Yes, that is what I was trying to say before. No matter <b>how</b> you're moving, as long as you're moving at a constant velocity, you'll measure the speed of light to be c. Period. If you witness a person moving at .5c turn on a light beam that travels away from them at c, from their frame of reference the speed of light is c. From your frame of reference, the speed is c. BTW, there is no absolute frame of reference. This was proven by the Michaelson-Morley (sp?) experiment. They shine light in different directions and measure the speed via and interference pattern, and guess what? It measures the speed of light to be c in any direction, even though the earth is moving at some thousands of miles an hour around the sun. Also, Prosoothus, the speed of light does not "speed up" in the light clock if it's moving horizontally, it stays the same. Hence time dialation, it takes longer for the light to travel between the two mirrors.
SpyFox_the_KMeson, Crisp, Thed, and James R, Unfortunately, nothing that you guys are saying is making any sense to me. Maybe I'm just stupid. Therefore I pose a question to all of you, so that I can understand what's really going on: There are to observers. Observer1 is (relatively) stationairy. Observer2 is traveling towards observer1 at .90 c(270,000 km/s). As soon as observer2 flys over over observer1, observer2 turns on his flashlight and shines the light in the same direction that he is travelling. Observer2 continues to travel at .90 c away from observer 1. One second after observer2 flew over observer1 where is the light?? a) The light is 300,000 km away from observer1. The light is 30,000 km away from observer2. Observer2 is 270,000 km away from observer 1. b) The light is 300,000 km away from observer1. The light is 300,000 km away from observer2. Observer2 is 270,00 km away from observer1. As you can see, if the answer is a), then observer2 is seeing the light travelling at only 30,000 km/s or only .10 c, because he is traveling at .90 c. However, all of you are claiming that the answer is b). This would mean that there are two beams of light: one beam which is 300,000 km from observer1, and the other which is 570,000 km away from observer1. Which is the correct answer??? Tom
Hi Tom, There is no point in using intuition here (as you did when formulating both your answers). I am pretty confident that both possibilities are wrong, but I'll do all the math in an hour or two and post it once I worked it out. Bye! Crisp
The maths Hi Tom, I've done the maths for the problem you mentioned, here are the results: Problem Consider two observers O and O', where O' moves horizontally with a speed v = 0.9c with respect to O. When the two observers meet, O turns on a flashlight in the direction of motion of O'. After one second, measured in the frame of reference of O, where is the light for O' ? Solution For O, the solution is simple: since the speed of light is c for him, after one second in his frame of reference, the light will be 3 * 10<sup>6</sup> meters in front of him. Symbolic notation: t = 1 s d = 3*10<sup>6</sup> m (distance covered by light) Hence, for observer O, after one second, the light is located at spacetime point ( t, d, 0, 0 ), i.e. a distance of d along his x-axis. We assume that the light is stopped there (eg. it is absorbed or something) This same event can be viewed from the perspective of O', by using the Lorentz transformation between the two observers. This transformation is given by: t' = gamma * ( t - V*d / c<sup>2</sup> ) d' = gamma * (d - V*t) V = 0.9c gamma = 1 / sqrt(1 - (V/c)<sup>2</sup>) = 1 / sqrt( 1 - 0.9*0.9 ) Hence, for O', the light will stop at (t', d', 0, 0) with: t' = 0,2294 s d' = 6,8824 * 10<sup>4</sup> m This difference arises because there are two important relativistic effects to consider when looking from the perspective of O': Lorentz contraction and time dilatation. The Lorentz transformation takes these two effects into account. As you can see, the speed of light from the frame of reference of O' is given by: speed = distance / time = d' / t' = c Explanation The speed of light is invariant and thus the same for the two observers. The distance the light covers and the time it requires to do so differ. You'll probably will question the Lorentz transformation at this point, but to be completely honest, that is something I cannot explain without writing a 20 page document and attaching it to a message. In case you are wondering: No, I will not do that Please Register or Log in to view the hidden image!. Bye! Crisp
Crisp, nice work! Prosoothus, if you want to read something (relatively accessible, but still quite hard to read) on the subject, maybe Roger Penrose's "The Emperor's new Mind" is good. Somewhere halgway the volumunous book is a piece on the subject. (Does anybody have an alternative?) Merlijn
Prosoothus, it would be answer B I believe. Now you understand why relativity is so weird, and why so many people refused to accept it.
