Existence

Hi NoNukes, I am not the one that is changing the direction of the pulse after it leaves the laser pen.I am insisting that the pulse travel in a straight line in a vacuum at 180 relative to the position the laser pen was in when the pulse was emitted from the laser diode.You on the other hand require the pulse to turn at a 63.43 angle relative to the position the laser pen was in when the pulse was emitted from the laser diode.If I an outdoors and point my laser pen at the moon and press the button on my laser pen and let it go, how or what can effect where that pulse will go?Will it go in a straight line until it scatters?Or will it go at an angle to the position the laser pen is in when the pulse is released until it scatters.?God Bless,

I think everyone of us agrees to exactly that. But you don't define what you mean by "the position the laser pen was in when the pulse was emitted", now do you? I've described one method of determining the position and alignment of the laser pointer. I simply look at the light path of laser beam inside of the laser pointer. That direction is the direction in which the laser beam is pointed. I agree that when that definition is used that the beam continues to travel in the same direction upon leaving the laser pointer.So how do you determine that position? Do you determine it by simply looking at the lengthwise dimension of the laser pointer? If so, what assurance do we have that the light beam travels along this direction even in a stationary frame. Are laser pointers perfectly aligned in this way? Or do you use some other method?The method used does matter, but you have yet to specify what pointing direction method you do use. Until you do, your question will contain an ambiguity that will produce different answers to your question.Provide your answer to your problem, and I will then demonstrate that your answer does indeed require a change in direction at the time of light emission as measured in one or more frames of reference. What method do you use to align the laser pointer with the moon? Do you simply sight down the length of the laser pointer, or is some other aiming technique used?Light is of course subject to diffraction, refraction, reflection, various scattering effects, reflection, absorption, and the photoelectric effect, among other things. What's your point?But for the short distances involved in your thought, the light beam simply continues along the path it had immediately before it left the laser pointer. I address this point in Message 939, and I look forward to your response.When might we expect to see your citation of Einstein's papers regarding postulate #2. It's been weeks since you promised that.Edited by NoNukes, : Address ICANTs second question.

How did you factor in the speed at which the Earth and Moon are moving relative to the center of the Milky Way? Or are you saying that there is something special about the Earth's frame of reference that is not found in the car's frame of reference?

HI ICANT,Just a quick response to a single point in your message. No ICANT, that is not what my calculations show. I did not show a change of direction within a single reference frame. Instead what was shown was that a 90 trajectory in one reference frame's coordinates would be measured as a 60 trajectory in the coordinate system of a different inertial reference frame. In each reference frame, the pulse was shown to continue along the same direction before and after leaving the laser pen.But as I predicted before starting the calculations, you continue to insist on confusing the answers for different inertial frames. There is in fact, no help for you at your current level of misunderstanding is there? There is literally no technique that can allow you to correctly picture an event from more than one reference frame. Pictures of coordinate systems confuse you and high school mathematics baffles you.My prediction from the past message:

Hi Taq, The same place it is if you had answered my question.

|           |           | these represents the pen
               E     P     E     P     E     P
               |           |           |
               |           |           | these represents the pulse
               |           |           |
               |           |           |
_________._____,_____._____,_____._____,_____,
         S     D     S     D     S     D     S

S = sensor
D = detector
E = pulse emitted
P = pen when pulse strikes the detector

Since the pen is pointed at the D when the pulse is emitted and the pulse will travel in a straight line in a vacuum at c independent of the motion of the emitter the pulse will hit the detector and the light will flash.At the same time the sensor attached to the frame on the car will pass over the sensor on the track and send the signal to the laser pen which will emitt a pulse directly above the second detector that will strike the detector when the laser pen reaches the position directly above the third sensor etc.Now for the pulse to hit the sensor which is directly under the P when the pulse hits the detector rather than the detector as you claim it would do the pulse would have to leave the laser pen at a 63.43 angle relative to the position the laser pen is in when the pulse is emitted.But what does the position the laser pen being in when the pulse strikes the detecetor have to do with where the pulse strikes? There is nothing special about the Salt Lake Flats.They do not grab the pulse and stop it.The pulse just continues on it's journey at 180 relative to the position of the pen when emitted from the pen. So if the Salt Lake Flats is all of a sudden receeding from the car so would the pulse be receeding from the car, as the car is traveling at 0.5 c horozontally relative to the Salt Lake Flats.The pulse is traveling zero horozontally to the Salt Lake Flats. It is also traveling c towards the Salt Lake Flats relative to the point it was emitted from the laser pen.God Bless,

