Robittybob1

If we put real figures to the formula DimaMazin gave us? http://www.sciencefo...ty/#entry922079. I think it will result in the OP seeing how the time dilation equation work and so will I. I just get confused how to put time in as a number. Maybe it just be a difference in time from [latex] T_0 {=} 0[/latex], then [latex] T_n [/latex] will either positive or negative or also zero.

How far away do you think the horizon is? If in the morning (at dawn) and you see the full moon you must be looking to the West. So for someone just below the horizon to the West the Moon will be also to their West. They too will be in the morning. Morning and dawn being somewhat similar. I don't think you have proved anything as yet.

Which of those three answers my question?

Is this suppose to be factual? It doesn't sound right to me. When there is a full moon it is full from the East to the West, so it is also full below the horizon as well in that phase.

If we were to check this against the real life speed of the satellites at these heights/velocities what value do we give to [latex]T_1, T_2..... T_N[/latex] etc. Like is it [latex]0 {=} T_0[/latex]? Or better [latex] T_0 {=} 0[/latex]

OK sounds like you'll be able to have rabbit with your goat milk cheeses one day.

[latex]\frac{t_n\gamma_n}{r_n}[/latex] That [latex]\gamma[/latex] is a velocity dependent term. For satellites surely that is related to the orbital velocity? The ISS and GPS satellites orbit at different [latex]{r}[/latex] values and different velocities accordingly so their time dilations are different. [latex]\gamma = \frac {1}{\sqrt{1-\frac{v^2}{c^2}}}[/Latex] #7

[latex]t_1\gamma_1 / (1- 2GM / r_1c^2)^{1 / 2} = t_2\gamma_2 / (1-2GM / r_2c^2)^{1/2} = t_3\gamma_3 / (1-2GM / r_3c^2)^{1/2}[/latex] Conclusion: seems to imply [latex]\frac{t_n\gamma_n}{r_n}[/latex] for any [latex]n[/latex] is a constant. Did I get that conclusion right? [Thanks Mordred for the Latex correction.]

What happens to the parts of the branches they don't eat?

Is it possible to describe "<T" in terms of atomic clock tick rates? Would this be equivalent to what you had said above: The tick rate of atomic clocks on the International Space Station are slower than the tick rate of clocks on the Earth and both are slower than the tick rate of clocks on GPS satellites?

@Permieguy - In your video you mention "tree hay". Could you please tell us what is that and how do you make it and use it?

I'm not knocking what you are doing, and I'm looking into Allan Savory's work as well. What I could see was a lot of quite risky work. Presumably you have to cut branches off trees and then cut them with that machete. A good wind could lift that goat tractor. Kneeling on damp ground to milk the goats. Will the really long grass be a fire risk in summer?

How many people will 3000 food calories fed per day?

Thanks Strange. I like the images of 3D spacetime as opposed to the rubber sheets. Like this http://cdnimg.visualizeus.com/thumbs/df/49/grid,planet,space,spacetime-df4956ec0ccee7ecf4f84ea9439b57d0_h.jpg?ts=93246 There seems to be the idea of frame dragging so space rotates with the spinning BH. OK (the Lorentz transform only applies in locally flat [Minkowski] space-time).

That was #76 Is the event horizon spinning in the case of a spinning BH? If the spin is in the same direction as the orbital motion would there need to be the need for the addition of velocities using the Lorentz factor? https://en.wikipedia.org/wiki/Lorentz_factor

You are right.

So does the "curvature" of a circle (the EH) relate to the curvature of spacetime? I've never heard that before. Have you got some reference to that? You might have thrown me into the deep end before I know how to swim. "Schwarzschild metric"!

But at the EH there would always be the same curvature (gradient) at that point only. It is the point where the curvature equals the escape velocity. So the tangent to the curvature will have the same slope I presume. Was that the point you were making?

As an aside: I was impressed with what New Zealander Roy Kerr had calculated I phoned him up and thanked him (about 16 years ago). He said "it was nothing", but recently it was suggested he might receive the Nobel prize for his work on rotating black holes. Do all circles have the same curvature? I'm tempted to say "no" but I'll accept correction. http://www.math.washington.edu/~lee/Books/Riemannian/c1.pdf That seems to suggest different curvatures are possible.

Gosh. So a larger BH (has more mass) and has a wider area (volume) of curvature. For would you say they all have the same curvature at the EH. Have you got a favourite image of the spacetime curvature round a black hole? Do you think of space itself falling into the BH?

No hair theorem - a BH can be characterised by 3 parameters: 1. Mass 2. Momentum 3. Electric charge In the list of where the mass could be. If it was a solid core then overlapping of the EH would be a big deal. Is there such a thing as gravitational mass so you don't need to worry about that? "P" what was that formula calculating?

I would love to be able to do that but what is the image? I don't want to think in terms of curved rubber sheets. I have wanted to put some values on the curvature so I can begin. How do you think of curvature? Do you just say "curvature" but have no mental image? Is it just the same as saying gravity for no one has an image of gravity either?

The LIGO team only had the trace to go from. From that they made list of deductions. So is that the type of observer you are talking about? I don't follow the going into a BH concepts very well. Just confirm for me please: a BH has mass beyond the EH and that mass produces/has gravity?