Existence

I think you're giving him too much credit that will confuse him later on. As this quote from him shows: he seems to believe in an absolute frame of reference and the aether theory.He won't understand why he's coincidentaly accurate to a large degree even though his reasoning is wrong and I doubt he understands the concept of accuracy and margin of error in physics. Even though you're essentially correct in what you said, saying it like this will confuse him even more and may make discussion even harder.Edited by Son, : No reason given.

No, that's not correct. Relative to its own reference frame the car is motionless and the Salt Flats are shooting by underneath at .5c. Relative to the reference frame of the Sun the car is shooting around the sun at .5 c plus the Earth's motion around the Sun. Relative to the local supercluster (the neighborhood that includes the Milky Way galaxy) the car is moving at .5 c plus the Earth's motion around the Sun plus the Sun's motion through the galaxy plus the galaxy's motion through space, and so on.When we translate between these different reference frames we have to add the velocity of the reference frame we've left. That's what it means for velocity to be relative to something. We've not made any claim that light always travels in a straight line in the direction it was emitted. Haven't you heard of lenses? Haven't you heard of mirrors? These are all ways to change the direction of light. But the pulse is not traveling at a 90 degree angle relative to the salt flats. It's traveling at a 90 degree angle relative to the car. Relative to the salt flats, the pulse is traveling at an angled path that intersects with the moving detector.By trigonometry, we know that the path of the light pulse relative to the salt flats is much longer than the path of the light pulse relative to the car. (The hypotenuse of a right triangle is always longer than either of the two other sides.) If it were a tennis ball and not a light pulse, we would resolve this difference by observing that the velocity of the tennis ball relative to the salt flats is much faster than the velocity of the tennis ball relative to the car, because the tennis ball is imparted with the car's velocity (because it starts out at rest relative to the car.)But light is different than tennis balls. While the speed of a tennis ball depends on the relative velocity of the observer, the speed of light does not. The speed of light is the same for all observers regardless of their velocity.Hence the paradox - relative to the salt flats, the light pulse has a longer path to cover than relative to the car, but relative to observers in both frames of reference it has to have the same speed. It can't go any faster to cover the longer path relative to the salt flats, so the observer on the salt flats observes a longer period of time for the pulse to be captured by the detector than the driver does. In other words, relative to the observer standing still on the salt flats, things happen slower for the driver of the car. Not just for the driver, but time is actually slower within the reference frame of the moving car. Chemical and nuclear reactions happen slower. Clocks run slower. At rest, the engine of the car puts along at a "chug-chug-chug." But as the car drives by, the salt flats observer hears it go "chuuuug......chuuug.....chuuuuug" - despite the high rate of speed, the RPM of the engine actually appears to be lower, much lower, than when the car is at rest. According to the driver, it is faster - but a "minute" for the driver is a much longer period of time than for the stationary observer. Time is actually slowing down for the driver. That's time dilation. No, you will not. Motionless relative to you would mean that the distance between us was not changing, but since you are getting closer to me and then farther away when you pass, I am clearly not motionless relative to you. Because relative to the emitter the detector has not moved. The light pulse has moved at c in a straight line towards the detector. Since the detector hasn't moved relative to the emitter, and the emitter has always been pointed at the detector, the light pulse hits the detector. In the reference frame of the car, nothing has moved at all but the light pulse and the salt flats outside. In the reference frame of the salt flats, the emitter has moved along with the detector and the light pulse emerges from the emitter on an intercept trajectory to where the detector is going to be - two feet ahead, to use your numbers.The difference is the angle of the trajectory. Relative to the car it is 90 degrees. Relative to the salt flats it is less than that. Yes. http://en.wikisource.org/...Earth_and_the_Luminiferous_EtherThis experiment occurred in 1887 and proved, with no room for doubt, that the speed of light is the same for all observers regardless of their velocity. This is an explanation of the experiment: http://en.wikipedia.org/...chelson%E2%80%93Morley_experimentNo, this is not correct. Einstein quite clearly says that the speed of light is the same for "all inertial observers." An "inertial observer" is an observer whose velocity is constant. Therefore it's another way of saying that the speed of light is the same for all observers regardless of their velocity, as I've been saying. I'm simply restating Einstein's words in a simpler way for your understanding. That's a very perceptive question! And the answer is - all inertial observers. Everyone who measures the speed of light in a vacuum observes it to be moving at C, regardless of their own velocity relative to anything else. Directly vertical (a "90 degree angle" you would say.) The "point emitted" is the emitter, which has zero velocity relative to the car. If you're talking about a point in coordinate space, then you need to say which coordinate space you're talking about - the coordinate space of the car, or the coordinate space of the salt flats? Because relative to the salt flats yes, the pulse has motion in the forward direction relative to where the emitter was when the pulse was emitted.

