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According to the equivalence principle the laws of physics remain the same in each reference frame. The reference frame that is travelling at 0.867c has the same laws of physics as the stationary frame. The stationary frame measures the moving frame as length contracted, but the moving frame measures itself as being the same length as if it were stationary. What is causing the differing measurements?
It's not that I wasn't taught the same thing you were its just that subjective thinking of physical contraction seems contrived.
Do you agree that a radio signal is light?
Do you agree the radio signal can pass through the trains walls?
Do you agree that if we used a radio signal instead of ''light'' that there is no scenario to discuss?
The thing is you are still not doing it correctly.
Lets restart this and try to go at a slow pace taking one issue at a time into consideration. You say ''Length of vehicle = d''Ok, are you happy at defining (d) to be 299 792 458 m in length at relative rest?
Do you agree that a round trip for light travelling then a return trip would take 2 seconds? [attachment id=0 msg=514855]
(This time we add the speeds together instead of subtracting, so it's a "closing speed" of 1.5c to cover distance d.)
Lets get back to the 0.866025 speed of light for a clock. Let's make the mirrors in the clock both perpendicular to and with the direction of motion. Obviously the same distance apart. The clock going 0.866025 we use the Gods eye again to start the photons from each mirror. We can divide 1 by 0.133 to get 7.5 of your centimeters for the light to reach the front mirror.
We can also do the Pythagoras by squaring 0.866025 for the same distance of 7.5 as the ratio to 1 as the speed of light in the forward direction with vector velocity.
The return light rounded off is about 0.57 relative to 1. 8.07 / 2 for the two way measurement of light. 4.035 cycle distance for the photon. Basically a 1/4 ratio 0.25 to 1. If we take the sq. rt. of 0.25 we get 0.5 vs. relative rest. This is what the Lorentz contraction represents.
Now when the forward direction of light hits the forward mirror the perpendicular light has not reached its mirror yet.
The photon has to follow the hypotenuse and has not reached the opposing mirror by the 7.5 forward ratio.
Explain why we need to contract the object physically? The visual contraction of the hypotenuse angle fits the contracted view. If it were also physically contracted it would not fit what is observed.It's not that I wasn't taught the same thing you were its just that subjective thinking of physical contraction seems contrived.
Quote from: GoC on 22/05/2017 18:12:48Lets get back to the 0.866025 speed of light for a clock. Let's make the mirrors in the clock both perpendicular to and with the direction of motion. Obviously the same distance apart. The clock going 0.866025 we use the Gods eye again to start the photons from each mirror. We can divide 1 by 0.133 to get 7.5 of your centimeters for the light to reach the front mirror.I assume you're still using 10cm as the clock length, so if light moves 10 cm from the rear, the front will move 8.67cm in the same length of time, meaning that it is still 8.67cm ahead of the light. That will happen in an amount of time that we can call "t" (and be aware that this t is not the same size as the t used in my discussions with TheBox - with him the t is a second, but here the t is a third of a nanosecond). Anyway, you will have to repeat this step quite a few times before the light can actually catch up with the front mirror. The closing speed of the light and mirror is 10cm/t minus 8.67cm/t, so that's 1.33cm/t. The gap to be closed is 10cm, so we divide that 10 by the 1.33 and get the time it takes for light to catch the mirror, and that will be 7.4641t. You have mistaken this time as the distance light has to go to get from the back of the carriage to the front, and you should have realised that it can't possibly cross the gap in such a short distance when that distance isn't even as big as the carriage length. To get the actual distance the light has to go before it catches the mirror, you need to multiply the distance that light goes in t (i.e. 10cm) by 7.4641t, so that's going to be a whopping great 74.641cm.QuoteWe can also do the Pythagoras by squaring 0.866025 for the same distance of 7.5 as the ratio to 1 as the speed of light in the forward direction with vector velocity.The same distance as what? The time that you mistook for distance (and whose value is not quite 7.5)? You've made a massive error which you're now trying to build upon.QuoteThe return light rounded off is about 0.57 relative to 1. 8.07 / 2 for the two way measurement of light. 