Existence

Hi NoNukes, Then show where the math is wrong. Well the one I chose was one that could bounce between the top mirror and bottom mirror 149,896,229 times per second once started, eternally or until the tube was removed. Enjoy yourself, just don't bust a gut. Figure anything you want to figure. It only takes 1 second for the light to flash at the first T from the time it flashed a the first B. Well, no I did not make the same mistake as you did.I did make one mistake in stating when the light came on at the second B 1 second had passed when really 2 seconds had passed. I will underline what I am refering to in the following quote.1 second takes place every 149,896,229 times the light pulse hits the top and bottom mirror, when 1 of the lights flashes.But when I give the breakdown of distances I did state the correct time in seconds.In Message 786 I laid out exactly what would happen.

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I really think a light pulse can be forced to travel in a straight line by confining it in a tube. You did propose it was in a vacuum enclosure, in the post I am replying too.

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I see you can't understand simple things as they are too non-complicated for your education.

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Lets modify my clock on the cycle and see.

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So let me try a few more modifications to the clock and see if I can force the pulse to go verticle at a 90 angle to the motion of the cycle.

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Lets make that tube a black metal vacuum tube so the light can't escape.

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Lets put a detector at the top and bottom that causes a light to flash.

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Lets program the detectors to flash the light at T and B alternating between them every 149,896,229 pulses.

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Lets assume this light pulse is similar to the atom and it will strike the top mirror 149,896,229 time's and flash the top light, once it is started. Then strike the bottom mirror 149,896,229 time's and flash the bottom light, which would equal 1 2 seconds. This continuing for eternity or until the tube is removed from between the mirrors.

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To clarify, 1 second will take place between the alternate flashes at the top and the bottom of the light clock.I stated the light pulse would have to travel between the two mirrors 149,896,229 times before the light would flash at T and then another 149,896,229 times before the light would flash at the second B.Get it through your thick skull that in this modification of the clock you can only see the light flash at the top or bottom of the light clock.Thus if you were standing at the first B when it flashes 1 second later the light at the first T would flash.The cycle with the light clock attached has traveled 149,896,229 meters from the time the light flashed at the first B.If you draw a straight line in the direction the cycle is traveling 149,896,229 meters and then extend a line at a 90 angle to that line at that point, 1 meter high you would be at the first T. Then draw a straight line from T to the first B, that line according to my angle caculator would be 149,896,229 metersWhen I convert meters to angstroms I get T to B as 1,498,962,290,000,000,000 angstroms.There is not 1 angstrom difference in the straight line from B to T as there is for the line from B to a point 90 below the first T.Thus the angle at the first B is 3.8223629703908244e-7 according to my angle calculator.If the light could travel in a straight line from the first B to the first T it would take 499,999,999.9999999 nanoseconds which is identical to the 499,999,999.9999999 nanoseconds for the cycle to travel the 149,896,229 meters. Show how much more time would be required for the light to travel in a straight line from the first B to the first T than would be required for the light to flash at the first T by the pulse bouncing between the two mirrors. I will wait for your details showing the math is wrong.Just because you can't read and understand the modified experiment does not make my math wrong.So let me clarify my example further.The light clock has 2 mirrors that are 1 meter apart.There is a black metal vacuum tube connecting these two mirrors.The light will pulse at the first B.The light pulse will travel between the 2 mirrors 149,896,229 times and cause a light to flash at the first T. 1 second has passed since the flash was observed at the first B.The light pulse will travel between the 2 mirrors another 149,896,229 times and cause a light to flash at the second B. 2 seconds has passed since the flash was observed at the first B.The light pulse will travel between the 2 mirrors another 149,896,229 times and cause a light to flash at the second T. 3 seconds has passed since the flash was observed at the first B.The light pulse will travel between the 2 mirrors another 149,896,229 times and cause a light to flash at the third B. 4 seconds has passed since the flash was observed at the first B.The time interval between light flashes is 1 second.You or anyone should be able to determine the distance from the first B to the first T using the baseline of the bottom mirror which is 1 meter from the top mirror.You have a triangle with 1 90 angle that has a baseline of 149,896,229 meters. Name that baseline b and the verticle distance of the mirrors of 1 meter as h you should be able to solve for a straight line from B to T. I did and I don't get 1 angstrom of difference between the b line and the BT line.1 angstrom is equal to 1/10,000,000,000 of a meter.So you don't have to go back to Message 786 I will present the diagram again. I will put the top and bottom of the light clock closer together to represent a better picture.

fig 5
                 T                               T
B                               B                                B
1 2 The #1  obsever is 100 meters from B and the #2observer is 100 meters to the right of #1
observer.

