Relativistic Lagrangian and Hamiltonian for a free particle

[dyn]

Hi.
I am working through a QFT book and it gives the relativistic Lagrangian for a free particle as L = -mc²/γ. This doesn't seem consistent with the classical equation L = T - V as it gives a negative kinetic energy ? If L = T - V doesn't apply relativistically then why does the Hamiltonian H = T + V as the total energy still apply ? Thanks

 

[pervect]

Momentum is the partial of the Lagrangian with respect to $\dot{x}$, which in this case would be velocity. So if you want positive momentum for a positive mass and a positive velocity, you want the negative sign in the Lagrangian - as v increases, gamma increases, and the Lagrangian becomes less negative as written, meaning the momentum is positive.

 

[Orodruin]

It does not give a negative kinetic energy. In the non-relarivistic limit you have $-1/\gamma = -\sqrt{1-v^2}\simeq -1+v^2/2$ which is just the regular classical Lagrangian up to an additive constant. L = T-V does not hold relativistically, the kinetic term is not quadratic in $\dot x$.

 

[pervect]

The energy function (which is the total energy, the Hamiltonian in different v) is $v\frac{\partial L}{\partial v} - L$)

Hi.
I am working through a QFT book and it gives the relativistic Lagrangian for a free particle as L = -mc²/γ. This doesn't seem consistent with the classical equation L = T - V as it gives a negative kinetic energy ? If L = T - V doesn't apply relativistically then why does the Hamiltonian H = T + V as the total energy still apply ? Thanks

How are you calculating the energy? Given the Lagrangian, $L(x,\dot{x},t)$ one usually uses the legendre transformation

$$H = v\,\frac{\partial L}{\partial \dot{x}} - L$$

which comes out positive and equal to $$\frac{mc^2}{\sqrt{1-(v/c)^2}}$$

 

[dyn]

Thanks for your replies. So L = T - V does not apply relativistically. Does H = T + V apply ? If not , does H still give the total energy ?

 

[stevendaryl]

Thanks for your replies. So L = T - V does not apply relativistically. Does H = T + V apply ? If not , does H still give the total energy ?

The general relationships for Lagranian mechanics are these:

$p_\mu = \frac{\partial L}{\partial U^\mu}$
$H = p_\mu U^\mu - L$

where $U^\mu = \frac{dx^\mu}{ds}$, for $s$ the "evolution parameter" (not necessarily time--relativistically, it is often proper time). Only in the special case where $L$ is quadratic in $U^\mu$ (that is, $L$ involves only zeroth, first and second powers of $U^\mu$) does $L = T - V$ or $H = T+V$.

 

[pervect]

Thanks for your replies. So L = T - V does not apply relativistically. Does H = T + V apply ? If not , does H still give the total energy ?

No, H in the relativistic case is given by the Legendre transformation as in my post #4. (Which works for both relativistic and non-relativistic physics).

Basically, the least-action principle when applied relativistically results in the principle of maximal aging, see for instance http://www.eftaylor.com/leastaction.html.

[add]In other words, the expression for L is proportioanl to $d\tau$, because $d\tau = dt / \gamma$ and $L = -mc^2 / \gamma$ and $mc^2$ is constant for a free particle.

 

[dyn]

Thanks. So neither L = T-V or H = T+V apply relativistically. But does the Hamiltonian always give the total energy ?