newtonian gravity - How is a result of no time variation in the gravitational constant $G$ related to a measurement of no local expansion?

In this answer @PhillS linked to the paper Progress in Lunar Laser Ranging Tests of Relativistic Gravity which has given me a great start. While its main focus is on the Equivalence Principle (EP) they also mention that no local expansion was seen. The last sentence in the abstract is:

" The search for a time variation in the gravitational constant results in G/G ˙ = (4 ±9) ×10 −13 yr − 1 ; consequently there is no evidence for local ( ∼1 AU) scale expansion of the solar system."

Briefly, the data here is 34 years of laser ranging from Earth of a retroreflector array on the moon. While the distance between a given measurement site and the reflector on the moon can vary by as much as 50,000km (due mostly to the Moon's elliptical orbit and the size of the earth) these can be and have been painstakingly modeled. The amazing result is that in 34 years the residual scatter is only about 2cm!!

What I'm looking for is a midrange answer - not (exclusively) high level cosmology, but more than just balloon and raisin cake analogies. Something that will help towards understanding the relationship between the two.

Question: How is a result of no time varying $G$ related to a measurement of no local expansion?

Bonus mini-question: If I understand correctly and the null measurement of expansion is from the earth-moon ranging data, why does the sentence say "local (~1 AU) scale" when the earth-moon distance is only 0.0027 AU?

Lunar Libration image from here

Lunar Laser Ranging images from here

Answer

If cosmological expansion applies on the scale of the earth moon system, then in some short period of time $\delta t$ the distance between the earth and moon increases from $r$ to $r+\delta r$. So the force of gravity between the bodies changes to:

$F+\delta F=\frac{GMm}{(r+\delta r)^2}\approx\frac{GMm}{r^2}\left(1+\frac{\delta r}{r}\right)^{-2}\approx\frac{GMm}{r^2}\left(1-\frac{2\delta r}{r}\right)=F\left(1-2\frac{\delta r}{r}\right)$

The same change in gravitational force could also come about via change in $G$:

$F+\delta F=\frac{(G+\delta G)Mm}{r^2}=F\left(1+\frac{\delta G}{G}\right)$

Equating the $\delta F$ terms shows you that the change in the force due to increasing distance is the same as the change in the force due to decreasing $G$ if

$\frac{\delta G}{G}=-2\frac{\delta r}{r}$

Now if our $\delta r$ is due to cosmological expansion being applicable to this distance scale, then for an expansion velocity of $v$, $\delta r=v\delta t$ and by using Hubbles law $v=H_0 r$ where $H_0$ is Hubble's constant, which is around $70kms^{-1}Mpc^{-1}$, which we want to converft to SI units ($ms^{-1}{m^-1}=s^{-1}$) which gives us $H_{0}=2.26\times10^{-18}s^{-1}=7.1\times10^{-11}yr^{-1}$

So plugging that in to our equation gives $\frac{\delta G}{G}=-2\frac{H_{0}r\delta t}{r}$. Simplifying, and turning $\frac{\delta G}{\delta t}$ into $\dot{G}$ we end up with

$\frac{\dot{G}}{G}=-2H_{0}$

This is a simplistic way to do it, but the basic idea is on the right track I believe. A cosmologically-caused increase in the earth moon distance would give a reduction in the mutual force between them the same as a decrease in $G$ would if they stayed at the same distance. Rearranging the maths as above gives you a a direct proportionality between $\frac{\dot{G}}{G}$ and $H_0$, although the constant of proportionality for general relativity is probably not exactly what I've come up with here (and other theories of gravity may give different values).

The paper you mention puts an experimental limit on that constant of proportionality of (roughly) $0.006 \pm 0.012$ (using the value $H_0$ above), as opposed to the 'predicted' value of $-2$

The reason for expressing the result in terms of $\frac{\dot{G}}{G}$ idea that that is directly derived from what they measure independent of any theory. I.e. it is as far as you can go towards putting numerical limits on the expansion without committing to a particular theory or value of $H_0$

(I believe that John Rennie's comment is correct that the are using '$AU$ scale' to indicate that they are talking about effects on the order of the solar system, rather than galactic or cosmological distances.)

