# Tag Info

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In the simplest form the saddle point method is used to approximate integrals of the form $I \equiv \int_{-\infty}^{\infty} dx\,e^{-f(x)}$ The idea is that the negative exponential function is so rapidly decreasing -- $e^{-10}$ is $10000$ times smaller $e^{-1}$ -- that we only need to look at the contribution from where $f(x)$ is at its minimum. Lets say ...

7

In the standard model (as in all traditional relativistic quantum field theory), particles are pointlike. All experimentally available facts about microphysics seem to be consistent with the standard model. This is the (fully sufficient) reason for believing that particles in Nature are pointlike. Pointlike is a technical term that refers to the fact that ...

7

You are right about exact results, these depend on your definition of "exact". The best definition of an exact is if you have a fast algorithm to calculate the result in a reasonable time. The faster the algorithm, the more exact the result. For Helium atoms, the answer is yes--- you can use the variational method to produce a result to as good a precision ...

5

In real life, the current can't jump instantaneously because there is always some finite inductance in a circuit. However, this is just a typical idealized textbook problem where the inductance is assumed identically zero, so the current can jump instantaneously according to the assumptions of the problem. Note the current also jumps in their solution for ...

5

BebopButUnsteady has explained the mathematics behind it and I'll provide you with some references I've found useful and quite like that get into the more technical mathematical details although they are still very readable. These deal more concretely with the complex analysis required and how to properly pick the correct contour so you don't get divergences ...

4

No, these semiclassical formulas are no good. The first formula just gives the free energy as the volume of classical phase space, and this is incorrect, since, for instance, it predicts the specific heat of a cold gas is 1.5k independent of temperature, and this vanishes for cold quantum gasses as the discrete energy levels of the box the gas is contained ...

4

Not exactly, since the total wave function is not factored as $\psi_j(q)\phi(R)$ (where $q$ are the electronic coordinates and $R$ the nuclear coordinates) but as $\psi_j(q;R)\phi(R)$. This means that whenever $R$ changes the electronic wave function instantaneously adapts to remain in the same quantum state. That necessarily implies that the electronic ...

4

The modeling used in the GPS is based the static weak field metric $$ds^2 = -(1+2\Phi)dt^2 + (1-2\Phi)dS^2,$$ where $dS^2$ is the Euclidean metric and $\Phi$ is the gravitational potential of the Earth, though only the monopole and quadrupole terms are used. The proper time of a clock is therefore determined by $$d\tau = ... 3 Wikipedia has the following text: Newton's laws were verified by experiment and observation for over 200 years, and they are excellent approximations at the scales and speeds of everyday life. Newton's laws of motion, together with his law of universal gravitation and the mathematical techniques of calculus, provided for the first time a unified ... 3 Elementary particles don't really have a shape or a size, these are emergent qualities that stem from interactions between particles. In quantum physics a particle is represented by its quantum state, and if you want to describe that in space you get a wave function which tells us how much of the particle is present at any given point in space. Because there ... 3 Strictly speaking, light rays from distant stars are not perfectly parallel, but the typical angle (in radians) between them can be estimated as the diameter of the star divided by the distance from the star to the Earth; this value may equal 10^{-7} or less, so the rays are parallel to high accuracy. 3 There are many physical intuitions often presented in various texts on fluid dynamics. I won't mention those here. I will, however, mention that mathematically the passage from a particle point of view to a continuum point of view is still a largely un-resolved problem. (With suitable interpretation, this problem was already posed by Hilbert as his 6th of 23 ... 3 The theory of fluids introduces material parameters in the stress tensor, which help model the substance. "The viscosity coefficient is the proportionality constant relating a velocity gradient in a fluid to the force required to maintain that gradient. The thermal conductivity is the proportionality constant relating the temperature gradient across a fluid ... 3 Whenever two quantum systems are united, the Hilbert space of the wavefunctions of the resulting system is always the tensor product of the Hilbert spaces of the wavefunctions of the source systems. This does not mean that the total wavefunction is always a product of two wavefunctions from the source spaces. In general case it looks like this:$$ \Psi = ...

3

first, I don't know exactly whta "sequence of infinitesimal Lorentz boosts" you're referring to. A boost of the flat Minkowski space gives you a Minkowski space back, so you can't get a curved space by any sequence of boosts that act globally on the spacetime. Also, the adjective "infinitesimal" could be pretty much inconsequential. Transformations in ...

