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Natural Science Forum / Physics / Relativity / June 2005GR for Roberts, Carlip, Daryl, Bilge/ Tucker
RE: Discussion of Roberts, Carlip, Daryl, Bilge & Tucker. This refs to Weinberg's "Grav&Cosmo", pg 148, "Geodesic Deviation"... I would like to comment on the equation of motion, given in ascii by, 0 = d^2x^u/ds^2 + GAMMA^u_vw(x) U^v U^w. (1) Without doubt, the x^u requires a relation involving TWO particles to define x^u. Using GR it is *by choice* we set either particle to be at rest to calculate the d^2x^u/ds^2, in order to compute relative coordinate acceleraction. OTOH, each of those particles is in classical geodesical motion, (classical meaning there is no exchange of photons to create inertial forces and reactions). Shifting to Eq.(6.10.1), we'll find those same TWO particles separated into distinct geodesics whose finite range I'll re-denote from the finite delta x^u used in the (6.10.1) to the finite x^u used in the geodesic (1). We have two expressions to describe a field, Eq(1) is one single *absolute* geodisic that presumes an FoR, but (6.10.1) requires relative geodesics. It appears, from point of view of GR, Eq.(6.10.1) needs no presumption of either geodesic being at rest in the computation, but instead only the relativity of two arbituary geodesics. Let's do an example. Consider the orbit of Mercury. Using Eq.(6.10.1) both the Sun and Mercury are on geodesics, and we should find the *precession* accountable by, R^a_bcd =/=0, without any specific centalization of a ref that Eq.(1) appears to require, in the spirit of GR. Regards Ken S. Tucker Ken S. Tucker says... >I would like to comment on the equation of motion, >given in ascii by, [quoted text clipped - 3 lines] >Without doubt, the x^u requires a relation involving >TWO particles to define x^u. >Using GR it is *by choice* we set either particle >to be at rest to calculate the d^2x^u/ds^2, in order >to compute relative coordinate acceleraction. I don't understand what you mean by that. Equation (1) is the equation of motion for a test particle described in curvilinear coordinates. What second particle are you referring to? It is true that the R_ijkl can be computed in terms of the deviation of two nearby geodesics, but equation (1) is only the equation for a single geodesic. -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > [quoted text clipped - 9 lines] > >to be at rest to calculate the d^2x^u/ds^2, in order > >to compute relative coordinate acceleraction. Hi Daryl, > I don't understand what you mean by that. Equation (1) > is the equation of motion for a test particle described > in curvilinear coordinates. What second particle are > you referring to? That 2nd particle maybe regarded as the gravitator producing the curvilinear coordinates, if you choose that particle as a ref, then Eq(1) becomes relevant. > It is true that the R_ijkl can be computed in terms of > the deviation of two nearby geodesics, but equation (1) > is only the equation for a single geodesic. That's right, in GR you can use the R_ijkl (6.10.1) should you select two gravational particles that are relatively geodesical, and that's my point. Any field is defined by the relative relation of particles, independent of any preferred reference. > Daryl McCullough Thanks Daryl Regards Ken S. Tucker i cant see the idea in putting moron bilge together in the subject line lets take the moron away from the subject line > RE: Discussion of Roberts, Carlip, Daryl, Bilge > & Tucker. [quoted text clipped - 47 lines] > Regards > Ken S. Tucker > This refs to Weinberg's "Grav&Cosmo", pg 148, > "Geodesic Deviation"... [quoted text clipped - 3 lines] > Without doubt, the x^u requires a relation involving > TWO particles to define x^u. Merely quoting an equation out of a book is NOT sufficient to understand what is going on -- you also need to understand what the symbols in the equation actually mean. Please read the text in that section. Quite clearly the x^u(\tau) describe the trajectory of ONE of the two particles under discussion, where the x^u(\tau) are determined by referring to some specific (but unspecified) coordinate system. These 4 values are functions of the particle's proper time \tau, as indicated by Weinberg's notation. The second particle has a trajectory x^u(\tau)+dx^u(\tau) [my ASCII d is Weinberg's \delta], referred to THE SAME coordinate system. > Using GR it is *by choice* we set either particle > to be at rest to calculate the d^2x^u/ds^2, in order > to compute relative coordinate acceleraction. You have fallen completely of the rails, and are wandering around in fantasy land here. The x^u are referenced to a specific (but unspecified) coordinate system. EXPLICITLY neither particle is "at rest", because they follow trajectories x^u(\tau) and x^u(\tau)+dx^u(\tau) -- neither of those is zero (they are _arbitrary_ functions of \tau; i.e. specific but unspecified functions of \tau). And Weinberg is most definitely NOT computing "relative coordinate acceleraction". He is, as the title of the section says, computing geodesic deviation. That is, his (6.10.1) relates the behavior of nearby geodesics to the curvature tensor. > OTOH, each of those particles is in classical > geodesical motion, Yes. You got something right. But the adjective form of the word is "geodesic". And "classical" is redundant, as Weinberg's entire book discusses non-quantum gravitation and cosmology. > (classical meaning there is > no exchange of photons to create inertial forces > and reactions). This is nonsense -- no "photons" are required to "create inertial forces and reactions". The presence of "photons" would explicitly indicate the presence of electromagnetic interactions, implying explicitly NON-inertial (non-geodesic) motion. And, of course, photons are quantum objects and Weinberg's book discusses only classical (non-quantum) gravitation and cosmology. > Shifting to Eq.(6.10.1), we'll find those same TWO > particles separated into distinct geodesics Please actually _READ_ what Weinberg wrote -- these two particles were always on separate geodesic paths, as that is the context of this entire section. > That 2nd particle maybe regarded as the gravitator > producing the curvilinear coordinates, No. Both particles are following geodesic paths in an arbitrary Lorentzian manifold characterized by a metric and its associated connection and curvature tensor. _NEITHER_ particle is a "gravitator", as in order for them to follow geodesic paths they must be of negligible mass. No "gravitator" is mentioned in this section, but the influence of any and all "gravitators" are contained in the metric, connection, and curvature tensor. > Any field is defined by the relative relation of > particles, independent of any preferred reference. Nonsense. A field is a function on the manifold. It is utterly unrelated to any "particles" or any "preferred reference". > [... further fantasies utterly unrelated to Weinberg's book] Tom Roberts tjroberts@lucent.com Roberts and Bilge and to some extent Carlip, tend to retain the late 60's idea of GR, aka classical GR, as I did as a student. In the intervening 35 years there has been a progession, called a 2nd look. In doing that Tucker predicts the LIGO experiment to be null, and in the period 1999-2005 (6 years) no g-wave has been detected as Tucker predicted in 1996. Nevertheless Roberts is almost polite in this post so I suppose we should reply. > > This refs to Weinberg's "Grav&Cosmo", pg 148, > > "Geodesic Deviation"... [quoted text clipped - 10 lines] > Please read the text in that section. Quite clearly the x^u(\tau) > describe the trajectory of ONE of the two particles under discussion, motion is relative, there is no reason in GR to choose either particle as the coordinate origin, that was the point of my post. Get a handle on GR. > where the x^u(\tau) are determined by referring to some specific (but > unspecified) coordinate system. These 4 values are functions of the [quoted text clipped - 9 lines] > You have fallen completely of the rails, and are wandering around in > fantasy land here. Stupid Roberts comment. > The x^u are referenced to a specific (but > unspecified) coordinate system. EXPLICITLY neither particle is "at [quoted text clipped - 6 lines] > geodesic deviation. That is, his (6.10.1) relates the behavior of nearby > geodesics to the curvature tensor. Back to square one <sigh> which of those "nearby" geodesics does the R^a_bcd in (6.10.1) apply to? ((don't that question again!!!!)) > > OTOH, each of those particles is in classical > > geodesical motion, > > Yes. You got something right. But the adjective form of the word is > "geodesic". And "classical" is redundant, as Weinberg's entire book > discusses non-quantum gravitation and cosmology. You have formed an unsupported opinion, evidentally without consideration of the advanced concepts. > > (classical meaning there is > > no exchange of photons to create inertial forces [quoted text clipped - 6 lines] > objects and Weinberg's book discusses only classical (non-quantum) > gravitation and cosmology. Provide an inertial force. > > Shifting to Eq.