208
votes
Accepted
Is $\pi^2 \approx g$ a coincidence?
The differential equation for a pendulum is
$$\ddot{\phi}(t) = -\frac{g}{l}\cdot\sin{\phi(t)}$$
If you solve this, you will get
$$\omega = \sqrt{\frac{g}{l}}$$
or
$$T_{1/2}=\pi\sqrt{\frac{l}{g}}$$
$...
54
votes
Is $\pi^2 \approx g$ a coincidence?
It's annoyingly unclear how far it's a coincidence, but at any rate it isn't completely a coincidence.
As you can see in e.g. the Wikipedia article about the metre, a unit almost equal to the metre ...
49
votes
Accepted
How is it possible to differentiate or integrate with respect to discrete time or space?
Let's say space is really a lattice with spacing $\Delta x$. It turns out that this idea has more trouble with experiment than you might think, but we can plow ahead for the purposes of this question.
...
41
votes
Far away from a charged conductor, the field is like a point charge. Where's the point located?
The answer is that it doesn't matter.
The distance at which fields resemble that from a point charge is also the distance at which it does not matter where that point is located within the structure. ...
40
votes
Accepted
Why doesn't physics like infinity (or does it)?
There's no rule against infinity, only a rule against being empirically wrong. Historically, the infinite has hinted we're missing something, but so have plenty of other things too. Let's discuss some ...
38
votes
Accepted
The reasoning behind doing series expansions and approximating functions in physics
The key reason is that we want to understand the behavior of the system in the neighborhood of the state rather than at the state itself.
Take the equation of motion for a simple pendulum, for ...
37
votes
Accepted
Is there a rigorous definition of 'much greater than'?
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 ...
36
votes
Accepted
Far away from a charged conductor, the field is like a point charge. Where's the point located?
Based on some of the back-and-forth I see, I think you're asking the wrong question. I think the question you want to ask is "Given a charge distribution $\rho(\mathbf{r})$, where should I place a ...
36
votes
Accepted
What is the point of a voltage divider if you can't drive anything with it?
Oh, but you can. You can drive an high impedance input with it...including a buffer, which can then in turn be used to drive whatever you want. The more current you draw the more the voltage will ...
34
votes
Accepted
Why does a simple pendulum or a spring-mass system show simple harmonic motion only for small amplitudes?
A simple pendulum does not strictly show simple harmonic motion unless you allow some approximations and uncertainties. It approximately behaves as a harmonic oscillator for small amplitudes.
An ...
33
votes
Accepted
What is the true nature of gravity?
The job of physics is to construct models that are able to explain and predict empirical observations. You can never be completely sure that a given model is the "true" description, only ...
32
votes
Accepted
Elliptical Trajectory, or Parabolic?
A parabola and an ellipse are both conic sections, which can be constructed in a plane as all the points where the distances from some reference point (the "focus") and some reference line (the "...

rob♦
- 70.2k
30
votes
Accepted
Why is a particle non-relativistic when its kinetic energy is small compared to its rest energy?
When we say a particle is non-relativistic we mean the Lorentz factor $\gamma$ is close to one, where $\gamma$ is given by:
$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$
So saying $\gamma$ is close to ...
27
votes
Accepted
Why is the Newtonian expression for kinetic energy called the "first order" approximation of the relativistic expression?
The way I see it, there are four possible answers. You can pick the one you like the most, because in the end it doesn't matter very much.
You're right, it's a second order approximation and those ...
24
votes
Accepted
Is the existence of a sole particle in an hypothetical infinite empty space explicitly forbidden by QM?
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, ...
24
votes
Accepted
Near Earth vs Newtonian gravitational potential
Your equation (2) is the change in potential energy when the object moves vertically by a distance $h$ i.e. when the object moves from $r$ to $r+h$. Let's use equation (1) to calculate this:
$$ \...
23
votes
Why do physicists say that elementary particles are point particles?
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 ...
22
votes
Why is a particle non-relativistic when its kinetic energy is small compared to its rest energy?
'Non-relativistic' means $v\ll c$.
That is effectively the same as $\gamma \approx 1$ as $\gamma={1 \over \sqrt{1-v^2/c^2}}$.
But also $\gamma={E_{tot}\over E_{rest}} \equiv 1+{E_{kin} \over E_{...
22
votes
Accepted
If charges is quantised, how can we use integrals in electrostatics?
You may also wonder why we use the concept of density of a material despite that material is made of molecules, or atoms, or in general quantized entities. This is the basis of hydrodynamics and solid ...
22
votes
Why doesn't physics like infinity (or does it)?
Infinity is a shorthand for unbounded. When we say that $\frac 1 x$ goes to infinity as $x$ approaches $0$, what we really mean is that we can make $\frac 1 x$ as large as we like by using a value of $...
21
votes
Accepted
Understanding this quote by Feynman
[...] the wave theory was unable to explain things like the Photoelectric Effect and Compton Scattering [...]
This is true. As a result, we know that classical electrodynamics is only an ...
19
votes
Approximating an expression for a potential
There is nothing wrong with your first approximation. You got the leading term $2k/l$ correct. You just did not expand out far enough in powers of $x/l$ to see the $2k x^2/l^3$ term. If you were to ...
19
votes
Is projectile motion an approximation?
Just as the motion of body around the earth is ellipse (1st Kepler law replacing sun by earth), so is the motion of a projectile. Notice that almost everything we deal is an approximation, the earth ...
18
votes
Elliptical Trajectory, or Parabolic?
If gravity is uniform - the force has the same magnitude and direction everywhere, the trajectory is a parabola. This is a very good approximation for trajectories that don't go very far.
But in ...
18
votes
Accepted
Accuracy of physics laws
Accuracy can mean different things. While the question asks about the statistical accuracy, what immediately comes when talking about the Newton's laws is that they are non-relativistic, i.e., they ...
17
votes
Accepted
Should not we apply pseudo-forces all the time as the Earth is a non-inertial frame of reference?
The way to properly answer your question is, for lack of a better word, by enumeration. Consider all the ways the Earth moves, compute the fictitious forces corresponding to them and check that they ...
16
votes
Accepted
Why don't the Navier-Stokes equations simplified for hydrodynamics contain gravitational acceleration?
If you go through the process of non-dimensionalizing the equations, the math becomes more clear. If you start with the momentum equation (ignoring viscous forces because they aren't important for the ...
15
votes
Accepted
Calculating the total time elapsed until two pendulums "stop colliding" gives a divergent result
Your calculation is perfectly correct, under the standard idealizations in mechanics.
From a mathematical point of view this isn't that surprising; divergent times are pretty common. For instance, ...
15
votes
Is there an approximation for the Lorentz factor for very large velocities?
This is roughly the simplest you can get it:
$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-v/c} \sqrt{1+v/c}} \approx \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1-v/c}}.$$
In other words, if $\Delta ...
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