Tag Info

Hot answers tagged

34

The error is that you assume that the density distribution is "nearly spherically symmetric". It's far enough from spherical symmetry if you want to calculate first-order subleading effects such as the equatorial bulge. If your goal is to compute the deviations of the sea level away from the spherical symmetry (to the first order), it is inconsistent to ...


23

Arnold Neumaier's comment about statistical mechanics is correct, but here's how you can prove it using just thermodynamics. Let's imagine two bodies at different temperatures in contact with one another. Let's say that body 1 transfers a small amount of heat $Q$ to body 2. Body 1's entropy changes by $-Q/T_1$, and body 2's entropy changes by $Q/T_2$, so ...


20

Nature has no preferences, and therefore entropy tends to increase. Sounds paradoxical? The point is that each microscopic state (describing the exact position and velocity of each atom in the system) is equally likely. However, what we typically observe is not a micro state, but a course-grained description corresponding to incredibly many micro states. ...


17

To make it fall you need a torque. This torque is provided by the weight force acting on the center of mass of the object and by the offset between the center of mass and the edge of the object. Imagine your domino standing upright then tilt it. You are moving the center of mass. When the center of mass (blue) is on the right of the edge (red) then you have ...


16

From a fundamental (i.e., statistical mechanics) point of view, the physically relevant parameter is coldness = inverse temperature $\beta=1/k_BT$. This changes continuously. If it passes from a positive value through zero to a negative value, the temperature changes from very large positive to infinite (with indefinite sign) to very large negative. ...


13

Here we would like to calculated analytically Lubos Motl's solution to the first order in the flatness parameter $f$, $$0<f:=1-\frac{b}{a}\approx\frac{a}{b}-1 \ll 1, $$ where $a$ and $b$ are the equatorial and polar radius of the Earth, respectively, and $a>b$. (The $\approx$ symbol will from now on mean equality up to higher-order terms in $f$.) We ...


10

There was some doubt about Lubos' answer (which I've accepted), so this is just a verification. I copied the method Lubos described and found the potential difference for an ellipsoid with different eccentricities. Sure enough, for an oblate spheroid, if you make the center-equator distance a fraction $e$ larger than the center-pole distance, the ...


10

If you made the most perfect cone possible, so that its tip was a single atom, and stood it on the most perfect surface possible (a perfectly smooth, perfectly hard sheet of atoms), and completely removed all forces other than gravity, it would still topple. This is because those atoms are all jiggling around due to thermal motion. This effect fundamentally ...


7

The forces are never balanced, as there is only ever one force - gravity. The key is to remember Newton's second law: $F = ma$. Force and acceleration are paired, not force and velocity. Knowing just an object's current velocity tells you nothing about what forces are acting on it. There are two ways to see how the velocity goes to zero. Either the initial ...


7

The excerpt from the text forgets to mention that you assume Local Thermodynamic Equilibrium, and not full Thermodynamic Equilibrium, so to make it possible to define point to point (or from region to region) an EoS. If there is no sense of being 'close' to thermodynamical equilibrium, it is simply impossible to talk about EoS, pressure and the like from ...


6

Some engineering texts use "moment" and "couple" to talk about forces that tend to rotate an assembly (what physicist mean when they say "torque", but the engineers sometimes have a slightly different meaning for that word). A roughly translation guide is... A "couple" is a pair of opposite forces whose points of action are not co-linear. A couple is ...


6

If a plane is flying without any rudder input, then the banner will always fly straight behind the plane, with nose-tail-banner in a straight line, no matter what the speed or direction of the plane and/or wind. The only thing that affects the plane and banner is the flow of air over the control surfaces. How would the banner know that there was wind ...


5

Other than User58220's answer, I'm reading a lot of nonsense here. When you fly an airplane (I and many other people on this site do), when you are cruising in the air, you center the rudder. The plane has no awareness of the movement of the air mass over the ground (wind). The plane has a vertical stabilizer (tail) which causes it to point into its ...


5

It takes a lengthy proof, but Lyman Spitzer shows in the second chapter of Physical Processes in the Interstellar Medium (the standard text in interstellar matter studies) that the velocity distribution of interstellar gas particles (which is what forms nebulae) is very nearly Maxwellian - the deviation is less than 1%. Other larger systems, probably not so ...


5

You need to read this paper by Jaynes. I can't explain it as well as him, but I will try to summarise the main points below. The first thing is to realise that the entropy is observer-dependent: it depends on what information you have access to about the system. A finite temperature means that you don't have access to all the information about the state of ...


