[ExI] physics

[John Clark]

On Mon, Apr 25, 2016  Anders Sandberg <anders at aleph.se> wrote:

​>> ​
>> ​If "real" means something that's invariant then curved spacetime is not
>> real because it looks different for different observers.
>
>
> ​> ​
> Nope. I was using curvature here to denote the metric tensor (or, if you
> want to go deep, the Riemann tensor). Tensor equations are
> observer-invariant.
>

​Yes, tensors measure how far a curve is from a geodesic for a given
manifold and all would agree on that, but they won't agree on what manifold
they want to use, many different ones and their associated tensors can
correctly describe how things move. Everyone will agree on what the minimum
length of a spacetime interval between 2 points will be but lots of
different manifolds are consistent with that. Equations are just
descriptions of physical reality and a gravitational force in flat
spacetime is equivalent to curved spacetime and no gravitational force, so
just use the manifold that's most convenient.


> ​> ​
> Geodesic curves for one observer are geodesics for all others.


​A ​
geodesic
​ is the shortest path between 2 points and all observers will agree about
that, and any object that is not accelerating will follow a geodesic
through 4D spacetime, and
a straight line is not the shortest path between 2 points in curved
spacetime.
​ So​
if you're
 in a closed box and feel a 1g acceleration
​pushing your feet to the floor of the box ​then
you know that you can't be on a Geodesic curve
​; but how can you know if you're in a rocket in intergalactic space in a
flat spacetime manifold​ following a curved timeline or if you're in a
curved spacetime manifold sitting on the surface of the earth following a
straight timeline? Both will produce the same correct description of the
way things move so just use the one you find easiest to calculate.


Well OK, you really could tell the difference because the guy on the
surface of the earth would feel tidal forces and the guy in the rocket
accelerating at 1g in deep space would not, but by making the man and the
box arbitrarily smaller the tidal force will be arbitrarily smaller.

​> ​
> So while you might measure using a different coordinate system from me and
> get different numbers, there is a simple rescaling between our results.
>

​I agree and that rescaling factor is gravity
, although​

​I've never heard ​
​​
tensor equations
​ being referred to as simple before.


> ​> ​
> How observers see things is irrelevant:
>

​Not to the observers it isn't.​

​

 John K Clark​
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