Existence

Hi ICANT You believe as you do because you don't know what a reference frame is. And nobody has said anything about time having its own reference frame. Most of the above makes no sense. The pulse is in every reference frame, so that's fine. But reference frames are not contained in other reference frames. As Justice Scalia might say, that paragraph is almost 100% goobledygook. Any doubts about that are yours. The laser pen and the light pulse are in every reference frame. Is a pulse an object?Your statement makes no sense. Colloquially, we might identify an object with a reference frame if the object is at rest in said reference frame. But since a light pulse is never at rest in any inertial frame, even that colloquial use is not applicable. Your statement has no meaning that has any physical import.But let's be frank about what you do when you say the light pulse is in its own reference frame. You couple that statement with other statements indicating that such a reference frame is the only one that counts. You talk about objects leaving a reference frame and entering their own frame and insist that we cannot see light beams unless they are in our own reference frame. You ask me how a man and his wife can be in the same reference frame when they are facing eye to eye. Well, that kind of talk is completely asinine.Reference frames are simply coordinate systems used to measure, record, and analyze events. Coordinate axes associated with reference frames extend indefinitely. Reference frames exist independently of objects, and they are usually associated with observers rather than objects. When analyzing a physics problem, we select a particular reference frames based on convenience. We may know the velocity, position and time for a space-time event in a given reference frame, so it makes sense to start there. If necessary, we can translate those observations to other coordinate systems, with the goal being to obtain knowledge about reality in the reference frame that is important to a particular observer. That observer is never at rest on a photon pulse. Given that the premise of the question is based on your misunderstanding of what a reference frame is, I'll respond to it later if necessary. But yes there is a reference frame in which Earth is at rest. But is that the Earth's own frame given that every other object in the universe has coordinates and velocity in that frame?Added by Edit All wrong. You are just making stuff up here. You aren't getting this stuff from professional relativity deniers. The above is your own folly. And it is not even close to being right.Your predictions about what could be observed from a reference frame moving at 1.0 c are all completely wrong. Length contraction effects from traveling at c would reduce all distances in the direction of travel to zero. Thus no object moving less than "c" would be observed to have any movement in that direction from such a frame. You have no idea how weird things would appear in such a frame. Fortunately, we never make measurements or observe from such a reference frame.Regardless of what is 'observed' in the pulse frame of reference, things appear differently in "ordinary" reference frames in which observers are at rest. And those frames do matter.Edited by NoNukes, : Add response to pulse reference frame assertions.

Hi NoNukes, 1. I know that reference frame, may refer to a coordinate system.2. I know that reference frame, may refer to a set of axes wherein the position, orientation, and other properties of objects can be measured.3. I know that reference frame, may refer to an observational reference frame tied to the state of motion of an observer.4. Reference frame, may also refer to a combination of #1 and #3 as a unit. In Message 882 you said:

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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.

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If the pulse can not leave the reference frame of the car and have a reference frame of its own, how can the car have its own reference frame when it was created in the Earth's reference frame?In Message 874 you said:

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The only way objects enter or leave a reference frame is when they are created or destroyed.

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So when does the pulse begin to exist?
Does it exist before it is emitted?
Or does it begin to exist when contact is made and the pulse is emitted. Well you don't agree that the pulse can be in a reference frame where the pulse is 0.0847252802667504 nanoseconds which is 1 inch above the roof of the car and 1 1/2 inches from the pole attached to the roof of the car while being 47 inches from the detector on top of the pole.The pulse can be observed at those coordinates by any observer that is not inside the car.So you are the one saying the pulse can not be in all reference frames in the universe.So according to your definition of a reference frame does the pulse have a reference frame? The pulse can be observed. Therefore it is not a concept.
Some say it is a particle and others say it is a wave. Some say it is one part of the time and the other part of the time. Why can't a light pulse be at rest in it's own reference frame?You tell me the car that is traveling "0.5 c" is at rest in it's frame.You tell me my cycle that is traveling "0.5 c" is at rest in it's frame.So why can't the photon in the example you were trying to prove that I was looney because you said I was saying the photon was doing "1.5 c" when it was coming towards me in the cycle discussion.Why couldn't the photon be at rest in it's frame and observe the earth receeding at "c" and me approaching at "0.5 c" on my cycle? I thought the whole thing was a thought experiment.First point of mine. "The pulse would observe the car going at a 90 angle to it's own travel."What is wrong with this point?
Does the pulse travel in a straight line from the point emitted?
Is the laser pen mounted at a 90 angle to the motion of the car?Second point of mine. "The pulse would observe there was distance between it and its source."Does the pulse travel distance when emitted?
Is there distance increasing at "c" between the pulse and its source?Third point of mine. "The pulse would observe there was distance between itself and the pole that was moving away from it's path of travel."From the frame of the pulse the car would be moving at a 90 angle to its motion at the speed of "0.5 c".From any observer outside of the car they will observe the car traveling at "0.5 c" relative to the Salt Lake Flats.The same observer would see the pulse (if it were possible to view light from the side) traveling at "c" at a 90 angle to the motion of the car.You said these were wrong, so where are they wrong? Is this an admission that the pulse does have a frame of reference?God Bless,

