How much energy is stored in the charge on each electron?

The energy of electron formation becomes mass, charge, spin and momentum. How much becomes the charge? We know how much becomes mass: E = mc^2. What is the equation for the charge?

quantum mechanics - Why do we obtain classical physics by taking the limit of Planck's constant to zero?

Why if we specifically set Planck's constant equal to zero (the limit of it) do we sometimes get classical physics? I mean, what does it mean physically to set the constant equal to zero? Or to say it in another way, what did the first human to do this think of in order to come up with it?
I have seen in various situations somebody taking h-->0, but what is making anybody who does this to take h-->0 in the first place? Why would somebody take h-->0 to take the classical limit in the first place?

Answer

In classical physics there exists no h, and everything was going happily along until the data/measurements showed that classical electrodynamics + classical mechanics could not explain a number of data.

The datum that introduced the otherwise unknown and undefined h_bar was black body radiation. It was not possible to reconcile the fact that no excess ultraviolet radiation was observed from the experimental study of the radiation from bodies in the lab. Planck had the brilliant idea to assume that the electromagnetic radiation was not continuous but came in quanta of energy h*nu, where nu is the frequency, and could fit the data. Thus the constant took his name. It became obvious that a new framework for physical observations was emerging.

When the photoelectric effect, and then the spectra of atoms appeared in the labs the new framework became consistent by using this h*nu identification for photons that make up light. These data are not classical physics. Quantum mechanics was developed to explain it, and it soon became clear that all classical observations emerged from an underlying quantum mechanical framework, and this can be proven with advanced mathematical methods. For example, the way the classical electromagnetic radiation is built up by photons.

A second foundation stone for quantum mechanics is Heisenberg's Uncertainty Principle. , HUP . This is what sets the stage for deciding whether quantum mechanical methods are needed or whether classical theories are adequate. For example

this inequality does not hold for classical physics, since h is a very small number. The constraints of the inequality appear at the level of particle physics, for micrometers and lower values. The statement that one gets classical physics when h is zero comes from the observation that classical momenta times classical dimensions multiplied have no lower bounds up to dimensions of micrometers and smaller . The handwaving statement "one gets classical physics if h is zero" is based on this relation.

The HUP is in the foundation of the mathematics of quantum mechanics, it appears in the commutator and anticommutator relations of operators representing measurable observables, like position and momentum. If h is zero, one has to fall back on the classical theories. , the consistency check also needs advanced mathematics to see the correspondence.

riddle - I am an expression of a deer mint - Clue One

You wake up one morning, and immediately look over at the letter on your bedside table. It reads:

Dear [redacted]:
You have been chosen to assemble the most powerful thing in the world - the [redacted] serum. It is composed of [redacted] clues. Your task: Decode the clues, and then assemble them. Good luck!

P.S. If you cop out, we'll know. And you'll regret it.

The Riddle:

I am the color of first.
I am always good – I'll never be a brat.
I am made by number 2.
I am an expression of a deer mint.

What am I?

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_{Next clue --->}

Answer

The answer is

Honey

I am the color of first.

The 'color of first' being gold (e.g. gold medal), which is the color of honey. (@Brent Hackers for this one)

I am always good – I'll never be a brat.

Honey is sweet, and sweet could be considered synonymous with 'good' and antonymous with 'being a brat'.
Honey doesn't spoil or 'go bad', i.e. it's always good.

I am made by number 2.

Honey is made by a bee; 'B' is the 2nd letter.

I am an expression of a deer mint.

'Honey' is a common term of endearment (assuming the 'a deer mint' = 'endearment' theory ala @seasnake).

mathematics - Smallest PRIME containing the first 11 primes as sub-strings

In Smallest number containing the first 11 primes as sub-strings, @Alconja successfully found the smallest number which contains the first eleven primes (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31) as concatenated sub-strings. This inspired me to propose the following followup:

What is the smallest prime which contains each of the first eleven primes as a sub-string?

Obviously the answer is at least

113,171,923,295,

but that's not prime. How much further do we need to go?

