Is there an easy (aka intuitive) way to understand that the conserved quantity from time translation symmetry is just what we call energy?

In other words, we use two definitions of energy. One is with Noethers theorem, and I've been told this is the fundamental one. The other is what is you learn in school and is mentioned in the examples below. The question is how to connect this two definitions.


  • I can lift weights, so they get more energy.
  • I can boil water for tee, so it gets more energy.
  • I can burn $CaO$ with carbon, so I get carbide, that has more energy.
  • ..

If I define energy as conserved quantity, how do I arrive at my examples..?

(Well, "easy/intuitive" is in the eye of the beholder. Thank you nevertheless.)

Side question: We have energy conservation in thermodynamics. I have never seen a Lagrangian formulation of thermodynamics. Can I only hope for an answer of my main question in theories that have such a formulation?


Answer to your side question:

Energy conservation indeed follows from symmetry properties of Lagrangian. The first law of thermodynamics is a little more than energy conservation. It says that although the heat change $\delta Q$ in the system and the work done $\delta W$ on it are inexact differentials, their sum $dU$ is an exact differential. That is, although the total heat change and work done depend on the path taken by a thermodynamic transformation, their sum is the same for all paths. Their sum $U$ is therefore a state function, called the system's internal energy.

The first law of thermodynamics guarantees the existence of an internal energy function.


Falko, if I understand your question right, then this would be the following: how is it that getting some of that particular conserved quantity (called "energy") transferred into a (mechanical, chemical etc.) system may have such dramatic consequences as a phase transition (boiling water) or chemical bond making and breaking.

Intuitively speaking: how can you realize that it is exactly the Noether-current associated with time translation invariance, the one quantity necessary to boil your water?

I would try to attack the problem by pointing out that it comes to your (arbitrary) partitioning of the system under consideration, and whether you apply the symmetry (i.e. conservation law) to the entire system or to an arbitrarily out singled part of it. Let us take as an example the boiling of water for tea during an airliner flight.

As long as you consider the entire kitchen in the airplane, with the heater and its batteries, it is time translation invariant, the "energy" Noether current is conserved and nothing happens (Here we considered the entire kitchen/plane as a system A). If you define the water in the boiler as a system B in its own, then you break time translation invariance for B once you want to get a pot of tea (and you switch the button at a given time, so the hamiltonian of B becomes time-dependent!) This transfers some of the Noether conserved current (energy) between the battery and your boiler and effectively boils your water.

Now let us consider as system C a second plane flying in line of sight of your plane A. Pilot of C can see plane A but he neither is aware that you have left your seat and have put up a cup of tea, nor he sees any flashes or auras surrounding A as a result of yours having done this. For the pilot of C, system A is time translation invariant and its energy is conserved. Noether's theorem is thus corroborating his own empirical observations.

In summary: your breaking or not breaking of the symmetries of a system (and of the transfer of associated conserved current) depends very much on how accurately you delineate or circumscribe (within a fictitious separating membrane) your system under consideration. This is very similar to the method in Thermodynamics.

  • $\begingroup$ The question is "if energy is conserved, why can I put energy into things". Naively, you can only move energy from one place to another, so how can you heat water, since this requires taking energy out of something else. It's a thermodynamic question. $\endgroup$ – Ron Maimon Aug 25 '12 at 10:28
  • $\begingroup$ @ Maimon: conservation of Noether currents means indeed that you can only move the conserved charge from one place to the other. If it is a thermodynamic or a mechanical question reflects only points of view. These only have to be mutually compatible. $\endgroup$ – Lupercus Aug 25 '12 at 10:54
  • $\begingroup$ "how can you realize that it is exactly the Noether-current associated with time translation invariance, the one quantity necessary to boil your water?" - yes that was what I intended to ask. If you see a way to get my question more to the point, feel free to edit - I'd like to learn where I misled Ron. (This really irritates me.) $\endgroup$ – Falko Aug 25 '12 at 19:03

I'm not sure if this will answer your question but I thought I'd give it a shot nonetheless!

