quantum mechanics - Is the single slit experiment a practical example of Heisenberg's uncertainty principle?

I saw some videos where a person points a laser through a slit. As they reduce the width of the slit, the diffracted image spreads out, like this:

Can this pattern be viewed as a consequence of Heisenberg's uncertainty principle, applied to photons?

Answer

Yes, light diffraction may be viewed both as a classical phenomenon and as a quantum mechanical consequence of Heisenberg's uncertainty principle. However, since both explanations work equally well, it doesn't provide any direct evidence for quantum mechanics.

Let me explain why the two explanations are equivalent. I'll do the classical uncertainty bound first.

You may have noticed that in far-field diffraction, the product of the width of the aperture and the size of the pattern on the screen is a constant: a slit half as wide makes a pattern twice as big. In fact, plugging in the formulas for any kind of diffraction whatsoever will give something like $$\sigma_{\text{slit}} \sigma_{\text{screen}} \gtrsim D \lambda$$ where $D$ is the distance to the screen. This is the "uncertainty principle" for classical diffraction.

Now let's view this on the quantum level. The position uncertainty is simply $$\sigma_x = \sigma_{\text{slit}}.$$ The momentum is given by the de Broglie relation $$p = \frac{h}{\lambda}$$ but we want the uncertainty in the $x$-component of the momentum, $p_x = p \sin \theta \approx p\theta$. Now, the uncertainty in the angle $\theta$ is just $\sigma_{\text{screen}}/D$ by trigonometry, so we have $$\sigma_{p_x} = p \sigma_{\theta} = \frac{h}{\lambda} \frac{\sigma_{\text{screen}}}{D}.$$ Putting this all together, $$\sigma_x \sigma_{p_x} = \frac{h}{\lambda D} \sigma_{\text{slit}} \sigma_{\text{screen}} \gtrsim h.$$ Up to a constant, this is the usual Heisenberg uncertainty principle; the two pictures are equivalent.

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From a mathematical perspective, both uncertainty principles listed above special cases of a more general fact: the product of the width of a function $f$ and the width of its Fourier transform is bounded.

In the quantum case, the Fourier pair is position and momentum. In the classical far-field diffraction case, the pair is the screen and source, as proven here. This result also has applications in signal processing.

geography - Devoted to high culture

_{This puzzle is part 4 of Gladys' journey across the globe. Each part can be solved independently. Nevertheless, if you are new to the series, feel free to start at the beginning: Introducing Gladys.}

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Dear Puzzling,

Today I visited a rather unusual religious site. The people here seem highly spiritual, yet at the same time extremely tolerant. Unfortunately I was not allowed to witness their ceremonies myself, but I did get to take a tour of the church during the day.

Wish you were here!
Love, Gladys.

Across
1. From the creators of Ice Age | Unpaid trainee
2. Irish singer | Joker actor
3. Small image | Gemstone
4. Politician who would be sober tomorrow
5. Ballad's Chevy | English county
6. Couch | Piece of artillery

7. Frantic | Stitch up
8. Mediterranean island | Funeral container

Down
1. German empire | Fantasy monster
2. 2.54 centimetres | Ethiopian language
3. Happy | Japanese politician
4. Author Fleming | Repeat aloud
5. Open with a key | City with its own gulf
6. Name spelt with PH instead of V | Signified by Σ
7. Basque separatists | Central, interior

8. Small bread | Urban area

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_{Gladys will return in "Gladys spins the wheel".}

Answer

Gladys is at

the INTERNATIONAL CHURCH OF CANNABIS (yes, that's a real thing) in Denver, CO

The gray numbers

are meant to be interpreted as a gridless Slitherlink where blanks are considered 0. The solution to this Slitherlink encloses some of the letters of the crossword, which when read from top to bottom and left to right give us Gladys's location.

Completed grid:

Additionally, the answer is confirmed by

the title mentioning "high" culture and the letter in the flavortext saying that the people there are spiritual but tolerant.

homework and exercises - Equations of motion for bob-on-a-string -- am I missing some terms?

The dynamics of a type of physical system I am currently working on are modeled in most of the literature by replacing the moving parts of that system with an equivalent set of pendulums. Parameters and initial conditions for this approximate pendulum model are chosen such that the total mass and the center of mass of these pendulums describe those of the real system as accurately as possible.

For one particular system however, I need to take this one step further because that system is losing mass rapidly. (Also, it is spinning about an axis that lies outside the system, but let's leave that out of consideration here).

Therefore, I ended up modeling this particular system with a series of "specialized" pendulums:

• the pendulum masses are free to move in any direction (i.e., not constrained to move in any single plane)


• the point masses are attached to the pendulum hinge points by means of massless strings of infinite strength but finite elasticity (spring constant $k$)



• The pendulum mass $m$, the height of the hinge point $h$, the restoring force $m\mathbf{g}$ and the nominal string length $L$ are all functions of time.

In this figure, $h(t) = \left|\mathbf{h}(t)\right|$ and $\left|\mathbf{s}(t)\right| = L(t)$ when the string is just stretched. $\dot{\mathbf{h}}$ and $\ddot{\mathbf{h}}$ are constrained to only the pure vertical direction.

Leaving the fictitious forces out of consideration, I have come up with the following equation of motion for this kind of pendulum:

$$ m\ddot{\mathbf{r}} = -m\mathbf{g} + \begin{cases} 0 & \text{if } |\mathbf{s}| < L\\ m\left(\frac{\mathbf{s}\times\mathbf{\dot{r}}}{|\mathbf{s}|^2}\times\mathbf{\dot{r}}\right) - \gamma_1\left(\frac{\mathbf{s}\times\mathbf{\dot{r}}}{|\mathbf{s}|^2}\times\mathbf{r}\right) & \\ \ \ \ \ \ \ \ \ \ - k\left(|\mathbf{s}|-L\right)\frac{\mathbf{s}}{|\mathbf{s}|} - \gamma_2\left(\frac{\mathbf{s}}{|\mathbf{s}|}\cdot\mathbf{\dot{r}} + \dot{L}\right)\frac{\mathbf{s}}{|\mathbf{s}|}& \text{if } |\mathbf{s}| \geq L \end{cases} $$

where $\gamma_1$ is the torsional damping ratio, $k$ the string constant of the (flexible) string and $\gamma_2$ its linear damping ratio.

The first two terms of the second case lie in the current plane of rotation of the mass $m$ about the hinge point. They describe the centripetal force towards the hinge point, and torsional damping, respectively. The last two terms are in the radial direction, and describe the stretching and damping of the stretched string, respectively.

Numerical results of this equation are encouraging. However, this is all a bit messy (especially with the terms for the ficticious forces included), so I'd really appreciate if someone else took a look at it.

Why I ask this is because I can't figure out why the $\dot{\mathrm{h}}$, $\ddot{\mathrm{h}}$, $\ddot{L}$ and $\dot{m}$ do not show up in this equation...

Am I missing any terms anywhere?

Answer

Since $\mathbf{h} + \mathbf{s} = \mathbf{r}$, there's no need to use $\mathbf{h}$ or any of its time derivatives.

Physics doesn't care about something like $\dot{m}$. It cares about things like, "What's the energy?" and "What's the force?". But the kinetic energy of the bob is $\frac{1}{2}mv^2$ regardless of $\dot{m}$. The energy in the stretched spring is $\frac{1}{2}k(s-L)^2$ regardless of $\dot{L}$, etc.

experimental physics - How to guess the correct fitting function to some data?

Imagine we are plotting some date points $\left(x_i,f(x_i)\right)$ that we obtained experimentally, and that we want to know what $f(x)$ is. The way to do this is to use some software and try to fit the data to some guessed function. For example, if the behavior of the data points looks like exponential decay we then choose an exponential decaying function ..etc.

My question is: sometimes the data points are perfectly fitted to the exponential decaying trial function only on a certain region, but then the rest of the points show deviation away from the trial function.

1. How to proceed in this case? is there a catalog (something like mathematical tables) for functions and their plots that one can use as a guide?


2. Is there a systematic way to get the best fit instead of that trial and error method?

strategy - What frequent patterns can be used to overcome Minesweeper obstacles?

Minesweeper, to those who do not know the strategies involved, is a difficult game.

What patterns exist in the game which I can take advantage of?

Meta-context: this question is an offshoot of the current original question, for which the scope has been changed to be simply pattern-based. Feedback indicated this was somewhat too broad. This question has now been edited to address these concerns. Please contribute to discussion on Meta if you feel inclined.

Answer

Small notes

• Remember that the number of remaining mines is the clue number minus the number of surrounding mines. This is important in patterns like 1-2-1; 3-2-3, for instance, also works if there are two mines next to each 3, since they both need 1 mine to be solved.


• A one by itself on a corner always indicates that there is a mine


• A two by itself next to a convex corner always indicates two mines


• Patterns that work on horizontal walls obviously work on vertical walls, and vice versa.



• Patterns do not work when there is interference from adjacent clues or unknowns; i.e. when a pattern is in a corner or has another unknown near it.

The double-click trick

On the Windows version of Minesweeper, if you click on a clue with both your left and right mouse buttons, if all that clue's mines have been identified, then the game will remove all the other surrounding unknowns. You should always use this to clear spaces, unless you are guessing.

Additionally, you can use this to check visually if all the spaces surrounding a clue are mines. If you need two mines, and two squares are highlighted when you use this trick, you know they're both mines.

