argument about fallacy of diff(M) being a gauge group for general relativity

I want to outline a solid argument (or bulletpoints) to show how weak is the idea of diff(M) being the gauge group of general relativity.

basically i have these points that in my view are very solid but i want to understand if there are misconceptions on my part that i'm simply not getting and if its so, i ask for help to make the case more solid, or understanding why it doesn't apply (to gravity):

• 
  

  gauge groups are not the same as a symmetry group (thanks to Raymond Streater for making that point completely clear)

  
  
  


• 
  

  gauge invariance in electrodynamics is an observation that physical observables are unchanged after a gauge transformation without changing coordinate frame ( we are ask to believe that in gravity someone did the same? that is, someone made the observation that physical observables are unchanged after a diffeomorphism-gauge-transformation, only to later argue that because of this, that there are no physical observables to begin with, that doesn't make a lot of sense, to not say that its just plain stupid circular argument )

  
  


• 
  

  classic electrodynamics is also invariant (as in symmetry invariant, not as gauge-invariant) under Diff(M). The invariance is of course broken when the theory is quantized and $\hbar$ makes an appearance, because it assumes a preferred scale for certain energies. The key point here is: classical gravity is not special regarding having diff(M) as a symmetry group

  
  


• 
  

  from bulletpoints 2 and 3, if i cannot infer that Diff(M) is a gauge-invariance of electrodynamics, the same should apply to gravity

  
  

For this question, i would say that a valid answer would either disprove any of the arguments as fallacies themselves (hence showing a solid argument why gravity is special and diff(M) is without a doubt its gauge group), or improve the argument for making it bullet-proof (sorry for the pun)

Answer

I will mostly talk about the classical physics as this is complicated enough already (I might mention something about quantum stuff at the end). So, let us first get all the relevant terms straight, so that we avoid any further confusion. In particular, we need to be precise about what we mean by invariance because already two different notions have been thrown into one bag.

• 
  

  A symmetry group of a physical system is a group of transformations that leave the system invariant. E.g. the electric field of a point charge is invariant under rotations w.r.t. to that point. In other words, we want the group to act trivially. But that means that this immediatelly rules out any equations that carries tensorial indices (i.e. transforms in a non-trivial representation of the rotation group). For these equations, if you perform a rotation, the equation will change. Of course, it will change in an easily describable manner and a different observer will agree. But the difference is crucial. E.g. in classsical quantum mechanics we require the equations to be scalar always (which is reflected in the fact that Hamiltonian transforms trivially under the symmetry group).

  
  


• 
  

  Carrying a group action is a broader term that includes tensorial equations we have left out in the previous bullet point. We only require that equations or states are being acted upon by a group. Note that the group action need not have any relation to the symmetry. E.g. take the point charge and translate it. This will certainly produce a different system (at least if there is some background so that we can actually distinguish points).

  
  
  


• 
  

  A gauge group of a system is a set of transformations that leave the states invariant. What this means is that the actual states of the system are equivalence classes of orbits of the gauge group. Explicitly, consider the equation ${{\rm d} \over {\rm d} x} f(x) = 0$. This has a solution $g(x) \equiv C$ for any $C$. But if we posit that the gauge group of the equation consists of the transformations $f(x) \mapsto f(x) + a$ then we identify all the constant solutions and are left with a single equivalence class of them -- this will be the physical state. This is what gauge groups do in general: they allow us to treat equivalence classes in terms of their constituents. Obviously, gauge groups are completely redundant. The reason people work with gauge groups in the first place is that the description of the system may simplify after introduction of these additional parameters that "see" into the equivalence class. Of course historical process went backwards: since gauge-theoretical formulation is simpler, this is whan people discovered first and they only noticed the presence of the gauge groups afterwards.

  
  

Now, having said this, let's look at the electromagnetism (first in the flat space). What symmetries are the Maxwell equations invariant under? One would like to say under Lorentz group but this actually not the case. Let's look at this more closely. As alluded to previously the equation $$\partial_{\mu} F^{\mu \nu} = J^{\nu}$$ can't really be invariant since it carries a vector index. It transforms in the four-vector representation of the Lorentz group, yes -- but it is certainly not invariant. Contrast this with the Minkowski space-time itself which is left invariant by the Lorentz group.

We also have ${\rm d} F = 0$ and therefore (in a contractible space-time) also $F = {\rm d} A$ which is obviously invariant w.r.t. to $A \mapsto A + {\rm d} \chi$. In terms of the above discussion the equivalence class $A + {\rm d} \Omega^0({\mathbb R}^{1,3})$ is the physical state and the gauge transformation lets us distinguish between its constituents

Let's move to a curved spacetime now. Then we have $$\nabla_{\mu} F^{\mu \nu} = J^{\nu}$$ Again, this is not invariant under ${\rm Diff}(M)$. But it transforms under an action of ${\rm Diff}(M)$. The only thing in sight that is invariant under ${\rm Diff}(M)$ is $M$ itself (by definition).

In the very same way, GR is not invariant under ${\rm Diff}(M)$ but only transforms under a certain action of it (different than EM though, since GR equations carry two indices). Also, ${\rm Diff}(M)$ can't possibly be a gauge group of any of these systems since it would imply that almost all possible field configurations get cramped to a single equivalence class possibly indexed by some topological invariant which can't be changed by a diffeomorphism. In other words, theory with ${\rm Diff}(M)$ as a gauge group would need to be purely topological with no local degrees of freedom.

speed of light - Special Relativity and $E = mc^2$

I read somewhere that $E=mc^2$ shows that if something was to travel faster than the speed of light then they would have infinite mass and would have used infinite energy.

How does the equation show this?

The reason I think this is because of this quote from Hawking (I may be misinterpreting it):

Because of the equivalence of energy and mass, the energy which an object has due to its motion will add to its mass. This effect is only significant to objects moving at speeds close to the speed of light. At 10 per cent of the speed of light an objects mass is only 0.5 per cent more than normal, at 90 per cent of the speed of light it would be twice its normal mass. As an object approaches the speed of light its mass rises ever more quickly, so takes more energy to speed it up further. It cannot therefore reach the speed of light because its mass would be infinite, and by the equivalence of mass and energy, it would have taken an infinite amount of energy to get there.

The reason I think he's saying that this is as a result of $E = mc^2$ is because he's talking about the equivalence of $E$ and $c$ from the equation.

Answer

I read somewhere that $E=mc^2$ shows that if something was to travel faster than the speed of light then they would have infinite mass and would have used infinite energy.

Nope, not true. For a couple of reasons, but first, let me explain what $E = mc^2$ means in modern-day physics.

The equation $E = mc^2$ itself only applies to an object that is at rest, i.e. not moving. For objects that are moving, there is a more general form of the equation,

$$E^2 - p^2 c^2 = m^2 c^4$$

($p$ is momentum), but with a little algebra you can convert this into

$$E = \gamma mc^2$$

where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$. This factor $\gamma$, sometimes called the relativistic dilation factor, is a number that depends on speed. It starts out at $\gamma = 1$ when $v = 0$, and it increases with increasing speed. As the speed $v$ gets closer and closer to $c$, $\gamma$ approaches infinity. Armed with this knowledge, some people look at the formula $E = \gamma mc^2$ and say that, clearly, if a massive object were to reach the speed of light, then $\gamma$ would be infinite and so the object's energy would be infinite. But that's not really true; the correct interpretation is that it's impossible for a massive object to travel at the speed of light. (There are other, more mathematically complicated but more convincing, ways to show this.)

To top it off, there is an outdated concept called "relativistic mass" that gets involved in this. In the early days of relativity, people would write Einstein's famous formula as $E = m_0 c^2$ for an object at rest, and $E = m_\text{rel}c^2$ for an object in motion, where $m_\text{rel} = \gamma m_0$. (The $m$ I wrote in the previous paragraphs corresponds to $m_0$ in this paragraph.) This quantity $m_\text{rel}$ was the relativistic mass, a property which increases as an object speeds up. So if you thought that an object would have infinite energy if it moved at the speed of light, then you would also think that its relativistic mass would become infinite if it moved at the speed of light.

Often people would get lazy and neglect to write the subscript "rel", which caused a lot of people to mix up the two different kinds of mass. So from that, you'd get statements like "an object moving at light speed has infinite mass" (without clarifying that the relativistic mass was the one they meant). After a while, physicists realized that the relativistic mass was really just another name for energy, since they're always proportional ($E = m_\text{rel}c^2$), so we did away with the idea of relativistic mass entirely. These days, "mass" or $m$ just means rest mass, and so $E = mc^2$ applies only to objects at rest. You have to use one of the more general formulas if you want to deal with a moving object.

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Now, with that out of the way: unfortunately, the passage you've quoted from Hawking's book uses the old convention, where "mass" refers to relativistic mass. The "equivalence of energy and mass" he mentions is an equivalence of energy and relativistic mass, expressed by the equation $E = m_\text{rel}c^2$. Under this set of definitions, it is true that an object approaching the speed of light would have its (relativistic) mass approach infinity (i.e. increase without bound). Technically, it's not wrong, because Hawking is using the concept correctly, but it's out of line with the way we do things in physics these days.

With modern usage, however, I might rephrase that paragraph as follows:

Because energy contributes to an object's inertia (resistance to acceleration), adding a fixed amount of energy has less of an effect as the object moves faster. This effect is only significant to objects moving at speeds close to the speed of light. At 10 per cent of the speed of light, it takes only 0.5 per cent more energy than normal to achieve a given change in velocity, but at 90 per cent of the speed of light it would take twice as much energy to produce the same change. As an object approaches the speed of light, its inertia rises ever more quickly, so it takes more and more energy to speed it up by smaller and smaller amounts. It cannot therefore reach the speed of light because it would take an infinite amount of energy to get there.

