Posted on 02/11/2006 4:31:06 PM PST by PatrickHenry
ah i see - that is a difference I hadn't thought of
Apprently, you have to attend the conference to find out....
At some point the mass is ejected from the field, IMHO.
Don't tell anyone, but you multiply it by itself and the answer is three.
I dare you to tell that to "turtle-face" in person!
Only if we need to clear some asteroids from our path, or perhaps for unsportsmanlike conduct.
Well, we're off for some yummy sushi and hot sake in celebration of my 10th SkyVista system sold today.
The customer's place isn't far from here, but he's up and over a ridge on a road that is no foolin' scary in places. Ruts deep enough to high-side a Suburban. Right about now, that's a nasty combo of water, ice and mud. I think I'm going to tack on a hazard surcharge for this one.
congrats
This whole thing is as queer as the square root of a three-dollar bill.
The reciprocal of which is "57.7" ents.....
The fix is in. Damn NFL refs!
Congrats my friend.
L
Will I be rioted upon? If you get rioted on by other than conservatives, you did a wrong thing and must have said something intolerant. (If you get rioted on by conservatives, you're a courageous free thinker.)
The correct solution to the twins "paradox" has been known all along. It's only ever been a source of mystery to those who don't know the math.
The answer is that the axis of simultaneity is frame dependent. The twin who remains in one inertial frame is the "object of reference". The one who changes frames (by turning around and coming back) is the one who experiences less time.
On each leg of the traveller's journey, time is passing more slowly on Earth than on the traveller's ship, as measured from the traveller's point of view. At the same time, time is passing more slowly on the ship as seen from the Earth's point of view.
Those statements seem irreconcilable until you understand one key fact: that on the ship, the time spent on the outbound leg is simultaneous with a (short and slowly passing) period of Earth time shortly after the traveller's departure, and on the inbound leg is simultaneous with a (short and slowly passing) period of Earth time shortly before the traveller's arrival. At the turn-around, there is a "simultaneity gap": a large stretch of the Earth time that was never (or rather, only very briefly) simultaneous with the ship time, during the turn-around.
Let's put it another way. Suppose the traveller isn't coming back, but zooms out past Betelgeuse, and passes an Earth-bound ship at very close range. At the moment the ships pass, the time on Earth is extremely different for both ships, even though they're at the same place at the same time.
Why? Because they're in different inertial frames.
I understood that better before I read your explanation.
Precisely where anti-deflationary forces become detectable...
Sattelite => Satellite.
Imagine you're moving towards a blk hole. As you approach it, your speed goes to c. Folks on Earth see you decelerate and hang at the event horizon. The guy on the ship sees the blk hole accelerating towards him.
To an observer on Earth, the clocks on the ship slow down and the mass of the ship goes huge. All the guy on the spaceship sees is the black hole accelerating towards him and eventually he becomes one with the hole. Like a guy crashing into the wall.
On Earth, it seems the guy driving the ship has slowed down for some reason and parked his ship at the event horizon. This is the nature of the apparent repulsive force. The guy on the ship doesn't believe in that force.
This makes a lot more sense when you wash it down with a very fine single malt scotch and it justs gets better as the keybord loose's focus.
I don't think that's what Felber is talking about.
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