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All the fundamental laws of physics can be written in terms of an action principle. This includes electromagnetism, general relativity, the standard model of particle physics, and attempts to go beyond the known laws of physics such as string theory. For example, (nearly) everything we know about the universe is captured in a Lagrangian where the terms carry the contributions of Einstein, Maxwell (or Yang and Mills) and Dirac respectively, and describe gravity, the forces of nature (electromagnetism and the nuclear forces) and the dynamics of particles like electrons and quarks.

Source: David Tong

Me: I am a second year undergrad and have a nice familiarity with the -1/4 F_ij F_ij term (electromagnetism) and how it results to the Maxwell's Equations, but am still curious to know how it seems like there is always a Lagrangian. Apart from one obvious advantage that the Euler Lagrange equations hold in any coordinate system and the Lagrangian holds the key to the symmetries of the system.

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    $\begingroup$ It's a good question, but always might be a little too strong. For some insight about classical models, see How do I show that there exists variational/action principle for a given classical system?. For quantum models, see Tachikawa's slides "What is Quantum Field Theory?" (link to pdf), and also How general is the Lagrangian quantization approach to field theory? $\endgroup$
    – Chiral Anomaly
    Commented Apr 9, 2021 at 20:30
  • $\begingroup$ There is a self-answer by me on mathoverflow: derivation of Hamilton's stationary action from the work-energy theorem (Hamilton's stationary action as a formulation of classical mechanics.) I'm confident the reasoning presented generalizes to application of variational calculus in other dynamics fields, so that answer may well address your question. (I posted that derivation on mathoverflow because the derivation is application of mathematics only; in terms of physics content the work-energy theorem and Hamilton's stationary action are identical.) $\endgroup$
    – Cleonis
    Commented Apr 9, 2021 at 20:55
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    $\begingroup$ Most interesting quantum systems do not have a Lagrangian description. At least quantum mechanically, theories which have a weak coupling limit are the only theories for which Lagrangians exist. So any strongly coupled system (of which there are many) do not have a Lagrangian. $\endgroup$
    – Prahar
    Commented Apr 9, 2021 at 21:20
  • $\begingroup$ About derivation: stackexchange contributor knzhou has made some very incisive statements about derivation in physics "[...] in physics, you can often run derivations in both directions: you can use X to derive Y, and also Y to derive X. That isn't circular reasoning, because the real support for X (or Y) isn't that it can be derived from Y (or X), but that it is supported by some experimental data D. This two-way derivation then tells you that if you have data D supporting X (or Y), then it also supports Y (or X)." $\endgroup$
    – Cleonis
    Commented Apr 9, 2021 at 21:39
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    $\begingroup$ @ChiralAnomaly thanks.. that was an awesome summary by Tachikawa. looking forward to taking this attitude in my future research $\endgroup$
    – Gravity_CK
    Commented Apr 24, 2021 at 14:51

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Why does it seem like there is always a Lagrangian?

Typically we study forces of nature using differential equations, and these differential equations can typically be re-written in terms of other formalisms such as Lagrangian or Hamiltonian formulations.

In many cases (not all), we are able to understand the forces of nature using a Lagrangian formulation. This is probably "why [] it seem[s] like there is always a Lagrangian."

There is likely no other really satisfying answer appropriate to this forum--this forum is focused on main stream physics, not philosophy--or perhaps any forum.

All the fundamental laws of physics can be written in terms of an action principle.

Not necessarily.

In fact, even all the forces we know of are not immediately susceptible to the usual Lagrangian way. For example: Quantum gravity. We have yet to make sense of such a theory in the usual sense.

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