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 $B_\mu u^\mu$.

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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:

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  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.

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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.

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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.

differential geometry - Are contravariant basis vectors and basis 1-forms identical?

The reason I'm asking this is because I am trying to develop a set of notes from my reading of MTW (and Wrede, Menzel, Bergman, etc.).

I represent covariant basis vectors with $\mathfrak{e}_{i}$, or if they are orthonormal $\hat{\mathfrak{e}}_{i}$. In cases where the author is explicitly using the terminology "contravariant basis vector", I use, $\mathfrak{e}^{i}$ and $\hat{\mathfrak{e}}^{i}$ for the dual entities. My desire to represent basis 1-forms using identical with contravariant basis vectors started as a software issue. I don't have a pretty way to consistently represent bold-faced Greek letters with the editors I use. But it has become a matter of principle.

I believe I can continue to use my contravariant basis notation for the basis 1-forms appearing in MTW with impunity. That is, I won't have to go back and change my notes because it turns out that basis 1-forms and contravariant basis vectors are geometrically distinct entities.

I have no way of concisely summarizing my understanding of these concepts. Every author has a different approach to developing the fundamentals of tensor analysis and differential geometry pertinent to applications in physics.

There are two prima facie different "schools".

For example, Menzel and Wrede use dual covariant/contravariant basis vector sets. The MTW school uses covariant basis vectors and their dual basis 1-forms. I understand that basis vectors and basis 1-forms are intended to induce conceptually different notions in the student's mind. As I understand these, the (covariant) basis is represented as an arrow because that depicts the distance along one coordinate curve between points of intersection with other coordinate curves. That is, the distance between "lines" of the coordinate mesh.

Basis 1-forms represent the "density" of intersecting coordinate curves. So an arrow representation is misleading. By this reasoning, depicting the gradient with an arrow is misleading because it is actually a 1-form which represents the density of surfaces of constant value. (There is, however, a signed normal direction to each such surface.)

Contravariant basis vectors are typically concocted in some way to support transformation invariance.

But, to me, basis 1-forms and contravariant basis vectors appear to represent the same commodity with different labeling. For example: if we denote the dual basis 1-forms by $\omega^{j}$, then

$$\mathfrak{e}_{i}\cdot\mathfrak{e}^{j}=\left\langle \mathfrak{e}_{i},\omega^{j}\right\rangle =\delta_{i}^{j} \, .$$

Is this a reasonable assessment? Is there an example situation where identifying basis 1-forms (dual to covariant basis vectors) with contravariant basis vectors (dual to covariant basis vectors) fails?

Answer

Gravitation and Inertia Ignazio Ciufolini & John Archibald Wheeler" pages 416-417. 1-forms and covariant vectors are defined as synonymous. of the transformation law for 1-forms they state ”This is just the coordinate dependent definition of a covariant vector."

The answer is: yes, the basis 1-forms and contravariant basis vectors typically used in physics are identical objects. Any distinctions involve notational differences, particularly in how they are used in binary operations. Most significantly; when the basis 1-forms are written as coordinate differentials $dx^i$ and are used as the projection operators mapping vectors to their individual components, we have the defining characteristic of a basis 1-form:

$$\left\langle dx^{i},\mathfrak{v}\right\rangle =\mathfrak{e}^{i}\cdot\mathfrak{v}=v^{b}\mathfrak{e}_{b}\cdot\mathfrak{e}^{i}=v^{i}$$.

In addition, both types of object transform by the same contravariant transformation law.

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Edit to add the following references:

Compare Mathematical Physics, by: Donald H. Menzel $\S\S$ 2-31 Tensor Analysis, to Gravitation, by Charles W. Misner, Kip S. Thorne & John Archibald Wheeler $\S\S$ 8.4 Tensor Algebra in Curved Spacetime, Box 8.4 Tensor algebra at a fixed event in an arbitrary basis.

See also the lecture notes Vectors and Dual Vectors PDF

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Edit to add graphics.

The first image depicts a vector (blue) resolved onto a covariant basis (red and green), and onto a contravariant basis (cyan and magenta). The basis 1-forms dual to the covariant basis are likewise depicted in cyan and magenta. They are "surfaces" of constant coordinate value. So, if the red line is tangent to the coordinate inducing the red basis vector, the cyan lines are tangent to the subsequent coordinate curves "parallel" to the red line, etc. Notice that the dual (contravariant) basis vectors are orthogonal to their corresponding covariant basis vectors. Also observe that the contravariant basis vectors are orthogonal to their corresponding 1-form surfaces. The same geometric object was used to draw both the 1-forms and the contravariant basis vectors. They were just passed to different functions.

The second image depicts a vector "resolved" onto the covariant basis in terms of both covariant and contravariant components. This requires some scaling which is described in Theoretical Physics, By: Georg Joos, Ira M. Freeman

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Edit to add: The following image shows equivalent forms of a the component matrix associated with a change of coordinate induced bases. It is by no means exhaustive. It would be too much effort to translate it from Mathematica to LaTeX where I produced it. Of course, one needs context in order to expand the single row and single column forms to show their equivalence with the other forms.

radiation - Can you speed up radioactive decay of plutonium?

We all know the issue of deep geological repositories for fuel rods. Is there a currently feasible way to speed up the rod's decay to render them harmless in less than 10 years?

nuclear physics - What is an intuitive picture of the motion of nucleons?

