- #31
shoehorn
- 420
- 2
Unfortunately, I find that I've lent my copy of Lanczos to someone and don't have another to hand. I could walk to the library to get one but, well, it's bloody freezing here at the moment. As a result, I'll try to quote the general line of thinking from memory and get back to you later with some references.
Let me begin by looking at a simpler case than the one the OP asked about. Suppose that we have a non-relativistic particle of mass m which moves in a potential V(x^i,t). The canonical action for this theory is
<br /> S[x^i(t),p_i(t)]<br /> = \int dt\, \left( p_i\frac{dx^i}{dt} - H(x^i,p_i,t) \right),<br />
where p_i is the momentum conjugate to x^i. The Hamiltonian is
<br /> H = \frac{p_ip^i}{2m} + V(x^i,t).<br />
Note that this form of the Hamiltonian makes the implicit assumption that we are dealing with Newtonian theory because of the form of the inner product p_ip^i=\delta^{ij}p_ip_j.
Now suppose that we introduce a parametrization of the path in phase space. Let me introduce some arbitrary parametrization of this path, and label it by \tau. Moreover, suppose that we look at a larger phase space given by
<br /> x^\alpha = (t,x^i),\,\,<br /> x^\alpha = x^\alpha(\tau),\,\,<br /> p_\alpha = p_\alpha(\tau).<br />
(By the way, my memory of Lanczos is that this is covered within the context of a discussion of Jacobi's principle and is explained in detail there.) We can then re-express the canonical action as
<br /> S[x^\alpha(\tau),p_\alpha(\tau)]<br /> = \int d\tau\, (p_i\dot{x}^i - H\dot{t}),<br />
where dots denote differentiation with respect to the parameter \tau. If one varies with respect to x^i(\tau) and p_i(\tau), one obtains the standard Hamiltonian equations of motion, while if one varies with respect to t(\tau) one obtains what (I believe) is widely called the "energy balance equation"
<br /> \dot{H} = \dot{t}\frac{\partial H}{\partial t}.<br />
However, we have still to vary with respect to p_0(\tau), the momentum conjugate to t(\tau). It's easy to do this: we find that p_0 = -H and since H=H(x^\alpha,p_\alpha) we can vary p_0 independently in the action. Thus, one obtains the constraint equation
<br /> \mathcal{H} = p_0 + H(x^\alpha,p_\alpha) = 0.<br />
So far, so bog standard. If we want to be able to vary all of the (x^\alpha,p_\alpha) freely, we just follow Dirac's method and add the constraint into the action together with a Lagrange multiplier, N. We find that
<br /> S[x^\alpha(\tau),p_\alpha(\tau);N(\tau)]<br /> = \int d\tau\, (p_\alpha\dot{x}^\alpha - N\mathcal{H}).<br />
If we vary N we obtain the constraint \mathcal{H}=0 again. If we vary p_0 we get
<br /> N = \frac{d(\textrm{absolute "Newtonian" time})}{d(\textrm{label time})}<br /> = \dot{t},<br />
an expression which assigns physical meaning to the Lagrange multiplier. Again, such occurrences are common all over physics (think about the status of the Lagrange multipliers in general relativity for example).
We can also recover the ordinary Hamiltonian equations of motion and an energy balance equation from the remaining variations. The form of the action is also seen to be invariant under arbitrary reparametrizations of the form
<br /> \tau\to \tau'(t)<br />
whenever
<br /> N \to N' = N\frac{d\tau}{d\tau'}<br />
and the (x^\alpha,p_\alpha) are invariant under this reparametrization. Then, and this is the central point, the Hamiltonian for a non-relativistic point particle of mass m can be taken as N\mathcal{H}. What's more, on a classical solution one has \mathcal{H}=0, the claim I made earlier in the thread.
All of this depends crucially on the theory being invariant under such (essentially arbitrary) reparametrizations. For the fifth or sixth time in this thread let me repeat it: this is, or at least should be, emphasised ad nauseum in a course on analytical mechanics. To me, it's right up there with the Hamilton-Jacobi equation in importance.
So far I've dealt only with the non-relativistic case. To see how the relativistic case works out, let's take the usual Lagrangian:
<br /> L = -m\sqrt{1-\dot{x}_i\dot{x}^i}.<br />
One can go through essentially the same procedure as in the non-relativistic case, recovering the idea that the Hamiltonian vanishes on-shell. The proof is left as an exercise for anyone who's interested.
Finally, in case you're not convinced of the ubiquity of this method, let me note that it pops up in loads of places. It happens, as a well known example, in bosonic string theory, where heavy use is made of world sheet reparametrization invariance in order to write the Hamiltonian as a quantity which vanishes on shell. Similarly, (vacuum) general relativity has an identically vanishing Hamiltonian of the form
<br /> H = \int d^3x\, (N\mathcal{H} + N^i\mathcal{H}_i) = 0.<br />
where (N,N^i) are the lapse function and shift vector, respectively, and \mathcal{H},\mathcal{H}^i are the Hamiltonian and momentum constraints, respectively.