(6.10.1), we'll find those same TWO > > particles separated into distinct geodesics [quoted text clipped - 13 lines] > any and all "gravitators" are contained in the metric, connection, and > curvature tensor. Wrong, both particles are gravitators, thus Guv = Tuv is required, Guv=0 does not apply physically. > > Any field is defined by the relative relation of > > particles, independent of any preferred reference. > > Nonsense. A field is a function on the manifold. It is utterly unrelated > to any "particles" or any "preferred reference". Wrong again, you have no idea of what relativity is do you. It is a relation, that's all. That's why I stressed the importance for you to learn (6.10.1) prior to puking on GR, two particles, two geodesics and one R^a_bcd. If you intend to advance, you'll need to understand that and deal with the question I posed above, or resign. > > [... further fantasies utterly unrelated to Weinberg's book] One of us knows why g-waves can't be detected, and it it ain't Roberts! > Tom Roberts tjroberts@lucent.com Yes of course, bye Ken Ken S. Tucker says... >> No. Both particles are following geodesic paths in an arbitrary >> Lorentzian manifold characterized by a metric and its associated [quoted text clipped - 5 lines] > >Wrong, both particles are gravitators, No, that's not true. A geodesic is the path that a *test* particle takes, where a test particle is one whose mass is so small as to not affect the spacetime metric. So geodesic deviation is definitely *not* about the relative motion of two gravitators. Actual particles with sizable mass *don't* follow geodesics. >thus > >Guv = Tuv > >is required, Guv=0 does not apply physically. Well, yes. T_uv is never perfectly equal to zero, because there is always stray energy in the form of photons and cosmic rays. But the approximation of T_uv = 0 is a very good approximation for motion in the vacuum of space. >> Nonsense. A field is a function on the manifold. It >> is utterly unrelated to any "particles" or any "preferred >> reference". > >Wrong again, you have no idea of what relativity >is do you. Of course he does. Tom understands it about as well as anyone who posts to this newsgroup other than maybe Steve Carlip. -- Daryl McCullough Ithaca, NY | Ken S. Tucker says... | [quoted text clipped - 13 lines] | *not* about the relative motion of two gravitators. Actual particles | with sizable mass *don't* follow geodesics. Hmm... I think the only "test" particle we know that could possibly fit that bill is a photon. http://en.wikipedia.org/wiki/Geodesic_%28general_relativity%29 "Thus, for example, the orbital path of a planet around a star is the projection of a geodesic of the curved 4-D spacetime geometry around the star onto 3-D space." FrediFizzx http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf or postscript http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps FrediFizzx says... >| No, that's not true. A geodesic is the path that a *test* particle >| takes, where a test particle is one whose mass is so small as to not [quoted text clipped - 4 lines] >Hmm... I think the only "test" particle we know that could possibly fit >that bill is a photon. Well, it's a matter of scale. If you are calculating the path of a space probe around the sun, then the mass of the probe is completely negligible compared with the mass of the sun, so a geodesic gives a very good approximation to the path. -- Daryl McCullough Ithaca, NY > | Ken S. Tucker says... > | [quoted text clipped - 27 lines] > star onto 3-D space." > FrediFizzx Agreed, (My recent comments to Daryl, about relativity of geodesics notwithstanding). One important reason I'm interested in your (Freddi's) work on photon's (in the links below) regards a possible finite structure of a photon. If that's possible Weinberg's *tidal* equation (6.10.1) is more accurate than a simple geodesic. Of course that's nil in outer space, but when GR is considered in particle physics, like the way gamma rays interact, that tidal mechanism may be at the basis of pair production (gamma => e+ e-), the hypothetical concept being tidal force (the Roche limit is an analog), tear apart a photon. But if the photon is a structureless point, tidal forces (W (6.10.1)) will NOT apply. An additional test of that hypothesis is the relation of photons near the surface (but exterior to) of a neutron star. It would be a major achievement to determine if a photon has structure. Regards Ken S. Tucker > http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf > or postscript > http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps > | A geodesic is the path that a *test* particle > | takes, where a test particle is one whose mass is so small as to not > | affect the spacetime metric. > > Hmm... I think the only "test" particle we know that could possibly fit > that bill is a photon. Not true. For example, when studying the precession of the perihelion of Mercury, the planet Mercury is itself small enough to be considered a test particle when compared to the mass of the sun. The error in doing this is far less than the measurement resolution of its trajectory. Note, however, that in this case Jupiter cannot be considered to be a test particle; but the influences of the planets can be handled perturbatively (i.e. without a complete computation of their effects on the metric). Physics is a QUANTITATIVE science. When actually computing something one _MUST_ make approximations, so one must understand how good they are. For example, using a digital computer with a floating-point accuracy of 64 bits is good enough for most BUT NOT ALL computations, and some computations in GR are in that latter category. Tom Roberts tjroberts@lucent.com | > | A geodesic is the path that a *test* particle | > | takes, where a test particle is one whose mass is so small as to not [quoted text clipped - 17 lines] | 64 bits is good enough for most BUT NOT ALL computations, and some | computations in GR are in that latter category. Sure, however nature does not care about such things. Mercury is *not* a test particle to nature and you are merely speaking of our *present* limitations. Consider a system where you have two objects that are equal in mass (which is what I believe Ken was talking about), no matter how small, you have to consider that they are gravitators to each other. Does gravity shut off below the Planck mass like someone over in s.p.research was talking about? I don't think so. FrediFizzx http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf or postscript http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps Hi all... > | > | A geodesic is the path that a *test* particle > | > | takes, where a test particle is one whose mass is so small as to [quoted text clipped - 23 lines] > | 64 bits is good enough for most BUT NOT ALL computations, and some > | computations in GR are in that latter category. " Sure, however nature does not care about such things. " That's exactly the principle of covariance in words, that AE tried to explain. > Mercury is *not* > a test particle to nature and you are merely speaking of our *present* > limitations. Consider a system where you have two objects that are > equal in mass (which is what I believe Ken was talking about), no matter > how small, you have to consider that they are gravitators to each other. I have the right to use the test particle as the center of my CS. The Sun moves relatively to Mercury, by just reversing the geodesic used to make Mercury move on a geodesic relatively to the Sun. As Fredi says, "nature does not care". If I were to interpret that statement it would go like, "The laws of Nature are Generally Covariant". Kids, GC is a very powerful principle to discipline our thinking, and the math of tensors is the language, of GC and we're learning that new language. In view of Weinberg's (6.10.1), he used R^a_bcd to define a pair of geodisics...