[David Cooper]

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

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

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

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.

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?

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?

[guest39538]

I 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 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 0.6666666s
In 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.

[David Cooper]

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.

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. That means that light pulses will leave the laser twice as often on the stationary apparatus as they do on the one moving at 87%c.

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?

If your clock depends on something vibrating, its vibration rate will be slowed by its movement through space in the same way as light is slowed. The mechanism of all clocks involves communications at the speed of light, so they are all doing the same thing as a light clock - their mechanisms are merely disguised. If, for example, you have a piece of metal vibrating and producing a musical note, the ability of the material to slow, stop, accelerate the other way, slow, stop, accelerate back the way again, etc. is all governed by forces that apply at the speed of light, so if you increase the distance through space over which these forces are having to operate, you will slow their functionality to the same degree as with a light clock. The musical note will be an octave lower if the piece of metal is moving through space at 87%c, but anyone moving with it will have their functionality slowed to match, so they will not hear it as a lower pitch. So, you can wire up any kind of clock you like to the laser to govern the rate at which it pulses and it will tick half as often if it's moving through space at 87%c than if it's stationary.

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.

If you view the Earth from a frame of reference through which the Earth is moving at 87%c, you will see it take twice as long to rotate than if you view it from the frame in which the Earth is stationary. If the Earth is actually stationary relative to the fabric of space, what you see from the frame in which it appears to be moving at 87%c will be an illusion and it will actually be you that is living in slow motion, but if the Earth is actually moving at 87%c through the fabric of space, it really will be taking twice as long to rotate. The same thing applies to a pendulum - if the pendulum is sitting on a planet moving through space at 87%c, it will tick at half the rate of an identical pendulum sitting on an identical planet that's stationary.

The only way we could get a clock to tick at the same rate regardless of its speed of travel through space would be to make one that uses superluminal communications for the forces acting between its parts, but no such clocks have been invented, and they won't be invented until the speed of light barrier can be broken by something. Every clock has parts which influence each other by forces acting at the speed of light. The only things that might exist that have no separation between their parts cannot serve as clocks.

[David Cooper]

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 0.6666666s
In 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.

If the first part of your trip is the part where the light is moving from the front of the train to the rear, then that's the 2/3 needed for (7), so you now appear to have agreed with it. You also appeared to be in agreement with all the points (1) to (6) that come before it. 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.

(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.

[GoC (OP)]

What 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.

 

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. 

How did you know I had a tree house? 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.

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]

Hint, The front mirror had a head start.

[David Cooper]

What 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.

A cycle involves movement or change which is carried out over time. A distance doesn't.

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.

It will catch it in 0.7461 what? Are you seriously telling me that the light moves 7.461cm from the back of the carriage and somehow manages to catch the front of the carriage at that point even though the carriage started 10cm ahead of it and has moved further on ahead during the time the light has moved 7.461cm? I don't know how you get it to pull off that trick.

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?

Hint, The front mirror had a head start.

Yes, and so does the runner (representing the mirror) that you (representing the light) are chasing on the track. You run 7.461m and are still short of where the other runner started, but he's run further on ahead during that time. It is not credible that you could be so stupid as to think you could catch him so soon, so the only rational explanation of what's going on here is that you're playing avoidance games as you don't like where the maths is leading you, and if you're going to go on doing that, I'm not going to waste any more of my time discussing this stuff with you.

[guest39538]

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 0.6666666s
In 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.

Oh you mean 199861638.667m  and 49965409.6667m

added - I am not happy the second amount is correct so have re-done it below

(299792458/3)*2=2/3 ? = 199861638.667m?

199861638.667/2=99930819.3333   m?

p.s I quite like learning and have to work things out, thank you for the mental exercises.

(8 ) Distance vehicle has moved during the time the light was coming back = 1/3d = 99930819.3333   m

(9) Distance light has moved during second part of trip = 2/3d = 199861638.667m


8+9=299792458m   

Sorry for taking my time with the correct answer, it did not sink in right away what you was asking.

added- no, my second answer is still wrong....
 (8 ) Distance vehicle has moved during the time the light was coming back = 1/3d =49965409.6667m   

My first answer was correct after all.

/2=.5c d1/3

added- lol now I am unsure, perhaps I should clear my mind and start again .   

added - sorry I was seeing your question ambiguously , I think you are just asking for one third of the distance which is 99930819.3333m.

[David Cooper]

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.

[guest39538]

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.

10 ) We now have a round trip for the light completed in 2 2/3t = 2.6666666666s

The light has moved 2 2/3d = 799446554.667m through space

The vehicle has moved a total of 1 1/3d = 399723277.333m

[David Cooper]

So, the final version of assignment 1 (the light clock aligned with its direction of travel) is as follows, and we appear to be in agreement on it (for now, at least):-

(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 [= 2.6666666666s]. The light has moved 2 2/3d [= 799,446,554.667m] through space. The vehicle has moved a total of 1 1/3d [= 399723277.333m], which is half the distance the light travelled, and that's no surprise as the light was moving twice as fast as the vehicle.