Hi ICANT,I think you've bitten off too big a chunk this time. The statements quoted above are not consistent with each other.If the salt flats were to suddenly recede away from the car, why would that cause a change in the direction of a light that was emitted from the car? After all there is no connection between the flats and the massless light beam. Why wouldn't the light continue on in the direction in which it was currently moving regardless of any motion by the salt flats? As you describe things there certainly does appear to be something very special about the salt flat's frame. According to you, the motion of the car cannot affect the direction of the light beam in anyway, but if the salt flats recedes, then according to your statement above, the light beam must also recede.On the other hand, when the moon recedes, nothing particular happens to the light beam. So yes, that sounds pretty special.It's one thing to insist that the direction of the light beam is independent of the source, something which is not correct, but it is a completely different idea to suggest that the light beam must maintain a fix relationship to the salt flats regardless of how the salt flats moves. ICANT, even you ought to be able to see the problem with your statement above.Can you provide a not yet discredited source other than yourself that supports your position. No version of postulate #2 of which I am aware suggests that reality works this way. It's pretty easy for any of us to find support for the speed of light in a vacuum being c as measured in any reference frame. In fact, you yourself have quoted that fact numerous times here, and I've already mentioned a couple of places supporting the same position in Einstein's 1905 paper.I challenge you to find and cite support for your own interpretation of postulate #2.Feel free to cite quacks as support. But stand by if you do.By the way. Other than not liking the result, I did not note any criticism of my demonstration regarding the path of a light beam in two different reference frames. Was there a particular step of my presentation or line of reasoning that went awry? I'm not going to ask you if the math is wrong, because we both know that it is possible to can add and subtract numbers correctly and yet get an irrelevant result.The biggest problem with your view point is the absolute lack of symmetry. While you accept that the source of light cannot affect the speed of light in a vacuum, you are perfectly willing to allow the receiver of the light beam to dictate the direction of a light beam. Unfortunately, that does make the receiving reference frame special. Relativity allows us to use any inertial frame as the stationary frame. But when we try that using your distortion of postulate #2, the symmetry is completely broken.Here's something to think about. Referring again to your original car and sensor experiment, surely it is possible to aim the light pen at some angle so that it would strike the sensor. What angle would that light beam take from the point of view of an observer sitting on the roof of the car? Since the 4 foot pole is vertical, how would the angle traveled by the light beam seem to the roof observer be at any angle other than vertical? How is that possible if the light pen is not mounted vertically? How do you reconcile that with the fact that a vertical beam (as determined by a salt flat's bound observer) would miss the sensor?Edited by NoNukes, : No reason given.

I haven't taken a serious look at this problem yet, but I don't think the problem is as simple to analyze as it appears. Unlike the situation with the roof mounted sensor, the potential targets in this hypothetical are all at rest in a different inertial frame from the one in which the laser device is at rest.In the frame of reference in which the car is at rest (referred to hereinafter as the car frame), the spacing between the sensors is not 4 feet, but is less due to length contraction. (I think gamma is 1.1547 for this problem). So the separation between sensors is 3.464 feet in the car frame.Apparently the sensor on the car has only a tiny clearance above the track sensor, so we can neglect time delays due to that.The light beam clearly travels vertically in the car frame of reference, so it makes sense to solve the problem using that frame. It also seems clear that the propagation delay for the trigger signal to the laser is 4 feet divided by c. Once triggered, the light beam then takes an additional delay of the same amount to reach the ground. Total delay 8 feet divided by c. (No length contraction in the vertical direction) During that time, all of the sensors move 4 feet to the left (0.5c * 8ft/c), in the car reference frame. By the time the beam reaches the ground both the detector and the second sensor will be well beyond the point at which the beam strikes the ground. (Remember length contraction) If the beam is on for some significant duration, it may hit some other sensor but we have not been provided with enough info to tell.ABE:It also occurs to me that the sensors, and detectors mounted on the ground provide uniformly spaced markers for a coordinate system in the salt flat reference frame. I am looking forward to ICANTs answer to this problem.Edited by NoNukes, : No reason given.