Let me begin by looking at a simpler case than the one the OP asked about. Suppose that we have a non-relativistic particle of mass m which moves in a potential V(x^i,t). The canonical action for this theory is
<br /> S[x^i(t),p_i(t)]<br /> = \int dt\, \left( p_i\frac{dx^i}{dt} - H(x^i,p_i,t) \right),<br />
where p_i is the momentum conjugate to x^i. The Hamiltonian is
<br /> H = \frac{p_ip^i}{2m} + V(x^i,t).<br />
Note that this form of the Hamiltonian makes the implicit assumption that we are dealing with Newtonian theory because of the form of the inner product p_ip^i=\delta^{ij}p_ip_j.
Now suppose that we introduce a parametrization of the path in phase space. Let me introduce some arbitrary parametrization of this path, and label it by \tau. Moreover, suppose that we look at a larger phase space given by
<br /> x^\alpha = (t,x^i),\,\,<br /> x^\alpha = x^\alpha(\tau),\,\,<br /> p_\alpha = p_\alpha(\tau).<br />
(By the way, my memory of Lanczos is that this is covered within the context of a discussion of Jacobi's principle and is explained in detail there.) We can then re-express the canonical action as
<br /> S[x^\alpha(\tau),p_\alpha(\tau)]<br /> = \int d\tau\, (p_i\dot{x}^i - H\dot{t}),<br />
where dots denote differentiation with respect to the parameter \tau. If one varies with respect to x^i(\tau) and p_i(\tau), one obtains the standard Hamiltonian equations of motion, while if one varies with respect to t(\tau) one obtains what (I believe) is widely called the "energy balance equation"
<br /> \dot{H} = \dot{t}\frac{\partial H}{\partial t}.<br />
However, we have still to vary with respect to p_0(\tau), the momentum conjugate to t(\tau). It's easy to do this: we find that p_0 = -H and since H=H(x^\alpha,p_\alpha) we can vary p_0 independently in the action. Thus, one obtains the constraint equation
<br /> \mathcal{H} = p_0 + H(x^\alpha,p_\alpha) = 0.<br />
So far, so bog standard. If we want to be able to vary all of the (x^\alpha,p_\alpha) freely, we just follow Dirac's method and add the constraint into the action together with a Lagrange multiplier, N. We find that
<br /> S[x^\alpha(\tau),p_\alpha(\tau);N(\tau)]<br /> = \int d\tau\, (p_\alpha\dot{x}^\alpha - N\mathcal{H}).<br />
If we vary N we obtain the constraint \mathcal{H}=0 again. If we vary p_0 we get
<br /> N = \frac{d(\textrm{absolute "Newtonian" time})}{d(\textrm{label time})}<br /> = \dot{t},<br />
an expression which assigns physical meaning to the Lagrange multiplier. Again, such occurrences are common all over physics (think about the status of the Lagrange multipliers in general relativity for example).
We can also recover the ordinary Hamiltonian equations of motion and an energy balance equation from the remaining variations. The form of the action is also seen to be invariant under arbitrary reparametrizations of the form
<br /> \tau\to \tau'(t)<br />
whenever
<br /> N \to N' = N\frac{d\tau}{d\tau'}<br />
and the (x^\alpha,p_\alpha) are invariant under this reparametrization. Then, and this is the central point, the Hamiltonian for a non-relativistic point particle of mass m can be taken as N\mathcal{H}. What's more, on a classical solution one has \mathcal{H}=0, the claim I made earlier in the thread.
All of this depends crucially on the theory being invariant under such (essentially arbitrary) reparametrizations. For the fifth or sixth time in this thread let me repeat it: this is, or at least should be, emphasised ad nauseum in a course on analytical mechanics. To me, it's right up there with the Hamilton-Jacobi equation in importance.
So far I've dealt only with the non-relativistic case. To see how the relativistic case works out, let's take the usual Lagrangian:
<br /> L = -m\sqrt{1-\dot{x}_i\dot{x}^i}.<br />
One can go through essentially the same procedure as in the non-relativistic case, recovering the idea that the Hamiltonian vanishes on-shell. The proof is left as an exercise for anyone who's interested.
Finally, in case you're not convinced of the ubiquity of this method, let me note that it pops up in loads of places. It happens, as a well known example, in bosonic string theory, where heavy use is made of world sheet reparametrization invariance in order to write the Hamiltonian as a quantity which vanishes on shell. Similarly, (vacuum) general relativity has an identically vanishing Hamiltonian of the form
<br /> H = \int d^3x\, (N\mathcal{H} + N^i\mathcal{H}_i) = 0.<br />
where (N,N^i) are the lapse function and shift vector, respectively, and \mathcal{H},\mathcal{H}^i are the Hamiltonian and momentum constraints, respectively.
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