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The geodesic equation can be derived using the action $$S_0 ~=~ \int d\tau \sqrt{-\dot{x}_\mu\cdot \dot{x}^\mu}\tag{1}$$ (I am using the (-+++) convention and $\dot{x} = \frac{dx}{d\tau}$). To simplify calculations one then chooses an explicit parametrization namely the arc length $\tau$ i.e. $$\dot{x}_\mu\dot{x}^\mu = -1.\tag{2}$$ From my point of view this means that I minimize the action with the constraint: $$\dot{x}_\mu\dot{x}^\mu + 1 = 0.\tag{3}$$ So the resulting equation should be the same if I use the following action instead $$S = \int d\tau \sqrt{-\dot{x}_\mu\cdot \dot{x}^\mu} + \lambda(\dot{x}_\mu\dot{x}^\mu + 1)\tag{4}$$ where $\lambda$ is a Lagrange multiplier.

Let's find the eom in Minkowski space: $$0 = \dot{p}_\mu = \frac{d}{d\tau}\left(\frac{-\dot{x}_\mu}{\sqrt{-\dot{x}_\mu\dot{x}^\mu}} + 2\lambda\dot{x}_\mu\right)\tag{5}$$ $$\dot{x}_\mu\dot{x}^\mu + 1 = 0.\tag{6}$$

The square root in the first equation equals 1. So $$p_\mu = (2\lambda - 1)\dot{x}_\mu.\tag{7}$$ From the second equation I find $$\ddot{x}^\mu \dot{x}_\mu = 0.$$ Using this $$\frac{d}{d\tau} \dot{x}^\mu p_\mu = \ddot{x}^\mu p_\mu + \dot{x}^\mu \dot{p}_\mu = 0.\tag{8}$$ So $$\mathrm{const.} = \dot{x}^\mu p_\mu = 1-2\lambda \Rightarrow \dot{\lambda} = 0\tag{9}$$ Putting all together yields: $$ \dot{p}_\mu = (2\lambda - 1) \ddot{x}_\mu = 0.\tag{10}$$

In the case $\lambda \neq \frac{1}{2}$ this simply gives the old eom $\ddot{x} = 0$. However in the case $\lambda = \frac{1}{2}$ there is no restriction to $\ddot{x}$.

I don't understand where this case $\lambda = \frac{1}{2}$ comes from. How do I deal with it? Can I simply neglect it? Or have I forgotten something?

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  1. First of all, we should stress that what OP calls $\tau$ is not$^{\dagger}$ proper time off-shell but just some world-line (WL) parameterization. However, the constraint $$\dot{x}_\mu\dot{x}^\mu ~\approx ~-1\tag{A}$$ will imply that the WL parameter $\tau$ is the proper time on-shell.

  2. Since the EOM depends on the first derivative $\frac{d\lambda}{d\tau}$ of the Lagrange multiplier, we should specify a single condition, e.g. an inertial condition (IC) for $\lambda$. If we choose the IC different from $1/2$, we avoid the problem when $\lambda$ is $1/2$.

  3. The nature of the $\lambda=1/2$ pathology is a degeneracy of the constraint force/missing rank issue. To see this more clearly note that we can get an equivalent action $$ \tilde{S} ~=~ \int_{\tau_i}^{\tau_f}\!\mathrm{d}\tau \left(\sqrt{1} + \lambda(\dot{x}_\mu\dot{x}^\mu + 1)\right) \tag{B}$$ by inserting the constraint (A) into the first term in OP's action (4). If we repeat OP's calculation for the equivalent action $\tilde{S}$ we will see that the trouble has shifted to $\lambda=0$. Clearly, the case $\lambda=0$ corresponds to a degenerate case where the stationary action principle (B) is ill-defined.

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$^{\dagger}$ If the WL parameter $\tau$ is proper time off-shell as well, it would mean that OP's action (4) is just $S=\tau_f-\tau_i$, which is fixed by boundary conditions (BC). In other words, the action would not depend on the WL, i.e. the variational problem would be ill-defined.

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  • $\begingroup$ 2. ok that's a good point. But I guess this problem will also occur in other Lagrangians. Is there a way to find a non-critical value of $\lambda$ in general? 3. Is there any way to fix this degeneracy? Like adding another constraint? $\endgroup$
    – toaster
    Oct 15, 2018 at 20:59
  • $\begingroup$ 2. No, it will depend on the variational problem. 3. Yes. See point 2. $\endgroup$
    – Qmechanic
    Oct 16, 2018 at 1:55
  • $\begingroup$ I actually don't see where this degeneracy comes from. This does not happen when using Lagrange multipliers to minimize a function $f: \mathbb{R}^n \to \mathbb{R}$ with constraints. Why does it happen here? At what prevents me from choosing $\lambda = \frac{1}{2}$? $\endgroup$
    – toaster
    Oct 16, 2018 at 11:46
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    $\begingroup$ A similar phenomenon appears if we try to find stationary points for a function of the form $$f(g(x))+\lambda g(x),\qquad x\in \mathbb{R}^n. $$ $\endgroup$
    – Qmechanic
    Oct 16, 2018 at 12:19
  • $\begingroup$ Would it be possible to solve this by demanding that the Euler Lagrange equations should hold for every $\lambda$? $\endgroup$
    – toaster
    Oct 18, 2018 at 13:38

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