The observer is at rest with respect to the observer's inertial frame, assuming constant velocity.The pen laser, detector, and car are all at rest with respect to the observer, so the entire experiment is in the observer's frame of reference. It doesn't matter what the speed of the car is relative to the salt flats lake as far as the observations made by the driver of the car. The car and the Earth are different frames of reference in this experiment. One is NOT nested within the other. One is not the creation of the other. And all of those observers will observe that the light hits the detector dead center, including the observer in the car. Why can't you focus on the driver as the observer?So please tell me why the light pulse will not hit the detector when nothing is moving?

Hi ICANT, They don't really. But in order to label the reference frames, we call them by an object that is at rest in that reference frame. We have to name them something. Is any of that stuff the least bit relevant? If we want to go into detail about when a laser photon begins to exist, then we need to discuss how light is generated in a laser. That's way outside the scope of this discussion. I'm not going there.We care about the photon when it is emitted from the laser pen. It does exist before that point, but we don't care about it before it is emitted. I dunno. Which reference frame is that? Name the observer or object that is at rest in that particular reference frame and perhaps I'll know what you are talking about. I really haven't been paying much attention to the details of your car hypothetical. No ICANT. That's completely wrong. The observed coordinates will vary depending on the speed of the observer relative to the car or the salt flats, or whatever. So what?I'm not aware of any meaning behind saying that a photon or a light beam is "in its own reference frame". For objects traveling less than the speed of light, we might link the object and a reference frame in which the object is at rest as a convenient way to name the frame. But there is no inertial frame in which a photon is at rest. If you want to come up with some way of matching a reference frame with a photon, your going to have to invent one.But if your invented definition for reference frame does not match what is meant by reference frame in postulate #2, I'm going to object to it as not being useful for this discussion.You are great at repeating definitions of reference frames we've given you and at looking them up yourself. But you don't understand them. Nope. It's part of an explanation why your attempt to describe such a frame is wrong. Why don't you tell us what you mean when you talk about a photon's own frame of reference.My best guess is that you think a frame of reference is a location from which you can observe.

Hi ICANT,I am making a second post to address more of your questions. You get to make up the scenario, but you don't get to make up the physics/science. Photons cannot observe. Presumably you are asking me what an observer moving at the speed of light would observe/measure/calculate, but I'm not even sure that's what you mean. But to an observer moving at the speed of light, the entire length of the salt flats in the direction of motion would be length contracted to a single point. Is this really something we want to discuss? Yes. Yes, in at least one reference frame. But not in every reference frame. Certainly not in the frame in which the salt flats is at rest. See above. Yes in every reference frame. Nope. The light is going to hit that sensor atop the pole, so this is definitely wrong. Nope. A photon moves at speed c relative to any and all observers at rest in an inertial frame. Postulate #2 dude. An observer moving in the opposite direction at 0.5c relative to the salt flats would observe the car moving away from him at 0.8c. But I believe he would conclude that the car at 0.5c relative to the ground. Nope. Maybe it would be worthwhile to draw a couple of coordinate systems to show that this is not true. I tried to walk you through this once using the flat car example, but you couldn't or wouldn't keep up with me. Different observers will measure different angles of travel for the light beam.Here is a way to demonstrate that in at least some scenarios the observers must measure different angles. Let's do whatever it takes to aim the light beam so that it hits the sensor. Given that the sensor is directly over the light pen in the frame of reference in which the driver is at rest, what angle will the driver measure?But would not this same angle, if measured in the frame of reference in which the salt flats is at rest produce a miss? So the salt flat's observer must measure a different angle for the same photon flight because he must agree that the light hit the sensor.

This is due to length contraction, is it not?

