Existence

What you are failing to get is the meaning of a frame of reference.

Hi ICANT,You say the most puzzling things. Your question makes no sense. I made a statement that applies to any inertial reference frame. So you are free to pick any inertial reference frame. The photon can be moving vertically from any object at rest in the coordinate system of your choice.So the answer to your question is ICANT picks them. (Absent the ridiculous choice of a frame moving at speed c.)Given that choice, I can specify a second inertial frame in which the direction traveled by that same photon (or any other particle moving at finite speed) will not be vertical. In fact I'll tell you in advance what frame I'm going to pick. If the photon is traveling along the y axis, I'm going to pick an inertial frame that is moving horizontally at constant speed relative to the origin of the frame you choose.That statement is not the least bit controversial. I can prove it mathematically if necessary, but I won't bother doing so here because you would not be able to follow the algebra. However I have already provided several examples in this thread.Your circle dot stuff is irrelevant. Your diagram does not reflect what I am describing. There are a couple of things wrong with your drawing.First, only one coordinate system is provided on the drawing. I am describing how things would appear in a second coordinate system.Secondly, if as viewed from one coordinate system, the photon strikes point "D", the photon must reach point "D" in every reference frame, inertial or non-inertial. Simply changing the frame of reference does not change what events occur despite the fact that angles, durations, and speeds of objects ( i.e., speeds of objects other than photons) are different in different frames.So assuming that in track frame of reference, the photon does strike point 'D' (something which I am not disputing for now) then the photon must also strike point 'D' as viewed from an inertial frame in which the laser pen and sensor is at rest. So to correctly show things as I am arguing them, you would need to show a photon traveling at 26.5+ degrees but still striking point 'D'. You are trying really hard to find something to be right about here. I'll be careful to call this light segment a pulse in the future. But how does that change anything I've said about the path of a photon? The sensor inertial reference frame would not be "placed" on a drawing showing the track frame of reference because the origin of the sensor coordinate system would be moving relative to origin of the first coordinate system. The relationship between the two inertial frames might more easily be shown using a set of two or more time-sequenced drawings. I seem to recall someone providing just such a set of drawings for a different problem and a second someone failing miserably to understand them. It's no surprise to me that you failed. You are attacking the problem in a manner that cannot possibly produce success. Sigh...

Found a really nice animation that ICAN'T should check out:http://galileoandeinstein.physics.virginia.edu/...tclock.swfAnd especially this one:http://einstein.byu.edu/~masong/htmstuff/Clock2.htmlEdited by Taq, : No reason given.

ICAN'T,If you are having trouble imagining what the ball looks like as it is tossed between the two boys, plug in 0.5 into the animation at the top of this page:http://webphysics.davidson.edu/...ativity/illustration4.html

As we've explained, where we would place that frame such that the pulse leaves the emitter at a lower than 90° to the direction of travel of the car is stationary with regards to the salt flats.That is why the observer inside the car will see the pulse travel in a direction 90° to the car's direction of travel relative to the salt flats, but the observer on the salt flats will see the pulse travel at a shallower angle relative to the car's direction of travel relative to the salt flats. They reason that they don't agree on the angle of travel is because these two coordinate systems are in motion relative to each other. You have the origin of your coordinate axis pointed out the back of the laser pointer instead of the front. Could you explain why that is?

Nice find, even though I know ICANT disagree with what they show happening, maybe at least he'll understand what the current theories actually say.

Hi ICANT,The following statements of yours kept rattling around in my head so I took a second look at them. If you press and release the button on the side of the light pen in one microsecond, a feat that I know that humans are not capable of, you will generate a "pulse" of light about 1000 feet long. No you cannot change the trajectory of a beam of light. When you point the laser in a new direction, you send brand new photons in that new direction. The old ones continue to act just like the ones making up what you are calling a pulse. There is no sense in which the photons traveling in the new direction are part of the same "beam" as those traveling in the first direction. Your definitions are pure hogwash.

Hi ICANT,I believe your reading comprehension skills could use some honing. Perhaps spending some restful time with the family is in order.In the following excerpt, emphasis is added by me. Do you see the disconnect between what Taq said (correctly) and your attempted rebuttal?In short, Taq says that the guys' motion relative to the ground is at right angles to the path of the ball. That is certainly true. You respond by denying that the ball travels at right angles to the ground's motion.I plan to post a list of things that ought not to be said about reference frames along with an explanation of why those things make no sense. The above exchange won't make the list, but you'll likely see some familiar stuff. Really? How high did Taq say that hot air balloon was flying?If the observer thought that the boys were rolling the ball on the ground, that would make no difference with regards to Taq's point, so why even bother to say it?

