alan2here Posted June 10, 2007 Share Posted June 10, 2007 Im gona split this question up into two parts and ask the second part when I know I havn't messed up my resoning on the first. If a person was traveling inside something traveling at 0.99 times the speed of light that has a hollow inside and they had a lamp in there so they could see stuff then when they walked around as I understand it they could see all walls of there spaceship with no trobble. They could trow a ball up an catch it. They could walk around and wouldn't notice anything odd about how they light was working in the room? Link to comment Share on other sites More sharing options...
insane_alien Posted June 10, 2007 Share Posted June 10, 2007 yup as long as they are moving at constant speed Link to comment Share on other sites More sharing options...
abskebabs Posted June 10, 2007 Share Posted June 10, 2007 Alan2here this is the fundamental principle of relativity, the laws of physics are the same in all inertial frames of reference frames, and by consequence, light propogates at the same speed in all inertial reference frames. The situation you mentioned involves an inertial frame of reference travelling with respect to some other "rest" frame(presumably Earth's), so of coursea person in the ship would observe no difference in how light propogates. The bit about a tennis ball I'm a bit more cautious to answer, but I suppose if you're ship is generating its own gravitational field, there may be no observed difference with throwing a tennis ball up in the "air". The principle of special relativity, is really just a revision of Newtonian relativity. Think about travelling on a steady train at constnt velocity with respect to the ground. Is any difference observed by you when throwing a ball up in the air and catching it, to the doing the same thing on the ground? Link to comment Share on other sites More sharing options...
swansont Posted June 10, 2007 Share Posted June 10, 2007 They would have no clue that they were moving at 0.99c with respect to anything; or, that something was moving at 0.99c WRT them. They are at rest with respect to the ship Link to comment Share on other sites More sharing options...
alan2here Posted June 10, 2007 Author Share Posted June 10, 2007 Fantasic. This brings me onto the second part, which is that a smaller person inside a smaller spaceship witch was traveling round in circles inside the origanal spaceship at 0.5 times the speed of light would presumably also not know he was moving just by observing what was happening with the light created by the smaller lamp (The one in the new smaller ship). My question is then if both thease people open the blinds on there spaceships so that light can go in and out. Isn't an observer on the outside going to see the inner spaceship traveling at 1.49 times the speed of light? Link to comment Share on other sites More sharing options...
insane_alien Posted June 10, 2007 Share Posted June 10, 2007 1/ the little guy will know he is accelerating and therefore moving. 2/ velocities do not add linearly in relativity. it will be 0.9999 c or something. i forget how to do it. Link to comment Share on other sites More sharing options...
alan2here Posted June 10, 2007 Author Share Posted June 10, 2007 Maybe 0.99 + ((1 - 0.99) * 0.5) = 0.995 Thanks. Reality has an odd way of always working dosn't it. Link to comment Share on other sites More sharing options...
Janus Posted June 11, 2007 Share Posted June 11, 2007 Maybe 0.99 + ((1 - 0.99) * 0.5)= 0.995 Thanks. Reality has an odd way of always working dosn't it. Actually [math]\frac{0.99c+0.5c}{1+ \frac{0.99c(0.5c)}{c^2} }= .9967c[/math] Link to comment Share on other sites More sharing options...
alan2here Posted June 12, 2007 Author Share Posted June 12, 2007 Actually [math]\frac{0.99c+0.5c}{1+ \frac{0.99c(0.5c)}{c^2} }= .9967c[/math] lol, so my equation is not bad then :¬) Your's is slightly more accurate. Link to comment Share on other sites More sharing options...
Janus Posted June 12, 2007 Share Posted June 12, 2007 lol, so my equation is not bad then :¬)Your's is slightly more accurate. Actually your equation is quite bad. While it gives a fairly close answer for this particular situation, this not the case for all possible variations. For instance, if you make one velocity 0.5c and the other -0.5c, the correct formula gives [math] \frac{0.5c+(-0.5)c}{1+\frac{0.5c(-0.5c)}{c^2}}= 0c[/math] While yours gives [math] 0.5c +(1-0.5)(-.05c) = .25c[/math] Secondly, the proper equation is derived from the postulates of the theory, while yours was merely an attempt at a "curve fit". Link to comment Share on other sites More sharing options...
alan2here Posted September 25, 2007 Author Share Posted September 25, 2007 This seems very related http://www.scienceforums.net/forum/showthread.php?t=28606 Length contraction seems to remove additional paradoxes I previously couldn't get my head around in this example. I was going to ask but then found that thread, so im posting a link for future people reading this thread. Edit: It was a little confusing what would happen if 3 spaceships where having a race, it would seem they would be able to relative to each other keep getting faster and faster but relative to a stationary observer just keep getting closer and closer to C. Link to comment Share on other sites More sharing options...
CPL.Luke Posted September 27, 2007 Share Posted September 27, 2007 or tanh (tanh^-1 v + tanh^-1 v') easiest way of doing the relativistic velocity transform, and a useful way to remember hyperbolic tangent identities. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now