Hi NoNukes, But it does not exist in the form it is in when it leaves the laser pen.It is only electrons stored in batteries that is transformed into photons when the circuit is completed between the battery and the laser diode.This laser diode emits a laser beam through a lens on the head of the pen.If you hold the button you get a steady stream of photons but if you press and release the button you get a pulse of photons. The reference frame observed by an observer standing on the Salt Lake Flats that observes the car go by at the same time the pulse is released from the laser pen.That observer would observe the car moving across the Salt Lake Flats at 0.5 c. The observer would also observe the pulse leaving the laser pen at a 90 angle to the car roof traveling at c. Due to the fact the car moves 2 feet before the pulse can reach the detector the observer will observe that the pulse misses the detector and proceeds in a straight line from the point it was emitted.The pulse will continue in that direction until it is scattered or absorbed. Lets say I am traveling along side of Taq's car on my cycle at the speed of 149,896,229 meters per second relative to the Salt Lake Flats.That would mean that I am traveling at zero meters per second relative to Taq's car.I have mounted a super duper video camera on my cycle that can record the flight of the pulse when it leaves the laser pen mounted flush with the top of Taq's car.The pulse is released and the video camera records for 2 seconds.I peal off and return to the hanger to observe what was recorded.I show the recording on a big screen at a super duper slow speed.I observe the pulse as it leaves the laser pen at a 90 angle to the motion of the car.I then observe the distance from the point the pulse was released the pulse traveling in a straight line.I also observe that the distance between the pole extended from the roof of Taq's car and the straight line the pulse is traveling in is increasing due to the fact Taq's car is traveling at 149,896,229 meters per second relative to the release point of the pulse as the car traveles 149,896,229 meters per second relative to the Salt Lake Flats.I will observe due to the motion of Taq's car relative to the straight line the pulse is traveling in the pulse will miss the detector that is mounted to the top of the pole mounted to the roof of the car. Actually I thought a frame of reference was where an observer could locate items relative to other items.God Bless,

You will also notice that none of the apparatus is moving relative to the observer's car. Therefore, the light pulse hits the detector dead center. The car has no motion in the observer's frame of reference.

Edited by Admin, : ICANT - I'm no longer providing the free service of fixing your wide posts, I'm just going to hide the entire text. Preview your posts before posting from now on.

Hi ICANT, No ICANT. That is not what an observer who is at rest with respect to the salt flats would see/observer/measure/calculate or experience. In particular the angle you announce for the salt flat observer is wrong.First let me describe the two coordinate systems of interest here. For explanatory purposes, I will refer to the frame in which the salt flats is at rest as the stationary frame. But the choice of stationary frame is completely arbitrary. I could just as easily have chose the car frame to be the stationary frame.The drawings below show the relationship between the two coordinate systems. The upper system in drawings A and B is the reference frame in which the car is at rest, while the lower system represents the coordinate system in which the salt flats is at rest. Drawing A is taken at time zero in both coordinate systems, while Drawing B represents the coordinate system at some time later as measured in the coordinate systems. In the two systems, the X direction is the direction of motion of the car as observed from the stationary reference frame.Note that the x' coordinate axis is shown as being above the x coordinate axis only for purposes of illustration. In reality, the x and axis and the x' axis are at the same vertical level. We don't show the z and z' axes because they are not important to the discussion.So the coordinates of an event in the stationary frame are x, y, t, while the coordinates in the moving frame for the same event are x', y', t'.Let's imagine that the laser pen emits the laser beam at time = 0. I think it is pretty clear that at time zero, the two coordinate systems exactly coincide so that both the car observer and the salt flat observer agree that at the time of emission, the coordinate for the photon at time of emission are as follows:Moving frame coordinates x' = 0, y'=0, and t' = 0.
Stationary frame coordinates x=0, y=0, and t = 0.Note that there is no disagreement about where the photon started.So where will the photon be 3.3356 nanoseconds later? And what angle will the path of the photon take with respect to the horizontal?Let's first assume that we can somehow point direct the photon at what is vertical in the x'/y' coordinate system. I'm also going to use timing devices in the moving frame for this example. In the interest of equal time (and to avoid complaints from ICANT) I'll also work the problem for the case where the photon is directed vertically as observed in the stationary frame.The diagram below shows the result. Note that I've shown the coordinate systems aligned vertically as they should be. Light moves 1 meter, and no more in 3.3356 nanoseconds.Starting first in the x', y', t' frame, there should little if any any controversy about the fact that the photon has the coordinates x'=0, y'=1 meter, t' = 3.3356 nanoseconds. There should also be little controversy about the fact that as measured in the moving frame, the origin of the stationary frame coordinate system is 0.5 meters to the left of the origin of the moving frame coordinate system.So what are the coordinates of the photon in the stationary coordinate system?It does not take a college education in physics to see from the diagram that the photon does not travel in a vertical line as measured in the stationary coordinate system. The diagonal line connecting the starting and ending points for the photon as observed in the stationary frame is not vertical. It's also blindingly obvious that the length of the path as determined by the observer at rest in the stationary frame is greater than 1 meter.Without doing any math, two things are pretty clear.1) The photon cannot reach the point shown in the diagram in only 3.3356 nanoseconds as measured in the stationary frame.2) As measured/observed in the stationary frame, the path of light is not vertical as it is in the moving frame.For completeness, I draw the situation that exists if the photon is directed vertically as measured in the stationary frame. Again, the observers agree on the initial coordinates for the photon, so it is pretty easy to show that they must disagree on the angle and time required for the photon to reach the endpoint. Quite clearly, the observers in the stationary and the moving frame see the photon traveling at different angles. The observer in the car agrees that the photon misses the sensor because in his frame, the photon was actually directed towards the rear of the car. However, the observer standing in the salt flats will rightly insist that the photon was directed vertically.Edited by NoNukes, : Accidentally posted before finishing.Edited by NoNukes, : No reason given.

