Tres Juicy Posted January 30, 2012 Posted January 30, 2012 Hi all, So, Relativity says that the speed of light is the same in all frames of reference, but I can't get my head around it... Here's my question: 2 observers A an B A is riding a horse (why not?) at 1/2 c - B is stationary. A fires a laser at a distant target and measures the speed of light travelling away from him to be c B, our stationary observer also measures the speed of light travelling away from him to be c So lets say that the laser was fired while the two observers are at an equal distance away from the target and that from that dstance the light will take 10 seconds to reach the target from that point. Does that depend on the reference frame? Does the stationary B measure the time it takes to be 10 seconds, while the moving A measure it to be 5 seconds? How can the light move away from A (who is travelling at 1/2 c) at c and not reach the target sooner than it would if fired from a stationary position? Or do they both measure the same? If so, that can't be right can it? If they both measure the same then the speed of light is different in each frame(?!) That's enough for now I think.... Can someone enlighten me as to how this works?
Cap'n Refsmmat Posted January 30, 2012 Posted January 30, 2012 This is much like the old "relativistic flashlight" idea, which I have a simulation of here: http://www.refsmmat.com/jsphys/relativity/relativity.html#flashlight (Works best on Firefox and Chrome. Press Next to switch to the reference frame of the observer, instead of that of the flashlight.) And yes, the time it takes for the light to reach the target is dependent on reference frame. From the perspective of observer A, the distance to the target is slightly shorter, and the photon is able to get there faster; from the perspective of observer B, A's clock is running slow.
Janus Posted January 30, 2012 Posted January 30, 2012 There are three factors to be considered here: Time dilation Length contraction Relativity of simultaneity First we'll consider length contraction. Since B is at rest with respect to the target, I assume that the distance between target and B is 10 light seconds, as measured by B However since there is a relative velocity difference of 0.5c between A and B, A will measure that same distance as being 8.66 light seconds. Thus for B, the light will take 10 sec to travel to the target and for A it will take 8.66 seconds. Now consider time dilation According to B, A's clock runs at a rate of 0.866 that of his own, and thus will read 8.66 sec when the light reaches the target (so far so good, A agrees to this.) However, according to A, it is B's clock that is running at a rate of 0.866, and thus B's clock reads only 7.5 sec when the light reaches the target. This leads us to the third factor, the relativity of simultaneity. Events that are simultaneous for B are not always going to be so for A. In this case, B's clock reading 10 sec and the light reaching the target are simultaneous for B, but are not so for A (B's clock does not read 10 sec until some time after the light reaches the target.) For further illustration, let's put a clock at the target, which is synchronized to B's clock according to B (according to anyone at rest with respect to B and the target, the two clocks always read the same. This means that the light will reach the target clock when it reads 10 sec. As stated above, according to A, B's clock only ticks away 7.5 sec while the light travels from it to the target. However, A must also agree that the target clock reads 10 sec when the light reaches it(contradictions arise otherwise). Also, B's clock and the target clock run at the same rate, and B's clock has to read 0 when A passes it and the light is emitted. This leads to the conclusion that, according to A, the clock at the target must already read 2.5 sec at the moment the light was emitted. Again, no agreement of simultaneity between A and B, B says that the target clock always reads the same as his, and A says that it always runs 2.5 sec ahead of B's clock. The point is that frames in relative motion with respect to each other have different measurements of time and space.
