string theory - What if the LHC doesn't see SUSY?

A question in four parts.

1. 
   

   What are the main problems which supersymmetry purports to solve?

   
   


2. 
   

   What would constitute lack of evidence for SUSY at the proposed LHC energy scales (e.g. certain predicted superpartners are not in fact observed)?

   
   



3. 
   

   Are there alternative theoretical approaches which would address the SUSY problem set and which would still be credible in such an LHC no-SUSY-scenario?

   
   


4. 
   

   Where would LHC-disconfirmation of SUSY leave String Theory?

   
   

I would like to think that these four points could be taken together as one question.

Answer

First, let me emphasize something that is being covered by a thick layer of misinformation in the media these days: it is totally premature to conclude whether the LHC will see SUSY or not. The major detectors have only collected 45/pb (and evaluated 35/pb) of the data. The "slash pb" should be pronounced as "inverse picobarns".

The LHC is designed to collect hundreds or thousands times more data than what it has recorded so far, and it should eventually run at a doubled energy (14 TeV total energy instead of the current 7 TeV total energy). Each multiplication of the integrated luminosity (number of collisions) by 10 corresponds to the access of new particles whose masses are approximately 2 times larger or so. It means that the LHC will be able to decide about the existence of new particles at masses that are 4-16 times higher than the current lower bounds (16 also includes the likely upgrade from 2x 3.5 TeV to 7 TeV).

There are at least two "mostly independent" parameters with the dimension of mass in SUSY - I mean $m_0$ and $m_{1/2}$. So the number from the previous sentence should really be squared, and in some sensible counting and with a reasonable measure, the LHC has only probed about 1/16 - 1/256 of the parameter space that is accessible to the LHC over its lifetime.

So the only thing we can say now is that SUSY wasn't discovered at an extremely early stage of the experiment - which many people have hoped for but this possibility was never supported by anything else than a wishful thinking. Whether the LHC may see SUSY may remain an open question for several years - unless the LHC will see it much sooner than that. It's an experiment that may continue to 2020 and beyond.

We don't really know where the superpartner masses could be - but they may sit at a few TeV and this would still mean that they're accessible by the LHC.

Now, your questions:

What SUSY helps to solve

First, SUSY is a natural - and mostly inevitable - consequence of string theory, the only consistent quantum theory that includes gravity as well as the Yang-Mills forces as of 2011. See

http://motls.blogspot.com/2010/06/why-string-theory-implies-supersymmetry.html

In this context, supersymmetry is needed for the stability of the vacuum and other things, at least at a certain level. For other reasons, to be discussed below, it's natural to expect that SUSY should be unbroken up to LHC-like energy scales (i.e. that it should be visible at the LHC) - but there's no sharp argument that might calculate the superpartner scale. Some string theorists even say that it should be expected that supersymmetry is broken at a very high scale (near the GUT scale or Planck scale) - because this is a "generic behavior" in the stringy landscape (the "majority" of the minima have a high-scale SUSY breaking which would make SUSY unavailable to any doable experiments) - so these proponents of the anthropic reasoning don't expect SUSY to be seen at the LHC. However, more phenomenological considerations make it more natural for SUSY to be accessible by the LHC.

Why? There are several main arguments: SUSY may offer a very natural dark matter particle candidate, namely the LSP (lightest supersymmetric particle), most likely the neutralino (the superpartner of the photon or Z-boson or the Higgs bosons, or their mixture), that seems to have the right approximate mass, strength of interactions, and other things to play the role of the majority of the dark matter in the Universe (so that the Big Bang theory with this extra particle ends up with a Universe similar to ours after 13.7 billion years). See an article about SUSY and dark matter:

http://motls.blogspot.com/2010/07/susy-and-dark-matter.html

Also, SUSY with superpartners not far from the TeV or LHC energy scale improves the gauge coupling unification so that the strengths of the couplings get unified really nicely near the GUT scale (and maybe incorporated into a single and simple group at a higher energy scale not far from the Planck scale), see:

http://motls.blogspot.com/2010/06/susy-and-gauge-coupling-unification.html

The unification in the simplest supersymmetric models is only good if the superpartners are not too far from the TeV scale - but if they're around 10 TeV, it's still marginally OK. The same comment with the same value 10 TeV also holds for the dark matter job of the neutralinos discussed above.

Finally and most famously, SUSY with superpartner masses not far from the TeV or LHC scale stabilizes the Higgs mass - it explains why the Higgs mass (and, consequently, the masses of W-bosons and Z-bosons, among other particles) is not driven towards a huge energy scale such as the Planck scale by the quantum corrections (with loops of particle-antiparticle pairs in the Feynman diagrams). Those otherwise expected quantum corrections get canceled at the TeV accuracy if the superpartner masses are near a TeV - and the resulting Higgs mass may then be naturally in the expected 100 GeV - 200 GeV window with an extra 10:1 luck (which is not bad).

The lighter the superpartner masses are, the more "naturally" SUSY explains why the Higgs mass remains light. But there is no strict argument that the superpartners have to be lighter than 1 TeV or 10 TeV. It just "sounds strange" if they were much higher than that because a non-negligible portion of the hierarchy problem would remain. See a text on SUSY and the hierarchy problem:

http://motls.blogspot.com/2010/07/susy-and-hierarchy-problem.html

One may say that experiments already do disprove 99.999999999+ percent of the natural a priori interval for a conceivable Higgs mass in the Standard Model. SUSY changes this counting - the probability that the Higgs mass ends up being approximately as low as suggested by the electroweak observations becomes comparable to 100 percent according to a SUSY theory. To agree with other available experiments, SUSY needs to adjust some other parameters but at good points of the parameter space, none of the adjustments are as extreme as the adjustment of the Higgs mass in the non-supersymmetric Standard Model.

Can we decide whether SUSY is there at the LHC?

SUSY may hide for some time but the LHC is simply scheduled to perform a certain number of collisions at a certain energy, and those collisions may eventually be studied by the most up-to-date methods and the evidence for SUSY will either be there in the data or not. Some phenomenologists often want to stay very modest and they talk about numerous complex ways how SUSY may keep on hiding - or remain de facto indistinguishable from other models. However, sometimes the very same people are capable of reverse-engineering a randomly constructed man-made model (fictitiously produced collision data) within a weekend: these are the games played at the LHC Olympics. So I don't really expect too much hiding. With the data, the fate of the LHC-scale SUSY will ultimately be decided.

Obviously, if SUSY is there at the LHC scale, the LHC will eventually be discovering fireworks of new effects (SUSY is also the most attractive realistic possibility for the experimenters) - all the superpartners of the known particles, among other things (such as an extended Higgs sector relatively to the Standard Model). Their spins and couplings will have to be checked to (dis)agree with those of the known particles, and so on. All the masses may be surprising for us - we don't really know any of them although we have various models of SUSY breaking which predict various patterns.

Alternatives in the case of SUSY non-observation

The dark matter may be composed of ad hoc particles that don't require any grand structures - but such alternatives would be justified by nothing else than the simple and single job that they should play. Of course that there are many alternatives in the literature but none of them seem to be as justified by other evidence - i.e. not ad hoc - as SUSY. I think that in the case of no SUSY at the LHC, the LHC will remain some distance away from "completely disproving" SUSY particles as the source of dark matter because this role may work up to 10 TeV masses or so, and much of this interval will remain inaccessible to the LHC.

So the LHC is a great gadget which is stronger than the previous one - but one simply can't guarantee that it has to give definitive answers about all the questions we want to be answered. This fact may be inconvenient (and many laymen love to be promised that all questions will inevitably be answered for those billions of dollars - whether it's true or not) but it's simply a fact that the LHC is not a machine to see every face of God. There are various alternatives how to solve the hierarchy problem - the little Higgs model, the Randall-Sundrum models (which may be disproved at the end of the LHC, too - the LHC is expected to decide about the fate of each solution to the hierarchy problem although they may always remain some uncertainty), etc. - but I am convinced that even in the case that SUSY is not observed at the LHC, superpartners with slightly higher masses than those accessible by the LHC will remain the most well-motivated solution of the problems above.

Of course, if someone finds some better new models, or some amazing experimental LHC (or other) evidence for some existing models, the situation may change. But right now, away from SUSY, there are really no alternative theories that naturally explain or solve the three problems above at the same moment. This ability of SUSY to solve many things simultaneously is surely no proof it has to be the right solution of all of them - but it is a big hint. It's the reason why particle physicists think it's the most likely new physics at this point - a conclusion that may change but only if new (theoretical or experimental) evidence arrives.

While it is clear that the absence of SUSY at the LHC would weaken the case for SUSY and all related directions, I am convinced that unless some spectacular new alternatives or spectacular new proofs of other theories are found in the future, SUSY will still remain the single most serious direction in phenomenology. In formal theory, its key role is pretty much guaranteed to remain paramount regardless of the results of LHC or any conceivably doable experiments. The more formal portions of high-energy theory a theorist studies, obviously, the less dependent his or her work is on the LHC findings.

I don't have to explain that the absence of SUSY at the LHC would mean a sharper splitting of the particle physics community.