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Disclaimer: all I've said here applies to a fundamental particle or object moving in a straight line. When you start to consider particles with components which may be moving relative to each other, the idea of relativistic mass kind of makes a comeback... kind of. But that's another story.

thermodynamics - Have negative pressures any physical meaning?

Some cubic thermodynamical equations of state predict negative pressures, have negative pressures any physical meaning? Could they be related to negative mass?

lateral thinking - Liar liar pants on fire

Here is an easy and straightforward liar puzzle.

Four people are in a room, stack reader, boboquack, Rand al'Thor and Brent Hackers.
stack reader states that at least 2 of the other 3 are liars.
boboquack states that Rand al'Thor and Brent Hackers are the only 2 liars or are both truth tellers.
Rand al'Thor states that stack reader is the only liar.
Brent Hackers states that Rand al'Thor is the only liar or the only truth teller.

There is more than 1 liar!
Can you find who are the liars?
I can assure you that no one will find the answer!

Answer

I think the answer is:

Stack Reader is the only liar

I mean,

Here is an easy and straightforward liar puzzle is a lie - look at how many differing opinions there are

Four people are in a room, stackreader, boboquack, Rand al'Thor and Brent Hackers. is a lie

stackreader states that at least 2 of the other 3 are liars. is a lie (he says at least one of the others is lying, below)

boboquack states that Rand al'Thor and Brent Hackers are the only 2 liars or are both truth tellers. <- lie, bbq never said that. Same for "Rand al'Thor states that stackreader is the only liar." and "Brent Hackers states that Rand al'Thor is the only liar or the only truth teller."

There is more than 1 liar! Lie - stackreader is the only liar

I can assure you that no one will find the answer! lie, somebody will probably find the answer

mathematical physics - Discreteness of set of energy eigenvalues

Given some potential $V$, we have the eigenvalue problem

$$ -\frac{\hbar^2}{2m}\Delta \psi + V\psi = E\psi $$

with the boundary condition

$$ \lim_{|x|\rightarrow \infty} \psi(x) = 0 $$

If we wish to seek solutions for

$$ E < \lim_{|x|\rightarrow \infty}V(x), $$

I find textbooks like Sakurai and others assert that (without proof) set of all such eigenvalues $E$ that admits nontivial solutions is discrete, which gives us liberty to index them with natural numbers (principal quantum numbers). Can anyone help me with a little elaboration of this mathematical fact ? As a student of mathematics, I know that compact operators have discrete spectrums, so for the operator $ T : L^2 \rightarrow L^2 $given as

$$ T(\psi) = -\frac{\hbar^2}{2m}\Delta \psi + V\psi. $$

Are there methods to show that $T$ is compact? Or possibly other ways to establish discreteness of its eigenspectrum?

Answer

Consider the Sturm-Liouville regular problem in self-adjoint form $$(A-\zeta)\,v=f$$ whose explicit solution is $$v=\int_a^bG(x,s)\,f(s)\,ds$$ where the Green function $$G(x,s)=\begin{cases}\frac{\varphi_b(x)\,\varphi_a(s)}{W(s)},&a\leq{s}\leq{x}\\[0.1in]\frac{\varphi_a(x)\,\varphi_b(s)}{W(s)},&x\leq{s}\leq{b}\end{cases}$$ is built by the solutions $\varphi_a$, $\varphi_b$ of the equation $A\,\varphi=0$ which satisfy respective boundary conditions at $x=a,b$. It's a long way to get here, but you can verify this. The resolvent associated to $A$ is given by $$R(A,\zeta)\,f=\int_a^bG(x,s)\,f(s)\,ds$$ Now, showing that an operator $K$, whose kernel $k(x,s)$ satisfies $$\int_a^b\int_a^b\left|k(x,s)\right|^2ds\,dx<\infty\hspace{0.5in}(*)$$ is compact (meaning the functions $k$ are of square-integrable), may be used to show that if the Green function $G$ of the resolvent operator $R$ is continuous, then $G$ satisfies equation $(*)$ on a finite interval $[a,b]$, meaning that $R$ is a compact self-adjoint operator for regular Sturm-Liouville problems. You could just do this last thing in general, and just show that with the operator $$\hat{H}\equiv{T}=-\frac{\hbar^2}{2m}\nabla^2+V$$ (the Hamiltonian), the eigenvalue problem $\hat{H}\psi=E\psi$ (known as Schrödinger stationary -or time independent- equation) can be reduced to a regular Sturm-Liouville problem.

All of this because, complementing what you said, the eigenvalues of the problem $T\varphi_k=\lambda_k\varphi_k$ for a compact symmetric operator $T$ on a Hilbert space $H$ with interior product $\langle\cdot\mid\cdot\rangle_H$ are a bounded countably infinite set that converges to zero, $\displaystyle{\lim_{k\to\infty}\lambda_k=0}$. Other (pretty relevant) properties are that the multiplicity of each eigenvalue $\lambda_k$ is finite, and that the set of all eigenfunctions $\varphi_k$ define a complete basis of the space $H$, so that any element $f$ of $H$ can be expanded as $\displaystyle{f=\sum_{k=1}^\infty{f_k}\,\varphi_k}$.

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Update

Please note the relevance of the condition $\boxed{\displaystyle{E<\lim_{|x|\rightarrow \infty}V(x)}}$. Consider this next image as an illustration (I've taken it directly from the web, so take the potential $V=U$); as the math is pretty straightforward, I'll go a little more with physical picture this time

with the condition above, the results found are valid for the equation $\hat{H}\psi=E_i\psi$ with $i=0,1$. Indeed, both equations may presumably be reduced to a regular S-L problem because they would be ones with a finite or penetrable potential well, where we can know the energy spectrum just from $\psi$ inside the potential well, i.e. for a finite interval, whereas considering $i=2$ the interval for $\psi$ would inevitably be infinite and it is evident that $\hat{H}\psi=E_i\psi$ couldn't be reduced to a regular S-L problem and thus the energy spectrum presumably wouldn't be discrete. Feel free to consider any potential you like.

This corresponds to a beautiful analogy with classical mechanics when we find closed or open orbits, which here are bounded and unbounded states (hope I got the translations right) respectively. Think of a massive nucleus and an electron that interact by means of an atractive Columb potential. If the condition above is satisfied, the system may be a hydrogen-type atom that would have a discrete energy spectrum, as known. Although if someone shot the electron against the nucleus from very far away and with enough kinetic energy, the total energy would be positive and the nucleus would deflect the electron without capturing it and without changing it's energy, which evidently is free to take any value. The simplest case for a continuous energy spectrum in QM is that of the free particle, i.e. $V(\mathbf{r})=0,\,\forall\mathbf{r}\in\mathbb{R}^3$; there the normalization condition $\int\psi^*_n\psi_m\,dx=\delta_{nm}$ is not valid and may be generalized using the Dirac delta, also wavefunctions strictly belong to a Banach space.

homework and exercises - Force on a point charge $q$ inside a cavity in an uncharged conductor

This is problem 2.40 from Introduction to Electrodynamics by D. J. Griffiths:

A point charge $q$ is inside a cavity (not necessarily spherical or anything similarly regular) in an uncharged conductor. Is the force on $q$ necessarily zero?

What my understanding tells me is that if the charge on the inner surface of the conductor is just enough to cancel the field from the point charge from every direction, then the force from any two opposite pieces on the inner surface will produce equal but opposite forces. Thus, the net force is zero.

Is my reasoning correct?

event horizon - What happens if you let a cable roll slip into a black hole?

1. 
   

   Does the cable roll spin faster the more cable goes into the black hole in reference of a observer standing next to it?

   
   


2. 
   

   Can gravity pull the cable that it exceed the speed of light inside a black hole in reference to the center of the black hole?

   
   

Answer

Whether it's a black hole or some other more ordinary mass pulling on your rope isn't actually that interesting. Let's think about a cable unrolling above Earth to start with.

What we have is a pulley with a rope hanging off one side. The weight of the rope exerts some force on the edge of the pulley, causing it to undergo angular acceleration (starts to spin). This releases some more rope, so the force increases a bit, acceleration increases, pulley spins faster and faster. Some assumptions:

• Let's assume that our spool of rope is very big, so that there's always a lot of rope left in it, and the change in the mass of coiled rope is unimportant.


• Let's also assume the bearings or whatever that allow the pulley to rotate are frictionless.


• Also, let's not complicate anything with relativistic speeds, yet.

Once the end of the rope hits the ground, the weight of the portion of the rope lying on the ground is no longer felt by the pulley. Since there is always a constant amount of rope between the pulley and the ground, the force is constant. The pulley continues to spin faster, but with constant acceleration. Still assuming speeds stay non-relativistic for now. Now for the first consideration with the black hole. Instead of "hitting the ground" the end of the rope now "hits the singularity". The weight of the portion of the rope absorbed by the singularity is no longer felt by the pulley. We're adding a bit of mass to the black hole, but let's assume that the mass of rope that we're feeding in is small compared to the mass of the black hole (similar to how you'd assume the mass of rope lying on the ground is much less than the mass of the Earth). This means the gravitational force exerted by the black hole doesn't increase. So we really have exactly the same scenario. Don't worry, it will get more complicated in a minute.