timo Posted January 30, 2012 Posted January 30, 2012 (edited) The point is that frames in relative motion with respect to each other have different measurements of time and space. My gut feeling is that in this case the point could also be that frames in relative motion with respect to each other have different measurements of the speed that the target approaches them with. Edited January 30, 2012 by timo
D H Posted January 30, 2012 Posted January 30, 2012 Here's my question: 2 observers A an B A is riding a horse (why not?) at 1/2 c - B is stationary. A fires a laser at a distant target and measures the speed of light travelling away from him to be c B, our stationary observer also measures the speed of light travelling away from him to be c So lets say that the laser was fired while the two observers are at an equal distance away from the target and that from that dstance the light will take 10 seconds to reach the target from that point. Does that depend on the reference frame? The stationary observer will calculate the time needed for the light to reach the target as 10 seconds while to the person on the horse it's 8.660254 seconds. Your confusion arises from a number of misconceptions. #1, You can't measure the one-way transmission time (one-way speed of light). You can only measure round trip time. The assumption is that space is homogeneous. The speed of light is the same in all directions. (This is testable). #2, Simultaneity is also relative. That statement "while the two observers are at an equal distance away from the target" doesn't quite work in the relativity. So let's change the experiment so that the laser is triggered right as the rider goes by stationary observer. Here simultaneity does work because the rider and observer are (instantaneously) co-located. They are not however at the same distance from the target. #3, Distance is also relative. The stationary observer sees the target as 2,997,924.58 kilometers (10 light seconds) away. To the person on the 0.5 c horse, the target is only 2,596,278.84 kilometers (8.660254 light seconds) away. 1
imatfaal Posted January 31, 2012 Posted January 31, 2012 .../snipped This leads us to the third factor, the relativity of simultaneity. Events that are simultaneous for B are not always going to be so for A. In this case, B's clock reading 10 sec and the light reaching the target are simultaneous for B, but are not so for A (B's clock does not read 10 sec until some time after the light reaches the target.) For further illustration, let's put a clock at the target, which is synchronized to B's clock according to B (according to anyone at rest with respect to B and the target, the two clocks always read the same. This means that the light will reach the target clock when it reads 10 sec. .../ "which is synchronized to B's clock according to B (according to anyone at rest with respect to B and the target, the two clocks always read the same" A clock at target which is synchronized according to B is OK - but one that is synchronized to anyone at rest with respect to B and the target causes me problems. Someone at rest at the proposed target clock will see B's clock as 20 secs slow, anyone between B and the target will see varying difference in the readings of the two clocks - only someone at rest 10 lightseconds away (and exactly where B is) will have the clocks synchronized. 1
Tres Juicy Posted January 31, 2012 Author Posted January 31, 2012 #3, Distance is also relative. The stationary observer sees the target as 2,997,924.58 kilometers (10 light seconds) away. To the person on the 0.5 c horse, the target is only 2,596,278.84 kilometers (8.660254 light seconds) away. How can distance be realtive? Surely they are either at the same distance or they are not? With this in mind, if B (stationary) fires a laser past A while he is moving at 0.5c B measures the light to move away from him at c and A also observes the light going past him to move away from him at c (even though he is moving at 0.5c) Is that right? If it is surely one of these measurements is false? I mean the light cannot be doing different speeds for different observers can it?
imatfaal Posted January 31, 2012 Posted January 31, 2012 How can distance be realtive? Surely they are either at the same distance or they are not? aye, there's the rub. the distances are the same when measured from within the same frame - not when measured from frames in relative motion With this in mind, if B (stationary) fires a laser past A while he is moving at 0.5c B measures the light to move away from him at c and A also observes the light going past him to move away from him at c (even though he is moving at 0.5c) Is that right? If it is surely one of these measurements is false? I mean the light cannot be doing different speeds for different observers can it? neither is false, both are correct when viewed from their own frame of reference. the speed of light is the same - time and distance in a frame in relative motion vary. we like to think of distance as invariant - it is not. this is not instinctive and goes against much that seems incontrovertible - however is has a massive amount of experimental proof behind it and is mathematically as tight as a drum. 1