_________________________________________________________________


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.

[guest39538]

_________________________________________________________________


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.

Yes, at this stage you have been objective, you have not tried to add anymore than it is, the stationary clock ticks faster than the clock in motion.

added - however they are still synchronous. While the stationary clock measures 2 seconds, the light in the clock in motion is in the 2 second position although it as not reached the detector.

p.s my previous diagram

[guest39538]

ruler3.jpg (125.04 kB . 1765x505 - viewed 4816 times)

The stationary clock will read 2.6666666s when the clock in motion reads 2 seconds.   The 2.66666s would be the correct time. 2s=2.666666s

[Le Repteux]

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.

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.

[guest39538]

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.

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.

We can make a light clock by using a strobe emitting at a constant speed over a constant distance. Not even that difficult to be honest. 
Using the rate of detection to measure time, the rate would always be constant.
FLASH,DETECT,FLASH,DETECT,FLASH, DETECT, TICK AND TOCK


Also this is a way to physically measure the speed of light.


detect...............detect...............detect
..........time....................time..................

t=d

Imagine a device that is 10 meters long, at one end a strobe emitting 1 pulse every 10 seconds.

Now imagine at the other end of the device is a detector, it will detect 1 pulse every 10m travelled.

At this stage we do not know the speed of light.

However we have a graph running at detector end, the space between detects represents the 10 meters the light has travelled, of course we also have a time line on the graph so we can work out how long it took for the light to travel the 10 m.
We then can work out the speed.

[David Cooper]

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.

Before we can start crunching any numbers, we need to think about how light is going to get from one mirror to the other when movement of the train will mean that it has to travel at an angle if it's to hit the mirrors rather than going perpendicular to the train and landing where the mirrors used to be instead. 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.

You can easily see that the angle at which the light actually goes through space will not be perpendicular to the train, but the angle is also different for different speeds of travel by the train. If the train moves at 0.5c, the angle of the light's path will be 30 degrees to the perpendicular, whereas if the train moves at 0.866c the angle will be 60 degrees instead. However, you should insist on seeing proof that it has to go at those angles for those speeds, and you will then need to have an efficient mathematical method to apply to work out which angles go with which speeds.

You should maybe start with a trial-and-error method on a piece of paper as that will allow you to test different angles. Draw the two mirrors in their initial positions (which you can label as A and B), then draw parallel lines from them right across the paper running in the direction in which the train is moving. Then move the train the distance you think it might move in one tick of the clock at 0.5c and mark where the mirrors will be once it's moved that far (with these points labelled as A" and B"). Once you've done that, you should draw the mirrors again where they would be with the train half way between those two positions (labelling them as A' and B'). You can now draw a line at an angle from A to B' and another line from B' to A". Then get a ruler and measure those two lines. They should be the same length as each other every time you do this. Add them together. If the angle is right for that speed of travel of the train, the length you have just measured for the two sloping lines should be exactly twice the distance the train has moved. I've drawn a diagram of my own to give you an idea of what will happen, so see the attachment below. The red attempt failed because the distance light has to travel along on the sloping red lines is much longer than twice the length of the horizontal red line at the bottom. The black attempt also failed because the combined length of the sloping black lines is less than twice as long as the horizontal black line at the bottom. The blue attempt looks reasonably good though - it was drawn without measuring anything, so it'll be slightly out, but it's clearly giving us an angle close to the predicted 30 degree angle.

How can we actually calculate the correct angle mathematically then? If you start at the blue B' and draw a vertical line downwards from there until it hits the horizontal line that the A mirror moves along (and by chance it will hit it near to where I put the red A" mirror location), it will hit that line at 90 degrees, so we have a right-angled triangle (with the corners A, blue B' and a point next to or possibly right on top of red A"). We also know that the height of this new line is the width of the train, so that's a known length. Perhaps we can apply that funny soh cah toa trig stuff using sin, cos or tan on a calculator. We don't know the angle the light goes at yet because that's what we're trying to work out, so we need another length before we can apply trig. We don't know how far the train will go before the light can get back to the first mirror to complete a cycle, so we don't have the length of the horizontal side of the triangle, except that we do know that the light must go exactly twice as far as the mirror does if the train's moving at 0.5c.