You will primarily use this trick to identify patterns quickly through visual recognition of shapes.

Note: if you have a mine incorrectly marked, this will cause you to lose the game - but then again, you would have anyway.

The 1-2 pattern

Whenever you see a 1 next to a 2 against a wall, you know that the space opposite the 2 is a mine.

This is because there can only be one mine in the two spaces adjacent to the 1, which means the second mine must be adjacent to the 2 in the final space.

Note that in the second case, since there is a mine next to the 3 and 2, they need 2 and 1 mines respectively, matching the 1-2 pattern.

The 1-2-1 pattern

One of the frequently-encountered patterns along walls in Minesweeper is the 1-2-1 pattern. This is an extension of the 1-2 pattern, in that two of the pattern exist side-by-side. In this case, as an extension of 1-2, the mines must be underneath the ones.

Note again that in the second case, the 2-3-2 are all adjacent to a known mine, which means they need 1-2-1. In the third case, we know that there is a mine next to the two by the one and corner in olive-green. therefore, the 1-2-2 matches the 1-2-1 pattern, implying there are two mines in front of it.

The first position was honestly just dumb luck. Yes, the 1-2-1 pattern does chain.

Also note that it is very rare to actually encounter a 1-2-1 specifically. Usually, these are by reduction from surrounding mines.

The 1-2-2-1 pattern

Whenever you find a 1-2-2-1 pattern against a wall, you know that the two squares under the 2s are mines.

Unfortunately, I found myself unable to generate a true 1-2-2-1 case in a reasonable amount of time, but here is a valid position anyway that demonstrates the point.

The 1-1-X pattern

When a 1-1-X are sequential against a wall and against the edge of the grid, the space under the third clue, marked here with a question mark:

This is because the first two spaces must contain a mine by the first clue, which means both the first and second ones are already guaranteed to be satisfied. If there were a mine in the third space, it would conflict with the second one.

This pattern is, admittedly, a lot less useful. However, it has the possibility to save your game, especially when you're stuck.

thermodynamics - Counting the number of microstates that there are for a given configuration. How to prove this result?

I'm doing some statistical physics and I came across a result which I'm not sure how to derive. Any help?

The answer turns out to be:

Can anyone help with this derivation? Thank you :D

Answer

Can anyone help with this derivation?

There are $N!$ ways to arrange $N$ different objects, but you don't have $N$ different objects, you have $n_0$ indistinguishable objects in the ground state, $n_1$ indistinguishable objects in the first state, and so on. So that means, from your $N!$ different permutations (assuming all the objects are distinguishable) all the permutations where you mix around the $n_0$ objects in the ground state all look exactly the same. And there are $n_0!$ such permutations, so you have to divide by $n_0!$. The argument is the same for the objects in the first excited state, so you also have to divide by $n_1!$, and so on.

Here is an example. Suppose you have one letter "A" and two indistinguishable letters "B". But, suppose that somehow we actually have marked one of the "B"s with a "1" and one of the "B"s with a "2" so we actually can distinguish them. In that case there are $3!=6$ different configurations (see below)

A B1 B2
A B2 B1

B2 B1 A
B1 B2 A
B1 A B2
B2 A B1

But really, we can't distinguish between the "B"s, so these 6 "different" configurations really look like:

A B B
A B B
B B A
B B A

B A B
B A B

We can't really differentiate between the permutations that only mix up the "B"s. I.e., there are really only $3=\frac{3!}{2!}$ different configurations:

A B B
B B A
B A B

quantum mechanics - Eigenvalues and eigenfunctions of the exponential potential $ V(x)=exp(|x|) $

For $a$ being positive what are the quantisation conditions for an exponential potential?

$$ - \frac{d^{2}}{dx^{2}}y(x)+ ae^{|x|}y(x)=E_{n}y(x) $$ with boundary conditions $$ y(0)=0=y(\infty) $$ I believe that the energies $ E_{n} $ will be positive and real

I have read a similar paper:

P. Amore, F. M. Fernández. Accurate calculation of the complex eigenvalues of the Schrödinger equation with an exponential potential. Phys. Lett. A 372 (2008), pp. 3149–3152, arXiv:0712.3375 [math-ph].

However, I get this strange quantisation condition

$$ J_{2i\sqrt{E_{n}}}(\sqrt{-a})=0 $$

How can I handle this in the case $ a >0 $?

Answer

The paper you quote covers a similar case, which was solved previously by S.T. Ma (Phys. Rev. 69 no. 11-12 (1946), p. 668), but deals with the scattering problem on the tail of the exponential - hence the complex energies. What follows is partly inspired by that paper but is quite distinct from it. The tricky part is not getting scared by the Bessel functions, but that's why we have the theory of special functions.

For one, the exponential potential $e^{|x|}$ you ask for is an even function, which means that the corresponding eigenfunctions on $(-\infty, \infty)$ will be either even or odd. Therefore, they can be treated as eigenfunctions of the simpler potential $e^x$ on $(0,\infty)$, with boundary condition $\psi'(0)=0$ or $\psi(0)=0$, respectively. Since you ask about the latter condition, there is no point in keeping the absolute value.

The problem, then, is the eigenvalue problem $$-\frac{d^2}{dx^2}\psi+A e^x \psi=E\psi\text{ under }\psi(0)=0=\psi(\infty).\tag{problem}$$

(A word on dimensions: To get the equation down to this form, we've had to set $\hbar, m$ and the length scale of the exponential to 1, by taking appropriate units of time, mass and length respectively. This means that there is no more dimensional freedom, and the hamiltonian has one free parameter, $A$, which will affect not only the scale of the spectrum (which you might expect as $A$ is a scaling on the potential) but also its structure.)

Amore et al. treat this as a boundary-value problem in $\mathbb C$ and using a change to a complex variable. This complicates the issue more than is really necessary and for simplicity I will use only real variables, though this comes at the cost of dealing with modified Bessel functions instead of standard ones. The initial step is to change variable to $z=2\sqrt{A}e^{x/2}$, so that $Ae^x=z^2/4$ and derivatives transform as $$ \frac {\partial }{\partial x}=\frac {\partial z}{\partial x}\frac {\partial }{\partial z}=\frac {z }{2}\frac {\partial }{\partial z} \text{ so } \frac {\partial^2 }{\partial x^2} =\frac14\left( z^2\frac {\partial^2 }{\partial z^2}+z\frac {\partial }{\partial z} \right). $$ The final equation is thus $$ \left[ z^2\frac {\partial^2 }{\partial z^2}+z\frac {\partial }{\partial z}-(z^2+\nu^2) \right]\psi=0 \tag{equation} $$ where $\nu=i\sqrt{4E}$. (Yes. Some complexness is inevitable. No fear, it will eventually not matter.)

This equation is Bessel's equation in modified form with index $\nu$. This is exactly the same as Bessel's equation for more normal situations; the index is complex but that is all. Two linearly independent solutions are the modified Bessel functions of the first and second kind, $I_\nu(z)$ and $K_\nu(z)$, so the general solution of $(\text{equation})$ looks like $$ \psi(z)=aI_\nu(z)+bK_{\nu}(z). $$ We then only need to impose the boundary conditions $\psi|_{z\rightarrow \infty}=\psi|_{z=2\sqrt{A}}=0$:

• 
  

  The condition at infinity requires that we set the coefficient of $I_\nu$ to zero, since the First Kind function always explodes. We could have done this from the start: $K_\nu$ is, by definition, the exponentially decaying solution, while $I_\nu$ grows exponentially.

  
  



• 
  

  The condition at $x=0$ then simply requires that $K_\nu(2\sqrt{A})=0$. In terms of energies, then, $$ \boxed{K_{2i\sqrt E}(2\sqrt{A})=0,} $$ and this is your quantization condition.

  
  

As it happens, $K_\nu(z)$ is real for real $z$ and purely imaginary $\nu$. One way to prove this is via this integral representation: $$ K_{\nu}(x) =\sec( {\nu\pi}/{2})\int_{0}^{\infty}\cos(x \sinh\nolimits t)\cosh(\nu t)dt, $$ which is the analogue of Bessel's First Integral for $K_\nu$; alternatively, it also follows from the conjugation symmetry $$ K_{\nu^*}(z^*)=K_\nu(z)^*, $$ which follows from the integral representation but is a more fundamental property of the function.

Since $K_\nu$ is real here, for whatever reason, we can ask for its zeros. As with all Bessel zeros there is no chance of an elementary formula for them, but they can be found quite easily using numerical methods (for properties of the zeros, see this DLMF reference). For a taster, here are some graphs, in log-linear scale (so zeros show up as downward, log-like peaks), of $K_{2i\sqrt{E}}(2\sqrt{A})$ as a function of $E$, for a few different values of $A$.

While there isn't all that much to say about the energies from this, it is clear that there are a countable infinity of them, that they are bigger than $A$, and that their spacing increases with increasing $A$ and $n$ (why?) - but that's really all you'd really want to know!