I understand the "motion" of electrons within an ordinary atom (say argon at room temperature and pressure). They are moving in "orbits" defined by quantum mechanical wavefunctions where the "orbits" are smeared out in space with some locations more probable than others but they don't possess a definite trajectory. But in some sense, we know they are moving around the nucleus in some rough shape orbits even if we can't hope to describe the motion in a Newtonian sense.

My question is what is the similar "intuitive" picture of nucleons in an ordinary atom. Is the location of the nucleons fundamentally more "locked down" than for electrons? If so, does the nucleus look similar to a solid where the individual nucleons are vibrating (maybe w/ some QM smearing) around equilibrium positions? Also, recognizing the strong force will dictate what is possible, can nucleons "jump" around their neighbors to a different "location" in the nucleus (maybe say with a coincident jump of a similar particle exchanging positions)?

Answer

Shell models of the nucleus propose orbitals for nucleons much like those we see for electrons in an atom, but the "distances between them" (measured from regions of maximal likelihood or somesuch) are comparable to the size of the nucleons, so they can't really be thought of as separated in the way one can figure that atomic electrons have regions of maximal likelihood that are distinct from one another. That's why the liquid drop models that nate mentions were pretty useful.

The models require as inputs some potentials and form-factors that are not as well understood as those that affect electrons. Further the shell model must be solved numerically in all but the simplest cases so the results are only approximate.

My dissertation work included taking pictures of the shell distributions (of protons only) in momentum space. Here is one view of a carbon nucleus (after considerable manipulation to approximately remove the final state effects):

the horizontal scale is basically binding energy and you can see the p- (high peak on the left) and s-shells (low peak on the right). The vertical scale has arbitrary units. The lines are to guides the eye only as they have limited physics content.

This figure shows both the energy dependence and the momentum dependence

where the red diamond markers represent the (manipulated to approximately remove final-state interactions) data and the blue circular markers the result of a Monte Carlo calculation based on a shell model using earlier spectral functions and the de Forest prescription for the nuclear form-factors. As before the vertical scale is arbitrary and the horizontal scale is in GeV (natural units).

The double-bump in the momentum data are (in effect) the lobes of the p-orbitals, and the fact that the center is distinctly non-zero is the mark of the s-orbital.

Spatial distributions of bound objects are related to the momentum distributions by a Fourier transformation. You can't actually work that transformation on this figure because it is in energy rather than momentum space, but you might well guess that there would be a lot of spatial overlap and you would be right.

general relativity - Is there a binary black hole system in the middle of the galaxy?

We have observed gravity effects from black holes in the center of galaxies, but galactic centers are dusty so we can’t tell if it’s one black hole or two black holes in a binary system in there. A new paper is about the recent discovery of a binary black hole system, they zoomed in on a black hole with a radio telescope, and observed that it actually where two black holes in a binary system. The black holes were 250 light years away and had an orbital period of 4 million years,source. Binary black holes are then no longer just a hypothetical idea. Binary black holes are then no longer just a hypothetical idea.

The stars in the galactic disk have almost the same speed the center about 200-250 km/s.

This does not fit with Newtonian physics and Kepler’s third law. The most accepted theory why this doesn’t fit is the presence of dark matter. The dark matter has to be distributed in such a way that it pulls the stars so they get the same orbital speed, and most of it has to be placed at the outskirts of the galactic disk. Dark matter is currently quite mysterious and there are many proposals to what it can be, we don’t know what it is, but we know it is heavy and weighs more than the galaxy. We currently think the universe consists of about 23 % dark matter, 73 % dark energy, 3,6 % intergalactic gas and only 0,4 % stars:

We neither know why dark matter would be distributed in the way we think it is, and we might wonder if this almost constant orbital speed of stars is just a coincidence. Could there be alternative solutions to the dark matter problem? Is there another solution that would explain that the stars have almost the same speed around the center of the galaxy? If there were some sort of engine in the galactic center that pulled all the stars with equal force around the middle, we could do without dark matter. But such an engine might be an absurd thought, as we don’t know any wave or force that could transfer energy in such a way on this scale.

According to Einstein accelerated mass creates gravitational waves, and a binary black hole system has great acceleration of mass and is a huge gravitational wave source. The binary black hole system creates a gravity wave that travels radially outwards with the speed of light, and the rotation of the binary black hole system creates a spiral of gravitational waves:

Put together with our spiral galaxy the gravitational wave spiral will look something like this:

The binary black hole system probably has slow rotation, thousands or maybe even millions of years, while the gravitational wave goes radially outwards with the speed of light. This creates a pretty steep gravitational spiral that will almost perpendicularly intersect the spiral arms of the galaxy.

I pictured the gravitational wave spiral as moving gravity well ripples, that sweeps the galactic disk. And as stars that interact with the ripples, they get accelerated for a while.

But I found no scientific confirmation that gravitational waves can accelerate mass, only how gravitational waves squeeze mass and set it into some kind of vibration. I started to wonder if I had pictured it all wrong. Then I found some information on how Einstein looked at gravity waves:

“Gravity, according to Einstein's general theory of relativity, is how mass deforms the shape of space: near any massive body, the fabric of space becomes curved. But this curving does not always stay near the massive body. In particular, Einstein realized that the deformation can propagate throughout the Universe, just as seismic waves propagate in Earth's crust. Unlike seismic waves, however, gravitational waves can travel in empty space — and they do so at the speed of light." source.

