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20

Topology is of fundamental importance even to systems in classical mechanics. The configuration space (or phase space) of a generic classical mechanical system is a manifold and manifolds are topological spaces with some extra structure (e.g. a smooth structure in the case of smooth manifolds). At the very start of any classical mechanics problem, you need ...


13

As it is, differential forms don't tell you the whole story--strictly speaking, differential forms only deals with covectors and wedge products of covectors and then uses the hammer of the Hodge star to be able to clumsily do inner products. To me, it is too far removed from the vector calculus you may already know. Instead, I strongly urge you to look ...


13

This was something that confused me for awhile as well until I found this great set of notes: homepages.physik.uni-muenchen.de/~helling/classical_fields.pdf Let me just briefly summarize what's in there. The free Klein-Gordon field satisfies the field equation $(\partial_{\mu} \partial^{\mu} +m^2) \phi(x) = 0$ the most general solution to this equation ...


11

I would really recommend the book by Frankel, The Geometry of Physics. He deals with all the fundamental concepts of topology and differential geometry, but gives clear and detailed applications to classical mechanics, electromagnetism, GR and QM. He is not too formal, but develops really a lot of useful tools using differential forms. Another book, which ...


6

The particles that communicate the Weak interaction, i.e W Bosons and Z bosons are massive. So unlike Electromagnetism which is communicated by massless particles(Photons), the weak interaction has a very short range. For Massive particles the Potential of interaction falls as $V(x) = -K \frac{1}{r} e^{-m r} $ The range of this force is approximately ...


6

There is a very easy way to see this and it is through an $\hbar$ series. This claim can be traced back to Sydney Coleman and states that in the ultraviolet one is doing an expansion with $\hbar$ going to zero. A previous answer cited these lectures on classical fields but I would like to start from the generating functional of the scalar field theory and ...


6

There has to be a few assumptions. Let's assume we are talking about a linear plane wave in relatively deep water. Because the the case where the bottom comes into play the upward hydrostatic force distorts the wave. Picking deep water or insuring the relative depth of d to L (d is average water depth and L is the wavelength of the wave) is $d/L > ...


5

If you stick to gases then things are relatively straightforward because the temperature is related to the relative velocity of the gas molecules, that is the velocity of the gas molecules relative to each other. If you put your canister of gas in a fast moving (but non-relativistic) rocket moving at some velocity $v$ then you add the same velocity $v$ to ...


4

This may not be exactly what you are looking for, but I am going to recommend two specific texts. Misner, Thorne, and Wheeler, Gravitation, Chapters 4, 9, and end of 14 Solidly in the realm of physics but they have a lot of tidbits of interpretation in there. Choquet-Bruhat and DeWitt-Morette, Analysis, Manifolds, and Physics, Chapter IV.C I mean, this ...


4

This paper (http://www-stat.stanford.edu/~cgates/PERSI/papers/dyn_coin_07.pdf) shows that the probability distribution of getting a head, if I toss with the head side up is given by: $p(ψ, φ) =\frac{1}{2}+\frac{1}{\pi} \sin^{-1} (\cot(φ) \cot(ψ))$ if $(\cot φ)(\cot ψ) ≤ 1$, =1 if $\cot(φ) \cot(ψ) ≥ 1$ where $\phi, \psi$ are the Euler angles.


4

Well, is fruit really in a solid phase? Consider that fruit consists of a lot of water; that water is in a liquid phase prior to blending. Saying that fruit is a solid is like saying a water balloon is solid. When you blend the fruit, all you're really doing is slicing it up. What's left is fragments of fibres, membranes and so on in a liquid-ish suspension ...


3

Yes, of course, classical strings are just the $\hbar\to 0$ limit of "ordinary strings" and they do interact although the rate goes to zero in the limit, too. The local picture of the interactions used for "quantum strings" should be interpreted literally and it does allow strings to interact even if they're "classical": On the picture, you see the "type ...


3

I can give you one example. Topology plays an important role in chaos theory. http://www.scholarpedia.org/article/Chaos_topology


3

They're two different limits in which two different constants are sent to zero and the resulting limiting theory has different names. However, both of them are limits for dimensionful constants and the analogy is perfect. The properly derived $\hbar\to 0$ limit of a quantum mechanical theory is a classical theory – its classical limit – in the very same ...


3

As a perhaps "softer" book suggestion, I am finding "The Road to Realty" by Penrose to be quite nice in getting an overview of many mathematical interpretations of physics. Also, there are exercises in this book so although you feel like you're buying some sort of coffee table book, there is plenty to work through if you are willing.


