Newtonian force as a covariant or contravariant quantityby bcrowell Tags: contravariant, covariant, force, newtonian, quantity 

#91
Jan3013, 11:58 AM

P: 2,892

So it is clear that some extra structure is needed, it just seems natural to me that this connection needed to parallel transport is the canonical Riemannian connection in the vast majority of classical physical problems, be it in configuration space (configuration manifold), Euclidean space, Lorentzian manifolds, etc. My point was that I don't understand Burke's prejudice against metric tensors and what is it "not geometric" about them that might bother anyone. I agree with him that it should be more enphasized when the Euclidean metric is being implicitly used and when it is not. 



#92
Jan3013, 12:26 PM

C. Spirit
Sci Advisor
Thanks
P: 4,924





#93
Jan3013, 12:43 PM

P: 2,892





#94
Jan3013, 12:49 PM

P: 1,657

Anyway, I was aiming my comments at the level of someone who is familiar with vectors in the context of Euclidean space, but doesn't realize the full implications of lacking a metric. I think going into the full complexities of differential geometry is the wrong level for the discussion. 



#95
Jan3013, 12:53 PM

P: 2,892





#96
Jan3013, 12:58 PM

P: 2,892





#97
Jan3013, 01:11 PM

P: 1,657





#98
Jan3013, 01:31 PM

Emeritus
Sci Advisor
PF Gold
P: 5,500





#99
Jan3013, 02:00 PM

Mentor
P: 16,582





#100
Jan3013, 05:42 PM

Emeritus
Sci Advisor
PF Gold
P: 5,500

I actually made two examples. Example #1 is a billiard ball of unit mass in two dimensions, constrained by a diagonal wall to have y<x. The Lagrangian formalism just leads to the expected Newtonian expressions, [itex]p_x=\dot{x}[/itex], [itex]p_y=\dot{y}[/itex]. The force of constraint is [itex]F_a=dp_a/dt[/itex]. Let [itex]w^a[/itex] be a vector parallel to the wall. Since we do happen to have a metric in this example,we can say that [itex]F_aw^a=0[/itex]; most people would say that the force was perpendicular to the wall. Example #2 is a human arm with a heavy mass gripped in the hand. The upper arm is raised at an angle [itex]\theta[/itex] and the lower arm raised at an angle [itex]\phi[/itex] (both measured relative to the vertical). The arm's weight is negligible compared to the unit mass of the gripped weight, and both the upper and lower arm have unit length. Because of the construction of the elbow joint, we have a constraint [itex]\theta \le \phi[/itex]. The conjugate momenta (which are actually angular momenta) turn out to be [itex]p_\theta=\dot{\theta}+\cos(\phi\theta)\dot{\phi}[/itex] and a similar expression for [itex]p_\phi[/itex]. The force of constraint is [itex]F_a=dp_a/dt[/itex]. The surface of constraint can be represented by a vector [itex]w^a[/itex], which is on a diagonal line in the [itex](\phi,\theta)[/itex] plane. Since there is no metric, it doesn't make sense to say that [itex]F_a[/itex] is perpendicular to [itex]w^a[/itex]. Geometrically, Burke has a nice representation of a 1form as a pair of parallel lines, with one of the two lines marked with an arrowhead (see http://www.scribd.com/doc/37538938/BurkeDivGradCurl ). In this representation, the parallel lines representing the force of constraint are clearly parallel to the surface of constraint. But I'm not clear on how to notate this idea that in both examples, the force 1form is parallel to the surface. Do you represent the surface as, say, a 1form created by taking the (infinite) gradient of a step function across the wall? Apart from my notational confusion, does the rest of this seem right? 



#101
Jan3013, 06:04 PM

Sci Advisor
P: 8,005

Wouldn't one use a metric in writing cosine of an angle?




