• 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.


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.


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.

  • 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?'

  • 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.
  • 2
    $\begingroup$ Good question. I think the gist is that the missing constraints in 3D turn into conservation of angular momentum, but others here are certainly better qualified than I am to show that (if I'm not wrong) so I'll leave it to someone else to write an answer. $\endgroup$
    – David Z
    Commented Dec 16, 2013 at 3:47
  • $\begingroup$ That would be a very elegant resolution! I think it would lead to 'double-counting' constraints though. Suppose my particle accelerates in a circle... is that because the force is tangent to the circle at each point, or because there's a torque? I think the answer is 'either,' because the conditions are redundant. $\endgroup$ Commented Dec 16, 2013 at 8:13
  • $\begingroup$ Yeah, that was my thought as well. After all, at least in classical mechanics, the whole rotational dynamics formalism (angular momentum, torque, etc.) is a way of eliminating degrees of freedom with trivial behavior from certain kinds of systems, namely those that exhibit pure rotational motion. $\endgroup$
    – David Z
    Commented Dec 16, 2013 at 8:27

5 Answers 5


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.


(Not an answer as such, more an extended comment and suggestion).

You wrote "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. "

Actually the 1D elastic collision problem can be solved with (i) Conservation of Momentum (COM) (ii) application of symmetry (iii) application of galilean relativity where (ii) and (iii) replace COKE (Conservation of Kinetic Energy).

Thus:- state your 1D problem in original inertial frame then shift velocities to the frame of zero net momentum, conserving mass. Then the "after-collision" velocities are simply given by the inverse of the "before-collision" velocities. Then shift back to the original frame to find the "after-collision" velocities in that frame.

Perhaps this doesn't help you because it seems that I have replaced one constraint (COKE) with two new constraints (symmetry, galilean relativity) but maybe those are already "hidden axioms" in your original model.

Also any simple 3D elastic collision (e.g. between 2 spheres) can be considered a 1D collision along the line (axis) connecting the two centres. Then you can apply the rule $$dV1 = +/-2.MCV.m2/(m1+m2)$$ where m1,m2 are the sphere masses and MCV is the mutual closing velocity and dV1 is the change in along-axis component of velocity of sphere1 across the collision. With this rule you don't need COM or COKE.

Your tentative question "Where {does} Conservation of Energy come from, from a constraints perspective?".

I'm not sure about the "constraints perspective" but COKE can be derived from principles (i), (ii) and (iii). I have a derivation somewhere. The $$V^2$$ term in COKE comes from pythagoras $$(dVmag^2 = dVx^2 + dVy^2 + dVz^2)$$ COKE is a useful rule, but COM, symmetry and galrel are, in my view, more "primitive".


I interpret your question as

What type of equation is the equation $$\int_{t_0}^{t} \vec{F}(\vec{x}(t')) \cdot d\vec{x}(t') = \frac{1}{2} m ((\dot{\vec x}(t))^2 - (\dot{\vec x}_{0})^2) \; \; \;\; (1)$$ where $t$ is variable and $\dot{\vec x}_{0}$ fixed?

The answer is, it is a first order integro-differential equation, not a mere ordinary differential equation (ODE). The same is true for $$\int_{t_0}^{t} \vec{F}(\vec{x}(t')) dt' = m (\dot{\vec x}(t) - \dot{\vec x}_{0}) \; \; \;\; (2)$$

That is, neither of the stated are ODEs. However, equation (1) is not equivalent to Newton"s equations. Only the full set of the three equations (2) is equivalent to $$\vec{F}=m\ddot{\vec x}, \, \dot{\vec x}(t_0)=\dot{\vec x}_{0} \;\;\;\; (3)$$ You can get (3) from (2) by a derivative with respect to time $t$. To gain a set of 6 first order ODEs is much easier than you imagine: $$\vec{F}=m\dot{\vec v},\, \dot{\vec{x}}=\vec v$$ Where $\vec{v}(t)$ and $\vec{x}(t)$ are treated as independent functions. Exactly this kind of procedure is meant by the statements you have mentioned.

Note that none of (1,2,3) are conservation laws per se. For a particle in a potential the momentum is not conserved. Equation (1) tells you kinetic energy does change, otherwise your right hand side would be zero. Equation (2) does not tell you momentum is conserved for the very same reason. The stated are simply laws or "balances" as you say.

