csmyth3025

Welcome to scienceforums.net John J. Wright! I think you need to learn some basic science before you start posting white papers here. Bona fide science questions are answered with patience here. White papers based on voodoo science are not so kindly treated. Chris

If I understand your reply correctly, a dimensionless point (a singularity) cannot rotate and therefore cannot have any angular momentum. On the other hand. such a point possessing mass cannot rotate around a center unless there is some other point possessing mass to "counterbalance" it. Since two such points would merge almost instantaneously in the black hole's severe gravity environment, the only viable configuration is a one-dimensional line representing a non-zero circumference for the rotating mass. This circumferential line would have length (2*pi*r) but no volume. Is this approximately the description of a ring singularity? Also, Airbrush cited a maximum of 1150 rev/s for a 14 solar mass black hole singularity. What determines this limit and what happens if it's exceeded? Conversely, is there any known mechanism that would prevent this limit from being exceeded? Chris

As a follow-up to my last post, I take it that if the observable universe has negative curvature (-2% lower WMAP limit), then it would be hyperbolic and "open". The so-called "saddle" topography. If the observable universe has positive curvature (+2% upper WMAP limit), then it would be essentially spherical and "closed". In this case the universe would, indeed, have a radius of curvature. Is there is any way to calculate the radius of curvature of the observable universe based on this (+2%) upper limit of the WMAP data? Chris

I'm not sure where Widdekind got a present radius of curvature for the universe of 10's of Gly's. If the CMB is considered the limit of the present comoving distance to the edge of the observable universe, then the observable universe today spans a radial comoving distance from the Earth of about 45.7 Gly (about 91.4 Gly in diameter). (ref. http://en.wikipedia....rvable_universe ) According to WMAP data, the CMB indicates that the universe is "almost flat" (flat, +/- 2%): (ref. http://wmap.gsfc.nas...AP_Universe.pdf - "Shape of the universe") Chris

As far as I know, temperature has no effect on the half-life of nuclei that sponataneously fission and will not make a nucleus that has no tendency to fission aquire such a tendency. Chris

I'm not aware of any such proofs. Can you provide a reference? Chris

We do get realistc videos of the sun in UV light and other wavelengths - and astronomers combine infrared, visible, UV and x-ray images in various ways to make pictures of galaxies and other astronomical objects. I really don't have a clue about what kind of an image gravity waves or neutrinos could produce. I'm sure such images would have to be computer generated from a data stream. I doubt if any sort of "real time video" is possible. This is a good question, though. If anyone has any information on this please jump right in. Chris

I was actually being a bit of a smart-alec with my post. I probably should have put the words government project in bold letters. Better yet, I probably should have stuck to serious science. My inspiration was the proposed 398 million dollar "bridge-to-nowhere" which would have replaced two existing ferry operations (a seven minute ride) from the Alaskan mainland to Gravina Island - upon which a one-runway "international airport" and about 50 residents reside. The bridge hasn't been built. It was ostensibly canceled in 2005. In an ironic twist, a 28 million dollar federally funded road (~3 miles) was built on Gravina Island in 2007. It stops at the edge of the island where the bridge would have been. A gravity train on the Moon just sounded to me like the sort of "inspired" project a lunar government might relish. Chris

Although the gravity train idea is impractical for Earth-bound travelers, I think a good case might be made for such a (government) project on the Moon. This assumes, of course, that at some time in the future there is a substantial population on the Moon dispersed in several (or many) locally urbanized areas. It also assumes that this population would be willing to expend the tremendous amount of manpower and resources for all of the tunneling that would be required for this sort of point-to-point travel (translation: lots of taxes). Hmmm... Maybe they would rather just ride around in their Moon-buggies on their Moon-expressways (adorned with the requisite moon-billboards). Chris

The sun's coronal heating may be the result of contributions from all of these processes. Sort of like the Earth's weather, there are a lot of things going on in the sun's atmosphere and they all interact in complex ways. Regarding the portion about spicules you referenced: This information has been updated according to a Jan, 2011 Space.com article: http://www.space.com...lasma-jets.html Chris Edited to insert "Jan, 2011"

Although there are many isotopes of elements heavier than iron that decay spontaneously, there are also many that are not thought to be subject to spontaneous decay (even theoretically). I believe the heaviest of these is Zr92.There are an additional 55 heavier elements (up to atomic number 164) that are thought to spontaneously decay in theory but have never been observed to do so. (ref. http://en.wikipedia....observed_decays ) Chris

The Wikipedia article on the Solar Corona can provide you with a wealth of information. The portion of that article on spicules contains the following: (ref. http://en.wikipedia....s_.28type_II.29 ) Chris