ICANT did not do any dishonest editing. I found the statement exactly as ICANT quoted it on Wikipedia. Nonetheless, ICANT has himself quoted the formulation of postulate #2 that does include the more complete statement and we've discussed it at length here. He is now apparently claiming that "any inertial frame of reference" is a translation error, despite the fact that he cannot read German.That said, I do consider ICANT's question to be dishonest. He knows full well that where I got my statement of postulate #2 from, and I've called him on this exact point previously.There is no translation of Einstein's paper that does not support the concept that the speed of light is the same for observers at rest in different inertial frames. The concept is used in section three of Einstein's paper entitled "Theory of the Transformation of Co-ordinates and Times from a Stationary System to another System in Uniform Motion of Translation Relatively to the Former" to derive the coordinate transform equations. So regardless of ICANT's honesty, he's just wrong.

Harder??? ROFL!! It's already like explaining Chinese calculus to a 6th grader in Tupelo, Mississippi.I do see your point, but I cannot argue that way. Besides, didn't ICANT already make the point himself?Edited by NoNukes, : No reason given.

I found on this wikipedia page: from Theory of relativity - Wikipedia
I don't see this quote in the reply he wrote, it's still missing the reference to all observers regardless of their relative motion in ICANT's message.
Did you mean that he quoted it from another article or in an earlier reply?

Hi ICANT,I've addressed postulate #2 elsewhere. What problem is that? If the laser is fired from the car, isn't it going to hit the same distance behind the car no matter when the laser is fired?But I'll play along with your scenario once a few things are clarified. Do you believe that this matters? Why would it? Over what distance does this signal travel. If the signal is traveling over a wire, how does it get to a moving vehicle? I have no idea where you are saying this detector is located. Please clarify what "in the middle them" means. A diagram would be really helpful. Your description isn't very clear.

He quoted it from another article. ICANT quoted the abridged statement from the Wikipedia article titled "Special Relativity", rather than the "Postulates of Special Relativity" article.But any honest person would know that using a quote with ellipsis dots at the exact point of disagreement is a suspect thing to do. His claim to be using a correct translation is made up out of whole cloth.

Hi NoNukes, I seem to get the idea for the driver to fire the laser at the Salt Lake Flats as the car travels over the tracks he would need a hole in the floorboard of the car. Because as stated the driver just pointing a laser pen at the Salt Lake Flats and firing it the pulse would hit the inside bottom of the car and stop. The laser pen mentioned is mounted through the roof of the car and flush with the exterior of the car.The car is traveling in a vacuum at 0.5 c over the Salt Lake Flats.That means when the pulse is emitted from the laser pen it is on the exterior of the car and in a vacuum.There is nothing in that vacuum to alter the direction of motion of the pulse. According to everything I have read that pulse will travel in a straight line in the direction the laser pen was pointed at the moment the pulse was emitted. That pulse will not move the width of a hair in the direction the car is traveling. There is nothing to force the pulse to hit the detector as there is in the modified cycle where the pulse is traveling in a vaccum tube between the mirrors.Now can you explain how the pulse can not travel at a 180 angle relative to the mounted laser pen?If you can't then the pulse can not hit the detector as the detector has moved 2 feet relative to the point in the vacuum the pulse was emitted from the laser pen. The laser pen is mounted on a frame on the car 4 feet above the sensor that is mounted on the track.A sensor is mounted on the frame that passes over the sensor on the tracks which sends the signal to the laser pen to release the pulse.The sensor mounted on the frame on the car is conected to the laser pen with a wire so that the signal has to travel exactly 4 feet from the sensor on the tracks to cause the laser pen to emitt a pulse.

_________._____,_____._____,_____._____,
         S     D     S     D     S     D

There is 2 feet between the S and the D.
There is 4 feet between the S's and 4 feet between the D's.

S equals sensor and D equals detector.The sensor on the car passes over the sensor on the tracks the signal goes to the laser pen which emitts a pulse. At the time the pulse is emitted the laser pen is directly over the detector.My question was, will the pulse hit the detector that is below the laser pen when the pulse is emitted?Or will the pulse hit the second sensor?God Bless,

