Existence

Hi cavediver, And what has dropping a bag of peanuts from the roof of the car have to do with a light pluse released from a laser pen that is pointed at a 90 angle to the travel of Taq's car which is traveling at 149,896,229 meters per second?But in your example if you held the bag of peanuts up to the roof of the car above your lap they would fall into your lap.If you held the bag of peanut up to the roof of the car over the footwell they would fall into the footwell.If you held them above your face and you were looking up at them when you dropped them they would hit you in the face.But if you were in Taq's car which is in a vacuum traveling at 149,896,229 meters per second, wouldn't the bag of peanuts stay right where they were when you released them?If you opened the sunroof and pitched them 4 foot into the air above the roof of the car they would not return to the inside of the car. They would fall somewhere towards the rear of the car.But if you were in Taq's car the bag of peanuts would continue in the direction you tossed them, at the speed you tossed them, wouldn't they since the car is traveling in a vacuum?But why don't you take the example that is being discussed and show me how Taq can be right when he says the pulse will hit the detector dead center.The car has a pole in the roof of the car that extends at least 4 feet above the car, with a detector mounted exactly 4 foot from the laser pen that is mounted flush with the roof at a 90 angle 1 inch from the pole. The detector protrudes 9 inches from the pole. The car is traveling at 149,896,229 meters per second.My argument is that since the light pulse can not add the forward motion of the car to its verticle speed the light pulse will miss the detector as the detector will have moved 2 feet by the time the light pulse reaches the location the detector was when the light pulse was released.Taq is trying to convince me that even though the detector has moved 2 feet from a 90 angle from the location the light pulse was released, the light pulse will hit the detector dead center.When the light pulse leaves the laser pen it is in free vacuum without anything to cause the pulse to alter the direction of it's travel.The detector has to only move 10 inches for the pulse to miss the detector.cavediver will the pulse of light hit the detector?If so, why?God Bless,

---

"John 5:39 (KJS) Search the scriptures; for in them ye think ye have eternal life: and they are they which testify of me."

Hi NoNukes, I know you do not advocate that anything can travel faster than light.What I can't figure out is why you think you have to stop my cycle that is traveling at 149,896,229 meters per second relative to the distant stars when I turn the headlight on to keep the speed of the light from exceeding 299,792,458 meters per second relative to the distant stars. I can't find, "as measured in any and every reference frame." in the quote of postulate #2.I do find "any inertial frame of reference", which I have asked you to present an inertial frame of reference in the universe that is not affected by an outside force.I also see, "light is always propagated in empty space with a definite velocity c". To borrow your words "Don't skip over that vital point."There is no such thing as empty space. I have no problem with the speed of light being observed from any frame traveling at the speed it travels due to the restrictions of the medium it is passing through, and that speed to be very near to 299,792,458 meters per second.I do have a problem when you tell me when I am travling in a vacuum at 149,896,229 meters per second and I turn on my headlights and the light beam is traveling at 299,792,458 meters per second when you combine the speeds you get 299,792,458 meters per second.You can't combine the two speeds for anything. You can substract the speed of the cycle from the speed of light to determine how fast the distance between the lead edge of the light beam and the cycle is increasing at.The speed of the light can't be affected by the speed of the cycle according to postulate #2.

quote:

---

light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body.

---

It makes no difference what the cycle is doing the light beam from the headlight will travel at 299,792,458 meters per second. I don't see in Message 418 any indications that the clock is enclosed in a vacuum chamber.In fact when I enclosed it in a tube you said it would bounce around and mess up, not working. Where does that take place especially since postulate #2 says:

quote:

---

light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body.