Aha! It turns out that we don't even need to use the length of the vertical line on the triangle running from the blue A' to the red B" because we already know the ratio of the other two sides to each other: the length from A to the red B" is half as long as the length from A to the blue B', and so long as we use numbers related to each other by the right ratio for these two lines, we are guaranteed to get the correct angle out of this. So, let's call the horizontal line on the triangle 0.5 and the sloping line 1. We can then apply the soh (sine of the angle = opposite over hypotenuse) or cah (cosine of the angle = adjacent over hypotenuse) rules to calculate the other angles of the triangle. With soh we can get the angle at blue A' (between the sloping side of the triangle and the vertical line running down from there): sine of that angle = opposite/hypotenuse, so that's 0.5/1, and that equals 0.5. So, we simply use arcsin(0.5) on a calculator to get the angle, and the calculator spits out the number 30. That's our 30 degrees. To find the function arcsin on a scientific calculator, you usually have to press a key with shift, inv (inverse) or 2ndf (second function) followed by the key with "sin" on it. With some calculators you would type the 0.5 in first and then type inv and sin, but on other calculators you have to type shift sin first and then type in the 0.5.

[The cah route gives you the other angle: arccosine of the angle at A = 0.5/1, so again it's just a matter of typing 0.5 inv cos or shift cos 0.5, and the calculator gives us the answer 60 degrees.]

This method works for any speed of the train, so if it's going at 0.866c, we know the ratio of its speed to the speed of light and the distances they both travel will be 0.866 to 1. Again we can do arcsin 0.866/1, and again the number we're dividing by is 1, so it makes no difference to the 0.866. The result is that all we need do is type 0.866 inv sin, or shift sin 0.866 and the calculator will give us the answer 60 degrees for the angle at the top of the triangle.

If you don't have a scientific calculator and don't know how to find the one that's likely included with your operating system, you can use my JavaScript one online instead: http://www.magicschoolbook.com/maths/3.html - it has an inv/2nd button, but it also has direct asin, acos and atan buttons which you can use instead. So, type 0.5 or 0.866 or any other speed for the train into it and then click on the asin button to get the angle the light goes at relative to a line perpendicular to the direction of travel of the train.

[guest39538]

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.

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



tri.jpg (5.28 kB . 421x458 - viewed 4828 times)


You do also realise light does not do this?

My answer at a glance , 

at rest

5.s one way trip

10.s round trip

in motion
3.s
7.s return

[guest39538]

Yea that is my answers


tri1.jpg (8.75 kB . 421x458 - viewed 4817 times)

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.



size.jpg (15.26 kB . 421x458 - viewed 4860 times)

[Le Repteux]

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.

I think the only way to measure the speed of light one way would be to use atomic clocks, but to synchronize those clocks while they are traveling in aether, we would need to know the exact speed and direction they travel with regard to it, because the synchronization signal would take more time one way than the other, but we don't know our speed in the aether, so if it exists, or if light propagates as if it was the case, I'm afraid this experiment is impossible to make.

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.

I thought that we had an agreement on that on Facebook, but it seems that I did not understand properly what you meant. You said:

«As you say, a stationary observer to the side of the runway (and some way away from it) who caught a glimpse of the laser light would see it as coming from further back down the runway before the vehicles reach the point of closest approach, and he would see the vehicles back there at the same time, even if they are by that time at the point of closest approach and far ahead of where he sees them. That is quite different from an observer in one of the cars who sees the other car where it actually is even though the light he's seeing from it has come to his eyes from further behind.»

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.


Anybody knows why [nofollow] pops out after every link I put?

[David Cooper]

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.

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.

The speed of light is also the speed of force - the atoms are held together by forces which act at the speed of light, so the exact same delays apply to those. If you move an atom through space, its functionality is slowed. It is not a light clock, but a force clock, and the two are directly equivalent.

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.

There is nothing to stop light clocks being made. A one gigahertz processor has a clock rate of a billionth of a second. During that time, light only moves the length of a ruler. Also, the MMX is essentially a pair of light clocks, so it goes beyond thought experiments into actual measurements. We can also make radio clocks where we send signals to and fro between two points and use them as light clocks, and I do mean light, because if you were to run into radio waves fast enough, you would be able to see them as light.

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.

That frequency is dictated by the tick rate of a force clock, and a force clock behaves exactly like a light clock because they are governed by the same speed limit.

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

That won't work because they are force 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.

There is no need to point the laser anywhere other than perpendicular to the direction of travel - the angle the light actually follows is dictated by the movement of the laser while the light is moving through it (and indeed before that in the mechanism of how the light is emitted in the first place). This relates to the "headlights effect". If you have a light bulb in the middle of a square room and you move the room at relativistic speed, one of the walls may be moving directly away from the light that's chasing it from the bulb, and the opposite wall is racing directly towards the light coming at it from the bulb, but all four (or six) walls remain equally brightly lit because the movement of the bulb through space affects the way light is spread from it with more of the light being projected forwards. An attempt by an emitter to throw a photon directly sideways will fail because of the movement of the emitter - its own speed of movement in its direction of travel is automatically passed on to the photon, affecting the angle it leaves it in.

