Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result in accumulation of liquid inside the pipe which is not possible. Does streamline terminates or it changes its direction at the open end of pitot tube?

homework and exercises - Moment of inertia of a hollow sphere wrt the centre?

I've been trying to compute the moment of inertia of a uniform hollow sphere (thin walled) wrt the centre, but I'm not quite sure what was wrong with my initial attempt (I've come to the correct answer now with a different method). Ok, here was my first method:

Consider a uniform hollow sphere of radius $R$ and mass $M$. On the hollow sphere, consider a concentric ring of radius $r$ and thickness $\text{d}x$. The mass of the ring is therefore $\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi r\cdot\text{d}x$. Now, use $r^2 = R^2 - x^2:$ $$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \left(R^2 - x^2 \right)^{1/2}\text{d}x$$ and the moment of inertia of a ring wrt the centre is $I = MR^2$, therefore: $$\text{d}I = \text{d}m\cdot r^2 = \frac{M}{4\pi R^2}\cdot 2\pi\left(R^2 - x^2\right)^{3/2}\text{d}x $$ Integrating to get the total moment of inertia: $$I = \int_{-R}^{R} \frac{M}{4\pi R^2} \cdot 2\pi\cdot \left(R^2 - x^2\right)^{3/2}\ \text{d}x = \frac{3MR^2 \pi}{16}$$

which obviously isn't correct as the real moment of inertia wrt the centre is $\frac{2MR^2}{3}$.

What was wrong with this method? Was it how I constructed the element? Any help would be appreciated, thanks very much.

Answer

The mass of the ring is wrong. The ring ends up at an angle, so its total width is not $dx$ but $\frac{dx}{sin\theta}$

You made what I believe was a typo when you wrote

$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \left(R^2 - x^2 \right)\text{d}x$$

because based on what you wrote further down, you intended to write

$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \sqrt{\left(R^2 - x^2 \right)}\text{d}x$$

This problem is much better done in polar coordinates - instead of $x$, use $\theta$. But the above is the basic reason why you went wrong.

In essence, $sin\theta=\frac{r}{R}$ so you could write

$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \frac{r}{sin\theta} \ \text{d}x \\ = \frac{M}{4\pi R^2}\cdot 2\pi \frac{r}{\frac{r}{R}} \ \text{d}x\\ = \frac{M}{4\pi R^2}\cdot 2\pi R \ \text{d}x\\ = \frac{M}{2 R} \ \text{d}x$$

Now we can substitute this into the integral:

$$I = \int_{-R}^{R} \frac{M}{2 R} \cdot \left(R^2 - x^2\right)\ \text{d}x \\ = \frac{M}{2R}\left[{2R^3-\frac23 R^3}\right]\\ = \frac23 M R^2$$

newtonian mechanics - Under what conditions does the relation $vec{L} =I vec{omega}$ holds good?

if $\vec L=I\vec ω$ holds good in all the cases then directions of angular momentum and angular velocity must be parallel always which is not true in some of the situations. So under what conditions does the relation $\vec L=I\vec ω$ holds good ?

electromagnetism - How will SR EM Lagrangian change if we find a magnetic charge?

When we introduce electromagnetic field in Special Relativity, we add a term of $$-\frac e c A_idx^i$$ into Lagrangian. When we then derive equations of motion, we get the magnetic field that is defined as $$\vec H=\nabla\times\vec A.$$

If we now take divergence of both sides of this definition, we automatically get

$$\nabla\cdot\vec H=0,$$

which is equivalent to inexistence of magnetic charges.

But suppose we've found a magnetic charge. What will change in our Lagrangian or in definition of electric and magnetic fields in this case to make $\nabla\cdot\vec H=\sigma$?

In this Phys.SE answer it's asserted that magnetic field would get an additional term "gradient of a scalar potential". Is this "a" scalar potential instead "the" $A^0$ potential?

