This question arose because of my first question Interpreting Vector fields as Derivations on Physics. The point here is: if some force $F$ is conservative, then there's some scalar field $U$ which is the potential so that we can write $F = - \nabla U$.

That's fine, it says that force is a covector, but the point is: when we start thinking about curved spaces, in general instead of talking about gradients and covectors we talk about exterior derivatives and one forms.

My question then is: if a force is conservative with potential $U$ then it's correct do represent the force by the one-form obtained by the exterior derivative of the potential, in other words the form $F = -dU$ ?

In second place, if the force isn't conservative, is it correct to think of it as a one form yet ? But now what's the interpretation ?

I tried to give this interpretation: suppose we're dealing with some manifold $M$ and suppose that $(V,x)$ is a coordinate chart. Then $\left\{dx^i \right\}$ spans the cotangent space, and so, if we interpret some force at the point $p$ as some one form $F \in T^\ast_pM$ then we'll have $F=F_idx^i$ using the summation convention.

Now if i take some vector $v \in T_pM$ we can compute $F(v) = F_idx^i(v)$, however, $dx^i(v)=v^i$ and hence $F(v)=F_iv^i$ and so my conclusion is: if I interpret force at a point as a one-form at the point, then it'll be a form that when given a vector, gives the work done moving a particle in the direction of the given vector.

So if a force varies from point to point, I could represent it as a one-form field that can be integrated along some path to find the total work done.

Can someone answers those points and tell me if my conclusion is correct?

  • $\begingroup$ For a non-relativistic treatment of a charged particle on a Riemannian 3-manifold, see. e.g. this Phys.SE post. $\endgroup$
    – Qmechanic
    Apr 7 '13 at 21:24

To understand what a Newtonian force field is, let's take a look at Newton's second law $$ F = ma $$ This translates to the following differential-geometric relation $$ (m\circ\dot q)^\cdot = F\circ\dot q $$ where $m:\mathrm{T}M\to \mathrm{T}^*M$ maps from velocity to momentum space and $q:I\subset\mathbb R\to M$ is the trajectory.

The force field ends up being a map $$ F:\mathrm{T}M\to \mathrm{T}\mathrm{T}^*M $$

Let $$ \pi:\mathrm{T}^*M\to M \\ \Pi:\mathrm{T}\mathrm{T}^*M\to \mathrm{T}^*M $$ be the bundle projections.

Then $$ m = \Pi\circ F \\ \mathrm{T}\pi\circ F = \mathrm{id}_{\mathrm{T}M} $$

The latter equation is the equivalent of the semi-spray condition and tells us that we're dealing with a second-order field.

Because the bundles $\mathrm{T}\mathrm{T}^*M$ and $\mathrm{T}^*\mathrm{T}M$ are naturally isomorphic - in coordinates, we just switch the components $(x,p;v,f)\mapsto(x,v;f,p)$ - we can represent it as a differential form on $\mathrm{T}M$, which is just the differential $\mathrm dL$ of the Lagrange function (the Euler-Lagrange equation are Newtonian equations of motion).

Now, the space of Newtonian force fields doesn't come with a natural vector space structure, but rather an affine structure. You need to specify a zero force - a force of inertia - to make it into one. Such a force can for example be given by the geodesic spray of general relativity.

Once that's done, you can represent the force field as a section of the pullback bundle $\tau^*(\mathrm{T}^*M)$ where $\tau:\mathrm{T}M\to M$. This is a velocity-dependant covector field, which you can indeed integrate over or derive from a potential function (in case of velocity independence).

Now, for those who are uncomfortable with this level of abstraction, let's try a more hands-on approach:

Geometrically, the acceleration is given by $(x,v;v,a)\in\mathrm{TT}M$. However, that space has the wrong structure - if we add two accelerations acting on the same particle, we end up with $(x,v;2v,a+a')$, which is no longer a valid acceleration.

What we want instead are vectors $(x;a)\in\mathrm{T}M$ or $(x,v;a)\in\tau^*(\mathrm{T}M)$ in case of velocity-dependent accelerations, and a recipe how to get from these to our original acceleration as that's what occurs in our equation of motion.

So let's assume our acceleration is velocity-independent and represented by $(x;a)\in\mathrm{T}M$. By lifting the vector vertically at $(x;v)\in\mathrm{T}M$, we arrive at $(x,v;0,a)\in\mathrm{TT}M$. What's 'missing' is the horizontal component $(x,v;v,0)\in\mathrm{TT}M$.