Absence of SUSY and string theory

Clearly, if no SUSY were seen until 2012 or 2015 or 2020, the critics of string theory would be louder than ever before. Within string theory, the anthropic voices and attempts to find a sensible vacuum with the SUSY breaking at a high-energy scale would strengthen. But nothing would really change qualitatively. The LHC is great but it is just moving the energy frontier of the Tevatron at most by 1-1.5 order(s) of magnitude or so.

If there is some non-SUSY new physics found at the LHC, most particle physicists will obviously be interested in whatever models that can be relevant for the observations. If the LHC sees no new physics, e.g. if it only sees a single Higgs boson, and nothing else ever appears, the current situation will qualitatively continue and the tensions will only get amplified. Serious physicists will have to continue their predominantly theoretical and ever more careful studies (based on the observations that have been incorporated into theories decades ago) without any guidance about new physics from the available new experiments (simply because there wouldn't be any new data!) - while the not so serious physicists and people around science will strengthen their hostile and utterly irrational claims that physics is no longer science.

Sociologically, the situation would almost certainly become unpleasant for good physicists and pleasant for populist and uneducated critics of science who are not really interested in the truth about the physical world. But Nature works in whatever way She does. She is not obliged to regularly uncover a part of Her secrets.

A paper with the same question in the title

Amusingly, there exists a 2-week-old preprint by 8 authors:

http://arxiv.org/abs/arXiv:1102.4693
What if the LHC does not find supersymmetry in the sqrt(s)=7 TeV run?

You may see that the question in their title is almost identical to your question at Physics Stack Exchange. Their answer is much like my answer above: if the LHC is not found during the 7 TeV run (that should continue until the end of 2012), SUSY would still remain an acceptable solution to all the problems I mentioned above; just our idea about the masses of the strongly interacting superpartners (gluinos and squarks) would have to be raised above 1 TeV or so. It's pretty natural for those strongly interacting superpartners to be the heaviest ones among the superpartners - which automatically makes them harder to be seen at hadron colliders such as the LHC.

classical mechanics - Having trouble understanding spectral lines

In my notes I wrote that Rutherford's model of the atom could not explain spectral lines, because that is what my textbook says. I'm not really sure about the details of spectral lines though.

I know that when electrons interact with photons of specific frequency, they transition between energy levels (ie they could increase and become excited). After they become excited, they 'de-excite'. This allows them to release a photon and jump back to a lower orbit after they release a photon. This photon has frequency that corresponds between the difference in transition. This frequency correlates to a particular colour.

So how then does this colour produce spectral lines, and therefore balmer series? (is that the right word usage?). How does Bohr's model not explain spectral lines (does it not talk about electron transitions, electron orbits etc and thus cannot deduce further?). What current model explains this?

Answer

This is going to take a bit, so be prepared. Also, some of what I say will not be exactly correct, in order to get the larger point across, but bear with me.

First, the Rutherford model really did not speak to the question of emission lines. It simply noted that, on the basis of Rutherford's scattering experiments, all of the protons in an atom had to be concentrated in a dense clump. Before this, the fact that protons repel each other suggested that an atom was made of separate protons (which repelled each other) and electrons (which also repelled each other), while the attraction between the protons and electrons kept the whole assembly from flying apart. This was called the Plum Pudding Model.

In this model if you try to fire an $\alpha$ particle through something like gold foil, the $\alpha$ particle ought to hit a large number of small obstacles, and this would produce a specific scattering effect. When Rutherford actually did this, he discovered that this isn't so - the experiment behaved as if there were a small number of very large obstacles. Rutherford concluded that, although it made no immediate sense, the protons were somehow clumped together.

Shortly thereafter, Niels Bohr pointed out that this would make sense if the atom's electrons, being attracted to the clump (which shortly became known as the nucleus) of protons, could be thought of as orbiting the nucleus, just as planets orbit a star. This became the Rutherford-Bohr model. At it happens, it made all kinds of sense out of the known behavior of atoms, and was quickly adopted. It is still taught in grade schools and high schools, even though it has been overshadowed by the more accurate and powerful ideas of quantum mechanics. You can think of it as being like the difference between Newtonian and relativistic physics. For some cases, Newtonian physics works just fine. For some analyses the Rutherford-Bohr model works just fine, too.

Thinking of an atom as a little solar system, you can think about the energy of the planets/electrons. Let's take Mercury, for instance. If you put a really big rocket on Mercury, you could speed it up and cause it to move to a larger orbit. The increase in speed means an increase in energy, so you can say that larger orbits have more energy than smaller orbits.

This leads to the idea that, when an excited electron emits light, it loses energy and falls to a lower (smaller) orbit. With me so far? Now a problem comes up. If you look at the emission spectrum in anna v's answer, you may notice something: there are only a few sharp lines visible. This means that, in an excited gas which is emitting light, billions of electrons are all emitting exactly the same amount of energy when they "de-excite". Think about what this means in the Rutherford-Bohr model.

First: Each hydrogen has one proton in its nucleus, and all protons are identical, so for every hydrogen "solar system" the central sun is exactly the same - check.

Second: Each hydrogen has one electron as a planet, and all electrons are identical, so for every hydrogen "solar system" the planet is exactly the same - check.

Third: Since each "de-excited" planet gives off one of a small number of specific energies, the orbits found in each hydrogen "solar system" must be the same - che - wait, what?

Why in the world would this be true? In a macroscopic solar system, you can get any orbit you want by tweaking the speed of the moving body. So why would electron orbits be restricted to a small number of possibilities?

Well, that's what quantum theory is all about. The energy that a bound electron can have is restricted to integral multiples of some base energy - it is quantized.

And that is why the Rutherford-Bohr model does not explain emission lines. In the case of the hydrogen atom, an electron can switch between different energy levels as it emits light, and the relationships between the possible energy states of the electron produce a well-defined relationship between the different emission spectra. The Balmer and Lyman series are examples of different classes of orbit-switching.

Eigenvalues of a quantum field?

Fields in classical mechanics are observables. For example, I can measure the value of the electric field at some (x,t).

In quantum field theory, the classical field is promoted to an operator-valued function of space-time. But no one talks about eigenvectors of the quantum field! If I try to measure the field at one point in spacetime, I should get a real value which should be an eigenvalue of the quantum field, right? I guess the eigenvectors of the quantum field also live in Fock space?

Answer

If I try to measure the field at one point in spacetime, I should get a real value which should be an eigenvalue of the quantum field, right? I guess the eigenvectors of the quantum field also live in Fock space?

Yes, that's basically correct. If the value of the field at a point is observable, the eigenvalues of the operator representing it are the values the field can attain at that point. And the eigenvectors live in the Hilbert space of states, which you can think of (at least conceptually) as $L^2(\{\mbox{initial boundary conditions}\})$. This Hilbert space is a Fock space in free field theories.

There's a couple of subtleties worth mentioning:

The value of the field at a point might not be a physical observable. In electrodynamics, for example, you can't actually measure the value $A_\mu(x)$ of a component of the connection 1-form; instead, you can measure gauge invariant quantities like the curvature $F_A(x)$ and the holonomy $Hol_L(A)$ along a loop $L$. Likewise, in nonlinear sigma models, where the classical fields are maps $\phi: \Sigma \to X$ to some curved manifold, you can't measure the value $\phi(x)$. Eigenvalues are complex numbers, not points on a manifold. But you do get a real observable $\mathcal{O}_f(x)$ for every function $f: X \to \mathbb{R}$; measure the value of $f(\phi(x))$.

It's also not strictly correct to say that quantum fields are operator-valued functions on spacetime. The physical problem is that if you measure the value of the field at one point, you'll disturb the field near that point, affecting the values at other nearby points. The closer you look to the place where you made the measurement, the bigger the disturbance; even in free scalar field theory, the 2-point correlation function $\langle \phi(x) \phi(y) \rangle$ blows up as $x \to y$. This tells you that the fields aren't quite functions, because you can't multiply the 'value at a point' observables when they live at the exact point.

The mathematically correct thing to do is to think of the field (and more generally local observables constructed from fields) as an operator-valued distribution. Distributions are mild generalization of functions; they are objects which don't have values at a point, but which do have average values in an arbitrarily small (but finite) region. Basically, for any test function $f$ on your spacetime, you get an operator $\phi(f)$ which you can think of as measuring the value "$\int f(x) \phi(x) dx$" of $\phi$ sampled by a probe with resolution $f$. Distributions can only be multiplied when their singularities don't coincide; they exhibit the same obnoxious behavior that quantum field operators do.

Probably you don't have to worry about this too much. For one thing, even if you can't (strictly speaking) define an operator $\phi(x)$, you can still safely talk about the correlation function $\langle \phi(x)\phi(y)\rangle$. (It's the kernel function of the multilinear map $(f,g)\mapsto \langle \phi(f)\phi(g) \rangle$.)

Physicists don't spend a lot of time worrying about solving the eigenvalue problem for the field operators. Usually the spectrum is all of $\mathbb{R}$, and finding the eigenvectors isn't worth the trouble. There is one important exception though: In the Standard Model, it's pretty important that the vacuum vector is an eigenvector of the Higgs field operators, with non-zero eigenvalue.

cipher - Hidden message #2

One day I was going to a temple. On the path I met an old person who was giving cards to everyone. On card there is a strange message written.

