andrewgray Posted June 19, 2007 Report Posted June 19, 2007 Zanket, So sorry I haven't been more clear. All statements really need an observer associated with them. OK, here are the rocket equations: We will rewrite in LaTex for clarity: [math]sinh(x) = \frac{e^x-e^{-x}}{2}[/math] [math]cosh(x) = \frac{e^x+e^{-x}}{2}[/math] [math]tanh(x) = \frac{sinh(x)}{cosh(x)}[/math] Let's let c=1 for simplicity. [math]t = \left (\frac{1}{a} \right ) sinh(aT) = sqrt{(d^2 + \frac{2d}{a}}[/math] [math]d = \left (\frac{1}{a} \right ) cosh(aT) - \left (\frac{1}{a} \right ) = \left (\frac{1}{a} \right ) (-1+sqrt{1 + (at)^2} )[/math] [Latex]v = tanh(aT) = \frac{at}{ sqrt{1 + (at)^2}}[/Latex] [Latex]T = \left (\frac{1}{a} \right ) sinh^{-1}(at) = \left (\frac{1}{a} \right ) cosh^{-1} (ad + 1)[/Latex] [Latex]\gamma = cosh(aT) = sqrt{1 + (at)^2} = ad + 1[/Latex] So Zanket, these equations describe the rocket itself. They do not address the more complicated subject of the rocket's coordinate system. I will prepare more in my next post. (I am not in my office right now.) But while I am doing this, notice that in the third velocity equation, if t<0, then the velocity of the rocket is negative, i.e., it is backing up. Andrew A. Gray P.S. Finally, if we let [math]x=d+\frac{1}{a}[/math], then we get [math]t = \left (\frac{1}{a} \right ) sinh(aT)[/math][math]x = \left (\frac{1}{a} \right ) cosh(aT)[/math] and we see clearly the hyperbolic nature of the equations. (more to come)
Guest Zanket Posted June 19, 2007 Report Posted June 19, 2007 So Zanket, these equations describe the rocket itself.I would say that they describe the relationship between the rocket and its gantry, an object in free fall that is initially at rest with respect to the rocket. But while I am doing this, notice that in the third velocity equation, if t<0, then the velocity of the rocket is negative, i.e., it is backing up.Yes, the rocket would be decelerating relative to the gantry. The equations can be used to show that an object that is dropped from a rocket hovering at r=2M+ε can never again reach r=2M+ε once it has fallen to r=2M, or, put differently, the box in the OP cannot pass outward through the horizon.
Guest Zanket Posted June 19, 2007 Report Posted June 19, 2007 Finally, if we let [math]x=d+\frac{1}{a}[/math], then we get ...Well, the Rindler horizon is at d minus 1/(a*γ). Why add?
andrewgray Posted June 21, 2007 Report Posted June 21, 2007 Well, the Rindler horizon is at d minus 1/a. Why add? Zanket, notice that [math]x^2-t^2=\frac{1}{a^2}[/math]. This is a hyperbola, and a very simplifying coordinate system. Also, this allows us to go from your rocket equations, which you understand, to Gravitation's coordinate system. So d is the distance traveled, and x=0 is at the origin of the coordinate system: The final step to switch to Gravitation's notation is that you have used T as the proper time for the rocket, and thus X for distances inside the rocket. Gravitation has T → ξo and X → ξ1. So, t,x,y,z (or xo,x1,x2,x3) are the "stationary" inertial frame's coordinate of the center of mass of the rocket, and ξo, ξ1, ξ2, ξ3 are the coordinates that the astronauts use inside their rocket ship, distances from their center of mass. Are you with me? If so, then perhaps this will make more sense to you: http://www.modelofreality.org/accel.jpg (zoom in) Don't worry too much about Fermi-Walker transport. It is just the gyroscope transport of the coordinate basis. Are you still with me? If so, then we want to consider coordinates that are slightly away from the center of mass of the rocket. For example, if the astronauts measure 3 cm behind the center of mass (ξ1=-3), what is the x coordinate of this point? Another example: If the astronauts measure this point at 1000 seconds after takeoff according to their watches (ξo=1000), what will the corresponding t be? Andrew A. Gray
Guest Zanket Posted June 22, 2007 Report Posted June 22, 2007 Are you with me?I’m not actually. Your image above doesn’t make sense to me. You should note who measures what, like “So d is the distance traveled” in the gantry’s frame. The gantry is a free test object dropped from the rocket at t=0. The distance d between the gantry and the rocket, and the elapsed time t, are measurable in the gantry’s frame. You have x=d+(1/a). If 1/a is meant to be the distance between the rocket and the Rindler horizon, then note that you’re mixing frames, because this distance is measurable only in an MCIF of the rocket. The rocket is always at d, which is x-(1/a), so the rocket should be to the left of the hyperbola. For simplicity I’d treat the rocket as arbitrarily small. In any case there’s no need to go to Gravitation’s coordinate system. My point of wanting to using the relativistic rocket equations was to avoid the complexity of the diagram in that book. I’d like you to refute me in the simplest possible way. It’s a lot easier to use the gantry’s inertial frame. In the gantry’s frame the Rindler