Motor Daddy Posted June 19, 2008 Author Share Posted June 19, 2008 That is demonstrably wrong. Read this link: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html Again, why are you proving to me that clocks are inaccurate? Link to comment Share on other sites More sharing options...
mooeypoo Posted June 19, 2008 Share Posted June 19, 2008 You're just not READING what we're SAYING. A second is a specific duration, regardless what your watch says if it runs slow or fast. A second is a second is a second. Accurate is maintaining a constant duration, regardless of the acceleration or velocity, or distance traveled! You measure a second, and when that second has elapsed, the entire universe has also elapsed a second. A second is a second. Every time - again and again - we tell you that your assumptions are wrong. We explain what the theory says and where to get more information about it. You keep repeating the same mantra again and again. and again. and again. The theory of relativity is counter intuitive. No doubt about that. The fact you don't understand it (and you obviously not, as we pointed out, and explained despite your insistence to ignore us) does not mean it's false. Either talk science, or stop wasting this forum's precious database allocation. You're IGNORANT OF THE THEORY. You are claiming the ball is rectangular when it is proven to be spherical. How do you argue reality with someone who denies it? How do you debate reality with someone who refuses to learn bout it at all? Stop convincing us that you're ignorant of facts, and start fixing the situation and read and learn about the theory that multiple people in MANY YEARS proven it right. Over and over again. And over again. And over again. And again. and again. and again. Your insistence to deny facts is just silly. Seriously. Now please don't tell me a second is a second is a second. ~moo Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 19, 2008 Author Share Posted June 19, 2008 (edited) You're just not READING what we're SAYING. Every time - again and again - we tell you that your assumptions are wrong. We explain what the theory says and where to get more information about it. You keep repeating the same mantra again and again. and again. and again. The theory of relativity is counter intuitive. No doubt about that. The fact you don't understand it (and you obviously not, as we pointed out, and explained despite your insistence to ignore us) does not mean it's false. Either talk science, or stop wasting this forum's precious database allocation. You're IGNORANT OF THE THEORY. You are claiming the ball is rectangular when it is proven to be spherical. How do you argue reality with someone who denies it? How do you debate reality with someone who refuses to learn bout it at all? Stop convincing us that you're ignorant of facts, and start fixing the situation and read and learn about the theory that multiple people in MANY YEARS proven it right. Over and over again. And over again. And over again. And again. and again. and again. Your insistence to deny facts is just silly. Seriously. Now please don't tell me a second is a second is a second. ~moo I'm still waiting on you to tell me how my numbers are wrong, and what the correct numbers are? Or maybe you actually agree with the numbers in the example? If not, show me the numbers! How far does an object travel in 1 second if its initial velocity is zero, and it accelerates at the rate of 500 ft/sec^2? Basic, simple science and math, that precisely determines the distance traveled in a specific duration. These are facts, that are measurable, predictable, and just plain correct! Edited June 19, 2008 by Motor Daddy Link to comment Share on other sites More sharing options...
mooeypoo Posted June 19, 2008 Share Posted June 19, 2008 I'm still waiting on you to tell me how my numbers are wrong, and what the correct numbers are? How far does an object travel in 1 second if its initial velocity is zero, and it accelerates at the rate of 500 ft/sec^2? Basic, simple science and math, that precisely determines the distance traveled in a specific duration. These are facts, that are measurable, predictable, and just plain correct! I did tell you that your assumptions - that created those numbers - are wrong, therefore the numbers are wrong. Example: I claim the earth is rectangular. I supply math. To invalidate my claim, you don't even NEED to go to my math. It's irrelevant. Math can be 'true' and yet meaningless. Which it invariably is in my claim. All you need to do is prove that the earth is not rectangular, and you invalidate my claim, math and all. We showed you how your assumptions and how the way you produce your conclusion - including the way you produce the math - is wrong. Therefore your numbers, your math, and your conclusions are invariably wrong. Stop insisting silliness, and start talking science. Or perhaps reconsider if a scientific (with scientific rigor and analysis) forum is the right place for you to post in. ~moo Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted June 19, 2008 Share Posted June 19, 2008 Again, why are you proving to me that clocks are inaccurate? I could give the exact same results with a sufficiently accurate quartz clock, an amazing pendulum clock, a calibrated hourglass, or one of your revolving cylinder doohickies. I'd hardly call that inaccuracy. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 19, 2008 Author Share Posted June 19, 2008 (edited) I did tell you that your assumptions - that created those numbers - are wrong, therefore the numbers are wrong. Then show me the correct numbers. Example: I claim the earth is rectangular. I supply math. To invalidate my claim, you don't even NEED to go to my math. It's irrelevant. Math can be 'true' and yet meaningless. Which it invariably is in my claim. All you need to do is prove that the earth is not rectangular, and you invalidate my claim, math and all. Show me your correct math of the Earth being rectangular, then. We showed you how your assumptions and how the way you produce your conclusion - including the way you produce the math - is wrong. Therefore your numbers, your math, and your conclusions are invariably wrong. What assumptions? I gave the accelerations, the velocities, the distances, and the times. Are you suggesting that these maths are invalid? Then show me the correct math. You do have the correct math and numbers for the problem, don't you? You claim it's wrong, but don't give the correct answers. Is it a secret or something? Show me where my math is wrong? Stop insisting silliness, and start talking science. What silliness, this is mainstream calculations of mass, distance, and time. It's how it is done. Or perhaps reconsider if a scientific (with scientific rigor and analysis) forum is the right place for you to post in. ~moo I stated earlier that if you prove my numbers wrong I will acknowledge such and admit I am ignorant to mass, distance, and time. I could give the exact same results with a sufficiently accurate quartz clock, an amazing pendulum clock, a calibrated hourglass, or one of your revolving cylinder doohickies. I'd hardly call that inaccuracy. My constant velocity rotating shaft has no inaccuracies. Constant velocity shafts don't change their duration of revolution. Edited June 19, 2008 by Motor Daddy multiple post merged Link to comment Share on other sites More sharing options...
