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I'm reading An Introduction to Riemannian Geometry with applications to Mechanics and Relativity by Godinho and Natario, and they model mechanics as follows: a mechanical system is a triple $(M,\langle\cdot,\cdot\rangle, \mathscr{F})$, where:

  • $M$ is a smooth manifold (the configuration space);

  • $\langle\cdot,\cdot\rangle$ is a Riemannian metric on $M$, which defines a mass operator $\mu:TM\to T^*M$ by $\mu({\bf v})({\bf w})\doteq \langle {\bf v},{\bf w}\rangle$ and

  • the external force $\mathscr{F}$, which is a map $\mathscr{F}:TM\to T^*M$ satisfying $\mathscr{F}(T_pM)\subseteq T_p^*M$ for each $p\in M$.

Then they procceed to study motions $c:I\to M$ which are solutions to Newton's equation $$\mathscr{F}(c') = \mu\left(\frac{Dc'}{{\rm d}t}\right),$$among other things. I know that we can see $TM$ as the space of positions and velocities and $T^*M$ as the space of positions and momenta, and although I notice the striking resemblance of Newton's equation with his second law $F=ma$, I'm having trouble understanding

1) why the mass operator and the external force should be $T^*M$-valued;

2) the relation between $\mu$ and the musical isomorphism $\flat$ used to lower indexes using $\langle \cdot,\cdot\rangle$, because they're formally the same;

3) how to interprete the condition $\mathscr{F}(T_pM)\subseteq T_p^*M$.

I probably should point that I'm a graduate mathematics student trying to understand a bit of mechanics, so I'll likely understand more mathematical-oriented answers better.


It was difficult to choose an answer to accept. All the answers were great. I hope to understand physics well enough to contribute with the forum like this some day. Thanks to all!

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    $\begingroup$ are you sure inside $\mu$ its Dcprime/dt? $\endgroup$ – lalala Sep 23 '17 at 17:08
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    $\begingroup$ @lalala yes, I am sure. $\endgroup$ – Ivo Terek Sep 23 '17 at 17:10
  • $\begingroup$ So prime denotes the tangential map? $\endgroup$ – lalala Sep 23 '17 at 17:18
  • $\begingroup$ Yes. $c'(t) = {\rm d}c(t)(1)$ and $Dc'/{\rm d}t$ denotes the covariant derivative of $c'$ with respect to the Levi-Civita connection of the metric. $\endgroup$ – Ivo Terek Sep 23 '17 at 17:28
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Well,

1) why the mass operator and the external force should be $T^∗M$-valued;

For the mass operator, this is just the multiple ways you can represent tensor products. A Riemannian metric at point $x\in M$ is a bilinear map $T_xM\times T_xM\rightarrow\mathbb{R}$, so if you insert a vector into only one argument, it becomes a map $T_xM\rightarrow\mathbb{R}$, which is an element of $T^*_xM$. So it maps elements of $T_xM$ into elements of $T^*_xM$. When extended to all points, it becomes thus a vector bundle isomorphism $TM\rightarrow T^*M$ (this map respects the fibration in the sense that if $\pi_{TM}(v)=x$, then $\pi_{T^*M}(\mu(v))=x$ as well).

For the external force, this is a very clumsy formulation, imo. The force may depend on positions, velocities (and possibly on 'external' time, but I'm gonna ignore that), so it is a function $\mathcal F:TM\rightarrow X$ where $X$ is left undetermined for now. In a less general context, where the config. space is just $\mathbb{R}^3$, the force is understood to be a velocity-dependent vector field, $\mathbf{F}(x,v)$. It should be understood as a 1-form though because work is given by $W(c)=\int_cF=\int_c\sum_i F_idx^i$.

In the general context of configuration spaces, we rarely define forces though, since we prefer to work with Lagrangian or Hamiltonian formalism, where there are no forces. However, if we'd like to abstract forces, then it should be understood that $X$ must be a space that houses objects that can be integrated along curves (curves of $M$), so $\mathcal{F}:TM\rightarrow T^*M$, where $\mathcal F$ is one again a bundle morphism (but now no longer necessarily an isomorphism). However in this case $\mathcal{F}$ is not usually a fibrewise linear map, so it should not be understood as a type (0,2) tensor field $\mathcal F_{ij}$, but instead as a $T^*M$ valued function on $TM$, $\mathcal F(x,v)_i$ (as a mathematician, you might be annoyed by tensor index notation, but I believe here it offers clarity).

2) the relation between $\mu$ and the musical isomorphism $\flat$ used to lower indexes using $\langle\cdot,\cdot\rangle$, because they're formally the same;

They are the same.

3) how to interprete the condition $\mathcal{F}(T_xM)\subseteq T^∗_xM$

The thing to emphasize here is that the same $x\in M$ appears in both the initial space and the target space. This is to emphasize that $\mathcal F$ is a fibre bundle morphism, it preseves fibers, so $\mathcal F(x,v)$ takes values in $T^*_xM$ while $\mathcal{F}(y,u)$ takes its values in $T^*_yM$.