_{Disclaimer: I don't know the answer myself. I'm hoping it won't need a computer to find ...}

Answer

(Kind of) analytical solution that only requires small amount of calculations, (potentially) doable by hand.

First step: we can safely drop 2, 3 and 7 from the equation as those digits are used in 23 and 17. Now, we need to build a prime from: 5, 11, 13, 17, 19, 23, 29 and 31.

Second step: let's try to build the shortest number possible from these numbers. To do this we need to maximize the number of overlaps.

To do this, let's build a graph of possible overlaps:

An edge from number A to number B means that A and B can overlap (e.g. 11 and 13 can combine into 113). 5 and 29 can't overlap with other numbers. Maximum number of overlaps is equivalent to the (totally) longest possible set of paths in the "main" clique.

After going through all the possible starting points (11, 13, 31 and 23) we find that the maximum number of overlaps is 3 and there're 10 possible sets of paths with this number of overlaps:

• 11 -> 13 -> 31 -> 17 = 11317



• 11 -> 13 -> 31 -> 19 = 11319


• 13 -> 31 -> 11 -> 17 = 13117


• 13 -> 31 -> 11 -> 19 = 13119


• 23 -> 31 -> 11 -> 17 = 23117


• 23 -> 31 -> 11 -> 19 = 23119


• 13 -> 31 -> 17 = 1317, 11 -> 19 = 119


• 13 -> 31 -> 19 = 1319, 11 -> 17 = 117


• 23 -> 31 -> 17 = 2317, 11 -> 19 = 119


• 23 -> 31 -> 19 = 2319, 11 -> 17 = 117

Corollary 1: Any prime number that can be represented as a permutation of one of these 10 sets of numbers (let's call it a candidate):

• 5, 29, 11317, 19, 23


• 5, 29, 11319, 17, 23


• 5, 29, 13117, 19, 23


• 5, 29, 13119, 17, 23


• 5, 29, 23117, 13, 19


• 5, 29, 23119, 13, 17


• 5, 29, 119, 1317, 23


• 5, 29, 117, 1319, 23



• 5, 29, 2317, 119, 13


• 5, 29, 2319, 117, 13

will be the shortest possible prime that contains the first 11 primes. If al least one candidate exist, the smallest of them will be the solution.

Corollary 2: if there are candidates that start with 11317 then the smallest of them will be the solution, as 11317 is the alphabetically smallest sequence among all presented.

Step three: Let's sort the first set in alphabetical order and then go through permutations one by one in increasing order until we find a prime number:

• 11317, 19, 23, 29, 5 - not a prime, 5 * 22634384659


• 11317, 19, 23, 5, 29 - not a prime, 7 * 16167417647


• 11317, 19, 29, 23, 5 - not a prime, 5 * 22634385847



• 11317, 19, 29, 5, 23 - not a prime, 59 * 1918168297


• 11317, 19, 5, 23, 29 - not a prime, 337 * 335821817


• 11317, 19, 5, 29, 23 - bingo!

The answer is: 113171952923.

P.S. Now, all of this looks horrible, but the only step that requires truly obscene amount of calculations is a primality test for 113171952923. If we can use a computer for that, we're good. We kind of got lucky that the answer is so close to the start of the search, though.

electromagnetism - Can I transform electromagnetic tensors by matrix multiplication?

I know that the eletromagnetic field tensor $F^{\mu\nu}$, can be transfomed to another reference frame by $$F^{\alpha\beta} = \varLambda^{\alpha}_{\mu}\varLambda^{\beta}_{\nu}F^{\mu\nu}$$

