Existence

Oh, I think I see your problem - there is no light source, there's just a light pulse bouncing back and forth between two mirrors relative to the fixed position of the light clock relative to the space cycle, which is moving at .5 C.The light pulse has always been between these two mirrors, so it's not necessary for there to be a "source"; the light pulse has whatever initial trajectory is necessary for it to maintain direct up and down travel relative to the space cycle, because we define it into existence that way. They are moving. The train is moving, the mirrors are moving, and the light pulse is moving both up and down and laterally: /\/\/\/\/\/\/\/\/\/ and so on. Its all moving together, so there's never any instance where the light pulse "misses" a mirror, because the mirrors are always exactly where they need to be to reflect the light pulse.The result is that one of two things must be true. Either the speed of light is not the same for all observers, or time cannot be synchronized between two different frames of reference. It's a paradox. And the way that we resolve the paradox is by experimentation - experiments prove that the speed of light is the same for all observers.

You can assume anything you want and most probably be wrong.This does not alter the fact that Einstein's Section 2, principle 2, deals only with the measured velocity of the light not the path of the light beam.Your insistence that somehow Sec 2, p 2 is violated by some contrived circumstance dealing with the light path is wrong.Edited by AZPaul3, : No reason given.

Hi crash, Lets modify that open clock so we know for sure the pulse of light is going straight up to the top mirror by puting a glass tube over the head of the laser pen and extending it to the top mirror.We now know the light pulse will travel to the top mirror and return to the bottom Now it takes 3.33564095198152 nanoseconds for the light pulse to reach the top mirror, as it is 1 meter away. The pulse will strike the top mirror and return to the bottom mirror 149,896,229 times in 1 second.During which time the train or cycle with the clock attached has traveled 149,896,229 meters.So I see the light pulse go up and down in my frame as the clock is not moving in reference to me. Since the beam will take 6.67128190396304 nanoseconds to make the round trip I will see a solid white light.

fig 1
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Local observer       

The observer in a frame from the side will see the trip of the light pulse to the top mirror and return in the following manner.

fig 2
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       Observer at a distance 

The observer would see the light pulse strike the top mirror 3.33564095198152 nanoseconds after the light pulse strikes the top mirror.
You say that is time dilation.I say it is a mess.The first line at the moment the light strikes the top mirror is delayed due to the time delay of the observer receiving the light data to process due to the fact the cycle is traveling at 1 meter per 6.671281903963041 nanoseconds.The further the clock gets from the observer the more delayed the light data will be received until it is totally scattered and can not be observed. You like to use the pointer in your Power Point presentations also.I do most of mine at about 12' so I will figure on that basis.I will use 1 million miles per hour for the earths movement, although it is less than that.If I used the 67,000 mph of the earth around the sun it would be much less that the following. About 1/14.That equals 1,609,344,000 meters per hr which equals 1,609,344,000,000,000,000.00 nano meters per hr.
which equals 0.1998447616 nano meters per nanosecond.At 12' or 144" at 3.33564095198152 nanoseconds per meter it takes the laser light 0.0847254496312299 nano seconds to travel the 12 feet.At 1,609,344,000,meters per hr. the earth would move 0.0169319372830059 nano meters in the time it took for the laser beam to hit the targer 12' away, if my math is correct.To put that into perspective that is about 50 millionths of the width of a human hair.So I don't really think it is going to bother your aiming of your laser pen, or what you see the laser light do..My hand shakes a lot more than that. But Einstein says the w frame is a stationary inertial frame and the W frame is a moving non-inertial frame in his papers. Relative to what frame? I agree.That also means you can not add the forward motion of the emitter source of the light pulse on my train to the light pulse that is emitted at a 90 angle to the direction of the train. With the speed of a 12 lb cannon ball with a muzzle velocity of about 1769 Feet Per Second.The speed of light is 983,571,056.430445 feet per second.Yep that is a no brainer. The cannon ball could not catch up with the light, much less affect its speed.God Bless,

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"John 5:39 (KJS) Search the scriptures; for in them ye think ye have eternal life: and they are they which testify of me."

