I have studied high school physics but only have a pop-sci kind of understanding of ideas like Noether's theorem so go easy.

As far as I know, Noether's theorem simply states that any kind symmetry of a physical system is accompanied by a particular conserved quantity.

The normal example of this is that if we consider a system to be time symmetric (the laws that govern the system are the same at all points in time), then energy conservation emerges. There are, however, supposed to be examples of systems where energy is not conserved, implying that the laws that govern the system isn't symmetric.

My problem is with this idea that laws are not symmetric over time. What does this really mean? Does this mean that the laws literally change over time? Is it not a very basic assumption for the study of physics that the laws that govern the universe never change? If we found systems where laws changed, how could we ever study the laws?

I believe that my confusion here is down to a misunderstanding of what it means for a system to be time symmetric but maybe it's something else. Thanks for any answers.

It's a little more involved than you say because the object that possesses the symmetry is the action for the system. The equations that describe the motion, i.e. the laws of physics for this system are then derived from this action using the Euler-Lagrange equation.

But we can describe a very simple example. Suppose you're in your spaceship and you plan a flyby of some large object e.g. the Sun. You start far from the Sun with some velocity $v$, and the Sun's gravity accelerates you inwards with the usual Newtonian equation for gravity:

$$ F = \frac{GMm}{r^2} $$

You swing in towards the Sun, accelerating as you go, then as you move away from the Sun again its gravity slows you back down and you depart with the same velocity $v$ that you started with. Your kinetic energy is unchanged.

However suppose it turns out that Newton's constant $G$ is actually time dependent and is decreasing with time:

$$ F(t) = \frac{G(t) Mm}{r^2} $$

That means $G$, and therefore the force, is lower for your outward journey than your inward journey. You are decelerated less on your outward journey and you leave with more energy than you started. Energy has not been conserved, and it's the time dependence of $G$ that is responsible.

  • Not a good example, as the question was to provide an example where energy is not conserved without the laws of nature changing over time. There plenty of such examples starting with the expansion of the universe. – safesphere Aug 17 at 15:43

When we consider a physical system, the environment denotes everything else that is not part of the system.

While the system has dynamical variables, which we are trying to solve, the variables of the environment are assumed given.

It often happens that the system is not isolated, and external forces from the environment act on the system. If the environment is changing, theses forces typically acquire explicit time dependence.

Example: Let the physical system be a child on a swing. The environment may e.g. be a parent pushing the swing in a time-dependent fashion. The energy of the child needs not be conserved.

  • So energy is not conserved in an open system. This is not what the question was about. – safesphere Aug 17 at 15:45

Simplest and most common system not symmetrical due to time shift is simply any mechanical system with dissipation. For example two bricks moving one on another with friction. In such circumstances part of energy is dissipated to heat, which is not taken into consideration in purely idealistic mechanical description. System as a whole of course conserves energy, but its mechanical part ( described via classical mechanics, involving action, Lagrangian, Jacobi equations etc) doesn't contain part of description which have to be carried via thermodynamics. In this case Noether theorem does not work. There is no need for Quantum General Relativity to understand this example and in fact I would suggest to focus rather on two bricks system to understand this stuff.

  • This is incorrect. The Noether theorem, as well as the energy conservation law, applies to closed systems. Your example is not a closed system and is not applicable here. If you make your two-brick system closed, e.g. by treating heat as potential energy or as mechanical molecular movement, the Noether theorem would work just fine. – safesphere Aug 18 at 5:32
  • Of course, that even included in the post above - please read 3rd and 4th sentence above once more. But this is exactly the causes of the effect to appear. You cannot give example of closed physical systems, described by Lagrangian mechanics, non conserved energy. Can you? – kakaz Aug 18 at 8:24
  • I believe you can. I can think of 3 conditions required for this to happen: spacetime is curved, the system is not static, and the motion is not reversible. The FLRW metric should comply. Does it conserve energy? It is debatable, but the consensus is that it does not since energy lost to the redshift is not recoverable, because the motion cannot be reversed. Although the time-energy symmetry relation remains (Fourier conjugates) and energy remains a Noether charge, it just does not conserve. – safesphere Aug 18 at 14:50
  • So basically you think of system with time shift symmetry of an action, completely described by this action, and not having constant of motion related to this symmetry? In this case you are expecting to find counterexample of Noether theorem. It would be great news... – kakaz Aug 23 at 19:58
  • It took me a minute to see where the confusion is coming from, it is from my statement that, "the time-energy symmetry relation remains". This does not refer to the time translation symmetry, but to the symmetry of time and energy being Fourier conjugates whether or not the time translation symmetry exists. In other words, energy is a Noether charge related to time whether or not the energy conserves. For example, the energy of a photon is still reverse proportional to the time period of EM oscillations, even if the energy is lost to the redshift with no time translation symmetry in place. – safesphere Aug 23 at 20:18

There is an amazing example which shows the effect of Noether's theorem: The non-conservation of energy in General Relativity. It actually requires the understanding of this theory but it is so clear that it is very useful as case of demonstration. Noether's theorem says (among others) that if in a physical system a translation in time (each moment in time is equivalent to any other) does not change the system, then the energy in this system is conserved. It works for million of cases, but not for the gravitation in the view of Einstein's theory.

