ravell Posted June 1, 2018 Posted June 1, 2018 1. In the laboratory on Earth, the 2 m long cylinder was filled with air at 20 atmospheres pressure. After opening the small measuring valve, the pressure in the cylinder drops to 10 atmospheres after 200 seconds. Questions: 1. What pressure will be present in this cylinder placed, along the direction of travel, in a spacecraft flying at 260,000 km /s? According to relativity theory at 260,000 km/s, the cylinder length will be contracted to 1 meter. 2. After what time, according to the light clock on the ship, the pressure in the cylinder will drop to half the value after opening the measuring valve? 3. What will be the above values when the ship will accelerate to 290,000 km /s ? The length of the cylinder will then be 50 cm.
Strange Posted June 1, 2018 Posted June 1, 2018 7 minutes ago, ravell said: In the laboratory on Earth, the 2 m long cylinder was filled with air at 20 atmospheres pressure. After opening the small measuring valve, the pressure in the cylinder drops to 10 atmospheres after 200 seconds. What is the relevance of opening the valve to reduce the pressure? How does that affect the following questions? 4 minutes ago, ravell said: 1. What pressure will be present in this cylinder placed, along the direction of travel, in a spacecraft flying at 260,000 km /s? According to relativity theory at 260,000 km/s, the cylinder length will be contracted to 1 meter. Is this before or after the pressure is reduced in the cylinder? As measured by what frame of reference? As far as the people on the spacecraft are concerned, there will obviously be no change. How would an observer moving relative to the spacecraft measure the pressure? Quote 2. After what time, according to the light clock on the ship, the pressure in the cylinder will drop to half the value after opening the measuring valve? As measured on the ship, nothing changes. The cylinder is 2 m long, the pressure is 20atm, the time taken is 200 seconds. Quote 3. What will be the above values when the ship will accelerate to 290,000 km /s ? The length of the cylinder will then be 50 cm. As measured on the ship, nothing changes. The cylinder is 2 m long, the pressure is 20atm, the time taken is 200 seconds.
ravell Posted June 3, 2018 Author Posted June 3, 2018 On 1.06.2018 at 2:46 PM, Strange said: What is the relevance of opening the valve to reduce the pressure? How does that affect the following questions? It is measurement of time of emptying the reference cylinder, to the half of the initial pressure, in the laboratory on Earth, in order to compare this time to the same experiments repeated in the spacecraft, at the speeds specified in the questions. Quote Is this before or after the pressure is reduced in the cylinder? On the spacecraft, before the launch, we install the same 2 reference cylinders as on the Earth, 2 m long and 20 atm pressure, to measure the time of pressure drop by half, after opening the valve, at the given speed of the spacecraft. The question concerns the pressure before opening the valve. Quote As measured by what frame of reference? As far as the people on the spacecraft are concerned, there will obviously be no change. Are you sure? Could you explain it? The clock on the spacecraft for 260 000 km /s runs 2 times slower, and for 290 000 km /s runs 4 times slower than the clock on Earth. Quote How would an observer moving relative to the spacecraft measure the pressure? I do not know. In the same way in which he sees and measures the relativistic length contraction of the spacecraft and the cylinder? Transmission by CNN? I am just curious about possible explanations, how will this whole experiment be seen by the observer remaining on Earth?
Strange Posted June 3, 2018 Posted June 3, 2018 Just now, ravell said: Are you sure? Could you explain it? The clock on the spacecraft for 260 000 km /s runs 2 times slower, and for 290 000 km /s runs 4 times slower than the clock on Earth. As seen from Earth. Meanwhile, the spaceship will see clocks on Earth run 2 (or 4) times slower than the clock on the spaceship. It's all relative. The people on the spaceship will always see their lengths and times unchanged; as we do on Earth even though we are moving at 50% or 90% or 99% of the speed of light relative to something. Quote I am just curious about possible explanations, how will this whole experiment be seen by the observer remaining on Earth? I think, the observer on Earth will see the tank being half as long and taking twice as long to empty. It looks like this means they will consider the pressure to be double. Which is an interesting result, if true.
