D H Posted November 13, 2006 Share Posted November 13, 2006 To avoid confusion I won’t refer to eq. 17 from my online reference; I’ll refer to eq. 4 in my paper, the same equation. Notice the circular logic in your statement. Looking at your paper, Fig. 1. An inconsistency of general relativity. The drawing at left depicts a ball free to move within a relativistic rocket. The drawing at right depicts a ball in free fall within a uniform gravitational field above a planet. The equations in section 8 show that the ball can traverse the rocket vertically in either direction in an arbitrarily short time in the frame of its crew. Try again. In the frame of the crew (fixed with respect to the rocket), the ball takes more than [math]h/c[/math] time to traverse the rocket, where [math]h[/math] is the height of the rocket. Your principal mistake is a misinterpretation of the equivalence principle: According to general relativity’s equivalence principle, the crew experiences the equivalent of a uniform gravitational field. This is not quite right; the equivalence principle holds no experiment performed in a closed system can distinguish between an accelerated reference frame or a reference frame in a uniform gravitational field. You explicitly propagate this mistaken interpretation of the equivalence principle in section 1, where you equate the non-uniform gravitational field induced by a point mass to a uniform gravitational field. This is invalid. Your entire analysis hinges on this fatal flaw. For a proper analysis of the problem of a particle of negligible mass falling from rest at [math]r=\infty[/math] toward a point mass, see Voracek, P., "Relativistic gravitational potential and its relation to mass-energy", Astrophysics and Space Science, vol. 65, no. 2, Oct. 1979, p. 397-413. The theory of relativity is 90+ years old. Don't you think that this blatant flaw, should it exist, would have been uncovered by now? Link to comment Share on other sites More sharing options...
Zanket Posted November 13, 2006 Share Posted November 13, 2006 Try again. In the frame of the crew (fixed with respect to the rocket), the ball takes more than [math]h/c[/math] time to traverse the rocket, where [math]h[/math] is the height of the rocket. No. See the equations for the rocket in section 8 in my paper. These equations come from the Usenet Physics FAQ that section 8 references. Let the ball remain at rest in the gantry’s frame, initially at the top of the rocket. The time required in the crew’s frame for the bottom of the rocket to reach the ball is given by eq. 19 in my paper, where d is the height of the rocket. The equation shows that the time elapsed in the crew’s frame can be arbitrarily short for any given d. Any decent book on SR tells you that the crew can reach the far side of a galaxy in an arbitrarily short time on their clock. Then of course they can reach a ball on the far side of their rocket in an arbitrarily short time on their clock as well. Your principal mistake is a misinterpretation of the equivalence principle: According to general relativity’s equivalence principle, the crew experiences the equivalent of a uniform gravitational field. This is not quite right; the equivalence principle holds no experiment performed in a closed system can distinguish between an accelerated reference frame or a reference frame in a uniform gravitational field. You have given one of many compatible definitions of the EP. What you quoted of mine follows from it, and is not intended to be a definition of the EP. You explicitly propagate this mistaken interpretation of the equivalence principle in section 1, where you equate the non-uniform gravitational field induced by a point mass to a uniform gravitational field. I do not equate those anywhere in the paper. Quote me. For a proper analysis of the problem of a particle of negligible mass falling from rest at [math]r=\infty[/math] toward a point mass, see Voracek, P., "Relativistic gravitational potential and its relation to mass-energy", Astrophysics and Space Science, vol. 65, no. 2, Oct. 1979, p. 397-413. That’s like trying to refute GR by referencing a paper on Newtonian mechanics. It proves nothing. The theory of relativity is 90+ years old. Don't you think that this blatant flaw, should it exist, would have been uncovered by now? One could have said the same thing about Newtonian mechanics in 1915, except change “90+” to “200+”. Section 2 has a related reader-author comment: Reader: Do you really expect us to believe that an inconsistency has existed undetected in general relativity ever since it was published? Author: It is known that general relativity predicts central singularities where its equations break down and where it is incompatible with quantum mechanics. Then it should not be a surprise that the theory is flawed in another way. Link to comment Share on other sites More sharing options...
