Some physical quantities like position, velocity, momentum and force, have precise definition even on basic textbooks, however energy is a little confusing for me. My point here is: using our intuition we know what momentum should be and also we know that defining it as $p = mv$ is a good definition. Also, based on Newton's law we can intuit and define what forces are.

However, when it comes to energy many textbooks become a little "circular". They first try to define work, and after some arguments they just give a formula $W = F\cdot r$ without motivating or giving intuition about this definition. Then they say that work is variation of energy and they never give a formal definition of energy. I've heard that "energy is a number that remains unchanged after any process that a system undergoes", however I think that this is not so good for three reasons: first because momentum is also conserved, so it fits this definition and it's not energy, second because recently I've heard that on general relativity there's a loss of some conservation laws and third because conservation of energy can be derived as consequence of other definitions.

So, how energy is defined formally in a way that fits both classical and modern physics without falling into circular arguments?


4 Answers 4


The Lagrangian formalism of physics is the way to start here. In this formulation, we define a function that maps all of the possible paths a particle takes to the reals, and call this the Lagrangian. Then, the [classical] path traveled by a particle is the path for which the Lagrangian has zero derivative with respect to small changes in each of the paths.

It turns out, due to a result known as Noether's theorem, that if the Lagrangian remains unchanged due to a symmetry, then the motion of the particles will necessarily have a conserved quantity.

Energy is a conserved quantity associated with a time translation symmetry in the Lagrangian of a system. So, if your Lagrangian is unchanged after substituting $t^{\prime} = t + c$ for $t$, then Noether's theorem tells us that the Lagrangian will have a conserved quantity. This quantity is the energy. If you know something about Lagrangians, you can explicitly calculate it. There are numerous googlable resources on all of these words, with links to how these calculations happen. I will answer further questions in edits.

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    $\begingroup$ I disagree, for the reasons given in my answer. (I'm not downvoting, because I think the answer still provides valuable insight, but I think the claims being made are too strong.) $\endgroup$
    – user4552
    Commented Apr 1, 2013 at 21:00
  • $\begingroup$ There is another definition of energy, which is consistent with the above one. It is defined as the negative of the change in the action per unit displacement of the end point of the trajectory. H=-dS/dt(both derivatives are partial). I find this definition particularly comfortable. en.wikipedia.org/wiki/Hamilton%E2%80%93Jacobi_equation $\endgroup$
    – Prathyush
    Commented Apr 2, 2013 at 6:33
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    $\begingroup$ Minor comment: The Lagrangian formalism is a bit of a red herring here. You can simply define energy to be the generator of time translations. This should work in any formalism you choose. $\endgroup$
    – user1504
    Commented Apr 2, 2013 at 15:10
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    $\begingroup$ ...and call this the Lagrangian, for the map $t\mapsto q(t)$ it should read action there, alternatively you can speak about the image of that map. $\endgroup$
    – Nikolaj-K
    Commented Feb 27, 2014 at 12:48
  • $\begingroup$ @Prathyush Your definition seems promising. Can you describe it extensively in an answer to my question(the link to the question is below)? Thank you in advance. physics.stackexchange.com/q/288773/126696 $\endgroup$ Commented Nov 2, 2016 at 10:09

The problem here isn't that energy needs to be defined more rigorously like everything else. The problem is that you're making an incorrect assumption that everything else can be rigorously defined for once and for all. For example:

"[...]using our intuition we know what momentum should be and also we know that defining it as p=mv is a good definition."

Actually this doesn't work. For example, a beam of light has zero mass and nonzero momentum, so p=mv is false for light. If intuition told you p=mv, intuition was wrong.

The general way to go about defining a conserved quantity is to pick something that is your standard amount of that quantity, and then use experiments to find out how much of various things can be converted into that standard. For example, if you pick a 1.00 kg mass moving at 1.00 m/s as your definition of the unit of momentum, then you will find through experiments that its momentum can be exchanged for 2.00 m/s worth of motion for a 0.50 kg mass. This naturally leads to the hypothesis that p=mv. Further experiments seem to verify that hypothesis. But then eventually you do experiments with electrons moving at 30% of the speed of light, or with beams of light, and you find out that p=mv was wrong. It was only an approximation valid under some circumstances. You're forced to revise your definition of p. It's a purely empirical process.

Same thing for energy. The only approach that fundamentally works is to define something as your standard unit of energy. This could be the energy required to heat 0.24 g of water by 1 degree C. Then experiments would show that you could trade that amount of energy for the kinetic energy of a 2.00 kg object moving at 1.00 m/s. Ultimately, all you can do is proceed empirically.

"[...]recently I've heard that on general relativity there's a loss of some conservation laws [...]"

Yes, and this is why I don't agree with Jerry Schirmer's answer. He says that energy is the conserved quantity that you get because of time-translation invariance. But this procedure doesn't work in GR. In technical terms, the relevant symmetry becomes diffeomorphism invariance, and that doesn't satisfy the requirements of Noether's theorem. The more fundamental reason it can't work in GR is that in GR, energy-momentum is a vector, not a scalar, and you can't have global conservation of a vector in GR, because parallel transport of vectors in GR is path-dependent and therefore ambiguous. What you can do in GR is define local (not global) conservation of energy-momentum. Even if the technical details are mysterious, I think this counterexample shows although Noether's theorem does provide a deeper insight into where conservation laws come from, the ultimate definition of conserved quantities is still empirical.