From the slingshot effect we know that one moving gravity well can accelerate another. Like when Voyager 1 was accelerated by the moving gravity-wells of Jupiter and Saturn. So could a binary black hole in the middle of the galaxy create moving gravity well ripples in spacetime that can accelerate stars? So far gravitational radiation has not been observed and we can’t even be sure it exists, and its interaction with matter is on a theoretical basis. But if there is an engine in the middle of the galaxy that can replace dark matter, and accelerate the stars to the same velocity, gravitational radiation might be a part of the solution. The black hole also accelerate mass to relativistic speeds and have huge relativistic mass, which have relativistic moving gravitational fields, and as gravitation act at the speed of light, this might give different gravitational effects than ordinary mass, especially in a binary black hole system. Steven Hawking has proposed Hawking radiation from black holes due to quantum effects. We also have the black hole information paradox, if information is lost in a black hole it violates the principle of quantum mechanics “–that complete information about a physical system at one point in time should determine it state at any other time.”

So there might be some unknown radiation or force that may accelerate the stars in the galactic disk, at least we should hold this possibility open, as it might give a more elegant solution, without the use of the mysterious dark matter.

refraction - Do electromagnetic waves always move in straight lines?

When we send an electromagnetic short wave to the sky, it reflects due to the ionosphere effects. But if we send it horizontally, is it correct that it moves around the surface of the earth, and if it has enough energy, it can return to its first position?

If yes, then how could that happen?

Answer

Due to the refraction index dependence on the air density, optical (and radio) rays bend in the atmosphere. It turns out that an atmospheric layer with temperature inversion (temperature increasing with height) can create conditions where the curvature of the ray would match the curvature of the Earth surface. This is called "circulating rays" and these web-pages contain explanations and references on this phenomenon: http://mintaka.sdsu.edu/GF/explain/simulations/ducting/duct_intro.html, http://mintaka.sdsu.edu/GF/explain/atmos_refr/bending.html#circulating

electromagnetism - How do we prove that the 4-current $j^mu$ transforms like $x^mu$ under Lorentz transformation?

Given that the position vector $\textbf{r}$ to be a vector under rotation, we mean that it transforms under rotation as $\textbf{r}^\prime=\mathbb{R}\textbf{r}$. Now, taking two time-derivatives of it, one can easily see that the acceleration $\textbf{a}=\ddot{\textbf{r}}$ transforms as $\textbf{a}^\prime=\mathbb{R}\textbf{a}$ i.e., also behaves as a vector under rotation.

Now a four-vector is something which transforms under Lorentz transformation as $x^\mu$ does. Given the transformation of $x^\mu$: $$x'^\mu=\Lambda^{\mu}{}_{\nu} x^\nu\tag{1}$$ how can one show that the four-current density $j^\mu$ also transforms like (1) preferably from the definition $j^\mu=(c\rho,\textbf{j})$?

visible light - What is Gray, from a physics POV?

Quora explains how white and black colors fit into the spectrum of visible light. It explains that white is all colors together while black is the lack of color.

So, where is Gray? Gray is the mix of all-colors and no-color!! What does that mean? Can somebody explain Gray, the mysterious color?

Answer

All colors are only in the mind. Light has a mix of wavelengths, but it doesn't have color until someone sees it.

When light enters the eye, it hits rods and cones in the retina. Cones are color receptors. There are three kinds. Each kind is sensitive to a range of wavelengths. Color is the result of stimulation of the cones, and additional processing in the brain.

The image is from The Color-Sensitive Cones at HyperPhysics. Copyright by C. R. Nave, Georgia State University. A good starting link is the Light and Vision page.

Loosely, the sensors are sensitive to long, medium, and short wavelengths. The ranges overlap. Most light, even single wavelength laser light, stimulates more than one. The graph shows which are stimulated by single wavelength light at different wavelengths. The colors we see are determined by the mix of stimulations. The bottom of the graph gives names of colors for single wavelength light.

Grey is not on the list. Grey requires a mix of wavelengths that stimulate the three types more or less equally. So do black (very little stimulation) and white (more).

There is more to it than that. The perception of color is affected by colors around it. There are photographs where two different patches reflect the same light. But the colors we perceive are different because of the surroundings. For example, see the Checker Shadow illusion.

By Original by Edward H. Adelson, this file by Gustavb [Copyrighted free use], via Wikimedia Commons

Also no wavelength will stimulate only the "Green" cones. They are always stimulated in combination with other cones. I once read it is possible to stimulate them with a probe. The person saw a color he had never seen before. I wish I could find a link. Quora might be a good place to start.

humidity - Why doesn't the moisture freeze out of the air when the air temperature is below freezing?

I would think that there would be zero humidity outside right now and it's 21 degrees. The weather site says there is 71% humidity.

Answer

Even at the height of summer, it's well below the boiling point of water, and yet there's still water vapor in the air. The reason that water vapor can exist in the atmosphere below the boiling point is the same reason that water vapor can exist in the atmosphere below the freezing point. Water can exist in vapor form even at very low temperatures.

