Another action for the relativistic particle

[Heidi]

Hi Pf
i am accustomed with the action $$mc \int ds $$ for relativistic particle.
i found a paper by Andrew Wipf with another lagrangian. please look at the beginning of chapter seven (7.1) Is it possible to deduce from it the shape of its trajectory? I have a lot of question about this chapter...

 

[vanhees71]

I'm not familiar with the formalism provided in the paper, but I guess it's about the fact that you can derive the geodesic equation for the motion of a test particle in a given spacetime by the Lagrangian
$$L=\frac{1}{2} g_{\mu \nu} \dot{q}^{\mu} \dot{q}^{\nu},$$
where the dot refers to an arbitrary world-line parameter, $\lambda$ which is in using this particular form of the Lagrangian automatically affine.

This is easy to see: Because the Lagrangian does not explicitly depend on $\lambda$ the "Hamiltonian",
$$H=p_{\mu} \dot{q}^{\mu}-L=L, \quad p_{\mu}=\frac{\partial L}{\partial \dot{q}^{\mu}}$$
is conserved along the trajectory.

If you have a massive test particle then you can make $L=\frac{1}{2}$. Then the affine parameter is proportional to the proper time of the particle, but this form of Hamilton's principle also works for massless particles.

You can also use this trick for the covariant dynamics of relativistic particles in special relativity. For this case, the alternative form for the free part of the Lagrangian is derived in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Heidi]

Do you thind that in the gauge transformation $$t --> t - \xi$$
it is a constant or a function of t?

 

[formodular]

It's just the einbein form of the action for a point particle, e.g. eq. (2.1.5) of GSW (note the action you posted is technically zero for massless particles, one adds in the constraint p^2 = - m^2 as a Lagrange multiplier then eliminates momentum to get the action as in (7.1)).

 

[Heidi]

Yes and the action i knew is an einbein-free equivalent to this action.
i discovered it in https://physics.stackexchange.com/questions/4188/whats-the-point-of-having-an-einbein-in-your-action

 

[samalkhaiat]

Hi Pf
i am accustomed with the action $$mc \int ds $$ for relativistic particle.
i found a paper by Andrew Wipf with another lagrangian. please look at the beginning of chapter seven (7.1) Is it possible to deduce from it the shape of its trajectory? I have a lot of question about this chapter...

Let (\mathcal{M}^{1} , e) be a 1-dimensional “space-time” with coordinate \tau and metric d\sigma^{2} = e(\tau) d\tau d\tau , with e(\tau) \equiv e_{\tau \tau}(\tau) being the component of the (obviously symmetric) metric tensor. Under a general coordinate transformation (or, which is the same thing, diffeomorphism or gauge transformation) on \mathcal{M}^{1}, \tau \to \tau^{\prime} = \tau^{\prime}(\tau), the (component) of the metric tensor transforms as e^{\prime} (\tau^{\prime}) = \left( \frac{d\tau}{d \tau^{\prime}}\right)^{2} \ e(\tau) . This means that d\tau \sqrt{e(\tau)} is a diffeomorphism-invariant measure on \mathcal{M}^{1} (prove it). Next, we think of \left(\mathcal{M}^{1}, e(\tau)\right) as the world-line x^{\mu}(\tau) of particle in the 4-dimensional (pseudo Riemannian) space-time (\mathcal{M}^{(1,3)}, g_{\mu\nu}(x)). In other words, we define e(\tau) to be a metric on the world-line x^{\mu}(\tau), or (equivalently) define x^{\mu}(\tau) to be 4 scalar fields on the 1-dimensional “space-time” \left(\mathcal{M}^{1}, e(\tau)\right).

