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Other application of Liouville's theorem besides thermodynamics Are there any other important practical and theoretical consequences of Liouville's theorem on the conservation of phase space volume besides the calculation of the microcanonical potential in Thermodynamics?
| 1) On a symplectic manifold $(M,\omega)$, Liouville's theorem is often stated as that every Hamiltonian vector field $X_f=\{f,\cdot\}$ is divergence-free
$$ {\rm div}_{\rho} X_f~=~0 ,$$
where the volume density $\rho$ comes from the canonical volume form
$$\Omega~=~\rho dx^1 \wedge \ldots \wedge dx^{2n}.$$
Here the canonical volume form
$$\Omega~:=~\omega^{\wedge n}$$
is top exterior power of the symplectic 2-form $\omega$.
Equivalently, the Lie derivative of the canonical volume form
$$ {\cal L}_{X_f}\Omega~=~0, $$
wrt. a Hamiltonian vector field, vanishes.
2) It is interesting to note that Liouville's theorem generalizes to symplectic supermanifolds $(M,\omega)$ with Grassmann-even symplectic structure.
3) However Liouville's theorem fails for Grassmann-odd symplectic manifolds, also known as antisymplectic manifolds, or Batalin-Vilkovisky (BV) manifolds. Such manifolds do not have a canonical volume density $\rho$ either. Nevertheless, let us assume that there at least exists some volume density $\rho$. Then the failure of Hamiltonian vector fields to be divergence-free is measured by the odd Laplacian
$$ \Delta_{\rho} f ~=~\frac{(-1)^{|f|}}{2}{\rm div}_{\rho} X_f, $$
where $|f|$ denotes the Grassmann parity of the function $f$.
| {
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Quantum field theory, particle interpretations and path integrals? I am trying to find some names or models of a particle interpretation of quantum field theory which isn't a literal path integral approach? Are there any particle interpretations of quantum field theory which don't use path integrals?
| In some sense 2nd quantization is equivalent to path integral approach when introduce quantized fields. But the point is, even if you use 2nd quantization you still need to calculate things like cross sections, which is more related to generating functional of path integral. Also, many things of more advanced QFT are based on path integral. I think you would like Ron's comment on path integral which is more professional: What is the fundamental probabilistic interpretation of Quantum Fields?.
| {
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Why planets are orbiting only in one plane? Since gravity is three dimensional why planets are orbiting only in one plane around sun.
| I would probably say it has something to do with the gravity or rather fabric of space if you want. E.g. take a look of the shape of our galaxy:
Do you see the similarity?
How the galaxy resembles our solar system?
It's about the fact that after a while the biggest "rotators" like the Sun or biggest planets win and other "slower rotators" will adjust to their way of rotation.
After than they cannot change their rotation from left to right or stop. Other force like collision of other planet or asteroid etc. can cahnge the direction of rotation.
A lot of things are involved. And one of them is gravity too.
Btw. even strange things are pulsars:
http://en.wikipedia.org/wiki/Pulsar
This next few lines are off topic just to take in consideration ;):
*To tell you the truth the scientists have troubles to detect some meteorites because they are moving too fast etc. So, be very sceptical to all scientists who claim that nowadays science know how pulsars and galaxy works. And the "hypo-theories" changes all the time. Take e.g. how Big Bang was the coolest thing 15- 20 years ago, now it's overcome with multiverse and other things. Another example is Big Freeze (brrr, slow ;) or her majesty recollapsing Big Crunch - These were the hypothesis popular among most of the scientists not more than 2 years ago. Now it seems that our Universe will pop like a balloon when it will get older. Who is right? Hard to say but...
...keep trying ;)
| {
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How is parity relevant to determining angular momentum? Question:
Particle A, whose spin $\mathbf{J}$ is less than 2, decays into two identical spin-1/2 particles of type B.
What are the allowed values of the orbital angular momentum $\mathbf{L}$, the combined spin $\mathbf{S} = \mathbf{s}_1+\mathbf{s}_2$ (where $\mathbf{s}_1$ and $\mathbf{s}_2$ are the spin vectors of the B particles), and the total angular momentum $\mathbf{J}$ of the decay products? Give the results from two cases: first if particle A has odd parity and then if A has even parity.
My thoughts:
Particle A can be spin-1/2, spin-1, or spin-3/2. Since $\mathbf{J}<2$, we see that there are four possibilities for A:
$$
\begin{align*}
&(1): \;\;\mathbf{S}_A = 1/2 \quad\quad \mathbf{L}_A = 0 \quad\Rightarrow\quad \mathbf{J} = 1/2 \\
&(2):\;\;\mathbf{S}_A = 1/2 \quad\quad \mathbf{L}_A = 1 \quad\Rightarrow\quad \mathbf{J} = 3/2 \\
&(3):\;\;\mathbf{S}_A = 1 \quad\quad\quad \mathbf{L}_A = 0 \quad\Rightarrow\quad \mathbf{J} = 1 \\
&(4):\;\;\mathbf{S}_A = 3/2 \quad\quad \mathbf{L}_A = 0 \quad\Rightarrow\quad \mathbf{J} = 3/2 \\
\end{align*}
$$
The total spin of the B particles can be either $1$ or $0$, and each particle can individually have an orbital angular momentum, along with the angular momentum of the particles as a system. With this thought, cases 1,2, and 4 are impossible because the orbital angular momentum of the B particles is an integer, as is their total spin (and therefore their total angular momentum too). Thus we find that only case 3 is allowed, so the total angular momentum of the B particles is $1$ and their orbital angular momentum is $0$ (so $\mathbf{J}=1$).
I have a strong feeling that this is incorrect, because the question asks for the cases when A has odd parity and even parity (what does that even mean?!) so I suspect there should be more than one possible answer. Where did I go wrong?
| I actually searched for a deeper understanding of how parity is linked with angular momentum for my own, but I know from Griffiths "Introduction to Elementary Particles" that the Parity for relative angular momentum l is given by (-1)^l. All particles have parity, either + or -1, which one is determined by QFT. (creation and annihilation operator which are dependent of momenta, which change sign under parity transformation). Parity is multiplicative (you multiply the parity of your particles and the relative angular momentum) and for strong and electromagnetic processes conserved.
| {
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Electrical conductivity of an intrinsic semiconductor On which factor does the electrical conductivity of an intrinsic semiconductor depend? It doesn't have an excess of charge carriers in fact, does it?
| Conductivity of intrinsic semiconductor is due to their own internal charge carriers. The bonding between between two electrons of two neighboring atoms is covalent, therefor at NTP, there is no free charge carrier for conduction. When it is heated, some covalent bonds break due to heat and thus some electron get free for conduction. As soon as one electron gets free, there is a deficiency of electrons at its preceding position which acts as a positive charge or a hole, The number of holes is equal to number of electrons. At normal temperature, only $1$ ou of $10^9$ bonds break and therefore, conductivity is very low about few milli amps.
| {
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Will observers moving on a sphere experience time dilation? A single source of light exists at a fixed point in space relative to two observers. The two observers move on the surface of a shell with a fixed radius with the light source at its centre. They move along a circular path on the sphere. They move at different velocities u and v. Will they experience time dilation relative to each other?
If relativity says that there will be time dilation on the shell of a sphere with a fixed radius with a light source at its centre, then there is a problem. Let's say the light blinks once every 10 secs. Then that light will reach all observers on the shell at the same time no matter what their relative velocity to each other. All observers would agree the light blinks once every 10 secs, no matter what their velocity is as long as they are on the shell.
| You write: "All observers would agree the light blinks once every 10 secs, no matter what their velocity is as long as they are on the shell."
Using their own stationary clock, no they wouldn't, because the light source is moving relative to them. Each would record the light blinking at a rate slower by a factor $\gamma$
| {
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How does the solar sailing concept work? Wikipedia describes solar sailing as
a form of spacecraft propulsion using a combination of light and high speed ejected gasses from a star to push large ultra-thin mirrors to high speeds.
I understand the part where ejected gasses bump into the sail pushing the spacecraft. On the other hand, I don't understand how light can do this, since light has no mass.
How does that work? Does this mean that if I have a mirror balancing on a needle I would be able to push it over with my flashlight?
| solar sailing work on the principal impulse applied on the body by a moving photon on a reflective surface. If the ratio of the force applied by the photon which is directly proportional to the area of the reflecting surface and intensity of the light to the weight of the space craft is large enough it would accelerate the spacecraft and further you can go to following link it would help you through interesting videos
(http://www.youtube.com/results?search_query=solar+sail+spacecraft%2F+discovery&oq=solar+sail+spacecraft%2F+discovery&gs_l=youtube.3...1334.10855.0.11007.34.27.1.6.7.1.497.5586.7j7j5j3j4.26.0...0.0...1ac.1.BqfM0pGzoz8)
| {
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Initial separation of neutron star/black hole binaries? How would I go about finding the distribution of initial separations (i.e. the lengths between the centres of mass) of stars that make up binary systems. I am interested in neutron stars and stellar black holes.
This paper uses an assumption that
"The initial separation of a binary is chosen from a distribution that is assumed to be logarithmically flat: $\Gamma(A) \propto 1/A$."
but it neither defines $\Gamma$ or $A$, so yeah... helpful.
| The paper is rather lacking in definitions, I'll give you that.
As Murphrid points out, $\Gamma$ is just the PDF for separation, and $A$ is the variable used for the separation itself. The constant of proportionality is set by requiring the total probability to be unity.
Of course, as defined $\Gamma$ is not normalizable - it diverges logarithmically for both $A \to 0$ and $A \to \infty$. Presumably they put in a cutoff (much as they have a mass distribution $\Psi(M) \propto M^{-2.7}$, but they cut it off at $4~M_\odot$ and $100~M_\odot$), but I couldn't find any reference to the values.