If you have a modern refrigerator/freezer, you've probably seen ice cubes that have become smaller and riddled with holes when left in the freezer for a long time. This is because your freezer operates at an extremely low humidity to keep ice from depositing on the walls of the freezer. This low relative humidity makes the cubes gradually sublimate (transform directly from ice to water vapor without passing through the liquid state), resulting in a decrease in size and a pockmarked surface.

I would think that there would be zero humidity outside right now and it's 21 degrees.

The answer to your question is found in the phase diagram for water, depicted below.

_{Source: https://en.wikipedia.org/wiki/File:Phase_diagram_of_water.svg}

Note that water vapor, marked in tan in the above diagram, exists well below freezing. Below the triple point, ice sublimates directly into water vapor without transitioning to liquid water. (Similarly, water vapor at temperatures below the triple point temperature undergoes deposition to directly transition from vapor to ice, without an intermediate transition to liquid water.)

The reason water can exist in vapor form below the boiling point is that atmospheric pressure is irrelevant. All that matters is the partial pressure of the water vapor in the air. Liquid water will evaporate and ice will sublimate when the partial pressure is below the saturation level. The partial pressure of completely saturated air can be found as the point at which a vertical temperature line intersects the upper boundary of the tan region in the above phase transition diagram.

The weather site says there is 71% humidity.

Relative humidity is the ratio of the amount of water vapor that is present in the air versus the amount of water vapor that would be present in completely saturated air at the same temperature. The saturation pressure at -6° C (21° F) is 364 pascals. That represents about 3 grams of water vapor for every cubic meter of air. 71% humidity means that the air instead held about 2 grams of water per cubic meter.

forces - What happens when two bodies collide in a dynamics point of view?

I'm a high school student learning about energy and momentum. What confuses me is the things with elastic and inelastic collisions.

I completely (in my opinion) understand those concepts in energy and momentum point of views. The problem is I can't reconcile these concepts with what I learned about forces.

Suppose we have two masses $m_1$ and $m_2$ traveling with velocities $v_1$ and $v_2$ and collides each other. To simplify let it be in 1-dimensional case. Here is my picture of what is going on. After the collision there is a change in velocity, so there is an acceleration. Since there is an acceleration there is also force. But how do we calculate that force? The velocity function is a constant function which has a discontinuity at the time of collision. So it is not differentiable at that time, which means the concept of force and acceleration is meaningless at that time!? I have always thought that energy is just a convenient computational tool and we can do everything with classical mechanics based solely on the concepts of forces and Newton's law.

If my suspect is right, we should go to molecular level to understand what is going on? Because in my understanding collision is the result of some repulsive forces between the atoms and without them there would be no collision as the bodies just go through each other. So if possible, can someone elaborate on this and also explain what causes the two different types of collisions (elastic and inelastic) ?

Answer

You are correct that this is a difficult problem. Energy methods are required to determine the contact force because kinetic energy is stored in strain ('spring') energy during the collision, when both bodies deform.

If the bodies are assumed rigid, the contact force is undefined. As you correctly conclude from $a=\frac {\mathrm{dv}}{\mathrm{dt}}$, bodies cannot decelerate instantly (ie. $t=0$). In reality, the bodies decelerate during the time (and distance) of deformation. Without deformation, $d =0= \frac {v'-v}{2}t$, and $F = \frac {\mathrm {dP}}{\mathrm {dt}}$, require $t = 0$ and $F \to \infty$. Therefore, to determine the contact force between colliding bodies, deformation must be solved.

Impact Mechanics have been studied extensively for engineering applications. Complex collision problems are solved numerically via Continuum Mechanics and Finite Element Methods, as shown below.

In practice, analytical approximations are possible with simple geometry (and conservative assumptions)- this is discussed in the following.

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Think of the problem in three time steps:

1) Before contact:

Both bodies have mass ($m_{1}, m_{2}$) and velocity ($v_{1,i}, v_{2,i}$).

2) During contact:

The time interval begins at contact between bodies and ends at the instant $v_{f} = v_{1,f} = v_{2,f}$. The final velocity is solved by Conservation of Momentum: $$m_{1}(v_{f} - v_{1,i}) = m_{2}(v_{f} - v_{2,i}) \; \Rightarrow \; \therefore v_{f} = \frac {m_{1}v_{1,i}-m_{2}v_{2,i}}{m_{1}-m_{2}}$$ Next, apply Conservation of Energy and assume all energy is stored in strain energy (where no energy is lost to heat, sound, etc.). Strain energy is analogous to energy stored in a spring, where the effective spring constants, $k=\frac {EA}{L}$, depend on the geometry and material of the bodies. $$\Delta E_{Kinetic} = E_{Strain}$$ $$[\frac {1}{2}m_{1}v_{1,i}^2 + \frac {1}{2}m_{2}v_{2,i}^2]_\text{i} - [\frac {1}{2}(m_{1} + m_{2})v_{f}^2]_\text{f} = \frac {1}{2}k_{1}\delta_{1}^2 + \frac {1}{2}k_{2}\delta_{2}^2$$ While it is not possible to analytically solve for the displacement of both bodies ($\delta = \delta_{1} + \delta_{2}$), engineers simplify the problem and solve for the $\delta$ they care about¹; this is a conservative assumption because it predicts a higher contact force than if both bodies deformed. $$[\frac {1}{2}m_{1}v_{1,i}^2 + \frac {1}{2}m_{2}v_{2,i}^2]_\text{i} - [\frac {1}{2}(m_{1} + m_{2})v_{f}^2]_\text{f} = \frac {1}{2}k\delta^2$$ $$\therefore \delta = \sqrt {\frac {[\frac {1}{2}m_{1}v_{1,i}^2 + \frac {1}{2}m_{2}v_{2,i}^2]_\text{i} - [\frac {1}{2}(m_{1} + m_{2})v_{f}^2]_\text{f}}{\frac {1}{2}k}}$$ Finally, the average contact force is determined from: $$\boxed {\therefore F_{contact} = k\delta}$$ Similarly, the same result is obtained from the kinematic equations: $$t = \frac {2\delta}{v_{f}-v_{i}} \;\Rightarrow \; a=\frac {v_{f}-v_{i}}{t} \; \Rightarrow \; \therefore F_{contact}=ma$$

