The most commom derivation I've seen of the geodesic equation of a massive particle is by the use of the Variational Principle. My problem is that I can't realize what the meaning of find a spacetime path (the geodesic) such that the proper time is extremized. (If the signature is $(+,-,-,-)$ it should be a maximum as some textbooks say.)

I understood that the action integral must be proportional to the line element $ds$ because we need that all the observers compute the same value of action to obtain the same equations of motion.

What I don't understand is the physical meaning of finding a maximum proper time instead a minimum, and what physical implications it leads to. How can I conclude that what I need to find a geodesic equation is maximize proper time of the massive particle? If possible, make an analogy with the Minkowski Space.

  1. Let us for simplicity consider Minkowski space although the generalization to curved spacetime is straightforward. Lorentz invariance suggests that the Lagrangian one-form for a massive point particle should be $$\mathbb{L}~=~ f(\dot{x}^2)\mathrm{d}\lambda, \qquad \dot{x}^2~:=~\eta_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}~>~0, \qquad \dot{x}^{\mu}~:=~\frac{dx^{\mu}}{d\lambda}, \qquad x^0~\equiv~ct, \tag{1}$$ for some function $f$. Here $\lambda$ denotes a world-line (WL) parameter, and a dot denotes differentiation wrt. $\lambda$. We chose Minkowski signature $(+,-,-,-)$ so that timelike vectors have positive length.

  2. WL reparametrization invariance implies that the function $$f~\propto ~\sqrt{\cdot}\tag{2}$$ is proportional to a square root.

  3. In other words, the variational principle finds stationary paths for timelike arc-length (or equivalently, proper time multiplied with $c$): $$c\tau~=~\int_{\lambda_i}^{\lambda_f}\!\sqrt{\dot{x}^2}\mathrm{d}\lambda. \tag{3}$$ The corresponding Euler-Lagrange (EL) equations are the geodesic equations. In Minkowski space the geodesics are just straight lines.

  4. Let us impose boundary conditions (BCs) $$ x(\lambda_i)~=~x_i\qquad\text{and}\qquad x(\lambda_f)~=~x_f. \tag{4}$$ By changing coordinate system, we may assume that ${\bf x}_i={\bf x}_f.$ Finally let us choose static gauge $\lambda=t$.

  5. Then eq. (3) becomes $$c\tau~=~\int_{t_i}^{t_f}\!\sqrt{c^2-\dot{\bf x}^2}\mathrm{d}t. \tag{5}$$ Eq. (5) is clearly maximal for $\dot{\bf x}={\bf 0}$, i.e. a particle at rest, which is also what a free particle would do with the given BCs. This answers OP's question.

  • $\begingroup$ Thank you so much, your explanations completely clarifiedy mind about this topic and made me understood where's my logical mistakes. $\endgroup$ Feb 8 '20 at 15:10
  • $\begingroup$ is this phenomenon directly associated with time dilation or is it another thing suitable thing? $\endgroup$ Feb 8 '20 at 17:43
  • $\begingroup$ I found a document with a similar explanation, like yours and at after obtained the result the author states that: "Don’t confuse this result with the time dilation principle, since they are considering different cases. Time dilation is an effect that occurs when two different observers (one moving at a speed v relative to the other) measure the time interval between two events on the same world line. In the current case, we’re looking at the difference in proper time between two different world lines, as analyzed in the same inertial frame. " $\endgroup$ Feb 8 '20 at 17:51
  • 1
    $\begingroup$ The particle's trajectory is independent of coordinate system. Time dilation depends on the observer's coordinate system/reference frame. $\endgroup$
    – Qmechanic
    Feb 8 '20 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.