Hi ICANT and fellow physics enthusiasts! I have to admit that trifling responses of the type above annoy me. Taq asked for a diagram of events as they occur in the driver's frame of reference. Taq did not ask for a drawing of what the driver could see from his vantage point in the car. Yet ICANT, in typical fashion insists on denying that the driver can see anything and refuses to answer the question.In fact, if we ignore the effects of special relativity, there is a very simply mathematically relationship the coordinate systems of the two inertial reference frames involved in this problem. We can exploit that relationship to show what happens in one of the inertial reference frames given the knowledge of what happened in another inertial reference frame. The equation works even if the observer's view is blocked. Taq's question can be answered regardless of what a driver in the car can see. Does anyone seriously think a different set of events occurs when the driver turns his back on the action?The diagram below illustrates the relationship between the relevant coordinate axes. In the above drawing, the red coordinate axes represent the salt flat's coordinate system, with O representing the origin of the salt flat's coordinate system. The blue coordinate axes represent the position of the car coordinate system at time = t. At time = 0, the origins of the O and O' axes would have coincided. However, the origin O' moves to the right relative to the O origin at a velocity V. So at time t, the separation between the O origin and the O' axis is V*t as shown in the drawing. Each coordinate system has a third axis Z, which is perpendicular to the axes shown. In the case of this problem, nothing interesting occurs in the Z axis, and the Z axes are not shown.The black dot represents an event in space at time t. The coordinate axes for each frame extend to infinity. Accordingly the event has coordinates in both coordinate systems. In fact every possible event that has or will occur has coordinates in every coordinate systems regardless of whether that frame is inertial or non-inertial. Statements like "particle X has left coordinate system O' and is in its own coordinate system are clearly seen to be without meaning." Particle X has coordinates in every reference frame. A photon outside of a car has coordinates in the car reference frame.The horizontal coordinate for the event in the O' frame is X', while the horizontal coordinate for the event in the O frame is X. It should also be clear from the drawing that X' = X - V*t'. On the other hand the vertical coordinate for the event is the same in the two coordinate systems.Using these equations,Y' = Y
X' = X-V*tallows us to calculate the coordinates of events in the O' frame whenever we know the time and location of the events in the O frame. People familiar with physics will recognize the equations above as the Galilean transform equations. The equations do not take into account any aspect, whatsoever, of special relativity.Let's apply the equations to the problem Taq asked ICANT to solve. Let's even use ICANTs version of how events occur in the salt flat frame (O). The velocity V for this problem is 0.5c. ICANT says the the photon travels vertically from point E to point D. I presume that this occurs at the speed of light, although that scarcely matters. Accordingly, the coordinates and times for point E and D are as follows.For the O frame
E 0 feet horizontal, 4 feet high at time 0.00 nanoseconds
D 0 feet horizontal, 0 feet high at time 4.0668 nanosecondsWe now have all of the information we need to calculate what the coordinates in the O' reference frame. Note that the vertical (y) coordinates are the same in both reference frames.X' at time zero = 0 - 0.5c * 0 = 0 feet
X' at time 4.0668 = 0 - 0.5c * (4.0669 nanosecs) * 186,282 miles/sec * 5280 feet/mile = -2 feetFor the O' frame.
E 0 feet horizontal, 4 feet high at time 0.0
D -2 feet horizontal, 0 feet high at time 4.0669 nanosecondsSo we arrive at a familiar answer once again. In order to provide the answer requested by Taq (well at least an ICANTized version of that answer), let's plot those coordinates on the O' axes. Note the complete lack of any mirrors in the diagram. ABE:Exercise for the intuitive student: Where would the location of sensor S be on the above diagram.Edited by NoNukes, : No reason given.Edited by NoNukes, : Grrr... Fix plot of point D.Edited by NoNukes, : No reason given.

Hi NoNukes, When the laser pen reaches the horizontal point that the sensor sends the signal to the laser pen to cause it to emit the second pulse the laser pen is directly over the S sensor.The question is what did the first photon that was emitted from the laser pen strike.If the laser pulse was 1 attometer long when it is emitted from the laser pen, where would it strike?Would it strike the D or the S?God Bless,

D and S are further away than 1 attometer, so it wouldn't strike either of them. However, the laser pulse would be directly below the pen laser in both frames of reference, and it will be there for the entire flight of the photon. I linked several websites in my posts above that have nice animations showing just this. Also, this is exactly what Michelson and Morley saw in their experiment. This phenomenon has been directly observed in experiments.

Hi ICANT.Congrats on making post #1000. That is not "The" question. It is one of several questions.As I've demonstrated in detail, the answer depends on the frame in which the aiming is done. In my opinion, the set up of the thought experiment suggests that the laser was aimed vertically in the car frame of reference. You disagree.Not only do you disagree, you don't even understand what a frame of reference is, which makes discussing them with you incredibly difficult if not impossible.The answer to my question is that sensor S would be at the origin of coordinate system O' at time 4.00669 nanoseconds. I've already answered your question about where the photon would strike. Ignoring SR effects, the photon would strike S under my understanding of the problem, and D under your understanding. But if we require the photon to travel at "c", neither answer is correct.But using either understanding case, the photon would not take a vertical path in one of the car and track inertial reference frames. You have yet to address any of my demonstrations that such is the case.

I don't understand this answer. I can agree that the pulse would miss both D and S, but only by taking into account SR. How does the length of the pulse establish that the pulse will miss both D and S?

The internet is fully of excellent, free reference material on relativity. I highly recommend Leonard Susskind's lecture series on Special Relativity on iTunes and Youtbe. In fact, Susskind has free semester length lectures on a wide variety of physics topics including classical mechanics, statistical mechanics, QM, string theory, general relativity and cosmology. For Quantum Mechanics I'd recommend the Oxford QM lectures on iTunes over Susskind's lectures.I've linked to Lecture 1 on special relativity. The first few minutes of Susskind's lecture 1 on special relativity cover inertial reference frames and coordinate systems. The math level escalates a bit after that. 1000 NoNukes to anyone who watches the entire 110 minutes.Edited by NoNukes, : Fix video link

Perhaps I misunderstood what ICAN'T was trying to say. When the light pulse is one attometer away from the end of the pen laser it is still directly below the pen laser, as it is the entire transit from the pen laser to the spot where it strikes the tracks.A light beam is nothing more than a collection of singular photons, so his description of a one attometer long pulse really doesn't make sense anyway. It would be much more productive to look at a single photon, wouldn't you agree?