3) After contact.

Elastic collisions are idealized so that all kinetic energy is conserved. Inelastic collisions describe real collisions- some kinetic energy is permanently lost to heat, sound, plastic deformation, etc. and the resulting final kinetic energies are lower (than predicted by the idealized elastic model). A coefficient of restitution ($CoR$) is applied to the Conservation of Momentum to reflect the bodys' final velocities. Note that for elastic collisions, the $CoR=1$. While inelastic collisions are more accurate, the $CoR$ is often unknown; therefore, collisions are assumed elastic to provide a reasonable estimate.

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¹ "Machine Design: An Integrated Approach (2nd Ed.)"- Robert L. Norton

fluid dynamics - How do I calculate the Reynolds number in multiphase flows?

I am modeling a gas flowing through a liquid. How do I calculate the Reynolds number in multiphase flows? And, at what Reynolds number should I consider the flow to be turbulent?

The problem is of a typical drilled wellbore in the oil and gas industry. Fresh cement is poured into an annulus between the steel casing and the drilled formation. Before the cement is set it is vulnerable to formation gas over coming its pore pressure (or hydrostatic head) and flowing through the cement, often times forming channels for more gas to flow through in the future, rendering the cement sheath useless.

EDIT:

I am currently using a Bingham fluid to describe the viscosity of the cement slurry. The yield stress will be in the range of 300 to 700 kPa. The plastic viscosity will be in the order of 0.05-0.10 Pa.s.

It is hard for me to estimate gas velocity but in preliminary models I have seen velocities in the range of 5-10 mm/s in preliminary models.

electrostatics - Electric field outside a capacitor

I know that the electric field outside of a capacitor is zero and I know it is easy to calculate using Gauss's law. We create cylindrical envelope that holds the same amount of charges (of opposite signs) in each plate.

My question is why can't I pick an envelope which include only part of one of the plate? Gauss's law states, specifically, that I can pick any envelope I want.

Note: I have encountered this question a couple of years a go and I got an answer which I was not completely happy with.

Answer

Outside two infinite parallel plates with opposite charge the electric field is zero, and that can be proved with Gauss's law using any possible Gaussian surface imaginable. However, it might be extremely hard to show if you don't choose the Gaussian surface in a smart way.

The usual way you'd show that the electric field outside an infinite parallel-plate capacitor is zero, is by using the fact (derived using Gauss's law) that the electric field above an infinite plate, lying in the $xy$-plane for example, is given by $$ \vec{E}_1=\frac{\sigma}{2\epsilon_0}\hat{k} $$ where $\sigma$ is the surface charge density of the plate. If you now put another plate with opposite charge, i.e. opposite $\sigma$, some distance below or above the first one, then that contributes its own electric field, $$ \vec{E}_2=-\frac{\sigma}{2\epsilon_0}\hat{k} $$ in the region above it. Since the electric field obeys the principle of superposition, the net electric field above both plates is zero. The same happens below both plates, while between the plates the electric field is constant and nonzero.

Your way of doing it is a little more tricky, but again gives the same answer. For example, if you choose the Gaussian surface to have an hourglass shape with different radii for the two sides, then indeed the net charge enclosed is not zero. However, when you calculate the total electric flux through that surface, you have to be careful to realize that there is nonzero electric field between the two plates, and therefore there is a nonzero flux through the part of the Gaussian surface that lies between the plates. That flux, of course, has to be accounted for. Assuming that you know the electric field inside the capacitor, $\vec{E}_\text{inside}$, you can do the integral $\oint\vec{E}_\text{inside}\cdot d\vec{A}$ for such a Gaussian surface (it's not that hard actually), and you find that the flux through the part of the surface that lies between the plates is exactly equal to $q_{\text{enclosed}}/\epsilon_0$. Thus, the net flux through the part of the Gaussian surface that lies outside the plates has to be zero, proving, after a little thought, that the electric field outside the capacitor is zero.

The final answer for $\vec{E}$ never depends on the Gaussian surface used, but the way to get to it always does. That's why the Gaussian surface has to be chosen in a smart way, i.e. in a way that makes the calculation of $\oint\vec{E}\cdot d\vec{A}$ easy.

fluid dynamics - What is going on in front of and behind a fan?

