Spyman Posted March 2, 2005 Share Posted March 2, 2005 . This is a mindboggling question about Speed of Light and Theory of Relativity. It involves 5 theoretical, (but fully doable), simple experiments. I will begin with explaining the setup and this figure: C1 A1 O = Lightsource / / S2 [ and ] = Shutters ¤ o / o = Light detectors o----+----[O]----+----o ¤ = Computers \ / o ¤ \ - = light Path A2 S1 \ \ B2 + = Midways B1 C2 There are Three towers in a straight line, (just to get above all obstacles). Distance between the towers are 300 kilometers. The center tower has a powerful lightsource with Two shutters called S1 and S2. Also close to the lightsource are Two lightsensors called A1 and B1. The Two outer towers has lightsensors pointed at the center tower called A2 and B2. Midways between the center and outer towers are Two computers called C1 and C2. Lightsensors A1 and A2 are connected to computer C1. Lightsensors B1 and B2 are connected to computer C2. The computers measures the time taken between sensor 1 and sensor 2, detects light. Now follows the experiments, (number 3 is the important one). All are executed in the night without interfearing lightsources. (1 & 2 are for calibrating the computers.) (4 & 5 are for rechecking calibrations.) 1. Computer C1 On, Shutter S1 Open. Computer C2 Off, Shutter S2 Closed. -> Turn lightsource On and Off. Computer C1 measures the lightspeed from A1 to A2. Roughly the time taken should be 1 millisecond. 2. Computer C1 Off, Shutter S1 Closed. Computer C2 On, Shutter S2 Open. -> Turn lightsource On and Off. Computer C2 measures the lightspeed from B1 to B2. Roughly the time taken should be 1 millisecond. 3. Computer C1 On, Shutter S1 Open. Computer C2 On, Shutter S2 Open. -> Turn lightsource On and Off. Computer C1 measures the lightspeed from A1 to A2. Computer C2 measures the lightspeed from B1 to B2. 4. Computer C1 On, Shutter S1 Open. Computer C2 Off, Shutter S2 Closed. -> Turn lightsource On and Off. Computer C1 measures the lightspeed from A1 to A2. Roughly the time taken should be 1 millisecond. 5. Computer C1 Off, Shutter S1 Closed. Computer C2 On, Shutter S2 Open. -> Turn lightsource On and Off. Computer C2 measures the lightspeed from B1 to B2. Roughly the time taken should be 1 millisecond. Finally comes the important question: What is the time measured by both the computers in No:3 ? If it is roughly 1 millisecond then the photons going in opposite directions, (from center tower to Left tower and from center tower to Right tower), have had an relative speed between eatch other of twice the lightspeed. Which according to Theory of Relativity is impossible. (Nothing can go faster than the Speed of Light.) If the time doubles to roughly 2 milliseconds then the only thing needed to lower the Speed of Light is measuring its speed in opposite direction. Which according to Theory of Relativity also is undoable. (The Speed of Light is the same to all observants.) . Link to comment Share on other sites More sharing options...
swansont Posted March 2, 2005 Share Posted March 2, 2005 .If it is roughly 1 millisecond then the photons going in opposite directions' date=' (from center tower to Left tower and from center tower to Right tower), have had an relative speed between eatch other of twice the lightspeed. Which according to Theory of Relativity is impossible. (Nothing can go faster than the Speed of Light.) [/quote'] Your setup does not measure the relative speed of one photon to the other. It measures the speed of two separate photons by some observer. To measure the speed of one photon to the other, you'd have to be in one of the photon's frames of reference, which you can't do. Link to comment Share on other sites More sharing options...
syntax252 Posted March 2, 2005 Share Posted March 2, 2005 If two cars were traveling towards one another at their top speed, which was 50 MPH, then they would be approaching each other at a rate of 100 MPH, even though neither of them was capable of that speed--correct? Link to comment Share on other sites More sharing options...
Severian Posted March 2, 2005 Share Posted March 2, 2005 If two cars were traveling towards one another at their top speed, which was 50 MPH, then they would be approaching each other at a rate of 100 MPH, even though neither of them was capable of that speed--correct? Correct, but if two space ships where moving towards each other at 2x10^8 m/s the speed of the second to someone in the first would not be 4x10^8 m/s. Link to comment Share on other sites More sharing options...
syntax252 Posted March 2, 2005 Share Posted March 2, 2005 Correct, but if two space ships where moving towards each other at 2x10^8 m/s the speed of the second to someone in the first would not be 4x10^8 m/s. OK, but in the posted experiment, the time measurement was taken from the ground and not from inside the photons, so the anology is correct--correct? Link to comment Share on other sites More sharing options...