mooeypoo Posted June 19, 2008 Share Posted June 19, 2008 Look back at the posts and follow the links and explanations people worked on putting for you, the answers are there. Trolling reported. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 19, 2008 Author Share Posted June 19, 2008 Look back at the posts and follow the links and explanations people worked on putting for you, the answers are there. Trolling reported. Trolling? I'm not the one trolling here. I gave facts, you are evading the facts and trying to turn this discussion into a pissing match. Show me your numbers. Give me a similar example of mass, distance, and time, so that I can can compare the differences on a apples to apples comparison. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted June 19, 2008 Share Posted June 19, 2008 My constant velocity rotating shaft has no inaccuracies. Constant velocity shafts don't change their duration of revolution. Cesium atoms don't decide to resonate at different frequencies, and quartz crystals don't start producing different frequencies at random (within a certain margin of error.) But I can absolutely guarantee that if you put your shaft on a plane and fly it at high speed, and compare it to a separate shaft on the ground, it'll show a slightly different answer. I defer the maths to someone with more experience in special relativity, as I am likely to screw it up. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 19, 2008 Author Share Posted June 19, 2008 Cesium atoms don't decide to resonate at different frequencies, and quartz crystals don't start producing different frequencies at random (within a certain margin of error.) But I can absolutely guarantee that if you put your shaft on a plane and fly it at high speed, and compare it to a separate shaft on the ground, it'll show a slightly different answer. I defer the maths to someone with more experience in special relativity, as I am likely to screw it up. But if the shaft changes rotational velocity it is no longer 300 revolutions per second. In other words, the shaft may only rotate 299 revolutions per second if it changes rotational velocity. That means the shaft is no longer accurate. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted June 19, 2008 Share Posted June 19, 2008 So don't change the rotational velocity of it then. Just change the linear velocity. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 19, 2008 Author Share Posted June 19, 2008 So don't change the rotational velocity of it then. Just change the linear velocity. The linear velocity is irrelevant, as the constant rotational velocity is determining a duration. It was already established that a second was 300 revolutions of the constant velocity shaft. If the shaft changes ROTATIONAL VELOCITY a second will no longer be 300 revolutions, and I will have NO IDEA how much time has actually elapsed during an event. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted June 19, 2008 Share Posted June 19, 2008 Exactly. So I'm saying that you should put your shaft in a spaceship and send it off at a high linear velocity and bring it back, never changing its rotational velocity (from the perspective of someone sitting next to it in the spaceship). It'll still disagree with one sitting on the ground. Let me describe a scenario that you can visualize a little better. Suppose I take a pair of twins born eight minutes apart and put one of them on my latest new spaceship. I send him off at an incredibly high speed for five years; after five years, he'll turn around and come blazing back at the same high speed, for a total elapsed time of ten years from departure to return (according to him). So I send him off on my mission when he's aged twenty five. He comes back, aged thirty five, and sets off to find his twin to tell him how the journey was. Much to his shock, his twin died years ago -- in the time he was away, the twin had children, grandchildren, great-grandchildren, died, and so on. Four hundred years have elapsed on Earth, but the twin has only aged ten. That scenario is possible (assuming we could build such a spaceship). All of modern physics, and all of our experimental evidence, agrees that it could happen. What you are doing is denying all of that evidence and all of that physics. I suggest you take a course in special relativity to better understand it and learn why it has been demonstrated to be correct. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 19, 2008 Author Share Posted June 19, 2008 Exactly. So I'm saying that you should put your shaft in a spaceship and send it off at a high linear velocity and bring it back, never changing its rotational velocity (from the perspective of someone sitting next to it in the spaceship). It'll still disagree with one sitting on the ground. Let me describe a scenario that you can visualize a little better. Suppose I take a pair of twins born eight minutes apart and put one of them on my latest new spaceship. I send him off at an incredibly high speed for five years; after five years, he'll turn around and come blazing back at the same high speed, for a total elapsed time of ten years from departure to return (according to him). So I send him off on my mission when he's aged twenty five. He comes back, aged thirty five, and sets off to find his twin to tell him how the journey was. Much to his shock, his twin died years ago -- in the time he was away, the twin had children, grandchildren, great-grandchildren, died, and so on. Four hundred years have elapsed on Earth, but the twin has only aged ten. That scenario is possible (assuming we could build such a spaceship). All of modern physics, and all of our experimental evidence, agrees that it could happen. What you are doing is denying all of that evidence and all of that physics. I suggest you take a course in special relativity to better understand it and learn why it has been demonstrated to be correct. What year were they born, and what year is it when he returns? Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted June 19, 2008 Share Posted June 19, 2008 Let's suppose they're both born in 2050. He departs on his mission in 2075. When he returns, it's 2475 -- but the twin on the spaceship is 35. (Well, he has only aged 35 years, but he's clearly 425.) Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 19, 2008 Author Share Posted June 19, 2008 Let's suppose they're both born in 2050. He departs on his mission in 2075. When he returns, it's 2475 -- but the twin on the spaceship is 35. (Well, he has only aged 35 years, but he's clearly 425.) I don't doubt the aging difference, or the differentiation of the clocks. The trip took 400 years. The brother is 400 years older, regardless of how long his beard is, how many wrinkles he has, or the state of physical condition. Those are symptoms, not distance, and time. If the brother took a 400 year journey at .5c, he traveled 200 light years in 400 years. What's the problem? Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted June 19, 2008 Share Posted June 19, 2008 Both his clocks and his body attest to him only spending ten years in space. Who's right? The people on Earth or the twin and his clock? Link to comment Share on other sites More sharing options...
Rakdos Posted June 19, 2008 Share Posted June 19, 2008 I don't doubt the aging difference, or the differentiation of the clocks. The trip took 400 years. The brother is 400 years older, regardless of how long his beard is, how many wrinkles he has, or the state of physical condition. Those are symptoms, not distance, and time. If the brother took a 400 year journey at .5c, he traveled 200 light years in 400 years. What's the problem? All physical tests performed on the trip-taking brother should show his age to be 35 though. Link to comment Share on other sites More sharing options...
D H Posted June 19, 2008 Share Posted June 19, 2008 (edited) Then show me the correct numbers. Phase 1: The rocket accelerates from rest with respect to the Earthbound brother at 93000 miles/second2 as sensed by the brother on board the rocket for one second as measured by the brother on board the rocket. The relevant equations of motion here are those of the relativistic rocket. By the numbers, The coordinate time duration of phase 1 is 1.0420612 seconds, not 1 second. The distance covered in the Earthbound brother's frame is 47,473.8749 miles, not 46,500 miles. The velocity in the Earthbound brother's frame is 85,973.2076 miles/sec, not 93,000 miles/sec. Phase 2: The rocket coasts for ten second as measured by the brother on board the rocket. The relevant equations of motion are the Lorentz transformation. By the numbers, The velocity in the Earthbound brother's frame is 85,973.2076 miles/sec, not 93,000 miles/sec. (Velocity at end of phase 1) The coordinate time duration of phase 2 is 11.2723131 seconds, not 10 second. The distance covered in the Earthbound brother's frame is 969,116.913 miles, not 930,000 miles. Phase 3: The rocket decelerates for one second as measured by the brother on board the rocket, coming to a rest with respect to the Earthbound brother. The time and distance in the Earthbound brother's frame are the same as in phase 1. Outbound leg totals as sensed by the Earthbound brother: Time: 13.3564355 seconds (rather than 12 seconds) Distance: 1,064,064.66 miles (rather than 1,023,000 miles) The return leg has identical numbers. Edited June 19, 2008 by D H Fix link for velocity calculation Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 19, 2008 Author Share Posted June 19, 2008 Both his clocks and his body attest to him only spending ten years in space. Who's right? The people on Earth or the twin and his clock? He spent 400 years in space, regardless of the physical state of his condition or what the clock says. I just happen to have a birth certificate and calender to prove it. His birth certificate says he was born in 2050, and the calendar says it's 2475. He is 425 years old. He departed when he was 25, so he spent 400 years in space. If he traveled at .5c he traveled a total of 200 light years in 400 years. There is no disagreement with the numbers of distance and time. Who cares how old he looks, you just told me he is still alive, and the numbers prove he spent 400 years in space. Link to comment Share on other sites More sharing options...
