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Muon lifetime determination My colleagues and I performed several experiments to determine the lifetime of the muon (from secondary cosmic rays) using scintillator detectors coupled to multi-channel analysers. The results invariably showed a muon lifetime lower than the standard 2.2 microseconds. Apart from poor statistics,and assuming no faults in the equipment used, what other factors could be responsible for the discrepancy?
| As you suggested in your comment, the $\mu^-$ and $\mu^+$ that stop in matter do not have the same lifetimes. The $\mu^+$ come to rest between the atoms of your stopper (eg: scintillator?) and decay into $\nu_{\mu}e^+\nu_e$ with the standard 2.2 usec lifetime. However, the $\mu^-$ get captured into Bohr orbits about the stopper nuclei. The $\mu^-$ then transistions down to n=1 L=0 orbit by emitting Auger electrons and x-rays. In this closest orbit there is an overlap between the $\mu^-$ wave function and the nucleus, and therefore some rate to interact with the protons and neutrons. So, the seen decay rate of the $\mu^-$ is a sum of the nuclear interaction rate and the natural decay rate of the muon.
$$
\frac{1}{\tau_{Seen}}=\frac{1}{\tau_{Nuclear}}+\frac{1}{\tau_{Natural}}
$$
The nuclear interaction rate increases with the Z of the nucleus because the orbital radii are smaller and there are more nucleons as Z increases. The lifetimes $\tau_{Nuclear}\approx\tau_{Natural}$ for $Z\approx 10$. There is an extensive review of all this in Physics Reports 354 (2001) 243-409. Table 4.2 shows some $\tau_{Seen}$ for $\mu^-$ stopping in different elements.
In summary, your number of decays versus time is the sum of two exponentials. One for $\mu^+$ with a 2.2 usec lifetime, and one for $\mu^-$ with a lesser lifetime that depends on the elements in your stopper. The ratio of the number of positive muons to negative muons at sea level in cosmic rays is about 1.2 . It is reasonable that you are measuring <2.2 usec for the overall lifetime, but for quantitative sense you will have to fit two exponentials and use the $\tau_{Nuclear}$ for your stopper elements.
| {
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Coherence length of a single photon If I pass individual photons through a M-Z interferometer with equal arms I will observe interference (eg only one detector will respond). As I increase the path length of one arm I will observe the two detectors responding alternately as I pass through each phase cycle. Eventually I suspect that at a certain point, the interference will disappear and the two detectors will respond with equal probability. What determines this point and what does this tell us about the 'length' of an individual photon. What does QM predict when the path difference is greater than this?
| CuriousOne makes a good point in comments and I will elaborate on that.
While you are asking about the coherence length of a single photon, in the experiment that you describe you will have to detect many photons to judge if they are detected in both arms with equal probability. Where do this photons come from?
If you assume that the photons are identical, then you would need some ideal light source with infinite coherence length. If you take a real light source, the difference between individual photons will define your decoherence length.
You can see, then, that what you are actually measuring with your experiment is the coherence length of the light source. A single photon does not have such property.
| {
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What is a stroboscopic map? I have an assignment where I'm supposed to generate a "stroboscopic map" of some orbits of a dynamical system. I have a hard time finding information about exactly what this kind of map is on the Internet and in my course book, but I suspect it might be the same thing as a Poincaré map or a special case of it? Can anyone explain this type of map and how it differs from the Poincaré map?
| A stroboscopic map is indeed a special case of a Poincaré map for driven systems. The distinguishing feature is that a given phase of the driver’s period is used for mapping (instead of some other marker event like a local maximum or a zero crossing). If, e.g., your driving term is $\sin(t)$, you would obtain the stroboscopic map by regarding the state of the system in intervals of $2π$ time units.
For example from Scholarpedia:
For a non-autonomous vector field $\frac{dx}{dt} = f(x,t)$
with $f(x,t) = f (x,t+τ)$ for some $0<τ<∞$, the calculation of the stability properties of a periodic orbit with period $T = \frac{p τ}{q}$, where $p$ and $q$ are integers (see Arnold tongues),
can be done by considering a stroboscopic map which takes
$$x(t) \rightarrow x\left(t + \frac{p τ}{q} \right).$$
The stability properties follow from the eigenvalues of this map, as above.
| {
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Does the sign of imaginary part of complex permittivity have any physical meaning? I have noticed some papers having written complex permittivity as $e' + je'' $
and others as $e' - je''$. The data in literature does not specify the sign. What should I use and does the sign of $e''$ implies anything?
| Work out how a plane wave propagates. If its intensity grows in the direction of propagation, then you need to switch the sign of the imaginary part! It depends on whether you define $e^{+i\,k\,z}$ or $e^{-i\,k\,z}$ as your diffraction operator in the direction of propagation. If it's $e^{+i\,k\,z}$, then a lossy dielectric always has a positive imaginary part to its refractive index. If your diffraction codes (or similar) blow up and give you a whole lot of NaNs as output, then the sign of imaginary parts of permitivities / diffraction operators and so forth is the first thing to look at - usually someone's slipped up on this convention to cause this symptom.
| {
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Meaning of the phase space in statistical physics I have a silly question about the phase space. I am confused with the meaning of points in phase space. Does the each point in phase space represent concrete particle of the system, or does it represent the whole state of the system? Our teacher told us, that we use the phase space to describe the development of each particle. It is not right, isn't it?
| It represents the "space occuped" by the hypervolume momentum-position of a particle.
If you integrate the three components of position you get volume; if you integrate the three components of momentum you get kind of a momentum volume. If you integrate both together you get the space phase volume
| {
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Why is oil a better lubricant than water? How come mineral oil is a better lubricant than water, even though water has a lower viscosity?
When two surfaces slide over each other with a gap filled with a fluid, the different layers of the fluid are dragged at different speeds. The very top layer touching the top metal surface will have the same speed as the surface itself, while the bottommost layer is stationary. The speed in the layers between is distributed linearly and there exist friction forces between those layers that slow the movement. Those frictional forces should be reduces however, if a fluid with a lower viscosity is chosen.
How come this is not so?
Does it have to do with water's polarity, so that it sticks to surfaces in a different way than oil?
| Your derivation is composed of correct statements and indeed, if something is known to act as a lubricant, we want the viscosity to be as low as possible because the friction will be reduced in this way. For example, honey is a bad lubricant because it's too viscous.
However, your derivation isn't the whole story. The second condition is that the two surfaces must stay apart. If you use a lubricant with too low a viscosity, the surfaces will come in contact and the original friction will reappear.
So the optimum lubricant is the least viscous liquid that is viscous enough to keep the surfaces apart. Which of them is the optimal one depends on the detailed surfaces and other conditions. For example, there exist situations in which water is a better lubricant than oil – for example when ice slides on ice. Some of the ice melts and the water is why the ice slides so nicely.
| {
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Is a spacetime of constant positive curvature just a 4-hypersphere? In discussions of basic cosmological models, I don't see "spacetime of constant positive curvature" described more simply as a "4-hypersphere". What am I missing?
| You are presumably thinking of the FLRW metric for a universe with greater than critical density i.e. a closed universe.
We normally use comoving coordinates to describe this, in which case the time coordinate is not curved and at every point along this time coordinate the three spatial coordinates have the topology of a 3-sphere. That is, if we draw a straight line in any direction and continue it indefinitely the line will eventually return to it's starting point.
This isn't a 4-sphere because this is not true of the time dimension. The time dimension starts at the Big Bang and ends at the Big Crunch so it is just a line not a loop. In fact it is geodesically incomplete at both ends since both the Big Bang and Big Crunch are singular points.
| {
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How was the half-life of Uranium 235 determined and by whom? Wikipedia says that the half-life of Uranium-$235$ is $7.038 \times 10^8$ ($\text{703 800 000}$) years.
This is very long. Therefore, on a human time scale, the decay is very small, posing difficulty for determining the half-life.
So Who and how was the half-life calculated?
| The earliest reference I've been able to find on the half-life of 235U is in The Uranium Half-Lives: A Critical Review, by Norman Holden, which reviews various early studies of each of the common isotopes of uranium (232U, 233U, 234U, 235U, 236U, and 238U). The earliest study he cites is Nier (1939) (A. 0. Nier, The isotopic constitution of uranium and the half-lives of the uranium isotopes, Phys. Rev. 55 150 (1939))1.
Nier looked at ratio of 206Pb to 207Pb in 21 samples of radiogenic lead - that is, lead that has been formed by radioactive decay, in this case, from Uranium. 206Pb is a decay product of 238U, while 207Pb is a decay product of 235U. He then determined that the actinium series - the decay chain of 235U to 207Pb - is 4.6$\pm$0.1% as active as the uranium series - the decay chain of 238U to 206Pb. Then, using the ratio of 238U to 235U in natural samples, he was able to calculate the half-life of 235U. It is not mentioned as to how he was able to calculate the half-life of 238U.
Interestingly enough, Nier's value is closer to the accepted value than the half-lives found in many later studies using $\alpha$-counting, which looked at the number of $\alpha$ particles emitted during the decay of 235U.
1 I was able to find Nier's abstract in Report of the Committee on the Measurement of Geologic Time (pages 45-46).
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Product of two Pauli matrices for two spin $1/2$ In the lecture, my professor wrote this on the board
$$
\begin{equation}
\begin{split}
(\vec{\sigma}_{1}\cdot\vec{\sigma}_{2})|++\rangle &= |++\rangle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\blacktriangledown)\\
(\vec{\sigma}_{1}\cdot\vec{\sigma}_{2})(|+-\rangle+|-+\rangle) &= (|+-\rangle+|-+\rangle)\\
(\vec{\sigma}_{1}\cdot\vec{\sigma}_{2})(|+-\rangle-|-+\rangle) &= -3(|+-\rangle+|-+\rangle)
\end{split}
\end{equation}
$$
but I don't get how these are correct. I know that
$$
\begin{equation}
\begin{split}
|1\;1\rangle &= |++\rangle \\
|1\;0\rangle &= \frac{1}{\sqrt{2}}(|+-\rangle+|-+\rangle) \\
|0\;0\rangle &= \frac{1}{\sqrt{2}}(|+-\rangle-|-+\rangle)
\end{split}
\end{equation}
$$
I will work out equation $(\blacktriangledown)$ in the usual matrix representation of the eigenstates of $S_z$ basis:
$$
|+\rangle=\begin{pmatrix}1\\
0
\end{pmatrix},\;\;\;\;\;\;\;\;\;\;\;\;\;\;|-\rangle=\begin{pmatrix}0\\
1
\end{pmatrix},
$$
So we have
$$
\begin{equation}
\begin{split}
(\vec{\sigma}_{1}\cdot\vec{\sigma}_{2})|+\rangle_{1}\otimes|+\rangle_{2}&=&\vec{\sigma}_{1}|+\rangle_{1}\otimes\vec{\sigma}_{2}|+\rangle_{2}\\&=&\begin{pmatrix}1 & 1-i\\
1+i & -1
\end{pmatrix}_{1}\begin{pmatrix}1\\
0
\end{pmatrix}_{1}\otimes\begin{pmatrix}1 & 1-i\\
1+i & -1
\end{pmatrix}_{2}\begin{pmatrix}1\\
0
\end{pmatrix}_{2}\\&=&\begin{pmatrix}1\\
1+i
\end{pmatrix}_{1}\otimes\begin{pmatrix}1\\
1+i
\end{pmatrix}_{2}
\end{split}
\end{equation}
$$
but this is not $|++\rangle=|+\rangle\otimes|+\rangle$. What did I do wrong here? What have I misunderstood?
| Your expression for:
$$(\vec \sigma_1 \cdot \vec \sigma_2) |+\rangle_1 \otimes |+\rangle_2=\vec \sigma_1 |+\rangle\otimes \vec \sigma_2 |+\rangle_2$$
Is wrong. It sould read:
$$(\vec \sigma_1 \cdot \vec \sigma_2) |+\rangle_1 \otimes |+\rangle_2=\sigma_{1x}|+\rangle_1\otimes \sigma_{2x}|+\rangle_2+\sigma_{1y}|+\rangle_1\otimes \sigma_{2y}|+\rangle_2+$$ $$\sigma_{1z}|+\rangle_1\otimes \sigma_{2z}|+\rangle_2$$
$$=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\otimes\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}$$
$$+\begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\otimes\begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}$$
$$+\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\otimes\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}$$
$$=\begin{pmatrix} 0 \\1 \end{pmatrix}\otimes\begin{pmatrix} 0 \\1 \end{pmatrix}$$
$$+\begin{pmatrix} 0 \\i \end{pmatrix}\otimes \begin{pmatrix} 0 \\i\end{pmatrix}$$
$$+\begin{pmatrix} 1 \\0 \end{pmatrix}\otimes \begin{pmatrix} 1\\0\end{pmatrix}$$
$$=\begin{pmatrix} 0 \\1 \end{pmatrix}\otimes\begin{pmatrix} 0 \\1 \end{pmatrix}$$
$$-\begin{pmatrix} 0 \\1 \end{pmatrix}\otimes \begin{pmatrix} 0 \\1\end{pmatrix}$$
$$+\begin{pmatrix} 1 \\0 \end{pmatrix}\otimes \begin{pmatrix} 1\\0\end{pmatrix}$$
$$=\begin{pmatrix} 1 \\0 \end{pmatrix}\otimes \begin{pmatrix} 1\\0\end{pmatrix}$$
i.e. I think you have to do the dot product between the Pauli matrices vectors first then put them through the tensor product.
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Would an HI or HII region have a color? That is, if we were approaching an HI or HII region in a spaceship, would the cloud have a color visible to the naked eye? Of the HI region, Wikipedia says:
These regions do not emit detectable visible light (except in spectral
lines from elements other than hydrogen) but are observed by the 21-cm
(1,420 MHz) region spectral line. This line has a very low transition
probability, so requires large amounts of hydrogen gas for it to be
seen.
But I'm not sure what it means by "so requires large amounts of hydrogen gas for it to be seen." What does "except in spectral lines from elements other than hydrogen" entail?
| "Requires large amounts of hydrogen to be seen" means simply that the light is extremely weak, so you need lots of hydrogen available to get a reasonable number of photons created the 21-cm line. That's a microwave line, by the way, so wouldn't be visible to the eye.
As far as "except in spectral lines from elements other than hydrogen" is just a reminder that clouds of hydrogen gas are generally not composed exclusively of hydrogen. There will be a mix of hydrogen, helium, oxygen, carbon, neon, etc. Your typical cloud of hydrogen will be something like 74% hydrogen, 24% helium, 1% oxygen, 0.5% carbon, 0.1% neon, 0.1% iron, 0.1% nitrogen, etc. This is by number of atoms, not by mass. So even though you won't see any emission lines from the hydrogen itself, you might see emission lines from other elements.
| {
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How to pour water from a bottle as fast as possible? When one pours water out of a bottle, it first flows smoothly but then a pressure 'blockage' develops and the pouring becomes interrupted and turbulent, so that the water comes out in splashes. This seems to slow down the flow of water from the bottle.
What is the optimal way to pour the water so that it completely empties fastest?
Possible strategies:
*
*Holding the bottle at a certain angle
*Wildly shaking the bottle
*Squeezing the bottle
*Other...
It probably depends on the shape of the opening and/or the bottle itself, but we shall assume this beautiful example of a standard water bottle:
CLARIFICATION
The question is asking how to pour the water the fastest, so no straws, hole insertion and evaporating lasers allowed...
| The question is: what is the optimal way to pour the water so that it [the bottle] completely empties fastest? I conclude the aim is to have the empty bottle, not the water in another container.
Solution: Create a centrifuge-like setup, bottle opening to the outside. The setup will generate artificial gravity for the water in non-inertial frame of reference associated with the bottle.
You may hold the bottle in your hand, stretch the arm and spin fast enough. Example videos below. There is no bottle in any video, so it looks like you have the chance to be the pioneer.
*
*Basic setup here.
*Use two bottles in two hands to create symmetric setup for balance. Unbalanced centrifugal setup may fail like this one.
*More technically advanced setup is possible: link.
*The setup is prone to miniaturization (notice the renewable energy source): link.
This solution may be combined with squeezing the bottle.
Enhancement: do it in vacuum (but watch the temperature and don't let the water freeze before it leaves the bottle). It is obvious and marvelously impractical thing to do. You may be picky again and expand your clarification now. :)
| {
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What is meant by rest in rest-mass? As far as I know only photons are considered to have no rest-mass. In common words when it doesn't move it 'disappears'.
*
*Electrons and quarks should have a rest-mass. But are they really at rest?
In atoms and molecules is always a kind of zero-point energy left which implies that there is still some 'movement' in the particles.