Janus Posted June 3, 2018 Posted June 3, 2018 (edited) 1 hour ago, Strange said: I think, the observer on Earth will see the tank being half as long and taking twice as long to empty. It looks like this means they will consider the pressure to be double. Which is an interesting result, if true. They will consider the pressure as being the same. Pressure is approximately found by P = 1/3 (nmv^2) where n is the number of molecules per unit volume, m is the mass of each molecule, and v the average velocity of the molecules. For our cylinder moving at 0.866c relative to our observer: the cylinder has 1/2 the volume but contains the same number of molecules, so n doubles to 2n The "effective" mass of each molecule doubles to 2m Time runs at 1/2 the rate so v is v/2 1/3 ((2n)(2m) (v/2)^2) = 1/3(4nmv^2/4)= 1/3(nmv^2) 1 hour ago, ravell said: Are you sure? Could you explain it? The clock on the spacecraft for 260 000 km /s runs 2 times slower, and for 290 000 km /s runs 4 times slower than the clock on Earth. I do not know. In the same way in which he sees and measures the relativistic length contraction of the spacecraft and the cylinder? Transmission by CNN? I am just curious about possible explanations, how will this whole experiment be seen by the observer remaining on Earth? All measurements are relative. While the spacecraft is moving at 0.866 c relative to the Earth, the Earth is moving at 0.866 relative to the spaceship. So while the Earth measures the spaceship clock running 1/2 as fast as its own, and the cylinder contracted to 1/2 the length of the Earth cylinder, an observer in the spacecraft would measure the Earth clock as running slow and the Earth cylinder as being contracted to 1/2 the length of the ship cylinder. Neither of them would measure any difference in their own clock or cylinder. Edited June 3, 2018 by Janus
Strange Posted June 3, 2018 Posted June 3, 2018 3 hours ago, Janus said: They will consider the pressure as being the same. Pressure is approximately found by P = 1/3 (nmv^2) where n is the number of molecules per unit volume, m is the mass of each molecule, and v the average velocity of the molecules. For our cylinder moving at 0.866c relative to our observer: the cylinder has 1/2 the volume but contains the same number of molecules, so n doubles to 2n The "effective" mass of each molecule doubles to 2m Time runs at 1/2 the rate so v is v/2 1/3 ((2n)(2m) (v/2)^2) = 1/3(4nmv^2/4)= 1/3(nmv^2) Excellent explanation. But doesn’t v also depend on the length contraction (in one dimension)?
ravell Posted June 9, 2018 Author Posted June 9, 2018 On 1.06.2018 at 2:46 PM, Strange said: As measured on the ship, nothing changes. The cylinder is 2 m long, the pressure is 20atm, the time taken is 200 seconds. On the spacecraft the cylinder is 2 m long, the pressure is 20atm and this is obvious because the length contraction in reality does not exist. But the light clock on the spacecraft at 260 000 km /s runs 2 times slower, and it stems from classical physics. So as I understand it, emptying the cylinder up to ½ should take 100 local seconds and not 200s? Quote All measurements are relative. While the spacecraft is moving at 0.866 c relative to the Earth, the Earth is moving at 0.866 relative to the spaceship. So while the Earth measures the spaceship clock running 1/2 as fast as its own, and the cylinder contracted to 1/2 the length of the Earth cylinder, an observer in the spacecraft would measure the Earth clock as running slow and the Earth cylinder as being contracted to 1/2 the length of the ship cylinder. Neither of them would measure any difference in their own clock or cylinder. The observer on the Earth is able to see theoretically, only the transverse diameters of the spacecraft and the cylinder, but in no way he is able to see and measure the length of the spacecraft (and the cylinder) and its contraction , during its move away with the speed of 0.866c. Therefore, the discussion about observed apparent contraction of the spacecraft at the speed of 0.866c (or other) is abstract and frankly amazing for me. Observed on Earth, the run of the light clock in the spacecraft moving away at the speed of 0.866c will be not 2 times slower , but almost 4 times slower, after taking into account the Doppler effect.
