Newtonian force as a covariant or contravariant quantity

[bcrowell]

I recently came across a very cool book called Div, Grad, and Curl are Dead by Burke. This is apparently a bit of a cult classic among mathematicians, not to be confused with Div, Grad, Curl, and All That. Burke was killed in a car accident before he could put the book in final, publishable form, but it can be found in various places online, e.g., on scribd.com. At the beginning of section 12, in the context of Newtonian mechanics, he says,

We now come to the first surprise. Force is not a vector, but a 1-form. The most direct way to see this is to think of the work done by a force. Force is the operator that takes in a displacement, a vector, and tells you how much work was done. This makes forces dual to vectors, i.e., 1-forms.

In the language of Einstein-style index gymnastics, applied in a nonrelativistic context, this amounts to a statement that energy is a scalar, and displacement is a contravariant (upper-index) vector, so force should naturally be considered as a covariant (lower-index) vector.

The first thing I'm unsure about here is whether energy is really a scalar in the context of nonrelativistic mechanics, if "scalar" is taken to have its full Einstein-style interpretation of "invariant under any change of coordinates." Doesn't nonrelativistic energy change when you rescale your coordinates, implying that it's a scalar density rather than a true scalar? (After all, in relativity, rescaling coordinates changes all the components of the energy-momentum tensor, which means it changes the mass-energy.)

The second thing that bugs me is that if you were to reason from Newton's second law, it seems like you would "naturally" conclude the opposite, that force is a contravariant vector.

If we follow the usual but arbitrary convention of saying that upper indices are used for distances measured on a ruler, then this breaks the otherwise perfect symmetry between vectors and their duals. It then seems clear that things like velocity, which can be obtained by differentiation with respect to a scalar, should also take upper indices (be contravariant). But I'm less convinced that this then breaks the duality symmetry in the case of Newtonian force, or a relativistic quantity like the stress-energy tensor...?

 

[WannabeNewton]

I think here when we talk about W({x}) = \int_{{x_{0}}}^{{x}}F_{i}dx^{i} being a scalar, it just means it is a real valued function defined on the image of the curve that the particle is traveling on and not necessarily a coordinate invariant quantity. Note, regarding your second point if I understood the question correctly, that when we are dealing with \mathbb{R}^{n}, the dual vectors and vectors are naturally identified.

 

[bcrowell]

A similar but slightly different example...

This WP article http://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors says,

Examples of vectors with contravariant components include the position of an object relative to an observer, or any derivative of position with respect to time, including velocity, acceleration, and jerk.

But if I follow Burke's philosophy that work is obviously a scalar, then power should be a scalar as well. If there is only one force acting on a particle of mass m, then the power is given by P=m a\cdot v. This implies that a and v can't both naturally be contravariant as claimed by WP.

 

[WannabeNewton]

Yes if you write it in that form i.e. P = ma_{i}v^{i} then the acceleration shows up as a co - vector but if we are talking about particles moving in \mathbb{R}^{3}, the dual space of \mathbb{R}^{3} IS \mathbb{R}^{3} so it isn't wrong to also say it is a vector if we are just talking about euclidean space with the canonical basis.

 

[bcrowell]

I think here when we talk about W({x}) = \int_{{x_{0}}}^{{x}}F_{i}dx^{i} being a scalar, it just means it is a real valued function defined on the image of the curve that the particle is traveling on and not necessarily a coordinate invariant quantity. Note, regarding your second point if I understood the question correctly, that when we are dealing with \mathbb{R}^{n}, the dual vectors and vectors are naturally identified.

What you're saying seems sensible, but it would seem to completely contradict the statements by Burke and in the WP article to the effect that this quantity or that quantity "is" covariant or contravariant. That is, they seem to be claiming that the natural identification provided by duality is not so natural.

As a side issue, I'm not sure what you mean by "when we are dealing with \mathbb{R}^{n}." As far as I can see, we're always dealing with a tangent space on a manifold, which is always isomorphic to \mathbb{R}^{n}. Whenever you have a vector space, you can make duals of vectors.

 

[robphy]

I think power is F_a v^a.
I think that to write Newton's Second Law with acceleration,
one needs a metric: F_a =mg_{ab}a^{b}.

Using momentum (thought of as a covector), the Second Law is
F_a= \frac{d}{dt} p_a

 

[bcrowell]

the dual space of \mathbb{R}^{3} IS \mathbb{R}^{3}

I don't think this is true. They're isomorphic, but not the same. An element of \mathbb{R}^3 is a 3-tuple of real numbers. An element of its dual space is a linear function. Physically, elements of the two spaces have different transformation properties.