As the pulley keeps spinning faster, its edge (the part that's spinning fastest) grows to an appreciable fraction of the speed of light. The rope is coming out at the same speed that the edge of the pulley is moving, so it is also now moving relativistically. Now Newtonian intuition starts to break down. Another way to look at what has been going on is in terms of energy. The rope starts with some gravitational potential energy (from having a height above the Earth/black hole/whatever). As it falls it loses potential energy, which goes into the kinetic energy of the rope and rotating pulley. Newtonian considerations would tell you something along the lines of: $$-\Delta U = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$ $v$ is the speed of the rope, and $\omega = v/r$ is the angular speed of the pulley (equal to the speed of the rope divided by the radius of the pulley). Anyway. This breaks down once speeds start getting relativistic. Energy is still conserved, but the change in speed obtained for a given energy input starts to decrease. You can put in more and more energy, and the speed will always keep increasing, but the increase for a given amount of energy gets smaller and smaller. The speed keeps increasing, approaching the speed of light, but never reaches it. You'll get to $.9c$, then $.99c$, then $.999c$, eventually $.9999999999999...999999c$, but never reach $1.0c$.

In practice, a lot of other things prevent you from getting to relativistic speeds in the first place. You might have a nearly frictionless pulley, but nearly isn't zero, and as speeds pick up the heat production from friction will destroy your equipment. Or air friction on the rope will increase with the speed until equilibrium is reached and the rope is falling at a constant speed. There is no air above a black hole of course... You need a lot of energy to get to relativistic speeds, which means to need to drop a LOT of rope. And the more rope you need to drop, the more needs to be coiled. But this increases the mass of the coil, which increases the energy needed to accelerate it. So more rope is needed. But this increases the mass of the coil... it's likely your coil will get so massive that it will collapse and form it's own black hole before you can get to relativistic speeds by unrolling it.

There are also worries about the tensile strength of the rope. You can't hang your spool arbitrarily high, at some point the weight of the rope below a given segment will be so much that the rope will break. This gets even worse around a black hole. As the rope approaches the singularity, the tidal force on the rope increases dramatically. Basically the BH is pulling harder on a nearby piece of rope than a more distant piece. As the singularity is approached, the tidal force tends to infinity, so even the most durable imaginable rope will be torn apart.

So far this has all been classical mechanics and special relativity. I won't even try to start bringing general relativity into this.

That's some ideas to get you thinking a bit. You can play with assumptions and adding in new considerations for different effects more or less ad infinitum. But the big take-away message is:

• Nothing with mass can move faster or equal to $c$, even if black holes are involved (massless particles like the photon can move at $c$, and in fact cannot move any faster or slower than this).


• The faster an object is moving, the more energy it takes to increase its speed by a fixed amount. This becomes important when speeds are a significant fraction of $c$. At much lower speeds the effect is safely ignored.

rotational dynamics - Why is the Earth so fat?

I made a naive calculation of the height of Earth's equatorial bulge and found that it should be about 10km. The true height is about 20km. My question is: why is there this discrepancy?

The calculation I did was to imagine placing a ball down on the spinning Earth. Wherever I place it, it shouldn't move.

The gravitational potential per unit mass of the ball is $hg$, with $h$ the height above the pole-to-center distance of the Earth (call that $R$) and $g$ gravitational acceleration.

Gravity wants to pull the ball towards the poles, away from the bulge. It is balanced by the centrifugal force, which has a potential $-\omega^2R^2\sin^2\theta/2$ per unit mass, with $\omega$ Earth's angular velocity and $\theta$ the angle from the north pole. This comes taking what in an inertial frame would be the ball's kinetic energy and making it a potential in the accelerating frame.

If the ball doesn't move, this potential must be constant, so

$$U = hg - \frac{(\omega R \sin\theta)^2}{2} = \textrm{const}$$

we might as well let the constant be zero and write

$$h = \frac{(\omega R \sin\theta)^2}{2g}$$

For the Earth,

$$R = 6.4*10^6m$$

$$\omega = \frac{2\pi}{24\ \textrm{hours}}$$

$$g = 9.8\ m/s^2$$

This gives 10.8 km when $\theta = \pi/2$, so the equatorial bulge should be roughly that large.

According to Wikipedia, the Earth is 42.72 km wider in diameter at the equator than pole-to-pole, meaning the bulge is about twice as large as I expected. (Wikipedia cites diameter; I estimated radius.)

Where is the extra bulge coming from? My simple calculation uses $g$ and $R$ as constants, but neither varies more than a percent or so. It's true the Earth does not have uniform density, but it's not clear to me how this should affect the calculation, so long as the density distribution is still spherically-symmetric (or nearly so).

(Wikipedia also includes an expression, without derivation, that agrees with mine.)

Answer

The error is that you assume that the density distribution is "nearly spherically symmetric". It's far enough from spherical symmetry if you want to calculate first-order subleading effects such as the equatorial bulge. If your goal is to compute the deviations of the sea level away from the spherical symmetry (to the first order), it is inconsistent to neglect equally large, first-order corrections to the spherical symmetry on the other side - the source of gravity. In other words, the term $hg$ in your potential is wrong.

Just imagine that the Earth is an ellipsoid with an equatorial bulge, it's not spinning, and there's no water on the surface. What would be the potential on the surface or the potential at a fixed distance from the center of the ellipsoid? You have de facto assumed that in this case, it would be $-GM/R+h(\theta)g$ where $R$ is the fixed Earth's radius (of a spherical matter distribution) and $R+h(\theta)$ is the actual distance of the probe from the origin (center of Earth). However, by this Ansatz, you have only acknowledged the variable distance of the probe from a spherically symmetric source of gravity: you have still neglected the bulge's contribution to the non-sphericity of the gravitational field.

If you include the non-spherically-symmetric correction to the gravitational field of the Earth, $hg$ will approximately change to $hg-hg/2=hg/2$, and correspondingly, the required bulge $\Delta h$ will have to be doubled to compensate for the rotational potential. A heuristic explanation of the factor of $1/2$ is that the true potential above an ellipsoid depends on "something in between" the distance from the center of mass and the distance from the surface. In other words, a "constant potential surface" around an ellipsoidal source of matter is "exactly in between" the actual surface of the ellipsoid and the spherical $R={\rm const}$ surface.

I will try to add more accurate formulae for the gravitational field of the ellipsoid in an updated version of this answer.

Update: gravitational field of an ellipsoid

I have numerically verified that the gravitational field of the ellipsoid has exactly the halving effect I sketched above, using a Monte Carlo Mathematica code - to avoid double integrals which might be calculable analytically but I just found it annoying so far.

I took millions of random points inside a prolate ellipsoid with "radii" $(r_x,r_y,r_z)=(0.9,0.9,1.0)$; note that the difference between the two radii is $0.1$. The average value of $1/r$, the inverse distance between the random point of the ellipsoid and a chosen point above the ellipsoid, is $0.05=0.1/2$ smaller if the chosen point is above the equator than if it is above a pole, assuming that the distance from the origin is the same for both chosen points.

Code:

{xt, yt, zt} = {1.1, 0, 0};

runs = 200000;
totalRinverse = 0;
total = 0;

For[i = 1, i < runs, i++,
 x = RandomReal[]*2 - 1;

 y = RandomReal[]*2 - 1;
 z = RandomReal[]*2 - 1;
 inside = x^2/0.81 + y^2/0.81 + z^2 < 1;
 total = If[inside, total + 1, total];
 totalRinverse = 
  totalRinverse + 
   If[inside, 1/Sqrt[(x - xt)^2 + (y - yt)^2 + (z - zt)^2], 0];
]

res1 = N[total/runs / (4 Pi/3/8)]

res2 = N[totalRinverse/runs / (4 Pi/3/8)]
res2/res1

Description

Use the Mathematica code above: its goal is to calculate a single purely numerical constant because the proportionality of the non-sphericity of the gravitational field to the bulge; mass; Newton's constant is self-evident. The final number that is printed by the code is the average value of $1/r$. If {1.1, 0, 0} is chosen instead of {0, 0, 1.1} at the beginning, the program generates 0.89 instead of 0.94. That proves that the gravitational potential of the ellipsoid behaves as $-GM/R - hg/2$ at distance $R$ from the origin where $h$ is the local height of the surface relatively to the idealized spherical surface.

In the code above, I chose the ellipsoid with radii (0.9, 0.9, 1) which is a prolate spheroid (long, stick-like), unlike the Earth which is close to an oblate spheroid (flat, disk-like). So don't be confused by some signs - they work out OK.

Bonus from Isaac

Mariano C. has pointed out the following solution by a rather well-known author:

http://books.google.com/books?id=ySYULc7VEwsC&lpg=PP1&dq=principia%20mathematica&pg=PA424#v=onepage&q&f=false

cosmology - Why should I believe that matter does not expand along with the space between it

This question has been asked before under other guises. I am not a scientific profesional however i have some schooling in pyhsics and mathematics and have a keen interest in these subjects.

It seems plausible to me and was a thought i had my self many years ago whilst at college, that the universe could possibly be expanding from within itself. Ie matter and space were expanding at the same rate as each other.

I observe there is no relative distance increase between the objects i see on a day today basis and as such this type of expansion wouldn't change our perception of this relatively speaking. Everything we observe would remain the same and we would be unaware of the expansion..

Where does energy come from in a nuclear fission if nucleon count doesn't change?

In a nuclear fission the total number of protons and neutrons are conserved. Then the mass converted to energy $E=mc^2$. From where this mass come from? Does that mean that all protons and neutrons are not identical in mass.

electromagnetism - Electrons and Magnetism

Electron at rest generates Electric field. Electron moving without acceleration produces electric and magnetic field. Electron moving with acceleration produces electromagnetic waves.