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Just for completeness: the eigenfunctions themselves, then, are of the form $$\psi_n(x)=C_nK_{2i\sqrt{E_n}}\left( 2\sqrt{A}e^{x/2} \right).$$ It is interesting to note that the dependence in $n$ comes through the index instead of a coefficient before $x$. This is partly to ensure the very strict decay $\psi\sim e^{-\exp(x/2)}$, which is required by the very hard exponential wall of the potential. For some information on how these Bessel functions behave, try the Functions of Imaginary Order subsection in the DLMF; particularly important results are asymptotics on $K_{i|\nu|}$ at large $x$ and for the oscillatory region. The latter is $$ {K}_{{i\nu}}\left(z\right)=-\left(\frac{\pi}{% \nu\mathop{\sinh}\left(\pi\nu\right)}\right)^{{\frac{1}{2}}}% \mathop{\sin}\left(\nu\mathop{\ln}\left(\tfrac{1}{2}% z\right)-\gamma_{\nu}\right)+\mathop{O}\left(z^{2}\right), $$ so the asymptotic for the wavefunction is of the form $\psi(x)\sim\sin\left(\sqrt{E_n}x\right)$, as it should be. (Note, though, that this holds little physics beyond the standard: the information on the potential's variation is encoded in the change of the instantaneous frequency as in e.g. these formulas, and would require beefier maths.)

optimization - A perfect metro map

You are working for a company and asked to create a perfect metro map where there will be as many stops as possible. But there are two constraints which limits the number of tracks (railroads) and maximum number of stations between one station to another one:

• There are at most $3$ tracks (railway) from one stop to others since it is not wanted to dig the underground too much and have the least amount of railways.


• The maximum number of stations between destination and departure stations is limited as $1$. So you can go from one station to another one with only seeing other $1$ different stations. In other words, if you move from one station (departure), there has to be at most $1$ stations between departure to destination.


• The tracks on the map can overlap to each other, since it is underground you can adjust the depth of the tunnels accordingly.

For example, if this question is asked for at most $3$ tracks with the maximum number of stations between stations as $0$ (no station between destination and departure stops), the answer would be 4 stations at most as seen below:

Answer

[EDITED: the original puzzle has been amended from "at most 4 stations between any two" to "at most one station between any two" which makes it no longer a research problem because the answer is already known. See below.]

I think this is a pure mathematical research problem in the guise of a puzzle. (Hence, no spoiler markings.)

This question asks for the largest graph with all degrees $\leq3$ and diameter $\leq5$. There is an easy upper bound on this, called the "Moore bound". Start at any vertex and start walking. You can reach 3 vertices after one step, then at most another 6 after the next, at most another 12 after the next, and so on. Even if these are all different -- i.e., the only way to repeat a vertex by this process, when walking as far as the diameter of the graph, is to retrace your steps exactly -- we can find at most 1+3+6+12+24+48=94 vertices this way.

It's been known for ages that this bound (call it $M$) can't be attained for cubic graphs. The answer can't be exactly $M-1$ for parity reasons. It was proved by Leif Jørgensen that it can't be $M-2$ either. Nor can it be $M-3$, again for parity reasons. So the most we could hope for is 90 vertices.

(That's for actual cubic graphs. Perhaps one can do a little better when, as here, we are allowed graphs of degree at most 3 rather than exactly 3. I suspect one can't, and I bet that if one can it's by only one or two vertices, but I haven't tried to prove that.)

The best actually found as of 2009 has 70 vertices. I had a brief look on the web and didn't find any descrption of the actual graph, but the paper you need is by Alegre, Fiol, and Yebra. From another table on that page it appears that no upper bound better than the one in the previous paragraph is known, but maybe I'm misunderstanding because the upper bound in the table is just the Moore bound, which is known not to be optimal for cubic graphs.

[EDIT: Here is what happens for diameter 2, the case now in the puzzle statement.]

When we allow at most one station between two others -- diameter 2, in graph-theory lingo -- the solution is the famous

Petersen graph (with 10 vertices)

which looks like this:

Why does the Stern–Gerlach quantum spin experiment conflict with classical mechanics?

My understanding of the Stern–Gerlach experiment is that neutral (0 total charge) particles are sent through a non-homogeneous magnetic field, with the expectation that the field will push that particle's path higher or lower on a detector because of the collective spin of that particle. While the detector can detect particles in a two-dimensional surface, the results of the experiment are that particles appear in only two localized areas directly above the path of the particle stream, or directly below the path of the particle stream - spin up or spin down. The conclusions from these measurements are that the particle, when measured, will always have about the same magnitude.

Why does this not follow from classical mechanical theory related to magnetism? If you shot a magnet through a similar apparatus, I would expect the magnet to be rotated to align with the magnetic field in some way which, at high enough field strengths relative to the mass of the magnet, would cause us to measure basically the same magnitude as if the magnet entered the apparatus pre-aligned with the field.

How is my explanation incorrect?

Answer

The magnet has a finite moment of inertia. What would happen when the magnet with "wrong" orientation enters Stern-Gerlach apparatus? Of course, the magnetic field will exert torque on it. The magnet starts rotating. After it comes to the equilibrium orientation, i.e. is oriented along the field, the torque is zero, but angular velocity is at maximum, and the magnet overshoots — just as in motion of usual oscillator.

If you find average magnetic moment over all the motion time, i.e. multiple periods of oscillation, you'll find that it has smaller magnitude than actual magnetic moment of the magnet. This means that net displacement in the direction of field is smaller. Now if there're lots of such identical magnets with random initial orientations, they all will have random average magnetic moment, and thus their displacement will form a continuum instead of just two points.

pattern - What is a Useful Word™?

_{In the spirit of the What is a Word™/Phrase™ series started by JLee, a special brand of Phrase™ and Word™ puzzles.}

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If a word conforms to a special rule, I call it a Useful Word™.
Use the examples below to find the rule.

$$ % set Title text. (spaces around the text ARE important; do not remove.) % increase Pad value only if your entries are longer than the title bar. % \def\Pad{\P{0.0}} \def\Title{\textbf{ Useful }} % \def\S#1#2{\Space{#1}{20px}{#2px}}\def\P#1{\V{#1em}}\def\V#1{\S{#1}{9}} \def\T{\Title\textbf{Words}^{\;\!™}\Pad}\def\NT{\Pad\textbf{Not}\T\ }\displaystyle \smash{\lower{29px}\bbox[yellow]{\phantom{\rlap{rubio.2019.05.15}\S{6px}{0} \begin{array}{cc}\Pad\T&\NT\\\end{array}}}}\atop\def\V#1{\S{#1}{5}} \begin{array}{|c|c|}\hline\Pad\T&\NT\\\hline % \text{ ALL }&\text{ ANY }\\ \hline \text{ NEAR }&\text{ AT }\\ \hline \text{ BORROW }&\text{ BEG }\\ \hline \text{ CURSE }&\text{ BLESS }\\ \hline \text{ SNITCH }&\text{ BLUDGER }\\ \hline \text{ CANER }&\text{ CANNER }\\ \hline \text{ REBS }&\text{ DEBS }\\ \hline \text{ LIGHTEST }&\text{ GRAVEST }\\ \hline \text{ SUPPED }&\text{ LUNCHED }\\ \hline \text{ DIAMOND }&\text{ MOD }\\ \hline \text{ AND }&\text{ NOR }\\ \hline \text{ COVERT }&\text{ OVERT }\\ \hline \text{ QUAYS }&\text{ PIERS }\\ \hline \text{ GREEN }&\text{ RED }\\ \hline \text{ OTHERS }&\text{ SELF }\\ \hline \text{ GAMBLE }&\text{ WAGER }\\ \hline \text{ BLACKEST }&\text{ WHITEST }\\ \hline \end{array}$$

And, if you want to analyze, here is a CSV version:

Useful Words™,Not Useful Words™
ALL,ANY
NEAR,AT
BORROW,BEG
CURSE,BLESS
SNITCH,BLUDGER
CANER,CANNER
REBS,DEBS
LIGHTEST,GRAVEST
SUPPED,LUNCHED

DIAMOND,MOD
AND,NOR
COVERT,OVERT
QUAYS,PIERS
GREEN,RED
OTHERS,SELF
GAMBLE,WAGER
BLACKEST,WHITEST

The puzzle satisfies the series' inbuilt assumption, that each word can be tested for whether it is a Useful Word™ without relying on the other words.

These are not the only examples of Useful Words™; many more exist.

What is the special rule these words conform to?

—HINTS—

helpfulness level 0:

The name fits here, technically; a different name might have been more appropriate.

helpfulness level 1:

There are way more Useful Words™ than Not Useful Words™.

helpfulness level 2:

This is likely impossible to solve unless you start with the right words.

Answer

A Not Useful Word™ is a word that...

...can have "one" added somewhere within it to make a new word:

ANY => ANYONE
AT => ATONE
BEG => BEGONE
BLESS => BONELESS
BLUDGER => BLUDGEONER
CANNER => CANNONEER
DEBS => DEBONES
GRAVEST => GRAVESTONE
LUNCHED => LUNCHEONED
MOD => MOONED

NOR => NOONER
OVERT => OVERTONE
PIERS => PIONEERS
RED => REDONE
SELF => ONESELF
WAGER => WAGONEER
WHITEST => WHITESTONE

As for the name Useful Word™...

...Rand al'Thor figured that out in a comment below:

The idea behind the name seems to be based on the downvote tooltip on Stack Exchange. You give minus one for an answer that's not useful, hence "not useful" for a word which, when ONE is 'subtracted', is still meaningful.

language - Word connections

I am trying a new kind of puzzle so feel free to leave constructive comments and opinions! Thank you!

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The objective of this puzzle is to connect 2 words by using a chain of words that connects each other.

For example,

• 
  
  

  a 2 word chain solution for the following :

  
  

  
  

  Connect the words air and chart

  
  

  
  

  Could be :

  
  

  
  

  airflow > flowchart

  
  

  
  

Rules

• No spaces are allowed.