Why is it that when you drop paper behind a fan, it drops, and is not blown/sucked into the fan, whereas if you drop paper in front of a fan, it is blown away?

Answer

There is a YouTube video that visualizes the air flow around a propeller for various configurations. I caught a screen shot of a moment that more or less shows what is going on:

As you can see, this happens at 2:07 into the clip - this happens to be for a dual rotor configuration (two counter rotating blades) but the principle is the same.

Behind the rotor (above, in this picture) the air is moving slowly. Air over a wide range of area is drifting towards the rotor, where it is accelerated.

I will leave it up to others to describe the mathematics behind this contraction - but I thought visualizing the flow would at least confirm your observation that it is indeed slower behind the fan, and faster in front of it. In other words - it pushes, but doesn't suck.

A better image showing the flow lines around the propeller is given at this article about the mechanics of propellers

As the pressure is increased, the flow velocity goes up and the flow lines end up closer together (because of conservation of mass flow). This gives the flow the asymmetry you observed. But it's still more intuitive than rigorous...

AFTERTHOUGHT Hot Licks made an excellent observation in a comment that I would like to expand on. The air being drawn towards the fan is moving in the pressure differential between the atmosphere at rest, and the lower pressure right in front of the fan blades. The pressure gradient is quite small, so the air cannot flow very fast - and it has to be drawn from a wide region to supply the mass flow. After impact with the blade (or at least after "interacting" with the blade), the air has a LOT more momentum that is directed along the axis of the fan (with a bit of swirl...). This higher momentum gives the air downstream of the fan its coherence as can be seen in the diagram.

quantum mechanics - Bound States in a Double Delta Function Potential

Let $V(x) = −u \delta(x) - v \delta(x − a)$ where $u, v > 0$ correspond to a potential with two $\delta$ wells. Let $v > u$. If $a$ is very large, there is certainly a bound state: the particle sits in the $\delta$-well. As $a$ decreases to a certain critical value, the bound state disappears. I need help finding that value.

My idea was: Before the bound state disappears, its energy approaches $0$. I'm trying to assume that the energy $E$ is a very small negative number, solve the Schrodinger equation, and find the suitable value of $a$, but I'm having trouble doing this.

Would someone be able to help me with this problem?

newtonian mechanics - Why body starts moving when force is applied?

The recent question by m.buettner regarding self-inductance and its resistance to EMF Faraday's law - does the induced current's magnetic field affect the change in flux?, recalled me the great question that I always wanted to ask. The 3rd law of Newton says To every action there is always an equal and opposite reaction. This immediately means that action is exactly balanced by reaction force and, because of zero net force, no speed increase should be observed. How do you explain why our concerns with m.buettner are unfounded?

Answer

For each action of body A on body B, there is a reaction of body B on body A. The two forces do not apply on the same body and thus won't cancel each other when looking at the movement of body B only.

thermodynamics - Does a reversible heat engine exchanging heat with an ideal gas that does a transformation imply that transformation is reversible?

Consider a reversible heat engine working between two sources. Suppose that one of the sources is a thermostat, while the other is an ideal gas which follows a transformation and exchanges some heat with the engine.

Does the fact that the engine exchanging heat with the gas is reversible imply that the transformation of the ideal gas (no matter what that transformation is) is a reversible transformation?

In other words, does the following hold?

A reversible heat engine exchanges heat with an ideal gas that does a transformation $\implies$ the transformation of the gas is reversible

Said in another manner, can I have a reversible engine that takes heat from a sudden (but not adiabatic) and irreversible compression of some ideal gas and gives heat to a thermostat?

I'm confused about this beacuse in the definition of reversible heat engine nothing is said about an hypotetical transformation that happens in the source of the heat itself. The only necessary thing is that the cycle inside heat engine is made of reversible transformations. On the other hand, if the transformation in the ideal gas which is the source is not reversible, the heat engine cannot work backwards of course (and that's in contrast with the definition of reversible engine).

Answer

Does the fact that the engine exchanging heat with the gas is reversible imply that the trasformation of the ideal gas (no matter what trasformation is) is a reversible trasformation?

Yes. A reversible engine is an engine that only performs reversible transformations. A reversible transformation in which two systems interact (the engine and the gas-reservoir, in our case) cannot be reversible for one system and irreversible for the other.

The motivation is that reversibility is a property of the transformation as a whole. In a reversible transformation, the variation of entropy of the world (system + environment) is $0$. Another way to say this is that the two systems which are interacting must be in thermodynamic equilibrium throughout the transformation.

Take for example the Carnot cycle, which is the reversible cycle operating between two sources. It goes like this:

1. Reversible isothermal expansion ($1 \rightarrow 2$): The system absorbs a quantity $Q_1$ of heat and $\Delta S_1 = Q_1/T_1$ of entropy from the reservoir $1$.


2. Reversible adiabatic expansion ($2 \rightarrow 3$): This transformation is isentropic, so the entropy of the world is unchanged.


3. Reversible isothermal compression ($3 \rightarrow 4$): The system transfers a quantity $Q_2$ of heat and $\Delta S_2 = Q_2/T_2$ of entropy to the reservoir $2$.


4. Reversible adiabatic compression ($4 \rightarrow 1$): This transformation is isentropic, so the entropy of the world is unchanged.