horizon is at d-(1/(a*γ)). Plot the position of the rocket and a photon that is initially at the horizon and moves toward the rocket. You’ll see that the photon never reaches the rocket. In the plot above the gantry stays at d=0, so you can see it fall through the horizon in its own proper time t. You don’t need to take my word for it: Click on the image: The spaceship's world line is asymptotic to that of the flash of light, so the light never catches up with the spaceship. You’ve proven that there’s a Rindler horizon at r=2M for an observer hovering at r=2M+ε, where ε is zero in the limit. So you should agree that the box in the OP cannot pass outward through the horizon. If a photon initially at r=2M can’t reach r=2M+ε, then it can’t pass outward through the horizon. If a photon can’t do that then neither can the box. And any doubt that the box cannot pass outward through the horizon is removed by GR’s prediction that an infinite acceleration is required to remain at r=2M. All objects at r<=2M must fall, predicts GR, with no possibility of misinterpretation. In your LaTex you have d = (1/a) * (cosh(a * T) - 1) – (1/a). The bolded a shouldn’t be there.
andrewgray Posted June 22, 2007 Report Posted June 22, 2007 Ok, Zanket, I can work with your setup. These graphs: are essentially the same as Gravitation's. The only difference is that only times after t=0 are shown. If you extend the coordinates to t<0, you get the identical drawing that I was using before: where s is the distance variable for the rocket coordinates, and q is the time variable.So do you understand these drawings? Andrew A. Gray P.S. You say the Rindler horizon is 1/(a*γ) behind d . Where does the γ come from? P.S.S. The bolded a must have been a browser quirk, as it was not there in the LaTex.
Guest Zanket Posted June 22, 2007 Report Posted June 22, 2007 So do you understand these drawings?I think that’s beside the point until I understand why we should go beyond the refutation I have already made. What was wrong or suspect in my analysis that shows that nothing can pass outward through r=2M? It’s supported by the two charts I gave above and the Rindler site, and any doubt is removed by GR’s equation for the acceleration required to hover at r=2M. What compelling argument can you make for a more complicated counter-analysis, when mine is clear cut? P.S. You say the Rindler horizon is 1/(a*γ) behind d . Where does the γ come from?The horizon is a distance 1/a below the rocket measured in a frame that momentarily comoves with the rocket (an MCIF). Rulers in that MCIF are length-contracted in the gantry’s frame, which measures d. To account for the length contraction you divide by γ.
andrewgray Posted June 23, 2007 Report Posted June 23, 2007 OK, then we shall continue using your diagrams. Sorry if I am slow to address all of your logic. . . . any doubt is removed by GR’s equation for the acceleration required to hover at r=2M. Zanket, according to your site: We can imagine a flotilla of spaceships, each remaining at a fixed value of s by accelerating at 1/s. This statement is refering to this diagram: So this means that in this Special Relativity diagram it is impossible to "hover" at s=0 because of the infinite acceleration required. Therefore, according to you, an inertial frame X cannot "span" s=0 momentarily because the acceleration required to hover there is infinite. But we know this is to be false. Comments? Andrew A. Gray
Guest Zanket Posted June 24, 2007 Report Posted June 24, 2007 So this means that in this Special Relativity diagram it is impossible to "hover" at s=0 because of the infinite acceleration required. Therefore, according to you, an inertial frame X cannot "span" s=0 momentarily because the acceleration required to hover there is infinite. But we know this is to be false.We’re talking about two different equations here. You’re talking about SR’s equation for the acceleration a relativistic rocket requires to be a distance s above its Rindler horizon, where s is measured in an MCIF of the rocket: a = 1 / s I’m talking about GR’s equation for the gravitational acceleration at a given r-coordinate: a = M / (r^2 * sqrt(1 - (2M / r))) If SR contradicts the second equation—what you are implying—then GR is self-inconsistent, which would be GR’s problem, not mine. Then I can safely focus on the second equation and ignore SR’s predictions. (In fact you depended on the second equation in your proof.) Put differently, a self-inconsistency of GR does not save GR from being self-inconsistent! Note that I cover this type of argument in the OP: Another type of argument tries to use the self-inconsistency of GR against the proof of that. ...The second equation tells us that all material objects at r=2M must fall. An object cannot have a zero or positive upward velocity relative to r at r=2M, because the infinite gravitational acceleration there would reduce any such velocity to less than zero in zero time on the object’s clock.
