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16

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 ...


15

No, the value $9.8\frac{\mathrm{m}}{\mathrm{s}^2}$ is an approximation that is only valid at or near the Earth's surface. You can go a few miles up or down and it'll still be good enough, but once you get any significant distance away from the surface of Earth, you would need to use a different value for gravitational acceleration. You can calculate the ...


14

To expand a little on David's point assume we move from the nominal "surface" where $g$ is $9.8\text{ m}/\text{s}^2$ to another point at radius $r + \Delta r$. How much does the acceleration of gravity change? $$ g = \frac{GM}{(r+\Delta r)^2} = \frac{GM}{r^2(1 + \Delta r/r)^2} $$ and as long as $\Delta r$ is small compared to $r$ we can reasonably ...


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 $ma$ ...


9

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 ...


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

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 ...


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

$g$ becomes $ g \approx 9.7 \frac{m}{s^2}$ at a height of about 35km, so it would be ok to use the value $9.81$ for "down to earth" problems. The relevant wikipedia article has lots of useful information, like for example the following approximation formula for different heights: $$ g_h=g_0\left(\frac{r_e}{r_e+h}\right)^2 $$ Where $g_h$, is the gravity ...


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 ...


6

Is the Coulomb potential also used to solve the hydrogen atom in relativistic quantum mechanics? Yes, the Coulomb potential is there in the solution of the hydrogen atom with the Dirac equation, which is formulated in the relativistic framework. Now it is time to specialize to the hydrogen atom for which $$\frac{V}{\hbar c}=-\frac{Z\alpha}{r}$$ ...


6

You say: The Coulomb potential comes from classical electrodynamics but actually the Coulomb potential is predicted by quantum electrodynamics as a low energy limit. Quantum field theory describes the interactions between charged particles as the exchange of virtual particles, and it's not immediately obvious that it would lead to an inverse square ...


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 ...


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 ...


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 think the issue here is that you need to keep a consistent level of approximation in your "small angle approximation." By small angles, we typically mean $\theta_1$ and $\theta_2$ are both of order $\epsilon$, where $\epsilon \ll 1$. Then the question is - to what order in $\epsilon$ do you want to write down the equations of motion? When you neglect ...


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

"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

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

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 = ...


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

The other answers are correct. I would like to add to them with an example. Take a spring, with spring constank $k$, with a mass, $m$, at one end and fixed to a large immovable object at the other. Let the only force acting on the mass be due to the spring and the difference from the equilibrium position to be $x$, which can be positive and negative. This ...


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

This is a quite general question. Whether or not one should use an approximation depends on several things. Disclaimer: my answer is not restricted to differential equations and contains examples from perturbation theory, but the general idea still applies. The most important question is whether an approximation makes sense from a physical point of view. ...



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