Zolobrine Posted May 20, 2017 Share Posted May 20, 2017 This is my first post and I'm glad to join this awesome community. So yeah lately I've been wondering about this and I thought of a lot of possibilities this could end in, I discussed this in some other forums but none of them really could theorize a solution. So let's say we have two men in two spaceships in a vacuum, they both have clocks which show the same time at the start of the experiment. One of them is "A" and the other is "B". They are facing apart from each other (180^{o}) and as the experiment starts they both start traveling at 75% the speed of light (away from the starting point). If you look at this from a Classical Physics perspective technically they should travel at 150% the speed of light apart from each other. Questions: 1. Is the Classical Physics perspective correct and if not explain the answer. 2. Will the clocks in both spaceships differ from each other? 3. Explain what will happen and why. Thank you, and sorry if I'm posting this in a wrong sub-forum. Link to comment Share on other sites More sharing options...

KipIngram Posted May 20, 2017 Share Posted May 20, 2017 I'll take a shot. First off, welcome to the community. Ok, so let me add an observer C who remains at the launch point of the two rockets. That observer © will measure a 150% light speed separation rate for the two ships. That doesn't violate the theory fo special relativity, because C is not seeing either one of the rockets exceed light speed - the 150% comes from adding together two separate observations. A and B will each see C moving away from them at 75% of light speed, just as C sees them moving with 75% light speed. How A and B observe one another, though, requires the relativity formula for velocity addition: u' = (u + v) / (1 + uv/c^2) where c is the velocity of light. Here u' is the speed A sees B moving away at, u is the speed A sees C moving away at, and v is the speed C sees B moving away at. Taking c = 1 and plugging in u = v = 0.75, you get u' = 0.96 So A and B will see each other moving away at 96% of the speed of light. As far as clocks go, the observer left at home © will see both A's clock and B's clock running at a reduced rate. The formula for rate reduction is (1-v^2/c^2)^0.5. So C will see A's clock and B's clock running at 66.1% of the normal rate, and A and B will see each other's clocks running at 28% of the normal rate. I would assume that if A and B maintain completely symmetric motion patterns (i.e., if they eventually slow down, turn around, and come back home, while mirroring each other's velocity pattern at all times) then when everyone is back together again A's clock and B's clock will agree with one another, but both of them will show that less time has elapsed than A's clock shows. Why? That's just what the theory of special relativity predicts. Einstein assumed that 1) the laws of physics are the same for observers moving at constant velocity with respect to one another, and 2) all observers will agree that an object moving at the speed of light is, in fact, moving at that speed. They will not get different results when measuring such a velocity. The rest is just algebra. 4 Link to comment Share on other sites More sharing options...

Zolobrine Posted May 20, 2017 Author Share Posted May 20, 2017 Awesome xD now I get it, thanks so much, last night I spent 3 hours of sleep time on this problem and couldn't figure it out, this is why I find the internet to be humanities biggest invention. Link to comment Share on other sites More sharing options...

Delta1212 Posted May 20, 2017 Share Posted May 20, 2017 The answer is that the classical solution has a major failing that is addressed in relativity. In the frame of someone at the starting point, the separation rate of the two spaceships would indeed be 1.5c. It turns out, however, that velocities don't add linearly. This only becomes really noticeable at appreciable fractions of the speed of light, though. For A and B, the velocity that one would observe the other as moving away at would be given their respective velocities moving away from c would be given by the following formula: v = (A + B)/(1+AB/c^2) So in this case 1.5c/1.5625 or they will each see the other moving away at 96% of the speed of light. Now, the reason that we don't notice this difference in our daily lives is that, at very low speeds, the above formula very closely approximates the classical formula for addition of velocities. So, for example, if you plug in the numbers for two cars traveling away from C each at 20mph, each car will I arc the other as traveling away at approximately 39.999999mph Which is a difference of about an inch and a half per 1,000 miles travelled off of what the classical formula gives you. Edit: Missed it by that much Link to comment Share on other sites More sharing options...

Mordred Posted May 20, 2017 Share Posted May 20, 2017 (edited) I'll take a shot. First off, welcome to the community. Ok, so let me add an observer C who remains at the launch point of the two rockets. That observer © will measure a 150% light speed separation rate for the two ships. That doesn't violate the theory fo special relativity, because C is not seeing either one of the rockets exceed light speed - the 150% comes from adding together two separate observations. A and B will each see C moving away from them at 75% of light speed, just as C sees them moving with 75% light speed. How A and B observe one another, though, requires the relativity formula for velocity addition: u' = (u + v) / (1 + uv/c^2) where c is the velocity of light. Here u' is the speed A sees B moving away at, u is the speed A sees C moving away at, and v is the speed C sees B moving away at. Taking c = 1 and plugging in u = v = 0.75, you get u' = 0.96 So A and B will see each other moving away at 96% of the speed of light. As far as clocks go, the observer left at home © will see both A's clock and B's clock running at a reduced rate. The formula for rate reduction is (1-v^2/c^2)^0.5. So C will see A's clock and B's clock running at 66.1% of the normal rate, and A and B will see each other's clocks running at 28% of the normal rate. I would assume that if A and B maintain completely symmetric motion patterns (i.e., if they eventually slow down, turn around, and come back home, while mirroring each other's velocity pattern at all times) then when everyone is back together again A's clock and B's clock will agree with one another, but both of them will show that less time has elapsed than A's clock shows. Why? That's just what the theory of special relativity predicts. Einstein assumed that 1) the laws of physics are the same for observers moving at constant velocity with respect to one another, and 2) all observers will agree that an object moving at the speed of light is, in fact, moving at that speed. They will not get different results when measuring such a velocity. The rest is just algebra. Excellent and accurate shot. +1. Edited May 20, 2017 by Mordred Link to comment Share on other sites More sharing options...

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