A gravitational field is no longer invariant under translations in time because the gravitational field is considered as curvature of space-time, so space and time are in general curved, in particular the curvature can change in time. So moments in time are then no longer really equivalent. So then Noether's theorem tells us that the energy is not necessarily conserved.

That might be a rather shocking result, but it can easily be demonstrated in another way: By a coordinate transformation the gravitational effect can be locally "transformed away" i.e. locally eliminated, so if in one coordinate system an energy budget (including the gravitational energy) of the system is set up, in the other system, where the gravitational effect is locally "transformed away", the gravitational energy has gone too (so no conservation of the total energy). This is an ultimate consequence of the equivalence principle.

  • While this is true, in almost all cases this effect is renormalizable by introducing the "gravitational potential energy" for any time reversible process. The only true effect of this non-conservation is when the process cannot be reversed in time, such a in the expansion of the universe. Even in this case energy remains the Noether charge, that is not conserved, because we cannot introduce the global potential energy of the universe due to the expansion not being reversible. – safesphere Aug 17 at 15:52

Yes, violating the uniformity of the laws in time (time translation invariance) indeed means that the laws are NOT the same at different times.

While it's common in physics to assume that the laws are constant in time, this need not be the case in principle. If the laws change in particular ways, we could still describe the laws and calculate their ramifications and hence do science.

John Rennie gave a good example of this situation - if Newton's universal law of gravity actually read $F = G(1+t) \frac{m_1 m2}{r^2}$, with $t$ the time since the Big Bang (say), then we will have a very-specific law of physics and we could determine its consequences and test them against experiments and so on. We could do physics, even though the law is not time-invariant.

Now, things get a bit complicated in General Relativity. In GR what "time" is is a bit subtle so you end up with a theory whose basic laws do not explicitly depend on time (like the $t$ above) but that still allows for the effective-laws to change in time when space is expanding. The result is a bit of a mess as to whether energy is or is not conserved. It kinda depends on what you define as 'energy', and, well, it's a mess.

My problem is with this idea that laws are not symmetric over time. What does this really mean?

It is supposed to mean that Lagrangian function, which encodes interactions between different degrees of freedom of the system and the environment, does not depend explicitly on time. That is often the case and one can view it as thanks to the fact that basic physical laws of motion and interaction of bodies do not change in time. For example, a Lagrangian of a body orbiting Earth can be written as

$$ L = \frac{1}{2}m(v_x^2 +v_y^2 +v_z^2 ) + \frac{GMm}{\sqrt{x^2+y^2+z^2}}. $$ The fact that gravity does not change in time implies that $G$ does not change in time either; it is constant, the same number for any time we use the formula. If gravity did change in time, the above Lagrangian would not work. Another one, where some part depends on time could work.

But there may be other reason for time to appear in the Lagrangian.

For example, particle acted upon by an EM wave can be modeled by a Lagrangian of type

$$ L(x,\dot x) = \frac{1}{2}m\dot x^2 + q x E_0\sin(\Omega t). $$

There is a time-dependent term due to harmonic EM wave, because the wave is given as a process in time. But nothing about physical laws of EM theory changing in time is implied.

  • It is simple not true. Symmetric over time, means exactly what is said: when shifting time, action functional doesn't change. It is far from saying that Lagrangian doesn't depend on time! Lagrangian independent on time is just particular example, but it can be, in general. Especially as action is integral over time it is possible that boundary terms cancel, even if Lagrangian explicitly contains function of t. Think of it as continuous group of symmetry, just like continuous translation symmetry, but just in t. – kakaz Aug 25 at 10:31
  • kakaz, perhaps you're right. Do you have some example where Lagrangian depends non-trivially on time (in a way that is not possible to eliminate by subtracting a function of time and coordinates), but still action does not depend on time? – Ján Lalinský Aug 25 at 19:10
  • It is not easy. First, if Lagrangian doesn't depend explicitly on time, the same is true for equations of motions. Such type of equations are called autonomous. So if we're looking for example, it should be nonautonomous system of pde. Second, it is hard to say if conserved quantities for such systems, can be called energy or momentums, however they may be related to various lie groups of symmetry which for autonomous systems gives us such invariants. In other words we are talking about bold generalisation. – kakaz Aug 25 at 19:55
  • Explicit example of such system, and invariant, you may find at the end of this paper: fismat.unizar.es/~jfc/pdfs/pub62.pdf As you may see, and probably expected, it is hard to read.. Maybe other important remark here is that if we replace oryginal, time independent Lagrangian £, by £' = £ + π where π is complete time derivative of some function of (p,q,t), then action integral remains not changed, so probably by smart trick, you can produce time dependent Lagrangian which in fact would be kind of fake example... – kakaz Aug 25 at 19:55

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