Background:
- Starting from $F = ma$, integrating with respect to time, and using basic calc, one can derive $\int Fdt = m (v_f – v_i)$
- Starting from $F = ma$, integrating with respect to distance, and substituting $a\ ds = v\ dv$ (from calculus), one can derive $\int Fdx = KE_f – KE_i$
(Obviously I care about these results because, when combined with $F_{ab} = -F_{ba}$, they give conservation of momentum and a starting point for conservation of energy, but for this question I'm not going to consider this part of the derivation.)
In 1-D, the two bulleted results make sense to me: the starting point is a 2nd-order ODE, and the two results form a coupled system of two 1st-order ODEs. This is precisely what math says should be possible – the number of mathematical constraints has been conserved.
Question:
A similar (slightly more involved) derivation is possible in 3 dimensions, but it's harder for me to classify the resulting mathematical constraints and reassure myself that the transformation doesn't add or remove constraints:
- Starting Point: $\vec{F}=m\ddot{\vec{s}}\ \rightarrow$ 3 coupled second-order ODES
- Result 1: $\int \vec{F}dt = m(\dot{\vec{s}_f} -\dot{\vec{s}_i})\ \rightarrow$ 3 coupled first-order ODEs (viewing initial state as a boundary condition)
- Result 2: $\int \vec{F}\cdot d\vec{s} = \frac{1}{2}m(\dot{\vec{s}_f} \cdot \dot{\vec{s}_f}) – KE_i\ \rightarrow $ Unsure how to classify this
- Rewriting this as $\int \vec{F}(x,y,z)\cdot d\vec{s} = \frac{1}{2}m(\dot{x}_f^2 +\dot{y}_f^2 +\dot{z}_f^2 ) – KE_i $ makes it clearer that this is a single non-linear ODE with the first derivatives of 3 different dependent variables in it.
Math says that a system of 3 coupled second-order ODEs can be rewritten as a system of 6 coupled first-order ODEs, but I obviously only have 4 equations. What type of ODE is Result 2? A single first-order ODE? A 'triple' first-order ODE with 3 dependent variables?
I'm ultimately looking to reassure myself that commuting the problem 'solve $F=ma$' to the problem 'solve conservation of momentum and conservation of energy' doesn't add or remove mathematical constraints. If anyone can refer me to a textbook which addresses this idea, I'd appreciate that as well.
Edit:
On further consideration, I think I've made a mistake. $F=ma$ is a differential equation with time as the independent variable, so integrating with respect to time (as is necessary to derive conservation of momentum) changes the order, and it's not actually reasonable to expect two coupled first-order differential equations as a result any more. In fact, we could integrate Conservation of Momentum wrt $t$ again, and then we'd no longer have a differential equation at all.
Since integrating is a valid technique for solving differential equations, it now kind of appears as though Conservation of Momentum could be viewed as a [/the] solution to $F=ma$, and could thus be expected to describes all the information which is present in the original ODE.
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That would make the $F=ma$-to-Conservation of Momentum derivation logical (from a Conservation of Constraints perspective) in both 1-D and 3-D, and then make the real question 'Where Conservation of Energy come from, from a constraints perspective?'
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On the other hand, the 1-D elastic collision problem can't be solved with Conservation of Momentum alone – Conservation of Energy is also required. Since Conservation of Energy can be derived from $F=ma$ and calculus alone, it must be representing a constraint which is in the original equation. Thus $F=ma$ holds more information than Conservation of Momentum.
Other things I've been thinking about:
- This question may actually relate to integro-differential equations, which I know nothing about.
- There may be some subtleties related to $F$ which I'm missing, and writing it as $\vec{F}$ instead of $\vec{F}(\vec{s},\dot{\vec{s}},t)$ may be sweeping them under the carpet.
Best Answer
In theory, the Momentum, Force and Energy pictures may each contain the same amount of information. In practice, the given constraints and available mathematical tools determine which picture(s) to use.
To 'derive' Energy from $ \mathbf{F} = m \mathbf{a}$, we integrate along the path $\cdot d\mathbf{s}$ which means that Energy is a scalar (multi-variable, non-linear, differential) equation with potentially many unknowns and units of $ml^2/t^2$ where '$m$' is mass, '$l$' is length, and '$t$' is time. If you cannot integrate the forces, or if the force is non-conservative and you cannot integrate the path, the equation remains integro-differential. We would recover the 'missing constraints' that result from the dot product with vector equations for the Energy picture.
As you correctly guessed, Energy admits a system of 6 equations where 3 momenta each get an independent equation and an operation on an Energy quantity. The Lagrangian and Hamiltonian formulations, which are related by Legendre transformation, both suffice (see backward and forward derivations from Forces). They each require knowledge of Momentum, Energy and more complicated mathematics. Force, Energy and Momentum solve most Newtonian problems more quickly (counterexamples: forces on constrained objects and canonical momenta), but the vector equations have more theoretical value and application to the other physical sciences.
Emmy Noether, for example, formalized the relationship between conserved quantities and physical symmetries using the Lagrangian picture. She showed that the conservation of Momentum, Energy and Angular Momentum result from physical laws being constant through space, time and orientation, respectively.
Some of the answers/discussions claim that the missing constraints may be recovered with the angular version of each of these pictures. That view is patently false. We may derive $\mathbf T = \mathbf I \ddot{\boldsymbol \theta}$ by taking $\mathbf r \times \left(\mathbf F = m \mathbf a \right)$ which means Torques are a consequence of Forces. When we take $\left( \mathbf T = \mathbf I \ddot{\boldsymbol \theta} \right) \cdot d\boldsymbol \theta$, we recover the scalar Angular Energy which means rotation suffers the same limitation (from which it also recovers with the Hamiltonian and Lagrangian formulations).
I used Thornton and Marion for intermediate mechanics and it was great.