No, this is not correct. The observer from the side will see the light pulse do the same thing you see it do - reflect back and forth between the two mirrors - but because the mirrors are moving, he sees the path as

\    /\    /\
\  /  \  /  \
\/    \/    \

And the point is that this is a longer path than you observe - it's a simple function of trigonometry to prove that it is - and so one of two things must be true - either the light pulse is observed moving at a different relative speed, relative to the two of you; or else the two of you can't agree on when the light pulse is actually bouncing off of the mirrors.Because we know that the speed of light is the same for all observers, we know that the second consequence must be the one that is true. That's called "time dilation" and it's been experimentally verified, but this simple thought experiment proves that it happens regardless. But even if we fire lasers as far as the moon, we never observe any deflection of their path due to the Earth's movement. Doesn't happen. Relative to any frame. Regardless of your velocity relative to any frame whatsoever, you will never find any photon traveling through a vacuum at less than C. That's why it's called "C" - it's always the same, and so we can give a name to that speed. Again, experimentally verified since 1886. I can't add the speed but I can add the velocity. Remember when I explained the difference? I can't accelerate light to a speed greater than C, but the velocity of an emitter can be added to it with the result that the light experiences a change in velocity; the change not being to the magnitude of the velocity (which will always be C) but to the direction. It's not a matter of speed; even a .5 C super-cannon couldn't produce light that traveled any faster than C.

This is the incorrect part. To the outside observer the beam of light appears to move with the direction of the train.Take one photon of light. Just a single photon. To you standing next to the clock the photon as it travels moves up only:*
*
*
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*To the outside observer the photon also moves in the direction of travel:

*
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         *
      *

This does not violate Einstein's principle since the measured velocity, the speed of the photon, is always c for both observers.Edited by AZPaul3, : No reason given.

Fine. So paper the entire top of the frame with photographic film (except where the mirrors are and provide a thin polyester film at the bottom of the frame that the laser can make a hole in when it punches through. We can then determine the path of the laser beam by lining up the hole in the bottom with the mark on the film top frame. I agree, sorta. But to be pedantic "c" is only the speed of light and not the velocity of light. No it does not say light is independent, whatever the heck that means. Only that the speed is always c, regardless of the speed of the source or the receiver.Yes, only the speed is independent. For example, we know that the frequency of light at the receiver depends on the relative motion between source and destination. Were this not true, then doppler radar would not work. Perhaps you can use this defense in court if you get a traffic ticket. I'd advise my own client not to try it. That's the part that you've made up out of whole cloth. Unless I'm mistaken, you seem to accept that observers in the train won't measure a light beam as moving vertically despite the fact that observers on the ground will. That's is almost too funny. You should consider doing physics stand up comedy. You'd wow audiences on the Cambridge Mass., Georgia Tech, Cal Berkeley circuit.You just snap your fingers and suddenly you can see what everyone else knows without seeing. Grrrr... What did you do? Did you add some cigar smoke within the mirror frame you constructed? But when you want to deny something, you just don't look?Central to your reasoning is that we can determine the absolute velocity of observers so that we can identify a truly non-moving frame. But Einstein meant nothing of the sort when he described stationary frames. Any inertial frame can be designated as stationary and generally the stationary frame is picked for convenience sake. If you are doing experiments in a non accelerating space ship, then the logical stationary frame for making measurements and applying the laws of physics is likely to be the space ship rather than earth.Further, we don't really need to "see" things in a thought experiment. In the case of a light clock, if we know what it is doing in one frame, we can infer what an omniscient observer in another inertial frame would see. For example, we know that if the light beam hits the top mirror in one frame, then observers in other frames cannot receive a contrary indication no matter how fast they whiz by.