Hi ICANT,Here's an answer to what I consider the most important of your questions. I'm not going to bother with the rest of them in this post. This post is already way too long. Maybe someone else will take them on. The pulse will travel along the axis of the laser pen in every conceivable inertial reference frame. However, it is a simple matter to show that the axis of the laser pen will not measured as being along the same direction in every reference frame.Your question does not specify the reference frame for which I'm supposed to provide an answer.Before I embark on this demonstration, I do want to point out that your own explanations never address the difference in coordinate systems between different inertial reference frames. When I say that the car is at rest in one particular reference frame, you respond with "so when did the car stop", failing to consider that the car is still moving as measured in another reference frame. If I suggest that a photon path is not vertical in the flat car reference frame, you claim that the laser on the ground was mounted vertically. Well it was mounted vertically, but only in the ground frame of reference.The types of responses I've illustrated above do not even attempt to rebut the arguments I and others post. If someone describes an action in the coordinate system of a first inertial reference frame, it is not a rebuttal to say, "but that's not how it appears in some second reference frame" in which the origin is moving as measured in the first reference frame. You should expect that the measurements are different.I predict that you will continue to make those same non rebuttals in response to this post and that for you my efforts will be water off a ducks back. Hopefully, someone else will learn something.With that in mind, I'm going to address the question of the direction of the emitted photon. What I intend to do, is to calculate the angle of the emitted photon in two different coordinate systems. I'll present a non-SR calculation first, show it's problems, and then I will present the SR calculation. I will be careful to note the reference frame in which each measurement is made. In other words, I'm not going to be making vague statements about velocities, coordinates, durations, etc. without nailing down the frame in which those measurements/parameters apply.The first question to resolve is "How do I determine which direction the laser pen is pointing/mounted?" You never tell how this is done. I'll include a method for doing so in this response.For the purpose of doing the calculations, I will make an assumption about the emission path of the photon within the laser pen. I will assume that the length of that path is 3 millimeters (as measured in the reference frame in which the laser pen is at rest). Any finite length will yield the same results. Picking a value simply allows me to use arithmetic rather than algebra.The internal light path is also the direction along which the laser beam will be directed/pointed/mounted.I will also assume that the car is moving along the tracks to the right as measured in a coordinate system in which the tracks are at rest.Still with me ICANT?Let's start by considering the coordinate system in which the laser pen (and the car) is at rest. For convenience, I will arbitrarily refer to this frame as the stationary frame. If someone does not like my choice of stationary frame, let them redo the work. Galileo, Newton, and Einstein all say the choice is arbitrary. If you think you know better, then let's hear it now.We will locate emission point of the laser pen at the origin of the stationary frame. Of course, we should expect that the emission point will remain at the origin of the stationary frame for the entire duration of ICANT's thought experiment because the velocity of the laser pen is zero in this stationary frame of reference. If you don't understand why this must be, then ask and I'll respond politely.For ICANT in particular, note that the path of the photon as described above is completely within the light pen. Apparently this matters to you. It does not matter one iota to me or anyone else. The only reason for mentioning it is to foreclose a inevitable and ridiculous "in its own frame" argument.In a vacuum, light travels 3 millimeters 0.010 in nanoseconds (your choice of inertial frame). Assuming that the laser beam is pointed downwards as measured in the stationary frame, then the beginning of the emission path is at:x=0, y=3millimeters, t =0.The end point of the emission path is atx=0, y=0, t=0.0100 nanosecond.So what is the direction of the photon in the stationary frame? Since light travels along a straight line, the direction is simply the direction of the path along the straight line connecting the starting and ending point of the photon. In this case anyone can determine that the direction of the path is exactly vertical by inspection. But let's apply the trigonometric equation we are going to need later. Theta, which is the angle of the path from vertical is given by the following. Theta = zero implies a vertical direction.Let us now estimate the angle for the photon in the moving reference frame by considering things without applying special relativity. The moving reference frame is the one in which the tracks and detector are at rest. Of course those items are take no part in determining the angle of travel for the photon.In the moving frame of reference, the emitting point of the photon is again at the origin at the time the photon leaves the laser pen, but quite obviously, the emitting point of the laser pen is not fixed at the origin of the moving frame of reference. In fact, at the time the photon was at the start of the emission path, which according to our calculation for the stationary frame was 0.010 nanoseconds prior to the time the light beam reaches the point for leaving the laser pen, the axis of the light pen must have been to the left of the origin of the moving frame by some amount. Ignoring SR, we would calculate the axis of the light pen to have been 0.010 nanoseconds * 0.5c * 299,792,458 m/second = 1.5 millimeters to the left of the origin.Are you still with me ICANT??Completely ignoring SR, the coordinates for the starting point of the photon are x'=-1.5 millimeters, y'= 3.0 millimeters, t'= -0.010 nanoseconds. The coordinates of the photon at the point of emission are x'=0 milimeters, y'=0.0 millimeters, t' =0.0.Still there ICANT??Now for the trigonometry to calculate the deviation from vertical. Well that number looks ought to look familiar. Notice that we have yet to apply the requirement that light travel at c as measured in any inertial reference frame. Instead we've used sixth grade physics with some trigonometry. It is not special relativity that requires that the angle of a light beam be different in two different reference frames. As I've explained to others, this inescapable conclusion is simply the result of light moving at a finite speed. If light moved at an infinite speed, it would appear to move in the same direction in the moving frame and in the stationary frame of our example.Now, let's take a SR review of our result. We see that our non-relativistic calculation is wrong because it predicts a speed of light greater than c as measured in the moving frame of reference. In particular:The length of the path within the light pen, using the coordinates calculated above (from the Pythagorean theorem) is 3.3541 millimeters and light cannot travel that distance in 0.010 nanoseconds. It would instead take 0.011188 nano seconds. Of course the laser pen would have moved a small additional amount in this slightly increased time. If continue this iterative process by recalculating the hypotenuse using the calculated time and then recalculating the required time to travel the hypotenuse, the result converges. It turns out that as measured in the moving frame of reference, the photon actually traverses the length of the travel path within the light pen in 0.011547 nanoseconds. This value corresponds to the start of the photon path being 1.732 millimeters to the left of the origin of the moving frame. Recalculating theta provides the following result. Note that regardless of whether you accept postulate #2, the angle of the path of the photon is different in diverse inertial reference frames. But only the SR result is consistent with postulate #2. It's also the one that agrees with experiment.Edited by NoNukes, : Premature postingEdited by NoNukes, : No reason given.Edited by NoNukes, : Clarify moving frame coordinates