D H Posted November 13, 2006 Share Posted November 13, 2006 Any decent book on SR tells you that the crew can reach the far side of a galaxy in an arbitrarily short time on their clock. Then of course they can reach a ball on the far side of their rocket in an arbitrarily short time on their clock as well. Wrong. A sufficient velocity relative to the galaxy makes the galaxy shrink to the size of a pinhead in the crew's frame. The rocket has zero velocity in the crew's frame. You explicitly propagate this mistaken interpretation of the equivalence principle in section 1, where you equate the non-uniform gravitational field induced by a point mass to a uniform gravitational field. I do not equate those anywhere in the paper.I do not equate those anywhere in the paper. Quote me. 1 Equations of Motion for a Uniform Gravitational FieldAccording to the equivalence principle, the crew of a relativistic rocket experiences the equivalent of a uniform gravitational field. [ text elided ] 2 A Flaw of General Relativity Let a test particle fall radially from rest at infinity toward a large point mass while measuring the velocity v of objects fixed at each altitude as they pass directly by. Section 1 shows that directly measured free-fall velocity asymptotes to c in a uniform gravitational field. The particle always falls through a uniform gravitational field (a succession of them, each applying only locally) since a nonuniform gravitational field is everywhere uniform locally. This is the crux of your mistake. The equivalence principle talks about the inability to distinguish between uniform acceleration and a uniform gravitational field. You are applying the principle erroneously. It is not surprising that you are getting erroneous results. The nonuniform gravitational field produced by a point mass is everywhere nonuniform. For a proper analysis of the problem of a particle of negligible mass falling from rest at [math]r=\infty[/math] toward a point mass, see Voracek, P., "Relativistic gravitational potential and its relation to mass-energy", Astrophysics and Space Science, vol. 65, no. 2, Oct. 1979, p. 397-413.That’s like trying to refute GR by referencing a paper on Newtonian mechanics. It proves nothing. People like you are dangerous. People who don't know math very well may see you spouting nonsense and not see that what you spout is nonsense. You showed us how to solve the problem incorrectly. I posted the link to show how the problem can be solved correctly. Link to comment Share on other sites More sharing options...
Farsight Posted November 13, 2006 Share Posted November 13, 2006 D H: I don't understand Zanket's paper. But I've seen this before. There was a guy called lalbatros who argued blue in the face that Zanket was wrong, then conceded he was right. Fine, talk with the guy, debate. Knock the argument down. But please resist the temptation to say stuff like "people like you are dangerous". One day people like Zanket will be right. And then the dangerous people are the other guys. http://forum.physorg.com/index.php?showtopic=9591&st=122 Link to comment Share on other sites More sharing options...
Zanket Posted November 13, 2006 Share Posted November 13, 2006 Wrong. A sufficient velocity relative to the galaxy makes the galaxy shrink to the size of a pinhead in the crew's frame. Yes, and the ball along with it. That’s how the ball (or any other object in the galaxy that is at rest with respect to the galaxy) traverses the rocket in an arbitrarily short time in the crew’s frame. The rocket has zero velocity in the crew's frame. Irrelevant. The ball moves independently of the rocket. The equivalence principle talks about the inability to distinguish between uniform acceleration and a uniform gravitational field. You are applying the principle erroneously. Nothing you boldfaced of mine violates the EP. “The crew of a relativistic rocket experiences the equivalent of a uniform gravitational field” is implied by the EP. They must experience the equivalent of a uniform gravitational field, if they are unable to distinguish between what they experience and what someone in a uniform gravitational field experiences. The second statement you boldfaced is not based on the EP. “A nonuniform gravitational field is everywhere uniform locally” is implied by the definition of “local”. From that it follows that the “particle always falls through a uniform gravitational field (a succession of them, each applying only locally)”. Deem the gravitational field between your head and feet as uniform. Then a particle that falls between your head and feet falls in a uniform gravitational field. If it fell from higher up, then it fell through a succession of uniform gravitational fields, each applying only locally. The nonuniform gravitational field produced by a point mass is everywhere nonuniform. Except locally, where by definition it is uniform. People like you are dangerous. People who don't know math very well may see you spouting nonsense and not see that what you spout is nonsense. You showed us how to solve the problem incorrectly. I posted the link to show how the problem can be solved correctly. It’s the other way around. People who see only what they want to see and favor a rush to judgment are dangerous. I gave you the equation for the ball, I showed you that it comes from the Usenet Physics FAQ, but still you ignore it in favor of your own viewpoint. You think SR is inapplicable despite its experimental confirmation. And basic logic says that you cannot refute a paper that purports to refute GR by simply referencing a paper on GR; that ignores my evidence to the contrary. Link to comment Share on other sites More sharing options...