BTW, there is a good exposition of this philosophical position in the Feynman Lectures. He discusses conservation of energy using the metaphor of a bishop moving on a chess board and always staying on the same color. Although that treatment is aimed at people who don't know anything about Noether's theorem or general relativity, I think his philosophical position holds up very well in the full context of what is currently known about all of physics.

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    $\begingroup$ This procedure DOES work in GR, when you have a global timelike killing vector in your spacetime, which is the analogue of having a time translation symmetry. Alternately, it will work if you have asymptotic flatness, etc. $\endgroup$ Commented Apr 1, 2013 at 20:45
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    $\begingroup$ I don't think these issues are obscure at all. For example, undergrads often ask what happens to conservation of energy in cosmology. The answer is that energy simply isn't conserved. "I discussed energy specifically." Sorry to have mischaracterized what you said. However, you made a claim about energy, and I provided a counterexample concerning energy. $\endgroup$
    – user4552
    Commented Apr 1, 2013 at 21:03
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    $\begingroup$ @BenCrowell: you can still define energy as the Noether charge corresponding to time translations even if it's not conserved $\endgroup$
    – Christoph
    Commented Apr 1, 2013 at 21:21
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    $\begingroup$ @JerrySchirmer: I believe you're missing Ben's point: if you're defining energy as the conserved quantity given by Noether's theorem, you're left with no definition at all if energy conservation does not hold; but see my other comment for the resolution of that issue... $\endgroup$
    – Christoph
    Commented Apr 1, 2013 at 21:40
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    $\begingroup$ @Christoph: that's exactly my point. If you don't have a special time, there's no hope in getting an energy. Newtonian physics grants this from on high. Special cases of general relativity also have special notions of time. The general case does not. It's meaningless to chug forward, then, since we can always just arbitrarily change coordiantes. $\endgroup$ Commented Apr 1, 2013 at 21:58

Precise definitions only ever apply to specific models. One of the more instructive ones for energy comes from special relativity, where space and time are not independent, but rather part of a single entity, space-time.

Directional quantities in SR not only have components in the spatial x-, y- and z-directions, but also a fourth (or by convention zeroth) component in time direction. For momentum, that component is (up to a constant factor) the energy.

It turns out that $\vec p=m\vec v$ is less fundamental than $\vec p=\gamma m\vec v$ which is the spatial part of the 4-vector $$p^\mu=\gamma m\begin{pmatrix} c \\ v_x \\ v_y \\ v_z \end{pmatrix}$$ where $\gamma = 1/\sqrt{1-(v/c)^2}$.

As $$\gamma = 1 + \frac 12 (\frac vc)^2 + \mathcal O((\frac vc)^4)$$ the relativistic definition of $\vec p$ agrees with the classical one for $v\ll c$.

Inserting our approximation up to second order into the definition of 4-momentum yields $$ p^0\approx mc + \frac{mv^2}{2c} = \frac 1c \left( mc^2 + \frac 12 mv^2 \right) = \frac 1c \left( E_\mathrm{rest}^\mathrm{Einstein} + E_\mathrm{kin}^\mathrm{Newton} \right) $$

Setting $\vec v = 0$, this also shows that rest energy and mass are essentially the same thing (they only differ by a constant factor). It's important to note that rest energy includes internal binding energy, which leads to the mass defect in nuclear reactions.


Definitions of physical quantities in physics are dependent on context. For example the definition of energy in classical general relativity is different from the definition used in the quantum physics of the standard model. We don't yet have the most "fundamental" theory of physics so we don't know what the fundamental definition of energy will be like, or even if there will be one. Perhaps energy is emergent at the level of quantum gravity so it does not have a fundamental definition. We wont know until we understand quantum gravity better than we do now.

However, there is a general theory about energy and its relation to time translation invariance that is embodied in Noether's theorem. The theorem says that there is a conserved quantity associated with any symmetry of nature. Energy is related to time symmetry while momentum is linked to space translations, angular momentum is linked to rotations, charge is linked to electromagnetic gauge invariance etc.

Noether's theorem was originally stated and proved for classical systems but there is also a version that works for quantum physics, so it could be said that energy is defined as the quantity that comes out of Noether's theorem that is linked to time invariance. This may be the most fundamental definition we can give now, but it depends on the context of currently known physics and we have no idea if it will survive in some form at more fundamental levels of theory than those currently known.

When we speak of time invariance in Noether's theorem we are talking about the fact that the complete laws of physics do not change with time. The early universe may have been very different from the one we live in now, but the laws of physics were the same. This means that Noether's theorem works perfectly well in general relativity for example. The universe may expand and cosmology may evolve but Einstein's field equation for gravity is always the same, so energy is conserved. Many people, especially those in this forum dispute this, but they are wrong. The arguments to this effect given in other answers here are flawed. Energy is conserved in GR without caveats about special cases or global meaning. For detailed refutations of individual claims see http://vixra.org/abs/1305.0034

In case anyone thinks I sound like a lone voice contradicting the mainstream view, that is not the case. When I wrote in a recent FQXi essay about how energy is conserved in GR despite claims to the contrary, Carlo Rovelli responded by writing "I do not see anything in what you say that goes beyond what is written in all GR books about energy conservation in GR. There is a vast literature on this." He is mainly right. You will find conservation of energy explained in books on gravitation by Weinberg, Dirac, Landau and Lifschitz, etc. It is covered well in Wikipedia and there was even a Nobel prize awarded for the application of energy conservation in GR to binary pulsars. The idea that energy is not conserved in GR is a meme perpetuated in some blogs and forums like this one. It stems from an article written on the subject in the physics FAQ some years ago which unfortunately I have been unable to get changed. Do not be fooled.


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