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Hamiltonian for a free particle (Special relativity)

  1. Oct 8, 2013 #1
    I've seen that the lagrangian for a relativistic free particle is

    [tex] -mc \sqrt{\eta_{\mu\nu} \dot{x^{\mu}}{\dot{x\nu}} [/tex] but when I construct the hamiltonian as

    [tex] p_{\mu} \dot{x^{\mu}} - L [/tex]

    I seem to get zero. Im not really sure what I'm doing wrong. I find that if in the first term of the hamiltonian, I only sum over spatial indices, I end up with γmc^2 which is what I would expect.
     
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  3. Oct 8, 2013 #2
    Actually, you forgot to square the speed of light

    [itex]L = - m c^2 \sqrt {\frac{\eta_{\mu\nu} \dot{x^{\mu}}{\dot{x^{\nu}}}}{c^2}}[/itex]
     
    Last edited: Oct 8, 2013
  4. Oct 8, 2013 #3
    Thanks for the reply, I see that but couldn't you just pull a factor of 1/c out of square root and get what I wrote? Also, can you show how to construct the hamiltonian from here?
     
  5. Oct 8, 2013 #4
    Note that the dot operation in the Lagrangian doesn't mean the same thing as the dot operation in the construction of the Hamiltonian. in the Lagrangian it means derivative with respect to proper time while in the second equation it means derivative with respect to coordinate time. These are not the same thing within relativity. Try rewriting the lagrangian as

    [itex]L = - m c^2 \sqrt {1 - \frac{v^2}{c^2}}[/itex]
     
    Last edited: Oct 8, 2013
  6. Oct 8, 2013 #5
    Forget what I said on my previous post about the dot operation. What I said is true in a different context, but not here because the Lagrangian isn't a Lorentz invariant. The dot represents derivative with respect to coordinate time in both equations here
     
  7. Oct 8, 2013 #6
    Your expression for the Hamiltonian isn't correct. It should be
    [itex]H= \Sigma_i p_i \dot x^i - L[/itex] ,
    where the index i runs over spacial coordinates alone, not the time coordinate.
     
  8. Oct 8, 2013 #7

    PeterDonis

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    As dauto pointed out, that is in fact what you're supposed to do. But that does raise an obvious question: why?

    The answer is that, in order to derive the Hamiltonian at all, we have to pick a reference frame; in other words, we have to pick a particular split between space and time. We then treat the space coordinates as the generalized coordinates, but treat time (the coordinate time in our chosen frame) as a parameter, the parameter with respect to which we take derivatives.

    You implicitly did this when you wrote the Lagrangian using ##\dot{x}##; the dot is a derivative with respect to coordinate time, but that means coordinate time can't be one of the generalized coordinates. And *that* means that, when we derive the Hamiltonian, we don't include coordinate time in the sum ##p_i \dot{x}^i##; we only include the generalized coordinates, which here are the spatial coordinates.
     
  9. Oct 8, 2013 #8

    PeterDonis

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    Of course, this statement now raises the question, why is the ##\dot{x^0} = \dot{t} = \partial t / \partial t = 1## term included in the Lagrangian to begin with? One way of looking at this is to see what the Lagrangian looks like in the non-relativistic approximation:

    $$
    L = - m c^2 \sqrt{ 1 - \frac{v^2}{c^2} } \approx - m c^2 \left( 1 - \frac{1}{2} \frac{v^2}{c^2} \right) = \frac{1}{2} m v^2 - m c^2
    $$

    In other words, in the non-relativistic approximation, the rest energy ##m c^2## acts like a potential energy. We normally gloss over that by effectively subtracting it out, which amounts to adjusting the zero point of energy. But that doesn't mean it isn't there; it just means we can ignore it in most non-relativistic situations.

    Relativistically, though, we can't leave out the rest energy in the Lagrangian (which is what leaving out the ##\dot{x^0}## term would amount to), because if we did, the quantity left under the square root would be negative, and the Lagrangian is supposed to be a real number. Physically, what this means is that in relativistic situations, we can no longer treat the rest energy as something constant that we can just ignore by adjusting the zero point of energy. It's kind of hard to see this in the case of a free particle, because free particles can't interact with anything and so they can't change their rest energy. But relativistic particles in general can undergo interactions that change them into particles with different rest mass--they can even be converted into particles like photons with *zero* rest mass. The relativistic Lagrangian forces you to acknowledge that by taking rest energy into account from the start.
     
  10. Oct 8, 2013 #9

    dextercioby

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    Of course you get a 0 Hamiltonian, since you're dealing with a constrained system. A way to get simpler calculations is to put an einbein there.
     
  11. Oct 8, 2013 #10

    rubi

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    That's not the exact reason for why the Hamiltonian is 0. There can also be constrained systems that have a non-zero Hamiltonian.

    The reason for why the Hamiltonian is zero is the invariance of the action under reparametrization of the proper time. One can show that every system that is invariant under this kind of reparametrization has a vanishing Hamiltonian.

    It's true however, that we're dealing with a constrained system here. The constraint you're getting is ##p_\mu p^\mu = m^2##. (Quantization of this leads directly to the Klein-Gordon equation.)

    You can remove the reparametrization invariance by setting ##\tau = x^0##. You will get a non-zero Hamiltonian then.
     
  12. Oct 8, 2013 #11

    dextercioby

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    I see your point, what I wrote is a wrong cause-effect statement. I stress for the OP the most important part of your comment:

    The reason for why the Hamiltonian is zero is the invariance of the action under reparametrization of the proper time. One can show that every system that is invariant under this kind of reparametrization has a vanishing Hamiltonian.
     