D H Posted November 13, 2006 Share Posted November 13, 2006 Irrelevant. The ball moves independently of the rocket. So what? The rocket traverses the galaxy in an arbitrary period of time because the galaxy shrinks due to length contraction. The rocket itself does not shrink in the frame of the crew. You are mixing frames, a very bad thing to do, particularly when dealing with relativistic effects. The ball cannot move faster than the speed of light as seen by the crew. In particular, the point at center of the ball, as seen by the crew, cannot traverse the length of the rocket in time [math]h/c[/math] or less. The black hole at the center of the galaxy (or any other point) is subject to the same constraint. They must experience the equivalent of a uniform gravitational field, if they are unable to distinguish between what they experience and what someone in a uniform gravitational field experiences. This is the misapplication of the equivalence principle that gets you into trouble. From http://relativity.livingreviews.org/Articles/lrr-2001-4/: The outcome of any local non-gravitational experiment is independent of where and when in the universe it is performed. You are not performing a non-gravitational experiment here. You are asking how a point mass accelerates a body with negligible mass. I gave you the equation for the ball, I showed you that it comes from the Usenet Physics FAQ, but still you ignore it in favor of your own viewpoint. You misapplied those equations. You think SR is inapplicable despite its experimental confirmation. The second postulate (the speed of light is the same to all observers) is very applicable. Think about what you are saying in your paper in light of this postulate. By saying the ball can traverse the spacecraft in an arbitrarily small amount of time, you are saying the ball can exceed the speed of light. Link to comment Share on other sites More sharing options...
Zanket Posted November 14, 2006 Share Posted November 14, 2006 The rocket traverses the galaxy in an arbitrary period of time because the galaxy shrinks due to length contraction. The rocket itself does not shrink in the frame of the crew. You are mixing frames, a very bad thing to do, particularly when dealing with relativistic effects. No, I am not mixing frames. For the sake of argument drop your insistence that the ball needs to move faster than c to traverse the rocket in an arbitrarily short time in the crew’s frame; otherwise you’ll never see it. Keep an open mind. Now consider this situation: [CXXXXXB]XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXS [CXXXXXB] = rocket, initially at rest with respect to the galaxy C = crew, who remain in a fixed position with respect to the rocket X = empty space B = ball, which remains at rest with respect to the galaxy S = star at far side of galaxy As shown, the ball is initially at the top of the rocket. Let the rocket begin accelerating toward S. Then the ball, which remains at rest with respect to the galaxy, moves towards the crew. You agree that the crew can get to S in an arbitrarily short time in their frame. But the crew will get to the ball first. Then they can get to the ball in an arbitrarily short time in their frame too. Nothing need move faster than c. Eq. 19 in my paper returns the time elapsed in the crew’s frame, for them to reach either the ball or S. Eq. 16 shows that the ball does not move faster than c. The ball, because it remains at rest with respect to the galaxy, can be treated as part of the galaxy. The ball shrinks due to length contraction along with the galaxy of which it is part. That the rocket itself does not shrink in the frame of the crew is irrelevant. We’re discussing what happens to the ball in the crew’s frame, not what happens to the rocket in that frame. This is the misapplication of the equivalence principle that gets you into trouble. You are defying logic there. My statement follows logically from the definition of the EP you gave. From http://relativity.livingreviews.org/Articles/lrr-2001-4/: The outcome of any local non-gravitational experiment is independent of where and when in the universe it is performed. You are not performing a non-gravitational experiment here. You are asking how a point mass accelerates a body with negligible mass. The definition you give here is one of the many definitions of the strong EP. Wikipedia says:The strong equivalence principle states that the results of any local experiment, gravitational or not, in an inertial frame of reference are independent of where and when in the universe it is conducted. Then my usage of the EP is fine. You don’t fully understand it. You misapplied those equations. No, the ball is treated the same as the star. If you want the crew’s elapsed time to reach the ball, then you put the height of the rocket into eq. 19 for d, which is the initial distance between the crew and the ball as measured in the galaxy’s frame (or the crew's frame; doesn't matter because they are initially at rest with respect to the galaxy). If you want the crew’s elapsed time to reach S (the star), then you put the initial distance between the crew and the star as measured in the galaxy’s frame. The second postulate (the speed of light is the same to all observers) is very applicable. Think about what you are saying in your paper in light of this postulate. By saying the ball can traverse the spacecraft in an arbitrarily small amount of time, you are saying the ball can exceed the speed of light. No, as shown by eq. 16. The velocity of the ball (v) is always less than c, just as it is for S. Link to comment Share on other sites More sharing options...
