# 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

$\boxed{E=\sqrt{m^2c^4+c^2p^2}=H}$

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:

$\boxed{L=pv-H}$

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

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

Then,

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

and finally

$\boxed{L=-mc^2\sqrt{1-\dfrac{v^2}{c^2}}=-\dfrac{mc^2}{\gamma}=-mc\sqrt{-\dot{X}^2}}$

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

$ds^2=d\mathbf{x}^2-c^2dt^2$

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

$ds^2=-c^2d\tau^2=d\mathbf{x}^2-c^2dt^2$

Thus, dividing by $dt^2$

$-c^2\dfrac{d\tau^2}{dt^2}=\mathbf{v}^2-c^2$

and

$d\tau^2=\gamma^{-2}dt^2=\dfrac{1}{\gamma^2}dt^2=\left(1-\dfrac{\mathbf{v}^2}{c^2}\right)dt^2$

so

$d\tau=\sqrt{1-\dfrac{\mathbf{v}^2}{c^2}}dt$

$cd\tau=ds=\sqrt{c^2-\mathbf{v}^2}dt=\sqrt{-\dot{X}^2}dt$

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

$\boxed{L=-mc^2\sqrt{-\dot{X}^2}}$

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):

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

since

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

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

$H=E=\sqrt{\mathbf{p}^2c^2+(mc^2)^2}$

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

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

$\boxed{L=-mc^2\sqrt{-\dot{X}^2}}$

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

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:

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