So I take it you're familiar with Noether's Theorem: If there exists a symmetry of the mechanical system that leaves the mechanical properties unchanged, then Noether's Theorem says that there also exists a conserved quantity of any such system.

Okay, let's look at an example! (I'm also going to presume you've encountered Langrangian and Hamiltonian mechanics). Consider the Hamiltonian of a system of N-particles, all interacting:

$H = \sum_{i=1}^N \frac {1} {2m_i}p_i^2 + U(q_1,...,q_N)$.

Here $t$ does not appear explicitly in $H$

$H = H(q_i,p_i) \Longrightarrow H(t+\Delta t) = H(t)$

So now the time translation $t+\Delta t$ is related to a conservative quantity, which is $H$. To check if the quantity is conserved we take the time derivative of $H$:

$\frac {dH} {dt} = \frac {\partial H} {\partial q_i}\dot{q_i} + \frac {\partial H} {\partial p_i}\dot{p_i} + \frac {\partial H} {\partial t}$.

Because there is no explicit dependence on $t$, $\frac {\partial H} {\partial t}=0$. So we are left with:

$\frac {dH} {dt} = \frac {\partial H} {\partial q_i}\frac {\partial H} {\partial p_i} - \frac {\partial H} {\partial p_i}\frac {\partial H} {\partial q_i}$ $=0$

The Hamiltonian is the conserved quantity associated with the time translation so therefore we can say that the energy conservation and the invariance of the system under time translation are linked. Hope that helps!

  • 5
    $\begingroup$ This is not really a correct demonstration--- the fact that H is time independent doesn't mean that H is the conserved quantity corresponding to time translation. For this, you need to show that H acting on the phase space produces an infinitesimal time translation. But this is the definition of H, so perhaps you can be forgiven for a vacuous demonstration. The Hamiltonian formalism assumes that H generates time translations. $\endgroup$ – Ron Maimon Aug 25 '12 at 10:17

Your examples involve an external agent which produces entropy in order to do work. hen you put heat in water, you are moving energy from a state where it is low entropy (like in the chemical bonds of petroleum) into a state where it is high entropy (hot water). This is a separate question from energy conservation and Noether's theorem. It's thermodynamics.

But given a Lagrangian that is time translation invariant

$$ S = \int_0^T L(q,\dot{q}) dt $$

If you shift the path in time infinitesimally by $\epsilon$, you get an action which is the same, but the endpoints are translated. The variation in the action is

$$ \delta S = - \epsilon( H(T)-H(0) ) = - \epsilon (p\dot{q} - L)|_0^T $$

You can see this from doing the variation. The L term comes from moving the endpoints, the $p\dot{q}$ comes from varying L inside the integral: the change in q at any time t is $\epsilon \dot{q}$, while the change in $\dot{q}$ is $\epsilon \ddot{q}$. So the change in the action is

$$ \delta S = \epsilon \int {\partial L \over \partial q} \dot{q} - {\partial L \over \partial \dot{q}} \ddot q = - \epsilon \int {d\over dt} (p \dot q) $$

The time translation invariance tells you that H is conserved.

  • $\begingroup$ "When you put heat in water, you are moving energy .." - yes, and I want to understand why this energy is connected to Noethers theorem. $\endgroup$ – Falko Aug 25 '12 at 18:53
  • $\begingroup$ I've ever only seen formulae like yours, containing abstract generalised coordinates (something I once learned for point masses, but never understod the way to generalise it to my example 2 and 3). I'm just still not educated enough to see why the quantity I move around to have my steam engine going is just the same as the one you describe here. Thats what I want to understand. Thank you for the response, Ron. $\endgroup$ – Falko Aug 25 '12 at 18:59
  • $\begingroup$ @Falko: The reason is that energy is conserved, and it is positive, so it partitions equally among different modes. So when you do macroscopic work, and increase the energy of the piston, the macroscopic kinetic energy of the piston distributes itself into the kinetic energy of the atoms. You don't find it from Noether, but from thermodynamics. $\endgroup$ – Ron Maimon Aug 25 '12 at 20:32

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