D H Posted November 14, 2006 Share Posted November 14, 2006 A quick question: Let the height of the rocket be [math]h[/math]. Suppose the ball can traverse the rocket in an arbitrarily small amount of time in the frame of the crew. In particular, suppose the crew observes the ball to traverse the rocket in time [math]0.1 h/c[/math]. What is the velocity of the ball in the frame of the crew? Edited to add I looked at your diagram. You are indeed mixing frames. This time, you be the one to keep an open mind. [ CXXXXXB ] XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXS CXXXXXB is fixed in the frame of the crew and is not subject to length contraction. You even used braces to show that this is in the frame of the crew. XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXS is in the frame of the star S and is subject to length contraction. It is the length contraction that makes the crew able to traverse the length of the galaxy in an arbitrarily short period of time. The CXXXXXB length is not contracted. Suppose our spacecraft is 10 meters long and that we are moving at 99.99999999999999999999999999999999995% the speed of light relative to some galaxy that is about 10<sup>5</sup> light years in diameter. Our velocity shrinks that galaxy down to an apparent diameter of 1 millimeter. It does not take long at all (about 3.34 picoseconds) for a particular point on the spacecraft to traverse the length of that apparently miniscule galaxy. It takes a bit longer (33.4 nanoseconds) between the time the nose of the spacecraft passes over the first arm of the galaxy and the tail passes over the last arm. The spacecraft is still 10 meters long. Add a whole bunch more 9s and our spacecraft can traverse tip-to-tail an entire galactic supercluster in 33.4 nanoseconds. Add a whole bunch more 9s and we can make that 33.4 nanoseconds to traverse the known universe, tip-to-tail. But it still takes 33.4 nanoseconds for the rocket to pass tip-to-tail over any one particular star in the known universe. Or, for that matter, 33.4 nanoseconds to pass tip-to-tail over a ball tossed into the air on a planet orbiting that one particular star. Mods, please put this thread in Speculations. Link to comment Share on other sites More sharing options...
D H Posted November 14, 2006 Share Posted November 14, 2006 Suppose our spacecraft is 10 meters long and that we are moving at 99.99999999999999999999999999999999995% ... My bad. There are not enough nines here to make the galaxy 1 millimeter in diameter. I want [math]\gamma=\frac 1{\sqrt{1-(v/c)^2}} = \frac{10^5\ \text{ly}}{10^-3\ \text{m}} = 10^8 \frac{\text{ly}}{\text{w}} \approx 10^{24}\text{\ using\ } 1\ \text{ly} \approx 10^{16}\ \text{m}[/math] The correct velocity is v = 99.99999999999999999999999999999999999999999999995% c. Link to comment Share on other sites More sharing options...
Farsight Posted November 14, 2006 Share Posted November 14, 2006 D H, check out the link I posted at #54. This was a guy who said what you said: Now, from equation [16] one gets: [elapsed time]² = (T1-T2)² = 1/sh(aT)² = 1/th(aT)² - 1 = 1/v² - 1 Indeed, as v approaches the speed of light the time elapsed can be zero. Link to comment Share on other sites More sharing options...
D H Posted November 14, 2006 Share Posted November 14, 2006 All it takes is a little thought to see that this cannot be the case and remain consistent with special relativity (with which zanket claims to agree). If the ball can traverse the rocket in an arbitrarily short interval of time in the frame of the crew, the ball can exceed the speed of light in the frame of the crew. Something is obviously wrong here! Most reasonable people would think, "Oops, I screwed up somewhere. Now where did I make my mistake?" Crackpots do not think that way. They cannot be make mistaken. They charge forward, piling one mistake on top of another. They use non sequitur, rhetoric, false analogy, and argue till you are blue in the face. They refuse to admit errors, just like happened over at the website to which you referred in post 54 (this one: http://forum.physorg.com/index.php?showtopic=9591&st=122). Just because zanket pulled the wool over the eyes at some other website does not mean zanket is right. Here is another website to which zanket posted his garbage: http://www.sciforums.com/showthread.php?t=58421. What happened there is a different story. One of the users knows general relativity. There has been no reply from zanket after his "theory" was shown to be rubbish. Typical crackpot reaction. No apology; no response at all. Untangling the web of errors in a crackpot theory is a chore and leads to little of any use. This is why a lot of reasonable people on this website don't bother addressing crackpot theories. This is why dead-serious physics websites (those that attract a lot more physics professionals than does SFN) simply ban the crackpots. Link to comment Share on other sites More sharing options...