(aka tidal force) to define R^a_bcd. I understand what W is doing there, the text is an introduction. That equation requires the metric tensor "g_uv" to be defined by a finite length and NOT at a point but instead across the delta x^u. Classical GRist's fall back on Riemann Geometry as some sort of fuckin god, without understanding it's initial limits, especially where AE used it. I think (6.10.1) challenges the differential field, and produces a conceptual field derived using finite relations. Regards Ken S. Tucker Ken S. Tucker says... >> Mercury is *not* >> a test particle to nature and you are merely speaking of our *present* [quoted text clipped - 4 lines] >I have the right to use the test particle as the >center of my CS. This discussion has *nothing* to do with the choice of the origin of your coordinate system. The geodesic equation is *covariant*, you can use any coordinates you like. That isn't the issue. The issue is whether an object follows geodesics. In general, a massive, extended object does *not* follow a geodesic. The primary equations of motion for GR is not the geodesic equation, but the *Einstein* equations G_uv = T_uv To figure out the motion of a massive object, you have to solve those equations, subject to boundary conditions. -- Daryl McCullough Ithaca, NY FrediFizzx says... >Sure, however nature does not care about such things. Mercury is *not* >a test particle to nature and you are merely speaking of our *present* >limitations. Consider a system where you have two objects that are >equal in mass (which is what I believe Ken was talking about), no matter >how small, you have to consider that they are gravitators to each other. That's true. But in that case, the geodesic equation (d/ds)^2 x^u = Gamma^u_vw U^v U^w is *not* the correct way to compute the trajectories of the two objects. Real objects don't move on geodesics except as an approximation, and that approximation is only valid if one object's mass is very small compared with the dominant masses affecting the spacetime metric. -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > [quoted text clipped - 11 lines] > takes, where a test particle is one whose mass is so small as to not > affect the spacetime metric. That's a good definition. But a zero mass/momentum test particle does NOT exist. (I added the stipulation, about momentum in view of Freddi's recent post about photons). How far will you simplify the theortical context of a problem for "all practical purposes", shall we wind back 3 century's to Newton? OTOH, going forward, and studying the sublties of GR may yield unexpected results when new questions are allowed. > So geodesic deviation is definitely > *not* about the relative motion of two gravitators. > Actual particles > with sizable mass *don't* follow geodesics. And how will you define the x^i in the geodesic, D(dx^i/ds) = 0 (absolute derivative of U^i). Why would you choose for example Mercury to follow a (approximate) geodesic relative to the Sun, but not the Sun following a geodesic relative to Mercury? The whole point of GR is to express physical laws without choosing arbituary origins or states of motion. Therefore the Sun follows a geodesical path *relative* to Mercury, and thus the size of the test mass is immaterial. It is the relation that matters. In practical applications young GRist's think in terms of using the Sun as the center to calculate orbits, then somehow move that to something important in the theory, and they're back to Newtonianism. ((I like that term Newtonianism)). > >thus > > [quoted text clipped - 6 lines] > cosmic rays. But the approximation of T_uv = 0 is a very > good approximation for motion in the vacuum of space. The motion of Sun is not a very good vacuum. [snip gossip] > -- > Daryl McCullough > Ithaca, NY Ken S. Tucker says... >> ...A geodesic is the path that a *test* particle >> takes, where a test particle is one whose mass is so small as to not >> affect the spacetime metric. > >That's a good definition. But a zero mass/momentum >test particle does NOT exist. Yes, that's true. Real particles do not obey the equation (d/ds)^2 x^u = Gamma^u_vw U^v U^w That equation is only an approximation that is valid to the extent that the mass of the particle is small compared with the masses of nearby sources of gravitation. >How far will you simplify the theortical context of >a problem for "all practical purposes", shall we wind >back 3 century's to Newton? I'm only arguing that your interpretation of the geodesic equation is incorrect. It isn't the path of a "gravitator", except as an approximation. And that approximation is only valid to the extent that you can *ignore* the gravitational effects of the test particle. >> So geodesic deviation is definitely >> *not* about the relative motion of two gravitators. [quoted text clipped - 9 lines] >but not the Sun following a geodesic relative to >Mercury? Neither the Sun nor Mercury follows a geodesic, exactly, but since the mass of Mercury is very small compared with that of the Sun, the path of Mercury is approximately described by a geodesic. But the mass of the Sun is not small compared with the mass of Mercury, so approximating the path of the sun by a geodesic around Mercury would make no sense. The sun probably does follow an approximate geodesic through the Milky Way. >The whole point of GR is to express physical laws >without choosing arbituary origins or states of >motion. Therefore the Sun follows a geodesical >path *relative* to Mercury, and thus the size >of the test mass is immaterial. No, it doesn't. And a geodesic isn't *relative* to another particle or object. A geodesic is a path through spacetime. It can be *described* using coordinates, and that's what the geodesic equation does. >It is the relation that matters. I think you are mixing up GR with some other, purely relational theory. In GR, spacetime is a physical entity in its own right. Objects don't directly interact with each other gravitationally; instead, they interact via spacetime. Objects distorts spacetime, and their motion is affected by these distortions. But the motion of one particle makes no reference to the location of any other particle; the only relevant quantity is the metric. >In practical applications young GRist's think in >terms of using the Sun as the center to calculate >orbits, then somehow move that to something important >in the theory, and they're back to Newtonianism. >((I like that term Newtonianism)). Using the geodesic equation to calculate the motion of Mercury has nothing to do with whether your coordinate system puts the Sun in the center. You can use any coordinates you like to compute the geodesic, it doesn't have to be centered on the Sun. >> Well, yes. T_uv is never perfectly equal to zero, because >> there is always stray energy in the form of photons and >> cosmic rays. But the approximation of T_uv = 0 is a very >> good approximation for motion in the vacuum of space. > >The motion of Sun is not a very good vacuum. I don't understand what that sentence means. The space between the Sun and Mercury is a vacuum to a very good approximation. The Einstein tensor G_uv is equal to zero in this region, to a good approximation. >[snip gossip] How is it gossip to say that Tom Roberts and Steve Carlip understand General Relativity better than most posters? -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > [quoted text clipped - 8 lines] > > (d/ds)^2 x^u = Gamma^u_vw U^v U^w Again I'll differ. That equation simply reiterates DU^u =0 (absolute derivative), meaning there is no absolute acceleration. > That equation is only an approximation that is > valid to the extent that the mass of the particle > is small compared with the masses of nearby sources > of gravitation. As I explained the mass of the particle is immaterial in GR. > >How far will you simplify the theortical context of > >a problem for "all practical purposes", shall we wind [quoted text clipped - 5 lines] > the extent that you can *ignore* the gravitational effects > of the test particle. Careful, what you're doing is setting G_uv=0 to some sort of law, well it isn't cause G_uv isn't invariant. You need to unlearn the GR introduction. > >> So geodesic deviation is definitely > >> *not* about the relative motion of two gravitators. [quoted text clipped - 11 lines] > > Neither the Sun nor Mercury follows a geodesic, Now that's contradictory. The geodesic specifies absolute acceleration vanishes, (DU^=0), what you attempt to do is reverse that and claim the Sun or Mercury follow some course specified by absolute acceleration. >exactly, > but since the mass of Mercury is very small compared with [quoted text clipped - 3 lines] > the path of the sun by a geodesic around Mercury would make > no sense. That's reversing the application onto the theory. > The sun probably does follow an approximate geodesic > through the Milky Way. Let's confine ourselves to the Sun and Mercury. > >The whole point of GR is to express physical laws > >without choosing arbituary origins or states of [quoted text clipped - 4 lines] > No, it doesn't. And a geodesic isn't *relative* > to another particle or object. So what you're doing is essentially producing an absolute FoR like an aether. Your attempting to regard geodesics as something absolute instead of relative. > A geodesic is > a path through spacetime. It can be *described* > using coordinates, and that's what the geodesic > equation does. > >It is the relation that matters. > > I think you are mixing up GR with some other, purely > relational theory. GR is a "purely relational theory" that's why we call it "General Theory of Relativity". > In GR, spacetime is a physical entity in its own right. > Objects don't directly [quoted text clipped - 4 lines] > to the location of any other particle; the only > relevant quantity is the metric. What you've written in that last paragraph is a good explanation of G_uv=0 which is an easy linear approx. But GR is really about G_uv =T_uv, an in the example of a binary star system, using G_uv =0 is out of the ?. > >In practical applications young GRist's think in > >terms of using the Sun as the center to calculate [quoted text clipped - 19 lines] > approximation. The Einstein