Since these tensors can be represented by matrices so I thought that I could represent the electromagnetic field tensor in another inertial reference frame by doing matrix multiplications, but I ended up with:

$$F^{\alpha\beta}= \begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & -\frac{1}{c}E_x & -\frac{1}{c}E_y & -\frac{1}{c}E_z \\ \frac{1}{c}E_x & 0 & -B_z & B_y\\ \frac{1}{c}E_y & B_z &0 & -B_x\\ \frac{1}{c}E_z & -B_y & B_x & 0 \end{bmatrix} =$$ $$=\begin{bmatrix} -2\frac{\gamma^2\beta}{c}E_x & -\frac{\gamma^2}{c}E_x(1+\beta^2)& -\frac{\gamma^2}{c}E_y(1+\beta^2)+2\gamma^2\beta^2B_z & -\frac{\gamma^2}{c}E_z(1+\beta^2)-2\gamma^2\beta B_y \\ \frac{1}{c}E_x & 2\frac{\gamma^2\beta}{c}E_x & 2\frac{\gamma^2\beta}{c}E_y-\gamma^2B_z(\beta^2+1) & 2\frac{\gamma^2\beta}{c}E_z+\gamma^2B_y(\beta^2+1)\\ \frac{1}{c}E_y & B_z &0 & -B_x\\ \frac{1}{c}E_z & -B_y & B_x & 0 \end{bmatrix} $$ But as it can easily be seen, this matix is not anti-symmetric as an eletromagnetic field tensor should be, by definition, so does this mean that I cannot use matrix multiplication on tensors or does it mean that I made a mistake somewhere in the calculations of the matrix products?