Will you help me to read the message?

Hint

Hidden message #1

Answer

This is a very partial answer. I used brute force to get this. Just hope it’s correct.

---

I think the answer is

Nice to meet you

When you decipher the pigpen, you get

RMGIUQQIIXULR

Then you do rot-22 (equivalent to taking each letter four letters back) and get

NICEQMMEETQHN

I used rot-3 on QM (basically taking each letter in UQ one letter back) to get

TP

And rot-7 on QHN (that’s the same as taking each letter in ULR three letters ahead) to get

XOU

Did a little change (P to O & X to Y) to give me

Nice to meet you

homework and exercises - Does conservation of angular momentum break conservation of momentum?

Say we have a spinning ring of mass $M$, rotating at $W_0$, at a radius $r$ from some pivot point. This ring has massless spokes extending out to a length of $2r$.

From this, we can calculate the initial angular momentum:

$L_0 = IW_0$

$I_0 = Mr^2$

$L_0 = Mr^2W_0$

While it's spinning, we drop on top of the massless spokes another ring of twice the radius, $2r$, but the same mass, $M$. Friction between the spokes and the ring makes sure the two rings spin together.

Conservation of angular momentum should allow us to find the final angular velocity of the two rings together:

$L_0 = L_f$

$I_fW_f = I_0W_0$

$I_f = Mr^2 + M(2r)^2 = 5Mr^2$

$5Mr^2W_f = Mr^2W_0$

$W_f = \frac{W_0}{5}$

However, I've always learned that conservation of angular momentum doesn't break conservation of momentum. That is, conservation of momentum should still hold.

However, if I try to solve this same problem using only conservation of linear momentum, I get a different result. Let $P$ stand for momentum, $V_{inner}$ stand for the final linear velocity of little segments of the inner disk, and $V_{outer}$ stand for the final linear velocity of little segments of the outer disk:

$P_0 = MV = MrW_0$

$P_0 = P_f$

$MV_0 = MV_{inner} + MV_{outer}$

$MrW_0 = MV_{inner} + MV_{outer}$

$MV_{inner} = MrW_f$

$MV_{outer} = M(2r)W_f$

$MrW_0 = MrW_f + M(2r)W_f$

$MrW_0 = 3MrW_f$

$W_f = \frac{W_0}{3}$

As you can see, conservation of linear momentum gives me a different result...what am I doing wrong?

Answer

Momentum is a vector quantity. The total momentum of a uniform spinning disc is zero, so the momentum begins at zero and ends at zero.

Behavior of propagator of quantum field as function of spin

Can we write down a generic expression for propagator for any arbitrary spin? At the least, about the ultraviolet behavior of these propagator. Especially, I would like to know whether there is a specific dependence on spin of the field.

Answer

It is possible to write an expression for the propagator of a massive spin $j$ particle whose ultraviolet limit can be shown to vary as $q^{-2+2j}$.

To see why this holds one has to find a natural generalization of Pauli matrices for $(j,0)$ and $(0,j)$ and of gamma matrices for the one where you deal with parity conserving interaction which involves fields transforming as $(j,0)\oplus (0,j)$ representation the details to which can be found in the reference mentioned.

Momentum space propagator of a particle of spin $j$ (eqn. 5.13 of reference which need not be mentioned in the context) has the structure involving a matrix $\Pi(q)$ written in terms of $2j+1$ dimensional matrix of finite Lorentz transformation $D^j[L (\bf {p})]$ where $L (\bf{ p})$ is lorentz boost (eqn. 4.3 of reference) in the numerator. Explicit calculation of matrix $\Pi(q)$ is given in the appendix of the reference.

In ultraviolet limit, the propagator behaves as $\Pi(q)/q^2$ where the value of monomial $\Pi(q)$ is given in table 1 of the reference. From the table it is easily seen that the monomial’s leading term in ultraviolet region goes as $q^0,q,q^2,q^3$ and so on for respectively $0,1/2,1,3/2$ spin particle and hence contributes a factor of $q^{2j}$ and in total propagator behaves as $q^{-2+2j}$ for a spin $j$ particle.

The case of $(j,0)\oplus (0,j)$ yields similar behavior for massive particle. This analysis actually breaks down for massless particles because of the non semi-simple structure of the little group associated with $m=0$ case and there is not a definite relation between massless particles propagators and the spin as photon and graviton both will behave as $j=0$ in the above relation.

S. Weinberg , Feynman rules for any spin. Phys. Rev. 133,1964.

thermodynamics - What is thermal radiation? How does it move through space?

Assume that by some mysterious(at the lack of a better word) way I was able to make a bonfire on the moon and was able to sit next to it without a space suit.

I will not be able to feel the heat form the fire as a result of convection as there is no atmosphere

Now my question is will I be able to feel the heat from the fire as a result of radiation? As far as I know a wood fire does not produce electromagnetic waves or does it?

In short what is thermal radiation?

Answer

I will not be able to feel the heat form the fire as a result of convection as there is no atmosphere

True, but no atmosphere also means no oxygen for the fire, so this example is kind'a off.

Now my question is will I be able to feel the heat from the fire as a result of radiation?

Yes.

If the fire was at Earth (in an atmosphere with oxygen) the strong heat you feel when holding your hands above the fire is from natural convection. But standing beside the fire, you still feel the heat. Convection cannot be the cause of this, if no wind is moving hot air your way - instead this is due to thermal radiation.

As far as I know a wood fire does not produce electromagnetic waves or does it?

Yes, it does.

Any material with a non-zero temperature produces thermal radiation - this is simply a type of electromagnetic wave. You own body radiates heat, and anything else you see around you radiates heat. Stefan-Boltzmann's law describes this:

$$I=\epsilon \sigma T^4$$

($I$ is heat from each square meter $[W/m^2]$ as thermal radiation, $\epsilon$ the emmisivity (how well the object radiates heat), $\sigma$ the Boltzmann's constant, $T$ the temperature.)

In short what is thermal radiation?

The thermal radiation is electromagnetic waves as any other kind of radiation. If you can see fire, sparks, lightning etc. the temperature is so that the electromagnetic waves are in the visible spectrum. As a red-hot iron bar. If not, then you can't see the radiation - but it still is there outside of the visible spectrum (in the infrared part most likely).

One can from Wien's displacement law see that higher temperature apart from radiating more strongly, also shifts the wavelenghts of the radiation towards lower values - towards the visible spectrum:

What effects could/can gravitational waves have on us?

Are gravitational waves something a human beeing could notice if it were powerful enough? Or would it be more like a higher dimensional force, since it would alter us in the sameway as our surroundings so as my eye deforms in the same relation to the enviroment it would anyway the enviroment capture like we are used to?

To calrify what I'm interested in:

Let us assume our earth wouldn't be orbiting sun, but instead be orbiting (or what ever it would be) 2 merging black holes with each of mass ~50¹ solar masses with earth having a distance of lets say 5000 kilometers when the merging occurs.

Please just ignore all other none GW caused effects that would have extinguished earth allready anyway.

So when this merging event occurs:

Would we be able to notice² the shapeshifting caused by the waves or would it be in an higherdimensional sense not be notable for us as it is not changing shape in any irregular manner?

---

¹_{I used this figure in hope it would cause an effect strong enough to express what I'm interested in.}

²_{May it be pain caused by the deforming of our body or altered vision by seeing the objects stretching happen.}

homework and exercises - Free Particle Path Integral Matsubara Frequency

I am trying to calculate $$Z = \int\limits_{\phi(\beta) = \phi(0) =0} D \phi\ e^{-\frac{1}{2} \int_0^{\beta} d\tau \dot{\phi}^2}$$ without transforming it to the Matsubara frequency space, I can show that $Z = \sqrt{\frac{1}{2\pi \beta}}$. However, I have a problem in obtaining the same result in the Matsubara frequency space: \begin{equation} \phi (\tau) = \frac{1}{\sqrt{\beta}} \left( \sum_{n} \phi_n \ e^{i\omega_n\tau} \right), \end{equation} with $\sum_n \phi_n =0, \omega_n = \frac{2\pi n}{\beta}$. And \begin{equation} Z = \int \prod_n D\phi_n\ \delta\left(\sum_n \phi_n\right)\ e^{-\frac{1}{2} \sum_n \phi_n \phi_{-n} \omega_n^2 } \end{equation} which, I think, vanishes.

I guess the problem lies in the measure. Any comments?

Info: I write the Schulman's derivation in imaginary time here. \begin{eqnarray} Z &=& \int\limits_{\phi(0) =\phi(\beta) = 0} D\phi(\tau) e^{-\frac{1}{2}\int_0^{\beta}d\tau\dot{\phi}^2}\\ &=& \text{lim}_{N \rightarrow \infty} (\frac{1}{2\pi \epsilon})^{(N+1)/2} \int d\phi_1 \dots d\phi_N e^{-\frac{1}{2\epsilon} \sum_{i =0}^N (\phi_{i+1} -\phi_i)^2} \end{eqnarray}

Then, we can use the identity \begin{equation} \int_{-\infty}^{\infty} du \sqrt{\frac{a}{\pi}} e^{-a(x-u)^2}\sqrt{\frac{b}{\pi}} e^{-b(u -y)^2} = \sqrt{\frac{ab}{\pi(a+b)}} e^{-\frac{ab}{a+b}(x-y)^2} \end{equation} to evaluate the sum to be \begin{equation} Z = \sqrt{\frac{1}{2\pi \beta}}. \end{equation}

gravity - Potential well for gravitational waves

Can one consider the gravitational field of a gravitating body such as a planet or a star as a potential well for gravitational waves? In other words, would it be possible for such a gravitating body to capture gravitational waves in some bound state, similar to the way electrons exist in bound states around the nucleus in atoms or to the way light can be captured in resonant cavities?