  13. Oct 8, 2013 #12

    PeterDonis

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    Doesn't this correspond to doing the computation in the rest frame of the particle and setting the zero point of energy to the particle's rest energy?

    And doesn't this correspond to doing the computation in an arbitrary frame, in which the particle does not have to be at rest, and *not* setting the zero point of energy to the particle's rest energy?
     
  14. Oct 8, 2013 #13

    rubi

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    I don't think so. The action is Lorentz-invariant, so ##x^\mu## could be coordinates in an arbitrary inertial frame. If you write ##L=-m c \sqrt{-g_{\mu\nu}(x)\dot x^\mu \dot x^\nu}## instead, ##x^\mu## could even be coordinates for a completely arbitrary frame and still the Hamiltonian would be zero.

    The parameter ##\tau## is actually not related to a choice of frame. It only parametrizes the points on the world line of the particle. It's really a completely unphysical parameter (it doesn't even need to be the proper time; that's just an arbitrary choice that is usually made. Setting ##\tau = f(\sigma)## would yield the same equations of motion. Using ##\tau = x^0## is also an arbitrary choice). Reparametrizing ##\tau## doesn't change the coordinate system you're in. You have already made this choice when you wrote down ##x^\mu## in the first place. The coordinate-independent form of the action would look something like ##S=-m c \int_c \mathrm d s## instead.
     
  15. Oct 8, 2013 #14

    PeterDonis

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    Yes, I see that. Repameterizing amounts to changing ##x^{\mu}## and also ##g_{\mu \nu}##, in such a way that the action remains invariant.

    If you include all four coordinates, yes. But don't you have to pick a frame for the Hamiltonian to be meaningful? In other words, the Lagrangian can be written in coordinate-independent form (as you do later in your post), but can the Hamiltonian? The Hamiltonian is supposed to describe "time evolution", but that requires picking a particular split between space and time, correct? (Essentially, that would mean that in order to derive the Hamiltonian, one would need to set the parameter ##\tau = x^0 = t## where ##t## is the coordinate time in some particular frame.)
     
  16. Oct 9, 2013 #15

    rubi

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    There is a coordinate-free formulation of Hamiltonian mechanics. In general, a Hamiltonian system is given by a Hamilton function on a symplectic manifold ##(M,\omega)##, that is a map ##H:M\rightarrow\mathbb R##. But that kind of math is overkill for this situation, so it's better to stay with the coordinate formulation here.

    The formulation doesn't require "time" to be physical time as measured by clocks. It's a purely mathematical framework after all. The purpose of the "time" variable is to parametrize the solutions. It needn't be an observable quantity. In our case, the "time" variable is the ##\tau##.

    The space/time split is needed for field theories. However, a relativistic particle isn't a field theory.
     
  17. Oct 9, 2013 #16

    PeterDonis

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    Ok, I've seen references to this before, but it's been a while. :redface:

    I see that this can work mathematically, but physically, isn't the Hamiltonian, as an operator, supposed to govern time evolution? (More precisely, the operator ##\exp ( i H t )## governs time evolution.) Or does saying that amount to choosing a particular parametrization (the one you refer to as setting ##\tau = x^0##) among many possible ones?

    In the canonical formalism, yes, I agree, but I'm not sure this is true of the path integral formalism.

    Agreed.
     
  18. Oct 10, 2013 #17

    rubi

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    If the Hamiltonian is non-zero, then it generates time-evolution. But that needn't be the case. In the case of the free relativistic particle, you get a constraint equation ##p_\mu p^\mu=m^2## instead. If you quantize this, you get the Klein-Gordon equation, which isn't a Schrödinger type equation and thus doesn't have solutions of the form ##\mathrm e^{i t H}\Psi(0)##. It's rather a restriction on the set of admissible states in the physical Hilbert space. (This is also the root of many misconceptions about the KG equation.) The same thing happens in the case of general relativity, by the way.

    I was talking about the classical Hamilton formalism. If you do path integrals, you usually don't need a split, that's true.
     
  19. Oct 10, 2013 #18

    WannabeNewton

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    Well ##| \psi,t \rangle ## is the trajectory of an initial state ##| \psi,0 \rangle## in an appropriate Hilbert space much like ##\gamma(\tau)## is the "trajectory" in space-time of a time-like particle so in the same way ##\tau## is just a choice of parametrization of said "trajectory", ##t## is just a parametrization of the trajectory ##| \psi,t \rangle ## (I put "trajectory" in quotes for ##\gamma(\tau)## because of course it isn't a trajectory in the literal sense since time-like particles don't move in space-time). There needn't be any operational definitions attached. The same goes for the global "time" function one chooses when working with the Hamiltonian formulation of classical GR on globally hyperbolic space-times.
     
  20. Oct 10, 2013 #19

    PeterDonis

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    So basically, it comes down to whether we want to view the system as a particle with a "location in space" that "evolves in time", or as a worldline in spacetime that doesn't change. The latter view means we treat all four spacetime coordinates as generalized coordinates, in which case the Hamiltonian vanishes; the former view means we treat the time coordinate as a parameter and only treat the space coordinates as generalized coordinates, in which case we get a non-vanishing Hamiltonian. Both views appear to me to be valid (although the former view requires choosing a space-time split while the latter does not), though one may be more useful than another for a particular problem.

    Yes, I understand this. Deriving the Schrodinger equation from the K-G equation requires making a non-relativistic approximation, which amounts to choosing a space-time split and assuming that the second time derivative of the field can be ignored.

    I assume this refers to the Wheeler-DeWitt equation?
     
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