*
*So when an electron or quark is really at rest does it still exist?
*Or does it 'disappear' just like a photon?
*And hence is there really a difference in mass-property of particles or is the equation $E=mc^2$ already suggesting that there is no real property difference in mass between a photon and electron? Or there is?
| Rest mass means the mass which would appear if a paricle were at rest. Do not confuse between particles and photons. These particles are metarialistic particles behaving as energy in some circumstances. Being matter they possess rest mass. While photon is a bundle of energy behaving sometimes as a particle and hence can not possess a rest mass or more precisely, invariant mass.
| {
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How the data or models given in physics or mathematics problems can be obtained in real life? When ever there is physics or mathematics problem. we are already given a mathematical model. Can't we make own models from given diagrams or data collected from the problem? sometimes angle is given in the problems like a plane makes an angle of theta= pi radian etc... how they measure angle of a plne and how can we measure angle in real life... suppose if I want to calculate the length of a ladder that is leaning against a wall & ladder makes an angle with respect to ground , how that angle can be measured..? .. in physics books if we want to find position by integration we already are given acceleration as a function of time..why is it so..? can't we obtained these functions; models; or angles etc ourself?
| There are essentially two ways in which we obtain functions and models ourselves: through the use of an appropriate measuring instrument (ruler, protractor, scale, clock...) or via mathematical logic.
Measuring instruments can give us an excellent but imperfect approximation, while mathematical logic provides us an exact value based upon input values that are either theoretical or approximate.
Most school mathematics assumes values that are exact in order to encourage us to work through the mathematical logic. In practice, however, we may be working with measurements that rather poorly known (especially in sciences such as astronomy). For this reason, we calculate the uncertainty of our results based upon how well we were able to make our measurements.
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If the speed of light is constant, why can't it escape a black hole? When speed is the path traveled in a given time and the path is constant, as it is for $c$, why can't light escape a black hole?
It may take a long time to happen but shouldn't there be some light escaping every so often?
I'm guessing that because time is infinite inside a black hole, that this would be one possible reason but wouldn't that mean that we would require infinite mass?
What is contradicting with measuring black holes in solar masses, what means they don't contain infinite mass.
So how can this be?
| The explanation I like is thus:
In GR, all things, from planets to photons, travel in straight lines through curved space bent by mass. Black holes bend and distort spacetime so severely that the curvature captures the photon.
Scale things down and it behaves much the same way passing asteroids can be captured by a star. For us, the speed of the asteroid(photon) is only relevant up until the point it crosses the threshold of capture, the point of no return and tips into the star's(black hole) gravity well. The asteroid(photon) will never escape, the "walls" of the well are too high. Does it matter if it is me or the worlds best high jumper who's trapped at the bottom of a pit, if the walls are 50 feet high? Neither one has any chance of escaping, the fact that one of us can jump the highest a person can possibly jump is irrelevant.
The fact the light goes the the fastest anything can go is a similar red herring. What matters is that it has gotten into a situation from which there is no escape. The speed of light is not a get out of jail free card any more than the high jumpers.
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What are some resources for learning about x-ray powder diffraction? I am looking for up-to-date, applied resources to learn about x-ray powder diffraction. There is a lot of jargon with which I am unfamiliar. I am trying to refine theoretical curves to collected data using the program Maud using the Rietveld Method but I am unsure about which parameters to adjust.
| I'd prefer myself to start from an introductory book or a lecture before diving into research papers.
The most often recommended book on the rietveld_l mailing list (which is the biggest mailing list about powder diffraction) is:
Fundamentals of Powder Diffraction and Structural Characterization of Materials by Vitalij K. Pecharsky & Peter Y. Zavalij.
Note that rietveld_l is a low-traffic list and most-often recommended there means a few times per decade.
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Rolling motion of a rigid object I have the situation described in this picture
I know the speed of the ball at the top of the loop ($v_{top} = 2.38 m/s$), and I have to demonstrate that the ball does not fall from the track at the top of the loop.
From what I understand, the two forces acting on the ball are the gravitational force ($F_g = m g$) and the normal force of the track, that provides the centripetal acceleration of the ball ($F_n = m \cdot a_c$, where $a_c = \frac{v^2}{r}$). Since the centripetal acceleration balances the tangential velocity $v_{top}$, that is, forces the ball to follow the circular path, why does not the ball fall under the effect of the gravitational force $F_g$?
| At the top of the loop if the normal reaction on the ball due to the track is $F_n$ down and the weight of the ball is $F_g$ down then using Newton's second law
$$F_n + F_g = m\frac {v^2}{R}$$
where $v$ is the speed of the ball, $m$ is the mass of the ball and $R$ is the radius of the loop.
This equation tells you that the faster the ball is moving the larger is the value of the normal reaction.
However as the speed of the ball at the top gets less the normal reaction $N$ gets smaller until there comes a time when the normal reaction is zero and $$F_g = m \dfrac {v_{\text{minimum}}}{R}$$
where $v_{\text{minimum}}$ is the minimum speed that the ball can have and still keep in contact with the track.
If the speed of the ball is less than this minimum speed it will lose contact with the track before reaching the top of the loop.
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What is the color of a group of trillions of electrons, protons, and neutrons Since an electron is smaller than visible light, then what what color would a group of electrons (trillions of electrons) be if there were enough of them to be seen by the eye? What color would a group of trillions of protons be? Color of trillions of neutrons? I don't mean a group of electrons, protons, and neutrons mixed together into atoms, I mean a group of each of them separately. Would they be an actual color (red, black, green, etc), clear but visible (the color of water, glasses's lenses), or invisible?
| I cant comment on large groups of electrons, but we do actually have two real world examples of groups of neutrons, and one of groups of protons.
Neutrons
*
*In labs, we have Bose-Einstein condensates. But since they evaporate with the slightest motion or exposure to light, I dont know how we could ever see them.
*Possibly more helpful could be neutron stars. I think they have a ~4 inch thick atmosphere of some gas and a crust of iron in the way though, so seeing bare neutrons might still be impossible.
Protons
*
*There is an isotope of hydrogen called protium that has no neutrons. Cationic protium (positively charged protium, ie stripped of the electrons) is nothing but protons. To find out the color of pure protons, just find out what color cationic protium matter is.
This is just a conjecture, but, it may be possible that isotope and/or charge is irrelevant to the color of hydrogen, so maybe you just have to look at liquid hydrogen to find out what color protons are.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is conservation of momentum and energy valid for non-inertial frames? Conservation laws of momentum and energy are said to be the most basic principles of physics. Are they also valid for non-inertial frames, and in what way?
| Regarding total momentum conservation, the point is that in non-inertial reference frames inertial forces are present acting on every physical object. Momentum conservation is valid in the absence of external forces.
However, if these forces are directed along a fixed axis, say $e_x$, or are always linear combinations of a pair of orthogonal unit vectors, say $e_x,e_y$, (think of a frame of axes rotating with respect to an inertial frame around the fixed axis $e_z$ with a constant angular velocity), conservation of momentum still holds in the orthogonal direction, respectively. So, for instance, in a non-inertial rotating frame about $e_z$, conservation of momentum still holds referring to the $z$ component.
Mechanical energy conservation is a more delicate issue. A general statement is that, for a system of points interacting by means of internal conservative forces, a notion of conserved total mechanical energy can be given even in non-inertial reference frames provided a technical condition I go to illustrate is satisfied.
Let us indicate by $I$ an inertial reference frame and by $I'$ the used non-inertial frame. Assume that our physical system is made of a number of points interacting by means of conservative true forces depending on the differences of position vectors of the points, so that a potential energy is defined and it does not depend on the reference frame.
If the origin of $I'$ has constant acceleration with respect to $I$ and the same happens for the angular velocity $\omega$ of $I'$
referred to $I$ (it is constant in magnitude and direction), then only three types of inertial forces take place in $I'$ and all them are conservative but one which does not produce work (Coriolis' force). In this case, the sum of the kinetic energy in $I'$, the potential energy of the true forces acting among the points and the potential energy of the inertial forces appearing in $I'$ turns out to be conserved in time along the evolution of the physical system.
| {
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Can interference occur between two waves that are parallel but separated by a small distance?
This is a image of diffraction in crystal. My doubt is how the parallel waves coming out interfere if they are seperate?
| In answering your question, a lot could be said about the art of mathematical modeling, but, the short answer is: They don't.
But, the rays in the scheme are only an approximation, and one that fails at the atomic scale - a beam of light, no matter how laser like or faint, is never exactly a 1-D mathematical line, it spreads sideways. That's why they can interfere.
| {
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Specific heat of the classical ferromagnetic Heisenberg model I have simulated the classical ferromagnetic Heisenberg model on a cubic lattice using Monte Carlo and I get a finite specific heat near zero temperature. My understanding is that from the magnon theory,
we should get a specific heat that behaves like ~ $T^{3/2}$. I have looked everywhere to find an answer to this question, but I have found none. I have read somewhere about the number of ground states not being necessarily the same for a finite system as for an infinite one, but I can't quite see
the connection. Any help with this question will be much appreciated. The included figure is a sample of the results that I have obtained.
| This is essentially a result of the equipartition theorem where each degree of freedom contributes $k_B T/2$ to the energy. Given that the specific heat in this context is just ${\partial E}/{\partial T}$ then each degree of freedom contributes $k_B/2$ to the specific heat. For the classical model of lattice vibrations in solids this leads to the Dulong-Petit Law; i.e. a constant specific heat.
For the classical Heisenberg model of interest the analogous harmonic modes are the spin waves about the ferromagnetically ordered state. Since there are as many spin wave modes as sites, and each contributes two degrees of freedom (roughly: the two directions transverse to the ordered moment) you would expect an energy per site from these modes of $E = 2 (k_B T/2) = k_BT$ and thus a contribution to the specific heat of $C \sim k_B$. In more natural units this would be $C \sim 1$, as you find in your simulations. This should only occur at low temperature (well below the ordering transition) where this spin-waves are well defined.
In regard to the $C \sim T^{3/2}$ law for magnons; this is a result of quantum mechanics, in the same way the low temperature Debye specific heat, $C \sim T^3$, is for phonons. To find this result one needs both the ferromagnetic dispersion that goes as $\omega \sim k^2$ at long wave-lengths, as well as Bose statistics for the magnons.
| {
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Could a "living planet" alter its own trajectory only by changing its shape? In Stanislaw Lem's novel Solaris the planet is able to correct its own trajectory by some unspecified means. Assuming its momentum and angular momentum is conserved (it doesn't eject or absorb any mass), would this be possible (in Newtonian mechanics) and how? If not, can it be proven? The assumption is that the planet orbits a star (or perhaps a binary star) system.
Intuitively this seems possible to me. For example, tidal forces result in a planet losing its rotational energy, so it seems possible that by altering its shape, a body could alter at least its rotation speed.
My ideas go as follows: Assume we have an ideal rod consisting of two connected mass points. The rod rotates and orbits around a central mass. When one of the points moves towards the central body, we extend the rod, getting it closer to the center. thus increasing the overall gravitational force that acts on the rod. When one of the points is getting away from the center, we shrink the rod again, thus decreasing the combined gravitational force. I haven't run any simulations yet, but it seems this principle could work.
Update: An even more complex scenario (conserving momentum and angular momentum) would be if the planet ejected a piece of matter and absorbed it again after some time.
| By conservation of momentum and energy, the only possible way to change a planet's trajectory is to eject some (large) mass at high velocity in specific direction, like rockets do. But you are also correct that by increasing the moment of inertia, the rotational speed can be changed. But this cannot influence the movement of center of mass.
Edit2: Other answers capture what I missed while looking for a fast solution. The interplay between rotational and orbital angular momentum can indeed produce some effect (credit to @WetSavannaAnimalakaRodVance and @valerio92).
Let's assume that the rotational axis of the planet and it's orbit are aligned. Then, we have 2 invariants:
$$E = \frac12 I \omega^2 + \frac12 m R^2 \Omega^2 - G \frac{m M}{R} $$
$$ L = I \omega + m R^2 \Omega $$
where $I$ is a moment of inertia of a planet and $\omega$ is the rotational frequency while $\Omega$ is the orbiting frequency. $m$ and $M$ are masses of the planet and a star, respectively. Now, let's exclude $\omega$:
$$ \omega = \frac{1}{I} (L-M R^2 \Omega) $$
$$ E = \frac{1}{2 I} (L-M R^2 \Omega)^2 + \frac12 m R^2 \Omega^2 - G \frac{m M}{R} $$
For $\Omega$ we have a condition of staying on orbit:
$$ \Omega^2 R = G \frac{M}{R^2} $$
$$ \Omega^2 = G \frac{M}{R^3} $$
Then,
$$ E = \frac{1}{2 I} \left(L-M R^2 \sqrt{G \frac{M}{R^3}} \right)^2 + \frac12 G \frac{M m}{R} - G \frac{m M}{R} = \frac{1}{2 I} \left(L-M R^2 \sqrt{G \frac{M}{R^3}} \right)^2 - \frac12 G \frac{M m}{R} $$
There might be a mistake somewhere, but we can solve this for $R$ and, keeping $L$ and $E$ constant, we can vary $I$ changing the orbit radius.
Edit: Not directly related to the question formulated in the title. Okay, among futuristic options would be destruction of some nearby objects like closest planets or the hosting star. If this won't destroy our planet, it's course will definitely change. But to do so, one needs to accurately disperse the mass comparable or much bigger than the planet.
Basically, everything boils down to changing distribution of the mass.
| {
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What is the angular velocity of the electron? An electron has angular momentum. Shouldn't it also have angular velocity?
Ignoring the g-factor (just for the order of magnitude approximation) and the fact that an electron is not a sphere the electron's angular velocity should be around:
$$ \omega \approx \frac{\mu}{er^2} $$
or about 0.01 to 10^17 rad/s depending on whether the radius is the classical radius, the compton wavelength, or the planck length.
Is there some "average" angular velocity that can be assigned to the electron?
| You can't generate spin 1/2 from motion in space, so no, there is no way to assign an angular velocity to the electron. Orbital angular momentum only comes in integer multiples of h-bar.
This situation actually doesn't change very much even if we do discover substructure to the electron. Google "preon" and "confinement problem." You're still going to need a preon with half-integer spin, and that spin still can't come from orbital angular momentum.
| {
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Can I recirculate water from an open reservoir to the bottom of a bigger, closed one, without a pump? A fountain head pumps water out of the main tank into a 'pond' reservoir. Can the water recirculate back into the main tank without the help of another pump?
I'm sorry if this a dumb question. I'm guessing it would not function as the diagram shows, as the pressure of the water in the main tank would not let any water in at the bottom,right? Any solutions? (not requiring additional pumps)
| The set up shown will work just fine, though to be safe I'd put the check valve on the pump so as to prevent water and or air from flowing back into the tank from above.
I also made a few assumptions:-
*
*The free space in the top of the tanks is small enough
*The tank is not too tall (less than about $9~\text{m}$ should suffice)
*The volume of the pond is large enough
*Water cannot flow back through the pump
*The diameter of the outflow pipe is small compared to the tank diameter.
At first, the pressure of the air gap in the tank is at atmospheric pressure: this means it has a gauge pressure (pressure above atmospheric) of 0.
Check valve must be placed on the outlet pipe as close to the pump as possible.
Before the pump is turned on, the weight of the water will cause it to flow out into the pond through the bottom pipe connecting the two. As this does, the volume of the air gap will increase, and thus it's pressure will drop. Eventually the pressure will reach a point such that the force it exerts on the water will cancel out the gravitational force pulling the water out. (Negative gauge pressure)
When the pump turns on, it will remove water from the main tank again increasing the air gap volume. This will mean there is an overall negative gauge pressure at the tank side of the inlet pipe, thus causing water to be drawn into the tank to replace the water removed by the pump.
The reason this all works is due to the difference of the external pressure and the internal pressure, and works on a similar principle this.
| {
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System rigidity What is the meaning of system rigidity in mechanics? I can't understand how to classify the system as rigid or not, and what is the effect of rigidity on the whole system.
If you know anything about system rigidity... Please let me know
| It simply means that the object is so stiff that it can only move by proper Euclidean isometries. At any time, the whole body can only be a combination of rotations and translations of itself at any other time. In particular:
This means that the distance between every pair of points within the body stays the same at all times and all angles between lines in the body are invariant
Any combination of rotations and translations can be represented by a single translation followed by rotation (or rotation followed by translation - it doesn't matter as long as the order is consistent in a given discussion).
Can you represent the body at any time as a single translation followed by a rotation of itself at any other time? Can you translate to match up the centers of mass, then rotate into the correct orientation to find a perfect match? It is rigid if and only if the answer is yes.
