CraigD Posted March 15, 2008 Report Posted March 15, 2008 You brought up an interesting point about the center of gravity mathematically causing vector canceling. Although the directional force vectors cancel, the gravity scalar is at a maximum in the center. Does that scalar addition correspond to GR? The reason I ask that is the GR of a planet is often explained, in simple terms, as analogous to a bowling ball making a well in the fabric of space-time. If the vector addition is important than the bottom of that well should contain a peak since the gravity vector addition decreases toward the center. But it is never explained with a peak in the center, but appears to assume scalar addition for only a well.The gravitational time dilation predicted by GR for a specific point depends simply of the gravitational potential of the point. As the graph in preceding linked wikipedia section depicts, the gravitational potential for any body single body – Sol, Earth, etc. – is zero at 2 radii from its center of mass, zero and infinity. Though not a pretty graphic, here’re results of the two gravitational time dilation equations given in the wikipedia article “gravitational time dilation”, from roughtly the orbit of Mercury to the center of Sol.r/rS t0/tf 100 .999999978790432605 99 .999999978576194548 98 .999999978357584286 97 .999999978134466596 96 .99999997790670062 95 .999999977674139572 94 .999999977436630415 93 .999999977194013535 92 .999999976946122375 91 .999999976692783058 90 .999999976433813978 89 .999999976169025367 88 .999999975898218834 87 .999999975621186863 86 .999999975337712289 85 .999999975047567723 84 .999999974750514954 83 .999999974446304287 82 .999999974134673848 81 .999999973815348829 80 .999999973488040686 79 .999999973152446259 78 .999999972808246848 77 .999999972455107191 76 .999999972092674386 75 .999999971720576706 74 .999999971338422332 73 .999999970945797975 72 .999999970542267385 71 .999999970127369737 70 .999999969700617869 69 .999999969261496382 68 .999999968809459557 67 .999999968343929095 66 .99999996786429165 65 .999999967369896128 64 .999999966860050747 63 .999999966334019798 62 .999999965791020108 61 .999999965230217149 60 .999999964650720758 59 .999999964051580421 58 .999999963431780072 57 .999999962790232342 56 .999999962125772193 55 .999999961437149856 54 .999999960723022987 53 .999999959981947934 52 .999999959212369994 51 .999999958412612526 50 .99999995758086476 49 .999999956715168103 48 .999999955813400752 47 .999999954873260321 46 .999999953892244219 45 .9999999528676274 44 .999999951796437087 43 .999999950675423968 42 .999999949501029271 41 .999999948269347027 40 .999999946976080668 39 .999999945616492956 38 .999999944185347994 37 .999999942676843842 36 .999999941084533902 35 .99999993940123482 34 .999999937618918141 33 .999999935728582266 32 .999999933720100396 31 .999999931582039045 30 .999999929301440266 29 .999999926863558806 28 .999999924251542951 27 .999999921446044431 26 .999999918424738324 25 .999999915161727719 24 .999999911626799551 23 .999999907784486311 22 .999999903592871851 21 .999999899002055992 20 .999999893952158524 19 .999999888370692872 18 .999999882169064333 17 .99999987523783239 16 .999999867440196398 15 .999999858602875533 14 .999999848503080163 13 .999999836849469994 12 .999999823253591293 11 .999999807185734406 10 .999999787904305802 9 .999999764338114781 8 .999999734880375224 7 .999999697006137375 6 .999999646507151345 5 .999999575808566619 4 .999999469760680159 3 .999999293014177733 2 .999998939521079163 1 .999997879041033708 .99 .999997921248161006 .98 .999997963031094316 .97 .99999800438983369 .96 .99999804532437918 .95 .99999808583473084 .94 .99999812592088872 .93 .999998165582852873 .92 .999998204820623347 .91 .999998243634200193 .9 .999998282023583461 .89 .999998319988773199 .88 .999998357529769456 .87 .999998394646572278 .86 .999998431339181715 .85 .999998467607597812 .84 .999998503451820615 .83 .999998538871850171 .82 .999998573867686523 .81 .999998608439329717 .8 .999998642586779797 .79 .999998676310036806 .78 .999998709609100787 .77 .999998742483971783 .76 .999998774934649835 .75 .999998806961134984 .74 .999998838563427271 .73 .999998869741526738 .72 .999998900495433422 .71 .999998930825147363 .7 .999998960730668601 .69 .999998990211997173 .68 .999999019269133116 .67 .999999047902076468 .66 .999999076110827264 .65 .999999103895385541 .64 .999999131255751334 .63 .999999158191924679 .62 .999999184703905609 .61 .999999210791694158 .6 .999999236455290359 .59 .999999261694694245 .58 .999999286509905848 .57 .999999310900925199 .56 .99999933486775233 .55 .999999358410387271 .54 .999999381528830052 .53 .999999404223080703 .52 .999999426493139253 .51 .999999448339005728 .5 .999999469760680159 .49 .999999490758162571 .48 .999999511331452991 .47 .999999531480551447 .46 .999999551205457962 .45 .999999570506172563 .44 .999999589382695274 .43 .99999960783502612 .42 .999999625863165122 .41 .999999643467112305 .4 .99999966064686769 .39 .999999677402431301 .38 .999999693733803157 .37 .999999709640983281 .36 .999999725123971691 .35 .999999740182768408 .34 .999999754817373451 .33 .999999769027786839 .32 .999999782814008588 .31 .999999796176038719 .3 .999999809113877246 .29 .999999821627524187 .28 .999999833716979558 .27 .999999845382243373 .26 .999999856623315649 .25 .999999867440196398 .24 .999999877832885635 .23 .999999887801383373 .22 .999999897345689626 .21 .999999906465804403 .2 .999999915161727719 .19 .999999923433459583 .18 .999999931281000007 .17 .999999938704348999 .16 .99999994570350657 .15 .999999952278472728 .14 .999999958429247482 .13 .999999964155830839 .12 .999999969458222808 .11 .999999974336423394 .1 .999999978790432605 .09 .999999982820250445 .08 .999999986425876919 .07 .999999989607312033 .06 .99999999236455579 .05 .999999994697608194 .04 .999999996606469247 .03 .999999998091138953 .02 .999999999151617313 .01 .999999999787904329 0 1Notice that gravitational time dilation peaks at an unspectacular 0.999997879041033708 at Sol’s surface, r/rS=1, and is non-existent at its center, r/rS=0. Also note that these calculations assume constant density and no rotation, so far more approximate than their number of significant digits imply, but are still useful for illustrative purposes. To get a dramatic, or even infinite (t0/tf=0) gravitational time dilation, a body’s radius must be nearly less than its Schwarzschild radius. With its low density, Sol doesn’t come close to this: its mean radius is about 696,000,000 m, its Schwarzschild radius 2,952 m. Other stars don’t do much better, as even super-giant ones more than 100 times as massive as Sol have even lower densities, with radii several hundred times Sol’s, which their Schwarzchild radii increase only proportional to their masses. I just can’t imagine any measurement that could detect the effect of a time dilation factor on the order of 0.9999999 in a Star’s fussion, over the many turbulent and only approximately calculable effect that can effect the rate dramatically. Quote
modest Posted March 16, 2008 Report Posted March 16, 2008 Like an Earth-like planet’s, the gravitational field strength that determines the gravitational time dilation factor actually decreases at some point as it approaches the center of the star, to nearly zero at the center. So the time dilation factor in the core may be less than in a star’s outer layers, and the time dilation in the core of a more massive star may be less than that of a less massive one. A good and succinct, if necessarily ideal description of this can be read in the “outside and inside a non rotating sphere” section of the wikipedia article “gravitational time dilation”. The gravitational time dilation predicted by GR for a specific point depends simply of the gravitational potential of the point. As the graph in preceding linked wikipedia section depicts, the gravitational potential for any body single body – Sol, Earth, etc. – is zero at 2 radii from its center of mass, zero and infinity. I believe the wikipedia article: “outside and inside a non rotating sphere” section of the wikipedia article “gravitational time dilation”If one is inside the sphere, the sphere can be split in two parts: a hollow sphere above and a solid sphere below. One is weightless anywhere in the interior of a uniform hollow sphere. With respect to one's gravitational potential, it is as if the hollow sphere is not there[1][2]. What is left is the solid sphere below, and its mass is:... The implication is that the gravitational time dilation reaches its maximum at the surface of the non-rotating massive spherically-symmetric object, and that the gravitational time dilation reaches its minimum at the center of the sphere. is wrong. They are ignoring the outer shell or hallow sphere part when calculating time dilation which is not correct. It is true that the spacetime in the hallow shpere would be flat or that only the mass beneath you (when you are descending) would contribute toward the curvature or the local force of the gravitational field. However, gravitational time dilation is not a measure of gravitational force. It is rather proportional to gravitational potential phi or it's positive GR counterpart U. For this, the "outer shell" must be considered. The potential in a hollow sphere would be constant - indicating no gravitational force - but it would not be zero. This is a distinction the wiki authors didn't seem to realize. The gravitational potential outside a spherical body is -GM/r and inside the sphere it is [imath]-.5\times GM/R \times [1+r^2/R^2][/imath] where r is the distance to center and R is the radius of the body So, if the gravitational potential is zero at infinity and it's taken to be positive (which is a conversion you have to make for the following equation) then it gets larger as you approach the sphere and continues to get larger inside the sphere approaching the center. Gravitational time dilation is proportional to the potential. Therefore, gravitational time dilation is greatest at the center of a massive object. EDIT: Using U, you can calculate the gravitational time dilation inside and outside the sphere:[math]T = \frac{T_0}{\sqrt{1-U/c^2}}[/math] -modest Quote