Please explain the mechanism or the process how this happens. How magnetic field is generated by moving charges? (A moving charge produces a current, or rather a time varying electric field in space.)

Also link https://www.physicsforums.com/threads/why-does-moving-electron-produce-magnetic-field.184619/ said

A moving electron alone actually does not produce a magnetic field. It requires electrons and protons to produce a magnetic field. As the electrons move relative to protons (ions) there is a relativistic charge per unit volume difference between the positive and the negative charges. This causes any external charges to feel a force we know as the magnetic field.

The magnetic field is a relativistic correction to the electrostatic field.

I could not understand these lines. And I don't know much about quantum physics. So, can you please explain in a simpler way? Please use quantum physics also, if it works here.. I may understand it sooner or later.

Also how electromagnetic waves are produced by accelerating charges? An electron with more energy gives it out in the form of photons i.e., E.M. waves. Is it that electric and magnetic fields produced by electron superpose? That could not have possibly happened because varying electric and magnetic field give rise to each other independently..

What happens there actually? And how magnetic field is relativistic correction to electric field?

Thank you.

logical deduction - Five professors and nine dishes

Here's yet another puzzle adapted from a puzzle book (in my case, with edits to make some of the specifications of the puzzle more clear because I didn't really understand them the first time I read it).

Five professors are spending three days on a retreat, and every day they eat dinner at the same restaurant. This restaurant has a special "roulette menu" with nine items, in which you choose a menu item from 1 to 9 that corresponds to one of the nine available items, but you don't know which number corresponds to which item. When the items are ordered, they are brought to the table in no particular order, so if a group orders multiple items at a time, they cannot figure out individually which item was ordered by whom.

These professors have tasked themselves with figuring out which number corresponds to which item in three days regardless. Each professor will order one item per day. How should they plan their ordering of the items in order to find out which number corresponds to which item for all items?

Answer

Okay, 15 orders, 9 dishes, that means 6 dupes. So 3 dishes will only be ordered once, obviously never on the same day. Now to split the six dupes so they can all be detected.

The first 3:

Day 1: 1   3
Day 2: 1 2
Day 3:   2 3

The dish that shows on day 1 and 2 is #1, and so on.

Now add in same-day pairs:

Day 1: 1   3 4 4
Day 2: 1 2   5 5 
Day 3:   2 3 6 6

So the dish that's doubled on day 1 is #4, etc.

And now the last three:

Day 1: 1 7 3 4 4
Day 2: 1 2 8 5 5 
Day 3: 9 2 3 6 6

The dish that only shows up on day 1 is #7, and so on.

And then, secure in their triumph, they can leave five bad Yelp reviews warning people away from this crazy place.

Sudoku false positive (wrong move)

I'm new to Sudoku puzzles.

---

I tried solving the one above but my last move was flagged as a wrong move(highlighted red). I believe it is a false positive since the $6$(highlighted red) I inserted is unique on the rows and columns as well as within the small square. Can someone justify this for me?

logical deduction - When is Cheryl's Birthday?

I saw this question on Facebook:

Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.

May 15 16 19

June 17 18

July 14 16

August 14 15 17

Cheryl then tells Albert and Bernard separately the month and the day of her birthday, respectively.

Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know, too.

Bernard: At first, I didn't know when Cheryl's birthday is, but I know now.

Albert: Then I also know when Cheryl's birthday is.

When is Cheryl's birthday?

There is an official solution which is explained in the link below, but I don't understand the logic.

July 16, as explained here. https://www.facebook.com/kennethjianwenz/posts/386479228197631

What is wrong with the logic in the answer I posted below (I put it in spoilers if others want to try)?

Answer

I will try to help you understand the answer. It is important to note that Albert and Bernard both know the answer before we do.

With Albert's first statement, it is clear to us that Albert was told either July or August. Had he been told May or June, he would not be able to state definitively that he knew Bernard didn't know. (May 19 and June 18 both could be uniquely identified immediately by Bernard without Albert's help, so in order for Albert to know that Bernard doesn't know, the month he was told must not be May or June.)

Because Bernard has been able to identify Cheryl's birthday after Albert's statement, he must not have been told 14. Because Albert's statement revealed July and August, had Bernard been told 14, he would still be unclear on the date. (If he said he didn't know, then Albert would know the birthdate, but Bernard would never be able to deduce it.)

We still don't know whether it is July 16, August 15, or August 17.

However, since Albert (who was only told the month) is able to state that he also knows the birthday, he must not have been told August. Had he been told August, he would be unable to decide if Bernard was told the 15th and 17th.

It is only after the third statement that we (as outsiders) can deduce the birthday. Bernard knew after the first statement and Albert knew after the second.

atoms - Is there any significance of atomic orbitals?

We have been taught that the atomic orbitals we read about are probability density region of finding electrons of particular energies which are designated by the various quantum numbers.

Since, there is high degree of uncertainty when we talk about the position of merely the detection of an electron as well as the energy that we "observe" and which may pretty much not be the actual energy of the observed electron, what then is the significance of any atomic orbital?

I am not very well read in the various techniques and theories that we have developed over time, but still I think that atomic orbitals are at best a high probability region of finding certain electrons. Since we must not be able to do that with enough certainty how then can we check the validity of the existence of orbitals?

Is it possible, that the idea of atomic orbital is now outdated in terms of modern development and only used for simplistic explanations of complex electron motions and phenomenons?

Overall, I am interested in learning the significance of atomic orbitals. Google provides various links explaining the theory and listing its uses in modern chemistry but I could not find anything which properly explains the significance or even tries to explain the actual presence of something like an atomic orbital other than saying that we find so and so probability regions while working with Schrödinger equations.

Added after answers and comments: The answers are listing the uses of the contruct only, I tried to say that I am interested in finding out whether something like an orbital is actually there or not.

Maybe an example will help, there is no boundation for an electron of certain energy to move in an orbital that is dumbled or double dumbled shaped! I agree with electrons having certain energies described by quantum numbers, but how can one definitively prove that an electron belonging to say 2p_x resides in a dumbled shaped orbital?

Again, are orbitals true? Are they actually significant and not just a helpful mathematical construct? Can it be definitively proven?

quantum field theory - How long can a virtual particles exist in a vacuum state?

As much as I have understood, virtual particles keep appearing and disappearing in a vacuum state. This leads to zero-point energy, Casimir effect etc. Now, I want to know if there is a lifetime (i.e. a time limit) for which these virtual particles can exist before disappearing. Also can there be a mechanism by which they can be made to stay longer?

Answer

The concept of virtual particles is simply an analogy which resembles the underlying phenomena and which is often used to explain the QFT effects such as Casimir or Hawking radiation to beginners.

Reality is a bit different.

Classical vacuum

In classical field theories we have a well-defined notion of "vacuum": a background spacetime with no field content. For instance, for a scalar field $\phi(t, \vec{x})$:

$$ \phi(x^{\mu}) = \phi(t, \vec{x}) = 0 $$

defines the classical vacuum configuration. Equivalently, it can be defined as a point on the phase space:

$$ \phi(\vec{x}) = 0, \quad \pi(\vec{x}) = 0, $$

where $\pi(\vec{x})$ is the conjugate momenta density field (canonically conjugate to $\phi(\vec{x})$). Note how the two definitions of the vacuum are equivalent: you can take the phase space point defined by $\phi(\vec{x}) = \pi(\vec{x}) = 0$ and evolve it using Hamilton's equations to get $\phi(t, \vec{x}) = 0$.

Uncertainty relation

In quantum mechanics, however, everything is much more complicated. Remember the renowned uncertainty relation? For any two canonically conjugate variables $x$ and $p$ we have $$ \Delta x \cdot \Delta p \sim \hbar. $$

Lets apply this to the theory of a scalar field. Suppose that we know with absolute certainty that $$ \phi(\vec{x}) = 0. $$

But this means that we cannot be sure of the value of $\pi(\vec{x})$! Moreover, Hamilton's equations will use this undefined $\pi(\vec{x})$ to take the field variable $\phi(\vec{x})$ to an undefined value in an arbitrary short interval of time.

This observation lies in the heart of the QFT treatment of vacuum: the uncertainty relation does not permit us to have a classical vacuum state, where the value of the field is zero at all spacetime points.

QFT vacuum

Now to the mathematics of QFT. For free field theories we are able to define the QFT vacuum state, which we usually denote by $\left| 0 \right>$. As follows from the previous section, it does not have the properties of the classical vacuum state, especially, we can measure nonzero values of $\phi$ in this state!

The actual wavefunction for $\left| 0 \right>$ is the multi-dimensional generalization of the Gauss wavepacket:

$$ \left| 0 \right> [\phi(\vec{x})] = \exp \left[ i \cdot \int d^3 x \, \alpha \, \phi(\vec{x})^2 \right], $$

where $\alpha$ is a dimension-full parameter which depends on the mass of the field and on $\hbar$.

It can be seen from this integral that the expectation (average value) of the field variable observable is $$ \left< \phi(\vec{x}) \right> = \left< 0 \right| \phi(\vec{x}) \left| 0 \right> = 0, $$

which is what we expect from the vacuum state. However, the expectation of the field observable squared is nonzero:

$$ \left< \phi(\vec{x})^2 \right> = \left< 0 \right| \phi(\vec{x})^2 \left| 0 \right> \neq 0. $$

This is proportional to $\hbar$ and thus extremely small in the classical approximation (I am ignoring the divergence issues here for simplicity). That is why we can say that in the classical limit $\left| 0 \right>$ is observationally equivalent to the classical vacuum.