We now try to formulate “general relativity”- type theory on the world line, i.e., we try to construct a diffeomorphism-invariant action-integral on \mathcal{M}^{1} using the metric e(\tau) and the scalar fields x^{\mu}(\tau). From the invariant measure on \mathcal{M}^{1}, we form the following cosmological-constant-type (gauge invariant) action S[e] = - \frac{m^{2}}{2} \int d \tau \ \sqrt{e(\tau)} , where m is a constant of mass dimension. For the scalar fields x^{\mu}(\tau), we seek a scalar (Lagrangian) that plays the part of the Ricci scalar in General Relativity, i.e., we would like to find an action of the form S[x] = \int d\tau \sqrt{e(\tau)} R(\mathcal{M}^{1}), where R(\mathcal{M}^{1}) is some scalar depending on the fields e(\tau) and x^{\mu}(\tau). Let us define the function \mathcal{L} (x) = \frac{1}{2} g_{\mu\nu}(x) \frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}, and examine its behaviour under the diffeomorphism \tau \to \bar{\tau} = \bar{\tau}(\tau) (infinitesimally, this is written as \bar{\tau} = \tau + \epsilon (\tau), but we don’t need this in here). Since \bar{x}^{\mu}(\bar{\tau}) = x^{\mu}(\tau), we find (I invite you to do the simple algebra)\mathcal{L}(\bar{x}) = \mathcal{L}(x) \left(\frac{d\tau}{d\bar{\tau}}\right)^{2} . From this, we can identify the Ricci scalar on \mathcal{M}^{1} by R(\mathcal{M}^{1}) \equiv e^{-1} (\tau) \mathcal{L}(x) = \bar{e}^{-1}(\bar{\tau}) \mathcal{L}(\bar{x}) . Thus, we have the following “Einstein-Hilbert” action on the world-line S[x] = \int d\tau \sqrt{e} \ R(\mathcal{M}^{1}) = \frac{1}{2} \int d\tau \sqrt{e(\tau)} e^{-1}(\tau) g_{\mu\nu}(x)\dot{x}^{\mu}(\tau)\dot{x}^{\nu}(\tau) . Therefore, for the total action, we have S[e,x] = \frac{1}{2} \int d\tau \left( \frac{1}{\sqrt{e}}g_{\mu\nu}(x)\dot{x}^{\mu}\dot{x}^{\nu} - m^{2} \sqrt{e}\right) . This action is suitable for massive and massless particles. Let us find the equations of motion for the fields (Please, fill in the algebraic details!): For e(\tau) \frac{\delta S}{\delta e} = 0 \ \Rightarrow \ e^{-1}g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu} + m^{2} = 0 . This is a constraint equation. For m \neq 0, we can write e = - \frac{g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}{m^{2}} . Substituting this back in the total action, we find the familiar action for massive particle S_{m}[x] = -m \int d\tau \ \sqrt{g_{\mu\nu}(x)\dot{x}^{\mu}\dot{x}^{\nu}} = -m \int \ ds.

Exercise(1): Calculate the conjugate momentum p_{\mu} = \frac{\partial L}{\partial \dot{x}^{\mu}} = \frac{1}{\sqrt{e}} g_{\mu\nu}(x)\dot{x}^{\nu}, and show that the constraint equation is nothing but the dispersion relation g_{\mu\nu}p^{\mu}p^{\nu} + m^{2} = 0. Choose \tau to be the proper time and show that p^{\mu} = m\dot{x}^{\mu}.

Exercise(2): Show that \frac{\delta}{\delta x^{\mu}}S[e,x] = 0 \ \Rightarrow \ \sqrt{e} \frac{d}{d\tau} \left( \frac{1}{\sqrt{e}}\dot{x}^{\mu}\right) + \Gamma^{\mu}_{\rho \sigma}(x) \dot{x}^{\rho}\dot{x}^{\sigma} = 0. This can be written as \ddot{x}^{\mu} + \Gamma^{\mu}_{\rho\sigma} \dot{x}^{\rho}\dot{x}^{\sigma} - \frac{1}{2} \frac{\dot e}{e} \dot{x}^{\mu} = 0. What is the geometric meaning of the factor \frac{1}{2}e^{-1}\dot{e} in the last term of the above equation of motion? Why is it always possible to choose \tau such that e(\tau) = 1? Of course, in this case you will have p_{\mu} = g_{\mu\nu}\dot{x}^{\nu}, p_{\mu}\dot{x}^{\mu} + m^{2} = 0 and \ddot{x}^{\mu} + \Gamma^{\mu}_{\rho \sigma}\dot{x}^{\rho}\dot{x}^{\sigma} = 0.