Answer

In the absence of magnetic monopoles, Maxwell's equations are

$$ \begin{align} \text d F &= 0 ,\\ \text d{\star F} &= J_e , \end{align} $$

where $J$ is the 4-current 3-form due to electric charges (assuming a metric with signature $(-,+,+,+)$). For cohomological reasons, from the first equation one can asserts that there exists a 1-form $A$ such that $F = \text d A$, and $A$ is the interpreted as the 4-potential $(\phi,\mathbf A)$ (up to the musical isomorphism between tangent and cotangent bundle to Minkowski spacetime). In the presence of magnetic monopoles (or charge, to even symmetrise terminology) the above equations would become

$$ \begin{align} \text d F &= J_m ,\\ \text d{\star F} &= J_e , \end{align} $$

where $J_m$ is the 4-current for magnetic charges. Therefore in this extended theory of electrodynamics both the Faraday tensor $F$ and its Hodge dual $\star F$ (sometimes also denoted by $G$) figure in constitutive equations.

Since $F$ is no longer a closed form, its expression must be modified by the introduction of a non-exact part, say $C$, so that

$$F = \text d A + C.$$

Since the equations are symmetric in $F$ and $\star F$ we can postulate there exist 1-forms $B$ and $D$ such that

$$\star F = \text d B + D,$$

and assume that $C$ depends on $B$, while $D$ depends on $A$. But since $\star\star = -1$ in special relativity, we conclude that

$$F = \text dA - \star\text dB,$$

which can be related to the Helmholtz decomposition into polar and axial part for twice differentiable vector fields.

The Lorentz force for a particle with electric charge $q_e$ and magnetic charge $q_m$ would be $$K = \iota_u(q_e F + q_m G),$$ where $u$ is the particle's 4-velocity vector and $\iota$ denotes the interior product. The extra term can then be reproduced with a Lagrangian containing the extra term $B_\mu u^\mu$.

---

To make contact with the usual vector notation, observe that the Faraday tensor has the covariant matrix representation $$F = \begin{bmatrix}0&-~\mathbf E^T\\\mathbf E&\star\mathbf H\end{bmatrix}$$ where $\star\mathbf H$ is the Hodge dual of the magnetic field $\mathbf H$, and can be thought as the linear map $(\star\mathbf H)\mathbf v = \mathbf v\times\mathbf H$ for any $\mathbf v\in\mathbb R^3$. Skew-symmetric tensors as the one above are then represented by a polar vector $\mathbf E$ and an axial vector $\mathbf H$, and can be denoted as $F=(\mathbf E,\mathbf H)$. Having defined this notation, the action of the Hodge dual is then $\star(\mathbf E,\mathbf H) = (\mathbf H,-\mathbf E)$ (up to a sign which I can't be bothered remembering). The exterior derivative of the 4-current $A$ is a tensor of the form above, and it turns out that $$\text dA = \left(\nabla A^0+\frac{\partial\mathbf A}{\partial t},\nabla\times\mathbf A\right),$$ where the first component is the polar part and the second one is the axial part. Hence with no magnetic charges we recover the electric and magnetic fields. Now for the extra potential $B=(B^0,\mathbf B)$ we have, using the rule for the Hodge dual discussed a few lines above, $$\star\text dB = \left(\nabla\times\mathbf B, - \nabla B^0 - \frac{\partial\mathbf B}{\partial t}\right)$$ Remark Here $\mathbf B$ is an extra vector potential, not to be confused with the magnetic induction.

Reconstructing the Faraday tensor according to the prescription $F=\text dA - \star\text dB$ given above we then have, in terms of polar and axial parts $$F = \left(\nabla A^0 + \frac{\partial\mathbf A}{\partial t} - \nabla\times\mathbf B, \nabla\times\mathbf A + \nabla B^0+\frac{\partial\mathbf B}{\partial t}\right),$$ whence $$\mathbf E = \nabla A^0 + \frac{\partial\mathbf A}{\partial t} - \nabla\times\mathbf B$$ and $$\mathbf H = \nabla\times\mathbf A + \nabla B^0 + \frac{\partial\mathbf B}{\partial t}.$$

cosmology - How can a quasar be 29 billion light-years away from Earth if Big Bang happened only 13.8 billion years ago?

