# Tag Info

14

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 ...

12

No, it's not exact, and it's not a definition either. Consider that acceleration has a definition that no one will dispute. It is the time derivative of velocity. More than likely, he had in mind the relativistic generalization of the equation. The more general form of the equality is: $$F = \frac{dp}{dt}$$ You can easily see how this results in $ma$ ...

8

You cannot use the second kinematical equation because it is valid only when the acceleration due to gravity, $g$ , is constant. This is incorrect for distances comparable to the radius of the earth, and velocities comparable to the escape velocity. The first correctly assumes a $\frac{1}{R^2}$ fall-off of the gravitational attraction on the body due to ...

8

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 ...

8

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

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 ...

6

From the context of the statement within Feynman's lectures, it is evident that what he had in mind was the idea that mathematical statements such as $F=ma$ are just an idealization of nature. In the text, he goes on to explain that for example the precise mass of a physical object is not known. He gives the example that the mass of a chair is just an ...

6

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 ...

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

A small portion of any smooth curve looks the same as a small piece of a parabola in the limit. Choose a coordinate system so that the tangential direction in the middle of the segment is along the $x$ axis and choose a translation for the middle of the segment to sit at $(0,0,0)$, the origin of the coordinates. Then $y,z$ on the curve (ellipse etc.) may be ...

5

In physics and engineering, we often abstract and idealize a physical problem to gain insight into the physics, e.g., infinite plane of charge, infinite line of charge, point charge, etc. Now, it goes without saying that if these idealizations didn't represent good approximations of relevant physical systems, they wouldn't be used. With regards to your ...

4

Occam's razor suggests that in the simplest explanation is the most probable. Physicists will assume that elementary particles are point-like, until they have evidence to suggest otherwise.

4

I) In this answer we discuss a systematic approach to linearization and stability analysis. Imagine that the physical system under consideration is described by an autonomous Lagrangian $L=L(q,\dot{q})$ of $n$ generalized coordinates $$\tag{1} q~=~(q^1, \ldots, q^n)~\in~ \mathbb{R}^n.$$ One of the first questions one would like to ask is, if a specific ...

4

Assuming that $m_1$ and $m_2$ take up a finite amount of space (e.g., two spheres of mass with radius $r_0$), that equation isn't even valid for $r < r_0$, so there's no inconsistency. The derivation follows from Gauss' law; it is analogous to the application of Gauss' law in electrostatics; the $m_1$ and $m_2$ are the mass enclosed at some distance $r$. ...

4

Yes, the continuity of the electric charge is an approximation that is valid whenever the relevant charges are much greater than the elementary charge (of the electron, or the proton). When we deal with numbers like $1,234,567\,e$, it doesn't really matter that it should have been $1,234,567.8\, e$: very large numbers may be approximated by a nearby integer ...

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

"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 ...

4

Of course Newton's three laws of motion are correct, because they were verified several hundred of years ago and they continue working today, for such systems. Science is accumulative. What modern physics has done is to constraint the range of validity of those laws. Although some 18th century physicists believed that the laws were valid elsewhere, we know ...

4

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 ...

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 $$d\tau = ... 3 Strictly speaking, light rays from distant stars are not perfectly parallel, but the typical angle (in radians) between them can be estimated as the diameter of the star divided by the distance from the star to the Earth; this value may equal 10^{-7} or less, so the rays are parallel to high accuracy. 3 The theory of fluids introduces material parameters in the stress tensor, which help model the substance. "The viscosity coefficient is the proportionality constant relating a velocity gradient in a fluid to the force required to maintain that gradient. The thermal conductivity is the proportionality constant relating the temperature gradient across a fluid ... 3 Whenever two quantum systems are united, the Hilbert space of the wavefunctions of the resulting system is always the tensor product of the Hilbert spaces of the wavefunctions of the source systems. This does not mean that the total wavefunction is always a product of two wavefunctions from the source spaces. In general case it looks like this:$$ \Psi = ...

3

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

There are several methods, and several reasons. Sepration of variables is not The method of "separation of variables" is a misnomer in this exact context. What separation of variables really means in quantum mechanics is that you solve the eigenvalue problem for the linear operator of the Hamiltonian, and this gives the general solution by superposition. ...

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 ...

3

The formula: $$\Delta U = mgh$$ is an approximation that applies when the distance $h$ is small enough that changes in $g$ can be ignored. As you say, the expression for $U$ is: $$U= -\frac{G M m}{r}$$ So the change when moving a distance $h$ upwards is: $$\Delta U = \frac{GMm}{r} - \frac{GMm}{r + h}$$ We rearrange this to get: \begin{align} ... 3 Hints: Define difference \delta:=\Delta-\Delta_0. Deduce from |\delta|\ll |\Delta_0| that the lhs. of eq. (1) is\tag{A}\text{lhs}~\approx~ -\frac{\delta}{\Delta_0}.$$Substitute \xi=x\Delta  in the integral on the rhs. of eq. (1). Deduce using \hbar \omega_D \gg \Delta that the rhs. is$$\tag{B} \text{rhs}~\approx~ \int_{\mathbb{R}} \! ...

3

As it turns out, you can use it for the vast majority of the time. The ideal gas law is considered "ideal" because it assumes interactions between the gas molecules only occur due to collisions and no long-range forces are present. If the fluid you are modelling is non-polar and doesn't have things like van der Waals forces between molecules, then the ideal ...

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