• Every words in the chain must also be connected by an intermediary word.(ex. flow in the above example)

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Here is the real one, good luck!

Connect the words play and out

Can you find

• a 2 word chain?


• a 4 word chain?

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NOTE

I am aware that using some internet tools could make this puzzle drastically easier so if possible, please refrain from doing so.

Answer

2-word chain:

playtime $\rightarrow$ timeout

3-word chain:

playable $\rightarrow$ ablest $\rightarrow$ stout

4-word chain:

playtime $\rightarrow$ timeout $\rightarrow$ outlay $\rightarrow$ layout

• 
  

  with spaces:

  
  
  
  
  • 
    

    
    

    playtime $\rightarrow$ timetable $\rightarrow$ tabletop $\rightarrow$ top out (not sure if this last one is valid)

    
    

    
    
  
  
  
  • 
    

    
    

    playboy $\rightarrow$ boyfriend $\rightarrow$ friendzone $\rightarrow$ zone out

    
    

    
    
  
  
  
  

5-word chain:

playoff $\rightarrow$ offbeat $\rightarrow$ beatable $\rightarrow$ ablest $\rightarrow$ stout

Could we Bose-condense Higgs bosons?

Apart from the obvious problem that we'd have to create enough of them and somehow not let them decay, is there anything that would prevent a Higgs boson condensate?

Interesting topics to research in mathematical physics for undergraduates

I'm planning on getting into research in mathematical physics and was wondering about interesting topics I can get into and possibly make some progress on.

I'm particularity fond of abstract algebra and topology and if possible any topics that involve abstract algebra would/topology/calculus of variations would be especially appreciated

spacetime - Time resulting of a thermodynamics effect?

I heard in a documentary that : Irreversibility of time is linked to Heat Dissipation in thermodynamics. The theorist says "passage of time" is only a macro sensation,

the underlying effect is the heat dissipation (loss in energy) and 2nd law of thermodynamics.

I feel this theory looks very interesting and quite un-common, maybe some specialist may highlight their views and any papers on it.

1) Am referring to the relation between time and amount of heat dissipation. Time == Qty of Heat Dissipation

2) EDITED: Additional questions from the answer.

Arrow of time seems showing irreversibility....

1) Suppose we are at zero kelvin, no atoms move, nothing.

Is there any existence of time in this local environment ?

2) Is there any Equation linking Time variation and Entropy variation, like :

P[ dS/dt > 0 ] == 1

Answer

You are talking about the thermodynamic arrow of time; I'll talk about this in reference to entropy. It's actually not an uncommon idea, especially in cosmology.

Let's think about a video of a teacup being dropped to the floor and shattering. We know (one might say instinctively) when the video is being played whether it is being played forwards or backwards. We know that teacups do not leap into our hands after lying shattered on the ground; instead, they fall from our hands and shatter - in other words, entropy increases. That's the second law: that most of the time, entropy will increase (as formulated by Boltzmann, $S = k*log W$, with k being Boltzmann's constant, S being the amount of entropy in a system and W being the number of indistinguishable arrangements of a system).

That's the thermodynamic arrow of time: we know the direction of time because entropy increases most of the time, so we know something is played backwards when entropy decreases, like a broken teacup becoming intact as it leaps into our hand. Now, the formulation of entropy is statistical. What this means is that for 99.99999999999999999999999999999% of the time (not an exact number) entropy increases, but occasionally, there'll be fluctuations. This could lead you into a whole description of the beginning of the universe and the question of why it started at low entropy (for more information on that, look at this question).

Dissipation is the result of an irreversible process. That is, it doesn't obey time symmetry; it looks different from the direction of past to future as opposed to the direction of future to past. Heat transfer is dissipative because it is the transfer of energy from a hotter body to a colder one. This dissipation obeys the second law of thermodynamics (aka, entropy). This is where the connection to the thermodynamic arrow of time comes in, of course. (If you'd like the equation, the entropy change of a system at temperature $T$ is $dS = δQ/T$, though that's simplifying it dramatically).

You can look at this website for more information.

Hope this helps!

riddle - While I have six eyes, I don't need an optician

I thought I might try writing a riddle. Maybe this is too simple for some of the adept riddle solvers, but oh well. Here it is:

While I have six eyes, I don't need an optician.

There's just one factor to my primary condition.
Emerge from the inn, then head off to the right.
End with tea, where you can trust your eyesight.

What am I?

Answer

You are:

Indivisibility

While I have six eyes, I don't need an optician.

Indivisibility contains 6 of the letter i, but doesn't have anything to do with optometry

There's just one factor to my primary condition.

Something with 1 factor is indivisible by anything else

Emerge from the inn, then head off to the right.

The word indivisibility starts with in (sounds like "inn"), with "head off" to the right - this could either mean div in the sense of separating from the inn, or in the sense of

having similarity to as a valid HTML tag

End with tea, where you can trust your eyesight.

The word indivisibility ends with visibility, which itself ends with ty (sounds like "tea").

thermodynamics - Will a cloud form in a depressurized room?

If a room is depressurized will a visible cloud or mist form?

I would expect it would, but when I apply a vacuum to a flask I see no cloud form inside the flask. Would a room behave differently than the flask?

I am more interested in the case of sudden evacuation than a gradual depressurization.

cosmology - Big bang and time

I heard Carl Sagan talking about the Universe 15 Billion years ago, and the Big Bang. He made the statement that it was the biggest explosion of all time (at first I thought this a subtle pun). This leads me to my question. What would time have been like at +1 "moment" after the big bang? What I'm trying to ask is, and I hate to say it because I'm afraid I'll sound foolish, did time flow at the same rate? Wouldn't all that mass in one place have distorted space/time (and why didn't it "rip" it)?

If I were inside of that mass with a stop watch, I'm guessing I wouldn't have been able to measure a difference because time would have effected me the same way as it effected all the other space/mass in the area. I'm guessing I would have to have something inside (the initial Big Bang mass) and something outside measuring time and see if there was a difference (intuitively this feels weird to consider, could I actually place something "outside" the "Big Bang mass").

Maybe I've said too much, or made the question too complicated. I apologize if this is the case.

Update

A black hole is a lot of Mass collapsed into a small space. I believe that as mass increases time dilation increases. I remember hearing that if you fell into a black hole, you'd never experience the last second of your life...

If this is true of black holes, how did time pass in the mass/energy that is responsible for the big bang? As the big bang occurred, did time speed up with the expansion of the universe? I'm trying to explain what I'm getting at by asking more questions related to what I was asking. I'm trying to understand what time itself looked like. As I understand it time prevents everything from happening at once. If time was a line, were the ends smashed together into a point before the Big Bang ? Maybe as mass/energy expanded the "time line" expanded too?

Answer

Here is a sense in which this can be answered a bit unambiguously--it is a known effect that gravitational fields both dilate time, by a factor $\sqrt{1-\frac{2\,G\,M}{c^{2}\,r}}$ and redshift light waves by that same factor.

It is also known that cosmological effects redshift gravitational waves. This time, it is done by a factor of $a(t)$, the so-called 'radius of the universe'. For example, the cosmic microwave background radiation was believed to have been radiated from a surface whose temperature (and therefore, emitted wavelength) is roughly equivalent to the surface of a hot star. It is a matter of simple algebra to find a value of $\frac{M}{r}$ for which the two effects are roughly equivalent, and, if you wish, you can think of this as describing a "different rate of time flow."

To my knowledge, there really isn't a useful reason to do this, though.

measurements - If time standard clocks and any memories about the time standard are destroyed, can we recover the time standard again?

Assume the time standard clocks and any memories about the time standard are destroyed. Can we recover the time standard again exactly?

Recovering the time standard again means we can determine the date and time that are exactly the same as the destroyed time standard clocks will show (if they are not destroyed).

Answer

It depends on your definition of "any memories". If you don't remember what a second is, there is no solution.

If you remember the "old" definition (a day has 24 h on 60 minutes, each of it is 60 s), and live on Earth not to far from now, you can rebuild an approximate time standard. If you remember the modern definition, i.e.

(the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.)

you can reconstruct it as good as we can now, even if you are in a spaceship far away from Earth.

That would give you a second duration, from which you can define a time-scale. The next-step is to synchronize your new time-scal with the old time-scale. This is basically what historian do when they translate Egyptian or Mayan date in our modern system of datation. To have something precise, you basically need some record of an astronomical event.

The best analogue I can see is the work discussed here and here, where the difference between the solar time and atomic time is extrapolated in the past, before the atomic clock era. There are 10 s error bars around 1700, and 3h error bar for the year 1000 BC.

Wave equation in classical mechanics!

We represent the wavefunction of any wave on the string as $$y=f(x-vt),$$ where $v$ is velocity of the wave and $x$ is distance from origin and $t$ is time taken to reach the given point and $y$ is the displacement from the the $x$-axis.

but my question is why do we represent wave by this equation (i have clearly understood that the vertical displacement is the function of distance from the mean origin and time), why can't we represent a wave function by something like $$y=f(x^2-v^2t^2)$$ or any other function and why is there negative sign between $x$ and $vt$. i think negative sign doesn't make any sense.

on referring other book i saw wave equation was represented as $$y=f(t-\frac{x}{v}),$$ are they both same?

Please try to give answer with simple explanation, i am not expert in differentiation or differential equations, so i can't understand their meanings!