I was reading through the Wikipedia article on Quasars and came across the fact that the most distant Quasar is 29 Billion Light years. This is what the article exactly says

The highest redshift quasar known (as of June 2011[update]) is ULAS-J1120+0641, with a redshift of 7.085, which corresponds to a proper distance of approximately 29 billion light-years from Earth.

Now I come to understand that the Big Bang singularity is believed to be around 13.8 Billion years ago.

So how is this possible? Does the presence of such a quasar negate the Big Bang Theory?

I'm not a student of Physics and was reading this out of (whimsical) curiosity. Is there something I'm missing here or the "proper distance" mentioned in the fact is a concept that explains this?

Edit: My Bad! Here's how..

A simple google search led me to this article which says the farthest quasar found is 12.9 billion LYs and not 29 billion.
So in the end we have just proven that wikipedia needs more moderation.

quantum field theory - What does it mean to say that "the fundamental forces of nature were unified"?

It is said that immediately after the Big Bang, the fundamental forces of nature were unified. It is also said that later they decoupled, becoming separate forces.

Indeed, if we look at the list of states of matter on Wikipedia we see:

• 
  

  Weakly symmetric matter: for up to $10^{−12}$ seconds after the Big Bang the strong, weak and electromagnetic forces were unified.

  
  



• 
  

  Strongly symmetric matter: for up to $10^{−36}$ seconds after the Big Bang, the energy density of the universe was so high that the four forces of nature — strong, weak, electromagnetic, and gravitational — are thought to have been unified into one single force. As the universe expanded, the temperature and density dropped and the gravitational force separated, a process called symmetry breaking.

  
  

Not only is it said that the forces were once unified, but this is also somehow related to the states of matter.

I want to understand all of this better. What does it truly mean, from a more rigorous standpoint, to say that the forces were unified and later decoupled? How this relate to the states of matter anyway?

Answer

When we say that the forces were unified, we mean that the interaction was described by a single gauge group. For example, in the original grand unified theory, this group was $SU(5)$, which spontaneously broke down to $SU(3) \times SU(2) \times U(1)$ as the universe cooled. These three components yield the strong, weak, and electromagnetic forces respectively.

I'll try to give a math-free explanation of what this means. To do so I'll have to do a decent amount of cheating.

---

First, consider the usual strong force. Roughly speaking, the "strong charge" of a quark is a set of three numbers, the red, green, and blue color charges. However, we don't consider the strong force three separate forces because these charges are related by the gauge group: a red quark can absorb a blue anti-red gauge boson and become blue. In the case of the strong force, we call those bosons gluons, and there are 8 of them.

At regular temperatures, the strong force is separate from the electromagnetic force, whose charge is a single number, the electric charge, and whose gauge boson is the photon. There is no gauge boson that converts between color charge and electric charge; the two forces are independent, rather than unified.

When we say all the forces were unified, we mean that all of the Standard Model forces were described by a common set of charges, which are intermixed by 24 gauge bosons. These gauge bosons are all identical in the same way that the 8 gluons are identical. In particular, you can't point at some subset of the 24 and say "these are the gluons", or "this one is the photon". They were all completely interchangeable.

---

As the universe cooled, spontaneous symmetry breaking occurred. To understand this, consider slowly cooling a lump of iron to below the Curie temperature. As this temperature is passed, the iron spontaneously magnetizes; since the magnetization picks out a specific direction, rotational symmetry is broken.

In the early universe, the same process occurred, though the magnetization field is replaced with an analogue of the Higgs field. This split apart the $SU(5)$ gauge group into the composite gauge group we have today.

The process of spontaneous symmetry breaking is closely analogous to phase transitions, like the magnetization of iron or the freezing of water, which is why we talk about 'strongly/weakly unified' matter as separate states of matter. Like the iron, which state we are in is determined by the temperature of the universe. However, a exact theoretical description of this process requires thermal quantum field theory.

newtonian mechanics - How is a running man able to Accelerate himself?

