Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Einstein's Hamiltonian?

  1. Nov 30, 2015 #1
    Hello! I have recently bought the book The Principle of Relativity by Einstein (Along with Minkowski, Lorentz and Weyl). This book is simply a collection of papers published by Einstein (along with the other three scientists mentioned) concerning the development of Special and General Relativity. To the best of my knowledge it is in chronological order. Now, don't worry, I am not attempting to study the Theories of Relativity from the original papers published on them. I have already self-studied SR and GR by watching the excellent video lectures on YouTube by Leonard Susskind, and reading Sean Carroll's free notes on GR (Which have been published into a graduate textbook that is unfortunately too expensive for me at the moment). Now to my question.

    In the paper by Einstein titled The Foundation of the General Theory of Relativity. Section 15 (page 145 in the book), he mentions the Hamiltonian Function for the Gravitational Field. Which he says, "To show that the field equations correspond to the laws of momentum and energy, it is most convenient to write them in the following Hamiltonian form..." He then goes on to define the Hamiltonian for a gravitational field in the absence of matter as ##H=g^{\mu\nu}{\Gamma}^{\alpha}_{\mu\beta}{\Gamma}^{\beta}_{\nu\alpha}## along with the convention that ##(-g)^{1/2}=1## where ##g## is the determinant of the metric. To the best of my knowledge, the modern way to derive the Field Equations is to vary the action of the gravitational field, having the Lagrangian density for the gravitational field be ##R## the Ricci Scalar. I have looked everywhere to see if this form of the Hamiltonian for a non-quantum gravitational field (in the absence of matter) is used, and how valid is it. When looking up the Hamiltonian for Gravity I often get the ADM Formulation for Quantum Gravity, which isn't what I'm looking for as I would like a non-quantum formulation. My question(s), I guess, are;

    1) Is this a correct Hamiltonian for the Gravitational Field?

    2)If yes, where did Einstein get this form of the Hamiltonian from? Guess and Check? I doubt that, though.
     
  2. jcsd
  3. Nov 30, 2015 #2

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is a side issue, but the notion of constraining the determinant of the metric is extremely old-fashioned. It also occurs in Schwarzschild's original paper on the Schwarzschild spacetime. It's not consistent with how anyone does relativity today, because it amounts to a constraint on the permissible coordinate systems. Today we think of all coordinate systems as being equally valid. In general this is an example of why IMO it really doesn't make a lot of sense to spend a lot of time studying centuries-old papers. If there is something interesting in such an old paper, OK, fine -- but if it's simply archaic and requires a lot of effort to reinterpret in terms that anyone today can understand, then it becomes a waste of time.
     
  4. Nov 30, 2015 #3
    bcrowell, I actually didn't know that the constraint on the determinant was a hindrance, so thank you for that info. I guess you could say these papers are very archaic (interpreting the notation is really difficult being so different from what is used now). I'm reading them mostly out of curiosity as to how Einstein was able to develop GR. When I came across his "Hamiltonian" I got a bit excited, because I always wondered if anyone had written a Hamiltonian for classical GR (classical as in non-quantum), and had tried to find one on the internet with no luck.
     