Answer

We generally calculate the motion of gravitational waves using an approximation called linearised gravity. With this approach gravitational waves behave just like light does so it can't be bound in a gravitational potential well any more than light can.

Just like light, gravitational waves cannot escape from behind an event horizon, and they could in principle be captured in a circular orbit (called the photon sphere) though this orbit is unstable. But neither of these really count as a bound state.

When you say light can be captured in resonant cavities I'd guess you're thinking of waveguides. These work because EM waves interact very strongly with the conduction electrons in the metal, but gravitational waves interact so weakly with matter that a gravitational waveguide isn't possible.

The linearised gravity approximation I mentioned above ignores the gravitational field produced by the energy of the waves themselves. If instead we use a full calculation it has been suggested that sufficiently intense gravitational waves can form a bound state called a geon. It has been proven that such states can exist but it is currently unknown if they are stable.

quantum mechanics - Wave equations for two intervals at Potential step

Lets say we have a potential step as in the picture:

In the region I there is a free particle with a wavefunction $\psi_I$ while in the region II the wave function will be $\psi_{II}$.

Let me take now the schrödinger equation and try to derive $\psi_I$:

\begin{align} &~~W \psi = -\frac{\hbar^2}{2m}\, \frac{d^2 \Psi}{d\, x^2} + W_p \psi ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\nonumber \\ &~~W \psi = -\frac{\hbar^2}{2m}\, \frac{d^2 \Psi}{d\, x^2}\nonumber \\ &\frac{d^2 \Psi}{d\, x^2} = -\frac{2m W}{\hbar^2}\,\psi \nonumber\\ {\scriptsize \text{DE: }} &\boxed{\frac{d^2 \Psi}{d\, x^2} = -\mathcal L\,\psi}~\boxed{\mathcal{L} \equiv \sqrt{\tfrac{2mW}{\hbar^2}}} \nonumber\\ &~\phantom{\line(1,0){18.3}}\Downarrow \nonumber\\ {\scriptsize \text{general solution of DE: }} &\boxed{\psi_{I} = C \sin\left(\mathcal{L}\, x \right) + D \cos \left(\mathcal{L}\, x \right)}\nonumber \end{align}

I got the general solution for the interval I, but this is nothing like the solution they use in all the books: $\psi_{I} = C e^{i\mathcal L x} + D e^{-i \mathcal L x} $ where $\mathcal L \equiv \sqrt{{\scriptsize 2mW/\hbar^2}}$. I have a personal issue with this because if $x= -\infty$ part $De^{-i \mathcal L x}$ would become infinite and this is impossible for a wavefunction! I know that i would get exponential form if i defined constant $\mathcal L$ a bit differently as i did above:

\begin{align} {\scriptsize \text{DE: }} &\boxed{\frac{d^2 \Psi}{d\, x^2} = \mathcal L\,\psi}~\boxed{\mathcal{L} \equiv -\sqrt{\tfrac{2mW}{\hbar^2}}} \nonumber\\ &~\phantom{\line(1,0){18.3}}\Downarrow \nonumber\\ {\scriptsize \text{general solution of DE: }} &\boxed{ \psi_{I} = C e^{\mathcal L x } + D^{-\mathcal L x} }\nonumber \end{align}

This general solution looks more like the one they use in the books but it lacks an imaginary $i$ and $\mathcal L$ is defined with a - while in all the books it is positive. Could anyone tell me what am i missing here so i could understand this?

optics - If refraction slows down light, isn't it possible to hold light still?

I have a quick question about the refraction of light, and I'm sorry if it seems a bit simplistic or even stupid, but I'm still learning.

We know that when light passes through a denser medium, it slows down. However, is there a limit to that slowing down? If no, wouldn't it be possible, with some type of medium we make ourselves (like a crystal), to decrease the light's speed down to a point where we would be able to "see the photons moving forwards"?

I know it is way more complicated than that and that you can't see light unless some other light bounces off it, what I mean is slowing down light to an extreme and unusual speed). That sounds completely absurd but I can't seem to find what is wrong. Can anyone can explain it to me? :)

Answer

Yes, light can be brought to a complete halt under the right conditions.

To understand how this happens you need to understand what is going on when light slows down in a medium. Light is an oscillating electromagnetic field, and when it passes though anything that contains charged particles (i.e. any matter made from electrons and protons) the electric fied of the light interacts with those charges.

When the light interacts with the charges we have to describe the light/matter system by a new wavefunction that includes all the interacting components. This means the light is not longer purely light - we have a quantum system that mixes up the light with the charged particles. If this mixing is very strong it produces a quasiparticle called a polariton. The polariton is a massive particle so it moves at less than the speed of light.

Under normal circumstances, e.g. light passing through glass, the interaction of the light and the medium is relatively weak and we probably wouldn't use a polariton description. Instead we'd use a classical description as described very nicely in the answers to Why do prisms work (why is refraction frequency dependent)?.

However it is possible to make systems that interact very, very strongly with the light. Specifically Bose-Einstein condensates. This interaction can be so strong that the polaritons become essentially motionless i.e. the light is brought to a stop. This was first achieved in 2001 by a group at the Harvard-Smithsonian Center for Astrophysics, who managed to bring light to a stop in a gas of rubidium atoms. At about the same time a group in Harvard achieved the same result using a Bose-Einstein condensate made from sodium atoms.

The experiments mentioned above require rather extreme conditions to work, but they involve basically the same physics that causes light to slow down by a factor of about 1.5 in glass.

mathematics - Coin weighing with a single weighing device

You have 12 coins which each weigh either 20 grams or 10 grams. Each is labelled from 1 to 12 so you can tell the coins apart. You have one weighing device as well. At each turn you can put as many coins as you like on the weighing device and it will tell you exactly how much they weigh.

What is the minimum number of weighings that will always tell you exactly which coins weigh 10 grams and which weigh 20?

Clearly you can do it in 12 weighings by just weighing each coin separately.

---

There is now a solution using $7$ static weighings by Julian Rosen!

Current record found by computer search is $7$ (dynamic) weighings by Victor.

There is a hand developed record of $8$ (dynamic) weighings by Joel Rondeau.

---

Hint

Here are the first three lines of a solution that uses $7$ weighings.

0 0 1 0 1 1 1 0 0 0 0 0
1 0 0 1 0 1 1 1 0 0 0 0
1 1 0 0 1 0 1 1 1 0 0 0

---

Bounty awarded to Victor for the first correct answer. Julian Rosen gave a static solution with $7$ weighings but without an explanation yet (just external references).

I will award the "win" to the first static solution with $7$ weighings and an explanation.

Answer

I decided to try a different approach: Brute force.

So, I created a Java 8 program to brute-force the creation of a decision tree in order to find how to weight the 12 coins in 7 measurements, and it worked. Further, I could use it to prove that this is the optimum, because it said that with 6 there is no solution.

The program found a solution in 35 seconds on my computer. There are many solutions, most due to permutations.

The program creates a tree of nodes, where each one iterates through all 4096 possibles forms of choosing the coins, but it do variable renaming to eliminate permutations, i.e, this eliminates all but a tiny part of the permutations.

Then, for each measurement form, it simulates all the possibles combinations of light-or-heavy coins, already excluding possibilities eliminated by previous nodes in the tree. Then it checks if the generated node is able to separate different combinations, putting those in child nodes.

When a node passes the maximum depth (i.e. the maximum number of measurements), it is considered bad and is pruned. If it results in a single possibility, no further measurements are needed. If the node created only one child node, then the coin combination for the measurement is useless and the node is considered bad. A node that has bad nodes as children is considered bad and is pruned, even if not all of its child nodes were checked.

If a node tries every possible combination of coins to measure and is unable to determine all the possibilities uniquely, it is considered bad.

In the end, after encountering the solution, it is outputted.

I uploaded the human readable output to THIS LINK.

I also uploaded a JSON-formatted output IN THIS OTHER LINK.

I couldn't post either output here because they are too large. If you don't like the fact that the solution is provided in links, know that I don't like either, but again, it is simply too large to post here. You may just compile and run the program instead.