 

[bcrowell]

I think power is F_a v^a.
I think that to write Newton's Second Law with acceleration,
one needs a metric: F_a =mg_{ab}a^{b}.

Sure. If we assert complete symmetry between vectors and their duals, then the whole issue becomes a non-issue. But both Burke and the WP article seem to be asserting that the symmetry is not complete, and that there is some mode of reasoning that tells you which form is preferred, covariant or contravariant, for a given physical quantity.

 

[WannabeNewton]

I don't think this is true. They're isomorphic, but not the same. An element of \mathbb{R}^3 is a 3-tuple of real numbers. An element of its dual space is a linear function.

Hi ben yes strictly they are not equal but their vector space structures are identical so for all practical purposes they are the same. For vector spaces over the reals, of with which we are concerned with, we can readily go from a vector to a corresponding dual vector via the Riesz representation so why should there be some a priori preference for saying say force must be co - vector as opposed to a vector: is that what is being asked?

 

[Bill_K]

I recently came across a very cool book called Div, Grad, and Curl are Dead by Burke.

Bill was a remarkable person, and one of my best friends. His biography is here.

 

[micromass]

Hi ben yes strictly they are not equal but their vector space structures are identical so for all practical purposes they are the same. For vector spaces over the reals, of with which we are concerned with, we can readily go from a vector to a corresponding dual vector via the Riesz representation so why should there be some a priori preference for saying say force must be co - vector as opposed to a vector: is that what is being asked?

If that is the question, then the answer is easy: because you want to define line integrals.

The entire reason for defining 1-forms is to be able to take line integrals. A line integral should only ever be defined on 1-forms (or differential forms). Taking line integrals of vector fields is done in practice, but is not a good standard.

So there is a reasonable process to decide whether something should be a 1-form or a vector. The question you should ask: do you intend to take line integrals of your object? If the answer is yes, then you are dealing with a 1-form.

Since taking line integrals of force fields is something that is very common, this is an explanation of why force should be a 1-form and not a vector.

 

[PAllen]

Just to add to this, Bergmann, in his 1942 book on SR+GR, treated force as a covariant vector, and velocity and 4-acceleration as contravariant vectors. 4-momentum was treated inconsistently, sometimes contravariant, sometimes covariant.

 

[WannabeNewton]

Just to add to this, Bergmann, in his 1942 book on SR+GR, treated force as a covariant vector, and velocity and 4-acceleration as contravariant vectors. 4-momentum was treated inconsistently, sometimes contravariant, sometimes covariant.

Just to further add on to this PAllen if you don't mind =D, Soper in his text on classical field theory also defines force as a one form and you can see it in the tensorial nature of the euler lagrange equations which are one - form equations.

 

[bcrowell]

Bill was a remarkable person, and one of my best friends. His biography is here.

I really like the book, and it's a shame that it seems to have been condemned to this sort of dusty junk-shop existence on the internet. Do you have any idea whether his heirs (Violet John?) would be interested in seeing it spiffed up and distributed for free under a creative commons license? I'd been thinking of making inquiries, but didn't know who to contact initially, was thinking maybe his colleagues at UCSC.

 

[robphy]

This point of view about force as a 1-form appears in Burke's
Applied Differential Geometry and in Spacetime, Geometry, Cosmology.
A related article is a J.Math Phys article on Twisted Differential Forms for Electromagnetism ( http://dx.doi.org/10.1063/1.525603 )

see also
http://www.ucolick.org/~burke/home.html

Earlier sources include Schouten's Tensor Analysis for Physicists.

Momentum as a covector seems natural from a Hamiltonian Mechanics viewpoint.


...all of this talk is nudging me to get back to a project of mine to visualize tensors and differential forms.

 

[WannabeNewton]

Momentum as a covector seems natural from a Hamiltonian Mechanics viewpoint.

Here I think it is more strictly required than seen as one possible viewpoint in order for the phase space to be the cotangent bundle to the configuration space manifold. Unlike this case that you mentioned, in GR I find it hard to keep track of quantities that are naturally taken to be one - forms or vectors in the context of index based computations because one freely uses the metric tensor to go form one to the other.

