|
Register | Sign In |
|
QuickSearch
EvC Forum active members: 65 (9164 total) |
| |
ChatGPT | |
Total: 916,423 Year: 3,680/9,624 Month: 551/974 Week: 164/276 Day: 4/34 Hour: 1/0 |
Thread ▼ Details |
|
Thread Info
|
|
|
Author | Topic: Existence | |||||||||||||||||||||||||||||||||
ICANT Member Posts: 6769 From: SSC Joined: Member Rating: 1.5 |
Hi Son,
Son writes: What Taq actually asked was how much the car moved in the reference frame of the car. Does the car still move two feets away in this frame reference? The car moves 2 feet relative to the Salt Lake Flats. I think that means the frame the car is in moves 2 feet relative to the Salt Lake Flats. So how would the car move 2 feet in its own reference frame? 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."
|
|||||||||||||||||||||||||||||||||
crashfrog Member (Idle past 1488 days) Posts: 19762 From: Silver Spring, MD Joined: |
For the pulse to travel the way you have it drawn the vacuum tube in the modified light clock would have to be 1.1180 meters long and swing back and forth from strike point to strike point. No, because the tube is moving as well, because it's attached to the car. The image you're confused about is the path of the light pulse relative to an observer not moving with the car; by trigonometry, the path is longer, yet the light pulse has the same rate of travel to that observer as to the one in the car. Therefore the light clock "ticks" slower for the stationary observer than for the one in the car, proving that time dilates as a function of velocity. Edited by crashfrog, : No reason given.
|
|||||||||||||||||||||||||||||||||
fearandloathing Member (Idle past 4166 days) Posts: 990 From: Burlington, NC, USA Joined: |
Hi,
The pulse is dragged sideways as it makes its journey between the mirrors. This dragging is caused by the forward motion of the cycle relative to the Earth and PlanetX. You seem to be on the verge of enlightenment, so let me ask if the pulse is moving up and down and dragged forward how could the path it takes not be longer than 1 meter? It really is just simple geometry as has been pointed out previously by many others.
ICANT writes: Hi fearandloathing,
fearandloathing writes:
Please explain to me why you feel like light behaves differently than say...anything you drop in a moving vehicle? The pulse released from the pen on my cycle is in an open clock, the same as the one from the car. The pulse is released in a vacuum. It is not released inside of anything, other than a vacuum. The only one that is in a vacuum tube is the one NoNukes and I are discussing. God Bless, If there is an answer to my question in there I don't see it. Would you like to try again? "No sympathy for the devil; keep that in mind. Buy the ticket, take the ride...and if it occasionally gets a little heavier than what you had in mind, well...maybe chalk it off to forced conscious expansion: Tune in, freak out, get beaten." Hunter S. Thompson Ad astra per aspera Nihil curo de ista tua stulta superstitione.
|
|||||||||||||||||||||||||||||||||
hooah212002 Member (Idle past 823 days) Posts: 3193 Joined:
|
So how would the car move 2 feet in its own reference frame? So you admit and realize that the car is not moving within it's own reference frame, yes? Since you just admitted it, how would the light pulse do anything BUT go straight up and down? How would it miss whatever is directly above it? "Why don't you call upon your God to strike me? Oh, I forgot it's because he's fake like Thor, so bite me" -Greydon Square
|
|||||||||||||||||||||||||||||||||
Taq Member Posts: 10038 Joined: Member Rating: 5.3 |
The car moves 2 feet relative to the Salt Lake Flats. We are not using the lake's frame of reference. We are using the car's frame of reference. In the car's frame of reference there is no relative motion between the pen laser, detector, and car. None.
|
|||||||||||||||||||||||||||||||||
Taq Member Posts: 10038 Joined: Member Rating: 5.3 |
What is what in reference within the reference frame of the car?
Relative to the car, how far do the pen laser and detector move during the transit of the light pulse? The answer is zero. They don't move at all, correct?
So the 2 feet you are asking about in this post is the 2 feet the car moves in a vacuum at 149,896,229 meters per second relative to the Salt Lake flats . . . As I stated above, we are not using the lake's frame of reference. We are using the car's frame of reference.
|
|||||||||||||||||||||||||||||||||
Son Member (Idle past 3851 days) Posts: 346 From: France,Paris Joined: |
Since neither the car, nor the pen laser moves in the car's frame of reference, why then would the laser miss it's target?