This is the program:

import java.util.Arrays;
import java.util.Set;
import java.util.TreeSet;

/**
 * @author Victor
 */
public class BruteForce {
    // Change this to false to output in pretty-print mode.

    private static final boolean COMPACT = true;

    private final int coins;
    private final int maxDepth;
    private final int possibilities;

    public final class Node {
        private final boolean[] combs = new boolean[possibilities];
        private final boolean[] shouldWeight = new boolean[possibilities];
        private Node[] children = null;

        private int counts = 0;
        private final int level;
        private int lastCount = -1;
        private final Node parent;
        private Set coinRenames;
        private int whichCoinsToWeight = -1;
        private int tag; // For writing it in some human readable format. Not used for creating and processing the tree.

        public Node() {
            this(0, null);

        }

        private Node(int level, Node parent) {
            this.parent = parent;
            this.level = level;
            if (parent != null) return;
            for (int i = 0; i < possibilities; i++) {
                count(i);
            }
        }


        public void count(int c) {
            combs[c] = true;
            counts++;
            lastCount = c;
            if (coinRenames == null) coinRenames = new TreeSet<>();
            int renamed = renameCoins(c);

            // Not really important, but if you want a solution in a few seconds instead of
            // waiting for some hundreds of years, keep this check here.

            if (coinRenames.add(renamed)) {
                shouldWeight[c] = true;
            }
        }

        // Convert an int to an array of boolean.
        private boolean[] asArray(int coinSet) {
            boolean[] array = new boolean[coins];
            for (int i = 0; i < coins; i++) {
                 array[i] = ((coinSet >> i) & 1) != 0;

            }
            return array;
        }

        private void summarizeCoinsUsedForWeighting(boolean[][] weightingCoinsPerLevel) {
            weightingCoinsPerLevel[level] = asArray(whichCoinsToWeight);
            if (parent != null) parent.summarizeCoinsUsedForWeighting(weightingCoinsPerLevel);
        }

        private int renameCoins(int thisLevelCoins) {

            // This arrays represent, for each level of the tree that is an ancestor to this node
            // the set of coins used for weighting, where each int corresponds to a sequence of
            // 0 and 1 bits representing which coins are being used.
            boolean[][] weightingCoinsPerLevel = new boolean[level + 1][coins];
            if (parent != null) parent.summarizeCoinsUsedForWeighting(weightingCoinsPerLevel);
            weightingCoinsPerLevel[level] = asArray(thisLevelCoins);

            // Transpose the array, so we have a new array, where for each coin, we have an int corresponding
            // to a sequence of 0 and 1 bits where the 1 bits are the levels in which the coin is used for weighting.
            // But since we do want that the levels near the root of the tree are represented by the most

            // significative bits, we reverse the bit order of the levels.
            int[] usingLevelsPerCoin = new int[coins];
            for (int i = 0; i <= level; i++) {
                for (int j = 0; j < coins; j++) {
                    if (weightingCoinsPerLevel[i][j]) usingLevelsPerCoin[j] |= 1 << (level - i);
                }
            }

            // Rename the coins to ensure that the most firstly-used coins are the first ones.
            // Coin-renaming serves to avoid recalculating combinations that are simply permutations to some

            // previously calculated position. Further, this coin renaming provides a single canonical representation
            // for each group of equivalent permutations. To ensure that, we simply sort the array, so the most used
            // coins with higher bits sets will be the LAST ones, I.E, the resulting array would still be in the reverse order.
            Arrays.sort(usingLevelsPerCoin);

            // We could untranspose the sorted array, so we have again coins per level, but with the coins are renamed.
            // However, now we are interested only in the last level, so we just get the last bit from each row on the array.
            // Further, we reverse the order of the bits that we get because the array is sorted in reverse order. This way it
            // will be in the right order again.
            int sortedCoinsOnLastLevel = 0;

            for (int j = 0; j < coins; j++) {
                sortedCoinsOnLastLevel |= (usingLevelsPerCoin[j] & 1) << (coins - j - 1);
            }

            return sortedCoinsOnLastLevel;
        }

        public boolean separate() {
            if (counts == 1) return true; // Just one possibility, it is in the lastCount variable.


            // If is recursing too deply and there is still at least 2 coins left in this possibility,
            // so there was a bad choice somewhere up in the decision tree.
            if (level >= maxDepth) return false;

            // The coins choosen to weight are the 1 bits. Start counting with 1 because there is no sense to weight zero coins.
            // This means that it will choose every possible combination of coins to weight, until it founds a good one or resigns.
            // Save it in an instance field, so it can be remembered when the solution is found.
            a: for (whichCoinsToWeight = 1; whichCoinsToWeight < possibilities; whichCoinsToWeight++) {

                // If this is just a permutation of some previously tested combination, just skip it.

                if (!shouldWeight[whichCoinsToWeight]) continue;

                children = null;
                int childCount = 0;

                // Iterate every possible valuation for the coins.
                b: for (int selectedCoinsCombination = 0; selectedCoinsCombination < possibilities; selectedCoinsCombination++) {

                    // If the choosen valuation never happens in this place, skip it.
                    if (!combs[selectedCoinsCombination]) continue;


                    // Weight the coins. This works by setting 0 to coins which
                    // are not being weighted, leaving 0 for light coins and 1 for heavy ones.
                    int bits = Integer.bitCount(whichCoinsToWeight & selectedCoinsCombination);

                    // Now start to populate the children. If it does not have any, create the array.
                    if (children == null) children = new Node[coins + 1];

                    // Choose the appropriate children node in the decision tree.
                    Node child = children[bits];


                    // If the selected node still do not exist, create it.
                    if (child == null) {
                        child = new Node(level + 1, this);
                        children[bits] = child;
                        childCount++;
                    }

                    // Add the selected coins combination as possible to the child node.
                    child.count(selectedCoinsCombination);

                }

                // If we have just one child, then the selected coins to weight did not gave any information,
                // so discard this choice and proceed to the next one.
                if (childCount <= 1) continue;

                // Now, perform the separation recursively in each children.
                // If some of them fails, then the selected coins to weight combination is useless and should
                // be discarded, even if just some of the nodes were tested.
                for (Node n : children.clone()) {

                    if (n == null) continue;
                    boolean r = n.separate();
                    if (!r) continue a;
                }

                // Yeah, found some useful combination!
                return true;
            }

            // We tried every combination of coins to weight and none of them was useful.

            return false;
        }

        // Output a JSON!
        public void reportResult(StringBuilder sb) {
            boolean first = true;
            sb.append("{");
            if (counts == 1) {
                sb.append("\"HEAVY\": [");
                for (int i = 0; i < coins; i++) {

                    if (((lastCount >> i) & 1) != 0) {
                        if (first) {
                            first = false;
                        } else {
                            sb.append(", ");
                        }
                        sb.append(i + 1);
                    }
                }
                sb.append("], \"LIGHT\": [");

                first = true;
                for (int i = 0; i < coins; i++) {
                    if (((lastCount >> i) & 1) == 0) {
                        if (first) {
                            first = false;
                        } else {
                            sb.append(", ");
                        }
                        sb.append(i + 1);
                    }

                }
                sb.append("]}");
                return;
            }
            if (!COMPACT) {
                sb.append("\n");
                ident(level + 1, sb);
            }
            sb.append("\"TO WEIGHT\": [");
            for (int i = 0; i < coins; i++) {

                if (((whichCoinsToWeight >> i) & 1) != 0) {
                    if (first) {
                        first = false;
                    } else {
                        sb.append(", ");
                    }
                    sb.append(i + 1);
                }
            }
            sb.append("]");

            for (int i = 0; i <= coins; i++) {
                if (children[i] == null) continue;
                sb.append(",\n");
                ident(level + 1, sb);
                Integer weighting = Integer.bitCount(whichCoinsToWeight);
                sb.append("\"").append((i + weighting) * 10).append(" grams\": ");
                children[i].reportResult(sb);
            }
            if (!COMPACT) {
                sb.append("\n");

                ident(level, sb);
            }
            sb.append("}");
        }

        private void ident(int level, StringBuilder sb) {
            for (int i = 0; i < level; i++) {
                sb.append(COMPACT ? " " : "  ");
            }
        }


        // Output a human readable algorithm!
        public void reportAlgorithm(StringBuilder sb) {
            this.tag = 1;
            for (int i = 0; i < maxDepth; i++) {
                reportAlgorithm(new int[maxDepth], i, sb);
            }
        }

        private void reportAlgorithmDone(StringBuilder sb) {

            if (counts != 1) return;
            boolean first = true;
            sb.append("Heavies = [");
            for (int i = 0; i < coins; i++) {
                if (((lastCount >> i) & 1) != 0) {
                    if (first) {
                        first = false;
                    } else {
                        sb.append(", ");
                    }

                    sb.append(i + 1);
                }
            }
            sb.append("]; Lights = [");
            first = true;
            for (int i = 0; i < coins; i++) {
                if (((lastCount >> i) & 1) == 0) {
                    if (first) {
                        first = false;
                    } else {

                        sb.append(", ");
                    }
                    sb.append(i + 1);
                }
            }
            sb.append("].\n");
        }

        private void reportAlgorithm(int[] levelCounters, int desiredLevel, StringBuilder sb) {
            boolean first = true;

            if (counts == 1) {
                reportAlgorithmDone(sb);
                return;
            }
            if (level < desiredLevel) {
                for (int i = 0; i <= coins; i++) {
                    if (children[i] != null && children[i].counts != 1) children[i].reportAlgorithm(levelCounters, desiredLevel, sb);
                }
                return;
            }

            sb.append(level + 1);
            if (level > 0) sb.append(".").append(tag);
            sb.append(". Weight the coins [");
            for (int i = 0; i < coins; i++) {
                if (((whichCoinsToWeight >> i) & 1) != 0) {
                    if (first) {
                        first = false;
                    } else {
                        sb.append(", ");
                    }

                    sb.append(i + 1);
                }
            }
            sb.append("]\n");
            for (int i = 0; i <= coins; i++) {
                if (children[i] == null) continue;
                Integer weighting = Integer.bitCount(whichCoinsToWeight);
                sb.append("  ").append((i + weighting) * 10).append(" grams: ");
                if (children[i].counts == 1) {
                    children[i].reportAlgorithmDone(sb);