2

In this case "perfectly parallel" means "more closely parallel than can be detected". How parallel is that? Consider the geometry. The biggest angular deviation possible is between a photon originating from the left side of the star impinging on the right side of the detector and vice versa. By the small angle approximation we can write this difference as ...

2

This is an extremely comprehensive review of electronic properties in two-dimensional electron systems (2DESs): http://rmp.aps.org/abstract/RMP/v54/i2/p437_1 but, as you can imagine, it covers almost everything there is to cover in 2DESs. For areas (in transport) you're focusing on you will find only sections IV C and D useful; it involves computation of ...

2

Even a physical quantity which changes by discrete amounts can often be well approximated by a continuous function of time. The derivative is a property of a mathematical function. Any differentiable function must necessarily be continuous, and a continuous function will change by arbitrarily small values for an arbitrarily small change in inputs. The ...

2

Pointlike and point are entirely different concepts. The planet Jupiter is pointlike to likely 6 or more decimals of precision when studying the dynamical evolution of the Solar system. Does not mean that Jupiter is a point! Just because something behaves pointlike has always meant that we just don't know enough yet. String theory is one theory about a ...

2

"Newton's Laws" are - like most physics - a mathematical model that describes how the world - or the universe - works. All models are wrong, in that they don't describe the complete complexity of the physical world, but some models are useful, in that they let us make predictions. Newton's Laws, as a model, work well, unless you are dealing with things ...

1

Your question is based on an assumption that the vacuum is empty, and matters (including particles) are things we placed in the empty vacuum. But the Casimir effect shows that the vacuum is not empty but a dynamical medium. This led to an emergence point of view of elementary particles: they are quantized collective motions of the vacuum medium. In one ...

1

Let me take a slightly different perspective to Ron Maimon and say that the answer depends on whether you're after an exact solution of some mathematical model, or whether you want to calculate the exact physical behaviour. Any method for calculating the physical properties of a system rely upon an approximation. If you choose some model you can certainly ...

1

When talking about infinitesimals, you need to specify how close to zero it is. The entropy loss is $\delta Q\over T$ and the gain is $\delta Q \over T-\delta T$, so the net gain is $\delta Q \delta T \over T^2$ to leading infinitesimal order, and it vanishes linearly in $\delta T$ and $\delta Q$ both. For iterated infinitesimal processes that approximate a ...

1

This question is working within the realm of 'circuit theory', which is an idealization useful for introductory teaching of electromagnetism. It is really a simplification of electrodynamic field theory, just a special case making useful assumptions. A lot of conceptual problems in circuit based questions come from forgetting that you are dealing with a ...

1

The electron spin can only enter through magnetic effects, and these are usually smaller by about an order of magnitude than electrostatic effects by about an couple of orders of magnitude. It is therefore a relatively safe assumption when one is trying out the problem for the first time to ignore magnetic, and therefore spin, effects. Of course, there is ...

1

Physics is all about making the right approximations, in the hope that we can gain some actual physical insight into our problem and make verifiable predictions. For example, say you wanted to calculate the trajectory of a cannonball that has been fired from a cannon. It would be a Sisyphean task to account for all the possible variables that could affect ...

1

Yes, you are right that you should keep all terms to a given order. However if you look at the full expression that is being calculated: $\zeta= \left<0|(1+iH_1+H_2)(1-iH_1+H_2)|0\right>-\left<0|(1+iH_1+H_2)|0\right> \left<0|(1-iH_1+H_2)|0\right>$ You will find that the $H_2$ terms cancel out identically. Surprisingly, this answer is ...

1

Good question! When I started writing this answer I couldn't think of an example, but then I realised that Brownian motion fits the bill, as I'll explain below. So please forgive the somewhat tangential introduction: To a reasonable degree of approximation, temperature can be thought of as energy per degree of freedom. I say approximation because in quantum ...

1

The answer is in general No, for various reasons. The WKB approximation is generally not exact for a finite energy-level $N$. Point 1 is true even if we include the metaplectic correction from the Maslov index. OP's sought-for formula is not exact for a finite energy-level $N$ even for the simple quantum harmonic oscillator (SHO). (This is despite the ...

1

I think I worked it out myself. What I found is that trying to arrive at the Hartree-Fock Hamiltonian via this mean-feald procedure isn't really that appropriate. Instead, one says: Given a two-particle operator $\hat O = c_1^\dagger c_2^\dagger c_3 c_4$, how can I find a single-particle operator O^{MF}\$ that is a good approximation? If we have a complete ...

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