  5. Nov 30, 2015 #4

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  6. Nov 30, 2015 #5

    samalkhaiat

    User Avatar
    Science Advisor

    This is not what we now call Hamiltonian. Back then “Hamiltonian function” used to be the name for the integrand of the action. Now we call that the Lagrangian density. In his original work, Einstein observed that the Lagrangian density (his “Hamiltonian”)
    [tex]
    \begin{equation}
    \mathcal{L} = g^{\mu\nu}\Gamma^{\sigma}_{\mu\rho}\Gamma^{\rho}_{\nu\sigma} ,
    \end{equation}
    [/tex]
    which is a function of [itex]g_{\mu\nu}[/itex] and [itex]\partial_{\sigma}g_{\mu\nu}[/itex] only, together with the additional constraint [itex]\sqrt{-g} = 1[/itex] leads to the free field equations as its Euler-Lagrange equations and defines a conserved gravitational energy-momentum pseudo-tensor
    [tex]
    \begin{equation}
    \partial_{\mu} (\sqrt{-g} \mathcal{T}^{\mu}{}_{\nu}) = \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}g_{\alpha \beta})} \partial_{\nu}g_{\alpha \beta} - \delta^{\mu}_{\nu} \mathcal{L} \right) = 0 .
    \end{equation}
    [/tex]
    Clearly, the constraint [itex]\sqrt{-g} = 1[/itex] is not consistent with general covariance for it restricts the allowed coordinate systems. However, this shortcoming can be removed by writing (what we now days call the Einstein’s Lagrangian)
    [tex]
    \begin{equation}
    \mathcal{L}_{E} = \sqrt{-g} g^{\mu\nu}\left( \Gamma^{\sigma}_{\mu\rho}\Gamma^{\rho}_{\nu\sigma} - \Gamma^{\sigma}_{\mu\nu}\Gamma^{\rho}_{\rho \sigma} \right) .
    \end{equation}
    [/tex]
    Notice that [itex]\mathcal{L}_{E}[/itex] reduces to [itex]\mathcal{L}[/itex] when [itex]\sqrt{-g} = 1[/itex]. However, as well as producing the continuity equation (2), the Euler-Lagrange equations of [itex]\mathcal{L}_{E}[/itex] are now generally covariant equations, i.e., they hold true in any coordinate system. In fact, it is easy to show that the Einstein Lagrangian, [itex]\mathcal{L}_{E}[/itex], is identical with the Hilbert-Einstein Lagrangian, [itex]\mathcal{L}_{HE}=\sqrt{-g}R[/itex], up to a total divergence:
    [tex]
    \sqrt{-g}R = \mathcal{L}_{E} + \partial_{\sigma} \left( \sqrt{-g}( g^{\sigma\mu}\Gamma^{\rho}_{\mu\rho} - g^{\mu\nu}\Gamma^{\sigma}_{\mu\nu}) \right) .
    [/tex]
    As for the Hamiltonian formulation of pure gravity, well it is a bit complicated because GR describes constraint system. But it has been done. Basically, one considers a (3+1) foliation of space-time by space-like hyper-surfaces defined by [itex]t = \mbox{constant}[/itex] and write the metric as [itex]g_{\mu\nu} = h_{\mu\nu} + n_{\mu}n_{\nu}[/itex], where [itex]n^{\mu}[/itex] is a time-like unit normal to the 3-surface [itex]t = \mbox{constant}[/itex]. Then, one can express (up to total divergence) the 4-curvature scalar [itex]R^{(4)}[/itex] in terms of the 3-curvature [itex]R^{(3)}[/itex] and the extrinsic curvature [itex]K_{ij} = (1/2)\partial_{t}h_{ij}[/itex] of the 3-surface [itex]t = \mbox{constant}[/itex]:
    [tex]\int d^{4}x \ \sqrt{-g}R^{(4)} = \int d^{4}x \ \sqrt{-g} \left( R^{(3)} + K_{ij}K^{ij} - K^{n}_{n}K^{m}_{m}\right) .[/tex] From this you obtain the canonical momentum [itex]\pi^{ij}[/itex] conjugate to the gravitational field [itex]h_{ij}[/itex] by the usual definition
    [tex]
    \begin{align*}
    \pi^{ij} &= \frac{\partial(\sqrt{-g}R)}{\partial (\partial_{t}h_{ij})} \\
    &= \frac{\sqrt{-h}}{2} \frac{\partial}{\partial K_{ij}} \left( K_{mn}K^{mn} - K^{m}_{m}K^{n}_{n}\right) \\
    &= \sqrt{-h} \left( K^{ij} - K^{n}_{n}h^{ij}\right) .
    \end{align*}
    [/tex]
    This now can be solved for the “velocity” [itex]\dot{h}^{ij}[/itex]
    [tex]\dot{h}^{ij} = 2K^{ij} = \frac{2}{\sqrt{-h}} \left( \pi^{ij} - \frac{1}{2}\pi^{n}_{n} h^{ij}\right) .[/tex]
    Now, the Hamiltonian of the gravitational field is calculated by the usual definition in field theories
    [tex]
    \begin{align*}
    H(\pi , h) &= \int_{t = \mbox{const.}} d^{3}x \left[ \pi^{ij}\dot{h}_{ij} - \sqrt{-h} \left( R^{(3)} + K_{ij}K^{ij} - K^{m}_{m}K^{n}_{n} \right) \right] \\
    &= \int_{t = \mbox{const.}} d^{3}x \ \sqrt{-h} \left( K^{ij}K_{ij} - K^{m}_{m}K^{n}_{n} - R^{(3)} \right) \\
    &= \int_{t = \mbox{const.}} d^{3}x \left[ - \sqrt{-h} R^{(3)} + \frac{1}{\sqrt{-h}} \left( \pi^{ij} \pi_{ij} - \frac{1}{2} \pi^{i}_{i}\pi^{k}_{k}\right)\right] .
    \end{align*}
    [/tex]
    The equations of motion derived from this Hamiltonian are indeed equivalent to the Einstein’s equations.
     
  7. Dec 1, 2015 #6

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    @samalkhaiat - Is what you're describing in the second half of #5 the ADM formalism?
     
  8. Dec 1, 2015 #7
    Thank you for the link bcrowell, I'll have to look over that paper for the next couple of weeks as it does seem interesting. Also thank you samalkhaiat, I didn't know Hamiltonian in those papers was what we now consider the Lagrangian Density. I guess that is another way in which the notation is very outdated.
     