This is because 1) to calculate work along a path $c$ you take $W(c)=\int_c\mathcal{F}(c(t),c'(t))$, which kind of makes sense only when $\mathcal{F}(c(t),c'(t))\in T^*_{c(t)}M$, 2) to make sense of Newton's equation, because $\mu(Dc'/dt(t))$ is an element of $T^*_{c(t)}M$ since $\mu$ is a vector bundle morphism, so $\mathcal {F}(c(t),c'(t))$ should also be in $T^*_{c(t)}M$.

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  • $\begingroup$ It's beyond me how could I forget about the work integral. That makes a lot of sense, perfect! $\endgroup$ – Ivo Terek Sep 24 '17 at 19:13
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I'll answer these in an order that makes the exposition a little more concise.

2) The Isomorphism Induced by Riemannian Structure

You'll have to say more about the definitions of you were given to explain why you're confused, They're the same thing, and since I'll be using them throughout this answer I will write it out to fix notation:

\begin{equation} \mu: TM \to T^*M ,~~ v \mapsto \langle v, \cdot\rangle \end{equation}

3) The Bundle Morphism:

This is the statement that the following commutes $$ \newcommand{\ra}[1]{\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} TM & \ra{\mathcal F} & T^*M \\ \da{\pi_M} & & \da{\pi_{T^*M}}\\ M & \ra{Id} & M \\ \end{array} $$

There's a nicer way to say this. The category of smooth vector bundles, $\hat V$, admits a functor $Forget$ to the category $\hat{M}$ of smooth manifolds, the functor acts on objects by projecting to the base space, and on morphisms by composing with the projection maps. This functor $\hat{V} \xrightarrow{Forget} \hat {M}$ is a groupoid fibration $M$. The above then says $\mathcal F \in Hom_{\hat V}(TM, T^*M)$ is a morphism that lies over the identity in the groupoid fibration. $Forget^{-1}(Id_M)$ is said to be the fiber over $M$, and it's a groupoid. The statement is then simply $\mathcal F$ lives in the fiber over $M$ in this forget functor. (Exercise: all this is fairly obvious but if you're unfamiliar with these constructions it's a cute exercise to check it).

This geometrically says you have a smooth map that's "vertical" as in it acts as the identity when restricted to see the base space. The following section should answer why you want this property.

1) The Differential Operator

We take hints from the more physical answers in this thread, and note that in general you want to describe your system by some smooth function

\begin{equation} \mathcal L \in C^\infty(T^*M, \mathbb R) \end{equation}

Now the tangent space $T^*T^*M$ admits a natural splitting given by the exact sequence \begin{equation} 0 \to T^*M \xrightarrow{z} T^*T^*M \xrightarrow{p} V(T^*M) \to 0 \end{equation}

where $z$ is the 0-section and $p$ is given locally by projection to the fiber, the exactness defines the bundle $V$. There's a natural bundle isomomrphism

$$ \varphi: V(T^*M) \to T^*M $$

(Exercise: check this, e.g. by comparing tangent spaces at each point.)

Now your force is supposed to the section given by the composition of the following sequence

$$ T^*M \xrightarrow{d\mathcal L} T^*T^*(M) \xrightarrow{p} V(M) \xrightarrow{\varphi}T^*M $$

Geometrically it's the "vertical/fiber" component of the differential of $\mathcal L$, naturally viewed as a vector field in the cotangent bundle. Now the version you have is a version that acts on the tangent bundle, you can build it by pre-composing the natural isomorphism. $$ TM \xrightarrow{\mu} T^*M \xrightarrow{d\mathcal L} T^*T^*(M) \xrightarrow{p} V(M) \xrightarrow{\varphi}T^*M $$

You want this because in your Newton formula you want to take a vector field as an argument.

To summarize, your force is supposed to be given by the following, for some smooth function $\mathcal L$ \begin{equation} \mathcal F \equiv \varphi \circ p \circ d\mathcal L \circ \mu: TM \to T^*M \end{equation}

This answers the question 1). Now back to 3), simply observe we've projected ourselves to vertical sections when we applied the projection $p$.

Edit/Appendix: takeaways

As I was walking home from the pub where I wrote this I started wondering why (aside from Bourbon) I wrote this much about something that appears to be a minor notation/definition confusion. Here's what I think:

We're staring at a second order differential equation on a Riemannian manifold. A priori the differential operator that defines it should live on some jet. However we all know we can write it down as an equation of vector fields on the tangent bundle. Why?