Exercise(3): Investigate the case when m = 0.

 

[Heidi]

I found an article that explains where this einbein may come from:https://www.ellipsix.net/blog/2010/08/the-origin-of-the-einbein.html

 

[formodular]

Your article is a discussion of the GSW equation (2.1.5) I mentioned above, however GSW introduce it out of thin air and the article introduces the einbein basically via magic.

The fool-proof way to arrive at the einbein form of the action, the way that generalizes, is discussed from equations (2.13) to (2.16) here

http://www.damtp.cam.ac.uk/user/examples/3P6.pdf

with this method generalizing to the Nambu-Goto action to give the Polyakov action via equations (3.10) to (3.19) and then equations (3.48) to (3.50).

 

[Heidi]

In his answer Samarkhaiat gave the constraint $$C=p^2 + m^2=0$$ `associated to the eom of the invariant action by a gauge transformation. I read that the associated constraints are the generators of the gauge transformations. and that they are exp (i e C). is it the case here?

 

[formodular]

You are analyzing a constrained dynamical system, and it is commonly said that 'first class constraints are generators of gauge transformations', e.g. section 1.4 of Henneaux 'Quantization of Gauge Systems'. On a classical level, they are not commonly written in exponential form like that, however on a quantum level they are, e.g. one can take C to be the Hamiltonian and one ends up with the Klein-Gordon equation.

Section 2.1 of these notes http://www-th.bo.infn.it/people/bastianelli/ch6-FT2.pdf treat the relativistic point particle as a constrained system, derive the constraint and derive the einbein action, and also treat the constraint C as the Hamiltonian on a quantum level (thus it is in exp (i C) form) but the Hamiltonian vanishes.

 

[Heidi]

I still have a question about the gauge fixing in the relativistic free particle model.
When we consider the electromagnetisme hamiltonien using the potential A, we know that gauge transformations leave the electric and magnetic filelds unchanged.
Here we have the lapse e and a four verctor representing the position of the particle in space time. the space positions are measurable.
I read this in another paper:
the gauge transformation reads
$$x(\tau) -> x(f(\tau))$$
$$e(\tau) -> f'_\tau e(f(\tau)) $$
then the author writes fixing the gauge with tau = x0 ...

does this leave the positions x,y and z unchanged?

 

[haushofer]

I still have a question about the gauge fixing in the relativistic free particle model.
When we consider the electromagnetisme hamiltonien using the potential A, we know that gauge transformations leave the electric and magnetic filelds unchanged.
Here we have the lapse e and a four verctor representing the position of the particle in space time. the space positions are measurable.
I read this in another paper:
the gauge transformation reads
$$x(\tau) -> x(f(\tau))$$
$$e(\tau) -> f'_\tau e(f(\tau)) $$
then the author writes fixing the gauge with tau = x0 ...

does this leave the positions x,y and z unchanged?

Well, yes. I guess you mean the static gauge. This only restricts the temporal coordinate $$x^0$$, not the spatial ones.

I'm not sure what you mean by "the space positions are measurable". You can also measure the time coordinate, but by its very nature these are coordinate dependent.

By the way, if you refer to papers or books, please give references. We don't have magic abilities here at PF and context is often crucial in giving some aid ;)

 

[Heidi]

These lines are in a french thesis written by David Louapre.
I attached the paragraph about the subject.
in electromagnetism the gauge group transforms A -> A' A' - A obeying a linear differential law. it is easy to see that we have gauge group and to find the generators.
but here when f is not infinitesimal , have we also a gauge group, and what are the lie generators?
if the primed values denote transformed value we have $$ x' ( \tau') = x(\tau), e'( \tau') d\tau' = e( \tau) d\tau$$
Does if help for the group question?