Hi ICANT,It is certainly possible for a photon to be emitted from the laser pen in such a way that it takes the paths shown in your drawing. So let's just accept the drawing you provided as the reality at least for the purpose of this post.Of course your drawing shows the path of the photon pulse in the reference frame of the tracks, but what would the path of the photon be in the reference frame of the car? Notice that I did not ask you what the path would look like to an observer in the car, because you would probably insist that the observer could not see the path. But we can calculate that path even if we cannot see it. And, for the time being, let's ignore special relativity completely. That should not bother you, right?Referring to the drawing you provided, at time the light was emitted, the laser pen was at point E and the car sensor was directly below that point at the first D. At least I would expect that you would agree with that.Let's chose the car sensor as the origin (0, 0 coordinates) of the car reference frame. Of course any arbitrary point fixed to the the car can serve the origin point. Initially the car sensor is directly below the point at which the photon was emitted. The coordinates of the photon at the time of emission are 4 feet high, and zero feet horizontally away from the car sensor which is our origin point. So x,y coordinates are (0, +4 feet).But at the time that the photon reaches point D, according to your diagram the car sensor is now at the second point S and the light beam is two feet behind the origin of the car reference frame (the sensor) at the first D a height of zero. Surely you would agree with this description. So relative to the car sensor the pulse is now two feet behind the car sensor at height zero. Second x, y coordinates (-2 feet, 0 feet).Given the above pair of coordinates, what is the path of the light beam as calculated in the reference frame of the car? Is there any doubt that one and only one line can pass through the the two points determined above, and that the line must represent the path of the photon in the car reference frame?I'm not going to do the calculation, but it should be pretty clear to any SD who understands 9th grade math, that the path of the photon in the car reference frame is not vertical despite a path of the photon in the track reference frame that is vertical. The result is not because the car has affected the motion of the light beam, but simply because the coordinate system of the car sensor moves relative to the track reference frame.But more to the point when you say that the path is vertical, without stating in which reference frame you mean, your problem specification is ambiguous. You might well mean that the photon travels vertically in the car reference frame. In fact the latter assumption makes the most logical sense given that the light pen is mounted to the car. Most of us would reach the opposite conclusion if the laser pen were mounted to the ground. In that case I probably would draw something quite similar to your drawing.Added By EditI need to apologize for an error. Apparently I overestimated the math skills required to understand the path of the light beam. According to this, eight graders are supposed to know how to perform coordinate translations and to determine the slope of a line that passes through two points.Edited by NoNukes, : Some humor.

Last I checked, the angle between P and the second S is 90 degrees. The angle between P and the first D is not 90 degrees. You are the one stating that it shoots off at an angle in the driver's frame of reference, not I. That angle needs to be 90 degrees, as you have stated numerous times. When the light pulse strikes the second S, not the first D, the angle between the second S and the P is 90 degrees as it should be. Then why do you have the light going backwards at an angle in the driver's frame of reference? No, it goes backwards according to you. So how is the salt flats grabbing the light and dragging it backwards? Is there some kind of aether wind or something? Why wouldn't the light pulse continue at a 90 degree angle in the driver's frame of reference? Why does it have to go backwards? How is the salt flat grabbing the light?

For the sake of clarity, let's just assume that the space between the sensors and detectors is equal to the distance that the car will travel in the time it takes light to travel the distance between the pen laser and detector/sensor, whatever that length may be. I think we will confuse ICAN'T with the inclusion of length contraction and time dilation in the car's frame of reference, although it is a fun math problem nonetheless.

I sometimes do that, but only after I noisily announce that I'm ignoring SR. I expect ICANT to leap onto any inconsistency, so I don't intend to leave any unexplained ones.But I intended that last message for you. I'm not going to talk down to you in order to avoid confusing ICANT.

Hi NoNukes, Well actually the only place the Salt Lake Flats would be receeding from, is the car frame you declared as the stationary frame that is traveling at 0.5 c relative to the Salt Lake Flats.When you declare the car's frame of reference as stationary the car does not stop traveling 0.5 c relative to the Salt Lake Flats.It can only be said that the Salt Lake Flats is receeding from the reference frame of the car due to the fact the car is traveling at 0.5 c relative to the Salt Lake Flats.The earth does not stop its rotation.
The earth does not reverse its rotation and begin to rotate backwards at 0.5 c.The car is still moving relative to the Salt Lake Flats at 0.5 c.The car is still moving relative to the pulse that was emitted from the laser pen at 0.5 c.The pulse is still moving relative to the point it was released at c. The Salt Lake Flats do not change their movement they are attached to the earth and move with the earth relative to the core of the earth. Relative to what? The laser pen would have to be installed through the roof of the car at a 26.57 angle relative to the motion of the car. An observer on the roof would observe the pulse to travel at 180 relative to the point the pulse was emitted. Both observers would observe the pulse to travel at 180 relative to the point the pulse was emitted, and miss the detector.God Bless,

From the car's frame of reference, it is the salt lake flats that are moving at 0.5 c.

ICAN'T,Just so that we are are on the same page . . .You are the driver in the car. You position yourself so that you are directly above the pen laser and looking directly down at the tracks, and the tracks are moving from your left to your right. When the pen laser fires, 1) do you "see" (assuming you can see photons in flight) the light pulse move straight down at 3E8 m/s at the tracks until it strikes the track? Or are you saying 2) that at the instant the pen laser fires the photons move at an angle to your right so that they can hit the second detector when the light pulse travels the distance between pen laser and the tracks?Is it 1) or 2) ?Edited by Taq, : No reason given.

I think ICANT doesn't properly understands the difference from the salt flats frame and the frame of the car? In one frame it is the car moving, in the other it is the salt flats.