 

[robphy]

Here I think it is more strictly required than seen as one possible viewpoint in order for the phase space to be the cotangent bundle to the configuration space manifold. Unlike this case that you mentioned, in GR I find it hard to keep track of quantities that are naturally taken to be one - forms or vectors in the context of index based computations because one freely uses the metric tensor to go form one to the other.

In typical presentations of GR or other subjects, there is little attention paid
to what fundamental structures (e.g. metric, volume-form, connection, etc...)
are being used in a construction or calculation.

So, we have this blurring of whether something is naturally a vector, covector (one-form), or maybe even a bivector or two-form. This is especially true when one deals with a space (e.g. Minkowski space or Euclidean space) with lots of symmetries.

The references by Burke and Schouten (and others) try to tease out what could be the fundamental structures... For example, there is work on metric-independent formulations of electromagnetism... where, e.g., the magnetic field is fundamentally a two-form. Then, only with additional structure, can it be thought of as an axial-vector [in three spatial dimensions].I actually like the [abstract-]index computations because they help me keep track of
what kinds of objects I am dealing with. Ideally, it would be nice to a have a geometrical picture
to go along with the computations to suggest what is physically going on.

 

[Ben Niehoff]

Using momentum (thought of as a covector), the Second Law is
F_a= \frac{d}{dt} p_a

I think this is the most sensible answer, and eliminates the discrepancy. Force is a covector in every place it appears. Momentum is naturally a covector, being canonically conjugate to position.

What is missing from this picture is how you relate velocity and momentum. Incidentally, this is exactly the same question as relating Lagrangian to Hamiltonian dynamics. At some point you are forced to identify the tangent and cotangent bundles. Luckily there is a canonical way to do so.

 

[PAllen]

What is missing from this picture is how you relate velocity and momentum. Incidentally, this is exactly the same question as relating Lagrangian to Hamiltonian dynamics. At some point you are forced to identify the tangent and cotangent bundles. Luckily there is a canonical way to do so.

Wondering if you know of anything better than: Oh, we want momentum to be a covector derived from 4 velocity, so take m times 4-velocity and make it a covector with the metric. In one variation or another, that's what I've seen, and it has seemed a bit artificial to me.

 

[WannabeNewton]

I actually like the [abstract-]index computations because they help me keep track of
what kinds of objects I am dealing with. Ideally, it would be nice to a have a geometrical picture
to go along with the computations to suggest what is physically going on.

I find this is true in the symplectic geometry based classical mechanics class I'm taking because here the index free approach really has more geometric insight and is extremely elegant. In the GR books I've seen the index based computations tend to cloud it but these calculations have a certain flair of their own: who doesn't like writing down a large mess of co variant derivatives ;)

 

[stevendaryl]

This is a very interesting topic. In Newtonian physics, there are two different ways to define "force":

1. F^i = m \dfrac{dU^i}{dt}, where U is velocity/
2. F_i = -\partial_j \Phi, where \Phi is potential energy.

The first would lead you to think of force as a vector, and the second would lead you to think that it is a co-vector. From experience with General Relativity, one learns to suspect that if there is a confusion between vectors and co-vectors, then that means the metric tensor is secretly at work. Using the metric tensor g_ij you can certainly resolve the tension by writing:

m g_{ij} \dfrac{dU^i}{dt} = F_j

But that's a little unsatisfying, because there is a sense in which there is no metric tensor for Newtonian physics. Why do I say that? Well, if you formulate Newtonian physics on Galilean spacetime, there are two different notions of distances between points:

1. The time between events.
2. For events taking place at the same time, the distance between events.

The latter notion of distance between events is undefined in Galilean spacetime for two events that are not simultaneous. So that makes me wonder: what is the covariant notion of spatial distance in Galilean spacetime? It's not a tensor, so what is it?

 

[stevendaryl]

This is a very interesting topic. In Newtonian physics, there are two different ways to define "force":

1. F^i = m \dfrac{dU^i}{dt}, where U is velocity/
2. F_i = -\partial_j \Phi, where \Phi is potential energy.

The first would lead you to think of force as a vector, and the second would lead you to think that it is a co-vector. From experience with General Relativity, one learns to suspect that if there is a confusion between vectors and co-vectors, then that means the metric tensor is secretly at work. Using the metric tensor g_ij you can certainly resolve the tension by writing:

m g_{ij} \dfrac{dU^i}{dt} = F_j

But that's a little unsatisfying, because there is a sense in which there is no metric tensor for Newtonian physics. Why do I say that? Well, if you formulate Newtonian physics on Galilean spacetime, there are two different notions of distances between points:

1. The time between events.
2. For events taking place at the same time, the distance between events.

The latter notion of distance between events is undefined in Galilean spacetime for two events that are not simultaneous. So that makes me wonder: what is the covariant notion of spatial distance in Galilean spacetime? It's not a tensor, so what is it?

