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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.

Qmechanic
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Prajith Velicheti
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1 Answers1

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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.

A very similar division into 3 parts happens for point charges and E&M fields, cf. e.g. my Phys.SE answer here.

Qmechanic
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