The take-away message is that our understanding of the initial distributions of star properties still needs a lot of work. We can hardly claim to understand the distribution of initial masses beyond rough empirical measurements, and binary mass ratios (assumed to be uniform on $[0, 1]$ in this paper) and binary separations are at least as elusive. The assumptions in this paper were made so as to have something to feed into the simulations - they should by no means be taken to be the actual distributions of parameters.
| {
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Why does Planck's law for black body radiation have that bell-like shape? I'm trying to understand Planck's law for the black body radiation, and it states that a black body at a certain temperature will have a maximum intensity for the emission at a certain wavelength, and the intensity will drop steeply for shorter wavelengths. Contrarily, the classic theory expected an exponential increase.
I'm trying to understand the reason behind that law, and I guess it might have to do with the vibration of the atoms of the black body and the energy that they can emit in the form of photons.
Could you explain in qualitative terms what's the reason?
| Joshua has beaten me to an answer, but I'll still post this since it's written at a simpler level.
The reason you get a maximum because there are two effects that oppose each other. The number of modes per unit frequency rises as frequency squared, so as long as the energy of the modes is well below kT the energy is proportional to frequency squared. This is why the black body spectrum initially rises approximately as frequency squared.
However the probability that a mode is excited falls exponentially as soon as the energy of the mode is greater than kT, so as the frequency goes to infinity the emitted radiation falls to zero.
The net result of the two effects is that the emission first rises then falls again, and that's why there is a maximum in the middle.
| {
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What does it mean that an inspiral is 'adiabatic'? Binary systems emit gravitational radiation. This causes the system to lose energy, which results in a shrinking of the semi-major axis. I have read on countless occasions that this 'inspiral' is adiabatic (here for example). What does this mean that the shrinking is adiabatic?
| I believe it's an approximation applied to that phase of the inspiral, stating that the relative change in orbital frequency over time is small with respect to the orbital frequency itself. This defines what is known as the adiabatic parameter:
$$
\xi = \frac{\dot\omega}{\omega^2}
$$
Why adiabatic? No loss or gain of heat energy. Much like the heat death of the universe.
| {
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Relating generalized momentum, generalized velocity, and kinetic energy: $2T~=~\sum_i p_{i}\dot{q}^{i}$ According to equation (6) on the first page of some lecture notes online, the above equation is used to prove the virial theorem. For rectangular coordinates, the relation
$$
2T~=~\sum_i p_{i}\dot{q}^{i}
$$
is obvious. How would I show it holds for generalized coordinates $q^{i}$?
| The equation $2T=\sum_{i=1}^n p_i\dot{q}^i$ holds both in Lagrangian and Hamiltonian formalism for large classes of systems.
*
*In Lagrangian formalism, it holds for Lagrangians of the form $L(q,\dot{q},t)=T(q,\dot{q},t)-V(q,t)$, where the kinetic energy is of the form $T=\frac{1}{2}\sum_{i,j=1}^n \dot{q}^i m_{ij}(q,t)\dot{q}^j$. Now use the Lagrangian definition of momentum $p_i = \frac{\partial L}{\partial \dot{q}^i}$.
*In Hamiltonian formalism, it holds for Hamiltonians of the form $H(q,p,t)=T(q,p,t)+V(q,t)$, where the kinetic energy is of the form $T=\frac{1}{2}\sum_{i,j=1}^n p_i m^{ij}(q,t)p_j$. Now use Hamilton's equations $\dot{q}^i = \frac{\partial H}{\partial p_i}$.
In both cases, we are secretly using the Euler homogeneous vector field $\sum_{i=1}^n\dot{q}^i\frac{\partial}{\partial \dot{q}^i}$, which counts the number of $\dot{q}$s, as Emilio Pisanty also points out in his answer.
| {
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Future light cones inside black hole In Caroll's Spacetime and Geometry, page 227, he says that from the Schwarzschild metric, you can see than from inside a black hole future events all lead to the singularity. He says you can see this because for $r<2GM$, t becomes spacelike and r becomes timelike. I don't understand his reasoning though. Why does r being timelike mean you can only travel towards the singularity?
| Objects that follow timelike trajectories always have a unique timelike coordinate associated with each point on the trajectory. This is not true for spacelike trajectories, where a Lorentz transformation can render two points on the path simultaneous.
Hence, an object on a timelike trajectory within the event horizon can only cross a specified radius once along its trajectory. This means that it must either fall all the way inward without stopping or turning back; or it can be completely ejected without stopping or turning back. These two possibilities are mutually exclusive--if one is possible, the other is not (because time only flows in one direction) for a given system. The former describes a black hole; the second a "white" hole.
| {
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Deriving equations of motion of polymer chain with Hamilton's equations This is related to a question about a simple model of a polymer chain that I have asked yesterday. I have a Hamiltonian that is given as:
$H = \sum\limits_{i=1}^N \frac{p_{\alpha_i}^2}{2m} + \frac{1}{2}\sum\limits_{i=1}^{N-1} m \omega^2(\alpha_i - \alpha_{i+1})^2 $
where $\alpha_i$ are generalized coordinates and the $p_{\alpha_i}$ are the corresponding conjugate momenta. I want to find the equations of motion. From Hamilton's equations I get
$\frac{\partial H}{\partial p_{\alpha_i}} = \dot{\alpha_i} = \frac{p_{\alpha_i}}{m} \tag{1}$
$- \frac{\partial H}{\partial {\alpha_i}} = \dot{p_{\alpha_i}} = -m \omega^2 (\alpha_i - \alpha_{i+1} ) \tag{2}$
, for $i = 2,...,N-1$. Comparing this to my book, (1) is correct, but (2) is wrong. (2) should really be
$- \frac{\partial H}{\partial {\alpha_i}} = \dot{p_{\alpha_i}} = -m \omega^2 (2\alpha_i - \alpha_{i+1} - \alpha_{i-1}) \tag{$2_{correct}$}$
Clearly, I am doing something wrong. I suspect that I'm not chain-ruling correctly. But I also don't get, where the $\alpha_{i-1}$ is coming from. Can anybody clarify?
| Instead of using the chain rule (although it of course gives the same answer) expand the square of the $i^\mathrm{th}$ term in the sum parentheses to obtain
$$
\alpha_i^2 - 2\alpha_i\alpha_{i+1}+\alpha_{i+1}^2
$$
differentiating this with respect to $\alpha_i$ gives
$$
2\alpha_1 - 2\alpha_{i+1}
$$
Now, from the $(i-1)^\mathrm{th}$ term
$$
\alpha_{i-1}^2 - 2\alpha_{i-1}\alpha_i + \alpha_i^2
$$
you get an additional
$$
-2\alpha_{i-1} + 2\alpha_i
$$
when you take the $\alpha_i$ derivative. Putting these results together gives the answer in the book.
| {
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What will happen if we place salt water on a induction cooker? As we know that induction cooker works on the principal of induction of current in a conducting plate. So I just wanted to know what will happen if we place salt water in a plastic container on the induction cooker will it get hot as it is also a conductor or nothing will happen?
| The induction cooker works by passing a rapidly changing magnetic field through the thing to be heated.
That rapidly changing magnetic field creates an EMF (Electro Motive Force) field. (Faraday's law). This is like a voltage. Current will flow, proportional to the conductivity. No conductivity, no current. Low conductivity, low current. High conductivity, high current. Heat will be produced according to I^2 R.
So there will be a current in the salt water, and it will be heated. But not very much. Its conductivity is much less than copper or iron, so it will not have a very large current or amount of heating.
As for Iron or other magnetic materials, they might be part of the design, required to concentrate the magnetic field, but in general, induction heating is from currents induced in a conductor.
Many weeks later: Those darn induction cookers are advertising everywhere. Sure enough, their writeup says you need some magnetic properties to the cookware. Their explanation as to why, something like the first line in JJ Fleck's answer, is garbled and not exactly correct. The addendum of Fleck's answer has it spot on, and with references too.
| {
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Why do we hear a higher pitched sound outside the water when we smash two stones in water? The observer is outside the water; the stones are in water (say, 1 m below the surface). This produces a higher-pitched sound for the observer than if both the observer and the stones are in air.
Is this because it takes more energy for the sound waves to travel through water than through air, so that the ones that we hear from outside are the ones that had higher frequencies after the collision to begin with?
Does the density of the medium that is disturbed by a rigid body collision have any effect on the frequency distribution of the sound waves that are generated? For example, does the higher "stiffness" of the cage of water molecules surrounding the stones that are vibrating mean higher frequency normal-modes?
Finally, does the refraction at the water-air interface play any role?
| The higher pitched frequencies have higher energy. The path of minimal time is to the surface.
Lower frequencies naturally travel further in water despite their lower energy value. Hence the use of infrasound by whales for the purpose of long distance communication.
| {
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Pendulum axes confusion Suppose I have a simple pendulum and I want to calculate its acceleration when the bob reaches the maximal angle. I usually choose my axes such that the y-axis will be parallel to the rope. Then the acceleration would be $g \sin\alpha$ (Because: $mg\sin\alpha=ma$). However, if I choose the y-axis to be parallel to the gravitational force ($mg$), the acceleration equals to $g\tan\alpha$ (Because: (1): $T\cos\alpha=mg$ and (2): $T\sin\alpha=ma$ where $T$ is the tension force (centripetal force) and $\alpha$ is the given angle). Obviously, these two accelerations aren't the same. So my question - which one is right? And why the other one is wrong?
| The total gravitational force is $mg$ and this force is the hypotenuse of a triangle whose sides are $mg\cos\alpha$ in the direction of the rope and $mg\sin\alpha$ in the orthogonal direction. So the latter, $mg\sin\alpha$, is the force (mass times acceleration) along the circular orbit because the centripetal part of the force is cancelled by the string's tension.
