- #36
WannabeNewton
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Thanks for the link robphy I'll read that and keep thinking aloud, it is interesting to hear your thoughts on this =D.
robphy said: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.
robphy said: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
Urs Shreiber said:However, the canonical momentum associated with [itex]v^j[/itex] is
[tex]p_j = \frac{\partial L}{\partial v^j}[/tex]
where L is the Lagrangian and where [itex]p_j[/itex] carries a lower index because the right hand side transforms covariantly. Hence [itex]p_j[/itex] are the covariant components of a covector that lives in cotangent space.
DaleSpam said:In any case, I am pretty sure that the topic of the thread is 3D vectors and covectors with their corresponding 3D metric tensors.
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 said:Generalized force is a 1-form on an n-dimensional configuration space, which does not have a metric tensor.
WannabeNewton said: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.
robphy said: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 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.bcrowell said:This setup has a lot of objectionable features, such as the degeneracy of the metric
DaleSpam said: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.
bcrowell said: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?
bcrowell said: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 said:I think they're right, in a sense. Momentum is naturally a co-vector because, if you have a Lagrangian, then
[itex]P_\mu = \dfrac{\partial L}{\partial U^\mu}[/itex]
and force is naturally a co-vector because:
[itex]F_\mu = \dfrac{\partial L}{\partial x^\mu}[/itex]
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 [itex]L[/itex] 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.
DaleSpam said: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.
stevendaryl said: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 [itex]L[/itex] 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 said: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.
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").stevendaryl said:This is a very interesting topic. In Newtonian physics, there are two different ways to define "force":
- [itex]F^i = m \dfrac{dU^i}{dt}[/itex], where [itex]U[/itex] is velocity/
- [itex]F_i = -\partial_j \Phi[/itex], where [itex]\Phi[/itex] 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 [itex]g_ij[/itex] you can certainly resolve the tension by writing:
[itex] m g_{ij} \dfrac{dU^i}{dt} = F_j[/itex]
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:
- The time between events.
- 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 said: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.
TrickyDicky said: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").
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.atyy said: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.
TrickyDicky said: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").
TrickyDicky said: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
stevendaryl said: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.
TrickyDicky said:I'd say you need 3 numbers for each parametrized hypersurface.
It is just [itex]\mathbb{R}^{4}[/itex] 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.stevendaryl said: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.
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.stevendaryl said: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 said: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.
stevendaryl said:And you need one number to say which hypersurface. That's 4.
TrickyDicky said: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 said: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 [itex]R^n[/itex].
stevendaryl said: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.
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.stevendaryl said: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 [itex]R^n[/itex].
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.pdfstevendaryl said:No, it doesn't. A manifold is any object that is locally like [itex]R^n[/itex].