CraigD Posted July 10, 2016 Report Share Posted July 10, 2016 (edited) So you [A-Wal] are claiming while the Universe continues forward in time that the object travelling at c does not experience time and time stops for the object...Yes, I believe that’s what A-Wal is saying, and I agree, as would nearly anyone with an adequate academic introduction to physics, such as a scientifically inclined US high school graduate or first year college student. This is simply one of the obvious geometric conclusion that one is exposed to (and unlikely to reach without being exposed to it in school or from conversation with someone familiar with physics), like the conclusion that the Earth is a spheroid because distant objects appear to sink beneath the horizon. ....so how do you assume that speed in any way affects time?That the passage of time, as measured by any clock – an ordinary wristwatch, a jar containing a fly that dies in a day or two, etc – differs for objects in motion relative to one another by the relationship [math]t_1 = t_2 \sqrt{1 - \left( \frac{v}{c} \right)^2}[/math] is not an assumption, but a conclusion based on 2 assumptions:The laws of physics are the same no matter what an objects velocity relative to other objectsThe speed of light is the same as measured by any experiment, regardless of the relative velocity of the source and detectors of the light.Once you’ve accepted these assumption, which are conventionally known as the postulates of Special Relativity, and have been experimentally confirmed, the conclusion that time is affected by speed as described above can be reached through simple geometry. The simplest derivation I know is the light clock. It can be found in many books and websites, including the Wikipedia article link here. At about the 6:15 mark, this Youtube video has a nice animation of it. Once you understand the example of the light clock, it should be obvious how time dilation is a consequence of the postulates of SR. If you’re unaware of experiments confirming their truth, there’s nothing stupid or illogical about not accepting the postulates. Until around 1900, this is what most physicists believed: that the speed of light depended on the relative velocity of its source and receiver (a “ballistic” theory of light), or of the receiver relative to some medium (a “wave” theory of light). If one of these were true, clocks could run at the same rate regardless of their relative motion. A great scientific surprise of the late 19th and early 20th century is that light doesn’t behave as predicted by ballistic and wave theories of light, like tiny bullets or longitudinal vibrations in air or transverse ones in a stretched string, but as postulated by the Special Theory of Relativity. It’s essential in science to accept theories that are supported by experiments over ones that are not, even when they are more difficult to understand. IMHO every thoughtful person should understand the light clock illustration of time dilation. I got one. Two observers with their watches in synch (from the perspective of an observer half way between them), one observer at each station, one takes the instantly accelerating train to meet the other observer. The traveler's watch is 6 minutes behind the other observer's watch after the journey. How fast did the train move in the station's reference frame?[math]t_1 = t_2 \sqrt{1 - v^2}[/math] [math]t_2 = \frac{8}{v}[/math] [math]t_2 - 6 = t_2 \sqrt{1 - v^2}[/math] [math]\frac{8}{v} -6 = \frac{8}{v} \sqrt{1 - v^2}[/math] [math]\left( \frac{8}{v} -6 \right) \frac{v}{8} = \sqrt{1 - v^2}[/math] [math]1 -\frac{6 v}{8} = \sqrt{1 - v^2}[/math] [math]1 -\frac{3 v}{2} + \frac{9 v^2}{16} = 1 - v^2[/math] [math]-\frac{3 v}{2} +\frac{9 v^2}{16} +v^2 = 0[/math] [math]-\frac{3 v}{2} +\frac{25 v^2}{16} = 0[/math] [math]\frac{25}{16}v = \frac{3}{2}[/math] [math]v = \frac{48}{50} = 0.96 c[/math] Edited July 12, 2016 by CraigD fixed yet another mistake! Quote Link to comment Share on other sites More sharing options...