In Newton-Cartan theory, which is a covariant formulation of Newtonian gravity, I am not sure what happens to the spatial metric tensor. How is F = ma expressed in Newton-Cartan theory, or is it only a theory of pure gravity, no non-gravitational forces?

 

[Dale]

But that's a little unsatisfying, because there is a sense in which there is no metric tensor for Newtonian physics.

There is always a metric tensor; it is just a 3D tensor for doing 3D vectors and covectors. Consider doing Newtonian physics in spherical coordinates. You still need the metric to tell you how to translate your coordinates into distances. The only difference is that time is not a coordinate but a parameter.

 

[stevendaryl]

There is always a metric tensor; it is just a 3D tensor for doing 3D vectors and covectors. Consider doing Newtonian physics in spherical coordinates. You still need the metric to tell you how to translate your coordinates into distances. The only difference is that time is not a coordinate but a parameter.

In Newtonian physics, time is both a coordinate and a parameter. It's a coordinate, because just as in SR, it takes 4 numbers to locate something in Galilean spacetime.

 

[PAllen]

In Newtonian physics, time is both a coordinate and a parameter. It's a coordinate, because just as in SR, it takes 4 numbers to locate something in Galilean spacetime.

But I've never seen velocity, momentum, or force treated as anything but 3-vectors in Newtonian mechanics; and there is no 4-dim metric at all (and can't be). However, the despite that, the general issue remains of a good set of arguments for velocity as a vector, momentum and force as covectors. Cartesian coordinates in flat space it doesn't matter, so is generally ignored. But in polar coordinates, it would matter even in Newtonian mechanics.

 

[stevendaryl]

In Newtonian physics, time is both a coordinate and a parameter. It's a coordinate, because just as in SR, it takes 4 numbers to locate something in Galilean spacetime.

It's actually interesting to formulate force-free motion in Galilean spacetime covariantly:

\dfrac{dU^i}{dt} + \Gamma^i_{jk} U^j U^k = 0

This works in all coordinate systems, including accelerated, rotating and curvilinear. But to work, you need a U^0, which is, the time component of velocity. That is, U^0 = \dfrac{d x^0}{dt}, where x^0 is coordinate time, and t is Newtonian universal time. Of course, you can always choose your time coordinate to be equal to t, in which case, U^0 = 1. But U^0 plays a role in the equations of motion, if you use accelerated or curvilinear coordinates.

• \Gamma^i_{00} U^0 U^0 gives rise to the "g" force due to using an accelerated coordinate system.
• \Gamma^i_{0j} U^0 U^j gives rise to the Coriolis force.
• \Gamma^i_{jk} U^j U^k gives rise to the Centrifugal force.

Using time as a coordinate unifies these three types of "fictitious" forces, and all three can be seen to be manifestations of the connection coefficients \Gamma^\mu_{\nu \lambda}

 

[PAllen]

It's actually interesting to formulate force-free motion in Galilean spacetime covariantly:

\dfrac{dU^i}{dt} + \Gamma^i_{jk} U^j U^k = 0

This works in all coordinate systems, including accelerated, rotating and curvilinear. But to work, you need a U^0, which is, the time component of velocity. That is, U^0 = \dfrac{d x^0}{dt}, where x^0 is coordinate time, and t is Newtonian universal time. Of course, you can always choose your time coordinate to be equal to t, in which case, U^0 = 1. But U^0 plays a role in the equations of motion, if you use accelerated or curvilinear coordinates.

• \Gamma^i_{00} U^0 U^0 gives rise to the "g" force due to using an accelerated coordinate system.
• \Gamma^i_{0j} U^0 U^j gives rise to the Coriolis force.
• \Gamma^i_{jk} U^j U^k gives rise to the Centrifugal force.

Using time as a coordinate unifies these three types of "fictitious" forces, and all three can be seen to be manifestations of the connection coefficients \Gamma^\mu_{\nu \lambda}

Well, you can have a connection without a metric, not something I'm very used to. I maintain that a 4-metric for Newtonian physics is not logically possible. The geometric structure has completely independent spatial invariants; and physical clock time is taken to be invariant.