From the discussion above, the crucial property that makes this possible is that the differential operator $\mathcal F$ gives a section everywhere vertical in the bundle. Similarly if the section was everywhere horizontal, the analysis is again easy because we again have an isomorphism from the horizontal subbundle (alternatively just observe the inclusion is just the 0-section so solving this differential equation should be easy). Anything deviating from these cases are obstructions to our ability to write the differential equation in the tangent bundle. This is the important takeaway here.

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    $\begingroup$ I sure didn't expect category theory to pop up here, but that did help, thanks! $\endgroup$ – Ivo Terek Sep 24 '17 at 19:12
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1) why the mass operator and the external force should be $T^*M$-valued;

Arguably, this has more to do with physics than geometry (contracting momentum with a direction yields the 'amount of motion' in that direction, the curve integral over the relativistic momentum covector yields the action, ...).

But note that geometrically, this allows us to derive forces from potentials (which opens up the pathway towards the Hamiltonian and Lagragian formulations).

2) the relation between $\mu$ and the musical isomorphism $\flat$ used to lower indexes using $\langle \cdot,\cdot\rangle$, because they're formally the same;

In this particular model, they are indeed the same, though that's not necessarily the case in general: If you understand the Euler-Lagrange equations as Newtonian, the mass operator would be given by the fibre derivative instead.

3) how to interprete the condition $\mathscr{F}(T_pM)\subseteq T_p^*M$.

That's just compatibility with the fibration, ie the force acts at the same base point.


As an aside, I'd also like to point out what we need the covariant derivative for: Without a connection, Newton's first law makes no sense (no notion of straight lines or constant velocities without additional structure).

Geometrically, if you're dealing with a second order system, the equation

$$ c'' = 0 $$

is incompatible with initial conditions $$ c' \not= 0 $$

due to the structure of ${\rm TT}M$, whereas $$ {{\rm D}c' \over {\rm d}t} = 0 $$

works fine.

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Ok, let me try a clumsy answer. I won't regard velocity depending forces like your text suggest.

There are basically two ways to formulate classical mechanics, Newtons equation and Lagrange formalism. The Lagrange formalism is suitable for extending to curved spaces, so here we go with flat space first:

Define $$ L(q,\dot{q})= \frac{m}{2}\dot{q}^2 - V(q)$$ with $m$ the mass of the particle, $q$ the position vector, $\dot{q}$ the velocity vector and $V$ the potential. It has been observed if you take a path $q$ with fixed beginning and end $q(t_1)=q_1$ and $q(t_2)=q_2$ and calculate $$S[q] = \int^{t_2}_{t_1} L(q(t),\dot{q}(t))\, dt$$ $S$ being a function mapping trajectories to numbers; that the real trajectory a particle takes is given by the trajectory at which $S$ attains an extremum. (with fixed beginning and end points) The corresponding equations can be obtained by a method called calculus of variations. This leads to the Euler-Lagrange equations which are: $$ \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} = \frac{\partial L}{\partial q_i}$$ Right side is called force and you can see it is actually an element of cotangential space (maps directions to numbers) which are of course Newton's equation: $$ \frac{d}{dt} (m\dot{q}_i) = - \frac{\partial V}{\partial q_i}$$ So far so good. Now how can one generalize the Lagrange function to a manifold, so that it scalar under coordinate transformation. For the potential $V$, no problem, for the kinetic term you need a scalar product, so you would end up with (here $(q, \dot{q})$ parametrize the tangetial space) $$ L(q, \dot{q})= \frac{m}{2} \langle\dot{q},\dot{q}\rangle -V(q)$$ Now you can choose to absorb the factor of $m$ also in the scalar product. This explains what $\mu$ has to do with mass.

Now doing the same kind of variation calculus, (and using $c$ for $q$), you should end up with something along the line of

$$\frac{d}{dt}{\langle\dot{c}, \epsilon\rangle} = - dV(c) \cdot \epsilon$$ for all variations $\epsilon$. Since it is for all $\epsilon$, and the scalar product is time independent, this is equivalent to $$\mu\left(\frac{d}{dt} \dot{c}\right) = F(c)$$ with $F(c):= -dV(c)$. This already looks very similar to what you have, except for $\dot{c}$ in the $F$ in your formula. I guess they are extending this to also velocity depending forces (since the tangential map contains the position, this is a generalization)

Yes it is a bit sketchy, but maybe you can see where they are coming from.

EDIT: This is the single particle case (as I am sure your question talks about), if you go multi-particle then $\langle\cdot,\cdot\rangle$ has to map from $TM^n \times TM^n$ to $R$ instead (also sometimes called the 'mass-matrix')

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  • $\begingroup$ Why the downvote? $\endgroup$ – lalala Sep 24 '17 at 7:44
  • $\begingroup$ Rest assured that I'd ask for clarification instead of hitting a downvote (perhaps the downvoter thought that the formatting could be improved?). Anyway, thanks! $\endgroup$ – Ivo Terek Sep 24 '17 at 19:15

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