Tres Juicy Posted January 31, 2012 Author Posted January 31, 2012 aye, there's the rub. the distances are the same when measured from within the same frame - not when measured from frames in relative motion neither is false, both are correct when viewed from their own frame of reference. the speed of light is the same - time and distance in a frame in relative motion vary. we like to think of distance as invariant - it is not. this is not instinctive and goes against much that seems incontrovertible - however is has a massive amount of experimental proof behind it and is mathematically as tight as a drum. Could you point me towards some of the experimental proofs? I think I'd like to read a little more on the subject Thanks
Schrödinger's hat Posted January 31, 2012 Posted January 31, 2012 How can distance be realtive? Surely they are either at the same distance or they are not? With this in mind, if B (stationary) fires a laser past A while he is moving at 0.5c B measures the light to move away from him at c and A also observes the light going past him to move away from him at c (even though he is moving at 0.5c) Is that right? If it is surely one of these measurements is false? I mean the light cannot be doing different speeds for different observers can it? The cause of all the weirdness is that there is a speed that is the same for all observers. If B sees the light travelling at c (relative to him because he's at rest in his frame), and A sees the light travelling at c relative to him, and speed is distance/time, then something has to change with your definitions of distance and time. B's seconds and A's seconds are not the same. B's metres and A's metres are not the same. 1
D H Posted January 31, 2012 Posted January 31, 2012 Could you point me towards some of the experimental proofs? B.Rossi and D.B.Hall, "Variation of the Rate of Decay of Mesotrons with Momentum", Phys. Rev. 59, 223–228 (1941) Abstract: http://prola.aps.org/abstract/PR/v59/i3/p223_1 The Rossi-Hall experiment and a number of follow-ons looked at various short-lived particles created in the upper atmosphere by collisions of cosmic rays with the atmosphere. The Rossi-Hall experiment focused on muons (which the called mesotrons). Suppose you measure the flux (number of particles per second) of these short-lived particles at the top of a mountain and at sea level. You should observe more of those particles atop a mountain compared to at sea level because that increased elevation at the mountaintop gives those particles less time to decay. Measuring this particle flux was one of the key tests of special relativity. Ignoring relativistic effects, the time spent by those particles as the move from the mountaintop elevation to sea level is simply the altitude difference divided by the velocity. Divide this time by the half life of the particle and you get the expected number of half lives. This number of half lives should in turn dictate the ratio of the flux observed atop the mountain to that observed at sea level. To make this concrete, suppose the mountaintop elevation is 6300 feet (1920 meters), the particles are moving at 99.94 c, and the particles have a half life of 2.197 µs. A non-relativistic calculation would say that it takes the particles 6.409 µs to go from 6300 feet to sea level, or 2.917 half lives. A mountaintop flux of 538 particles per hour should mean a sea level flux of 71 particles per hour. That is not what was observed. The sea level flux was much higher than that predicted ignoring relativistic effects, about 500 particles per hour. So how to explain this from a relativistic perspective? One way is to look at it from the perspective of the Earth-bound observer. Time is dilated for those relativistic muons. The 2.197 µs muon half life becomes a time dilated 63.43 µs, so that 2.197 µs is only 0.101 half lives. Given a mountaintop flux of 538 particles per hour, the sea level flux should be about 502 particles per hour. Another way to look at it is from the perspective of the muon. From this perspective it is the Earth that is moving toward the muon at 0.9994 c. That 6300 foot difference from mountaintop to sea level becomes a paltry 218.2 feet. At 0.9994 c, it takes the Earth only 0.222 µs to cover this distance, or 0.101 half lives. That's the same as the Earth-bound observer explanation. Tests of time dilation are tests of length contraction. The two concepts go hand in hand. 1
Schrödinger's hat Posted January 31, 2012 Posted January 31, 2012 Could you point me towards some of the experimental proofs? I think I'd like to read a little more on the subject Thanks Missed this somehow before my other post. I gather from seeing your other posts around, that you are reasonably mathematically literate. So you may be able to derive the Lorentz transforms from some simple thought experiments. Here's one to get you started (it should yield a time dilation formula). Consider a pair of paralell plates or mirrors with a photon or laser beam (or other speed of light object) bouncing between them. Simulation of this situation here: http://www.refsmmat....tml#light-clock The object/photon/whatever (let's call it a ball to avoid confusion with wave stuff) is bouncing between the plates which are (for convenience) 1 light second apart. The ball takes 1 second per bounce, (and the local clocks will tick once for every time it bounces) Now consider a frame in which the whole contraption is moving up. The ball bouncing off of a plate is an event, the fact that it happens cannot change. But in a different frame, the time and place of an event might change. The ball has moved a different distance (to meet the moving plate). Some assumptions which should help: 1. The ball is moving at the same speed in both frames (that is the magnitude of its velocity is constant). 