 

[stevendaryl]

But I've never seen velocity, momentum, or force treated as anything but 3-vectors in Newtonian mechanics;

That's what Newton-Cartan theory does. Treating velocity as a 4-vector is very nice, as I said in another post, because it unifies three types of "fictitious forces" that appear in Newtonian physics when you use accelerated or curvilinear coordinates: "g" forces, Centrifugal forces, and Coriolis forces. All three can be seen as manifestations of connection coefficients.

But I'm not sure how non-gravitational forces fit in.

and there is no 4-dim metric at all (and can't be).

Yes, that was my point. So I wonder what becomes of the spatial metric in Newton-Cartan theory. It can't be a metric, so what is it? How are non-gravitational forces and momentum treated in Newton-Cartan theory, I wonder.

However, the despite that, the general issue remains of a good set of arguments for velocity as a vector, momentum and force as covectors. Cartesian coordinates in flat space it doesn't matter, so is generally ignored. But in polar coordinates, it would matter even in Newtonian mechanics.

 

[stevendaryl]

Well, you can have a connection without a metric, not something I'm very used to. I maintain that a 4-metric for Newtonian physics is not logically possible. The geometric structure has completely independent spatial invariants; and physical clock time is taken to be invariant.

A connection is more general than a metric. You're right, there is no 4-metric in Newtonian physics. That's what I said in my first post.

 

[robphy]

One could say that the analogue of the Lorentz-signature 4-metric in Newtonian physics is degenerate (non-invertible) with signature (+000) ... applicable for timelike components. So, one needs a second (also-degenerate) contravariant (indices-up) metric with signature (0---) or (0+++) for the spacelike components, as well as specification of a connection that is compatible with this pair of degenerate 4-metrics.Check out Malament's formulation
www.socsci.uci.edu/~dmalamen/bio/papers/GRSurvey.pdf#page=37

 

[Dale]

In Newtonian physics, time is both a coordinate and a parameter. It's a coordinate, because just as in SR, it takes 4 numbers to locate something in Galilean spacetime.

That certainly doesn't obviate the need for a metric. Whether you are doing a typical space+time or a more exotic Galilean spacetime, either way you still need a metric.

In any case, I am pretty sure that the topic of the thread is 3D vectors and covectors with their corresponding 3D metric tensors.

 

[WannabeNewton]

But I've never seen velocity, momentum, or force treated as anything but 3-vectors in Newtonian mechanics; and there is no 4-dim metric at all (and can't be). However, the despite that, the general issue remains of a good set of arguments for velocity as a vector, momentum and force as covectors.

It is done implicitly because when you have a line integral like this W = \int_{\gamma }F_{i}dx^{i} you must integrate with a one - form otherwise the line integral won't make sense. On top of that if we have that dF = 0 i.e. \partial _{i}F_{j} = \partial _{j}F_{i} we conclude as always that F = d\varphi i.e. F_{i} = \partial _{i}\varphi for some smooth scalar field (of course in general this relationship between closed and exact one - forms is local but this is physics =D) so it makes sense to talk about force as being a co - vector naturally. As for momentum, outside of the symplectic geometry formulation of classical mechanics I cannot see immediate why it would naturally be associated with a co - vector in say GR.

 

[bcrowell]

That certainly doesn't obviate the need for a metric. Whether you are doing a typical space+time or a more exotic Galilean spacetime, either way you still need a metric.

In any case, I am pretty sure that the topic of the thread is 3D vectors and covectors with their corresponding 3D metric tensors.

I actually think the interface between relativistic and nonrelativistic physics is an interesting way of trying to get to the bottom of these issues. But reasoning from relativity to the nonrelativistic limit can be surprisingly subtle. For instance, you can get weird stuff like "Carroll kinematics" http://arxiv.org/abs/1112.1466 , and the fact that there are two different classical limits of electromagnetism http://arxiv.org/abs/physics/0512200 . I'm used to thinking about tensors in the GR context where the theory has an extremely high degree of symmetry (between time and space, and with respect to any change of coordinates), so it's hard to adjust to the Galilean situation, which lacks these symmetries (or lacks them unless you go to some formalism like Newton-Cartan, which I'm not familiar with).