Z Observer is at a distance of 149,896,229 meters from the light pulse at the first B. 

God Bless,Edited by Admin, : Fix width.

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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."

Sorry to butt in, ICANT, but I wsa wondering if you could just answer the question I posed? It is very trivial compared to all the hard work you are putting in, but it does actually lie at the heart for everything you are talking about. If you get a moment, let me know...Edited by cavediver, : No reason given.

Hi ICANT, I agree with all of that. 'Tis as I understood it to be.Let me show a representation (not to scale, we don't have that much screen space) of the triangle we are talking about. The triangle will represent only the portion from the flash of the top mirror to the subsequent flash at the bottom mirror. First observe the direction of the line segment T-B2. You are trying to claim that the line is essentially vertical. Well does it appear vertical to you? Or does the line appear to be nearly horizontal? Is it that difficult to picture how it would appear if the drawing could be drawn to scale. That small angle at B2 is the angle you are calculating to be 3.8223629703908244e-7. Not the angle at T.Secondly, no photon, in particular not the one photon we are interested in, actually travels that path between T and B2. Instead, as you've stated, a photon completes 299,792,458 trips from one mirror to another in the interval between B1 (not shown) and B2, regardless of whether or not you can actually see them. Why don't you attempt to draw the triangle for one such path either from bottom mirror to top mirror to bottom mirror, ICANT. I'll even give you a hint. The height of the triangle is one meter.If you insist that it cannot be done, I'll do if for you.Added by Edit:Just to show how hokey your analysis and claim of "No Angle" are, let's imagine that the space ship is moving at only 3 mph relative to the observers. Point B2 would be a mere 4.4 feet (1.3411 eters) distant rather than about 150 million meters. The angle at B2 would then be [latex] tan^-1 (\frac{1m}{1.3411m}) = 36.71 degrees.Of course, arguing about the angle at B2 is pointless. (Well, other than for the purpose of pointing out your errors).We know the photon did not travel the path directly from T to B2 because the length of that path is only 149,896,229 meters (or, tee-hee, 4.4 feet when the ship is moving at 3 mph). Light travels 299,792,458 meters in 1 second.And you say my skull is thick? Edited by NoNukes, : grammarEdited by NoNukes, : ABE: Finish stomping a mud hole in this inane argument.

Hi NoNukes, In Message 796 I stated:

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Thus the angle at the first B is 3.8223629703908244e-7 according to my angle calculator.

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So I never said the angle at T was 3.8223629703908244e-7The angle at the first T would 89.999999617763790 on both sides of the T.The angle at all the B's will be 3.8223629703908244e-7.The angle at all the T's will be 89.999999617763790. I can not draw a triangle that the light will travel as you insist it would.I can draw a triangle of where the vacuum tube will be when the pulse strikes the top mirror and the bottom mirror and back to the top mirror.But that is irrelavant.The vaccum tube will move 1/2 meter while the pulse travels from the bottom mirror to the top mirror. The vaccum tube will move another 1/2 meter while the pulse returns to the bottom mirror.The pulse is always traveling verticle at a 90 angle relative to the bottom mirror at the same time that the vacuum tube is moving in the direction of the travel of the cycle.The light does not travel at an angle as you assert. That is why I enclosed it in a vacuum tube so there was no possibility of the pulse traveling at an angle. If it did it would travel 1.0307764064044151 meters instead of the 1 meter it actually travels between the top mirror and the bottom mirror.Since the pulse is in a vacuum tube that is forcing the pulse in the direction of the travel of the cycle the pulse is in an accelerated frame, as the pulse has outside force exerted upon it.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 cavediver, Your question. Taq never said relative to what.I assumed since he was using a car he was talking about relative to the earth, and the car traveling on something like the salt lake flats.But what does that have to do with the dropping of the bag of peanuts from the inside roof of the car to do with the light pulse released from a laser pen that was mounted flush with the top of the roof of the car to the exterior. Which was 1" from a pole mounted on the roof of the car that 4 foot up on the pole a detector was mounted to catch the pulse when released from the laser pen.You said the pulse would strike the detector.Could you explain how this pulse that is in free vacuum with nothing to force it in the direction of the travel of the car could strike the detector which has traveled 2 feet from where it was when the pulse was released from its source?The pulse would have to travel 4.47213595499958 feet at a 26.56505117707799 angle from the point of release to strike the detector.How is that possible when the pulse is released at a 90 angle to the travel of the car?As I understand postulate #2 the pulse can not add the forward motion of the car to the direction of the light pulse.Is that true?The bag of peanuts does add the forward motion of the car and thus will strike you in your face if you hold them directly above your face and drop them from the ceiling of the car.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."

