Existence

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 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,

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

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

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

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

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

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

Hi NoNukes, If the laser pen is installed at a 90 angle relative to the travel of the car on the tracks will the pulse hit the detector as drawn in Message 950? Lets see if I understand what you are trying to say.

y = pen
x = sensor
| = direction pen pointed when pulse emitted.
> = direction of car traveling at 0.5 c.
                |          |         |          |
                y          y         y          y

                                     >>>>>

     x          x          x          x          x
     S          D          S          D          S  

The signal is sent when the sensor passes over the first S.
This signal causes the laser pen to release a pulse at the first y.
At this point the sensor mounted on the car just above the tracks is over the first D.

y
.
.
.
.
.
x     x  

Are you saying because the sensor mounted on the car and has moved 2 feet from the first x to the second x the pulse can not travel at a 180 relative to the position the laser pen was in when the pulse was emitted?You do realize the laser pen is directly over the second x when the sensor gets there as they are attached to the same frame.Is Newton's first law which is

quote:

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Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

---

incorporated and operational in postulate #1 of SR?If so what external force do you apply to cause the pulse to travel at an angle? What difference does it make whether the laser pen is mounted on the frame added to the car or mounted on the ground and trigered by a sensor on the car.The pulse when emitted from the laser pen will travel in a vacuum in a straight line in which the laser pen is pointed at the moment the pulse was emitted.That means the location the sensor on the car is after it has moved 2 feet will have no effect on the pulse that was emitted from the laser pen directly above the detector on the track. The pulse will hit the detector and cause the light to flash.God Bless,

Hi Taq, Do you ever read anything posted?Are do you just put your fingers in motion and allow them to type anything without first putting your brain into gear?The experiment is set up with the car traveling over the Salt Lake Flats at 0.5 c relative to the Salt Lake Flats.There is a pole mounted on top of the car at a 90 angle relative to the travel of the car. On top of that pole is exactly 4 feet from the exterior surface of the roof of the car. The detector extends 9 inches towards the rear of the car.The laser pen is mounted through the roof of the car flush with exterior of the roof at a 90 angle relative to the travel of the car.The laser pen is not mounted at an angle other than 90 relative to the travel of the car.Now according to Newton the pulse will travel in a vacuum in a straight line at a constant c unless an external force is applied.This law of Newton is a part of postulate #1. Yes. Yes the pulse will travel in a straight line relative to the pole.The problem is the pole is moving as it is attached to the car that is traveling at 0.5 c relative to the emitted pulse. So the pulse will travel in a straight line path from the point it was emitted which will be 2 feet from the pole when the pulse reaches the 4 foot height the detector is from the roof of the car.Therefore the pulse will miss the detector. But the pulse is not traveling parallel to the vertical pole, because the pole is moving at 0.5 c horozontal relative to the vertical motion of the pulse. The laser pen has been mounted through and flush with the exterior of the roof of the car since Message 827.God Bless,

Hi Taq, Yes. Yes. The pulse does not travel in a straight line from the laser pen.So no the pulse will not strike the second S.The pulse is not traveling verticle relative to the position of the pen.The pulse is traveling verticle relative to the position the pulse was emitted at Y.At the moment the pulse is emitted from the laser pen at Y which is mounted at a 90 angle to the motion of the car the pulse is in a vacuum with nothing to exert force on it to change it's direction.Since it was emitted at the point it was directly over the detector aimed at the detector the pulse will hit the detector and cause the light to flash.For the pulse to somehow change its direction and hit the S as you claim there would have to be external force placed upon the pulse to change it's straight line journey to the detector.What do you propose to exert that force.In the modified light clock on my cycle where the pulse is in a vacuum tube the tube exerts external force on the pulse as it travels between the mirrors and forces the pulse in the direction the cycle is traveling.But in the open vacuum there is nothing to force the pulse to not travel at 180 relative to the laser pen which is mounted at a 90 angle to the travel of the car.God Bless,

Hi crash, How do you place the pulse in the frame of the car?The pulse has it's own coordinate system. It has distance from the laser diode and distance from the pole and distance from the detector.So the car is traveling relative to that coordinate system at 0.5 c. The pulse is traveling at c in a straight line relative to the position of the laser pen at the point the laser pen was activated and the pulse was emitted, which is at a 90 angle relative to the motion of the car.So the statement the car is traveling at 0.5 c relative to the point the pulse was emitted at is true. And if you want to say the point the pulse was emitted at is traveling at 0.5 c relative to the car that is also true. Maybe I did not make that clear. If you disagree explain why. To be correct you would have to say the observer will observe the pulse moving at c at a 90 angle relative to the motion of the car. I was talking about the car with the laser pen mounted on a frame on the rear and you are talking about the one with the laser pen in the roof with the pole on top. So I will make you a diagram.