If we look at the entropy of the world, $S_w$, we see that

1. $\Delta S_{w,1} = \frac{Q_1}{T_1}-\frac{Q_1}{T_1} = 0$


2. $\Delta S_{w,2}=0$


3. $\Delta S_{w,3} = \frac{Q_2}{T_2}-\frac{Q_2}{T_2} = 0$


4. $\Delta S_{w,4} =0$

So that $\Delta S_{w,total}=\sum_{i=1}^4 \Delta S_{w,i}$ is identically $0$. But if we make the first transformation irreversible (and remember: it doesn't make sense to say that it is reversible for the engine but irreversible for the reservoir...), then $\Delta S_{w,1} \neq 0$ and it follows that $\Delta S_{W,total} \neq 0$, so the engine is not reversible.

double slit experiment - Shooting a Single Photon

How is it even possible that a single photon can be isolated and "shot" through a slit? What tool or mechanism or setup allows such an incredible feat, especially since the photon has no mass?

In mechanics, is shock really better expressed as jerk instead of acceleration?

Some expensive electronics or mechanical devices are designed to be shock-resistant. However, the manufacturers often market the level of shock-resistance in units of g-force (I know g-force is really a measure of acceleration). I'm not really convinced that that's the proper unit.

In fact, the Wikipedia article for mechanical shock describes shock as a sudden acceleration or deceleration. Here, the term "sudden" seems to imply that the acceleration or deceleration is not constant during a shock, which would mean that there should be a jerk component to the equation that describes the movement or position of the object as a function of time.

So here are my three related questions:

1. Is shock better expressed as g-force per second? If not, why (i.e. why is g-force a better unit)?


2. When you bang a smaller object that is reasonably rigid (e.g. a wristwatch with stainless steel case and bracelet) against another object that is reasonably massive, immovable, and rigid (e.g. a brick wall), how does the plot of position as a function of time actually look like, supposing we can record time and distances with extreme precision?


3. Do common mechanical devices suffer mostly from high acceleration or from high jerk?

Update The ISO 1413 shock resistance standard seems to give some clues. The testing procedure consists of letting a 3 kg hard plastic hammer traveling at 4.43 m/s hit a watch. Which suggests that we really care about the instantaneous transfer of energy or of momentum. But how fast does the transfer happen? Is it in the millisecond or nanosecond granularity?

experimental physics - How to prove that we are living in a 3+1D world?

Is there any scientific experiment that can lead us to conclude we live in 3 spatial dimensions without the premise of the conception of limited dimensions?

Thank you all who helped in the improvement of this question (which was not clear at first).

EDIT:

I know that this can be a little philosophical, but it is also a scientific question.

Let's consider the scenario where the mankind was not ever able to see.

Let's also consider that this limitation could be surpassed thus not limiting us to reach a scientific and technological knowledge "similar" to what we have today.

Would this civilization of blind people reach the conclusion that they are living in a 3D spatial world?

Is the sense of touch enough to reach that conclusion? Is there any scientific experiment that can lead us to that conclusion without the premise of the conception of limited dimensions?

Would it be easier, harder, or just different to reach a conclusion predicted by the M-Theory? (please do not focus only on this last question)

Answer

This question has changed in such a way that my answer (previously here) didn't seem even related anymore. I therefore came up with something new, gladly inheriting 4 upvotes, but much less confident. In fact, I can plainly state that I'm fully incompetent in these matters.

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With that out of the way, another introductory remark. Science doesn't prove things. Dimensions (in this case) are part of mathematical models that aim to describe nature. Whether dimensions "exist" or not is a philosophical question.

Now, let's see...

Anthropic arguments are frowned upon by many, but that is when they are used to explain something. Here one can use anthropic arguments just to reason our way to, not why, but rather, that something is the case. (I guess these arguments aren't really anthropic, but rather anthropic-like.)

My main source, as usual, is Wikipedia. The thing in which we are interested is spacetime (note: a mathematical model). From there we pick up this:

How many dimensions are needed to describe the universe is still an open question. Speculative theories such as string theory predict $10$ or $26$ dimensions (with M-theory predicting $11$ dimensions: $10$ spatial and $1$ temporal), but the existence of more than four dimensions would only appear to make a difference at the subatomic level.

Then the section "Privileged character of 3+1 spacetime" gives us a whole bunch of reasons of the following form:

"if $N\neq3$, then the world (or stuff in it) as we know it wouldn't exist",

where $N$ does not include compactified dimensions invoked by string theory and undetectable to date. (See, e.g., Calabi-Yau manifolds.)

Planets wouldn't have stable orbits. Stars wouldn't have stable orbits. Electromagnetism wouldn't work (at all). Electrons would either fall into the nucleus or disperse. Nerves cannot cross without intersecting. (Some of these arguments apply to $N<3$, $N>3$, or both.)

Hence anthropic and other arguments rule out all cases except $N = 3$ and $T = 1$ [...]—which happens to describe the world about us.

