I was trying to figure out the situation in which energy is conserved and momentum is not and it was quite easy to find out one case which is of a stone tied to a string moving in a uniform circular motion.

Then I thought to consider the reverse situation in which momentum is conserved but energy is not. To me it seems that as soon as we chose a system in which momentum is conserved then it automatically is implied that in such system energy is conserved too. But looking at the case how these two conservations come into existence due to two different symmetries (one related to the invariance in physical laws due to translation in space and the other in time).

So it would be quite helpful

  • if someone can point out to a case in which momentum is conserved but energy is not.

  • otherwise if the above is not possible then explain why that is the case?

[Note that I am considering every form of energy of the given system.]

This is to summarise my question so that no further confusion occurs to future visitors. The question in short is:

  • can we lose time symmetry and retain translation symmetry? (Give an appropriate example for the case)
  • $\begingroup$ Related: How to explain independence of momentum and energy conservation in a 2-body collision in elementary terms? $\endgroup$
    – Qmechanic
    Commented Jan 17, 2020 at 20:30
  • $\begingroup$ Be careful when choosing the system in which you want to observe conservation laws. They generally are only true for closed systems -- or if you want, closed systems are defined as those which do not "leak" (energy, momentum). Obviously such systems do not exist, short of the entire universe. But in your case, because e.g. the Moon's, the Sun's, the planet's and the galaxy's gravity can be neglected, we can consider just the Earth and the stone and the string. In that case, of course, both momentum and energy are conserved, as always. It's just that Earth's $\Delta v$ is so small. $\endgroup$ Commented Jan 18, 2020 at 11:43
  • $\begingroup$ If you want to disprove the proposition "momentum conservation implies energy conservation", then what you want is a situation in which momentum conservation holds, but energy conservation doesn't. Not the other way around. Maybe your question title is backwards: Does energy conservation imply momentum conservation? $\endgroup$
    – Kaz
    Commented Jan 18, 2020 at 19:55
  • $\begingroup$ Your question doesn't cover the question body. $\endgroup$ Commented Jan 19, 2020 at 1:06
  • $\begingroup$ @descheleschilder why not? $\endgroup$
    – user249968
    Commented Jan 19, 2020 at 1:13

9 Answers 9


While there are some good answers, no one has yet mentioned Noether's theorem. Fundamentally, the conservation of momentum and conservation of energy have different characters. Notably conservation of linear momentum is linked with the laws of physics being independent of where something happens and conservation of energy is linked with the laws of physics being independent of when something happens.

Simply: Linear momentum conservation is linked to position invariance. Energy conservation is linked to time invariance.

More rigorously, Noether's theorem demonstrates that conservation of linear momentum arises from a Lagrangian that has no explicit dependence on position and conservation of energy from a Lagrangian that has no explicit dependence on time. Furthermore, other symmetries imply other conservation laws. A lack of explicit dependence on the angle in the Lagrangian gives conservation of angular momentum, suggesting that angular momentum can be considered fundamentally different than linear momentum (which is backed up by spin in quantum mechanics not being interpretable as linear motion). As an aside, purely (Newtonian) 2-body gravitational systems have another conserved vector owing to symmetry in the Lagrangian of the inverse square force law.

Hence, it is easy to write down models where energy is conserved, but momentum is not, and systems where momentum is conserved and energy is not. Furthermore, these models can be good approximations of many real situations as described in other answers. You can have models of collisions where momentum is conserved but energy is "lost" to internal degrees of freedom or radiation. You can have thermodynamic models where you have a container of effectively infinite mass (which leads to momentum not being conserved when gas molecules collide with the walls) but with conserved energy. In the end, however, linear momentum, angular momentum, and energy seem to be conserved in fundamental interactions and maybe in the universe as a whole.

As an aside, Special Relativity treats position and time in a similar way and likewise wraps up momentum and energy in the 4-momentum, which shows energy as the time component of a relativistic 4-momentum. However, although relativity shows some symmetry between position and time and consequently momentum and energy, they still have different characters owing to different signs in the signature ($\mathrm{d}s^2 = -c^2\,\mathrm{d}t^2 + \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2$).