So if a body is composite: e.g. a hinged system, or if it is squidgy like a jellyfish so that its surface deforms when pushed, then it does not move as described above and is not "rigid".
| {
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How to find entropy production of opening the windows in the winter? Every time you open the windows in the winter (to bring in healthy and fresh air), room's warm air goes outside into the atmosphere.
How to find entropy production of opening the windows in the winter?
The room's volume is denoted by $V_r$ and its temperature by $T_r$, and atmospheric volume is denoted by $V_a$ and its temperature by $T_a$.
| Note: We need one property of the room's air for example mass or pressure of it.
From "THERMODYNAMICS An Engineering Approach, Fifth Edition, by YUNUS A. CENGEL and MICHAEL A. BOLES":
Irreversibilities such as friction, mixing, chemical reactions, heat transfer through a finite temperature difference, unrestrained expansion, nonquasiequilibrium compression or expansion always generate entropy, and anything that generates entropy always destroys exergy. The exergy destroyed is proportional to the entropy generated:
$$X_{\textrm{destroyed}}=T_0S_{\textrm{gen}}\ge 0$$
So, assuming whole of the room's air is replaced with outside air, current question is reduced to:
Calculate the exergy of a room that contains air by the volume of $V_r$ and temperature of $T_r$ while the environment is at temperature of $T_a$ and pressure of $P_a$ (I assume we know the pressure, if not we can calculate it by using ideal gas law $P_a=\frac{RT_a}{V_a}$)
Because according to equation above we have: $S_{\textrm{gen}}=\frac{X_{\textrm{destroyed}}}{T_a}$ and if we assume that no useful work is done, then we have $X_{\textrm{destroyed}}=X_{\textrm total}$ (I.e. whole of exergy is destroyed)
Now let's calculate the exergy of the room before opening the window:
From the same book we have:
$$X=m\left[(u-u_0)+P_0(v-v_0)-T_0(s-s_0)+\frac{\mathscr V^2}2+gz\right]$$
So that:
$X$ is exergy of closed$^1$ system
$m$ is the mass of the system
$u$ is the internal energy of unit mass
$P$ is the pressure
$T$ is the temperature
$s$ is the entropy of unit mass
$\mathscr V$ is the speed of the system (Whole air of the room)
$z$ is the height of the center of mass of the system
The subscript of $0$ refers to the environment state.
In your case:
The room is fixed. So, we have: $\mathscr V=0$
Also we can ignore the potential energy of the air, i.e. $gz=0$
So, we will have:
$$X=m\left[(u-u_a)+P_a(v-v_a)-T_a(s-s_a)\right]$$
As we have two independent intensive properties of inside and outside air (we know temperature and pressure of them), we can determine all quantities of latter equation.
Once we found $X$, we can say $S_{\textrm {gen}}=\frac X{T_a}$ so that $S_{\textrm {gen}}$ is entropy generation.
$^1$ Our system is a closed system because we want to calculate the exergy of the system before opening the window.
| {
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Why does the frequency of a wave remain constant? They say the frequency of a wave is its fundamental character, thus remain constant throughout its propagation regardless the medium through which it travels. Could anyone explain why frequency of wave is fundamental character but its wavelength isn't?
| When source and receiver are moving away or towards each other, the observed frequency changes. This effect is called a "doppler shift".
The frequency becomes lower ("red shift) when the source moves away. The frequency becomes higher ("blue shift") when the source moves towards the observer.
So yes, one can change the frequency of light. With light this effect is noticible by looking at stars.
The same phenomenon can be observed / heard when an emergency car, with sirens on, is passing by.
| {
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looking for a technique for conducting experiment I have used a servomotor and a controlling circuit to produce oscillatory motion for an experiment that involves flapping of a tiny metal plate (~few mm) inside liquid. I could not be very successful in controlling the amplitude and the frequency precisely.
Is there any technique or method like magnetic or electric piezo crystals or robotic technology I can use to produce intended motion?
| I would reach for an Arduino or similar microcontroller to do the job. You can set the amplitude and frequency of the vibration in code which, in my experience, is always easier than tweaking an electronic circuit.
| {
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The analytical result for free massless fermion propagator For massless fermion, the free propagator in quantum field theory
is
\begin{eqnarray*}
& & \langle0|T\psi(x)\bar{\psi}(y)|0\rangle=\int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}+i\epsilon}e^{-ik\cdot(x-y)}.
\end{eqnarray*}
In Peskin & Schroeder's book, An introduction to quantum field theory
(edition 1995, page 660, formula 19.40), they obtained the analytical
result for this propagator,
\begin{eqnarray*}
& & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}+i\epsilon}e^{-ik\cdot(x-y)}=-\frac{i}{2\pi^{2}}\frac{\gamma\cdot(x-y)}{(x-y)^{4}} .\tag{19.40}
\end{eqnarray*}
Question: Is this analytical result right? Actually I don't know
how to obtain it.
| As alluded to in the other answer here, the integral can basically be evaluated in several ways. One of them is the one that OP has himself followed. (PS - OP, congratulations on completing that feat!)
Let me present here a way of computing this using Schwinger parameterization. We will use
$$
\frac{1}{a} = \int_0^\infty d\tau e^{- \tau a} ~, a > 0~.
$$
We want to compute
$$
I = \int \frac{d^4k}{(2\pi)^4} \frac{i e^{- i k \cdot x}}{k^2+ i \epsilon} = \int \frac{dk^0}{2\pi} \frac{d^3k}{(2\pi)^3} \frac{i e^{- i k^0 t + i \vec{k} \cdot \vec{x} }}{(k^0)^2 - \vec{k}^2+ i \epsilon}
$$
Do a Wick rotation $k^0 \to i k^0_E$, $t \to - i t_E$. Then,
$$
I = \int \frac{dk_E^0}{2\pi} \frac{d^3k}{(2\pi)^3} \frac{ e^{ k_E^0 t + i \vec{k} \cdot \vec{x} }}{(k_E^0)^2 + \vec{k}^2+ i \epsilon} = \int \frac{d^4k}{(2\pi)^3} \frac{ e^{ - i k_E^0 t_E + i \vec{k} \cdot \vec{x} }}{ k^2 }
$$
where in the last equation, we now have a Euclidean $k^2$ that is always positive over the range of integration. Now, we may use the Schwinger parameterization so that
$$
I = \int_0^\infty d\tau \int \frac{d^4k}{(2\pi)^4} e^{- k^2 \tau - i k_E^0 t_E + i \vec{k} \cdot \vec{x}}
$$
Now, we can do the integral of $k$ quite easily since $k^2 = \sum_i k_i^2$. This gives
$$
I = \int_0^\infty d\tau \frac{1}{(4\pi)^2} \frac{e^{ - \frac{1}{4\tau} ( t_E^2 + \vec{x}^2 ) } }{\tau^2}
$$
Now to perform the integral over $\tau$, define new integration variable $y = \frac{1}{4\tau} $. Then
$$
I = \int_0^\infty dy \frac{1}{(2\pi)^2} e^{ - y ( t_E^2 + \vec{x}^2 ) }
$$
This last integral is again the Schwinger parameter one. It converges and is nice so we compute it and find
$$
I = \frac{1}{4\pi^2 ( t_E^2 + \vec{x}^2 ) } = - \frac{1}{4\pi^2 ( t^2 - \vec{x}^2 ) } = - \frac{1}{4\pi^2 x^2 } ~.
$$
where in the last step, we have performed the inverse Wick rotation to go back to Lorentzian time.
| {
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Should the complex conjugate of a derivative of a Grassmann number include a sign? Take a real Grassmann variable, by which I mean $\theta=\theta^*$. We have
$$\int d\theta~ \theta =1,\qquad \frac{\partial}{\partial\theta}\theta=1$$
If I define the conjugation of Grassmann variables to invert their order, $$(\eta\theta)^*= \theta^*\eta^*,$$ should I then have $$(d\theta\theta)^*=\theta^*d\theta^*~?$$ But this means
$$(\int d\theta~\theta)^*=1 \quad \Rightarrow \quad\int \theta^* d\theta^*=-\int d\theta^* ~\theta^*=1,$$
so if $\theta=\theta^*$, I should have $$d\theta^*=-d\theta.$$ The same can be found for the derivative of $\theta$.
Is this the usual convention to take, or should I instead choose $$(d\theta \theta)^*=d\theta^* \theta^*~?$$
| Yes. OP is right. There is a minus. Since by convention the complex conjugation obeys
$$ (z w)^{\ast} ~=~ w^{\ast}z^{\ast}~=~(-1)^{|z|~|w|} z^{\ast}w^{\ast} \tag{1}$$
for any two supernumbers $z$, $w$ (of definite Grassmann parities $|z|$,$|w|$), we should also have
$$ (A f)^{\ast} ~=~(-1)^{|A| ~|f|} A^{\ast}f^{\ast} \tag{2}$$
for complex conjugation of an operator $A$ and a function $f$, cf. e.g. Refs. 1 & 2. Eq. (2) reduces to eq. (1) if $A$ is a left multiplication operator. It is easy to check that eq. (2) implies that$^{1}$
$$ \left(\frac{\partial_L}{\partial z}\right)^{\ast}~\stackrel{(2)}{=}~ (-1)^{|z|} \frac{\partial_L}{\partial (z^{\ast})}.\tag{3}$$
Since Berezin integration is the same as left differentiation
$$ \int \!d\theta ~=~\frac{\partial_L}{\partial \theta}, \qquad
\int \!d\theta^{\ast} ~=~\frac{\partial_L}{\partial (\theta^{\ast})},\tag{4} $$
we derive that complex conjugation of Grassmann-odd differentiation produces a minus
$$ \left( \int \!d\theta \right)^{\ast}
~\stackrel{(4)}{=}~\left(\frac{\partial_L}{\partial \theta}\right)^{\ast}
~\stackrel{(3)}{=}~-\frac{\partial_L}{\partial (\theta^{\ast})}
~\stackrel{(4)}{=}- \int \!d\theta^{\ast}.\tag{5} $$
References:
*
*B. DeWitt, Supermanifolds, Cambridge Univ. Press, 1992; eq. (2.2.19).
*S.J. Gates, M.T. Grisaru, M. Rocek & W. Siegel, Superspace, or One thousand and one lessons in supersymmetry, arXiv:hep-th/0108200; eq. (3.1.9).
--
$^{1}$ The subscript $L$ ($R$) denotes left (right) differentiation, i.e. acting from left (right), respectively. For completeness, let us mention that
left and right differentiation are connected via the formula
$$ \frac{\partial_L f}{\partial z}~=~(-1)^{(|f|+1)|z|}\frac{\partial_R f}{\partial z},\tag{6} $$
so that complex conjugation satisfies
$$\left(\frac{\partial_L}{\partial z}\right)^{\ast}f~\stackrel{(3)+(6)}{=}~ (-1)^{|z||f|} \frac{\partial_R f}{\partial (z^{\ast})}, \qquad \left(\frac{\partial_R}{\partial z}\right)^{\ast}f~\stackrel{(3)+(6)}{=}~ (-1)^{|z||f|} \frac{\partial_L f}{\partial (z^{\ast})}.\tag{7} $$
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Why do thin films need to be thin? No matter what thickness a piece of glass is wouldn't its optical thickness be close to an integer multiple of a wavelength such that it could create interference effects? I feel like I am missing something here.
| You are completely right in stating that the same effect should occur for thicker slabs. There are however at least 3 practical reasons why the effect is more easily observed in thin films.
*
*Light sources are typically not completely monochromatic. They emit slightly different colours at the same time. Imagine a light source whose wavelength varies by 0.1 percent. For a layer which is a couple of wavelengths thick, all colours will interfere destructively under the same angle. However when the layer is 1000 wavelengths thick, one colour will interfere constructively, while the other interferes destructively. The interference pattern will thus be lost. BTW, the length over which a light source can interfere is called the coherence length.
*Light sources are typically not infinitesimally small. Instead of being a point source, the source has a certain width. This means that the light effectively arrives under various angles at a certain point. For thin samples this does not matter as all these angles will interfere constructively at the same time. However for a thick layer, light under certain angles will interfere constructively, while other angles interfere destructively and the interference pattern is lost.
*Light sources are typically not completely spatially coherent. This means that the wave front is not completely neat and flat. The distance over which the wavefront is still flat (~up to half a wavelength) is called the coherence width. The result is that when a part of the beam interferes with another part of the beam, which is further away than the coherence width, the interference pattern is lost. For a thick layer, when light is applied under an angle, the light typically interferes with another part of the beam further than the coherence width. Note that mathematically reasons 2 & 3 are actually the same.
*It is difficult to make thick flat films. However not as impossible as it might seem. Wafers are typically polished up to atomic flatness over micrometer distances and vary only hundreds of nanometers over millimeter distances.
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Why must $v$ be $< c$ in the Lorentz transformations? Do these equations not apply to light? I was trying to understand how things look from the perspective of light. Looking at the Lorentz transformations, it seems that the universe would contract along the direction of movement into a plane, and time would stop. But I have heard that these equations cannot be applied to light speed, when $v=c$.
Why don't the Lorentz transformations apply when $v=c$?
| Lorentz transformations apply to objects with nonzero mass. For an object with mass, it would require an infinite amount of energy to reach light speed.
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Is there any effect on mechanical waves by electromagnetic waves (and vise versa)? Do electromagnetic waves like light and gravitational waves (due to moon for instance) affect on mechanical waves like sound?
Can sound change the path of light?
| I can answer half your question in that a sound can change the path of light.
A change in the density of the air produces a change in the refractive index of the air and so a Schlieren photograph can make this visible.
Here is a YouTube video to show a sound wave produced by clapping.
| {
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Force of water hitting a wall If you had a 8" pipe with 500psi stream of water exiting it and hitting a wall at 90 degrees 8 feet away, what would the force of the water on the wall be?
Thank you all.
Non-mathematician.
|
Using Bernoulli's equation and the momentum conservation equation, we can show that water flowing out of a pipe with cross-section $A$ at speed $v$ exerts a force $F$ on a wall (at 90 degrees), acc.:
$$F=\rho Av^2$$
With $\rho$ the density of the water.
But your specification of "8" pipe with 500psi stream of water exiting it and hitting a wall at 90 degrees 8 feet away" does not allow to compute a value for $v$. So the problem is undefined.
| {
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Torque on electric dipole placed in non uniform electric field When electric dipole placed in non uniform electric field, what is the approach to calculate torque acting on it? Can it be zero?
| The torque $ \tau $ on an electric dipole with dipole moment p in a uniform electric field E is given by $$ \tau = p \times E $$ where the "X" refers to the vector cross product.
Ref: Wikipedia article on electric dipole moment.
I will demonstrate that the torque on an ideal (point) dipole on a non-uniform field is given by the same expression.
I use bold to denote vectors.
Let us begin with an electric dipole of finite dimension, calculate the torque and then finally let the charge separation d go to zero with the product of charge q and d being constant.
We take the origin of the coordinate system to be the midpoint of the dipole, equidistant from each charge. The position of the positive charge is denoted by $\mathbf r_+ $ and the associated electric field and force by $\mathbf E_+$ and $ \mathbf F_+$, respectively. The notation for these same quantities for the negative charge are similarly denoted with a - sign replacing the + sign.
The torque about the midpoint of the dipole from the positive charge is given by
$$ \mathbf \tau_+ = \mathbf r_+ \times \mathbf F_+ $$
where
$$ \mathbf F_+ = q\mathbf r_+ \times \mathbf E_+(\mathbf r+) $$
Similarly for the negative charge contribution
$$ \mathbf \tau_- = \mathbf r_- \times \mathbf F_- $$
where
$$ \mathbf F_- = -q\mathbf r_- \times \mathbf E_-(\mathbf r-) $$
Note that
$$ \mathbf r_- = -\mathbf r_+ $$
We can now write the total torque as
$$ \mathbf \tau_{tot} = \mathbf \tau_- + \mathbf \tau_+ =q\mathbf r_+ \times (\mathbf E(\mathbf r_+)+\mathbf E(\mathbf r_-))$$
It is clear that in taking the limit as the charge separation d goes to zero, the sum of electric fields will only contain terms of even order in d.
Noting that $$ \mathbf |r_+| = \frac{d}{2} $$
and defining in the usual way $$ \mathbf p = q\mathbf d = q(\mathbf r_+ - \mathbf r_- ) $$
We can write that $$ \tau_{tot} = \mathbf p \times \mathbf E(0) + \ second \ order \ in \ d $$
As we take the limit in which d goes to zero and the product qd is constant, the second order term vanishes.
Thus, for an ideal (point) dipole in a non-uniform electric field, the torque is given by the same formula as that of a uniform field.