4.035 cycle distance for the photon. Basically a 1/4 ratio 0.25 to 1. If we take the sq. rt. of 0.25 we get 0.5 vs. relative rest. This is what the Lorentz contraction represents.For the return trip, the carriage moves 53.5898 and the light moves 4.641cm before they meet. I can't make sense of what you're trying to do there with any of what you've done there.QuoteNow when the forward direction of light hits the forward mirror the perpendicular light has not reached its mirror yet.Not possible - light takes 2t for the round trip on both clocks with the carriage at rest. With the carriage moving at 0.867c, it takes 4t on the perpendicular clock, and 2t for each half of that, so it reaches the far perpendicular mirror in 2t and reaches the front mirror of the other clock in 7.4641t.QuoteThe photon has to follow the hypotenuse and has not reached the opposing mirror by the 7.5 forward ratio.It reaches the mirror long before the 74.641cm point which is the distance you should be using.QuoteExplain why we need to contract the object physically? The visual contraction of the hypotenuse angle fits the contracted view. If it were also physically contracted it would not fit what is observed.It's not that I wasn't taught the same thing you were its just that subjective thinking of physical contraction seems contrived.My interactive diagrams at the top of http://www.magicschoolbook.com/science/relativity.html should already have shown you how the maths relates into reality, so why am I having to point you to them again? Do you not trust your own eyes? The MMX apparatus is like a pair of light clocks perpendicular to each other. The way I've arranged things in those diagrams sends the light from the front mirror to the rear one first, so it's the second part of the light's journey that takes a long time. Study it carefully. How fast is the apparatus moving across the screen and how fast are the red dots moving across the screen? Have I cheated in some way with the diagrams? No - you can see the speeds, lengths, distances and angles by eye and tell that they are correct. On the first interactive diagram you can see what happens without length-contraction. On the second interactive diagram you can see how length-correction produces the null result.
Quote from: David Cooper on 22/05/2017 19:18:43(This time we add the speeds together instead of subtracting, so it's a "closing speed" of 1.5c to cover distance d.)Before I answer your left out results, can you please explain where the hell you are getting 1.5c from? Not once in the entire scenario does anything travel at 1.5c. Why are you adding the speeds together? maybe its just me and I need to think about that one.
The light travels , b contracts the distance while c also contracts the distance. The light hits b after 2/3rds of second because there is less distance travelled by c.
The light then returns while (a) is moving forward at 0.5c expanding the distance.
C then takes more time to catch up. I do not even ''see'' a contraction. Where is your contraction?
Quote from: Thebox on 22/05/2017 20:08:09Quote from: David Cooper on 22/05/2017 19:18:43(This time we add the speeds together instead of subtracting, so it's a "closing speed" of 1.5c to cover distance d.)Before I answer your left out results, can you please explain where the hell you are getting 1.5c from? Not once in the entire scenario does anything travel at 1.5c. Why are you adding the speeds together? maybe its just me and I need to think about that one.It's a mathematical method which I explained that in the final paragraph, so I'll repeat it here:-"If you have difficulty understanding step seven, we work out the closing speed by adding the speed of light to the speed of the train because that tells us how long either of them would take to hit the other if the other was stationary. This is a short cut which is useful because it avoids doing multiple additions as we move the light a bit towards the rear of the train and the back of the train half as far towards the train - if you try to work out the answer that way it becomes an ordeal of trial and error as you don't know how far to move either of them to get them to the point where they will collide. Feel free to use trial and error though with multiple moves of light and train until you can see how far they have to go before them meet."I invited you to check the validity of the mathematical method if you wish to by using trial and error instead - that may involve doing the calculation through many steps with guessed distances until you happen upon the right numbers for the light to hit the rear of the train.QuoteThe light travels , b contracts the distance while c also contracts the distance. The light hits b after 2/3rds of second because there is less distance travelled by c.What do you mean by "b contracts the distance" and "c contracts the distance"? When light is moving from a to b, b is moving away from it while the light chases it down. It takes 2 seconds for the light to catch b, as described in step (4). The separation distance is closed over time by the relative movement of the light and the front end of the train, but it's not a good idea to use the word "contracted" in this context.QuoteThe light then returns while (a) is moving forward at 0.5c expanding the distance.It's also not a good idea to use the word "expanding" there. This time we have a separation distance being closed over time by the relative movement of the light and rear of the train, and it takes 2/3 of a second for them to meet.QuoteC then takes more time to catch up. I do not even ''see'' a contraction. Where is your contraction?There is no contraction involved yet. We only apply length-contraction later on when we try to make this light clock tick at the same rate as a perpendicular light clock which will tick faster than the uncontracted lengthways light clock.