andrewgray Posted June 24, 2007 Report Posted June 24, 2007 Zanket, I am not explaining this well enough, I guess. Apologies. We’re talking about two different equations here. You’re talking about SR’s equation for the acceleration a relativistic rocket requires to be a distance s above its Rindler horizon, where s is measured in an MCIF of the rocket: a = 1 / s I’m talking about GR’s equation for the gravitational acceleration at a given r-coordinate: a = M / (r^2 * sqrt(1 - (2M / r))) I am not talking about SR's equation for the acceleration for a rocket. I am talking about SR's equation for the hyperbolic acceleration at a given s-coordinate. You are missing the point. There is a one-to-one analogy between the two cases. So let me change directions completely and come at this problem for a different viewpoint. Take for example, an object launched from r=2M+ε at a certain coordinate velocity. It rises to r=2M+h, and falls back to r=2M+ε, reaching the same coordinate velocity in the opposite direction. This would be like a rocket ship "hovering" with its tail at r=2M+ε, while an astronaut "in the back" tossed a ball up to r=2M+h and the ball dropped back down to him. This process is time reversal symmetric around the time when the object reaches "it peak" at r=2M+h, I think you would agree. And, as you take the limit as ε→0, the process remains time reversal symmetric, correct? Next, an object is launched from s=0+ε at a certain coordinate velocity. It rises to s=0+h, and falls back to s=0+ε, reaching the same coordinate velocity in the opposite direction. This would be like a rocket ship "hovering" with its tail at s=0+ε, while an astronaut "in the back" tossed a ball up to s=0+h and the ball dropped back down to him. This process is also time reversal symmetric around the time when the object reaches "it peak" at s=0+h, I think you would also agree. And, as you take the limit as ε→0, this process also remains time reversal symmetric. Are you with me? If not, I guess you are going to force me to derive the hyperbolic coordinates in the local vicinity of r=2M. Andrew
Guest Zanket Posted June 24, 2007 Report Posted June 24, 2007 And, as you take the limit as ε→0, the process remains time reversal symmetric, correct?Yes, I agree. This process is also time reversal symmetric around the time when the object reaches "it peak" at s=0+h, I think you would also agree. And, as you take the limit as ε→0, this process also remains time reversal symmetric. Are you with me?Yes, I agree. But GR's equation above still shows that all material objects at r=2M must fall, and that makes my case. How does your analysis change that?
andrewgray Posted June 24, 2007 Report Posted June 24, 2007 OK, so on we go. There are no discontinuities in spacetime near s=0 or r=2M. We know this for a fact near s=0, since this is flat spacetime. So in both cases, we can extend this experience a little further, say to [math]r=2M-\frac{\epsilon}{1000000}[/math] , and [math]s= 0-\frac{\epsilon}{1000000}[/math] So we know that it is possible for an object to be launched from s=0-ε/1000000, rise up to s=0+h, and return to s=0-ε/1000000 in a time reversal symmetric manner. Here is a plot of the geodesic: The point is that the object has "passed upward through a coordinate worldline that has infinite acceleration associated with it." It has started at a coordinate value less than the horizon, and made it to a coordinate value greater than the horizon. Yet the coordinate value for the horizon has an infinite acceleration associated with it. Not only this. We can "pass upwards through s=0, and get to any large value for s desired (get out to infinity). Here is a plot of a geodesic to do this: The point is that the object has again "passed upward through a coordinate worldline that has infinite acceleration associated with it." It has started at a coordinate value less than the horizon, and "made it to infinity". Yet the coordinate value for the horizon has an infinite acceleration associated with it. The region of spacetime near r=2M is locally flat, and observers at fixed r experience constant acceleration. The constant acceleration Schwarzchild observers experience the very same thing here in this locally flat region of spacetime as the constant acceleration hyperbolic observers experience in their flat region of spacetime. Thus, it is possible to start from an r-coordinate value less than the horizon, and make it out to an r-coordinate value greater than the horizon in this local region of spacetime. Perhaps the the key to understanding this is that if something has a constant acceleration, "then sometime in the past, it stopped and went the other way". If you still do not agree, I guess you will force me to derive it mathematically for you. Andrew A. Gray
