browndn Posted April 12, 2008 Posted April 12, 2008 Hypotherical Q: Lets say a spaceship A was travelling at .8c (C=speed of light) and space ship B was travelling at .8C too but in the opposite direction. So from the frame of reference A, doesn't it look like spaceship B is travelling at 1.6C? am i missing something here? EDIT: it was just a typo to type 1.8C. and why would it look like it is .9somethingc?
YT2095 Posted April 12, 2008 Posted April 12, 2008 Hypotherical Q: Lets say a spaceship A was travelling at .8c (C=speed of light) and space ship B was travelling at .8C too but in the opposite direction. So from the frame of reference A, doesn't it look like spaceship B is travelling at 1.8C? am i missing something here? yes, 2 things: 1) UTFSE! 2) .8 + .8 = 1.6 Not 1.8!
insane_alien Posted April 12, 2008 Posted April 12, 2008 if velocities added linearly it would only appear to be doing 1.6c your maths is flawed there. and velocities don't add linearly in the real world, it would only appear to be doing 0.9something c
browndn Posted April 12, 2008 Author Posted April 12, 2008 yes, 2 things: 1) UTFSE! 2) .8 + .8 = 1.6 Not 1.8! if i were to use the search engine, then what hell is this site for if anyone can search their question on google? please dont reply if you cant answer please. u seem to be a regular member to this site so please act mature.
YT2095 Posted April 12, 2008 Posted April 12, 2008 oh I`m sorry, I guess I missed the part where I said Google? Our Search Engine Fool! considering this same question has been done to death at a sum greater than the Planck constant reciprocal! 1
ydoaPs Posted April 12, 2008 Posted April 12, 2008 if i were to use the search engine, then what hell is this site for if anyone can search their question on google? We have a search engine for a reason. We've answered this question before. Velocities don't add linearly. However, at low speeds, linear velocity addition is a close enough approximation.
Sayonara Posted April 12, 2008 Posted April 12, 2008 YT, there is no need to be rude to new members. Someone with about 10 posts under their belt probably hasn't spent enough time on the forums to realise how frequently the same questions get asked. Your first reply is like being told by the receptionist to "use the ****ing floor plan". 1
thedarkshade Posted April 13, 2008 Posted April 13, 2008 You don't add velocities like that in relativity. You do that by [math]V=\frac{v_1 + v_2}{1 + \frac{v_1v_2}{c^2}}[/math]
Cap'n Refsmmat Posted April 13, 2008 Posted April 13, 2008 Or, in other words, relativity defies common sense.
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