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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?

Qmechanic
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Yildiz
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    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
  • You are right, I said wrong things in my last message: now everything is clear, thanks :) – Yildiz Dec 24 '16 at 17:12
  • Today I have thought to a particular case: in the Freedman-Robertson metric the Lagrangian for a massless particle is not invariant for time translation because of the scale factor, so how is it possible to talk about the energy of a photon if there is no Noether current? – Yildiz Dec 25 '16 at 13:51
  • It seems to me that in the context of general relativity is not possible to talk in general of a "energy conservation" because metric usually doesn't allow for coordinate translation, am I right? – Yildiz Dec 25 '16 at 13:53

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Comments to the post (v3):

  1. The notion of Hamiltonian and the notion of total energy do not need to coincide, cf. this Phys.SE post and links therein. Total energy is the Noether charge associated with time translations. In relativity theory, the notion of time (and thereby the notion of energy) depend on the chosen coordinate system. In particular the notion total energy (unlike the notion of rest energy) is not an invariant. See also above comments by ACuriousMind & JamalS.

  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.

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Qmechanic
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  • Thanks Qmechanic, but I can' t use the static gauge in my case. If I try to use it I also define the factor scale, and this is inconsistent. – Yildiz Dec 27 '16 at 13:47
  • Could you show what you mean by using formulas? – Qmechanic Dec 27 '16 at 14:17
  • The Lagrangian is a Lagrange parameter times the constraint with a flat FRW metric. If you calculate the Euler-Lagrange equations for the dynamics of the time coordinate you' ll find a simple relation: the derivative of the time coordinate respect to the evolution parameter is equal to the inverse of the factor scale times a constant. At this point if you set the time coordinate t as the free parameter, you will fix the factor scale to a constant: this is my problem. – Yildiz Dec 27 '16 at 17:58
  • Forgive if can't post formulas, but I don' t know how to do it: anyway I gave my best in order to explain the problem. – Yildiz Dec 27 '16 at 17:59
  • You have to learn how to type formulas. It is really quite simple. Click the 'edit' button on other posts to reverse-engineer how it is done (without actually making an edit). Or see here and links therein. – Qmechanic Dec 27 '16 at 21:41
  • The Lagrangian is $L = g_{\mu\nu}(x)\dot{x}^{\mu}\dot{x}^{\nu}\lambda$ where $\lambda$ is a Lagrange multipler and $g_{\mu\nu}$ is the FRW metric with ${a}$ as a factor scale. The dynamics in $x^{0}$ is given by $\partial x^{0}/\partial {t}= 1/ {a}$ and as you can see, If I choose the static gauge $x^{0}=ct$, $a$ becomes a constant. Does it mean that I am using a wrong Lagrangian? – Yildiz Dec 28 '16 at 13:10
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    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
  • It's not the static gauge and you are right, anyway if you set $x^{0}=t$, $a$ will be uniquely defined and this pose a conflict with reparameterization invariance, do you agree with me? – Yildiz Dec 28 '16 at 23:09
  • Prior to application of EFE (i.e. Friedmann equations), the scale factor $a$ is in principle arbitrary. – Qmechanic Dec 29 '16 at 10:48
  • Ok, but I would like to couple this Lagrangian with the Einstein-Hilbert one, and in this case you can' t say that the factor scale is arbitrary. – Yildiz Dec 29 '16 at 11:14