Note that it is not correct to start with the expression for a force on an ideal/point dipole in a non-uniform field and then calculate torque from this force. To derive this expression one ends up first taking the limit of a point dipole (on which there is zero force in a uniform field) and then one finds a torque of zero, which is incorrect. One must start with the case of a finite dipole, calculate torque and only then pass to the limit.
When p and E are parallel and anti-parallel, the torque is zero, so yes zero is possible. But the case in which p and E are anti-parallel is one of an unstable equilibrium, and a small angular perturbation will cause the dipole to experience a torque which attempts to align the dipole with the electric field.
| {
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Relation between entropy and internal energy I am confused as to what is the relation between entropy and internal energy. Entropy is always presented as a measure of the randomness in a system. So when we supply heat to a well insulated system say ideal gas in a container with fixed boundary, the internal energy and temperature increase, which implies that the motion of gas particles increases and hence the system becomes more chaotic and thus entropy increases. But if we take the same system, and supply heat isothermally and reversibly, the defnition of entropy change ΔS=Q/T , says that the entropy of system would increase(at the cost of equal entropy drop in surroundings). But for an ideal gas, internal energy is only a function of temperature and so internal energy remains constant here,no change in average kinetic energy of gas particles takes place, so where does the chaos come from to increase entropy of the system.
How are the two related?
| Entropy is the order of disorderness of a system, which means greater will be the irreversibility of a process.
Internal energy is the sum of kinetic and potential energies of particles.
Entropy increases only if there is enough energy in particles. Thus, if there is no internal energy there won't be any entropy.
| {
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Is the energy needed for a current through a straight and a coiled wire different? When you add current to a straight piece of wire does it use less electricity than if it was coiled? The power wire on telephone pole's are curved while buried cables are pretty strait in comparison. Does the curve of the wire and proximity to itself make a difference in energy consumption?
| I may be wrong, but I think Lenz's Law might provide an answer.
The circuit with the straight wire takes in current i(suppose) once the switch is closed.
The one with the looped wire, will having a changing flux through it once the switch is closed. Since any change is to be opposed, the current drawn this time will be less,(assuming the dimensions of the wire loops do not change).
NOTE: In both cases, we observe and measure the current very soon after the switch is closed.
The situation is quite like an inductor. The current grows slowly(compared to a normal scenario). However, as the equations show, after a sufficiently long time, both should draw the same current.
| {
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Does the use of $\gamma=\left(1-v^{2}/c^{2}\right)^{-1/2}$ automatically assume a (+ - - - ) metric? In Special Relativity, does the use of $\gamma=\left(1-v^{2}/c^{2}\right)^{-1/2}$
automatically assume a (+ - - - ) metric convention? For introductory textbooks, the Lorentz factor is is always defined the same. For a spacial velocity v:
$$\gamma=c\left(c^{2}-v^{2}\right)^{-1/2}$$
$$\frac{c}{\gamma}=\sqrt{c^{2}-v^{2}}$$
$$c=\gamma\sqrt{c^{2}-v^{2}}$$
which can be written as the four velocity $U^{\alpha}=\gamma v^{\alpha}$
(taking $v^{0}=c$
):
$$c=\sqrt{U^{\alpha}U_{\alpha}}$$
Which utilizes a (+ - - - ) convention. I thought this was strange since relativistic texts generally use the (-+++) convention. It may lead to confusion
| This definition doesn't depend on the metric signature convention. Note that in definition of $\gamma$ the metric doesn't appear anywhere. It is defined purely in terms of "3-vectors" and "3-scalars" measured by particular observer. So it is impossible for metric to appear here explicitly.
| {
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Quantum master equation and off-diagonal terms I have a couple of related questions
*
*What is exactly the difference between the quantum master equation and the regular master equation? My understanding is that the normal master equation is used to find a "vector" of state probabilities (like in a regular Markov chain), whereas in the quantum master equation one finds the density matrix. Is this correct? If this is the case, how does the "transition matrix" look, in the quantum case?
*I'm also a bit confused about the off-diagonal elements in a density matrix. As the density matrix is self-adjoint it can be diagonalizable in some orthonormal basis. So why do we speak about off-diagonal elements? Is it because the density matrix $\rho$ can be time dependent (and orthonormal basis stop being so as time evolves $\Rightarrow$ off-diagonal elements appear)?
| The off-diagonal terms appear when you analyze the measurement problem. Say you have a system S with some observable $\hat{S}$ such that it haves eigenvalues $s_i$ and you wish to measure the state. In order to do that you consider an apparatus $\hat{A}$ wich is initially in a pointer state $| a_0 \rangle$ so the initial state of your system is given by the tensor product:
$|\Psi \rangle = |s \rangle \bigotimes |a_0 \rangle$
The act of measurement is the interaction between the apparatus and your system so schematically your Hamiltonian will be:
$H = \hbar \Omega \sigma_z \bigotimes z$
But in general that interaction will form an EPR like state between the system and the apparatus:
$|\Psi \rangle = |s \rangle \bigotimes |a_0 \rangle \mapsto \sum c_i |s_i \rangle |a_i \rangle $
So now your density matrix has off diagonal terms wich in general you don't see. So how do we kill them? Via decoherence from the environment
| {
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Reciprocal Time Dilation in Special Relativity I'm trying to understand theory of special relativity, but there is one thing that really makes me confused which is reciprocal time dilation in special relativity.
In special relativity, the time dilation effect is reciprocal: as observed from the point of view of either of two clocks which are in motion with respect to each other, it will be the other clock that is time dilated. (This presumes that the relative motion of both parties is uniform; that is, they do not accelerate with respect to one another during the course of the observations.) - Wikipedia http://en.wikipedia.org/wiki/Time_dilation
This paragraph tells us that: as your friend flying on a high-speed moving rocket passes around you (who is at the rest in space), you see her age more slowly than yourself. She in turn will see you age more slowly.
This conclusion seems rather contradicting to me. What does this conclusion gives as the result of age-relationship between you and your friend?
Does it mean if the rocket were forever in the uniform motion going away from you, you will be always older than your friend, and friend will be always older than you depending on which frame of reference you take?
| Yes, Einstein's postulates entail symmetrical (reciprocal) time dilation, but in 1905 Einstein deduced, invalidly (in the sense that this does not follow from the postulates), asymmetrical time dilation - when the moving clock passes the stationary one, the former lags behind the latter. Nowadays you can often hear the same incorrect conclusion: "Moving clocks run slower than stationary ones", "Time slows down for you if you start moving" etc.
| {
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Electric field dependence on distance How can it be proved that for a point charge, $E$ is proportional to $$1/r^2$$ using the concept of Electric field lines (or lines of force)? I tried to show that if field lines are close, then magnitude of Electric field is higher. But, I couldn't show the given dependence.
| You can prove it using the concept of electric flux. For instance. If you surround a point charge with a sphere if r=1, or a sphere with r =10, you know that the electric flux ( field strength times area) must be the same. A sphere is easy because every point is equidistant to the charge.
| {
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Mass dropped on a spring I thought this would be a particularly simple problem but it is turning out to be quite the opposite. I am sure I am doing a very simple mistake.
The problem statement is that there is a mass which is just barely kept on the spring (help by the force equal to the weight of the spring) such that the spring is uncompressed. As soon as I let the object go, it will compress the spring and come to rest at some height. The energy stored in the spring will be equal to the difference in the potential energy at the two height of the spring (compressed and uncompressed),
$$mgh_i -mgh_f= 1/2 kx^2$$
Furthermore, the compression of the spring would just be the difference in the $h_f -h_i$. This gives me two roots for compression,
$$
x=0
$$
And the second would be,
$$
mg=1/2kx
$$
But then by this, $kx=2mg$ and $kx$ is force by Hooks law, but then does it means that the force compressing the spring is twice the weight of the object? That sounds odd. I was expecting I would just recover Hooks law but I guess I am doing something wrong here. But I am not sure what that is.
Any help will be much appreciated.
P.S. This is not a homework problem. We are designing a project for our school.
Thanks for your time.
| When the spring reaches maximum compression, the mass is instantaneously at rest but it is not in static equilibrium. The net force on it is not zero : $kx \ne mg$. Like a pendulum at the end of each swing, there is a net force on the mass causing it to accelerate towards the equilibrium position - at which the net force on it is then (for an instant) zero. Your calculation is correct. It is your interpretaion of the result which is at fault.
At the lowest point the compression force in the spring is $kx=2mg$, acting upwards on the mass, while gravity is still pulling down with force $mg$. There is a net force of $mg$ acting upwards.
| {
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How did physicists know that there are two kind of charges? Problems
The question I am about to make is either too stupid or hasn't bothered anyone because its obvious because I can't really find the answer anywhere.
I am currently studying electricity and magnetism and my book starts by telling that matter consists of atoms that are made of a heavy positively charged nucleus and a cloud of light negatively charged electrons orbiting around it and it bases the rest of the book on these facts.
However electricity, positive and negative charges were studied before the atomic theory was confirmed. Actually at that time it wasn't even remotely supported. Benjamin Franklin for example thought that electrical charge was some kind of cloud/gas. When there was abundance of this cloud the charge was positive while otherwise it was negative.
Questions
Firstly, I suppose that classically we define charge as the physical property that objects have to have in order to interact electrically - at least for now.
*
*How did physicists back then know about the existence of two charges, positive and negative? Sure, if you bring two glass rods close after rubbing them with silk they repel each other, while if you bring glass rod and one plastic rod they attract each other, but is that really enough?
*Moreover, how did they know that opposite charges attract while same charges repel each other? You can't arbitrarily choose to be so, since electrons really do repel each other. Was there a way to tell that a glass rod had abundance of that cloud after the rubbing while the plastic had a deficit?
In my book it is said that Franklin, after a series of experiments, determined that there are two kinds of charges without elaborating. I cannot find those experiments anywhere. All I get is about the famous kite experiment.
| Among competing hypotheses, the one with the fewest assumptions should be selected.
Some electrified objects repel, some attract. This can be explained by two kinds of charge. Nothing that cannot be explained by two charges can be explained by adding a third kind of charge. So we continue to describe electricity as occurring in two kinds.
| {
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Sign mistake calculating electrostatic potential energy formula! I have a problem calculating the electrostatic potential energy.
I rely on these equations coming from mechanics:
\begin{equation}
U_{B}-U_{A} = -W_{A \ \rightarrow \ B} (done\ by \ the \ field \ force)
\end{equation}
\begin{equation}
U_{B}-U_{A} = W_{A \ \rightarrow \ B} (done\ by \ the \ opposite \ force)
\end{equation}
According to the next picture
Work done by the coulomb force (field force) is:
\begin{equation}
W= \int_{A}^{B} \! \vec{F}.\,\vec{dr}
\end{equation}
According to the picture
\begin{equation}
F = \frac{q_{1}q_{2}}{4\pi e_{o} x^{2}} \vec{i}
\end{equation}
\begin{equation}
\vec{dr} =- dx \vec{i}
\end{equation}
Therefore:
\begin{equation}
W= \int_{A}^{B} \! \vec{\frac{q_{1}q_{2}}{4\pi e_{o} x^{2}} \vec{i}}.\,(- dx \vec{i})
\end{equation}
let $B=r$ and A=$\infty$ be
\begin{equation}
W= -\int_{\infty}^{r} \! \frac{q_{1}q_{2}}{4\pi e_{o} x^{2}} \, dx
\end{equation}
Let $B=r$ and A=$\infty$ be
\begin{equation}
W= \frac{q_{1}q_{2}}{4\pi e_{o} } (\frac{1}{x} from\ \infty \ to \ r )
\end{equation}
Then:
\begin{equation}
W= \frac{q_{1}q_{2}}{4\pi e_{o} r}
\end{equation}
When I put this result into equations at the top:
\begin{equation}
U_{B}-U_{A} = -\frac{q_{1}q_{2}}{4\pi e_{o} r}
\end{equation}
As $U_{A} =0$
Finally:
\begin{equation}
U_{B} = -\frac{q_{1}q_{2}}{4\pi e_{o} r}
\end{equation}
It turned out the potential energy is negative, but it is suppose to be positive since a external force is putting energy into the system. I don't know where my mistake is!
| You are doing a "backwards" integration. From higher to lower rather than from lower to higher x-axis values.
Integrating "backwards" from $\infty$ to $r$ (backwards because $\infty>r$) is the "flipped" and "opposite" version of the one from $r$ to $\infty$,
$$\int_\infty^r \dots dx=-\int_r^\infty \dots dx\,.$$
$\int$ is a generalized summation symbol. It sums up all the small bits. $\int_\infty^r$ and $\int_r^\infty$ should cover the exact same thing. Mathematically, the same area under the drawn graph; physically the same displacement. One is just columns summed left-to-right, and the other columns summed right-to-left; or physically summations over the same path, just starting from the right and then from the left. They should be exactly equal.
But writing them out will give a sign issue, because we always say "final situation" minus "starting situation":
$$\int_\infty^r \dots dx=\underbrace{F_r}_\text{final}-\underbrace{F_\infty}_\text{start}\qquad\quad \int_r^\infty \dots dx=\underbrace{F_\infty}_\text{final}-\underbrace{F_r}_\text{start}$$
$F_r-F_\infty$ and $F_\infty-F_r$ are not the same - their signs are opposite. But we know that they are exactly equal - just summed from different starting points. This mathematical issue means that integrating "backwards" requires us to add a minus sign. Otherwise we do not get the same area under the graph or the same path summation.
| {
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Why can we see the moon when it is between the Earth and the Sun? A rather stupid question, why can we see the moon when it is between the Earth and the Sun?
| The premise of this question is wrong.
If the moon is in between the earth and the sun (as shown on your diagram), and you can see the moon, then it is day, not night:
If on the other hand, you are on the opposite side of the earth during that configuration (so that it is night), then you can't see the moon because the earth is blocking your view of it:
| {
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Extension of Schrödinger's cat thought experiment My question is quite simple. In the thought experiment of Schroedinger's cat: When the scientist measures the state of the cat, its wavefunction collapses into either the alive or dead state. But wouldn't then the scientist in turn be in a superposition of measuring dead respectively alive until someone opens the door to the laboratory and asks the scientist about the outcome of the experiment (and therefore measures the state of the system)?
| It depends of what interpretation of quantum mechanics you are using. By interpretation it is meant that the mathematical predictions of the quantum mechanics formalism are the same, but the philosophical meaning of each is what differs. In the copenhagen interpretation that you seem to describe, the wave function collapses when a conscious observer makes a measurement. Before the measurement the system can be in a superposition of states. It is supposed that the cat is not conscious thus, in that interpretation the wave function does not collapse and the cat is in a superposition of states. If a human opens the box then the function collapses to either live or death. The human observer will not be in a superposition of states.
But this is an old interpretation and the are many others. In particular, in the Bohm interpretation there is no wave function collapse and the cat is in a specific state (either alive or death) regardless of he being observed or not.
| {
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Plotting hydrogen wave functions This may sound a bit dumb but how do I plot the hydrogen wave functions? For example, what is exactly being represented in this image? Is it just the norm-squared of the wave function and is the z-axis sticking out of the page?
I'm not sure how to use any other application but I'm using the Mac grapher tool. Say I wanted to plot $\psi_{200}$. The combined radial and angular equation for this state is $$\psi_{200}=\frac{1}{4\sqrt{2\pi}a_0 ^\tfrac{3}{2}} \left(2-\frac{r}{a_0}\right)e^{-\tfrac{r}{2a_0}}$$ where $a_0$ is the Bohr radius. As I said, i'm pretty clueless. I'm not sure how to plot this in spherical coordinates so I just converted $r=\sqrt{x^2+y^2+z^2}$. Basically, I plotted $$z=\frac{1}{4\sqrt{2\pi}a_0 ^\tfrac{3}{2}} \left(2-\frac{\sqrt{x^2+y^2+z^2}}{a_0}\right)e^{-\tfrac{\sqrt{x^2+y^2+z^2}}{2a_0}} $$When I do plot this, I get a flat plane with a sort of half-sphere in the centre, which is not what the image I linked above shows. I also tried graphing $\psi^2$ but I still did not get it. I feel like i'm missing something big.
| The plots you see in the Wiki images, are, as their title suggests, probability density plots ($\psi^2$, for Real $\psi$). Light shaded areas represent high probability areas, darker areas lower probability density.
Furthermore, they've been sliced, e.g. with an $x,y$-plane. For radially symmetric functions like $\psi_{2,0,0}$ (aka the $2s$ orbital) the specific slicing plane doesn't even matter: the $\psi^2$ value is the same in all directions as it only depends on $r$.
There's little to be gained from these plots for radially symmetric wave functions with regards to the simpler radial distribution plots like these.
