BurtJordaan
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Why is relativity so hard to learn?[10X]
May 5th, 2017, 12:13 am

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OK, back to Alice's mission. Her plan has changed a little and she will now first stop in the region of Alpha-C (AC) for some observations. It is way too late to achieve the rocket braking at the original very low (0.075g) acceleration, so Alice asked her INS computer to get them into a 'parking' orbit at a more comfortable 1g retro-burn. Her INS and my trusted spreadsheet 'know' that at -1g, it will take 0.69 years to get the coasting ship to zero velocity relative to Bob, and that it must start to brake 0.2 lyrs (on their rotated map) short of the parking position. At just over 11 years into her mission, Alice comes to rest in Bob's spacetime structure again.

During the 0.69 years of deceleration, Alice's tilted structure has gradually rotated back to align with Bob's structure once again. While their structures were at an angle to each other, space and time were different for them. Here is a summary of the space and time accounting columns:

Stage/event _______ T_Bob yr __ D_Bob lyr...

[ continued ]

Last edited by BurtJordaan on May 6th, 2017, 12:43 pm, edited 3 times in total.
0 Comments Viewed 145 times

This is a "whiteboard dump" for a post following on part [9] of a Philosophy of Science thread with the above name. I have thought to move things deemed too technical here for reference (As sort of Appendices).

BurtJordaan » 04 May 2017, 11:44 wrote:That said, in the next piece we will make Alice turn around en head home, and provided that her long-playing rocket has enough fuel left, make it little swifter.

As Mitchell has indicated above, there is a slight possibility that a future super-dooper technology could perhaps bring Alice back home with the amount of fuel that she has left. As we also discussed, if you have the technology, a higher acceleration would be better, so let us look at a one earth gravity (1g) setting ...

[ continued ]

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In the last installment I attempted to cross the bridge from Special relativity to General relativity by means of things that are accelerated. We will now let gravity do the acceleration and see what it implies in simple situations.

Relativistic gravity is mostly about spacetime geodesics - the shortest path (or rather path of least action) that particles, objects or whatever, take through spacetime. When gravity is present, those paths are curved in most coordinate systems. Pictured below is a two-dimensional cut of a spacetime geodesic ($c\Delta t'$) in quasi-Newtonian spacetime.[1]

At the origin sits a large mass M and a small object is released from rest at position $r = r_0$. The object is moving upward at a "speed $c$",...

[ continued ]

Last edited by BurtJordaan on December 28th, 2014, 6:30 am, edited 3 times in total.
1 Comment Viewed 1309 times

After wrapping my head around the biggest issues of special relativity theory, I obviously had to move on to general relativity, where gravity takes the center stage. Einstein used uniformly accelerated frames to 'build a bridge' between special relativity and gravity. He established that a uniformly accelerated frame is precisely equivalent to a hypothetical 'uniform gravitational field' - a thing that does not really exist. However, reasonably small stretches of space and time (in a 'real gravitational field') will very, very closely resemble a uniformly accelerated frame.

Engineers normally understand gravity pretty well in terms of how to calculate it and use it in their designs, be it a tall building or the trajectory of a missile or a rocket. They understand that gravity is 'stronger' at ground level than at altitude, and they understand the equivalence between acceleration and gravity in broad terms. But, tell them that the tail end of a uniformly accelerated frame in free space...

[ continued ]

0 Comments Viewed 887 times

I'm trying to understand Lincoln's values for frame C. So I added some values from GrayGhost's Minkowski diagram for essentially the same (ABC) scenario:

With $\large v= 0.866c$, $\large L/v =4$ and $\large \gamma = 2$:

Lincoln's transformations with Gray's values calculated:
$\large (x,t)_{1A} = (0, 0)$
$\large (x,t)_{2A} = (L, \frac{L}{v})=(3.464,4)$
$\large (x,t)_{3A} = (0, \frac{2L}{v})=(0,8)$

$\large...$

[ continued ]

Last edited by BurtJordaan on December 23rd, 2014, 12:45 am, edited 4 times in total.
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