Jacques Posted March 3, 2005 Share Posted March 3, 2005 Correct, but if two space ships where moving towards each other at 2x10^8 m/s the speed of the second to someone in the first would not be 4x10^8 m/s. You would apply the gamma factor to get the relative speed of the space ship. But in that experiment you have two photons are going in opposites directions and I am not sure what would be the gamma factor. Can some one how knows relativity better than me make the calculations ? Link to comment Share on other sites More sharing options...
Spyman Posted March 3, 2005 Author Share Posted March 3, 2005 To measure the speed of one photon to the other, you'd have to be in one of the photon's frames of reference, which you can't do. Why can't a third party observer calculate the relative speed between Two moving objects ? If measuring their individually speed against "standstill" in the same time frame is possible and the direction of their movement is known. Link to comment Share on other sites More sharing options...
swansont Posted March 3, 2005 Share Posted March 3, 2005 Why can't a third party observer calculate the relative speed between Two moving objects ? If measuring their individually speed against "standstill" in the same time frame is possible and the direction of their movement is known. You can, unless it's for a photon. You can't be in the photon's frame of reference. Link to comment Share on other sites More sharing options...
Saint Posted March 3, 2005 Share Posted March 3, 2005 Correct, but if two space ships where moving towards each other at 2x10^8 m/s the speed of the second to someone in the first would not be 4x10^8 m/s. OK - I'm assuming that the speed you used is referenced to a common inertial frame? If your spaceships are moving toward each other at 2x10^8 m/s as referenced to some common inertial frame, where would the measurement need to be made from in order to find the relative velocity between the two spaceships as 4x10^8 m/s? From the common inertial frame. It seems like people are simply discounting the valididty of measurements taken from that frame. I thought that one of the tenents of relativity was the idea that no reference frame would be preferred over another. My point is that if two spaceships each travel in opposite directions from a given origin at .999 c, regardless of what they see on their individual ships, there must be some frame that will describe their relative speed as being greater than c. Otherwise, we cannot say that each is traveling from the origin at .99c. It's nonsensical. Link to comment Share on other sites More sharing options...
Spyman Posted March 3, 2005 Author Share Posted March 3, 2005 You can, unless it's for a photon. You can't be in the photon's frame of reference. Why have not the photons from the third party observers point exceeded the lightspeed in relative to eath other ? They start at the same time, travels at lightspeed in opposite directions, and arrives at the target in the same time. What if the experiment used some other particle with almost the speed of light ? Link to comment Share on other sites More sharing options...
Spyman Posted March 3, 2005 Author Share Posted March 3, 2005 My point is that if two spaceships each travel in opposite directions from a given origin at .999 c, regardless of what they see on their individual ships, there must be some frame that will describe their relative speed as being greater than c. Otherwise, we cannot say that each is traveling from the origin at .99c. It's nonsensical. But if there is another observer and the origin is already travelling with .999c from that observer ? Repeat and repeat... You will finally end up in infinite speed ! Link to comment Share on other sites More sharing options...
swansont Posted March 3, 2005 Share Posted March 3, 2005 What if the experiment used some other particle with almost the speed of light ? Then the relativistic velocity addition applies, and the final answer will still not exceed c. Link to comment Share on other sites More sharing options...
ydoaPs Posted March 3, 2005 Share Posted March 3, 2005 OK - I'm assuming that the speed you used is referenced to a common inertial frame? If your spaceships are moving toward each other at 2x10^8 m/s as referenced to some common inertial frame, where would the measurement need to be made from in order to find the relative velocity between the two spaceships as 4x10^8 m/s? holy cow those are fast spaceships! light goes roughly 3x10^8 m/s. Link to comment Share on other sites More sharing options...
Spyman Posted March 3, 2005 Author Share Posted March 3, 2005 Then the relativistic velocity addition applies, and the final answer will still not exceed c. Then We are back to the Moment 22 catch and You have still not explained why and how... Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted March 3, 2005 Share Posted March 3, 2005 Why have not the photons from the third party observers point exceeded the lightspeed in relative to eath other ? They start at the same time' date=' travels at lightspeed in opposite directions, and arrives at the target in the same time. What if the experiment used some other particle with almost the speed of light ? [/quote'] From this perspective the photons can be considered to be 2c in relative velocity. This does not violate the "c" speed limit. Similarly 2 masses can approach 2c relative velocity as you have described, from the third party perspective. (although it's not usually viewed that way) From either of their own it would still be less than c. Link to comment Share on other sites More sharing options...