Hi Taq, It would be hard for something that is attached to observer's car to
move relative to said car.The car is moving 149,896,229 meters per second relative to the salt lake flats.Which means when the pulse is emitted and travels straight up the car is moving relative to the point the pulse was emitted at 149,896,229 meters per second. Relative to what?

Of course it will. The pen laser, detector, and pole are all in the same inertial frame. At this point, it doesn't matter what the car's velocity is with respect to the Salt Lake Flats since none of the pieces of the equipment are in the inertial frame of the Salt Lake Flats. If the observer is in the inertial frame defined by the pen laser, pole, car, and detector the observer will observe that the light travels straight up and nails the detector without any deviation from 90 degrees. The driver is the observer. The observer is the one doing the measurement. Therefore, the observer uses their inertial frame of reference. For the driver, neither the pen laser, the pole, the car, nor the detector are moving regardless of the relative velocity of the car to the Salt Lake Flats. The only reason that you would need to factor in the velocity with respect to the Salt Lake flats is if the pen laser, pole, or detector were stationary on the Salt Lake Flats. Since this isn't the case, the relative velocity between the car and Salt Lake Flats can be ignored for the driver's calculations.

Then why are you adding in a 0.5c velocity? That is irrelevant. What matters is the observer's frame of reference, and in this frame the velocity is zero for all of the pieces. The car isn't moving in the driver's frame of reference. It is at rest.

Hi NoNukes,I assume you had these drawings already prepared as you have reverted to the light clock on the cycle, where the pulse is traveling up and down in a vacuum tube between the two mirrors which are 1 meter apart. In the car experiment or the cycle experiment we can agree as the start point of the pulse.In the cycle experiment we can agree where the end point after the pulse has traveled 1 meter is. The only problem we have there is how the pulse reached that destination.Now lets return to the car experiment which is the one we have been discussing lately. I will still refere to the pictures you have drawn just using 1 meter = 4 feet and 1/2 meter = 2 feet.In the A B picture you have drawn 90 angles for the pulse to travel relative to motion of the source at the point emitted.In the other diragram you did not label you have drawn a line the pulse would have to travel to hit the detector after it had moved 2 feet.My question is, what causes the pulse to travel at the angle you have drawn?In Message 890 you said:

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Does the pulse travel in a straight line from the point emitted?

Yes.

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Since the pulse will travel in a straight line from the point emitted independent of the motion of the source emitted, why did you draw the line at an angle to the point emitted?Or have you changed your mind about the pulse going in a straight line from the point emitted? That is no problem if we use the 4 foot 2 foot measurements of the car experiment you would simply mount the laser pen through the roof of the car at a 26.56505117707799 angle to the forward motion of the car.If you did that the pulse will travel 4.47213595499958 feet and strike the detector.The problem is the laser pen is mounted at a 90 angle to the travel of the car.If the pulse travels in a straight line from the point emitted independent of the motion of the laser pen the pulse will miss the detector as it will have moved 2 feet since the pulse was emitted.Now if you can somehow add the motion of the laser pen to the pulse to cause it to travel at an angle instead of a straight line relative to the point emitted then you could get it to hit the detector.The problem is that does not satisfy postulate #2. I have never argued otherwise.I just argue the pulse does not travel at the angle you have drawn. But if it did it would take longer than 3.3356 nanoseconds to reach that destination.But since the pulse only travels the 1 meter distance from the bottom mirror to the top mirror in a vacuum tube it only takes 3.3356 nanoseconds for the pulse to travel between the two mirrors.Any forward motion of the pulse is caused by the force exerted up the pulse by the movement of the vacuum tube. This force being exerted upon the pulse causes the pulse to cease being in an inertial frame of reference, and places it in a non-inertial frame of reference.God Bless,

Hi Taq, At rest relative to what?God Bless,