I Will a spring extend more when going at relativistic speeds?

[jartsa]

OK, here's the super simple version involving the notion of a "dropped mass" style of accelerometer. These are commonly used as "absolute" gravimeters.. It gets the right answer, but it's not very rigorous, I have some concenrs in that regard.

Before we get into this, though, I want to point out that the "dropped mass" in this case is just an instantaneously co-moving inertial observer. The observer on the elevator, and also Superman, are accelerating. The "dropped test mass" is moving inertially.

If we consider an instantaneously co-moving observer who starts out "at rest" with respect to the elevator floor, they will see the elevator accelerate away from them , to a high degree of precision obeying the Newtonian law s = 1/2 a t^2, where a is the acceleration of the elevator. If they watch long enough, eventually the elevator will reach relativistic velocities and we'd need to do a relativistic analysis, but if we consider short time periods this is not necessary.

So if the acceleration, a, of the elevevator is 32 feet/sec^2 (1 gravity), at t = 1 second, s = 16 feet.

The accelerating observer at rest on the elevator floor will see essentially the same thing as their associated instantaneously co-moving inertial observer.

Now let's consider Superman's point of view. Superman also has an instantaneously co-moving observer. Superman and that observer both share essentially the same inertial frame of reference, which is different from the frame of reference of the elevator floor. Due to time dilation, the moving observer's clocks will run slower by the factor gamma. So 1 second on the stationary observers clock is only 1/2 a second on the moving clock, if we assume the gamma factor is 2. This means the elevator accelerates away from the moving observer by 16 feet in 1/2 second. Plugging that into s=1/2 a t^2, the moving observer computes that the elevator is accelerating away at 4 gravities. When we generalize this to a general factor of gamma , we see that the acceleration went up by a factor of gamma^2.

Thus superman (and the instantaneously inertial observer co-moving with him) sees the elevator's acceleration as $\gamma^2 a$.

Lorentz contraction doesn't have any effect on the analysis, the distances are measured perpendicularly to the direction of motion.

The issue that I'm concerned with is to justify how we can ignore the relativity of simultaneity.

Shouldn't the vertical motion of the elevator be slowed down according an observer that observes an elevator that has lot of horizontal speed? (Compared to an observer that is glued on the elevator floor)

And shouldn't the elevator floor still be curved, according to that observer?(I'm not sure what is discussed here, but I'm guessing that we want to know what superman flying through an accelerating elevator observes )

 

[PeterDonis]

The issue that I'm concerned with is to justify how we can ignore the relativity of simultaneity.

If the acceleration is independent of time, as it is in this scenario, then relativity of simultaneity is not an issue because it doesn't matter where the different spacelike slices of the two observers (Green Lantern and Superman) intersect the world tube of the elevator; the curvature of the world tube is the same everywhere, and so are the curvatures of the worldlines of the two observers.

If the acceleration is not independent of time, then you will indeed have to take relativity of simultaneity into account.

 

[Idunno]

Yes, but it's important to understand why. It's not because the mass of the object on the end of the spring increases (by a factor $\gamma$). It's because the proper acceleration of the object on the end of the spring increases (by a factor $\gamma^2$). In other words, the moving object has to accelerate harder than the stationary object. (Note also that this all assumes that both objects stay at a constant height above the floor of Green Lantern's elevator, as seen in Green Lantern's frame.)

To put it another way, for both objects, $W = m_0 a$, where $W$ is weight, $m_0$ is rest mass, and $a$ is proper acceleration. But for the stationary object, $a = g$, while for the moving object, $a = \gamma^2 g$.

Thanks a bunch everyone, I think I understand all this much better now, for whatever that's worth. :)

 

[pervect]

Something I'd like to clear up. Calculating the 4-acceleration from the 4-velocity is a very straightforwards procedure that can be found in most texts, in the context of either special relativity or general relativity.

The tricky part of the problem problem (IMO) is finding the 4-velocity of Superman, aka the sliding block. I do this in https://www.physicsforums.com/threads/the-notion-of-weight-in-relativity.701257/#post-4445483, but it's helpful to eliminate the factor K I used in the derivation, and replace it with $\beta_0$, the block's normalized velocity, and $\gamma_0 = 1/\sqrt{1-\beta_0^2}$

The formulae to do the conversion are in the original post, but the converted results are not written down explicitly. Doing so now, we find

$$\frac{dt}{d\tau} = \gamma_0 \sqrt{1+g^2\,t^2} \quad \frac{dx}{d\tau} = \beta_0 \, \gamma_0 \quad \frac{dz}{d\tau} = \gamma_0 \, g \, t$$

This express the 4-velocity in terms of t, not $\tau$. My original intent was to stop here, but as I wrote this I realized it'd be helpful to go further and additionaly re-write the proper velocity as a function of $\tau$ rather than t. To do this we make use of the relation between t and $\tau$ derived in the original post:

$$\tau = \frac{\sinh^{-1} gt}{ \gamma_0 g} \quad t = \frac{1}{g} \sinh \gamma_0 g \tau $$

We can then substitute to find the 4-velocity as a function of $\tau$

$$\frac{dt}{d\tau} = \gamma_0 \sqrt{1+\sinh^2 \gamma_0 g \tau} = \gamma_0 \cosh \gamma_0 g \tau \quad \frac{dx}{d\tau} = \beta_0 \gamma_0 \quad \frac{dz}{d\tau} = \gamma_0 \, \sinh \gamma_0 g \tau $$

While I'm at, I realized I can write the 3-velocity of Superman as a function of time:

$$\frac{dx}{dt} = \frac{\beta_0}{\sqrt{1+g^2\,t^2 }} \quad \frac{dz}{dt}= \frac{g\,t}{ \sqrt{1+g^2t^2}} $$

We can see that the velocity $dz/dt$ is the same function of time t as for the relativistic rocket, as it should be.

 

[PeterDonis]

Thread closed for moderation, since it looks like the OP question has been answered and the thread is now bogged down in an off topic tangent.

 

[PeterDonis]

A number of off topic posts have been deleted. Since the thread has run its course and the OP has expressed satisfaction with the answers given, the thread will remain closed.