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Newtonian force as a covariant or contravariant quantity

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dx
#145
Feb1-13, 02:19 AM
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Quote Quote by bcrowell View Post
But I'm not clear on how to notate this idea that in both examples, the force 1-form is parallel to the surface. Do you represent the surface as, say, a 1-form created by taking the (infinite) gradient of a step function across the wall?
The constraint surface can be represented by an equation of the form f(Xi) = 0. The constraint surface in your second example is

f(φ, θ) = θ - φ = 0

A constraint force which is parallel to the constraint surface can be written as a multiple of df, since df is parallel to the constraint surface.

F = λdf = λ(dθ - dφ)
stevendaryl
#146
Feb1-13, 05:55 AM
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Quote Quote by bcrowell View Post
No, I'm pretty sure you're wrong about this. The operator [itex]\partial_a=\partial/\partial x^a[/itex] is written with a lower index precisely because it's a covector and transforms like one. The notation isn't misleading. The notation shows exactly how the thing transforms. For instance, you can convince yourself of this by writing down a rescaling of the coordinate and applying the tensor transformation law.
Yeah, these conventions are confusing, but they make some sense if you think about it.

In differential geometry, a vector [itex]V[/itex] is defined via its effects on scalar fields. Specifically, if we choose a particular coordinate basis, we can write:

[itex]V(\Phi) = V^\mu \partial_\mu \Phi[/itex]

with the implicit summation convention. Now, if [itex]V[/itex] happens to be a basis vector [itex]e_\sigma[/itex], then that means that [itex]V_\sigma = 1[/itex], and all the other components are equal to 0. So we have, in this case:

[itex]e_\sigma(\Phi) = \partial_\sigma \Phi[/itex]

Since this is true for any [itex]\Phi[/itex], it's an operator equation:

[itex]e_\sigma = \partial_\sigma[/itex]
stevendaryl
#147
Feb1-13, 06:10 AM
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Quote Quote by micromass View Post
The notation [itex]\frac{\partial}{\partial x^i}\vert_p[/itex] represents a tangent vector,and not a 1-form. A 1-form would look like [itex]dx^i\vert_p[/itex]. See "Introduction to Smooth Manifolds" by Lee.
Yeah, this stuff is pretty confusing. Even though [itex]\frac{\partial}{\partial x^\mu}[/itex] is a vector, the prototypical covector is [itex]\nabla \Phi[/itex] for some scalar field [itex]\Phi[/itex], which has components [itex]\frac{\partial}{\partial x^\mu} \Phi[/itex]

To complete the confusion, even though [itex]dx^\mu[/itex] is a 1-form, the prototypical vector is a tangent vector to a parametrized path [itex]P(\lambda)[/itex], which has components [itex]\dfrac{dx^\mu}{d\lambda}[/itex].
TrickyDicky
#148
Feb1-13, 07:34 AM
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Quote Quote by stevendaryl View Post
Yeah, this stuff is pretty confusing. Even though [itex]\frac{\partial}{\partial x^\mu}[/itex] is a vector, the prototypical covector is [itex]\nabla \Phi[/itex] for some scalar field [itex]\Phi[/itex], which has components [itex]\frac{\partial}{\partial x^\mu} \Phi[/itex]

To complete the confusion, even though [itex]dx^\mu[/itex] is a 1-form, the prototypical vector is a tangent vector to a parametrized path [itex]P(\lambda)[/itex], which has components [itex]\dfrac{dx^\mu}{d\lambda}[/itex].
It is confusing, but even more if one doesn't distinguish the geometrical object from its component representation. [itex]dx^\mu[/itex] is a 1-form, but transforms contravariantly, however if it had components other than unity they would transform covariantly.

So I'm still waiting for someone to explain why the rate of change of momentum is not naturally a tangent vector.
stevendaryl
#149
Feb1-13, 09:07 AM
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Quote Quote by TrickyDicky View Post
It is confusing, but even more if one doesn't distinguish the geometrical object from its component representation. [itex]dx^\mu[/itex] is a 1-form, but transforms contravariantly, however if it had components other than unity they would transform covariantly.

So I'm still waiting for someone to explain why the rate of change of momentum is not naturally a tangent vector.
If momentum is defined to be [itex]\vec{P} = m \frac{\vec{dU}}{dt}[/itex], then it is naturally a vector. If it is defined via a Lagrangian [itex]L[/itex] by [itex]P_\mu = \frac{\partial}{\partial U^\mu}L[/itex], then it is naturally a covector.

Another way to see that momentum should be considered contravariant might be:

In curvilinear coordinates, the equations of motion for a free particle is

[itex]\dfrac{d}{dt} U_\mu = 0[/itex]

not

[itex]\dfrac{d}{dt} U^\mu = 0[/itex]
bcrowell
#150
Feb1-13, 09:36 AM
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I think the basic cause of the ugly inconsistency in Einstein notation is not that there's anything wrong with Einstein notation but that it's being mixed inappropriately with Sylvester's notation. Sylvester was the one who originally introduced terms like "contravariant" and "covariant." The pure Sylvester notation is described in this WP article: http://en.wikipedia.org/wiki/Contravariant_vector Unfortunately Sylvester's notation and terminology are much more cumbersome than Einstein notation, and it doesn't correspond to the way physicists customarily describe the objects themselves as having transformation properties, rather than the objects staying invariant while their representations transform.

In Einstein notation, a sum over up-down indices represents a tensor contraction that reduces the rank of an expression by 2. In Sylvester notation, a vector is expressed in terms of a basis as a sum over up-down indices, which is not a tensor product. Mixing Einstein and Sylvester notation produces these silly-looking things where it looks like we're expressing a covector as a sum of vectors or a vector as a sum of covectors.

The abstract index form of Einstein notation is carefully designed so that you *can't* inadvertently refer to a coordinate system; anything written in abstract index notation is automatically coordinate-independent. This means that in abstract index notation, you can't express a vector as a sum over basis vectors. You don't want to and you don't need to.

There is also an issue because in Einstein notation we refer to vectors as covariant and contravariant, whereas in Sylvester notation a given vector or 1-form is written in terms of components and basis vectors that are covariant and contravariant. Neither is right or wrong, but mixing them leads to stuff that looks like nonsense.
atyy
#151
Feb1-13, 09:50 AM
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Ok, I actually can't understand the Lewis notes I linked to in #127.

These guys http://www.google.com/url?sa=t&rct=j...41867550,d.b2I do it differently. In proposition 7.8.1 they do define a force via contraction with a vector field. However they contract with a symplectic 2 form, not a metric. I think the relevant definition of the 2 form is on p161.

Incidentally the Lewis notes say ad hoc "motivation" is that the Euler-Lagrange equations transform as the components of a one form. He say this is hand-waving because the EL equations are not a one-form.
atyy
#152
Feb1-13, 06:23 PM
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I guess the Lewis notes http://www.mast.queensu.ca/~andrew/t...f/439notes.pdf are fine after all. Force is a one-form. What I was confused by is their notation F: R X TQ -> T*Q. All they mean is that a force is an assignment of a one form to every velocity at every position. Just as they say a vector field is V: Q -> TQ by which they mean a vector field is an assignment of a vector at every position.


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