My guess is that instead of dividing the gravitational force into the two orthogonal components, you are trying to divide the tension force into two components and identify one of these two components with the gravitational force.
The plan wants to divide the tension $T$ into two directions – I suppose it's the horizontal and vertical direction. In the vertical direction, you think to have the cancellation between $mg$ and $T\cos\alpha$ and in the horizontal direction, between $T\sin\alpha$ and $ma$.
However, you forgot that in the vertical direction, there's a contribution to the force from the acceleration, too. So the right conditions in this coordinate system are actually
$$T\cos\alpha = mg-ma\sin\alpha, \quad T\sin\alpha = ma\cos\alpha$$
Note that instead of $ma$ from your formulae, I had to write $ma\sin\alpha$ and $ma\cos\alpha$ because the acceleration is a vector in a direction that is neither vertical nor horizontal. Multiply the first (correct) equation by $\sin\alpha$, the second one by $\sin\alpha$, and subtract them in order to eliminate $T$. You will get
$$ 0 = T(\sin\alpha\cos\alpha-\cos\alpha\sin\alpha) =\\ = mg\sin\alpha - ma\sin^2\alpha-ma\cos^2\alpha=mg\sin\alpha-ma $$
which gives you $a=g\sin\alpha$ again. In other words, you have forgotten to realize that the acceleration $a$ is in a direction tilted by $\alpha$ so if you rewrite the vectors in the horizontal and vertical components, you have to rotate it into $a\sin\alpha$ and $a\cos\alpha$ with the right signs.
| {
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Eye sensitivity & Danger signal Why are danger signal in red, when the eye is most sensitive to yellow-green?
You can check luminosity function for more details...
| My hypothesis is that the use of red for danger signals is because we instinctively recognize red things as potentially dangerous. Aposemitism (Wikipedia) is the use of bright colors by prey to signal the presence of secondary defense mechanisms, such as toxicity. Aposematism only functions as a defense mechanism because predators, such as humans, instinctively recognize and avoid brightly-colored markings.
Take a look at the yellow-banded poison dart frog, and the granular poison frog. Does their coloration look familiar?
Of course, this kind of "pushes back" the question.
Q: Why are danger signs red or yellow? | {
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Will molten iron stick to a magnet? I've known for a long time that if you heat a magnet, there is a point it loses its magnetism (the Curie temperature). It isn't clear to me if this applies to induced magnetism like iron sticking to a magnet.
Will molten iron behave like a ferrofluid and be attracted to a magnet or will it just have a very weak paramagnetic attraction?
| The loss of magnetism at the Curie temperature applies primarily to "induced magnetism like iron sticking to a magnet".
Ferrofluid doesn't really include a molten iron; ferrofluid is a collection of many small but mezoscopic particles, "sawdust", and its magnetism doesn't differ so much from magnetism of normal pieces of iron except that it's easier for the particles to change the orientation. Ferrofluid normally have lots of "normal fluid" in it, like water or organic liquids, and some "coating", so the ferromagnetic material is a relatively minor component and it's surely not melted.
When a ferrofluid gets heated to the Curie temperature, it loses its magnetism, too.
The Curie temperature of all sensible materials is and has to be below the melting point. When you approach the melting point while heating the material, the magnetism has been lost for quite some time.
Yes, molten iron is paramagnetic, much like every ferromagnetic material above the Curie temperature.
| {
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Does quantum computing rely on particular interpretations of quantum mechanics? It is my understanding that quantum computing relies on quantum superposition and entanglement to work--qbits must exist in all states simultaneously before giving a particular result when observed.
Would this mean that quantum computing is impossible in interpretations of quantum mechanics in which qbits are not in reality existing in all states simultaneously until observed? Thus, wouldn't quantum computing be incompatible with non-local hidden variable interpretations (deBroglie-Bohm, for example), or with other interpretations in which the underlying reality is deterministic such as 't Hooft's?
| Generally, when you make a quantum calculation, you have to make some sort of measurement of the qubits at the end of the algorithm where the result you're looking for is a very probable (but not necessarily certain) result. In any interpretation that actually agrees with the basic results of quantum mechanics, these probabilities will still hold and the algorithm will still work.
If an interpretation is ruled out by the possibility quantum computing, then it's (probably) wrong because it contradicts quantum mechanics. To the best of my knowledge all of the interpretations you mentioned, while deterministic, still give results in agreement with quantum mechanics and can't be ruled out by the existence of a quantum computer.
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Why do objects accelerate as they fall? Most importantly, what must change in order for the falling object to change its speed? Is it the distance to the centre of the planet? If you pull the earth away from the object as the object falls, will the object slow down or will it keep accelerating?
| An object accelerates when a force is acting on the object. This given by the Newton's second law $F=ma$, where $F$ is the net force act on the object, $m$ is the mass of the object and $a$ is the acceleration of the object. The reason why objects accelerate as they fall is because the gravity of earth acts on the object. If you pull the earth away from the object as the object falls, the distance between the object and the earth increases. Now the garavitattional varies inversely to the square of the distance.So, the gravitatational force acting on the object due to earth will decrease and therefore the acceleration of the object will also decrease.
what must change in order for the falling object to change its speed
Nothing. As long as there is a net force forcing on the object, the object will accelerate.The acceleration will be given by the Newton's second law.
| {
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Proof that a spherical lens is stigmatic In geometric optics, we generally allow that, for example in the case of a convex lens, rays coming from a particular point get refracted towards another particular point on the opposite side of the lens.
How is this proven from Snell-Descartes' law? Do we need to use the paraxial approximation in order for this to be true, or is it actually exactly true?
| According to an article by the Optometric Science Research Group titled Stigmatic optical systems:
"There would appear to be little disagreement on what constitutes an astigmatic system in the case of a thin lens: the cylinder is not zero. A spherical thin lens is stigmatic or not astigmatic. The issue is less clear in the case of a thick system. For example, is an eye stigmatic merely because its refraction is stigmatic (spherical)?"
According to Snell's law, (from Wikipedia) the largest possible angle of incidence which results in a refracted ray is called the critical angle. When something exceeds this angle no there is no refraction point for the object in the image, meaning the result is astigmatic.
Also according to Stigmatic optical systems:
"An eye may be astigmatic despite having a stigmatic refraction."
| {
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Why doesn't fusion contradict the 1st law of thermodynamics? I was reading up on the 1st law of thermodynamics for my Chemistry exam and I was wondering why doesn't fusion contradict the 1st law of thermodynamics?
The 1st law states that
The energy of an isolated system is constant
or that whatever is put into the system, you get out of it, but in fusion you get more out of the initial reactions than you put in
Hydrogen + Hydrogen = Helium etc.
I am still a bit confused...
Thanks for any help!
| By that logic, a battery violates the law also.
"Look, all I did was put in enough energy to flip a switch, and now an LED keeps shining and shining! I got more energy out than I put in!"
In nuclear fusion, we are releasing some latent energy which is present in the materials, by changing the nuclear structure into other materials that contain less energy.
The energy we are getting out was "put in" to those nuclear reagents.
| {
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(Re-)use of a space elevator (basic mechanics and potential energy source) It's said that if a space elevator were made then it would be much more efficient to put objects in orbit. I've always wondered about the durability of a space elevator though. I don't mean the material strength but rather what affect using the elevator has on the elevator.
To put some massive object in orbit requires increasing its potential energy by a lot. Where is this energy coming from? Is the energy 100% from the fuel used to power whatever climbs the elevator? Is energy sapped from the Earth's rotation? Does climbing the elevator move the counterweight at all and does the position of the counterweight have to be adjusted after each climb?
I assume that a space elevator can be used over and over but I'd like to understand what the ultimate source of energy is and what allows for elevator re-use.
| I suppose that for a properly designed space elevator, the counterweight is placed far enough above geosynchronous orbit height to be able to withstand both the weight of the cable and the weight of the climber (i.e. the elevator cabin and the cargo inside it). If so, there is no reason why going up in the elevator would move the counterweight. In principle, Earth and the counterweight would together move ever so slightly in the opposite direction of the climber, since the center of mass of the whole system must remain the same, but the position of the counterweight relative to Earth would not change. This effect is of course negligible due to the magnitude of Earth's mass, just as the corresponding effect when travelling in a normal elevator.
The increase in the potential energy of the climber should be supplied from a source on the ground. Different methods for delivering this energy to the climber is discussed here.
| {
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Galaxies moving away at the speed of light As an arts student, I really find those cosmological questions hard to understand and hence come here to seek your kind help.
The Hubble constant $H_0$ is estimated to be about 65 km/s/Mpc, where 1 Mpc (megaparsec) is around 3.26 million light-years. At what distance would galaxies be moving away at exactly the
speed of light? (I found that there is something called Hubble Radius, but is this the same as Hubble Radius?) If there were galaxies farther than the Hubble radius, how would they appear to us?
| That's a very, very good question! Actually, the point you are addressing is the reason why physicists coined the term "observable universe". Those galaxies moving away from us with a speed bigger than the speed of light will never be visible (in the light they emit right now) to us and are outside the so-called "Hubble sphere". The distance you are looking for is $\frac{c}{H_0}$.
So as sad as it may sound, every second more and more galaxies and astronomical objects are leaving the observable universe, never to be seen again.
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The rule breaker, emissivity + reflectivity = 1 If emissivity and reflectivity are inversely proportionate, why does glass have a high emissivity of around 0.95-0.97 as well as being very reflective for IR Radiation?
normally it works but not with glass!
Can anyone explain this?
| One has to be careful with reflection. If you take a car rear-view mirror, it might seem to be close to 100% reflectivity, but in practice may be around only 60% (this is helped partly because the inside of the car is usually darker than the outside).
