Existence

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.

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.

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 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 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.

Hi ICANT, Then forget the angles. Explain how the red dot gets from start to finish in 3.3356 nanoseconds when the distance between the two points is 1.1180 meters. Show me the shortcut. Which is exactly what is shown in the picture. Only one tube is present, but the tube is shown in different places. Apparently, you need me to explain things in baby step. No ICANT. There is only one tube in the drawing. The tube is shown in a number of the locations it will occupy as the space cycle moves along in observer #1's coordinate system. And why do your other remarks in your post acknowledge that you were well aware of this fact? (For example your discussion of the clear tube)Your water example is irrelevant. The physics of a water molecule traveling in a tube full of water is nothing like the physics of a photon traveling in a vacuum. I'm not going to bother sorting out whether you're even right about the water results. Sure ICANT. What you've just said is that I've accurately described reality, but because the tube is not transparent, you can deny that the events you admit happened actually occurred. Nice work my man.And given the fact that the tube really does not do anything anyway, you've all but admitted that you are wrong.But forget the obscured path of the photon and forget the angles. The bottom line is this. The photon starts at a first point and somehow ends up at a second point which you say is one meter below and 0.5 meters to the right of the first point. The distance between those points as measured in the observer's frame is inescapably 1.1180 meters and there is no route between start to finish that is smaller than that distance.I challenge you to draw a path for the photon that starts at the first point shown in the drawing, ends at the last point shown in the drawing, but that is less than that 1.1180 meters. If yd.ou don't like my path, draw your own. Use as many or as few tubes as you like. Use whatever angles you want to use. Draw in a stair case or a zig-zag bath bouncing of the side of the tube, but somehow you need to get shrink the bath down to 1 meter or you've failed.Or not. Your call.Let me help you out of the silly dilemma that causes you to say such ridiculous things. Simply deny that postulate #2 is correct. It is your insistence that postulate #2 is correct that is the albatross around your neck. It should be increasingly clear to you that postulate #2 is completely consistent with SR. Holding unto it is completely inconsistent with denying SR, so just let the thing go. You were fine with doing just that 400 messages ago.

No ICANT, you did not draw the path of a photon between the top and the bottom mirrors. You said that you could not draw such a path.Instead you drew that large triangle with the 150 million meter base. But we know that hypotenuse of that triangle cannot be the path of a photon because if it were, the photon would only be traveling 1/2 of the speed of light as measured observer #1's frame.We also know that the large triangle does not show the path of the photon because it does not show the photon returning to the top and bottom mirrors just short of 150 million times for each mirror. Yet you've insisted that this number of round trips is exactly what occurs during a one second time interval. Those up and down excursions that you fail to show are what makes the actual distance traveled by the photon consistent with a speed "c".So far, you have yet to draw or describe any path for a photon that is both consistent with postulate #2, and that allows the photon to return to a moving top and bottom mirror 150 million times. I don't expect you to the path for 150 million round trips, but surely asking you to draw one or two of them (or even a single trip from top to bottom mirror) is not unreasonable.We should be able to take whatever horizontal advance you show to occur in a single round trip path, and multiply it by the number of round trips you admit occurs and the result should match the distance you show in your large triangle diagram. So far, you refuse to admit that a single trip up and down covers any horizontal extent at all. That produces an inconsistency with your claim that 149+ million meters horizontal meters were traversed. How can the bottom mirror flash be 149+ million meters away form the top mirror flash if none of the up down movements covers any horizontal distance.You seem to be incapable of understanding how coordinates and events happen in different inertial frames. Now that your unneeded tube is in place, you've "locked in" at least partially on events as they appear in the space cycle/tube inertial frame of reference. Yes, in that frame of reference the photon does move strictly vertically, and no horizontal distance is covered. But in that frame of reference, the space cycle also does not move, and the clock flashes do tick off 1 second intervals. But why are you showing a 149+ million meter horizontal excursion for the photon. That measurement is not consistent with a space cycle/tube frame of reference. Yes, ICANT, but as measured in observer #1s reference frame, the photon must simultaneously move horizontally with the tube and the cycle or it would be left behind.The short answer: Simultaneous horizontal and vertical motions constitute diagonal motion. It is just that simple.Simple question. Can you move forward one full meter without getting out of your car seat? Yes you can if the car moves that full meter. Well the photon "trapped" in the metal tube accomplishes the same thing in the same way.Imagine bouncing vertically on a trampoline on a flat car moving at 60 mph. Keep yourself in a vertical tube if you like. Is it too hard to imagine that from a ground observer's frame of reference, you move horizontally, while also moving vertically. Well the trapped photon accomplishes the same thing.The mistake you make here is to combine observations in the space cycle frame with observations in the observer's frame. In the space cycle frame, the light beam does travel perfectly vertically. But if the photon failed to move horizontally in observer 1s frame of reference, while the space cycle does move as measured in that frame, then the photon would be left behind. We know that does not happen. I understand that perfectly. But in observer #1s frame of reference the tube also moves. The photon moves along with the tube while moving up and down along the tube. I did not merely hint at the answer your question. I've expounded on it at length. I've just made yet another attempt to explain things up for you. The attempt will fail, of course. You will never admit to understanding.I can make other attempts using different illustrative examples. Others have done so.In the water experiment, no water molecule is observed to travel the length of a horizontally moving tube. If we did observe the water molecule from a frame in which the piping is moving horizontally, then the water molecules will also have to move horizontally or they cannot stay in the tube. However, the water molecules are not required to move at C as measured from an inertial frame In fact they cannot move at that velocity in any frame of reference.I don't care how cheap the experiment is to perform. If it does not model the situation under discussion, then the experiment is not relevant.Edited by NoNukes, : No reason given.