Suppose a man is running and he gradually speeds himself up . For this he applies a force on the ground backward and the ground pushes him forward . This is probably due to to friction between the shoes of the man and the ground . The friction acting in this case is Kinetic friction which has a constant magnitude. Since the magnitude of frictional force is constant, how is the man able to accelerate himself ? Who is providing this force?

Answer

Don't think kinetic friction, just because some part of him is moving. Is it also kinetic friction if I stand still but swing my arms? Many particles in him or elsewhere might or might not move. They are irrelevant.

Only the particles in contact with the ground are relevant.

And they are not moving. His foot is not moving during the step. It is stationary and not sliding while touching. There is static friction here.

And static friction can vary easily.

quantum mechanics - How does light oscillate?

Why do we say that electromagnetic wave is oscillating? Or does light propagate really in a wavy form like this image?

What is making the photons oscillate and how is it oscillating is it oscillating back and forth (and if oscillating back and forth how does it move if its moving back and forth at the same point?)?

Why do we say that the EM field is constantly changing when in the above picture its constant? (we know that a compass is not able to follow an electromagnetic radiation because the EM field is constantly changing)

Answer

In light propagation, oscillation does not mean any movement in space. It is the value of the electromagnetic field, at one given point in space, that oscillates.

The picture that you quote does not represent the movement in space, but the electromagnetic field value as a function of time.

Compare to waves in water: if you put a little boat on the water, the boat will actually go up and down when a wave passes by, showing that the water is indeed going up and down. For electromagnetic waves, there is no matter or photons that go up and down.

Instead, you have to imagine that there is a little arrow associated to each point in space: this little arrow is the electric field direction. Another arrow, at the same point, is the magnetic field. These two arrows change size and direction with time, and in fact they oscillate. But remember they only are related to one point in space. The next point in space will have different arrows. The whole space is filled with many arrows, one at each point, and they interact with each other. This interaction allows for the oscillation of one of them to be transmitted to the next one, and the next , etc.

newtonian mechanics - Would a sneeze by a cosmonaut in a spacesuit affect his movement?

Naive question; feel free to shoot me down

It is a truism that any motion in space would continue indefinitely unless it is opposed by an external force. If a cosmonaut were to sneeze within his/her spacesuit, would it have any impact upon their movement? I assume the suit and cosmonaut would, between them, totally absorb the force exerted by the 100mph sneeze air velocity. This would leave the cosmonaut unaffected ... probably.

Answer

When the cosmonaut sneezed they would start moving, and rotating, in the opposite direction, but when the sneeze hit their faceplate (ugh!) this would stop the motion. The net result is that the velocity of the cosmonaut would not have changed, but their position and angle would have.

According to Wikipedia a typical breath is 500cm$^3$ and a sneeze velocity is around 15m/s. If the density of air is about 1.2kg/m$^{3}$ the momentum of a sneeze is about 0.009kg.m/s. I weight about 70kg, so the sneeze would leave me moving at about 0.0013m/s. Lets say the faceplate is 5cm away from my mouth, then with the sneeze moving at 15m/s I'd only move for 0.0033s before the sneeze hit my faceplate (ugh again!) and stopped me. In that time I'd have moved about 4 microns.

I must admit that's less than I thought when I started this.

nuclear physics - Why is there a scarcity of lithium?

One of the major impediments to the widespread adoption of electric cars is a shortage of lithium for the batteries. I read an article a while back that says that there is simply not enough lithium available on the entire planet to make enough batteries to replace every gasoline-powered car with one electric car. And that confuses the heck out of me.

The Big Bang theory says that in the beginning, there was a whole bunch of hydrogen, and then lots of hydrogen started to clump together and form stars, and those stars produced lots of helium through fusion, and then after helium, all the rest of the elements. That's why hydrogen is the most common element in the universe by far, and helium is the second most common.

