michel123456 Posted March 6, 2011 Author Share Posted March 6, 2011 (edited) The velocity addition equation works for objects in a single frame of reference. You are mixing frames, so the formula does not apply. The solution is simply this: v = d/t. How far does each object move in a time t, relative to the observer. The relative speed is the difference in the distance traveled divided by that time. Which just turns out to be the difference in their speeds. If a rocket travels at c/2, then an observer will see the light outpacing it by c/2. If the rocket goes in the other direction, the separation occurs at 3c/2. 3c/2 = superluminal. Is that allowed? Use u+v=w if u, v and w are all measured by C. Use the relativistic formula if u and w are measured by C and v is measured by B. They are solutions to different problems so you can't substitute the solution of one into the other. that's what i have done in point 3: C is the muon's FOR The muon measures u=0,98c w is measured in the muon's FOR w=c v is measured in Earth's FOR v=0,02c (on your assumption) the result is inaccurate. Edited March 6, 2011 by michel123456 Link to comment Share on other sites More sharing options...
Iggy Posted March 6, 2011 Share Posted March 6, 2011 v=.02c if it is measured by the same frame of reference that measures u and w. unless u, v, and w are all measured by the same frame of reference you cannot use u+v=w. They are not all measured by the same frame here, C is the muon's FOR The muon measures u=0,98c w is measured in the muon's FOR w=c v is measured in Earth's FOR v=0,02c (on your assumption) the result is inaccurate. so you should not expect to be able to use v=.02c Link to comment Share on other sites More sharing options...
michel123456 Posted March 6, 2011 Author Share Posted March 6, 2011 v=.02c if it is measured by the same frame of reference that measures u and w. unless u, v, and w are all measured by the same frame of reference you cannot use u+v=w. They are not all measured by the same frame here, so you should not expect to be able to use v=.02c Look I am getting completely nuts. In BOTH Frames, w=c, u=0,98c because it is totally symmetric. The result of the relativistic equation gives the result that is measured in the other frame. The result is c, not 0,02c. Link to comment Share on other sites More sharing options...
Iggy Posted March 6, 2011 Share Posted March 6, 2011 yeah, the relativistic equation is probably complicating things needlessly. The distinction can be summed up by saying that light goes 1c faster than the muon in the muon’s frame and that light goes .02c faster than the muon in earth’s frame. In the same fashion, light goes 1c faster than the earth in earth’s frame and light goes .02c faster than the earth in the muon’s frame. Link to comment Share on other sites More sharing options...
swansont Posted March 6, 2011 Share Posted March 6, 2011 3c/2 = superluminal. Is that allowed? Yes. It's moving objects measured with respect to each other, in a different frame of reference. They are not causally connected. Link to comment Share on other sites More sharing options...
Iggy Posted March 6, 2011 Share Posted March 6, 2011 They are not causally connected. How do you mean? (best diagram I could find) A and G are both in the other's past lightcone. Link to comment Share on other sites More sharing options...
michel123456 Posted March 6, 2011 Author Share Posted March 6, 2011 yeah, the relativistic equation is probably complicating things needlessly. The distinction can be summed up by saying that light goes 1c faster than the muon in the muon’s frame and that light goes .02c faster than the muon in earth’s frame. In the same fashion, light goes 1c faster than the earth in earth’s frame and light goes .02c faster than the earth in the muon’s frame. You are confusing v and w IMHO. The muon's frame calculates v in Earth's frame. Please show me on the first diagram where is v in Earth's frame. On this one Link to comment Share on other sites More sharing options...