---

That does not say sometimes it says 'always', independent of the state of motion of the emitting body.So it does not make any difference what the observer does or does not do the light is 'always' independent of the motion of the emitting body. So we can agree that the speed of light is not always 299,792,458 meters per second, as the medium through which the light must travel can reduce the speed of light, as it travels through that medium.So saying that the speed of light is always 299,792,458 meters per second is a constant speed is a mistatement. IOW it is false, as postulate #2 refers only to the speed of light in a vacuum independent of what the source is doing. If the train is stopped at the station the speed of the light would still be 299,792,458 meters per second in the vacuum the thought experiment was set up in.If the train is traveling at 74,948,114.5 meters per second the speed of the light pulse would still be 299,792,458 meters per second in the vacuum the thought experiment was set up in.If the train is traveling at 149,896,229 meters per second the speed of the light pulse would still be 299,792,458 meters per second in the vacuum the thought experiment was set up in.If the train is traveling at 99% of the speed of light, the speed of the light pulse would still be 299,792,458 meters per second in the vacuum the thought experiment was set up in. So we are in agreement that cycle can travel at 149,896,229 meters per second and when the headlights are turned on the light pulse will travel at 299,792,458 meters per second in the vacuum. In any frame where the clock is sitting on the bottom of the clock frame with the mirrors parallel with the travel of the cycle or train. The train thought experiment was presented at both speeds, as well as with the laser pens attached to the track with a sensor to cause it to send a light pulse towards the center of the top mirror when a sensor triger passed over the pen sensor. Since the top mirror is only 9 inches from center to the edge the pulse will miss the to mirror by .8 of an inch. I don't remember attaching any car to the train other than a flatcar with a clock attached to it. Why?The clock is on the flatcar.Now if I did attach a passenger car between the engine and the flatcar, anyone in that car would observe the clock from the front and would not know if the light pulse hit the top mirror or not unless they look up at the ceiling of the tunnel and saw the reflection from the light pulse hitting the ceiling of the tunnel.So no they would not observe the light beam travel at a diagonal from the train. Apparantly you have changed my thought experiment and took my clock off the flatcar and placed it in a passenger car on my minature train.Lets go back to the cycle thought experiment where you can place an observer on a plane parallel to the travel of the cycle.I will present my post shortly on that topic.

---

"John 5:39 (KJS) Search the scriptures; for in them ye think ye have eternal life: and they are they which testify of me."

Hi ICANT,I've answered some of the questions in your post several times. You do see the word "any" in postulate #2, don't you? That means that the choice of inertial reference frame is up to us. No matter which one we pick, the postulate holds. Yes, I did add "every" for emphasis. I could further add, "each and every inertial reference frame, without any exception, including a space cycle traveling in interstellar space at constant speed" without changing the meaning of postulate #2. First, the space cycle cannot be traveling at 0.5c relative to "the distant stars" because the distant stars each have their own different motions. As postulated, the cycle is traveling at 0.5c relative to earth and planetX. No one has ever said anything differently up until now. There is no aether frame of reference or "distant stars" frame of reference.Secondly, I never required that the space cycle stop. I should not have to keep denying this every post. Stop attributing this foolishness to me.In different inertial frames of reference, the space cycle's velocity might be 0.5c, 0.4c, or 0.0000c without the space cycle slowing or coming to a stop. Is that too difficult for you to understand?As you drive on the highway at constant, in one frame of reference your car's speed might be 60mph, in yet another frame of reference, your car's speed is 70,000 mph during the "same moment". This true regardless of whether the frames are even inertial. When we say that you are going 60 mph, does that mean that the earth has stopped revolving or does it simply mean that we are using a terra firma frame of reference? For all intents and purposes, the space cycle is traveling inertially up until time to turn around. At the midway point in the journey, the space cycle is a light year away from any other objects in space. Really? I thought you believed that the speed of light in a vacuum was a constant and that it was unaffected by the speed of the source. I have to admit that I cannot understand your issue with this. You don't seem to the like the "as measured in any inertial frame" part. Somehow your eyes seem to glide right over those words. As measured in any inertial reference frame. Let's never for get that as is your wont. Yes I agree that postulate #2 refers to the speed of light propagating in a vacuum. So what? Interplanetary space and interstellar space are close enough. In fact, air at standard temperature and pressure is very nearly close enough. What clock are you talking about? Is there a light clock on the rail car, flatcar or whatever? I thought we were talking about a laser shooting a beam upwards at a mirror on the top frame mounted to a flat car.If that isn't the correct thought experiment, then perhaps I did provide an irrelevant answer.

Everything. And what do you mean by Taq's car travelling at 149,896,229 ms^-1? With respect to what? Of course!