Crisp, First of all, light travels at 3*10^8 m/s and not 3*10^6 m/s as you indicated. Secondly, your formulas provide no proof. You assumed that time slows down, and you calculated the distance light travelled so that the speed of light would be constant to the second observer. That's very bad logic. To prove you wrong, let's look at distance of the light travelled compared to the stationary observer: If you change c in your formulas to the correct value, then: d2=6.8824 * 10^6 m However, the total distance the light would pass in that relative time period would be: Total distance=distance of the moving observer from the stationary observer+distance light travelled from the moving observer Which is: Total distance = .90c + 6.8824 * 10^6 =2.768824 * 10^8 That means that if you add the distance of the moving observer from the stationary observer, with the distance that the light travelled from the moving observer, you get 2.768824 *10^8 m. However, you said that the light moved 3 * 10^8 meters from the stationary observer. There is a discrepency of 8% in your formulas. As you can see, your formulas give two different results in the stationary observer's frame of reference. Tom
SpyFox_the_KMeson, It can't be b. Answer b gives two different results, just like Crisp's formulas. Logic dictates that the same photon can't be at two different places at the same time. Tom
Tom, Here's your problem: <i>There are to observers. Observer1 is (relatively) stationairy. Observer2 is traveling towards observer1 at .90 c(270,000 km/s). As soon as observer2 flys over over observer1, observer2 turns on his flashlight and shines the light in the same direction that he is travelling. Observer2 continues to travel at .90 c away from observer 1. One second after observer2 flew over observer1 where is the light??</i> The first thing to ask is: according to whom? Who is trying to determine the position of the light? Observer 1 or observer 2? It makes a difference, since we need to know who the one second time interval is measured by. Since you have not specified, I will do both for you. First, <b>observer 1's frame of reference, assuming that observer 1 sees observer 2 switch on the light and then looks at its position 1 second later as measured by observer 1's clocks.</b> After 1 second, the light is 300,000 km away from observer 1 by this measure. Observer 2 is 270,000 km away from observer 1. The light is therefore 30,000 km from observer 2 <b>as measured by observer 1</b>. If observer 1 calculates the speed of light using the measured distance and time he finds it is 300,000 km/1 second = 300,000 km/s. Now, in the 1 second that observer 1 measures, observer 2 observes a time interval of 0.44 seconds due to time dilation. In observer 2's frame of reference, in that amount of time observer 1 travels 270,000 km/s × 0.44 s = 117690 km. In the same time, the light travels 130767 km, but in the opposite direction. The distance between the light and observer 1 is therefore 130767+117690 = 248457 km, as measured by observer 2. If observer 2 calculates the speed of light by dividing the distance by the time he gets 130767/0.44 = 300,000 km/s. Second, <b>observer 2's frame of reference, assuming that observer 2 switches on the light and then looks at its position 1 second later as measured on observer 2's clocks.</b> Observer 2, from his own frame of reference, is stationary, so he sees observer 1 moving at 0.9c relative to him in the opposite direction to the beam of light. He sees the light moving at speed c in front of him. So, after 1 second as measured by him, he is 270,000 km away from observer 1, the light is 300,000 km away from him, and the light is 570,000 km away from observer 1. Observer 2 measures the speed of light to be the distance divided by the time, again giving 300,000 km/s. Now, in the 1 second that observer 2 measures, observer 1 observes a time interval of 2.29 seconds due to time dilation. In observer 1's frame of reference, in that amount of time observer 2 travels 270,000 km/s × 2.29 s = 619422 km. In the same time, the light travels 688247 km, in the same direction. The distance between the light and observer 2 is therefore 688247 - 619422 = 68825 km, as measured by observer 1. If observer 1 calculates the speed of light by dividing the distance by the time he gets 688247/2.29 = 300,000 km/s. Notice that in both cases, observers 1 and 2 measure different travel distances and times for each other and for the light, but their calculations of the speed of light always give 300,000 km/s. In other words, the speed of light is the same in both reference frames, but their ideas of distances and times change.