The problem is that in your scenario, your pulse moves in the direction the salt flat lakes are traveling if it misses the detector.

The light pulse has not travelled across the distance between the pen laser and the tracks yet. When it does transit this distance it will strike the second S from the left in your diagram. At that very instant the pen laser will be directly above the second S from the left. The light pulse moves 90 degrees downard relative to the CAR the whole time.Edited by Taq, : No reason given.Edited by Taq, : No reason given.

Hi Son, Well in my scenario the car is traveling at 0.5 c relative to the Salt Lake Flats.I did not declare the car frame as stationary.But if you do declare the car frame as stationary and say the Salt Lake Flats are receeding from the car at 0.5 c the point the pulse is released into the vacuum will also receed from the car at 0.5 c.So let me modify my scenario and ask you a question.

_________._____,_____._____,_____._____,
         S     D     S     D     S     D

              Son

There is 2 feet between the S and the D.
There is 4 feet between the S's and 4 feet between the D's.
There is 25 feet between Son and the first detector.

S equals sensor and D equals detector.The sensor on the car passes over the sensor on the tracks the signal goes to the laser pen which emitts a pulse. At the time the pulse is emitted the laser pen is directly over the detector.My question was, will the pulse hit the detector that is below the laser pen when the pulse is emitted?Or will the pulse hit the second sensor?My question for you is what will Son observe?Will Son observe the detector to flash a light?Or will Son observe the pulse to hit the second sensor instead?God Bless,

Where is the pen laser in relation to the spot on the tracks where the light strikes at the moment it does strike? That is a much better question.added by edit: Why? What is so special about the salt flats that they can remotely grab light and stop it?Edited by Taq, : No reason given.Edited by Taq, : No reason given.

I will reiterate Taq's question and ask you what's so special about those salt Flat Lakes? And are you aware that by using an absolute frame of reference, you are disagreeing with relativity? Relative and absolute are in fact opposites. If from the start you were disagreeing with the fact that the speed of light is constant in a vacuum as mesured in every referential frame, you should have said so from the start. If moreover you do believe there's one absolute frame of reference from which we can determine everything else, say so. Of course, both positions have already been disproven by experiment...As for your question, the pulse will hit the point on the tracks that is directly below the car in such a way that the pulse will have no horizontal speed in the inertial frame from which it originated and where the laser pen is at rest.

Hi ICANT,I could not pass up responding to this even though it was addressed to Son. Nobody cares about your declaration or the lack thereof.The choice of which reference frame is stationary belongs to the person attempting to solve the problem. It is not something you as the proposer would get to declare when you set up the problem. If you do make a declaration, the problem solver should ignore the declaration and chose his own coordinate systems. If necessary, the problem solver can use a coordinate transform to specify the answer in any other coordinate system.Generally speaking the person working a physics problem should pick an initial reference frame based on one or more facts given in the problem.In this case, it appears to me that we know the laser pen's orientation in the reference frame in which the car is at rest from your statement of the problem. So it makes sense to choose that particular reference frame at least initially.Your own preferred answer to this problem appears to require that the laser beam make a substantial change in direction from its path within the laser pen to the path after the laser beam leaves the pen. I'd sure like to know how and why that happens.