Even though such a horizontal lift looks trivial in coordinates, it is not a 'natural' operation in differential geometry. You can fix this in two obvious ways by either providing a connection (it's trivial to see how this works out if you take the geometric approach due to Ehresmann) or by manually specifying the 'zero' acceleration due to inertia.

The question that's left to answer is why we're using forces instead of accelerations, or formulated another way, why do we move to the cotangent space?

From the point of view of differential geometry, one answer to that question is because we want to work with potentials, which are less complicated objects, and the differential yields covectors instead of vectors.

Another point of consideration is that $\mathrm{TT^*}M$, $\mathrm{T^*T}M$ and $\mathrm{T^*T^*}M$ are naturally isomorphic, whereas $\mathrm{TT}M$ is not. These isomorphisms lead to several (more or less) equivalent formulations of analytical mechanics, including the Newtonian, Lagrangian and Hamiltonian approach.

Apologies for expanding the scope of the question - feel free to ignore these ramblings ;)

  • $\begingroup$ Wow, very in-depth. Didn't really consider this generality. Are there any applications of this? $\endgroup$
    – user21299
    Feb 24 '13 at 14:20
  • 1
    $\begingroup$ @Christoph, very good answer indeed, it's the kind of approach to physics that I like. Can you recommend me some book that covers more of those topics ? I mean, studying physics using rigorous math ? Thanks for your answer again. $\endgroup$
    – Gold
    Feb 24 '13 at 16:02
  • $\begingroup$ I like the approach you are taking but things really seem strange to me. Shouldn't the force naturally be a covector field $F:M\rightarrow T^*M$ and Newton's 2nd Law is $g(F,\cdot)=ma$ or something? Is is not natural to find a spray in this manner? $\endgroup$
    – levitopher
    Feb 25 '13 at 1:51
  • $\begingroup$ @alexarvanitakis: I'm not aware of applications; the 'interesting stuff' is normally done using the Hamiltonian or Lagrangian approach, which come with well-developed generalizations $\endgroup$
    – Christoph
    Feb 25 '13 at 7:36
  • 1
    $\begingroup$ @cduston: let's forget about dual spaces for now; the second-order equation $\ddot q = Y\circ\dot q$ tells us that the acceleration field $Y$ needs to be a semi-spray; the problem is that there is no natural way to get such a second-order vector field from a first-order vector field without additional structure (eg a connection) $\endgroup$
    – Christoph
    Feb 25 '13 at 7:44

There was a rather lengthy discussion about whether force is naturally a vector or a covector over at physicsforums: http://www.physicsforums.com/showthread.php?t=666861 .

If you define momentum as "that which is conjugate to position," then momentum is a covector. I.e. if you have a Lagrangian, then:

$$p_\mu =\frac{\partial L}{\partial \dot{x}^\mu}$$

Force can then be interpreted as $d p_\mu / d \tau$. Or, you can define force directly from the Lagrangian as:

$$F_\mu=\frac{\partial L}{\partial x^{\mu}}$$

Combined with the argument about work that you provided, where $W=\int F_{i} dx^{i}$, it seems very compelling that force should naturally be interpreted as a covector.

  • $\begingroup$ It it necessary to denote spacetime point by $x^\mu$? If I use $x_\mu$ to denote spacetime point, then both the momentum and force would be a vector rather than a covector. The only thing is that it can be shown that if $x^\mu$ is a vector $\frac{\partial}{\partial x^\mu}$ of some scalar function is a covector. @Jold $\endgroup$
    – SRS
    May 26 '17 at 6:38
  • $\begingroup$ @SRS $x_\mu$ doesn't have any meaning in the absence of a metric, which is why $x^\mu$ is the more "natural" representation. $\endgroup$
    – Jold
    Jun 12 '18 at 4:12

Everything you said is correct. If a force is not conservative then it still makes sense as an 1-form, albeit one that is not exact.

Note also that the condition $\vec{\nabla} \times \vec{F} =0$ for a force to be locally determined by a potential can be written as $d F=0$ so that $F=-d U$ for some function $U$ by the Poincare lemma.

More generally we have p-form potentials $A$ to which we associate p+1-form field strengths $dF$. E.g in electromagnetism (again!) we can combine the vector and scalar potentials into a 1-form on spacetime (3+1=4) and the resulting field strength tensor is this one.


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