tensor G_uv is equal to zero > in this region, to a good approximation. That doesn't make it right. This entire post has argued for G_uv=0 as ok. Why is there so much resistance to Einstien's Law expressed by the generic Guv=Tuv, I find that's easier. > >[snip gossip] > > How is it gossip to say that Tom Roberts and Steve Carlip > understand General Relativity better than most posters? Evaluate the post. Ken > -- > Daryl McCullough > Ithaca, NY Ken S. Tucker says... >> Yes, that's true. Real particles do not obey the equation >> [quoted text clipped - 5 lines] > >meaning there is no absolute acceleration. Yes, and that's not literally true for a real, massive, extended object. >> That equation is only an approximation that is >> valid to the extent that the mass of the particle [quoted text clipped - 3 lines] >As I explained the mass of the particle is immaterial >in GR. That's not true. The mass of the particle modifies the metric. The greater the mass, the greater the modification. The geodesic equation describes the path of a particle in a *external* metric (one that doesn't include modifications due to the test particle). >> I'm only arguing that your interpretation of the geodesic >> equation is incorrect. It isn't the path of a "gravitator", [quoted text clipped - 4 lines] >Careful, what you're doing is setting G_uv=0 to some sort >of law, well it isn't cause G_uv isn't invariant. No, the discussion of geodesics has nothing to do with G_uv. >> >Why would you choose for example Mercury to follow >> >a (approximate) geodesic relative to the Sun, [quoted text clipped - 4 lines] > >Now that's contradictory. No, it's not. The geodesic is the path that would be taken by a particle of negligible mass. The sun and Mercury do not have zero mass, and they don't follow geodesics. >The geodesic specifies absolute acceleration vanishes, >(DU^=0), what you attempt to do is reverse that and >claim the Sun or Mercury follow some course specified >by absolute acceleration. There was a discussion about this a few months ago. The correct way to figure out the motion of Mercury (or the Sun) is to solve the Einstein equations of motion. Those equations don't reduce to the geodesic equation except in the limiting case of the motion of a test particle of negligible mass. >> exactly, >> but since the mass of Mercury is very small compared with [quoted text clipped - 5 lines] > >That's reversing the application onto the theory. I don't know what that sentence means. The correct way to compute the motion of Mercury or the sun is to solve Einstein's equations of motion. Those equations are much more complicated than the geodesic equation. However, you can get an approximate solution to the Sun/Mercury system by using the following approach: 1. Compute the metric for the sun in the absence of any planets. 2. Compute the paths of the planets under the assumption that they move along geodesics of the metric found in 1. >> No, it doesn't. And a geodesic isn't *relative* >> to another particle or object. > >So what you're doing is essentially producing an >absolute FoR like an aether. No, I'm not. I don't know why you got that impression. >Your attempting to regard geodesics as something absolute >instead of relative. The path itself is a geometrical object, independent of any coordinate system. The coordinate description of the path is, of course, relative to a coordinate system (*not* relative to another object). >> I think you are mixing up GR with some other, purely >> relational theory. > >GR is a "purely relational theory" that's why we >call it "General Theory of Relativity". No, it's not. >> In GR, spacetime is a physical entity in its own right. >> Objects don't directly [quoted text clipped - 10 lines] >But GR is really about G_uv =T_uv, an in the example >of a binary star system, using G_uv =0 is out of the ?. When people write that G_uv = 0, they aren't denying that G_uv = T_uv. They are, instead, saying that T_uv = 0 (to a good approximation). >> I don't understand what that sentence means. The space >> between the Sun and Mercury is a vacuum to a very good [quoted text clipped - 3 lines] >That doesn't make it right. This entire post has >argued for G_uv=0 as ok. Yes, and in the region between the Sun and Mercury, G_uv = 0, to a good approximation. > Why is there so much resistance to Einstien's Law >expressed by the generic Guv=Tuv, I find that's easier. Nobody's denying that. We're talking about the case in which T_uv = 0 (to a good approximation). >> How is it gossip to say that Tom Roberts and Steve Carlip >> understand General Relativity better than most posters? >Evaluate the post. Which post? Anyway, it is still true---Tom Roberts and Steve Carlip understand GR better than most posters, and frankly, that includes both you and I. -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > [quoted text clipped - 154 lines] > Carlip understand GR better than most posters, and frankly, > that includes both you and I. Speak for yourself, you are less than informed. > -- > Daryl McCullough > Ithaca, NY Ken S. Tucker says... >> Which post? Anyway, it is still true---Tom Roberts and Steve >> Carlip understand GR better than most posters, and frankly, >> that includes both you and I. > >Speak for yourself, you are less than informed. I know enough to know that you know less about GR than Tom or Steve --- by a large margin. -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > [quoted text clipped - 6 lines] > I know enough to know that you know less about > GR than Tom or Steve --- by a large margin. That's gossip, can you sustain a point, otherwise I need not waste time with you. Daryl, get OFF THE POT and consider (6.10.1). I know your brain, and your friends...boring. Ken > -- > Daryl McCullough > Ithaca, NY Ken S. Tucker says... >> I know enough to know that you know less about >> GR than Tom or Steve --- by a large margin. > >That's gossip No, it's not. Gossip is rumor or talk of a personal, sensational, or intimate nature. Making a judgement about a person's competence based on *public* behavior isn't gossip. -- Daryl McCullough Ithaca, NY > How far will you simplify the theortical context of > a problem for "all practical purposes", shall we wind > back 3 century's to Newton? Physicists simplify problems in order to be able to obtain solutions to the equations that describe them. And to do so requires knowing the validity of the simplification involved. For instance, in considering Mercury to be of negligible mass compared to the sun involves an approximation more accurate than the resolution for measuring Mercury's orbit, so such an approximation is clearly warranted. > OTOH, going forward, and studying the sublties of GR > may yield unexpected results when new questions are > allowed. Making such an approximation as above does not "disallow" any "questions". > And how will you define the x^i in the geodesic, > D(dx^i/ds) = 0 (absolute derivative of U^i). Using _ANY_ valid coordinate system you like. I repeat: you _REALLY_ need to learn the basics -- you keep quoting formulas without having the slightest notion of what they mean or what the symbols in them represent. That is not physics, that is not math, that is just plain stupid. > Why would you choose for example Mercury to follow > a (approximate) geodesic relative to the Sun, > but not the Sun following a geodesic relative to > Mercury? Because Mercury is of negligible mass compared to the sun, but the sun is not of negligible mass compared to Mercury. Sheesh! > The whole point of GR is to express physical laws > without choosing arbituary origins or states of > motion. No. The point is to describe physics without having to choose any particular coordinate system. It often happens that the PHYSICAL SITUATION includes some specific "origin or state of motion" -- when that happens the theory must obviously conform to the PHYSICAL SITUATION. For instance, the sun is much larger than any planet. This imposes a constraint on any model of the solar system thgat expects to be valid. > Therefore the Sun follows a geodesical > path *relative* to Mercury, and thus the size > of the test mass is immaterial. Not true. In GR only TEST PARTICLES follow geodesic paths through spacetime, and this is really only approximate (but an excellent approximation to the extent that one can neglect the influence of the test particle on the geometry). BTW geodesic paths are not "relative" to any object, they are determined completely by the _GEOMETRY_. In GR, of course, the geometry is determined by all matter and energy EXCEPT test particles. > It is the relation that matters. In practical > applications young GRist's think in terms of > using the Sun as the center to calculate orbits, > then somehow move that to something important > in the theory, and they're back to Newtonianism. > ((I like that term Newtonianism)). Only "young GRist's" that don't understand basic physics, and how one properly performs approximations. Such as yourself. >>Well, yes. T_uv is never perfectly equal to zero, because >>there is always stray energy in the form