If you answer that it's because it moves relative to the salt lakes, why then is the salt's lake frame of reference more important than the car's frame of reference in determining light's behaviour?
|
|||||||||||||||||||||||||||||||||
NoNukes Inactive Member |
Hi ICANT
I need more coddling. Okay. I'll serve up some coddling. I apologize in advance for any thing I say that under estimates your knowledge. I expect that I'll probably overestimate.
If I am in my van on a oval race course and set my cruise control at 100 mph I will be traveling at 100 miles per hour and yet I will be accelerating due to the force exerted upon the car and my body will want to go toward the passenger side of the car. Exactly right, ICANT. But the point is that you cannot cover increased distance without increasing your speed. Your explanation of acceleration is fine. But it is unhelpful to explaining your way out of the contradiction you've created. You cannot travel further than 100 miles in an hour without increasing your speed. You say that the photon is dragged. So the question remains how does the photon follow the dragged path without exceeding the speed "c". You have yet to answer that question. Draw the dragged path, or just acknowledge that I've already done so. Then show that the dragged path is 1 meter or less. But you cannot do that, can you?
Reality is the vacuum tube is 1 meter long between two mirrors once the pulse is started it will travel back and forth between the mirrors striking the top mirror 149,896,229 times the light on top of the vacuum tube will flash. Yes. I agree with what you say above. But being constrained to stay in a one meter tube does not prevent the photon from traveling more than one meter during the trip from top to bottom of the tube. For example if the tube is moving in a given reference frame, the photon must travel along with the tube while also moving up and down the tube.
ICANT writes: For the pulse to travel the way you have it drawn the vacuum tube in the modified light clock would have to be 1.1180 meters long and swing back and forth from strike point to strike point. This must be among the more ridiculous things you've written in this thread. It's surely in the top twenty. I'm seriously contemplating making a list. No ICANT, the tube would not be required to be longer than 1 meter or to swing back and forth. As you've indicated, the tube is vertical. But the tube does not stretch from the lower mirror to a future position of the top mirror. In order to contain the photon while the photon moves along a diagonal, all that is required is that the vertical tube move horizontally at constant speed as the photon moves up and down along the tube. No need for any to and fro tube swinging or stretching. This should be actually an easy thing to picture, so I do not accept that understanding this is beyond you. I'll keep trying. I've got some alternate methods of explaining the phenomenon in mind. I'll try them if I see further denials on your part. Seriously, forget the photon. As you've explained it, the tube must swing back and forth simply to stay attached to the two mirrors. You are simply ignoring the time component. The mirror stretches vertically from the position of the bottom mirror at one given instant in time to the position of the top mirror at the same instant in time.
I don't ever remember saying the interior of the tube was black. I did say a black tube. Explain how being black helps contain the photon in the tube. If the interior of the tube is not black, what difference does the color of the tube make to the photon. Why not just admit that the choice of color was hastily made? Edited by NoNukes, : No reason given.
|
|||||||||||||||||||||||||||||||||
Taq Member Posts: 10038 Joined: Member Rating: 5.3 |
The pulse will be going up and down in the vacuum tube and yet the motion of the cycle the light clock is attached too will exert force on the tube which in return exerts force upon the pulse. The cycle is going at a constant velocity, therefore there is no force being exerted on the tube. If you were to sever all physical contact between the tube and the cycle it will happily float along with the cycle. Just think of the shuttle putting a satellite into orbit. When the astronauts sever contact between satellite and the shuttle does the satellite suddenly loose all of the velocity of the shuttle compared to the launching pad? Does the satellite shoot away from the shuttle at 25,000 mph as soon as all physical contact is lost? Why do you need a tube to begin with? To block out the Aether wind?
|
|||||||||||||||||||||||||||||||||
ICANT Member Posts: 6769 From: SSC Joined: Member Rating: 1.5 |
Hi Taq,
Taq writes: We are not using the lake's frame of reference. We are using the car's frame of reference. In the car's frame of reference there is no relative motion between the pen laser, detector, and car. None. But once the pulse is emitted from the laser pen it is no longer attached to the laser pen nor the car. The car that is moving at ".5 c" relative to the Salt Lake Flats is also moving at ".5 c" relative to the pulse as the pulse is moving at zero horizontally relative to the Salt Lake Flats. The pulse is moving at "c" at a 90 angle relative to the Salt Lake Flats. 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."
|
|||||||||||||||||||||||||||||||||
crashfrog Member (Idle past 1488 days) Posts: 19762 From: Silver Spring, MD Joined:
|
But once the pulse is emitted from the laser pen it is no longer attached to the laser pen nor the car. Why would it matter if it is attached or not? How does not being attached to the car remove it from the car's reference frame? Does the light pulse undergo any kind of acceleration? Does its velocity change in any respect? No. Therefore it remains in the same reference frame it started in - the same frame as the rest of the light apparatus.
