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35

There is a consistent definition, but it involves a couple of arbitrary thresholds, so I doubt you'd consider it rigorous. The construction $X \gg Y$ means that the ratio $\frac{Y}{X}$ is small enough that subleading terms in the series expansion for $f\bigl(\frac{Y}{X}\bigr) - f(0)$ can be neglected, where $f$ is some relevant function involved in the ...


31

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 $...


23

Scattering experiments can be used to determine the size of a particle. The results for an extended object are different than that of a point particle. But all of these scattering experiments depend on getting the probe particle "close" to the scattering object. In the case of electrons, that means launching the probe with enough energy to overcome the ...


21

That the eigenfunctions of the free Hamiltonian $H\propto p^2$ are not actually normalizable due to its completely continuous spectrum and therefore cannot be actual quantum states is well-known, although rarely suitably emphasized. (See e.g. Why are eigenfunctions which correspond to discrete/continuous eigenvalue spectra guaranteed to be normalizable/non-...


17

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 ...


13

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

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 ...


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 $ma$ ...


12

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 t)...


9

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 ...


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

The concept of infinity as used in calculus arises mathematically as an equivalence class of divergent series to compactify the reals (or complex numbers), so infinities are, even in a purely mathematical (calculus) setting, in some way approximations of large numbers. There are rich theories of other concepts related to infinity not directly related to the ...


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 10-...


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

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.


6

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 ...


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

Huygens's Wave Theory is what you call a first order scalar diffraction theory of light. So what does it describe and what does it fail to describe? First order means that electromagnetic effects like induced currents in surfaces etc. are ignored. These can be described by solving Maxwell's equations for the same system instead of working with the wave ...


6

The explanation is not a very full one. As you correctly note, you're taking a limit, so the assumption $\sin\theta \to\theta$ as $\delta z\to0$ becomes exact. So Eq 16-23 contains no approximation. The assumption creeps in subtly when one assumes that the force calculated in Eq 16-23 is at right angles to the $z$ axis. That is, that $\mathrm{d}y/\mathrm{d}...


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

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

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 ...


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

Yes, the term "effective action" has the same meaning in particle physics and condensed matter physics. After all, the discovery of the concepts of the Renormalization Group by Ken Wilson and others was being done simultaneously in both disciplines and the exchange of ideas was fruitful in both directions. On the other hand, what the effective theories ...


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 ...



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