of photons and >>cosmic rays. But the approximation of T_uv = 0 is a very >>good approximation for motion in the vacuum of space. > > The motion of Sun is not a very good vacuum. "Motion" cannot possibly be "vacuum" -- they are completely incommensurate concepts. The sun itself is definitely not vacuum, and nobody ever said it is. What is vacuum to excellent approximation is the region of the solar system outside the sun and the planets. And in the approximation that Mercury is of negligible mass compared to the sun, then Mercury follows a geodesic path through a vacuum region of the solar system. If one neglects the rotation and oblateness of the sun, and all the other planets, then one can compute Mercury's orbit in the Schwarzschild manifold with M corresponding to the mass of the sun; doing so gives excellent agreement with the observed orbit of Mercury (after subtracting away those influences omitted in the model) -- in particular the precession of its perihelion is well matched. I grow weary of trying to get you to see you own shortcomings. Good bye. Tom Roberts tjroberts@lucent.com > > How far will you simplify the theortical context of > > a problem for "all practical purposes", shall we wind [quoted text clipped - 89 lines] > I grow weary of trying to get you to see you own shortcomings. > Good bye. Tucker will classify new GR data. Ken On the meaning of senility, [...] > For instance, the sun is much larger than any planet. > This imposes a constraint on any model of the solar system > thgat expects to be valid. > Tom Roberts... Wow, Roberts has decided General Relativity is now superseded by the Copernicus model, what is that the Roberts theory of FAT planets? If Roberts is a *state of the art* GRist as some think, he makes Jim Jones look like an amateur. The stupidity of Robert's statement is the apitamy of the morass of logic. If any other professional GRist agrees with him what the f.ck does "constraint" mean???? His post gets a FUF grade, f.cked UP and Failed, that's the lowest I know, but Roberts might invent lower grades, to his credit. Regards Ken S. Tucker Ken S. Tucker says... >On the meaning of senility, > [quoted text clipped - 7 lines] >superseded by the Copernicus model, what is that the >Roberts theory of FAT planets? Ken, you say things that don't make any sense, with no purpose other than being insulting. -- Daryl McCullough Ithaca, NY Ken S. Tucker says... >On the meaning of senility, > [quoted text clipped - 3 lines] >> thgat expects to be valid. >> Tom Roberts... >The stupidity of Robert's statement is the apitamy >of the morass of logic. You disagree that the sun is much larger than any planet? You disagree that the sun has a disproportionate effect on the spacetime metric of the solar system? What in the world are you disagreeing with? It seems to me that you were just in a mood to lash out, and it didn't really matter what Tom wrote. This is in contrast with Tom's criticism of your posts. He criticized *specific* claims and assumptions that you made, and he explained why you were wrong. >If any other professional >GRist agrees with him what the f.ck does "constraint" >mean???? If you didn't know what Tom meant, why didn't you ask *before* you conclude that it is stupid? A constraint means the same thing as a condition: a constraint on a model is something that must be true in order for the model to be acceptable. In GR, mass distorts the spacetime metric. An object with higher mass has a larger effect than an object with smaller mass. So the metric of the solar system is mostly determined by the largest mass around, the sun. The effects of the planets are minor corrections. That's what Tom was saying, and I don't see how you can disagree with it, much less consider it the "apitamy of the morass of logic". -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > > [quoted text clipped - 15 lines] > were just in a mood to lash out, and it didn't really > matter what Tom wrote. Bull, Roberts earned that lash, and so will you if you insist GR has anything to with the FAT planet theory Roberts f.cked your brain with. He and now you puke on GR by suggesting there is any preferred ref because one mass is FATTER than another. That's the idiocy this thread is taking. [snip a load of gossip] > In GR, mass distorts the spacetime metric. An object with > higher mass has a larger effect than an object with smaller [quoted text clipped - 3 lines] > and I don't see how you can disagree with it, much less > consider it the "apitamy of the morass of logic". Daryl, what happened to your brain? We are in the GENERAL THEORY OF RELATIVITY, not some pussy assed flower crap you and Roberts are pushing in some ding-bat application called solar system. GTR is true in the whole universe, no exception. Mass does NOT make a preferred ref. You guys are forming a weird cult. THE GENERAL THEORY OF RELATIVITY will survive U. Ken S. Tucker > -- > Daryl McCullough > Ithaca, NY This post concerns invariants. The Kronecker delta has two values, 0 and 1, and to offset that 1 to something like .999 is like offsetting 0 to .001, doesn't make sense, so the (0,1) is logical. Commonly, authors use c=1, q=1 and h=1, as speed of light, fundamental charge and Plancks, and those are recognized invariants. In view of the above, we will consider the GR invariant, ds^2 = g_uv dx^u dx^v and the integral of that, $ds = s. Because "s" is an invariant, we either obtain s=0 or s=1. Using ds=0 means we have an integration constant. Suppose hypothetically, by calculus, we obtained, s^2=g^uv x_u x_v as the finite invariant "s". IMO that would not make good sense. Instead I would need s^2 = g^uv a_u b_v where the finite interval "s" spans from location "a" to "b". And it follows the g_uv is defined by the relativity of "a" and "b". So, in view of the integral, GR demands ds^2 = g_uv da^u db^v, to produce, s^2 = g^uv a_u b_v wherein g_uv is defined by the relation of points "a" and "b". I think the last definition of ds^2 is necessary in order to realize GR is a theory of relations. Historically, the Reimann Geometry was developed in the mind frame of absolute motion and the fields that provide that, such as analysing a point on a curved surface. The General Theory of Relativity applied by AE that definition as an introduction, but we can clearly see that as definition of a continuum at a point disrepects the meaning of relativity of relation. Ken S. Tucker Ken S. Tucker says... >This post concerns invariants. > [quoted text clipped - 18 lines] >Because "s" is an invariant, we either obtain >s=0 or s=1. No, you don't. s is proper time. It can be any real number. -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > > [quoted text clipped - 22 lines] > > No, you don't. s is proper time. It can be any real number. Of course and so can "c, q or h". I presume one could set those to any non zero scalar, but my point is, it's arbituary in relativity, so effectively s=0 or 1. The only reality occurs when a measurement is made w.r.t a basis, that basis defines the g_uv. Check this out, let me set c=q=h=1 to c=g_uv c^uv = q=g_uv q^uv = h=g_uv h^uv QUESTION: Does it follow c^uv = q^uv = h^uv ? Well IMHO it does not. As an example, in a very simple universe g_00=1 , g_11=1 , g_01 =0. Use c^00=1 , c^11=0 and q^00=0 , q^11=1, (c^01=q^01=0 , and everything is symmetric). Clearly c^uv =/= q^uv , but some measure like c^uv and q^uv produces the SAME invariant "1". Any difference of an invariant from "1" is only a scale factor that is arbitrary. A unified field theory in a covariant form would relate the tensors, c^uv , q^uv , h^uv GR , EM , QT which may help to see the problem and the solution. Regards Ken S. Tucker > -- > Daryl McCullough > Ithaca, NY Ken S. Tucker says... >> >Because "s" is an invariant, we either obtain >> >s=0 or s=1. >> >> No, you don't. s is proper time. It can be any real number. > >Of course and so can "c, q or h". No, they can't. Once you've chosen a system of units, c q and h are *fixed* by that choice. That isn't true of s. s is invariant, but it isn't a *constant*. s is proper time. If an ideal clock moves from point A to point B, then s along the path will be the elapsed time (the time at B minus the time at A). You can certainly choose a time scale so that the elapsed time at B is exactly 1, but then there will be points between A and B at which s is a number different from 0 or 1. >Check this out, let me set c=q=h=1 to > [quoted text clipped - 3 lines] > >c^uv = q^uv = h^uv ? No, of course not. What does that have to do with s? Daryl McCullough Ithaca, NY > Ken S. Tucker says... > [quoted text clipped - 6 lines] > > No, they can't. Once you've chosen a system of units See, thats where you're misconcieving GR/GC (General Covariance) and the purpose of tensors. The point of of GC is it applies to all systems of units, it's NOT pre-chosen! >c q and h are *fixed* by that choice. That isn't true of s. > s is invariant, but it isn't a *constant*. Well, if light-rays are the basis of our survey of spacetime and ds=0 for light-rays it follows "s" is a constant. > s is proper time. If an ideal clock moves from point A to point B, > then s along the path will be the elapsed time (the time at B > minus the time at A). You can certainly choose a time scale so > that the elapsed time at B is exactly 1, but then there will be > points between A and B at which s is a number different from 0 or 1. Look, that's queer, you're setting an invariant to have different scalar quantities because of a transformation. There is no such thing as proper time measured in seconds! > >Check this out, let me set c=q=h=1 to > > [quoted text clipped - 5 lines] > > No, of course not. What does that have to do with s? Tensor technology allows two choices, s^2 = g_uv x^u x^v or s^2 = g_uv a^u b^v. How do you choose? And I want you guys to render an opinion on the question you're prepared to support, and if not don't snip it, that sucks. ((Roberts, Carlip, Bilge haved snipped the hard stuff, and went fluffy, then complain they don't understand unified field theory...duh)). Now go back to Weinberg's Eq.