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My background is in maths, but I have been studying some basic physics with occasional input from a friend who is studying for a physics PhD. Due to my background, I am keen to visualize things geometrically, and find the idea that the Lagrangian might behave as a (pseudo-Riemannian) metric very appealing. The question "Can Lagrangian be thought of as a metric?" discusses this, but the answers there only address the case where potential energy terms can be neglected so that the corresponding metric is positive definite. I am specifically interested in the more general case of a mixed-signature metric.

Let $M$ be a pseudo-Riemannian manifold with metric $g$. Geodesics $\gamma$ : $[\tau_0,\tau_1] \rightarrow M$ can be characterized as being stationary for the functional $E(\gamma) = \int g\left(\frac{d\gamma}{d\tau},\frac{d\gamma}{d\tau}\right)d\tau$ over curves with fixed endpoints $\gamma(\tau_0),\gamma(\tau_1)$ where $\frac{d\gamma}{d\tau}$ is never zero. The co-ordinates on $M$, and hence the components of $\frac{d\gamma}{d\tau}$, can be labelled however we like: for example, some of the latter could represent velocities (or momenta) of particles while others represent quantities like 1/distance (or wavenumber). We can think of the coefficients in $g$ as scaling factors which ensure that the corresponding components of $g\left(\frac{d\gamma}{d\tau},\frac{d\gamma}{d\tau}\right)$ have the appropriate weightings (and "units").

If the integrand in $E$ represents a Lagrangian, then the components of $\frac{d\gamma}{d\tau}$ must split into those which contribute to kinetic energy and those which contribute to potential energy. Depending on the choices of units for the scaling factors in $g$, the former look like velocity or momentum ($n$-dimensional), while the latter look like 1/distance or wavenumber (1-dimensional). The corresponding coefficients in the metric must have opposite signs; alternatively, one set of components of $\frac{d\gamma}{d\tau}$ can pick up a factor of $i$, to make all coefficients positive. Note that velocity and momentum are quantities which "involve time" while 1/distance and wavenumber are quantities which do not. Finally, momentum and wavenumber are connected via Planck's constant (the quantum of action) in the de Broglie relations, while $E$ itself corresponds to action.

To me, all of this seems very suggestive of a connection with relativity. Could any of what I've said be meaningful? Is any aspect of it discussed in the literature in any way?

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    This question appears to be asking for opinions, which makes it off-topic. – Kyle Kanos Apr 26 '15 at 16:49
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    I'm not quite sure what your question really is - the energy functional you write down is precisely what we take as the action of a test particle. What exactly do you want to know about that? – ACuriousMind Apr 26 '15 at 17:20
  • A metric is a defining mathematical property of a special linear space, which in themselves are special topological spaces. You can study the properties of these spaces all by themselves without ever encountering a physics problem. The only physical question would be why macroscopic physics seems to invariably converging on equations of motion that have these properties. Nobody can answer that for you, at the moment, because we do not know what microscopic physics (if any) causes this layer of reality. – CuriousOne Apr 26 '15 at 18:20
  • StackExchange has informed me that I can only notify one user per comment, so I'll respond to each comment separately.

    @KyleKanos: the question was meant as a request for known information and/or references on related material, not for opinions.

    – Robin Saunders Apr 26 '15 at 18:39
  • @ACuriousMind: M is not spacetime, g is not the usual metric tensor on spacetime, γ is not the path of a single particle and τ is not proper time. I will rewrite my question to try to make that more clear. – Robin Saunders Apr 26 '15 at 18:39
  • @CuriousOne: I am not asking where the equations of motion ultimately come from, but rather whether there is a known connection (via geodesics) between the "mixed signature" metrics which feature in both the Lagrangian - a classical concept - and relativity. – Robin Saunders Apr 26 '15 at 18:39
  • @Robin: You are proposing a theory/relation and then asking Could any of what I've said be meaningful; this is asking for an opinion, plain & simple. If you want something different, ask that instead. – Kyle Kanos Apr 26 '15 at 18:43
  • The Lagrangian is not a classical concept. It exists in relativity and quantum mechanics as well. That's exactly what I meant to express. We do, to the best of my knowledge, not know "why" nature seems to preserve the concept of the Lagrangian in all theories. That question is, without having observations of a non-Lagrangian theory, not even a meaningful physical question. In my opinion, and that is all this is, Lagrangians AND metrics occur as artifacts of the statistical mechanics of a deeper layer of nature that we haven't seen, yet. – CuriousOne Apr 26 '15 at 18:48
  • @KyleKanos: I apologize. Perhaps my poor choice of wording comes from my being used to mathematicians, for whom there are only facts and not theories - so when I asked "could this be meaningful", I suppose I was using that as shorthand for "is there a known mathematical framework that this fits into or relates to in some way?" I admit there's some leeway in interpreting "in some way", but I'm certainly interested in answers which talk about facts, not opinions. I am currently rewriting the original post. – Robin Saunders Apr 26 '15 at 18:52
  • @CuriousOne: Sorry, I meant "classical" in the sense of "not-necessarily-non-classical": in other words, you don't need relativity to define a Lagrangian, but when you interpret it geometrically then this mixed-signature metric still shows up. I would then be interested to know whether this mixed metric is mathematically connected to the mixed metric of relativity, a question that I hoped theoreticians might address without necessarily referencing physical interpretations. I appreciate your comments about deeper "physical meaning", but that was not what I had intended to ask about. – Robin Saunders Apr 26 '15 at 18:58
  • @RobinSaunders: I have no idea what you mean by "not-necessarily non-classical". The existence of a Lagrangian in physical systems is an observation, not a free choice or even consequence of theory. The mixed sign of the metric of spacetime is also an observation on a certain scale (the scale on which c is not large), not a choice. My opinion is that both can be derived from statistical properties of the theory of everything... which is obviously something that we don't have, yet. You could turn the question around and ask how that assumption limits the structure of the TOE. – CuriousOne Apr 26 '15 at 19:06
  • When I say that the Lagrangian is a classical or "not-necessarily-non-classical" concept, I just mean that it exists even in models which do not take relativity into account. Those models exist independently of the phenomena they are supposed to describe, and it's the relationship between certain aspects of the models that I am interested in. – Robin Saunders Apr 26 '15 at 19:59
  • All of that said, however, I've noticed that the question as I've formulated it rests crucially on a false assumption: the momentum terms appearing in the expression for kinetic energy are quadratic, and I was writing under the assumption that the 1/distance terms appearing in the expression for potential energy were likewise quadratic, but of course they are actually linear. I believe something resembling my question can still be salvaged, but it will require a reworking of the setup. So until I've done that, this question should be ignored. – Robin Saunders Apr 26 '15 at 20:06

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