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I Difference between 1-form and gradient

  1. Feb 6, 2017 #1
    I have seen and gone through this thread over and over again but still it is not clear.

    https://www.physicsforums.com/threads/vectors-one-forms-and-gradients.82943/


    The gradient in different coordinate systems is dependent on a metric

    But the 1-form is not dependent on a metric. It is a metric independent mapping. Is that correct first of all ?

    But I also see that the 1-form is defined in term of a gradient. Is that accurate ?

    So is the 1-form dependent on a metric ? I am really confused. So any clarifications would help.
     
  2. jcsd
  3. Feb 6, 2017 #2

    stevendaryl

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    Without a metric, there is a big distinction between vectors and co-vectors. With the conventions that I'm familiar with, components of a vector are written with raised indices, [itex]V^\mu[/itex], while components of a co-vector are written with lowered indices, [itex]\omega_\mu[/itex]. Without a metric, you can still combine a vector with a co-vector, [itex]\sum_\mu V^\mu \omega_\mu[/itex], but you cannot take a scalar product of two vectors or two co-vectors; you need one of each type of object.

    So without a metric, you can define a kind of derivative of a scalar field [itex]\Phi[/itex], [itex]d \Phi[/itex], which has components [itex](d \Phi)^\mu = \frac{\partial \Phi}{\partial x^\mu}[/itex]. That looks like a gradient, but it's a co-vector, not a vector.

    If you have a metric [itex]g[/itex], then you can convert a co-vector into a vector using the inverse of the metric:

    [itex]\nabla \Phi = g^{-1}(d\Phi)[/itex]

    In terms of components:

    [itex](\nabla \Phi)^\mu = \sum_\nu g^{\mu \nu} (d \Phi)_\nu = \sum_\nu g^{\mu \nu} \frac{\partial \Phi}{\partial x^\nu}[/itex]

    So you don't need a metric to get a gradient-like object that is a co-vector, but to get a gradient-like object that is a vector requires a metric.
     
  4. Feb 6, 2017 #3
    Many thanks Steve for your prompt response.

    So if I were to summarize your response -

    1) There is no difference between a vector and a co-vector in flat space. The distinction between covariant and contravariant vectors is irrelevant
    2) However in curved space or the usage of curvilinear coordinates that describe the space the idea of a metric slips in and now the distinction between covector and vector is very much on.

    I am still not clear on this fact
    -Why a covector alone can be defined independent of a metric whereas to get a gradient like object that is a vector requires a metric ? What makes the covector special ? Is that just to do with the definitions ?
     
  5. Feb 7, 2017 #4

    lavinia

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    The 1 form one gets from a differentiable function can be defined without a metric. If ##X## is a vector at a point ##p## then ##df(X)## is just the derivative of ##f## along a curve whose derivative equals the vector ##X## at ##p##. No metric is needed.

    If one has a metric ##<,>## then there is a vector field ##grad(f)## (not a 1 form) that satisfies the relation ##df(X) = <grad(f),X>##. If one changes the metric, then one gets a different vector field ##grad(f)##. So the vector field ##grad(f)## is dependent on the metric.

    However the 1 form ##ω(X) = <grad(f),X>## is the same no matter what the metric is. So even though one can represent the 1 form using a metric the 1 form is not dependent on the metric.
     
    Last edited: Feb 7, 2017
  6. Feb 7, 2017 #5
    Let us have a frame. Is there a way to picture a one-form by its components in that frame?
     
  7. Feb 7, 2017 #6

    lavinia

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    Not sure what you mean by "picture". And do you mean by "frame" any set of vector fields that from a basis for the tangent space at each point?
     
    Last edited: Feb 7, 2017
  8. Feb 7, 2017 #7

    stevendaryl

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    I would say, rather, that in flat space, using Cartesian coordinates, the conversion between them is trivial.

    No, I wouldn't say that's correct. The metric is already there in flat space; it's just that it's a very simple metric. A metric is what allows you to convert between covectors and vectors. If you don't have a metric, you can't convert between them. So it's the lack of a metric that makes a distinction between vectors and covectors.

    Let me illustrate the distinction between vectors and covectors in a setting where there is no possible way of converting between the two.

    Suppose we're studying weather, and I tell you that air pressure is a function of temperature and altitude (I don't know if that's true, but suppose that it is). So if you are driving around in your car, both your altitude and your temperature might be changing as a function of time.

    Abstractly, we can represent your current state as a point on a 2D graph, where the x-axis is the temperature and the y-axis is the altitude. We can use a function [itex]P(r)[/itex] to represent the pressure corresponding to point [itex]r[/itex]. We can use a function [itex]R(t)[/itex] to represent your position on the graph as a function of time. Technically, [itex]P(r)[/itex] is a scalar field on the abstract space of temperature x altitude. [itex]R(t)[/itex] is a parametrized path on that abstract space.

    Now, we can associate a one-form [itex]dP[/itex] with the pressure, which tells how the pressure is varying with position on the graph. It has components [itex](dP)_x = \frac{\partial P}{\partial x}[/itex] and [itex](dP)_y = \frac{\partial P}{\partial y}[/itex].

