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Originally Posted by Zanket
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Originally Posted by papageno
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Originally Posted by Zanket
The equivalence principle implies that the laws of motion for a uniform gravitational field are the same as the laws of motion for a relativistic rocket.
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Can you show us how this implication works? |
There’s a ton of explanations for this on the web. Just google for “equivalence principle” or “principle of equivalence”. For example, this site has some mpeg movies showing what you quoted from the paper. |
The Theory of General Relativity is based on two postulates:
1. General principle of Relativity;
2. Principle of Equivalence,
Now, "
the laws of motion for a uniform gravitational field are the same as the laws of motion for a relativistic rocket" looks like a consequence of the first postulate.
The link you gave talks about the consequence of the second postulate.
Can
you explain how the Equivalence Principle implies that "
the laws of motion for a uniform gravitational field are the same as the laws of motion for a relativistic rocket"?
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Originally Posted by Zanket
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Originally Posted by papageno
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Originally Posted by Zanket
To get to another galaxy within their lifetimes, the crew of a relativistic rocket must effectively exceed the speed of light.
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Why? |
Because another galaxy is further away in light years than the maximum years in a human lifetime, and the crew cannot attain or exceed the speed of light. |
Why does it have to
exceed the speed of light? (See below.)
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Originally Posted by Zanket
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Originally Posted by papageno
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Originally Posted by Zanket
This is doable thanks to length contraction.
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I thought time was the problem.
Isn't it sufficient that the relativistic rocket's speed is close enough to c, so that the travel time measured by the crew is less than their lifetimes? |
Length contraction is what reduces their proper time to another galaxy, as described in section 2. Instead of traversing 2 million light years, say, to get to another galaxy, they might traverse only 20 light years. The 2 million light years reduced to 20 light years thanks to length contraction. |
So, you are looking at it from the point of view of the crew.
Again, why does the crew need to
exceed the speed of light?
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Originally Posted by Zanket
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Originally Posted by papageno
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Originally Posted by Zanket
For instance, when the rocket has accelerated to a velocity of sqrt(0.5) ≈ 0.7071c the crew observes length contraction to a percentage given by eq. 8.13, ≈ 0.7071.
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What happens to the time? |
The crew also observes time dilation, to the same percentage. Passing clocks run at about 70.71% of the rate of the ship’s clocks, as the crew measures. Length contraction and time dilation go hand-in-hand in relativity. |
So, how do yo exceed the speed of light?
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Originally Posted by Zanket
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Originally Posted by papageno
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Originally Posted by Zanket
One light year as they measured before they began accelerating now measures as 0.7071 proper light years along their axis of motion. At 0.7071c a distance of 0.7071 light years is traversed in one year.
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0.7071 ly is the distance as measured by the crew.
The speed of the rocket as measured by the crew is zero.
How do you get one year travel time, and in what frame of reference is this time measured? |
The “as they measured” indicates the crew’s frame. The velocity of the rocket relative to the crew is zero, yes, but the velocity that’s being used here is the rocket’s (or the crew’s) velocity relative to the galaxies in question. That velocity is 0.7071c. At 0.7071c they traverse every formerly-measured light year, now length-contracted to 0.7071 light years, every proper year (a year on their clock).
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Originally Posted by papageno
In the frame of reference of the gantry, wouldn't they take ~1.4142 years to cover 1 ly at 0.7071c speed?
Wouldn't it take ~0.293 years from the point of view of the crew?
(Of course I could have made mistakes in my calculations.)
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In the gantry’s frame, yes, it’ll take the rocket (1 light year / 0.7071c) = ~1.4142 years to traverse a light year. How did you get ~0.293 years for the crew’s frame? |
I got confused in the calculations.
I should be more careful.
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Originally Posted by Zanket
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How did you obtain those numbers?
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In the crew’s frame, moving at 0.7071c relative to the galaxies in question, one formerly-measured light year (that is, before they began accelerating, or in the gantry’s frame) is contracted to a percentage given by eq. 8.13, = sqrt(1 – (0.7071c)^2) = ~0.7071. 1 formerly-measured light year * 0.7071 = 0.7071 light years. The crew traverse 0.7071 light years in (0.7071 light years / 0.7071c) = 1 proper year (a year on their clock). |
Hang on.
Shouldn't you get:
[x^2 - (ct)^2](gantry's frame) = [x'^2 - (ct')^2](rocket's frame) ?
left: (1 ly)^2 - (c * x/v)^2 = {1 - (1/0.7071)^2 } ly^2 = 1 ly^2,
right: (0.7071 ly)^2 - (c * 1 y)^2 = 0.5 ly^2;
I don't think you are comparing the correct intervals.
If we assume that left equals right, we obtain t' = 1.225 y, the time interval to pass each signpost at one lightyear separation (in gantry's frame).
I hope posters with more experience in Relativity can help.