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    $\begingroup$ can you provide an off-shell example where momentum is conserved and energy is not? $\endgroup$
    – lurscher
    Commented Jan 17, 2020 at 19:07
  • $\begingroup$ @lurscher That's getting outside of my expertise. My research is mostly in the Newtonian realm. Maybe someone else can help. $\endgroup$ Commented Jan 17, 2020 at 19:32
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    $\begingroup$ @lurscher You could take the Lagrangian $L=\frac12m\dot{x}^2-V(t)$ [sic]. $\endgroup$
    – J.G.
    Commented Jan 18, 2020 at 21:00
  • $\begingroup$ My research is mostly in the Newtonian realm. Then why you bring in relativity? $\endgroup$ Commented Jan 19, 2020 at 7:59
  • $\begingroup$ @descheleschilder I've taken graduate level classes that include Special Relativity, so I'm not a novice in relativity. But I'm not an expert either since currently that's not what I work on. $\endgroup$ Commented Jan 20, 2020 at 17:58

The total energy (internal + kinetic) of hot body isotropically emitting radiation while moving in the vacuum far from other bodies, will decrease, while momentum is conserved.

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    $\begingroup$ In terms of newtonian mechanics, yes. (And indeed the context of the question is newtonian mechanics). Still, let's see how things work out in terms of relativistic physics. If we assume the source of the energy is nuclear fusion then the heat generation goes at the expense of inertial mass. In the absence of nuclear fusion: in terms of relativistic physics the kinetic energy of the criss cross motion of the particles that the emitting body consists of has a corresponding inertial mass, which deceases as the body emits thermal radiation. $\endgroup$
    – Cleonis
    Commented Jan 17, 2020 at 12:24
  • $\begingroup$ @GeorgioP Your answer is not an answer to what is asked by the OP, in the question above the question body. It's not even true. Momentum and energy are both conserved in your example. $\endgroup$ Commented Jan 19, 2020 at 1:10
  • $\begingroup$ @descheleschilder look I had a misconception that if momentum is conserved then so is energy (in total) which I wanted to know if can be proved wrong. So I asked the answerers to give a situation in which this isn't the case. And GiorgioP provided this. But I really feel sad that I can't simultaneously accept other good answers (Though I have upvote them). $\endgroup$
    – user249968
    Commented Jan 19, 2020 at 1:12
  • $\begingroup$ Alright then!!! $\endgroup$ Commented Jan 19, 2020 at 2:49
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    $\begingroup$ @descheleschilder Energy is not conserved in his example because the system the example was looking at is the sphere by itself, not the sphere + emitted photons. This kind of solution (looking at a closed system rather than the whole universe) must be within the scope of what the OP was looking for because their example for a system where momentum is not conserved but energy is only has non-conservation of momentum if you're looking at just the rock on the string, and not the body the string is attached to as well. $\endgroup$ Commented Jan 19, 2020 at 14:53

Yeah, it's interesting to contemplate the relation between energy and momentum.

Energy can morph from form to form, with kinetic energy being only one of them. Momentum, on the other hand, deals with a subset of all forms of energy: motion.

So let's set up a system with the energy oscillating between potential energy and kinetic energy: let's say a vibrating string (for example a guitar string).

The momentum of the string is oscillating, while the total energy is conserved.

However, looking at it that way is cheating. The string is part of a system: the system as a whole is the guitar with the string. To aid this thought demonstration, take a guitar with one string, and make that string really heavy. Have that system be suspended, such that it cannot exchange momentum with surroundings. Then the string and the body are continuously exchanging momentum with each other, and the momentum of that system is conserved.

In your question, you submitted the case of an object connected to a central axis that is infinitely rigid, with the object moving in a circular motion.

That example doesn't actually offer what you want.
You need to consider the system as a whole. This means that you also need to consider the momentum of the support that maintains the circular motion of the circumnavigating object.

Consider for example the motion of the planet Jupiter around the Sun. Jupiter and the Sun are both orbiting the common center of mass of the Sun/Jupiter system. That common center of mass lies somewhat outside the Sun, not even inside the surface of the Sun.

About the title of your question:
"Does momentum conservation imply energy conservation?"

In my opinion: from an axiomatic point of view: no.

I think that in terms of formal logic the two are independent axioms. In other words, I think that in terms of formal logic neither can be derived from the other.

However, if both energy conservation and momentum conservation are laws that hold good universally, and presumably they do, then in any thought demonstration we come up with we will always see that energy and momentum are both conserved.

I think: in terms of formal logic the fact that both hold good universally does not logically imply that they are interdependent (but thought-provoking for sure).

