Stephen A. Lawrence wrote: Here's another cute example: A spherical chamber cut out of a uniformly dense planet which was _offset_ from the center would have a _uniform_ (but non-zero) G-field inside it.
Horace Heffner wrote: It should have a g field due to the sphere having the radius from the hole to the center. Stephen A. Lawrence wrote: If you work it out, it's a completely uniform field. Very strange. (Easiest way to analyze it is to pretend the chamber is a separate sphere of "negative mass" and just sum its "negative" field with the field of an intact planet.) Horace Heffner wrote: I think any object held in that chamber would experience a gravitational red shift proportional to the g at its location, not to the gravitational potential. Stephen A. Lawrence wrote: I don't know if I understand you. Light should be redshifted as it crosses the chamber, in proportion to the intensity of the field in the chamber, right? Horace Heffner wrote: Since the amount of red shift is a function of g, the change in red shift is a function of the change in g as movement occurs. Stephen A. Lawrence wrote: If that's what you're saying, I agree. Horace Heffner wrote: I'm not sure we even agree on the g field in the bubble being uniform. As you move across the bubble you become "outside" a larger and larger sphere of material. Stephen A. Lawrence wrote: Note again that since the field strength is the gradient of the potential, that's equivalent to saying the degree of redshift varies with the potential. Horace Heffner wrote: I'm not sure we are using the same terminology. I'm looking at red shift as the difference between how a photon, or the energy of an atomically emitted photon, is observed in a zero gravity situation vs in the presence of a gravitational field. Horace Heffner wrote: If what you were saying were true then objects in the center of the universe (assuming here a big bang) should all be massively red shifted, instead of vice versa. Stephen A. Lawrence wrote: Only if there's a gravitational field filling the universe, pointing to the center. That's the only way you'll get a lower gravitional potential at the center of the universe. And if there is such a field, then there must be a redshift associated with it, too. Horace Heffner wrote: No matter how you cut it, clock rate is a function of gravitational field. If the effects of the gravitational field differ from the effects of acceleration (this difference at any point) then Einstein's fundamental assumption for GR is violated and GR disappears in a flash! 8^) I also have to question the validity of the tangential straight rod approach you use. I could be missing something, but it doesn't seem to account for how we would see the clock advance as it passes behind the earth in the opposite direction. Stephen A. Lawrence wrote: You can't synchronize all the clocks on a rotating disk. Horace Heffner wrote: I didn't mention synchronization. Stephen A. Lawrence wrote: You can't synchronize all the clocks on the Equator. If you try, you find there is a "date line" where two adjacent clocks are out of sync. It's crossing the "date line" which causes the hiccup. Horace Heffner wrote: Here again you are talking about how things appear in motion. I just want to figure out in an intuitive way what accounts for differences when clocks are brought back together. Stephen A. Lawrence wrote: Hmmm ... Consider again the laser-ring gyro. What causes the fringe shift when you rotate it? Horace Heffner wrote: A change in distance travelled. Stephen A. Lawrence wrote: Signal velocity relative to the rim of the disk can be measured and is constant. Horace Heffner wrote: If there is no velocity effect which does this, then what remains except acceleration? Retardation is out of the picture. Now that I can see some real data it would be good to look at the effects of gravimagnetism, because these should modify the expected values. It is probably going to take an FEA program to do this right, and I just do not have the time right now. What I *can* see from the airplane data is it can not be fully analysed using only the earth's gravimagnetic field. It requires quantifying the solar and/or galactic gravimagnetic field. Stephen A. Lawrence wrote: But would such effects not be swamped by the local influence of the Earth? Horace Heffner wrote: More to come on that. On Jan 22, 2006, at 5:05 AM, Stephen A. Lawrence wrote: Acceleration doesn't affect clocks. That's been verified (can't cite references, sorry). A clock in a centrifuge slows only as a result of the speed at which it's traveling, not as a result of the centripetal force. Horace Heffner wrote: This can not be consistent with relativity, Stephen A. Lawrence wrote: But it is. It's built into GR from the get-go. Horace Heffner wrote: I thought Einstein's equivalence principle was built into relativity. On Jan 23, 2006, at 5:12 AM, Stephen A. Lawrence wrote: The equivalence principle is built in. So is the principle of relativity, and, as a consequence of the assumption that you can change to any arbitrary coordinate system without affecting the results, the lack of any local effect due to acceleration is built in, too. [snip enormous amounts, after reading -- thanks for the additional explanation of the retardation comments] Stephen A. Lawrence wrote: The rate at which a clock is observed to tick does not depend on whether the clock is _currently_ undergoing acceleration. That has been both predicted and observed to be true, to the limits of the experiments which have been done. Horace Heffner wrote: Then you have conclusively proved GR is based upon a false assumption. Stephen A. Lawrence wrote: No, I haven't, because, as stated elsewhere, clocks in GR are