Excellent animation of the moon changing through the month BTW.

general relativity - Could there be any alternative to a supermassive black hole that might explain Sgr A*?

I am aware of How do we know the stars orbiting Sgr A* are orbiting a supermassive black hole and not just the center of mass of the Milky Way galaxy?, which asks why the motions of stars near the Galactic centre suggest a 4 million solar mass black hole.

But are there any theoretical ideas to explain these observations, that avoid the conclusion of a black hole, that remain unfalsified?

And will subsequent observations (e.g. with the Event Horizon telescope) be able to falsify these alternatives or indeed the hypothesis of a supermassive black hole?

newtonian mechanics - What is an intuitive explanation using forces for the equatorial bulge?

The earth is not a sphere, because it bulges at the equator.

I tried fiddling with centripetal force equations and gravity, but I couldn't derive why this bulge occurs.

Is there

(a) a mathematical explanation using forces (not energies) and

(b) a simple intuitive explanation to explain to others why the bulge occurs?

Answer

Equatorial bulging of a planet is caused by the combination of gravity and centrifugal force. To show this I will first make a few assumptions:

• The planets is assumed to be made up of a liquid of constant density.


• All liquid is at rest relative to itself, which means that there are no shear stresses within the liquid, since this would induce a flow.


• The equatorial bulging is small, such that the acceleration due to gravity, $\vec{a}_g$, at the surface can be approximated with: $\vec{a}_g=-G\frac{M}{\|\vec{x}\|^3}\vec{x}$, where $G$ is the gravitational constant, $M$ the mass of the planet and $\vec{x}$ the position on the surface relative to the center of mass of the planet.


• The planet is axis symmetric and rotates around this axis with a constant angular velocity $\omega$.

A small volume, $dV$, experiences two volumetric accelerations, namely gravitational and centrifugal, and normal forces by the neighboring liquid in therms pressure. The sum of all accelerations on $dV$ should add up to zero to comply with the second assumption (the centrifugal acceleration already accounts for the fact that the reference frame is rotating). At any point on the surface there is a constant pressure, because above it there would be the vacuum of space. This means that the neighboring liquid, also at the surface, has the same pressure and therefore can not exert any force on each other in the plane op the surface. The only direction that liquid can exert force on each other is in the normal direction to the surface. However the sum of all accelerations still should add up to zero and therefore the sum of the gravitational and centrifugal acceleration should also point in the normal direction of the surface.

The magnitude of the gravitational acceleration, $a_g$, is defined by assumption three and its direction is always radially inwards. The magnitude of the centrifugal acceleration, $a_c$, is equal to:

$$ a_c = \omega^2 \sin\phi\ \|\vec{x}\|, $$

where $\phi$ is equal to $\pi/2$ minus the latitude; its direction is always parallel to the plane of the equator and its line of action always goes through the axis of rotation. These accelerations are illustrated in the figure below.

For the next part I will define local unit vectors $\vec{e}_r$ and $\vec{e}_t$, where $\vec{e}_r$ points into the local radial outwards direction and $\vec{e}_t$ is perpendicular to it, lies in the plane spanned by the axis of rotation and $\vec{x}$ and faces the direction closest to the equator. The direction of vectors also correspond with the grey vectors in the figure above. Using these unit vectors, the vector sum of the gravitational and centrifugal acceleration can be written as

$$ \vec{a}_g + \vec{a}_c = \vec{e}_r \left(\omega^2 \sin^2\!\phi\ \|\vec{x}\| - G\frac{M}{\|\vec{x}\|^2}\right) + \vec{e}_t\ \omega^2 \sin\phi\ \cos\phi\ \|\vec{x}\|. $$

If there would be no bulging then the normal vector should always point radial outwards. However the normal vector has to point in the opposite direction as the equation above, which means that for $\omega>0$ it will not point in the same direction as $-\vec{e}_r$ for all values of $\phi$. This means that the surface will have a small slope, $\alpha$, relative to $\vec{e}_t$

$$ \alpha = \tan^{-1}\left(\frac{\omega^2 \sin\phi\ \cos\phi\ \|\vec{x}\|}{G\frac{M}{\|\vec{x}\|^2} - \omega^2 \sin^2\!\phi\ \|\vec{x}\|}\right). $$