Conservation laws arise only for systems with particular properties such as symmetries.

The solution of Newton's equations (3) can be put equal to finding integrals of motion only in special cases. Generally you need $N$ integrals of motion for a system with $N$ degrees of freedom to be Liouville integrable (e.g. a particle moving in 3D without a constraint has 3 degrees of freedom). This means that in such a case the equations can be rearranged into quadratures (the relation between positions and time can be expressed explicitly with integral expressions involved). But in such a process you actually use Newton's equations.

The only partial escape is when you find more than $N$ integrals of motion for an $N$ degree of freedom system. That is the case of the motion in Newtons gravitational potential. You have a particle in 3D, but we have 4 integrals of motion hidden in conserved energy, conserved angular momentum and the Laplace-Runge-Lenz vector. Just by requiring these to be constant you get the unique shape of the orbit. But to solve for time evolution, you have to go back to Newton's equations.

However, not every system has such numbers of integrals of motion. Sometimes the integrals simply do not exist and the motion becomes chaotic.

A great book to get a better understanding of classical mechanics is Goldstein Classical Mechanics or the more mathematical Arnold Mathematical Methods of Classical Mechanics.



I'm the original asker of this question (9 months ago); thanks to the comments and answers I've gotten here, I think I've pieced together an answer that I'm happy with.

Short forms used in this answer:

  • CoLM = Conservation of Linear Momentum
  • CoAM = Conservation of Angular Momentum
  • KEB = the Kinetic Energy Balance
  • CoE = Conservation of Energy


The equation derived in the question is actually KEB. KEB doesn't provide as much information as CoLM does, but because of the symmetries of some problems, specific results are easiest to extract using KEB. Full CoE, however, is a different equation, and does incorporate new information: it accounts for the fact that work, heat, and internal energy can inter-convert, connecting the motion problem to internal energy and heat transfer. If the force laws are all known, then CoLM is all that is required to determine the motion, but CoE provides information about the sources/sinks in the of heat and internal energy - information which can't be extracted from CoLM.

To address a few of the specific issues raised in the question:

  • CoLM contains exactly as much information as F = ma does
  • KEB is not a new constraint; it's sort of redundant on CoLM (in the sense that the two can never be incompatible), but [in two or more dimensions] it contains less information than CoLM and therefore can't be used in its place. In many cases, the symmetries of the problem are such that KEB alone can be used to extract useful results. In those cases, the discarded information (which was in CoLM but isn't in KEB) pertains to aspects of the problem which aren't of physical interest.
  • CoE is not redundant on CoLM; a new constraint has been introduced, but so have new variables (heat transfer and changes in internal energy)
  • The 1-D elsatic collision problem actually ties everything together quite well. Firstly, when solving with CoLM, adding KEB doesn't actually impose new constrains - as mentioned by steveOw, the new information comes from an assumption that the kinetic energy lost during the compression stage of the collision is equal to the kinetic energy gained during the extension stage of the collision, i.e., that the loss to heat/internal energy is zero, or equivalently the force at a given compression level is the same during both compression and extension. This assumption is applied in KEB by setting the source/sink term to zero, but it could have also been substituted in CoLM (without KEB) to derive the same result (although the integration required would essentially re-derive KEB).

    Secondly, this is an example of how KEB can leverage symmetries of the problem to make it easier to extract specific results. The symmetry here is the fact that the force at any point during compression is the same as the force at the same point during expansion. Per KEB, the full force law is actually somewhat irrelevant so long as it respects this symmetry - it doesn't impact the final speeds, at least. A full force law would be required to generate a full result, however; we know the velocities of the two objects, but we don't know how long they spent interacting, so we don't know what the starting point for the post-interaction paths was and thus don't know either object's position. A full solution would require a full force law (and could also admit a force law without the compression/expansion symmetry), and the only equation used to fully solve the motion would be CoLM.

