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There is no such thing as time. There is only one distance measured by another distance that cycles.

Light will hit the front mirror in 7.461 cm in 10 cm length.

It will hit the front mirror in 0.74671 n a 1 cm length. It is a ratio. A 100 cm cell length will be 74.61 ratio. There is no fixed time only a fixed ratio to c for energy of motion. In a frame time measurement is the same ratio for a cm as a km.I am confused to my monumental error?

Quote from: Thebox on 24/05/2017 00:08:32I do not think you viewed my latest last post. My numbers are correct.If they're correct, you should be able to insert them into the right places [between square brackets] in the following list:-(1) Length of vehicle = d [= 299,792,458m](2) Time for light to travel distance d = t [= 1s](3) Time for light to make round trip lengthways when vehicle at rest = 2t [= 2s](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.)(5) Distance vehicle has moved by this point = d [= 299,792,458m](The light moved 2d and the vehicle moved half that.)(6) Distance light has moved by this point = 2d [= 2 x 299,792,458m](7) Time for light to make second part of trip = 2/3t [= 2/3s](This time we add the speeds together instead of subtracting, so it's a "closing speed" of 1.5c to cover distance d.)(8 ) Distance vehicle has moved during the time the light was coming back = 1/3d [= ...](9) Distance light has moved during second part of trip = 2/3d [= ...](10 ) We now have a round trip for the light completed in 2 2/3t [= ...s]. The light has moved 2 2/3d [= ...] through space. The vehicle has moved a total of 1 1/3d [= ...], which is half the distance the light travelled, and that's no surprise as the light was moving twice as fast as the vehicle.You say that your latest numbers are correct, but I haven't seen you produce the right answers yet to put into (8 ) and (9), and those are the numbers you should be calculating. Also, you still haven't told me whether you agree with the number in square brackets for (7). Until these [= ...] parts remain unfilled with your numbers, you have not completed your assignment.

I do not think you viewed my latest last post. My numbers are correct.

Of course, if we wait till the dots hit the detector to send a new one, then there will be more time between the dots on the right diagram, but if we send them at one second interval on both diagrams, it seems to me that they will be detected at the same frequency on both detectors, because there will be no doppler effect to alter the frequency.

Time depends on the frequencies of vibrating or rotating bodies, not on the distance they travel through the fabric of space during their rotation or vibration, so why would it depend on the distance light travels instead of depending on its frequency?

It is not the distance the earth's surface travels in space that determines the sidereal day length, it's when the same star passes at the zenith. It is not the distance traveled by a pendulum that determines the tics of a clock either, it's when it passes in the middle of its course.

I am finding it hard to fill in the brackets because I am finding it hard to change to your way of doing it and still trying to understand so I put the answers in the correct places. I am getting close to understanding you, I understand you are saying it takes 2 thirds of second for one one event in which I replied 0.6666666sIn my last diagram I stopped the carriage at the next station allowing me to retain synchronous time by taking the delay out by being at rest in the station. ''(7) Time for light to make second part of trip = 2/3t [= 2/3s]''0.6666666s however that is the first part of my trip.

What is a distance that cycles? Without time, how can it cycle?

If you think light can go from the rear of a 10cm to the front while the front is racing away from it and can catch up with it while only covering 7.461cm, you must be out of your tree.

Quote from: David Cooper on 24/05/2017 17:22:31What is a distance that cycles? Without time, how can it cycle? A cycle is the distance light moves between mirrors and back. They are interchangeable.

Anyway if you are chasing a mirror with light and the mirror is moving at 0.866 then the light will catch the front mirror in the direction of travel in 0.7461 to 1 cm. 7.461 in 10 cm and 74.61 in 100 cm. Its just a ratio to one. Can you explain the problem I must be thick.

QuoteImagine you're on an athletics track, 100m from the finish line. You can run at 10m/s (and you have a remarkable gift of being able to reach that speed in an instant from a standing start). There's another runner who will start 10m ahead of you in the race, but he can only run at 8.66m/s (a speed which he too can reach in an instant). How far will you have to run before you catch up with him? Can you really catch him after only 7.641m?Hint, The front mirror had a head start.

Imagine you're on an athletics track, 100m from the finish line. You can run at 10m/s (and you have a remarkable gift of being able to reach that speed in an instant from a standing start). There's another runner who will start 10m ahead of you in the race, but he can only run at 8.66m/s (a speed which he too can reach in an instant). How far will you have to run before you catch up with him? Can you really catch him after only 7.641m?

Quote from: Thebox on 24/05/2017 17:40:23I am finding it hard to fill in the brackets because I am finding it hard to change to your way of doing it and still trying to understand so I put the answers in the correct places. I am getting close to understanding you, I understand you are saying it takes 2 thirds of second for one one event in which I replied 0.6666666sIn my last diagram I stopped the carriage at the next station allowing me to retain synchronous time by taking the delay out by being at rest in the station. ''(7) Time for light to make second part of trip = 2/3t [= 2/3s]''0.6666666s however that is the first part of my trip.. So, what you now need to do is calculate the numbers for (8 ) and (9). To do that, you need to work out how far the train travels in 2/3 of a second at 0.5c, and how far light travels in 2/3 of a second at c. You can do that. Maybe we'll be able to move onto the perpendicular clock some day soon too, because step (10 ) is easy to do - it's just a matter of adding together some of the numbers that you've already collected in the earlier points.