Hi ICANT, Already answered. The photon pulse moves up and down while also moving horizontally along with the tube. The combined motion is along a diagonal. Yes, it does, but only as measured by observer's moving along with the space cycle. Other observers might determine a different duration and path for the pulse movement. I'll allow you to give my answer to this question. I've played this game long enough given that you did not respond a single one of my questions. You have no response, because your position is untenable; it is not even self consistent.

It should be abundantly clear by now that you are being asked to draw the path of the photon including the 'dragging' you mention above. Since you've acknowledged that photons cannot be accelerated such that the speed of the photon is increased above c, you are also being asked to show that the resultant speed of the 'dragged' photon does not exceed "c".Is that clear, or do you need yet more coddling? No ICANT, things are not independent of what is observed in a given reference frame. In fact there no preferred frames. Your choice of the space cycle frame of reference is completely arbitrary.The argumentation has already been given to you at least a dozen times. Speaking of events as measured in the inertial frame in which observer 1 is at rest, the photon must move horizontally while moving up and down the tube. Postulate 2 requires that this net motion be exactly 'c' given a vacuum in the tube. Yet, the net distance traveled by the photon is clearly greater than 1 meter.Why is the interior of the tube black? Are you ever going to explain that ridiculous choice. That's what we are supposed to be doing. But you ducking questions as you 'prefer' not to answer is not conducive to having a conversation. Your own version of events requires the speed of light as measured/observed/calculated in observer #1s frame of reference to greater than c, meaning that you are not accurately describing said events. I mean whichever object under discussion that is moving vertically while also moving horizontally such that it even makes sense to talk about a combined motion. The obvious 'dragged' object of interest that you deliberately avoided mentioning in your short list. Why don't you take a shot at answering your own question?This kind of silly behavior is of course what I refer to when I denounce playing stupid games. Edited by NoNukes, : No reason given.