Rakdos Posted June 19, 2008 Share Posted June 19, 2008 He spent 400 years in space, regardless of the physical state of his condition or what the clock says. I just happen to have a birth certificate and calender to prove it. His birth certificate says he was born in 2050, and the calendar says it's 2475. He is 425 years old. He departed when he was 25, so he spent 400 years in space. If he traveled at .5c he traveled a total of 200 light years in 400 years. There is no disagreement with the numbers of distance and time. Who cares how old he looks, you just told me he is still alive, and the numbers prove he spent 400 years in space. Numbers in one Reference Frame. The Reference Frame (an atomic clock) aboard the ship would say otherwise. http://en.wikipedia.org/wiki/Hafele-Keating_experiment << should help explain it. Link to comment Share on other sites More sharing options...
Motor Daddy Posted June 19, 2008 Author Share Posted June 19, 2008 Phase 1: The rocket accelerates from rest with respect to the Earthbound brother at 93000 miles/second2 as sensed by the brother on board the rocket for one second as measured by the brother on board the rocket. The relevant equations of motion here are those of the relativistic rocket. By the numbers,The coordinate time duration of phase 1 is 1.0420612 seconds, not 1 second. The distance covered in the Earthbound brother's frame is 47,473.8749 miles, not 46,500 miles. The velocity in the Earthbound brother's frame is 85,973.2076 miles/sec, not 93,000 miles/sec. Phase 2: The rocket coasts for ten second as measured by the brother on board the rocket. The relevant equations of motion are the Lorentz transformation. By the numbers, The velocity in the Earthbound brother's frame is 85,973.2076 miles/sec, not 93,000 miles/sec. (Velocity at end of phase 1) The coordinate time duration of phase 2 is 11.2723131 seconds, not 10 second. The distance covered in the Earthbound brother's frame is 969,116.913 miles, not 930,000 miles. Phase 3: The rocket decelerates for one second as measured by the brother on board the rocket, coming to a rest with respect to the Earthbound brother. The time and distance in the Earthbound brother's frame are the same as in phase 1. Outbound leg totals as sensed by the Earthbound brother: Time: 13.3564355 seconds (rather than 12 seconds) Distance: 1,064,064.66 miles (rather than 1,023,000 miles) The return leg has identical numbers. I was just getting ready to call it quits tonight, it's 1:30 AM. I am going to bed, but I am going to tell you why your numbers are wrong in the morning, and yes, they are wrong, and I can prove it. Sorry, but I'm seriously tired. I'll reply in the morning. Link to comment Share on other sites More sharing options...
swansont Posted June 19, 2008 Share Posted June 19, 2008 I was just getting ready to call it quits tonight, it's 1:30 AM. I am going to bed, but I am going to tell you why your numbers are wrong in the morning, and yes, they are wrong, and I can prove it. Sorry, but I'm seriously tired. I'll reply in the morning. The proof had better not be along the lines of "a second is a second is a second." Seriously. Link to comment Share on other sites More sharing options...
timo Posted June 19, 2008 Share Posted June 19, 2008 That's funny, because on another board, the "expert" SR guy didn't agree with this example, ... Perhaps you said something different there or he understood something you didn't explicitly state. If you are not familiar with Relativity then you easily omit potentially important information, e.g.: You haven't stated in whose reference frame the numbers are measured. If one assumes (I did) that all numbers are given in the same frame of reference (there is no reason to believe otherwise, I think) then the numbers are fine. If on the other board you somehow used these numbers in a different frame of reference or mixed the frames without noticing then what you said there was probably wrong. Effectively, you have (till the point where you wrote above quote; I have not followed the thread any further) not said much. Only set up a scenario with a few numbers that, assuming they are all meant in some fixed frame and taking the kind of special definition of average speed you used, do not violate any constraints set by nature (v<c). It is well possible that the magic word "merlon" ... errr ... "twins" made people believe you had said something that you haven't said (namely having spoken about eigentimes). 1 Link to comment Share on other sites More sharing options...
D H Posted June 19, 2008 Share Posted June 19, 2008 The proof had better not be along the lines of "a second is a second is a second." Seriously. Nor along the lines of this either. Link to comment Share on other sites More sharing options...
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