If you couple this with the usual compressed, pseudo-color look up table that most thermal cameras use, the apparent reflectivity of objects can be 'accentuated'.
| {
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Special Relativity - speed of light question Just a basic question:
I know that if you are traveling at $x$ speed the time will pass for you slower than to an observer that is relatively stopped. That's all just because a photon released at the $x$ speed can't travel faster than the $c$ limit.
I want to know what happens if you have two bodies, $A$ and $B$ moving towards each other. If $A$ releases a light beam, and $B$ measures it (the speed of the photons), the speed measured is still the same? The only difference will be the wave length?
And if we have the opposite case, $A$ and $B$ are moving away from each other, we get the red shift, but the speed measured will be still the same?
I just want to know if I got it right...
| The speed of light will be the same, yes. This is the fundamental tenet of special relativity - that all inertial observers see the same laws of physics, including universal constants like the speed of light.
And yes, the wavelength $\lambda$ will change. The frequency $\nu$ will also change, since after all we still must have
$$ \lambda \nu = c. $$
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Is lattice enthalpy positive or negative? I've learnt that the lattice enthalpy (defined as the energy change from a solid ionic lattice to separate gaseous ions) is always positive, obviously. However, I've seen it explained as the opposite other places, so it's negative.
What is correct?
| the lattice dissociation enthalpies are always positive and the lattice formation enthalpies are always negative. when 1 mole of lattice breaks down into its gaseous ions then energy is gained here hence +ve enthalpy but when gaseous constituent ions come together to form a lattice energy is released and hence enthalpy is -ve as exothermic reactions have negative enthalpy and endothermic reactions have positive enthalpies.
| {
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How does the freezing temperature of water vary with respect to pressure? I know when the pressure is reduced, the boiling temperature of water is reduced as well. But how does the pressure affect the freezing point of water?
In a low-pressure environment, is water's freezing temperature higher or lower than $0\sideset{^{\circ}}{}{\mathrm{C}} \, ?$
| If you decrease the pressure, the freezing point of water will increase ever so slightly. From 0° C at 1 atm pressure it will increase up to 0.01° C at 0.006 atm. This is the triple point of water. At pressures below this, water will never be liquid. It will change directly between solid and gas phase (sublimation). The temperature for this phase change, the sublimation point, will decrease as the pressure is further decreased.
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Higher order covariant Lagrangian I'm in search of examples of Lagrangian, which are at least second order in the derivatives and are covariant, preferable for field theories. Up to now I could only find first-order (such at Klein-Gordon-Lagrangian) or non-covariant (e.g. KdV) ones. Also some pointers to the literature about general properties of such systems are welcome.
Thanks
| You can look at the Lagrangian for the galileon particles for instance in this paper. It has the property that the equations of motion remains 2nd order in the derivatives and covariant.
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How does the wavelength change in relativistic limit? In the text, it reads that the momentum of a particle will change if it is moving at speed close to light speed. In the general case, the wavelength is given as
$$
\lambda = \frac{h}{p}
$$
and
$$p = \frac{mv}{\sqrt{1-v^2/c^2}}$$
when $v \to c$, $p\to\infty$, so is it say that the wavelength is ZERO? I don't understand why the wavelength will change to zero if it is moving at speed very close to light speed?
| Yes. The energy-momentum equation $E^2=p^2c^2+m^2c^4$ says that a massive object's mass (relativistic mass), momentum and energy approaches $\infty$ as a particle is accelerated towards $c$. There's nothing obvious about the fact that it requires infinite energy to accelerate it to $c$. This strictly means that you can't accelerate the object to speed-of-light...
Substituting Planck's wave equation $h\nu$ in $E=pc$ for photons, we get $$\lambda=\frac{hc}{pc}$$
Or in case of particles with rest energy where we can substitute for total energy, $$\lambda=\frac{h}{\sqrt{E^2-{(m_0c^2)}^2}}=\frac{h}{\sqrt{p^2-2mE_0}}$$
This equation says that matter waves are observable only for matter (massive) particles which always travel at $v<c$. In other words, there is no matter wave for such objects.
| {
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Could a planet sized bubble of breatheable atmostphere exist? I'm reading a book (Sun of Suns by Karl Schroeder) that the main location is a planet called Virga, which contains air, water, and floating chunks of rock, and has no or a very small amount of gravity. There is a main 'sun' at the center of the planet, which provides the heat for weather.
Could a 'planet' of this type exist?
| A sun or a star is not possible to exist on this scale; to be as massive as a core of a planet, it's just not massive enough. But you didn't mention the size of it.
So if we put that aside, first of all there's no such thing as no gravity. Where there's mass there's gravity, and that gravity has to be strong enough to hold gas (atmosphere). And the rocks will have to sink into the core since they are the denser objects.
If however we compared this to an existing example, where the sun is in the center of the solar system and holding planets (floating chunks of rocks), there's still vacuum in between. Because at the distances these planets are from the sun, the sun's gravity isn't strong enough to hold gas. Where the sun's gravity is strong enough there's gas, and that ends as far as the outer atmosphere of the sun itself. Which mainly doesn't extend to the planets. Therefore it's not possible.
| {
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How do you tell if a metric is curved? I was reading up on the Kerr metric (from Sean Carroll's book) and something that he said confused me.
To start with, the Kerr metric is pretty messy, but importantly, it contains two constants - $M$ and $a$. $M$ is identified as some mass, and $a$ is identified as angular momentum per unit mass. He says that this metric reduces to flat space in the limit $M \rightarrow 0$, and is given by
$$ds^2 = -dt^2 + \frac{r^2 + a^2 \cos^2\theta}{r^2 + a^2}dr^2 + \left(r^2 + a^2 \cos^2\theta \right)d\theta^2 + \left(r^2 + a^2 \right)\sin^2\theta d\phi^2 $$
and $r$, $\theta$ and $\phi$ are regular spherical polar co-ordinates.
But I don't understand why this space is obviously flat. The Schwarzschild metric also contains terms involving $dt^2$, $dr^2$, $d\theta^2$ and $d\phi^2$ but is curved. I always thought that a metric with off diagonal elements implied a curved space, but clearly I was very wrong.
Question: How do you tell if a metric is curved or not, from it's components?
| The flat space time refers here to the spacetime of Minkowski written with the spherical coordinates (I think one of your sign is wrong in your equation)
$$
ds^2 = -dt^2 + dr^2 + r^2 d\theta^2 + r^2 sin^2 \theta d\phi^2.
$$
where the metric is diagonal and has constant coefficients $g_{\mu \nu} = ( -1,1,1,1)$.
I would say that the conditions for a flat space-time, regarding only its metric, refer to its diagonal shape and constant (at least constant)
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Whats the anti-torque mechanism in horizontal take-off aircraft? In most helicopters there is the anti-torque tail rotor to prevent the body from spinning in the opposite direction to the main rotor.
What's the equivalent mechanism in horizontal takeoff single engine propeller, and jet aircrafts, where the air or the jet coming out back from the turbine or propeller is spinning and will cause such aircraft to roll?
| If you place two engines spinning in opposite directions equidistantly from the center of gravity, the torque is cancelled. Hence the old Air Force song:
"Don't give me a P-38, with props that counter-rotate
They'll loop, roll and spin but they'll soon auger in
Don't give me a P-38!"
Single engine planes can compensate with a bit of rudder . Check Martin Beckett's answer below.
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What is baryon loading in the context of gamma ray bursts (GRBs)? I've read that with short hard gamma ray bursts (shGRBs) associated with the coalescence of NS-NS and NS-BH binaries are expected to be beamed along the axis of the orbital angular momentum (i.e. perpendicular to the orbital plane) because the "baryon loading is lowest there". Here, NS = neutron star, BH = (stellar mass) black hole.
*
*What is meant by baryon loading?
*Why is it lowest along the direction of the orbital angular momentum?
| The baryon load is either the number or the density of baryons. It is the amount of baryons the burst must either clear out or carry along.
It is low at the poles and high in the orbital plain because there is nothing but empty space at the poles, whereas in the orbital plane there can be an accretion disk as well as the two possibly disrupted orbiting bodies and their remnants, which can be a very significant amount of matter, and thus smother (or obscure) the burst in these directions.
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Is it possible to "add cold" or to "add heat" to systems?
Amanda just poured herself a cup of hot coffee to get her day started.
She took her first sip and nearly burned her tongue. Since she didn't
have much time to sit and wait for it to cool down, she put an ice
cube in her coffee and stirred it with a metal spoon. After a moment,
she felt the spoon warm up, but when she took another sip, the coffee
was cooler. She was pretty sure, the ice added cold to her coffee,
and the coffee added heat to her spoon.
Would you agree?
| This is not an opinion based exercise. Thermal energy is generated through conversion. It is not inherent. The natural state of empty dark space is absolute zero. Stars and other celestial bodies generate thermal energy which is distributed through space and absorbed and retained by other objects. It is not possible to add cold. It is only possible to add or remove heat. This is most easily observed in the concept of computer hat sinks. They cool the cpu by drawing heat into the metal and using air movement to remove it from inside the chassis
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If the moon was rapid enough would it be able to orbit the earth from a close distance? If the moon was close in orbit that it's surface was like 100 km away from the earth's surface. And it had a large enough angular velocity will it be able to hold orbit?
If this was possible, is something similar possible to exist in the universe (the very close orbit)?
If this was hypothetically in effect (ignoring the damage that would cause to life on earth), would a person on earth, or a person on the moon tend to levitate, or gain weight on the far sides?
| You should do some hard assumptions, as no atmosphere, but the real problem is the so called Roche Limit, that states that at this limit, the tidal forces , difference of the gravitational potential between the face facing the Earth and the one opposite, is so large that rips the body (Moon) apart.Wiki
With the Wikipedia formula:
$
d = 2.44 R_{Earth} \left(\frac{\rho_{Earth}}{\rho_{Moon}}\right)^{\frac{1}{3}} \sim
2.44\cdot R_{Earth}km \left(\frac{5.5}{3.3}\right)^{\frac{1}{3}} \sim 2.89R_{Earth}
$
That would be well beyond the Moon in your scenario.
