# Tag Info

23

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

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 ...

13

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 ...

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 ...

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

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 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$to ... 4 Well, is fruit really in a solid phase? Consider that fruit consists of a lot of water; that water is in a liquid phase prior to blending. Saying that fruit is a solid is like saying a water balloon is solid. When you blend the fruit, all you're really doing is slicing it up. What's left is fragments of fibres, membranes and so on in a liquid-ish suspension ... 4 This paper (http://www-stat.stanford.edu/~cgates/PERSI/papers/dyn_coin_07.pdf) shows that the probability distribution of getting a head, if I toss with the head side up is given by:$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$are the Euler angles. 4 This may not be exactly what you are looking for, but I am going to recommend two specific texts. Misner, Thorne, and Wheeler, Gravitation, Chapters 4, 9, and end of 14 Solidly in the realm of physics but they have a lot of tidbits of interpretation in there. Choquet-Bruhat and DeWitt-Morette, Analysis, Manifolds, and Physics, Chapter IV.C I mean, this ... 3 I can give you one example. Topology plays an important role in chaos theory. http://www.scholarpedia.org/article/Chaos_topology 3 They're two different limits in which two different constants are sent to zero and the resulting limiting theory has different names. However, both of them are limits for dimensionful constants and the analogy is perfect. The properly derived$\hbar\to 0$limit of a quantum mechanical theory is a classical theory – its classical limit – in the very same ... 3 Yes, of course, classical strings are just the$\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 No, it is not possible, and the argument is simple--- there is no dimensional parameter with unit of length, so if there were a stable equilibrium at one radius, there would be many such equilibria obtained by rescaling the original solution to a one-parameter family of solutions. In fact, it is easier to see that the stable solution is for the electron to ... 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 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 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. 3 A closed system can not speed itself up, that's the momentum conservation law which is also the key to your problem. As far as I can see you are implicitly supposing the following three equalities to hold $$E_i+A=E_f\\p_i=p_f\\m_i=m_f$$ where subscripts$i$and$f$stays for the initial (before acceleration) for the final (after acceleration) states ... 3 In short: you have calculated the maximum displacement of the stone, and your friend has calculated the equilibrium displacement. If you place the stone of mass$m$on the spring (which I assume to be of negligible mass), the stone ends up oscillating around the equilibrium point. So the sum of the potential energy stored in the spring and the kinetic ... 2 Short answer: The distance the rock falls in your thought experiment is not the same as the spring compression described in the actual question. Long answer: In your thought experiment, the rock is released from rest and it falls some distance before turning around. So the question arises: For an object of mass$m$released from rest just touching a ... 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 The charge of an atom is defined by its constituent number of protons/electrons and local fluctuations in their density distribution which cause instantaneous dipoles, unless we are talking about ions which have a permanent charge. Charge is a classical concept that has real meaning in classical physics and can be described in various fields ... 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 ... 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 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 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 ... 1 1st: you must take care that friction is a impulsive force and momentum never remains conserved when$f$changes. 2nd:the normal reaction will become$mg+I/\Delta t$where$I$is the impulse imparted on the ball ie:$mv(1+e);v=\sqrt{2gh}\ and\ \Delta t .$is the small time of impact.So, $$N=m(g+v(1+e)/\Delta t)$$ As,the momentum and energy both are not ... 1 The kinetic energy the ball has when it hits the block is$mgh$. We have the relation: $$\frac{mv^2}{2}=mgh$$ This can be rewritten: $$m^2v^2=2m^2gh$$ Which means the downward momentum of the ball is: $$p=mv=m\sqrt{2gh}$$ The block will excert a force on the ball to cancel this momemntum and give it a momentum of$e\cdot p$in the opposite ... 1 Equilibrium will be reached when the net torque on the armature is zero. Since, as we will see below, it will be impossible to have the net torque vanish over an extended interval of time, we'll look for the situation when the torque averaged over time vanishes. The Lorentz force law tells us that for a wire of length vector$\vec{l}$carrying current$I\$ ...

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