lagrangian formalism - Constraints in classical mechanics

I am self-studying classical mechanics and I have a couple of questions about constraints. Goldstein in his book Classical Mechanics writes in the beginning of Section 1.3 that:

It is an overly simplified view to think that all problems in mechanics are reduced to solving the set of differential equations: $$m_i \ddot{\bf r}_i ~=~ {\bf F}_i^{(e)} + \sum_j {\bf F}_{ji} \tag{$\dagger$}$$ where ${\bf F}_i^{(e)}$ denotes the net external force on particle $i$ and ${\bf F}_{ji}$ denotes the force exerted by particle $j$ on $i$ because one may need to take into account the necessary constraints for the system.

Then, he says that the constraints introduce two types of difficulties in solving mechanical problems:

$(\text{I})$ $r_i$ are no longer independent.

$(\text{II})$ constraint forces are not known in general.

My questions are:

$(1)$ IF it is possible to identify all the constraint forces, then all the problems would be reduced to solving $(\dagger)$, where ${\bf F}_i^{(e)}$ includes all the constraints forces. Wouldn't it? If not, is there a constraint that cannot be translated into a corresponding constraint force?

$(2)$ Isn't $(\text{II})$ in fact the only difficulty with solving mechanical problems? It seems to me that $(\text{I})$ is not a "difficulty" because provided we can identify all the constraint forces the fact that $r_i$ are not independent would be incorporated into the constraint forces that would appear in the equations of motion. Isn't the fact that $r_i$ are not independent nothing but that the equations of motion are coupled ODEs?

mathematics - Standard Sudoku Specialized with just 2 Primes

A standard 9x9 Sudoku uses the digits 1 to 9.

You are only allowed two distinct primes to represent 1 to 9.

Find out the minimum number of characters (digits + signs) needed to construct a Sudoku puzzle with a unique solution.

Allowed signs are plus, minus, division, multiplication, factorial, and exponentiation.

Answer

The minimal Sudoku contains 17 digits: 2 each of seven of the nine digits, 3 of another digit, and none of the last one. (By symmetry, it doesn't matter which two digits are selected to appear 3 and 0 times in the Sudoku.)

If we select our two primes as

2 and 3, then we can represent the digits as $$3-2,2,3,2+2,2+3,3!,?,2\wedge3,3\times3.$$ (We omit 7 because we only need to represent all but one of the digits for the minimal Sudoku.)

So we choose the minimal Sudoku with

seventeen digits filled, two each of $1,2,4,5,6,8,9$ and three of $3$.

How many symbols do we need to use then?

Each of $1,4,5,8,9$ requires three symbols, $2$ and $3$ require only one, and $6$ requires two. So the total is $$5(2\times3)+(2\times1)+(2\times2)+(3\times1)=30+2+4+3=39.$$

mathematics - The Erasmus polyhedron

Professor Erasmus has constructed a special convex polyhedron from perfectly homogeneous material, which he modestly calls the "Professor-Erasmus-polyhedron". The professor claims that he can put the polyhedron into the lake next to his house, so that the polyhedron floats

• with 90% of its volume below the water level, and


• more than 50% of its surface area above the water level.

Has the professor once again made a mathematical blunder, or does such a polyhedron indeed exist?

Answer

Ok, i'll take a stab at it.

Take a cone of radius 100 and height 10. place it point first into the water, so that .34 units stick out of the water. At that point, 90% of the volume (94397.4 of 104720) is underwater, and less than half of the surface area (29462 of 62988.5) is under water.

Oh fine, polyhedron, i'll change it to a

pyramid of base 100x100 height 10. Point first, if .34 units stick out of the water. Volume (30048 of 33333), surface (9516 of 20198)

I came to this idea from the idea that a really flat pyramid would have the surface area of the base be close to 1/2 of the total surface area. Then I just had to work out where 90% of the volume would be, and check if that area is under 50%.

Let's take height as a ratio of width (xw). So, we know that the height underwater is .966 of the total height, which means the width underwater is also .966 of the original width.
So, the surface area of the part underwater is $2 * .966w * \sqrt{\frac{(.966w)^2}{4} + (.966xw)^2}$
And of course the full surface area is $w^2 + 2 * w * \sqrt{\frac{w^2}{4} + (xw)^2}$

Plugging that all in gives an X < .2883

fluid dynamics - What is the shape of a stream of water coming out of pipe?

As shown in the photograph below, when water comes out of a pipe (located on the surface of the earth) it travels downwards in an arc.

What is the shape and size of this arc as a function of D, the inner diameter of the pipe and, P, the pressure of the water? I suppose the shape is either a parabola, or maybe a catenary. Assume the pipe is horizontal and the water is exiting the pipe perpendicular to gravity.

(Although a thrown object will have a parabolic trajectory, I don't think I can assume that a stream of water will necessarily be parabolic because the water stream has viscosity.)

quantum chemistry - Why are do neutral atoms shrink as their valence shells approach 8 electrons?

Why do neutral, unbonded atoms shrink in size as they approach having 8 electrons in their valence shells? A good example is elements 3 through 10 in this table, that is, lithium (1 valence electron) through neon (8 valence electrons).

