Ok, so I remember reading that every conservation law has a corresponding symmetry (i.e. conservation of momentum is translational symmetry, conservation of angular momentum is rotational symmetry).

Now conservation of energy is temporal symmetry (you can rewind the tape and it looks exactly the same, but in reverse--you don't get a different "movie" running the tape in reverse).

But I just saw an article here...


...where researchers found violations of time-reversal symmetry.

Now wouldn't that therefore mean that they found processes that create or destroy energy?? Does T-Symmetry violation mean that energy is not being conserved?


2 Answers 2


Conservation of energy follows from invariance under translation in time, not inversion. This symmetry states that no matter when you do your experiment, it will give the same results. All isolated systems obey this symmetry (and therefore conserve energy) and no violation of it has ever been detected. (Needless to say, it would be a huge event if it were.)

In classical physics, only continuous symmetries - that is, symmetries that can be continuously connected to the identity transformation - have a corresponding conservation law. Quantum physics does permit conservation laws for discrete symmetries but these laws are far harder to visualize.

An example of this is conservation of parity, $P$, which corresponds to invariance under inversion in space, and which gives the parity - even or odd - of wavefunctions. Temporal inversion, $T$, is even harder to turn into a physical quantity because it requires a full relativistic treatment in which time is a coordinate like space and not a parameter (as it is in non-relativistic quantum mechanics). A third discrete symmetry is charge conjugation, $C$, which exchanges particles for their antiparticles.

It turns out that any consistent field theory must be invariant under all three operations when taken together - i.e. under $CPT$. Thus violation of parity - an experiment and its mirror image behaving differently - is possible, for example, if it comes together with violation of $C$ - i.e. the mirror experiment behaves like the original one if it is made of antimatter -, as was discovered in the sixties.

Violations of $C$ and $P$ together have also been discovered in recent years, which means that in some situations violations of $T$ must occur. The recent $B$-meson experiments confirm this. Since the $T$ symmetry does not correspond to energy but to a far more abstract quantity (which is not conserved), this does not lead to a nonconservation of energy.

  • $\begingroup$ Isolated local systems, that is. Energy conservation seems not to apply to the universe as a whole, though. $\endgroup$
    – David Z
    Commented Nov 20, 2012 at 2:50
  • 2
    $\begingroup$ So what is the "far more abstract quantity" that T-Symmetry refers to? $\endgroup$
    – John
    Commented Nov 20, 2012 at 16:30
  • $\begingroup$ For a feel of the maths, try this WP section. $\endgroup$ Commented Nov 20, 2012 at 18:18
  • 1
    $\begingroup$ The "far more abstract quantity" is, in the context, time. Your confusion stems from the extreme degree of its abstraction, which he'd made a remark about earlier: Properly, an "amount" of time should be referred to as a "duration", but, as in the probable linguistic derivation of time's "flow" from the medieval use of water clocks, comparisons between time and other subjects of study in physics are usually made with terminology more appropriate to those other subjects. Even "duration" is related to material things that "wear out". The answer's as math-free as the question was. $\endgroup$
    – Edouard
    Commented Jan 26, 2020 at 20:09

Let us here just consider the classical case ($\hbar=0$).

I) Noether's theorem does not work for discrete symmetries like time reversal symmetry,

$$\tag{1} T: t ~\longrightarrow~ -t, $$

cf. e.g. this Phys.SE post.

II) Instead, energy conservation follows from time translation symmetry

$$\tag{2} t ~\longrightarrow~ t + a, \qquad a~\in~ \mathbb{R}, $$

which is a continuous symmetry.

III) However, if one has time reversal symmetry around every instant $t_0$ in time

$$\tag{3} t-t_0 ~\longrightarrow~ t_0- t, \qquad t_0~\in~ \mathbb{R}, $$

then it is not difficult to show that one also have translation symmetry (2), and therefore also energy conservation.

  • $\begingroup$ Disclaimer: I'm not a math guy. If you could provide a conceptual explanation, I'd very much appreciate it, though. $\endgroup$
    – John
    Commented Nov 19, 2012 at 22:04
  • $\begingroup$ @John: Conceptual explanation of which part? $\endgroup$
    – Qmechanic
    Commented Nov 19, 2012 at 22:20

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