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I'm trying to get the relativistic action (or Lagrangian) for a free particle in the case of violation of Lorenz invariance. Suppose we have the modified dispersion relation:

$$ E^{2}=\Omega^{2}(p^{2}) $$

Here $E$ and $p$- energy and momentum of the particle, $\Omega^{2}$ is a function that takes the form $\Omega^{2}(p^{2})=m^{2}c^{4}+p^{2}c^{2}$ if we have Lorentz invariance.

In the other word, we have equation $$ \left(v\frac{\partial L}{\partial v}-L\right)^{2}=\Omega^{2}\left(\left[\frac{\partial L}{\partial v}\right]^{2}\right) $$

We want to define the Lagrangian.

For example, if $E^{2}=m^{2}+(1+\xi)p^2$, we can obtain(overriding $\tilde{v}=v(1+\xi)^{-1/2}$, we present the equation written above to the standard relativistic form, above which we know everything): $$ L=-m\sqrt{1-\frac{v^{2}}{1+\xi}} $$

I want to find lagrangian and action for the following case:

$$ E^{2}=p^{2}c^{2}+m^{2}c^{4}+\frac{p^{4}}{M^{2}} $$

Here $M\gg m$.

So, we have equation:

$$ \left(v\frac{\partial L}{\partial v}-L\right)^{2}=\left(\frac{\partial L}{\partial v}\right)^{2}c^{2}+\frac{1}{M^{2}}\left(\frac{\partial L}{\partial v}\right)^{4}+m^{2}c^{4} $$

-I tried to find the solution in the following way:

$L(v)=L_{0}+L_{1}v$, where ${{ L_0}}^{2}={{ L_1}}^{2}{c}^{2}+{\dfrac {{{ L_1}}^{4}}{{M} ^{2}}}+{m}^{2}{c}^{4}$, and $|L_0| \ge mc^2$. I set it in the original equation, but it didn't give significant results.

Here (pages $6$-$7$) they obtaining the Lagrangian for the case I described above, and they say that in the general case Lagrangian can be defined as (equation $18$) $$ L=-m\sqrt{1-v^{2}}F\left(\frac{1}{\sqrt{1-v^{2}}}\right) $$

I tried to find a solution in this form (by substituting in the equation) but I didn't succeed.

Can you help me to obtain Lagrangian (action) for the relation I wrote? Also it would be great if you gave the usefull references devoted to the topic (for example, where they obtain Lagrangians, etc)

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2 Answers 2

up vote 2 down vote accepted
+50

In mechanical problems where the Hamiltonian does not depend explicitly on time, then the Hamiltonian is equal to the total energy $H = E$. These problems can be solved using the (inverse) Legendre transform, which allows the computation of the Lagrangian from the Hamiltonian, without the need to solve a differential equation.

The Lagrangian can be computed from the Hamiltonian $H = E$ as follows:

The velocity is given by the Hamilton equation of motion (Here $c=1$):

$v = \dot{x} = \frac{\partial E}{\partial p} =\frac{p+\frac{2p^3}{M^2}}{\sqrt{p^2+m^2+ \frac{p^4}{M^2}}}$

The Lagrangian is given by the (inverse) Legendre transform

$ L = p\dot{x}-H = pv-E = \frac{-m^2+\frac{p^4}{M^2}}{\sqrt{p^2+m^2+ \frac{p^4}{M^2}}}$

The only remaining problem is to express the Lagrangian as a function of $v$. Now, $p$ is given implicitly in terms of $v$ in the Hamilton equation of motion. In the special case at hand, this equation can be solved in a closed form because it is of the third degree in $p^2$:

$ 4 \frac{(p^2)^3}{M^4} + 4 \frac{(p^2)^2}{M^2} +p^2(1-v^2) - m^2v^2=0$

Now, please see the following wikipedia page for the exact form of the solution, which is cumbersome and will not be given here. Moreover, one must be careful in the selection of the correct root, which will depend on the values of $\frac{m^2}{M^2}$ $\frac{p^2}{m^2}$ .

In the following an approximate solution will be given for the case $p^2\ll M^2$ and $m^2\ll M^2$. In this case we substitute:

$ p^2 = m^2\frac{v^2}{1-v^2} + \frac{\delta }{M^2}$

Please, notice that the first term is the exact solution of the case $M^2 \rightarrow \infty$.

Substituting this solution in the Hamilton equation and keeping only the leading terms in $\frac{1}{M^2}$, we get the approximate solution:

$ p^2 = m^2\frac{v^2}{1-v^2} + \frac{m^4 v^4}{M^2(1-v^2)3}$

Which can be substituted in the Lagrangian

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Thanks for your answer! –  xxxxx Mar 5 '13 at 16:57

In the article you cited they get to their form of the Lagrangian from the action

$$S_{pp}=-m\int ds\,F(u_\mu v^\mu),$$ where F is an arbitrary function with $F(1)=1.$ We get to the desired form by changing to a frame of reference where $u_0=1$ and $u_i=0$. With the relations

$$d\tau=ds=\sqrt{1-v^2}dt$$

and

$$v^0=\frac{1}{\sqrt{1-v^2}},$$

where $v^2=v^iv^i,$ we arrive at

$$S_{pp}=\int dt\,L_{pp},$$

with

$$L_{pp}=-m\sqrt{1-v^2}F\left(\frac{1}{\sqrt{1-v^2}}\right).$$

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