Changing sign of Lagrangian & Hamiltonian: how to interpret energies then?

Question

In Lagrangian mechanics, it is possible to multiply the Lagrangian by a constant $a$. Let's assume I take $a=-1$.

Then, the Hamiltonian will have its sign changed as well. And it will represent the good dynamic as well.

What I don't understand is the possible energy interpretation we can have of the Hamiltonian. If $H$ and $-H$ are physically equivalent, how to know what is the lowest energy of a system?

How to make connection with statistical mechanics for example (canonical probabilities).

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Let's say everything is classical, consider for example a classical Ising Hamiltonian: $H=J S_1 S_2$ with $J>0$. In one description the fundamental would be $++$ or $--$, in the other, $+-$ or $-+$. The physical description are thus different in term of statistical physics.

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I think that @knzhou is indeed the good one but something is still perturbing me. If someone gives you those two equivalent hamiltonian: $H=J S_1 S_2$ and $H'=-J S_1 S_2$

You want to know what is the occupied state at $0K$. Conceptually would you agree it is absolutely impossible to know what it will be, you could state ($++$,$--$), or ($+-$ or $-+$) because the formalism is symmetric via $H \rightarrow -H$.

Then to know what in practice you will see you must do an experiment. This experiment will tell you which sign to choose so that it matches statistical physics description. It is impossible to know theoretically what it will be.

If no experiment are needed it necesserally mean that I need an extra mathematical axiom in addition to hamiltonian formalism to do this mapping. If so, which one ?

Answer 1

Let's say that L' = -L then indeed one can see that H' = -H, but L' doesn't longer equals to T - V but to -T + V which is T' - V'. So you can see that although you get the same dynamical solutions your representation of the question is different and in this representation H' = E' = T' + V' = -(T + V) = -E = -H.

So in other words H' is just another representation of the same problem with the same solutions, we don't care that the energy is not the same because we care only about the difference in the energy which is the same (zero in this case).

Note that we assumed that H = T + V, which is not trivial as one sometimes think. (See discussion in Goldstein 339).

Answer 2

Let's call your weird definition the schmenergy $\tilde{E} = -E$. Everything about physics works the same, as long as you are consistent with flipping the signs everywhere.

• The temperature of an object is defined by $T = dE/dS$. Note that $\tilde{E}$ appears nowhere in this definition.
• If you want an alternative definition that involves $\tilde{E}$, you could define the schemperature $\tilde{T} = d \tilde{E} / dS$. Then you could have a whole theory of schmermodynamics based on $\tilde{T}$.
• The crucial point is that you need to focus on unambiguous physical predictions, and not those tied to your weird choice of words. For example, is it "paradoxical" that normal objects will have negative schemperatures? No, because schemperature is a different thing from temperature, so we shouldn't expect it to have the same intuitive properties.
• As an example of an unambiguous physical statement, you can carry on your derivations in schmermodynamics, in exactly the same way as a standard textbook but with extra minus signs, to show that as the schemperature approaches zero, the schmenergy approaches a maximum. But this is exactly the same thing as the energy approaching a minimum, so your results aren't any different from the standard ones.

More generally, our intuition for energy applies to quantities that are of the form $$E = mv^2/2 + \text{other stuff}.$$ You are free to define an alternative quantity of the form $$\tilde{E} = -mv^2/2 + \text{other stuff},$$ there's nothing mathematically illegal about it. It just won't obey the same intuition. I mean, you are also free to call the number $2$ "three", but then you shouldn't be surprised by the fact that $\text{"three"} + \text{"three"} = 4$.