Answer

^{\begin{align} F^{\alpha\beta} & = \begin{bmatrix} \hphantom{-}\gamma & -\gamma\beta & \hphantom{-} 0 & \hphantom{-}0\hphantom{-} \vphantom{\dfrac12}\\ -\gamma\beta & \hphantom{-}\gamma & \hphantom{-}0 & \hphantom{-} 0 \hphantom{-} \vphantom{\dfrac12}\\ \hphantom{-}0 & \hphantom{-}0 & \hphantom{-}1 & \hphantom{-} 0 \hphantom{-} \vphantom{\dfrac12}\\ \hphantom{-} 0 & \hphantom{-} 0 & \hphantom{-} 0 & \hphantom{-} 1 \hphantom{-} \vphantom{\dfrac12} \end{bmatrix} \begin{bmatrix} 0 & -E_x & -E_y & -E_z \vphantom{\dfrac12}\\ E_x & \hphantom{-} 0 & -cB_z & \hphantom{-} cB_y \vphantom{\dfrac12} \\ E_y & \hphantom{-} cB_z & \hphantom{-} 0 & -cB_x \vphantom{\dfrac12} \\ E_z & -cB_y & \hphantom{-} cB_x & \hphantom{-} 0 \vphantom{\dfrac12} \end{bmatrix} \begin{bmatrix} \hphantom{-}\gamma & -\gamma\beta & \hphantom{-} 0 & \hphantom{-}0\hphantom{-} \vphantom{\dfrac12}\\ -\gamma\beta & \hphantom{-}\gamma & \hphantom{-}0 & \hphantom{-} 0 \hphantom{-} \vphantom{\dfrac12}\\ \hphantom{-}0 & \hphantom{-}0 & \hphantom{-}1 & \hphantom{-} 0 \hphantom{-} \vphantom{\dfrac12}\\ \hphantom{-} 0 & \hphantom{-} 0 & \hphantom{-} 0 & \hphantom{-} 1 \hphantom{-} \vphantom{\dfrac12} \end{bmatrix} \nonumber\\ & = \begin{bmatrix} \!\!-\gamma\beta E_x &\!\! -\gamma E_x & -\gamma(E_y-\beta cB_z) & -\gamma(E_z+\beta cB_y)\vphantom{\dfrac12} \hphantom{-} \\ \hphantom{-}\gamma E_x & \!\gamma\beta E_x & \hphantom{-}\gamma(\beta E_y- cB_z) & \hphantom{-}\gamma(\beta E_z+cB_y)\vphantom{\dfrac12}\hphantom{-}\\ \hphantom{\gamma\beta} E_y & \hphantom{-} cB_z &0 & -cB_x\vphantom{\dfrac12}\hphantom{-}\\ \hphantom{\gamma\beta} E_z & -cB_y & cB_x & \hphantom{-}0\vphantom{\dfrac12}\hphantom{-} \end{bmatrix} \begin{bmatrix} \hphantom{-}\gamma & -\gamma\beta & \hphantom{-} 0 & \hphantom{-}0\hphantom{-} \vphantom{\dfrac12}\\ -\gamma\beta & \hphantom{-}\gamma & \hphantom{-}0 & \hphantom{-} 0 \hphantom{-} \vphantom{\dfrac12}\\ \hphantom{-}0 & \hphantom{-}0 & \hphantom{-}1 & \hphantom{-} 0 \hphantom{-} \vphantom{\dfrac12}\\ \hphantom{-} 0 & \hphantom{-} 0 & \hphantom{-} 0 & \hphantom{-} 1 \hphantom{-} \vphantom{\dfrac12} \end{bmatrix} \nonumber\\ & =\begin{bmatrix} \hphantom{-}0 & -E_x & -\gamma(E_y-\beta cB_z) &-\gamma(E_z+\beta cB_y)\hphantom{-}\vphantom{\dfrac12} \\ \hphantom{-} E_x & 0 & \hphantom{-}\gamma(\beta E_y- cB_z) & \hphantom{-}\gamma(\beta E_z+cB_y)\hphantom{-}\vphantom{\dfrac12}\\ \hphantom{-}\gamma(E_y-\beta cB_z) & -\gamma(\beta E_y- cB_z) &\hphantom{-} 0 & -cB_x\hphantom{-}\vphantom{\dfrac12}\\ \hphantom{-}\gamma(E_z+\beta cB_y) & -\gamma(\beta E_z+cB_y) & \hphantom{-} cB_x & \hphantom{-}0\hphantom{-}\vphantom{\dfrac12} \end{bmatrix} \nonumber\\ & =\begin{bmatrix} \hphantom{-}0 & -E'_x & -E'_y & -E'_z \hphantom{-}\vphantom{\dfrac12}\\ \hphantom{-}E'_x & \hphantom{-} 0 & -cB'_z & \hphantom{-} cB'_y \hphantom{-}\vphantom{\dfrac12} \\ \hphantom{-}E'_y & \hphantom{-} cB'_z & \hphantom{-} 0 & -cB'_x \hphantom{-}\vphantom{\dfrac12} \\ \hphantom{-}E'_z & -cB'_y & \hphantom{-} cB'_x & \hphantom{-} 0 \hphantom{-}\vphantom{\dfrac12} \end{bmatrix} \tag{01} \end{align}} Since $\:\beta=\upsilon/c$ \begin{align} E'_x & = \hphantom{-}E_x \tag{02.x}\vphantom{\dfrac12}\\ E'_y & = \gamma(E_y-\upsilon B_z) \tag{02.y}\vphantom{\dfrac12}\\ E'_z & = \gamma(E_z+\upsilon B_y) \tag{02.z}\vphantom{\dfrac12}\\ B'_x & = \hphantom{-}B_x \tag{03.x}\vphantom{\dfrac12}\\ B'_y & = \gamma(B_y+\dfrac{\upsilon}{c^{2}} E_z) \tag{03.y}\\ B'_z & = \gamma(B_z-\dfrac{\upsilon}{c^{2}} E_y) \tag{03.z} \end{align}

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^{For Your Information :}

^{The equations of a more general Lorentz Transformation between two systems $\:\mathrm S(\mathbf{x},t)\:$ and $\:\mathrm S'(\mathbf{x}',t')\:$, the latter translating with constant velocity $\:\mathbf{v}\!=\!\upsilon\mathbf{n}\,,\Vert\mathbf{n}\Vert=1\,, \upsilon \in (-c,+c)$, with respect to the former, are : \begin{align} \mathbf{x}'& \!=\!\mathbf{x}\!\boldsymbol{+}\!(\gamma\!\boldsymbol{-}\!1)(\mathbf{n}\boldsymbol{\cdot}\mathbf{x})\mathbf{n}\!\boldsymbol{-}\!\gamma\mathbf{v}t \tag{ft-01a}\\ t' & \!=\! \gamma\left(t\!\boldsymbol{-}\!\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{x}}{c^{2}}\right) \tag{ft-01b}\\ \gamma & \!=\!\left(1\!\boldsymbol{-}\!\dfrac{\upsilon^{2}}{c^{2}}\right)^{\boldsymbol{-}1/2} \tag{ft-01c} \end{align} see Figure.(1)}