A slope means a change of height, and thus radius, when displacing horizontally. To simplify the expression, $r$ will substitute $\|\vec{x}\|$. For a slope $\alpha$ the change of the radius, $dr$, for a small change in $\phi$, $d\phi$, will be equal to:

$$ dr = \tan\alpha\ r\ d\phi. $$

By substituting in the equation for $\alpha$ the following differential equation can be obtained

$$ \frac{dr}{d\phi} = \frac{\omega^2 \sin\phi\ \cos\phi\ r^2}{G\frac{M}{r^2} - \omega^2 \sin^2\!\phi\ r} . $$

When $\phi$ is equal to $0$ or $\frac{\pi}{2}$, the poles and the equator respectively, this equation will be zero, however for any value in between, it will be positive, since when denominator would become negative it would mean that the centrifugal force will be bigger than gravity and the liquid would be flung into space. So this planet would have the smallest radius near the poles after which the radius will increase with $\phi$ until you reach the equator.

electromagnetism - Possiblity of predicting behavior of system when properties described as functions over space?

Suppose I am given a system that consists of a distribution of charged particles(which are all over space and are point-charges). They are described by a set of functions instead of variables. These functions are:

• $C(x)$ - a function describing the amount of charge at each point in space. For example, take the one dimensional example of $C(x) = \sin(x)$.


• $M(x)$ - a function describing the amount of mass at each point in space.


• $V(x)$ - the function describing what the sum of all the velocities of the particles at a point is.

Given these are the functions to describe the system at time $t = 0$, is it possible to predict the state of this system (the charge distribution, velocities and mass distributions) at a later time, say, $t = 2$, using Newtonian dynamics?

My system can be thought of as a fluid, where distribution of mass and charge vary. I ignore charges if that simplifies things. I want to take into account gravity, though, at least. How do I exactly proceed to find the distribution functions at $t = 2$?

pattern - What is a Re-Tileable 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 has a certain property, I call it a Re-Tileable Word™.

You can use the examples below to find the property:

$$\begin{array}{|c|c|} \hline \bbox[yellow]{\textbf{Re-Tileable Words™}}&\bbox[yellow]{\textbf{Un-Re-Tileable Words™}}\\ \hline \text{KALE}&\text{CHARD}\\ \hline \text{ROOSTER}&\text{COCKEREL}\\ \hline \text{KARMA}&\text{FORTUNE}\\ \hline \text{SALMON}&\text{MACKEREL}\\ \hline \text{TEST}&\text{QUIZ}\\ \hline \text{DREAM}&\text{NIGHTMARE}\\ \hline \text{FAST}&\text{QUICK}\\ \hline \text{CRAFT}&\text{ORIGAMI}\\ \hline \text{VIEW}&\text{SCENERY}\\ \hline \text{ORGAN}&\text{GLAND}\\ \hline \text{SCORE}&\text{MARKS}\\ \hline \text{EROTIC}&\text{SEXUAL}\\ \hline \end{array} $$

For those without MathJax, or if you want to pop this into a spreadsheet, here is a CSV version:

Re-tileable Word™, Non-re-tileable Word™  

KALE, CHARD  
ROOSTER, COCKEREL  
KARMA, FORTUNE   
SALMON, MACKEREL  
TEST, QUIZ  
DREAM, NIGHTMARE  
FAST, QUICK  

CRAFT, ORIGAMI  
VIEW, SCENERY  
ORGAN, GLAND  
SCORE, MARKS  
EROTIC, SEXUAL  

Answer

A Re-Tileable Word™ seems to be ...

... an allowable Scrable word that can be anagrammed to one or more other allowable Scrabble words. Some anagrams are obscure e.g. makar, monals, tercio, but they are all in the Scrabble dictionary. The Un-Re-Tileable words have no useful anagrams.

These words are Re-Tileable™, because ...

... to make an anagram of a word on a Scrabble board just means to rearrange the tiles.

special relativity - Euclidean geometry in non-inertial frame

Refer, "The classical theory of Fields" by Landau&Lifshitz (Chap 3). Consider a disk of radius R, then circumference is $2 \pi R$. Now, make this disk rotate at velocity of the order of c(speed of light). Since velocity is perpendicular to radius vector, Radius does not change according to the observer at rest. But the length vector at boundary of disk, parallel to velocity vector will experience length contraction . Thus, $\dfrac{\text{radius}}{\text{circumference}}>\dfrac{1}{2\pi}$ , when disc is rotating. But this violates rules of Euclidean geometry.