Iggy Posted March 6, 2011 Share Posted March 6, 2011 You are confusing v and w IMHO. I'm not confused, Michel. The muon's frame calculates v in Earth's frame. The muon can calculate v in Earth's frame or its own frame. In the muon's frame the muon is at rest and the earth has a velocity of .98c. v is the difference between earth's velocity and light's velocity. In earth's frame v=1c and in the muon's frame v=0.02c. When Janus said, You're doing an improper frame switch. So far, this whole thread has been about the Earth frame... And in the Earth frame, the ship does chase after its own light. he is saying that you should not switch frames for what you are doing. Light goes .5c faster than the ship in earth's frame just like light goes .02c faster than the earth in the muon's frame. Please show me on the first diagram where is v in Earth's frame. On this one C is the muon. It is 'at rest'. B is the earth. The earth moves to the right at u=.98c compared to the muon. A is a ray of light. The ray of light moves to the right at v compared to the earth and w compared to the muon. If u, v, and w are all in the muon's frame then use u+v=w so that u=0.98c, w=1c and v=0.02c. If u and w are in the muon's frame and v is in earth's frame then use w=(u+v)/(1+uv/c2) so that u=0.98c, w=1c and v=1c. In other words, in earth's frame the distance between the earth and light changes one light-second per second. In the muon's frame, on the other hand, the distance between the earth and light changes .02 light-seconds per second. The value of v depends on which frame you are talking about. In the twin 'paradox' the distance between the ship and a ray of light changes 0.5 light-seconds per second in earth's frame and the distance changes 1 light-second per second in the ship's frame. Does this make sense? Link to comment Share on other sites More sharing options...
swansont Posted March 6, 2011 Share Posted March 6, 2011 How do you mean? (best diagram I could find) A and G are both in the other's past lightcone. They are not causally connected in the present. The limitation on the speed of light refers to single objects. This is a version of the lighthouse paradox. Two unrelated events are not restricted to occurring within d/c of each other. Link to comment Share on other sites More sharing options...
michel123456 Posted March 6, 2011 Author Share Posted March 6, 2011 (edited) I'm not confused, Michel. I am. The muon can calculate v in Earth's frame or its own frame. In the muon's frame the muon is at rest and the earth has a velocity of .98c. v is the difference between earth's velocity and light's velocity. In earth's frame v=1c and in the muon's frame v=0.02c. (...) C is the muon. It is 'at rest'. B is the earth. The earth moves to the right at u=.98c compared to the muon. A is a ray of light. The ray of light moves to the right at v compared to the earth and w compared to the muon. If u, v, and w are all in the muon's frame then use u+v=w so that u=0.98c, w=1c and v=0.02c. If u and w are in the muon's frame and v is in earth's frame then use w=(u+v)/(1+uv/c2) so that u=0.98c, w=1c and v=1c. In other words, in earth's frame the distance between the earth and light changes one light-second per second. In the muon's frame, on the other hand, the distance between the earth and light changes .02 light-seconds per second. (...) Now take C as the Earth, B the muon, and A the ray of light. Where upon the diagram is v that has been calculated from the muon's FOR ? I don't know if we are communicating properly. What is measured from one FOR must correspond to the other, it is a single phenomena, either observed from earth, either from the muon. When the muon calculates what is experienced in Earth's FOR, it must be what is calculated by Earth's FOR. v is v. Unless you tell me that v in one FOR is w in the other, and vice-versa. That is the reason why I 'll ask you to show me upon the diagram where is v in Earth's FOR (the value you say is 0,02c), as calculated by the muon (=1c), when C is the Earth. Edited March 6, 2011 by michel123456 Link to comment Share on other sites More sharing options...