modest Posted March 16, 2008 Report Posted March 16, 2008 It occurs to me that the equation in my above post for potential in a ball is either wrong or weird or perhaps both. This website:Gravitational potential due to rigid bodyI just found which has the formula:[math]\phi = -\frac{GM}{2a^3}(3a^2-r^2)[/math]where a is the radius of the mass and r is distance from center. It shows the derivation for this formula and shows the formula for r=0:[math]\phi = -\frac{3GM}{2a}[/math] and says this about it: This may be an unexpected result. The gravitational field strength is zero at the center of a solid sphere, but not the gravitational potential. However, it is entirely possible because gravitational field strength is rate of change in potential, which may be zero as in this case. which is basically what I was saying above. In any case, somebody should change the wiki article. -modest Quote
Erasmus00 Posted March 16, 2008 Report Posted March 16, 2008 I believe the wikipedia article: “outside and inside a non rotating sphere” section of the wikipedia article “gravitational time dilation” is wrong. Then lets solve for it. Without proof, I posit that the spherically symmetric static space-time solution in a vacuum is the Schwarzchild metric, which has a t component given by [math] ds^2 = \left(1-\frac{C}{r}\right)dt^2 +... [/math] Now, C is a constant to be set by boundary conditions. Inside a hollow sphere, one condition is that at [math]r=0[/math] everything should be finite, so C =0. So inside a hollow sphere, everything is minkowski, and there is no gravitational time dilation. Hence, for gravitational time dilation you need only consider the part of a sphere that is below you. So, I'm pretty sure wiki got it right. Edit: On second thought, I take that back. The above argument assumes superposition is valid (inside a sphere= hollowsphere above + solid sphere below), but GR's eqs. are nonlinear. -Will Quote
CraigD Posted March 16, 2008 Report Posted March 16, 2008 I believe the wikipedia article: “outside and inside a non rotating sphere” section of the wikipedia article “gravitational time dilation” is wrong. I believe you’re right. The wikipedia discussion page does, too, but nobody has corrected the main page yet :(Using U, you can calculate the gravitational time dilation inside and outside the sphere:…As I don’t know a formula for the gravitational potential [math]U = \frac{GM}{r}[/math] inside a sphere of uniform or other density, I’ve fallen back to a numeric approximation, representing the sphere as a cloud of 26,252 point masses. I get this data (r/rS is distance in solar radii, t0/tf is time dilation factor, U is gravitational potential, a is acceleration of gravity):r/rS t0/tf U a 100 .999999989395200034 1906223767.813480345 .02738835683753690541 90 .999999988216884665 2118027173.356373618 .03381282283334155303 80 .999999986743988548 2382781772.589150796 .04279441867358078436 70 .999999984850261461 2723181173.04637664 .0558948742980524389 60 .999999982325284976 3177048304.152280841 .0760793931197597338 50 .999999978790302062 3812465131.93427698 .1095549435136705927 40 .999999973487785805 4765597898.447833819 .1711813736583099953 30 .999999964650117036 6354177963.942459629 .3043292452352190608 20 .99999994697404736 9531469655.77949344 .6847843307467345995 10 .999999893936075265 19065099315.48882054 2.740056558554051093 9 .999999882147080632 21184183073.96256291 3.383131871617272929 8 .999999867409051173 23833358802.36715171 4.282379482834933189 7 .999999848456927781 27240022127.71027448 5.594436683759973575 6 .999999823181128431 31783372494.31814896 7.61693168208359074 5 .999999787781443652 38146501260.77327098 10.97353725131199335 4 .999999734648802308 47697146293.75077 17.15950419178133069 3 .999999646000573906 63631752231.64106418 30.54568677801467849 2 .999999468448974921 95546821916.539406 68.79537428188544505 1 .999998952223784709 188338761261.5377773 236.2785547840037095 .9 .999998860377357958 204848233543.4009131 232.8268030348554124 .8 .999998774380242872 220306285770.2400017 209.8060072617887853 .7 .999998698623743462 233923577798.4557373 180.9041294609779052 .6 .999998634623462806 245427679195.0222742 149.3471828252405965 .5 .999998583165326013 254677319714.0637629 116.2546377940258245 .4 .999998544699015786 261591668889.2030174 82.36314408431174265 .3 .99999851937075551 266144443224.168515 48.59539333978102965 .2 .999998506800302837 268403991752.1595836 17.02233713424564801 .1 .999998505013092641 268725244156.0610172 4.77347197305522623 .09 .999998505215889535 268688791258.7991411 5.64477991360693624 .08 .999998505445717968 