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

AI Thread Summary
The action S=S(a,b,c) describes a functional relationship among fields a, b, and c, where the solution for field c is expressed as f(a,b). When f(a,b)=0, it indicates that c vanishes on-shell, meaning that c's behavior is entirely determined by fields a and b when they satisfy the equations of motion. This on-shell condition signifies that field c lacks independent dynamics, acting as a constraint on the system. Additionally, such conditions may arise from underlying symmetries or conservation laws within the action S. Overall, the vanishing of field c on-shell reveals important insights into the system's dynamics and symmetries.
Baela
Messages
17
Reaction score
2
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?
 
Physics news on Phys.org


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.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top