Zanket Posted November 14, 2006 Share Posted November 14, 2006 CXXXXXB is fixed in the frame of the crew and is not subject to length contraction. No, the ball is not fixed in the frame of the crew. The ball remains at rest with respect to the galaxy. The crew moves with respect to the ball and the galaxy. The ball is subject to length contraction in the crew's frame. My drawing is the initial condition only; that was clear from the text. XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXS is in the frame of the star S and is subject to length contraction. It is XXXXXXBXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXS that is subject to length contraction in the crew's frame. The ball is in the frame of the star S. It is the length contraction that makes the crew able to traverse the length of the galaxy in an arbitrarily short period of time. The CXXXXXB length is not contracted. No, it is [CXXXXXX] that is not length contracted in the crew’s frame. The ball moves relative to the crew. Mods, please put this thread in Speculations. You are illogical. Above you say “suppose the crew observes the ball to traverse the rocket in time [math]0.1 h/c[/math]”, yet you say that the ball does not move relative to the crew with “CXXXXXB is fixed in the frame of the crew”. The ball cannot both move and not move relative to the crew. Link to comment Share on other sites More sharing options...
Zanket Posted November 14, 2006 Share Posted November 14, 2006 If the ball can traverse the rocket in an arbitrarily short interval of time in the frame of the crew, the ball can exceed the speed of light in the frame of the crew. No, you agreed that the star S need not exceed c to traverse the rocket in such time. Then the ball, which is in the frame of the star S, need not exceed c either. They use non sequitur, rhetoric, false analogy, and argue till you are blue in the face. A classic case of the pot calling the kettle black. More on that below. There has been no reply from zanket after his "theory" was shown to be rubbish. I haven’t replied after one day! I must be a crackpot! Untangling the web of errors in a crackpot theory is a chore and leads to little of any use. Let’s recap your “logic” displayed in this thread: You think SR is not applicable in the presence of a gravitational source. No word from you on how it could then be experimentally confirmed. You think the equivalence principle is limited to non-gravitational experiments only. And you think it applies to gravitational experiments. You think a ball can simultaneously both move and not move relative to someone. Link to comment Share on other sites More sharing options...
D H Posted November 14, 2006 Share Posted November 14, 2006 No, the ball is not fixed in the frame of the crew. The ball remains at rest with respect to the galaxy. The crew moves with respect to the ball and the galaxy. The ball is subject to length contraction in the crew's frame. My drawing is the initial condition only; that was clear from the text. The ball is subject to length contraction. Big deal. We can shrink its diameter down to the Planck length with a high enough relativistic velocity. The ball still has some location as perceived by the crew. This location is some fixed point on the ball. A point is not subject to length contraction. It remains a dimensionless point regardless of the observer's velocity relative to the point. The speed of light is the same to all observers. How can some object (e.g., the ball) be observed to exceed the speed of light by the crew? It cannot (unless you reject special relativity also). One can get a superuminal velocity by mixing frames. Case in point: You agree that the crew can get to S in an arbitrarily short time in their frame. Sure. The crew can traverse multiple light years (as measured in the galaxy frame) per picosecond (as measured by the crew). Do you see the mixing of frames going on here? If you want to measure the velocity of some object as perceived by an observer in some frame, the distances must be measured with rulers fixed in the observer's frame and times must be measured with clocks fixed in the observer's frame. You are not doing this. It is XXXXXXBXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXS that is subject to length contraction in the crew's frame. The ball is in the frame of the star S. In other words, you want to measure the ball's crew-frame velocity using a ruler in star frame. CXXXXXB is fixed in the frame of the crew and is not subject to length contraction. You even used braces to show that this is in the frame of the crew. I'm sorry for the poor wording here. Obviously the ball is moving in the crew frame. What I meant was that this vector is expressed in the crew frame. I used fixed here meaning not subject to length contraction. Correcting that poor choice of words, CXXXXXB is expressed in the frame of the crew and is not subject to length contraction. You even used braces to show that this is in the frame of the crew. It had better be expressed in the crew frame or the resultant velocity will be invalid. You are illogical. No, I just choose my words poorly. Right back at ya, you are a pedant and a crackpot. Link to comment Share on other sites More sharing options...
