gravitational waves - Redshift of merging black holes

How did they found that the gravitational waves where emitted at redshift $z=0.09$?

I understand the measurement of redshift for an electromagnetic wave where we have measured in a lab various transitions and therefore we can make a comparison with the wavelength we receive.

But how can they manage to get the redshift for the emitter of gravitational waves, since we have no reference?

Answer

As stated in the LIGO discovery paper (pdf), the event is placed at $410^{+160}_{-180}\ \mathrm{Mpc}$ luminosity distance, equivalent to a redshift of $z = 0.09^{+0.03}_{-0.04}$. This gives a clue as to how one measures the distance for this event.

If we know how intrinsically luminous an object (like a star, or a supernova) is, we can compare this to how bright it seems and recover a distance via the standard inverse-square law. The distance we get is by definition the luminosity distance. For this detection, the same principle applies, since the simulations predict the intrinsic strength of the signal.

Actually, we also can leverage frequency information. Again, we have numerical simulations that predict waveforms, and the waveform itself will be redshifted in the same way as any other signal propagating at the speed of light.

In practice, one takes the entire waveform and a bank of numerical simulations, and does a statistical analysis to see how well the signal matches models, and what self-consistent distance/redshift make it fit. This is detailed in Veitch et al. 2015 Phys. Rev. D 91 042003.

Note there is some degeneracy with inclinations. The detectors are not monopole antennae, but at least with two of them we can sort of localize the source on the sky to figure out what fraction of the power is actually absorbed. A more stubborn degeneracy lies in the orientation of the astrophysical system with respect to our line of sight. Since gravitational waves are (at least) quadrupolar in order, an edge-on system nearby will be hard to distinguish from a face-on system further away. This is at least part of the reason for the large uncertainties.

fluid dynamics - Why doesn't water come out of tap/faucet at high pressure when I turn it on?

(tap=faucet) When I turn a tap on full and then put my thumb over the spout covering, say, 90% of it, then the water spurts out. If I turn it on to, say 10%, then the water dribbles out.

What's the essential difference between the way that my thumb covers most of the hole and the way the tap valve opens that makes the water come out at a different pressure in either case?

Answer

It's the shape of the tap - in fact the very reason for their being a tap rather than just a valve on the end of the pipe.

When you open the valve slightly the water does come out at high speed, hits the inside of the tap spout and is stopped - it then runs out of the tap at slow speed instead of spraying you.

Imagine if you put your finger over the end of the tap to create a spray but then collected that spray in another pipe and looked at the water coming out of the other end of the pipe - it would be a slow trickle, just like the tap.

mathematics - What's the largest number you can spell?

If I give you one of each letter in the alphabet what's the largest you can spell (in word form)?

Bonus: What's the smallest?

Bonus 2: What if you can use the words "minus", "plus" and "times"?

Answer

For a number immensely bigger than $\omega$, consider the uncountably infinite number hidden below. (Note, by the way, that $\omega$ is countably infinite, and rather than being the biggest something, it is in fact “the smallest infinite ordinal ... as it is the least upper bound of the natural numbers” [1]). So omega is a good candidate for the first bonus, the smallest number one can spell if given one of each letter in the alphabet.

Answer:

The transfinite number aleph sixtyfour appears to be the biggest aleph ($\aleph$) number one can spell if given one of each letter in the alphabet.
Note that $\aleph_{64} > \aleph_{63} > ... \aleph_1 = 2^{\aleph_0} > \aleph_0 = \omega$.
(For a big number that doesn't quite work because it has two a's and e's, see wikipedia's Aleph-ω article; aleph omega is the least upper bound of ${\aleph_n : n\in\{0,1,2,\dots}\}$. But if we use five Roman and one Greek letter, or one Hebrew and one Greek letter, aleph $\omega$ or $\aleph_{\omega}$ work ok.)