^{Under (ft-01) the vectors $\:\mathbf{E},\mathbf{B}\:$ of the electromagnetic field in empty space are transformed as follows :
\begin{align} \mathbf{E}'& \!=\!\gamma\mathbf{E}\!\boldsymbol{-}\!(\gamma\!\boldsymbol{-}\!1)(\mathbf{n}\boldsymbol{\cdot}\mathbf{E})\mathbf{n}\boldsymbol{+}\:\gamma\left(\mathbf{v}\boldsymbol{\times}\mathbf{B}\right) \tag{ft-02a}\\ \mathbf{B}'& \!=\!\gamma\mathbf{B}\!\boldsymbol{-}\!(\gamma\!\boldsymbol{-}\!1)(\mathbf{n}\boldsymbol{\cdot}\mathbf{B})\mathbf{n}\!\boldsymbol{-}\!\dfrac{\gamma}{c^{2}}\left(\mathbf{v}\boldsymbol{\times}\mathbf{E}\right) \tag{ft-02b} \end{align} Equations (02),(03) are a special case of (ft-02) for $\:\mathbf{n}=(1,0,0)$.}

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^{(1) See a 3D version of this Figure here : Figure 3D version}

homework and exercises - Wave speed of a hanging rope

Let us consider a homogeneous rope hanging from the ceiling. I will call the vertical direction $x$ and the horizontal displacement $y$. When we apply the second Newton's Law to a portion of mass $\Delta m$ and proceed in the same way we do for a horizontal string we get $$\mu\frac{\partial^2 y(x,t)}{\partial t^2}=\frac{\partial}{\partial x}\left[T(x)\frac{\partial y(x,t)}{\partial x}\right].$$ The difference now is that the tension is not constant. Defining $x=0$ at the free end of the rope and orienting it upwards we have the tension $T=\mu gx$. Therefore, $$\frac{\partial^2 y}{\partial t^2}=gx\frac{\partial^2y}{dx^2}-g\frac{\partial y}{\partial x}.$$

So, my issue is that I have seen in a couple of physics books (for instance) that the wave speed of this rope is simply $$v=\sqrt{\frac{T}{\mu}}=\sqrt{gx}.$$ In my opinion these books are either:

i) Cheating the students: they know it is wrong but assume it is right just to show some nice features (the speed would increase as the wave goes upwards).

ii) Had misconception assuming that the usual wave equation are still valid.

iii) Is doing some obscure approximation, which is another cheat if you do not reveal it.

My questions are:

1) Is there an expression for the wave speed for the "wave equation" above?

2) Is there an approximation leading to the wave speed given above?

newtonian mechanics - Why is the tension between two masses connected by a rope and undergoing a force along the direction of the rope less than that force?

Two different known masses at rest on a frictionless surface are connected by a rope. A known force F is applied horizontally to mass 2 so that the boxes begin to accelerate in that direction. The acceleration of both masses together is given by a=F/(m1+m2) (I think they can be considered the same object because of the rope and the lack of friction). The only horizontal force acting on mass 1 is the pull coming from the tension in the rope. Therefore the tension T is given by T=m1*F/(m1+m2) (again, the acceleration is the same because of the rope and the lack of friction, I think).

Intuitively, this makes absolutely no sense to me, because the tension in the rope ends up being less than the force acting on mass 2. What's going on? Are you supposed to not count both masses in the calculation of acceleration? If so, why?

Answer

It is best to draw free body diagrams for the two masses.

$F$ is the applied force and $T$ the tension in the massless and inextensible rope joining the two masses.
There is no friction and both masses have the same acceleration $a$.

Applying Newton's second law for each of the masses:

$T = m_1\;a$ and $F-T= m_2\; a \Rightarrow F = (m_1+m_2)\;a$ so $F>T$

You can think of it as the force $F$ is accelerating both masses whereas the force $T$ only has to accelerate mass $m_2$.