Iggy Posted March 7, 2011 Share Posted March 7, 2011 They are not causally connected in the present. Sorry, I guess I don't follow. I certainly agree that two present events cannot have a causal relationship if they are not co-located, but this would be true regardless of the velocity of the world lines the events populate. I'm sure what you're saying makes sense, I'm just not getting it. The limitation on the speed of light refers to single objects. This is a version of the lighthouse paradox. Two unrelated events are not restricted to occurring within d/c of each other. No doubt--I agree. I am. I animated some frames that might help. I don't know if we are communicating properly. What is measured from one FOR must correspond to the other, it is a single phenomena, either observed from earth, either from the muon. When the muon calculates what is experienced in Earth's FOR, it must be what is calculated by Earth's FOR. v is v. Unless you tell me that v in one FOR is w in the other, and vice-versa. Here is earth's FOR where a rocket (going 0.5c) and a ray of light (the red arrow going 1c) are moving away from the earth in the same direction. This is what the frame looks like at sequential moments in time: And here is the rocket's FOR: V is the velocity (or the change in distance per time) between the rocket and the ray of light. In earth's frame it is .5c and in the rocket's frame it is 1c. In earth's frame the speed of light compared to the rocket is .5c. The rocket 'chases' the light in earth's frame. That is the reason why I 'll ask you to show me upon the diagram where is v in Earth's FOR (the value you say is 0,02c), as calculated by the muon (=1c), when C is the Earth. The astronaut can solve the value of v in earth's frame by finding u and w in earth's frame and solving w=u+v. 1 Link to comment Share on other sites More sharing options...
michel123456 Posted March 7, 2011 Author Share Posted March 7, 2011 (edited) Quoting Here is earth's FOR where a rocket (going 0.5c) and a ray of light (the red arrow going 1c) are moving away from the earth in the same direction. This is what the frame looks like at sequential moments in time: And here is the rocket's FOR: V is the velocity (or the change in distance per time) between the rocket and the ray of light. In earth's frame it is .5c and in the rocket's frame it is 1c. In earth's frame the speed of light compared to the rocket is .5c. The rocket 'chases' the light in earth's frame. The astronaut can solve the value of v in earth's frame by finding u and w in earth's frame and solving w=u+v. Thank you for the animations. They show what I was afraid of when saying I don't know if we are communicating properly. What is measured from one FOR must correspond to the other, it is a single phenomena, either observed from earth, either from the muon. The 2 diagrams are different in length. At t=10, the distance between the Earth and the ray of light are different, small in the first diagram, long in the second. Since we are talking about the same and only one phenomena, how is it possible? Can you make some other diagrams showing at the end the ray of light at the same distance no matter the FOR? Edited March 7, 2011 by michel123456 Link to comment Share on other sites More sharing options...
swansont Posted March 7, 2011 Share Posted March 7, 2011 Q The 2 diagrams are different in length. At t=10, the distance between the Earth and the ray of light are different, small in the first diagram, long in the second. Since we are talking about the same and only one phenomena, how is it possible? They are from two different frames of reference. Lengths are contracted in moving frames, but we weren't focusing on that. Link to comment Share on other sites More sharing options...
michel123456 Posted March 7, 2011 Author Share Posted March 7, 2011 They are from two different frames of reference. Lengths are contracted in moving frames, but we weren't focusing on that. These are 2 diagrams representing only one phenomena. Iggy may use lorentz transformations at will. I would like to see that. Link to comment Share on other sites More sharing options...
Iggy Posted March 7, 2011 Share Posted March 7, 2011 (edited) The 2 diagrams are different in length. At t=10, the distance between the Earth and the ray of light are different, small in the first diagram, long in the second. That's because the earth is at rest in the first frame and the rocket is at rest in the second. Both things are moving in the same direction in earth's frame and they are moving in opposite directions in the rocket's frame. Also, the speed of light compared to the earth in the first fame has to be the same as the speed of light compared to the rocket in the second frame. Can you make some other diagrams showing at the end the ray of light at the same distance no matter the FOR? The diagrams are correct. I can't change them. Edited March 7, 2011 by Iggy Link to comment Share on other sites More sharing options...
michel123456 Posted March 7, 2011 Author Share Posted March 7, 2011 I can't change them. Sure you can. Link to comment Share on other sites More sharing options...