apparently affected by gravitational _potential_ but not by the local intensity of the gravitational _field_. When you accelerate, in SR, you find that distant clocks are apparently affected by _your_ acceleration. _THAT_ is equivalent to the GR clocks being affected by the gravitational potential. The effects are identical. You cannot separate the observations from the observer, and the concept of observable properties of external things being affected by changes within yourself (such as your acceleration) is a consequence of that. Horace Heffner wrote: Agreed - but only if you agree that clocks involving mass actually change due to velocity alone. In other words, if m = m0*gamma is purely due to appearances, i.e. due to retardation, then the only effect left to cause a time difference upon rejoining the clocks is acceleration. I don't think it is generally accepted an more that m=m0*gamma is a real effect. I definitely read that in some text. Stephen A. Lawrence wrote: Whether an effect is "real" or not is so slippery that I don't think there's any definitive answer. IMHO the Sagnac effect proves that time dilation is "real". In the opinion of lots of other people it does not. If you spin a rigid disk it will crack due to Fitzgerald contraction. I think that proves the contraction is "real". Many other people think it does not. The trouble with m0*gamma is it's the total energy of the object, and that's frame-dependent. Your point of view determines how big it is. But does that mean it isn't "real"? I don't think so -- that's like saying kinetic energy isn't "real" because it's frame dependent. Stephen A. Lawrence wrote: What acceleration _DOES_ do is affect _DISTANT_ clocks. When YOU accelerate, clocks that are far, far away and toward which you are accelerating seem to you to run _faster_. Horace Heffner wrote: Irrelevant. Who cares how the clocks appear during the journey. I only care what happens when they come back together in the same location in the same reference frame. Then all retardation effects have cancelled because there is no retardation remaining. Stephen A. Lawrence wrote: Retardation explains what you see as you watch the other party, right? By using a powerful telescope, you can actually watch the other party's clock throughout the whole trip. By looking at the _size_ of the image, and the rate at which it's changing, you can see how fast the other party is moving (relative to you) and how far away the other party is. That picture-show which you can watch _MUST_ agree with the physical effects observed when you get home again and put the clocks next to each other. If you can explain how that happens you're probably as close to "understanding" this as you can get. The weird thing is that all the "effects of retardation" do _not_ cancel when you get home, and it's very hard to draw a line between what was "real" and what was an illusion. The weirdness you apparently see comes directly from the (assumed) fact that the light signal travels at C relative to _both_ the stationary and the moving parties. * * * In the "stationary" frame you can explain it all by using time dilation -- you can, with the help of extra (stationary) observers spaced out along the route, actually observe the traveler's clock running _slow_. (Just assume time dilation is real and you're done!) In the "moving" frame you've got a much bigger problem; just exactly when does the stationary clock run _fast_? The answer is: while you are accelerating. Stephen A. Lawrence wrote: If the traveler accelerates in a blazing flash lasting a few microseconds, and then IMMEDIATELY decelerates again, he'll experience negligible time skew. On the other hand, if he accelerates, coasts a long time, and then decelerates, he'll experience a lot of time skew. Horace Heffner wrote: OK, but then this implies the clock is mass related, and m=m0*gamma is a real, not a retardation effect. Stephen A. Lawrence wrote: Relativistic Doppler shift includes a term for gamma. The emitter's motion changes the apparent frequency, _and_ the emitter's different time base changes the apparent frequency, and the two effects must be combined to obtain the total observed effect. The "extra" mass of a moving body is (gama-1)*m0. At relativistic speeds that's where most of the body's energy is. If that's not "real", then most of the energy isn't "real", either. Stephen A. Lawrence wrote: The gravitational time dilation is due to the gravitational potential, _not_ the local acceleration of the field. Horace Heffner wrote: I think this is not the only possible explanation. An alternative explanation is the red shift is due to the effect of gravity on the photon. Gravitons exchange momentum with photons, but not virtual photons. If this were not true black holes would not exist. In the case of a spherical shell object with a hole in it, I think the red shift would occur at the surface as light goes through the hole. Stephen A. Lawrence wrote: That would make sense. The redshift "happens" in regions where the field is non-zero, of course, which is exactly where the photons would be interacting with it. If the field is the gradient of the potential, then the places where the field is strongest are also the places where the potential, and degree of redshift, are changing most rapidly. But I'm just talking about where and when you would "observe" a red- shift. The "observed" redshift is a function of the gravitational potential, but that's not the same as saying it's "caused" by it -- I should be more careful about how I say things... Horace Heffner wrote: It sounds like you are attributing a "real" effect to static gravitational potential that should be matched by an equivalent "real" effect from