Epimenides paradox

Epimenides (a Cretan) once wrote a poem, in which he stated that all Cretans are liars.

Since he is a Cretan, and therefore a liar, Cretans are veracious.

But then again he wouldn't be a liar!

This paradox can be solved, i.e. it can be shown that it's not an actual paradox. How?

Edit: In this puzzle's "universe", liars always lie. I don't know how to precicely explain it, but "black and white logic" applies.

Additionally, there are at least 52 Cretans.

Answer

The negation of "All x are y" is "There is at least one x which is not y".

So, Epimenides is a liar. Therefore his statement "All Cretans are liars" is false. This means that not all Cretans are liars. This means that at least one Cretan tells the truth. He can still be a liar, there just has to be at least one Cretan who's not a liar.

Now, if Epimenides is the only Cretan, we'd have a bigger problem.

aerodynamics - Bicycle Wheel Drag in Slipstream

I was recently driving behind a car that had a bicycle mounted on a carrier over the rear bumper.

The bicycle wheels were not bound so they were rotating in the slip-stream of the car. I wonder, the fact that the wheels are turning; does this theoretically increase or decrease the drag on the car?

lateral thinking - 100 Prisoners and a Light 2: Electric Boogaloo

You may recall the prison experiment done along the following lines:

One hundred prisoners have been newly ushered into prison. The warden tells them that starting tomorrow, each of them will be placed in an isolated cell, unable to communicate amongst each other. Each day, the warden will choose one of the prisoners at random, and place him in a central interrogation room made of concrete walls 7ft on all sides and containing only a light bulb affixed to the ceiling with a toggle switch on one wall. The prisoner will be left alone for 1 minute in this room. The prisoner will be able to observe the current state of the light bulb. If he wishes, he can toggle the light bulb. He also has the option of announcing that he believes all prisoners have visited the interrogation room at some point in time. If this announcement is true, then all prisoners are set free, but if it is false, all prisoners are executed. The warden leaves, and the prisoners huddle together to discuss their fate. Can they agree on a protocol that will guarantee their freedom.

The solution to this puzzle (I surprisingly cannot find it on PSE, is found here)

Since that experiment's success, a new, crueler warden has taken over. He does not want the prisoners to succeed in obtaining their freedom. He is bound by law to offer the prisoner's the same opportunity, but he is allowed to change the rules to make it harder for them. To that end, he comes up with a plan that he thinks makes it impossible to succeed.

He removed the lightswitch that controls the bulb from the room. In addition, after each prisoner's time in the central room, he will peer in the room himself and see if anything has been altered or left behind; if he sees any difference, all prisoners will be executed.

Under these new rules, what plan can the prisoners devise to guarantee their freedom?

CLARIFICATION: To help keep the possibilities from being too broad, I want to specify that the answer I have in mind does not involve anything besides the room itself (nothing like 'prisoners banging on their doors if they've visited the room', etc) and can yield a strategy that can guarantee with 100% certainty when all prisoners have visited the room (nothing like 'wait x years and they've probably all been').

Answer

Using the same logic in the original answer (of switching the light bulb on and off), you can still use that algorithm by

slightly loosening/tightening the bulb in it's socket. (about a quarter turn or so). The light bulb would still be lit/unlit, but the next prisoner can easily tell the "on/off" state from the "tightness" of the light bulb. And the warden sees no visible difference in the room.

Link to answer by Yisong Song https://www.math.washington.edu/~morrow/336_11/papers/yisong.pdf

vibrations - Could it be that all energy is seeking ground?

Edited version.

If an electrical charge like a lightning bolt seeks ground (think of a lightning's rod's purpose), and if all matter is seeking ground (gravitational pull), and if all energy and all matter are various forms of vibration (electromagnetic spectrum), what is the likelihood that all forms of energy are directly or at least indirectly also seeking ground?

I'm just looking for an energy's basic common behaviors with other energy (and matter) Specifically I'm thinking of mechanical (vibrational) energy once it attaches itself to an object.

Thanks,

optics - How to correctly calculate the colour of the sky?

It is well-known that the sky is blue due to Rayleigh-scattering.

What bothers me though is the question how to actually calculate the spectrum of the scattered sunlight. Yes, use one of the formulas involving

$$I\propto \frac{1-\cos^2\theta}{\lambda^4},$$

that is clear, and the solar black-body spectrum has to be convoluted with the three human eye sensitivity curves to obtain an RGB image. But since our visual perception system (a.k.a. eyes) evolved such that the VIS-range is not strongly absorbed, I assume multiple scattering needs to be considered. So the formula above has to be applied to all paths involving no, single, double, triple etc. scattering, in other words there'll be a path integral involved.

Is there a closed form solution for this multiple scattering? Or at least an established routine to do the calculation?

visible light - Is a single photon emitted as a spherical EM wavefront?

If yes, could the same photon hit multiple targets as it expands? If not, how does the photon acquire the wave-ness if it is not born as a spherical wave? Also, in second case, how can multiple photons synchronize to make up a single wave front?

Answer

The only way to induce electromagnetic radiation is to disturb subatomic particles. For the emission of a photon it's enough that in an atom an (excited) electron falls back into a lower level. Once emitted the photon travels through empty space as a quanta of energy. The photon is indivisible during its travel. Hence a single photon couldn't have a spherical wavefront.

The sum of the emitted photons - say from an electric bulb - is called electromagnetic radiation. The emission from a laser is very strong directed, from a bulb it is much more spherical directed. So the radiation could be spherical, no matter would this be from a bulb or a star. Being far enough away from the source one would receive single photons. But this is not a wavefront.

how does the photon acquire the wave-ness?

A wavefront one could produce with radiation waves. Radio waves are produced by periodical acceleration of electrons in the antenna rod.

how can multiple photons synchronize to make up a single wave front?

Since this acceleration happens nearly synchronously for all involved electrons the number of emitted photons follows the frequency of the antenna generator. So for radio waves one really could measure wave properties (which is not possible for a bulb powered by a DC current). But again, being far enough from the antenna one will receive single photons.

electromagnetic induction - Secondary current in a loosely coupled transformer

How does the secondary current depend on the resistive load placed on the secondary winding of a loosely coupled transformer, in which the primary winding is supplied with a linear current ramp ?

Answer

The easiest way to analyze such problem is to write Kirchhoff's equations for a pair of coupled coils, the way EEs do it: $$V_1 = L_1 \frac{dI_1}{dt} + M \frac{dI_2}{dt}\\ V_2 = M \frac{dI_1}{dt} + L_2 \frac{dI_2}{dt}$$ with the termination condition : $$V_2=-I_2R_2$$

You have also specified that the input current is a ramp $I_1=k_1t$ so that the coupled equations become $$V_1 = L_1 k_1 + M \frac{dI_2}{dt}\\ -I_2R_2 = M k_1 + L_2 \frac{dI_2}{dt}$$ and after rearrangement $L_2 \frac{dI_2}{dt}+I_2R_2 + M k_1=0$ for $t>0$ the solution of which is an increasing exponential $I_2(t)=-\frac{Mk_1}{R_2} (1-e^{-t\frac{R_2}{L_2}})$. Notice that this approaches a constant current $I_{max}=-\frac{Mk_1}{R_2}$ as $t \to \infty$

homework and exercises - Quantum Stat-Mech Proof of an Inequality for the Partition Function

I have the following problem that I was unable to solve for class, but I had a couple first steps that I started with that I am unable to finish. I know I can't get this since it's already been turned in, but I would like to see if this was even a viable option for doing this proof in the first place.

``Given an orthogonal set of states, $\{\phi_{n}\}$, and a Hamiltonian, $\hat{H}$, show that the partition function, $Q_{\beta}$, satisfies the following $$ Q_\beta \geq \sum_{n}\exp\{-\beta \langle \phi_n|\hat{H}|\phi_n\rangle \}$$ with equality holding when the $\phi_n$ states are eigenstates of the Hamiltonian.''

I started by dropping in the identity in the exponential (as eigenstates of the Hamiltonian) $$ \sum_n \exp\{ -\beta\sum_k \langle \phi_n|\psi_k\rangle\langle \psi_k |\hat{H}|\phi_n\rangle\}=\sum_n \exp\{ -\beta \sum_k E_k |c_{nk}|^2\}$$ Then I am left with showing that $$\sum_n \exp\{ -\beta \sum_k E_k |c_{nk}|^2\} \leq \sum_k \exp\{ -\beta E_k\}$$ with equality showing up again the same way, with a Kronecker delta $\delta_{nk}$ collapsing the sum in the exponential.

I realize I didn't make it very far, so this might not be the best way to show this, but it seems manifestly true just by looking at it, but I can't actually show it. Does anyone have any hints about how to continue with this?

