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Suppose a Galilean transform with non-constant velocity $u$. Then, the equation of motion $\frac{d^2 r}{dt^2} = F$ is transformed to $\frac{d^2 r'}{dt^2} = F - \frac{d u}{dt}$. In other words, $\frac{d u}{dt}$ appears on the right hand side of equation of motion as an inertial force.

I want to do the same thing for the Lorentz transform with non-constant velocity $u$. How is the inertial force look like?

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Lorentz transform is used to convert the vectors from one inertial reference frame into another reference frame.

The problem is that you cannot associate an inertial reference frame with accelerating observer. Let me give you a physical example. If you and I start from rest, me at position $z=-L/2$, you at $z=L/2$, and then we both start accelerating at the same acceleration ($a$), in z-direction, then if $\frac{aL}{c^2}$ ($c$ is the speed of light) is large enough our clocks will loose synchronization. Now, Lorentz transform, amongst other things, tells you the time-dilation between the clocks in the current interial frame and the in the 'new' inertial frame, so you can see that concept of intertial frame is incompatible with acceleration. You can associate an instaneneous inertial frame with an accelerating observer, but this will make sense only locally.

Globally, you must allow for spacetime-dependent transformations, which brings you to Rindler coordinates https://en.wikipedia.org/wiki/Rindler_coordinates

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There is no such thing as the Lorentz transform with non-constant velocity. By definition the Lorentz transform is a transform between inertial frames, so the velocity is constant.

However, your question could be broadened to ask for arbitrary transforms, especially non inertial ones. A couple of people have mentioned the Rindler coordinates, which is the simplest such transform. Under any generic transform you can write:

$$m\frac{d^2 x^{\mu}}{d\tau^2}=f^{\mu}-m\Gamma^{\mu}_{\nu\lambda}\frac{dx^{\nu}}{d\tau}\frac{dx^{\lambda}}{d\tau}$$

This is the equivalent of your expression for any coordinate system. The $\Gamma$ terms are called the Christoffel symbols and can be calculated from the metric in any coordinates of interest.

For example, the Christoffel symbols are given for a rotating coordinate system using polar coordinates in section 2.1.5 here:

https://arxiv.org/abs/0904.4184

They are $\Gamma^r_{tt}=-\omega^2 r$, $\Gamma^{\phi}_{tr}=\Gamma^{\phi}_{rt}=\omega/r$, $\Gamma^r_{t\phi}=\Gamma^r_{\phi t}=-\omega r$, $\Gamma^{\phi}_{r\phi}=\Gamma^{\phi}_{\phi r}=1/r$, $\Gamma^r_{\phi\phi}=-r$

So, for example, calculating the radial component of the acceleration we get:

$$m\frac{d^2 r}{d\tau^2}=f^r + m \left( \omega^2 r \frac{dt}{d\tau}^2 + 2\omega r \frac{dt}{d\tau}\frac{d\phi}{d\tau} \right)$$

Which we recognize as being the relativistic versions of the centrifugal force and the radial component of the Coriolis force

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  • $\begingroup$ Thank you for the answer. What I was thinking is a situation where the observer is rotating with almost speed of light, and what the centrifugal and Coriolis force would look like. Can such a thought be done via Rindler coordinates? $\endgroup$ Commented Sep 5, 2018 at 7:06
  • $\begingroup$ No, the Rindler coordinates are for a linearly accelerating observer, not a rotating observer. $\endgroup$
    – Dale
    Commented Sep 5, 2018 at 11:04
  • $\begingroup$ I will add some information to the answer specifically about rotating coordinates $\endgroup$
    – Dale
    Commented Sep 5, 2018 at 11:26

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