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Representing forces as one-forms
 Asked 8 years, 11 months ago Active 3 years, 11 months ago 3k timesViewed
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This question arose because of my first question 
. The point here is: if some force is conservative, then there's some scalar field 
which is the potential so that we can write .
Interpreting Vector fields as Derivations on
Physics F U
F = −∇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 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 ?
U
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 and suppose
that is a coordinate chart. Then spans the cotangent space, and so, if we
interpret some force at the point as some one form then we'll have 
using the summation convention.
M
(V , x) {d }xi
p F ∈ MT ∗p F = dFi x
i
Now if i take some vector we can compute , however, 
and hence 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.
v ∈ MTp F (v) = d (v)Fi x
i d (v) =xi vi
F (v) = Fiv
i
 
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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?
mathematical-physics differential-geometry vectors
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edited Feb 14 '18 at 14:12 asked Feb 24 '13 at 2:01
Gold
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For a non-relativistic treatment of a charged particle on a Riemannian 3-manifold, see. e.g. Phys.SE
post.
this
– Qmechanic ♦ Apr 7 '13 at 21:24
3 Answers Active Oldest
18
To understand what a Newtonian force field is, let's take a look at Newton's second law
This translates to the following differential-geometric relation
where maps from velocity to momentum space and is the
trajectory.
F = ma
(m ∘ = F ∘q̇ )⋅ q̇
m : TM → MT∗ q : I ⊂ R → M
The force field ends up being a map
F : TM → T MT∗
Let
be the bundle projections.
π : M → MT∗
Π : T M → MT∗ T∗
Then
m = Π ∘ F
Tπ ∘ F = idTM
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 and are naturally isomorphic - in coordinates, we just
switch the components - we can represent it as a differential form
on , which is just the differential of the Lagrange function (the Euler-Lagrange
T MT∗ TMT∗
(x, p; v, f) ↦ (x, v; f, p)
TM dL
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, j g g ( g g
equation Newtonian equations of motion).are
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 
 where . This is a velocity-dependant covector field, which you can
indeed integrate over or derive from a potential function (in case of velocity independence).
( M)τ ∗ T∗ τ : TM → M
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 . However, that space has the
wrong structure - if we add two accelerations acting on the same particle, we end up with 
, which is no longer a valid acceleration.
(x, v; v, a) ∈ TTM
(x, v; 2v, a + )a′
What we want instead are vectors or 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.
(x; a) ∈ TM (x, v; a) ∈ (TM)τ ∗
So let's assume our acceleration is velocity-independent and represented by . By
lifting the vector vertically at , we arrive at . What's 'missing'
is the horizontal component .
(x; a) ∈ TM
(x; v) ∈ TM (x, v; 0, a) ∈ TTM
(x, v; v, 0) ∈ TTM
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 , and are naturally isomorphic,
whereas is not. These isomorphisms lead to several (more or less) equivalent
formulations of analytical mechanics, including the Newtonian, Lagrangian and Hamiltonian
approach.
T MT∗ TMT∗ MT∗T∗
TTM
Apologies for expanding the scope of the question - feel free to ignore these ramblings ;)
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edited Feb 25 '13 at 10:42 answered Feb 24 '13 at 11:03
Christoph
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.6 3 56
 
Wow, very in-depth. Didn't really consider this generality. Are there any applications of this?
– user21299 Feb 24 '13 at 14:20
1
 
@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. –  Gold Feb 24 '13 at 16:02
 
I like the approach you are taking but things really seem strange to me. Shouldn't the force naturally be
a covector field and Newton's 2nd Law is or something? Is is not natural
to find a spray in this manner?
F : M → MT ∗ g(F , ⋅) = ma
– levitopher Feb 25 '13 at 1:51
 
@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 – Christoph Feb
25 '13 at 7:36
1
 
@cduston: let's forget about dual spaces for now; the second-order equation tells us that the
acceleration field 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)
= Y ∘q̈ q̇
Y
– Christoph Feb 25 '13 at 7:44
8
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:
=pμ
∂L
∂ẋμ
Force can then be interpreted as . Or, you can define force directly from the
Lagrangian as:
d /dτpμ
=Fμ
∂L
∂xμ
Combined with the argument about work that you provided, where , it seems
very compelling that force should naturally be interpreted as a covector.
W = ∫ dFi xi
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Jold
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It it necessary to denote spacetime point by ? If I use 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 is a vector of some scalar function is a covector. @Jold
xμ xμ
xμ
∂
∂xμ
– SRS May 26 '17 at 6:38
 
@SRS doesn't have any meaning in the absence of a metric, which is why is the more "natural"
representation.
xμ x
μ
– 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-
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3
y g y
form, albeit one that is not exact.
Note also that the condition for a force to be locally determined by a potential can
be written as so that for some function by the Poincare lemma.
× = 0∇⃗  F ⃗ 
dF = 0 F = −dU U
More generally we have p-form potentials to which we associate p+1-form field strengths 
. 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 one.
A
dF
this
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