Virtual particle lifetimes

As suggested by @Frederic Thomas (in the comments), the time-energy uncertainty relation

$$ \Delta t \cdot \Delta E \sim \hbar $$

is useful to estimate the average time scale relevant to the effects connected to virtual particles with energy scale $\Delta E$. However, this is as far as it goes: a useful tool for estimating the average time scale. It is wrong to interpret $\Delta t$ as a lifetime of the virtual particle.

Concerning your last question: how can we extend the lifetime of a virtual particle? As you probably already guessed from the spirit of this answer, virtual particles are just an analogy and don't exist in reality. The only way we can "extend" the lifetime is to make a real particle, that is - an excitation of the quantum state. The physical properties of these excitations are described by the mathematics of QFT.

geometry - Triangles tiling on a hexagon

Each number means the numbers of blue tiling surrounding the number

example

Answer

The blue-filled hexagon in below:

How I solved this:

First, There had to be blue on both triangles between the 4's. That gives us 4 blues triangles already done. The rest was obvious.

error analysis - How to Interpolate When Reading a Scale

Suppose, I am taking measurement with a scale whose least deviation is 1mm. the length of something seems to lie between 44mm to 45mm. But my instinct says, it's 44.7mm. So, the question is I should report 44.7+-0.5mm or 44.5+-0.5mm or something else?

Answer

There are a few different concepts we need to cover here.

First - the difference between precision and accuracy. Precision tells me the number of digits I can write down: in your case, you might be able to estimate that the number is close to 44.7 and you might even believe your ability to interpolate is such that the answer is 44.7 ± 0.1 mm. Of course, when you are reading an analog scale, you usually need to worry about parallax: depending on where your eye is with respect to the alignment of the measurement with the scale, you may get an error. Old analog meters often addressed this by putting a mirror behind the scale: you had to line up your eye such that the reflection of the needle fell right behind the needle - this is how you would minimize the parallax error.

The second issue is accuracy: when the supplier of your analog device (a ruler, in this case) put markings on the ruler, they did so with a certain tolerance. In a rare case, this tolerance might be noted on the ruler - most of the time, you assume it is "as good as the divisions on the ruler". In other words, when a ruler is marked in mm, you can reasonably assume that a measurement of 44 mm is closer to 44 mm than it is to 45 mm or 43 mm. But the manufacturer usually doesn't claim that the line that is drawn at 44 mm is actually drawn at 44.0 mm

When you try to interpolate, you are in effect assuming that 44 = 44.0 ; while that may sometimes be true (it would have to be independently confirmed), it won't always be. And if you start quoting numbers like 44.7 you are in fact claiming that your measuring instrument has accuracy beyond its markings.

In summary - there are actually multiple sources of error that affect your measurement. There is error aligning the ruler (including parallax), error reading the ruler (especially interpolating between markings), and errors in the ruler itself. Proper error estimation accounts for all these things.

If you cannot fully analyze all these sources of error, it is a useful shortcut to round to the nearest whole marking. When you add digits without knowing the accuracy of your ruler, you may be falsely claiming a precision that is not there.

To clarify: I recommend in your case to quote 45 +- 0.5 mm

speed of light - Special relativity and missing factors of $c$

I am doing problems in a textbook called 'Introduction to Classical Mechanics' by David Morin. In one of the questions it says the following:

In the lab frame, two particles move with speed $v$ along the path shown

The angle between the trajectories is $2\theta$. What is the speed of one particle, as viewed by the other?

I know how to do this question but what has confused me in the answers he has left out all of the c's. Here is his method

Consider the frame S' that travels along the point P midway between the particles. S' moves at speed $v cos\theta$, so the $\gamma$ factor relating it to the lab frame is: $$\gamma =\frac{1}{\sqrt{1-v^2cos^2\theta}}$$

From what I have learned there should be a $v^2/c^2$ instead of just $v^2$. He has repeated a similar method 4 times in the book (from what I can see) all without the $c^2$, which makes me think that I have got something wrong thinking their should be a c. Please explain thanks.

He then goes onto use the velocity addition formula like this:

$$V=\frac{2u'_y}{1+u^{'2}_{y}}$$

Answer

The beginning of the chapter, the author does use $c$ for the Lorentz transformations (cf., Equation (13.1)). \begin{align} A_0&=\gamma(A_0'+(v/c)A_1') \\ A_1&=\gamma(A_1'+(v/c)A_0') \\ A_2&=A_2'\\ A_3&=A_3' \end{align}

Shortly after Equation (13.1), the author lists several enumerated remarks. In particular is #4:

4. Lest we get tired of writing the $c$'s over and over, we'll work in units where $c=1$ from now on.

And then he re-writes the Lorentz transformations in the new units in Equation (13.2), \begin{align} A_0&=\gamma(A_0'+vA_1') \\ A_1&=\gamma(A_1'+vA_0') \\ A_2&=A_2'\\ A_3&=A_3' \end{align}

homework and exercises - Photoelectric effect intensity

I understand the PE effect quite well but I'm failing to understand one thing. Intensity is the amount of energy per second incident to a given area.

So can you can increase the intensity by either increasing the electrical energy or making the area smaller? By increasing the electrical energy, what exactly is happening? I assume that by increasing intensity you are increasing the number of photons incident to an area of the metal thus increasing the rate of photoelectric emission. But why doesn't increasing intensity via more electrical power lead to greater energy of the electrons, if by increasing intensity you are increasing the energy of the incident waves?

I understand the results but I'm trying to understand why the intensity works the way it is thanks

quantum mechanics - Non-relativistic limit of complex scalar field

In page 42 of David Tong's lectures on Quantum Field Theory, he says that one can also derive the Schrödinger Lagrangian by taking the non-relativistic limit of the (complex?) scalar field Lagrangian. And for that he uses the condition $\partial_{t} \Psi \ll m \Psi$, which in fact I suppose he means $|\partial_{t} \tilde{\Psi}| \ll |m \tilde{\Psi}|$, otherwise I don't get it. In any case, starting with the Lagrangian:

$$\mathcal{L}=\partial^{\mu}\tilde{\psi} \partial_{\mu} \tilde{\psi}^{*} -m^{2}\tilde{\psi}\tilde{\psi}^{*}$$

Using the inequation I think it's correct, I can only get to:

$$\mathcal{L}=-\nabla\tilde{\psi} \nabla \tilde{\psi}^{*} -m^{2}\tilde{\psi}\tilde{\psi}^{*}$$

And from that I've tried relating $\tilde{\psi}$ or $\psi$ (as we can write the above Lagrangian with both, as it's invariant under multiplying by a pure phase), to $\dot{\psi}$

resource recommendations - What is a good non-technical introduction to theories of everything?

I'm not a physicist but I'm interested in unified theories, and I do not know how to start learning about it. What would be a good book to read to start learning about this topic?

Answer

As a theory of everything includes a theory of all particular things, it would be good if you start by learning about the theories that need to be unified. This means first

• some quantum mechanics,


• something about classical electromagnetism,


• something about special anf general relativity, then


• some quantum field theory,


• something about quantum electrodynamics,


• something about the standard model.

So you should look at the recommendations for introductions to these subjects available at our Book recommendations.

Unless you are content with such books as ''The elegant universe'' http://en.wikipedia.org/wiki/The_Elegant_Universe where you learn the buzzwords without a deeper understanding.

pattern - Bongard n°2: Shoelace

As with my previous puzzle I made the 12 "rooms" larger but kept the same rules. I'm also adding the link to the rules for those that don't know the puzzles Click me!

And have fun!

Hint:

the black circles are only cosmetic, please ignore them.

Edit:

There was a minor mistake that made the puzzle a bit inelegant (the solution was not changed). Now it is fixed, I also removed the incorrect solution that was found; this is the last edit, I will eventually add the answer if no one can find it.

Hint 2:

Some numbers don't play well with others.

thermodynamics - Heat rejected by a refrigerator?

What is the amount of heat rejected by the condenser of a vapour compression refrigeration system, typically those found in households? Something in the nature of a 25W compressor, for example

Answer

If you have a 25 W compressor, then the net heat flux into the environment will be 25 W. This is made up of the heat extracted from the inside of the refrigerator, plus the heat of losses in the compressor, minus the heat that travels from the environment back to the inside from the fridge (which is why the compressor needs to keep running).

If we concentrate just on the compressor as a heat pump, we can make some simple assumptions about the COP (coefficient of performance) of the compressor to come up with an estimate.

Assuming the inside of the fridge is at 5°C and the environment at 22°C, the compressor is pumping against a 17°C gradient. A perfect Carnot engine operating between 278 K and 295 K would have an efficiency of

$$\eta = 1 - \frac{T_l}{T_h} = 5.7\%$$

Conversely, a perfect heat pump can pump more heat than the energy you put into it (that's why they are sometimes used for heating homes, when a suitable source of nearby "heat" is available, e.g. groundwater). The ratio of heat moved vs work done is $\frac{1}{\eta} = 17.4$.

That number is the ratio between the heat rejected and the power consumed - so for an "ideal" refrigerator compressor operating between the limits given, the heat rejected would be about 430 W. Now a refrigerator pump is rarely as efficient as you would like - certainly it doesn't get close to the efficiency of the Carnot cycle.

There's a good paper on the measurement of refrigerator efficiency that probably has more information than you ever wanted to know about the subject. It includes measurements of the thermal insulation of a typical refrigerator cabinet (1.21 W/K for the freezer, 0.88 W/K for the refrigerated compartment). From Table 3 of that publication, I find a COP of about 1.7 - suggesting that for our example 1.7*25 W = 42 W of heat is extracted, and 42W+25W =

You can compare these numbers to the ones from the article, and find they are comparable - although the refrigerator in their example had a given energy consumption of 28 kWh/month which suggests about 39 W average power used. Still, that's the same ball park as 25 W.