2. The dimensions of space are independant. Change in velocity upwards will not cause any non-classical or otherwise unexpected change in sideways distances or positions. 3. Constant velocity. Something moving inertially (at constant speed) in one frame will be moving at constant speed in other frames. From this and other similar situations, you can derive special relativity logically with no need for further experimental results (other than the constancy of the speed of light). Once you have a working formula for time dilation, you can derive length contraction as well (a good thought experiment is to consider a car moving on a rail and the observations of the moving and stationary observer -- add beacons, laser beams (a pair of lasers fired from the center of the car simultaneously at either end of the car is useful) etc as you like to see what will happen). Other notes: Keep careful track of signs, probably the easiest thing to muck up. Velocity of [math]\frac{\sqrt{3}}{2}c[/math] leads to convenient numbers. You may see factors of [math] \sqrt{1 - \frac{v^2}{c^2}}[/math] pop up a lot, it helps to give them a name. Traditionally we set [math]\gamma = \frac{1}{ \sqrt{1 - \frac{v^2}{c^2}}}[/math] You don't really need to keep track of three directional dimensions, you can happily set z=0 v_z =0 and leave it out. Taking derivatives and keeping track of some calculus identities can sometimes sidestep a bit of algebra (but you only need algebra to do the calculations). Feel free to post your progress/requests for further help here if you decide to have a go at it (or if you want to try, but are still a bit lost) 1
Tres Juicy Posted January 31, 2012 Author Posted January 31, 2012 Missed this somehow before my other post. I gather from seeing your other posts around, that you are reasonably mathematically literate. So you may be able to derive the Lorentz transforms from some simple thought experiments. Here's one to get you started (it should yield a time dilation formula). Consider a pair of paralell plates or mirrors with a photon or laser beam (or other speed of light object) bouncing between them. Simulation of this situation here: http://www.refsmmat....tml#light-clock The object/photon/whatever (let's call it a ball to avoid confusion with wave stuff) is bouncing between the plates which are (for convenience) 1 light second apart. The ball takes 1 second per bounce, (and the local clocks will tick once for every time it bounces) Now consider a frame in which the whole contraption is moving up. The ball bouncing off of a plate is an event, the fact that it happens cannot change. But in a different frame, the time and place of an event might change. The ball has moved a different distance (to meet the moving plate). Some assumptions which should help: 1. The ball is moving at the same speed in both frames (that is the magnitude of its velocity is constant). 2. The dimensions of space are independant. Change in velocity upwards will not cause any non-classical or otherwise unexpected change in sideways distances or positions. 3. Constant velocity. Something moving inertially (at constant speed) in one frame will be moving at constant speed in other frames. From this and other similar situations, you can derive special relativity logically with no need for further experimental results (other than the constancy of the speed of light). Once you have a working formula for time dilation, you can derive length contraction as well (a good thought experiment is to consider a car moving on a rail and the observations of the moving and stationary observer -- add beacons, laser beams (a pair of lasers fired from the center of the car simultaneously at either end of the car is useful) etc as you like to see what will happen). Other notes: Keep careful track of signs, probably the easiest thing to muck up. Velocity of [math]\frac{\sqrt{3}}{2}c[/math] leads to convenient numbers. You may see factors of [math] \sqrt{1 - \frac{v^2}{c^2}}[/math] pop up a lot, it helps to give them a name. Traditionally we set [math]\gamma = \frac{1}{ \sqrt{1 - \frac{v^2}{c^2}}}[/math] You don't really need to keep track of three directional dimensions, you can happily set z=0 v_z =0 and leave it out. Taking derivatives and keeping track of some calculus identities can sometimes sidestep a bit of algebra (but you only need algebra to do the calculations). Feel free to post your progress/requests for further help here if you decide to have a go at it (or if you want to try, but are still a bit lost) Cool, I will have a look at this properly when I'm not at work Thanks SH!
DrRocket Posted January 31, 2012 Posted January 31, 2012 Cool, I will have a look at this properly when I'm not at work Thanks SH! You might want to look at a couple of books on special relativity. Introduction to Special Relativity by Wolfgang presents the usual physics approach in which the Lorentz transformations are derived from thought experiments and the two postulates of special relativity. The Geometry of Minkowski Spacetime, an Introduction to the Mathematics of the Theory of Special Relativity by Gregory Nabers derives special relativity by considering transformations that preserve the Minkowski metric. Both are useful perspectives, dependiing on the specific issue at hand. 2
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