Existence

We can use salt flats, that is fine with me.So we have many different frames of reference that we can choose from. We can choose the driver's frame of reference, the person standing on the salt flats, or someone standing on the surface of the sun. Let's see what each person observes.The driver observes that the pen laser, car, and detector are not moving or accelerating. An accelerometer does not measure any acceleration (other than the gravity of the Earth) in any direction. The person standing on the salt flat would observe that the Earth is at rest, with no acceleration in any direction. The person standing on the Sun would observe that both the driver (and the rest of the stuff on the car) are moving as is the person standing on the salt flat.For now, we are talking about the driver. In his frame of reference, neither the pen laser, detector, or car are moving relative to each other. If nothing is moving in this frame of reference, why should the light miss the detector? How is it possible for the pen laser to miss when the pen laser and detector are not moving relative to each other? The car is not moving relative to the pen laser and detector. There is no motion to add to begin with.ABE:Here is another way to look at the problem. Let's pretend that the Earth is not revolving. We put the car on the salt flat with the driver and the equipment described above and put it in neutral. Next, we start spinning the Earth so that it achieves 0.5c relative to it's previous motion. Since the car is in neutral, the driver will see the Earth spinning by at 0.5c.ICAN'T, in this scenario, will the laser light hit the detector?Edited by Taq, : No reason given.

ICANT my man. Are you kidding me?You are spouting nonsense. The triangle I've drawn shows a single round trip for a photon between mirrors. It is one of hundreds of millions of triangles actually traversed by a photon between flashes emitted by the bottom mirror.In other words, you don't stretch out the triangle I presented. You replicate it millions of times per second.On the other hand, no photon of interest travels any portion of your 149,896,229 meter triangle, so why should I care about its dimensions? What relevance does your big triangle have? Does light travel only 149,896,229 meters in a second. Nope! So stop showing me triangles with those dimensions as though they meant something. Is it really? I think I've demonstrated that a photon traveling that path would not comply with postulate #2. As viewed by observer #1, the distance from B1 to B2 is 1.1547 meters which does comply with postulate #2. Really? So when I show mirror T as not being above where mirror B was at the time the photon hit mirror B, what effect am I accounting for? That's right, I'm accounting for the relative motion between the mirrors and observers #1 and #2. Wrong. I did the calculation for you in my last response. The correct answer is 1.1180 meters. Your answer is for x=0.25 meters.Edited by NoNukes, : tags and shorten responseEdited by NoNukes, : Tweak

Hi NoNukes, The problem is you used B1 and B2 shortly after I had a very lengthly post using B1 and B2. My B1 and B2 was 299,792,458 meters apart. Which in Message 798 you presented a nice triangle not to scale but a fair picture of what the triangle would look like.You then presented another triangle in Message 804 and did not supply any distances.You then obfuscate by saying: You say we will use the 0.5 meters for at least a bit.In the next sentence you say: Here you lay out what you think my claim is.Then you follow that with: Then you explain to me it takes 1 second for the trip from B1 to T.You then further obfuscate by adding, "(149,896,229 round trips) * 1 half trips/2 round trips = 3.3356 nano seconds".I think maybe here I should call upon Einstein for one of his quotes.

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If you can't explain it simply, you don't understand it well enough

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What makes you think you understand what you are presenting when you revert to such obfuscation? Do you ever read anything before you preach your sermons?In Message 796 I presented the details of what the modifications would produce.

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So let me clarify my example further.

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The light clock has 2 mirrors that are 1 meter apart.

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There is a black metal vacuum tube connecting these two mirrors.

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The light will pulse at the first B.

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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.

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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.

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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.

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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.

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The time interval between light flashes is 1 second.

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In Message 786 You can find the modifications to the light clock.I would mention one thing here that you questioned.

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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 second. This continuing for eternity or until the tube is removed from between the mirrors.