Well, lithium is #3 on the periodic table. By extrapolation, there ought to be several times more lithium around than, say, iron or aluminum, which there is definitely enough of for us to build plenty of cars with. So why do we have a scarcity of lithium?

No-slip boundary condition for viscous fluids

When dealing with fluid mechanics of viscous fluids, both theoretically and numerically, I've always been told that the boundary condition applied at solid walls has to be a no-slip one. My teachers or textbooks never really explained why, except sometimes "well, the fluid's viscous, so it sticks to solid" which is far from an explanation to me. Therefore I'm reading a little bit to understand the real origin of this condition, and so far there is on thing that I don't understand in what I've found: in Volume II of Modern Developments in Fluid Dynamics by S. Goldstein, it is written that:

"[...]; finally he [Navier] decided on the first [hypothesis on the behaviour of a fluid near a solid body], on the grounds that the existence of slip would imply that the friction between solid and fluid was of a different nature from, and infinitely less than, the friction between two layers of fluid, and also that the agreement with observation of results obtained on the assumption of no slip was highly satisfactory."

I do not understand what is in bold: what does the "different nature" of friction means, and why would it be "infinitely less"?

NB: After this statement, Goldstein refers to a bibliographic entry, but I don't know if it's possible to find such archive on the Internet. Here it is anyway: Trans. Camb. Phil. Soc. 8 (1845), 299, 300; Math and Phys. Papers, 3, 14, 15.

Plasma and Plasma Globes

Plasma is the fourth state of matter. Wikipedia says, that all the gaseous atoms will be ionized into positive ions and free electrons at extremely high temperatures. But, what does this explain in a plasma-globe and other applications. Do all these artificial apps have a relation with the plasma in stars...? If so, how is this high temperature maintained in such a small area?

If this relation is perfect, then plasma could be used for conduction of electricity, as it has extra-ordinary number of free electrons. Am I correct in stating that plasma conducts electricity very well?

Answer

The ions and the electrons don't necessarily have the same temperature (non-thermal plasma), but if you leave them for a while, they will undergo equilibration. I would not overestimate the value in Kelvins of different degrees of freedom of subsystems. The temperature is associated with a mean kinetic energy. If you tackle an electron, you can accelerate it easily because of its low mass. Conversely, even a fast electron will not give raise to the same momentum transfer as a heavy particle. So a fast electron is "not as powerful" as an equally fast ion.

If you have an application like the ball, there the effect is mainly generated by accelerating of electrons in the electric field. If you go away from the electrodes, the field gets weaker and there the electrons lose their kinetic energy due to collisions with heavier particles. This is why a too high particle density (or pressure) is not the friend of open corona discharges - the glow effect can't extend too far away without an opposing charge somewhere else, such that there is a relevant electric field in between. Of course, if the temperature is generally high (thermal plasma) as in the sun, then you will have charges flying around in any case. But for the earthly applications you have in mind, the area containing free electrons/ions doesn't extend forever and the temperature will not kill you unless the electric field that produced it is super strong.

Then as Shaktyai pointed out, plasmas are not always totally ionized, usually the opposite is the case. For some cases the Saha equation holds and there you get an idea about the functional dependence of the ionization degree with temperature. For high $T$, the factor goes against 1 (graph exp(-1/x) in wolfram alpha or so).

quantum mechanics - Translation Operator and Position Basis

In Modern Quantum Mechanics by Sakurai, at page 46 while deriving commutator of translator operator with position operator, he uses $$\left| x+dx\right\rangle \simeq \left| x \right\rangle.$$ But for every $\epsilon > 0$ $$\langle x+ \epsilon \left| x \right\rangle = 0.$$ Therefore this limiting process $$\lim_{\epsilon \rightarrow 0} \left| x+ \epsilon \right\rangle = \left| x \right\rangle$$ does not make sense for me. I couldn't derive commutator relation without using these.

