tag:blogger.com,1999:blog-2322970904637850676.post6352450754204721280..comments2017-05-13T12:03:58.000-07:00Comments on No Black Holes: Why black holes with permeable event horizons are perpetual motion machinesEditorhttp://www.blogger.com/profile/13969164309873348346noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-2322970904637850676.post-11935329932227282742009-05-11T15:50:00.000-07:002009-05-11T15:50:00.000-07:00Anon:
You have stated: “The main problem with pa...Anon:<br /><br />You have stated: “The main problem with paper is its interpretation… you need to use a coordinate system in which this observer is at rest, and a coordinate time which corresponds to this observer's proper time. This calculation has been done and it has been found that the infalling observer will cross the event horizon in a finite amount of time as measured by his clock.”<br /><br /><br /> The paper “Journey toward the Schwarzschild radius of a mass” clearly shows the infalling observer will not cross the event horizon in a finite amount of time as measured by his clock. This should be readily apparent based on the second law of thermodynamics. <br /> <br /> That is, as a result of time dilation, the finite passage of proper time necessary to reach the event horizon requires (i.e., is equal to) an infinite passage of Schwarzschild time. Well before this infinite passage of Schwarzschild time, and therefore before the infalling observer can reach the event horizon in proper time, everything (including the infalling observer and the black hole) will disintegrate. Even if the infalling observer were hardy enough to last an infinite amount of Schwarzschild time, Stephen Hawking has shown that the black hole is not.<br /> <br /> The disintegration that occurs in Schwarzschild time before the infalling observer reaches the event horizon, by necessity, also occurs in proper time before the infalling observer reaches the event horizon.Editorhttps://www.blogger.com/profile/13969164309873348346noreply@blogger.comtag:blogger.com,1999:blog-2322970904637850676.post-32347690829108991042009-05-11T15:45:00.000-07:002009-05-11T15:45:00.000-07:00Anon:
You have stated: “Energy is the 4th compon...Anon: <br />You have stated: “Energy is the 4th component of the 4-momentum vector, and this component will be different for different observers.”<br /><br /> If you define the variable “E” as representing the 4th component of the 4-momentum vector—your statement about energy is correct. But that is not how the variable “E” is defined or used in the present paper.<br /><br /> In the present paper, total energy “E” of mass m is specifically defined in equation (2) to be equal to mc^2. Based on that definition, the total energy E of a mass m is shown in equation (6)—and therefore in the Schwarzschild metric—to be equal to mc^2 in every reference frame!<br /> <br /> For example, in equation (6), mc^2, represents the total energy for mass m as measured in the reference frame of the Schwarzschild coordinates. In equation (6), the time component mc^2 (d tau/dt)^2 represents the energy for mass m that is neither gravitational nor kinetic energy. In the reference frame of the proper coordinates this time component of energy is equal to mc^2, i.e., it is the total energy for mass m as measured by the proper coordinates. However, because of how time dilation affects the speed of light, the time component of energy when measured in the reference frame of the Schwarzschild coordinates is equal to mc^2 (d tau/dt)^2.<br /> <br /> The paper “Conservation of momentum and energy in the Schwarzschild metric” goes into additional detail on conservation of both momentum and energy.Editorhttps://www.blogger.com/profile/13969164309873348346noreply@blogger.comtag:blogger.com,1999:blog-2322970904637850676.post-23785845241783231992009-05-11T15:30:00.001-07:002009-05-11T15:30:00.001-07:00The main problem with the paper is its interpretat...The main problem with the paper is its interpretation. It is well understood that an observer at infinity will observe that it takes an infinite amount of his time (i.e. the Schwarzschild coordinate time t) for the mass to reach the event horizon; i.e. it will never do so. But since you have only analyzed the process using this metric, this tells you nothing about what the infalling observer will experience. For that you need to use a coordinate system in which this observer is at rest, and a coordinate time which corresponds to this observer's proper time.<br />This calculation has been done and it has been found that the infalling observer will cross the event horizon in a finite amount of time as measured by his clock.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2322970904637850676.post-52657332554199147952009-05-11T15:30:00.000-07:002009-05-11T15:30:00.000-07:00You have asserted that in the Schwarzschild metric...You have asserted that in the Schwarzschild metric, the total energy associated with a particle of mass m is E=mc^2. This statement is incorrect, if by 'm' you mean the rest mass at infinity; in the Schwarzschild metric, the total energy is given by, e.g., eq. (25.14) or (26.16a) in the text "Gravitation" by Misner, Thorne and Wheeler. Energy is the 4th component of the 4-momentum vector, and this component will be different for different observers. E=mc^2 is the total energy that an observer who is in free fall with the mass would observe. The Schwarzschild metric corresponds to an observer at infinity, and that observer will see an energy different from E=mc^2.Anonymousnoreply@blogger.com