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This means that the light clock was modified in such a way this light pulse could strike the top mirror exactly 149,896,229 time's and flash the top light and then strike the bottom mirror 149,896,229 time's and flash the bottom light. Alternating in this fashion forever.There can be no deviation from that process for eternity unless the tube is removed from between the two mirrors.So during the time required for the cycle to travel 149,896,229 meters the pulse will strike the top mirror 149,896,229 times and cause the light to flash.If you disagree please present your argumentation. Do you need to get your glasses changed?I can't see the fixed mirrors in a light clock not being verticle to one another at all times. You are correct, that is what I get for copying and pasting instead of doing the math again. Sorry about that.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 Taq, When the light pulse leaves the top of the laser pen it is not moving relative to the car, laser pen, or detector.The pulse is moving at a 90 angle to the travel of those items when it leaves the laser pen in the direction of the position the detector is at, when the pulse is released.It is just that when the pulse reaches the position the detector was when the pulse was released from the laser pen, the detector is not there as it has moved 2 feet. The car, laser pen, and detector is moving at 0.5 c relative to the ground underneath the car.The reason the pulse misses the detector is that the pulse is not traveling c relative to the car, laser pen, or detector. It is traveling in the direction the laser pen was pointed when the light pulse was released. Just as soon as the light pulse is released at a 90 angle to the travel of the car, laser pen, and detector, which are traveling 0.5 c relative to the ground underneath the car, the light pulse is in free fall.For the light pulse that is attached to neither the car, laser pen, or detector to strike the detector it has to add the forward motion of the car, laser pen, and detector. Is the car parked on earth doing zero meters per second relative to the earth?If the car is parked on earth the driver will not see the earth spinning no matter how fast you make it go.If the car is parked in empty space above the earth the driver would see the earth spinning. If the car is parked on earth the pulse released from the laser pen will not strike the detector.If the car is parked in empty space above the earth the pulse will strike the detector every time it is released from the laser pen.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, OK so in this frame will the pulse released in free fall by the laser pen reach the detector or will it miss the detector?If you disagree as your statement implies then explain:Explain how the pulse is absorbed by the detector when it is 2 feet from where the light pulse passes it.What causes the pulse which is in free fall, attached to nothing to travel at a 63.43494882292201 angle in order to strike the detector?The distance the pulse would have to travel to strike the detector is 4.47213595499958 feet.Even though the car would move several inches during this added time the pulse would still strike the detector as the detector was 9 inches long extended towards the laser pen in relation to the pole the detector is mounted on. So now you have a frame that passes the car moving at 0.5 c + 30 mph. Is that correct, as you did not explain what you were talking about? Could you explain this one as I can find no example that covers this situtation. In the frame where an observer sees the car move 2 feet backwards due to the forward motion of that observer the pulse will miss the detector as the car has moved two feet from the position it was in when the light pulse was emitted into the vacuum at a 90 angle the car is traveling 0,5 c in.The same applies to the frame where the observer sees the car traveling 0.5 c + 30 mph relative to the observers frame. The pulse will still miss the detector.If it is possible for the car to travel at 0.5 c and be observed to go sideways at 100,000 mph the pulse would still miss the deteftor as the detector would move 2 feet in the forward direction of the travel of the car as well as the distance sideways the car would travel.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 you are driving down the road and shine a laser beam perpendicularly out of the side window, do you really excect the beam to curve backwards a little bit because you're moving forward!?Or can you realize that the laser beam will remain perfectly straight?Now if you had a dector on a pole...

It's not two feet from where the light pulse passes it; it's directly in the path of the light pulse because when the light pulse was emitted, it was emitting towards the detector and the detector hasn't moved relative to the light pulse's origin. Because that origin is on a moving vehicle, same as the detector. They're all moving at the same velocity so they don't move relative to each other. Imagine that the light clock is stationary and you are observing it from a car. The light clock sits there, happily ticking away, the light pulse bouncing back and forth between two perfectly aligned mirrors. It never misses because the mirrors are always where they're supposed to be.Now, you drive past it in your car right as the light pulse leaves the bottom mirror. Does the pulse suddenly escape and break the clock? After all, relative to you the top mirror is suddenly two feet (or whatever) from where it was when the pulse left the first mirror, because the entire light clock has suddenly gained a velocity relative to your perspective. Of course, relative to me standing near the clock, you're the one who is moving - the clock is stationary.How can you observe the light pulse escape the clock while I observe it continue to tick? That's nonsense. Obviously, regardless of your velocity you continue to observe a ticking light clock, because relative to you the light pulse has whatever velocity it needs to gain to keep up with the rest of the clock.