Some thoughts that may or may not be helpful:

In 1+1-dimensional SR, with the usual (t,x) coordinates, expressed in units with c\ne1, we have, up to an arbitrary constant, g_{\mu\nu}=\operatorname{diag}(1,-c^{-2}) and g^{\mu\nu}=\operatorname{diag}(1,-c^{2}). In the limit c\rightarrow\infty, we get a degenerate lower-index metric g_{\mu\nu}=\operatorname{diag}(1,0), and since that's not invertible there's nothing to stop us from saying that the upper-index version is not g^{\mu\nu}=\operatorname{diag}(1,-\infty) but g^{\mu\nu}=\operatorname{diag}(0,-1). So the lower- and upper-index metrics represent completely separate timelike and spacelike metrics. The result of measuring a vector with the lower-index metric is invariant under galilean transformations, but the result from the upper-index one isn't invariant unless the timelike part is zero. Some types of vectors, such as displacements along the world-line of an observer, are always timelike and therefore never have a frame-independent "upper-index magnitude" (i.e., spatial length). Raising and lowering of indices are in general noninvertible operations.

This setup has a lot of objectionable features, such as the degeneracy of the metric, but if you are willing to take it seriously, then it definitely doesn't make sense to say that coordinate displacements are naturally upper-index vectors. Time and mass-energy want to be upper-index vectors (so that you can measure them), while distance and momentum want to be lower-index vectors.

 

[robphy]

I actually think the interface between relativistic and nonrelativistic physics is an interesting way of trying to get to the bottom of these issues. But reasoning from relativity to the nonrelativistic limit can be surprisingly subtle.

Yes, this is the approach I use.
With a loss of symmetry of some kind, see what you can still accomplish.
It can then be argued that those accomplishments are in some sense more fundamental when formulated in a way that does not use that symmetry.

Yes, one has to take care with how one takes a non-relativistic limit.
As you probably know, it's not sending every "c" to infinity.

[slightly off topic] Here is an interesting paper
http://dx.doi.org/10.1119/1.12239
"If Maxwell had worked between Ampère and Faraday: An historical fable with a pedagogical moral" by Jammer and Stachel (1980)
...which discusses how you can re-order the history of EM to see a Galilean-invariant theory along the way... then be forced into a Lorentz-invariant one. The idea is based on the paper by Le Bellac and Levy-Leblond.

 

[robphy]

As for momentum, outside of the symplectic geometry formulation of classical mechanics I cannot see immediate why it would naturally be associated with a co - vector in say GR.

Borrowing ideas from quantum mechanics,
maybe we should think of momentum
like the wave-[co]vector k_a=\nabla_a \phi, which is the gradient of a phase [i.e. the exterior derivative of a scalar field].

Here's another possible way to think about things.
(I don't have an answer to this way of thinking about it... I'm just thinking out loud.)

It seems natural to think of an observer as t^a, a tangent vector to a future-timelike curve.
Given a particle somehow described by its 4-momentum, what is the energy of that particle according to our observer?
Is it t^a p_a or t^a g_{ab} p^b?
From a physical / theory-of-measurement point of view,
do we need to use a metric to determine the energy of a particle
(apart from possibly normalizing the observer's tangent vector t^a)?Finally... here's is an ancient thread on this question of the 4-momentum as a covector
https://www.physicsforums.com/showthread.php?t=135193
which didn't resolve the issue.

 

[WannabeNewton]

Thanks for the link robphy I'll read that and keep thinking aloud, it is interesting to hear your thoughts on this =D.

 

[bcrowell]

Finally... here's is an ancient thread on this question of the 4-momentum as a covector
https://www.physicsforums.com/showthread.php?t=135193
which didn't resolve the issue.

This usenet post by John Baez might be enlightening
http://www.lepp.cornell.edu/spr/2003-02/msg0048957.html

another set of usenet posts
http://www.lepp.cornell.edu/spr/2002-01/msg0038319.html
http://www.lepp.cornell.edu/spr/2002-01/msg0038342.html

Alas, poor Usenet. I knew him, Robphy.

However, the canonical momentum associated with v^j is

p_j = \frac{\partial L}{\partial v^j}

where L is the Lagrangian and where p_j carries a lower index because the right hand side transforms covariantly. Hence p_j are the covariant components of a covector that lives in cotangent space.

[My interpretation of Schreiber's asciified math.]

Ahhh...now things make a little more sense. So now we get the same answer about the force vector by two different methods.

Method #1: Work is a scalar, and displacement is an upper-index vector, so if we want W=F_i \Delta x^i, we have to make force a lower-index vector.