  9. Dec 1, 2015 #8

    samalkhaiat

    User Avatar
    Science Advisor

    Yes, It is the ADM formulation but in the so-called synchronous gauge, i.e., the time [itex]t[/itex] is taken to be the proper time along geodesic lines on the normal vector field [itex]\mathbf{n}[/itex]. This is why the final form of [itex]H[/itex] contained no constraints. If we repeat the calculation in an arbitrary (time-foliation) gauge, the so-called lapse function [itex]N = \mathbf{g}(\mathbf{n},\partial_{t})[/itex] and the shift vector [itex]\mathbf{m}= \partial_{t} - N\mathbf{n}[/itex] will enter into the Hamiltonian as Lagrange multipliers
    [tex]
    H = \int_{\Sigma}d^{3}x \sqrt{-h} \left[\left(K_{ij}K^{ij} - K^{2} - {}^3R \right) N - m_{i}\nabla_{j}\left(K^{ij} - K h^{ij}\right)\right] .
    [/tex]
    Since [itex](N,m_{i})[/itex] are non-propagating fields, we therefore have the following constraints
    [tex]\frac{\delta H}{\delta m_{j}} = 0 \ \ \Rightarrow \ \nabla_{i}\pi^{ij} = \nabla_{i}(K^{ij} - K h^{ij}) = 0 ,[/tex]
    [tex]\frac{\delta H}{\delta N} = 0 \ \ \Rightarrow \ K_{ij}K^{ij} - K^{2} - {}^3R = 0 .[/tex]
    Together, these constraints imply that [itex]H(h,\pi) = 0[/itex] which is the characteristic feature of any generally covariant theory.
    I should also say that the Gauss-Codazzi equation was the starting point of my derivation: A surface [itex]\Sigma[/itex] embedded into a space-time with metric [itex]\mathbf{g}[/itex] receives the induced metric [itex]{h} = {g}|_{T\Sigma}[/itex] which can be used to define the induced Levi-Civita connection [itex]{}^3\nabla[/itex] and calculate the corresponding 3D Riemann tensor [itex]{}^3R(\mathbf{x},\mathbf{y},\mathbf{z},\mathbf{t})[/itex] for tangent vectors [itex]\mathbf{x},\mathbf{y},\mathbf{z}[/itex] and [itex]\mathbf{t}[/itex]. The Gauss-Codazzi equation is the following relation between the reduced 3D Riemann tensor and the original 4D Riemann tensor
    [tex]
    {}^3R(\mathbf{x},\mathbf{y},\mathbf{z},\mathbf{t}) = {}^4R(\mathbf{x},\mathbf{y},\mathbf{z},\mathbf{t}) + K_{\Sigma}(\mathbf{x},\mathbf{t}) K_{\Sigma}(\mathbf{y},\mathbf{z}) - K_{\Sigma}(\mathbf{y},\mathbf{t}) K_{\Sigma}(\mathbf{x},\mathbf{z}) ,
    [/tex]
    where [tex]K_{\Sigma}(\mathbf{x},\mathbf{y}) = h( \nabla_{\mathbf{x}}\mathbf{n}, \mathbf{y}) , \ \mbox{or} \ K_{\mu\nu} = h_{\rho \nu} \nabla_{\mu}n^{\rho} ,[/tex] is the so-called extrinsic curvature tensor of the surface [itex]\Sigma[/itex], i.e., the distortion tensor [itex]g( \nabla_{\mathbf{x}}\mathbf{n}, \mathbf{y})[/itex] of the vector field [itex]n^{\mu}[/itex] with [itex]g[/itex] replaced by the induced metric [itex]h[/itex].
    The 3D curvature scalar [itex]{}^3R[/itex] is defined by the 3-trace, [itex]\mbox{Tr}^{(3)}_{(. , .)}[/itex], of the induced Riemann tensor [itex]{}^3R(\mathbf{x},\mathbf{y},\mathbf{z},\mathbf{t})[/itex] with respect to [itex](\mathbf{x},\mathbf{z})[/itex] and [itex](\mathbf{y},\mathbf{t})[/itex]. Since [itex]n[/itex] is normal to [itex]\Sigma[/itex], the 3-trace of any tensor [itex]T[/itex] can be given in terms of its 4-trace as
    [tex]\mbox{Tr}^{(3)}_{(\mathbf{x},\mathbf{y})} \left( T(\mathbf{x},\mathbf{y}) \right) = \mbox{Tr}^{(4)}_{(\mathbf{x},\mathbf{y})} \left( T(\mathbf{x},\mathbf{y}) \right) - T(\mathbf{n},\mathbf{n}) .[/tex]
    Now, if we trace the Gauss-Codazzi equation and use the above relation together with the symmetries of the Riemann tensor, we can show that
    [tex]{}^3R = {}^4R - K_{ij}K^{ij} + K^{2} + \mbox{ total divergence} .[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Einstein's Hamiltonian?
  1. Einstein Hamiltonian (Replies: 1)

  2. Newton Or Einstein (Replies: 22)

  3. Einstein's Twin (Replies: 4)

  4. Einstein Today (Replies: 1)

Loading...