Newtonian force as a covariant or contravariant quantity

[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.

 

[WannabeNewton]

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 is just $\mathbb{R}^{4}$ with the standard topology and with a preferred set of coordinates (because of the preferred frames). The standard euclidean metric tensor is valid. If you are describing the same set but with some other topology then you have to show that it is a topological manifold with respect to that topology before even proceeding.

 

[robphy]

By "metric", I think stevendaryl is referring to the physically-motivated metric that will model the situation.

 

[TrickyDicky]

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.

Might be a manifold but I was referring (I added the qualification in my post now to make it more clear) to the word "spacetime" as defined in Wikipedia: "For physical reasons, a spacetime continuum is mathematically defined as a four-dimensional, smooth, connected Lorentzian manifold (M,g)." Which seems to imply it needs a Lorentzian metric to qualify as a Spacetime.
In any case my point wasn't exactly about terminology, see my answer to your other post below.

 

[TrickyDicky]

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.

The fact is what you say is true for general manifolds. But I think in Euclidean or in general flat manifolds there exists a canonical identification between vectors and covectors even in the absence of a metric, just as long as one uses a canonical basis, by virtue of the equivalence of their tangent and cotangent spaces with the manifolds itself:the galilean spacetime manifold is a vector space R^4 even in the absence of a metric tensor, right?

 

[TrickyDicky]

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

Well by that formula any parametrized object is a manifold with n+1 dimensions because you need one more number to specify where you are looking at. That IMO goes against the concept of manifold as an intrinsically defined object with no need to go to an ambient space.

 

[stevendaryl]

Well by that formula any parametrized object is a manifold with n+1 dimensions because you need one more number to specify where you are looking at.

Yes, any parametrized object, you can treat the parameter as another dimension. But there is a much more intimate connection between the dimensions in the case of time, which is that equations of motion explicitly relate what's happening on one time slice to what's happening on another time slice---that's what an "equation of motion" is.

That IMO goes against the concept of manifold as an intrinsically defined object with no need to go to an ambient space.

No, it doesn't. A manifold is any object that is locally like $R^n$.

 

[stevendaryl]

Yes, any parametrized object, you can treat the parameter as another dimension. But there is a much more intimate connection between the dimensions in the case of time, which is that equations of motion explicitly relate what's happening on one time slice to what's happening on another time slice---that's what an "equation of motion" is.

No, it doesn't. A manifold is any object that is locally like $R^n$.

I've been trying to think of some sense in which Galilean spacetime is less of a real manifold than Minkowsky spacetime, but I don't think that there really is one. If you're thinking that in Minkowsky spacetime, there is some kind of symmetry (Lorentz transforms) mixing space and time, I don't see why the Galilean transform doesn't count.

 

[TrickyDicky]

I've been trying to think of some sense in which Galilean spacetime is less of a real manifold than Minkowsky spacetime, but I don't think that there really is one. If you're thinking that in Minkowsky spacetime, there is some kind of symmetry (Lorentz transforms) mixing space and time, I don't see why the Galilean transform doesn't count.

But I'm not saying that Galilean "spacetime" is not a manifold, I was objecting to calling it spacetime if by spacetime we understand a manifold with Lorentzian metric.

 

[TrickyDicky]

Yes, any parametrized object, you can treat the parameter as another dimension. But there is a much more intimate connection between the dimensions in the case of time, which is that equations of motion explicitly relate what's happening on one time slice to what's happening on another time slice---that's what an "equation of motion" is.
No, it doesn't. A manifold is any object that is locally like $R^n$.

Sure, again, I'm not arguing here that the Galilean model is not a manifold, here I was talking about the parametrized 3-space versus the 4-manifold and which looks more natural. I guess both views are ok, I just find it more natural to use the former, which is the one we are used to from classical mechanics, the 4-manifold representation is more of a modern retelling in the light of what we know about relativity and Lorentzian manifolds.

 

[WannabeNewton]

No, it doesn't. A manifold is any object that is locally like $R^n$.

There are more conditions than just that for a topological space to be a manifold. Anyways it seems like it is just turning out to be an issue of semantics regarding whether galiliean space - time is a "space - time". From what I'm seeing it seems to be a commonly used terminology and I don't see any reason to object it regardless. Also one can take a look at this: http://ls.poly.edu/~jbain/spacetime/lectures/11.Spacetime.pdf