(6.9.1) and (6.10.1) for some background, that's tough stuff, physics isn't for fluzzy's. In view of Eq.(6.9.1), (6.10.1) describes a relative relation between geodesics, but uses a single metric tensor for both. Hence that tensor spans the geodesics alluded too in (6.9.1), where "a" and "b" are *introduced*, but formalized in (6.10.1). So now we need to choose a solution to those equations, which in my view, use separate geodesics (tidal), but in consequence, the invariant, s^2 = g_uv a^u b^v. where "a" and "b" are distinct. By distinct I mean "a" and "b" have differing spacetime coordinates. In juxtaposition, consider our usual, s^2 = g_uv x^u x^v and find x=x, and the "x" are NOT distinct. Enough Ken S. Tucker re > Daryl McCullough > Ithaca, NY > > Ken S. Tucker says... > > [quoted text clipped - 11 lines] > of GC is it applies to all systems of units, it's NOT > pre-chosen! Ken fans the air, wildly, on a slow pitch. Well, if the ball had been where Ken thought it was, that might have been a hit. > >c q and h are *fixed* by that choice. That isn't true of s. > > s is invariant, but it isn't a *constant*. > > Well, if light-rays are the basis of our survey of > spacetime and ds=0 for light-rays it follows "s" > is a constant. Ah, the mluttgens school of physics. Nothing but lightlike intervals allowed. > > s is proper time. If an ideal clock moves from point A to point B, > > then s along the path will be the elapsed time (the time at B [quoted text clipped - 6 lines] > a transformation. There is no such thing as > proper time measured in seconds! Look, indeed. :-) I'm guessing that Daryl hasn't responded yet because he hasn't picked himself up from the floor. And he's too polite to say ROFL himself. I can't resist filling the vacuum, so, at risk of a spate of invective from you know who, here goes: Ken has no clue. Oh sure, he can throw tensor notation and associated buzzwords around like Zeus throwing thunderbolts, and that's an admirable talent of sorts, one I can't hold a candle to. But what's the good of all that if he can't grasp a simple statement about s, but instead, comes up with a gem like the above? [I snip the rest because I am so fluzzy, I don't even own a copy of Weinberg. Yeah, yeah, I know.] Please see my post to Daryl it covers the snip. [...] (enjoyed the ha-ha's:-). > [I snip the rest because I am so fluzzy, I don't even > own a copy of Weinberg. Yeah, yeah, I know.] If you would like to recommend a secure on-line on GR, where the tidal forces are described, we could analyze that with a common ref. Some of those equations are tough to display in ascii. Ken PS: Oh-no another fluzzyist, then comes fluzzyism and then of course, anti-fluzzyism. Ken S. Tucker says... >> s is proper time. If an ideal clock moves from point A to point B, >> then s along the path will be the elapsed time (the time at B [quoted text clipped - 5 lines] >to have different scalar quantities because of >a transformation. No, that's not what I'm doing. I thought that was what *you* were doing when you said that s is either 0 or 1. >There is no such thing as proper time measured in seconds! I don't know what in the world you are talking about. Of *course* proper time can be measured in seconds! As I said, proper time is the time that shows on an ideal clock. You can just look at your watch to find out the proper time for your particular path through spacetime. At the beginning of your "trip", you note the time on your wristwatch, say 12:00 am. At that point, s=0 seconds. Later, you look at your watch, and it says 12:30. That means that at that point, s = 1800 seconds. Even later, you look at your watch and it says 1:15. That means that at that point, s = 4500 seconds. s changes continuously from 0 to however long your trip lasts. Saying s = 0 or 1 makes as much sense as saying that you have a watch that only shows two possible times: 12:00 and 1:00. That's not a very good watch. -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > [quoted text clipped - 15 lines] > I don't know what in the world you are talking about. Of *course* > proper time can be measured in seconds! Seconds is as arbituary as minutes or hours or meters or cubits, all those are *relative* to a specific CS and are *components*, but "s" is a scalar with no units. It's the same number in all CS's. > As I said, proper time > is the time that shows on an ideal clock. You can just look at [quoted text clipped - 6 lines] > s = 4500 seconds. s changes continuously from 0 to however long > your trip lasts. Good example let's use it... What you're actually doing is recording the time in seconds between events in your CS by using components "x^0", such that x^0 = 1800 seconds. I happen to use minutes, so the time between events is different from yours, (I'm sitting in your back- seat on the trip), I read x'^0 = 30 minutes on my ideal clock. Add a bit more misery, when we get home, your brother who stayed home(twin paradox) figures the the trip took 2:00 hours to your 4500 seconds. Now you need to define "s" (say the trip duration from home to home which defines two events) to be the same for all CS's. > Saying s = 0 or 1 makes as much sense as saying that you have > a watch that only shows two possible times: 12:00 and 1:00. That's > not a very good watch. Ha, study above. From the PoV of events, the invariant spacetime difference when s=0 happens when the events occurs at same place and time otherwise s=1, or any arbituary non-zero scalar. It's really seems simple to me if you're careful. Perhaps review the meaning of the Kronecker Delta and scalars in relativity. Regards Ken S. Tucker Ken S. Tucker says... >> I don't know what in the world you are talking about. Of *course* >> proper time can be measured in seconds! [quoted text clipped - 4 lines] >a scalar with no units. It's the same number in >all CS's. No, that's not correct. The only number that is the same in all coordinate systems is a *dimensionless* number. s is not dimensionless, it has the dimensions of seconds (or hours, or whatever units of time you are using). >What you're actually doing is recording the time in >seconds between events in your CS by using components [quoted text clipped - 8 lines] >brother who stayed home(twin paradox) figures the >the trip took 2:00 hours to your 4500 seconds. >Now you need to define "s" (say the trip duration >from home to home which defines two events) to be >the same for all CS's. Sure. It's this: s = Integral of square-root(g_ij dx^i dx^j) where dx^i is the 4-vector displacement of the clock in question. The answer for the particular trip will be 4500 seconds. >It's really seems simple to me if you're careful. The concept of proper time is simple enough, but what you said about it is completely wrong. -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > [quoted text clipped - 10 lines] > the same in all coordinate systems is a *dimensionless* > number. Well then how would define that? >s is not dimensionless, it has the dimensions > of seconds (or hours, or whatever units of time you are [quoted text clipped - 24 lines] > in question. The answer for the particular trip will be > 4500 seconds. I'm sorry, your stay at home brother obtained 7200 seconds for the events interval...you leaving and returning, which I asked you to account for, you refused to do that. > >It's really seems simple to me if you're careful. Ken Ken S. Tucker says... >> Sure. It's this: >> [quoted text clipped - 6 lines] >I'm sorry, your stay at home brother obtained 7200 seconds >for the events interval Then he made a mistake. His answer should be 4500, also. That's what it means to be an *invariant*. From my point of view, for me to calculate the proper time for my path is simple: it's just the elapsed time on my clock, which is 4500 seconds. So I compute s = 4500. From the point of view of my stay-at-home brother, things are more complicated. There are three important events to take into account: Event A: I leave. Event B: I turn around. Event C: I return. In my brother's coordinate system, I'm travelling at .7806c (to make gamma = 