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14

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 ...


12

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 ...


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

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 > ...


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 ...


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

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

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

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 ...


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

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

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 ...


2

You did not carry out your integration quite correctly. We have: $$ a(t)= -B_0+B_1t $$ $$ v(t) = \int\! a(t)dt=-B_0 t+\frac{1}{2}B_1t^2+C$$ Then plugging in conditions to solve for $C$ we get: $$ v(t_s)=0 $$ $$ 0=-B_0t_s+\frac{1}{2}B_1t_s^2+C $$ $$ C=B_0t_s-\frac{1}{2}B_1t_s^2 $$ Now we can plug in $t=0$ and solve for $v(0)$ $$ ...


1

I think as nothing explanatory is given: $A$ is amplitude of 1 mass or both have seperate amplitudes A/2 which cause net extention as $A$. We should do this by equivalent mass. Equivalent/Reduced mass of system is $$(m*m)/(m+m)=m/2$$ So, $$w=\sqrt{K/(m/2)}$$ we'll get frequency and $K$ So, energy is $$E=kA^2/2$$ Maybe it's wrong....


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

The relevant equation to use for a standing wave string under some tension is: $$ f_n = \frac{n}{2 L} \sqrt{\frac{F_T}{\rho}}$$ where $\rho$ is the mass density of the string and $F_T$ is the tension force. For a fixed frequency we now have, after rearranging the above equation: $$ F_{T,n} = \rho \left(\frac{2 L f_{\mathrm{fixed}}}{n}\right)^2 $$ When you ...


1

When looking at phase in a sine wave, for example when you are interested in wave interactions such as comb filtering, values are between 0deg and 360deg (you can normalize between 180 and -180). You are looking for the distance between peaks. When looking at phase on a real source, such as a mono recording playing on two speakers, phase can easily exceed ...


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

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

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 ...


1

You can't use the formula in Coulomb's law to compute the force between two magnets because that would describe magnetic monopoles, which do not exist in nature (one of Maxwell's equations, $\vec{\nabla}\cdot \vec{B}=0$ expresses this fact), and more importantly, because this formula is incorrect. The force between two magnets should look like the formula ...



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