Hi ICANT,I wanted to comment on this picture. You cannot have thought through the implications of what you claim. I think we all agree that the distance between those vertical lines as measured by an observer for whom the ship is moving at 0.5c must be more than 93,000 miles. (Actually about 15.5% larger by my calculations)So how does the photon make those 90,000+ mile horizontal trips at the bottom and top of the light clock shown in fig 2. When I say that the photon as observed from the earth/planet X frame travels a path having an angle of a mere 30 degrees from vertical you balk. But then you show the light making a 90 turn in space and traveling horizontally?Edited by NoNukes, : No reason given.

I have to admit that I find this explanation a bit confusing. Part of the confusion results from people saying that the velocity of light is "c" when they clearly mean that the speed is "c". We know that the use of the term "velocity" is inexact in this context because c is a scalar quantity, while velocity is a vector. ICANT has latched onto this sloppy language because he thinks it helps his argument.I'm not sure that "adding velocity" makes much sense either. The way I like to think of things is that there are no frames that are stationary in the absolute sense, but all inertial frames are perfectly good coordinate systems for investigating the laws of physics. When observers at rest in different inertial frames conduct experiments, they measure different durations of events, different speeds of objects (other than photons), different angles, etc., but they all agree that momentum and energy are conserved in their own coordinate systems despite measuring differing values for those things.

Hi NoNukes, As you just learned even in peek mode there are somethings that can't be copied and pasted.Your fig 2 looks nothing like mine.If it did it would look like this:

fig 2
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Observer at a distance 

If you were to take a piece of cardboard 1 meter high and 1/2 meter wide and place 149,896,229 marks with a very fine point pen on the cardboard without placing one on top of the other you would have what would be produced to the observer at a distance.Now here is a different couple of figures for you.reference to me.

fig 3
              M
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Local observer       

This is what I would observe with the light pulse going to the top mirror and back to the bottom mirror from any angle on the cycle.

fig 4
                                        M             M             M
                                       F  |          F
                                      E    |        E
                                     D      |      D
                                    C        |    C
                                   B          |  B
                                  P             P             P
                           Observer at a distance