Afterthought - reading my solution again, I would not normally make an estimate like this and quote a number that seems to be precise to two figures. It would be more reasonable to say "roughly three times the power use is rejected". 1+1.7=2.7 which is approximately 3. Probably a better approach if you are estimating and making tons of assumptions.

newtonian mechanics - Proving that the angular velocity vector is equal to a limit involving the rotation vector

The angular velocity vector of a rigid body is defined as $\vec{\omega}=\frac{\vec{r}\times\vec{v}}{|\vec{r}|^2}$. But I'd like to show that that's equivalent to how most people intuitively think of angular velocity.

Euler's theorem of rotations states that any rigid body motion with one point fixed is equivalent to a rotation about some axis passing through the fixed-point. So let's consider a rigid body undergoing some motion with one point fixed, and for any times $t_1$ and $t_2$ let $\vec{\theta}(t_1,t_2)$ denote the "rotation vector" of the rotation that's equivalent to the rigid body's motion between time $t_1$ and time $t_2$. For those who don't know, the rotation vector of a rotation is a vector whose magnitude is equal to the angle of the rotation and which points along the axis of the rotation; see this Wikipedia article.

Now my question is, how do we prove that the limit of $\frac{\vec{\theta}(t_1,t_2)}{t_2-t_1}$ as $t_2$ goes to $t_1$ exists, and that it's equal to the angular velocity vector?

This would all be much simpler if rotations were commutative since then the angular velocity would just equal the derivative of $\vec{\theta}(t_0,t)$ with respect to time. But since rotations are non-commutative, $\vec{\theta}(t_1,t_2)$ does not equal $\vec{\theta}(t_0,t_2)-\vec{\theta}(t_0,t_1)$ and thus the relation between angular velocity and the time derivative of $\vec{\theta}(t_0,t)$ is considerably more complicated; see this journal paper for details.

Note: This is a follow-up to my question here.

EDIT: Note that what this journal paper calls $\vec{\alpha}(t)$ would in my notation be written as $\vec{\theta}(t_0,t)$. The paper discusses the fact that the angular velocity vector $\vec{\omega}(t)$ is not equal to the time derivative of $\vec{\alpha}(t)$. This means that the limit of $\frac{\vec{\theta}(t_0,t_2)-\vec{\theta}(t_0,t_1)}{t_2-t_1}$ does not equal $\vec{\omega}(t_1)$. But my question is about proving a slightly different statement, which is that the limit of $\frac{\vec{\theta}(t_1,t_2)}{t_2-t_1}$ as $t_2$ goes to $t_1$ DOES equal $\vec{\omega}(t_1)$. Note that the expressions $\frac{\vec{\theta}(t_0,t_2)-\vec{\theta}(t_0,t_1)}{t_2-t_1}$ and $\frac{\vec{\theta}(t_1,t_2)}{t_2-t_1}$ are not equal, because $\vec{\theta}(t_0,t_1)+\vec{\theta}(t_1,t_2)$ does not equal $\vec{\theta}(t_0,t_2)$ due to the non-commutativity of rotations. So none of what I'm saying contradicts or seeks to disprove the journal paper.

Answer

If you take $t_0 = t_1$ in the EDIT part of your question, with $\vec{\theta}(t_0,t_1) = \vec{\theta}(t_1,t_1) = 0$, you are in the special case $\omega = \dot{\alpha}$ from Asher Peres's paper you have mentioned, which then proves your statement according to your observation from the EDIT part, because in that case, you have $$\frac{\vec{\theta}(t_0,t_2)-\vec{\theta}(t_0,t_1)}{t_2-t_1} = \frac{\vec{\theta}(t_1,t_2)}{t_2-t_1},$$ whose limit when $t_2 \to t_1$ is equal to $\dot{\alpha}(t_1) = \omega (t_1)$.

Indeed, according to Asher Peres, we have: $$ w = \dot{\alpha} + \frac{1 - \cos \alpha}{\alpha^2} (\alpha \times \dot{\alpha}) + \frac{\alpha - \sin \alpha}{\alpha^3} (\alpha \times (\alpha \times \dot{\alpha})),$$ which, for $\alpha(t) = \vec{\theta}(t_1,t)$ and $t = t_1$, and using $\alpha(t_1) = \vec{\theta}(t_1,t_1) = 0$ (see above), reduces to $$ w(t_1) = \dot{\alpha}(t_1) + 0.$$ Note that luckily $1 - \cos \alpha = \mathrm{O}(\alpha^2)$ and $\alpha - \sin \alpha = \mathrm{O}(\alpha^3)$, hence there is not problem when passing to the limit when $\alpha \to 0.$

Why is the gauge potential $A_{mu}$ in the Lie algebra of the gauge group $G$?

If we have a general gauge group whose action is $$ \Phi(x) \rightarrow g(x)\Phi(x), $$ with $g\in G$.

Then introducing the gauge covariant derivative $$ D_{\mu}\Phi(x) = (\partial_{\mu}+A_{\mu})\Phi(x).$$

My notes state the gauge potential $A_{\mu} \in L(G)$, $L(G)$ being the Lie Algebra of the group $G$.

What's the connection between the Lie Algebra of the group and the gauge potential?

Answer

The gauge potential is an object that, when introduced in the covariant derivative, is intended to cancel the terms that spoil the linear transformation of the field under the gauge group. Every gauge transformation $g:\Sigma\to G$ (on a spacetime $\Sigma$) connected to the identity may be written as $\mathrm{e}^{\mathrm{i}\chi(x)}$ for some Lie algebra valued $\chi: \Sigma\to\mathfrak{g}$. The derivative of a transformed field is $$ \partial_\mu(g\phi) = \partial_\mu(g)\phi + g\partial_\mu\phi = g(g^{-1}(\partial_\mu g) + \partial_\mu)\phi$$ and it is the $g^{-1}(\partial_\mu g) = \partial_\mu\chi$ that we want to cancel here by adding the gauge field so that $D_\mu(g\phi) = gD_\mu\phi$. Since $\partial_\mu\chi$ is Lie algebra valued, so must the gauge field $A$ we add, and it has to transform as $$ A\overset{g(x)}{\mapsto} gAg^{-1} - g^{-1}\mathrm{d} g$$ to cancel the terms we want to cancel.

visible light - Why do we not see stars during the day?

As stars always propagate light, I was thinking that we should see them even during the day. I searched a lot and I find three questions related to the current one.

1. 
   
   

   Why do two beams of light pass through one another without interacting?

   
   


2. 
   

   Can photons pass through each other?

   
   


3. 
   

   Star visibility in outer space even during the day?

   
   

I couldn't get the answers given to these questions, but I could find some clues!

I guess beams of light pass through one another without interacting (and so we should see stars during the day) because of

1. 
   

   This sentence of Anna's answer to the first question: "Thus two light beams have no measurable interactions when crossing"

   
   


2. 
   

   The "Because" in the beginning of John Duffield's answer to the first question.

   
   


3. 
   

   This sentence of Anna's answer to the second question: "Thus we can say that for all intents and purposes photons scatter on each other without interacting"

   
   
   

In other hand, I guess we cannot see stars during the day because of this sentence of udiboy1209's answer to the third question: "if you can sustain the heat and the blinding radiation from the sun, you should be able to see stars when you are facing the sun"

May someone please clarify me by (as much as possible) simple explanations?

Answer

The daylight sky has a brightness of about magnitude 3 per square arcsecond. The brightest stars have an integrated intensity of about zeroth magnitude.

If your eyes had an angular resolution approaching 1 arcsecond then you would easily be able to see bright stars in the daylight sky - they would be about 10 times as bright as the sky. Unfortunately, the resolution of the eye is more like 1 arcminute. That means when comparing the starlight to the sky, the star is blurred over an area such that the contrast ratio with the sky is no longer large enough to discern it. However, even with this, if you knew exactly where to look, you could make out the very brightest stars, if your eyesight were good and this is obviously the case of bright objects like Venus, which are visible in the daytime sky.

If you look through a telescope (which increases collection of both starlight and daylight equally) then you can easily see stars. This is because the angular resolution of the telescope is around $1.22\lambda/D$, where $\lambda$ is the wavelength and $D$ the telescope diameter. A 10cm telescope can give you an angular resolution approaching 1 arcsecond (atmospheric conditions permitting) and thus a 3rd magnitude star has a similar brightness to the daytime sky through such a telescope.

astrophysics - Are Neutron stars transparent?

Neutrons have no charge so they would not, I think, interact with photons. Would a neutron star be transparent?

story - Clue 18 - Should this be on Stack Overflow?

_{<<---First clue}

_{<---Previous clue}

---

As you sit at the computer that you used to solve the last puzzle, a popup window opens. It appears to be a Stack Overflow question, with a very generic name: Why isn't my code working?

As you look closer, however, you notice several things - the URL has been blocked out, along with the name and avatar of the OP. The question score is at -18. Based upon this information, you assume that it must be your next Clue.

!(function(fivetimesthree){
va one = parsent('5D0', Math.og2(65536));
var two = fivetiesthree * 101 - 1;
var  three = Date.parshe('1/1/1970 0:00:01.489 GMT+0');
var four = one + two - three;
et first_two = String.fromCharCode(one, two);
cnst last = eval(`'\\u0${three.toString(16)}'`) + JSON.parse(JSON.stringify(our.toString(16)).replace(/"/,'"\\u0'));
console.lo([last, first_two].reerse().join(''));
})(5*3);

---

_{Next clue--->}

Answer

The

"wrong" letters I have found (NB these are the same ones as listed by stack reader, plus the extra space, but as it happens I found them independently)

when transformed by

A->Z, B->Y, ..., Z->A,

yield the phrase

"iron solute".