swansont Posted June 9, 2018 Posted June 9, 2018 4 hours ago, ravell said: On the spacecraft the cylinder is 2 m long, the pressure is 20atm and this is obvious because the length contraction in reality does not exist. But the light clock on the spacecraft at 260 000 km /s runs 2 times slower, and it stems from classical physics. So as I understand it, emptying the cylinder up to ½ should take 100 local seconds and not 200s? The clock on the spacecraft only runs slower as measure by an observer in another frame. Nothing is different for an observer on the spacecraft. As far as they are concerned, the ship is at rest. Quote The observer on the Earth is able to see theoretically, only the transverse diameters of the spacecraft and the cylinder, but in no way he is able to see and measure the length of the spacecraft (and the cylinder) and its contraction , during its move away with the speed of 0.866c. Therefore, the discussion about observed apparent contraction of the spacecraft at the speed of 0.866c (or other) is abstract and frankly amazing for me. Seeing and measuring a spacecraft moving at 0.866c is pretty abstract and amazing, so, really, that's not an issue with such a gedanken experiment. Quote Observed on Earth, the run of the light clock in the spacecraft moving away at the speed of 0.866c will be not 2 times slower , but almost 4 times slower, after taking into account the Doppler effect. No, that's not true. "What is measured" and "what is observed" are not synonymous. (Janus has multiple posts explaining this) You have to consistently apply your chosen measurement protocols. (similarly, a clock that is 1 light-minute away from you in your own frame is not a minute slow — you take the distance into effect)
ravell Posted June 12, 2018 Author Posted June 12, 2018 On 9.06.2018 at 5:07 PM, swansont said: The clock on the spacecraft only runs slower as measure by an observer in another frame. Nothing is different for an observer on the spacecraft. As far as they are concerned, the ship is at rest. I will repeat my earlier statement : The light clock on the spacecraft moving at the speed of 0,866c, must runs 2 times slower than the clock at rest, and this is not a relativistic effect, but the real effect which results from classical physics. If it were not so, it would not occur the twin paradox, and would not be needed the offset for the clocks in GPS satellites. The observer on the spacecraft can be able to measure this delay, by comparing the cycle of the clock on board with the cycle of the external reference clock at rest, such as the transversally distant reference pulsar. If for the observer at rest the pulsar makes 60 flashes per minute, then for the observer on the spacecraft moving at a speed of 0.866c , the pulsar will hold 120 flashes per the minute of the clock on the board. Quote Seeing and measuring a spacecraft moving at 0.866c is pretty abstract and amazing, so, really, that's not an issue with such a gedanken experiment. Right. Seen by an outside observer the alleged contraction of the spacecraft , at such unreal speeds close to the speed of light, is an extreme abstraction even in the field of science fiction. A stationary observer observing at the back the movement of a 100-meter long spacecraft, will see nothing, beside a smaller and smaller point. A stationary observer standing aside, at a very large distance, of the moving spacecraft at a speed of 0.866 c (or other), will see not the contraction of the spacecraft, but its elongations to several hundred km, due to the inertness of the eyes or the measuring camera. Thus, the discussion about the alleged contractions of observed objects in motion, lies outside the area of serious science. It can only be treated as a kind of useless intellectual entertainment.
Mordred Posted June 12, 2018 Posted June 12, 2018 (edited) You can repeat your earlier statement all you want under relativity all observers in the same reference frame as a reference clock will not notice any time dilation it is only an observer seeing a clock in a different reference frame that will notice the difference. regardless of whether or not you feel length contraction is involved is irrelevant compared to the success of relativity. It is one of the most tested theories out there and has been found to be incredibly accurate. The length contraction of geometry is certainly not outside of science despite your opinion lol Edited June 12, 2018 by Mordred
Endy0816 Posted June 13, 2018 Posted June 13, 2018 You could fly the ship past a series of closely spaced lasers. Would cut a different number of beams than a ship at rest.