If you really do want to plot radially non-symmetric probability density functions (e.g. for $2p$, 3$d$ or $4f$ orbitals) you'll need to find a tool to generate contour plots. I imagine advanced functions of Mathlab allow to do that.
Here's a *contour plot* of $\psi^2_{200}$ for $z=0$ (the $x,y$-plane):
The white areas are those of high probability density. The plot clearly shows the 'shell-like' structure of the hydrogen $2s$ orbital.
The plot was obtained by means of Wolfram alpha, by plotting the square of:
$$\psi=\frac{1}{4\sqrt{2\pi}a_0 ^\tfrac{3}{2}} \left(2-\frac{\sqrt{x^2+y^2}}{a_0}\right)e^{-\tfrac{\sqrt{x^2+y^2}}{2a_0}}$$
For $a_0=1$.
Here's a excellent resource for visualised wave functions: the Orbitron
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Is there a prohibited region in $P-V$ plane? Polytropic process generalize the particular thermodynamic processes with
$$P V^{n}= \mathrm{constant}$$
Where, if $n$ changes, the curve on $P-V$ plane changes, as shown in the diagram.
The orange region is not touched by any curve, so there is no value of $n$ for which the gas goes directly in to the orange region.
Why is that? I do not see any particular reason why there should not exist a process to make the gas go into the orange part.
| In a polytropic process other than adiabatic, you are controlling the temperature in tandem with P and V in such a way that n is constant. You can certainly achieve negative values of n by controlling the temperature appropriately.
From the ideal gas law, if T and P are expressed parametrically in terms of V, then:$$\frac{P}{P_0}=\left(\frac{V_0}{V}\right)^n$$
$$\frac{T}{T_0}=\left(\frac{V_0}{V}\right)^{n-1}$$
Just substitute a negative value of n.
| {
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Why don't high pressure gases stored in containers lose energy? Containers holding gas at a high pressure don't slowly lose the internal energy of the gas. It seems like the high speed particles would collide with the metal walls and slowly transfer their energy to the slower particles outside the container.
Even if the pressure is from more particles in the container, they can do work when released so they have energy. Shouldn't that energy dissipate over time?
| re: "Why don't high pressure gases stored in containers lose energy?"
They can gain & lose energy:
Energy (heat) is lost from a gas as the gas is compressed (whether thru mechanical compression or thru cooling compression (e.g. passing a gas thru a tube that is immersed in a very cold liquid -- like liquid nitrogen).
Energy (heat) is gained by a gas when a very cold gas is transferred to a warmer tank (e.g. especially from a high pressure tank to a lower pressure tank that isn't super insulated).
Energy (heat) can also be lost or gained while a gas is stored in a tank as ambient external temperatures fluctuate.
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If change in position over time is average velocity, why doesn't change in position over time squared equal average acceleration? For example, let's say a car is experiencing an acceleration of $1$m/s$^2$, for $6$ seconds so it goes $18$m. Now the average velocity is found through dividing $18$m by $6$s which is in line with the formula $v_\text{avg} = \frac{\Delta x}{\Delta t}$. And indeed, the average velocity is $3$m/s.
Acceleration has units of distance divided by time squared, however the average acceleration is not $18/6^2 = 0.5$m/s$^2$, the average acceleration is $1$m/s! So I have two questions from this:
*
*What exactly is that $.5m/s$ signifying? I know the kinematic equations including $\Delta x = v_0t+\frac{1}{2}at^2$ and this would allow us to find acceleration.
*Aren't the units a bit deceiving on acceleration? Maybe I'm just not super comfortable visualizing second derivatives yet but if I have an $m/s^2$ I feel like I should be able to plug in meters and seconds and get the average acceleration. And I feel like it's part how we define the units also. Because:
$$v = \frac{\Delta x}{\Delta t}$$
$$a = \frac{\Delta v}{\Delta t}$$
Substitution then gives us:
$$a = \frac{\Delta\frac{\Delta x}{\Delta t}}{\Delta t}$$
Which is different than the units seem to imply, from my perception.
| The glitch in your logic is that you supposed acceleration to be distance/time squared while using your formula: $$ \Delta x = v_0t+\frac{1}{2}at^2 $$
The above formula gives $$ a= \frac{2\Delta x}{t^2}$$ supposing that vo is equal to zero.
This makes the acceleration equal to the average acceleration in your case since the car is moving in one direction and with a constant acceleration.
| {
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Problem on Emissivity and absorptivity I have read the radiation chapter, where I have been introduced with the terms emissivity and absorptivity. emissivity tells about the ability to emit heat energy as thermal radiation compare to a black body. and absorptivity is the amount of heat absobed by body devided by the incident heat energy .but my question is ,are these emissivity and absorptivity constant or depends on temperature?
Actually I had a misconception that these quantities are intrinsic property of body
If I have really wrong concept and they depends on temperature then tell me and
Tell me how they depends on temperature
I mean I need The mathematical expression
| The figures below support Jim's statement.
Figure 1: The specific spectral radiation of a black (Schwarzer Strahler), grey (grauer Strahler) and real body (realer Strahler) at the same temperature and different wavelengths $\lambda$.
Figure 2: Specific spectral radiation of black bodies ($\varepsilon=1$) at different temperatures and wavelengths $\lambda$. According to Wien's displacement law. Note the wavelengths of sunlight are roughly at $380-780\ \mathrm{nm}$.
Source: Wikipedia.
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Harmonic oscillator in quantum mechanics I have brief questions regarding the attachments, which are notes from the book Introduction to Quantum Mechanics by Griffiths which explains the harmonic oscillator case. Any assistance would be appreciated. The attachments don't look healthy but the questions are quite simple.
How do we get [2.77], isn't $A$ a constant in the general solution [2.75], why does it become a function of $\xi$?
Lastly, why is this method of terminating the power series taken (which involves letting $a_{n + 2} = 0$ for some $n$ and letting either the odd or even terms all be zero) Surely there are other ways to define the power series so that they terminate (maybe for some $n$ let $a_j = 0$ for all $j \geq n$)?.
Thanks a lot for any assistance.
| *
*$A$ is constant in the approximate solution at large $\xi$. So Griffiths makes the ansatz that for general $\xi$, the solution is of the form for some function $A(\xi)$ that becomes "constant" compared to the exponentation at large $\xi$. It's an ansatz, it is not derived.
*Look at the condition on $K$ that terminating the sequence starting from a non-zero term imposes: If the sequence terminates at $n$, you have $K=2n+1$. If both odd and even terms were non-zero from the start, then you would have them terminating at $n_\text{even},n_\text{odd}$ with $2n_\text{even}+1=K=2n_\text{odd}+1$, which is impossible.
| {
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Does the speed of light in vacuum define the universal speed limit?
*
*Is light the thing causing the universal speed limit to be $299\,792\,458\,\mathrm{m/s}$? So the universal speed limit would be different if light travelled faster or slower?
*Or, is $299\,792\,458\,\mathrm{m/s}$ the universal speed limit anyway and light just goes that fast? Light is just something we commonly associate with it because it goes super fast.
| It's the second one: the reason the speed $299792458\ \mathrm{m/s} = c$ is special is because it's the universal speed limit. Light always travels at the speed $c$, whatever that limit may be.
The reason there is a "universal speed limit" at all has to do with the structure of spacetime. Even in a universe without light, that speed limit would still be there. Or to be more precise: if you took the theoretical description of our universe, and remove light in the most straightforward possible way, it wouldn't affect $c$.
There are many other things that depend on the speed $c$. A particularly important one is that it's the "speed of causality": one event happening at a particular time and place can't affect another event unless there's a way to get from the first event to the second without exceeding that speed. (This is sort of another way of saying it has to do with the structure of spacetime.)
| {
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Why decrease in velocity will increase pressure? According to the Bernoulli's equation, if velocity decreases, then pressure increases.
I am trying to understand the Bernoulli's effect based on a situation.
Suppose we have a stream of water. Let's assume it is an ideal fluid. Imagine the water flows out from a wider pipe to a narrower pipe. Since the area decreases, according to the Continuity equation, velocity of water molecules increase. This causes an decrease in pressure.
I don't understand the last part. If water molecules' velocity increase, then their kinetic energy also increases. Wouldn't this causes more collision between pipe's wall and water molecules, thus giving higher pressure?
| Pressure is momentum transfer due to molecular collisions once you have subtracted out their average motion. So decrease in pressure due to increase in average speed may be construed as transfer of kinetic energy from random molecular motion to mean motion. This means that random molecular motion (by which I only mean molecular motion with average subtracted out) now contains less energy, less momentum, and thus results in lower pressure reading.
Recall how pressure is measured in a pipe, for example. Pressure gauge is fitted on the wall such that flow does not directly impinge on it; otherwise you would be measuring total energy which manifests itself as a pressure head (called stagnation pressure), and in an ideal fluid (in which there is no viscous dissipation) this latter pressure would be constant everywhere.
| {
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Standard convention for $x$ error bars What is the standard convention regarding the error bars of the independent quantity in a graph? In what situations should I show the $x$ error bars? In case both $x$ and $y$ uncertainties are comparable or neither can be disregarded, should I show both or only the greater one?
I know this matter is more related to aesthetics and convenience, but I wonder if there is more straight rule for that.
| If you are measuring y at some value x, and both quantities have uncertainty, then in principle you should show the uncertainties on both axes.
In some circumstances you might omit the x error bars. This would be the case if the y value depends on x such that
$$\Delta y \gg |dy/dx| \Delta x,$$
where $dy/dx$ is your best estimate of the gradient of $y(x)$. In other words, where y changes by much less than its error bar over a change in x equal to the x error bar.
| {
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Large-scale antimatter production From what I can find, presently the only known means of producing antimatter consist of directing particle accelerators at various targets, and only infrequently getting positrons or anti-protons as byproducts of particle interactions.
Assuming a sufficiently large source of energy: Have more efficient means of producing antimatter been conceived?
And based on these is there a known upper bound on antimatter production efficiency? For example:
*
*If all known theory requires that an anti-particle be produced with its particle pair, then efficiency will always be under 50%.
*Are there known processes for "transmuting" matter into anti-matter? If so, do these allow for a theoretical upper-bound on production greater than 50% of input energy?
| Magnetic field containers can be used for the storage of antimatter. Also, increasing the size of the particle accelerator will increase the volume output of antimatter, the key fuel in warp drives. The size of particle accelerator can be 100 miles around by diameter and 30 stories high.
| {
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Water droplet lensing Suppose I have a camera that is looking straight down at a single droplet of water resting on a flat surface. The droplet is small enough that surface tension forms it into a rounded shape. Inside the water, there is suspended an object that I'd like to image and measure. Is it possible to somehow calibrate for the lensing effect of the drop and make accurate measurements of objects suspended inside it?
I know how to deal with calibration of the camera itself and measurement of objects in a flat focal plane, with no water surface in the way. So I think the core problem here is figuring out how to 'dewarp' the image seen through the water droplet into an equivalent flat plane image.
If I know the shape of the droplet, I think I can create a model of the optics and use it to calibrate using known reference features positioned under droplets, but I'm uncertain what parameters determine the shape of a water droplet on a flat surface. I'm planning to do some experiments, but I'm interested in identifying prior research to review on the subject.
| The curvature of the drop is influenced by the liquid-solid, solid-air and liquid-air interface forces. It can be determined by the Young-Laplace equation. For more details see this article
| {
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Does a wave function describe the motion of electrons or atoms? I took the course of quantum mechanics a while ago. I do not quite remember all the detail on how to derive the wave function for hydrogen but I still remember the general picture. I think the text always start the discussion with hydrogen because this atom contains 1 electron only. It is a simple 2-body model so we could solve the problem in the frame of center of mass. In other situation for atom has more than 1 electron but the outermost shell contains only 1 electron, we could reduce it as a hydrogen-like ion with 1 electron. Since the nucleus or the hydrogen-like ion is much heavier than the electron, we consider it a rest ion so the wave function we derive is mostly describing the motion of the electron? But in the book, they always mention the wave function of an atom. Is it the same thing to say the wave function for the atom or the electron (in the atom)?
As I still remember, if the wave function is used to describe the atom, the modulus square of the wave function is interpreted as the probability to find the atom in space; but if it is for electron, we should say it is interpreted as the probability of finding the electron in space. It is confusing.
| Atoms.
We sometimes use somewhat loose language when we speak of the wave function of the electron. We typically either have chosen the center of mass frame, or pretended that the nuclei are fixed (as you point out). It's the atom that has energy levels.
We can get away with the loose language because if we really were able to fix the positions of the atoms.
| {
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What is the measure of distance in higher dimensions? In our world we are using kilometers to measure distance. What measurement is used to measure distance in higher dimensions?
| Distance measurements in $n$ dimensional flat space follows the same pattern for $n$ equal 1,2,3, or higher values.
I'm going to assume a straight line, change in position to simplify the math (that is we're measuring what a introductory book would call the "displacement" $s$ rather than distance. But then distance is just an accumulation of many magnitudes of displacement, so the full case follows.
You should already be familiar with the one, two and three-dimensional cases (writing $s_{(n)}$ for the magnitude of displacement in a $n$ dimensional space):
\begin{align*}
s^2_{(1)} &= (\Delta x)^2 \\
s^2_{(2)} &= (\Delta x)^2 + (\Delta y)^2\\
s^2_{(3)} &= (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \,,
\end{align*}
and the extension to higher dimension follows the same pattern (allowing for the choice to label the usual dimensions with the last three letters of the alphabet):
\begin{align*}
s^2_{(4)} &= (\Delta w)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \\
s^2_{(5)} &= (\Delta v)^2 + (\Delta w)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \,.
\end{align*}
For generality, however, we might prefer to label our coordinate directions with a number, so that $x$ is named $x_1$, $y$ is named $x_2$ and so on. That lets us write
$$ s^2_{(n)} = \sum_{i=1}^n (x_i)^2 \,. $$
From this you also get an explanation of why the small, compact dimensions posited in some candidate next generation theories don't effect day-to-day life. Differences of position in those dimensions contribute so little to the total displacement between two points that we can't detect the effects.
This approach can be extended to a Minkowski space (the space of special relativity, by measuring interval in the form
$$ s^2_{(n)+(m)} = \left(\sum_{i=1}^n (ct_i)^2\right) - \left(\sum_{i=1}^m (x_i)^2\right) \,, $$
for a space of $n$ time-like dimensions and $m$ space-like dimensions.
Extension to general relativity requires the introduction of a metric.
| {
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Jones formalism for calculating quarter waveplate angle for circular polarized light I am planning to change the polarization of a vertically linear polarized laser to circular polarized light with the help of a quarter waveplate.
I know the final result: I have to rotate the fast axis of the waveplate 45° to the incident polarization axis in order to achieve left handed circular polarized light. A rotation of -45° results in right handed polarization.
My task is know to use the Jones formalism to calculate this result.
Vertically linear polarized light can be written in Jones matrices as
\begin{pmatrix}1 \\ 0\end{pmatrix}
right handed circular polarized light can be written in Jones matrices as
$$\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -i\end{pmatrix}$$
left handed circular polarized light can be written in Jones matrices as
$$\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ i\end{pmatrix}$$
Now I don't know how to proceed from here. Has anyone a good advise for me?
| Hints:
*
*The Jones matrix for e.g. right circular polarized light is: $$\frac{1}{2} \begin{pmatrix}
1 & i \\ -i & -1
\end{pmatrix}$$ To see this try applying this matrix to your linearly polarized Jones vector. It will give a right circularly polarized Jones vector.
*The Jones matrix for a quarter wave plate is:$$\begin{pmatrix}
1 & 0 \\ 0 & e^{i\pi/2}
\end{pmatrix}$$
You can see this e.g. by applying it to the vectors $\begin{pmatrix}
1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix}
0 \\ 1 \end{pmatrix}$ separately and observing the phase difference that is introduced between the two.
*Furthermore you can always build in rotation matrices, which corresponds to physically rotating e.g. your quarter waveplate relative to the incoming polarization.
Now you can try to build the required matrix (see first point above) from a quarter waveplate and rotations, which will solve the problem.
| {
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How do I calculate a upward-pulling force on a swinging pendulum? I'm trying to implement a simple pendulum using a 2d physics system that can model rigid bodies with gravity. The problem is that I don't know how to calculate the upward-pulling force of the rope, as in this image:
I've only found equations for calculating velocity of the pendulum, as shown on wikipedia, but the problem is that I can't change the velocity directly, as I need to be applying a force.
The problem is probably just the magnitude of the force vector. Since in the idle position the upward force is equal to negative gravity, I thought I could simply take something like $cos(\theta) \cdot v \cdot G$ where $v$ is the pendulum vector and G is gravity magnitude, but that doesn't work in my simulation.
| The rope pulls just enough that the pendulum doesn't fall to the ground, but follows an arc.