Answer

The derivation by Sakurai is by no means mathematically rigorous, so you should expect something like your argument about the scalar product. Indeed, we have everything more or less fine until $$ [x,\mathcal{T}(\epsilon)]|z\rangle=\epsilon|z+\epsilon\rangle $$ where we want to replace $|z+\epsilon\rangle$ by $|z\rangle$ and claim that it is ok in the first order in $\epsilon$. As soon as position eigenstates are non-normalizable, there is no measure of 'smallness' to use in our reasoning about orders. However, what makes sense is to deduce $[x,\mathcal{T}(\epsilon)]=\epsilon\mathcal{T}(\epsilon)$, which is true for any finite $\epsilon$. Here the reason why everything works nice is that $\mathcal{T}$ is a good bounded(=continuous) operator which is defined on the whole Hilbert space of states, and is easily understood even on the generalized vectors like $|x\rangle$. In fact, if you work in coordinate representation, you can deduce this commutator working only with normalizable wavefunctions, on which the action of $x$ is defined (they remain normalizable after this action), giving completely rigorous mathematical sense to your calculation.

What is different when you try to deal with Sakurai's $K$ (which you are trying to do every time when talking about infinitesimal translations) rigorously, is that it is a bad (unbounded, discontinuous) operator. Indeed, in a sense, $$ K=i\left.\frac{d}{d\epsilon}\mathcal{T}(\epsilon)\right|_{\epsilon=0}. $$ But the only way to give sense to this formula is to define the action of $K$ on states: $$ K|\psi\rangle=i\lim_{\epsilon\to0}\frac{\mathcal{T}(\epsilon)|\psi\rangle-|\psi\rangle}{\epsilon} $$ But this limit exists only for certain good states, which we say are in the domain of $K$. In fact, if you look at $K$ in the coordinate rep, it is just $-i\frac{d}{dx}$, which is defined on the (everywhere dense) subspace of differentiable functions of the space $L_2$ of square-integrable functions. When you deal with $K$ rigorously, you have to restrict yourself to the domain of $K$ (for example, if you consider joshphysics answer, where every formula with $K$ is restricted to the domain, it is almost a rigorous proof).

However, due to some reason, which is surely related to the fact that the domain $D(K)$ of $K$ is everywhere dense -- any state can be approximated by a state from $D(K)$ to any desired accuracy, a careless treatment like that of Sakurai works.

classical mechanics - Water bottle moment of inertia

I've noticed that I can make a full water bottle spin about its short axis easier than I can make it spin when it is 1/4 or 1/2 full. Also, when it is spun and is not full, the geometric center of the water bottle moves in a bizarre way (kind of circular).

In general the moment of inertia is directly proportional to the mass. So why is this the case?

I can see that the water in the bottle has more freedom to move when it is not full than when it is full, and this affects the spin. My idea is that it maybe has something to do with the fact that the center of mass of the system is not fixed to a certain point relative to the bottle when this bottle is not completely full.

But I'd like to see a mathematical explanation of this phenomenon.

Answer

This is a late answer; a recent question was marked as a duplicate of this.

The phenomenon discussed in the question goes under the general concept of the dynamics of sloshing liquid. Sloshing liquids can overturn tank trucks, derail railroad tanker cars, capsize ships at sea, crash aircraft, and cause spacecraft to lose controllability. This makes this a very important concept for economic and safety reasons and hence is the subject of many journal articles and entire technical books.

The dynamics of slosh are nonlinear, rather complex (particularly so if the sloshing is extreme and creates bubbles), and are highly dependent on container geometry. As a general rule, a container that is nearly full or nearly empty of fluid doesn't slosh much, and sloshing is at its worst when the container is close to half full.

Excitations from vehicle suspension, from vehicle acceleration and braking, from ocean waves, and from the control systems of aircraft and spacecraft can turn low amplitude sloshing into high amplitude sloshing, something that is best avoided.

But I'd like to see a mathematical explanation of this phenomenon.

You are inadvertently asking me to write a lot. Go to scholar.google.com and books.google.com and search for "slosh dynamics" and you'll see how much has been written on understanding and mitigating slosh. I'll instead provide an overview.