Actually, ICANT's position is more wrong than you suggest here Taq. Objects do not enter or leave a reference frame. The idea of objects leaving a reference frame (inertial or not) is foolish. It's all part of ICANT's "didn't see it, doesn't matter" wishful thinking.Objects may be at rest, at constant velocity or accelerating as observed from an inertial reference frame, but they do not enter or leave a reference frame even if they are light years distant from any observer or if they simply get put behind the milk in the fridge.Light propagating in empty space is never at rest in any inertial frame. It always moves at speed c as measured in any reference frame.

Hi NoNukes,In Message 425 You presented the following light clock showing the pattern my wife would see the light travel.I pointed out to you that you had mounted the clock parallel to my handle bars and that pattern could not be seen by my wife as she would see exactly whatever I saw as she would be looking over my shoulder.I then rearranged the light clock you had installed on my cycle where the example you showed above would have the possibility of being seen.I then pointed out that my wife was still behind me and would see exactly what I saw.To have an opportunity to see the light pulse go in a sawtooth pattern my wife would have to observe my cycle trip from the side.I then gave you an experiment you can do at your desk to prove that you could not see the light beam between the bottom mirror and the top mirror.If you have forgotten or ignored that experiment you can aim your laser pen at the ceiling and prove to yourself that you can not see the beam between your hand and the ceiling. You can observe the red dot on the pen and ceiling but you can not observe any light between the pen and ceiling.In your last post you still assert that the pulse is traveling at an angle thus it takes longer for the the pulse to reach T from B.I will see if I can punch a couple of holes in that assertion.In Message 710 You said But the light pulse does not travel at an angel.The light pulse travels at a 90 angle to the travel of the cycle.But since you keep assuring me that the light has to take a longer time to reach the top because it has to travel at an angle I will try one more time to make my point and refute yours.In Message 705 you said:

quote:

---

ICANT, this is just plain silly. You cannot force light to travel in a straight line by confining it to a tube. If the light is not originally directed down the axis of the tube, then the light would instead bounce off the sides of the tube. (If not on the first pass, after a few reflections between the mirrors). The actual light path would be quite a bit different then you describe here.

---

I really think a light pulse can be forced to travel in a straight line by confining it in a tube. You did propose it was in a vacuum enclosure, in the post I am replying too.I see you can't understand simple things as they are too non-complicated for your education.Lets modify my clock on the cycle and see.So let me try a few more modifications to the clock and see if I can force the pulse to go verticle at a 90 angle to the motion of the cycle.Lets make that tube a black metal vacuum tube so the light can't escape.Lets put a detector at the top and bottom that causes a light to flash.Lets program the detectors to flash the light at T and B alternating between them every 149,896,229 pulses.Lets assume this light pulse is similar to the atom and it will strike the top mirror 149,896,229 time's and flash the top light, once it is started. Then strike the bottom mirror 149,896,229 time's and flash the bottom light, which would equal 1 second. This continuing for eternity or until the tube is removed from between the mirrors.T = top flash
B = bottom flash
X = me
Z = observer at a distance
. is for spacer to keep program from putting letters close together.Now we have what I would see below.

T
.
.
.
.
.
              B
.
.             X local observer

I would see one of the light's flash every 500,000,000 nanoseconds.

fig 5
                  T                     T
.
.
.
.
.
      B                     B                     B
      1 2 The #1  obsever is 100 meters from B and the #2observer is 100 meters to the right of #1
.     observer.
.
      Z Observer is at a distance of 149,896,229 meters from the light pulse at the first B. 

