, the metric must be defined in some way. The manifold R^3 is turned into normal Euclidean space by defining a metric for which the basis vectors of a Cartesian coordinate are orthonormal. This is the definition of the metric.

    Defining the basis vectors associated with spherical coordinates to be orthonormal results in a completely different metric. In this case, the relationship between the two metrics is a (variable) change of scale. For this metric, e.g., a line of constant r and theta and changing phi is a geodesic.

    Usually, metrics that are useful, either mathematically or physically, are chosen. In the case of R^3, the first metric above is almost always chosen, and this is the metric that Baez and Munian chooses implicitly on the top of page 74.

    Just divide a vector by its metric-given length.

    You might not like my answer to this. The basis vectors associated with the {x, y, z} coordinate system are orthonormal ... by definition.

    This defines the usual metric.
     
    Last edited: Jun 29, 2006
  20. Jun 29, 2006 #19

    nrqed

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    Thanks again George. I have to leave shortly unfortunately but I hope you will keep helping me out!

    I understand the above, yes.
    And, unles I am mistaken, vector fields also exist without a metric. This is my main source of confusion. We have a vector field that we write using [itex] \partial_r...[/itex] and then we want to write it using [itex] {\hat e_r}...[/itex].

    If we go from the [itex] \partial_r...[/itex] to [/itex] \partial_x ...[/itex], there is no need to invoke the metric (unless I am mistaken).

    What confuses me is the role of the [itex] {\hat e_r}...[/itex] basis (or the i,j,k or any other unit vector basis. In *that* case, a metric is needed to define them, am I right? (I guess the fact that they are of unit length is a giveaway!). So they are fundamentally different than the partial derivative basis in that regard. This is the first point I want to clarify because I used to see all those basis as just corresponding to a differnt choice of basis without invoking a metric.


    My second point is that I am still wondering about scalar product between partial_x and partial_y, say. Again, these basis are defined independently of a metric, right? But the scalar product involves a metric. So it seems to me that there is a nontrivial step there that involves introducing a metric before the scalar product is taken. Am I correct? What I mean is that I could have a 3-D manifold with coordinates x,y,z on which no metric is defined and I still would work with partial_x, etc. But then I would never be able to get to a basis with the usual i,j,k or e_r, e_theta, r_phi, is that right??

    I guess that I want to see the "big picture".

    I hope this makes sense. I really appreciate yoru help. I have to run but I am looking forward to reading your comments!!
     
  21. Jun 29, 2006 #20

    mathwonk

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    sighhh... i try to explain these things in down to earth terms, and yet the students seem to prefer computational details and technical symbols to my plain explanations of what is really going on. [voice in wilderness]
     
  22. Jun 30, 2006 #21

    nrqed

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    Maybe it's because I am too stupid but I need to see both explicit calculations and discussions of the general concepts when I learn a new framework.

    I don't know why but it seems as if asking questions about differential geometry always bring comments of the type "this is obvious, you are wasting my time" :mad:

    EDIT: I want to add that I have appreciated immensely your comments in the past and that it's quite possible that you have already explained this to me before. I would not be surprised if that has happened. But I find myself spending several months at a time without having the opportunity to think about this subject (unfortunately) because I sometimes have over 20 contact hours of teaching per semester. This, in combination with the fact that I never reached a point where everything seemed crystal clear to me and the fact that I also spend my free time time learning other stuff (I spent the last month learning Python and deciphering code written in Fortran 90 and Python that performs calculations in lattice QCD) explains why I may find myself asking the same or similar questions a few months apart. Maybe that makes me stupid, I don't know. But it does not mean that I was not paying attention the previous time.

    Regards
     
    Last edited: Jun 30, 2006
  23. Jun 30, 2006 #22

    George Jones

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    I agree that post #10 by a certain mathwonk was very nice, and we will eventually get back to it; in the meantime, cover your eyes.
     
  24. Jun 30, 2006 #23

    George Jones

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    Right - no metric required.

    Right.

    Right, saying, e.g., which bases are orthonormal requires the use of a metric. In general relativity, a tetrad is an orthonormal basis, and, in many important spacetimes, tetrads are not coordinate bases.

    Right.

    Right.

    Right.

    Since a vector space is closed under scalar multiplication, you, without a metric, could work with, e.g., the vector

    [tex]\frac{1}{r} \partial_\theta,[/tex]

    but you couldn't assign a length to this vector, and you coudn't say which vectors are orthogonal to this vector.

    In our example, Cartesian coordinates are used to define the metric. In other words, define [itex]g[/itex] by:

    1) [itex]g[/itex] is bilinear;

    2) [itex]g \left( \partial_{i} , \partial_{j} \right) = \delta_{ij}.[/itex]

    This defines the scalar product of any two vectors.
     
  25. Jun 30, 2006 #24

    nrqed

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    Ok, then it all fits perfectly together. Thank you George!

    Now I realize what the source of my confusion was. What I had not realized is that some bases for vectors require the introduction of a metric and some other bases don't. *This* is really the reason for my confusion. In my mind, all bases were metric independent . Of course, in hindsight, the very name "orthonormal basis" is a dead giveaway. But my problem was that I knew that vector fields exist independently of a metric and I was (subconsciously) assuming that any representation of a vector field (in the sense of giving its components) had to be independent of a metric.

    Now this is much clearer!

    In regard to Mathwonk's comment of post # 10. Yes, that WAS *very* instructive and useful to me!!!! I finally understood that the gradient of introductory math was in fact a vector and I understood, because of Mathwonk, that more knowledgeable people are used to switch between the form df and its "associated" (through the metric) vector nabla f. And that cleared up a major source of confusion (and frustration) for me so I highly appreciate that post! (my sincere apologies for not replying with a thank you post, I really should have :frown: :frown: ).

    But this then gave me the impetus to actually go beyond discussing the concepts and actually sitting down and working out all the details in a specific case (which is the true test of understanding a new formalism, imho). And this is when I ran into a problem which led me to my post #11.

    I hope that this did not give the impression that I had ignored Mathwonk's post!!!!
    Notice that my question of post #11 is not the same as the question that led to Mathwonk's post #10! Mathwonk clarified for me the disticntion between the one-form df and the vector nabla f. In post #11 I started to use that knowledge to explicitly calculate a *directional derivative*. This was no longer an issue of distinguishing df and nabla f. It was an attempt at working out explictly df(v) and understanding why this comes out independent of a metric.

    Again, if Mathwonk felt that I had ignored his very appreciated post (and there were other posts in other threads that were illuminating), I apologize sincerely!

    Patrick
     
  26. Jun 30, 2006 #25

    mathwonk

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    thank you george and nrqed, for the feedback and encouragement. i am very insecure and appreciate it a lot.

    best,

    mwonk.
     
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