| From: Mik | 19/08/99
11:50:05 |
| Subject: Special Theory of Relativity | post id:
31502 |
| Can someone please prove or
disprove this example. Person A is standing on the earth. Person B launches in a Space Ship and travels at the speed of Light for 1 year, 6 months out (1/2 light year) and 6 months back (1/2 light year) and returns to earth. From Persons B point of reference he is one year older. From Person A point of view many more years have pased and he is an old man. Is this a correct example of how the theory works? Mik | |
| From: Dave | 19/08/99
12:01:04 |
| Subject: re: Special Theory of Relativity | post id:
31531 |
| No. The persons (both) will only
be 3 years older. Light speed is relative to how fast they travel, not the time frame. Dave! | |
| From: Chris (Avatar) | 19/08/99
13:04:27 |
| Subject: re: Special Theory of Relativity | post id:
31556 |
Þ Dave: ??? Þ Mik: You're close! :o) The big problem with your scenario is that your person can't travel at the speed of light, SR forbids any massive object from accelerating to light speed. So let's be technically correct (since this is a learning exercise). Both person A and person B start out in the rest frame of the earth (for the purists, consider the earth for very small Dt). Now person B accelerates quickly away from the earth until they reach a velocity of 0.99c with respect to A. Let's look at what is happening in B's frame of reference. B barrells along at 0.99c for one year, quickly turns around and heads back to earth. According to B's subjective time frame, 2 years have passed. However, for A things have changed dramatically!! Because B accelerated away from A to near light speed, B's time intervals have been dilated for A. A will measure that more time has passed since B left than B does. If we call t the time interval that B measured from departure to return, and we call A's measurement of the time between B's departure and return t', then SR relates the two intervals in this way: t' = t / g g is the SR dilation factor g = Ö (1 - v2/c2) So we have t' = t / Ö (1 - v2/c2) t' = t / Ö (1 - 0.992) t' » 7t So 14 years pass back on earth while only 2 pass in B's spacecraft. Hope this helps! Chris | |
| From: Mik | 20/08/99
13:48:18 |
| Subject: re: Special Theory of Relativity | post id:
31879 |
| Thanks Chris it helps me, however
one of my team (all IT people not Physicist(wrong spelling?)) also called
Mik is STILL not convinced that this is the case. In the following example
does the hands of the clock move. Given a person that travels away from a clock (in a fixed position) at infinetly close to the speed of light, does that person see the hands on the clock move?. If so why, and by what amount. Then if the person travels back towards the clock at infinetly close to the speed of light, does that person see the hands on the clock move twice as fast. If so why, and more importantly why not? If this is the case then please explain your answer in the previous response. Mik and Mike | |
| From: Cam (Avatar) | 20/08/99
14:18:43 |
| Subject: re: Special Theory of Relativity | post id:
31889 |
| Hi Mik I'm going to have a stab at this before Chris gets all excited about it! While travelling away, the clock hands appear to the mobile observer to move very slowly. While travelling towards the clock, its hands again appear to move very slowly. However, while the observer is doing his U-turn, the clock hands appear to go berserk, and crank away the time like there's no tomorrow (hmm... sounds like a mixed metaphor there). That's because the observer's frame of reference is not inertial during the turn-around. I'll leave Chris to fill in the gaps in my explanation, do the sums, and generally tidy it up! | |
| From: Chris (Avatar) | 20/08/99
16:14:42 |
| Subject: re: Special Theory of Relativity | post id:
31927 |
Ok. First replace the clock with a light pulse that flashes every minute (say). That way your traveller has a mechanism for "seeing" the clock. The flashes are timed in the frame of the earth, and they will flash religiously every minute until our traveller returns. The traveller races off at the same speed as before (0.99c), watching for the light pulses from the earth-clock to catch him. Now because he is moving away from the earth we have two effects in play: 1) From the spacecraft's point of view the earth-clock is moving away at high speed, so it will see time dilated in the earth-clock's frame. The pulses will appear further apart. 