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From basic principals, how does one prove that energy is conserved? Or a little more specifically - Why does this hold:

$$\Delta \mbox{ PotentialEnergy} + \Delta \mbox{ KineticEnergy} = 0 $$

Or, for extra credit, why does this hold:

$$\Delta \mbox{ PotentialEnergy } + \Delta \mbox{ KineticEnergy} + \Delta \mbox{ ThermalEnergy } = 0 $$


marked as duplicate by jinawee, BMS, Kyle Kanos, Qmechanic Apr 12 '14 at 18:25

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An energy conservation law only arises when the system studied has a Lagrangian which is invariant under time translations up to total derivatives, due to Noether's theorem. More generally, a quasi-symmetry under spacetime translations gives rise to an entire host of conserved quantities, encoded in the stress-energy tensor. Consider a scalar field; under translations

$$x^\nu \to x^\nu - \epsilon^\nu$$

the variation of the field is $\delta \phi = \epsilon^{\nu}\partial_\nu \phi$, as it is an active rather than passive transformation. Hence the conserved currents are,

$$T^{\mu\nu} = \partial^\mu \phi \partial^\nu \phi - \eta^{\mu \nu}\mathcal{L}$$

The corresponding conserved Noether charge is given by,

$$Q = \int \mathrm{d}^3 x \, \, T^{00} = \int \mathrm{d}^3 x \, \, \mathcal{H} = H$$

The Hamiltonian is equivalent to the total energy of the system, including kinetic and potential, and as Noether's theorem implies,

$$\frac{\partial}{\partial t} Q = 0$$

we may conclude that energy is conserved locally, i.e. energy conservation applies.

Caveat: It is somewhat incorrect, or at least not mathematically rigorous to speak of transformations of spacetime points, one should think of a change of frame by a Lorentz transformation instead, c.f. the answer by R. Ekman at: Lorentz transformations of fields evaluated at a point.

  • $\begingroup$ No, you are correct. Conserved quantities like energy and momentum arise from symmetries of the Lagrangian, like translation invariance, and translation is most certainly an operation that can be formulated rigorously, and transforms spacetime points. With a translation you parallel transport all the fields from the starting point and compare to the value they have at the end point, infinitesimally this is exactly (covariant) differentiation, so $\delta \phi = \epsilon^\nu \partial_\nu \phi$. $\endgroup$ – Robin Ekman Apr 12 '14 at 16:57

Well, it depends on what way you want it 'proved'. That is, in what context?

If you loo at the first chapter of Landau Vol. 1 it will give you some lovely derivations of the proof.

Energy is conserved due to the homogenity of time.

That is, because it should not matter if I measure the energy of some mechanical system today, or tomorrow say, or at any point in time, we have a 'Conservation Law'.

So if you know Lagrangian mechanics, we can say that the lagrangian

$$ L(q(t),\dot{q}(t),t) = T(\dot{q}(t)) - V(q(t)) $$

where $T(\dot{q}(t)) $ is the kinetic energy and $V(q(t)) $ is the potential energy, is invariant under transformations

$$ t \rightarrow t' = t + \epsilon $$

Then doing a bit of maths, which you should work through all the steps yourself, you will get an expression of the form

$$ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \dot{q} - L \right) = \frac{d}{dt} \left( (\dot{q}) \dot{q} - (\frac{1}{2}\dot{q}^2 - V) \right) = \frac{d}{dt} \left( \frac{1}{2}\dot{q}^2 + V \right) = 0 $$


$$ L = \frac{1}{2} \dot{q}^2 - V $$ say

Hence the term inside the brackets, which is our 'Energy' is conserved.

  • $\begingroup$ You can edit your own answers. $\endgroup$ – JamalS Apr 12 '14 at 16:17
  • $\begingroup$ What does 'homogenity of time' mean? $\endgroup$ – YoTengoUnLCD May 17 '16 at 3:20

If you want, you can pick a theory like e.g. Newton's mechanics and show, that these equation obey the principle of energy conservation.

But that's not how energy conservation is typically looked at. Energy conservation is(!) one of the basic principles theories are build on. Energy conservation is more fundamental than any single theory and is not derived from more basic principles.

  • $\begingroup$ If you pick General Relativity, you'll see that energy is not conserved globally. $\endgroup$ – jinawee Apr 12 '14 at 16:10
  • $\begingroup$ Fair enough, GR is a special case. But to my knowledge it's not that globally energy is not conserved. It is, however, not at all clear how to formulate the principle of energy conservation in several context. See here for a discussion by Baez: (math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html) $\endgroup$ – taupunkt Apr 12 '14 at 16:22

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