                } else {
                    children[i].tag = ++levelCounters[level + 1];
                    sb.append("Go to ").append(level + 2).append(".").append(children[i].tag).append(".\n");
                }
            }
            sb.append("\n");
        }
    }

    public BruteForce(int coins, int maxDepth) {

        this.coins = coins;
        this.maxDepth = maxDepth;
        this.possibilities = 1 << coins;
    }

    public void run() {
        Node n = new Node();
        boolean ok = n.separate();
        if (!ok) {
            System.out.println("Impossible");

            return;
        }

        // Uncomment this to output as JSON.
        //StringBuilder sb = new StringBuilder(1024 * 1024);
        //n.reportResult(sb);
        //System.out.println(sb.toString());

        StringBuilder sb2 = new StringBuilder(1024 * 1024);
        n.reportAlgorithm(sb2);

        System.out.println(sb2.toString());
    }

    public static void main(String[] args) {
        int coins = 12; // You may change the number of coins if you want.
        int measures = 7; // You may change the number of measures, if you want.
        BruteForce b = new BruteForce(coins, measures);
        b.run();
    }
}

^{I am sorry that this site does not have syntax-coloring enabled.}

You may modify the program as you wish, specially to brute-force different combinations of number of coins and number of measurements. This is cool because you may also use this program to solve any number of desired coins and measurements (as long as you have enough time and memory).

If you change the int measures = 7; near the end of the program to 6, it will output that it is impossible, proving that 7 are indeed the minimum.

You may uncomment the code in the run() method to get a compact JSON output. You may also change the value of the COMPACT variable in the beginning to false and then the JSON output will be a better looking one, but will be still larger. You may get that JSON and use it to work with something else. Or you may simple keep just the human readable output.

You may argue that this is a very inelegant, ugly and stupid way of solving this problem, as the output of the program is basically a bunch of random-looking arbitrary measurements without any clear, simple or intuitive pattern. I agree, but nevertheless, the problem is solved: Just follow the measurements show in the output and you will find the weight for any combination of 10 or 20 grams coins in only 7 measurements.

measurements - Can a ultracentrifuge be used to test general relativity?

With today's ultracentrifuge technology, they can spin so fast that the sample can be subjected to accelerations of up to 2 millions Gs. That is equivalent to two solar masses. Has someone tried to measure the time dilation in a radioactive sample? How calculate that time dilation respect to the time outside the ultracentrifuge, for example one week. I think that that time dilation will be significant enough to be measured.

physical constants - Is there a limit to acceleration?

As we all know the speed of light is the limit at which energy/matter can travel through our universe.

My question being: is there a similar limit for acceleration? Is there a limit to how quickly we can speed something up and if so why is this the case?

Short question, not sure what else I can elaborate on! Thanks in advance!

pressure - What exactly happens to an exposed human body in space?

I know that space is a vacuum, has no air, and is very, very cold. From what I have seen in movies and TV shows when a space suit is exposed the oxygen will escape like a visible gas and the person would explode.

In the movie Event Horizon one of the characters is blown out of an airlock and one of his comrades tells him to expel all the air in his lungs and to curl up, as he does he seems to be bleeding a lot.

In the movie The Hitchhikers Guide to the Galaxy it says you can hold your breath for a little while before anything happens. Of course all these examples are science fiction so they may or may not be accurate.

I am wondering what would happen if a human body was exposed in space? Would it blow up like so many shows seems to hint? Would the person be frozen to death because of the temperature in space?

NOTE: I ask this as in a fan fiction I am writing, in one scene the main character's daughter is sucked out through a hull breach of the ship before her friend willingly jumps out and holds her as she teleports the two through space to the nearest planet. I want to explain how the adverse effects on an exposed human body are negated, but I can't do that if I don't know what I am negating.

Answer

Unlike most depictions that you see in movies or heard of maybe, the human body can actually maintain its stability for a short while. I am not sure of exactly how long it takes for a permanent injury to occur or swellings to start appearing, but what is known is that there's no immediate injury.

Meaning you do not explode, your blood doesn't boil, nor does it freeze, and you don't lose consciousness instantly. Why? there are two important factors in play:

• Containing effect of your skin and your circulatory system allow us to withstand the instantaneous drop of pressure for a while.


• You do not instantly freeze because, although the space environment is extremely cold (close to 0 Kelvin), heat does not transfer away from your body quickly. In fact space is mostly a vacuum and can hardly transfer heat, so you can imagine that the main temperature worry for space suits is how to get rid of naturally generated body heat, not keeping heat inside!

Finally, surprisingly one actually dies of asphyxiation. Air would immediately leave the lungs due to the enormous difference of pressure between your body and the vacuum around you. Then any oxygen dissolved in the blood would empty into the lungs to try to equalize the partial pressure difference, and your brain starts to die.

In a nutshell, our skin and lack of matter(no atoms/molecules: exchange kinetic energy with what? of course radiation still takes place) in space make the instantaneous effects impossible.

For further reading: Human body in a vacuum

Are there examples in classical mechanics where D'Alembert's principle fails?

D'Alembert's principle suggests that the work done by the internal forces for a virtual displacement of a mechanical system in harmony with the constraints is zero.

This is obviously true for the constraint of a rigid body where all the particles maintain a constant distance from one another. It's also true for constraining force where the virtual displacement is normal to it.

Can anyone think of a case where the virtual displacements are in harmony with the constraints of a mechanical system, yet the total work done by the internal forces is non-zero, making D'Alembert's principle false?

waves - Why we don't use gamma rays, x-rays or ultraviolet to transmit data?

The greater the frequency range of a transmission medium, the greater the number of bits per second it can transmit. In other words, the bigger the bandwidth in hertz available, the bigger the bandwidth in bits per second that can be transmitted.

Given that, why we don't use gamma rays, x-rays or ultraviolet to transmit data instead of e.g. visible(light) in optical fiber or microwaves(cell phones)?

pattern - What is an R-complete Word™?

If a word conforms to a special rule, I call it an R-complete Word™.

Use the examples below to find the rule.

$$ % set Title text. (spaces around the text ARE important; do not remove.) % increase Pad value only if your entries are longer than the title bar. % \def\Pad{\P{0.0}} \def\Title{\textbf{ R-complete }} % \def\S#1#2{\Space{#1}{20px}{#2px}}\def\P#1{\V{#1em}}\def\V#1{\S{#1}{9}} \def\T{\Title\textbf{Words}^{\;\!™}\Pad}\def\NT{\Pad\textbf{Not}\T\ }\displaystyle \smash{\lower{29px}\bbox[yellow]{\phantom{\rlap{rubio.2019.05.15}\S{6px}{0} \begin{array}{cc}\Pad\T&\NT\\\end{array}}}}\atop\def\V#1{\S{#1}{5}} \begin{array}{|c|c|}\hline\Pad\T&\NT\\\hline % \text{TAX} & \text{TRIBUTE}\\\hline \text{NEEDLE} & \text{PIN}\\\hline \text{SUN} & \text{MOON}\\\hline \text{BEST} & \text{WORST}\\\hline \text{PUZZLE} & \text{RIDDLE}\\\hline \text{COFFEE} & \text{TEA}\\\hline \text{SCIENCE} & \text{PHILOSOPHY}\\\hline \text{COLUMN} & \text{ROW}\\\hline \text{CARTOONS*} & \text{SERIES}\\\hline \text{HEAVEN} & \text{HELL}\\\hline \text{TRUCK} & \text{LORRY}\\\hline \end{array}$$ * This is a Double R-complete Word™

CSV Version:

R-complete Words™,Not R-complete Words™
TAX,TRIBUTE
NEEDLE,PIN

SUN,MOON
BEST,WORST
PUZZLE,RIDDLE
COFFEE,TEA
SCIENCE,PHILOSOPHY
COLUMN,ROW
CARTOONS,SERIES
HEAVEN,HELL
TRUCK,LORRY

What is the rule?
There are many more R-complete Words™

Hints: A hint will be given every 50 views for the first three hints. After that, 1 hint every 100 views or 15 votes.

---

Hint #1 (50 views):

- It is also called 3R word because it is based on a 3-component score (Bonus: "Truck" is R-Complete and "Lorry" is not)

Hint #2 (100 views):

The R-score is defined as $$R = (R_1, R_2, R_3)$$ and a word is R-complete if $$ \min(R) > 0 $$

Hint #3 (150 views):

- Here is a hint in the form of mini-puzzle
- Bonus: Answering @rand-althor question, Yes, an anagram of the word will keep its property, it's still a 3-complete word.