268647479421.5523449 6.167159494655981265 .07 .999998505688868986 268603772839.607712 6.33303746024827012 .06 .99999850593166612 268560129868.6037794 6.15193642753531008 .05 .999998506161135347 268518882599.0099152 5.65030422326076242 .04 .999998506365607828 268482128513.5120911 4.868435598279108475 .03 .999998506535173245 268451648999.7996966 3.856129723603217361 .02 .999998506661966524 268428857810.7520283 2.668634036372359829 .01 .999998506740317896 268414774091.3511006 1.363715835633614655 0 .999998506766814506 268410011305.4346802 .0000000000000000105 Instead of a time dilation factor (t0/tf) of zero at the center of Sol, I get a slightly greater dilation (by a factor of about 1.0000005) than at the surface. Note also that, due to the granularity and accumulated calculator errors, this above-surface t0/tfs differs slightly from the pervious, more accurate calculations. Still, the gravitational time dilation in Sol, or even the most massive stars, remains very small, far below what I think could be detected by observation. Quote
HydrogenBond Posted March 16, 2008 Report Posted March 16, 2008 If you look at the vector addition of gravity in a planetary sphere, the vectors cancel in the center to give zero gravity, based on Newtonian gravity. One thing nobody talks about is where is the energy going? In other words, if we took apart a planet, each piece of matter could generate gravity. If place them all in a sphere we lose gravity in the center. What this implies is potential energy is being lost. If we assume the conservation of energy, then the lost gravity potential needs to go somewhere else. In this case into GR. Or GR appears to be a result of the lost energy within the Newtonian gravity vector addition. If we plot Newtonian gravity, starting at distant space, relative to an isolated planet, it starts close to zero. As we approach the planet the gravity will get higher and reaches a max at the planet. As we enter the planet the vectors begin to cancel until at the center, we get zero gravity just like at vast distance. So the Newtonian gravity-distance curve starts at zero reaches a max and then decays back to zero. If you look at this in term of a GR space-time well, the fabric of space-time starts flat at far distance, it begins to form a well that gets deeper as we approach the planet. As we enter the planet the well should start to form a peak as it attempts to reach the top of the fabric at zero gravity, i.e., same mathematical zero value of gravity as distant space in this example. But this peak is not expressed in GR, but is a virtual peak, with the potential energy of that Newtonian peak providing the energy that makes GR possible. The more massive the object, the more vector canceling and lost potential one will get inside the object. This higher amount of lost potential means a larger virtual peak in the GR well, and more space-time affect. I realize this is unorthodox, but where does the potential energy within the vector canceling go, within the Newtonian center of gravity? This analysis sort of makes gravity the sum of Newtonian plus GR, with GR due to Newtonian vector canceling and the induction of virtual energy. Some of the first experiments to show the zero potential in a sphere involved a shell with charge on the surface. The vector canceling or the wave canceling implies lost potential energy in the center. The affect is the surface charge becomes dynamic due to energy conservation. Quote
Majik Posted March 16, 2008 Report Posted March 16, 2008 If you look at the vector addition of gravity in a planetary sphere, the vectors cancel in the center to give zero gravity, based on Newtonian gravity. One thing nobody talks about is where is the energy going? There is also a vector cancelling of gravity out in empty space at Lagrange points between massive bodies. For example, we park satelites at a particular distance between the sun and the earth because the gravitational pull of the sun is balanced by the earth, and it takes less fuel to maintain its orbit there. Where does the potential energy go at those points? Quote
CraigD Posted March 16, 2008 Report Posted March 16, 2008 If you look at the vector addition of gravity in a planetary sphere, the vectors cancel in the center to give zero gravity, based on Newtonian gravity. One thing nobody talks about is where is the energy going? In other words, if we took apart a planet, each piece of matter could generate gravity.The energy – or, specifically, the gravitational potential ([math]U = \frac{G m}r[/math]) which give the difference in potential energy due to gravitational force for a test body of unit mass for any pair of points – is precisely what the preceding approximations have addressed, by doing literally what HBond describes – taking a planet – or rather, the Sun – apart into several thousand evenly-spaced pieces. The gravitational potential energy doesn’t go anywhere mysterious. The