A-wal Posted July 10, 2016 Report Share Posted July 10, 2016 [math]t_1 = t_2 \sqrt{1 - \left( \frac{v}{c} \right)^2}[/math][math](8-6) = 8 \sqrt{1 - \left( \frac{v}{c} \right)^2}[/math][math]\left ( \frac{1}{4} \right )^2 = 1 - \left ( \frac{v}{c} \right )^2[/math][math]\sqrt{\frac{15}{16}} = \frac{v}{c} [/math][math]v = \frac{\sqrt{15}}{4} \dot= 0.9682 c[/math]That's not the answer that I was expecting. :) Would you mind taking me through that calculation step by step? It doesn't matter if it's too much of a pain. How much time would the traveler's watch be behind the other observer's watch if the traveler moves at .8c? Quote Link to comment Share on other sites More sharing options...
xyz Posted July 11, 2016 Author Report Share Posted July 11, 2016 (edited) Yes, I believe that’s what A-Wal is saying, and I agree, as would nearly anyone with an adequate academic introduction to physics, such as a scientifically inclined US high school graduate or first year college student. This is simply one of the obvious geometric conclusion that one is exposed to (and unlikely to reach without being exposed to it in school or from conversation with someone familiar with physics), like the conclusion that the Earth is a spheroid because distant objects appear to sink beneath the horizon. That the passage of time, as measured by any clock – an ordinary wristwatch, a jar containing a fly that dies in a day or two, etc – differs for objects in motion relative to one another by the relationship[math]t_1 = t_2 \sqrt{1 - \left( \frac{v}{c} \right)^2}[/math]is not an assumption, but a conclusion based on 2 assumptions:The laws of physics are the same no matter what an objects velocity relative to other objectsThe speed of light is the same as measured by any experiment, regardless of the relative velocity of the source and detectors of the light.Once you’ve accepted these assumption, which are conventionally known as the postulates of Special Relativity, and have been experimentally confirmed, the conclusion that time is affected by speed as described above can be reached through simple geometry. The simplest derivation I know is the light clock. It can be found in many books and websites, including the Wikipedia article link here. At about the 6:15 mark, this Youtube video has a nice animation of it. Once you understand the example of the light clock, it should be obvious how time dilation is a consequence of the postulates of SR. If you’re unaware of experiments confirming their truth, there’s nothing stupid or illogical about not accepting the postulates. Until around 1900, this is what most physicists believed: that the speed of light depended on the relative velocity of its source and receiver (a “ballistic” theory of light), or of the receiver relative to some medium (a “wave” theory of light). If one of these were true, clocks could run at the same rate regardless of their relative motion. A great scientific surprise of the late 19th and early 20th century is that light doesn’t behave as predicted by ballistic and wave theories of light, like tiny bullets or longitudinal vibrations in air or transverse ones in a stretched string, but as postulated by the Special Theory of Relativity. It’s essential in science to accept theories that are supported by experiments over ones that are not, even when they are more difficult to understand. IMHO every thoughtful person should understand the light clock illustration of time dilation. [math]t_1 = t_2 \sqrt{1 - \left( \frac{v}{c} \right)^2}[/math][math](8-6) = 8 \sqrt{1 - \left( \frac{v}{c} \right)^2}[/math][math]\left ( \frac{1}{4} \right )^2 = 1 - \left ( \frac{v}{c} \right )^2[/math][math]\sqrt{\frac{15}{16}} = \frac{v}{c} [/math][math]v = \frac{\sqrt{15}}{4} \dot= 0.9682 c[/math] I do not understand your maths sorry but I do understand your video link which I have watched lots of times before your link, but thank you anyway. Supposing I could quite clearly show the demonstration of the light clocks to be wrong? Would you be remotely interested? probably not. Pause the video at approx 8.39mins. Firstly we would not observe the beams without a medium such as smoke. Secondly the mirrors are observer effect a normal surface does not reflect beams when considering a laser thirdly you are viewing the entire experiment through the ''gin-clear'' , Forthly any light you are observing is travelling perpendicular to you from your screen or from the beam or from the clocks if you was observing this in real life. Edited July 11, 2016 by xyz Quote Link to comment Share on other sites More sharing options...
A-wal Posted July 11, 2016 Report Share Posted July 11, 2016 Supposing I could quite clearly show the demonstration of the light clocks to be wrong? Would you be remotely interested? probably not.We would be very interested if you could do that but you can't. You only think you can because you have no understanding of what you're trying to disprove but instead of realising that it's your understanding that's at fault, you think it's the model. Firstly we would not observe the beams without a medium such as smoke.Wrong! Secondly the mirrors are observer effect a normal surface does not reflect beams when considering a laserWrong! thirdly you are viewing the entire experiment through the ''gin-clear'' ,Meaningless! Forthly any light you are observing is travelling perpendicular to you from your screen or from the beam or from the clocks if you was observing this in real life. What? No it isn't. Wrong! Quote Link to comment Share on other sites More sharing options...