#102
Jan3013, 07:01 PM

Emeritus
Sci Advisor
PF Gold
P: 5,500





#103
Jan3013, 07:23 PM

Sci Advisor
PF Gold
P: 1,806

While searching for information relative to this thread, I stumbled across this on Google Books (Richard A Mould, Basic Relativity, p.258). It claims that, under some circumstances, "covariant energy" (time component of covariant 4momentum) is globally conserved, but "contravariant energy" (time component of contravariant 4momentum) is conserved only locally but not globally. Unfortunately the next page is missing from the preview, so it's not clear what the "some circumstances" actually are, or how you prove the claim.
Does anyone know what is the correct statement and its proof? Well, in the case of Rindler coordinates [tex] ds^2 = g^2 z^2 \, dt^2  dx^2  dy^2 dz^2 [/tex](c=1) the contravariant 4momentum of a particle at rest is [tex] P^\alpha = \begin{bmatrix} \frac{m}{gz} \\ 0 \\ 0 \\ 0 \end{bmatrix} [/tex]whereas the covariant 4momentum is [tex] P_\alpha = \begin{bmatrix} mgz && 0 && 0 && 0 \end{bmatrix} [/tex]Of course [itex]mgz[/itex] is the correct formula for "gravitational" potential energy in Rindler coordinates, whereas [itex]m/(gz)[/itex] has no significance that I know of. Another nonrelativistic example I can think of is cylindrical polar coordinates in Euclidean 3space [tex] ds^2 = dr^2 + r^2 \, d\theta^2 + dz^2 [/tex]For an arbitrary particle [tex] x^i = \left( r(t), \theta(t), z(t) \right) [/tex]we have a contravariant 3momentum [tex] p^i = \begin{bmatrix} m \dot{r} \\ m \dot{\theta} \\ m \dot{z} \end{bmatrix} [/tex]whereas the covariant 3momentum is [tex] p_i = \begin{bmatrix} m \dot{r} && m r^2 \dot{\theta} && m \dot{z} \end{bmatrix} [/tex]Here again we see that the covariant component [itex]m r^2 \dot{\theta}[/itex] is the conserved angular momentum, whereas the contravariant component [itex]m \dot{\theta}[/itex] is not conserved. So all of the above seems to be more evidence to suggest that momentum and its timederivative, force, are naturally covariant rather than contravariant. 



#104
Jan3013, 07:37 PM

C. Spirit
Sci Advisor
Thanks
P: 4,924

Is the statement that the 0 component of contravariant 4  momentum is NEVER globally conserved and can only ever be locally conserved whereas the 0 component of covariant 4  momentum CAN be globally conserved under the appropriate conditions?




#105
Jan3013, 09:44 PM

Emeritus
Sci Advisor
PF Gold
P: 5,500

Here's another thing I have to try to wrap my head around: I think Killing vectors are naturally lowerindex vectors, and even though you do have a metric, it doesn't make sense to try to raise their indices. The reason is that the Killing vector is a field, not just a vector defined at a point, and the metric is varying...Does that make sense!? 



#106
Jan3013, 11:54 PM

Sci Advisor
P: 8,005





#107
Jan3113, 12:02 AM

C. Spirit
Sci Advisor
Thanks
P: 4,924





#108
Jan3113, 06:38 AM

P: 1,657

[itex]g(\nabla_Y (X),Z) + g(Y , \nabla_Z (X)) = 0[/itex] which is an equation on vector fields [itex]X^\mu[/itex]. However, it also says that in "local coordinates", this is equivalent to [itex]\nabla_\mu X_\nu + \nabla_\nu X_\mu = 0[/itex] which is an equation on covector fields [itex]X_\mu[/itex]. I don't quite understand this, because it seems that the latter equation doesn't even involve the metric (except indirectly, through the covariant derivative). It would seem to me that the latter could be defined even for a manifold without a metric, with just a connection, as a definition of a Killing covector field [itex]X_\mu[/itex]. 


Register to reply 
Related Discussions  
When is something covariant or contravariant?  Special & General Relativity  7  
Covariant vs contravariant  Differential Geometry  7  
covariant vs contravariant  Differential Geometry  62  
covariant vs. contravariant  Special & General Relativity  6  
covariant and contravariant ?  General Physics  15 