whirlwind Posted January 9, 2004 Posted January 9, 2004 (edited) and that is why gravity does not pull it back less you reatach the string or "turn gravity on" Edited January 9, 2004 by wirlwind Quote
iain Posted January 9, 2004 Posted January 9, 2004 my other geek question for the day is how long can you hang out in the NMR room before the machine sucks the iron in your hemoglobin through your skin? Quote
iain Posted January 9, 2004 Posted January 9, 2004 am I mistaken or did you just out-pagetop yourself? Quote
whirlwind Posted January 9, 2004 Posted January 9, 2004 yes i am refering to myself, for even trying to talk phisics with a bunch of engineers Quote
iain Posted January 9, 2004 Posted January 9, 2004 did you realize you are not actually sitting in your chair right now but hovering above on a cushion of magnetism? Quote
catbirdseat Posted January 9, 2004 Posted January 9, 2004 It's not yet clear whether protons decay, but if they do, it's at such a slow rate that you won't need to worry about it, Dru. Quote
Dru Posted January 9, 2004 Posted January 9, 2004 there is no chair  i am typing this with one hand and clapping with the other in a silent forest Quote
Dru Posted January 9, 2004 Posted January 9, 2004 It's not yet clear whether protons decay, but if they do, it's at such a slow rate that you won't need to worry about it, Dru.  hmmm.... all the protons were created in the big bang so they are all the same age  what if they all hit the best before date at the same time, and get yanked? Quote
iain Posted January 9, 2004 Posted January 9, 2004 what if they all hit the best before date at the same time, and get yanked? you finally get laid. BOOYAH! or maybe it's me. nevermind. Quote
Dru Posted January 9, 2004 Posted January 9, 2004 wtf iain the ladies are all over me when they see my ride  Quote
skykilo Posted January 9, 2004 Posted January 9, 2004 All right dudes, I'm gonna make one serious post here, and then I'm gonna go help put an experiment together. Â Yo Iain, you are right, speed affects the rate at which time passes for an observer, that is special relativity. In a relativistic theory of gravity, it is logically necessary that the passage of time is also affected by local spacetime geometry, which in turn is affected by the distribution of mass and energy. The obvious example of this phenomenon is gravitational redshift. Â Wirlwind, we can measure photons because they have momentum and energy. By mass, we mean rest mass, and photons don't have any rest mass. They don't exist at rest. Now, electromagnetic fields are made of photons, and the fields carry energy, which in turn affects the geometry of spacetime, so they have all the privileges of a massive object, except for one crucial thing-there is no photon at rest. A photon at rest is a nothing, which does not have mass. You could say something cool, like 'Photons travel along null geodesics, objects with mass propagate through spacetime on time-like geodesics.' An object with mass can never go the speed of light. Light can propagate at a net speed less than the speed of light by scattering off a the atoms of a material, which is why refraction happens. Â Just to be extra confusing and start another three pages of misinformed spray about physics, I'll mention Cerenkov radiation, which is what happens when something is moving faster than the speed of light in a given material. This is one way we detect neutrinos. The neutrinos are moving faster than the local speed of light (but still less than the speed of light in vacuum), and they radiate energy away until this is no longer true. Â Sorry to be so serious. I'm grumpy because I kept waking up during Nuclear Astrophysics today, so I neither learned from the lecture nor got a good nap. Â Quote
catbirdseat Posted January 9, 2004 Posted January 9, 2004 We can measure the energy of a photon, which is just hv. And we know that m=E/c^2. Thus the "mass" of a photon is equivalent to hv/c^2, but it is rather meaningless, isn't it? Quote
mr.radon Posted January 9, 2004 Posted January 9, 2004 Back on topic: Mars review. Â http://www.gamespy.com/fargo/january04/marsrover/ Quote
catbirdseat Posted January 9, 2004 Posted January 9, 2004 "My face reddened with rage. 'No LASERS!? What kind of a robot probe doesn't have lasers? What are you guys doing on this planet, anyways?' " Quote
cracked Posted January 9, 2004 Posted January 9, 2004 We can measure the energy of a photon, which is just hv. And we know that m=E/c^2. Thus the "mass" of a photon is equivalent to hv/c^2, but it is rather meaningless, isn't it? But that's just pointless. Take a 1kg rock and have it move at .5c. Now it's rest mass is greater than 1kg? Energy doesn't imply (rest) mass. Quote
whirlwind Posted January 9, 2004 Posted January 9, 2004 if photons don't exist at rest then of cource there would be no rest mass thats like saying the force of gravity is not a force because it does not exist when the earth stops rotating. oh and how can nothing have momentum and energy and isn't energy the transfer of electrons, and electrons have mass and also don't exist at rest Quote
mr.radon Posted January 9, 2004 Posted January 9, 2004 My favorite: When I suggested we try some whack-ass Halo-Warthog-esque grenade jumping, the guys at NASA got all wiggy and dragged me away from the joystick. Quote
Dru Posted January 9, 2004 Posted January 9, 2004 But that's just pointless. Take a 1kg rock and have it move at .5c. Now it's rest mass is greater than 1kg? Energy doesn't imply (rest) mass. Â Hmmmmm young cracked u still suck.... how is moving at 0.5c at REST? Quote
skykilo Posted January 9, 2004 Posted January 9, 2004 We can measure the energy of a photon, which is just hv. And we know that m=E/c^2. Thus the "mass" of a photon is equivalent to hv/c^2, but it is rather meaningless, isn't it? Â No, it's not meaningless. You're misapplying the formula. E=m*c^2 is the formula for rest mass energy. To get the rest mass of the photon, first measure its energy. Now, that was in your rest frame, in which the light was moving at c. Lorentz boost the light's energy-momentum 4-vector to the frame in which the light is at rest, and it's energy and momentum will be ZERO. The scalar magnitude of the E-p 4-vector E^2-p^2=m^2 is always zero. Â We can reaffirm if we compare the photon's quantum mechanical energy and momentum. p=h/lamda E=h*f f*lamda=c implies E=h*c/lamda Now check the invariant Lorentz product E^2-(p*c)^2=(h*c/lamda)^2-(h*c/lamda)^2=0 I was ignoring c's in the first paragraph above this. Forgive me, it's a very convenient habit to use c=1. Â You were just trolling, right? Quote
Sabertooth Posted January 9, 2004 Posted January 9, 2004 Let's go to Mars. All we need is one of these  and a setup like this... Quote
Dru Posted January 9, 2004 Posted January 9, 2004 HOW is it that no one has mentioned "quintessence" and "dark energy" or even "MOND" yet in this uninformed geek smackdown? Quote
cracked Posted January 9, 2004 Posted January 9, 2004 But that's just pointless. Take a 1kg rock and have it move at .5c. Now it's rest mass is greater than 1kg? Energy doesn't imply (rest) mass. Â Hmmmmm young cracked u still suck.... how is moving at 0.5c at REST? It isn't...and that's the flaw in CBS's argument. Sky explained it better than me, though, but that's to be expected. He's more of a than me, apparently. Quote
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