By Jaynes E.T.
We notice the details of historical past? as a framework on which to hold many history feedback in regards to the nature and motivation of Bayesian/Maximum Entropy equipment/ event has proven that those are wanted with a purpose to comprehend contemporary paintings and difficulties. A extra whole account of the background? with many extra info and references, is given in Jaynes ( 1978).The following dialogue is largely nontechnical: the purpose is barely to show a bit introductory "feel" for our outlook, function, and terminology, and to alert newbies to universal pitfalls of bewilderment.
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Extra resources for Bayesian methods
1 1µm 1Å ➤ A body executing Brownian motion. 1 Calculate, for the above data, the velocity which a molecule of typical velocity can impart to the body in a head-on elastic collision. Solution on page 88 mean free path Another scale that characterizes the situation is the distance that the body can traverse between two collisions. This distance is called the mean free path, or simply the free path (see also Chap. 3). In order to ﬁnd the orders of magnitude of the free path, we apply the following consideration: if the body were to move, parallel to itself, a distance L (Fig.
Cross section 52 Ch. 3 Transport Coeﬃcients v ➤ θ A ➤ –z Fig. 2 A molecule with velocity v will strike an area A of the side of the container. Actually we almost calculated this quantity on our way to Eq. 3). The number of molecules striking an area A during time ∆t and whose velocity component normal to the surface is vz , was found in Sec. 3) where n(vz ) is the density of molecules of velocity vz . Notice that what we called the x direction in Sec. 1 we here call z. In order to ﬁnd the total number of molecules that strike the surface, we have to sum over all the values of the velocity vz such that vz > 0, since molecules for which vz < 0 are moving away from the surface and will not strike it.
P(z+dz) ➤ ➤ ➤ ➤ g }∆z P(z) ➤ z Fig. 5 Gas inside a container residing in a gravitational ﬁeld. 10 Calculate the change in height at the earth’s surface that will cause the gravitational ﬁeld to change by a thousandth of a percent. What is the ratio between this distance and the average intermolecular distance in a gas at standard conditions? Solution on page 81 Thermal equilibrium guarantees that the velocity distribution is identical in all the layers; however, due to the gravitational force the pressure diﬀers at each height, since the pressure of the gas is determined by the weight of the gas above it.