    We can associate a tangent vector [itex]v = \frac{dR}{dt}[/itex] with your path. It has components [itex]v^x = \frac{dx}{dt}[/itex] and [itex]v^y = \frac{dy}{dt}[/itex].

    Note: With Cartesian coordinates, you can define a length of a vector [itex]v[/itex]: by: [itex]|v| = \sqrt{(v^x)^2 + (v^y)^2}[/itex]. But in our case, if we measure [itex]P[/itex] in atmospheres, and measure [itex]x[/itex] in degrees C, and measure [itex]y[/itex] in meters, then [itex](dP)_x[/itex] has dimensions [itex]\dfrac{atmospheres}{degree}[/itex] and [itex](dP)_y[/itex] has dimensions [itex]\dfrac{atmospheres}{meter}[/itex]. It doesn't make any sense to square them and add them together. So there is no meaningful way to compute the length of [itex]dP[/itex]

    Similarly, the components of the tangent vector [itex]v^x[/itex] has dimensions [itex]\dfrac{degrees}{sec}[/itex] and [itex]v^y[/itex] has dimensions [itex]\dfrac{meters}{sec}[/itex]. There is no way to compute a meaningful length of the vector [itex]v[/itex].

    Without a metric, there is no way to take the "dot product" of two vectors, or of two covectors. But you can take a meaningful dot-product of one vector and one covector:

    [itex]\langle v, dP\rangle = v^x (dP)_x + v^y (dP)_y[/itex]

    The first term has dimensions [itex]\frac{degrees}{second} \cdot \frac{atmospheres}{degree} = \frac{atmospheres}{second}[/itex]. The second term has dimensions [itex]\frac{meters}{second} \cdot \frac{atmospheres}{meter} = \frac{atmospheres}{second}[/itex]. So you can add those two together to get the total rate of change in [itex]P[/itex] as a function of [itex]t[/itex].

    So one way to think about a metric is that it provides a "conversion" between different components of a vector or covector so that they can be compared. Without a metric, it's as if the different components of a vector or covector are in different units.
     
  9. Feb 7, 2017 #8

    lavinia

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    The difference between a vector and a linear map on a vector space is that one is a vector the other is a function on a vector space. Flat or curved space has absolutely nothing to do with it. It is a question of linear algebra not of physics.

    When one has an inner product then for each linear function one can find a vector ##V## so that the value of the linear function is the same as the inner product with ##V##. That does not make ##V## the same as the linear function. It makes the inner product with ##V## the same linear function.

    It turns out that the association of a linear function with ##V## via an inner product is an isomorphism of vector spaces. But a different inner product will give you a different ##V##.
     
  10. Feb 7, 2017 #9
    I mean we have a basis and a metrics. We draw it in cartesian coordinates. How can we picture an arbitrary one-form there knowing its components?
     
    Last edited: Feb 7, 2017
  11. Feb 7, 2017 #10

    lavinia

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    A good way to visualize it is by the field hyperplanes in each tangent space that it sends to zero.
     
    Last edited: Feb 7, 2017
  12. Feb 7, 2017 #11
    In the fluid dynamics book I am reading the gradient is defined as the following

    $$q^n \partial / \partial q^n $$

    Is there a reason that it cannot be expressed as the following


    $$q_n \partial / \partial q_n $$

     
  13. Feb 8, 2017 #12
    Lavinia

    What is the difference between


    $$\omega(X) = <grad(f),X>$$


    and

    df(X) = <grad(f),X>. From the notation they are both scalar products am I wrong ?
     
  14. Feb 8, 2017 #13

    lavinia

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    They are the same. I used ##ω## to indicate that ##df## is a 1 form. ##ω=df##.

    It is more generally true that any 1 form whether or not it is the differential of a function can be expressed as a inner product with some vector field. Again this vector field changes if the inner product changes.
     
  15. Feb 8, 2017 #14
    In grad(f) I am presuming f is a scalar field. But I guess you can have grad(A) where is A is a vector field as well. Correct ?
     
  16. Feb 8, 2017 #15

    lavinia

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    No. The gradient is the gradient of a scalar field. The gradient of a scalar field is a vector field. The differential of a scalar field is a 1 form. The differential can be calculated by the inner product with the gradient vector field. The gradient field depends upon the metric. A different metric gives a different gradient field.

    Here is an example. Let ##f## be the scalar ##f(x) = x## defined on the real number line. Let the inner product be the dot product.

    The differential ##df## is ##df = dx##. The gradient ##grad(f)## is the vector field ##∂/∂x##.
    If ##a∂/∂x## is a tangent vector then ##df(a∂/∂x) = dx(a) = a##. This may be calulcated as the dot product ##∂/∂x⋅a∂/∂x## since ##∂/∂x⋅∂/∂x = 1##. Now change the inner product to be twice the dot product. With this new inner product ##<∂/∂x,∂/∂x> = 2##. So ##df(a∂/∂x) = dx(a)= a = <1/2∂/∂x,a∂/∂x>## so the gradient of ##f## under this new dot product is ##1/2∂/∂x##.
     
    Last edited: Feb 8, 2017
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