Regarding your immediate question, as I understand it the energy "decay" of photons due to the expansion of space is not a loss of energy globally, but rather the simple relationship that energy density/pressure is inversely proportional to volume. I'm not an expert on cosmology, so if I'm mistaken please feel free to correct me. Regarding your original post, the kind of universe your describing (with little or no mass) is, I believe, called a de Sitter universe: (ref. http://en.wikipedia....Sitter_universe ) Chris Edited to add the word "globally"

Swansont provided one possible theoretical answer to your question (edited by me to emphasize that this is a question about scientific theory). Neutrinos can handily pass through our atmosphere and buildings without interference. Unfortunately, they can also pass very easily through people, the Earth, and whatever type of telescope you may be using as if they aren't even there. For these reasons Swansont describes using them to see anything as "exceedingly difficult". I would go one step further and say that there is no known way, at present, by which this can be done. Another possible candidate might be gravitational waves. They, also, can pass unhindered through things. Like neutrinos, there is no known way, at present, by which they can be used to see anything. There are real scientific conjectures, however, that we might one day learn how to use neutrinos and gravity waves in somewhat the same way we use light. I would put these conjectures loosely in the same category as Jules Verne's 1865 novel "From the Earth to the Moon" in which the protagonists are shot to the moon from a huge cannon. The correct details are missing, but the basic idea might be possible one day if our scientific knowledge and engineering expertise becomes sufficiently advanced. In 1865 I'm sure everyone thought that the idea of actually going to the moon was pure fantasy. I suspect that Mr. Verne viewed the possibility a bit more seriously. Chris

If all you need is the slope, then the suggestion from Bignose is all the help you need. Post your equation with the slope denoted by a variable. Is there a reason why you cannot do this? Chris

Swansont is much more patient than am I. In all of the posts you've made I have yet to read a single statement that can even remotely be considered scientific, predictive or even coherent. I offer you the following quote from the Wikipedia article on Global Positioning Error Analysis: (ref. http://en.wikipedia....itioning_System ) Please explain in specific mathematical detail how you reconcile your notion that time runs faster close to the sun by virtue of the sun's consumption of energy with the verified Special and General relativistic effects of clocks in orbit around the Earth vs clocks on the ground. Chris ---------------------------------------------------------------------------------------------- As Klaynos said: "What equation?" You haven't presented a single equation. How on Earth can you claim to be a "logic savant" in one sentence and then turn around and claim that we haven't disproven an equation that doesn't exist. That is not logical. Chris Edited to correct spelling errors

Is there a reason why the singularity of a Kerr black hole is thought to be a ring with zero thickness but non-zero radius rather than a disk with zero thickness but non-zero radius? Chris

Thanks for the math Iggy. I can follow your calculation. It's obvious that these sorts of relativistic calculations can get very complicated very quickly. I tip my hat to anyone with the math skill to wade through them and not get lost. Chris

Just to keep the record straight, in my previous post (#79) I initially indicated that we could use an obital radius slightly larger that the innermost stable circular orbit as a simplifying condition. I subsequently picked an orbital radius that is twice the Schwarzschild radius. The innermost stable circular orbit for a black hole must be greater than three times the Schwarzschild radius, however. (ref. http://en.wikipedia....child_geodesics ) With rs=2.95 km, I should have picked a larger orbital radius like 10 km (3.39 rs) or even 9 km (3.05 rs) in order to be consistent. I don't think this changes the math of the calculations we made, though. I think the solution for resolving the orbiting observer's calculation of his orbital velocity with the distant observers calculation lies somewhere in the cited Wiki article on Schwarzschild geodesics. Unfortunately, my math skills are too poor to even understand what most of the equations in the article mean. Chris

Thanks Iggy. I forgot that to = tftimes sqrt.... If my (further) calculations are correct, a point object will orbit this black hole once every 2.471x10-4 sec. (ref. http://www.wolframal...w.m1---&x=5&y=9 ). The inverse of this would be about 4047 revolutions per second. A distant observer will count about 4047 orbits for this object every second according to his clock. An observer on the orbiting object will count about 4047 orbits every 0.5 seconds according to his watch. For the same event (4047 orbits) the distant observer will think the orbiting observer's watch is running slow (0.5 seconds) while the orbiting observer will think the distant observer's clock is running fast (1 second). For an orbital radius of 5.9 km, the circumference of the orbit would be 2*pi*5.9 km=37.1 km. 4047 orbits would make the distance traveled about 150,144 km. The distant observer would see the orbiting observer traveling at an orbital velocity of about 150,144 km/s. Because the orbiting observer is going so fast, to him each meter of his orbit will be about 0.8657 m rather than the "real" meter of the distant observer (ref. http://www.wolframal...9%5E2%29+meters ) . He will think that his velocity is about 260,000 km/s according to his calculation (ref. http://www.wolframalpha.com/input/?i=%28150144000+meters%29%280.8657%29%2F%280.50s%29 ) Is there an additional shortening of the orbiting observer's perceived orbital length due to the strong gravitational field that will make his calculation match the distant observer's value? Chris Edited to correct math errors