CraigD Posted July 11, 2016 Report Share Posted July 11, 2016 (edited) Would you mind taking me through that calculation step by step? It doesn't matter if it's too much of a pain.I live to take people through calculations step by step! :) [math]t_1 = t_2 \sqrt{1 - v^2}[/math] <- this is the usual formula for time dilation, [math]t_2[/math] the duration measured by the stationary clock, [math]t_2[/math] the duration measured by the moving clock, speed [math]v[/math] in units of c to simplify it. [math]t_2 = \frac{8}{v}[/math] <- calculate t_2 is the given distance of 8 light minutes divided by speed [math]v[/math] [math]t_2 - 6 = t_2 \sqrt{1 - v^2}[/math] <-replace [math]t_1[/math] with [math]t_2 -6[/math] from the given “The traveler's watch is 6 minutes behind the other observer's watch”. [math]\frac{8}{v} -6 = \frac{8}{v} \sqrt{1 - v^2}[/math] <- replace [math]t_2[/math] with[math] \frac{8}{v}[/math] [math]\left( \frac{8}{v} -6 \right) \frac{v}{8} = \sqrt{1 - v^2}[/math] <-algebra [math]1 -\frac{6 v}{8} = \sqrt{1 - v^2}[/math] <-algebra [math]1 -\frac{3 v}{2} + \frac{9 v^2}{16} = 1 - v^2[/math] <-algebra [math]-\frac{3 v}{2} +\frac{9 v^2}{16} +v^2 = 0[/math] <-algebra, rearrange to get 0 on one side of equation. [math]\frac{25}{16}v - \frac{3}{2}[/math] = <-algebra, divide both sides by v [math]\frac{25}{16}v = \frac{3}{2}[/math] <- rearrange and solve [math]v = \frac{48}{50} = 0.96 c[/math] How much time would the traveler's watch be behind the other observer's watch if the traveler moves at .8c?[math]t_1 = t_2 \sqrt{1 - \left( \frac{v}{c} \right)^2}[/math] [math]t_1 = 10 \sqrt{1 - \left( 0.8 \right)^2}[/math] [math]t_1 = 10 \cdot 0.6 = 6[/math] [math]t_2 - t_1 = 10 - 6 = 4 \, \mbox{minutes} [/math] Edited July 12, 2016 by CraigD redid for fixed math from previous post Quote Link to comment Share on other sites More sharing options...
A-wal Posted July 11, 2016 Report Share Posted July 11, 2016 Hmm. I thought the velocity would have been .8c if the traveler's watch ended up six minutes behind.Here's my thinking, please don't laugh. An object moving at .8c is moving through time at half speed, therefore length contraction also halves the distance. If the object has to go half as far as does it twice as fast on their watch then the journey takes a quarter of the time. Obviously it's relative so the observer at the station is length contracted and time dilated by the same amount from the traveler's frame of reference while they're making the journey but I look at it as acceleration making the station's frame of reference the one that applies once the traveler has accelerated back into that frame. Please don't tell me that thinking of it like that doesn't work.I appreciate the equation description but to really understand it I'd need to know why it's done exactly like that. I understand that it's to keep the speed of light consistent in every inertial frame and I can visualise the Lorenz transformations, roughly but can't quantify it well enough to follow why each step of the equation is what it is. I'm going to keep going through it and hopefully something will click. Quote Link to comment Share on other sites More sharing options...