But despite everything, the Moon and the Earth survives, and people also, then you should simply equate the gravitational pulls of each body.
$Earth_{Surface} = m_{individual}g = 70 \cdot 9.8m/s^2$.
$Moon_{100km above} = G\frac{M_{Moon}m_{individual}}{(d+R_{Moon})^2} \sim m_{individual}1.44 m/s^2$
where $m_{individual}$ is your mass. Earth pull would still be $6.8$ times stronger. You would not levitate, but you wouldn't need to follow a diet plan.
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Dimensional Regularization involving $\epsilon^{\mu\nu\alpha\beta}$ Is it possible to dimensionally regularize an amplitude which contains the totally antisymmetric Levi-Civita tensor $\epsilon^{\mu\nu\alpha\beta}$?
I don't know if it's possible to define
$\epsilon^{\mu\nu\alpha\beta}$ in e.g. $$d-\eta$$-dimensions where $\eta$ is considered small and which we set to zero in the end.
So what are your thoughts?
| this fails since the tensor $ \epsilon ^{a,b,c,d} $ is diemnsion dependent
however in the case of zeta regularization of integrals $ \int_{a}^{\infty}dx x^{m-s} $ with 's' a regulator we can overcome this problem
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Vector and Spinor Representation in Ramond-Neveu-Schwarz Superstring Theory I am learning Ramnond-Neveu-Schwarz Superstring theory (RNS theory). I often find the following notation, especially in the closed string spectrum etc.:
$$\mathbf{8}_s,\mathbf{8}_v $$
And it is noted that these are vector and spinor representations of something. I have two questions regarding these.
*
*What are these representations of? Are they representations of $SO(8)$?
*What do they actually mean? How do you represent something in vector/spinor notation.
| Yes, they're representations of $SO(8)$, more precisely $Spin(8)$ which is an "improvement" of $SO(8)$ that allows the rotation by 360 degrees to be represented by a matrix different from the unit matrix, namely minus unit matrix.
${\bf 8}_v$ transforms normally as
$$ v\mapsto M v$$
where $MM^T=1$ is the $8\times 8$ real orthogonal $SO(8)$ matrix. The spinor reps ${\bf 8}_s\oplus {\bf 8}_c$ label the left-handed and right-handed spinor, respectively. People usually learn spinors well before they study RNS string theory.
The spinor representation transforms under $SO(8)$ in a way that is fully encoded by the transformation rules under infinitesimal $SO(8)$ transformations, $1+i\omega_{ij} J^{ij}$ where $\omega$ are the angle parameters and $J$ are the generators.
In the Dirac spinor representation, $J_{ij}$ is written as
$$ J_{ij} = \frac{\gamma_i \gamma_j - \gamma_j\gamma_i}{4}$$
where $\gamma$ are the Dirac matrices that may be written as tensor products of Pauli matrices and the unit matrix and that obey
$$\gamma_i\gamma_j+\gamma_j\gamma_i = 2\delta_{ij}\cdot {\bf 1}$$
Each pair of added dimensions doubles the size of the Dirac matrices so the dimension of the total "Dirac" representation for $SO(2n)$ is $2^n$. For $n=4$ we get $2^4=16$.
This 16-dimensional spinor representation is real and may be split, according to the eigenvalue of the $\Gamma_9$ chirality matrix, to the 8-dimensional chiral (=Weyl) spinor representations labeled by the indices s,c.
For $SO(8)$, there are 3 real 8-dimensional irreducible representations that are "equally good" and may actually be permuted by an operation called "triality". This operation may be seen as the $S_3$ permutation symmetry of the 3 legs of the Mercedes-logo-like $SO(8)$ Dynkin diagram. I just wrote a text about it last night:
http://motls.blogspot.cz/2013/04/complex-real-and-pseudoreal.html?m=1
If you really need to explain what a representation of a group is, you should interrupt your studies of string theory and focus on group theory – keywords Lie groups, Lie algebras, and representation theory. Without this background, you would face similar confusion too often.
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How does a star ignite? I remember reading that X-Rays are generated by 'braking' electrons in a Coolidge tube.
Is it fundamentally a matter that the extreme gravity immediately before a star ignites is so strong that it affects the hydrogen atoms to the point the velocity of it's components must be let-off in the form of heat & light?
How does a star ignite?
| The nuclear fusion that powers stars has little to nothing to do with electrons. In the cores of stars, temperatures are high enough that all the electrons are stripped from the nuclei, leaving a pure plasma.
As stars contract and condense out of interstellar dust, their gravitational potential energy is converted to heat faster than this heat can be radiated away. Once the temperature reaches roughly $10^7\ \mathrm{K}$, protons (hydrogen nuclei, stripped of their electrons) have a nonnegligible chance of sticking together when they colide, with one of them converting to a neutron along the way:
$$ {}^1H + {}^1H \to {}^2H + e^+ + \nu_e. $$
This is the first step of the PP chain, and it releases energy. There are more steps that ultimately turn four protons into a helium-4 nucleus. In more massive stars than the Sun, there are other ways (e.g. the CNO cycle) to catalyze this process with the help of carbon, nitrogen, and oxygen.
In any event, there is nothing extreme about the gravity. It just happened to pull matter from a huge distance close together. If you took infinitely spread apart particles totaling mass $M$ and formed a uniformly dense sphere of radius $R$, the gravitational potential energy released would be
$$ \frac{3GM^2}{5R}, $$
about half of which you expect to go into heating the material. Once hot, hydrogen naturally forms helium in exothermic processes.
Stellar reactions are self-regulating in the sense that if the rate of fusion increases, the additional luminosity would push the outer layers of the star, causing the star to expand and cool, thus reducing the reaction rate. Thus as long as there is hydrogen in the core, stars more or less burn at a steady rate once ignited.
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Does Earth produce metallic elements in its core? Does Earth produce metallic elements in its core?
| By metal-production, I think you're mentioning nuclear fusion (like those that take place in hydrogen bombs and core of stars) by which elements can be produced. If you look at the Wiki article I've linked, you can see a quote:
to be about the same temperature as the surface of the Sun - approximately 5700 K
I don't think such a low (relative to sun) temperature is sufficient enough to fuse the nuclei of elements (not even hydrogen can fuse at such a low temperature) to form new elements. Nuclear fusion requires to about several million (or sometimes a billion) Kelvin temperature.
So roughly, it's NO - our Earth doesn't..! It lacks a few zeros at its heat value...
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Interval and proper time Is the definition of $$d s^2=-d \tau^2$$ assuming that $c=1$, so that we always have $$\left({ds\over d\tau}\right)^2=-1$$?
Is there a reason for this definition?
Don't we get an imaginary ${ds\over d\tau}$?
| It depends on what convention you're using for the metric's signature. Some people use the metric signature (-+++), which is what you have there. The interval is then:
$$ds^2=-dt^2+d \mathbf{r}^2$$
On the other hand, some people use the (+---) convention:
$$ds^2=dt^2-d \mathbf{r}^2$$
In this signature $ds^2=d \tau^2$. Which one you use is a matter of preference. Special/General Relativity textbooks tend to prefer (-+++), though a lot of Quantum Mechanics textbooks prefer (+---).
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Waveguides (in the ocean?) The speed of sound in the ocean is given by
$$c_s(\theta,z) = 1450 + 4.6\theta - 0.055\theta^2 + 0.016z$$
$\theta$ is the temperature in degrees celcius, and $z$ is the depth. In a simplified model, $\theta$ is constant at 10$\,^\circ $C for the part of the ocean above the "themocline". The thermocline is an interface at depth 700 m over which the temperature drops to 4$\,^\circ $C almost instantly.
The question: It claimed that the water below the thermocline can act as a waveguide. Why and what is the extent (in depth) of this waveguide?
My thoughts: Evaluating some relevant speeds: $c_s(10,700) = 1501.7 m\,s^{-1}$ and $c_s(4,700) = 1478.7 m\,s^{-1}$. As the speed changes at the thermocline, there will be refraction and reflection of incident waves from both sides (above and below).
So waves incident from below will be reflected back. However I don't understand what makes the waves reflect on the lower side of this "waveguide". As far as I can see, the speed of sound will increase with depth. If there is no interface with a sudden discontinuity like the thermocline, how does this situation work?
| I wonder if you're mixing up the thermocline and the SOFAR channel. The speed of sound is a minimum at the SOFAR depth, so water at this depth acts as a waveguide.
I imagine sound will reflect off the thermocline, but I don't see how this would act as a waveguide unless the sea bottom acts as the lower reflector.