Guest Zanket Posted June 25, 2007 Report Posted June 25, 2007 We can "pass upwards through s=0, and get to any large value for s desired (get out to infinity).I agree. (But let’s be clear that in an idealized gravity-free universe, while you can get to any large value for s, if you’re in free fall then you must come down eventually, because the rocket will reach you eventually.) The point is that the object has again "passed upward through a coordinate worldline that has infinite acceleration associated with it." It has started at a coordinate value less than the horizon, and "made it to infinity". Yet the coordinate value for the horizon has an infinite acceleration associated with it.I agree. The region of spacetime near r=2M is locally flat, and observers at fixed r experience constant acceleration. The constant acceleration Schwarzchild observers experience the very same thing here in this locally flat region of spacetime as the constant acceleration hyperbolic observers experience in their flat region of spacetime. Thus, it is possible to start from an r-coordinate value less than the horizon, and make it out to an r-coordinate value greater than the horizon in this local region of spacetime.You’re assuming here that GR is self-consistent. That’s a mistake in logic when evaluating whether GR is self-consistent. Let’s make that more obvious with an extreme example. Suppose Einstein added one more prediction to GR: any object that reaches r=2M ceases to exist. Do you see that you could use the equivalence principle to make your same argument above, yet now GR would clearly be self-inconsistent? GR’s equation that I gave above refutes that “it is possible to start from an r-coordinate value less than the horizon, and make it out to an r-coordinate value greater than the horizon in this local region of spacetime”. Such is impossible according to that equation. That you can use the equivalence principle to show otherwise doesn’t mean that I’m wrong. It means that GR is self-inconsistent. Note that GR’s equation that I gave above applies to any object, even one in free fall, whereas “SR's equation for the hyperbolic acceleration at a given s-coordinate” applies only to the rocket. If you still do not agree, I guess you will force me to derive it mathematically for you.Anything you derive that does not directly refute that “such is impossible according to that equation” would not refute me. You cannot ignore that equation in your refutation. Showing that something in GR contradicts it would only make my case that GR is self-inconsistent. You need to show that the equation itself does not predict that all material objects at r=2M must fall. I gave logic above to show that it does predict that. So in both cases, we can extend this experience a little further, say to [math]r=2M-\frac{\epsilon}{1000000}[/math] , andWhy didn’t you extend it all the way to r=2M? You did not, because if you did, the rocket would begin to fall, and eventually fall all the way to r=0. It would fall because GR says that the gravitational acceleration is infinite at r=2M, and the rocket’s acceleration is finite. If the rocket can never again rise above r=2M once it reaches there, then why do you think some other material object could do that? What makes the rocket special to explain why only it has to fall?
andrewgray Posted June 25, 2007 Report Posted June 25, 2007 OK Zanket, I understand what you are saying. So let us attempt to prove this mathematically, but as simply as possible. According to your website, the s coordinate experiences an acceleration of 1/s. For starters, let the acceleration at s=1 be a=1. This simplifies things. Thus the coordinate transformation from (t,x) to (q,s) coordinates becomes: [math]t = s\sinh(q)[/math][math]x = s\cosh(q)[/math] Notice that we have: [math]x^2-t^2=s^2[/math] The flat spacetime metric becomes in terms of these new coordinates: [math]-d\tau^2 = -s^2 dq^2 + ds^2 +dy^2+dz^2[/math] Notice that the horizon is at s=0. Again, for simplicity, we will work in the equatorial plane and drop references to z. In the equatorial plane, the Schwarzchild metric becomes: [math]-d\tau^2 = -(1-\frac{2M}{r})dt^2 + \frac {1}{(1-\frac{2M}{r})}dr^2 + r^2d\phi^2 [/math] In a small neighborhood of r=2M, we can revert to a coordinate system as shown: R will be a "measurement coordinate from the horizon", and y will be perpendicular to R. So we let [math]dy = rd\phi [/math] and [math]\frac{R^2}{2} = r-2M[/math] [math]RdR = dr[/math] So we must calculate (1-2M/r) to first order: [math]1-\frac{2M}{r} \;\; =\;\; 