Tangential Extra: CoLM and CoAM

A related issue, which was touched on in the comments by David Z, is redundancy of CoLM and CoAM, and how this connects to the "conservation of constraints" idea of this question. CoLM and CoAM are kind of redundant, but kind of not, depending on whether you take the real-life perspective (that you can't have a force or moment which acts at a point) or the simplified perspective (that distributed loads can be simplified as a resultant point force and resultant point moment). In the real-life perspective, there is a one-to-one correspondence between forces and moments - every moment is the result of a force and a moment arm - and CoLM and CoAM are redundant. Continuum mechanics takes the real-life perspective (all forces are resolved as stresses - no point forces or moments), and so in continuum mechanics only CoLM is solved (CoAM would provide no extra information). In the simplified perspective, the one-to-one force-moment correspondence is lost - there are point moments which don't have an associated force and moment arm - and CoLM and CoAM become distinct conditions. Statics usually involves the simplified perspective, so in statics CoLM and CoAM (or, as they're more often applied, Fnet = 0 and Tau_net = 0) are distinct conditions.

  • $\begingroup$ Each formulation of mechanics makes a tacit assumption about the conditions of the interaction. The (linear / angular) momentum picture assumes that (physical laws are space invariant) there are no external forces (torques) acting on the involved bodies. The Energy picture assumes that (physical laws are time invariant) the system in question accounts for all of the energy. If both of these are true, we actually have 7 (13!?!) equations available for use. $\endgroup$
    – user121330
    Commented Sep 23, 2014 at 16:35
  • $\begingroup$ I would be hesitant to make a distinction between conservation of energy and any sort of kinetic energy 'balance' because rotational and vibrational energies are also kinetic. $\endgroup$
    – user121330
    Commented Sep 23, 2014 at 16:44
  • $\begingroup$ Lastly, any body force may apply a torque. For example, one side of the moon is more massive and it points at Earth through tidal locking. $\endgroup$
    – user121330
    Commented Sep 23, 2014 at 16:54
  • $\begingroup$ Re comment 2: I don't think rotational energy breaks the KEB, as long as you consider the KE of a solid body to be \int \rho v^2/2 dV. This would give the same result as KE_{rot} = 1/2 I \omega^2 for solid-body rotation. Vibration is fine too - macroscopic vibration is a cyclic conversion of kinetic energy to and from potential energy, and microscopic vibration is considered as part of internal energy. $\endgroup$ Commented Sep 23, 2014 at 20:32
  • $\begingroup$ Re comment 3: I used the wrong phrase; by body torques'' I meant point torques.'' I corrected that above $\endgroup$ Commented Sep 23, 2014 at 20:33

One can consider the quantities

  • $\int F_x\,dx=\int m\ddot{x}\,dx=\frac{1}{2}m(\dot{x}_f^2-\dot{x}_i^2)$
  • The $y$ version of above
  • The $z$ version of above

Are these what you're after? These three quantities aren't usually considered in standard problems, but they seem valid to me. Your "Result 2" is the sum of the three bulleted equations here.

  • 2
    $\begingroup$ I thought this way as well at first. I don't think that those are appropriate integrals to perform on their own though. Result 2 comes from taking the line integral along the specific path that the object takes from A to B. Your result comes from integrating along one axis at a time, which physically represents moving the object along just one axis at a time. This will only coincide with the path that the object actually takes if the lateral forces are zero everywhere - which they aren't in general. $\endgroup$ Commented Dec 16, 2013 at 8:23
  • 1
    $\begingroup$ Yeah, BMS, I considered this line of thought as well but I came to the same objection user1476176 did. $\endgroup$
    – David Z
    Commented Dec 16, 2013 at 8:25
  • $\begingroup$ I don't think it necessarily corresponds to movement in just one dimension at a time. For path $C$, $\int_C \vec{F}\cdot d\vec{s}=\int_C F_x\,dx + \int_C F_y\,dy +\int_C F_z\,dz$, which is how one usually performs line integrals. Then, consider one of the terms at a time. $\endgroup$
    – BMS
    Commented Dec 16, 2013 at 16:12
  • $\begingroup$ When you equate the line integral you've written with the change in kinetic energy, the result is a scalar equation. You can't divide it into components because the contributions from the three dimensions are all on the same footing (as opposed to a vector equation where they are linearly independent and you can separate into components) $\endgroup$ Commented Dec 16, 2013 at 18:32
  • 1
    $\begingroup$ I agree that one can't automatically extract component information from $\int_C \vec{F}\cdot\,d\vec{s}=\Delta K$, and for the reason that you specify. However, it seems valid to write out $\int \vec{F}\cdot\,d\vec{s}=\sum_i\int F_i\,ds_i$. After writing this, can one separately consider what each of the three terms will mathematically evaluate to. $\endgroup$
    – BMS
    Commented Dec 16, 2013 at 23:37

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