---

For posterity, below is my earlier incorrect version of my question. It's incorrect because it based on a table that used the atomic radii of bonded atoms to describe their sizes. That is a very different measurement from the sizes of neutral atoms, since for bonded atoms you are tossing in additional electrons via the bonds. Very small atomic radii for halogens are easy to explain in that case, since the incomplete shells must fight very hard over the shared electrons to complete their own octets, making their bonding radii anomalously short. This question had an interesting back-and-forth, and I thank the two contributors both who both caught my chart interpretation error and answered the real and interesting question that lurked beneath my incorrect one.

---

If you look here at the relative sizes of charge-neutral atoms, you can see something I never noticed or even thought about until I tried to think through this question in the new Chemistry beta.

Why are nearly-complete atomic shells, such as the seven valence electrons of fluorine, so compact in comparison to the bloated-but-very-stable situation created by adding one more electron to make neon?

I thought I could answer that in terms of Pauli exclusion and wound up completely baffled. Can anyone help on this one? Why does "almost complete" equal "very small" in a charge-neutral atom?

Answer

I'm not sure how easy it would be to be to give a rigorous explanation of this. Here's an explanation based on the atomic orbital approach to the electronic structure of atoms. This is only an approximation, but I think it gives a good flavour of what is going on.

To clarify Terry's question: if you take the atomic orbital approximation to electronic structure you can ask what happens when e.g. the $p$, $d$, or $f$ orbitals are filling up. For example, along the first row of the periodic table from Boron to Neon the $2p$ orbital is filling up, or along the first row of the transition metals from Scandium to Zinc the $3d$ orbital is filling up, and in both cases the radius of the atom decreases.

The $p$, $d$ and $f$ (and higher if we ever make atoms that heavy) orbitals are generally expressed as a product of a radial and angular part, and for this question it's the radial part we have to consider. The atoms are shrinking because the radial part shrinks towards the nucleus, and as you've probably guessed it's because of the increasing nuclear charge attracting the electrons.

Consider the first row from Boron to Neon, i.e. the $2p$ shell is filling. You can get six electrons into this shell. They all share the same radial part of the atomic orbital and they're distributed between the three angular orbitals, $2p_x$, $2p_y$ and $2p_z$. As you add more electrons to the $2p$ orbital (and increase the nuclear charge) you get some mutual screening of the nuclear charge. However the electrons avoid each other as much as possible, so although they do screen each other from the increased nuclear charge the increase in screening is less than the increase in the nuclear charge. The result is that the effective potential felt by the electrons deepens and the radial part of the $2p$ orbital shrinks towards the nucleus. That's why Neon is the smallest atom in this row. When you add another electron it has to go into a new atomic orbital, the $3s$, and the radial part of the $3s$ extends further than the radial part of the $2p$ so the atom suddenly grows.

The reason all this is an approximation is because the splitting of the electron configuration into separate orbitals only works if there are no electron correlation terms i.e. if all the electrons feel a central averaged field, and of course this isn't true. However it's a pretty good approximation for light atoms and I think gives a reasonable explanation of Terry's question.

mathematics - Spreading Gossip

Initially, each of 50 Puzzling Stack Exchange users have a single distinct juicy bit of gossip not known to the others.

1. If $A$ sends an email to $B$, that email can include all the bits of gossip $A$ currently knows. What are the fewest number of emails that need to be sent so that each of the 50 users knows all 50 bits of gossip? Why can't it be done with fewer emails? Assume multiple recipients / carbon copies are not allowed.


2. If $A$ calls $B$, $A$ and $B$ can exchange all the bits of gossip that both know. What is the fewest number of calls required so that each of the 50 users knows all 50 pieces of gossip?

Note: Addressing why (2) can't be done with fewer calls is a little too difficult for a puzzle in my opinion. But then again I believe Bollobás included it in his puzzle book "Coffee Time in Memphis", so if you want to have a go at it knock yourself out =).

Answer

Problem 1:

As leoll2 illustrates, it can be done in

98 emails (everyone emails Alice, then she emails everyone).

Here's why it can't be done in fewer.

First, we show that after the $48$th email, there won't be anyone who has heard all of the gossip. Consider the undirected graph where the vertices are people, where two people are connected with an edge if one of them emailed the other. Since there are at most $48$ edges, and $50$ vertices, this graph is disconnected (every connected $n$ vertex graph has at least $n-1$ edges). This means that people in one connected component could not have heard any of the tidbits of gossip from the other components, so no one could have heard all the gossip.

Furthermore, the number of people who have heard all the gossip increases by at most one after every email. Thus, since there are 0 fully informed people after $48$ emails, there are at most $49$ fully informed people after $48+49=97$ emails have been sent, so $97$ emails is not enough.

gravity - How can the core of the Earth be younger than the crust due to gravitational time dilation if the crust experience more force than the core?

Gravitational force decreases with depth under the surface of Earth, and at centre, the surrounding forces cancel each other out effectively making it zero and hence a body at the center of gravity of the Earth (which would be very close to the centre of the Earth, the core) would experience weightlessness. Now the famous study in 2016 (https://arxiv.org/abs/1604.05507) says that the core is 2.5 years younger than the crust even though the former was formed first, due to gravitational time dilation causing the core to age slower even though the gravity acting on the core should be lower than what is on the crust. I understand that I might be confusing myself between two scientific principles but I want to understand both and know where my understanding is wrong.