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    $\begingroup$ What does this mean: "in my opinion, from an axiomatic point of view, no"? It's not a matter of opinion, it either is (with a proof) or isn't (with a counterexample). The answer is no, for example, as per the accepted answer. $\endgroup$
    – andrepd
    Commented Jan 17, 2020 at 15:11
  • $\begingroup$ @andrepd It seems to me that "in my opinion" here means "according to my interpretation of the question." $\endgroup$ Commented Jan 17, 2020 at 16:09
  • $\begingroup$ @andrepd In general: it's my understanding that it's not a good idea to even try axiomatization. Most physics theories aren't axiomatized because when you try it just gets messy. Theory of motion: it appears as if one can actually pare things down to just a couple of axioms. However, my understanding is that a treatment with mathematical rigor would be quite a bit more involved. My guarded 'in my opinion' is a reflection of that notion. With that said: just the attempt at formulating axioms forces one to examine one's own assumptions, which is important and instructive in itself. $\endgroup$
    – Cleonis
    Commented Jan 17, 2020 at 16:34

Due to the principle of energy conservation, energy is always conserved for an isolated system in an inertial reference frame. Therefore, considering the conservation of all forms of energies (the sum of them), the question does not make much sense.

It makes sense if referring to the mechanical energy only, whose conservation is a theorem and not a principle.

Here it is possible to show elementary counterexamples where the angular momentum is conserved whereas the mechanical energy is not.

A simple example is a point of the matter of mass $m$ attached to one endpoint an ideal stick with negligible mass and the other endpoint of the stick is fixed to the origin $O$ of the Cartesian axes and the point can freely rotate in the plane xy (without friction) of an inertial reference frame.

The trick is supposing that the length of the stick varies in time with a given function $d=d(t)$.

The motion of the point is described in terms of the angle $\theta(t)$ determined by the stick with respect to the axis $x$. The equation of motion are $$\frac{d}{dt}\left(m d(t)^2 \dot{\theta}(t)\right)=0$$ The mechanical energy is $$E= \frac{1}{2} md(t)^2 \dot{\theta}^2$$ and it is easy to see from the equation above that it is not conserved on every non-trivial solution of the equation of motion if the assigned function $d=d(t)$ is not constant.

Conversely, the angular momentum is directed along the axes $z$ and has a numerical value which exactly coincides with $$L=\left(m d(t)^2 \dot{\theta}(t)\right).$$ Therefore it is conserved (no matter how you choose the function $d$) on every solution of the equation of motion.


If you’re just considering mechanical energy, consider the case of a firework or other exploding object. The total momentum is conserved, but the kinetic energy is increased from the chemical energy of the explosion.

More formally, momentum conservation is associated with isotopic space while energy conservation is associated with no changes over time: the explosion happens at a moment and energy changed; you pull the circular motion in different directions and momentum changes. So you can develop different situations to do different things.

  • $\begingroup$ I was considering total energy of the system. $\endgroup$
    – user249968
    Commented Jan 17, 2020 at 6:19
  • $\begingroup$ In that case, when would the total energy not be conserved? $\endgroup$ Commented Jan 17, 2020 at 7:35
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    $\begingroup$ @user3518839 when time symmetry is violated. So the question is, can we lose time symmetry and retain translation symmetry or vice-versa. Noether's Theorem. $\endgroup$
    – rghome
    Commented Jan 17, 2020 at 8:36
  • $\begingroup$ Not in Newtonian or quantum mechanics $\endgroup$ Commented Jan 18, 2020 at 9:48

In Newtonian theory, you're dealing with only gravity and EM forces which are conservative and central so there is no example of a closed system where either energy or momentum are not conserved. So energy and linear and angular momentum conservation are always implied by Noether's theorem. Your example is wrong even if you only consider linear momentum for the whole system.

In QM Hamiltonian systems are always space- and time-symmetric so again the same thing.

In general( relativity) this symmetry doesn't have to be there. one can derive some similar conserved quantities but they don't necessarily equate to momentum or energy. So they are not conserved

For example, two laser light sources/receivers are some large distance (10 mpc or more) apart shooting light at each other as they are receding from each other. A distant observer will see the energy of light drop, in time at each receiver because of respective redshifts but momentum will be conserved. Or simply if they were two point masses as the space between them is increasing potential energy is lost while momentum stays the same.


Example of momentum conservation, while energy is not conserved (sticky collision):

Just let two pieces of sticky stuff with opposite momenta (so in total zero momentum), collide frontally. The end result is a piece of this stuff that has also zero momentum but the kinetic energy has conversed into internal heat (energy is alway conserved, just like total momentum, which in this case enters into the internal of the stuff). It's just an inelastic encounter.

So, also in the example you give, both energy and momentum (angular momentum) are always conserved.

So to answer your question: No. Momentum conservation doesn't imply energy conservation (when potential energy is not considered). More generally they don't imply each other both ways (again, not considering potential energy).

