BurtJordaan
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Why is relativity so hard to learn?[10X]
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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 $c$", but at the same time it is following a curve with its center of curvature on the opposite side of the mass, labelled $cc$.

The distance from the object to $cc$ is completely determined by mass $M$ and the distance of the object from the mass, together with constants $G$ and $c$, of course. It determines the radius of curvature |$R$|. As the object moves closer to the mass (in the -x direction), the curvature center cc will also move closer to the mass (in the +x direction). The radius of curvature will decrease and the curve will become 'tighter' (more curvature) as it gets closer to mass M. The 'speed' along the curve stays at $c$.

The object will move faster and faster in the spatial (-x) direction and at the same time it will be moving slower in the time (+ct') direction. Because of the the curved path, it can be loosely said that it "trades time-movement for space-movement". Quasi-Newtonian gravity sports gravitational- and velocity time dilation. It thus has spacetime curvature, but it lacks the curved space of General Relativity.

Before taking that extra step, we will do a few calculations for the gravity of Earth, using the above principles. If we take for Earth: mass as M=6E24 kg, surface radius r0=6.378E6 meters, G=6.67E-11 m3 kg-1 s-2, we get a radius of curvature |R| ~ 9.15E15 meters.[2]. Now all we need to do to get Earth's surface gravity is to divide that number into c2, i.e. $g = c^2/ 9.15E15$ = 9.83 m/s2.

This looks super-complicated just to get Earth's surface gravity and in a way it is. We do not need 'quasi-Newtonian' gravity for anything around Earth. The truth is that for Earth's low gravity, that factor $(1-GM/(r_0c^2))$ only differs from unity by one part in a billion, so it can be ignored. In the next Blog entry we will see where it does make a significant difference.

--
Regards
Jorrie

[1] Quasi-Newtonian spacetime has flat space, but curved spacetime. The definition used here is my own, but I guess there are many different meanings of the term - anything that's not Newton, not Einstein, but something in-between. Figure 4-4 is from Relativity 4 Engineers, but there is a unit-inconsistency in the equation shown in the diagram, where the first M should also have been multiplied by the factor G/c2. The equation should rather have read:

$|R|=r_0^2c^2/(GM) \times (1-GM/(r_0c^2))$

[2] Paste into Google search, without the quotes:
'6.378E6^2*3e8^2/(6.67E-11*6E24)*(1-6.67E-11*6E24/(6.378E6*3e8^2)) in meters'
and you should get the answer |R| ~ 9.15E15 in meters. To get the final answer 'g', just paste 'c^2/9.15E15 meters' without the quotes into Google search. One can sometimes get away without having to enter the values of constants G and c, but just the symbols. However, it does sometimes not work so nicely, especially if the constants appear multiple times in the formula.

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