2) Because the traveller is moving away from the light source the light will be doppler shifted, meaning the frequency will drop. The pulses will appear further apart. On the outbound leg, then, the traveller sees the earthclock running slow - in fact the outbound traveller will "see" a light pulse about once every 14 minutes! (While in the earth frame they're shooting off once a min). At the turn-around event time dilation will vary between factors of one and seven as the traveller accelerates to turn back towards earth. If we make the turnaround event last a day in the frame of the traveller (he's a hardy fella!) then it will have minimal effect in this analysis (it's SR after all, Cam! ;o). On the inbound event is when things get interesting. Now travelling towards the source, the relativistic doppler effect means that our traveller will see light pulses arriving at the rate of 14 per minute - the earth-clock now appears to be running very fast. Nearly 14 years will appear to pass (according to the pulse-clock) during the return trip home. The sum time for the journey: * the traveller watches the pulse clock on earth count out 1/14 of a year on the outbound leg while a year passes onboard the spaceship. * the pulse clock counts out about 1.6days during the 1 day turnaround event. * the traveller watches the pulse clock on earth count out 14 years during the return journey of one year. Total - just over 2 years pass on the space-craft, just over 14 years pass on earth as counted by the pulse clock. Just as in the example above. Hope this helps! Chris | |
| From: mike | 20/08/99
18:10:20 |
| Subject: re: Special Theory of Relativity | post id:
31942 |
| Cris Thanks for you
help BUT, some more questions.... Mik's actual example is, A Traveller taking a trip for 2 years in one direction and looking BACK at a clock (granted this is not really possible) on earth would see the clock slow and nearly stop. On the return trip the clock would speed up. By the time the traveller arrived at earth again, the clock would show 2 years has past. I believe if you apply Newtonian Mechnical theory this would be indeed be the case. (yes?) However, because of the effect of time dilation via a doppler shift, time does NOT appear the same at all points in the universe dependant on velocity. OK so how do I explain time dilation and relativistic doppler effect to young Mik!!! This started as a lunch time discussion and is really starting to get interesting! also if the doppler effect increases as velocity increases then a person flying at the speed of sound would appear to be "less old" than a person whom never flys at the speed of sound (if even by milli-seconds) Am I taking the theory too far? Mike D | |
| From: Chris (Avatar) | 20/08/99
18:37:02 |
| Subject: re: Special Theory of Relativity | post id:
31944 |
Mik's actual example is, A Traveller taking a trip for 2 years in one direction and looking BACK at a clock (granted this is not really possible) on earth I don't understand your objection here. Looking at something is receiving photons of light in your eyes. These photons have been reflected or emitted from what you're looking at and then must travel to your eyes. Since the speed of light is finite, you can't separate the travel time of the light image out of the problem. This is why I used the light pulse clock - so you can better appreciate what is actually going on. To suggest that the traveller can "magically" see the earth-frame clock without the light from the clock having to travel to the traveller is unphysical, and defeats the purpose of the problem. would see the clock slow and nearly stop [on the outbound leg]. On the return trip the clock would speed up. No, not slow, or speed up. If the relative velocity is constant, then the time dilation factor (which depends on v - see equations in my first post) is also constant. The traveller sees time dilate in the earth frame on the outward journey, and sees time run "fast" in the earth frame on the inbound leg. The trick is - and this will really test whether you can see what's going on - that the earth bound observer sees the same behaviour in clocks onboard the space-ship! By the time the traveller arrived at earth again, the clock would show 2 years has past. I believe if you apply Newtonian Mechnical theory this would be indeed be the case. (yes?) Yes. This is entirely Newtonian. Newton arives at this result because the speed of light is infinite in newtonian physics. Experimental evidence supports the relativistic result, so newton does not apply in this case. However, because of the effect of time dilation via a doppler shift, time does NOT appear the same at all points in the universe dependant on velocity. No. Time dilation is not dependant upon doppler shifting. Time dilation is a relativistic effect which is quite separate. It arises solely from the postulates of special relativity - ie that the speed of light is the same for all