Answer

Here is a sketched proof of the inequality. The problem is to show that

$$ \sum_n\langle \phi_n|e^{-\beta \hat{H}}|\phi_n\rangle ~\stackrel{?}{\geq}~ \sum_n e^{-\beta\langle \phi_n|\hat{H}|\phi_n\rangle} ,\qquad\qquad (1)$$

where the Hamiltonian $\hat{H}$ is a selfadjoint operator, and $|\phi_n\rangle $ denote orthonormal basis vectors in the Hilbert space of states. The lhs. of eq. (1) is the partition function ${\rm Tr}(e^{-\beta \hat{H}})$. By scaling $\hat{H}$, we may assume that $\beta=-1$. The inequality (1) would follow if we can show the inequality for each and every summand

$$ \langle \phi_n|e^{ \hat{H}}|\phi_n\rangle ~\stackrel{?}{\geq}~ e^{ \langle \phi_n|\hat{H}|\phi_n\rangle} ,\qquad\qquad (2)$$

or equivalently, in a simplified notation for fixed $n$,

$$ \langle e^{ \hat{H}}\rangle ~\stackrel{?}{\geq}~ e^{ \langle \hat{H} \rangle} .\qquad\qquad (3)$$

But eq. (3) is just Jensen's inequality for a convex function (with the exponential function playing the role of the convex function).

newtonian mechanics - Friction in Slipping and Rolling

I'm solving a rotational motion problem that involves slipping and rolling. The problem's basic setup has a sphere rotated and then placed onto the ground, at which point it begins to acquire translational velocity. A solution in my textbook assumes the following:

"Once the ball starts rolling without slipping, there is no more frictional sliding force, and so the velocity will remain constant."

Why does friction stop exerting a force? Am I mistaken that friction exerts a force whenever an object and its surface might be distanced from each other?

fluid dynamics - Patterns in laminar flow of tap water

This is a simple experiment that anyone can do at home. Open your tap so that the water maintains a laminar flow, and the cross section of flow is considerably thin. Place your finger 3-4 cm below the flow. A beautiful wave-like pattern is observed on surface of flowing water, near the region of your finger. Changing the position of finger changes the 'wave-length' of the wave.

The attached picture shows how to do this experiment, though the resolution is not clear.

What is the explanation for this phenomena? Probably a harder question would be if there is an intuitive explanation for it. Thanks!

${\bf Edit:}$ I observed a similar phenomena in the swimming pool...when the water is still, move your hand slowly along the surface of the water. It causes a 'bump' in the water, which decays away in a wave-like pattern in front of your hand. Clearly they are the same phenomena: water falling on the hand or hand moving against the water.

The answers below are very interesting (Thanks!). And as far as I can see, they are still being debated, suggesting that the phenomena is more non-trivial that it appears to be. I had accepted an answer earlier, but I am now taking it back, as I did not completely understand it. Looking forward to further discussion on it. It would be most exciting to find an answer that is elementary: without resorting to complex terminologies of fluid mechanics.

quantum mechanics - Using Wien's Law to show spectral distruibution function of one temperature represents all temperatures

This is a exercise question from Quantum Mechanic textbook by Bransden:

Using Wien's Law to show that if the spectral distribution function of black body radiation, $\rho(\lambda,T)$ is known at one temperature then it can be obtained at any temperature (so that a single curve can be used to represent black body radiation at all temperatures.

Wien's law being: $\rho(\lambda, T)=\lambda^{-5}f(\lambda T)$

I understand this question as that temperature $T$ only acts as a scalar to the $\lambda$ variable, so that the differentiation of Wien's Law with respect to $T$ will give a constant multiple to the whole function, but in fact it is not clear that this is true by examining only on the Wien's Law.

Any help would be appreciated.

Answer

Suppose that, for a temperature $T_1$, you know $$ \rho(\lambda,T_1) = \lambda^{-5}f(\lambda T_1) $$ for every value of $\lambda$. Now, for a temperature $T_2$, let's introduce a variable $$\bar{\lambda} = \lambda T_2/T_1. $$ Then $$ \begin{align} \rho(\lambda,T_2) &= \lambda^{-5}f(\lambda T_2)\\ &= (T_2/T_1)^5 \,\bar{\lambda}^{-5}f(\bar{\lambda} T_1)\\ &= (T_2/T_1)^5\rho(\bar{\lambda},T_1)\\ &= (T_2/T_1)^5\rho(\lambda T_2/T_1, T_1) \end{align} $$ so that you also know $\rho(\lambda,T_2)$ for every value of $\lambda$.

homework and exercises - Using Jacobians to prove an identity

I have to prove the identity $$ \left(\frac{\partial U}{\partial V}\right)_T=T\left(\frac{\partial p}{\partial T}\right)_V-p $$ I know how to do this using Maxwell's relations, but the assignment specified that I have to use Jacobian transformations. Reading many websites, pdf's, etc. has brought me no closer to the solution.

I think that I have to start from the equation for internal energy and then differentiate it using $dV$ - this produces almost the end result, but when I work with the Jacobian I'm always lost.

Any explanations on how to use the Jacobian would be great. I know the basic rules, but cannot solve this.

Could you help me please?

Answer

In a reversible cyclic process the heat absorbed is the work done, hence $ \int_{C} TdS = \int_{C} pdV$ where $C$ is a closed contour. But it is also true that for the enclosed area $A$ within $C$ we have $ \int_{C} TdS = \iint_{A} dTdS$ and $ \int_{C} pdV = \iint_{A} dpdV$, so $ \iint_{A} dTdS = \iint_{A} dpdV$, and if we go from the variables $T,S$ to $p,V$ one has $ \iint_{A} dTdS = \iint_{A} dpdV \frac{\partial(T,S)}{\partial(p,V)}$. This implies that the Jacobian must be 1: $$\frac{\partial(T,S)}{\partial(p,V)}=1$$, in other words $$\frac{\partial T}{\partial p}\frac{\partial S}{\partial V}-\frac{\partial T}{\partial V}\frac{\partial S}{\partial p} = 1$$ This implies that in general $$\frac{\partial(T,S)}{\partial(x,y)}= \frac{\partial(p,V)}{\partial(x,y)}$$ Now if $x=T,y=V$ as new variables then you get $$ \frac {\partial S}{\partial V} = \frac {\partial p}{\partial T}$$ from which and from $dU=TdS-pdV$ your equality follows.

radioactivity - Radio-dating and the age of the earth

I understand there are a few radio-dating methods to determine the age of the earth, uranium-lead to name one (maybe not the best though). The ratio Pb₂₀₆ + Pb₂₀₇ to U allows you to find when the uranium came into existence. What I don't understand is how this relates to the age of the earth. Can't the uranium have been rambling for 500 million years in space before lumping together with other elements to form the earth? Which would make the age of the earth in this case 4 billion years instead of 4.55 billion.

Or is it the same as with C₁₄ dating: the C₁₄ is formed in the higher atmosphere through interaction of cosmic rays with nitrogen atoms. This stops when the carbon enters living material, so we know then the decay starts. Is the radioactive uranium created by cosmic rays until it hits earth? But the early earth didn't have an atmosphere which would stop the cosmic rays, did it?

So I'm a bit confused. How do we know the uranium formed at the same time as the earth?

Applying Navier Stokes to Fluid on Inclined Plane

I am trying to solve the well known problem of identifying the velocity profile of a fluid on an inclined plane. The assumptions we know are that the flow is Newtonian, incompressible, with a low reynolds number. I have been following my professor's notes on the problem, but ran into an assumption that I do not understand.

In evaluating Navier Stokes along an axis in line with the incline, we assume that the change in pressure along the incline $$\cfrac{\partial P}{\partial z}$$ is 0, since the flow is gravity driven. I do not understand why it is reasonable for us to make this assumption.

By evaluating the continuity equation, and Navier stokes along the other two axes, we discover that the velocity along the incline does not change (no acceleration). In addition, there is no pressure change out of the page and perpendicular to the incline axis (duh). However, I don't see any mathematical expression that leads to this conclusion.

statistical mechanics - Is the energy of the Ising model random or deterministic?

I'm reading about the Ising model and I'm confused by the following point. It says that the probability of finding the model in a given state $s$ is proportional to $e^{-E(s)/T}$. Where $E(s)$ is the energy of the state and $T$ is the temperature.
On the one hand, if the state is random, then, since the energy is a function of the state, so is the energy. But on the other hand isn't the energy also determined by the non-random temperature $T$?

electromagnetism - In a simple battery + resistor circuit, what form of energy is lost from the electrons upon exiting the resistor?

I will give this question a little context. Firstly, as I understand it, as soon as I "close the switch" on a circuit, electric current pretty quickly establishes a steady state where, at any given cross section along a wire, the average kinetic energy of that slice is both constant and equal anywhere along the wire. (Please correct me if I am misunderstanding).

If the average kinetic energy is constant anywhere throughout the wire in the circuit (which confuses me, as I would think that the electrons should all accelerate after exiting the resistor since they are no longer being impeded as much), that means that the average velocity is constant. So, in the example of a simple circuit with a battery of v Volts and a resistor of o Ohms, my question is the following:

Because a 'voltage drop' is known to occur across the resistor, what type energy is being "traded" to generate the heat that radiates from the resistor? I would think that the reflexive answer is "electrical energy"...hence the VOLTAGE drop (the sacrificed energy is clearly not kinetic, as the velocity is constant everywhere). However, I find this confusing. Does this mean that if I had a positive test charge, it would be easier for me to bring it to the beginning of the resistor as compared to bringing it to the end of the resister?

Further, if the correct answer IS "electrical energy", why exactly DOES electrical energy get "dissipated" as the electrons pass through the resistor? I always here the comment Oh! It's because all the electrons are running into densely packed crap which impedes their flow but that to me sounds like a reason for their KINETIC energy to be reduced. However, clearly that's not the case. So, ultimately, I guess the real question is:

**What goes on in the resistor that is literally removing electrical energy **. I am sure that this is a quantum mechanics question, but my quantum mechanics knowledge is not terribly strong. If I could get an answer that is devoid of crazy wave function equations, I would greatly appreciate it. Thanks!

homework and exercises - Coriolis deflection:- is it 4 times if the object is thrown up, compared to a drop?