7200/4500 = 1.6). The coordinates of A are: t_A = 0 x_A = 0 Event B takes place at coordinates t_B = 3600 seconds x_B = 3600 * .7806 = 2810 light-seconds Event C takes place at coordinates t_C = 7200 seconds x_C = 0 To compute s for the entire trip, my brother breaks the trip up into two constant-velocity segments: segment 1: Start = (0,0), End = (3600,2810) delta-x = 2810 delta-t = 3600 ds = square-root(delta-t^2 - 1/c^2 delta-x^2) = square-root(12960000 - 7896100) = 2250 seconds segment 2: Start = (3600,2810), End = (7200,0) delta-x = -2810 delta-t = 3600 ds = square-root(delta-t^2 - 1/c^2 delta-x^2) = 2250 seconds s (entire trip) = 2250 + 2250 = 4500 seconds So my stay-at-home brother calculates the same value for s. Proper time along a given path is an invariant, but it isn't constant, and it isn't dimensionless. -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > [quoted text clipped - 11 lines] > Then he made a mistake. His answer should be 4500, also. That's > what it means to be an *invariant*. No, KISS, he's measured the TIME between two events you leaving and you return. He gets x'^0 = 7200. > From my point of view, for me to calculate the proper time for my > path is simple: it's just the elapsed time on my clock, which is 4500 [quoted text clipped - 14 lines] > t_B = 3600 seconds > x_B = 3600 * .7806 = 2810 light-seconds > Event C takes place at coordinates > [quoted text clipped - 21 lines] > So my stay-at-home brother calculates the same value for > s. That approach to the problem will not be a solution because it's ambiguous where the basis vectors transform and thus the components are ill defined. We need to agree on things far more basic. Let S be an arbitrary vector. I'll use capitals to designate a vector. In an arbitrary CS the basis vectors are, E^u or E_u then S = E_u x^u and S.E^u = x^u (dot product). The Kronecker delta = E_u . E^v = 0 or 1, without units, you all know that I hope! In Tucker's *specialized* CS, he uses, E_0 = 1/(1 minute) and E^0 =1 minute in accord with Kronecker, and x^0 = number of minutes (between events). where g_00 = E_0.E_0 = 1/minutes^2 . Then the invariant s^2 = S.S is, s^2 = g_00 x^0 x^0 = unitless scalar, (when Tucker is at rest, as usual :-). > Proper time along a given path is an invariant, but it > isn't constant, and it isn't dimensionless. Let's stop calling it "proper time" which is an SR hang-over and embrace GR and call it an invariant or scalar that has no units. Suppose Daryl uses seconds and he and I are at rest then the transformation to Tucker time is, x'^0 = (&x'^0/&x^0)*x^0 . Using x'^0 = 60*x^0 yields x'^0 = 60*x^0 because 60 seconds = 1 minute. Whatever you do, keep an eye on that Kronecker I prefaced about invariants. Regards Ken S. Tucker Ken S. Tucker says... >> Ken S. Tucker says... >> [quoted text clipped - 14 lines] >No, KISS, he's measured the TIME between two events >you leaving and you return. He gets x'^0 = 7200. But that's not the definition of s. s is defined to be s = Integral of square-root(g_ij dx^i dx^j) which is an invariant, the same value in every coordinates system. >> From my point of view, for me to calculate the proper time for my >> path is simple: it's just the elapsed time on my clock, which is 4500 >> seconds. So I compute s = 4500. [Computation deleted] >> So my stay-at-home brother calculates the same value for >> s. > >That approach to the problem will not be a solution >because it's ambiguous where the basis vectors >transform and thus the components are ill defined. What in the world are you talking about? >We need to agree on things far more basic. > [quoted text clipped - 10 lines] > >E_0 = 1/(1 minute) and E^0 =1 minute That's not the way things are usually done. The *basis* vectors don't carry any units. It's the coefficients x^u that carry the units. It's a little strange to give units to the basis vectors, because the same basis vectors must be used for quantities with different units. For example, 4-velocity U is dimensionless (in units with c=1), but if we write U = E_j U^j with your choice of units, it will turn out that U^j has dimensions of minutes, which is very strange. >in accord with Kronecker, and > [quoted text clipped - 5 lines] > >s^2 = g_00 x^0 x^0 = unitless scalar, You are saying that x^0 has units minutes. That means that the vector S = E_j x^j is dimensionless. In that case, of course s^2 is dimensionless. In general, if S has units d, then s^2 has units d^2. So for instance, if S is the momentum 4-vector, then s^2 will have units of kg^2 (in units where c=1, and where kg is the unit for mass). If S is the displacement 4-vector, then s^2 will have units of meter^2. >(when Tucker is at rest, as usual :-). > [quoted text clipped - 4 lines] >an SR hang-over and embrace GR and call it an >invariant or scalar that has no units. But it is proper time, and it *does* have units. Where did you get the idea that scalars in GR are unitless? That's not true. -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > >> Ken S. Tucker says... [quoted text clipped - 17 lines] > > But that's not the definition of s. That should have been x'^0 = 7200 seconds. > s is defined to be > > s = Integral of square-root(g_ij dx^i dx^j) > > which is an invariant, the same value in every coordinates > system. True, but to start, for simplicity let's approximate to constant g_uv over a small section, then consider the case where g_uv vary. > >> From my point of view, for me to calculate the proper time for my > >> path is simple: it's just the elapsed time on my clock, which is 4500 [quoted text clipped - 10 lines] > > What in the world are you talking about? Did you read my post prior to typing that? I thought I included an example. > >We need to agree on things far more basic. > > [quoted text clipped - 13 lines] > That's not the way things are usually done. The > *basis* vectors don't carry any units. What!? Draw a line on a blank piece of paper, Trace it onto 1 cm graph paper or one inch graph paper. All 3 of those lines are the same, and so that line is invariant. The *basis* vectors on the graph paper have direction AND magnitude, the magnitude being the unit, cm or inch. > It's the > coefficients x^u that carry the units. Yes, that too. > It's a little strange to give units to the basis > vectors, because the same basis vectors must be [quoted text clipped - 6 lines] > with your choice of units, it will turn out that U^j > has dimensions of minutes, which is very strange. No, U^j = dx^j/ds Both sides have the same unit as they should. > >in accord with Kronecker, and > > [quoted text clipped - 9 lines] > means that the vector S = E_j x^j is dimensionless. > In that case, of course s^2 is dimensionless. Yup. > In > general, if S has units d, then s^2 has units d^2. No S can't have units without a specialized CS, as I explained above using a line on a blank paper. > So for instance, if S is the momentum 4-vector, then > s^2 will have units of kg^2 (in units where c=1, and > where kg is the unit for mass). > If S is the displacement 4-vector, then s^2 will > have units of meter^2. Here's where you'll get into trouble. A invariant is a tensor of rank 0. But the units like meter^2 are relative and are not invariant, it's that simple. > >(when Tucker is at rest, as usual :-). > > [quoted text clipped - 8 lines] > Where did you get the idea that scalars in GR are > unitless? That's not true. For the 3rd time learn to use the Kronecker Delta prior to using the metric tensor, for example solve, ds^2 = Delta{^u_v} dx_u dx^v = dx_u dx^u and explain the units you use. Regards Ken S. Tucker Ken S. Tucker says... >> That's not the way things are usually done. The >> *basis* vectors don't carry any units. [quoted text clipped - 7 lines] >direction AND magnitude, the magnitude being >the unit, cm or inch. That's not the way it's usually done. As I said, the basis vectors are usually unitless. >> In general, if S has units d, then s^2 has units d^2. > >No S can't have units without a specialized CS, as >I explained above using a line on a blank paper. Where did you get the idea that invariants must be unitless? >> So for instance, if S is the momentum 4-vector, then >> s^2 will have units of kg^2 (in units where c=1, and [quoted text clipped - 5 lines] >is a tensor of rank 0. But the units like meter^2 >are relative and are not invariant, it's that simple. No, that's not correct. A scalar can have any units whatsoever. The fact that it is invariant has *nothing* to do with what its units are. As I said, rest mass is the invariant associated with the 4-momentum. Rest mass has dimensions of mass. The quantity E^2 - B^2 (where E = the electric field, and B = the magnetic field) is an invariant. It has dimensions "energy per unit volume". >> But it is proper time, and it *does* have units. >> Where did you get the idea that scalars in GR are [quoted text clipped - 6 lines] > >and explain the units you use. The Kronecker Delta has nothing to do with it. The Kronecker Delta is unitless, but that doesn't imply anything about whether ds^2 is unitless. It implies that the units for ds^2 