kxmode wrote:
I wrote this epiphany at 2 in morning and I've never had it since. I don't understand half of what I wrote, but at the time is seemed crystal clear, like a diamond of pure clarity.
I do, though I would like to know what you were doing that you somehow understood then but not now. Did you read up on special relativity?
kxmode wrote:
Figuratively time is a wheel.
We know time is visible through the method of past, present and future. Rather than one rigid, unchangeable shape, this time wheel itself has the ability to change and form a complete system where varied speeds of light could be achieved.
No, not quite. Rather, the speed of light is measured to be the same in all inertial reference frames regardless of how fast one reference frame is moving with respect to another.
To illustrate what that means, let's for now think up situation that doesn't involve special relativity but only requires known physics before Einstein (called Galilean relativity). Suppose you're on a racetrack. You're opponent's car is travelling at 50 km/h and your car is travelling at 20 km/h, both at constant velocity. According to Galilean relativity, if you could measure the velocity of your opponents car with respect to your own car while you were driving it, you would find it was travelling at 30 km/h with respect to you (meaning of course, it was travelling 30 km/h faster than you from your point of view). Now lets replace your opponents car with a light beam. According to special relativity, you would that light beam would be travelling at exactly the same speed with respect to you regardless of how fast your car would be travelling.
kxmode wrote:
Normal time is represented as a wheel with three rings and an inner core. Each successive ring rotates by a small amount compared to the preceded ring (illustrative examples: the ring animation from the movie Contact, or watch runners during a Track and Field event). While the inside ring rotates slowly the outermost ring moves at a high rate of speed. As the wheel rotates, the coordinates (x, y) of a point on the wheel to its center are dynamic, but the distance (r) between the point and the center are static. The following formula is supposed:
r2 = x2 + y2 = constant
The coordinates (x, y and z) of the interval between two points in three-dimensional space (a vector) change when the coordinate system is rotated in three dimensions. However the separation (r) of the two points remains constant. Thus we arrive at the formula:
r2 = x2 + y2 + z2 = constant
I don't no what you're trying to get at with that wheel analogy, it's much simpler than that. That's just the distance between two points in space in rectangular coordinates (in that case, one point was chosen to be the origin) and comes from Pythagoras' Theorem. The point is, no matter how you choose your coordinate system, the distance between those two points will work out to be the same. It's also not only true if the coordinate system was rotated. For example you can choose a new coordinate system with the origin at 2 units in the z direction with respect to old. If you calculate the distance between those two points with the same formula but using the numbers from the new coordinate system rather than the old one, you will still get the same answer.
kxmode wrote:
This again is what could be classified as normal time.
Here's where things get interesting…
Let's go into different warped times otherwise known as a space-time wheel. A space-time wheel is shaped like the letter "X". A vectored shape becomes horizontal where time (t) is vertical and space (x) is horizontal.
Like the normal three-ringed circular shape, a small velocity compared to the preceded ring, or branch there boosts each successive ring. As the space-time wheel boosts, the space-time coordinates (t, x) of a point on the wheel to its center change. Yet the space-time separation(s) among the point and the center remains constant. This formula could best represent this:
s2 = - t2 + x2 = constant
Moreover, the coordinates (t, x, y, z) of the interval between two events in four-dimensional space-time (a 4-vector) change whenever the coordinate system is boosted or rotated, but the space-time separations of the two events remain constant. This would result in formulating:
s2 = - t2 + x2 + y2 + z2 = constant
The invariant space-time separation(s) between two events is a rock in the sea of relativity (a quantity that remains the same for all observers). Though time and space they change for different observers. The space-time separation(s) is of basic importance in relativity.
Again, you don't need the wheel analogy. It just overcomplicates the matter. It's the same situation as before. If you represent events in space-time as points, then the space-time separation is a "distance" between those two points in four dimensional space-time. The reason for the minus in front of the time coordinate is due to the geometry of space-time being non-euclidean and so Pythagoras' Theorem is not obeyed.
kxmode wrote:
Let's suppose you took a trip across the Universe in a spaceship, constantly accelerating at one-Earth gravity (g). How far would you travel in how much time? The space-time wheel offers a unique way to solve this problem. Because the rotating space-time wheel might be regarded as representing space-time frames, that undergoing constant acceleration, points on the right quadrant of the rotating space-time wheel would represent world lines of persons whom accelerated with constant acceleration in their own frame.
If the units of space and time are chosen so that the speed of light and the gravitational acceleration are one, (c = g = 1), then the proper time experienced by the accelerating person is the boost angle, and the time and space coordinates of the accelerating person (to a person whom remains idle), are those of a point on the space-time wheel:
(t, x) = (sinh, cosh)
In the case where the acceleration is one-Earth gravity (g = 9.80665 m/s2), the unit of time is [c / g = (299,792,458 m/s) / (9.80665 m/s2) = 0.97 years], just short of one year.
After a slow start, you cover ground at an ever-increasing rate, crossing 50 billion light-years, the distance to the edge of the observable Universe, in just over 25 years of your own time.
Does this mean you go faster than the speed of light? No. From a person’s view on Earth, you never go faster than the speed of light. From your own view, distances along your direction of motion are Lorentz-contracted. So distances that are vast from Earth's view appear much shorter to you.
As the Universe rushes by it never goes faster than the speed of light. This is why stars that we see as being vibrant burnt out millions of years ago. Simply stated it took the light that long to reach us.
This rosy picture of being able to flit around the Universe has drawbacks. First, it would take too much energy to keep you accelerating at Earth's Gravity. Second, you would use too much Earth-time traveling around at relativistic speeds. For example, let’s say you took a trip to the edge of the Universe. By the time you got back not only would all your friends and relations be dead, but the Earth would probably be gone, swallowed by the Sun in its red giant phase. The Sun would have exhausted its fuel and shriveled into a cold white dwarf star. The Solar System, having orbited the Galaxy thousand times would be lost somewhere in its milky ways.
Also worth noting the Universe is expanding. The distance to the edge of the observable Universe is increasing. So it would actually take longer than expected to reach the edge of the observable Universe. Moreover if the Universe is accelerating, as recent evidence from the Hubble diagram of Type Ia Supernovae suggests (
Supernova Cosmology Project and
High-Z Supernova Search), then you will never be able to reach the edge of the observable Universe regardless of speed.
All of that is just talking about the time dilation and the Twin's Paradox. Here is a more detailed treatment of The Twin's Paradox:
http://en.wikipedia.org/wiki/Twins_paradox.
24 Jul 2010, 4:57 pm
