Relativistic generalization of Newton’s equation

[Zak]

If say you have some scalar field, θ(x^u), where x^u represents the 4-vector coordinates of spacetime, and then the typical classical equation of motion, a = -∇θ, how would one go about 'generalizing' this to a relativistic version? Since F = ma, would you have to write it as d/dt (P^u) where P^u is the relativistic 4-momentum? But since P^0 = m, this means that the 0 component of the 4-vector simply vanishes, and this seems unsatisfactory to me since you're essentially just keeping everything Newtonian. Can anybody shed some light on this?
(sorry for the bad notation)

Thanks in advance

 

[Simon Bridge]

Newtonian physics applies as long as you stay inside one inertial reference frame.
Relativity is all about the differences between observers.

For the general results, you want to look up "relativistic mechanics" and hunt down "equation of motion".

 

[Orodruin]

Since F = ma, would you have to write it as d/dt (P^u) where P^u is the relativistic 4-momentum?

Yes, this is how you define the 4-force used in special relativity. However, note that the usual thing to do is to define the 4-force as the derivative of the momentum with respect to the proper time of the object, not the coordinate time - which is frame dependent.

But since P^0 = m

This is incorrect. By definition, $P^0 = m\gamma$, where $\gamma$ is the Lorentz factor. You also have to use the relativistic momentum for the spatial part of the 4-momentum.

 

[Orodruin]

Newtonian physics applies as long as you stay inside one inertial reference frame.

This is not really true. Newtonian mechanics breaks down at speeds close to the speed of light regardless of whether you consider only one inertial frame or not.

It is also unclear what you mean by "stay inside one inertial reference frame". Physics applies equally to all inertial frames and any object that exists in one frame exists also in another.

 

[vanhees71]

The most intuitive way to guess equations of motion is to use the manifestly covariant action principle. For free particles you have
$$S_0=-m c^2 \int \mathrm{d} \lambda \sqrt{\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}},$$
where $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$ is the Minkowski-pseudometric and $\lambda$ an arbitrary scalar parameter to parametrize the world line. The Lagrangian is parametrization invariant, and that's why in fact only 3 equations are independent.

To get the interaction with an external scalar field, the most simple way to get a consistent equation of motion is to use again a parametrization-invariant scalar, and the most simple one is
$$S_{\text{int}}=-\int \mathrm{d} \lambda \Phi(x) \sqrt{\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}.$$
Then the Euler-Lagrange equations lead to the equations of motions,
$$(m+\phi) \ddot{x}^{\mu}=(\eta^{\mu \nu}-\dot{x}^{\mu} \dot{x}^{\nu})\partial_{\nu} \phi.$$
As you see, thanks to the manifest covariance and the parametrization independence you get the correct constraint condition enabling the on-shell condition for the four-momentum,
$$\dot{x}_{\mu} \ddot{x}^{\mu}=0.$$

 

[Simon Bridge]

This is not really true. Newtonian mechanics breaks down at speeds close to the speed of light regardless of whether you consider only one inertial frame or not.

Unfortunately I was called away from the computer before I could finish the thought... but that's why the suggestions for what to look for.

It is also unclear what you mean by "stay inside one inertial reference frame". Physics applies equally to all inertial frames and any object that exists in one frame exists also in another.

All quantities should be measured in the same frame... no using measurements from different frames in the same equation without the appropriate transformation.

I won't complete the thought left dangling since (a) good search terms are given, and (b) vanhees has a nice starter.