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Poles bit in a propagator Hi I am trying to derive the K-G propagator and am stuck on the bit where Cauchy's Integral formula is needed i.e evaluating from $$\int \frac{d^{3}p}{(2\pi)^3}\left\lbrace\frac{1}{2E_{p}}e^{-ip.(x-y)}|_{p^{o}=E_{p}}+\frac{1}{-2E_{p}}e^{-ip.(x-y)}|_{p^{o}=-E_{p}}\right\rbrace $$ to $$\int \frac{d^{3}p}{(2\pi)^3}\int \frac{dp^{0}}{i2\pi}\frac{-1}{P^{2}-m^{2}}e^{-ip.(x-y)} $$
I understand that the formula $g(z_{0})=\frac{1}{2\pi i}\int \frac{g(z)}{z-z_{0}}dz$ must be used but I just don't see how the solution can be found,
| For Feynman prescription, the poles are located at $p^0=\pm(E_p-i\epsilon)$. When $x^0>y^0$, we close the counter below the positive pole such that $\Re(-ip^0(x^0-y^0))<0$; When $x^0<y^0$, we close the counter above the negative pole such that $\Re(-ip^0(x^0-y^0))<0$. According to Jodan lemma, we know that
$$\int_{|p^0|=+\infty}\frac{dp^0}{2\pi i}\,\frac{-1}{P^2-m^2}e^{-ip^0(x^0-y^0)}=0$$
Notice that $(p^0)^2-E^2_p=(p^0-E_p)(p^0+E_p)$ and the counter we choose only has one pole. For $x^0>y^0$, we have
$$z_0=E_p,\quad g(z)=\frac{-1}{p^0+E_p}e^{-ip^0(x^0-y^0)}$$
and for $x^0<y^0$, we have
$$z_0=-E_p,\quad g(z)=\frac{-1}{p^0-E_p}e^{-ip^0(x^0-y^0)}$$
Then, with the residue theorem
$$\frac{1}{2\pi i}\int dp^0\,\frac{g(z)}{z-z_0}=g(z_0)$$
we can obtian that
$$\frac{1}{2\pi i}\int dp^0\,\frac{-1}{(p^0)^2-E^2_p+i\epsilon}e^{-ip^0(x^0-y^0)}=\frac{1}{2E_p}e^{-iE_p(x^0-y^0)}\theta(x^0-y^0)+\frac{1}{-2E_p}e^{iE_p(x^0-y^0)}\theta(y^0-x^0)$$
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In terms of the Doppler effect, what happens when the source is moving faster than the wave? I'm just trying to understand this problem from a qualitative perspective. The Doppler effect is commonly explained in terms of how a siren sounds higher in pitch as it is approaching a particular observer. I understand this is because the velocity of the wave is constant and so the frequency of the waves increase as they are bunched together. What would happen if a siren was mounted on say a plane traveling at a supersonic speed? To clarify what would the observer observe/hear? Apologies if my question is not phrase very well my knowledge of physics is very rudimentary.
|
The first image shows an object traveling at Mach 1 ($v=c$). The second one shows the object traveling at some supersonic velocity ($v>c$). For both the cases, the longitudinal pressure waves pile up. Say the observer is standing in the ground and the object is traveling at $c$. The observer can't hear the pitch of sound because, the waves reach him all at once and hence, he'd hear a loud "bash". The most necessary thing is that he had to wait until the source arrives. When the source is directly overhead, he hears the shock waves.
When the object breaks the sound barrier (supersonic), it's somewhat worse. The same loud "thump" is produced here. But, the observer would notice a delay in sound (i.e) he has to wait for the shock waves to reach him. There's also this Mach cone produced by these waves since the waves group so fast behind the object. And so, there's a region of high pressure at first followed by a low pressure zone. Thus, if the object passes by in some comparable distance, it makes a lot of disturbance, "breaking things", etc...
The comic thing is, for someone inside the aircraft, he can still speak with his partner, can hear the bump of a ball on the plane, etc. The problem is only for the distant observer who suffers...
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How long does a supernova last? Is a supernova over instantaneously? Or, does the (for want of a better word) explosion continue for a while? What is/are the order of timescales involved? What is the duration for which the supernova continues to release copious amounts of energy?
| I love theoretical physics, I am not capable of the math, but here's a neat comparison. This is one of many proposed solutions to the mean free path of a photon produced in the core of the sun, this one says 4000 years to travel to emission surface...pretty wild drunken walk indeed! This is due to the assumed density of the core and various assumed layers making up the solar structure.
http://image.gsfc.nasa.gov/poetry/ask/a11354.html
SN1987A was several solar masses. The equations relevant to this phenomenon come up with...somehow, a kinetically powered explosion lasting a few days, which manages to propegate through several solar masses of matter.
All things being equal it would seem that a core collapse supernova explosion for a several solar mass star should take alot longer to manifest visually.
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What happens when a star undergoes gravitational collapse? Immediately prior to becoming a supernova the core of some types of stars may suffer gravitational collapse.
*
*What happens to any planets in orbit around the star at the instant the mass is fully collapsed?
*Assuming this sudden change would cause some perturbation ; how large/distant would a planet in the system have to be to be relatively immune to such perturbation?
*Could unexplained perturbation serve as an indicator of a historic supernova in the vicinity?
| It depends on the nature of the system, and the explosion. If more than about half the mass of the system is lost from the central star, the planet will become unbound (interesting National Geographic article on the subject). This can be relevant even before the actually supernova - as massive stars lose a lot of mass through winds. In any case, if the surviving remnant is massive enough, the planet will stay in orbit.
Nothing happens at the instant of collapse, because the planet doesn't 'know' about it until the changes in gravity become important.
The dynamic effects of supernovae ejecta can be important. In particular, if the planet is a gas giant - significant amounts of mass can be blown off by the ejecta. If I recall correctly, even close in rocky planets can absorb enough energy to become disrupted themselves.
There are some additional perturbative effects that are important, especially possible 'supernova kicks' (see for example this astrobites article). Kicks are sometimes considered in binary star systems to explain large eccentricities (for example).
Currently it's hard to find planets around non-main-sequence stars (despite the first exoplanet being found orbiting a pulsar---which also shows that planets can remain bound after supernovae), so it's a bit early to seriously examine what types of thing to look for in those cases.
| {
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Parabolic motion (experiment) We performed a laboratory, performing six releases of a sphere with angles $15^\circ,30^\circ,45^\circ,60^\circ,75^\circ,40^\circ$ a parabolic movement, took five distances for each angle, the initial velocity was calculated $3.025~\text{m/s}$.
Then doing 5 tosses of the sphere with an angle of $90^\circ$ times were recorded and calculated initial velocity was $3.33~\text{m/s}$, the question is: Why are these speeds almost the same?
| I'm a little confused by one part of your question. You ask,
Why are these speeds almost the same?
If you are asking why they are almost the same, and not completely different, then the answer is that you expect them to be the same because they are caused by the same launch mechanism. But I don't think you are asking that.
I think you are asking why they are almost the same and not exactly the same. The answer is that you can't see that they are different until you take the uncertainties into account. You don't report the uncertainties, but my guess is that the uncertainty for the second part is much larger than the uncertainty for the first part. And my guess is that the two numbers are within the uncertainties of your experiment.
I agree with Gugg that you should check to see what time you should have in the second part. I solved for the intial speed in terms of the measured time for that part of the experiment, and I got that an error of $0.1 \mathrm{s}$ in the time will produce an error of $9.8 \mathrm{m/s}$ in the speed. So the difference you see in the speeds suggests that you are over-estimating the time by about $0.03 \mathrm{s}$ in that second part. That is actually smaller than normal human reaction time, so I would guess that your two values are well within the uncertainty you have in the experiment.
If you still have the apparatus, you might try repeating one of the angles from the first set of experiments, and measure both the time and the range. You could then compare the speed you get from those measurements to see how well they agree. In this case, you know for certain that the speed is the same (because it's the same launch), so any disagreement is due to the measurements themselves.
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Can the effects of a person's mass upon the local gravitational field be detected and measured remotely? As the title suggests, Can the effects of a person's mass upon the local gravitational field be detected and measured remotely?
I am aware any mass produces and effects gravity but couldn't find anything in my searching if it is possible or theoretically possible to detect this effect remotely.
| According to this site
http://en.wikipedia.org/wiki/Gravimetry
Gravimeters can have a accuracy of up to 0.002 mGal (= 2*10^-8 m/s^2)
The gravity of a person of lets say 100kg at a distance of one meter is approximately
G * m/r^2 = 6.7*10^-9 m/s^2
So that person would need to weigh at least 300kg to be detectable.
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Joule heating due to the (slow) electron drift velocity? I understand the concept of why the signal speed is higher than the electron drift velocity, but I can't understand the concept of joule heating. If electrons move slow then how do they produce a lot of heat when they hit the nucleus. Besides my friend once told me that the drift velocity is the net movement and electrons move fast in all directions, if that is the case why do they move like that?
| The absolute velocity of the electrons actually doesn't matter for joule heating. Think about it this way, if there is no current flowing there wouldn't be any joule heating. So, even if electrons are moving quickly and randomly when no current is flowing, we know no joule heating would occur and that joule heating is really about the net change in effect caused by the current. That is, the base electron velocity doesn't have an effect. All that matters is the $\Delta V$ over the base electron velocity which is given by the drift velocity.
Joule heating is really about electrical energy lost to heat due to resistance. Even if the average drift velocity of an electron is tiny, there are so many electrons moving that the tiny energy loss to heat for each electron adds up. As you know, current is the result of a huge number of moving electrons. It's a numbers game. The more electrons losing a tiny bit of energy there are, the more total heat is generated.
Via Ohm's Law you can see that $P_{ower} = I^2 R$ so it's no wonder that heat generation is proportional to $I^2 R$.
Also, in your question you mentioned an electron bumping into a nucleus. That is not what's happening. Electrons are colliding with the electron cloud of atoms and via electromagnetic repulsion are pushing the whole atom a bit, increasing its kinetic energy. It's just free electrons interacting with bound electrons.
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When does Thevenin's theorem not apply (modelling a power source with a ohmic internal resistance) Most physics text books say that a power source can be modelled as an EMF with a internal resistance. This is also know as Thevenin's theorem or Norton's theorem. However I have read in some sources that this is not always the case.
When does this model not apply/break down and for what reasons?
| It is doubtful if Thevenin's eqv also supports transients/dynamics. Suppose the ïnternals of a battery include a resistance as is usual and also an inductance in series with it. Concept of ïmpedance does not hold as supply is DC, and the same cannot also be converted to an eqv Norton's.
If the supply is AC and the "source" internals include a series resistance + an inductance ALONG with a capacitance across the output terminals of this "practical AC source" will the Thevenin's eqv hold for transients? No.
