Lagrange Equations in Special Relativity

The basis for Special Relativity is that the laws of physics are independent of which inertial coordinate system we write them in. We wish to write equations in terms of scalars, 4-vectors, and tensors, with both sides of the equation transforming the same way under rotations and boosts. Such equations are said to be covariant, because both sides of the equation transform in the same standard way. The Lagrangian is based on the principle of least Action. The Action as we know it so far is clearly not a vector or a tensor so our obvious choice is to make it a scalar. Our definition of the Action so far is.

\bgroup\color{black}$\displaystyle S=\int\limits_{t_i}^{t_f} L dt=\int\limits_{t_i}^{t_f} L\gamma d\tau $\egroup

We should then have \bgroup\color{black}$ \gamma L$\egroup be a scalar quantity. But what scalar do we have?

For a particle, the mass is the only scalar we can use. Clearly this is not related to external potential applied to the particle.

$\displaystyle \gamma L$ $\displaystyle = -mc^2$    
$\displaystyle L$ $\displaystyle ={-mc^2\over \gamma}$    
$\displaystyle p_i$ $\displaystyle ={\partial L\over\partial \dot{q}_i}=-mc^2{\partial \sqrt{1-\beta^2}\over\partial v_i} ={v_i\over c^2}\gamma mc^2=\gamma m v_i$    

This agrees with our previous result \bgroup\color{black}$ \vec{p}=\gamma m \vec{v}$\egroup indicating that we have picked the right kinetic term for the Lagrangian.

The (velocity independent) potential can be subtracted.

\bgroup\color{black}$ \displaystyle L={-mc^2\over\gamma}-V$\egroup
This is an important relativistic basis for our use of \bgroup\color{black}$ L=T-V$\egroup in nonrelativistic problems. To have fully relativistic problems we need to have potentials, for example, that are relativistically correct. We will take a look at the relativistic equations of electromagnetism, but this is really beyond the scope of this course. In field theory, there will be scalar interaction terms in the Lagrangian, like the interaction between the electric current due to particles with the vector potential, a dot product between 4-vectors.

The Hamiltonian can be computed from the Lagrangian.

\bgroup\color{black}$\displaystyle H=v_ip_i-L ={p_ip_i\over \gamma m}-L={p^2c^2\...
...r\gamma}+V={1\over\gamma mc^2}(p^2c^2+m^2c^4)+V={E^2\over\gamma mc^2}+V $\egroup

\bgroup\color{black}$\displaystyle H=E+V=T+mc^2+V $\egroup

It is the total energy.

It is quite interesting to look at this Action which is to be minimized in Minkowski space and to try to formulate the principle of least Action in a covariant way.

\bgroup\color{black}$\displaystyle S=\int\limits_{\tau_i}^{\tau_f} -mc^2 d\tau $\egroup

What Action should we use for particles in Special Relativity? The proper time is directly related to the space-time interval.

\bgroup\color{black}$\displaystyle ds^2=-c^2 d\tau^2 $\egroup

So we can write our action to be minimized in terms of the spacetime interval.

\bgroup\color{black}$\displaystyle S=\int\limits_{\tau_i}^{\tau_f} -mc^2 d\tau = -mc\int\limits_{\tau_i}^{\tau_f} \sqrt{-ds^2}$\egroup

That is the particle follows a path through 4-space that minimizes the space-time interval. In curved space-time, this path is called a geodesic.

We may take this back to an integral over proper time.

\bgroup\color{black}$\displaystyle S= -mc\int\limits_{\tau_i}^{\tau_f} \sqrt{-ds...
...u^2} d\tau
= -mc\int\limits_{\tau_i}^{\tau_f} \sqrt{-v_\mu v_\mu} d\tau$\egroup

While the factor of \bgroup\color{black}$ mc$\egroup may seem useless, it becomes useful if an external potential (other than gravity) is applied.

Jim Branson 2012-10-21