5

Strictly speaking there are no reversible processes in Nature; it is an idealization that enables one to get bounds on efficiency of nonequilibrium processes by using techniques of equilibrium thermodynamics only. A reversible process is therefore primarily a theoretical concept for discussing what would happen in a process if dissipation were absent. It ...


5

Using the method of images, you can calculate the force between the ring of charge and the sphere. Assume the sphere is on the z axis with it's center on the point $z$, a radius of $R_s$ and the ring's radius is $R_r$ with a charge density $\lambda$. So $z$ denotes the center of the sphere. To calculate the force, you can replace the sphere with a charged ...


5

If you initially give to the bob a velocity $\sqrt{4rg}$, it will actually take an infinite time for the bob to reach the top! A little lesser velocity will cause the bob to stop earlier and come back toward the initial point, while a little greater one will take the bob over the top (the motion will continue, with increasing velocity, to the other side). ...


4

To answer your question, you should first understand when is a system most stable. Firstly it shouldn't have a tendency to move or change state, thus it should be under equilibrium conditions, i.e. the net Force should be zero. We know that $$F = - \frac{dU}{dx}$$ Putting $F=0$, we get $$\frac{dU}{dx}=0 \tag{1}$$ Secondly, it should be able to maintain ...


4

"Equilibrium" means thermal equilibrium. The solid has one well defined temperature, and a constant Fermi energy. The Fermi energy is an energy value against which energy levels are compared to determine how fully occupied (or not) an energy level is. Generally when the Fermi level is constant throughout a solid electrons diffuse equally in all ...


4

No. You are wrong. Even if we exclude drag, the force of gravity does not suddenly drop to zero. The force acts to continuously accelerate the ball downwards. Because it starts off with a velocity upwards this will be reduced because of this force. At some point it will be zero as the velocity reverses sign, but there is no equilibrium as you can prove by ...


4

People don't immediately compress because the body is more or less a pressure vessel. It's not a very good pressure vessel for dealing with vacuum, but it's something. It's your body's resistance to pressure that lets you do things like spray bodily fluids (from your mouth, or from your bladder, or from your arteries). When my wife was in labor with my son, ...


4

First of all, let me note that it is misleading to say that we're not even dealing with a particle We are still dealing with a particle, but the state of the particle at a given moment in time is no longer described by a pair $(x(t),p(t))$ of position and momentum in a classical phase space, it is instead a state $|\psi(t)\rangle$ in a Hilbert space. ...


4

Start with the unperturbed gravitational potential for a uniform sphere of mass M and radius R, interior and exterior: $$ \phi^0_\mathrm{in} = {-3M \over 2R} + {M\over 2R^3} (x^2 + y^2 + z^2) $$ $$ \phi^0_\mathrm{out} = {- M\over r} $$ Add a quadrupole perturbation, you get $$ \phi_\mathrm{in} = \phi^0_\mathrm{in} + {\epsilon M\over R^3} D $$ $$ ...


4

Think of the layer of skin on the finger tip which is in contact with the object that has a lower temperature than the body. If the object has a high heat capacity and a high thermal conductivity (like metal), then the skin of the finger will come to an equilibrium temperature that is lower than if it is in contact with an object that has a lower heat ...


4

You gave an expression for the force, whereas Prathyush's answer treats it as the potential. If that expression really is the force, then the answer is different. You don't really need to do any calculation in this case to see what is going on. Clearly $F(0) = 0$, so this is indeed a point of equilibrium (the question should probably read something like ...


3

In this answer, I will present a framework to use, and then I will frame the prior answers within that framework. Let me sum up the values we have here. I'll use the same notation (as best as possible) as everyone else and Wikipedia for an oblate spheroid where $a$ is the large, equatorial, radius. Mark1, method in the question, $2 (a-b) = 21.6 km$ ...


3

You may demonstrate numerical friction by applying a Runge-Kutta method (or even the Euler method if RK is too advanced) to a conservative system (such as Sun - Jupiter - Saturn) and notice the dissipative effect of stochastic perturbances (aka discretization errors): Ultimately, the planets will fall into the sun, though this takes many revolutions. ...


3

Since at higher altitudes, the air pressure is lower, the boiling point of water decreases, since it's easier for the energy insde the water to get free. When A liquid starts to boil, you reach a critical point where the liquid loses a lot of heat, much more than when not boiling, thus requiring much more energy for the same increase in temperature, and ...


3

First, if ${\rm d}S\neq 0$, then the entropy will change, and because something is changing, it's obviously not an equilibrium. If the physical system doesn't maximize the entropy and it's composed of many parts that may interact with each other, directly or indirectly (it's not disconnected), then any path that allows the entropy to be increased (given ...



Only top voted, non community-wiki answers of a minimum length are eligible