swansont Posted June 13, 2018 Posted June 13, 2018 15 hours ago, ravell said: I will repeat my earlier statement : The light clock on the spacecraft moving at the speed of 0,866c, must runs 2 times slower than the clock at rest, and this is not a relativistic effect, but the real effect which results from classical physics. If it were not so, it would not occur the twin paradox, and would not be needed the offset for the clocks in GPS satellites. How would you go about showing this? What classical physics predicts that this would happen? Quote Thus, the discussion about the alleged contractions of observed objects in motion, lies outside the area of serious science. It can only be treated as a kind of useless intellectual entertainment. No, that's not true. It's a common practice in physics to analyze situations where some variable gets very large or very small in order to clearly understand certain effects. (e.g. you get to use factors of 2, as in this example, instead of much smaller modifications which require more effort to calculate and analyze) Length contraction is an effect that manifests itself in certain particle accelerator experiments, in order to properly model the collisions where the incident nucleus has has relativistic speed. It's not "intellectual entertainment"
ravell Posted June 13, 2018 Author Posted June 13, 2018 8 hours ago, swansont said: “The light clock on the spacecraft moving at the speed of 0,866c, must runs 2 times slower than the clock at rest, and this is not a relativistic effect, but the real effect which results from classical physics.” How would you go about showing this? What classical physics predicts that this would happen? Let us answer a question first, which concern the very theoretical example below. What time will show the light clock in the spacecraft after its return to Earth? The spacecraft with the light clock on board, took off from Earth and after a short acceleration period, reached the target speed of 0.866c . With this speed flew certain distance in space and, after 1000 hours on the clock on the Earth, returned to the Earth. Assume here, that the time of acceleration of the spacecraft to 0,866c, as well as decreasing to the speed of zero is the same and is 1 hour. Quote 19 hours ago, Mordred said: You can repeat your earlier statement all you want under relativity all observers in the same reference frame as a reference clock will not notice any time dilation it is only an observer seeing a clock in a different reference frame that will notice the difference. Sorry, Your statement is unclear to me. Do you think that the Earth (where our spacecraft took off ), the reference pulsar, and the spacecraft already moving at a rate of 0.866 c, constitute one and the same reference frame? If not, could you answer how many pulsars flashes per minute of the clock on the spacecraft will see the observer on the board, if for the observer on Earth this number is 60 flashes per minute of the clock on the ground.
swansont Posted June 13, 2018 Posted June 13, 2018 2 hours ago, ravell said: Let us answer a question first, which concern the very theoretical example below. What time will show the light clock in the spacecraft after its return to Earth? The spacecraft with the light clock on board, took off from Earth and after a short acceleration period, reached the target speed of 0.866c . With this speed flew certain distance in space and, after 1000 hours on the clock on the Earth, returned to the Earth. Assume here, that the time of acceleration of the spacecraft to 0,866c, as well as decreasing to the speed of zero is the same and is 1 hour. Or we could assume there was no change in speed, and the clocks were compared as the ship passed by the earth. We stll see time dilation. Now, how about answering my question? What classical physics predicts that this scenario would happen?
Mordred Posted June 14, 2018 Posted June 14, 2018 (edited) 7 hours ago, ravell said: Do you think that the Earth (where our spacecraft took off ), the reference pulsar, and the spacecraft already moving at a rate of 0.866 c, constitute one and the same reference frame? If not, could you answer how many pulsars flashes per minute of the clock on the spacecraft will see the observer on the board, if for the observer on Earth this number is 60 flashes per minute of the clock on the ground. they would not be in the same reference frame despite being at the same velocity as they both have different gravitational potentials. A pulsar for one has far more mass than a rocket ship. Here is a little trivia on just how complex this can get. There is time dilation however miniscule between your head and your feet as they reside at different gravitational potentials as you stand on Earth. So is your goal to keep coming up with scenarios hoping to find some mistake ? by the way they will still see the same number of flashes, the wavelengths will vary accordingly to the gravitational redshift formula, however the number of flashes will remain the same. Which by the way happens to involve length contraction. Edited June 14, 2018 by Mordred
ravell Posted June 14, 2018 Author Posted June 14, 2018 20 hours ago, swansont said: Or we could assume there was no change in speed, and the clocks were compared as the ship passed by the earth. We stll see time dilation. Now, how about answering my question? What classical physics predicts that this scenario would happen? Perhaps my question was not very precise, because you did not give a specific, numerical value in your answer. Let me repeat this example in a simplified, more clear form: Two identical A and B light clocks will be used to clarify the matter.Clock A remains on Earth as the clock at rest, and the clock B is placed in a measuring probe, which was sent from Earth in space for a 4 year travel at the speed v = 0.866c. At the start of the probe, both clocks A and B indicated the date of 1 January 2020. What date will the clock B show, when the probe returns to Earth after 4 years of travel? Clock A on Earth will then indicate the date 1 January 2024. The specific answer to the above example is necessary to clarify your questions. 16 hours ago, Mordred said: by the way they will still see the same number of flashes, the wavelengths will vary accordingly to the gravitational redshift formula, however the number of flashes will remain the same. Which by the way happens to involve length contraction. Are you sure? If the observer on Earth and the observer on the spacecraft moving at the speed 0.866c will see exactly the same 60 flashes / min. of the observed pulsar, which is treated here as a reference clock, it would prove that the clock on the spacecraft has exactly the same rate of run as the clock on Earth (!). Is that so?