Iggy Posted March 7, 2011 Share Posted March 7, 2011 Now that you see the frames, Janus' post #32 can't help but make sense. Link to comment Share on other sites More sharing options...
michel123456 Posted March 7, 2011 Author Share Posted March 7, 2011 (edited) No. It doesn't make sense that the 2 diagrams are different either. IMHO what happens is the following: _in the first animation, you made a velocity addition, a substraction actually. It is not different than a statement in which you say that 0,6c+0,6c=1,2c: it is wrong. You should have used the relativistic formula, although remaining in the same FOR. _the same goes for the second animation. You made there a clearer velocity addition in a newtonian way. The result is that light escapes from Earth at superluminal speed: it must be wrong. You should have used the relativistic formula in which no speed can attain SOL. If you do that for both animations, the 2 diagrams will give the same result, and this thread come to an end. The other way is to send me to the asylum. Edited March 7, 2011 by michel123456 Link to comment Share on other sites More sharing options...
Sisyphus Posted March 7, 2011 Share Posted March 7, 2011 Both animations are totally correct. In both frames, light moves at C, as it must, always. Since it is moving at C, it has a velocity of C relative to anything that is at rest: Earth in the first animation, the ship in the second. And clearly, it won't have a velocity of C relative to anything that is not at rest. You should have used the relativistic formula, although remaining in the same FOR. No, that's for translating between frames. In one frame, velocities add normally. It's just that no velocity can be greater than C. The result is that light escapes from Earth at superluminal speed: it must be wrong. Not so. The speed limit is C. That makes the maximum closing or separation speed 2C, for objects moving in opposite directions. 1 Link to comment Share on other sites More sharing options...
swansont Posted March 7, 2011 Share Posted March 7, 2011 Sure you can. He can't change them and have them be correct. You can make incorrect drawings quite easily, as evidenced by a number of threads in speculations that declare SR to be wrong. No. It doesn't make sense that the 2 diagrams are different either. IMHO what happens is the following: _in the first animation, you made a velocity addition, a substraction actually. It is not different than a statement in which you say that 0,6c+0,6c=1,2c: it is wrong. You should have used the relativistic formula, although remaining in the same FOR. _the same goes for the second animation. You made there a clearer velocity addition in a newtonian way. The result is that light escapes from Earth at superluminal speed: it must be wrong. You should have used the relativistic formula in which no speed can attain SOL. If you do that for both animations, the 2 diagrams will give the same result, and this thread come to an end. The other way is to send me to the asylum. The velocity addition formula gives you the speed object A measures measures B to be moving at, when the numbers come from frame C, i.e. it's the value of B's velocity in A's rest frame. That can't exceed c. It does not give the relative speed of two objects, when they are both moving. That value, as Sisyphus has said, is limited to 2c. Link to comment Share on other sites More sharing options...
Iggy Posted March 7, 2011 Share Posted March 7, 2011 (edited) No. It doesn't make sense that the 2 diagrams are different either. Different frames of reference are different. IMHO what happens is the following: _in the first animation, you made a velocity addition, a substraction actually. It is not different than a statement in which you say that 0,6c+0,6c=1,2c: it is wrong. You should have used the relativistic formula, although remaining in the same FOR. _the same goes for the second animation. You made there a clearer velocity addition in a newtonian way. The result is that light escapes from Earth at superluminal speed: it must be wrong. You should have used the relativistic formula in which no speed can attain SOL. The relativistic formula isn't used when placing multiple objects with different velocities onto a frame of reference. Many people have corrected you and I've given a source: This is due to a misunderstanding of the question which can easily be confused with the following one: Suppose the object B above is an experimenter who has set up a reference frame consisting of a marked ruler with clocks positioned at measured intervals along it. He has synchronised the clocks carefully by sending light signals along the line taking into account the time taken for the signals to travel the measured distances. He now observes the objects A and C which he sees coming towards him from opposite directions. By watching the times they pass the clocks at measured distances he can calculate the speeds they are moving towards him. Sure enough he finds that A is moving at a speed v and C is moving at speed u. What will B observe as the speed at which the two objects are coming together? It is not difficult to see that the answer must be u+v whether or not the problem is treated relativistically. In this sense velocities add according to ordinary vector addition. But that was a different question from the one asked before. Originally we wanted to know the speed of C as measured relative to A not the speed at which B observes them moving together. This is different http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/velocity.html In the rocket's frame of reference light does not travel c compared to the earth. In earth's frame of reference light does not travel c compared to the rocket. If you do that for both animations, the 2 diagrams will give the same result, and this thread come to an end. This is earth's reference frame: The distance between the earth and the ray of light changes at a rate of c in this frame (notice the earth is at rest in this frame). The distance between the rocket and the ray of light changes at a rate of w-u in this frame. This is the rocket's reference frame: The distance between the rocket and the ray of light changes at c in this frame (the rocket here is at rest). The distance between the earth and the ray of light changes at w-u in this frame. [edited: I forgot to make u negative in the diagram] Why do you assume this is wrong? Edited March 7, 2011 by Iggy Link to comment Share on other sites More sharing options...