1\;-\; \frac{2M}{(\frac{R^2}{2}+2M)} [/math] [math]\text{ }= \;\; 1 \;-\; \frac{1}{1+\frac{R^2}{4M}}[/math] [math]\text{ }\approx\;\; 1\;-\;(1-\frac{R^2}{4M}) \text{ where we have used 1/(1+x)}\approx\text{1-x}[/math] [math]\text{ }=\;\;\frac{R^2}{4M}\text{ valid in regions near r=2M}[/math] So to first order, the Schwarzchild metric becomes: [math]-d\tau^2\;=\; -\frac{R^2}{4M}dt^2 \;+\; \frac{1}{(R^2/4M)}R^2dR^2 \;+\; dy^2 [/math] Simplifying: [math]-d\tau^2\;=\; -\frac{R^2}{4M}dt^2 \;+\; 4MdR^2 \;+\; dy^2 [/math] For simplicity's sake we have let the acceleration be a=1 at s=1 and R=1. So we arbitrarily let 4M =1. (Zanket, if you want to see the complete proof, just say so. I have tried to keep it simple.) So finally, for the Schwarzchild metric, to first order we have: [math]-d\tau^2\;=\; -R^2dt^2 \;+\; dR^2 \;+\; dy^2 [/math] And indeed we see that to first order in the neighborhood of r=2M, there is no difference between a Rindler hyperbolic horizon and a gravitational horizon. This is because the Rindler Horizon (flat spacetime) metric: [math]-d\tau^2 \;=\; -s^2 dq^2 \;+\; ds^2 \;+\;dy^2[/math] Is of exactly the same form, and thus: To first order, the neighborhood of r=2M is a flat inertial frame and the equivalence principle is verified. Andrew A. Gray CraigD and InfiniteNow 2
Guest Zanket Posted June 26, 2007 Report Posted June 26, 2007 I understand what you are saying.I don’t think you did, because you ignored my key points. I said: Anything you derive that does not directly refute that “such is impossible according to that equation” would not refute me. You cannot ignore that equation in your refutation. Showing that something in GR contradicts it would only make my case that GR is self-inconsistent. You need to show that the equation itself does not predict that all material objects at r=2M must fall.But what you did, it seems, is ignore that equation and show that something in GR contradicts it, thereby helping make my case that GR is self-inconsistent. I’m talking about GR’s equation for the gravitational acceleration at a given r-coordinate: a = M / (r^2 * sqrt(1 - (2M / r)))
andrewgray Posted June 26, 2007 Report Posted June 26, 2007 Sorry Zanket, I did not mean to ignore your equation. What I showed is that in a local neighborhood around r=2M, spacetime is flat and the gravitational horizon is exactly equivalent to a Rindler horizon. That means that two objects in an inertial frame can stay equidistant with no restrictions on where they go. In other words, they do not have "to fall", and they do not have to become separated if they "span" the infinitely accelerated coordinate worldline R=0 (r=2M). Yes, the acceleration is infinite at R=0, just like the acceleration is infinite at the s=0 coordinate world line. To me, everything has been shown. Please continue to explain what is missing. I do not understand. Andrew A. Gray
Guest Zanket Posted June 26, 2007 Report Posted June 26, 2007 What I showed is that in a local neighborhood around r=2M, spacetime is flat and the gravitational horizon is exactly equivalent to a Rindler horizon.I don’t doubt that you showed this. Yes, the acceleration is infinite at R=0, just like the acceleration is infinite at the s=0 coordinate world line.It’s not “just like”, at least not when the acceleration at r=2M is given by GR’s equation that I gave above. The infinite acceleration at r=2M applies to all material objects, even those in free fall. The infinite acceleration at s=0 applies to only the rocket. To me, everything has been shown. Please continue to explain what is missing. I do not understand.Suppose a theory about birds makes the following predictions: Some birds can fly.No birds can fly.Obviously this theory is self-inconsistent; it contradicts itself. GR makes the following predictions: All material objects at r=2M must fall.Material objects at r=2M need not fall.Do you see that the theory contradicts itself? This equation of GR: a = M / (r^2 * sqrt(1 - (2M / r))) shows the first prediction. You’ve shown that GR also makes the second prediction, and the OP shows that as well. Suppose somehow you could dramatically increase the mass of the Earth while somehow maintaining its diameter. A bowling ball on the ground would become harder to lift, right? What would happen if you increased the gravitational acceleration at the ground to infinity, like it is at r=2M? What would happen is that the Earth including the ball would implode to r=0. The r-coordinate velocity of any object at ground level would become negative (downward) in zero time on the object’s clock. There’s no way that the object could reach a higher r-coordinate.
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