It's possible because the emitter is also attached to the car.I continue to be mystified by how you don't seem to understand this, and I wonder if it has something to do with your ignorance of calculus. I know I couldn't think about continuous processes like motion in a really continuous way until I learned calculus. I had to think about it the way you seem to be doing - first this moves a little bit, then that moves a little bit, then back to this - in discreet steps. Calculus helps you think continuously. That has not ever been what postulate 2 says.

Ah, and here-in lies the issue. It does. Without question. Travelled 2 feet relative to what? I don't believe you. Relative to me, it hasn't moved at all. Relative to my cousin it has actually gone backwards 3 feet. And my sister swears blind that the car has actually moved sideways. No it wouldn't. I told you, as far as I am concerned, the car isn't even moving.

Sigh.I assume by your handle that you dive. But fathoming these depths...?

Hi ICANT, Why can't you draw the picture? The picture as you describe is very easy to draw (at least schematically), and I'll draw it to scale for you. Unfortunately, the drawing won't reflect reality as presented in the reference frame of observer #1 (or in any other reference frame). That is true in the reference frame of the cycle rider. But if your statement were true in the reference frame of observer #1, the photon could never advance horizontally along the flight path of the cycle. The photon cannot advance horizontally by making hundreds of millions of exactly vertical motions.More to the point, you can now see that your analysis showing that tiny angle at B2 was completely pointless, right? It has no relevance to anything. Even you now admit that the photon does not travel that path. Since you last looked at my previous post, I've shown what happens when the space cycle is moving only 3 mph relative to earth. As it turns out, the slower the space cycle moves, the larger is the angle at B2.As promised, here is the picture based on your description. If the software here renders properly, the picture should be to scale. I've shown the path of the photon from B1 to T and from T to B2 as straight lines. Perhaps you disagree with that proposition that the photon travels in a straight line along those paths. Perhaps you think the photon is banging around in the tube. If so, I'm wondering why you made the walls of the tube black. A black interior will absorb the photons. Why not give the tube a white interior or a mirror finish?In any event, regardless of the path you, think the photon takes to get from B1 to T, we seem to agree that the photon does somehow get from B1 to T. The straight line segments shown are the absolute, shortest possible path that the photon could take. If you propose that the path is somehow different, you're simply going to be further along the wrong end of the analysis. Any other path will be longer, thus compounding the problem I'm going to discuss below.According to you, the distance between x and B2 is 0.5 meters. I did not label the picture with that value because I know that the number is wrong. But we'll work with 0.5 meters for at least a bit.So how long does the photon take to get from B1 to T. You claim that one second passes from the time the of a flash at the top mirror to the subsequent flash at the bottom mirror, and that the photon travels between the mirrors 149,896,229 times during that interval. So apparently the trip from B1 to T (which is half of a round trip) must take 1 second/(149,896,229 round trips) * 1 half trips/2 round trips = 3.3356 nano seconds.But what is the distance between B1 and T. The minimum value, (which assumes that the photon travels in a straight line between B1 and T) is: Of course you and I both know that light cannot travel that distance (or any longer distance based on whatever goofy path you think the photon takes) in only 3.3356 nsecs. The speed of light would work out to be at least 1.1180meters/3.3356 nanoseconds = 333,134,684 meters per second in that scenario, said speed being impossibly greater than c.Rework the problem with distances B1-X and X-B2 being 0.57735 meters and we can confirm that the photon can (at least potentially) reach point T from B1 without exceeding the speed of light. Of course the time between clock flashes (corrected for light travel time if necessary) will no longer be 1 second in the observer's reference. That's what we call dilation.Uh-oh... Inanity ahead... So you now believe you can change the speed of a photon by accelerating it? Since you have no science education to speak of, I guess your mind is free from constraints and you can just make stuff up. Sorry ICANT, but you are grossly misreading postulate #2 as well as doing bad arithmetic.Postulate 2 says, in essence, that the speed of light in a vacuum is a constant "c" as measured in any inertial reference frame.Whatever you think is going on inside that tube, to anyone outside of the tube, and presumably that's all of the observers in the problem, if we are in an inertial reference frame, we will measure the speed of light in a vacuum to be "c". Always. That's what postulate #2 requires.It is irrelevant whether or not the inside of the tube is an inertial frame, because it is the speed as measured in the inertial frame of the observer that counts.Plus this is yet another example of your failure to understand inertial reference frames. The photon is "in" every frame, inertial or not. Nothing is ever solely in its own frame. That idea is just mule stupid. So is the idea that you can add speed to a photon or any other object traveling at C by applying force to it. You might be able to change the observed value of the speed of light by accelerating yourself, but not by applying force to a photon. Sorry ICANT, but there is no possible path between B1 and T that will be as short as 1 meter given that B1 and T1 are separated by at least 0.5 meters horizontally and 1 meter horizontally.Further, 1.03078 meters corresponds to a speed of 0.25c and not 0.5c. Your 149,896,229 distance between top and bottom mirror flashes implies 0.5c.