Why that transformation? Because

if you fix the bugs in the code, what it puts in the console log is "Atbash" in Hebrew letters.

At Scimonster's request, here is the code with the bugs taken out:

!(function(fivetimesthree){
var one = parseInt('5D0', Math.log2(65536));
var two = fivetimesthree * 101 - 1;
var three = Date.parse('1/1/1970 0:00:01.489 GMT+0');
var four = one + two - three;
let first_two = String.fromCharCode(one, two);
const last = eval(`'\\u0${three.toString(16)}'`) + JSON.parse(JSON.stringify(four.toString(16)).replace(/"/,'"\\u0'));
console.log([last, first_two].reverse().join(''));
})(5*3);

On my browser, this actually still doesn't work and I have to fix up the date to something like this:

var three = Date.parse('1970-01-01 00:00:01.489+0000');

at which point I get the following in the console:

אתבש

homework and exercises - Varying an action (cosmological perturbation theory)

I am stuck varying an action, trying to get an equation of motion. (Going from eq. 91 to eq. 92 in the image.) This is the action

$$S~=~\int d^{4}x \frac{a^{2}(t)}{2}(\dot{h}^{2}-(\nabla h)^2).$$

And this is the solution,

$$\ddot{h} + 2 \frac{\dot{a}}{a}\dot{h} - \nabla^{2}h~=~0. $$

This is what I get

$$\partial_{0}(a^{2}\partial_{0}h)-\partial_{0}(a^{2}\nabla h)-\nabla(a^{2}\partial_{0}h)+\nabla^{2}(ha^{2})~=~0.$$

I don't really see my mistake, perhaps I am missing something. (dot represents $\partial_{0}$)

It is this problem (see Lectures on the Theory of Cosmological Perturbations, by Brandenburger):

quantum mechanics - Does entanglement have a speed or is it instantaneous

The phenomenon of observing one entangled particle and noticing the other take on corresponding values... Does this take a finite speed at all or is it instantaneous?

Answer

You must distinguish 2 concepts:

Transmitting an information.

The protocol is strict: An observer, at some point of space-time, emits an information (and any information carries energy), and this information is received by an other observer. The speed of transmission of this information cannot excess the maximum speed of information ("speed of light")

Correlations.

Quantum Entanglement is only a special kind of correlations. You may find quantum correlations, but also classical correlations (for instance, in probabilistic problems). A correlation has nothing to do with transmitting an information. It is only a set of numbers, which indicates a joint probability law, of the form $p(a,b)$, which corresponds to a joint measurement of aleatory variables $A$ and $B$

The illusion of "spooky action at a distance"

This illusion appears in the following case. Take a classical or quantum system made of 2 sub-systems, and create a correlation between these 2 sub-systems (so, at a given point of space-time). Then, move these 2 sub-systems so that they are spatially separated. If a measurement is done on A, there is certainly a correlation with the measurement done on B (this is the definition of correlations), but there is no signal transmitted.

dimensional analysis - Units INSIDE of a Dirac Delta function

I know that the units of a Dirac Delta function are inverse of it's argument, for example the units of $\delta(x)$ if $x$ is measured in meters is $\frac{1}{meters}$.

But, my question is what are the units inside the Dirac Delta?

For example if you have $\delta (x - 1)$ is the exact expression $x$ inside the dirac delta unitless? If not, that means that the $1$ would actually have to have units of meters. That seems odd to me, which is why I think that inside the units are unitless and as a whole you give the dirac delta inverse units of the argument but I'm not sure. If someone could clear this up for me I'd really appreciate it, thanks!

Answer

Let's clear up one thing first: you can never add (or subtract) two expressions with different units. For example, in $\delta(x - 1)$, the 1 is unitless, which means $x$ has to be unitless as well.

Then there's the separate issue of the units of the expression inside the Dirac delta. The expression inside the delta is its argument, and as you know, the argument does not have to be unitless. So there's your answer. You can write $\delta(x - 1\text{ m})$, for example, and since $x - 1\text{ m}$ has units of length, the delta function itself will have units of inverse length.

The reason the arguments of many functions have to be unitless is that those functions can be expressed as a power series,

$$f(x) = a_0 + a_1 x + a_2 x^2 + \cdots$$

where the $a_i$ are just numbers.¹ For example,

$$\begin{align} \exp(x) &= 1 + x + \frac{1}{2}x^2 + \cdots \\ \sin(x) &= x - \frac{x^3}{6} + \cdots \end{align}$$

If $x$ had units of, say, length, then you would be adding a number to a length to an area (length squared) to a volume etc., and as I said, that can't happen.

But the delta function is not one of these functions that can be expressed as a power series. In fact, it's not really a function at all. It's a distribution, defined implicitly by the integral

$$\int_a^b f(x) \delta(x - x_0)\mathrm{d}x = \begin{cases}f(x_0), & a \le x_0 \le b \\ 0,& \text{otherwise}\end{cases}$$

In order for this integral to come out with the same units as $f(x_0)$, the rest of the expression being integrated - that is, the combination $\delta(x - x_0)\mathrm{d}x$ - has to be unitless. And since $\mathrm{d}x$ has the same units as $x$, $\delta(x - x_0)$ has to have the unit that cancels that unit out.

---

¹One might argue that you can define a function $f$ as a power series where the coefficients $a_n$ do have units of the appropriate type, and in that case the argument $x$ would have units. But you can always generalize such a function to a different function of a unitless variable: just write $a_n = f_0 b_n/x_0^n$ where $b_n$ is unitless and $x_0$ has the same dimension as $x$, and then express the function as $$f(x) = \sum a_n x^n = f_0\sum b_n (x/x_0)^n = f_0 g(x/x_0)$$ where $g(y) = \sum b_n y^n$. The function $g$ expresses the same functional relationship as $f$, but in terms of a unitless variable, and is thus more generally useful.

pattern - What is a Fantastical Word™?

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

If a word conforms to a special rule, I call it a Fantastical Word™.

Use the following examples below to find the rule.

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

Fantastical Words™,Not Fantastical Words™
AKIN,DISSIMILAR
SOAR,FALL
DIARY,JOURNAL
LETHAL,BENIGN
RATER,GRADER
GRADE,CATEGORY
BASIN,PLATE

ZESTER,PEELER
GATORS,CROCS
AMOUR,HATRED
COULD,SHOULD
FIAT,DENIAL
BARTER,PURCHASE
NORIA,DRAIN
VINIER,BRANCH
DRAW,ERASE
LADE,UNLOAD

MARANTA,COCONUT
FAERY,BUTTERFLY
JAUK,HURRY
YUAN,DOLLAR
KUKRI,RAPIER
SHEA,PEANUT
NARF,CONQUER
SOLVERS,PUZZLERS
DREADS,WELCOMED
SHEAR,GROW

OWNS,DESIRES
LONE,GROUPED
TONICS,POISON
ISOLEAD,SQUARE
VARUS,REGULAR

Answer

Each Fantastical Word:

Is an anagram of a Final Fantasy Character.

Explanation of each:

AKIN -> KAIN, the Dragoon from FF4
SOAR -> ROSA, white mage from FF4
DIARY -> RYDIA, summoner from FF4
LETHAL -> TELLAH, mage from FF4
RATER -> TERRA, from FF6
GRADE -> EDGAR, from FF6
BASIN -> SABIN, from FF6
ZESTER -> SETZER, from FF6

GATORS -> STRAGO, from FF6
AMOUR -> UMARO, from FF6
COULD -> CLOUD, from FF7
FIAT -> TIFA, from FF7
BARTER -> BARRET, from FF7
NORIA -> RINOA, from FF8
VINIER -> IRVINE, from FF8
DRAW -> WARD, from FF8
LADE -> ADEL, from FF8
MARANTA -> AMARANT, from FF9

FAERY -> FREYA, from FF9
JAUK -> KUJA, from FF9
YUAN -> YUNA, from FF10
KUKRI -> RIKKU, from FF10
SHEA -> ASHE, from FF12
NARF -> FRAN, from FF12
SOLVERS -> VOSSLER, from FF12
DREADS -> REDDAS, from FF12
SHEAR -> SERAH, from FF13
OWNS -> SNOW, from FF13

LONE -> NOEL, from FF13-2
TONICS -> NOCTIS, from FF15
ISOLEAD -> IEDOLAS, from FF15
VARUS -> RAVUS, from FF15

angular momentum - If dark matter only interacts with gravity, why doesn't it all clump together in a single point?

I'm a complete layperson. As I understand, dark matter theoretically only interacts with the gravitational force, and doesn't interact with the other three fundamental forces: weak nuclear force, strong nuclear force, and electromagnetism.

Those are my understandings going in. If I'm wrong, please correct me. I've done some googling, and I haven't found anything confirming or denying that dark matter is affected by either of the fundamental nuclear forces.

So since dark matter only interacts with gravity, what causes any dark matter particle to be repelled from another? If they can pass freely through each other, and they are gravitationaly attracted to each other, why don't such particles clump together in a single 'point' in space?

It seems to me that particles occupying a single 'space' are philosophically not distinct particles, but I don't know how actual physics would play into this.

Edit This article, author's credentials unknown, but implicitly claims to be a physicist or astronomer, says "...[P]hysicists generally take all dark matter to be composed of a single type of particle that essentially interacts only through gravity."