The car that is moving at ".5 c" relative to the Salt Lake Flats is also moving at ".5 c" relative to the pulse as the pulse is moving at zero horizontally relative to the Salt Lake Flats. Incorrect. The light pulse has the lateral velocity of the car relative to the salt flats, because the pulse is in the reference frame of the car and not in that of the salt flats.
|
|||||||||||||||||||||||||||||||||
ICANT Member Posts: 6769 From: SSC Joined: Member Rating: 1.5 |
Hi Taq,
Taq writes: Relative to the car, how far do the pen laser and detector move during the transit of the light pulse? The answer is zero. They don't move at all, correct? The laser pen is attached to the car through the roof and is flush with the exterior. The pole with the detector on top is attached to the roof of the car. Thus they are one unit whatever one does the other does.
Taq writes: As I stated above, we are not using the lake's frame of reference. We are using the car's frame of reference. So you are using the car's frame of reference relative to what? You are not using it relative to the pulse flying throught the vacuum at "c" that was emitted from the laser pen attached to the car. You say you are not using the car's frame of reference relative the the Salt Lake Flats. The only thing you could insert relative to the car's frame of reference that would observe the car's frame at zero would be another car traveling along side of your car at ".5 c". 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."
|
|||||||||||||||||||||||||||||||||
Taq Member Posts: 10038 Joined: Member Rating: 5.3 |
But once the pulse is emitted from the laser pen it is no longer attached to the laser pen nor the car. It is not attached to the lake, either.
The car that is moving at ".5 c" relative to the Salt Lake Flats is also moving at ".5 c" relative to the pulse as the pulse is moving at zero horizontally relative to the Salt Lake Flats. So your are saying that the light pulse takes on the velocity of the salt lake flats in violation of your supposed postulate #2?
|
|||||||||||||||||||||||||||||||||
Taq Member Posts: 10038 Joined: Member Rating: 5.3 |
So you are using the car's frame of reference relative to what? Relative to the pen laser and detector.
You are not using it relative to the pulse flying throught the vacuum at "c" that was emitted from the laser pen attached to the car. I am measuring the light pulse within the reference frame of the car. Since neither the car, pen laser, or detector are moving relative to each other the light pulse should hit the detector dead on.
You say you are not using the car's frame of reference relative the the Salt Lake Flats. No, I am not, which makes me wonder why you keep referencing it. All frames are equally valid.
The only thing you could insert relative to the car's frame of reference that would observe the car's frame at zero would be another car traveling along side of your car at ".5 c". Since the other car would have zero velocity relative to the car's frame of reference the driver of the other car would also see the light pulse travel upwards at a perfect 90 degrees and strike the detector.
|
|||||||||||||||||||||||||||||||||
ICANT Member Posts: 6769 From: SSC Joined: Member Rating: 1.5 |
Hi Son,
Son writes: Since neither the car, nor the pen laser moves in the car's frame of reference, why then would the laser miss it's target? What laser? There is a pulse emitted from the laser pen so that would be a pulse of light that is traveling at "c" in a vacuum. It is not inside the car nor is it attached to the car. The pulse is traveling at a 90 angle relative to the Salt Lake Flats. The car is traveling at "0.5 c" relative to the Salt Lake Flats. The pulse is not traveling horizontally in the direction the car is traveling relative to the Salt Lake Flats. Therefore the car will move 2 feet relative to the pulse that was emitted at a 90 angle to the travel of the car. Making it impossible for the pulse to strike the detector.
Son writes: If you answer that it's because it moves relative to the salt lakes, why then is the salt's lake frame of reference more important than the car's frame of reference in determining light's behaviour? The only frame that matters is the one the pulse is in. Nothing affects the pulse's behaviour once it leaves the laser pen. The motion of the car with the pole and laser pen attached to it, nor the Salt Lake Flats. If will travel in the vacuum at "c" in a straight line from where it was emitted until it is scattered, or absorbed by something. 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."
|
|
|
Do Nothing Button
Copyright 2001-2023 by EvC Forum, All Rights Reserved
Version 4.2
Innovative software from Qwixotic © 2024