No. It's just general ignorance. I didn't gradudate HS and failed Algebra 3 times (only because I didn't attend class, which was also the reason I dropped out) and I have a much firmer grasp on the concept than ICANT seems to (thank you Nova Science Now and space documentaries!). Perhaps it is because physics/cosmology/astrophysics/science in general has always excited me and I actually want to learn....unlike some folks.Edited by hooah212002, : No reason given.

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"Why don't you call upon your God to strike me? Oh, I forgot it's because he's fake like Thor, so bite me" -Greydon Square

Yep, given enough helium, I'm happy dropping into the abyss...But you are spending pages and pages trying to explain all of this to ICANT, and he doesn't even appreciate relative motion. Do you not think that this lesson is vital before spending any time on anything else? The laser must shine vertically upwards by simple symmetry. Until you can get someone to appreciate this symmetry, the rest is pointless.

Hi cavediver, You quoted:

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Could you explain how this pulse that is in free vacuum with nothing to force it in the direction of the travel of the car could strike the detector which has traveled 2 feet from where it was when the pulse was released from its source ?

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Emphasis added.I take that to mean the car with the pole with the detector on top moves 2 feet relative to the position the car was in when the light pulse was released from the laser pen. If that was too complicated I apologize.

4' high
lp   D
|   /|
|  / |
| /  |
|/   P
E    C  

E = emitted
lp = light pulse
C = Car
P = Pole on car
D = DetectorLaser pen in the roof of the car is at position E when the light pulse is emitted.The car then moves 2 feet relative to that position by the time the light pulse reaches 4' from the position emitted.If the light pulse travels in a 90 angle to the travel of the car which the laser pen is pointed when the light pulse is emitted the light will miss the detector.The only way for the light pulse to hit the detector is for the light pulse to travel at the angle between E and D.Unless the car does not move the 2 feet from the position the light pulse is emitted.You can declare the car traveling at zero v in your frame all you want if the car moves 2 feet at 0.5 c relative to the position the light pulse was emitted the light pulse will miss the detector.

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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."

What position where the light was emitted? That's the same position it is now!Ok, yes, there is another frame where the car appears to be moving at 0.5c in the forward direction, where the car moves two feet between pulse emission and pulse absorption at the detector, but that is one frame out of the infinite number possible. Why does that frame dictate what the light pulse does?Why not the frame where the car is moving at 0.5c backwards, and the car moves 2 feet in reverse between emission and absorption?Or the frame where the car is moving at 30mph forwards? Or 100,000mph sideways to the left?Which view is the correct one so that you can say with certainty what the pulse will do from the perspective of someone in the car?

