MadSci Network: Physics |
I will answer what I interpret your question to be, and then show you the maths behind some of special relativity (it requires nothing more than Pythagoras's rule - honest!). Relativity is not quite as simple as some would make out. Imagine if we moved everything in the direction of the light at the speed the spaceship was going in the other direction. The space-ship is now stationary, and the stationary observer moving in the same direction as the light (suspiciously similar to when the spaceship shone a torch out the front window?). Now who's time goes slower? The whole principle of relativity states that it is actually impossible to tell which observer is staionary and which is moving. It is the process of turning the spaceship around and coming back which actually decides that it was the spaceship moving. There is a good explanation in 'The Emperors New Mind' by Roger Penrose (which is a discussion of quantum theory, relativity, and conscioussness). The Galilean principle of relativity says that it is not possible to determine an 'absolute' velocity. That is, if you are in a space-craft with no windows moving at a constant speed, you would have no idea what speed you were moving at. Newton's laws obey this principle. The problem came with the development of Maxwell's laws of electromagnetism (in 1865). These showed a finite speed of light, c (= 300,000,000 metres per second). This, together with Newton's laws, would imply that a car moving at speed u would see light to go at a speed c-u, and thus, by the speed of light in your window-less space rocket, you could determine your speed. However, experiments showed that this was not the case. Consider the following thought experiment, but when describing it, let us assume that the principle of relativity holds. That is, the speed of light is the same for all observers. MIRROR _____ C___________________ /\ | / \ | - Height L / \ | LIGHT / \ |_ SOURCE: A B: DETECTOR Length AB=d, Length AC=h. Speed of light=c If I was stationary (with respect to the apparatus) and watching this experiment, you would see the light to move a distance of 2h between A and B. Therefore, the time taken 9from my point of view) must be t=2h/c. But what if you were moving in the direction AB at speed v, such that you are at A when the light leaves A, and at B when the light reaches B. To you, the experiment would look like this: _____C_____ | | | | A,B So, you see the light to move a distance of 2L. So your time must be t'=2L/c. But from the Pythagorean theorem, (d/2)^2+(L)^2=(h)^2. BUT, t=2h/c => h=ct/2; t'=2L/c => L=ct'/2. And (from my point of view), you were moving at speed v, and covered the distance d in time t, therefore d=vt. So, (d/2)^2+(L)^2=(h)^2 => (vt/2)^2+(ct'/2)^2=(ct/2)^2 => (c^2-v^2)(t^2)=(ct')^2 => t=1/[[1-(v/c)^2]^.5] (I couldn't find the square-root or squared signs, ^2 means squared and ^.5 means square-rooted). This is the Lorentz transformation for time. It was developed so that Maxwell's laws remained unchanged when worked out for a moving observer, i.e. so that the speed of light appears the same to all observers. Mass, and distance, are also transformed in the same way, yielding results that quite a few people have heard of, e.g. E=m(c^2). Let's try out this formula on a problem: Imagine that an astronaut goes off at 0.8c, relative to the Earth, and returns after 30 years have elapsed on the rocket. We have v=0.8c, and t'=30 years. Therefore, t= 30/[[1-(0.8)^2]^.5] = 30/[1-0.64]^.5 =30/0.6 = 50 years have gone by on Earth! Special relativity is quit difficult to get your head round (it made Einstein's hair stand on end). Well done for trying. I reccommend the following books: 'Introducing Einstein' (formerly 'Einstein for Beginners') by Joseph Scwarz and Michael McGuiness. 'Six Not-So-Easy Pieces,' a collection of Feynman's brilliant lectures. 'E=mc^2, an Equation That Changed the World' (I can't remember who it's by, but it is very good).
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