Hint #4 (250 views):

More examples:

- R-Complete, Not R-Complete
- Ant, Bee
- Bull, Cow
- Camera, Picture
- Double, Triple

Answer

I think an R-complete word is one which:

Uses letters from all 3 rows on a standard QWERTY keyboard.

In each of the counter-examples above:

There is at least one row of the keyboard which is not used.

If we:

Label letters from each row as (T)op, (M)iddle or (B)ottom and count up how many of each are used in a word then you can produce the 3-part 'R-score', R(T,M,B) like so:

TAX = TMB = R(1,1,1)
NEEDLE = BTTMMT = R(3,2,1)
SUN = MTB = R(1,1,1)

BEST = BTMT = R(2,1,1)
PUZZLE = TTBBMT = R(3,1,2)
COFFEE = BTMMTT = R(3,2,1)
SCIENCE = MBTTBBT = R(3,1,3)
COLUMN = BTMTBB = R(2,1,3)
CARTOONS = BMTTTTBM = R(4,2,2)
HEAVEN = MTMBTB = R(2,2,2)
TRUCK = TTTBM = R(3,1,1)

As per the second hint, you can see that:

The minimum value in brackets in each of these is 1 or more.

In contrast, for the counterexamples:

TRIBUTE, PIN and MOON have no letters from the middle row,
WORST, RIDDLE, TEA, PHILOSOPHY, SERIES, HELL and LORRY have no letters from the bottom row,
ROW has no letters from the middle or bottom rows.

Hence, each of these will contain a '0' somewhere in their R(T,M,B) triplet, making min(R)=0 and ensuring these words are not 'R-complete'.

The 'R' in R-complete most probably stands for:

ROWS

cosmology - Dark Matter vs Modified Gravity

Why do cosmologists and astrophysicists assume that the reason for the higher velocities of outer stars in galaxies is due to matter at all? The name dark matter seems misleading. Couldn't gravity just work different on larger scales? After years of searching for a dark matter 'particle', we still have nothing. I'm a little puzzled at the constant search for a weakly interacting particle.

Topics in particle cosmology

I am interested in learning more about this interdisciplinary approach.

1) What are some of the top questions in particle cosmology (e.g nature of dark matter, inflationary structure, topological defects in the universe)?

2) Who are some of the top researchers in this field?

3) Recommended review papers (on arxiv, for instance) will be greatly appreciated.

astrophysics - What types of fusion reactions happened in population III stars?

I have read that, in smaller stars, such as our Sun, the fusion reaction that takes place is a proton-proton chain, or PP chain for short.

From what I have learned, in larger stars, a different process takes place, known as the CNO cycle, in which carbon, nitrogen and oxygen are utilized as a catalyst.

The first stars were many times more massive than the Sun, but pretty much the only elements they had to make use of were hydrogen and a little bit of helium. So, what types of fusion reactions went on in the first stars?

Answer

I think you already know the answer...

Pop III stars, by definition, are born from primordial gas that is basically Hydrogen, Helium with trace amounts of deuterium, tritium, lithium and beryllium; they initially contain almost no C, N, or O. Therefore the primary fusion in massive Pop III stars has to be (well, initially the deuterium is burned but this is not energetically important) the pp chain, which produces more helium from hydrogen. The lower temperature dependence of the pp chain means that the core can get hotter and denser than would be usual in more metal-rich massive stars.

It is not until He burning is initiated that C, N, and O can form, but this actually takes place before the red giant tip is reached in Pop. III stars.

Your recent questions all have a theme. You might find this presentation I came across interesting.

EDIT: Further to this: Current thinking is that CNO burning will take place on the main sequence for Pop. III stars more massive than $20 M_{\odot}$. This is because they cannot be supported by the pp chain alone; they contract and heat up sufficiently to start the triple alpha He burning phase that produce carbon; once the carbon concentration builds up to about $10^{-10}$ of hydrogen then the CNO cycle takes over (Ekstrom et al. 2008; Yoon et al. 2012).

quantum mechanics - Probability current density : Isn't there something wrong with this proof?

Is there something wrong with the following proof (see below)? To me, it seems like the third line should show $$\frac{dP_{ab}}{dt}=-\int_a^b \frac{\partial}{\partial t}J(x,t)dx$$ Am I missing something obvious?

---

Answer

If we look at line 2, we have an integral set equal to:

We then must note that $J(x,t)$ is defined as the negative of:

So the last equation on line 2 should be: $-\frac{\partial J(x,t)}{\partial x}$ as opposed to $-\frac{\partial J(x,t)}{\partial t}$.

Do photons have mass?

As a student in a highschool physics class, my teacher has repeatedly told me that photons are massless. Yet, I have also heard from other sources that photons have momentum. If photons were to have momementum, that would mean that they have mass as according to p = mv.

Do photons really have mass?

Also, how would this mass be calculated?

newtonian mechanics - How much thrust would be needed to turn a hobbyist weather balloon into a deep space probe?

I was reading the article Weather Balloon Space Probes that says you can put your own balloon probe at 65,000 ft temporarily.

Is it even remotely possible to raise the probe high enough using balloons, and then put it in orbit or escape the field using an on-board propulsion system?

Can it be done with easily purchased materials?

How much mass can be added for instrumentation?

How much fuel of is required? What kind?

I know it's kind of a broad question, but I just want to know a rough estimate.

electromagnetism - Classical explanation for volume conservation in magnetorestriction?

This is a follow-up to this question, on a more specific theoretical point. Magnetostriction is the phenomenon discovered by James Joule in which magnetic substances experiences stress when exposed to a magnetic field. Joule says:

About the close of the year 1841, Mr. F.D. Arstall, an ingenious machinist of this town, suggested to me a new form of electro-magnetic engine. He was of the opinion that a bar of iron experienced an increase of bulk by receiving the magnetic condition.

Joule then describes an experiment in which he detected an increase in length of an iron bar in the direction parallel to the field, and then a follow-up in which he immersed the bar in water and tested for a volume change using a capillary tube. He found no volume change, implying that each of the transverse dimensions changed by $-1/2$ the fractional change seen in the longitudinal direction.

The following graph (Great Soviet Encyclopedia, 1979) seems to show that the volume change is only zero for small fields. The slopes of the two curves seem differ by a factor of about $-1/2$ up to a field of about $1.5\times10^3$ oersteds. Assuming a permeability of about $10^4$, this is $B\sim10^7$ gauss, which I think is of the same order of magnitude as the nominal saturation field.

It seems to be a big deal or possibly controversial to see a volume change. (Chopra and Wuttig, Nature 521, 340–343 (2015)), and it only seems to happen in these very strong fields.

It seems clear that according to classical E&M we should expect a longitudinal compression and a transverse expansion, since the stress-energy tensor $T^\mu{}_\nu$ of a uniform field in the $x$ direction is proportional to $\operatorname{diag}(1,1,-1,-1)$ ($+---$ metric, $txyz$ coordinates). However, I would think that the volume would change, since the 3x3 stress tensor looks like $\operatorname{diag}(1,-1,-1)$ (as required by $T^\nu{}_\nu=0$), not $\operatorname{diag}(1,-1/2,-1/2)$ (which would give zero trace for the 3x3 part, but not for the 4x4 tensor, which is what's required).

So is there any classical explanation for the lack of volume change that is usually observed at small fields?

riddle - I am found in a pool, never sinking nor afloat; pass the squares that I rule, and a soldier must promote

Riddle me this:

Although I'm a cube, I do not have six sides;

Beneath me is green, although no grass resides.

My pockets are large enough — I can fall in!

But then you'll fall out, though at ends you might win.

I'm solid and black, but my pupil is white;

My letter and name are as tall as my height.

My value is tied, unlike my rotund form.

No tees are before you; no spots on me swarm,

But I despise stripes even when they're on cue.

My pale friend knows me, but what about you?

A classic riddle of mine. The title is also a clue, but it plays an entirely different game.

---

Hint:

I spin all the time, but will never get dizzy.
Don't count to my number when tipped — it ain't easy.

Answer

I think you are

The black ball in pool

Although I'm a cube, I do not have six sides;

The ball usually has the number '8' on it which is a cube

Beneath me is green, although no grass resides.

Pool tables often have a green baize (although it can be blue or red too)

My pockets are large enough — I can fall in!

Pool tables have pockets

But then you'll fall out, though at ends you might win.

If you pot the black ball before all others of your designated type, you will lose although you will need to pot at the end to win the game.

I'm solid and black, but my pupil is white;

The black ball has a small bit of white containing the number.

My letter and name is as tall as my height.

Not sure about this one, but possibly referencing the similarity between the number '8' and the letter 'B'.

My value is tied, unlike my rotund form.

The inherent value of the black ball in pool is the winning of the game and not the value '8'.

No tees are before you; no spots on me swarm,
But I despise stripes even when they're on cue.

The black ball is neither considered part of the 'spots' or 'stripes' in pool. You use a 'cue' to play the game.

My pale friend knows me, but what about you?

The cue ball

Hint

The black ball may spin a lot during play. The number '8' turned on its side is $\infty$, not easy to count.

