(Special) relativity final question; gyroscope at relativistic velocities

slaydez
Messages
1
Reaction score
0
This question on my final blew my mind and I have no clue where to even start: (verbatim, not exact wording and so 'theoretical' I almost posted it under relativity)

Given that an observer on Earth with acceleration due to gravity 9.8 m/s^2 has an indestructible frictionless gyroscope and a means by which to safely accelerate it to a rotational velocity of (some relativistic velocity, we'll say).85 c, how long will the gyroscope take to fall in a vacuum (I think the problem may have said neglecting air resistance) relative to the observer if dropped from a height of 15 meters? So to sum it up, g=9.8 m/s^2, and this 'indestructible' gyroscope with 'negligible friction' (Yes it's a crazy question) is dropped from a height of 15 meters through a vacuum while spinning at .85 c. How long will it take to fall relative to an outside observer?



Lorentz equations are necessary for sure, but it's not the calculations that really stumped me, it's the theory behind it. Part of me says (classically) it will fall just like anything else. If I'm not mistaken though special relativity applies to the gyroscope because it's moving uniformly. So would the gyroscope in effect fall "slower" to the observer due to time dilation occurring for an observer "on the gyroscope" such as an ant? As I said, I assumed it was a trick and wrote it off as simple mechanics. I don't even think it was a fair question to ask, mainly because it's slowly ruining my summer. Can anyone help me out?
 
Physics news on Phys.org
hi!
under the assumption that there is no component of the rotational velocity which points in the direction of free fall there would not occur any relativistic effect, since such effects only occur in the direction of movement.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top