Graduate If the solution of a field vanishes on-shell does it mean anything?

[Baela]

Let us consider an action $S=S(a,b,c)$ which is a functional of the fields $a,\, b,\,$ and $c$. The solution of the field $c$ is given by the expression $f(a,b)$. On taking into account the relations obtained from the solutions for $a$ and $b$, we find that $f(a,b)=0$. If the solution for the field $c$ vanishes on-shell, does it mean anything particular?

 

[member38644]

Yes, the fact that the solution for the field $c$ vanishes on-shell has a specific meaning in this context. It means that when the fields $a$ and $b$ satisfy the equations of motion, the field $c$ automatically satisfies the equation $f(a,b)=0$. This is known as the on-shell condition for the field $c$.

Physically, this on-shell condition indicates that the field $c$ does not have any independent dynamics and its behavior is completely determined by the fields $a$ and $b$. This can be interpreted as a constraint on the dynamics of the system, where the value of $c$ is completely determined by the values of $a$ and $b$ at every point in spacetime.

In some cases, the on-shell condition for a field may also arise due to symmetries or conservation laws in the system. For example, if the action $S$ is invariant under a certain symmetry transformation, then the on-shell condition for the corresponding field would be a consequence of this symmetry.

In summary, the vanishing of the solution for a field on-shell has a significant meaning and can provide insights into the dynamics and symmetries of the system.