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I am a math student and I have taken four-five lessons about special relativity in a course about Lagrangian and Hamiltonian mechanics, so be patient with me if my question is stupid. My teacher says that for a relativistic particle with an external vector potential, we can write the following Lagrangian (he says that's not the only possible choice)

$$L = -mc\sqrt{-\eta_{\mu\nu}u^\mu u^\nu} + eA_{\mu}(x^\lambda)u^\mu.$$

If we fix a particular parametrisation (proper time or relative time) we get the other two "classics":

$$L' = \frac{m}{2}\eta_{\mu\nu}u^\mu u^\nu + eA_{\mu}(x^\lambda)u^\mu.$$

$$L'' = mc\sqrt{c^2 - |v|^2} - e(cA_0 + A_i v^i).$$

From the last he said that we can see that "$A_0$ is the classical newtonian potential". So for example, I suppose that if I have a spring, I can set $A_0 = -\frac{1}{2}kx^2$ to study the motion.

But then he said that we cannot include gravity in special relativity (with fixed Minkowski metric $\eta_{\mu\nu}$) because it is an action at distance and that's clear to me.

What is not clear to me is why we cannot simply put $A_0 = \frac{1}{\rho}$ to include gravity. Here $\rho$ is the distance from the origin in polar/spherical coordinates.

I cannot see (at least formally) difference between gravity and elastic force.

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  • $\begingroup$ Where does the gravitational constant $G$ come into the Lagrangian then, if it's meant to describe Newtonian gravity? Actually, what do you see as the difference between gravity and elastic force in the classical case? $\endgroup$ – GodotMisogi Dec 23 '18 at 14:29
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    $\begingroup$ FWIW, there's a nice exercise (7.2 on page 179) in "Gravitation" where you are asked to show that a 'gravitational vector potential' yields gravitational waves that transport negative energy. $\endgroup$ – Alfred Centauri Dec 23 '18 at 14:33
  • $\begingroup$ @GodotMisogi is the constant important? I will set $G/\rho$ . Actually I don't see any difference between the two forces: for me are only different mathematical function and gravity has nothing special . What am I missing? $\endgroup$ – Marco Francesco Nervo Dec 23 '18 at 14:45
  • $\begingroup$ I think this conversation should be moved to chat, where some of the more basic conceptual difficulties can be addressed before this question is phrased clearly enough to be answered. (Comments are not for extended discussion.) $\endgroup$ – GodotMisogi Dec 23 '18 at 15:05
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    $\begingroup$ Excuse me, but what's not "phrased clearly"? I am asking why I cannot insert a gravitational potential while I can insert an elastic one. I don't see any (mathematical) difference bewteen them. If you have an answer, I'll please to read it; otherwise you just look arrogant $\endgroup$ – Marco Francesco Nervo Dec 23 '18 at 16:16
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One of the ground-lying concepts of the Lagrange formalism is the requirement of the invariance of the Lagrangian upon any kind of symmetry transformations. Once the Lagrangian of a physical theory checked on symmetry invariance (and showing that it is indeed invariant), we can be sure that the Lagrange-Euler equations (being equivalent with the equations of motion (or field equations if fields are studied) in that theory) also fulfill this symmetry (on which the considered theory is based upon).

In case of the Lagrangian given in your example, we observe that it is written in relativistic invariant (with 4-vectors) form. Actually the problem should be considered from a rather different point of view. We want to describe the motion of the particle in an electromagnetic (EM) field. Einstein actually found the theory of special relativity by consideration of the electromagnetical phenomena. If we want to find the equations of motion which describe the motion of a particle in an EM field, we have to require that they are compatible with the special theory of relativity. How to do this ? Setting up a Lagrangian which is does not change under changes of inertial reference systems (This is the symmetry invariance that we require). The 4-vector formalism, which is used in the given Lagrangian, guarantees this as long as $L$ is a 4-scalar which does not change if it is considered in another inertial reference system (for instance no change upon a change from rest to a train moving with constant velocity).

Including just the scalar potential of the Newton's gravity in the zero component of the 4-vector potential does not fulfill the requirement of invariance (of no change) of the Lagrangian $L$ if the inertial reference system is changed.

In order to understand this better, let's already look the electromagnetic theory. A quantity of 4 components $A_{\mu}$ is necessary to describe it. At rest a charged point particle could be described by a potential function $A_0\sim e/r$. However, if the charged point particle is observed from a moving inertial reference system, that's no longer the case. A magnetic field can be also observed with circular field lines around the moving particle. Due to the change of the reference system, the other 3 components of the $A_{\mu}$ have become non-zero. Aha, then may be we can also get additional contributions to $A_{\mu}$ from a gravitational field when observed from a moving reference system ?

The answer is no. Gravitation is not a theory which could be described by a 4-vector field simply because gravitation is only attractive, not repulsive. In EM-theory there are positive and negative charges, however, in gravitation there is only one type of matter which only exerts attractive forces on other matter. (In order to understand the latter thoroughly, field theory has to be studied which is out of scope of this answer.)

Actually, Einstein soon realized after his development of special relativity that gravitation cannot be described simply be a 4-scalar potential theory (neither a 4-vector theory), it would not fulfill the invariance under changes of inertial reference systems. Newton's theory of gravity considers instant action of forces between celestial bodies, something which strongly contradicts the special theory of relativity. No, it had/has to be much more complicated which eventually led him to the theory of General relativity, a tensor theory.

The last possibility would be to consider only a Lagrangian of classical Newton theory. However, the given Lagrangian would be a mixture of relativistic invariant parts and relativistic non-invariant parts. Such an approach is neither seriously considered in physics, therefore to be excluded.

So the gravitation cannot be included by simply adding a term $\sim M/r$ (or $\sim M/\rho$ if you prefer this notation) to $A_0$ in the Lagrangian. One has to accept that electromagnetism and gravity are very different theories, even if on the first sight it does not look like this.

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  • $\begingroup$ If I write my lagrangian "with gravity", is there an inertial change of coordinates such that an observer could see that force being repulsive? Or better: does this happen with Coloumb force? $\endgroup$ – Marco Francesco Nervo Jan 16 '19 at 16:20
  • $\begingroup$ Let me try... I have $L = -mc\sqrt{-(u^0)^2+(u^1)^2} - \frac{u^0}{x^1}$ and I make the change of frame $y^0 = -x^0$, $y^1 = x^1$. This is a Lorentz transform and the new Lagragian is $L = -mc\sqrt{-(v^0)^2+(v^1)^2} + \frac{v^0}{y^1}$, so the force became repulsive. Is this correct? $\endgroup$ – Marco Francesco Nervo Jan 16 '19 at 16:40
  • $\begingroup$ Oh no, that's not a Lorentz transformation, because it's not orientation-preserving.. $\endgroup$ – Marco Francesco Nervo Jan 16 '19 at 21:48
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My understanding is that $A_0$ in your analysis is the scalar φ component of the electromagnetic potential from which E and B fields and electromagnetic waves can be defined. One can form it as a four vector of special relativity.

You cannot mix a la cart the potentials of different forces. Newtonian gravitation has only the 1/r scalar potential because it is not Lorentz invariant and one cannot define a four vector for it. That is why you cannot treat the $A_0$ as just an independent potential, since Maxwell equations are Lorenz invariant.

Four vectors are inherent in General Relativity which describes Newtonian gravitation at the limit, but is a different story.

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