# Relationship between Lagrangians describing a particle interacting with a scalar field

In Susskind's Particles and Fields lecture, he considered the Lagrangian obtained by considering a particle and the effects of a scalar field $$\phi(t, x)$$ with coupling constant $$g$$ on the particle (20:45) as follows: $$\int L_P \ dt = \int -(m + g\phi)\sqrt{1 - \dot{x}^2} \ dt,$$ where the speed of light $$c=1$$ is one. Then, he considered the traditional Lagrangian of a field, and added the effect of a stationary particle (with $$\sqrt{1 - \dot{x}^2} = 1$$) on the field from the previous Lagrangian (41:37): $$\int L_F \ dx \ dt = \int \left[\frac{1}{2}\left(\frac{\partial \phi}{\partial t}\right)^2 - \frac{1}{2}\left(\frac{\partial \phi}{\partial x}\right)^2 + g\phi \delta(x)\right]\ dx \ dt.$$ I understand the use of $$\delta(x)$$ (the Dirac delta function) to rewrite the field in the context of the double integral, and that $$m$$ can be discarded for the purposes of determining the equations of motion, since it will vanish in the Euler-Lagrange equations. But I have two questions:

1. How did Susskind simply take the effect of the field on the particle in $$L_P$$ and insert it into $$L_F$$ to represent the effect of the particle on the field? I very vaguely understood it as a sort of "action-reaction" pair, but I would like to know a more rigorous treatment of this insertion and why it works for Lagrangians.

2. Susskind mentions that there is only one action for this system (29:09). Is he saying that since both equations describe the same system, both result in the same action? I would like a clarification about what he's referring to when he says that.

Susskind is essentially just saying that the total action of fields $$\phi$$ and point particles $${\bf r}$$ consists of 3 parts:
1. A free action $$S[\phi]= \int d^4x~{\cal L}$$ for the fields.
2. A free action $$S[{\bf r}]= \int dt~ L$$ for the point particles.
3. An interaction term of the form \begin{align}S_{int}[{\bf r},\phi] ~=~&-g\int dt~d^3{\bf x} ~\delta^3({\bf x}-{\bf r}(t))\phi({\bf x},t) \sqrt{1-\frac{\dot{\bf r}(t)^2}{c^2}}\cr ~=~&-g\int dt~\phi({\bf r}(t),t)\sqrt{1-\frac{\dot{\bf r}(t)^2}{c^2}}.\end{align}
When we then derive the 2 types of Euler-Lagrange equations for fields $$\phi$$ and point particles $${\bf r}$$, the interaction term will then lead to a source term for both types.