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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 ...

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 ...

15

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 ...

14

There's no doubt the solar system is accelerating. The milky way galaxy rotates, and we're quite on the outside. Hence, there's a permanent acceleration vector pointing to the center. However, this is a phenomenally small acceleration. If you'd try to measure it here on earth, you run into all kind of practical problems when you try to isolate it. For ...

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 $ma$ ...

10

There are two main reasons it is practical to ignore the pseudo forces due to the rotation of the earth/sun about the galaxy. First, the accelerations are pretty small, and second, they are pretty uniform. The sun moves around galactic center at about 800,000 kilometers per hour, but it takes around 250 million years to complete a single orbit of galactic ...

10

As was mentioned in the comments, the differential equation you give is not solvable analytically. What one can do is one can go beyond the small angle approximation in a controlled fashion by Taylor expanding the sine and cosine function and find (e.g. expanding up to order $x^3$) $$I \ddot x + b\dot x + mgl\left( x - \frac{x^3}{6}\right) = F \cos(\omega ... 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 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 ... 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 Shouldn't there be many psuedo-forces to account for planetary motion? In theory, yes. In practice, no. Consider the third body perturbations induced by Alpha Centauri (a two solar mass star system at a distance of 4.37 light years) on Voyager 1, which is currently about 130 astronomical units from the solar system barycenter. This is on the order of ... 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}$$... 6 You say: The Coulomb potential comes from classical electrodynamics but actually the Coulomb potential is predicted by quantum electrodynamics as a low energy limit. Quantum field theory describes the interactions between charged particles as the exchange of virtual particles, and it's not immediately obvious that it would lead to an inverse square ... 6 From the context of the statement within Feynman's lectures, it is evident that what he had in mind was the idea that mathematical statements such as F=ma are just an idealization of nature. In the text, he goes on to explain that for example the precise mass of a physical object is not known. He gives the example that the mass of a chair is just an ... 6 BebopButUnsteady has explained the mathematics behind it and I'll provide you with some references I've found useful and quite like that get into the more technical mathematical details although they are still very readable. These deal more concretely with the complex analysis required and how to properly pick the correct contour so you don't get divergences ... 6 g becomes  g \approx 9.7 \frac{m}{s^2} at a height of about 35km, so it would be ok to use the value 9.81 for "down to earth" problems. The relevant wikipedia article has lots of useful information, like for example the following approximation formula for different heights:$$ g_h=g_0\left(\frac{r_e}{r_e+h}\right)^2  Where $g_h$, is the gravity ...

5

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 ...

5

"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 ...

5

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 ...

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

The solutions for a forced/driven pendulum can be chaotic, in the sense of chaos theory, so the period may not even exist! For a tutorial on see these course notes for example. What you can do with an forced/driven pendulum is to simulate it and compute the various chaos-theory related parameters: Lyapunov exponents, Poincare map, etc. There are numerous ...

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

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 ...

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

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

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 ...

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