And then there is @Dilaton's comment to the question: "[M]easure how the gravitaional force depends on the distance". That may be expanded upon a little:

"[I]t is the three-dimensionality of space that explains why we see inverse-square force laws in Nature [...]" (Barrow 2002: 204). This is because the law of gravitation (or any other inverse-square law) follows from the concept of flux and the proportional relationship of flux density and the strength of field. If $N = 3$, then $3$-dimensional solid objects have surface areas proportional to the square of their size in any selected spatial dimension. In particular, a sphere of radius $r$ has area of $4\pi r ^2$. More generally, in a space of $N$ dimensions, the strength of the gravitational attraction between two bodies separated by a distance of $r$ would be inversely proportional to $r^{N-1}$.

For completeness' sake, I mention also i) the model with large extra dimensions and ii) causal dynamical triangulation, both of which I know even less of. The latter does apparently do something that might appeal to you:

It does not assume any pre-existing arena (dimensional space), but rather attempts to show how the spacetime fabric itself evolves. It shows spacetime to be two-dimensional near the Planck scale, and reveals a fractal structure on slices of constant time, but spacetime becomes $3+1$-d in scales significantly larger than Planck. So, CDT may become the first theory which does not postulate but really explains observed number of spacetime dimensions.

I would bet that the status of this approach (CDT) is controversial.

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Now, you stipulated: "without the premise [or] the conception of limited dimensions". I don't know what that really means. I hope that it doesn't mean "without the idea that mathematics can describe nature".

I also hope that I got at least something right. :)

cosmology - On Distant Lanterns and Relative Speed

It is said, that by measuring distant sources, we can gather information about the past state of the universe. The further we measure, the faster things appear to recede, which brings me to my question, which has puzzled me literally for years.

If distant sources tell something about the past, would that not mean the universe is slowing down today? If not, how come?

Answer

If distant sources tell something about the past, would that not mean the universe is slowing down today? If not, how come?

If in the past they are accelerating they should accelerate now too. Since we cannot expect that galaxies will slow down and come towards us at some point.

I want to talk about two things.

I-Redshift Observation

The EM spectrum of the galaxies shows us that, galaxies are receding from us And in cosmology, redshift observation is independent of time. For instance, we are observing a galaxy that is 100Mpc away. Which means we are observing 30 million years ago in time and we see that the galaxy is redshifted. Since in time it's not possible to expect that the galaxy will slow down and coming back to us, I think we can safely assume that it will be redshifted forever. Because from the observational data it seems that the further they are their velocity is higher. By observing the spectrum we can determine the speed of the galaxies or light sources. And it seems that they are accelerating.

II- The Deceleration Parameter

We can measure the density of the matter, and dark energy by modeling our universe. By using Friedmann Equations and by calculating density parameters we can understand the evolution of the universe. Now I want to talk about the deceleration parameter.

The deceleration parameter can be written as,

$$q_0=1/2\Omega_{m,0}-\Omega_{\Lambda}+\Omega_{r,0}$$

If $q_0>0$ it means that the universe is decelerating and if $q_0<0$ the universe is accelerating. By setting the current values of $\Omega_{m,0}=0.30$, $\Omega_{\Lambda}=0.70$ and $\Omega_{r,0}\approx0$

$$q_0\approx-0.55$$

And our observations and data best fit shows us that $q$ is negative and close to $-0.55$

So by looking at the CMBR data and observing the supernovas, we can see that the universe is accelerating.

The universe is accelerating because the dark energy density will become dominant over time. Since the matter density drops like $a^{-3}$ however the dark energy is independent of $a$ (where $a$ is scale factor). And this is currently what is happening.

spectroscopy - How to define the light "color" from a given spectral distribution?

The following question may be naive and incomplete in some way I don't know. I'm not a specialist on spectroscopy, colours and light curves, color spaces, etc.

Suppose you have a simple power-law function ; $f(\omega, \alpha) = \omega^{\alpha}$, which describes the spectral distribution of light angular frequencies as this : $$\tag{1} I = \kappa \int_0^{\omega_{\text{max}}} f(\omega, \alpha) \, d\omega, $$ where the exponent $\alpha$ is a given constant (a characteristic of the spectral distribution) and $\omega_{\text{max}}$ is another constant (the maximal value of the angular frequency allowed). $I$ is the total bolometric intensity of light at the detector's location, in watt/m^2 (the detector is a theoretical ideal device). $\kappa$ is just another arbitrary constant.

Then the question is this :

Assuming that $\omega_{\text{max}}$ is an angular frequency (rad/sec) which is in the visible spectrum or above it (i.e. ultra-violet), how can we define the color of the light described by the $\alpha$ index and the maximal value $\omega_{\text{max}}$ ?

By color, I mean something that could be compared in some way with the perception that we would have of that "$\alpha$-light", in the visible spectrum only.

For example, if $\alpha = 0$, the spectral distribution would be "flat" (i.e. uniform). What would be the color of light if $\omega_{\text{max}}$ corresponds to pure violet light, and $0 \le \omega \le \omega_{\text{max}}$ ? I guess white light !

If $\alpha = 2$, then the distribution would favour the violet and blue frequencies over the orange and red frequencies, so the light would look like blueish in some way, isn't ?

I hope the question is clear enough and doesn't bring me to the all messy/complicated problems of human/eye/brain/psychology perception ! I'm looking for something simple and "physical" only, if it exists ! In other words : is there a simple approximate "trick" to define a "color" from $\alpha$ and $\omega_{\text{max}}$ alone ? I'm just looking for some kind of approximation, to give an idea of what color the light might have.