event 1:The first distance contracts(rear), relative to light at 449688702m/s?0.5c+cevent 2: rear to front return, the distance contracts at 149896229m/s?added : much easier using cars cars.jpg (55.67 kB . 1445x505 - viewed 5118 times)
P.s : All you have explained thus far just explains that light takes more time to travel a further distance which I think we all know. I still await where you observe this physical contraction?
Quote from: Thebox on 22/05/2017 20:25:18event 1:The first distance contracts(rear), relative to light at 449688702m/s?0.5c+cevent 2: rear to front return, the distance contracts at 149896229m/s?added : much easier using cars cars.jpg (55.67 kB . 1445x505 - viewed 5118 times)I have no idea what your diagram's meant to be showing or how it is relevant to anything here. It appears to display contractions perpendicular to the direction of travel and has more to do with perspective and art.QuoteP.s : All you have explained thus far just explains that light takes more time to travel a further distance which I think we all know. I still await where you observe this physical contraction?You didn't appear to understand before that a light clock ticks slower if it's moving along because of the extra distance light has to travel through space for each tick. If you have now got your head around that, you should be able to produce numbers for this that fit with mine, and then we can move on to looking at the perpendicular light clock to see how much its ticking rate is slowed. After that, we can compare the two clocks to see whether they tick at the same rate as each other, and we'll find that they don't - the perpendicular light clock ticks more often than the light clock aligned lengthways along the train. The MMX shows us though that in the real universe the two clocks do tick at the same rate as each other, and that's why we conclude that there must be length-contraction in the real universe. At the moment though, you're a long way from being able to understand that, and it's not certain that your mind is capable of getting on top of it, even if it is the best mind on the planet.
Ok I do 'see'' your contraction now you mention the light clock, however you are ''playing'' with a ''parlour trick illusion'', twice the distance is twice the time, the clock is not broken in your scenario.
If you were to define geometrical points of position and took your measure from that, it removes all your error in the scenario. Changing origin points will change the length of the light second obviously because you are increasing or decreasing the distance the light has to travel. You are removing the constant time length by variance in the origin points creating this ''parlour trick''.
If the diagram is easier to understand for you this way, consider the 0.5c ''block'' is the front of the train and the c ''block'' is the chasing light.
There's no issue of a clock being broken. The longer the light takes to complete the round trip, the slower that clock will tick.
(4) Time for light to make first part of trip when vehicle moving at 0.5c = 2t [= 2s](Front of vehicle was ahead of light by d and moving at 0.5c while light is moving at c, so light is gaining on front of vehicle at 0.5c and will take 2t to catch it.)
Quote from: David Cooper on 22/05/2017 21:27:45There's no issue of a clock being broken. The longer the light takes to complete the round trip, the slower that clock will tick.Which I have been trying to tell you , means absolutely nothing. The clock is not ticking slower at all, the distance is being increased the light has to travel, you interpreting this as being a tick of a clock is totally unnecessarily. The subjective interpretation you are using is the parlour trick and you do not even realise why. It means nothing, it is babble that shows nothing except light has to travel less or more distance relative to the motion of the carriage. So why are you making it to be more than it actually is? I don't understand because it is simple laws of physics that needs no more interpretation than what I mentioned.
Quote from: David Cooper on 22/05/2017 21:46:09(4) Time for light to make first part of trip when vehicle moving at 0.5c = 2t [= 2s](Front of vehicle was ahead of light by d and moving at 0.5c while light is moving at c, so light is gaining on front of vehicle at 0.5c and will take 2t to catch it.)Is your starting point from the rear travelling to the front or from the front travelling to the rear ?I am starting from the front of the train, as the train moves the light is released in the rear direction. I have the shortest time firstly then the longer time secondly.
