Another trick question, right? If there really was an arithmetic equation that one could use to fill the orbitals, life would be much easier. There are some rules of thumb, but if you want it exactly, there is no way around looking it up.
Science and Technology Quiz 3
- Thread starter Aramazd
- Start date
- Status
Another trick question, right? If there really was an arithmetic equation that one could use to fill the orbitals, life would be much easier. There are some rules of thumb, but if you want it exactly, there is no way around looking it up.
thetrooper
Misanthrope
- Joined
- May 24, 2004
- Messages
- 8,759
Yo Perf - get on with it...
Perfection
The Great Head.
piss...
open floor
open floor
Olleus
Deity
A space ship, 10m long, is flying very fast just above the surface of an alien world. Very, very fast actualy, at about 0.866 times the speed of light. There are two probes attached to the space ship, one at the front and one at the back. At one point, an alien stands in the middle of the spaceship and presses a button that releases both probes at the same time.
A while later, another alien comes and looks at the two probes sticking in the ground and measures how far apart they are.
What is this distance?
A while later, another alien comes and looks at the two probes sticking in the ground and measures how far apart they are.
What is this distance?
ParadigmShifter
Random Nonsense Generator
I'll go for the obvious choice of 10 metres.
Olleus
Deity
But which length is being contracted?
And from the point of view of whom?
And from the point of view of whom?
Of course both lengths are contracted from the other point of view.
The question was a bit poorly worded, as in relativistic settings "at the same time" should never be said without saying for which observer it is at the same time. I assumed that it is at the same time for the alien pressing the button.
So the alien presses the button at (ct,x) = (0,0). The center of the spaceship also has the position (0,0). We know, that the front of the spaceship is 5m from the center in the system of the spaceship, so the button press event is at (ct, 5m). In the system of the alien, both probes are released at the time 0, so it's at (0, x). Using the Lorentz transformation, we get x=2.5m, and ct=-4.33m. The same considerations can be applied to the back, and there x=-2.5m and ct=4.33m
The answer, how far the pods are apart for the alien is thus 5m.
As I said, the length contraction goes both ways. So the distance between the two pods is only 2.5m for the pilot of the spaceship. This seems to be a contradiction to the fact that the spaceship is 10m long. But it's not, because the probes are not released at the same time in the spaceship system. The pilot sees the front probe released first, and then 8.66m / c later the back probe is released. In this time, the spaceship traves 7.5m. As the probes are 10m apart on the spaceship, that gives exactly the 2.5m distance between the released probes.
(And then there is the question, how much of the probes will be left for the second alien to measure, after they hit the ground at 0.866c. I guess he will have to measure the distance between the craters.)
Olleus
Deity
At the same time is defined for the spaceship in the alien as he is the one who presses the switch, apologies for not making that clearer. I must admit that I do not how to apply lorentz transforms in a coherent way, but I can assure you that the answer of 5m and 10m is wrong. Obviously, that only leaves one possible answer, but before I declare open floor I would like somebody to explain why this is.
(Of course not my dear friends, the probes are made of an infinitely strong material, and have duct tape at their ends at that they stick to the ground upon landing
)Ok, in that case, if the probes are released at the same time for the spaceship, then its even easier. The probes are at (0,+/-5m) in the spaceship and lorentz transforming that results in (+/-8.66m, +/-10m) which means that they land 20m apart. The reason is that it's the result of the lorentz transform
If you want an explanation with length contraction and time diletation: For the alien on the ground, the rear probe is released first (at ct=-8.66m) and the front probe is released later (at ct=8.66m). In the time between both releases, the spaceship travels 15m. With the size of the spaceship appearing to be a length-contracted 5m, this adds up to 20m.
Again you can also view the same thing from the spaceship. In that case the ground is length contracted and the probes land 10m apart in the spaceship's system, which corresponds to 20m in the ground system.
(Of course not my dear friends, the probes are made of an infinitely strong material, and have duct tape at their ends at that they stick to the ground upon landing
Reminds me of my theoretical mechanics lecture: "An infinitely long, massless steel bar..."
thetrooper
Misanthrope
- Joined
- May 24, 2004
- Messages
- 8,759
uppi said:
Wonderful!
Olleus
Deity
Yep uppi, your perfectly right.
Your turn.
Your turn.
Staying on the topic of relativity:
At the LHC they want to create the Higgs boson by colliding protons with the same speed but opposing directions. If one would want to do this by accelerating only one proton and colliding it with a stationary proton, roughly how much kinetic energy would the first proton need, if we want to cover Higgs masses up to 400 GeV?
The mass of a proton is roughly 1 GeV
At the LHC they want to create the Higgs boson by colliding protons with the same speed but opposing directions. If one would want to do this by accelerating only one proton and colliding it with a stationary proton, roughly how much kinetic energy would the first proton need, if we want to cover Higgs masses up to 400 GeV?
The mass of a proton is roughly 1 GeV
Perfection
The Great Head.
~1TeV?
~1TeV?
That would only produce masses up to 45 GeV.
Perfection
The Great Head.
~10TeV?
80 GeV?
It'd be 200 GeV of energy if they hit head on with the same speed. ymc^2=200 GeV -> y = 200 -> v=0.9999875 c in the original reference system. Then we Lorentztransform that with another 0.9999875 c. That gives 1 - 7.81261722 × 10^-11 c. That gives a y of 1 / sqrt(1 - (((0.9999875 + 0.9999875) / (1 + (0.9999875^2)))^2)) = 79 999.3998 = 80000, so the energy is 80000 GeV=80 TeV
It'd be 200 GeV of energy if they hit head on with the same speed. ymc^2=200 GeV -> y = 200 -> v=0.9999875 c in the original reference system. Then we Lorentztransform that with another 0.9999875 c. That gives 1 - 7.81261722 × 10^-11 c. That gives a y of 1 / sqrt(1 - (((0.9999875 + 0.9999875) / (1 + (0.9999875^2)))^2)) = 79 999.3998 = 80000, so the energy is 80000 GeV=80 TeV
- Status
Similar threads