D = detector
P = pulse emitted
| = direction pen pointed when pulse emitted.
> = direction of car traveling at 0.5 c.

     D          D          D

                  ^ 4 feet
                     direction of car travel  >>>>>

     ^          ^ direction pulse emitted
     P  > 2'    P
     |          | 

The laser pen is located below the P in the position it is in when the pulse is released.So the pulse is released from the pen at the first P in the direction the pen is pointed.
You say the pulse will travel in a straight line but will hit the second detector rather that the one directly above the point the pulse is emitted.What causes the pulse to change it's straight line direction so it hits the second detector instead of the one directly above the point the pulse is emitted?God Bless,

Hi NoNukes, No force needed to do what?In the following diagram for the pulse emitted from the pen above the second y when trigered by the signal from the first S to hit the second S the direction of the travel of the pulse has to be altered.

y = pen
x = sensor
| = direction pen pointed when pulse emitted.
> = direction of car traveling at 0.5 c.
     |          |          |         |          |
     y          y          y         y          y
                |                      \
                |                        \
                |                        >>\>>>
                |                            \
     x          x          x          x        \ x
     S          D          S          D          S

                           N
                        NoNukes 

You are standing on the Salt Lake Flats 100 feet from the track the car is traveling on.Will you see the pulse travel as depicted from the second y to the first D?This pulse travels at a 180 angle relative to the position of the laser pen when the pulse is emitted.Or will you see the pulse travel as depicted from the fourth y to the third S?This pulse travels at a 26.57 angle relative to the position of the laser pen when the pulse is emitted. I would agree that you could put a dot in the center of a piece of paper and draw a six inch line through the dot with 3 inches on each side of the dot and have the direction a photon or any other object could move in a straight line.You could then draw a line at any degree around a 360 degree circle intersecting the dot and still have a straight line for the photon to travel.So when you say a photon moves vertically at finite speed, what is the photon traveling vertical relative too?In the experiment we have been talking about it is specified that the pulse is emitted traveling at a 90 angle relative to the travel motion of the car on the tracks on the Salt Lake Flats.So it would not make any difference where the earth was in it's rotation cycle the pulse would be emitted traveling at a 90 angle relative to the travel motion of the car on the tracks on the Salt Lake Flats.Now where could you place an observer where said observer could see the pulse emitted that said observer would not see the pulse emitted at a 90 angle relative to the travel motion of the car on the tracks on the Salt Lake Flats. Where have I discussed a light beam in these experiments?A pulse of laser light from my laser pen occurs when I press and release the button on the side of the pen.I can not change the trajectory of this pulse of light.A beam of laser light from my laser pen occurs when I press and hold the button on the side of the pen.I can change the trajectory of the beam of light simply by changing the direction the laser pen is pointed or the position of the laser pen.Are you sure you are not getting the two confused? So where would you place an inertial reference frame that the pulse could be observed from by an observer when emitted from the laser pen that the pulse would not leave the laser pen at a 90 angle relative to the travel motion of the car on the tracks on the Salt Lake Flats?Playing around with my Chief Architect professional 3D architectural program I can find no place in a 3D picture I can place a 90 angle and turn and twist it that it ceases to be a 90 angle.God Bless,

Hi Taq, You did not use enough of the quote to ask a question.This is what brought about my comment you quoted.In Message 970 I said:

quote:

---

Taq writes:

​

If the pulse is travelling in a straight line from the pen laser in the car's frame of reference then the light pulse will strike the second S.

​

The pulse does not travel in a straight line from the laser pen.

​

So no the pulse will not strike the second S.

​

The pulse is not traveling verticle relative to the position of the pen.

​

The pulse is traveling verticle relative to the position the pulse was emitted at Y.