A particle is thrown up vertically with initial speed, reaches a maximum height and falls back to ground. Show that the Coriolis deflection when it again reaches the ground is opposite in direction, and four times greater in magnitude, than the Coriolis deflection when it is dropped at rest from the same maximum height.

I worked out this problem as 2 parts : going up and going down. I see that while going up, I get a deflection of factor 2/3 times $\omega\sin\theta v_{0}^3/g^2$ and while going down, I get -1/3 factor which is the same when dropped down from the same height. How is this wrong ? To get the correct answer, I have to integrate over twice the time required to reach the maximum height and then I get 4/3 for going up and coming down and -1/3 for being dropped down. How is the first method not correct since coming down part of the trajectory is same as being dropped down from the maximum height it reaches ? Should the deflection while coming down be not opposite to the deflection while going up ? I think I can derive it rigorously but it will be good to get an intuitive idea why the things work out by taking the total time of fall (ignoring the horizontal motion) but they don't if up and down motion are treated separately- Does the effect of horizontal motion cancel over going up and down together ?

special relativity - Connecting 4-velocity to equation for adding velocities

Is there a way to connect 4-velocity to equations for adding speeds? I know 4-velocity $U^\mu$ is derived like this:

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\begin{equation} \begin{split} P^\mu &= m U^\mu \Longrightarrow U^\mu = P^\mu \frac{1}{m} = \begin{bmatrix} p_x\\ p_y\\ p_z\\ \frac{W}{c} \end{bmatrix} \frac{1}{m} = \begin{bmatrix} \frac{mv_x \gamma(v)}{m}\\ \frac{mv_y \gamma(v)}{m}\\ \frac{mv_z \gamma(v)}{m}\\ \frac{mc^2 \gamma(v)}{c m}\\ \end{bmatrix} = \begin{bmatrix} v_x \gamma(v)\\ v_y \gamma(v)\\ v_z \gamma(v)\\ c \gamma(v)\\ \end{bmatrix} \end{split} \end{equation}

Or like this:

\begin{equation} \begin{split} U^\mu = \frac{d X^\mu}{d \tau} = \frac{d}{dt}X^\mu \gamma(v)= \frac{d}{dt} \begin{bmatrix} d x\\ d y\\ d z \\ c d t \end{bmatrix} \gamma(v) = \begin{bmatrix} v_x\gamma(v)\\ v_y\gamma(v)\\ v_z\gamma(v)\\ c \gamma(v) \end{bmatrix} \end{split} \end{equation}

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And i know how to derive Lorentz transformations for velocities $\perp$ to relative velocity $u$ (which is in direction of $x$, $x'$ axis) and $\parallel$ to $u$. It goes like this:

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a.) $\parallel$ to $u$: \begin{equation} \begin{split} v_y' &= \frac{dy'}{dt'}=\frac{d y}{\gamma \left(d t - d x \frac{u}{c^2} \right)} = \\ &= \frac{\frac{dy}{dt}}{\gamma \left(\frac{dt}{dt} - \frac{dx}{dt} \frac{u}{c^2} \right)}\\ &\boxed{v_y' = \frac{v_y}{\gamma \left(1 - v_x \frac{u}{c^2} \right)}} \end{split} \end{equation}

b.) $\perp$ to $u$: \begin{equation} \begin{split} v_x' &= \frac{dx'}{dt'}=\frac{\gamma (d x - u d t)}{\gamma \left(d t - d x \frac{u}{c^2} \right)} = \\ &= \frac{d x - u d t}{d t - d x \frac{u}{c^2}} = \frac{\frac{d x}{d t} - u \frac{d t}{d t}}{\frac{d t}{d t} - \frac{d x}{d t} \frac{u}{c^2}}\\ &\boxed{v_x' = \frac{v_x - u}{1- v_x \frac{u}{c^2}}} \end{split} \end{equation}

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QUESTION:

Where is the connection among 4-velocity and equations derived under a.) and b.)? How can i show the connection?

Answer

Yes, from the transformation law for the four-velocity, we can explicitly derive the transformations of three-velocities parallel to and perpendicular to a given boost.

First, we need to talk about what the transformation law is for the four-velocity with respect to a boost. Let the four-velocity be $U = (U^t, U^x, U^y, U^z)$. Let's boost this along the x-direction by a speed $\mu c$. Like any four-vector, the four-velocity transforms under Lorentz transformations like so:

$$\begin{align*} {U'}^t &= W(U^t - \mu U^x) \\ {U'}^x &= W(U^x - \mu U^t) \\ {U'}^y &= U^y \\ {U'}^z &= U^z\end{align*}$$

where $W = 1/\sqrt{1-\mu^2}$ is the Lorentz factor of the boost.

Now, break down the original four-vector $U$ as $U= \gamma c(1, \beta^x, \beta^y, \beta^z)$. We can find the components of $U'$ as

$$\begin{align*} {U'}^t &= W\gamma c(1 - \mu \beta^x) \\ {U'}^x &= W \gamma c (\beta^x - \mu) \\ {U'}^y &= \gamma c \beta^y \\ {U'}^x &= \gamma c \beta^z\end{align*}$$

You can then find the components of three-velocity in the primed frame by taking ${U'}^i/{U'}^t$.

$$\begin{align*} \frac{{v'}^x}{c} &= \frac{{U'}^x}{{U'}^t} = \frac{\beta^x - \mu}{1 - \mu \beta^x} \\ \frac{{v'}^y}{c} &= \frac{{U'}^y}{{U'}^t} = \frac{\beta^y}{W(1-\mu \beta^x)} \\ \frac{{v'}^z}{c} &= \frac{{U'}^z}{{U'}^t} = \frac{\beta^z}{W(1-\mu \beta^x)} \end{align*}$$

These are algebraically the same as the formulas you posted in (a) and (b). To me, this is much simpler than churning through velocity-addition. The transformation laws of four-vectors are simple to learn, and the amount of manipulation needed to find the new three-velocity is minimal.

*Note: your terminology on parallel vs. perpendicular to the boost seems confused. Nevertheless, I believe my results here capture what you intended. The boost velocity is in the direction of the x-axis.

rhyme - Riddle... pray that this is the last one

Stop worshipping me!
Away! you flea!
I'll stomp you with my boots,

You know I'll accept any substitutes,
At first, I change you,
Then I begin to arrange you,
And finally I end your precious heartbeat!

Who is yelling at you?

As usual, please explain all steps in deriving an answer :)

Answer

Yet again it's

a cat.

Stop worshipping me!
Away! you flea!

The cat is telling us, the riddlers of Stackexchange, to stop writing riddles about it :-p Also, in former times, Egyptians worshipped cats. Cats get and surely hate fleas :D

I'll stomp you with my boots,

This is a reference to Puss in Boots.

You know I'll accept any substitutes,

Cats are not known for loyalty.

At first, I change you,
Then I begin to arrange you,
And finally I end your precious heartbeat!

C is the first letter in change; then the letter A begins the word arrange, and finally the letter T ends the word heartbeat - spelling out CAT.

The title is also a hint, since two riddles with similar titles have both had the same solution.

general relativity - What determines which frames are inertial frames?

I understand that you can (in principle) measure whether "free particles" (no forces) experience accelerations in order to tell whether a frame is inertial. But fundamentally, what determines which frames are inertial (i.e. what principle selects in which frames free particles will not appear to accelerate)? I've been told that the cosmic microwaves determine the ultimate rest frame of the universe, but that doesn't make sense to me, since one can still ask why that frame is an inertial frame.

Also, I understand that there are no real inertial frames in general relativity, but it seems like there certainly exists approximate inertial frames and we can ask why those frames are approximately inertial and not others. For example, in the frame of a person riding a merry go round, free particles appear to greatly accelerate; while in the frame of someone standing next to the merry go round there are no such great accelerations. Why does the guy (or gal) on the merry go round see free particles accelerating while the other guy doesn't.

And if you're gonna tell me that it's "the rest of the stuff in the universe" that determines whether the person on the merry go round sees free particles accelerate, I'll ask how you know that all that stuff is not spinning.

I hope this question sort of makes sense, it's been bothering me for a while and my study of relativity (most of special relativity and just the outline of general relativity) hasn't really clarified it for me much.

general relativity - Redshift of distant galaxies: why not a doppler effect?

How can I explain to my 17 year old pupils that the observed redshift of distant galaxies cannot be interpreted as a doppler effect and inescapably leads to the conclusion that space itself is expanding?

I understand that this redshift is well explained in general relativity (GR) by assuming that space itself is expanding. As a consequence, distant galaxies recede from us and the wavelength of the light is "streched". Expansion, redshift and the Hubble law are explained coherently in GR, as well as many other phenomena (e.g. the cosmic microwave background), and the GR predictions about redshift agree with observations.

I understand that the redshift of distant galaxies cannot be explained as a doppler effect of their motion through space. Why exactly is a pupil's doppler interpretation wrong?

My first answer: "Blueshifted galaxies (e.g. Andromeda) are only seen in our local neighborhood, not far away. All distant galaxies show a redshift. At larger distances (as measured e.g. with Cepheïds) the redshift is larger. For a doppler interpretation of the redshift distant galaxies we must necessarily assume that we are in a special place, to the discomfort of Copernicus. In this view, space cannot be homogeneous and isotropic." Is this answer correct?