The following picture shows you how work out the force for the static case (no motion of the pendulum):
However, you need to take account of the fact that the pendulum is moving in an arc. When something moves in an arc, you need an additional force $F=\frac{mv^2}{r}$ to provide the centripetal force needed.
Combine these two forces, to obtain the total "mysterious force". Can you figure it out from here?
| {
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Principle of Least Action Question Let's say we have a particle with no forces on it. The path that this classical particle takes is the one that minimizes the integral
$$\frac{1}{2}m\int_{t_i}^{t_f}v^2dt.$$
So if we graph this for the actual path a particle takes it is a straight, horizontal line on the $(t,v^2)$ plane. But couldn't we lessen the integral if we first slow down and then speed up near the end to create a sort of parabolic line that has less area under the $(t,v^2)$ plane? So why doesn't the particle take this path? What am I missing in my thinking?
| Your proposed path has a VERY LARGE action. As @tparker pointed out, you have to minimize the path subject to the constraint that the average velocity doesn't change. Now, the action is quadratic in velocity. A little fiddling around with the math should convince you that to minimize the integral of $v^2$ subject to the constraint that the average velocity doesn't change, you want the velocity to be constant. Intuitively, going fast for a bit raises the action a LOT (because velocity is squared) while going slow for a bit doesn't do much to lower the action (because velocity is squared).
| {
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Does the universe expand in every direction evenly? I've heard that the universe is expanding constantly and that galaxies are moving further and further away from each other because of this. However, does the universe expand in every direction evenly or does it expand in one direction more than another direction?
| It is an assumption that the universe expands evenly in all directions, and the experimental evidence so far confirms the assumption.
Our mathematical description of the expanding universe is based on the assumption that on a very large scale the universe is homogeneous and isotropic, which basically means it's the same everywhere and in all directions. Since it's the same in all directions the expansion is the same in all directions.
I must emphasise the the homogeneity and isotropy is just an assumption that we make to make it easier to calculate how the universe expands. This assumption has to be checked by experiment to make sure it's correct. The main way this is done is by measuring the cosmic microwave background. If there were any anisotropies we would expect them to show up there, but we do not see anything that suggests an isotropy of the expansion. There are actually some weird things in the CBM e.g. the axis of evil. However opinion is divided about what exactly these obervations mean.
| {
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Spring behaviour under high atmospheric pressure We have a spring inside a chamber. We compress the spring and then let it decompress freely. Will its decompression (its speed and displacement) be the same if the air pressure of the chamber is $1\;\mathrm{atm}$ or $3000\;\mathrm{atm}$? If not, how will it be affected?
| Damping
The density of atmospheric air is approximately 1.225 kg/m$^3$. At 3000 atm, the density would be 3675 kg/m$^3$ compared to the density of water of 1000 kg/m$^3$. I don't know the viscosity of high density air, but as @DirkBruere answered, it could be a significant factor in the damping of the spring decompression.
The relative difference in decompression speed would depend on the spring stiffness. For example the motion of a slinky would be affected in a major way, but the motion of a car suspension spring might be similar in either environment.
Stiffness Change
Another factor to consider is the deformation of the spring due to the high pressures at its surface. 3000 atm is approximately 44,000 psi. That is likely enough to deform the steel cross section, making a difference in the spring stiffness.
| {
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Bell inequality violations evidence for 1935 EPR claims? Is it possible that Bell inequality tests provide experimental evidence in support for the EPR claims in their 1935 paper titled "Can Quantum-Mechanical Description of Physical Reality Be Considered Complete?", where Bell tests eliminate local realism as cause for entanglement, but allow for correlations which may be non-causal? In short, is it possible that quantum entanglement is just simply non-causal and therefore will never offer an opportunity to allow communication?
| Yes - these experiments have been conducted, most famously by Aspect et al., but also by others, see Wikipedia. They all observed violations of Bell's inequalities - Our world is therefore not local-realistic in the sense of Einstein.
An extension of Bell's inequalities by Legett (Legett inequalities) holds for non-local realistic theories. Their violation has been observed as well and therefore also ruled out this particular version of "realism".
Independent of that, it is impossible to send information via quantum entanglement by the No-communication theorem. In essence, the problem is that you need additional information in order to make sense of the observed states of the entangled particles, this information however can only be transmitted via classical channels.
| {
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Why is the resistance in a parallel circuit smaller than the resistance in a series circuit? So I was told in the physics class that the resistance in a parallel circuit is smaller than the resistance in a series circuit. Why does that happen?
Is this statement also true for circuits which have no resistors or resistance-offering devices connected to them?
And also from my calculations the total resistance in a parallel circuit is smaller than the resistances of any of the devices connected in the circuit. Now I really don't understand how this can happen.
| Question: Which of two pipes of equal length offers less resistance to the flow of water, one of which has twice the cross sectional area of the other?
Answer: The one of twice the cross sectional area.
But the one of twice the cross sectional area can be thought of as two of the smaller cross sectional area pipes in parallel.
This analogy gives an idea of the smaller resistance of a parallel arrangement of resistors although not completely.
A length $l$ of wire of cross sectional area $A$ has a resistance $R$ given by the equation $R = \dfrac {\rho l}{A}$ where $\rho$ is the resitivity of the wire.
Doubling the cross sectional area of a fixed length of wire decreases the resistance by a factor of two which is equivalent to having two wires of area $A$ in parallel.
However for a pipe the "resistance" to fluid flow is proportional to $\dfrac {1}{\text{area}^2}$.
So in the case of a pipe having the area increase by a factor of two decreases the resistance to fluid flow by a factor of four.
| {
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Plane waves in special relativity I don't understand how there can be plane waves that by definition are spread through all of space if nothing can travel faster than light.
Wouldn't every wave have to spread over time with at most the speed of light?
I could understand when they are only appearing in the mathematics of calculating waves, but isn't it common to talk about light or photons in terms of plane waves?
| Plane waves are not real, they are just a mathematical device.
In quantum mechanics, particles are represented by wave packets, which do not have infinite amplitude and allow a collection of plane waves to group together by interfering constructively within a certain area and destructively outside that area.
From Wikipedia Wave Packets
In physics, a wave packet (or wave train) is a short "burst" or "envelope" of localized wave action that travels as a unit. A wave packet can be analyzed into, or can be synthesized from, an infinite set of component sinusoidal waves of different wavenumbers, with phases and amplitudes such that they interfere constructively only over a small region of space, and destructively elsewhere.Each component wave function, and hence the wave packet, are solutions of awave equation. Depending on the wave equation, the wave packet's profile may remain constant (no dispersion, see figure) or it may change (dispersion) while propagating.
| {
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Principal moment of inertia for a rotating body My major is not in physics.
I am reading the following paper: (my problem is simple and not related with any optimization)
http://arxiv.org/abs/1410.2841 (p.5~p.6)
Suppose
*
*The body angular velocity is $\Omega(t_0) = [\omega \ \ 0 \ \ 0]^T$, $\omega$ is a constant.
*Torque free motion
My problem:
In this case, can I obtain the result: $I_1=I_2 = I_3$ since $\dot{\omega}_i=0$? It is bit odd since the paper assume $I_1\geq I_2\geq I_3$, so the only possibility is "$=$".
| No you cannot assume that. The initial rotation is about the major axis, and it will continue to be so (in the absence of torque, and since you were already rotating about the major axis). Instead, since $\omega_2=\omega_3=0$, your equations for the evolution of the angular momentum don't require the moments of inertia to be the same.
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Can we exit the event horizon of merging black holes? I have an intuitive scenario. Consider we have a spaceship just below the event horizon of a BH, which is merging with another black hole.
Finally, the singularities merge and we have a single black hole again.
But, in the transient stage, it is unclear to me if a timelike world-line would exist to leave the system.
I suspect, the metric is probably far too complex for an analytical solution, but in the worst case, it could be maybe solved numerically.
As far I know, black hole merges are examined mainly in an inspiral scenario. I suspect, maybe the escape is possible only if they have a hyperbolic-like orbit (i.e. there is no inspiral, but they simply collide).
Is it possible?
| No. When they merge their horizons will change shape, and eventually become the static or stationary shape of a BH horizon. Nothing inside either horizon while this is happening can escape. At all times the timelike curves stay inside, and the deformed horizons are where the lightlike curves end up. In each, and after they merge.
The area of each horizon right before they merge can not be smaller than before, as area is proportional to entropy which must increase or stay the same. All deformations will increase it (or be the same, but probably increase). At no times can lightlike Curves escape because of some deformation, and much less timelike curves.
I assume you meant you were right inside before. If you meant right outside anything can happen, now you'd have to take the ergospehereergosphere into account, and if inside it also probably no but I am a not sure.
There was similar question posted maybe 3 or so months ago, not in my saved list so I can't give you the reference. There were some answers.
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Mechanism for collapse of iron stars into black holes via quantum tunnelling In the wikipedia page "Future of an expanding universe" it refers to the scenario of a future without proton decay.
The page talks about how processes would lead to stellar-mass cold spheres of iron, calling these objects "iron stars":
In 101500 years, cold fusion occurring via quantum tunnelling should make the light nuclei in ordinary matter fuse into iron-56 nuclei. Fission and alpha-particle emission should make heavy nuclei also decay to iron, leaving stellar-mass objects as cold spheres of iron, called iron stars.
Under the heading "Collapse of iron star to black hole" it then says :
Quantum tunnelling should also turn large objects into black holes. Depending on the assumptions made, the time this takes to happen can be calculated as from 101026 years to 101076 years. Quantum tunnelling may also make iron stars collapse into neutron stars in around 101076 years.
How would quantum tunnelling lead to the collapse of an iron star to a black hole?
| Well iron stars will collapse due to quantum tunneling. Iron from the surface of the iron star over a really really really really really long time will go to the core. This will happen to all the iron atoms. Then the iron star will be so dense that it collapses into a neutron star. This neutron star then has the ability to turn into a black hole and by hawking radiation evaporate.
So in summary quantum tunneling tunnels iron nuclei from the surface of the iron star to the core forcing the density and gravity to go up until it turns into a neutron star and inevitably to a black hole.
https://en.wikipedia.org/wiki/Iron_star
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What frequency is the scratching of finger nails on a blackboard? This is the frequency/intensity that sets my teeth on edge.
Does anybody know what frequency (roughly) it is? I am guessing it is near the top of normal human hearing, 20kHZ, but I'm not sure if that's why it affects me.
I am sure the same frequency is played on some of the music I listen to, but somehow, it does not make me wince.
There is a related question here, with no answer Scratching on a Blackboard, but I just want a frequency value.
| From http://www.livescience.com/16967-fingernails-chalkboard-painful.html:
Interestingly, the most painful frequencies were not the highest or lowest, but instead were those that were between 2,000 and 4,000 Hz. The human ear is most sensitive to sounds that fall in this frequency range, said Michael Oehler, professor of media and music management at the University of Cologne in Germany, who was one of the researchers in the study.
No one knows all of the reasons why that sound is so painful to listen to, but some theorize that we evolved ear canals to amplify human speech as much as possible, and that sounds like this happen to have large portions of their energy in that frequency band.
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How does the LHC explore extra dimensions? The Large Hadron Collider (LHC) has been smashing particles for a long time and sometimes people say that it has found new dimensions. How is it even possible for a particle accelerator to find new dimensions?
| First, no evidence for other dimensions has been found.
However, there are ways for particle colliders to detect other dimensions. One of the main ones is to see if any energy "disappears" under very certain circumstances...then it could've possibly gone into another dimension.
Another way is to look for particles that can only exist if there are other dimensions. These particles would be around 100 times the mass of the W and Z bosons (these carry the electroweak force). Particles of this size could really only be detected by the LHC.
Other ways include examining the evaporation of mini black holes and seeing what particles are produced and looking for gravitons - gravitons would quickly disappear into extra dimensions, leaving an apparent loss of energy, which could be detected.
I'll be updating this as I find more information, though these are some of the main ways. Hope this helps!
This website has more information.
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Coleman-Mandula theorem and mass gap I had a couple of naive questions about Coleman-Mandula theorem.
*
*One of the assumptions of the theorem is the non-existence of massless particles in the spectrum. Since we do have massless photons in the standard model, how is the theorem relevant?
*Why aren't there examples of relativistic theories with hybrid symmetries and a massless particle in the spectrum (like some extension of QED)?
| The Coleman-Mandula theorem (CMT) does not rule out theories with massless particles. What it rules out is a theory with only massless particles. If you only have massless particles you either:
*
*End up with a free theory. This theory has trivial S-matrix, thus, the CMT does not apply here
*Have conformal symmetry. If you have conformal symmetry you cannot strictly speak about particles (there is no localized state) and there is no S-matrix, since there is no asymptotic free states.
This is the reason why they add a mass gap. But you can have theories with massless particles and massive particles.
The second question can be answered as follows: Suppose you have a symmetry that is not a direct product of Poincaré and an internal symmetry. This means that you can apply to a particle state of mass $M_1$ at position $x_1$ your symmetry and transform it in a state of mass $M_2$ (or just a different particle) at position $x_2$. This looks like a long range force. The CMT assumes local forces, otherwise you cannot assume asymptotic free states.
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Largest Mass Diffraction I have read "Matter-wave interference with particles selected from a molecular library with masses exceeding 10000 amu" which claims to observe diffraction patterns in objects of around 10'000 amu. What is the largest mass objects shown to have diffraction patterns and show wave-particle duality? I have heard a claims of this type have been shown for small amino acids, and possibly protein strands or even small viruses, but have struggled to find any references.
| According to this website, photons, electrons, atoms, and some molecules, including buckyballs, have displayed wave-particle duality. Here is a paper that talks about some of the experimentation for larger molecules. This paper talks about molecules with over 6,000 amu showing a wave-particle duality, and this article talks about the discoveries behind your paper. The article was written at the end of 2013, and it calls it a broken record. Everything I've looked at says this is the record. Interestingly, I did find a few sources (like this one) that said scientists have tested molecules with a mass over 10,000 amu; obviously, it could be that better experimental equipment will allow this to happen.
In reference to amino acids/DNA/viruses: The mass of the DNA strand Enterobacteria Phage T4 is roughly 40,000 amu. The mass of a DNA strand only makes up a small component of the mass of a virus, so I highly doubt that either DNA or viruses show their wave-particle duality (see this website). However, it should be mentioned that viruses are within an order of magnitude of 10,000 amu (100,000 amu).
I'll keep looking for information. Hope this helps!
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Satellite revolving problem gives two different answer Assume there's a satellite revolving about the Earth. If I would like to decrease its radius, should I increase or decrease its velocity?
I know the answer apparently should be decreasing its speed, but the following two formulas give different answers. Can someone explain why two formulas give two different answers?
$r = mv² / F$, where r and v are directly proportional
$v = √(GM / r)$, where v and r are inversely proportional
| You cannot simply pluck equations out of your textbook and apply them to any situation. Just because they both contain radius $r$ and speed $v$ does not mean that they are necessarily valid for your situation. You have to think about what you are doing.
The 1st equation applies for circular motion. It tells you that $r$ is proportional to $v^2$ providing that centripetal force $F$ and mass $m$ are held constant. If you change orbital radius, ie the distance between the satellite and the Earth, the gravitational force $F=GMm/r^2$ (which provides the centripetal force) also changes. So you cannot apply this equation.
However, substituting $F=GMm/r^2$ into the 1st equation gives you $v^2=GM/r$, which is the same as your 2nd equation. This is the equation you need to use. It tells you how the speed and radius of a satellite or planet are related for a circular orbit.
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How to find static friction coefficient and the applied force's angle of inclination with a horizontal plane? Okay so I have this weird problem in my statics book that wouldn't come to a solution no matter what I tried. It states the following:
A body of weight $4$ kg.wt is placed on a horizontal rough plane. If a force of magnitude $2$ kg.wt acts on the body with an inclination of $θ$ with the horizontal, or if a force of magnitude $4$ kg.wt acts on the body in the opposite direction of the previous force, the body is about to move. Find $θ$ and the coefficient of static friction.
I have tried to resolve the resultant of the two forces after they partially-cancelled each other, and I've tried to work with the first case alone, since in both cases the body is about to move and $F=μR$. But I could never reach a solution since in all the methods I've tried I always needed the normal reaction ($R$) and I can't get it without the relation $μR$ which equals $2cosθ$, because after resolving the force, the one pointing upwards was ($R+2Sinθ$) and I still had to get $θ$ in order to get mu or anything else. So basically my problem is with getting the theta. All I have so far is that $R=4-2sinθ$.