inertial observers. Explaining the concepts of relativistic space-times is difficult. My first suggestion is to make sure you understand it yourself, because you will surely be asked many many questions, and you need understanding to answer them. Basically the postulates of relativity and the finite speed of light mean that we must reassess our entire notions of absolute space and time. The measurements of intervals in space and in time made by two or more observers who are moving relative to each other will differ. That is probably the most concise, exact statement of what is going on in SR time dilation and length contraction. If you understand what that means, you're halfway there. And the math - surprisingly enough - is very easy. The concepts are tough! Hope this helps! Chris | |
| From: mike | 20/08/99
19:26:26 |
| Subject: re: Special Theory of Relativity | post id:
31957 |
| Chris, thanks once again,
Because I don't understand the finer points of the theory I have some trouble grasping all the concepts particually in relation to time dilation. I re-read your comments re the light being doppler shifted. Hence I have NO idea why I wrote the line re time dilation via doppler shifting (sorry brain is starting to hurt!). I have seen a number of examples over the past weeks on the subject which SHOW the results but do not really explain the WHY part. The statement re SR time Dilation does help somewhat. Ok, Therefore, if we stick with our example for a while longer....and at the risk of making a potentially dumb statement. The person that has returned to earth percieves that time has sped up once he see's his mate on earth has "aged" by 14 years. The person on earth see that for his mate that has just returned time has slowed. Therefore, the relative difference in time (is that calc'd by that formula in your first email?) is a result of speed and the effects of time dilation. This results in the "perception" of the rate at which time passes for the two observers. Or am I on the wrong track again!! Mike D | |
| From: Chris (Avatar) | 23/08/99
10:43:44 |
| Subject: re: Special Theory of Relativity | post id:
32180 |
Ok, Let's try an exercise in perception. We'll need a little basic math (Pythagoras) and some simple geometry, but the exercise should establish the "why" of time and space being dependent upon observer motion. Start with two boards, one above the other, and a ball bouncing between them: ¾ o ¯ ¾ If the boards are 1m apart, and the ball takes 1second to travel from the top board to the bottom board, then we can easily work out the ball's velocity: v = d/t = 1m/s. Now imagine the whole apparatus is moving past me at 1m/s. Our diagram now looks like this: ¾ / \ o o / \ ¾ ® ¾ What is the velocity of the ball with respect to me? Well, we know from above that it has a vertical (up and down) component of its velocity at 1m/s. We also know that the apparatus is moving with respect to me at 1m/s horizontally. We can make a right angled triangle of these velocities where the base of the triangle is the horizontal component, the vertical height of the triangle is the vertical component, and the total velocity with respect to me is given by the hypotenuse: x н 1 1 x = Ö(12 + 12) x = Ö 2 x » 1.41 Notice here that the speed of the ball with respect to me is greater when the apparatus is moving (1.41m/s) than when it is not moving (1m/s). This is all well and good - we intuitively expect this result every day. Now replace the ball in the above example with a photon (single "particle" of light) and the boards with reflective mirrors. Our photon bounces back and forward between the two mirrors just as the ball did in the first example. However this time there is a spanner in the works! According to the principles of Special Relativity the speed of light is measured to be the same for all inertial observers! This means that in both examples above we must measure the speed of the photon relative to us to be the same. That is when the mirror apparatus is stationary we measure the photon speed to be c, and when the mirror apparatus is moving with respect to us we measure the photon speed to be c. Now let's look at this. In the first case the mirrors are a distance d apart, and the velocity relation is c = d / t. In the second example the mirrors move, increasing the path distance to be travelled by the photon to d'. So we also have c = d' / t'. Since d ¹ d', we know that t ¹ t'. In words, the constancy of the speed of light forces us to re-evaluate our notions of absolute time and space. Hope this helps! Chris | |
| From: Mike | 24/08/99
15:08:26 |
| Subject: re: Special Theory of Relativity | post id:
32381 |
| OK Got it! Difficult concept easy
math! | |