CraigD Posted July 12, 2016 Report Share Posted July 12, 2016 (edited) Hmm. I thought the velocity would have been .8c if the traveler's watch ended up six minutes behind. ... An object moving at .8c is moving through time at half speed, therefore length contraction also halves the distance.You’ve made an arithmetic mistake. [math]\sqrt{1 – 0.8^2} = 0.6[/math], not 0.5. 0.8 c is a nice value to use for examples, because its Lorentz factor is a nice round 0.6. [math]\sqrt{1 – \left( \frac{\sqrt{3}}{2}\right)^2} = 0.5[/math]. [math]\frac{\sqrt{3}}{2} \dot= \, 0.866025403784438647[/math], is not a nice round number. If the object has to go half as far as does it twice as fast on their watch then the journey takes a quarter of the time.I think you’re getting confused here. As you know, the length contraction factor for a give speed is the same as the time dilation factor. Intuitively, this makes sense if you consider the experience of the traveler moving at 0.8 c. Their clock says they made the 8 light-minute long trip in [math]\frac{8 \,\mbox{light-minutes}}{0.8 \, c} \cdot 0.6 = 6 \,\mbox{minutes}[/math]. So the distance they calculate by multiplying their speed times the duration of the trip is [math]0.8 \, c \cdot 6 \,\mbox{minutes} = 4.8[/math] light-minutes, which is the same distance they would measure if they used a huge ruler (or a lot of small rulers put end-to-end along the track very quickly), which agrees with the length-contracted distance given by the transformation. Edited July 12, 2016 by CraigD fixed mistake Quote Link to comment Share on other sites More sharing options...
A-wal Posted July 12, 2016 Report Share Posted July 12, 2016 You’ve made an arithmetic mistake. [math]\sqrt{1 – 0.8^2} = 0.6[/math], not 0.5. 0.8 c is a nice value to use for examples, because its Lorentz factor is a nice round 0.6. [math]\sqrt{1 – \left( \frac{\sqrt{3}}{2}\right)^2} = 0.5[/math]. [math]\frac{\sqrt{3}}{2} \dot= \, 0.866025403784438647[/math], is not a nice round number.Oh it's 0.6, I thought it was 0.5. I didn't calculate it, I remembered it being a nice round number and thought it was .5, that would be too perfect. I think you’re getting confused here. As you know, the length contraction factor for a give speed is the same as the time dilation factor. Intuitively, this makes sense if you consider the experience of the traveler moving at 0.8 c. Their clock says they made the 8 light-minute long trip in [math]8 \cdot 0.6 = 4.8[/math] minutes. So the distance they calculate by multiplying their speed times the duration of the trip is [math]8 \cdot 0.6 = 4.8[/math] light-minutes, which is the same distance they would measure if they used a huge ruler (or a lot of small rulers put end-to-end along the track very quickly), which agrees with the length-contracted distance given by the transformation.But wouldn't that make that length contraction and time dilation two different ways of looking at the exact same thing rather than two seperate and equal effects? It would mean instead of this......you'd get this. Once the traveler gets to the station they can work out that the distance was length contracted in this frame because the time on their watch is behind the other observer's watch so if they made the trip in 4.8 minutes and they say the distance was 4.8 light minutes when it was length contracted then that's using the station's frame of reference to take length contraction into account but not time dilation. Their watch being behind the station observer's watch is a result of traveling through length contracted space in a time dilated duration of time, so 4.8 light minutes is using the time measurement of the station's frame and the distance measurement of the traveler's frame while they were making the journey. If length contraction made the distance 4.8 light minutes then at .8 c they should have completed the journey in 2.88 minutes on their own watch because in the station's frame they were traveling slower through time by the same amount as the distance was shortened in space so covering more space and doing it in less time. To make the the journey 8 light minute journey in 4.7 minutes at .8c the time dilation and length contraction factor should be just under under 0.775. Quote Link to comment Share on other sites More sharing options...