Most slosh models are a bit ad hoc. A simple approach is to model the fluid as being partitioned into a fixed part (one that moves with the container) and a sloshing part, with the sloshing part modeled as a spring/mass/damper system or as a damped pendulum system. The slosh wave slams into the container wall, and this has to be modeled as well. This works well for low amplitude slosh, not so well for high amplitude slosh. These low amplitude slosh models yield a natural slosh frequency. These models predate modern computing.

More recently, slosh has been studied using computational fluid dynamics, and even more recently, with smoothed-particle hydrodynamics originally developed by astrophysicists to model galaxy formation, star formation, supernovae, etc. The same techniques work quite nicely to model more mundane fluids such as sloshing in a container.

electromagnetism - Why is the B field of a solenoid equal to $mu_0 i n$ while that of a loop is $frac{mu_0 i R^2}{2(R^2+z^2)^{3/2}}$?

In the B field of the loop, if R is the radius and z is the distance along the axis perpendicular to the center of the loop, let z go to zero, and multiply by N loops. Starting with the B field of the loop axis: $$B=\frac{\mu_0 i R^2}{2(R^2+z^2)^{3/2}}$$ Becomes: $B_{\text{N loops}}=\frac{\mu_0 i N}{2R}$ and not $B_{\text {solenoid}}=\mu_0 i n$. What is the difference?

Answer

You can think of a solenoid as containing an infinite number of loops stacked one on top of the other. Thus, the expression for one loop becomes a small contribution to the net field of the solenoid: $$ B_{loop}\to dB_{solenoid}= \frac{\mu_0 (nidz) R^2}{2(R^2+z^2)^{3/2}} \tag{1} $$ where $n$ is the number of turns per meter so that $ndz$ is the number of current loops in a stack of thickness $dz$. Basically $n$ measures how densely you stack your loops.

Summing over all these loop contributions gives $$ B_{net}=\int_{-\infty}^\infty dB =\mu_0 ni \tag{2} $$ as in the solenoid.

This solution, which uses the superposition principle, is "easy" because the field on the symmetry axis of a loop is easy to compute.

A more general approach, using Ampere's law, shows that the field is constant inside the soleinoid, even for points that are off-axis. This latter result can also be shown using superposition but the integrations involved are a lot more technical.

quantum mechanics - Tunneling v. Hopping

Can someone explain the difference between hopping and tunneling? The context I'm considering is conduction in semiconductors, specifically between impurity states within the bandgap. It's always been my understanding that hopping is tunneling. Variable range hopping and nearest neighbor hopping, as I understand it, are both forms of tunneling between overlapping states (for example, see http://igitur-archive.library.uu.nl/dissertations/2002-0806-101243/c4.pdf). However, in papers such as DJ Thouless 1974 Electrons in Disordered Systems and the Theory of Localization, hopping and tunneling are described as two different processes. I guess I don't exactly understand what "hopping" is supposed to mean, with respect to charge transport.

electrostatics - What was discovered first - The Coulomb constant or Gauss law?

I checked out some resources on how the constant of proportionality of the Coulomb force was discovered and to my surprise, I found out that it was mathematically derived (unlike the Cavendish experiment for the gravitational constant) by using Gauss's law.

When I searched for the proof of Gauss's law, it used Coulomb's law WITH THE COULOMB CONSTANT AS $1/4πε$. Surely I am missing something, can you guide me?

Links - For Coulomb's constant, check out the 'Value of the constant' section https://en.wikipedia.org/wiki/Coulomb%27s_constant

For proof of Gauss's law, check out 'Deriving Gauss' law from Coulomb's law' in 'Relation to Coulomb law' section https://en.wikipedia.org/wiki/Gauss%27_law

Answer

You might want to read https://hsm.stackexchange.com/q/3553 for the history part.

