# Do length contraction affects accelerating objects?

1. Nov 24, 2013

### Trojan666ru

Suppose I'm standing on earth and i see an iron rod accelerating through space with a rate of 1m/s^2. Will it undergo length contraction after reaching a speed nearly to the c?
Will there be any deformations to a spinning disk?

2. Nov 24, 2013

### Meir Achuz

There will be the usual Lorentz contraction for the rod.

3. Nov 24, 2013

### Trojan666ru

Which means the rod appear foreshortened?

4. Nov 24, 2013

### TumblingDice

The rod will be length contracted in the direction of motion relative to your rest frame. Not foreshortened though, as LC is not an illusion.

5. Nov 24, 2013

### Meir Achuz

Yes.

6. Nov 24, 2013

### WannabeNewton

There are arguably many more subtleties involved with a spinning disk than with a uniformly accelerating rod. See here for a more or less detailed analysis: http://www.projects.science.uu.nl/igg/dieks/rotation.pdf [Broken]

Also, you have to be specific with the term "deformation". Take for example a disk spinning with constant angular velocity $\omega$. Each point on the disk has its own worldline and the worldline is an integral curve of the vector field $\xi^{\mu} = \gamma \partial_t^{\mu} + \gamma \omega \partial_{\phi}^{\mu}$ where $\gamma = (1 - \omega^2 r^2 )^{-1/2}$ and we are working in cylindrical coordinates. So I can represent the worldtube of the entire disk by $\xi^{\mu}$. If by "deformation" you mean shear and expansion stresses to the disk then in this case there will be none. The expansion tensor (which codifies both shear and expansion stresses) vanishes identically for this system: $\theta_{\mu\nu} = h_{\mu}{}{}^{\alpha}h_{\nu}{}{}^{\beta}\nabla_{(\alpha}\xi_{\beta)} = 0$ where $h_{\mu\nu}$ gives the spatial distances between neighboring points on the disk relative to $\xi^{\mu}$. In fact $\theta_{\mu\nu} = 0$ is equivalent to Born rigid motion so it is clear that there are neither shear nor expansion stresses in this system.

If, on the other hand, you are using "deformation" to mean a change in the "shape" of the disk then this is quite subtle. The "shape" of the disk is an aspect of its spatial (i.e. not space-time) geometry and as such has no frame invariant meaning. In fact, going back to the case of the uniformly rotating disk, there is no natural way of defining the spatial geometry of the disk relative to a family of observers each comoving with a respective point on the disk because the vector field $\xi^{\mu}$ above which describes the worldlines of these obeservers is not irrotational: $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\nabla_{\alpha}\xi_{\beta} \neq 0$ so what this means is that there is no natural notion of simultaneity between these observers and hence no natural notion of spatial geometry of the disk relative to these observers. See the following link: http://arxiv.org/pdf/gr-qc/0311058v4.pdf and take heed of the following quote from p.65 of the linked paper: "Every notion of simultaneity has associated a different notion of spatial length, and therefore a different radius and circumference length will a appear at each non-inertial observers, namely the disk 3-geometry will be simultaneity-dependent."

However there is a non-canonical notion of simultaneity that is well adapted to the above scenario that gets around the fact that the congruence of observers comoving with the disk is not twist-free. This is the simultaneity convention given by (local) radar distance. Under this the spatial geometry of the disk does become curved. See here: http://en.wikipedia.org/wiki/Born_coordinates#Radar_distance_in_the_small

Last edited by a moderator: May 6, 2017
7. Nov 24, 2013

### TumblingDice

Intended for accuracy - not to pick nits... I've read on PF that length contraction is not 'an illusion'. Any definition of foreshortened that concerns distance is all about illusion. The OP may be aware of this, so it's important to point out.

8. Nov 25, 2013

### nitsuj

Just to clarify, does "Any definition of foreshortened that concerns distance is all about illusion." refer to the scenario where a meter stick I calculate as contracted measures shorter distances then I do.

For example a ship that travels at 0.9c to a point one light year away from earth as measured from earth will itself measure/calculate that distance between Earth and the point as contracted compared to the original 1ly Earth frame value; while en route? (to the point where the speed of light is calculated to be c)

9. Nov 25, 2013

### TheBC

If you measure a rod at rest x meters long, then you measure the moving rod less than x meters long.

If you consider a rod at rest in front of you, you do not say the rod at rest 'appears' x meters long. You don't do this because you do not refer to any 'optical illusion'.
Stating that a moving rod 'appears' contracted insinuates there is some optical illusion involved. But there is no optical illusion invoved. Therefore there is no reason at all to state that a moving rod 'appears' contracted. If you state that the moving rod 'appears' contracted, then a rod at rest also only 'appears' measuring a certain length.

If you accept that you measure the rod at rest x meter long because the rod out there is 'really' (whatever that means) x meters long, then the moving rod with shorter length is also 'really' (whatever that means) out there. (Keep in mind that the moving rod in your frame refers to a collection of other events than the events in the frame of the rod). Therefore:

If you consider the simultaneous events of the rod at rest 'exist' (whatever that means), then the simultaneous events of the shorter moving rod in your frame (the moving rod is made of different events than in the rod's frame) also 'exist' (whatever that means).

The reason why people often think the contraction only 'appears' as such is because the contraction is reciprocal. How can the moving train be contracted for the station observer, but for the train observer the station is contracted, if its no optical illusion?
Consider the railway station and the train as 4D spacetime structures instead of evolving 3D objects, then the reciprocal length contraction and time dilation do make sense. They are different cuts with different 'direction' through the 4D spacetime structures (like cutting a loaf of bread in different directions result in different 'objects' -slices of bread- with its proper dimensions). Minkowski or Loedel diagrams visualize this very well.

If you are a 'realist', then these cuts are different but 'real' 3D spaces part of 'real' 4D spacetime 'existence'. Whatever your definition of 'real' and 'existence' is.
If you are a solipsist, then all the above is only a mathematical calculation wirh no reference to any 3D or 4D object 'really existing' out there. So be it.

But stating that a train at rest is 'real', but the contracted train only an optical or mathematical feature without 'real' content (and therefore only 'appears' contracted) does not make sense.