It isn't strictly for ICANTs benefit. Apparently others are interested in the discussion.I don't believe that ICANT will ever get it, because he does not want to do so. Once I accepted that, at least I stopped being frustrated.Edited by NoNukes, : No reason given.Edited by NoNukes, : No reason given.

Hi NoNukes, But when you streach out those triangles to their proper scale you will have 149,896,229 meters between B1 and x.You will also have 149,896,229 meters between x and B2.The straight line from B1 to T will not be 1 angstrom difference in the straight line from B1 to x. My calculator could be wrong but I don't think so.Now if you were trying to draw a triangle the pulse would travel according to NoNukes everytime the pulse hits the top mirror and the bottom mirror. The distance from B1 to x would be .5 meter.
From B1 to B2 would be 1 meter if you are trying to show the distances each pulse would travel between strikes on the mirrors.Your problem is that you do not account for the motion of the clock. But the pulse does not travel from B1 to T.The pulse travels in a vacuum tube mounted between two mirrors spaced 1 meter apart. The tube is fastened to the center of the mirrors. The pulse will strike the top mirror exactly 149,896,229 times and the light will flash at the first T. 1 second has passed since the light flashed at B1.While the light pulse is bouncing between the two mirrors, the mirrors will move 149,896,229 meters in the direction the cycle is traveling 0.5 c.It takes 1 second for the light pulse to reach T from B1. But the light pulse does reach T from B1 traveling in the vacuum tube bouncing between the two mirrors 149,896,229 times.Exactly 1 seconds passes. as it takes 3.33564095198152 nanosecs. for each of the
The light pulse striking the top mirror 149,896,229 times and the bottom mirror 149,896,229 times.We have 299,792,458 meters traveled at 3.33564095198152 nanoseconds. = 1,000,000,000 nanoseconds which equals 1 second.Can the pulse traveling traveling in the tube bounce between the two mirrors 149,896,229 times while the tube travels 149,896,229 meters and cause the light to pulse?Will that take 1,000,000,000 nanoseconds which equals 1 second?Anyway you slice it and dice it that tube will be under the T exactly 1 second after it was over the first B in my example.

fig 5

                 T                                T
B                                B                                B     

You don't have to increase the speed of the light pulse when it is accelerated by the forward motion of the cycle the light clock is attached to drags the tube the pulse is traveling up and down in is traveling at 149,896,229 meters per second.The light pulse is being accelerated sideways just like you are when you go around a curve in your car, or just like I would be in my experiment where I made a long u turn around planetx on my bike while maintaining 0.5 c speed. So what about when it is being forced sideways as it travels in a verticle motion bouncing back and forth between two mirrors in a vacuum tube? And in my thought experiment the light travels back and forth between the two mirrors striking each 149,896,229 times and then causing a light to flash.The speed of the light pulse is exactly 299,792,458 meters per second which is the speed of light or c if you prefer.The light will flash at the first B then the pulse will strike the top mirror 149,896,229 times and the light will flash. The pulse will then strike the bottom mirror 149,896,229 times and the light at B2 will flash. The pulse will then strike the top mirror 149,896,229 times and the light at T2 will flash. The pulse will strike the bottom mirror 149,896,229 times and the light at B3 will flash. Four seconds has passed since the light flashed at the first B.The cycle has traveled 599,584,916 meters.The light pulse has traveled 1,199,169,832 meters up and down in the vacuum tube between the two mirrors while being draged 599,584,916 meters. Check your assertions you are shortening up the distance between B1 and B2 which is 299,792,458 meters. If you take your triangle and use B1 and x then T for the top as you have drawn. If the distance between B1 and x = 0.5 meters the straight line beteen B1 and T would be 1.03078 meters. Which is what you say the pulse would have to travel between the bottom mirror and the top mirror.But the pulse in the vacuum tube fastened too the two mirrors will not allow the pulse to travel in that direction. It will only allow it to travel up and down in the vacuum tube while the tube is draged the .5 meter that you want the pulse to travel while going from the bottom mirror to the top mirror.Whenever you figure out how the light can travel up and down and get to the top mirror faster by going up and down between the mirrors while being draged alone by the movement of the light clock
than it could going at an angle.You would be able to answer many questions that are asked by many.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."