Hi ICANT Okay. I'll serve up some coddling. I apologize in advance for any thing I say that under estimates your knowledge. I expect that I'll probably overestimate. Exactly right, ICANT. But the point is that you cannot cover increased distance without increasing your speed. Your explanation of acceleration is fine. But it is unhelpful to explaining your way out of the contradiction you've created. You cannot travel further than 100 miles in an hour without increasing your speed.You say that the photon is dragged. So the question remains how does the photon follow the dragged path without exceeding the speed "c". You have yet to answer that question. Draw the dragged path, or just acknowledge that I've already done so. Then show that the dragged path is 1 meter or less. But you cannot do that, can you? Yes. I agree with what you say above. But being constrained to stay in a one meter tube does not prevent the photon from traveling more than one meter during the trip from top to bottom of the tube. For example if the tube is moving in a given reference frame, the photon must travel along with the tube while also moving up and down the tube. This must be among the more ridiculous things you've written in this thread. It's surely in the top twenty. I'm seriously contemplating making a list.No ICANT, the tube would not be required to be longer than 1 meter or to swing back and forth. As you've indicated, the tube is vertical. But the tube does not stretch from the lower mirror to a future position of the top mirror.In order to contain the photon while the photon moves along a diagonal, all that is required is that the vertical tube move horizontally at constant speed as the photon moves up and down along the tube. No need for any to and fro tube swinging or stretching. This should be actually an easy thing to picture, so I do not accept that understanding this is beyond you. I'll keep trying. I've got some alternate methods of explaining the phenomenon in mind. I'll try them if I see further denials on your part.Seriously, forget the photon. As you've explained it, the tube must swing back and forth simply to stay attached to the two mirrors. You are simply ignoring the time component. The mirror stretches vertically from the position of the bottom mirror at one given instant in time to the position of the top mirror at the same instant in time. Explain how being black helps contain the photon in the tube. If the interior of the tube is not black, what difference does the color of the tube make to the photon. Why not just admit that the choice of color was hastily made? Edited by NoNukes, : No reason given.

Not quite right. Although it seems counter-intuitive, light always travels at 'c' as measured in any inertial reference frame (at least when propagating in a vacuum, but approximately so even in air). Resolving the issues presented by this experimentally verified fact is what special relativity is all about.The only way objects enter or leave a reference frame is when they are created or destroyed.

I think it is important to be clear about this. The laws of physics are the same in every inertial frame as is the speed of light in a vacuum as measured in the coordinate system of any and every inertial frame. But the momenta, energies, etc. of objects (and photons for that matter) and the durations of events as measured/observed in different inertial reference frames may well differ.

Of course not. He'll never be ready. But I don't see any point in writing things that need to be corrected later. I also don't want to confuse other readers who are ready.ICANT does not understand what a reference frame is. His recent suggestion that some inertial reference frames don't matter and his request to know what some reference frame was "relative to" are clear evidence of that.But the fact that different measurements and observations are the reality for different inertial frames coupled with the fact that no inertial reference frame is special is pretty much central to understanding SR and to the discussion at hand. Discussing those things cannot be avoided.

Hi ICANT,I think you've taken up a topic that will summarize the differences you have with some of the SR supporters. Wrong, ICANT. No object is ever limited to being in only its own reference frame.Nothing in the definitions you've posted says anything about entering or leaving a reference frame. There is also nothing in the definitions that localizes a frame of reference to an object or particular point. That stuff is baggage you've invented. At most, the origin (0,0,0,0 point) of the origin can be fixed. Son and crashfrog's explanations are perfectly proper.The pulse (and every other object/event) is "in" every inertial frame conceivable during the entire time that it exists. Even if the pulse cannot be seen from some vantage point at rest in a particular inertial frame, the pulse still has coordinates and a velocity as measured using the coordinate system or the coordinate axes of that frame, and what's more to the point, every observer at rest in that frame will agree on the velocity and coordinates once they agree on the coordinate axes.The pulse never leaves the inertial frame of the salt flats, the car, or any other inertial or non-inertial frame no matter how distant it becomes from some object or observer.As an additional aside, those experienced with relativity know that it almost never helpful to talk about reference frames that are moving at speed c relative to any observers. After all, photons always move at speed c with respect to other inertial observers, so looking at things from the photon's perspective does not usually produce anything interesting. No sentient being ever moves at speed c. Instead it is the reference frame of the observer that is of interest.There is no way to accurately describe what postulate #2 means using your understanding of inertial frames of reference.Edited by NoNukes, : No reason given.