Spyman Posted March 3, 2005 Author Share Posted March 3, 2005 From this perspective the photons can be considered to be 2c in relative velocity. This does not violate the "c" speed limit. Similarly 2 masses can approach 2c relative velocity as you have described, [b']from the third party perspective.[/b] Your answer is much alike Saint's: My point is that if two spaceships each travel in opposite directions from a given origin at .999 c, regardless of what they see on their individual ships, there must be some frame that will describe their relative speed as being greater than c. Otherwise, we cannot say that each is traveling from the origin at .99c. It's nonsensical. Which sounds OK... But: But if there is another observer and the origin is already travelling with .999c from that observer ?Repeat and repeat... You will finally end up in infinite speed ! What I mean is: If that is correct then what is the maximum speed from an infinite number of observers with an infinite number of experiments ? Or in another way: Is the maximum speedlimit c relative to the "spacetime-web" ? (or ether) (So maximum is c of their own and from another observer maximum is 2c but not higher.) Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted March 3, 2005 Share Posted March 3, 2005 (So maximum is c of their own and from another observer maximum is 2c but not higher.) I think what you mean by this is correct. Link to comment Share on other sites More sharing options...
Meir Achuz Posted March 8, 2005 Share Posted March 8, 2005 Most of the errors in this thread arise because some people have implicitly assumed, as obvious, the simple nonrelativistic rule for addition of velocities, but it is wrong. Link to comment Share on other sites More sharing options...
Spyman Posted March 10, 2005 Author Share Posted March 10, 2005 Most of the errors in this thread arise because some people have implicitly assumed, as obvious, the simple nonrelativistic rule for addition of velocities, but it is wrong. Will you please enlighten me with some details why and how it is wrong. The reason for starting this thread was to learn from the answers... Link to comment Share on other sites More sharing options...
swansont Posted March 10, 2005 Share Posted March 10, 2005 Galilean transformation and velocity addition are incorrect. The speed of light is a constant for all inertial observers - this means that Lorentz transformations are correct. Link to comment Share on other sites More sharing options...
Meir Achuz Posted March 14, 2005 Share Posted March 14, 2005 Will you please enlighten me with some details why and how it is wrong.The reason for starting this thread was to learn from the answers... The simple nonrelativistic rule for addition of velocities is wrong, because it disagrees with experiment. It disagrees with conclusions from the Lorentz transformation, which has had numerous experimental verifications. More directly, if a fast moving elementary particle emits a photon in the forward direction, the photon velocity is still measured as c. Relativistic velocity addition has also been verified in decays like pi-->mu+nu. Link to comment Share on other sites More sharing options...
Spyman Posted March 15, 2005 Author Share Posted March 15, 2005 Galilean transformation and velocity addition are incorrect. The speed of light is a constant for all inertial observers - this means that Lorentz transformations are correct. The simple nonrelativistic rule for addition of velocities is wrong, because it disagrees with experiment. It disagrees with conclusions from the Lorentz transformation, which has had numerous experimental verifications. More directly, if a fast moving elementary particle emits a photon in the forward direction, the photon velocity is still measured as c. Relativistic velocity addition has also been verified in decays like pi-->mu+nu. swansont and Meir Achuz, I belive You are correct, I am not questioning GR or Lorentz transformations. I know the example is wrong because it contradicts itself, but I am trying to get at grip on GR so I made the example up to learn exactly where it flaws, how and why. So please have patience with a novice and follow me through the experiment once more, I will take it step by step this time and waiting for You to confirm or deny with an explanation, before next step. STEP1: I am the only observer standing inside the middle tower where the light switch is. When I toggle the light switch the Two photons starts from the same point at the same time, or at least so close it's neglibe. Very soon after, they passes the sensors, inside the middle tower, at the same time, or at least so close it's neglibe, at time 0 milliseconds. Confirm or Deny with explanation... Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted March 15, 2005 Share Posted March 15, 2005 In your reference frame 2 photons sent in opposite directions will "distance" themselves at a rate of 2c. This does not contradict relativistic velocity addition. Nothing is exceeding c in any inertial frame. Link to comment Share on other sites More sharing options...
mustang292 Posted March 15, 2005 Share Posted March 15, 2005 Well, 200 years ago, passing the speed of sound seemed impossible. I believe that passing the speed of light is atainable with out question. If not, then the human race is screwed. Link to comment Share on other sites More sharing options...
Spyman Posted March 15, 2005 Author Share Posted March 15, 2005 In your reference frame 2 photons sent in opposite directions will "distance" themselves at a rate of 2c. This does not contradict relativistic velocity addition. Nothing is exceeding c in any inertial frame. swansont and Meir Achuz seems to belive differently, or am I just misinterpreting their posts ? Link to comment Share on other sites More sharing options...
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