Quote from: Thebox on 22/05/2017 23:45:34Quote from: David Cooper on 22/05/2017 21:27:45There's no issue of a clock being broken. The longer the light takes to complete the round trip, the slower that clock will tick.Which I have been trying to tell you , means absolutely nothing. The clock is not ticking slower at all, the distance is being increased the light has to travel, you interpreting this as being a tick of a clock is totally unnecessarily. The subjective interpretation you are using is the parlour trick and you do not even realise why. It means nothing, it is babble that shows nothing except light has to travel less or more distance relative to the motion of the carriage. So why are you making it to be more than it actually is? I don't understand because it is simple laws of physics that needs no more interpretation than what I mentioned.If this is all a wind up, it's an extraordinary performance and you are indeed a comedy genius. However, I have to consider the possibility that you aren't that brilliant, in which case you really do think what you just said.A light clock (as used in thought experiments rather than the real world) works by sending out a pulse of light which travels along to a mirror and bounces back to a detector where the light was originally sent out from, at which point another pulse of light is sent out. A tick of this kind of clock is completed when the light pulse returns and a new one is sent out. If the light has to travel further through space due to the movement of the light clock, the light clock ticks less often. The light is a key part of the clock's mechanism and the time it takes to complete the round trip dictates the tick rate. [I don't know how else you imagine a light clock could be made to tick though - there's no possible way for it to measure how far the light's actually travelled and to tick once it's gone a set distance.]The whole point of using light clocks in thought experiments is that their mechanism is out in the open and we can see how the clock is slowed by its movement, but all clocks are slowed in the same way by their movement through space no matter how they're designed - they all have components which move in some way or other with delays introduced by increases in force communication distances caused by movement of the clock, and length-contraction also has a role in affecting their tick rate.
Quote from: Thebox on 22/05/2017 23:56:21Quote from: David Cooper on 22/05/2017 21:46:09(4) Time for light to make first part of trip when vehicle moving at 0.5c = 2t [= 2s](Front of vehicle was ahead of light by d and moving at 0.5c while light is moving at c, so light is gaining on front of vehicle at 0.5c and will take 2t to catch it.)Is your starting point from the rear travelling to the front or from the front travelling to the rear ?I am starting from the front of the train, as the train moves the light is released in the rear direction. I have the shortest time firstly then the longer time secondly.Throughout this thread I've worked with the rear-to-front direction first, so if you've switched it round to do the other direction first, it's a simple matter of swapping the numbers over.
I assume you're still using 10cm as the clock length, so if light moves 10 cm from the rear, the front will move 8.67cm in the same length of time, meaning that it is still 8.67cm ahead of the light. That will happen in an amount of time that we can call "t" (and be aware that this t is not the same size as the t used in my discussions with TheBox - with him the t is a second, but here the t is a third of a nanosecond). Anyway, you will have to repeat this step quite a few times before the light can actually catch up with the front mirror. The closing speed of the light and mirror is 10cm/t minus 8.67cm/t, so that's 1.33cm/t. The gap to be closed is 10cm, so we divide that 10 by the 1.33 and get the time it takes for light to catch the mirror, and that will be 7.4641t. You have mistaken this time as the distance light has to go to get from the back of the carriage to the front, and you should have realised that it can't possibly cross the gap in such a short distance when that distance isn't even as big as the carriage length. To get the actual distance the light has to go before it catches the mirror, you need to multiply the distance that light goes in t (i.e. 10cm) by 7.4641t, so that's going to be a whopping great 74.641cm.
The same distance as what? The time that you mistook for distance (and whose value is not quite 7.5)? You've made a massive error which you're now trying to build upon.
For the return trip, the carriage moves 53.5898 and the light moves 4.641cm before they meet. I can't make sense of what you're trying to do there with any of what you've done there.
Not possible - light takes 2t for the round trip on both clocks with the carriage at rest. With the carriage moving at 0.867c, it takes 4t on the perpendicular clock, and 2t for each half of that, so it reaches the far perpendicular mirror in 2t and reaches the front mirror of the other clock in 7.4641t.