xyz Posted July 12, 2016 Author Report Share Posted July 12, 2016 Oh it's 0.6, I thought it was 0.5. I didn't calculate it, I remembered it being a nice round number and thought it was .5, that would be too perfect. But wouldn't that make that length contraction and time dilation two different ways of looking at the exact same thing rather than two seperate and equal effects? It would mean instead of this...r_image-1.JPG...you'd get this.SR.JPG Once the traveler gets to the station they can work out that the distance was length contracted in this frame because the time on their watch is behind the other observer's watch so if they made the trip in 4.8 minutes and they say the distance was 4.8 light minutes when it was length contracted then that's using the station's frame of reference to take length contraction into account but not time dilation. Their watch being behind the station observer's watch is a result of traveling through length contracted space in a time dilated duration of time, so 4.8 light minutes is using the time measurement of the station's frame and the distance measurement of the traveler's frame while they were making the journey. If length contraction made the distance 4.8 light minutes then at .8 c they should have completed the journey in 2.88 minutes on their own watch because in the station's frame they were traveling slower through time by the same amount as the distance was shortened in space so covering more space and doing it in less time. To make the the journey 8 light minute journey in 4.7 minutes at .8c the time dilation and length contraction factor should be just under under 0.775.You have said I am wrong etc in the other post , you are not being objective or have considered any of what I said. Ok, for once in all my forum time let me try to get a bit serious . You are not considering the third observer at all in the scenario, Lets us have 3 observers, Observer (A) is at Shady grove Observer ( :cool: is at Tartaurus Observer © is on the Neo space station observing from a distance Shady grove and Tartaurus Observer (A) observes the rest length of distance between Tartaurus and Shady grove is X Observer ( :cool: observes the rest length of distance between Tartaurus and Shady grove is X Observer © observes the rest length of distance between Tartaurus and Shady grove is X All 3 observers agree on the rest length of X Do you Firstly agree with this? (Discounting velocities for now , an equilateral triangle). Quote Link to comment Share on other sites More sharing options...
A-wal Posted July 12, 2016 Report Share Posted July 12, 2016 You have said I am wrong etc in the other post , you are not being objective or have considered any of what I said. You are wrong in the other post, I am being objective, I have considered what you've said and you're just plain wrong. You are not considering the third observer at all in the scenario,There is no third observer in this scenario. If you want to introduce one that's a new, or at least a modified scenario. Lets us have 3 observers, Observer (A) is at Shady grove Observer ( :cool: is at Tartaurus Observer © is on the Neo space station observing from a distance Shady grove and Tartaurus Observer (A) observes the rest length of distance between Tartaurus and Shady grove is X Observer ( :cool: observes the rest length of distance between Tartaurus and Shady grove is X Observer © observes the rest length of distance between Tartaurus and Shady grove is X All 3 observers agree on the rest length of X Do you Firstly agree with this? (Discounting velocities for now , an equilateral triangle). If all three observers are at rest relative to each other then they're all in the same reference frame so of course they'll all measure the distance between the same two points in time or in space to be the same as the other two observers measurements. Quote Link to comment Share on other sites More sharing options...
xyz Posted July 12, 2016 Author Report Share Posted July 12, 2016 (edited) You are wrong in the other post, I am being objective, I have considered what you've said and you're just plain wrong. There is no third observer in this scenario. If you want to introduce one that's a new, or at least a modified scenario. If all three observers are at rest relative to each other then they're all in the same reference frame so of course they'll all measure the distance between the same two points in time or in space to be the same as the other two observers measurements.I mentioned the third observer earlier in the thread. So you agreed with me thus far, So if (A) travels to ( :cool: at the speed of light , what contraction does © observe of space or time? Both ( :cool: and © agree that L=Lx and remains constant with no contraction of space or time. See here for diagram- http://www.thenakedscientists.com/forum/index.php?topic=67594.0 Edited July 12, 2016 by xyz Quote Link to comment Share on other sites More sharing options...
A-wal Posted July 12, 2016 Report Share Posted July 12, 2016 So if (A) travels to ( :cool: at the speed of light , what contraction does © observe of space or time?That's really hard to work out because it uses coordinate acceleration. I'm having enough trouble with the standard frame conversion equations without any acceleration (instant acceleration). Put the third observer between the sun and earth (between the two stations) so that they can make instant measurements as the train/ship passes them. Then you can have the other two observers traveling at the speed of light in opposite directions so that they're traveling towards and then away from each other at twice the speed of light in the third observer's frame but at half that velocity in their own frames. Quote Link to comment Share on other sites More sharing options...