Tom, One more thing... <i>Secondly, your formulas provide no proof. You assumed that time slows down, and you calculated the distance light travelled so that the speed of light would be constant to the second observer. That's very bad logic.</i> It's exactly the same logic you are using but with different assumptions. Crisp and I are assuming that relativity is correct. You are assuming that Galilean relativity is correct, even though you probably don't realise that that's the name for your particular world-view in this example. The relativistic formulae give one set of answers. The "common-sense" Galilean formulae which you insist on using give another set of answers. This thought experiment you've put forward won't tell us which set of formulae are correct and which are wrong. The only thing that will tell us that is real experiments. Guess what? The real experiments show that the relativistic formulae are correct and Galileo's common sense is wrong. End of story.
Hi Tom, "First of all, light travels at 3*10^8 m/s and not 3*10^6 m/s as you indicated." LOL, indeed... This is the way you get punished for always putting c = 1 in your calculations Please Register or Log in to view the hidden image!. "It can't be b. Answer b gives two different results, just like Crisp's formulas. Logic dictates that the same photon can't be at two different places at the same time." You have to be careful saying that: in relativity both time and space are relative to the observer, so a photon can be at more places in the sense that two observers might not agree on the distance traveled. This gives two different results, but ofcourse does not mean that the photon has split into two and is at two places. Bye! Crisp
James R, You made the same mistake as Crisp, only more severe. The distance between observer 1 and the light should be equal for both observers. But 300,000 km does not equal 248,457 km. You cannot claim that the discrepancy is the result of time dilation either, since you included time dilation in your calculations already. Tom
Tom, <i>You made the same mistake as Crisp, only more severe. The distance between observer 1 and the light should be equal for both observers.</i> No, wrong again, I'm afraid. Length contraction is another well-known result of special relativity. Notice that the different distances are measured by <b>different</b> observers.
James R, It's kind of funny how the time dilation formula is so similiar to the length contraction formula that they balance each other out in the formula for velocity. As you pointed out in the "Galileo & Einstein - second thoughts" forum, time dilation and length contraction are superfluous, therefore time dilation and length contraction don't exist.Please Register or Log in to view the hidden image! Tom
Tom, <i>It's kind of funny how the time dilation formula is so similiar to the length contraction formula that they balance each other out in the formula for velocity.</i> Call it funny all you want. That's the way the world works. <i>As you pointed out in the "Galileo & Einstein - second thoughts" forum, time dilation and length contraction are superfluous, therefore time dilation and length contraction don't exist.</i> I don't like being misquoted. Please quote me where I said time dilation and length contraction are superfluous, provide a link to the relevant post, or admit that I said no such thing.