Farsight Posted November 14, 2006 Share Posted November 14, 2006 DH: Stop saying crackpot. People call other people crackpots far too quickly. Sure Zanket might be wrong, but I wouldn't be too sure about him being wrong about everything. The thing you're forgetting is that when the crew are travelling at a velocity of c, they experience no time. Zero. Zip. Nil. None. And that ball inside the rocket might be sitting still, with an absolute velocity of 0. So it traverses the rocket interior at c. And it takes no time at all as far as the crew are concerned. Jeez, read Time Explained, try to get a handle on it, and right or wrong try not to be so rude. Link to comment Share on other sites More sharing options...
D H Posted November 14, 2006 Share Posted November 14, 2006 Farsight, (1) You are still mixing reference frames here, and (2) the crew cannot go at the speed of light. The thing you're forgetting is that when the crew are travelling at a velocity of c, they experience no time. Zero. Zip. Nil. None. And that ball inside the rocket might be sitting still, with an absolute velocity of 0. So it traverses the rocket interior at c. And it takes no time at all as far as the crew are concerned. You, as an observer in some other frame, see the ball pass through the vehicle at c. The crew, moving at the speed of light, see nothing. They are frozen in time. There is no motion. For this crew, time and the ball freeze the instant the crew achieves light speed. Fortunately, the crew are not photons. They cannot go the speed of light. Let us examine the case where the crew is moving relative to the ball with all but the tiniest fraction of the speed of light. Assumptions: The ball passes directly through the rocket, from top to bottom, much to the dismay of the crew, The ball is quite small, thereby avoiding extreme damage to the vehicle and the crew, The vehicle has extremely precise collision sensors that record the times (crew clock, of course) when the ball punctures the tip and tail of the vehicle, and The vehicle measures 10 meters, tip-to-tail, in the crew frame. What is the difference between the recorded times of the tip and tail collision sensors for this event? My answer: about 33.4 nanoseconds. Anything shorter than that violates the second hypothesis of special relativity, the speed of light in vacuum is the same to all observers. Link to comment Share on other sites More sharing options...
Farsight Posted November 14, 2006 Share Posted November 14, 2006 It's a limit condition DH. OK, you've got this ball sitting there in space, and along comes this relativistic rocket. BANG! If the rocket is travelling at c, the stationary ball takes zero time to traverse the interior of the rocket as far as the crew are concerned. You can reduce the velocity of the rocket by as little as you like, and the stationary ball takes as little time as you like to traverse the interior of the rocket as far as the crew are concerned. Nowhere have we contradicted Special Relativity. Note that at c your apparent speed is infinite because Speed = Distance / Time and your time experience is zero. Thus the apparent speed of the ball can also be infinite. If you think I'm right you should say sorry to Zanket. English isn't his first language so try to cut him some slack. And read TIME EXPLAINED again. Link to comment Share on other sites More sharing options...
D H Posted November 15, 2006 Share Posted November 15, 2006 Wrong. If the crew is travelling at light speed, the crew cannot measure ANYTHING. Things blow up at [math]v=c[/math]. In particular, [math]\gamma\to\infty \text{\ as\ } v\to c[/math]. Time and motion stop. The crew sees nothing. You are mixing frames when you say the stationary ball takes zero time to traverse the interior of the rocket as far as the crew are concerned. The apparent speed of the ball to the crew is 0/0, indeterminate. It is not infinite. Forget the crew travelling at c. What does the crew see as the velocity of the ball when some other observer measures their relative velocity as something less than the speed of light? I don't care if you make [math]\gamma = 10^1\text{\ or\ }10^{1000}[/math]. But please do answer my question in light of a finite [math]\gamma[/math]. You can even take the limit as [math]v\to c[/math]. Just don't make [math]v=c[/math] when doing that. Link to comment Share on other sites More sharing options...