The verticle distance between two mirrors is 1 meter.The horozontal distance between B and T is 149,896,229 meters.The pulse travels back and forth between the mirrors 149,896,229 times and the light flashes at T 1 second has passed since the flash at the first B.If we draw a straight line on the bottom mirror line 149,896,229 meters and then draw a vertical line at a 90 angle to the bottom mirror and extend it 1 meter to the top mirror it will intersect with the point T that the light will flash at.If we solve for the distance from B to T using 10,000,000,000 angstroms as the distance from endpoint on B line to T and 1,498,962,290,000,000,000 angstroms as the distance the cycle has traveled in 1 second, we find, 1,498,962,290,000,000,000 angstroms.So when we convert meters into angstroms there is not 1 angstrom difference in the measurments.If we solve for the angle at B we find, 3.8223629703908244e-7.
If we solve for the angle at T we find, 89.9999996177637.So the angle at T is pretty close to 90.This is what the observer Z would see if parked at the first B.How could that observer Z tell the difference whether the light flash at T was on the top mirror or the bottom mirror at 149,896,229 meters?Now lets back the observer Z up and place observer Z at 149,896,229 meters from the point the light flashes at the first B.Lets use ZT for the straight line from Z to T.Solving for distance ZT we find it is, 211,985,280.00038323 meters.How long would it take for the light flash at the first T to reach the observer at Z?It would take the light flash 707,106,781.1865473 nanoseconds to reach the observer at ZWhy does it take 707,106,781.1865473 nanoseconds to reach the observer at Z when it only took 500,000,000 nanoseconds for the pulse to cause the light to flash after causing the flash at the first B?Since the light flash at the first B took 500,000,000 nanoseconds to reach the observer at Z, would it be due to the 149,896,229 meters distance the cycle has moved to the right of Z which would increase the distance from Z to the first T?Or is it caused by time dilation?Oh I forgot you said the light pulse had to travel at an angle, from B to T.Since there is not 1 angstrom difference in the measurments of the distance the cycle travels and the light flash has to travel, what would be the difference in time required for the light to travel at an angle going from B to T?The pulse travels back and forth between the mirrors 149,896,229 times and the light flashes at B₂ 2 seconds has passed since the flash at the first B.Distance from Z to B₂ is 335,178,157.6148753 meters.
Time required for light pulse to reach Z 1,118,033,988.749895 nanoseconds.The pulse travels back and forth between the mirrors 149,896,229 times and the light flashes at T₂ 3 seconds has passed since the flash at the first B.Distance from Z to TT₂ is 474,013,496.3101836 meters.
Time required for light pulse to reach Z is 1,581,138,830.08419 nanoseconds.The pulse travels back and forth between the mirrors 149,896,229 times and the light flashes at B₃ 4 seconds has passed since the flash at the first B.Distance from Z to B₃ is 618,037,985.048773 meters.Time required for light pulse to reach Z is 2,061,552,812.80883 nanoseconds.Adding the numbers for the #1 and #2 observer.Observer #1The light flashes at the first B.
Time required for light flash to reach o#1 is 333.564095198152 nanoseconds.The pulse travels back and forth between the mirrors 149,896,229 times and the light flashes at the first T, 1 second has passed since the flash at the first B.Distance from o1 to the first T is 149,896,229.00003335 meters.
Time required for light flash to reach o1 is 44,937,758.93685087 nanoseconds.The pulse travels back and forth between the mirrors 149,896,229 times and the light flashes at B₂ 2 seconds has passed since the flash at the first B.Distance from o1 to B₂ is 299,792,458.0000167 meters.
Time required for light flash to reach o1 is 1,000,000,000.000056 nanoseconds.The pulse travels back and forth between the mirrors 149,896,229 times and the light flashes at T₂ 3 seconds has passed since the flash at the first B.Distance from o1 to T₂ is 449,688,687.00001114 meters.
Time required for light flash to reach o1 is 1,500,000,000.000037 nanoseconds.The pulse travels back and forth between the mirrors 149,896,229 times and the light flashes at B₃ 4 seconds has passed since the flash at the first B.Distance from o1 to B₃ is 599,584,916.0000083 meters.
Time required for light flash to reach o1 is 2,000,000,000.000027 nanoseconds.If anyone is interested if you can check the time required for the light flash to reach observer #1 is decreasing the further the light pluse is from the observer.Observer #2The light flashes at the first B.
Distance from o2 to the first B is 141.42 meters.
Time required for light flash to reach o#2 is 471.7263434292266 nanoseconds.The pulse travels back and forth between the mirrors 149,896,229 times and the light flashes at the first T, 1 second has passed since the flash at the first B.Distance from o2 to the first T is 149,896,129.00003335 meters.
Time required for light flash to reach o2 is 499,999,666.4360158 nanoseconds.The pulse travels back and forth between the mirrors 149,896,229 times and the light flashes at B₂ 2 seconds has passed since the flash at the first B.Distance from o2 to B₂ is 299,792,358.0000167 meters.
Time required for light flash to reach o2 is 999,999,999.9999999 nanoseconds.The pulse travels back and forth between the mirrors 149,896,229 times and the light flashes at T₂ 3 seconds has passed since the flash at the first B.Distance from o2 to T₂ is 449,688,587.0000111 meters.
Time required for light flash to reach o2 is 1,499,999,666.435942 nanoseconds.The pulse travels back and forth between the mirrors 149,896,229 times and the light flashes at B₃ 4 seconds has passed since the flash at the first B.Distance from o2 to B₃ is 599,584,816.0000083 meters.
Time required for light flash to reach o2 is 1999999666.435932 nanoseconds.As you can see it only takes 4 seconds in all the examples for
The first T, the first B, the second T, and the third B to all flash a light.So the light does not travel at an angle as you assert.Now if I didn't mess up the math it should be correct.God Bless,Edited by Admin, : Fix width.