Method #2: Momentum is a lower-index vector, so if we want F_i=dp_i/dt, we have to make force a lower-index vector.

 

[dx]

In any case, I am pretty sure that the topic of the thread is 3D vectors and covectors with their corresponding 3D metric tensors.

Generalized force is a 1-form on an n-dimensional configuration space, which does not have a metric tensor.

 

[WannabeNewton]

Generalized force is a 1-form on an n-dimensional configuration space, which does not have a metric tensor.

Well technically the configuration space is a smooth manifold and you can always endow a smooth manifold with a riemannian metric but how useful it will be in the context of hamiltonian mechanics is a different story.

 

[dx]

Well technically the configuration space is a smooth manifold and you can always endow a smooth manifold with a riemannian metric but how useful it will be in the context of hamiltonian mechanics is a different story.

You could define such a tensor on the configuration space, but it would not represent anything physical, and is not technically a metric in the sense of physics because there is no notion of 'distance' between points of a configuration space.

 

[stevendaryl]

One could say that the analogue of the Lorentz-signature 4-metric in Newtonian physics is degenerate (non-invertible) with signature (+000) ... applicable for timelike components. So, one needs a second (also-degenerate) contravariant (indices-up) metric with signature (0---) or (0+++) for the spacelike components, as well as specification of a connection that is compatible with this pair of degenerate 4-metrics.


Check out Malament's formulation
www.socsci.uci.edu/~dmalamen/bio/papers/GRSurvey.pdf#page=37

That's a very interesting way of looking at it (although it seems a little long-winded--to end up with something as simple as F = ma, you have to go a huge distance with covariant and contravariant vectors, partial metrics, etc.)

But the upshot of it is this: In 4-D Galilean spacetime, you can, just as with SR, distinguish between "timelike" and "spacelike" vectors. IF a vector is spacelike, then it has a corresponding co-vector. So although acceleration is a vector, it is a spacelike vector, which means that it is associated with a covector, and that covector can be related to the force.

 

[Dale]

This setup has a lot of objectionable features, such as the degeneracy of the metric

That is my primary reason for not liking it. Because the metric expressed this way is actually a pair of degenerate metrics, to me it makes no sense to try to make a single Galilean spacetime. It seems much more natural to stick with a 3D space with time as a parameter and have a single non-degenerate metric.

The whole point of using a mathematical structure to model physics is because there is some close correspondence between how the physics behaves and the features of the mathematical structure. To me, the degeneracy indicates that the correspondence is not very close, so that choice of mathematical structure is not in close correspondence with the physics.

However, I have to admit that this did dissuade me from looking into it very deeply.

 

[robphy]

While it may have objectionable features,
I am actively trying to use it (Galilean spacetime [the simpler ideas, not the full machinery]) as a bridge from intro-physics to special-relativity (and general-relativity)
... in fact, a bridge from Euclidean Geometry to Special Relativity.

The non-euclidean geometry that underlies the "position-vs-time graph" in PHY 101 is a Galilean spacetime geometry... one of the Cayley-Klein geometries. (Draw a triangle of timelike lines in a position vs time graph... assign edge lengths according to the elapsed time (according to Galilean physics) for each worldline.)

By expressing intro-physics (at least kinematics and dynamics) in that context (thinking in spacetime concepts), one gets an earlier glimpse of special-relativity... making the transition to special relativity not-so-traumatic.

In addition, it is useful in clarifying the interpretation of features in special relativity and its non-relativistic limits.

 

[stevendaryl]

Because the metric expressed this way is actually a pair of degenerate metrics, to me it makes no sense to try to make a single Galilean spacetime.

The mathematics of manifolds is more general than the mathematics of manifolds with metrics. Some concepts about space require metrics, but others do not. It's actually informative to see where metrics are required, and where they are not.

Naively, a metric is required whenever you need a "dot" product of two vectors. However, in many cases (most?) you don't really have two vectors, but you have a vector and a covector, which can be combined to form a scalar without the need for a metric.

For example, if you have a parametrized path X^\mu(t) and you have a scalar function \Phi(X^\mu), then how do you compute the rate at which \Phi changes along the path?

\dfrac{D\Phi}{dt}

Well, it's the dot-product of the velocity vector with the gradient vector, but that doesn't mean that you need a metric, because the gradient is a co-vector:

(\nabla \Phi)_\mu = \dfrac{\partial}{\partial x^\mu} \Phi

while velocity is a vector:

V^\mu = \dfrac{d}{dt} X^\mu

 

[bcrowell]

I'm enjoying the stuff about the Galilean limit of relativity, but does anyone see a problem with #37 as a resolution of the original question?