Hi ICANT, Yes. I can see that the nomenclature would present some chances for confusion. But during my post, I clearly indicated why the base was unlabeled, and that I would assign it a starting value of 0.5 meters. What did you think that meant? I also explained exactly what B1 and B2 represented.None of that changes the fact that your big triangle is worthless. The line connecting the flash at the top mirror to the flash at the bottom mirror does not have any meaning and neither does the angle that line makes with the horizontal. You came up with that blather on your own.I also don't agree with the times and distances you give on that drawing. In fact there is no inertial coordinate system in which the flashes could be measured as 149,896,229 meters apart and in which the flashes occur at one second intervals.But that's actually the issue in dispute, so I'll leave it open until we've gotten further along in this triangle analysis. I apologize for the confusion. But you have some responsibility to use your wits.Seriously dude, when the difference is between 3.3356 nano seconds (or thereabouts) and 0.5 seconds (or thereabouts) I think we could easily sort out what refers to what with just a tiny bit of thought. I don't see the point of this rant. I completely agree with the above. I just happened to believe that the tube is unnecessary, and that a black interior is a rather silly choice given your (mis) understanding of what the tube is for.Of course I completely disagree with your assessment of the time interval between flashes, and of the distance between flashes. I have explained several times why the observer on the space cycle and observers #1 and #2 reach different conclusions. Postulate #2 leads directly to that result. We agree on that. But if T labels the mirror in its position at the time of a flash, the position of the lower mirror at that same instant is not of interest. I didn't draw it. Similarly, you don't show the 149, 896, 229 different locations for the top and bottom mirrors in your drawing. I hereby award you 500 NoNukes points for taking ownership of an error. I think you are up to 1500 points for the entire thread, which ain't so good given the number of errors you make and then silently ignore when you are called on them.Now that we've overcome my confusing notation, could you tell me how that photon gets from the bottom mirror to the top mirror (which has moved horizontally by at least 0.5 meters) without traveling at least 1.1180 meters as measured in observer #1s frame of reference? How is that distance covered in only 3.3356 nanoseconds?

Hi NoNukes, The answer to your question is very simple but since you are so indoctrinated into your beliefs I don't know if you can fathom the answer.But I will try to explain it in simple English.The pulse starts at the bottom mirror traveling at 299,792,458 meters per second.At the same time the vacuum tube is traveling at a 90 angle to the pulse at 149,896,229 meters per second.While the pulse is traveling 1 meter to the top mirror in the vacuum tube, the vacuum tube travels .5 meter at a 90 angle to the pulse.When this has transpired 149,896,229 times the vacuum tube has traveled 149,896,229 meters in the direction the cycle with the light clock is attached to has traveled. The pulse has traveled 299,792,458 meters between the mirrors, striking each mirror 149,896,229 times. 1 second has passed.I will give you an experiment you can use to prove to yourself what I am saying.Find someone who has a water level that has 12" tubes on each end.
If you don't have access to a water level you can use a clear plastic pipe with 2 elbows and a 12" tube on the top of each elbow. Fill the tube so that when level the water in each verticle tubes is a little less that half full.Lower the right tube until there is no water in the left tube.Draw a line 1' long on a table and mark it zero at the left end and at 6" and 12". Place the left tube at zero on the line.Have someone move the tube towards the 6" mark while you are raising right tube so the water rises in the left tube and reaches the top at the same time the left tube reaches the 6" mark.Have the person continue moving the tube while you lower the right tube so the water is out of the tube by the time the left tube reaches the 12" mark.Don't worry about the time it takes for the experiment only that the left tube reaches the 6" mark and is full of water and reaches the 12" mark and is empty.This is the same thing the pulse does in the vacuum tube, but it does not need any help as that is what it is designed to do.The water at no time goes at an angle.Neither does the pulse as it travels between the mirrors.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 crash, That is correct.The problem is that it has moved 2 feet in relation to where the pulse was released.If the pulse goes in a straight line from the point emitted it will miss the detector. Because the detector and the laser pen has moved 2 feet from where the laser pen released the light pulse into a vacuum in which the light pulse is in free fall.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, because where the pulse was released has moved 2 feet as well, relative to a stationary observer. Absolutely wrong. It'll hit the detector because relative to the pulse's point of origin, the detector hasn't moved at all.When you specify that something has moved or that something is stationary, you need to specify in relation to what that thing has moved or remained stationary. When you get into that habit you'll see that there's absolutely no problem with a moving light clock.