---

"John 5:39 (KJS) Search the scriptures; for in them ye think ye have eternal life: and they are they which testify of me."

Yes, and the light beam never contacts the chamber. What you really think is completely wrong. If you shine light down a tube, the light will not travel in a straight line down the tube. Are you familiar with fiber-optics? Do you know what path light follows in a light fiber. Yeah, sure ICANT. But because your mind is uncomplicated by knowledge, you can see pigs fly. So what? You have yet to cite any consequences of that. Are you suggesting that the path of a photon is something other than a straight line from source to destination? In fact all light pulses travel at some angle, even if that angle is 90degrees. The problem for you is that observers in different inertial reference frames will see different angles. I've demonstrated exactly that using a thought experiment in which you agree with the details but not the conclusion. The problem is that you have a blind spot regarding how moving coordinate systems work. I've explain it on a level that most high school students can grasp and its still miles over your head.Just to test your understanding, or lack thereof, Here is a simple question for you.Let's place the origin of a coordinate system (that moves along with the flat car) at the front edge of the flat car. Let's assume a laser beam directed vertically Tell me, in that coordinate system, the distance from the front edge of the flat car at which the leading photon (or any other photon) of the laser reaches the height of the lower frame, and also the distance from the front edge of the light car at which the light beam reaches the 1 meter high point above that lower frame. Now show me how a vertical line connects those two points in the flat car coordinate system.Hint: you cannot make the showing.I'm still reading your thought experiment analysis. I'll post detailed answers in another post. But I'll give the executive summary here: You don't know what an inertial frame of reference is, and your ignorance is on display for all to see.But what the heck does this mean:

Hi ICANT,I took a look at your calculations. As I anticipated, your calculations miss the point by a considerable margin, in those cases where they simply aren't wrong. And of course I pointed out to you that special relativity is not about what you see. Instead it is about reality as determined and measured in distinct coordinate systems. An observer at rest in a given inertial reference frame can be located anywhere in space. Remember when you asked me the off beat question of how two people in the same room facing each other could be in the same reference frame because "left" for one meant "right" for another.When you observer an object in the distance, the object distends a smaller angle, but is the object actually smaller? That's akin to the line of argument in your analysis above. Perspective effects from viewing things at an oblique angle or at a distance have nothing to do with length contraction. Sure ICANT. Pick any photon and tell me how being confined in a tube causes that photon to travel in a straight line. The purpose of the vacuum chamber is merely to keep out space dust. The vacuum chamber is large enough not to affect the path of the light beam. That ought to do it. The black tube will absorb the photon if it ever touches it. So I guess it won't escape. Problem solved. I suppose this is the benefit of not having a technical education.But if you think the tube helps, then let's see where it leads. Tee hee. I'm already giggling. Looks like you made two silly errors here.
---------------------------------------------------------------------
First using your numbers that tiny angle you calculated is the angle from horizontal, not from the vertical. Note that you've indicated a horizontal distance that is much larger than the vertical one. Using your numbers, the path of the light beam is nearly horizontal (180 degrees). Uh oh. If you want me to figure out the time dilation factor based on the 1 meter distance expanding into 149 million meters, we can do that, but you won't like the answer.IF you find it hard to accept that I'm right, try drawing a right triangle where the base is at only ten times the height, and you'll get the general impression. Using your numbers, the base is 149 million times the height. That makes the photon travel nearly horizontally and NOT nearly vertically.In any event the change from 90 to 180 degrees if correct is a substantial change in angle. Of course, as I show below, you've over estimated the angle.-------------------------------------------
Secondly, you've just made the same silly error you caught me making. The horizontal distance you've given is based on millions of photon round trips between mirrors and not on just a single trip by the photon. As we discussed, the non-relativistic estimate for the horizontal distance for a single trip is only 0.5 meters (but even that is wrong, because the length of that path would be 1.118 meters rather than 1 meter, and light cannot travel more than one meter in 3.33564 nsec). The corrected math (simple 8th grade algebra) gives a small angle of 30 degrees, but that is at least a lot smaller than the nearly 180 degree angle we get from using your numbers.Since you have the nerve to taunt me with your bad calculations, I'll indulge in the guilty pleasure of chuckling out loud. Yes I did say that. And unless you can reconcile your errors, your own presentation leads to the same result. Assuming your math is even correct, no that's not time dilation. That's simply light propagation delay. Special Relativity is about explaining time differences after taking light propagation delay into account.Once you redo your math based on my corrections above, you'll see that angle you are crowing about not existing is exactly 30 degrees. (Actually, you are more likely to claim that the angle is 26.5 degrees until I cajole you into understanding that answer is 30 degrees). I'm not going to bother checking that math. Assuming you mean the further the light pulse is from observer #1, I know that ain't right.About the only thing left to do is grade your paper. Why don't you assign the letter grade yourself.Edited by NoNukes, : No reason given.