                                  Z

This is what an observer from a distance to the right side of the direction of the travel of the cycle would see if it were possible to see the light pulse, and if the mirrors in the light clock was 299,792,458 meters apart.From P to M would seem as a angle line. I placed these two lines 1 meter apart but in reality it would be solid as the lines in fig 2 above.I did this math in a hurry so you might need to check it for accuracy.
But you should be able to get the idea.It takes the light 3.33564095198152 nanoseconds to travel 1 meter.
It takes the cycle 6.671281903963041 nanoseconds to travel 1 meter.I have divided this figure 4 up into 6 steps to the top.The light pulse is emitted at P and M is the mirror.The light source is observed at P by the observer at Z.When the observer sees the light pulse at B the light in the tube has travel distance =24,982,704.83333333 meters to the right of the observer which takes the light pulse 7,489,626.4894734800
nanoseconds longer to reach the eye of the observer than it did when emitted at P.When the observer sees the light pulse at C the light in the tube has travel distance =49,965,409.66666666 meters to the right of the observer which takes the light pulse 14,979,252.9789470000 nanoseconds longer to reach the eye of the observer than it did when emitted at P.When the observer sees the light pulse at D the light in the tube has travel distance =74,948,114.50 meters to the right of the observer which takes the light pulse 22,468,879.4684204000 nanoseconds longer to reach the eye of the observer than it did when emitted at P.When the observes sees the light pulse at E the light in the tube has travel distance =99,930,819.33333333 meters to the right of the observer which takes the light pulse 29,958,505.9578939000 nanoseconds longer to reach the eye of the observer than it did when emitted at P.When the observer sees the light pulse at F the light in the tube has travel distance =124,913,524.17 meters to the right of the observer which takes the light pulse 37,448,132.4473674000 nanoseconds longer to reach the eye of the observer than it did when emitted at P.When the observer sees the light pulse at M the light in the tube has travel distance =149,896,229 meters to the right of the observer which takes the light pulse 44,937,758.9368409000 nanoseconds longer to reach the eye of the observer than it did when emitted at P.The math says it is the fact that the clock has moved 149,896,229 meters to the right in relation to the position it was in when the light was seen to have been emitted by the observer causing the light pulse to have to travel an extra 44,937,758.9368409000 nanoseconds to reach the eye of the observer.Show me where the math is wrong.So the observer sees the light pulse strike the top mirror at 1,044,937,758.936841 nanoseconds after the observer saw the light pulse emitted.This extra 44,937,758.9368409000 nanoseconds and the angle is caused by the distance the light beam has been carried by the tube it is running up and down in.Does that mean light has exceeded 299,792,458 meters per second? No It means the distance has increased that the light data had to travel to reach the observer at Z.The second P is at 1 meter distance from the first P when released which is the same exact time the light pulse reaches the first M.Now by the time the pulse reaches the second M we have the same extra 44,937,758.9368409000 nanoseconds the observer will require to receive the light beam from the time the observer received the data that the pulse was emitted.Now the further the cycle gets from the observer the more acute the angles will become due to the observer's location in relation to the cycle. This will continue until the light pulse becomes a straight line to the observer if he/she has the capabilities to observe the light pulse. They would be 149,896,229 meters distance between them if the mirrors had 299,792,458 meters distance between them.That would be a big clock. What you been smoking, or drinking?The lines in my fig 2 is lines between two mirrors that are 1 meter apart. The light pulse is emitted from the middle of the bottom mirror and bounces off the top mirror because there is a tube the size of the beam placed over the laser pen and reaches to the top mirror.It takes 3.33564095198152 nanoseconds for the pulse to strike the top mirror and 3.33564095198152 nanoseconds to return to the bottom mirror.If another pulse is released at exactly 1 second then this will continue to be the case.God Bless,

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"John 5:39 (KJS) Search the scriptures; for in them ye think ye have eternal life: and they are they which testify of me."

Hi NoNukes, I use velocity because Einstein used velocity.I am not going by your interpertation of what Einstein said, or what you have been taught.From: The June 30, 1905 English translation of Einsteins paper titled:ON THE ELECTRODYNAMICS OF MOVING BODIES 2. On the Relativity of Lengths and Times

quote:

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1.The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of co-ordinates in uniform translatory motion.
2.Any ray of light moves in the stationary system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body. Hence

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SourceI will give you other papers of his that cover the 2 postulates later but now I have to take care of business and doctor appointments for a couple of days.God Bless,

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"John 5:39 (KJS) Search the scriptures; for in them ye think ye have eternal life: and they are they which testify of me."

Hi ICANT,I have to acknowledge that you've caught me in a really bad math error. As you have indicated, the distance between those seemingly parallel lines is on the order of 1/2 meter and not 90,000+ miles. But the distance it is not exactly 0.5 meters. Instead we can show that it is 0.57735 meters?But let me point out a tiny math error in your own calculations. Perhaps we can end this lemma. I'm not going to attempt to reproduce figure 4. But let me point out that the angle of the inclined path shown in figure 4 is indeed 30 degrees from vertical as I mentioned in my post, given a 0.5c relative motion between the the clock and the non-local observer.Can you calculate the length of the path between points P and M? Once you have done so, can you tell me how light manages to traverse that distance in 3.33 nanosecs? I think you'll find it quite impossible for light to traverse that distance in that amount of time at speed c. Here you are rambling on about a non-issue. When determining the time of an observed event, observer must correct his observed time by subtracting off the signal propagation time for the light to travel from the event to the observer. In the case of this thought experiment, the space ship travels a predictable path, and the signal travel time correction is quite easily done.

