In classical mechanics, the usual formula to translate the evolution of a quantity as seen from an inertial frame of reference to a rotational frame is: $$\frac{d \textbf{A} }{dt} \vert_{Inertial} = \frac{d \textbf{A} }{dt} \vert_{Rotational} + \boldsymbol{\omega}\times\textbf{A};$$

How do we make this relativistic?

  • $\begingroup$ The formula is relativistic as it stands as long as none of the velocities induced by the rotation exceedes the speed of light. $\endgroup$
    – user17116
    Dec 24 '14 at 14:52
  • 1
    $\begingroup$ @LittleBrownOne, is that true? For any ${\bf \omega}$ one can find a big enough distance from the rotating axis that everything there on is going to move faster than light according to that observer $\endgroup$
    – Hydro Guy
    Dec 24 '14 at 16:32
  • $\begingroup$ @Hydro Guy, you are right - that means that rotational frames should be used only locally in relativity. $\endgroup$
    – user17116
    Dec 24 '14 at 16:35
  • $\begingroup$ what about a general transformation? Not just local? $\endgroup$
    – SuperCiocia
    Dec 24 '14 at 22:57

The best way to describe a rotating reference system is via a simple change of coordinates. If you have a description of a phenomenon in some set of inertial coordinates $\{t, x, y, z \}$, then you can obtain a description of its motion in a rotating reference frame with coordinates $\{t', x', y', z'\} $ by making an appropriate substitution. For example, if $\vec{\omega} = \omega \hat{z}$, then we have \begin{align} t &= t' \\ x &= x' \cos \omega t' - y' \sin \omega t' \\ y &= x' \sin \omega t' + y' \cos \omega t' \\ z &= z' \end{align}

What's different is that now the metric is not as simple any more. If we take the differentials of all the above terms and substitute them into the Minkowski metric, we get \begin{align} ds^2 &= - dt^2 + dx^2 + dy^2 + dz^2 \\ &= -\left(1 - \omega^2 ({x'}^2 + {y'}^2) \right){dt'}^2 + 2 \omega (- y' dx' dt' + x' dy' dt') + {dx'}^2 + {dy'}^2 + {dz'}^2. \end{align} In particular, this means that an massive object's trajectory in these coordinates can easily have $|d\vec{x}/dt| > 1$; in other words, its coordinate velocity will be greater than $c = 1$. But if you calculate the four-velocity $\eta^\mu$ of this object, it will still be a timelike vector (or, equivalently, $ds^2 < 0$ for nearby points along its worldline.)

As far as the vector transformation law goes, that's a little more problematic. One of the main problems with the derivation of the result you cite is that splits up the derivative in the inertial frame into time derivatives of the coordinate components of the vector with respect to a set of rotating basis vectors, and derivatives of the rotating basis vectors themselves. Moreover, we assume that these basis vectors are constant with respect to space and time; in other words, a unit vector in the $x'$-direction at point A is the same as a unit vector in the $x'$-direction at point B. But you can see that defining a set of basis vectors is going to be problematic in a rotating reference frame; a vector that points in the $t'$-direction will change from timelike to spacelike as we cross the boundary $\omega^2 ({x'}^2 + {y'}^2) = 1$, so there's no way we could possibly make that into a constant vector.

This is not to say that we can't do particle dynamics in a rotating reference system, even for relativistic motion. The better way is to take the Lagrangian approach, in which a particle moving between two events in spacetime will extremize the proper time along its trajectory: $$ \tau = \int \sqrt{ - ds^2}. $$ In the rotating reference frame, this can be written as $$ \tau = \int \sqrt{ \left(1 - \omega^2 ({x'}^2 + {y'}^2) \right) + 2 \omega (- y' \dot{x}' + x' \dot{y}') - (\dot{\vec{r}'})^2 } dt' $$ and we can find a set of Euler-Lagrange equations for $\vec{r}'(t')$ that extremize this integral in the usual way. This would be the easiest way to generalize the Coriolis force and the centrifugal force into a relativistic context.

  • $\begingroup$ +1 especially for the second paragraph. Might be worth mentioning the Born co-ordinates for this kind of thing and also the Ehrenfest paradox - it emphasizes your second paragraph (about the nonpossibility of a constant $\omega\times$ split). Also it would be good to explain to the OP that a Newtonian rigid body in itself violates special relativity as it would have an infinite speed of sound and we replace the notion with Born rigidity - but even Born rigidity cannot be upheld whilst $\omega$ is time varying IIRC $\endgroup$ Jul 26 '15 at 12:00

The best start is to pull out a physics book and look at the various expressions for the wave equation. The simplest, and foundational for the invention of special relativity, is the form with no coefficients for the differentials. Now find the wave equation expressed in terms of spherical coordinates. We have coefficients corresponding the the transform required by the coordinate transform. Realize that the "objective observer" is never stationary but is moving along a time line himself. But any observer has his own "proper time" reference frame that he can consider stationary even though other observers don't see it that way.

This is the basis of General/Special Relativity and Tensor Analysis. First the quantity must be expressed in a coordinate free form (Tensor) and then evaluated in some frame of reference; typically "at rest". Then you apply coordinate transforms to bring the tensor to the frame of reference you want. The complication arises because you typically have to include "time" as a coordinate on a equal (more or less) footing with the space components. Take a rotating orbiting satellite: it has several reference frames; with respect to geostationary coordinates, with respect to it internal inertial retation, or with respect to the sun. But all of these can transformed and calculated by means of special relativity transforms consisting of spatial variations and time variations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.