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20

An orbit is stable because of conservation of angular momentum. Suppose we start with an object in an exactly circular orbit and slow it down slightly. That means it is moving at less than orbital velocity so it starts to fall inwards. However as its distance to the Sun decreases the tangential component of its velocity has to increase to conserve angular ...


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) The lowering of the ground state energy is a special case of the more general phenomenon of level repulsion (because the excited energy levels by definition must be larger than the ground state energy). II) Level repulsion is not just a quantum phenomenon. It also happens for purely classical systems, e.g. two coupled oscillators, as mentioned in the ...


11

The criterion you mention is roughly the threshold for the formation of the Coulomb gap in the Hubbard model or the local moment in the Anderson model. It is a common break-down region for many approaches starting from one of the limits (insulator/local moments versus conductor/mixed valence). For perturbation theory in $U$, see the PRB 36, 675 (1986) by ...


11

First, just to be sure about the answers to this particular problem: the eigenvalues of the $4\times 4$ matrix are $$0,\quad U\quad {\rm and}\quad U/2\pm \sqrt{(U/2)^2+4t^2}$$ When expanded to the first nontrivial order, the last two eigenvalues are $$ 0 - \frac{4t^2}U \quad {\rm and} \quad U+\frac{4t^2}U. $$ Note that the corrections to the energy arise ...


8

A nonperturbative theory means a theory where all the results can be calculated in principle to arbitrary accuracy on a computer. This is really just the same as a well-defined theory, a theory which is mathematically ok, and which makes sense, which isn't incomplete in some way at some high energy or when you do measurements at some high accuracy. An ...


8

The gravitational potential is what is known as a central force, which means that "how strong" the potential is only depends on how far away you are, and not on what angle you are relative to it. Having said that, gravitational systems are often treated in terms of an effective potential (full explanation provided on the Wikipedia page) which look like this ...


8

First: Scientific theories are never proven, only not falsified. Repeat that until it sinks in. Now, for the actual content of the question: That we only have perturbative ways to compute the S-matrix/scattering amplitudes for the Standard Model is not a reason to doubt its validity. Almost no physical system, apart from toy models, can be solved exactly, ...


7

No. There is nothing wrong with perturbation theory, or with theories with known, restricted accuracy. The point of theory is to explain the results of observation from as simple an initial theoretical standpoint as possible. Therefore: Since experiment always has a finite uncertainty, one can only ask that theory match the experimental value within its ...


7

I've been thinking about divergent series on and off, so maybe I could chip in. Consider a sequence of numbers (in an arbitrary field, e.g. real numbers) $\{a_n\}$. You may ask about the sum of terms of this sequence, i. e. $\sum a_n$. If the limit $\lim_{N\rightarrow\infty} \sum^N |a_n|$ exists then the series is absolutely convergent and you may talk ...


7

The reasons were given here. Essentially, at tree level you recover classical results. Loop corrections are proportional to powers of $\hbar$ and these are quantum terms.


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

Most of the difficulties of these calculations comes from the fact that people write the Lagrangian directly in terms of the metric perturbation $h_{\mu\nu}$. It is much simpler to write it in terms of the difference of connections tensor $F_{\mu\nu}{}^\beta$, i.e., $$(\nabla_\mu-\bar{\nabla}_\mu)A_\nu = \mathcal{F}_{\mu\nu}{}^\beta{}A_\beta,$$ where ...


6

Good question. You are correct in that without any restoring force, an object balanced precariously in an equilibrium position will be unstable. In physics, we use the scalar quantity of "potential" to find the equilibrium positions. These will be the maxima and minima in the potential field. The negative gradient of the potential gives the force. You've ...


6

Qualitative discussion (almost math free) The real key to understand orbits is the conservation of angular momentum. A two body orbit is nice this way insofar as it is a planar system and we get an easy expression for the angular momentum (we'll assume a satellite much, much less massive than the primary and not bother with the canonical transformation ...