In the Galileo & Einstein - second thoughts" thread I explained how absolute motion does exist, but that absolute motions cancel each other out in a frame of reference. Your response was: Therefore, since the relativistic effects of time dilation and length contraction cancel each other out in the velocity formula, I guess you would imply that they don't exist either??? I don't know about other people, but I get the impression that when Einstein tested his "invariant property of light" theory, he found that his results did not match reality. So instead of admitting that he was wrong, he made up length contraction so that his results would match reality. The fact is that if you assume that there is no time dilation and no length contraction you get the same result as if there is time dilation and length contraction. In layman's terms, Einstein made something up. Because what he made up didn't match reality, he made something else up that cancels the first thing that he made up. It's such a shame when scientists shape the universe based on their theories, instead of the other way around. Tom
Hi Tom, I am sure that 137 will pop into this thread and explain you nicely how the concepts of time dilatation and length contraction evolved. I'll give it a brief shot without too much details: It has for long been known that Newton's theory of mechanics does not work well for high speeds. Several experiments around the 1850's suggested that the speed of light was an upper limit for electrons that are accelerated in a high electric field. Around the 1880's, the idea had grown that light propagated through aether, and that perhaps that aether, at that time thought as the propagating medium for light, had something to do with the strange behaviour experiments brought forward (Newton simply didn't work out). The Michelson-Moreley experiment, performed around that time (1880's) hinted that there probably was no such thing as aether, but at that time, the aether was firmly believed in and the result of the experiment was explained by other means. One idea came from Lorentz, who, around the year 1900, proposed that the aether actually tampered with measurement devices, shortening them in the direction of movement. This idea could explain why the Michelson-Moreley experiment did not detect the presence of aether. The effect is still known today as length contraction or Lorentz contraction. However, that meant that aether was undetectable (it made itself undetectable by tampering with measurement equipment), and this idea annoyed physicists: there appeared to be some unknown "force" that played with the experiments. Einstein in 1905, after the concept of Lorentz contraction was introduced, proved that with the constancy of the speed of light, no mysterious force was required to explain Lorentz-contraction: it was a property of nature, a consequence of the invariance of the speed of light. (This also made the concept of aether superfluous, and there was more than just this reason). Hence, Einstein did not come up with the idea of Lorentz contraction. He did not tamper with some laws to get things right. Also, time dilatation and lorentz contraction do not nicely compensate in the formula's for moving observers. The proper relations are (if O' is moving with respect to O): t = gamma t' d = d' / gamma gamma > 1, hence t is larger than t' (moving clocks tick slower for the stationary observer) and d is smaller than d' (moving objects contract in length for the stationary observer). Where do the two effects compensate ? The fact that everything works out nicely for light is not because the formula's were tampered with, it is merely an indication that the theory of relativity is self-consistent: you start with the invariance of the speed of light, this induces the existance of time dilatation and lorentz contraction, and by taking those two effects into account (as James R nicely did), you get back to the invariance of the speed of light. Bye! Crisp
Tom, <i>In the Galileo & Einstein - second thoughts" thread I explained how absolute motion does exist, but that absolute motions cancel each other out in a frame of reference.</i> In other words, absolute motion (if it exists) has no observable consequences at all. Therefore, by Occam's razor it is superfluous, as I said. On the other hand, time dilation and length contraction is observable, so those concepts are not superfluous. <i>Therefore, since the relativistic effects of time dilation and length contraction cancel each other out in the velocity formula, I guess you would imply that they don't exist either???</i> No. They are measurable in other ways, unlike an absolute standard of rest. <i>I don't know about other people, but I get the impression that when Einstein tested his "invariant property of light" theory, he found that his results did not match reality. So instead of admitting that he was wrong, he made up length contraction so that his results would match reality.</i> 1. Einstein did not invent length contraction. You might want to look up Fitzgerald. 2. The invariance of the speed of light has been tested over and over again, not just by Einstein. It matches reality. <i>The fact is that if you assume that there is no time dilation and no length contraction you get the same result as if there is time dilation and length contraction.</i> Wrong. If you throw out time dilation and length contraction, you are forced to adopt the position that the speed of light is different in different reference frames That is contrary to observation. <i>In layman's terms, Einstein made something up.</i> I think it is you who is making something up. <i>It's such a shame when scientists shape the universe based on their theories, instead of the other way around.</i> It's such a shame when people comment on things they don't understand.
James R, First I'd like to thank you for attempting to explain the question I asked a few posts ago. It must have taken a lot of time. Unfortunately, after reading your post I still don't understand something: Where did you get the 130767 km, number from? I looked at results I got from the formulas for time dilation and length contraction but wasn't able to get that number in any way. Please explain the formula you used to get this number. Thanks Tom
The speed of light is 300,000 km/s. It is a fundamental postulate of special relativity that the speed is the same in all reference frames. You have previously agreed that that is the case. In a time of 0.44 seconds, light travels a distance given by s = vt = 300,000 × 0.44 = 130767 km.