9
$\begingroup$

A well known derivation of the free-space Lagrangian in Special Relativity goes as follows:

  • The action $\mathcal{S}$ is a functional of the path taken through configuration space, $\mathbf{q}(\lambda)$, where $\lambda$ is the path parameter

  • The action can be thought of as the total 'cost' of this path through configuration space. The path which is chosen is the 'cheapest' of these paths (i.e. the one which minimises the action)

  • 'Valid physics' can be retrieved by correctly assigning each point along the path a `cost', to do so we invoke a function called the Lagrangian, $\mathcal{L}$, such that:

$$\mathcal{S}[\mathbf{q}] = \int_{\lambda_1}^{\lambda_2} \mathcal{L}(\mathbf{q}(\lambda), \dot{\mathbf{q}}(\lambda), \lambda) \mathrm d \lambda \tag{1}$$

  • The extremal $\mathcal{S}$ is given when $\mathcal{L}$ satisfies the Euler-Lagrange equations.

  • In free space (assumed to be homogeneous and isotropic), the `cost' of each point along the path cannot be determined by either the position along the path, or the position in configuration space, as this would violate our free-space assumptions.

  • The only determining factor that can be allowed to influence the total cost of each point in space is the infinitesimal path length at each point, up to a dimensional constant $\alpha$. Therefore: $$ \mathcal{S}[\mathbf{q}] = \alpha \int_\mathbf{q} \mathrm d s \tag{2}$$

  • Using $\mathrm d s^2 = \mathrm d t^2 - \mathrm d \mathbf{x}^2 $, this gives: $$ \mathcal{S}[\mathbf{q}] = \alpha\int \sqrt{1 - \dot{x}^2} \mathrm d t\tag{3}$$
  • We choose $\alpha = - m c^2$ as the simplest invariant quantity that has the correct dimensions. Therefore if our path parameter is the coordinate time $t$, we have: $$ \mathcal{L} = - m c^2 \sqrt{1 - \dot{x}^2} \tag{4}$$

This proof is found in many different sources (probably most notably in Landau-Lifshitz Volume 2, Chapter 2). This idea generalises into General Relativity, where the free-space Lagrangian is: $$ \mathcal{L} \propto \sqrt{g_{\mu \nu} \dot{x}_\mu \dot{x}_\nu} \tag{5}$$ However. If we try to insert the Newtonian Euclidean 3-metric, it seems that we don't get the expected result: $\mathcal{L} = \frac{1}{2} m v^2 $. If we insert the Euclidean metric into the Landau-Lifshitz general definition, we find: $$ \mathcal{L} \propto |\mathbf{v}| \tag{6}$$ The equations of motion predicted by the normal Lagrangian are a statement of Newton's I axiom ($p = $ const in free space), but the result of this Lagrangian is: $$ \text{sgn}(v_i) = \text{const} \tag{7}$$ This isn't wrong, but it clearly does not contain all the information we expect the Lagrangian to contain!

Why does this approach (which has such resounding success in the relativstic case!) fail so badly when applied to the (supposedly simpler) Newtonian case? I know that under certain circumstances we can square the Lagrangian and retain the same equations of motion, but those proofs all relied on affine parameters etc., which seems like overkill for a Newtonian mechanics problem.

Am I missing something obvious? It seems like it should be trivial to recover classical mechanics from this method, when it is so 'easy' to get relativstic mechanics from it....

$\endgroup$
13
$\begingroup$

The problem with your approach is that your proposed action $$S = \int |\mathbf{v}| \, dt$$ is not invariant at all. While Landau's action is invariant under Lorentz transformations, and in fact completely coordinate independent, yours is not invariant under even Galilean transformations, which add a constant to $\mathbf{v}$. The space-only analogue of a Lorentz transformation is not a Galilean transformation, it's a rotation. Your result is invariant under rotations, but that's not enough.

Taking the nonrelativistic limit is not as simple as forgetting the time component. Parametrizing by time, the relativistic Lagrangian can also be written as $$\mathcal{L} = \sqrt{\dot{t}^2 - \dot{\mathbf{x}}^2} = \sqrt{1 - \mathbf{v}^2}.$$ We can arrive at your answer "getting rid of the one", but that's not the right way to take the limit. Instead the one becomes more important, and we have to keep it and Taylor expand about it, giving $$\mathcal{L} \approx 1 - \frac{\mathbf{v}^2}{2}$$ which recovers the usual nonrelativistic action. The argument for why we have to square $\mathbf{v}$ is also given in volume 1 of Landau and Lifshitz. Perhaps a simpler way to say this is that Galilean physics isn't simply relativity with the time dimension removed.

$\endgroup$
  • $\begingroup$ I of course agree with the second half of your post (regarding the Taylor expansion), and I agree that my Lagrangian is not symmetric. However, I guess my real question is why that is the case. It seems obvious that the metric should encode all symmetries (i.e. the Minkowski metric encodes the Lorentz symmetries), so why doesn't the Euclidean metric encode the correct symmetries? What line element should one use if one were to derive pure classical physics from this same argument (i.e. not in the limit of relativity) $\endgroup$ – almightyjack May 26 at 12:51
  • 1
    $\begingroup$ @almightyjack Nonrelativistic physics naturally separates space and time, so what we usually call a metric there is just the spatial metric. If you did want a geometric formalism, you might want to look into Newton-Cartan theory, which has two metrics, one to measure spatial distance and one to measure times. I'm not sure if that'll lead directly to what you want, though. $\endgroup$ – knzhou May 26 at 12:59
  • $\begingroup$ thanks! This is something that's been bugging me for a long time. It feels like it should be addressed somewhere, as it is a very natural thing to do, when presented with the elegant L&L proof.... The 'correct' way I guess is to enforce Galilean invariance by hand (i.e. expand in powers of |v| and throw away any terms that are not total derivatives), but I just wanted to produce a more elegant analogue. I will let you know if Newton-Cartan theory gets me anywhere :) $\endgroup$ – almightyjack May 26 at 13:10
  • $\begingroup$ @knzhou : Or, effectively, that space and time "pull apart" in the limit as $c \rightarrow \infty$. Suggesting another interpretation of $c$: it's the "extent to which space and time mingle" (or better, of $\frac{1}{c}$, which is also the per-distance minimum latency). $\endgroup$ – The_Sympathizer May 27 at 0:15
6
$\begingroup$
  1. A non-relativistic square root Lagrangian (6) has an issue with null-velocity vectors. It is not differentiable at zero velocity, which is unphysical from the perspective of Newtonian mechanics, cf. this related Phys.SE post.

  2. We should point out that the relativistic square root Lagrangian (5) has a similar shortcoming with null/lightlike vectors. It is only applicable to timelike vectors corresponding to massive point particles, cf. e.g. this Phys.SE post.

$\endgroup$

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.