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I can give a rough order-of-magnitude estimate, but a numerical simulation is required to give an actual answer. The sun is extremely well described as an ideal gas[1]. The total entropy is roughly [math]k_B[/math] per degree of freedom, where [math]k_B[/math] is Boltzmann's constant, giving a total entropy of roughly [math]S\approx 10^{57} k_B[/math] [2]. Now, the sun is in its main sequence core Hydrogen-burning phase, so the relevant timescale for entropy change is roughly the main sequence lifetime, which for a [math]1 M_\odot[/math] star is roughly [math]10^{10} yr\approx \pi\times 10^{17} [/math][math]s[/math][math][/math]. Putting this together gives a rough estimate of the rate of change of entropy in the sun at[math] \dot{S}\approx 10^{47} k_B yr^{-1} .[/math]This might seem large, but perhaps one should instead consider the fractional entropy change,[math] \frac{\dot{S}}{S}\approx 10^{-10} yr^{-1}.[/math]Again, let me reiterate that a proper stellar structure and evolution code is needed to get a real estimate.[1] Better, a bunch of layers of ideal gasses in hydrostatic equilibrium.[2] Since the total number of particles is roughly [math]N\approx M_\odot/m_H\approx 10^{57}[/math] where [math]M_\odot\approx 10^{33} g[/math] is the mass of the sun, and [math]m_H\approx 10^{-24} g[/math] is the mass of a Hydrogen atom.

Just that: the arithmetic seems right but the physics is wrong. An awesomely popular scenario.Problem:A material point of mass [math]m[/math] is attached to each extremity of a massless rigid rod of length [math]l[/math] which, initially, is standing still on a frictionless floor while leaning against a frictionless wall that is orthogonal with the floor (a mysterious, unspecified and irrelevant force is holding the contraption in place).Eventually the rod’s bottom edge, name it [math]B[/math], starts moving with a constant velocity [math]\vec{v}[/math] parallel to the floor (via the application of some mysterious and a very relevant and a very interesting force, name it [math]\vec{F_v}[/math]). Meaning that the rod’s bottom edge [math]B[/math] remains on the floor at all times.Question: (assuming that the entire system is at sea level, is subject to constant gravity [math]g[/math] and [math]v << c[/math]) what will the velocity of the rod’s top edge, name it [math]A[/math], be when it hits the floor?Incorrect Solution:More often than not students solve this problem by using the Pythagorean theorem. How hard could that be?We simply note that in the natural coordinate system for the rod’s bottom edge at [math]x[/math] and for the rod’s top edge at [math]y[/math] we have:[math]x^2 + y^2 = l^2 \tag{1}[/math]From where trivially:[math]y = \sqrt{l^2 - x^2} \tag{2}[/math]You have got to be kidding, right?Differentiating both sides of (2) over time once (hey, we know derivatives), we get:[math]\[/math][math]dot[/math][math] y = -\dfrac{x\cdot\dot x}{\sqrt{l^2 - x^2}} \tag*{}[/math]But[math]\[/math][math]dot[/math][math] x = v \tag*{}[/math]Hence:[math]\[/math][math]dot[/math][math] y = -\dfrac{x\cdot v}{\sqrt{l^2 - x^2}} \tag{3}[/math]Taking the limit of both sides of (3) as [math]x \to l[/math] (hey, we know limits), we find:[math]\displaystyle \[/math][math]dot[/math][math] {y_l} = -\lim_{x\to l}\dfrac{x\cdot v}{\sqrt{l^2 - x^2}} = -\infty \tag*{}[/math]Hm.Oh, OK. We know - so Einstein was wrong. Yep.See?The arithmetic is right but the physics is wrong. So wrong - it is not even funny.As we noted above this is a very popular scenario with the solutions of physics problems. Students often do not discern the physical essence of a problem at hand but plow with juggling some formulas right through anyway.The physical essence of the above problem, however, tells us that at some point the top edge of the rod - the point mass [math]A[/math] - will separate from the wall.Prior to that separation the rod’s center of mass (shown as red points below) will