Edit 2 The author is this Lisa Randall, "Professor of Science on the physics faculty of Harvard University."

Answer

Great question. Observations show that Dark Matter (DM) only noticeably interacts gravitationally, although it's possible that it may interact in other ways "weakly" (e.g. in the 'WIMP' model --- linked). Everything following has no dependence on whether DM interacts purely/only gravitationally, or just predominantly gravitationally --- so I'll treat it as the former case for convenience.

Observable matter in the universe 'clumps' together tremendously: in gas clouds, stars, planets, disks, galaxies, etc. It does this because of electromagnetic (EM) interactions which are able to dissipate energy. If you roll a ball along a flat surface it will slow down and eventually stop (effectively 'clumping' to the ground), because dissipative forces (friction) are able to transfer its kinetic energy away.

On the other hand, imagine you drill a perfect hole, straight through the center of the Earth, and you drop a ball down it. (Assuming the hole and the earth are perfectly symmetrical...) the ball will just continually oscillate back and forth from each side of the earth to the other --- because of conservation of energy. Just like a frictionless pendulum (no rubbing, no air resistance). This is how dark matter interacts, purely gravitationally. Even if there was no hole through the center of the earth, the DM will just pass straight through and continue to oscillate back and forth, always reaching the same initial height. To zeroth order, dark matter can only 'clump' as much as its initial energy (obtained soon after the big-bang) allows. One example of such a 'clump' is a 'Dark Matter Halo' in which galaxies are embedded. DM Halos are (effectively) always larger than the normal (baryonic) matter inside them --- because the normal matter is able to dissipate energy and collapse farther.

electromagnetism - Is there just one EM field for the whole universe?

Does our universe contain individual magnetic fields ?

For example two different magnets, one here on earth and one on mars. Do both of them have their own magnetic field? Or is there only one single field that stretches across the entire universe, but have different strength on different locations? For example around a magnet ?

Answer

Clear answer: yes, there is only one electromagnetic field for the whole universe.

Ontological answer: You can go to any spot in the universe and take a measurement. Even if you were to find a large chunk of space with "0" as your result (and for this part of the answer we do not care if this is physically possible or not), the field itself would still be there, displaying those "0"'s. Oh, let's ignore black holes as that's obviously not the idea you had in mind.

Physical answer: If people talk about the "field of an electron", making it sound as if that were some kind of localized thing, then that is just a shortcut to reduce incredibly complex maths to something they can handle. None of the interactions have a hard border or a defined cut-off radius. That quite literally means that every single charged particle in the universe influences/interacts with every single other charged particle in the universe, no matter how far away it is. If you wiggle your finger here on earth you will definitely change the EM field on Mars. In practice this influence is so utterly, unimaginably small that it obviously is neither measurable nor anything you need to keep track off. But it is still there.

Regarding the comment on this: Finger-Mars is obviously a half-joking example. The point is not the light cone, event horizon or the magnitude of the influence, but that it *is* the same *field*; the Finger-wiggling does propagate (at least in our theories), but obviously with such absurdly small numbers as to be totally irrelevant.

education - Doppler effect of sound waves

I am looking for interesting ways to introduce the Doppler effect to students. I want some situations in nature or every day life, where a student is possibly surprised and may ask "how could it be"?

The common example for Doppler effect of sound waves is the siren of a moving emergency vehicle. Clearly the siren example is one way to introduce it starting from a phenomenon. However are there other interesting, perhaps astonishing phenomena in nature or in every day life which can explained by Doppler effect of sound waves?

In this question I am only interested in Doppler effect of sound waves, not that one of light.

Answer

There is a list on Wikipedia.

Radar guns use an optical Doppler effect to measure speed. Their acoustic equivalent is used in medicine, where it's called Doppler ultrasound and used to measure blood flow or other sorts of motion in the body.

Animals that use echolocation can use the Doppler shift to gain information about the motion of their surroundings.

A sonic boom occurs when the Doppler shift shifts a frequency to infinity.

I guess one can continue to concoct scenarios. I wonder whether, when you drop a cat off a cliff, you can hear the pitch of its screaming drop as it accelerates. (It's not all that cruel - cats can usually survive a fall at terminal velocity.)

You could use it to determine which way a whale is swimming if you have two boats, both listening to the same whalesong. You could even use the Doppler shift to gain information about the position of the whale because both the whale's position and its velocity contribute to the observed Doppler shift at any given place. (I don't have any information about this actually being done, but it might be an interesting problem to work out the locus of possible whale locations for an observed Doppler shift.)

I was also curious about whether the Doppler effect gives us information on the motion of the crust that moves during an earthquake. I found this reference which suggests it does.

Strategy for solving Flow Free puzzles

There is an app on the app store called Flow Free.

Basically, there is a grid with a few sets of colored dots, and you need to connect each dot to the other dot of the same color, filling up the entire board.

For example:

It starts out pretty easy, but as you progress, it gets harder.

Is there a general strategy for solving these types of puzzles in the last moves? For example, this one:

Answer

It's quite funny actually. I was terrible at these puzzles until I started writing this answer. To demonstrate the strategies that I completely and suddenly winged out of nowhere, I will solve the given puzzle.

Firstly, I would like to note the following: Flow puzzles SOMETIMES have multiple solutions, and to my knowledge, these alternate solutions almost always involve a "2x2 pool", like so:

These pools can be used to fill space to get the full score for the puzzle (The entire grid should be filled by the end). The canonical solution to the puzzle will never include these. In fact, in the official rules of the puzzle genre Flow was derived from, these pools aren't even allowed! Much of our strategy will involve the abuse of this rule.

Strategy 1: WHY WOULD YOU EVER DO SOMETHING ELSE???

WHY WOULD YOU EVER DO SOMETHING ELSE??? is the most basic of strategies. Essentially, this strategy applies when no other line can even remotely be justified. We can immediately apply this:

Using WHY WOULD YOU EVER DO SOMETHING ELSE???, I have solved the purple line in the top right of the grid. If you are not convinced, notice how if any other line was drawn to connect them, either a 2x2 pool is created, or the line will close off part of the grid to the rest of the lines. By going out of our way to make this line more complicated, we make the rest of the puzzle harder and nothing else. Therefore, WHY WOULD YOU EVER DO SOMETHING ELSE????

Strategy 2: Who owns me?

In other words, look at an empty cell in the grid. Which line "owns it"? Which line can be the only one to pass through it?

I have already applied WWYEDSE??? to solve the blue line (Elaboration: You can consider the case in which this isn't the solution and notice a contradiction concerning closed loops). More importantly, notice the empty cell marked by the red X. It must be "owned" by the dark red line! This solves dark red. To further see this, consider the case in which this cell is NOT owned by dark red. Then, this cell cannot be owned by anyone! But we KNOW that in the intended solution, it must be owned by someone! Thus a contradiction.

Literally the exact same logic applies to the cell marked by the green X. This cell must be owned by dark blue. We repeat this logic to get a partial solution to dark blue:

I have applied WWYEDSE??? to solve the dusty color.

You may be saying, "GREENTURD! THE REST OF THE DARK BLUE LINE IS OBVIOUS! JUST DO IT ALREADY YOU FLIPPIN HALF COOKED PANCAKE!!!1!!1" Now, I'm sure with enough thinking and considering ALL possible cases, that you could apply WWYEDSE??? to solve dark blue... but it's hard to be certain. Instead, I would like to demonstrate...

Strategy 3: HYPOTHETICAL CORNER LINE PARTY!!!

Imagine that you start a new Flow puzzle. Now, imagine a corner of this puzzle... zoom in. See that corner cell? How it would be wonderful to know who owns it. But sometimes, it is unclear.

So, let's invent a new color that can be whatever the hell it wants to be. Let's call it Color X. We define it as the color of the line that passes through this corner cell.

How does this help? Think! If Line X owns this cell, what other cells must it own? Yes! The two adjacent edge cells! And wait: WE CAN APPLY THIS EVERYWHERE! Hence, we have created a HYPOTHETICAL CORNER LINE PARTY!!!

"But greenturtlez, that doesn't look like much of a party... the lines who were invited don't look like they are having fun! :("

It's just getting started! We can abuse the "no 2x2 pools" assertion to extend these lines, alot!

Now halt! Do... you notice anything? Look at the "orange" line at the top right. It can't connect to that weirdo pale color under't... so it must connect to... OMG "BROWN" AND "ORANGE" ARE PART OF THE SAME LINE!!!

And look! Their connection created a corner! Which set up another HYPOTHETICAL CORNER LINE PARTY!!! Corner lines create MORE CORNER LINES! Now, we use WWYEDSE??? and Who owns me? to deduce even more:

And CORNERS!

The cards are set, our hypotheticals and certainties in place... and then: BOOM!!!

Hm, that pink line... I do wonder. It has to own that "brown" line... but to what end? Wait a sec... is that a light I see? Oh lord. Everything that seemed strange, all that extra space we deemed so awkward, it all adds up to the one conclusion:

And with that, the puzzle is solved:

Summary

• 
  

  2x2 pools are not intended, so pretend they are illegal, and use this to your advantage.

  
  


• 
  

  If you see a possible way to solve a color, and all other ways to connect the dots have absolutely no justification whatsoever, just do it.

  
  
  


• 
  

  When you are lost, find a big mess in your puzzle, and look at an empty square... Think about which line MUST pass through this square. If you can point your finger at only one color that can "own" this square, you have made a valuable deduction.

  
  


• 
  

  Finally, when all is lost, draw hypothetical lines, that you know must exist as part of a line, but you do not know which.

  
  

Have fun solving these puzzles!