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Alright, and here's where I use the Feynman explanation:

The speed of light with respect to a given reference frame is constant and defined to be 299 792 458 meters per second. Note: defined. As in, the meter is defined to be 1/299792458 the distance that light travels in one second (the second is defined by atomic emissions; I don't know that one).

This is light in a vacuum of "uncurved" space. The way I learned relativity (from my dad) is like; take a piece of graph paper. Each grid is a square meter. Now put a solid object in it, and let it dip down. Depending on how heavy the object is, the more your "space" curves. The object may only be say, 5 grids in diameter, but now, to travel from one side to the other, it might take 9 or 10 grids.

This is how the "curvature" is defined. It's not special relativity; it's GTR, and you'll have to ask VorpalNeko to describe that. What I can say is what Feynman said: "Light always takes the shortest (least amount of time) path between any two points."

What this means is that; the speed of light in a vacuum is constant. What's changed is the space that light is travelling in; the "space" itself is longer. In equation terms, it's not that
distance=x=v*t=(c+k)*t
If this were the case, the time taken would be shorter. We could have
distance=x+delta x=(c+k)*t
But this is also not the case; the time taken is not the same. Rather,
distance=(x+delta x)=c*t
So the time taken is longer. c is still constant (in a vacuum); the only thing that's changed is the "distance" travelled.

Again, ask VorpalNeko for a better description of GTR. I am so unworthy! sweatdrop

I apologize, it was 1.7 seconds and it was referencing light passing by the sun(but not from the sun.) However the point I was trying to get to was not the number but rather the concepts it leads to, as well as the difficulty of truly measuring the speed of light.

And I see. What I am getting out of this so far is that matter causes space/time to be warped(or the converse, that space/time being warped causes matter, but as saying that matter is produced by gravity and not the opposite is not as sensible.) and as such perspectives of actions from different locations can very greatly.(Mostly in relative differences of gravity/velocity)

But that does bring up the problem of, what if one were to pass nearby a blackhole and overcome its gravity to reach the 'other side' which never saw the person/craft finish passing by the blackhole, but to the person passing by they are going at a normal rate. Would the object ever pass by it for those on the 'other side' of the blackhole, it would just take a long time? And for the person passing by the blackhole, a long time would pass without much effect?

Edit: Though I suppose this may be taking the generality that time is not relative and is instead something consistent, and that is something the theory addresses.

I am, and I was posing it theoretically, as obviously we couldn't do it now. However I suppose if you passed too closely, you might need to exceed c to pass by the blackhole, which is impossible sine it would require infinite energy.

Yes. If your speed is c and your velocity vector is tangent to the event horizon, then you will not fall in. On the other hand, you will be stationary relative to any external observer, so you will not make it out either. Additionally, this assumes that you are an idealized massless point-particle--this trick is impossible for extended bodies (every massive body is extended).

Edit: This is a bit counterintuitive, but the only "tangent to the horizon" direction possible is "in the t-direction only". It is tangent in the sense of spacetime, but non-tangent in the spatial section.

t doesn't orbit--it is stationary relative to external observers. As for light orbits, that's an interesting question. The full Schwarzschild metric is dλ² = (1-2m/r)dt² - dr²/(1-2m/r) - r²[dθ²-sin²θ dφ²]. By symmetry, it is obvious that any constant-speed orbits will be circular (dr = 0), and the coordinates can be chosen so that the orbit is in the equatorial plane (θ = π/2, dθ = 0), so in particular d²r/dλ² = d²φ/dλ² = 0. Under these conditions, the geodesic deviation for x^1 = r simplifies to [2m/r²][dt/dλ]² = 2r[dφ/dλ]². This essentially Kepler's third law Ω²r³ = m, except that reference frames are mixed. I can provide the details of the calculation if you're interested, but it's really just a lot of partial differentiation. Substituting this form of Kepler's third law back into the metric (along with dr = dθ = 0) gives dλ² = (1-3m/r)dt². Consequently, (1) there are no circular orbits below r = 3m and (2) r = 3m gives dλ = 0, i.e., a lightlike geodesic. This is the only closed orbit available for light.

According to Wikipedia:
I've always wondered about this: If the speed of light is defined as 299792458 metres per second, and a metre is defined as being 1/299792458 the distance that light travels in one second, then where does this terminate?

Prof. Verlinde was explaining this to us after the review session. The notion of a "pure vacuum" is still relatively undefined, and certainly not generally present on earth. Perhaps there are regions of space where, due to some nature of the space, the light travels even faster - an even purer vacuum, of sorts. Still, our measurements here on earth are pretty damn accurate, if you've ever seen the experimental setup.