a static electromagnetic potential A. No such effect exists to my knowledge. AFAIK, The only effects that manifest as real are the result of changes in A, i.e in @a/@t. Stephen A. Lawrence wrote: It seems that way but it's not. In relativity, the E field and the G "field" produce totally different kinds of "forces". Let's see if I can dredge this out of my memory.... The E field contains a heat-like component (heat is a force, too -- a candle increases the momentum of an object placed above it, so dP/ dt is nonzero in that case). Gravity is not a heat-like force, and I'm failing completely to recall just what difference that makes in this case. More mundanely, charge is conserved; you drop a charged rock down the hole, at the bottom of the hole it still has the same charge as it had at the top of the hole. If you turn it into a beam of light you need to figure out what to do with the charge -- you can't just throw it away. Gravitational mass is apparently _not_ conserved, not the same way; in particular, when you drop the rock down the hole, it gains gravitational mass. Oh well I'm just babbling at this point I should drop this line of reasoning until and unless I look it up again... Stephen A. Lawrence wrote: Here's another cute example: A spherical chamber cut out of a uniformly dense planet which was _offset_ from the center would have a _uniform_ (but non-zero) G-field inside it. It should have a g field due to the sphere having the radius from the hole to the center. Stephen A. Lawrence wrote: If you work it out, it's a completely uniform field. Very strange. (Easiest way to analyze it is to pretend the chamber is a separate sphere of "negative mass" and just sum its "negative" field with the field of an intact planet.) Horace Heffner wrote: I think any object held in that chamber would experience a gravitational red shift proportional to the g at its location, not to the gravitational potential. Stephen A. Lawrence wrote: I don't know if I understand you. Light should be redshifted as it crosses the chamber, in proportion to the intensity of the field in the chamber, right? If that's what you're saying, I agree. Note again that since the field strength is the gradient of the potential, that's equivalent to saying the degree of redshift varies with the potential. Horace Heffner wrote: If what you were saying were true then objects in the center of the universe (assuming here a big bang) should all be massively red shifted, instead of vice versa. Stephen A. Lawrence wrote: Only if there's a gravitational field filling the universe, pointing to the center. That's the only way you'll get a lower gravitional potential at the center of the universe. And if there is such a field, then there must be a redshift associated with it, too. Horace Heffner wrote: No matter how you cut it, clock rate is a function of gravitational field. If the effects of the gravitational field differ from the effects of acceleration (this difference at any point) then Einstein's fundamental assumption for GR is violated and GR disappears in a flash! 8^) I also have to question the validity of the tangential straight rod approach you use. I could be missing something, but it doesn't seem to account for how we would see the clock advance as it passes behind the earth in the opposite direction. Stephen A. Lawrence wrote: You can't synchronize all the clocks on a rotating disk. Horace Heffner wrote: I didn't mention synchronization. Stephen A. Lawrence wrote: You can't synchronize all the clocks on the Equator. If you try, you find there is a "date line" where two adjacent clocks are out of sync. It's crossing the "date line" which causes the hiccup. Horace Heffner wrote: Here again you are talking about how things appear in motion. I just want to figure out in an intuitive way what accounts for differences when clocks are brought back together. Stephen A. Lawrence wrote: Hmmm ... Consider again the laser-ring gyro. What causes the fringe shift when you rotate it? Signal velocity relative to the rim of the disk can be measured and is constant. Horace Heffner wrote: If there is no velocity effect which does this, then what remains except acceleration? Retardation is out of the picture. Now that I can see some real data it would be good to look at the effects of gravimagnetism, because these should modify the expected values. It is probably going to take an FEA program to do this right, and I just do not have the time right now. What I *can* see from the airplane data is it can not be fully analysed using only the earth's gravimagnetic field. It requires quantifying the solar and/or galactic gravimagnetic field. Stephen A. Lawrence wrote: But would such effects not be swamped by the local influence of the Earth? Horace Heffner wrote: Gravimagnetics, given the presence of the significant ambient gravimagentic field, should enhance the time difference on the *airplane* clocks. In other words it agrees qualitatively with the time differences, but may be too small to make any difference. It does mean the two east-west opposed orbit satellites would have their orbital parameters affected, thus their velocities, and thus their clocks. It is interesting that polar satellites should veer left going over the North pole and right going over the South pole, from the point of view of a person in the satellite oriented feet down and facing the direction of motion. Satellites going west-to-east should experience a lower g value than those going east-to-west, and the higher the velocity the lower the g value. This means the g value at the surface should be, due to gravimagnetism, slightly less at the equator than at the pole, and should cycle in value over a 24 hour period, due to