If you look back to post #75 there is an equation that I think demonstrates this effect. If we make the simplifying assumptions that the freely falling observer is in a circular orbit around a non-rotating, non-charged black hole at the distance slightly more than the innermost stable circular orbit then his (proper) time compared to a distant observer"s time will vary as described: To further simplify the problem we can say that the black hole has one solar mass and the circular orbit is twice the Schwarzchild radius. As it turns out, the Schwarzschild radius is about 2.95 km and the radius of the orbit is about 5.9 km in this case. Using the above formula, one second on the orbiting ("free-falling") observer's watch would equal: [1-(3/2 x 2.95 /5.9)]^0.5 = [1-(1.5 x 0.5)]^0.5 = [1-(0.75)]^0.5 = 0.25^0.5 = 0.5 seconds on the distant observers's clock. I have to admit that this result puzzles me - and I believe it has to do with how the term "proper time" is defined using Schwarzschild coordinates. For every second the distant observer's clock ticks off, the orbiting observer's watch will tick off two seconds. It seems from this that the orbiting observer's clock is running faster than the distant observer's clock. This is not what's supposed to happen for a clock that's in a gravitational field! I've clearly misused the formula or I'm misinterpreting the results. HELP! Chris

On the matter of clocks on satellites, I used a geostationary orbit as an example simply because it's a well defined orbit and easily referenced in Wikipedia. The main point is that the calculation shows the gravitational time dilation for orbiting clocks just as your previous post mentioned. Concerning the "frozen star" model of black holes, I can't present any definitive arguement for or against this model. I just don't know that much about it. To be sure - as Slinkey said - "there's something very heavy there...". Until a more well established "frozen star" model is presented, though, I'll have to follow the logic of this argument from my previous link: (ref. http://www.mathpages.../s7-02/7-02.htm ) Chris

Although this phenomenom doesn't have anything to do with length contraction, your idea isn't radically speculative. If a massive body is accelerating - in this case changing direction due to orbiting another body - it will produce gravitational waves: In the case of two neutron stars orbiting each other, the amount of energy carried off by gravitational waves can be quite large: (ref. http://en.wikipedia....orbiting_bodies ) PS - The energy carried off by gravitational waves in these systems comes from a loss of their angular momentum. In simple terms, they're spiralling in towards each other. In the case of the Earth, it will take about 10^13 times the age of the universe for Earth's orbit to decay to the point where the Earth will merge with the Sun. In the case of the binary pulsars with the closest known orbital separation (PSR J0737-3039), they're expected to merge in about 85 million years. Chris Edited to add PS

Iggy, I have to retract my pevious speculation about objects in circular orbits being in an inertial freame of reference indentical to that of a distant observer. I ran across this passage in Wikipedia's article on gravitational time dilation: (ref. http://en.wikipedia....l_time_dilation ) The Schwarzschild radius of any object (even the Earth) can be calculated as shown in the quote above. For Earth, this turns out to be 8.872x10^-3 m (about 0.35 in) (ref. http://www.wolframal...---.*--&x=5&y=7 ) A geostationary orbit has a radius of 42,164 km from the center of the Earth. (4.2164x10^7 m). (ref. http://en.wikipedia....ionary_altitude ) Using these values, one second on the geostationary satellite's clock would be: to= tf (1-(3/2)(8.872x10^-3 m)/(4.2164x10^7 m))^1/2 = ~0.9999999998 seconds on the distant observer's clock. Feel free to check my math, since it's very possible that my calculation isn't correct. By the way, the excellent reference you provided ( http://www.mathpages.../s7-03/7-03.htm ) and the previous sectin of this book ( http://www.mathpages.../s7-02/7-02.htm ) seem to argue in favor of the formation of black holes in a finite time from the distant observer's perspective. Chris

In the case of GPS satelites, the effects of both General and Special Relativity have, indeed, been demonstrated to a high degree of precision. I can certainly see how the atomic clock on the GPS satelite can be in an approximately inertial frame of reference because it's following the spacetime geodesic dictated by its distance from the Earth's center and its tangental velocity relative to the Earth. The atomic clock on the ground, however, is not in an inertial frame of reference - it's in an accelerated frame of reference. The fact that the device on the ground "feels" gravity (has weight) illustrates this fact. I wonder if any experiments have been performed that compare the atomic clock rates of two (or more) atomic clocks on satelites orbiting at different altitudes (11,000 and 22,000 km, for instance). I'll have to think about this for a while. I'm not ready to abandon the strong equivalence principle yet. Chris