What is wrong here?

Answer

What is wrong is the idea that one can actually make the disk rotate; and it will remain perfectly rigid.

In reality, what this correct argument shows is that relativity doesn't admit the existence of any perfectly rigid bodies. This is a perfectly basic, settled, and indisputable textbook material that every mature physicist knows. The first sentence of this paragraph contains a link to the Gravity Probe B website. The thought experiment is known as the Ehrenfest paradox and Ehrenfest himself already offered the right basic answer – no rigid objects exist in relativity – when he outlined the thought experiment in 1909.

When one takes a solid disk and makes it rotate, it will do all kinds of things resulting from the "imperfection of the material". It will tear apart by the centrifugal force, and if it won't, it will either tear basically along radial lines, or it will bend (the disk won't be planar anymore) because the circumference really shrinks by the Lorentz factor. If there existed a material that is perfectly rigid and cannot stretch or bend or tear, then it would be impossible to make it spin. In any world governed by relativity, the proper distances between the individual points/atoms of the objects simply have to change when the object is brought to motion. (The definition of rigidity using the constant proper distances between points/atoms of the object was given by Max Born in 1909 and is known as the Born rigidity.)

However, the non-existence of such a material may be shown even microscopically. It is not possible to "order" any solid object to keep the proper distances at every moment because the distance between two atoms (or points on the solid object) may only be measured with a delay $\Delta t = \Delta x / c$ simply because no information may move faster than light. That's why it's always possible to squeeze any rod on one end and the opposite end of the rod won't move at least for this $\Delta t = \Delta x / c$. This relationship between the "limited speed of signals by $c$" and "non-existence of rigid objects in relativity" was already pointed out by Max von Laue in 1911.

In fact, the delay will be much larger than that, dictated basically by the speed of sound, not by the speed of light. Whatever material you have, relativity guarantees that it can be squeezed as well as stretched as well as bent.

special relativity - Can relativistic mass be treated as rest mass?

In what real sense does the mass of an object increase with its speed? When we learn that the mass of an object increases according to the equation,

$$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

We can think of the mass of an object as its resistance to being moved (inertia). But suppose we devised a contraption that increased its mass by the motion of its internal parts. Would the inertia of the contraption be greater? In the following, I devise the simplest possible contraption that takes advantage of this fact and ask what happens when an object collides with it.

Consider a box with a negligibly light frame in which two balls each of mass $\frac{m_0}{2}$ start in the middle and move in opposite directions at velocity $v$. Call this box the "balls box".

The balls box has mass $m = \frac{m_{0}}{\sqrt{1 - \frac{v^2}{c^2}}}$.

Consider a box with rest mass $m$ when another box of rest mass $m$ hits it moving at velocity $v$. In an elastic collision, the first mass with move with velocity $v$ and the second will come to rest.

Now replace the box with rest mass $m$ with the balls box. Because the balls inside it are moving, it has rest mass $m_0$ but relativistic mass $m$. The balls box is at rest when a box with mass $m$ hits it at velocity $v$ in the direction along which the balls move. What happens?

According to a naive analysis, after the collision there will be some jiggling as information is transmitted down the length of the balls box. Eventually, the moving box will come to rest and the balls box will move with velocity $v$ to the right. Is this right?

What happens when we turn the balls box perpendicular to the motion of the moving ball?

Edit:

The rationale behind the naive analysis is that when the balls box is treated as a self-contained system—and when it's at rest—it has mass $m$. By that reasoning, it should behave the same as a box with rest mass $m$. The problem arises when we look inside the box after the collision. After information has propagated, one ball moves at $\frac{v - v}{\sqrt{1 + \frac{v^2}{c^2}}} = 0$. The other moves at $\frac{2v}{\sqrt{1 + \frac{v^2}{c^2}}}$. The box previously in motion is at rest. This leads to a momentum and energy different from the energy and momentum before the collision.