Zanket Posted November 15, 2006 Share Posted November 15, 2006 The ball still has some location as perceived by the crew. This location is some fixed point on the ball. The location of an object is not just one of many points on the object. Even if you choose one point arbitrarily and deem it the ball’s sole location, it only obfuscates the issue. How can some object (e.g., the ball) be observed to exceed the speed of light by the crew? I’ve shown you in multiple ways that the ball need not exceed c while traversing the rocket in an arbitrarily short time in the crew’s frame, including showing you the relevant equations from the Usenet Physics FAQ (in section 8 of my paper). You’re arguing against not just me, but also the Usenet Physics FAQ. If you want to measure the velocity of some object as perceived by an observer in some frame, the distances must be measured with rulers fixed in the observer's frame and times must be measured with clocks fixed in the observer's frame. You are not doing this. The equations implicitly do that. Do you see the mixing of frames going on here? Do you think the Usenet Physics FAQ mixed frames without anybody noticing for years? I'm sorry for the poor wording here. Obviously the ball is moving in the crew frame. What I meant was that this vector is expressed in the crew frame. I used fixed here meaning not subject to length contraction. Anything that moves relative to the crew is length-contracted in their frame. That’s basic SR. The relevant equation is eq. 23 in my paper, for the Lorentz factor. And you contradicted yourself again. Above you agreed that “The ball is subject to length contraction” in the crew’s frame. The ball cannot be simultaneously both length-contracted and not length-contracted in the crew’s frame. Right back at ya, you are a pedant and a crackpot. Despite your rudeness I will again help you to see that the Usenet Physics FAQ is correct. Consider this initial situation: [CXXXXXX] RRRRRRRRBRRRRRRRRRRRRRRRRRRRRRRRRRRRRRS [CXXXXXX] = rocket, initially at rest with respect to the rod C = crew, who remain in a fixed position with respect to the rocket X = empty space R = rod with ball and star affixed B = ball affixed to the rod S = star affixed to the rod As shown, the ball is initially near the top of the rocket. Let the rocket begin accelerating toward the star. Then the ball and the star move towards the crew. You agree that the crew can get to the star in an arbitrarily short time in their frame; the rod doesn’t change that. It should be ultra clear that the crew will always get to the ball before they get to the star. Then they must be able to get to the ball in an arbitrarily short time in their frame. In other words, the ball can traverse the rocket in an arbitrarily short time in their frame. Nothing need move faster than c. The entire rod, including the ball and star affixed to it, length-contracts in the crew's frame. They traverse it at less than c relative to it. Let the rod have a proper length of one hundred thousand light years. Let it be length-contracted to a length of one meter in the crew's frame (i.e. as measured by them), in which case the crew’s velocity relative to the rod is nearly c. Then they need move less than one meter at a velocity of nearly c, both measured in their frame, to reach either the ball or the star. That shows how they can get to either the ball or the star in a fraction of the time h/c on their clock (where h is the proper height of the rocket) while moving slower than c relative to those objects. Any decent text on SR explains this. Link to comment Share on other sites More sharing options...
Zanket Posted November 15, 2006 Share Posted November 15, 2006 You can reduce the velocity of the rocket by as little as you like, and the stationary ball takes as little time as you like to traverse the interior of the rocket as far as the crew are concerned. Correct. Nowhere have we contradicted Special Relativity. Correct. English isn't his first language ... Incorrect! Link to comment Share on other sites More sharing options...
Zanket Posted November 15, 2006 Share Posted November 15, 2006 You can even take the limit as [math]v\to c[/math]. That's what Farsight did when he said "You can reduce the velocity of the rocket by as little as you like". Link to comment Share on other sites More sharing options...
D H Posted November 15, 2006 Share Posted November 15, 2006 I’ve shown you in multiple ways that the ball need not exceed c while traversing the rocket in an arbitrarily short time/ No, you have not. You have mixed frames. Anything that moves relative to the crew is length-contracted in their frame. The rocket does not move with respect to the crew. The rocket remains a constant length in the eyes of the crew. That’s even more basic SR. And it is the key piece of information that you are ignoring. You agree that the crew can get to the star in an arbitrarily short time in their frame; the rod doesn’t change that. It should be ultra clear that the crew will always get to the ball before they get to the star. ... In other words, the ball can traverse the rocket in an arbitrarily short time in their frame. You are mixing frames. The ball cannot traverse the rocket in an arbitrarily short period of time. Doing so violates SR, and you claim no bones with SR. The tip of the rocket can get to the ball and star in an arbitrarily short period of time because the distance to the star can be made arbitrarily small in the crew's frame by making the relative velocity approach that of light. Suppose the ball is even with the tip of the rocket. The key question, and the one you refuse to answer, is how much longer does it take for the ball to traverse the length of the rocket as observed in the crew's frame? Certainly it is more than h/c, no? Remember, the length of the rocket is not subject to length contraction. Then they need move less than one meter at a velocity of nearly c, both measured in their frame, to reach either the ball or the star. In other words, the speed of the ball as measured by the crew, is less than one meter per some tiny fraction of a second. It is not a hundred thousand light years per fraction of a second. The crew must use its own ruler to measure time and distance. Link to comment Share on other sites More sharing options...