is the product of the units for dx_u and the units for dx^u. Typically, both dx_u and dx^u have dimensions of length (or time, depending on whether you include a factor of c or not) so ds^2 has dimensions length^2. Try actually *calculating* ds^2 for some simple example. In flat spacetime, it becomes (in Minkowsky coordinates) ds^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2) which clearly has dimensions length^2. In Schwarzschild geometry, it has this form: ds^2 = (1 - r_s/r) c^2 dt^2 - 1/(1 - r_s/r) dr^2 + angular part where r_s = the Schwarzschild radius. Again, it's clear that ds^2 has dimensions of length^2. Where did you get the idea that scalars in GR are dimensionless? That's not true. Rest mass is a scalar, and it has dimensions "mass". Proper time is a scalar, and it has dimensions "time". Proper length is a scalar, and it has dimensions "length". Total charge is a scalar, and it has dimensions "charge". -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > [quoted text clipped - 12 lines] > That's not the way it's usually done. As I said, > the basis vectors are usually unitless. Ah there is little I can add. If you decide cm = inches, I'm leaving, in a hurry! Ken > >> In general, if S has units d, then s^2 has units d^2. > > [quoted text clipped - 68 lines] > Daryl McCullough > Ithaca, NY Ken S. Tucker says... >> That's not the way it's usually done. As I said, >> the basis vectors are usually unitless. > >Ah there is little I can add. If you decide >cm = inches, I'm leaving, in a hurry! That response doesn't make any sense. I didn't say cm = inches. I said that basis vectors are usually unitless. That means that they are not measured in cm, and they are not measured in inches. -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > [quoted text clipped - 6 lines] > That response doesn't make any sense. I didn't > say cm = inches. Using graph paper I created both a cm and inch basis with unit vectors that have "unit magnitude". What part of the word "unit" is unclear? How do you define "unit" vector? > I said that basis vectors are usually unitless. What does "usually" mean? >That means that they are not > measured in cm, and they are not measured in > inches. Then tell us what you measure with, i.e what's that unit magnitude all them kooky mathematicians talk about? Ken Ken S. Tucker says... >Using graph paper I created both a cm and inch basis >with unit vectors that have "unit magnitude". A general vector can be written as a linear combination of basis vectors as follows: V = V^i e_i. In the case of a vector of length 1 inch that points in the x-direction, I would say that V^1 = 1 inch The *coefficients* V^j carry the units, not the basis vectors. Putting the units into the basis vectors is a very strange thing to do. The same basis vectors are used for displacement vectors (which have dimensions of length) and momentum vectors (which have dimensions of mass * length / time) and current (which has dimensions of charge/(length^2 * time)). It seems to me that it makes a lot more sense to let the basis vectors be dimensionless, and to put the dimensions into the coefficents. Anyway, getting back to the subject: You claimed that an invariant must be dimensionless. That is completely wrong. Here are some examples of invariants: Proper time : has dimensions of "time" Rest mass : (defined to be square-root(E^2/c^4 - p^2/c^2)) has dimensions of "mass" Proper length : (the proper length of an object is its length as measured in a frame in which the object is at rest) has dimensions of "length" Total charge : has dimensions of "charge" Electromagnetic energy density (defined by F^ij F_ij or E^2 - B^2) : has dimensions of "energy per unit volume" *All* these examples have dimensions. It's actually hard to think of an example from physics of an invariant that *doesn't* have units. Being a scalar has *nothing* to do with being dimensionless. -- Daryl McCullough Ithaca, NY > Ken S. Tucker says... > [quoted text clipped - 11 lines] > Putting the units into the basis vectors is a very strange > thing to do. Quote Daryl... "The same basis vectors are used for displacement vectors (which have dimensions of length)" > and momentum vectors > (which have dimensions of mass * length / time) and current > (which has dimensions of charge/(length^2 * time)). It seems > to me that it makes a lot more sense Quote Daryl again, "more sense to let the basis vectors be dimensionless," I'd like to bail, and let's take a break. Ken Ken S. Tucker says... >I'd like to bail, and let's take a break. Okay, but the following examples will be waiting for you: Invariant = proper time Dimensions = time Invariant = proper length Dimensions = length Invariant = total charge Dimensions = charge Invariant = E&M energy density Dimensions = energy per unit volume The claim that an invariant must be dimensionless is just incorrect. The dimensions of an invariant can be anything. -- Daryl McCullough Ithaca, NY > The claim that an invariant must be dimensionless is just > incorrect. The dimensions of an invariant can be anything. You guys are arguing over a PUN on "invariant" Once you select a unit for length, and 4 linearly-independent unit vectors you have determined all components of the metric tensor for the coordinate system that uses them as coordinate axes. From this point on, things like proper time, proper length, the metric tensor, etc. are all invariant UNDER COORDINATE TRANSFORMATIONS. But none of them are invariant UNDER A CHANGE IN UNITS. Daryl has been discussing invariance under coordinate transforms, and Ken has been discussing changes of units. Tom Roberts tjroberts@lucent.com > > The claim that an invariant must be dimensionless is just > > incorrect. The dimensions of an invariant can be anything. > > You guys are arguing over a PUN on "invariant" No. It's NOT a PUN, it's important. (Tom read your post). Let's discuss the simple 3D cartesian {i,j,k} unit vectors we all know and love, I hope. When a CS is selected, the magnitude of the unit of those vectors is calibrated by physics, cm, foot, cubit, etc, as the definition of length evolved. However none of those definitions affect the distance between two points. We can fart to eternity about units but in fact two physically distinct stationary points remain invariant, like 100 million years ago a Bronto- saurus said I pooped from "a" to "b" and that is invariant, i.e. 1 = a Bronto poop, because Bronto pooped once. Then comes Daryl using meters, and claims it's a 4 meter poop, and then Tom using inches it's a 160 inch poop. ((bilge would say it took 5 minutes for Bronto to poop, given bilges interest in things of that nature)) Tucker says it's one poop. Tucker's right, the Bronto pooped once and that's an invariant, poop=1. You guys can divide up the Bronto poop anyway you want, into meters, pounds, seconds, hours, cubits, but, it's only ONE BRONTO POOP! ((Knowing Carlip, he's likely to note the poop is steaming and hence it's mass is NOT invariant, and blow my argument)) Regards Ken PS: This is not shitty argument. > Once you select a unit for length, and 4 linearly-independent unit > vectors you have determined all components of the metric tensor for the [quoted text clipped - 8 lines] > > Tom Roberts tjroberts@lucent.com Ken S. Tucker says... >Let's discuss the simple 3D cartesian {i,j,k} >unit vectors we all know and love, I hope. [quoted text clipped - 3 lines] >foot, cubit, etc, as the definition of length >evolved. I wouldn't characterize it that way at all. If you set up a coordinate system in which the i-direction points from Buffalo, NY to New York City, then the displacement vector V from Buffalo to New York can be characterized variously as 1: (289 miles) e_i 2: (570 kilometers) e_i 3: (1.53 million feet) e_i 4: (18.31 million inches) e_i The *coefficient* V^i is equal to 289 miles in one system of units, and 570 kilometers in a different system of units. But e_i doesn't change. I suppose that if you are only interested in *displacement* vectors, then you could just as well put the units onto e_i, and consider the coefficient V^i to be dimensionless: 1: V^i = 289, e_i is a vector with magnitude 1 mile 2: V^i = 570, e_i is a vector with magnitude 1 kilometer 3: V^i = 1.53 million, e_i is a vector with magnitude 1 foot 4: V^i = 18.31, e_i is a vector with magnitude 1 inch However, that convention only works for *displacement* vectors. Would you have a different unit vector for momentum, and yet another unit vector for angular momentum? Letting the coefficient carry the units is much simpler, since you can use the *same* unit vector e_i for displacement, momentum, angular momentum, electric field, etc. The units would then be in the coefficients V^i. -- Daryl McCullough Ithaca, NY > When a CS is selected, the magnitude of the unit > of those vectors is calibrated by physics, cm, > foot, cubit, etc, as the definition of length > evolved. However none of those definitions affect > the distance between two points. Physics is a quantitative science. When we |
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