LOG#036. Action and relativity.

The hamiltonian formalism and the hamiltonian H in special relativity has some issues with the definition. In the case of the free particle one possible definition, not completely covariant, is the relativistic energy


There are two others interesting scalars in classical relativistic theories. They are the lagrangian L and the action functional S. The lagrangian is obtained through a Legendre transformation from the hamiltonian:


From the hamiltonian, we get the velocity using the so-called hamiltonian equation:

\dot{\mathbf{q}}=\mathbf{v}=\dfrac{\partial H}{\partial \mathbf{p}}=c^2\dfrac{\mathbf{p}}{E}


L=\dfrac{E}{c^2}\mathbf{v}^2-E=E\left(\dfrac{v^2}{c^2}-1\right)=-\dfrac{E}{\gamma^2}=-\dfrac{m\gamma c^2}{\gamma^2}=-\dfrac{mc^2}{\gamma}

and finally


The action functional is the time integral of the lagrangian:

\boxed{S=\int Ldt}

However, let me point out that the above hamiltonian in SR has some difficulties in gauge field theories. Indeed, it is quite easy to derive that a more careful and reasonable election for the hamiltonian in SR should be zero!

In the case of the free relativistic particle, we obtain

S=-mc^2\int \sqrt{1-\dfrac{v^2}{c^2}}dt

Using the relation between time and proper time (the time dilation formula):

dt=\gamma d\tau\rightarrow \dfrac{dt}{\gamma}=d\tau

direct substitution provides

-mc^2\int \sqrt{1-\dfrac{v^2}{c^2}}dt=-mc^2\int d\tau

And defining the infinitesimal proper length in spacetime as ds=cd\tau, we get the simple and wonderful result:

\boxed{S=-mc\int ds}

Sometimes, the covariant lagrangian for the free particle is also obtained from the following argument. The proper length is defined as


The invariant in spacetime is related with the proper time in this way:


Thus, dividing by dt^2







that is

\boxed{ds=cd\tau=\sqrt{-\dot{X}^2}dt=\sqrt{-\dot{x}^\mu\dot{x}_\mu}dt=\sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}dt}

and the free coordinate action for the free particle would be:

\boxed{S=-mc\int ds=-mc^2\int \sqrt{-\dot{X}^2}dt=-mc^2\int \sqrt{-\dot{x}^\mu\dot{x}_\mu}dt=-mc^2\int \sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}dt}

Note, that since the election of time “t” is “free”, we can choose t=\tau to obtain the generally covariant free action:

\boxed{S=-mc\int ds=-mc^2\int \sqrt{-\dot{X}^2}d\tau=-mc^2\int \sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}d\tau}

Remark: the (rest) mass is the “coupling” constant for the free particle proper lenght to guess the free lagrangian


Now, we can see from this covariant action that the relativistic hamiltonian should be a feynmanity! From the equations of motion,

P_\mu=\dfrac{\partial L}{\partial \dot{X}^\mu}=mc\dfrac{\dot{X}_\mu}{\sqrt{-\dot{X}^2}}

The covariant hamiltonian \mathcal{H}, different from H, can be build in the following way:

\mathcal{H}=P_\mu \dot{X}^\mu-L=mc\dfrac{\dot{X}_\mu \dot{X}^\mu}{\sqrt{-\dot{X}^2}}-mc\sqrt{-\dot{X}^2}=0

The meaning of this result is hidden in the the next identity ( Noether identity or “hamiltonian constraint” in some contexts):

\boxed{\mathcal{H}=P_\mu P^\mu+m^2c^2=0}


P_\mu P^\mu=m U_\mu mU^\mu=-m^2c^2

This strange fact that \mathcal{H}=0 in SR, a feynmanity as the hamiltonian, is related to the Noether identity E^\mu \dot{X}_\mu for the free relativistic lagrangian, indeed, a consequence of the hamiltonian constraint and the so-called reparametrization invariance \tau'=f (\tau). Note, in addition, that the free relativistic particle would also be invariant under diffeomorphisms x^{\mu'}= f^\mu (x)=f^\mu (x^\nu) if we were to make the metric space-time dependent, i.e., if we make the substitution \eta_{\mu\nu}\rightarrow g_{\mu\nu} (x). This last result is useful and important in general relativity, but we will not discuss it further in this moment. In summary, from the two possible hamiltonian in special relativity


\mathcal{H}=P_\mu P^\mu+m^2c^2=0

the natural and more elegant (due to covariance/invariance) is the second one. Moreover, the free particle lagrangian and action are:


\boxed{S=-mc^2\int d\tau=-mc\int ds=\int L dt}

Remark: The true covariant lagrangian dynamics in SR is a “constrained” dynamics, i.e., dynamics where we are undetermined. There are more variables that equations as a result of a large set of symmetries ( reparametrization invariance and, in the case of local metrics, we also find diffeomorphism invarince).

The dynamical equations of motion, for a first order lagrangian (e.g., the free particle we have studied here), read for the lagrangian formalism:

\boxed{\delta S=\delta \int L (q,\dot{q};t)dt =0\leftrightarrow\begin{cases}E(L)=0,\mbox{with E(L)=Euler operator}\\ E(L)=\dfrac{\partial L}{\partial q}-\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot{q}}\right)=0\end{cases}}

By the other hand, for the hamiltonian formalism, dynamical equations are:

\boxed{\delta S=\delta \int H (q, p; t)dt =0\leftrightarrow\begin{cases}p=\dfrac{\partial L}{\partial \dot{q}},\;\mbox{with}\; \det\left(\dfrac{\partial ^2L}{\partial \dot{q}^i\partial \dot{q}^j}\right)\neq 0\\ \;\\ \dfrac{dq}{dt}=\dot{q}=\dfrac{\partial H}{\partial p}\\ \;\\\dfrac{dp}{dt}=\dot{p}=-\dfrac{\partial H}{\partial q} \\ \;\\ \dfrac{dH}{dt}=\dot{H}=\dfrac{\partial H}{\partial t}=-\dfrac{\partial L}{\partial t}\end{cases}}


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