Test by shorting the outputs. The original with a capacitive termination will give an infinite current at t= 0+, while the eqv Zth, if inductive, will give Zero current at t = 0+.
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Is light red shifted in optical tweezers? This is a question I put to my supervisor during my PhD many years ago, and never really got a satisfactory answer to.
In an optical tweezers, assume that a beam of light is used to move a glass bead. My question is whether the outgoing light is red-shifted. If it is not I cannot reconcile how energy is conserved in the system, as work has been done to move the bead?
Answers appreciated as always...
| In theory, yes, the light will be redshifted. In practice, it sounds like the glass bead is too large for any measurable red shift. This is actually used in Mossbauer spectroscopy. What happens is that if your $\gamma$-ray source is a free atom, the recoil of the atom will cause the resulting radiation to be red-shifted relative to the natural frequency of the transition. In order to use the $\gamma$-ray usefully, you have to embed the source atom inside a crystal. Then it's the entire crystal that is recoiling, and that reduces the resulting red shift to something you can't measure.
This is the exact same problem as a collision between two billiard balls in classical mechanics. The photon has energy $E_\mathrm{ph} = hc/\lambda$ and momentum $|\vec p_\mathrm{ph}| = h/\lambda$. The glass bead has a kinetic energy $K = \frac{1}{2}mv^2$, momentum $\vec{p} = m\vec{v}$, and possibly some internal energy if we allow it to change state. The glass bead must change its velocity in order to conserve momentum. That will probably involve a change of speed, in which case the photon must change wavelength in order to conserve energy. It's easy to think of this as a doppler shift, if you think of the photon as being temporarily absorbed and then re-emitted. The magnitude of this goes down as the mass of the glass bead increases, in a manner that should be easy to work out.
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Voltage of open circuit
A battery with emf $\varepsilon$ and internal resistance $r$ is connected
with a resistor $R$ in the following open circuit. What is the voltage $V_{ab}=V_a-V_b$?
The answer is $- \varepsilon$. "No current. There is no voltage change across R and r.". But I don't really understand why ... I was thinking intuitively it should be $0$? Then thinking of how to get 0, I was thinking ... $V_a = - \varepsilon$ since its on the negative terminal, the $V_b = + \varepsilon$ since its on the positive terminal. But $V_a - V_b = -2 \varepsilon$ ... how do I make sense of this?
|
But Va−Vb=−2ε ... how do I make sense of this?
It's incorrect to write $V_a = \varepsilon$. The voltage $\varepsilon$ is across the battery.
Try this: place a ground symbol on the wire between the battery and the $a$ terminal; this is your zero node or the place you put the black lead of your voltmeter.
Now, if you place the red lead on the terminal of the battery connected to the resistor, you'll measure $\varepsilon$ volts.
If you place the red lead on the other side of the resistor, you'll measure $\varepsilon + V_R$ volts but $V_R = 0V$ so you still measure $\varepsilon$ volts.
But note that now you're measuring the voltage $V_{ba}$ since the read lead is connected to terminal b and the black lead is on terminal a.
So, $V_{ba} = \varepsilon = -V_{ab}$
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Purpose of Grover's algorithm? How is the output of Grover's algorithm useful if the result is required to use the oracle? If we already know the desired state, what's the point of using the algorithm?
So can you give me a concrete example of an oracle function. For example if the indexed items in a Grover search were, for example arbitrary patterns, what would the corresponding oracle function look like? Lets make the example more concrete. Each pattern is the image of a face and we want to see if an unknown face is located within pattern set. Classically our search algorithm is a correlation algorithm (e.g. Kendall-tau, rank correlation etc.). What would the analogue of this be for a quantum search?
| There is a difference between finding a solution and recognizing a solution. Oracle can recognize the solution or solve a particular instance of the problem but cannot give you the solution for complete problem. Or in other words, oracles gives you a part of solution and you may need to consult oracle a number of times to get the complete solution. Oracle also may be thought as a library function (as in programming languages) which will give you solution for one instance of a problem, and the real cost of computation is measured by how many times you call the function and not the inherent complexity of the function itself which is taken as black box. For example, lets say we have a oracle for a function $f(x) = x^2$, on presenting this oracle with a pair $(a,b)$ it will tell whether $b^2 = a$ or not. In this case time complexity is taken as how many time you need to consult the oracle to get the desired result.
More concrete example can be taken from oracle for verifying if the number if prime. Lets say we want to find the first prime number $p: a < p <b, a < b \in Z^+$. The problem has different complexity when you are given access to the oracle and when you are not given.
Physical example of an oracle: lets say our problem is to determine the angle between the floor and the wall which may not be necessarily $90^\circ$ to it. All you can do is throw a ball which will go elastic collision on the wall and return back. You have control of the angle you throw and you can note the angle it came back. Each throwing of a ball can be compared with the calling of oracle function and the constraint on the angle of the returning ball (reflection,which gives you a hint on the orientation of the wall) can be considered an oracle. The number of times you need to repeat the throwing of ball to get the orientation with desired accuracy may be considered as the complexity of the problem relative to the oracle.
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How can anything be hotter than the Sun? I've heard that if a space shuttle enters the atmosphere from a bad angle its surface will become so hot that it will be hotter than the surface of the Sun.
How can that be? It seems to an uneducated mind that Sun is really really hot, how could something seemingly minor such as a wrong angle of entrance to the earth's atmosphere could end up generating a heat hotter than the Sun?
| It depends on many factors such as the reentry velocity of the object, its shape (cone-spherical, etc.), what the planet's atmosphere is made of, whether it enters at some shallow angle and also the altitude where there's density variations in atmosphere, etc.
Googling on this, could return you a lot of results. And, all results matched a certain value. Here's the Wiki article on Thermal Protection which has its first phrase...
The Space Shuttle thermal protection system (TPS) is the barrier that protected the Space Shuttle Orbiter during the searing 1,650 °C (3,000 °F) heat of atmospheric reentry. A secondary goal was to protect from the heat and cold of space while on orbit
It's clear that it's nearly 4 times lower than the temperature of sun (5500 °C) When you hear things which you doubt, you should be skeptical and be sure to google it first...
And of course, a space shuttle entering at a bad angle (as you say) can have temperatures very much higher than $T_{sun}$. Here's a Wiki link (suggested by Gugg)
For example, a spacecraft entering the atmosphere at 7.8 km/s would experience a peak shock layer temperature of 7800 K.
But, it's still lower than the temperature of corona.
Your question's title is somewhat ambiguous. There are infinitely many number of celestial objects that are at a higher temperature than the sun (probably higher than its core).
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How can Ohm's law be correct if superconductors have 0 resistivity? Ohm's law states that the relationship between current ( I ) voltage ( V ) and resistance ( R ) is
$$I = \frac{V}{R}$$
However superconductors cause the resistance of a material to go to zero, and as I understand it, as $R \to 0$, $I \to \infty$. Does this present a problem for Ohm's law?
| Ohm's law works for ordinary conductors for a reason: the particles carrying the current (usually, but not always electrons) scatter incoherently and inelastically from features of the conductor. In the case of an electron current, at low temperature this scattering is caused by impurities in the conductor; at high temperatures, the dominant source of scattering is electron-phonon scattering (phonons are coherent vibrations of the conductors fundamental lattice). As long as those conditions pertain, you can expect Ohm's law to be a good approximation to the behavior of the current.
However, in a superconductor quantum mechanics rears its elegant head and generates a situation in which coherent effect dominate to a point where is effectively no inelastic scattering and as such no energy loss in the flow of current.
Nor are these the only possible situations, as Schlomo Steinbergerstein notes semiconductors exhibit a wide range of conduction behaviors.
The difference in macroscopic physics is down to a difference in microscopic physics.
This business where regimes dominated by coherent and incoherent interaction show very different behavior comes up a lot in various corners of physics and one could spend a long (and possibly enjoyable) time surveying those effects alone.
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Polarization of sound Sound can't be polarized because the vibration of such type can't be polarized i.e, it can't be limited or controlled by any barriers and so polarization is not possible in them.
This is what my teacher answered me when i asked the question. But i didn't understand what did he mean by "the vibration can't be controlled or limited." Does the word cant be limited or controlled make sense here? Moreover can anybody explain in details and more clearly to me?
| It would be difficult to imagine a polarisation of a longitudinal wave, but this certainly is true for transverse waves. The air couples weakly to transverse waves and so does your ear! Hence, we often consider sound waves (in the air) as longitudinal only, that would not be polarised.
Generally, though, sound waves can certainly be transverse and polarised. See for example a shear-stress mode in quartz.
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How can an asymptotic expansion give an extremely accurate prediction, as in QED? What is the meaning of "twenty digits accuracy" of certain QED calculations? If I take too little loops, or too many of them, the result won't be as accurate, so do people stop adding loops when the result of their calculation best agrees with experiment? I must be missing something.
| Suppose you're interested in computing some quantity $F(\alpha)$, like the excess magnetic moment $g-2$, which depends on the fine structure constant $\alpha \simeq .007$.
Perturbation theory gives a recipe for the coefficients $F_i$ of an infinite series $\sum_{i \geq 0} F_i \alpha^i$, which is expected to be asymptotic to $F$. This means that, if you add up the first few terms, you should get a decent approximation to $F$.
$F \simeq F_0 + F_1 \alpha + F_2 \alpha^2 + F_3 \alpha^3 + ... + F_N \alpha^N$
If you keep adding higher order corrections, making $N$ bigger, the approximation will get better for a while, and then, eventually, it will start to get worse and diverge in the limit $N \to \infty$. In QED, we don't know exactly where the approximation gets worse. However, one expects on general grounds that the series approximation will start to get worse when $N \simeq 1/\alpha \simeq 137$. (And this is indeed what we see in toy models, where we can make everything completely rigorous.)