The numbers you've put in for (8 ) and (9) are now correct...although you've subsequently edited in an alternative answer for (8 ) which is wrong. If you decide to go with the correct answer, then it'll just be number (10 ) left to deal with, and there are three numbers needed there:-The first one can be found by adding your answer to (4) to your answer for (7) because the time for the entire round trip is going to be the time taken for the first part of the trip plus the time for the second part of the trip.The second number you need to calculate is the total distance that the light has travelled through space during the round trip, so you need to add your answer for (6) to your answer for (9).The final number you need to calculate is the total distance that the train has travelled through space during the round trip, so you need to add your answer for (5) to your answer to (8 ). The number you produce for this should be half the size of the distance that your light has travelled.(1) Length of vehicle = d [= 299,792,458m](2) Time for light to travel distance d = t [= 1s](3) Time for light to make round trip lengthways when vehicle at rest = 2t [= 2s](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.)(5) Distance vehicle has moved by this point = d [= 299,792,458m](The light moved 2d and the vehicle moved half that.)(6) Distance light has moved by this point = 2d [= 2 x 299,792,458m](7) Time for light to make second part of trip = 2/3t [= 2/3s](This time we add the speeds together instead of subtracting, so it's a "closing speed" of 1.5c to cover distance d.)(8 ) Distance vehicle has moved during the time the light was coming back = 1/3d [= 99,930,819.3333](9) Distance light has moved during second part of trip = 2/3d [= 199,861638.667](10 ) We now have a round trip for the light completed in 2 2/3t [= ...s]. The light has moved 2 2/3d [= ...m] through space. The vehicle has moved a total of 1 1/3d [= ...m], which is half the distance the light travelled, and that's no surprise as the light was moving twice as fast as the vehicle.If you can fill in the missing three numbers in (10 ) and if they are compatible with my numbers, we will have reached agreement on how the non-perpendicular light clock behaves, and then we'll be able to move on to the second half of the process where we explore the performance of the perpendicular light clock.

_________________________________________________________________Before we move on to assignment 2, there's just important thing we should try to find agreement on now. When we compare the moving light clock on our train with an identical one that's stationary, we will see the stationary clock complete a tick in 2 seconds, but the moving clock won't complete its tick until 2 2/3 of a second have gone by, so the moving clock is ticking at a slower rate than the stationary one. Are you prepared to agree with that.

If you use a smaller light clock to govern the rate at which red dots are sent out from the laser, that rate will slow for the same reason the dots take longer to complete their journey through the MMX apparatus - the light clock governing their release will take twice as long to generate each tick.

Quote from: DavidIf you use a smaller light clock to govern the rate at which red dots are sent out from the laser, that rate will slow for the same reason the dots take longer to complete their journey through the MMX apparatus - the light clock governing their release will take twice as long to generate each tick.We don't really know what produces the light frequencies that we observe, but I suspect that atoms are not using light clocks to produce them. For the moment, light clocks are only mind experiments that help us to understand motion, they might not exist at all in nature, and we might not even be able to make one if we tried. What about a laser, can it make a light clock? Not really, because its frequency does not depend on the distance between the mirrors, it depends on the frequency emitted by its atoms. Can two hydrogen atoms forming a molecule constitute a light clock? If one of the wavelengths they emit was the same as their bond length, we could imagine that they use it to stay on sync, but could they do so while traveling through aether since light would take more time one way than the other? I think so, because there would be no doppler effect between them, because the length added or subtracted to the wave at emission would be exactly cancelled by the one subtracted or added to it at detection, which means that the two atoms could still use that light to stay on sync whatever their speed through aether and whatever their position with regard to the motion. By the way, that wavelength would be in the hard x-rays range, so it could be produced by the inner electrons, and it is also close to the bonding energy of such a molecule.

Okay then: it's time for Assignment 2 - the perpendicular light clock:-This time we have to put our clock sideways across the train so that it is aligned perpendicular to its direction of travel, but there's no reason why you shouldn't use a much bigger train and make this one as wide as the previous train was long. I will continue to work with my own numbers and you can see how well yours match up with mine. The maths involves very few stages this time, but it may use methods that you aren't yet comfortable with.

We can make a light clock by using a strobe emitting at a constant speed over a constant distance....Also this is a way to physically measure the speed of light.

If you look at the laser in my interactive diagrams of the MMX again ( http://www.magicschoolbook.com/science/relativity.html ) you will see that it is aligned perpendicular to the direction of travel of the apparatus, but its sideways movement steers the light inside it into following the correct angle of path needed to take it to where the mirror is going to be so that the light can hit it successfully.

Quote from: DavidIf you use a smaller light clock to govern the rate at which red dots are sent out from the laser, that rate will slow for the same reason the dots take longer to complete their journey through the MMX apparatus - the light clock governing their release will take twice as long to generate each tick.We don't really know what produces the light frequencies that we observe, but I suspect that atoms are not using light clocks to produce them.

For the moment, light clocks are only mind experiments that help us to understand motion, they might not exist at all in nature, and we might not even be able to make one if we tried.

What about a laser, can it make a light clock? Not really, because its frequency does not depend on the distance between the mirrors, it depends on the frequency emitted by its atoms.

I think the only way to measure the speed of light one way would be to use atomic clocks

I thought you meant that, for the two observers in the cars, aberration would avoid them to observe the real direction of the light they exchange, which to me, meant that it would still had to be aimed at their future position to hit them, thus that if they used lasers, they would have to be tilted towards that position. Of course, this possibility raises a problem: to tilt the laser towards the right direction, we need to know our direction in aether, and we don't know.

Ok, before I start , you do realise that if the train travels down a slope it makes it easier to measure the light?

Now I have given the answer to the angle I provided without knowing the speed of the carriage, however I am sure with it being maths, I can get the results to fit a speed of the carriage.

However before we move on from the last assignment , you still avoided my question, I am still waiting for the proof of the physical contraction you claim exists. I have not observed anything of the wow factor thus far, this is pretty basic stuff.

p.s the answer to assignment 2 is a variate relative to the dimensions of the carriage.