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Absolute zero

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About Absolute zero

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    Mesa, AZ
  1. Before all of the land on earth was mapped, map makers would shade in any areas not yet explored. This is where the phrases darkest Africa, the dark interior of North America and the dark side of the Moon come from. I think the person who coined the term dark energy was referring to the unexplored physics of dark energy. Dark matter does not release electro magnetic radiation so it is dark.
  2. The conversion of Newtonian orbital velocity to a Schwarzschild metric orbital velocity would use √(1+(rs/h)) or more familiar expression to me √(r/(r-rs)) or we could mix and match √(r/h). It would be odd for the same conversion factor to work for Newtonian weight to Schwarzschild metric weight. If you would be so kind, I would like to see the equation for Schwarzschild metric weight or where it can be found.
  3. I am not sure what you are referring to as being right. Is it v=√(GM/(r-rs)) is the correct equation for orbital velocity? Or could it be v=√(GM/r) is just a Newton approximation, good whenever r is much larger then rs? Until someone does a full GR evaluation, I will assume the equation, GM/(r-rs)^2 , is the correct one to answer the op's question with.
  4. From Wikipedia, (sorry I have not figured out linking here) Orbital velocity in general relativityEdit In Schwarzschild metric, the orbital velocity for a circular orbit with radius 'r' is given by the following formula: v=√(GM/(r-rs)) where rs=2GM/c^2 is the Schwarzschild radius of the central body. So when r=rs, the orbital velocity is infinite or undefined. At r=1.5rs, the orbital velocity is c. Script did not transfer correctly
  5. Force has both time and velocity components. Newtonian physics will give reasonable results when time is absolute and velocity is linear. GR must be used when a distant observer's rate of time differs from the local observer and\or where velocity is not linear. In the following, I am assuming both 'A' and 'B' are falling at negative escape velocity. Is 'A's clock running at the same rate as 'B's clock? Time slows down near a black hole by 1/√(1-(2GM/c^2R)). For every time unit 'A's clock shows, a distant observers clock would show 1.002 time units. Time also slows down do to velocity. 'A' has a velocity of .0547c. The time dilatation factor for velocity is √(1-v^2). For every time unit 'A's clock shows, a distant observers clock would show 1.001 time units. Time in this case is almost passing at the same rate as the distant observer so Newtonian physics should give reasonable results. For every time unit 'B's clock shows, the same distant observer would show 1732 time units. In this case, GR must be used to get reliable ansers. If a distant observer could do spooky action and apply a g of force to 'B' for 1732 seconds, 'B' would experience 1732g for 1 second. 'B's velocity would be .9999998c. The time correction would be 1581 seconds for the distant observer to 'B's 1 second. Multiple the two corrections times each other gives 273,800 as a correction factor for time. Which is the bigger change of velocity, .05470000c to .05470001c or .99999980c to .99999981c? (.05470001-.0547)/(1-.05470001•.0547)= .00000001 (.99999981-.9999998)/(1-.99999981•.9999998)=.0256 Close to Rs small changes in Newton velocity will be large changes in real velocity.
  6. Yes, the escape velocity at Rs is ,c, the velocity of light. In GR an orbital velocity of c is at R=1.5(Rs) while the orbital velocity at Rs is infinite. Question, how much force is required to accelerate a 1 gm mass to c?
  7. Astronaut B has the most force applied to him in a GR solution. One of the deferences between Newton's theory and GR is the location of infinite forces. In Newtons theory it is at R=0. The GR location is at R=(Rs), where all objects fall at the speed of light. This is why GR equations written in Newton form may have the term (R-Rs) or its equivalent.
  8. The main confusion with the expansion of the universe is the use of velocity in the expansion term. I wonder if the term was rewritten as distance/distance/time if some of the confusion would disappear.
  9. I think of it as a conservation of curvature term. Absolute zero
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