3

OP considers an equations of motion of the form $$\tag{1}\dot{\bf x}~=~{\bf B}({\bf x}),$$ where the vector field ${\bf B}$ is of the form$^1$ $$\tag{2} {\bf B}~=~{\bf \nabla}\times {\bf A}.$$ In other words, ${\bf B}$ is divergence-free $$\tag{3} {\bf \nabla}\cdot {\bf B}~=~0.$$ Eq.(3) is locally eqivalent to eq. (2), cf. Poincare's Lemma. Let ...


3

I think you can apply Euler Bernoulli beam theory. This means that the highest stress should take place closest to the wall. Why a tree branch does not break there is because it gets thicker closer to the trunk spreading the load over more material.


2

Yes, when people talked about the thermodynamic limit, they are always refer to the limit of large number of particle, or $N \to \infty$. The role of $k_B$ is to link the microscope quantities and macroscopic thermodynamic quantities. In Newtonian Mechanics, we have already defined the momentum and kinetic energy very clear. On the other hand, Thermodynamic ...


2

1) Usually, but not always, the word classical in physics refers to the limit $\hbar\to 0$. 2) Perturbative string interactions of open and closed strings are governed by the open and the closed string coupling constants, respectively, which are independent of Planck constant $\hbar$.


2

The classical analogue of quantum $\Phi^4$ theory is classical $\Phi^4$ theory, with the same action. There are no particles, but there is still scattering of waves! The correspondence between tree-level QFT and classical fields is on the level of fields only. (Particles make their appearance in classical field theory only in the limit where geometric optics ...


2

John Harrison's marine chronometer H4 did in fact solve the longitude problem with 18th century technology although the OP is I think wondering if it could be done without constructing a "sufficiently accurate" marine chronometer. H4 was first tested at sea, on HMS Deptford, from November 18 1761 to January 21 1762 and lost only five seconds, which is better ...


2

Screening sounds like it should help, but remember that screening, too, is a form of electron-electron interaction. I think ultimately it comes down to the remarkable results of Fermi liquid theory, which is that even once you take into account e-e interactions you still have electron-like quasiparticles moving in an electron-like way, and scattering off ...


1

I think photoelectric effect is a good example. Before formal quantum mechanics(e.g. Schrodinger's equation) was developed, it was believed the effect was due to the quantum nature of light. However, just using Schrodinger's equation+perturbation theory+classical EM wave it is sufficient to demonstrate the existence of photoelectric effect(the electron is ...


1

Your premise violates Newton's first law of motion: If there is no net force on an object, then its velocity is constant. The object is either at rest (if its velocity is equal to zero), or it moves with constant speed in a single direction. For an object (a body) to be accelerating there must be an external force applied. One of the reasons for ...


1

New answer What you've done here is just dimensional analysis. But you've gone a little too far. In particular, just because two things have the same dimensions doesn't mean that they are equal. If you want to expand $F/A$ a little more, you can choose your favorite from \begin{equation} F = \frac{d p}{dt} = \frac{d}{dt}(m\, v) = m\, \frac{d}{dt} v = m\, ...


1

I will give it a shot. Spoiler: I did this in the body frame so that the moment of inertia is time independent, before you get excited... Starting with Euler's equations: $$ I_i\dot{\Omega}_i+(I_j - I_k)\Omega_j \Omega_k = 0 $$ and taking cyclic permutations of $i,j,k$ to get the three of them; and in the absence of torques (I ignore air friction). It's a ...


1

There are classical systems without trajectories with the particles 'going through' all possible classical paths. Check for instance Poincaré resonances and the limits of trajectory dynamics. The concept of trajectory is an approximation both in quantum and classical mechanics (check above ref.); we recover trajectories when the states are localized $\sigma ...


1

The solution is not as simple as you wrote, it is a sum over discrete $\lambda_n$. To get the right solution, you first introduce a shifted $u$: $u'=u-T_0$. For $u'$ you will get $B=0$ and an equation for finding the discrete spectrum of $\lambda$. Then you make a superposition with different $A_n$ and make it obey the initial condition. This permits to ...


1

One way for finding the field of a magnet is to model it (as a polarized material inside volume $V$ ) with magnetic dipoles , as lots of dipoles near each other , and then sum the produced fields of all dipoles at the desired point. To find the field of a dipole, You can model it as two (to date, fictitious) magnetic monopoles and use coulomb force law ...



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