swansont Posted June 14, 2018 Posted June 14, 2018 1 minute ago, ravell said: Perhaps my question was not very precise, because you did not give a specific, numerical value in your answer. I did not answer, in part because it is you who owes me an answer. You own the burden of proof here, not me. Anyway, if gamma is two, the spaceship clock will read half of the elapsed time. It will read Dec 31 2021. (It'll be halfway through the spaceship day when the earth clock ticks over, since 2020 is a leap year. Earth will have experienced 1461 days, the ship 730.5) NOW will you tell me how Newtonian physics explains this?
Mordred Posted June 14, 2018 Posted June 14, 2018 (edited) I'm going to assume that English isn't your first lanquage. A pulse is a type of signal with regards to time dilation is is identical to signal propogation delay in an electronic circuit. The pulse RATE will vary not the number of received pulses. In other words the wavelength between pulses will vary as well as the duration of each individual pulse So no it does not prove no time dilation occurs but merely proves you don't understand what is involved with regards to pulse rate vs number of received pulses. Edited June 14, 2018 by Mordred 1
ravell Posted June 16, 2018 Author Posted June 16, 2018 On 15.06.2018 at 12:09 AM, Mordred said: I'm going to assume that English isn't your first lanquage. A pulse is a type of signal with regards to time dilation is is identical to signal propogation delay in an electronic circuit. The pulse RATE will vary not the number of received pulses. In other words the wavelength between pulses will vary as well as the duration of each individual pulse So no it does not prove no time dilation occurs but merely proves you don't understand what is involved with regards to pulse rate vs number of received pulses. You're right. I think I do not understand your explanations. Pulse rate or frequency (f) by definition is the number of pulses (flashes) per unit of time. Wave length (λ = 1 / f) by definition is the unit of time divided by the pulse rate. When motion of the spacecraft is perpendicular to the direction of the pulsar flashes, there is no Doppler effect for the observer in the spacecraft so we do not consider here the change in wavelength (pulselength). Thus, if you say that the observer in the spacecraft will see 60 pulses / local minute, that is the same as the stationary observer on Earth, then according to the physics the clock on the spacecraft must walk at the same rate as the clock on Earth? That's how I understand your explanation. Both observers are watching the same pulsar, yet! On 14.06.2018 at 8:49 PM, swansont said: Anyway, if gamma is two, the spaceship clock will read half of the elapsed time. It will read Dec 31 2021. (It'll be halfway through the spaceship day when the earth clock ticks over, since 2020 is a leap year. Earth will have experienced 1461 days, the ship 730.5) NOW will you tell me how Newtonian physics explains this? That's right! Thanks!Thus, you yourself confirm that the light clock in the spacecraft during its flight had to run twice as slowly as the clock on Earth. Indisputable proof of this fact is the indication of the time of the clock in the spacecraft, after its landing on Earth. The slowdown of the light clock during its motion, is not therefore a "relativistic illusion" for a stationary observer remaining on Earth, but a real physical event caused by the clock movement. Thus, in our example with the pulsar, the observer in the spacecraft moving at a rate 0,866c, will see twice more flashes per minute, than the observer on Earth. I suppose that a sufficiently clear explanation by classical physics, regarding the slowdown of the light clock during its motion, is presented on the link: https://www.dropbox.com/s/tdswmin3t6sl0tr/TimeDilationIsFalseProof.pdf?dl=0 (If you download the Excel file from Dropbox, you do not need to be logged in. You only have to choose the options: download> download directly> enable editing).
swansont Posted June 16, 2018 Posted June 16, 2018 1 hour ago, ravell said: That's right! Thanks!Thus, you yourself confirm that the light clock in the spacecraft during its flight had to run twice as slowly as the clock on Earth. Indisputable proof of this fact is the indication of the time of the clock in the spacecraft, after its landing on Earth. Landing on earth is not required, but were that to happen, the clocks would be running at the same rate. 1 hour ago, ravell said: The slowdown of the light clock during its motion, is not therefore a "relativistic illusion" for a stationary observer remaining on Earth, but a real physical event caused by the clock movement. It is relativistic, and ir is real, but nothing about relativity claims it's an illusion. 1 hour ago, ravell said: I suppose that a sufficiently clear explanation by classical physics, regarding the slowdown of the light clock during its motion, is presented on the link: https://www.dropbox.com/s/tdswmin3t6sl0tr/TimeDilationIsFalseProof.pdf?dl=0 The rules mandate that the discussion take place here. Let's have your explanation.
