# Problem conciliating relativistic momentum with Hamilton-Jacobi relations: massive object going at (imaginary) light-speed?

I can't seem to make sense of a strange paradox emerging from my attempts to conciliate the two physical statements described in the title. I'm sure it's some silly mistake I made in the process to cause it, but I can't identify why, and even my best guesses about the kind of mistake don't look to me as likely to generate it. I'd really appreciate any insight/explanation/correction/clarification.

1. In Special Relativity, assuming for simplicity a point-like, massive, free body moving along a single coordinate x (thus no quadri-potentials, no gravity, etc.), I have this equation for relativistic linear momentum along that coordinate in terms of gamma factor (depending in general on the velocity), rest mass and velocity: $$p_x=\gamma m v_x$$
2. Of course, in my reference frame the velocity is, trivially: $$v_x=\frac{\partial x}{\partial t}$$
3. I can use the mass-energy equivalence to replace the rest mass times gamma with the total energy, using the squared of the speed of light as proportionality factor: $$p_x=\frac{E v_x}{c^2}$$
4. If I want to solve for the velocity, I trivially get: $$v_x=\frac{p_x c^2}{E}$$
5. From (classical) Hamilton-Jacobi relations (which every single source I've found so far confirms can also apply to Special Relativity, provided that the Hamiltonian also includes the rest energy term) I can find the Hamiltonian $$H$$ as (minus) the partial time-derivative of the Hamilton-principal-function $$S$$ (analogous to action): $$H=-\frac{\partial S}{\partial t}$$
6. In a simple reference frame which doesn't depend explicitly on time, I can identify this Hamiltonian with the total energy of the body: $$E=-\frac{\partial S}{\partial t}$$
7. I can use the Hamilton-Jacobi relations for the momentum along $$x$$ as well, as a partial coordinate-derivative of the same $$S$$ (in the relativistic case, mechanical and canonical momentum are the same since I'm taking a simple case without potentials): $$p_x=\frac{\partial S}{\partial x}$$
8. If I try to match 4 with 6 and 7, I get: $$v_x=-\frac{\frac{\partial S}{\partial x}}{\frac{\partial S}{\partial t}}c^2$$
9. Which matching with 2 in "well-behaved enough" conditions (more on this later) should simplify as: $$v_x=-\frac{\partial t}{\partial x}c^2=-\frac{1}{v_x}c^2$$
10. This is quite alarming: while dimensionally the equation is still ok (the light-speed squared factor fixes the units), quantitatively speaking I'm equating a velocity with a negative reciprocal of a velocity, so much that if I try to solve I get: $$v_x=\pm \sqrt{-c^2}=\pm i c$$

I don't like the fact that massive objects can travel at the speed of life, let alone that they must always go at the speed of light, let alone that it's actually an imaginary speed of light! This seems quite evil.

SOME POSSIBLE (hints to) SOLUTIONS

Just to save some time to the kind answerers, I listed here, in order of increasing likeliness (according to me, that is), the things that I could possibly have got wrong:

• I could have messed up with rest/invariant vs relativistic/total quantities (I know that many people get $$E=mc^2$$ wrong, comparing total energy with rest mass without the gamma in non-stationary cases), but it really doesn't look like I did; also, I really struggle to see how a similar mistake could solve the "paradox", since it doesn't seem like multiplying or dividing for gamma once would improve much.
• I could have messed up with considering the Hamiltonian in 5 as the total energy in 3 (after all I'm admittedly using a classical result in a relativistic setup), but every source so far confirmed that in simple setups that should be exactly the case; also, I really struggle to see how a similar mistake could solve the "paradox", since it doesn't seem like adding or subtracting a rest-energy would improve much.
• I could have messed up in 9, "simplifying away" differentials and partial derivatives in a reckless way (It's not permitted, in general), but while on the one hand I think in this specific cases the way $$S$$ depends on $$x$$ and $$t$$ allows me to do that, on the other hand I could simply get rid of differentials integrating over a finite time interval, since for an insulated body the energy is a constant of motion (this is what I meant above with "well-behaved enough" conditions); also, I really struggle to see how a similar mistake could solve the "paradox", since some it doesn't seem like adding some integration constant would improve much.
• I could have messed up in 1 already, using the simple "relativistic mass" for the linear momentum (just like almost every source suggests), instead of the "longitudinal mass" (as opposed to "transverse"). Funny trivia: the linked source corrects the definition of momentum precisely to fix a similar "paradox" with the Lagrange formalism. This could be true (and most sources about relativistic momentum could be wrong), but still, another squared gamma factor doesn't improve the situation so much, since: $$p_x=\gamma^3 m v_x$$ $$v_x=\frac{p_x c^2}{E \gamma^2}=-\frac{c^2}{v_x \gamma^2}=-\frac{c^2}{v_x} (1-(\frac{v_x}{c})^2)=v_x-\frac{c^2}{v_x}$$ $$v_x^2=-c^2 v_x^2$$ $$c=\pm i$$ which is...well...not very reassuring (to the point that I really hope you will tell me to stick with the transverse mass instead)!

1. The Hamilton's principal function is \begin{align}S(x,t)~=~&p x -Et, \cr p~=~&\pm\sqrt{(E/c)^2-(m_0c)^2}, \end{align}\tag{1} for a relativistic free particle in 1+1D. The $$\pm$$ is the sign of the velocity/momentum.
2. From the triple product rule (TPR) we calculate $$\left(\frac{\partial x}{\partial t}\right)_S ~\stackrel{TPR}{=}~-\frac{\left(\frac{\partial S}{\partial t}\right)_x}{\left(\frac{\partial S}{\partial x}\right)_t} ~\stackrel{(1)}{=}~\frac{E}{p}, \tag{2}$$ which is the phase velocity.
3. The phase velocity (2) is not the velocity $$\frac{d x}{d t}~=~v~=~\frac{p}{\gamma m_0}~=~\frac{pc^2}{E}\tag{3}$$ of the particle. The latter is the group velocity.
Your issue has nothing to do with special relativity. It's just that the step $$\frac{\partial S / \partial x}{\partial S / \partial t} = \frac{dt}{dx}$$ doesn't make any sense. The quantity $$S(x, t)$$ means the total action for a path that travels a distance $$x$$ over a time $$t$$. This has no simple relation to $$dx/dt$$, which is the velocity for a given path at a given instant. To see that this step doesn't make sense, you can check it for any specific situation you want, such as a free nonrelativistic particle.