Hi ICANT, Of course there are. This question would not be asked by someone who understands the subject. In any reference frame, inertial or not, there is a system of coordinates in which an observer or some other object is stationary. You might be stationary at your desk right now.In its simplest form, relativity deals with inertial frames. In that sense, a "stationary frame" means any inertial frame so designated. The point is made clearly in Einstein's paper, and I don't think there is much dispute about the point even among SR doubters. The two statements are equivalent. And that is not just my version, it's everyone's.Yes, I'm sometimes a bit sloppy and leave of the "in empty space" part, but for the most part, the postulate is true for observations in air as well. I don't gloss over the "as measured in any inertial reference frame", or the "in the 'stationary' system of coordinates' as you like to do.

Hi NoNukes, So when I leave my wife and start my journey to planet X I can never leave my wife's frame of reference and be in my own frame of reference.Is that what you are saying?God Bless,

---

"John 5:39 (KJS) Search the scriptures; for in them ye think ye have eternal life: and they are they which testify of me."

Hi NoNukes, When I place my laser pen in a 12 inch tube and flash it, the flash covers the entire exit end of the tube upon exiting. The consequences of that is that your observer can not see the light travel at an angle in the light clock as you have asserted. Relative to what?In my thought experiment it goes in a 90 angel to the travel of the cycle or train. In Taq's it goes at a 90 angle to the car. You mean like the fellow in the car ahead of an accident would see one thing if he/she was looking in the rear view miror or actually facing the back of their automobile and the person driving the car behind the accident would see something different. The person standing by the roadside would even see something else. But they all would see an accident, just from different angles. As proposed you are correct I can not give you the distance from the front edge of the light car to the point the beam reaches 1 meter high. This is assuming the light can bounce back and forth between the mirrors 149,896,229 times in 1 second.Similar to the 9,192,631,770 cycles of radiation corresponding to the transition between two energy levels of the caesium-133 atom in 1 second.God Bless,

---

"John 5:39 (KJS) Search the scriptures; for in them ye think ye have eternal life: and they are they which testify of me."

Hi ICANTI wanted to answer your timely question. Your summary is partially correct. You will always be "in" your wife's frame of reference" to the extent that said phrase makes any sense at all. As measured in the wife's frame, your velocity will be 0.5c towards planet X, but as measured in your own frame of reference your velocity will be zero. That's why I say that being "in a frame of reference" is a confused man's concept, and leaving a frame is even more confused. I recommend avoiding those two expressions unless it is agreed what they meant.At any given point on the journey, you will have coordinates and a velocity in every conceivable frame of reference. The values for those parameters will be different depending on the chosen frame of reference. If the frames in question are inertial frames, then we can make additional statements about the velocities and coordinates and observed physics (postulates #1 and #2 for example) using the coordinate systems for those reference frames.If the frames are not inertial, then we can still apply SR, but our work is much more difficult. Most high school students cannot work relativity problems when the involved velocities are a significant fraction of c. Fortunately, in many situations involving short periods of time, we can treat earth as an inertial frame.In summary, when I say that you have a velocity of zero in the space cycle reference frame that does not mean that you aren't getting closer to planet X at 0.5c as measured in the earth/planet X coordinate system.In fact, inertial frame or not, when you are kicked back on the seat of your cycle, your rump's velocity is always zero in the cycle frame of reference. This observation is true regardless of whether your cycle is sitting in the hangar, racing down the highway, or traveling through interstellar space towards planet X.Hopefully, this will help understand what's being said during the discussion.