6

In order to do perturbation the expansion parameter needs to be small. Otherwise the the system will be strongly coupled and you're in the non-perturbative regime. It's the same as for instance in QM: for perturbative calculations the pertubation must be small.


6

The scattering states must be included in the perturbative calculations if the result is to be highly accurate. In particular, it is not justified to ignore the continuous spectrum at energies close to the dissociation threshold. The Hilbert space in the position representation is the space of square integrable functions on $R^3\setminus\{0\}$ with respect ...


5

My approach would be: first determine the time evolution of $\hat{x}(t)$ and $\hat{p}(t)$. For $\hat{x}$ you have $$ \frac{d}{dt}\hat{x}_H(t) = i[H_H,\hat{x}_H(t)] = \frac{i}{2m} [\hat{p}_H(t)^2,\hat{x}_H(t)] = \frac{\hat{p_H(t)}}{m} $$ and for $p$ you have (assuming $0\leq t \leq T$) $$ \frac{d}{dt}\hat{p}_H(t) = i[H_H(t),\hat{p}_H(t)] = -m\omega_0^2 ...


5

The idea is the following: Consider a series expansion in elementary charge squared, $e^2$: \begin{equation}S(e^2)=\alpha_0+\alpha_2 e^2+\alpha_4 e^4+\;\ldots\end{equation} Assuming finite convergence radius, $S(e^2)$ is analytic at $e=0$. This leads to the analyticity of $S(-e^2)$ (i.e. $e\rightarrow ie$). A theory with imaginary charges possesses an ...


5

Is there somewhere in the paper that they say that the third derivative vanishes, or invoke its vanishing as an approximation? In general, you can't make tensorial objects by differentiating the metric. To get a tensor by differentiating a tensor, you have to take a covariant derivative. But the covariant derivative of the metric vanishes identically. ...


5

What you do in perturbation theory is you assume the correct eigenvalue equation for a (hitherto) unknown correct wavefuction $|\psi_n\rangle$ and its associated eigenvalue $E_n$: $$ H|\psi_n\rangle = E_n |\psi_n\rangle,$$ where $H$ is the full Hamiltionian. What you have is a main contribution $H_0$ (e.g. the Coulomb potential) and a perturbation $H'$ ...


4

This is really a great question. Take a look at the figure attached here. The circle indicates a 2-dimensional degenerate subspace at $\lambda=0$. In red we indicate two possible basis states for the subspace. In blue we show another possible choice of basis states. Now look at the curvy green lines, these are the states as they evolve for $\lambda > ...


4

Firstly, I refer you to Prof. Binney's textbook (see below) which covers perturbation theory in quantum mechanics in explicit detail. When doing perturbation theory, we perturb the Hamiltonian $H^{(0)}$ of a system which has been solved analytically, i.e. the eigenstates and eigenvalues are known. Specifically, $$H^{(0)}\to H^{(0)} + \lambda H'$$ where ...


4

You're looking for quantum hadrodynamics. See, e.g. Serot and Walecka.


4

I thinnk the earliest papers where this was written down were by DeWitt. But for a reference easily available via the arXiv look at hep-th/9411092. Eq (2.17) has the expansion you want and eq. (2.18) even has it to fourth order in $h$.


4

I think "quasi-circular" is a misleading name for this problem. Perhaps "quasi-elliptical" would be better? I say this because this problem does, in fact, contain a closed circular orbit (the radius of which you have called $r_c$). For that orbit you can find the period using Kepler's second law, which gives the result you show. An interesting way to ...


4

I) This is discussed around eq. (23.7.1) on p. 462 in Ref. 1. The task is to perform the path integral $$\tag{1} \int_{BC} [d\phi]e^{\frac{i}{\hbar}S[\phi]} ~=~\sum_{\nu}\int\! du \int_{BC_0} [d\phi_q]e^{\frac{i}{\hbar}S[\phi_{cl}+\phi_{\nu,u}+\phi_q]} $$ over fields $\phi$ with some (possible inhomogeneous) boundary conditions $BC$. This is done by ...



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