As far as the confusion regarding Gauss law and Coulomb's law is concerned, you really can't prove either of them independently. And it makes sense, as Coulomb's law or Gauss law describe something related to the real world, and not something that could be logically or mathematically deduced. You can't tell whether masses attract just by logic. Similarly, you can't tell if Coulomb's/Gauss law is true by mere mathematics. You have to do experiments and draw conclusions from their results.

Then why do we even consider them true? That's because no experiment till date has given a result that defies them (that's pretty much the case with almost all physical laws which are considered true).

So, if you consider Gauss law to be more fundamental, then Coulomb's law is its consequence, and vice versa.

electrostatics - Charge Distribution On Hollow Sphere

Say we have a hollow conducting sphere (with some finite thickness). If this object has an excess charge amounting to +Q coulomb, and there is no extra electric field in the surroundings (due to other charges), how will the charge be distributed?

Intuitively it seems that the charge will be symmetrically distributed..(all other cases seem too ugly), but in this case how is charge inside the conducting surface zero? Does this have something to do with solid angle and the fact that two cones with vertex at a point in the conductor will have a special relation of the (charge/distance^2) factors? (more charge is distance is more...)

If this is the case can someone please prove that the effect of the two corresponding cones will cancel each other out?

Answer

Same charges repulse each other. So when they are confined in a system, they try to have stable distance among them as much as possible. For a hollow conducting sphere this stable maximum distance is equal distribution of charges in the outer surface. If any charge try to go the inside conducting surface it automatically decrease distance which increase repulsion. That's why charge inside the conducting surface zero.

quantum mechanics - Compactification of dimensions in string theory: Why our Universe has 3 large spatial dimensions?

Is the only way that string theory can respect the principles of quantum mechanics, and Einstein's special theory of relativity, is to formulate it in a hypothetical nine dimensional space?

You could use string compactification for $d$ number of non compact directions, all that is required to have $d$ non-compact directions is to choose an appropriate compact space of dimension $9-d$.

The question that arises from that is: Why does our Universe have 3 large spatial dimensions?

symmetry - The Ozma Problem

The "Ozma problem" was coined by Martin Gardner in his book "The Ambidextrous Universe", based on Project Ozma. Gardner claims that the problem of explaining the humans left-right convention would arise if we enter into communication (by radio waves, no images allowed) with life on another planet. We can ask the aliens to perform any experiment they want. It is claimed that the classical experiments with magnets, electrical currents, light polarization, gyroscopes etc. can't solve the problem. The only simple experiment that solves the problem is the beta decay in which the parity is not conserved and this can be used to distinguish humans left-right spatial relations. But all this are quite old, the parity violation in weak interactions was experimentally proved in 50s by Chien-Shiung Wu. I was wondering if something has changed since then and if there are some other experiments (maybe some thought experiments) and theories that can solve the Ozma Problem.

Answer

I haven't read that book, but I did read Feynman's discussion of (sounds like) exactly the same thing. Easy: Tell the aliens how to build a telescope, then describe the configuration of some galaxies near them. OK OK, but suppose we rule that out: We can't see any objects in common. Easy: Send them circularly-polarized radio waves (thanks @Anonymous Coward). OK OK, let's say our radio waves must be linearly polarized. Easy: Tell them to look at almost any phenomenon related to the weak force, for example the beta-decay of cobalt-60 in a magnetic field. But then there's one more catch--what if the aliens are made of antimatter and they actually were watching the beta-decay of antimatter-cobalt-60? In Feynman's discussion (if I recall correctly), that's where it ends: There's no way to be really sure that the aliens understand right and left correctly, because they may be made of antimatter.

But since 1964, when CP-violation was observed, we can even eliminate that possibility: We tell the aliens how to watch kaons decay (for example) and then the aliens can figure out whether they're made of (what we call) matter or antimatter, and therefore they can figure out which way is left and right without any more ambiguity. So I guess that aspect is a slight update from pre-1964 descriptions.

Watching atoms decay in a magnetic field is a pretty simple thing to do by the standards of particle-physics experiments. I don't know of any parity-violating experiments that are much simpler than that. It just has to involve the weak force.