quantum mechanics - Show that Propagator satisfies Schrödinger equation

I want to show that $$K=K(x,x',t-t')=\sum_{\beta}\exp\left[\frac{-iE_{\beta}}{\hbar}(t-t')\right]$$ satisfies the Schrödinger equation $$ H|\psi\rangle = i\hbar\partial_t|\psi\rangle$$ with respect to $x$ and $t$, where the $\beta$'s are the Eigenstates of the Hamiltonian and satisfy $$\sum |\beta\rangle\langle\beta|=1.$$ So I started calculating and got $$ i\hbar\partial_t K =...= \langle x\vert\exp\left[\frac{-iE_{\beta}}{\hbar}(t-t')\right]\vert x'\rangle. $$ But here I am stuck since I am not that familiar with QM and bra-ket notation/relations. What is the next step or which fact do I have to use to show the claim?

calculation puzzle - Number Sequence Series-Question 3

I want to share a series of questions that are created by myself.

I will give a hint in 24 hours and my answer in 3 days given that nobody could answer my question.

Here is my number sequence:

2020,1436,7575,?,23111311

If you guys want some extremely challenging questions. Please check these two questions posted by me.

Number sequences: 000, X00... and 6X000X9, 700XX08

What are the alphabets in the question mark?

Update

Hint1:

The logic is related to 1234

Hint2:

Separate the digits

Answer

The missing number is

$69811$

Procedure for getting from one step to the next

1. Separate out the digits (e.g, $2020 \rightarrow 2,0,2,0$).
2. Reverse the sequence (e.g, $2,0,2,0 \rightarrow 0,2,0,2$).
3. Add $1$ to the first digit, $2$ to the second, $\ldots n$ to the $n$th (e.g, $0,2,0,2 \rightarrow 1,4,3,6$).
4. Recombine the digits (e.g, $1,4,3,6 \rightarrow 1436$).

Other cases, step-by-step

$1436 \rightarrow 7575$
1. $1436 \rightarrow 1,4,3,6$
2. $1,4,3,6 \rightarrow 6,3,4,1$
3. $6,3,4,1 \rightarrow 7,5,7,5$
4. $7,5,7,5 \rightarrow 7575$

$7575 \rightarrow 69811$
1. $7575 \rightarrow 7,5,7,5$
2. $7,5,7,5 \rightarrow 5,7,5,7$
3. $5,7,5,7 \rightarrow 6,9,8,11$
4. $6,9,8,11 \rightarrow 69811$

$69811 \rightarrow 23111311$
1. $69811 \rightarrow 6,9,8,1,1$

2. $6,9,8,1,1 \rightarrow 1,1,8,9,6$
3. $1,1,8,9,6 \rightarrow 2,3,11,13,11$
4. $2,3,11,13,11 \rightarrow 23111311$

pattern - What is a Scalable Phrase™?

Now that you are familiar with Cyclone Phrases, let's look at another kind of phrase.

If a phrase adheres to a certain rule, then I call it a Scalable Phrase™.

Use the examples below to find the rule.

EDIT: I added the visual tag as a hint.

If you liked this puzzle, try others like it:
What is a Cyclone Phrase™?

What is a Triad Phrase™?

Answer

Here is my answer:

If you are coming from the left to the right, then "Scalable Phrases" are climbable while "Non-Scalables" are not climbable. Scaling a phrase refers to climbing it.

For a phrase to be climbable it has to be composed of climbable letters.

A climbable letter is any letter that, if you are on the "floor" (where the letter sits) then you can "climb" up the letter because no angle from the bottom to the top is greater than 90 degrees, or perpendicular from the floor. For example, V is not climbable because coming from the left the first obstacle (the \ of the V) is sort of an overhang. T is also not climbable even though the beginning is climbable, but then at the overhang it is not. H, F, E, M, etc are all climbable letter because they have no overhang.

Notice that I in the font chosen IS climbable, although in many fonts it is not, since it has an overhang.

Here is a full list of climbable letters:
ABDEFHIKLMNPR

Here is a full list of nonclimbable letters:
CGJOQSTUVWXYZ

Really nice puzzle!!

electromagnetism - What frequencies of em radiation can ionize air?

What frequencies of em radiation can ionize air at atmospheric pressure? Does it depend on the power of the transmitter/generator or just the frequency? Do some frequencies or power densities make air more conductive but not ionize it into a plasma? I know that temperature can make air into a plasma. Is there any way to heat air using em radiation other than with light?

