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Book: Classical Electrodynamics by J.D. Jackson (3rd ed.) - Chapter 12

Immediately after Eq. 12.8 he writes:

"The action (12.6) is proportional to the integral of the proper time over the path from the initial proper time $\tau_1$ to the final proper time $\tau_2$."

Eq. 12.6 reads:

$$ A=\int_{\tau_1}^{\tau_2} \gamma~L~d\tau , \tag{12.6} $$

and later in Eq. 12.7 he gives the Lagrangian of a free particle as

$$ L_{\rm free} = -m ~c^{2} \sqrt{1-\frac{u^2}{c^2}}. \tag{12.7} $$

I completely fail to understand his sentence in double quotes above (given these two equations). Please help.

Qmechanic
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Sashwat Tanay
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2 Answers2

5

  1. What Jackson is trying to convey (in his quote) is the fact that the action $S\equiv A$ of a relativistic massive point particle is $$S~=~ - E_0 ~ \Delta \tau, \tag{1}$$ where $$E_0~=~m_0c^2\tag{2}$$ is the rest energy, and $$\Delta \tau~=~\int_{\tau_i}^{\tau_f}\! \mathrm{d}\tau ~=~\tau_f-\tau_i \tag{3}$$ is the change in proper time. See also this Phys.SE post.

  2. Apart from the proportionality constant $E_0$, the formula (1) has a geometrical meaning via the principle of least action:$^1$ The massive point particle chooses the path that maximizes$^2$ its change in proper time, i.e. a timelike geodesic in spacetime.

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$^1$ As always, the devil is in the details: It does not make sense to use proper time $\tau$ as parametrization for the (possibly virtual) paths in the principle of least action because the corresponding boundary conditions $$ x^{\mu}(\tau_i)~=~x^{\mu}_i\qquad\text{and}\qquad x^{\mu}(\tau_f)~=~x^{\mu}_f\qquad\qquad(\leftarrow\text{Wrong!}) \tag{4} $$ would then make all paths have the same action value (1)! Instead we should pick another parametrization $\lambda$ of the paths with corresponding boundary conditions $$ x^{\mu}(\lambda_i)~=~x^{\mu}_i\qquad\text{and}\qquad x^{\mu}(\lambda_f)~=~x^{\mu}_f. \tag{5}$$ This means that we must relate the change $\mathrm{d}\tau$ in proper time $\tau$ to the change $\mathrm{d}\lambda$ in the new parameter $\lambda$. This relationship is mediated via the spacetime metric $$ c\frac{\mathrm{d}\tau}{\mathrm{d}\lambda}~=~\sqrt{g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}} , \qquad \dot{x}^{\mu}~:=~\frac{\mathrm{d}x^{\mu}}{\mathrm{d}\lambda}, \tag{6}$$ where we (as Jackson) have used the $(+,-,-,-)$ Minkowski sign convention. (We stress again that the parameter endpoints $\lambda_i$ and $\lambda_f$ are fixed, but that the change $\Delta\tau$ in proper time (3) is not fixed, so that the variational problem (1) is non-trivial.) In more detail, the action functional (1) therefore reads $$ S[x]~\stackrel{(1)+(2)+(3)+(6)}{=}~ -m_0c\int_{\lambda_i}^{\lambda_f}\! \mathrm{d}\lambda~\sqrt{g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}} .\tag{12.25} $$ The Euler-Lagrange (EL) equation for the action (12.25) is the geodesic equation, cf. e.g. this Phys.SE post. (The specific choice of the parametrization $\lambda$ will not matter because of parametrization invariance. Jackson's Lagrangian (12.7) corresponds to the so-called static gauge choice: $\lambda=t\equiv x^0/c$.)

$^2$ Notice the minus sign in eq. (1), which turns minima into maxima, and vice-versa.

Community
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Qmechanic
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4

In the action you quote, $A=\int_{\tau_1}^{\tau_2} \gamma L \, \mathrm d\tau$, the Lorentz factor $$ \gamma = \frac{1}{\sqrt{1-\frac{u^2}{c^2}}} $$ exactly cancels out the square root in the denominator coming from the Lagrangian $$ L = -mc^2\sqrt{1-\frac{u^2}{c^2}} $$ to give simply $$ \gamma L = -mc^2, $$ and therefore an action $$ A=\int_{\tau_1}^{\tau_2} \gamma L \, \mathrm d\tau =- mc^2\int_{\tau_1}^{\tau_2} \, \mathrm d\tau =-mc^2(\tau_2-\tau_1) $$ that's proportional to the ellapsed proper time interval.

Emilio Pisanty
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