xyz Posted July 12, 2016 Author Report Share Posted July 12, 2016 That's really hard to work out because it uses coordinate acceleration. I'm having enough trouble with the standard frame conversion equations without any acceleration (instant acceleration). Put the third observer between the sun and earth (between the two stations) so that they can make instant measurements as the train/ship passes them. Then you can have the other two observers traveling at the speed of light in opposite directions so that they're traveling towards and then away from each other at twice the speed of light in the third observer's frame but at half that velocity in their own frames.If we put the third observer between A and B at the half way point , we still have a triangle but you may now consider this a linearity. The third observer measures the distances between A and B and themselves, they measure the distance/length to be equal in either direction and state that L=X Observer A and B confirm the measurement and both concur L=X Now while anything travels between any of the points at any speed, none of the points move, in our imaginary ''stationary'' reference frame, L=X remains constant to all observers. So can we please ignore the journey times for now and can you tell me what you think actually contracts? because in this scenario everything is synchronous and constant, X is an invariant in time and space/length. Quote Link to comment Share on other sites More sharing options...
A-wal Posted July 12, 2016 Report Share Posted July 12, 2016 If we put the third observer between A and B at the half way point , we still have a triangle but you may now consider this a linearity. The third observer measures the distances between A and B and themselves, they measure the distance/length to be equal in either direction and state that L=XL=X Observer A and B confirm the measurement and both concur L=XYes, while all three are at rest relative to each other they all measure the same distance between the two stations. This is much easier to deal with because A moves at the same speed relative to C as A moves relative to B but you can then move B as well so they're all in different frames but it's easier to handle because they're in a straight line. Now while anything travels between any of the points at any speed, none of the points move, in our imaginary ''stationary'' reference frame, L=X remains constant to all observers.NO! You've been told time and again that this simply isn't true. It's very easily demonstrated, has a tone of evidence supporting it and couldn't possibly work any other way given that the speed of light is the same in all inertial frames. So can we please ignore the journey times for now and can you tell me what you think actually contracts? because in this scenario everything is synchronous and constant, X is an invariant in time and space/length.Of course it isn't! In the example where one observer travels from one station to the other, the third observer is at rest relative to the observer that stays at the platform so the time dilation and length contraction experienced by the traveler is the same because we're only dealing with two frames. At the speed of light they become infinite, the traveler measures no space at all between the two points and the trip also takes no time at all, they move instantaneously between the two stations from their own perspective but at c from the perspectives of the other two. Length contraction is also infinite so the traveler will appear strength out in space all the way between the two stations. This appears to be a contradiction of all frames being equal because the other observers (along with the stations) can be thought of as the ones moving at the speed of light so the trip should be instant from their perspective as well. Its not because they're not the ones who accelerate. This is why I don't like using the speed of light, it requires infinite acceleration. In B and C's frame (same frame for both if they're at rest relative to each other) the distance between the two stations doesn't change and A can only travel that distance at the speed of light but from A's perspective it accelerates into a frame where the distance between the stations shortens (length contraction) and it takes less time to cover the distance (time dilation). In the triangle example it's more complicated because A moves at the speed of light relative to B but slower than that relative to C and A's clock needs to end up behind C's clock by the same amount as it's behind B's clock because B's clock at A's clock are in synch. It works out because if you look at A's velocity relative to C it increases then decreases then increases again. This is coordinate acceleration and it should all add up to the same time lag behind C's clock. The interesting thing is when you take away observer B and the other station and regard A and C as the only objects in the universe. Now the coordinate acceleration becomes proper acceleration (acceleration that you feel), meaning that proper acceleration is defined by the a change in velocity relative to other objects. I think that's what Mach's principle is, acceleration is just as relative as velocity. I went a bit more in depth than I planned, you can just ignore the last paragraph, we're not using a triangle any more. Quote Link to comment Share on other sites More sharing options...