Hi ICANT,I did notice your sleight of hand here: And since the photon is in the tube, the photon also must travel the same 0.5 meters in the direction perpendicular to the tube axis while traversing the tube vertically even if you don't think that it does so in a straight line. Otherwise it would exit the tube.So being in the tube does not change the fact that the photon's path, as plotted in observer #1's coordinate system is greater than 1.1180 meters, tube or no tube.Consider the picture below. Since my nomenclature is confusing, I'm not going to supply any nomenclature and very little in the way of explanation . I suspect that you can figure it out yourself. Feel free to draw, or imagine in any additional light tube positions as you might find helpful. Your double talk was a nice try though. Or did it actually seem to make sense to you when you typed it? Can you describe, while looking at the above drawing, how the photon gets from the initial position shown in the drawing, to the final position shown without traveling at least 1.1180 meters? Because such a trip appears to be impossible.Imagine that you secretly had super speed, and that you ran home from work at 600 mph, covering a 10 mile (as the crow files) distance in 1 minute. If a friend called you from work 2 minutes after he saw you leave work, could you explain your super fast journey by claiming to take a short cut? Or would your friend rightly conclude that you must have traveled at least 300 mph?That's analogous the situation we have here. If the starting and ending points for the same photon are more than 1 meter apart as they are here, I don't care what path you claim that the photon took. The photon cannot traverse that distance in only 3.3356 nanoseconds without exceeding the speed C in a vacuum.And no I'm not going to consider that photons can be forced through a tube under hydrostatic pressure to exceed the speed of light.Unlike the case with photons, water is a essentially a non compressible fluid. When we put one water molecule in one end of the tube, a different water molecule flows out in short order. Photons on the other hand don't occupy space. We cannot stuff a light tube full of photons. Further, we are not discussing a frame of reference in which the water pipes themselves are moving as we are with the light clock. Postulate #2 says nothing important about the speed of water molecules in empty space. Water is not constrained to follow a geodesic path in space. I suppose I could come up with a few more differences if your experiment was even close to being relevant.That the light tube is a mere concession to your own inability to grasp reference frames in the first place. It's actually completely unnecessary, but you seem to like it...On the other hand, in the cycle frame of reference, the tube is at rest, and light does travel vertically up and down the tube. Of course, in that coordinate system, the tube, the photon, and the space cycle are all at fixed horizontal coordinates. Don't bother trying to understand all that.

Here you go again. Moved 2 feet according to whom? According to me it has not moved. According to my sister, it has moved but not by the amount and not in the same direction sa you are claiming. Great, so you have found someone (Nonukes was it?) saying that it has moved two feet forwards, but so what? What makes him God of determining the relative motions of objects?

Hi cavediver, According to the pulse that is free falling in a vacuum at a 90 angle to the travel of the car. And you are doing what relative to the car to declare that the car is stationary relative to you? And your sister is doing what relative to the car to declare that the car is not moving at 0.5 c in a forward direction relative to her? And what makes you God to decree that the light pulse will always hit the detector?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."

I don't understand. The pusle is emitted from a laser in the car - relative to the laser, the car is stationary. So from where does this idea about 2 feet come? I'm just a 1/2 mile away. And the car is not moving as far as I am concerned. When I first saw the car it was moving towards me quite quickly, but I just accelerated a bit to match speed. Well, if stating basic facts is all it takes to be God, I guess I've been missing a trick all these years...