Answer

Technically, one only needs photons with energies $\geq$13.6 eV to ionize a single monatomic hydrogen atom, which corresponds to ~3300 THz (i.e., ~3.3 billion MHz) or wavelengths of ~91 nm (i.e., ~$9.1 \times 10^{-8}$ m). Note that the energy of a photon is given by $E = h \ \nu$, where $h$ is Planck's constant and $\nu$ is the frequency. For electromagnetic radiation in vacuum, we also know that $c = \lambda \ \nu$, where $\lambda$ is the wavelength and $c$ is the speed of light. The index of refraction of air is close enough to 1.0 for the rough estimates I use below.

However, with air, assuming its Earth's atmosphere at STP, one would first dissociate the molecules before any ionization would occur. I wrote a detailed answer about the energies necessary to dissociate all of the molecules in Earth's atmosphere at https://physics.stackexchange.com/a/233126/59023.

You can easily scale these estimates down to a more practical lab setting. To dissociate $N_{2}$ (i.e., diatomic nitrogen) one needs ~945 kJ/mole or ~9.79 eV per $N_{2}$ bond (Note that 9.79 eV corresponds to a ~$2.36716 \times 10^{15}$ Hz or a ~126.7 nm photon). $N_{2}$ comprises ~78.08% of Earth's atmosphere by volume so it would occupy ~78.08% of a one meter cubed container. Thus, there would be ~$4.7 \times 10^{23}$ molecules of $N_{2}$ in a one meter cubed container or ~0.7808 moles. Thus, we would need ~737.86 kJ of energy to dissociate all of the $N_{2}$ molecules.

What frequencies of em radiation can ionize air?

It is generally stated that one needs at least UV light to ionize most atoms, thus why it is called ionizing radiation. The UV spectrum extends from ~10-400 nm or ~750-30000 THz, so you can see that both of the estimated wavelengths shown above fall in the UV spectrum range.

Does it depend on the power of the transmitter/generator or just the frequency?

I suppose the answer is yes, but not for the reasons your question seems to imply. The power of the transmitter would determine the number of photons per unit time generated while the frequency would determine the energy of each photon. If you only sent out one photon per second, my hand-wavy guess is that the recombination rate would swamp the ionization rate, thus you would not see a net charge. This is a good thing so we can live (i.e., the recombination rate is higher than the ionization rate from solar radiation).

Do some frequencies or power densities make air more conductive but not ionize it into a plasma?

I am not sure what you are asking. The air is an excellent insulator and really only conducts electricity when there is an arc, i.e., molecules are ionized along the path of the arc allowing electrons to flow, thus a current. This generally requires very large electric fields, like ~30 kV/cm (or ~76 kV/inch).

I know that temperature can make air into a plasma.

No, this is not really right. The temperature of a gas is a measure of the mean kinetic energy of the molecules in the center of momentum rest frame. If the gas is very tenuous, then there would be little-to-no particle-particle collisions. You can take a neutral particle and make it go as fast as you want without ionizing it (ignoring acceleration or the energy source) because its speed does not matter. In its rest frame, it does not know its moving unless it "looks" at something else.

High temperatures can lead to ionization if there is a sufficiently high particle-particle collision rate and if the kinetic energy of the collision exceeds the ionization energy of at least one of the atoms.

Is there any way to heat air using em radiation other than with light?

The short answer is yes. There are three basic energy transport methods: thermal conduction, radiation, and convection. A standard oven used for cooking food (i.e., not a microwave oven) uses a combination of thermal conduction and radiation (I think thermal conduction dominates here but have heard arguments for both sides).

Update
In the following question and answer Do conventional ovens heat by thermal conduction or radiation? I show that it is actually radiation that dominates in conventional ovens.

Regarding the last question, if the EM radiation can interact with the gas (e.g., similar to how microwave ovens vibrate/excite water molecules), one can deposit energy into the gas. If the gas is collisionally mediated, like Earth's atmosphere below ~10 km, then after enough energy is added through the radiation one could ionize the gas molecules.