I have a feeling that I'm doing something wrong. This is my first time studying this topic (friction) in statics, and the problem is a horizontal plane case, which is supposed to be too simple. However, for some reason I'm not finding it simple at all. Is there something wrong with the problem altogether? As in, should they have provided more or other givens? If not, would someone please tell me what I'm supposed to do?
| Here's how I interpret this question:
We have two equations: The horizontal projection of the 2 kg force equals $\mu$ times the sum of gravity and the vertical projection, i.e. $$2 \cos(\theta) = \mu(4-2\sin(\theta));$$ and similarly for the 4 kg force, except it's projection on the vertical direction is $+\sin(\theta)$, so $$4\cos(\theta) = \mu(4+4\sin(\theta)).$$
Taking a ratio of these two equations (latter over former) gives $2 = \frac{2+2\sin(\theta)}{2-sin(\theta)}$, which can be solved to get $\sin(\theta) = \frac{1}{2}$, i.e. $\theta=30^{\circ}$. Thus $\cos(\theta) = \frac{\sqrt{3}}{2}$, so the first of the original two equations becomes $$\sqrt(3) = 3 \mu \implies \mu = \frac{\sqrt{3}}{3}$$.
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What is the difference between the blackness of a Black Hole and the blackness of a Black Body?
*
*Light cannot escape the gravitational pull of the black hole and hence the 'black hole' is black.
*Any object that is black in color, absorbs all wavelengths of light and reflects none. So it appears black.
Could someone explain the key difference between these two phenomena ?
| A black body is a system that does not reflect photons. A black body can only emit photons, which is often fabricated as a hot cavity with a small hole. A razor blade can also approximate a black body if the sharp edge is smaller than the wavelength of photons of interest and is at a temperature so short wavelength photons are not significantly present. A black body then has a small reflectivity relative to the emissivity $r(\lambda)~<<~\epsilon(\lambda)$ for wavelengths of interest. An ideal black body has zero reflectivity for all wavelengths.
A black hole absorbs anything, including photons that reach it. It has zero reflectivity. Even if a photon entering a black hole reaches a mirror that is right at the event horizon, since the mirror has $z~\rightarrow~\infty$ the photon is not reflected back. The black hole also interacts with the quantum vacuum, in that virtual quanta across the horizon becomes an entangled pair of a negative energy quanta entering the black hole and a positive energy quanta that escapes to "infinity." This is a bit of heuristic, but it helps us see that a black hole losses a bit of mass and that mass escapes. The low energy form of this is of course photons, which have zero mass gap. As a result a black hole is about the most ideal black body in the universe.
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Are there higher genus equivalent of spherical harmonics? So the equation of a surface with topology S2 can be expanded out in terms of spherical harmonic functions. (I believe).
A torus T2 which is just S1xS1 can be expanded out in terms of ordinary harmonics which is just Fourier analysis.
But what about surfaces of higher genus, for example a double torus?
What is the equivalent harmonic functions, or Fourier series for these?
Edit: I am talking about how an arbitrary closed 1D curve can be represented as a Fourier series in terms of sines and cosines (as is done in string theory for example). So in terms of a 2D surface I guess the metric would be defined by its embedding in R3. Wouldn't the series expansion determine the shape of the surface and hence the metric not the other way round?
| If your surface has a metric, you can expand in eigenfunctions of the Laplace operator. This is kind of like the spherical harmonic expansion for g = 0 or 1, but it's not as nice. The sphere and torus are both homogeneous spaces, which allows you to use representation theory to organize the expansions. None of the higher genus Riemann surfaces are homogeneous spaces.
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Ground state of local parent Hamiltonians and invariance under local unitaries Assume that a finite-dimensional pure state $|\psi\rangle\in \mathcal{H}\simeq \mathbb{C}^m$, $m<\infty$, is the (unique) frustration-free ground state of a local parent Hamiltonian and suppose that the locality notion is given in terms of a connected set of neighbourhoods $\{\mathcal{N}_k\}$. My question is the following one: Is it true that any unitary $U$ satisfying $$U|\psi\rangle\langle \psi|U^\dagger=|\psi\rangle\langle \psi|$$ can be decomposed into a finite product of invariance-satisfying unitaries acting only on the neighbourhoods $\{\mathcal{N}_k\}$, that is $U$ can be written as $U=\prod_{i=1}^N U_{\mathcal{N}_{k_i}}$, where every $U_{\mathcal{N}_{k_i}}$ acts only on the neighbourhood $\mathcal{N}_{k_i}$ and it is such that $U_{\mathcal{N}_{k_i}}|\psi\rangle\langle \psi|U_{\mathcal{N}_{k_i}}^\dagger=|\psi\rangle\langle \psi|$
?
Any (partial) answer/comment/reference is very welcome.
Thanks in advance.
| Isn't simple translation symmetry an example?
E.g. suppose you have a one-dimensional ring of $L$ spins described by $|\psi\rangle = \sum_{\{\sigma_i\}} \psi_{\sigma_1,\sigma_2,\cdots,\sigma_L} \; |\sigma_1,\sigma_2,\cdots,\sigma_N\rangle$. Then this wavefunction could be invariant under the unitary transformation $\psi_{\sigma_1,\sigma_2,\cdots,\sigma_{L-1},\sigma_L} \to \psi_{\sigma_2,\sigma_3,\cdots,\sigma_L,\sigma_1}$. However you clearly cannot do this locally, unless I misunderstand your characterization.
An even more visceral counter-example would be (spatial) inversion symmetry.
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Why aren't trigonometric functions dimensionless regardless of the argument? Consider this equation :-
$$y = a\sin kt$$
where $a$ is amplitude, $y$ is displacement, $t$ is time and $k$ is some dimensionless constant.
My instructor said this equation is dimensionally incorrect because the dimension of $[kt] = [\text{T}^1]$ and since $\text{angles}$ are dimensionless, we can conclude that it is dimensionally incorrect.
I don't understand why it is so. Why do we need to check the dimension homogeneity of the term inside the $\sin$ to conclude whether the equation is dimensionally correct or not?
Why isn't the whole sine function is dimensionless $(\sin kt = \text{[T}^0]) $ regardless of the dimension of the argument inside as the range of sine function is $[-1, 1]$.
| The sine function $\sin(\theta)$ is defined as a function of an angle $\theta$ (measured in radians). So the question amounts to asking why an angle must be dimensionless. (Of course, the answers invoking properties of Taylor series are also correct.) The angle $\theta$ between two directions is defined as the ratio $\theta = a/r$, where $a$ is the arc length connecting the endpoints of a pair of lines of equal length $r$ pointing in those directions (see diagram below). An angle is therefore a ratio of two lengths, making the angle dimensionless. In particular, it doesn't matter if you measure $a$ and $r$ in metres, centimetres, inches or parsecs, the answer for their ratio $\theta$ will always be the same.
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Does an object float more or less with more or less gravity? This might be a stupid question, but I'm a newbie to physics.
An object less dense than water (or any other fluid, but I'm going to use water for this example) floats normally on Earth when placed in water. But if the object was placed in a hypothetical place where there is no gravity and there is air, it would not float on water. So if the object was placed in water on a planet with more gravity than Earth, would it float more or would it float less, or float the same as on Earth?
Would it float more because it doesn't float without gravity, but it does float with Earth gravity, therefore it'd float even more with more gravity.
Or would it float less because more gravity would pull the object down, so it won't float as much.
Or would it'd float the exact same as on Earth because the above two scenarios cancel each other out.
EDIT: By "float more," I mean it rises to the surface of the water faster, and it takes more force to push it down. By "float less," I mean it rises to the surface of the water slower, and it takes less force to push it down.
| Assuming both the water and the object are rigid and incompressible (pretty good approximation for water, may or may not be so good for the object) and that we can ignore surface tension (good approximation for large objects, not so good for tiny ones) then in equalibrium the same proportion of the object will be above the water regardless of the strength of gravity.
However stronger gravity will mean that the forces involved in the non-equalibrim state will be larger. Those larger forces will lead to faster movement.
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Do all rays of light from an object a being imaged by a camera pass through the focal point before being recorded on the image plane? Basically I'm trying understand the limitations of using the pin-hole model for a camera.
| In thin-lens optics, one often assumes that the center of the lens (whose faces by symmetry are locally perpendicular to the line of sight) does not refract light from distant objects, so one can ray-trace from the source to the image plane a straight line through that central point. It requires more calculation to ray-trace rays that hit the lens off-center, and it is only by doing that off-center ray tracing that one can determine the focus (focal point, and focal plane) of the lens.
For a pinhole "lens", it's all about the central point, of course. So there is no off-center light to be traced, no determination of a focal length (to the central focal point, for an image at infinity) or other focusing properties.
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How is potential energy lost when a water droplet is dropping down slowly on a wall? When a water droplet is on a vertical wall, it usually drops slowly, which is different from free falling. As the dropping speed is slower than free drawing, so I guess some energy must be lost.
I guess it is lost as internal energy, but if it is true, how a water droplet gains internal energy when dropping slowly in microscope view?
| The energy loss is frictive in nature. As the drop slides down the surface, there is a complex interaction at the interface between the droplet and the wall. This will go into changing the temperature of the water droplet, kicking some molecules off, etc...
Perhaps someone with a litter better skills in thermodynamics could help here, but I think you should be able to estimate the temperature change in the water droplet if you know its velocity as a function of time down the wall. The work done by friction should have some relationship to the amount of heat transferred into the droplet. My gut says those two are equal, but I am having trouble justifying that answer. Obviously in vacuum, there should be no heating.
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How increased current carrying capacity / ampacity affects wire gauge size In selecting a suitable wire size for manufacturing:
If I have a material that has the same resistivity and density as Copper, but has a higher current carrying capacity / ampacity for the same gauge (i.e. diameter and area), would I be able to use a thinner gauge wire? I am trying to understand how, all other things being equal, increased current carrying capacity would affect wire gauge size and why?
| It's all about heat buildup. A superconductor will, occasionally, lose coolant and quench, and you need enough conductivity of the windings to prevent that quench event from melting, straining or otherwise damaging your device. In non-superconducting wiring, the wire must not heat its surroundings (or itself) beyond the failure temperature, and the allowed currents for insulated house and appliance wiring reflect the thermal character of the surroundings.
One wants the time constant, L/R, for a quenching superconducting coil to be long enough that the dumped heat is spread over a large volume of the conductor, to limit the local temperature peak. Melting would be bad, evaporation worse.
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Spring force on both sides of spring I am a little confused about springs. I just wanted to know that if I pull an ideal spring of spring constant $k$ such that the spring has been symmetrically pulled and its elongation (total) comes out to be $x$ then would the force on one side by $$F=kx$$ or $$F=kx/2$$ I am a little bit confused and hence I resorted to ask it here.
| There is no such thing as a non-symmetrical pull of a string or some non-"total" elongation. $x$ is elongation, and that's it. $F=kx$ is the spring force, and that's it.
An object tied to one end of the spring experiences this spring force. The whole force. Something tied to the other end experiences by Newton's 3rd law the same force (in the opposite direction).
No need to half it.
Seen from each object at the ends of the spring, it would be:
*
*You draw the free body diagram of the object at one end, and it shows a spring force. This force is the push, the spring exerts on the object because it is compressed a bit and tries to return to the uncompressed state. And it is experimentally found to be proportional to the compression as $F=kx$. Which also feels intuitive: doubling the compression doubles the tendency of it to return to the relaxed state.
*You then draw the free body diagram of the object at the other end. It also feels a spring force. Same argument holds; the force from the spring is present, because the spring is compressed a bit, and the force appears to be proportional to the compression as $F=kx$.
We cannot talk about the "elongation $x$ caused at each end" and then say that "the total elongation is the sum of these". Because the elongation $x$ that happens is already caused by the two ends both being pulled - if only one was pulled (so only one end feels a force), there would be no elongation at all! (Assuming ideal mass less spring).
Force on one end does not contribute alone with one half of the total elongation; the forces on both ends cause it in collaboration.
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Heat pump intuition What is an intuitive explanation for the concept of heat pumps? I know that it is basically a reversed Carnot process. We can for example take an amount of heat $Q_1$ out of a warmer system and transform part of it into work W. The rest goes to the colder system. If we now reverse that process we need to take heat out of the colder system. For this to be done we need the same amount of energy W we got out of the process previously. But here my problem starts: How do you force the energy to come out of the colder reservoir? How can you explain that without just saying that it is an inversed Carnot process?
| You use a pump on a separate fluid (gas) system which is connected to the freezer (for example). The pump lowers pressure while the pipe system controls the volume of the fluid gas, and then this fluid can reach a smaller temperate than the internal of the freezer.
The freezer thus passes heat to the fluid system and cools down. The fluid cycles through the pipes to a part with higher pressure etc, which let it raise its temperature, so it can send away the heat to the surroundings.
The issue here is, that this pump performs an amount of work, and this is where $W$ enters the equation.
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Does the mass of water molecule increase when its just converted from ice to water? When we heat some object and its temperature raised then the K.E of molecules increased it means increase in velocity and according to $m=\frac{m०}{{(1-\frac{v^2}{c^2}})^{1/2}}$ mass is increased in very small amount.but when latent heat is gained by ice at $0°\text{C}$ and its converted to water at $0°\text{C}$, I think its K.E. is not increased, so I'm confused where has this energy of latent heat gone after separating molecules
| In the past, we might have said that any increase in molecular mass due to velocity increase would only be significant at speeds near that of light, but that is not how we view mass today at those velocities. No amount of heat we could put into water would ever result in molecular velocities anywhere near the speed of light.
The latent heat energy is used up in the phase change from ice to water, that is breaking the bonds that keep the water as crystal ice, and when liquid water is produced, it is (usually briefly) at the same temperature as the ice.
So that's where the latent heat energy goes, into breaking chemical bonds, with none left over to increase the k.e. of the water molecules
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Does empty space have energy? My physics friend suggested that "the answer to why matter exists in the universe" is because all massive particles are just the fabric of space excited into little packets. To illustrate, imagine a blanket on the ground. Then, pinch a small bit of the blanket and twist it. This is a particle that has mass.
It was intriguing to hear this (he's only studied up through Freshman year of college physics), but there are clear flaws (i.e. angular momentum of a "particle" tied to a "blanket"??).
Regardless, it made me wonder about vacuums. Is there any theory that suggests that a vacuum actually has energy in some form or another?
| Particles are not regarded as bits of curved spacetime, but rather as excitations of quantum fields.
It has been suggested that spacetime curvature can cause structures that behave a bit like objects (though not fundamental particles) and these objects are called geons. However it remains unclear whether these would be stable. It also isn't known what impact quantum gravity effects would have on the formation and stability of geons.
The vacuum has a precise definition in quantum field theory and it has a mean energy of zero by definition. However we can have gravitatinal waves propagating where no matter is present, and there is an energy associated with these gravitational waves. You could regard this as a vacuum having energy, though normally we wouldn't call it a vacuum if gravitational waves are present. This energy is in effect the energy of the spacetime curvature, though this turns out to be a rather elusive quantity.
| {
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What is the difference between these two ways to calculate average velocity? Average velocity:
$$v_{\rm avg,1}=\frac{v_{\rm final}+v_{\rm initial}}{2}$$
and average velocity:
$$v_{\rm avg,2} =\frac{\rm total\;displacement}{\rm time \;taken}=\frac{\Delta x}{\Delta t} $$
What is the difference between them and when do we use them?
| Taking the average of the initial velocity and final velocity is not necessarily, you are assuming a linear change in the velocity which is not the general situation. So only the second formule specifies the average velocity.
| {
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How many 'recursive' gravitational orbits are possible? I was thinking about how the moon orbits around Earth, which orbits around the Sun, which orbits around the center of the Milky Way.
I think of these kinds of orbits as recursive. For each body $B_n$, we can imagine adding another smaller body $B_{n+1}$ which orbits around $B_n$ such that the center of mass of the system remains within the radius of the larger body and the system as a whole is stable.
Is there a limit to how many times this can be repeated? Is there an ideal ratio of masses for these bodies? Assuming $B_0$ is the supermassive black hole at the center of the Milky Way, how small can we get before quantum effects take over?
| The other answers have nicely addressed the scale of increasing size. This answer provides detail on the scale of decreasing size. There are no known cases of moons having moons in the solar system. Two key things get in the way of moons having moonlets having even smaller moonlets (etc.).
One issue is the ever-decreasing volume surrounding an object about which orbits are stable. One way to express this volume is via the Hill sphere. which provides a rough estimate of the volume surrounding an object about which orbits are stable. As scale decreases, the Hill sphere will eventually become smaller than the moon itself.