Zanket Posted November 15, 2006 Share Posted November 15, 2006 The rocket does not move with respect to the crew. The rocket remains a constant length in the eyes of the crew. That’s even more basic SR. And it is the key piece of information that you are ignoring. This was not ignored above. The tip of the rocket can get to the ball and star in an arbitrarily short period of time because the distance to the star can be made arbitrarily small in the crew's frame by making the relative velocity approach that of light. Suppose the ball is even with the tip of the rocket. The key question, and the one you refuse to answer, is how much longer does it take for the ball to traverse the length of the rocket as observed in the crew's frame? Certainly it is more than h/c, no? Remember, the length of the rocket is not subject to length contraction. If you think the star can get to the tip of the rocket in an arbitrarily short period of time in the crew’s frame, why would you think the star cannot get to the crew in such time? After all, the star has traversed almost 100% of the distance by that point. If you read closely, I covered this above. But I’ll elaborate. In the initial condition where the rocket is at rest with respect to the rod, let the ball, which is attached to the rod, be even with the tip of the rocket. Let the crew be even with the tip of the rod opposite from the star. Let the proper height of the rocket (h) be 100 meters. Now let the rocket accelerate such that the entire rod (which has a proper length of one hundred thousand light years) length-contracts to a length of one meter in the crew’s frame (i.e. as they measure) in a negligible amount of time elapsed on the crew’s clock. At that moment the crew is beyond the tip of the rod and has moved partway along it. Then at that moment the distance in the crew’s frame between the crew and the star is less than one meter. The star has traversed over 99% of the rocket (> 99 meters / 100 meters). Meanwhile the ball, which is between the crew and the star, has traversed almost the entire rocket. Yet when the crew passes the ball, they will measure its velocity to be less than c. Same thing for the star. In the crew’s frame, even as the rod length-contracts from a length of one hundred thousand light years to a length of one meter in a neglible amount of time elapsed, decreasing the distance they measure to the star from one hundred thousand light years to less than one meter, they will always measure their velocity relative to any part of the rod passing directly by to be less than c. Any decent book on SR covers this. The equations in the Usenet Physics FAQ (section 8 of my paper) support this. The answer to your question is that it can take an arbitrarily short time for the ball to finish traversing the length of the rocket, as observed in the crew's frame. Link to comment Share on other sites More sharing options...
D H Posted November 15, 2006 Share Posted November 15, 2006 Can you not see the inconsistency in what you are saying? Yet when the crew passes the ball, they will measure its velocity to be less than c. It can take an arbitrarily short time for the ball to finish traversing the length of the rocket, as observed in the crew's frame. One way to measure the velocity is to note how much time passes when some object passes between two points that are a known distance apart. For example, the time of passage between (1) the ball crossing the tip of the rocket and (2) the ball crossing the tail of the rocket. If it takes an arbitrarily short time for this to occur, the ball's velocity can be arbitrarily high. Forget the star, forget the rod, make the ball a single point. You are making this overly complex and thereby not seeing your mistakes. Link to comment Share on other sites More sharing options...
Zanket Posted November 15, 2006 Share Posted November 15, 2006 One way to measure the velocity is to note how much time passes when some object passes between two points that are a known distance apart. For example, the time of passage between (1) the ball crossing the tip of the rocket and (2) the ball crossing the tail of the rocket. If it takes an arbitrarily short time for this to occur, the ball's velocity can be arbitrarily high. Forget the star, forget the rod, make the ball a single point. You are making this overly complex and thereby not seeing your mistakes. OK, make the ball a test particle. Measure the velocity using the method in your example. How can you tell how much time passes, when clocks all along the length of the rocket run at different rates due to gravitational time dilation? A clock at the tip of the rocket runs faster than a clock at the tail of the rocket. Measuring your way is mixing frames, which leads to invalid results. You can’t accurately tell how much time passes unless the clocks at both ends of the “known distance apart” run at the same rate. The only such distance is zero in the limit. When velocity is measured that way, it will always be less than c. While moving between two points zero distance apart in the limit, the test particle does not length-contract further. Then the increasing length contraction (increasing because the rocket is accelerating) that lets the particle traverse the rocket in an arbitrarily short time in the crew's frame does not affect the velocity measurement. Link to comment Share on other sites More sharing options...
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