execute the circular motion:The point in time when the rod’s center of mass hops off the circular arc (shown in red above) corresponds to the event of the rod’s top edge separation from the wall.If you can stomach it then the full and detailed solution of this problem is presented in this Quora blog post and the correct answer comes out to be a very finite number, namely:[math]\sqrt{v^2 + 3(glv)^{\frac{2}{3}}} \tag*{}[/math]Gee, Einstein was right after all?The separation angle [math]\theta_s[/math] - the angle which the rod forms with the floor at separation time - comes out to be:[math]\sin^3 \theta_s = \dfrac{v^2}{gl} \tag*{}[/math]and the rod’s internal tension force [math]\vec{T}[/math], the reactionary force of the top edge [math]\vec{N_a}[/math] and the mysterious constant velocity force [math]\vec{F_v}[/math] as the functions of the angle [math]\theta[/math] are given by:[math]T(\theta) = \dfrac{m}{\sin \theta}\Big(g - \dfrac{v^2}{l\sin^3\theta}\Big) \tag*{}[/math][math]N_a(\theta) = m\cot \theta \Big(g - \dfrac{v^2}{l\sin^3\theta}\Big) \tag*{}[/math][math]F_v(\theta) = m\cot \theta \Big(g - \dfrac{v^2}{l\sin^3\theta}\Big) \tag*{}[/math]and we see that when the angle [math]\theta[/math] reaches the magnitude of [math]\theta_s[/math] the forces [math]N_a, T[/math] and [math]F_v[/math] vanish - this time, however, in a mathematically sound and consistent fashion. Then, the force [math]N_a[/math] is gone for good but the vectors of [math]T[/math] and [math]F_v[/math] flip around and point in the opposite direction. The rod itself transitions from the state compressed, into neutral, into stretched and so on.Conversely.Just as often students may understand the underlying physics of the problem and get it right but - forget about the arithmetic being wrong - they may not even be equipped with enough tools to proceed mathematically.Problem (on, say, kinematics):A point executes a motion along a parabola[math]y(x) = kx^2, \; k > 0 \tag{4}[/math]in such a way that its acceleration vector [math]a[/math] remains parallel to the [math]y-[/math]axis at all times.Find: the point’s tangential and normal accelerations [math]a_t[/math] and [math]a_n[/math] as the functions of time.Solution:From (4) it follows that:[math]\[/math][math]dot[/math][math] y = 2kx\dot x \tag{5}[/math]And from (5) it follows that:[math]\ddot y = 2k\dot x^2 + 2kx\ddot x \tag*{}[/math]But we are told that[math]\ddot x = 0 \tag*{}[/math]Hence:[math]\ddot y = 2k\dot x^2 = a \tag{6}[/math]From where:[math]\[/math][math]dot[/math][math] x = \sqrt{\dfrac{a}{2k}} \tag{7}[/math]Integrating both sides of (6) once over time, we find:[math]\[/math][math]dot[/math][math] y = at \tag{8}[/math]if we agree to take the initial [math]y-[/math]velocity to be zero.Let [math]s[/math][math][/math] be the length of the arc travelled so far:[math]\[/math][math]dot[/math][math] [/math][math]s[/math][math] = v = \sqrt{\dot x^2 + \[/math][math]dot[/math][math] y^2} \tag{9}[/math]With (7) and (8) we have for (9):[math]\[/math][math]dot[/math][math] [/math][math]s[/math][math] = v = \sqrt{\dfrac{a}{2k}}\cdot\sqrt{1 + 2kat^2} \tag{10}[/math]Next we recover the radius of the curvature [math]R[/math] on this particular arc:[math]R = \dfrac{\dot s^3}{\dot x\cdot \ddot y - \ddot x \cdot \[/math][math]dot[/math][math] y} \tag*{}[/math]which with (7, 8, 10) becomes:[math]R = \dfrac{\sqrt[3]{1 + 2kat^2}}{2k} \tag{11}[/math]Lastly, with all the ducks in a row:[math]a_t(t) = \ddot [/math][math]s[/math][math] = t\sqrt{\dfrac{2ka^3}{1 + 2kat^2}} \tag{12}[/math][math]a_n(t) = \dfrac{\dot s}{R} = \dfrac{a}{\sqrt{1 + 2kat^2}} \tag{13}[/math]Problem solved.Compare the texture and the fabric of the two problems presented: in the first problem a fair amount of thought must be applied before we start juggling the formulas while in the second problem we are expected to whip out a solution in a small number of minutes without stopping much to think.

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