Gravitation
Notice: this page is actively under construction
Notice: this page is actively under construction
//Tampan,// Paminggir people. Lampung region of Sumatra, Kota Agung district circa 1900, 45 x 36 cm. From the library of Darwin Sjamsudin, Jakarta. Photograph by D Dunlop.
Tampan, Paminggir people. Lampung region of Sumatra, Kota Agung district circa 1900, 45 x 36 cm. From the library of Darwin Sjamsudin, Jakarta. Photograph by D Dunlop.

Let us describe particle P within a frame of reference F, and recall that for WikiMechanics reference frames are modeled by some enormous set of quarks. Consider some subset of these quarks located in the vicinity of P and generally call them the quarks surrounding P. We use $\mathbb{S}$ to represent this local environment, and employ the symbol $\mathbb{G}$ to note all of the other remaining quarks in F. Then the frame of reference is given by the union

$\sf{F} = \mathbb{S} \cup \mathbb{G}$

We usually assume that the surroundings are just a small part of the frame. Then $\mathbb{G}$ will contain almost the same quarks as F, and $\mathbb{G}$ will have almost the same characteristics as F. Let all these sets of quarks be described by $\ \overline{\kappa} \;$ their wavevectors. Then

$\overline{ \kappa }^{ \sf{F}} = \overline{ \kappa }^{ \mathbb{S}} + \overline{ \kappa }^{ \mathbb{G}}$

frame.png
We almost always assume that a frame of reference is inertial. Then since $\mathbb{G}$ is similar to F, we can reckon that $\mathbb{G}$ is big and heavy too. We could say that $\mathbb{G}$ has gravitas. Definition: $\mathbb{G}$ is called the gravitational component of the frame of reference. And we say that gravitational effects are perfectly negligible if

$\overline{\kappa} ^ { \mathbb{G} } = (0, 0, 0)$

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