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Relativistic rigid motion

Physics Asked by facenian on December 15, 2020

In Bryce DeWitt’s Lectures on Gravitation, in eq. 2.7 on page 25 when he describes the rigid motion of a continuum he states
$$x^mu(xi,tau)=x^mu(0,sigma)+xi^in^mu_i(sigma),,,(i=1,2,3) , .$$

In this equation $xi$ are the material coordinates and $x^mu(0,tau)$ is the origin of the labels $xi^i$, $sigma$ is the proper time for the particle taken as origin for material coordinates and $tau$ is the proper time for the particle with labels $xi$. I don’t understand why he assumes that as motion takes place the coordinates $xi$ do not change assuming that the three bases $n^mu_i(sigma)$ vary properly. I know this assumption is correct in Newtonian mechanics because you can imagine a rigid triad moving with the object and respect to it the object is static but this seem to be impossible according to relativity.

4 Answers

There is no such thing as a rigid body, but there is such a thing as rigid motion. It is that kind of motion when all the parts of a given body have worldlines such that, on a sequence of spacelike hypersurfaces that can be chosen in some reasonably natural way, the relative layout and proportions of the body are the same from one hypersurface to the next. In special relativity these spacelike hypersurfaces could, for example, be the planes of simultaneity for a sequence of inertial frames which are also instantaneous rest frames of some part of the body such as its centroid.

This kind of motion is highly constrained and therefore rare in practice, but merits study. It includes, for example, rigid rotation, and the Rindler metric (constantly accelerating frame).

Answered by Andrew Steane on December 15, 2020

In equation 2.7 the author is assuming there exists a reference frame with respect to which the continuum is at rest as is the case of newtonian mechanics. Apparently this is not in contradiction with SR as along as the dimensions of the medium are properly bounded as stated by the condition $(1+xi^ia_{0i})^2>xi^iOmega_{ik}xi^jOmega_{jk}$

However, the assumption is not explicitly stated by the author and I think this is not the general case of rigid motion if as such we accept the defintion $dot{gamma_{ik}}=0$

Answered by facenian on December 15, 2020

Page 25 is in the section on special relativity. In section 2.1 the $xi$ is basically a label based on the initial position. If you think of it as a function of $x^mu$ then on page 23 there are nonzero $xi^i_{,mu}$.

I'd you are referring to the equation after eq. 2.7 the author is tracking the motion if a particular label, and that is why there is no $xi^i_{,mu}$ in the equation after 2.7.

As for rigid bodies, there is no body that is rigid in all situations, no way to relativistically enforce rigidity in some situations in dynamics in a relativistic way.

But you brought up GR when that section of the book has no GR. And that section is also not talking about dynamics, there are no forces, just motion. Like if it rotated itself just because it always has been.

It's looking at the cases in SR that are most similar to Newtonian rigid motions.

Answered by Timaeus on December 15, 2020

I don't have the book in front of me so I can't say for sure, but I can guess: The formula for $x^mu$ is an assumption because it is a reasonable definition of the phrase rigid body. I would guess the author starts out with the assumption and then moves on to deduce what is really physically happening to achieve this.

Answered by user12029 on December 15, 2020

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