If you were to change the alignment of the moving laser, light would not leave it in line with the laser unless the laser's pointing directly forwards or backwards, so if you aim the laser at where the target is going to be rather than at where it currently is, it will hit a point far ahead of where the target will be.

[David Cooper]

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

No, I don't realise that, and your numbers don't appear to realise it either.

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.

How can you possibly calculate the angle without involving the speed of the carriage in the calculation?

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.

I've answered that many times before, but you clearly weren't ready to take it in. Remember the five points that I made in post #134:-

(A) A moving clock records fewer ticks than a stationary one in a given length of time, so it is not recording time, but is merely counting cycles. That is what all clocks do.

(B) A light clock aligned perpendicular to its direction of movement therefore records fewer ticks than a stationary clock.

(C) An uncontracted light clock aligned with its direction of travel (rather than perpendicular to it) will record fewer ticks than a light clock co-moving with it which is aligned perpendicular to their direction of travel.

(D) A correctly length-contracted light clock aligned with its direction of travel will record the same number of ticks as a light clock co-moving with it which is aligned perpendicular to their direction of travel.

(E) The null result of the MMX shows that the real universe length-contracts things in their direction of travel.

Once you have generated the right numbers for assignment 2, you'll be able to compare the two moving light clocks and see that the perpendicular one ticks more often than the other. The MMX shows that this is not how the universe works because real light clocks which move along together always tick at the same rate as each other regardless of their alignment. We also know that the slowing of real clocks matches up to the perpendicular clock's predicted behaviour and not to the predicted behaviour of uncontracted light clocks aligned with their direction of travel.

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

Your perpendicular light clock should be the same length as the one used in assignment 1 so that you can easily compare the numbers you get from the two assignments. Your job here is to work out how far the train moves at 0.5c while the light travels from mirror A to the far mirror (B') and back to the first one again (A"), and you need to know how to calculate the angle that the light moves away from the perpendicular. I told you how the angle can be calculated and I was hoping that you'd let me know whether you understand that method or not. If you want to find your own method for doing it, that's fine, but for it to be valid it will need to generate the same numbers. We will be stuck here until you understand the method I've spelt out to you, so I'll go through it again with a new diagram.

We want to calculate the angle x. We have a right angled triangle, but the only side of that triangle with a known length is the vertical line B'A' which has a length of d [= 299,792,458m]. However, we do know the ratio of the lengths of AA' to AB' because we know the ratio of the speed of the train to the speed of light, and that ratio is 0.5:1.

The trig rule soh tells us that sine(x) = AA'/AB'. We don't yet know the lengths AA' or AB', but we do know their ratio, so we can simply use the numbers from that ratio: sine(x) = 0.5/1. Because we're dividing a number by 1, the result will be unchanged, so we now have sine(x) = 0.5. All we need to do now is use a calculator to find x by typing in a sequence such as 0.5 inv sin or shift sin 0.5 (depending on the order the calculator wants you to press the keys in.

Having done that and got the number 30 degrees from the calculator, we can now set about working out the actual lengths of the lines AA' and AB'. The only real length we know for the triangle is the vertical line A'B', but that's enough to do the rest now as we also have an angle to work with. All we need to do is apply more trig: tan(x) = AA'/A'B', and cosine(x) =  A'B'/AB'. We can rearrange those a little before we pick up the calculator (and note that I'm going to use an asterisk for the multiplication symbol to avoid confusion with the angle x): AA' =  tan(x) * A'B', and AB' = A'B'/cos(x). Now the calculator is used: AA' = 0.57735, and AB' = 1.1547. Note that I have used 1 as the length of A'B' because my length unit is d, and d = 299,792,458m. To fit with all your previous calculations, you should use 299792458 instead of 1 so that your answers come out in metres.

If you double 0.57735, you'll get 1.1547, so these answers fit with the 0.5:1 ratio that we expect them to. The lower mirror has moved from A to A" by the time the light returns to it from the top mirror, so the train has moved 1.1547d and the light has moved 2.3094d. To convert those to metres, again you'd need to multiply them by 299792458. One tick of this moving clock will take 2.3094t, and as my time unit is a second long, that's 2.3094s.

We now have three tick rates. The stationary clock ticks once every 2 seconds. The moving clock from assignment 1 ticks once for every 2 2/3 seconds of the stationary clock. The moving clock from assignment 2 ticks once for every 2.3094 seconds of the stationary clock.

Now, lets see how many pages of posts it takes you to get up to speed with that. On the up side though, it looks as if you might reach the finish line before GoC, which no one reading this thread at the start would ever have predicted.