Title

I'm found in a pool - pool being the game rather than a swimming pool.

pass the squares I rule - not sure about this but several of the other balls have square numbers.
and a soldier must promote - an 8-ball is U.S. military jargon for a soldier often in trouble

quantum field theory - Compact QED and Non-compact QED - Polyakov textbook

This question is related with Polyakov, "Gauge Fields and Strings" section 4.3

Firstly, Polyakov define a QED on a lattice

1. 
   

   Compact QED \begin{align} S = \frac{1}{2} \sum_{x, \alpha, \beta} (1-\cos(F_{x,\alpha\beta}) ) \end{align} where $F_{x,\alpha\beta} = A_{x,\alpha} + A_{x+\alpha,\beta} - A_{x+\beta, \alpha} - A_{x,\beta}$ with $- \pi \leq A_{x,\alpha} \leq \pi$

   
   


2. 
   

   Non compact QED \begin{align} S = \frac{1}{4 e_0^2} \sum_{x,\alpha\beta} F_{x,\alpha\beta}^2 \end{align} here $- \infty \leq A_{x,\alpha} \leq \infty$

   
   

In the textbook, Compact version (periodic) is Natural version of QED which is related with charge quantization.

Here i have a few basic question.

1 : Why periodicity gives compactness?

2 : What is the physical difference or usefulness of compact or Non-compact QED? (Is this compact concept is only related with the lattice theory?)

3 : Is periodicity of $A_{x,\alpha}$ really related with charge conservation?

My guess from equation 4.32 in textbook, which is \begin{align} q_0 = \frac{1}{2\pi} \oint_{L} A_{x, \delta} \end{align} The periodicity of $A_{x,\alpha}$ gives some number after loop integral. $i.e$, (ration for certain loop)

Answer

1 : When $A$ is restricted to $[-\pi,\pi)$, the group of gauge transformations is the compact $\mathrm{U}(1)$ group. (Roughly speaking, to be compact, a group needs to be bounded.) In the nonperiodic theory, the gauge group is the group of real numbers under addition, which is noncompact.

2 : The most important distinction between the compact and noncompact theories is that the compact one allows ("Polyakov") monopoles, which are topological configurations of $A$. (For the same reason, the XY model has topological defects, called vortices, while a real scalar field does not.) Polyakov monopoles are very similar to Dirac monopoles, but Dirac's argument was that the Dirac string is unobservable because charges are quantized, while Polyakov's is that the string has no energy cost, because $F_{\alpha\beta} = 2\pi$ along its length.

You could define a compact $\mathrm{U}(1)$ gauge theory in the continuum, but monopoles are no longer possible. (Again, the same is true of vortices in the XY model; they have a core at which $\theta$ is undefined.) There could still be a distinction for the model on a space with nontrivial topology.

3 : How is compactness related to discreteness of charge?

Suppose we have a matter field $\psi$ with charge $q$ defined on the sites of the lattice. If you apply a gauge transformation $\psi \rightarrow \psi \mathrm{e}^{\mathrm{i} q \theta}$ on one site, then $A$ changes by $\pm\theta$ on all links connected to this site. If we choose $\theta = 2\pi$, this transformation of $A$ is redundant in the compact theory. Because $A$ is effectively unchanged, in order for the gauge transformation to be a symmetry, $\psi$ must also be unchanged, so we require $q$ to be an integer.

Another way to see this is to use the Dirac argument that the existence of magnetic monopoles requires discreteness of electric charges.

---

If you have access to it, you might want to look at J. B. Kogut, An introduction to lattice gauge theory and spin systems, Rev. Mod. Phys. 51, 659 (1979) for a detailed review of these ideas.

quantum field theory - Difference between 1PI effective action and Wilsonian effective action?

What is the simplest ay to describe the difference between these two concepts, that often go by the same name?

Answer

The Wilsonian effective action is an action with a given scale, where all short wavelength fluctuations (up to the scale) are integrated out. Thus the theory describes the effective dynamics of the long wavelength physics, but it is still a quantum theory and you still have an path integral to perform. So separating the fields into long and short wavelength parts $\phi = \phi_L + \phi_S$, the partition function will take the form (N.B. I'm using euclidean path integral)

$$ Z = \int\mathcal D\phi e^{-S[\phi]} =\int\mathcal D\phi_{L}\left(\int D\phi_{S}e^{-S[\phi_L+\phi_S]}\right)=\int\mathcal D\phi_{L}e^{-S_{eff}[\phi_L]}$$ where $S_{eff}[\phi_L]$ is the Wilsonian effective action.

The 1PI effective action doesn't have a length scale cut-off, and is effectively looking like a classical action (but all quantum contribution are taken into account). Putting in a current term $J\cdot \phi$ we can define $Z[J] = e^{-W[J]}$ where $W[J]$ is the generating functional for connected correlation functions (analogous to the free energy in statistical physics). Define the "classical" field as $$\Phi[J] = \langle 0|\hat{\phi}|0\rangle_J/\langle 0| 0 \rangle_J = \frac 1{Z[J]}\frac{\delta}{\delta J}Z[J] = \frac{\delta}{\delta J}\left(-W[J]\right).$$

The 1PI effective action is given by a Legendre transformation $\Gamma[\Phi] = W[J] + J\cdot\Phi$ and thus the partition function takes the form

$$Z = \int\mathcal D e^{-S[\phi] + J\cdot \phi} = e^{-\Gamma[\Phi] + J\cdot \Phi}.$$ As you can see, there is no path integral left to do.

electricity - Difference between current and voltage sources

I am confused about the current and voltage. My intuitive example would be that of a pipe of say water. The diameter of the pipe determines the amount of water flowing per second but the pressure is comparable to voltage. Am I right?

And what is the difference between voltage source and current source? In what class our electric sockets on the wall do fall?

speed - Why does the wheel of a car appear to be moving in opposite direction?

When a car is about to stop, the speed of its wheel reduces. Why do we see or feel that the wheel of the car is moving in the opposite direction? This can also be observed of a fan at home. So why is it?

Answer

I think that the reason for that is related to the fixed framerate of our eye which is 24 frames per second. In each next frame you'll see the wheel turned by some angle relative to the previous frame. If this difference is from 0 to pi radians (0 to 180 degrees) you'll see it in one direction. If during each frame, the angle will be from pi to 2pi radians (180 to 360 degrees) then you'll see it rotated in opposite direction in each frame. That's why the wheels of the car go through at least 3 stages of illusion: rotating in clockwise direction, slowing down up to non-rotating state and then rotate in opposite direction, while the car accelerates or stops. But in reality of course the rotation rate of the wheels in increasing or decreasing constantly.

quantum field theory - Why is $R^2$ gravity not unitary?

I have often heard that $R^2$ gravity (as studied by Stelle) is renormalisable but not unitary. My question is: what is it that causes the theory to suffer from problems with unitarity?

My naive understanding is that if the the Hamiltonian is hermitian then the $S$-matrix $$ \langle \text{out} \mid S \mid \text{in} \rangle = \lim_{T\to\infty} \langle \text{out} \mid e^{-iH(2T)} \mid \text{in} \rangle $$ must be unitary by definition. So why is this not the case for $R^2$ gravity?

I see that Luboš Motl has a nice discussion related to such things here, but I am not sure which, if any, of the reasons he mentions relate to $R^2$ gravity.

Are there other well known theories that have similar problems?

geometry - Now help me choose a new lock pattern for my phone

Well, thanks to Len, I managed to unlock my phone again, but now everyone knows my pattern, so I need to choose a new one.

_{For brevity, I won't include the explanation of how an Android pattern lock screen works again, but you still need to follow those rules (see previous puzzle).}

Here's the rules for my new pattern:

1. I still want to use all nine dots



2. I'm still obsessed with mirroring, so it must still have reflective symmetry


3. I don't like odd numbers, so the pattern's resulting line must have exactly two "ends"*

What should my new pattern be?

_{* by this I mean that visually, the final line formed by the pattern has exactly two dots which have a single line connecting to them. So, for example the pattern 2 1 3 has two line endings (at 1 & 3), but 5 2 1 3 has three (at 1, 3 & 5). All solutions found for the previous puzzle have three line endings.}

Answer

Woah... I think I got it after a whole lot of trying around. Here is my train of thought:

1. Since we have two open Ends and reflective symmetry, both ends must be mirrored.


2. Since we can never directly go to an already visited point, the last point must be an open end - so one of the two ends.



3. The two end segments (the last part, direct connection) cannot be directly connected, because we cannot leave the path and come back to a point.

Valid endpoints: 1-4 3-6 and 1-4 8-9

A . . .  B . . .              . . .   . . .   . . .   . . .
  | . |    | . .    Invalid:  . . .   .\./.   .\. .   . | .
  . . .    . ._.              ._._.   . . .   . .\.   . | .

Trying to connect the two valid positions for endpoints I realized I couldn't start at one of them, because I cannot leave the path and come back on the path - and without coming back on the path, it is no longer symmetric. So one of the endpoints had to be a hidden double-line (like 2-1-3) and the other the real endpoint.

So my path had to start on the mirroring point of this hidden path, since the mirrored path without the end-points would end there and has to be symmetric.

So the start-points for Figure A was 4 or 6 and for B was 4 or 8.

Some more trying around gave me this:

. ._. 4-5-6-3-2-8-9-7-1 |_|_| |_|_.