the earth's axis not aligning with the ambient gravimagnetic field. Stephen A. Lawrence wrote: Hmmm ... Interesting. On Jan 23, 2006, at 9:56 AM, Stephen A. Lawrence wrote: Time dilation is clearly real, then. I send a clock out to Pluto and back via a fast rocket, and check its time, and now it is slow. I do it again, and it's slower. Horace Heffner wrote: Well the time difference is the result of a real effect. hat effect is not necessarily time dilation, because time dilation (observed time differences) is at least in part due to retardation. Stephen A. Lawrence wrote: Mass increase -- m == m0*gamma -- seems real too, though you might disagree. Horace Heffner wrote: Yes, it seems that way to me. This effect can explain at least a portion of the "real" nature of the final time difference for the twins - depending on what kind of clock is used. If mass is used in the clock, then certainly real time differences can be expected. I am not sure there will be a real difference if the clocks used consist only of photons bouncing between mirrors. Which then brings up the "reality" of the Fitzgerald contraction. Is length contraction real? Certainly some of it is not real, because it is due at least in part to retardation. Stephen A. Lawrence wrote: I accelerate a clock to gamma=10, and let it collide with a clock which is "stationary". The energy given up by the traveling clock is consistent with its mass being m0*gamma; it makes a very real "bang", which involves locally observed forces that are far larger than those we would have observed had its mass been merely m0, at the speed at which it was traveling. I put a centrifuge into a (closed!!) box, and start it going. As it spins up I weigh it. It gets heavier, which again involves local measurement of a force. Once again, m == m0*gamma seems to me to be quite "real". Length contraction is far more dubious. As far as I know there is no way to observe it which doesn't involve making "simultaneous" measurements at separate locations which opens us up to all kinds of problems, though the "cracking spinning disk" experiment still bothers me. Horace Heffner wrote: I wonder if it is possible to make an accurate clock which doesn't depend on mass? Stephen A. Lawrence wrote: Finally, just for fun, I put a resistor into a centrifuge, and spin it up, and measure its resistance using a stationary meter....... WTF?? Horace Heffner wrote: Nothing like trying to iron out experimental artifacts! Stephen A. Lawrence wrote: No, I haven't, because, as stated elsewhere, clocks in GR are apparently affected by gravitational _potential_ but not by the local intensity of the gravitational _field_. Horace Heffner wrote: >From a QM point of view this is utter nonsense. Field potential is merely a calculation device. Stephen A. Lawrence wrote: Absolutely. I agree. It's not a "field" in relativity either, and the "gravitational potential" doesn't behave like a sensible "potential". It's just a convenient way to think of it, and it works pretty well in the low-curvature (Newtonian) limit. A "field" is something that can be represented as a mathematical tensor field and in GR, gravity certainly can't be. Einstein tried hard to make that work before he abandoned it, or so I've been led to believe. Stephen A. Lawrence wrote: [regarding that spherical hole cut in a larger uniform sphere:] I don't know if I understand you. Light should be redshifted as it crosses the chamber, in proportion to the intensity of the field in the chamber, right? Horace Heffner wrote: Since the amount of red shift is a function of g, the change in red shift is a function of the change in g as movement occurs. Stephen A. Lawrence wrote: If that's what you're saying, I agree. Horace Heffner wrote: I'm not sure we even agree on the g field in the bubble being uniform. As you move across the bubble you become "outside" a larger and larger sphere of material. Stephen A. Lawrence wrote: Yes but you're outside a larger and larger "bubble", too. It's a very cute example; I ran across it here: http://www.geocities.com/physics_world/gr/grav_cavity.htm Unfortunately it looks like the article has suffered an editing error (or six!) since the last time I saw it. It's completely illegible, at least in my browser, and the main illustration's gotten roached. Darn. (The page author's been having a rough time of it lately, I'm afraid, and hasn't had a lot of energy to worry about the state of his website.) Luckily I had stashed a copy of the (undamaged) page, to which I just referred to refresh my weak memory of the proof. The way to work this one out is to look at the field from an "intact" sphere, and _subtract_ the field due to the sphere we cut out to make the chamber. By the principle of superposition this is legit in Newtonian gravitation (doesn't quite work in GR, of course). The field at any point inside a uniform sphere of density rho is F = -(4/3)*pi*G*rho*R where "R" is the _radius vector_ from the center of the sphere to the point where we're finding the field. For the big sphere, let the radius vector be R1. For the small (cut-out) sphere let the radius vector be R2. (Note that they point from different origins, but that's OK, all we care about are the direction and length.) Then the net field anywhere inside the small (cut-out) sphere will be F(total) = -(4/3)*pi*G*rho*(R1 - R2) But (R1 - R2) is a _constant_, and is just the vector from the center of the big sphere to the center of the small sphere. So the force is also a constant, proportional to the distance between the spheres' centers, pointing along the line which connects the small sphere's center to the big sphere's center. According to GR the exact field won't be _quite_ uniform, I'm sure.