greater the distance of a test body from the center of mass of a large body (Sol in the previous examples), the greater its potential energy, which corresponds to a smaller value of U. U reaches its maximum at distance zero, while the net acceleration due to gravity reaches zero there. A potential source of confusion is the sign of U, which for gravitational time dilation calculations is positive, while for classical Newtonian calculations, is negative. So, using the values from post #39 a 1 kg body 90 solar radii from the center of Sol has about 2,118,027,173 - 1,906,223,767 = 211,803,406 Joules less, not more, gravitational potential energy than a 1 kg body at 100 solar radii. Intuitively, the “higher” something is, the greater its GPE. A body at a point with U=0 has the highest GPE it can have. To better illustrate this I’ve added a column for net acceleration to the table in post #39, and changed its scale to make it more readable. It still suffers from small calculator lack of precision and “granularity” relics, resulting in its gravitational potential actually maxing at 0.1 solar radii from center, where the lattice of pieces align, rather than at 0, but it’s less than a 1% inaccuracy.If place them all in a sphere we lose gravity in the center. What this implies is potential energy is being lost.No, it does not imply that. Energy is NOT equivalent to force, but to the sum of force times distance over which the force is applied. Assuming there are no other bodies in the universe, a body placed at and at rest relative to the center of mass of a second body has lost all of its gravitational potential energy. The energy has gone into whatever work was performed to place it there. Were no work done placing it there, or at any point less distant than some starting point – that is, has the body free-fallen there - the lost potential energy would be in the form of relativistic kinetic energy, which for low speeds can be accurately described by the classical formula [math]E = \frac12 m v^2[/math], or for any speed, an increase in the moved body’s mass given by [math]E = \Delta m c^2[/math]. Here’s an example illustrating the difference between force and work/energy: I push a car up a hill, giving it potential energy. The top of the hill is level, so the car experience no net force in its allowed direction of movement. If I release the car, it won’t do any work (roll down the hill). However, it still has the potential energy I gave it – if I push it over the edge of the hill, that potential energy, plus any I added getting it rolling, minus losses do to friction, will be transformed into kinetic energy. Quote
modest Posted March 16, 2008 Report Posted March 16, 2008 Still, the gravitational time dilation in Sol, or even the most massive stars, remains very small, far below what I think could be detected by observation. I agree. Unless you're modeling the collapse of a supermassive star, this is not a useful or significant variable in discussing the rate of fusion. -modest Quote
HydrogenBond Posted March 16, 2008 Report Posted March 16, 2008 Let me give an analogy for the virtual GR affect of Newtonian gravity. Say we had a generator with a single rope loop around a pulley. As you pull, the gear turns and the rope recycles within the loop. If we pull equally from opposite sides of the loop, i.e., rope top and rope bottom, there is no net force being applied to the generator, due to force vector addition. Yet we could generate energy constantly, using zero net force. The center of gravity, is zero net force, but it powers the GR generator. The might be explainable with a particle wave example. If gravity force was only expressed with particles, hypothetically, the center would be traffic central since it has the most combinations of particle interaction at the shortest distance. If we use the wave nature, only, the waves cancel in the center. Here is the paradox, traffic central should have tons of particles but the waves say there should be no particles. To maintain consistency, the particles can not be there. But over time, these particles are constantly being exchanged while constantly ending up in wave nothingness. Since the particles reflects force and distance or energy, this energy needs to appear elsewhere, to be conserved. Since it can not be expressed within the wave nothingness, it comes out as GR, as different type of wave. Quote
modest Posted March 16, 2008 Report Posted March 16, 2008 (As I don’t know a formula for the gravitational potential [math]U = frac{GM}{r}[/math] inside a sphere of uniform or other density, I’ve fallen back to a numeric approximation, representing the sphere as a cloud of 26,252 point masses. [math]\phi = -\frac{GM}{2a^3}(3a^2-r^2)[/math] derived on Gravitational potential due to rigid body gives equivalent answers (to yours). It assumes uniform density. -modest CraigD 1 Quote
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