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I haven't understood this thing: Physics is invariant for CPT trasform...But the Heat or diffusive equation $\nabla^2 T=\partial_t T$ is not invariant for time reversal...but it's P invariant..So CPT simmetry would be violed... What I haven't undestood?

Thank you

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The heat equation is a macroscopic equation. It describes the flow of heat from hot objects to cold ones. Of course it can not be time-reversible, since the opposite movement never happens.

Well, I say 'of course' but you actually have stumbled on something important. As you say, the fundamental laws of nature should be CPT invariant, or at least we expect them to be. The reason the heat equation is not CPT invariant is that it is not a fundamental law, but a macroscopic law emerging from the microscopic laws governing the motions of elementary particles.

There is however a problem here, how does this time asymmetry arise from microscopic laws that are themselves time reversal invariant? The answer to that is given by statistical mechanics. While the microscopic laws are time-reversible (I'll focus on T, and leave CP aside), not all states are equally likely with respect to certain choices of the macroscopic variables. There are more configurations of particles corresponding to a room filled with air than with a room where all the air would be concentrated in one corner. It is this asymmetry that forms the basis of all explanations in statistical mechanics.

I hope that clears things up a bit.

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  • $\begingroup$ Ok! A last question now: the Schrodingher equation isn't invariant for time reversal, for the first derivative in t. But is not that a microscopical law? $\endgroup$ – Boy Simone Dec 2 '10 at 12:54
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    $\begingroup$ Actually, the Schrödinger equation is invariant. But you have to take the complex conjugate of $\psi$. Since $\psi^*$ and $\psi$ have the same probability distributions $|\psi|^2$, the physics remains the same. $\endgroup$ – Raskolnikov Dec 2 '10 at 12:58
  • $\begingroup$ Nice summary. It worth noting, however, that this is a deep enough topic that multi-hundred page books have been written on the matter. $\endgroup$ – dmckee♦ Dec 2 '10 at 19:33
  • $\begingroup$ @dmckee: Of course, I didn't mean to give an exhaustive explanation. In fact, I left my explanation open to many attacks on purpose. I hope that Boy will think further and come to these questions by himself. But a thorough answer would indeed need a thorough course in statistical mechanics. $\endgroup$ – Raskolnikov Dec 2 '10 at 22:43
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CPT theorem is not a theorem for all of physics but only for a quantum field theory (QFT). Also CPT invariance doesn't mean that QFT is necessarily invariant with respect to any of C, P and T (or PT, TC and CP, which is the same by CPT theorem) transform. Indeed, all of these symmetries are violated by weak interaction.

Second, even if the macroscopic laws were completely correct it wouldn't mean that they need to preserve microscopic laws. E.g. most of the microscopic physics is time symmetric (except for small violation by the weak interaction) but second law of thermodynamics (which is universally true for any macroscopic system just by means of logic and statistics) tells you that entropy has to increase with time. We can say that the huge number of particles breaks the microscopical time-symmetry.

Now, the heat equation essentially captures dynamics of this time asymmetry of the second law. It tells you that temperatures eventually even out and that is an irreversible process that increases entropy.

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  • $\begingroup$ thank you for the answer! And why, in your example, huge number of particles breaks the microscopical time-symmetry? Why don't macroscopic effects preserve microscopical invariance CPT of quantum-field theory? $\endgroup$ – Boy Simone Dec 2 '10 at 12:39
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    $\begingroup$ @Boy: that has to do with statistical mechanics. You should really ask this as a separate question because answer is not completely simple. But in short: any given macroscopic state (given e.g. by energy and pressure) of the system can be realized by many microscopic states. Now your answer boils down to basic questions in probability theory: the more microscopic states there are, the more likely the resulting macroscopic state is. So system is more likely to move to move from the less probable state to more probable state and not in the other way. $\endgroup$ – Marek Dec 2 '10 at 12:46
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Maybe it's a little bit late to talk about this topic but I think equation $\alpha \nabla^{2} T = \partial_{t} T$ is not even a correct macroscopic equation. I mean it violates the fact that information (i.e. temperature change) needs some time to be transferred across the domain and as a result the information speed (i.e. wave speed) could not be infinite. In fact the correct equation which is invariant under Lorentz transformation is: $\frac{1}{c_{s}^{2}} \partial^{2}_{t} T + \frac{1}{\alpha} \partial_{t} T = \nabla^{2} T$. Now it looks like a Langevin equation with a viscous term $\frac{1}{\alpha}$ and if this viscous term vanishes for a hypothetical highly conductive material (i.e. $\frac{1}{\alpha} = 0$), finally you will end up with this wave equation as: $\frac{1}{c_{s}^{2}} \partial^{2}_{t} T = \nabla^{2} T$, which is indeed time reversal under T transformation. Obviously, it will not be again time reversal because of presence of viscous terms which may lead to increase entropy. I hope it helps to make it more clear after such a long time.

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