My second answer: "A doppler effect only occurs at the moment the light is emitted, whereas the cosmological redshift in GR grows while the light is traveling to us." My problem with this answer (if it is correct): what observational evidence do we have for a gradual (GR) increase of the redshift, disproving the possibility of an "instantaneous doppler shift at the moment of emission"?

My third answer: "For galaxies at $z>1$ you can only have $v

My fourth answer: "Recent observations of distant SN Ia show a duration-redshift relation that can only be explained with time dilation [see Davis and Lineweaver, 2004, "Expanding Confusion etc."]" My problem with this answer: does time dilation prove we have expanding space, in disagreement with a doppler effect?

My fifth answer would involve the magnitude-redshift relation for distant SN Ia [Davis and Lineweaver], but that's too complicated for my pupils.

Answer

Your first answer is the most correct one:

My first answer: "Blueshifted galaxies (e.g. Andromeda) are only seen in our local neighborhood, not far away. All distant galaxies show a redshift. At larger distances (as measured e.g. with Cepheïds) the redshift is larger. For a doppler interpretation of the redshift distant galaxies we must necessarily assume that we are in a special place, to the discomfort of Copernicus. In this view, space cannot be homogeneous and isotropic." Is this answer correct?

In other words, it is more likely that we are not in a special place and the universe is expanding than that everything in the universe is flying away from us. This is also supported by the fact that we cannot find anything else particularly special about our location in the universe: the galaxy we're in is typical, the group of galaxies our galaxy is in is typical (if a little low on the mass scale compared to clusters like Virgo or Coma), etc.

My second answer: "A doppler effect only occurs at the moment the light is emitted, whereas the cosmological redshift in GR grows while the light is traveling to us." My problem with this answer (if it is correct): what observational evidence do we have for a gradual (GR) increase of the redshift, disproving the possibility of an "instantaneous doppler shift at the moment of emission"?

We actually do have evidence for this. When the light passes through an especially massive cluster of galaxies on the way to us, the photons will gain energy as the fall into the cluster, and lose energy as they come out. If the universe is static, the photons would gain as much energy as they lose, only being deflected. With the expansion of the universe accelerating, though, the photons gain more energy when they fall into a well than when they come out, because the accelerating expansion of the universe has made the well more shallow while the photon was traveling through it. When this happens to a cosmic microwave background photon, it is known as the integrated Sachs–Wolfe effect.

"My third answer: For galaxies at $z > 1$[...]" The exception there is correct. The general Doppler shift in special relativity is given by:

\begin{align} f_r = \frac{1 - \frac{v}{c} \cos\theta_s}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} f_s \end{align} with $\theta_s$ the angle between the velocity $\vec{v}$ and the line of sight, as measured by the source. If you have a redshift of $z$, then your velocity is at least $$v_{\mathrm{min}} = c \frac{(z+1)^2 - 1}{(z+1)^2 + 1},$$ with any value up to $c$ allowable for the right choice of $\theta_c$. Fun fact: the rapidity $\phi_{\mathrm{min}}$ for $v_{\mathrm{min}}$ is defined by $v = c \tanh\phi$, leading to $\phi_{\mathrm{min}} = \ln(1 + z)$.

For what it's worth, we also see the effects of Doppler shifts of galaxy motion on redshifts when we study clusters of galaxies. Wikipedia discusses them in the Redshift-space distortions article. In particular, the "fingers of god" effect causes the redshifts of clusters of galaxies to be elongated along the line of sight, and the "pancakes of god" can elongate redshifts perpendicular to the line of sight.

My fourth answer: "Recent observations of distant SN Ia show a duration-redshift relation that can only be explained with time dilation [see Davis and Lineweaver, 2004, "Expanding Confusion etc."]" My problem with this answer: does time dilation prove we have expanding space, in disagreement with a Doppler effect?

This one does not carry information about whether the Doppler effect is relevant. The stretching out of a signal's wavelengths, with the speed of light held constant, will also cause the duration of the signal to increase, causing an apparent time-dilation. You can play around with slowing down and speeding up audio signals to see how this works - you need to do some extra work to keep the pitch the same if you do that. The converse is also true - if you just alter all of the pitches in an audio signal, you'll alter the duration, too, if you don't do extra work.

You can also throw in the observed existence of the CMB. It is very hard to explain using any model that doesn't have an expanding universe (I say "very hard" because I don't want to preclude the possibility of someone more clever than I figuring out a way in the future).

energy - Physics errors in the movie "Arrival"

After seeing the movie Arrival, though I enjoyed it, I nevertheless kept feeling that there were some serious violations of physics occurring. However, I'm not sure what the precise problems are (I am not a physicist).

Question: what, if any, are the physical laws broken in Arrival ?

Spoiler alert

In particular, I felt that the main character's seeming ability to not only see into the future, but also influence it, must have violated laws regarding the light-speed limit on information transfer.

But, maybe more importantly, doesn't it violate some kind of energy conservation? i.e. the movement of information and other influences over vast spans of spacetime, seemingly instantaneously, powered by...

her brain(?). Perhaps the movie implied that she was simply "remembering" the future, not influencing it, but I suppose that's a violation of relativity (future events influencing the past). And makes her future actions look inane.

Incidentally, I think that issues relating to the alien's technology are valid, but perhaps easier to explain away by e.g. "sufficiently advanced technology being indistinguishable from magic" arguments.

Also, I am aware that there are some media articles on this, but I do not find them to be very scientific (I always appreciate equations accompanying words).

riddle - Piece de Resistance - Introduction & Ace and A's

OP's Words

<sub>This puzzle is part of the "Piece de Resistance" series. Read the following section for the story

Ace Two Three Four Five ...</sub>

Hey guys (and gals), I have worked on this for weeks, took me even more effort than "A Masterpiece: A mirror or an ABC?". Hope y'all like it, and don't forget to press the upward button if you like it and/or appreciate my effort :D

Prologue

On Sunday night, you went to the bar to chill with your friend. After chatting for a while, your friend produced a deck of cards from his pocket.

"Here, happy birthday, mate! Bye for now!"

And then, without a word, he left.

You flipped through the cards, but they aren't the kind of playing cards you normally see. On the top-right corner of each of the cards are labels, from A through K. Strange though, there is no suit. You soon realised there is only one suit of 13 cards here.

Ace - and A's

You observed the Ace card in detail. It has a silver glossy surface. You tried rubbing a coin against the surface, and you realised there was another layer under the silver surface. The surface underneath was more like a typical playing card, with a white shiny facade. The black text on it says:

$\mathfrak{\text{Ace, I may be to the ear, without one letter.}}$
$\mathfrak{\text{A's, I may be to the eye, without two letters.}}$
$\mathfrak{\text{See? Three letters gone!}}$
$\mathfrak{\text{I am a victim.}}$

Answer

Case sounds like ace when you remove one letter (the first), looks like a's when you remove two letters (the first and last), is c (sounds like see) when you remove three letters (the last three), and is a victim of (falls to) arguments in a proof by cases. I don't, however, see a connection to the title ("Piece de Resistance - Introduction").

Constant mass flux in fluid dynamics?

OK so I am having a conceptual crisis about Hydrodynamics.

1) Since mass should be conserved, is $\frac{dm}{dt}=0$ ?

2) But I know that the formula $\rho_1u_1A_1 = \rho_2u_2A_2$ exists... and doesn't that mean the mass flux is conserved? Where does this formula come from?

3) How can we derive an expression for $\frac{dm}{dt}=0$ from the continuity equation $\frac{\partial\rho}{\partial t} + \nabla\cdot(\rho\mathbf{u})=0$

Answer

OK so in the end I found an answer:

$\textbf{Derivation}$

Mass is conserved, so $$\frac{dm}{dt} = 0$$

But $$m = \iiint \rho(\mathbf{r}, t)d^3\mathbf{r} $$ so that $$ \frac{dm}{dt} = \iiint \frac{\partial \rho(\mathbf{r}, t)}{\partial t}d^3\mathbf{r} = 0$$. From the continuity equation $$ \frac{\partial \rho(\mathbf{r}, t)}{\partial t} = -\nabla \cdot(\rho\mathbf{u})$$. So $$ \iiint \frac{\partial \rho(\mathbf{r}, t)}{\partial t}d^3\mathbf{r} = 0 \quad\text{implies} \quad -\iiint \nabla \cdot(\rho\mathbf{u}) d^3\mathbf{r}=0$$ which using Stokes' theorem becomes $$ \iint \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset (\rho\mathbf{u})\cdot d\mathbf{S} = 0$$

Applying this to a flow like the one in the diagram

so that the velocity is parallel to the cross sectional areas ($A_1$ and $A_2$) and there is no vertical component of the velocity in the top and bottom surfaces.

Assuming the density and the velocity are constants over that area (at a fixed $x$), the integral becomes: $$ \rho_1A_1\mathbf{v_1} = \rho_2A_2\mathbf{v_2} $$

$\textbf{Possible second derivation}$

There should also be another way of deriving this:

Starting once again from $$\frac{dm}{dt} = 0$$ and using the definition of a convective derivative (or Lagrangian derivative) $$ \frac{d}{dt} = \frac{\partial}{\partial t} + (\mathbf{u}\cdot\nabla)$$ where $\frac{\partial}{\partial t}$ is the Eulerian derivative, we get $$ \frac{\partial m}{\partial t} = -(\mathbf{u}\cdot\nabla)m $$

No idea how to get to the required equation though. Any pointers?