 

[stevendaryl]

I'm enjoying the stuff about the Galilean limit of relativity, but does anyone see a problem with #37 as a resolution of the original question?

I think they're right, in a sense. Momentum is naturally a co-vector because, if you have a Lagrangian, then

P_\mu = \dfrac{\partial L}{\partial U^\mu}

and force is naturally a co-vector because:

F_\mu = \dfrac{\partial L}{\partial x^\mu}

On the other hand, it's a little circular, because in order to even have a Lagrangian, you have to be able to form a scalar L from the dynamic entities like velocities, which are vectors. So you need some co-vectors to start with, or you need something like a metric tensor, before you can get a Lagrangian.

 

[robphy]

I'm enjoying the stuff about the Galilean limit of relativity, but does anyone see a problem with #37 as a resolution of the original question?

I think #37 is enough motivation.
What I would really like to see is a spacetime-formulation of Hamiltonian(Symplectic) Mechanics where the momentum as a covector in phase space leads to it being a covector in spacetime (Galilean or Lorentzian-signature)... then to the observer's [Euclidean] "space".

By the way, the notion of force as a covector/one-form works fine when it is derivable from a potential energy... however, for general forces (like friction), it's not so clear.

I think they're right, in a sense. Momentum is naturally a co-vector because, if you have a Lagrangian, then

P_\mu = \dfrac{\partial L}{\partial U^\mu}

and force is naturally a co-vector because:

F_\mu = \dfrac{\partial L}{\partial x^\mu}

On the other hand, it's a little circular, because in order to even have a Lagrangian, you have to be able to form a scalar L from the dynamic entities like velocities, which are vectors. So you need some co-vectors to start with, or you need something like a metric tensor, before you can get a Lagrangian.

In Mackey's Quantum Mechanics book,
he uses the kinetic energy to define a metric tensor on configuration space
http://books.google.com/books?id=qlpb2mWYmfYC&pg=PA102&dq=mackey+metric+"configuration+space"

 

[stevendaryl]

The whole point of using a mathematical structure to model physics is because there is some close correspondence between how the physics behaves and the features of the mathematical structure. To me, the degeneracy indicates that the correspondence is not very close, so that choice of mathematical structure is not in close correspondence with the physics.

Certain things are modeled very well by using a generally covariant theory on 4D spacetime. As I said in another post, it's nice that all the fictitious forces--"g" forces, Coriolis forces, Centrifugal forces--are seen to be aspects of the same thing, the connection coefficients. It's also nice how Newtonian gravity can be seen as just a dynamic modification of those connection coefficients. It all works pretty smoothly.

But the equations that are the Newtonian counterpart to the Einstein field equations, which describes how curvature works, are very messy and ad-hoc looking. I don't see that this implies that the mathematics of smooth manifolds is not a close match to Newtonian physics. What it implies is that, from a generally covariant perspective, Newtonian physics is an ugly theory. If someone had gone through the trouble of formulating a generally covariant version of Newtonian physics, they might have been led to something like GR for purely aesthetic reasons.

 

[bcrowell]

On the other hand, it's a little circular, because in order to even have a Lagrangian, you have to be able to form a scalar L from the dynamic entities like velocities, which are vectors. So you need some co-vectors to start with, or you need something like a metric tensor, before you can get a Lagrangian.

I don't think that indicates circularity. It just indicates that you need a foundation (including a metric) and at least one arbitrary choice that breaks the symmetry between objects and their duals (defining ruler measurements to be upper-index vectors).

By the way, the notion of force as a covector/one-form works fine when it is derivable from a potential energy... however, for general forces (like friction), it's not so clear.

Good point, but we clearly don't want some forces to be one type of mathematical object and other forces to be a different type.

 

[atyy]

Maybe one way of interpreting Burke's contention is that he is defining force by an the work done integral. Forms are primarily designed to be integrated. So whenever the primary definition is an integral, then the quantities are forms. There's a similar point of view which says that classically, one might suppose the Lagrangian viewpoint to be primary, so integration is primary, so the basic things in classical physics are forms, eg. the last paragraph of http://sophia.dtp.fmph.uniba.sk/~fecko/referaty/regensburg.pdf

Edit: seems similar to micromass's #11.