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You were saying the pulse had traveled at an angle to reach the second S when the pulse was emitted 2 feet to the left of that point.So you saying it was a straight line it was not correct as it would have to be at a 26.57 angle relative to the position of the laser pen when the pulse is emitted to strike the second S. The light pulse does travel straight down from the point the pulse is emitted.It does not travel straight down from the point the laser pen is above the second S. That would require instant travel. In any reference frame that the pulse can be observed when emitted from the laser pen the car will be moving at 0.5 c relative to the Salt Lake Flats. No aether is required.At the moment the pulse is created and emitted from the laser pen it is observed in the frame the tracks, sensors and detectors are motionless in and the car is traveling at 0.5 c relative too.The pulse is traveling at c relative to the point it was emitted toward the detector the laser pen was pointed at when the pulse was emitted.The point the pulse was emitted at and the detector are motionless in their frame. OK remove the vacuum tube from between the mirrors and the pulse will miss the top mirror.God Bless,

Hi Taq, Because the point the pulse is emitted is 2 feet to the left of where you say the pulse hits the second S.You would have to draw a straight line from the point emitted to the second S which would be at a 26.57 angle relative to the position of the laser pen when the pulse is emitted. OK.But when the pulse is created and emitted from the laser pen pointed at the detector the detector and the point the pulse was emitted is at rest in their frame.Therefore the pulse will travel in a straight line and hit the detector causing the light to flash.What you are failing to get is that the pulse when emitted is no longer in the laser pen and therefore no longer in the drivers frame of reference. From a seated position in the car there is no way the driver could observe the pulse being emitted from the laser pen which is mounted on the rear of the car pointed downard at a 90 angle to the motion of the car relative to the Salt Lake Flats.So no I can't draw a diagram for the driver's observations. But I have stated constantly that the pulse travels verticle at a 90 angle relative to the travel of the car relative to the Salt Lake Flats.So there should be no question of what I am talking about the pulse traveling relative too.

B1----------B2
     B1----------B2
     B1----------B2
     B1----------B2   

The boys are parallel and we will ignore the drop the ball will make between them.The ground is under both of their feet so the ball is travelling parallel to the ground as it travels back and forth between the boys.There is no place the ball is traveling at a 90 angle with respect to the motion of the ground below them.Now if one of them were to start dribbling the ball the ball would be traveling at or near a 90 angle to the motion of the ground below them. From a hot air balloon the observer would have a hard time telling whether the boys were rolling the ball on the ground or tossing it between themselves.That observer would not see anything other than the ball traveling parallel to the motion of the earth. Guess again.A 90 angle looks like this:
|
|
|
|
|_______
90 angle

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,

Hi NoNukes, For the pulse to strike S the laser pen would have to be mounted at a
26.57 angle to the travel of the car, as you would have to draw a straight line from the point emitted to the S.

y = pen
x = sensor
| = direction pen pointed when pulse emitted.
> = direction of car traveling at 0.5 c.
     |          |          |         |          |
     y          y          y         y          y
                |                      \
                |                        \
                |                        >>\>>>
                |                            \
     x          x          x          x        \ x
     S          D          S          D          S

                           N
                        NoNukes 

So for NoNukes to see the pulse hit the second S directly in from of him the pen at the second y would have to be mounted at a 26.57 angle to the travel of the car as the line from the fourth y to the third S shows.But the laser pen is mounted at a 90 angle to the travel of the car.Therefore for the 1 attometer long pulse to strike the S it could not travel in a straight line from the laser pen and strike the S.If the 1 attometer long pulse travels in a straight line from the point emitted where the laser pen is pointed directly at the D detector the pulse will strike the D.So does the pulse travel in a straight line in the direction the laser pen is pointed when the pulse is emitted?Or does the pulse acquire a 26.57 angle after emitted and travel in a straight line from that point to strike the S?I am trying to understand how the statement that the light will travel in a straight line independent of the motion of the emitter, can allow the pulse to travel at a 26.57 angle relative to the direction the pulse was traveling when emitted from the laser pen which is mounted at a 90 angle to the travel of the car.All the math in the world will not explain that.

y
|\
| \
|  \
|   \
|    \
D     S  

You claim the pulse will not hit the D but will hit the S instead so please explain how the pulse changes directions, when it is emitted pointed at D.God Bless,

Hi CS, Which reference frame is that?The drivers reference frame inside the car?The cars reference frame on the Salt Lake Flats?The Salt Lake Flats reference frame?The pulse's reference frame after it is emitted in the vaccum?Or NoNukes reference frame in my diagram?God Bless,