Newtonian force as a covariant or contravariant quantity

[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.

 

[robphy]

Yes, so with this perspective... regarding energy [and action] as primary

it is probably best to think of
force in units of "Joules per meter" (rather than Newtons or kg*m/s^2)
and
momentum in units of "action per meter" (rather than kg*m/s)

(Electric field in units of Volts/meter suggests that it is a covector.)
("g" "acceleration due to gravity" [or better "gravitational field" \vec g] should probably be "(Joules/kg) per meter"... is there a "gravitational volt"?)

 

[atyy]

With respect to the idea that integration is about forms, one of the things that I found confusing at first was the concept of integration being used. This article talks a bit about the various definitions, and the definition relevant to forms is the signed definite integral. http://www.math.ucla.edu/~tao/preprints/forms.pdf. Tao remarks "The concept of a closed form corresponds to that of a conservative force in physics (and an exact form corresponds to the concept of having a potential function)." And "Because Rn is a Euclidean vector space, it comes with a dot product ... which can be used to describe 1-forms in terms of vector fields (or equivalently, to identify cotangent vectors and tangent vectors) ... However, we shall avoid this notation because it gives the misleading impression that Euclidean structures such as the dot product are an essential aspect of the integration on differential forms concept, which can lead to confusion when one generalises this concept to more general manifolds on which the natural analogue of the dot product (namely, a Riemannian metric) might be unavailable."

 

[TrickyDicky]

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?

Well, it seems to me that calling either Newtonian or galilean models of space and time "spacetime" is not very rigorous and a bit of a stretch of the terminology if we look at the mathematical dfinitions, they are just collection of points rather than proper (pseudo)Riemannian manifolds like say Minkowski spacetime is. Sure nowaday everybody calls them spacetimes , but they simply look like parametrizations of either 3D Euclidean space (Newtonian "spacetime") or 3D affine space (galilean "spacetime").
Once one realizes this it seems straightforward (just by reading Tao's quote that atyy links above), that one can easily go from the covector to the vector form and viceversa, there'll be physical situations that mathematically naturally correspond to the vector or the covector form like have been commented for force, momentum, acceleration, etc, but it seems besides the point to try to decide whether a physical quantity is "really" a pure vector or a covector, it depends on how it mathematically acts or it's acted upon in the specific physical situation one uses it.

 

[TrickyDicky]

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.

I'd say you need 3 numbers for each parametrized hypersurface.

 

[robphy]

Well, it seems to me that calling either Newtonian or galilean models of space and time "spacetime" is not very rigorous and a bit of a stretch of the terminology if we look at the mathematical dfinitions, they are just collection of points rather than poper manifolds like say Minkowski spacetime is. Sure nowaday everybody calls them spacetimes , but they simply look like parametrizations of either 3D Euclidean space (Newtonian "spacetime") or 3D affine space (galilean "spacetime").

The Newtonian spacetime (as in the Malament article referenced http://www.socsci.uci.edu/~dmalamen/bio/papers/GRSurvey.pdf#page=37) certainly requires more work since there the metric tensor is degenerate (unlike in Minkowski spacetime and in GR-spacetimes). However, by providing additional structure [a connection] (which one gets for free with a nondegenerate metric), one is able to formulate Newtonian gravity in a 4-D spacetime setting.

Part of the philosophy behind such attempts is that
one is asking does one really need to regard the metric as fundamental? Or should it be the connection or its curvature tensor? Or the causal structure? ... Then, what else is needed to recover what you take for granted?
For example, there is the Ehlers-Pirani-Schild Construction
http://link.springer.com/article/10.1007/s10714-012-1352-5?LI=true
http://link.springer.com/article/10.1007/s10714-012-1353-4
which quotes an old theorem by Weyl "The projective and conformal structures of a metric space determine uniquely its metric."

A simpler case of the Galilean spacetime that I've been mentioning is actually well defined projectively as a Cayley-Klein Geometry, which includes Euclidean, Minkowski, and De-Sitter and Anti-Sitter spacetimes and their "Galilean limits", as well as the more-familiar non-euclidean hyperbolic and elliptical spaces. Yes, some stretching of definitions was needed to accept hyperbolic geometry as a legitimate geometry... and that's what projective geometry did. The Cayley-Klein formulation shows how similar those 9 geometries are to each other. In fact, one can relax definitions in order to describe all 9 in a unified way. That's what I am trying to do, with an eye to relativity.

Degenerate structures are actually not as bad as they sound.
From a projective-geometry viewpoint, our familiar Euclidean geometry has some degenerate structures. For example, there is no natural length scale in Euclidean geometry... one has to choose one... unlike in say elliptic/spherical geometry.

 

[bcrowell]

This article talks a bit about the various definitions, and the definition relevant to forms is the signed definite integral. http://www.math.ucla.edu/~tao/preprints/forms.pdf.

Thanks for the link to the Tao article --it's great! Unlike a lot of material on differential forms, it's not written in Martian.

 

[stevendaryl]

Well, it seems to me that calling either Newtonian or galilean models of space and time "spacetime" is not very rigorous and a bit of a stretch of the terminology if we look at the mathematical dfinitions, they are just collection of points rather than poper manifolds like say Minkowski spacetime is. Sure nowaday everybody calls them spacetimes , but they simply look like parametrizations of either 3D Euclidean space (Newtonian "spacetime") or 3D affine space (galilean "spacetime").

Galilean spacetime is absolutely a 4-dimensional manifold, so it isn't any kind of stretch to apply the term "spacetime". It's a manifold without a metric, but it's certainly a manifold.

 

[stevendaryl]

Once one realizes this it seems straightforward (just by reading Tao's quote that atyy links above), that one can easily go from the covector to the vector form and viceversa

But in the absence of a metric, you CAN'T go from one to the other. That's why to me 4-D Galilean spacetime is interesting, because it is an example of a manifold without a metric.

 

[robphy]

But in the absence of a metric, you CAN'T go from one to the other. That's why to me 4-D Galilean spacetime is interesting, because it is an example of a manifold without a metric.

It has a metric... but it's degenerate.
In terms of raising/lowering indices, you can go one way... but not the other.

 

[stevendaryl]

I'd say you need 3 numbers for each parametrized hypersurface.

And you need one number to say which hypersurface. That's 4.