They are separate conservation laws that can't be connected. Momentum by itself nevertheless is connected with (kinetic) energy by itself:

$$p=mv,$$ and E (leaving out the potential energy) $$E=\frac{1}{2} m v^2,$$ so $$\frac{dE}{dv}=mv$$

Taking potential energy into account. In this case yes. Both imply the other (momentum conservation and total energy conservation), as kinetic energy is transformed into internal energy (potential energy), like heat, chemical potential, etc.


For a single isolated particle (with constant mass) conservation of momentum does indeed imply conservation of kinetic energy (and also of mechanical energy, as if it is isolated there is no force applied to it and therefore there is no potential energy). This is easy to prove:

Let $\vec{p}=m\vec{v}$ be the momentum and $K=\frac{1}{2}mv^2$ the kinetic energy, then conservation of momentum reads:

$$\frac{d\vec{p}}{dt}=\vec{0}$$ expanding it $$\frac{d\left(m\vec{v}\right)}{dt}=\vec{0}$$ $$\frac{dm}{dt}\vec{v}+m\frac{d\vec{v}}{dt}=\vec{0}$$ $$m\frac{d\vec{v}}{dt}=\vec{0}$$ and multiplying it by $\vec{v}$ we have $$m\frac{d\vec{v}}{dt}\cdot\vec{v}=\vec{0}\cdot\vec{v}=0$$ $$\frac{d\left(\frac{1}{2}mv^2\right)}{dt}=0$$ $$\frac{dK}{dt}=0$$ $$\square$$

However, for a general isolated system of particles, this is not true in general. Say you have a completely inelastic collision(one where both particles end up stuck together) between to particles with masses $m_1$ and $m_2$ and velocities $\vec{v}_1$ and $\vec{v}_2$.

Before the collision the total momentum is: $$\vec{p} = \vec{p}_1 + \vec{p}_2 = m_1\vec{v}_1+m_2\vec{v}_2$$ and the total kinetic energy is $$K=K_1 + K_2 = \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$

After the collision both particles move with the same velocity, and the conservation of momentum dictates that this should be such that: $$\left(m_1+m_2\right)\vec{v} = m_1\vec{v}_1+m_2\vec{v}_2$$ $$\vec{v} = \frac{m_1\vec{v}_1+m_2\vec{v}_2}{m_1+m_2}$$

However, if we compute the kinetic energy after the collision we have

$$ K = \frac{1}{2}\left(m_1+m_2\right)v^2$$

which in general is not the same as before the collision, take for example $m_1=m_2=m$ and $\vec{v}_1=\vec{v}=-\vec{v}_2$ (that is, to equal particles moving with the same speed towards each other, which, after the collision remain still), and we have a constant momentum of $\vec{0}$ before and after the collision, however, the total kinetic energy goes from $K=mv^2$ to $0$.

Of course, inelastic collisions exist because energy dissipates in other forms of energy as acoustic, deformation, thermal... but the principle of conservation of momentum and energy talking about dynamics (that is kinetic and potential energy).


Mechanical energy is conserved in a system where only the conservative forces do work to the system..When you considered the vertical circular motion of an object tied with a string, the only forces acting on the object was the tension and gravity out of which tension does not do any work on the system as the motion is always perpendicular to the direction of tension and hence only gravity does work(which is a conservative force) and hence the total mechanical energy of the system is conserved...When you talk about conservation of momentum, no external force must be acting on the system.Here(vertical circular motion of an object) the gravitational force is an external force and it changes the momentum of the system. But in the case of collisions, we can say that momentum is conserved unless an external impulsive force acts on the system(Since collisions happen in an infinitesimally small time, non-impulsive force won't change the momentum of a system to an appreciable extent). So we can say that in a system, where the internal forces are non-conservative momentum will be conserved, but mechanical energy will not be conserved IF THE INTERNAL NON CONSERVATIVE FORCE DO WORK ON THE SYSTEM. And in a non-rigid system, internal conservative forces do work on the system and hence energy will not be conserved but since there is an absence of external force momentum will be conserved! Now if you want an example of such a system, it's very simple-just think of a block sliding over a rough plank and the system is placed on smooth ground. The system is non-rigid and kinetic friction will do work on the system,and momentum will be conserved and energy will not be conserved in this case.:D

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    $\begingroup$ Hi there! Make it somewhat more structuralized. In that way, it's easier to read. I made the same mistake when starting here. I've structuralized the book I'm writing thanks to this site! $\endgroup$ Commented Jan 21, 2020 at 16:06
  • $\begingroup$ Okay..I'm just a beginner here:D...What I meant was in a system where external force is absent and internal forces do work on the system..Mechanical energy will be dissipated,yet momentum will be conserved. $\endgroup$ Commented Jan 27, 2020 at 8:00