Hi ICANT, I really don't care why you use the term velocity. When you or anyone else says that the velocity of light is "c" then what is being described is a scalar quantity. It's simply not meaningful to say that the direction of a vector is "c" where c is a scalar.

Hi NoNukes, You attributed the statement I made to the wrong fig.What I would see is in fig 3, Not in fig 4.Fig 4 is what the observer at a distance would see. Why would I need to know that distance as nothing travels at a 60 angle?The light beam is in a tube just the size of the beam. It can go in no direction but straight up to the to mirror and back down to the bottom mirror.By enclosing the light pulse in a tube from the source to the top mirror I have now forced the light pulse to take on the motion of the cycle as the light clock is mounted on the handlebars.The light pulse goes up and down verticle at a 90 angle to the travel of the cycle.But since the cycle is traveling at 0.5 c the light pulse is dragged along with the cycle.This give the observer at a distance the optical illusion that the light pulse is going at an angle.The further the cycle gets from the observer at Z the more acute the angle will become and if the observer could see it long enough it would become a straight line at 180 because the light pulse only goes 1 meter from the bottom mirror before it strikes the top mirror and returns to the bottom mirror.It takes the light pulse exactly 3.33564095198152 nanoseconds to travel the 1 meter from the bottom mirror to the top mirror. I would agree if the tube was placed at a 60 to the source to the right of center of the top mirror.But then the distance the light pulse had to travel would be increased and the mirrors would no longer be 1 meter apart for the light pulse to travel between. You have not thought that through competely.Each lettter is going straight up in the tube that light pulse must travel in to reach the top mirror.The problem is that the cycle has moved the clock further from Z in B, C, D, E, F and M that it was when the light pulse was emitted.You see by enclosing the light pulse in a tube I have forced the light pulse to take on the forward motion (speed) of the cycle as the clock is attached to the handle bars.This invalidates postulate #2 as light can not take on the speed of the source.In the open clock I was discussing the light pulse could go at a 90 angle to the travel of the cycle as postulate #2 says it will. The problem with that is that the light pulse misses the top mirror.But that validates postulate #2.As my grandmother used to tell me, "you can't have your cake and eat it too". But the signal travel is not what you think it is.Think about it for a couple of days and I will be back home sometime late tomorrow.God Bless,

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"John 5:39 (KJS) Search the scriptures; for in them ye think ye have eternal life: and they are they which testify of me."

If it doesn't work on your bike then it shouldn't work on the Earth either since the Earth is hurtling through space at more than 1,500 km/h. The same for our whole solar system that is hurtling through space around the galactic center, and also the movement of your galaxy with reference to other galaxies. The Earth has a ton velocity, and yet you say the clock works fine on Earth. If it works on Earth then it should work just as well on the non-accelearting bicycle.We could also use a simpler example. Lets say that you are sitting in a window seat on a train travelling at a constant 60 mph. To pass the time you are tossing an apple up into the air. You happen to pass your wife who is standing at the side of the tracks. What do the two of you observe where it concerns the apple?You observe that the apple is travelling straight up and down. Your wife observes that the apple is travelling in a sawtooth pattern. The same applies to the light clock.

Hi ICANT, ABE: Actually, it invalidates the ICANT interpretation of postulate #2, which is quite distinct from the real postulate #2. Seriously, ICANT, where was the tube in the Michelson-Morley experiment which validates the real postulate #2?ICANT, this is just plain silly. You cannot force light to travel in a straight line by confining it to a tube. If the light is not originally directed down the axis of the tube, then the light would instead bounce off the sides of the tube. (If not on the first pass, after a few reflections between the mirrors). The actual light path would be quite a bit different then you describe here. Ooh. You are almost there except for your ridiculous denial. Well at least your denial here is on a basis that is easily seen as ridiculous. You probably cannot be convinced, but I doubt that any lurker would be confused about who is right here. Sure ICANT.Edited by NoNukes, : No reason given.Edited by NoNukes, : Respond to ICANT's postulate two silliness.