Nope. For entire existence of that photon it is travelling at c. It hasn't moved 2 feet in the inertial frame of the car. The pole keeps it directly above the pen laser the entire time. That's the point. Since all non-accelerating inertial frames can be considered the same, the same rules apply to all non-accelerating inertial frames. Stating that the car is moving relative to another inertial frame does not change the laws of physics in the car's inertial frame. This doesn't change the fact that the Michelson-Morley experiment falsifies your claims. You tried to explain it away by invoking some magical vacuum. Now you are backtracking on that as well. You forgot to move S along with D. Why do you need to park the car in order to have a non-accelerating inertial frame?Edited by Taq, : No reason given.

Right here:

quote:

---

I can be going towards the light and it will be traveling 299,792,458 meters per second, as long as it is in a vaccum.

​

I can be going in the same direction as the light and the light will be traveling 299,792,458 meters per second, as long as it is in a vaccum.

​

I can be waiting at the train station standing on the platform and the light will be traveling at 299,792,458 meters per second, as long as it is in a vaccum.

​

I can be dead and buried and the light will still be traveling 299,792,458 meters per second, as long as it is in a vaccum.

---

So if you are sitting on a train with a velocity of 0.5 c that is moving in the same direction as the light beam you will observe the light passing you by at c, correct?At that very spot along the tracks is someone who is standing stationary to the tracks. This person will also see the light passing them by at c, correct?So how can both observers, you on the train travelling at 0.5 c and someone standing stationary with respect to tracks, measure the same velocity for that light?

Hi ICANT,I thought I had posted a response to this message. Perhaps I neglected to submit. I apologize if the result is two responses. You have a gross conceptual error about what's happening in that tube. I'm not going to go into it here, but perhaps you need to think this through a bit. That does not matter one whit. What is important is not merely what is seen. The physics does not change if I turn my back or if I hide behind a rock or my own ignorance. What is important is what can be measured/calculated/determined using the coordinate system of the inertial frame of reference. I can determine the path of a photon if I know the coordinates (x, y, z, and t) of any two points on the path in the same reference frame. No, ICANT. That is decidedly NOT what I mean.I'm not talking about differences in perspective due to the different viewing angles, distances, perspective, and mental states of different observers. None of those things change the velocities observed/measured/calculated during the accident.I'll give you an example of what I mean.While traveling at excessive speed on the highway you observe a car in your rear view mirror, the car is maintaining a constant distance from your rear bumper. You might say that the car is closing at zero miles per hour.On the other hand, an observer on an overpass sees the two cars as the zoom away. The observer reports that both cars are moving at 120 miles per hour with you in the lead.A second observer traveling in your direction within the speed limit might say, "I was doing 55mph, but both cars shot by me as if I were standing still." Why can't you? Since you understand that the light will not hit the mirror surely you can at least agree that the point at a height of zero meters is closer to the front edge of the flat car than is the point of at one meter high. After all, the light is going to miss that mirror.In fact, the difference in distances is 0.2582 meters for a train moving at 0.25 c, and 0.57735 meters for a train moving at 0.5c. I haven't shown you how to obtain those answers, but I can do so if you are interested. But you should be able to check that those numbers are consistent with postulate #2. I have shown you how to do that.I find it amusing that you cannot make the calculation, yet you are sure than my answers are wrong because I am too well educated. Relative to what, indeed. As measured in a reference frame of your choosing, light can be directed in any direction. Not a problem. We're discussing different light beams and different observers. There is no contradiction. You can direct a light vertically in any inertial frame of reference.Edited by NoNukes, : grammar too vs two