xyz Posted July 12, 2016 Author Report Share Posted July 12, 2016 (edited) Yes, while all three are at rest relative to each other they all measure the same distance between the two stations. This is much easier to deal with because A moves at the same speed relative to C as A moves relative to B but you can then move B as well so they're all in different frames but it's easier to handle because they're in a straight line. NO! You've been told time and again that this simply isn't true. It's very easily demonstrated, has a tone of evidence supporting it and couldn't possibly work any other way given that the speed of light is the same in all inertial frames. Of course it isn't! In the example where one observer travels from one station to the other, the third observer is at rest relative to the observer that stays at the platform so the time dilation and length contraction experienced by the traveler is the same because we're only dealing with two frames. At the speed of light they become infinite, the traveler measures no space at all between the two points and the trip also takes no time at all, they move instantaneously between the two stations from their own perspective but at c from the perspectives of the other two. Length contraction is also infinite so the traveler will appear strength out in space all the way between the two stations. This appears to be a contradiction of all frames being equal because the other observers (along with the stations) can be thought of as the ones moving at the speed of light so the trip should be instant from their perspective as well. Its not because they're not the ones who accelerate. This is why I don't like using the speed of light, it requires infinite acceleration. In B and C's frame (same frame for both if they're at rest relative to each other) the distance between the two stations doesn't change and A can only travel that distance at the speed of light but from A's perspective it accelerates into a frame where the distance between the stations shortens (length contraction) and it takes less time to cover the distance (time dilation). In the triangle example it's more complicated because A moves at the speed of light relative to B but slower than that relative to C and A's clock needs to end up behind C's clock by the same amount as it's behind B's clock because B's clock at A's clock are in synch. It works out because if you look at A's velocity relative to C it increases then decreases then increases again. This is coordinate acceleration and it should all add up to the same time lag behind C's clock. The interesting thing is when you take away observer B and the other station and regard A and C as the only objects in the universe. Now the coordinate acceleration becomes proper acceleration (acceleration that you feel), meaning that proper acceleration is defined by the a change in velocity relative to other objects. I think that's what Mach's principle is, acceleration is just as relative as velocity. I went a bit more in depth than I planned, you can just ignore the last paragraph, we're not using a triangle any more.I am not sure you have the ability to think open minded enough to engage in discussion with me, you agreed with me earlier ''If all three observers are at rest relative to each other then they're all in the same reference frame so of course they'll all measure the distance between the same two points in time or in space to be the same as the other two observers measurements''' then contradictory xyz, on 12 Jul 2016 - 1:23 PM, said:Now while anything travels between any of the points at any speed, none of the points move, in our imaginary ''stationary'' reference frame, L=X remains constant to all observers.''NO! You've been told time and again that this simply isn't true. It's very easily demonstrated, has a tone of evidence supporting it and couldn't possibly work any other way given that the speed of light is the same in all inertial frames.'' I think you are confused. a,b,c, are stationary and do not move anywhere. Edited July 12, 2016 by xyz Quote Link to comment Share on other sites More sharing options...
A-wal Posted July 12, 2016 Report Share Posted July 12, 2016 (edited) God you're annoying. It's one thing if you don't have the capacity to grasp it but you're not even listening. I don't need a more open mind, you need a better one. And I didn't contradict myself. If they're AT REST relative to each other then all three measure the same distance the distance between the same two points in space or time to be the same. As soon as one of them accelerates they'll measure the distance differently through time dilation and length contraction so that the speed of light remains constant (as it's been proven to be many times) in all inertial frames of reference. This is proven fact. You can deny it all you want but it's just irrational, and you claim I need a more open mind. Edited July 12, 2016 by A-wal Quote Link to comment Share on other sites More sharing options...
xyz Posted July 13, 2016 Author Report Share Posted July 13, 2016 God you're annoying. It's one thing if you don't have the capacity to grasp it but you're not even listening. I don't need a more open mind, you need a better one. And I didn't contradict myself. If they're AT REST relative to each other then all three measure the same distance the distance between the same two points in space or time to be the same. As soon as one of them accelerates they'll measure the distance differently through time dilation and length contraction so that the speed of light remains constant (as it's been proven to be many times) in all inertial frames of reference. This is proven fact. You can deny it all you want but it's just irrational, and you claim I need a more open mind. I think you must be reading my post ambiguously because you are discussing something seemingly different. So to remove the ambiguity let me change the context. The great Pyramid, we have an exact copy of this ''floating'' around in space' with a difference of that it is solid throughout the structure. On one of the points sits an ant, the ant starts to walk along one of the axis's to another point, The length of the pyramids axis remains a constant L=Lx Are you ok with this ? do you understand thus far. Quote Link to comment Share on other sites More sharing options...
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