michel123456 Posted March 7, 2011 Author Share Posted March 7, 2011 (edited) Different frames of reference are different. (...) Of course they are different. But they must extract the same conclusions about the Universe. In both frames, light moves at C, as it must, always. Since it is moving at C, it has a velocity of C relative to anything that is at rest: Exactly. We agree. So why do you all believe that the rocket is "chasing" light? It is not, not even in the first animation. You cannot make the substraction c-v just like that, because c-v=c, always. Earth in the first animation, the ship in the second. And clearly, it won't have a velocity of C relative to anything that is not at rest. That's the point: it has a velocity of c no matter what. Why do you assume this is wrong? It is obvious. Edited March 7, 2011 by michel123456 Link to comment Share on other sites More sharing options...
swansont Posted March 7, 2011 Share Posted March 7, 2011 Of course they are different. But they must extract the same conclusions about the Universe. We have to agree on whether an event happened or not. But not in the ordering of those events. So why do you all believe that the rocket is "chasing" light? It is not, not even in the first animation. You cannot make the substraction c-v just like that, because c-v=c, always. That's the point: it has a velocity of c no matter what. No, that's not true, nor what he said. Light has a speed of c when measured against something at rest. In the earth's frame, the rocket is not at rest. Therefore the earth frame will not measure light to be moving at c relative to the rocket. Link to comment Share on other sites More sharing options...
Janus Posted March 7, 2011 Share Posted March 7, 2011 Of course they are different. But they must extract the same conclusions about the Universe. Exactly. We agree. So why do you all believe that the rocket is "chasing" light? It is not, not even in the first animation. You cannot make the substraction c-v just like that, because c-v=c, always. Try looking at this animation I made. It goes back to your original scenario, with a rocket going out and then returning. the animation shows events according to the Earth (green dot) frame. The ship is emitting light (the expanding circles) as it travels. The red and blue indicate the red or blue shift seen by an observer at any given point. As the ship starts on its journey, light is emitted in all directions at c, as shown by the first expanding circle. Note that this circle always remains centered on the Earth. As the ship continues on, it continues to emit light as shown by the subsequent circles. Each circles center in turn remains at the pint where the light was emitted, making each circle offset from the last, causing the waves to be closer together on one side and further apart on the other.(essentially, this is the Doppler shift). After 25 days, the light emitted when the ship left Earth reaches the point where the ship will turn around (the first pause in the animation). This is when a person located there will first see the Ship leave Earth. At 50 days, the ship arrives at the turnaround point, 25 days after the person there saw them leave Earth. The animation pauses at 51 days to show the light emitted at the moment of turnaround(the circle that is blue on the left) heading back toward Earth. The return trip is the same as the outbound trip with left and right switched. At 75 days (third pause) the light from the turnaround reaches Earth, while the ship is some 12.5 light days form Earth. 25 days later, at 100 days the ship returns to Earth. 1 Link to comment Share on other sites More sharing options...
michel123456 Posted March 8, 2011 Author Share Posted March 8, 2011 (edited) I understand the animation. Look. Someone said, if muons go at 0,98c in Earth's FOR, then the remaining c-0,98c=0,02c in Earth's FOR, because it is all measured in the same FOR. If a child is coming to you and ask you the result of 0,98c+0,02c What is your answer? Edited March 8, 2011 by michel123456 Link to comment Share on other sites More sharing options...
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