Hamiltonian for a massless particle - formal definition of energy

Given a Lagrangian, is possible to calculate momenta and from them the Hamiltonian, if the system is regular enough. Today, I have realized that the Lagrangian of a massless particle in a gravitational field is singular, and described by a constraint Hamiltonian. Here is my problem: given this Lagrangian, the Hamiltonian always vanishes; if it's always zero, how is it possible to talk of a "energy" associated with a massless particle?

• The energy of a massless particle is $E = pc$, what has the Hamiltonian to do with this? The Hamiltonian is not always the "energy", especially not when the system is constrained or time-reparametrization invariant. – ACuriousMind Dec 24 '16 at 16:26
• I agree with you, anyway I was looking for a formal definition of energy in the context of the Hamiltonian theory. For a free non relativistic system the answer is simple: the energy is the Hamiltonian, but in the relativistic case usually there is no Hamiltonian, if so where does the definition of energy come from? – Yildiz Dec 24 '16 at 16:45
• The energy is the Noether charge associated to translations in the time variable (which will usually be a phase space variable and not the evolution paramter (which is proper time) in the relativistic setting). – ACuriousMind Dec 24 '16 at 16:48
• I knew it, and here is my doubt: in the context of general relativity what do you mean by time variable? If it is the x-zero coordinate the Lagrangian is not invariant, so is it the evolution parameter, normally called lambda? – Yildiz Dec 24 '16 at 16:53
• @Yildiz As he said, the energy corresponds to the Noether charge associated with time translations. If the theory is invariant under Lorentz transformations, then you can extract the energy from the stress-energy tensor $T^{\mu\nu}$ which can be computed by taking a variation of the Lagrangian with respect to the metric, modulo factors of metric determinants and constants. However, when it comes to a system that is constrained or time-reparametrization invariant, then there are other complications as ACuriousMind pointed out. – JamalS Dec 24 '16 at 17:11

2. In the context of e.g. the Minkowski metric or the FLRW metric $$ds^2~=~\sum_{\mu,\nu=0}^3g_{(4)\mu\nu}\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}~=~-\mathrm{d}x^0\odot \mathrm{d}x^0+a(x^0)^2\sum_{i,j=1}^3g_{(3)ij}\mathrm{d}x^i\odot \mathrm{d}x^j ,\tag{1}$$ it is possible to make the Hamiltonian $H$ for a massless point particle equal to the total energy $$c|{\bf p}|~:=~c\sqrt{\sum_{i,j=1}^3 p_i g^{(4)ij}p_j}~=~\frac{c}{a(x^0)}\sqrt{\sum_{i,j=1}^3 p_i g^{(3)ij}p_j} \tag{2}$$ by choosing the static gauge condition $x^0=c\tau$, where $\tau$ is the world-line parameter (which is not the proper time). For details, see e.g. this Phys.SE post. Note that the total energy (2) is not conserved in the FLRW case due to the scale factor $a(x^0)$ with explicit time dependence.
• The relation $x^{0}=ct$ is a definition. It is not the static gauge condition $x^0=c\tau$, where $\tau$ is the world-line parameter (which is not the proper time). – Qmechanic Dec 28 '16 at 21:43