Mordred Posted June 16, 2018 Posted June 16, 2018 (edited) 5 hours ago, ravell said: You're right. I think I do not understand your explanations. Pulse rate or frequency (f) by definition is the number of pulses (flashes) per unit of time. Correct, the pulse rates vary between observers. aka gravitational redshift. This is precisely what I am talking about you receive the pulses at different rates due to time dilation. ie You can receive them at a different rate than the frequency they were sent..... time is the measure of rates of changing events aka pulses in this case. lol side note believe me if I had this wrong, there would be plenty of forum members questioning my response lol. Rightfully so if I make mistakes in a reply to any post. Edited June 16, 2018 by Mordred
ravell Posted June 24, 2018 Author Posted June 24, 2018 On 16.06.2018 at 9:45 PM, swansont said: It is relativistic, and ir is real, but nothing about relativity claims it's an illusion. The delay of light clocks in motion is a real phenomenon, having no relation to any observer and it can not be considered as a relativistic effect. In contrast, the following examples of claims of relativity theory have nothing to do with reality and are only a "relativistic illusion" or fiction: 1. For the observer in the moving spacecraft, the run of clocks on Earth is slowndown in relation to the clock in the spacecraft. 2. For the observer on Earth, the length of the spacecraft in motion is contracted, and for the observer in the spacecraft the objects on Earth have length contracted. 3. For the observer on Earth , the mass of the spacecraft in motion is increased and vice versa.
swansont Posted June 24, 2018 Posted June 24, 2018 9 minutes ago, ravell said: The delay of light clocks in motion is a real phenomenon, having no relation to any observer and it can not be considered as a relativistic effect. Since it's a consequence of c being invariant, how can it NOT be a relativistic effect? 9 minutes ago, ravell said: In contrast, the following examples of claims of relativity theory have nothing to do with reality and are only a "relativistic illusion" or fiction: 1. For the observer in the moving spacecraft, the run of clocks on Earth is slowndown in relation to the clock in the spacecraft. 2. For the observer on Earth, the length of the spacecraft in motion is contracted, and for the observer in the spacecraft the objects on Earth have length contracted. Length contraction and time dilation have been experimentally observed. 9 minutes ago, ravell said: 3. For the observer on Earth , the mass of the spacecraft in motion is increased and vice versa. Mass increase has likewise been experimentally observed, for a situation where it was predicted. Where is your promised explanation of these effects using Newtonian physics?
Strange Posted June 24, 2018 Posted June 24, 2018 11 minutes ago, ravell said: In contrast, the following examples of claims of relativity theory have nothing to do with reality and are only a "relativistic illusion" or fiction: As these are consequences of an effect that you consider “real” AND as they can be measured, I’m not sure on what basis you say they are not real. But then, science isn’t really concerned with “reality” (whatever that means) but only with can be measured.
Endy0816 Posted June 24, 2018 Posted June 24, 2018 1 hour ago, ravell said: The delay of light clocks in motion is a real phenomenon, having no relation to any observer and it can not be considered as a relativistic effect. In contrast, the following examples of claims of relativity theory have nothing to do with reality and are only a "relativistic illusion" or fiction: 1. For the observer in the moving spacecraft, the run of clocks on Earth is slowndown in relation to the clock in the spacecraft. 2. For the observer on Earth, the length of the spacecraft in motion is contracted, and for the observer in the spacecraft the objects on Earth have length contracted. 3. For the observer on Earth , the mass of the spacecraft in motion is increased and vice versa. The 'reality' is that all seperation, distances and times, can only be said to be greater than zero, not one specific value. There's no conflict if one person measures it as X and the other as Y. That's reality asserting itself, not the other way around.
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