Another issue is that as small objects tend to be less round than larger ones. Mimas at about 400 km in diameter is the smallest object rounded by self-gravitation in the solar system. Smaller objects look more like a lumpy potato than a sphere. This results in a non-spherical gravitational field. Even objects as large as our own Moon have a significantly non-spherical gravitational field. This means that orbits close to our Moon will be unstable. For example, read about the case of bizarre lunar orbits suffered by the PFS-1 and PFS-2 subsatellites released by Apollo 15 and 16.
| {
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Could a black hole pulling on a neutron star temporarily create a quark star? I believe a quark star is a hypothesized star that is composed of quark matter. If I'm correct then an even large gravitational pull than a neutron star has would be required to break down the individual neutrons forming a star made of quarks. For a neutron star to do this it would require so much more mass that it would make a black hole before coming a quark star.
What if a neutron star was caught in a black hole, could it become a quark star as it was being pulled in?
| No and maybe a yes. Quark stars formation from a neutron star in this is hard to conceive.
Yes it might be possible for black hole to collapse a neutron start into a quark star hybrid exotic star under different circumstances.If a small but significantly massive black hole collided and acted like a strangelet.
A shining example hybrid neutrons stars is the strange star that stars with strangelet collision. Stranglets are types of meta-stable particles that contain large numbers of quarks and strange quarks are believed to be in the majority.
These stranglets can bcan produce strange stars that are overdese(8 km or less) and have a core of quark matter.It is possible the strange quark matter is in more stable than nuclear matter and is favored at some point in colapse.In this phase it is called color superconductiviy.See article on strange hybrids https://arxiv.org/pdf/astro-ph/0407155.pdf.
Lattice Quantum chromodynamical simulations do not provide a conclusive guide yet as to the exact transition or behavior. So the state is still much debated but the color superconductivity state can be analogous to a type two superconductor with channels for the states of energy of the quarks in this phase.
So stellar cores that pass this phase can be hybrid stars.If stellar black hole of 3.8- 8 solar masses and has accretion disk.Along with it being Kerr-type it could produce a quark star in said disk.Kerr types may have sufficient rotational energy that it can produce strange quarks in majority in the disk(150 Mev per femtometer) and hence meta-stable quark matter.
See article on it https://arxiv.org/pdf/0908.2672.pdf.
| {
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Why is 3D stress tensor acting only on three surfaces? I'm trying to learn about the stress tensor (in 3D)
The tensors are said to have directions (the first subindex $i$ in $\sigma_{ij}$) and specify the surface upon which they act (the second subindex $j$ in $\sigma_{ij}$).
What confuses me is why is it defined only to act on three surfaces, even when the cube has six surfaces?
| We do consider other 3 faces, as we take into account (Sigma x), it means other opposite face of it is also taken but written only once which means the magnitude is same but direction is opposite, i.e. you will see 9 components but still there are 18 components, we only write 9 since other 9 will have same magnitude. So, think twice is it worthy to write twice ? answer is NO as if you know the magnitude, you can apply it on the element, if you are still confuse then think we take element under equilbrium condition.
| {
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Why doesn't Helium freeze at 0K? I have read that Helium does not freeze at absolute zero under normal pressures.
How could this be possible given that the absolute zero is the lowest attainable temperature and at that temperature, all random movements of the atom stop?
Shouldn't the atoms just stop vibrating and solidify instantly? Why do they possess kinetic energy at absolute zero?
| You have been misled by the idea that temperature is a measure of energy. While this is approximately true at high temperatures, it is not correct at low temperatures. Temperature is actually a measure of entropy; the derivative of entropy with respect to internal energy at constant particle number and volume is inverse temperature.
At very low temperatures, quantum mechanical effects become important, and even at absolute zero (0 K), the particles have energy, known as zero point motion. In helium, this zero point motion is large enough to prevent the atoms from sticking together as a solid - it remains a liquid.
Above roughly 3.2 MPa Helium-3 becomes solid at high pressure. For Helium-4 it will become solid above ~2.5 MPa.
http://ltl.tkk.fi/research/theory/helium.html
| {
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Unitary translation in phase space coordinate If we suppose that we can translate one point to another point in phase space $(x,p)$ through the following operators,
$$T(\Delta x) = \exp(-i p~\Delta x ) $$
and
$$T(\Delta p) = \exp(-i x~\Delta p ) ,$$
I want to see if there is any common point between these transformations and canonical transformation?
| Let us stick to Quantum mechanics, so, then, use P and Q for operators, and, to avoid confusion, s=Δx and t=Δp for classical shift parameters.
You then evidently wrote down the (conjugates of) celebrated U and V Weyl operators (1927), useful in the braiding form of the canonical commutation relations,
$$
e^{itQ} e^{isP} = e^{-i st} e^{isP} e^{itQ} .
$$
Their actions on functions of x are rephase ("clock") and shift,
$$
e^{itQ}\psi(x)= e^{itx}\psi(x),\qquad e^{isP}\psi(x)=\psi(x+s),
$$
and for p,
$$
e^{isP}\phi(p)= e^{isp}\phi(p),\qquad e^{itX}\phi(p)=\phi(p-t).
$$
They have particularly handy properties and serve as a bridge to finite dimensional Hilbert spaces (clocks).
I'm not sure how much you'd like to base your classical limit from quantum canonical transformations to classical ones (it is a truly daunting and treacherous area, approached with respect, if not trepidation!) on these; but, of course, they are both canonical as they preserve Heisenberg's commutation relation,
$$
e^{itQ} e^{isP} [Q,P] e^{-isP}e^{-itQ}= [Q,P]=i.
$$
If you really wished to get into quantum canonical transformations in phase space, where angels fear to tread, you may start from Bracken, Cassinelli, and Wood, "Quantum symmetries and the Weyl–Wigner product of group representations", Journal of Physics A: Mathematical and General 36.4 (2003) 1033 and pursue the groaning bibliography.
| {
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Does a higher water volume increase pressure? I am constructing a gravity flow water system. I have 100ft point where I can put my tank. My question is does the size of my tank matter? I am using a 1" pipe. Will I get more pressure if I use a bigger tank? For example what is the difference in pressure if I use a 10 gallon tank or a 50 gallon tank?
| The pressure experienced at the bottom of the pipe depends on the diameter of the pipe, the flow rate, the total height difference between the surface of the water and the point where you measure the pressure, and the density of the liquid.
When the flow rate is zero (no liquid flows: before you open the valve) the only thing left is the density and the height difference, and the pressure at the bottom will be higher than atmospheric pressure according to
$$\Delta P = \rho g \Delta h$$
Where $\rho$ is the density of water (1000 kg/m$^3$), $g$ is the gravitational acceleration (9.8 m/s$^2$), and $\Delta h$ is the height difference (100 ft).
Now if you have a small container at the top, then when water starts to flow the water level is likely to drop quickly - and this will give a small drop in pressure (small, because with 100 ft initial height there just isn't a lot of height to drop in a 10 gallon container). But otherwise there will be no difference between the two containers.
| {
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When should you jump off a falling ladder? If you stand on the top of a falling ladder you will hit the ground at a higher speed (and therefore presumedly sustain more injury) if you hold on to the ladder than if you jump off it.
This was solved
here.
Where is the "break even" height on the ladder, from where you will hit the ground with the same speed if you jump off it or if you follow it down? This question just makes an assumption that you would hit the ground more softly if you stay on to the ladder (compared to jumping off it), if you are located at the lower part of it.
I don't think the midpoint is the break even point. I quickly calculated and I think you should stay on the ladder if you are on its midpoint. (The following is just a quick computation, there could be errors in it.)
$$ v_{midpoint}^2 = \frac{1}{2}gl \frac{m + m_L}{\frac{1}{4}m + \frac{1}{3}m_L} $$
$m$ and $m_L$ are the respective masses of man and ladder, $l$ is the length of the ladder.
| You people are crazy with your useless calculations.
Doesnt take science to realise that if you hold on the ladder you will have a higher chances of falling on your back or ass and hit your head against the ground now I wouldnt know peoples preferences but yesterday I was at the top of the ladder which is about 3 meters I was trying to screw a piece of mdf sheet when I lost balance and gravity was pulling me backwards so I simply jumped off It and landed on my legs, now my left leg heel is killing me when I step on it but I rather have it this way than ended up in hospital with a concussion or broken back. Ps; the floor was concrete.
So yeah no matter your calculations it's better to jump and fall on your legs than hit any other part of the body, trust me.
| {
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What is a phase arrow? Feynman say's that a photon takes every path while reflecting off a mirror when going form A to B, but we only see the middle one(where incident angle = reflected angle) because all the others are cancelled out as they have longer routes and while following them their phase arrows cancel out each other.
If you can, please explain it in general terms without mathematics, as I am studying it the same way.
| This question is based, as pretty much everyone knows, on the book "QED, the strange theory of light and matter", by Feynman.
My answer is based on my limited understanding of the book.
Phase Arrows has a good description of Feynman's analogy.
As well the above source, if you visit Feynman Lecture Using Phase Arrows , you can watch RPF's description between times 29:41 and 36:27 of part He draws the "arrows" diagram on the chalkboard.
Similarly, if you visit Feynman Lecture On Photons, Feynman talks about this subject between times 59:33 and 60:32.
One of the differences between classical mechanics and quantum mechanics is that classically, objects have a definite path,or trajectory, so if you kick a soccer ball, you can be pretty certain where it will end up, and the path it will take.
Whereas in q.m.you can't say that a photon will definitely either follow a path or be found at a particular place. You can only assign a probability that a photon will be found at a particular place.
So Feynman sort of took this idea to the extreme, in a mathematical sense. Mathwise,you can say the photon took every available path, but all the paths that don't confirm to classical notions cancel each other out, by destructive interference. So you are left with just the paths that take the least time.
This is sometimes known as the sum over histories approach to quantum mechanics.
| {
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Why does the food in the microwave heat up but the bowl doesn't? I put a 1/4 inch thick clear glass container into the microwave with a plate on top and put it in for almost 5 minutes (there was lots of soup). When It came out the soup was really hot but I could still hold the top parts of the container. After measuring the soup's heat it was about 120-130 degrees but the top part felt cool to the touch. Why didn't it heat up like the soup did?
| The microwaves are primarily designed to vibrate/heat the water molecules in the food, as a way of ensuring that the foods gets cooked evenly. An aid to this process is the rotating plate within the machine.
Microwaves that cook your food pass through plastics, glass, and ceramics, with mimimal heating, as their water content is low and they are less prone to heating, explaining why you can pop your (almost ready to go) chicken curry and rice, along with its plastic packaging, straight into the microwave. It is also this feature of microwaves that makes them so energy efficient; they heat only the food and nothing more.
However, don't try to put eggs in a microwave, they will become minibombs as the water heats up, turns to steam and then blows the eggshell apart. Cups of water are not recommended, nor is just pressing the start button without food or liquid to absorb the microwaves, as the magnetron (which is what cooks your food), ends up absorbing the microwaves instead, which can damage your microwave and may even start a fire.
Metals, on the other hand, reflect these radio waves, a characteristic very cleverly put to use in the walls of the microwave such that no waves escape and cook anyone in the kitchen! However, you can see sparks from the edge of some decorative ceramic plates appearing now and again, from the microwave radiation.
Image Source: www.ccohs.ca
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Why doesn't the adiabatic reduction of first law of thermodynamics, $W = -\Delta U,$ hold for non-conservative forces? The first law of thermodynamics is$$
\Delta U~=~Q-W
\tag{1}
\,,$$where:
*
*$\Delta U$ is change in internal energy;
*$Q$ is the amount of heat supplied to the system;
*$W$ is the amount of work done by the system to the environment.
For a conservative force $F,$ work done by it is equal to
$$W = -\Delta U \,. \tag{2}$$
$\operatorname{Eq.}\left(2\right)$ is also the first law of thermodynamics (a version of the conservation of energy) with an adiabatic change $\left(Q=0\right),$ where positive work is done by the system.
Questions:
*
*Why isn't $\operatorname{Eq.}\left(2\right)$ applicable to a non-conservative force even though every object in this world follows conservation of energy?
*Is there any special case or condition where a non-conservative force, such as an external force by me on an object, follows the above equation?
| The above form is not the most general form of the 1st law of thermodynamics, which is:
$$ dU = \delta Q - \delta W $$
Here $\delta Q$ is the change in heat, i.e., the heat flowing into or out of the system. This is a statement of energy conservation.
Now, if the heat flow is zero, it reduces to your above equation. Non-conservative forces are usually forces, e.g., friction, which cause some kind of heat dissipation, and thus the formula is not applicable.
There can be no special case/condition where $dU = - \delta W$ for non-conservative forces because the internal energy $U$ is a state function. Any process starting and ending at the same state must have the same internal energy. But, the definition of a non-conservative force is one where going "in a closed loop" from one state back to itself, does not preserve energy - thus there must be processes in that cycle where there is dissipation in the form of heat (otherwise the force would not be non-conservative). For these, $dU=-\delta W$ does not apply.
| {
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Deceptively simple mass-spring problem? This question is inspired by two other, similar, so far unanswered questions (posed by different OPs).
Mass $m_2$ sits on a incline with angle $\theta$ that provides just enough friction for it not to start sliding down. It is connected by a massless string $S$ and perfect spring (with Hookean spring constant $k$) to mass $m_1$. Pulley $P$ is frictionless and massless. At $t=0$ the spring is not extended at all. Then $m_1$ is released.
Question:
What is the minimum $m_1$ to cause movement of $m_2$ up the incline?
Attempt:
Ignore the spring.
Determine static coefficient $\mu$ first.
\begin{align}m_2g\sin \theta &=\mu m_2g\cos \theta\\ \implies \mu &=\tan \theta\end{align}
To overcome the $m_2g$ component parallel to the inclined and the friction:
\begin{align}m_1g &\gt\mu m_2g\cos \theta+m_2g\sin \theta\\ \implies m_1 &\gt 2m_2\sin \theta\end{align}
But apparently this overestimates $m_1$. It has to be taken into account that $m_1$ starts accelerating before $m_2$ starts moving, because of the spring.
But how? Like several other members I can't see how the work done on the spring affects the minimum $m_1$. Conservation of energy?
| Let me try to do this without using any formulas. First consider what would happen if $m_2$ were glued to the incline. Then we would have a simple harmonic oscillator consisting of $m_1$ suspended from the spring $k$ (clearly it does not matter on which side of the pulley the spring is). In this situation the average tension force (over a full oscillation) of the spring must equal the weight of $m_1$, by conservation of momentum.
The initial position corresponds to the topmost position of the oscillation, and it is given that the spring has zero tension in this position. Then to get the predicted average over an oscillation, the tension must be twice the weight of $m_1$ halfway the oscillation, when $m_2$ is at its lowest position.
Since this is the largest tension that occurs during the cycle, the situation where $m_2$ is not glued to the surface will change as soon as this maximal tension is sufficient to make it start to slide upward. It follows that your original static computation overestimates the required mass of $m_1$ by a factor$~2$.
| {
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The meaning of covariant but not manifestly covariant What is the most general meaning of the expression covariant, but not manifestly covariant? Suppose I have a general (local) change of coordinates, $x^{\prime} = f(x)$, on an $(n+1)$-dimensional smooth manifold on which some classical fields are defined, say $A_{\alpha}(x_0,x_1,...,x_n)$, which transform into $A_{\alpha}^{\prime}(x^{\prime})$. Suppose the fields $A_{\alpha}(x)$ satisfy some equations of motion, where $x_0 = t$. How should these EOMs look like to be covariant with respect to the given change of coordinates, but not manifestly covariant? Could you explain in plain words the difference between the 2 forms of the EOMs?
If possible, can you write down a practical example of such a situation encountered in physics?
Thx.
| I'll expand on jjc385's answer with an example of how equations of motion can arise in a non-manifestly covariant form. In curved spacetime, the Lagrangian density can be written as $\mathcal{L}=\sqrt{\left| g\right|}\mathcal{L}_0$, with $\mathcal{L}_0$ a scalar called the scalar Lagrangian density. The action is then $S=\int d^4x\sqrt{\left| g\right|}\mathcal{L}_0$, also a scalar. The equation of motion $\frac{\partial\mathcal{L}}{\partial\phi}=\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}$ is not manifestly covariant, because the two sides of the equation aren't tensors in a general spacetime. However, you can derive the equation of motion from first principles in another form, $\frac{\partial\mathcal{L}_0}{\partial\phi}=\nabla_\mu\frac{\partial\mathcal{L}_0}{\partial\nabla_\mu\phi}$. (That's assuming $\phi$ has nothing to do with the metric tensor; if it does, we get a more complicated result, but the point I'm making remains the same.) This only uses tensors, so it is manifestly covariant.
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