In principle, then, we should add up all the Feynman diagrams of order $\leq 137$. In practice, we don't have the ability to do this. Computing all those diagrams is very time-consuming. Even with computer assistance, we have difficulty going beyond 4 or 5 loops.
| {
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What materials focus EM radiation in the 2.4GHz range If glass and similar materials refract visible light effectively, what materials would be best for focusing lower frequencies of EM radiation, if any? If not, what other methods exist for focusing these ranges?
The thought was inspired by wondering how you might build a camera that captures 'light' given off by Wifi routers
| Anything that 2.4Ghz EM waves (microwaves) will pass through that has a different index of refraction than air can be used as as a lens or prism. A really common material for this is wax.
Also, you can reflect microwaves using metal surfaces. This is how a satellite dish works.
For a simplified model of your wifi router, you can think about the microwaves as a bright light bulb flashing on and off very fast. An antenna is mostly designed to record the signal ("pattern of flashes") rather than the distribution of microwaves in space. To get an idea of how the signal propagates you'd need to take intensity samples in the space around the signal and generate a volumetric map.
| {
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Probability of position in linear shm? The problem that got me thinking goes like this:-
Find $dp/dx$ where $p$ is the probability of finding a body at a random instant of time undergoing linear shm according to $x=a\sin(\omega t)$. Plot the probability versus displacement graph. $x$=Displacement from mean.
My work:
$$v=dx/dt=\omega \sqrt{a^2-x^2}$$
Probability of finding within $x$ and $x+dx$ is $dt/T$ where dt is the time it spends there and T$$ is the total period.
Therefore
$$dp=dt/T=\frac{dx}{\pi \sqrt{a^2-x^2}}$$
because $t=2\pi /\omega$ and the factor 2 is to account for the fact that it spends time twice in one oscillation. The answer matches the answer and also the condition that integration $-a$ to $a$ of $dp =1$.
But when i try to find p as a function of x to plot the graph I get
$$p=\frac{1}{\pi}\arcsin(x/a)+C.$$
But then I get stuck as there is no way to find $C$ (except the fact that for $C=0$ the probability at the mean position is $0$ and hence $C$ cannot equal 0) which I know of. So how can I get a restraint on $C$ to find its value and hence to properly graph it with the condition that the probability from $-a$ to $a$ be 1?
| In your notation, $dp/dx$ is your probability density. $p(x)$ is your cumulative probability density, the probability that the particle is to the left of $x$ at a given time. Knowing this, there are at least three ways you can reason. As Emilio Pisanty pointed out in his answer, you can require that $p(-a) = 0$ (ie, that the particle cannot be to the left of $-a$. In addition to that, you can require that $p(a) = 1$ (ie, that the particle cannot be to the right of $a$), or you can require that $p(0) = 0.5$, because the system is symmetric about the origin. All of those should give you the same value for $C$.
| {
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Microscopic picture of an inductor I have a good understanding of how inductors behave in electrical circuits, and a somewhat rough-and-ready understanding of how this behaviour arises from Maxwell's equations. However, what I don't have a good mental picture of is how electromagnetic induction works on the microscopic level, i.e. in terms of forces experienced by individual electrons in the wire. Ideally I would like to be able to understand the operation of an inductor, at least qualitatively, in terms of the statistical mechanics of the electron gas in the coil and the electron spins in the core.
To clarify: in Maxwell's equations there is a term for $I$, the electric current. But current is a macroscopic quantity - it's the expected number of charges passing through a surface, with the expectation taken over an ensemble. So Faraday's law is a macroscopic relation, just like the gas equation. For the gas equation, we can understand how it arises from the microscopic motion of molecules. I want to understand how Faraday's law arises from the microscopic motion of electrons.
Would anyone be able to provide an explanation of how induction works in microscopic terms, or point me towards somewhere I can read up on it?
| You do not need neither a coil, nor core. The vacuum is a pretty good conductor of the magnetic field. Cores and coils only complicate the things. Consider a single loop of current. Surprise: It also has a magnetic field!
Individual electrons are susceptible to the electric field, that is induced by the changing magnetic field in the usual way: $F = q * E$. $\vec B$ circles around the wire, $E$ is in the direction of wire.
Might be you need to look at quantum electrodynamics?
| {
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Reaching the speed of light via quantum mechanical uncertainty? Suppose you accelerate a body to very near the speed of light $c$ where $v = c - \epsilon$. Although this would take an enormous energy, is it possible the last arbitrarily small velocity needed -- $\epsilon$ -- could be overcome with a minor bump in velocity due to the uncertainty principle?
| No, because the uncertainty principle operates between position and momentum rather than position and velocity. For speeds much less than $c$, momentum is just proportional to velocity: $p = mv$. But at relativistic speeds we have to use the relativistic version,
$$
p = \gamma mv,
$$
where $\gamma = 1/\sqrt{1-v^2/c^2}$. Substituting this in and squaring both sides we get
$$
p^2 = \frac{m^2v^2}{1-{v^2}/{c^2}},
$$
which we can rearrange a little to get
$$
v^2 = \frac{p^2}{ m^2 + p^2/c^2 },
$$
or
$$
v = \frac{p}{\sqrt{ m^2 + p^2/c^2 }}.
$$
Now, the limit of this as $p \to \infty$ is just
$$
v = \frac{p}{\sqrt{p^2/c^2 }} = c.
$$
The momentum $p$ can fluctuate due to the uncertainty principle, but now you can see that now matter how big $p$ gets, $v$ will always be less than $c$.
| {
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Evolution principle of the physical laws I wanted to know if there is a physical theory that considers that the laws of physics undergo an evolutionary process. That see the law of physics or the absence of them, as something dynamic, and that with time they slowly converge to something we know today. A kind of simulated annealing of the physical laws.
| This has nothing to do with theories of everything, but there is such a thing as quantum darwinism. I read about a long time ago and can not give a detailed answer, sorry. The link contains further reference. I remember the nature paper to be quite good.
Quantum Darwinism hopes to explain the collapse of wave functions in quantum mechanics as a dynamical process. Very roughly, the idea is that quantum information can somehow multiply in a thermodynamic environment. Only the states that are able to multiply efficiently can get amplified by the environment. This leads to the selection of eigenstates as observables.
| {
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Mathematical proof of non-negative change of entropy $\Delta S\geq0$ I understand that we can prove that for any process that occurs in an isolated and closed system it must hold that
$$\Delta S\geq0$$
via Clausius' theorem. My question is, how can I prove this in a mathematical way?
| Here's an enlightening special case: Take $n$ bodies with temperatures $T_1,\ldots T_n$ and bring them together until they reach a final temperature $T$. The first law of thermodynamics tells you that $T$ is the arithmetic mean of the $T_i$. The second law of thermodynamics tells you that the change in entropy is $n\log(T/G)$ where $G$ is the geometric mean. It's a standard theorem in pure mathematics that $T>G$, whence the change in entropy must be positive.
| {
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Are all points in the universe connected? Is it true that every point in the universe is connected or could be so theoretically? If so how is this mediated?
Is it through the quantum nature of the fabric of space or is it through the interrelationships of the gravity fields throughout the universe. From the gravity fields of single solar systems affecting each other to galaxies and galactic clusters as if they gravitational energy of each field has a knock-on effect on the next field?
| All points in the observable universe are "connected" in the sense that they can be acted upon by forces that have an infinite range (gravity and electromagnetism).
However, points that are outside of our cosmological horizon (due to the expansion of the universe) are no longer causally connected with points in our local vicinity, since they are receding from us faster than light. The same is true of points that are inside the event horizon of a black hole.
| {
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Lookback Time & Age of the Universe Calculations I try to calculate the age of the universe with the FLRW model:
$$ H(a) = H_0 \sqrt{\Omega_{\mathrm{R},0} \left(\frac{a_0}{a}\right)^4 + \Omega_{\mathrm{M},0} \left(\frac{a_0}{a}\right)^3 + (1-\Omega_{\mathrm{T},0}) \left(\frac{a_0}{a}\right)^2 + \Omega_{\Lambda,0}}. $$
I set $\Omega_{\mathrm{M},0} = 0.317$ (matter density) and $\Omega_{\Lambda,0} = 0.683$ (dark energy), as delivered by Planck 2013; $\Omega_{\mathrm{T},0} = 1.02$ (space curvature), according to this site; and $\Omega_{\mathrm{R},0} = 4.8\times10^{-5}$ (radiation density), according to this document.
For the time $t(a)$ I take the scale factor $a$ and divide it through the integrated recessional velocity
$$ t(a) = \frac{a}{\int_0^a{H(a')a'\ \mathrm{d}a'}/(a-0)} $$
and finally simplify to
$$ t(a) = \frac{a^2}{\int_0^a{H(a')a'\ \mathrm{d}a'}}. $$
But the problem is, I then get about $8\times10^9$ years for the age of the universe, but it should be around $12\times10^9$ years (which I get when I set $\Omega_{\mathrm{R},0}$ to zero):
$\Omega_{\mathrm{R},0} = 4.8\times10^{-5}$:
$\Omega_{\mathrm{R},0} = 0 \to 0.00001$:
Do I have to use some other models than FLRW/ΛCDM, or is one of my parameters outdated?
| The total energy density is by definition
$$ \Omega_{T,0} = \Omega_{R,0} + \Omega_{M,0} + \Omega_{\Lambda,0},$$
so with the values you cite ($\Omega_{R,0}=4.8\times 10^{-5}$, $\Omega_{M,0}=0.317$, $\Omega_{\Lambda,0}=0.683$), we get $\Omega_{T,0} = 1$, or in a more common notation $\Omega_{K,0}=1-\Omega_{T,0}=0$, i.e. a space with zero curvature.
It is also common to define the present-day value of the scale-factor as $a_0=1$, so that
$$
H(a) = H_0\sqrt{\Omega_{R,0}a^{-4} + \Omega_{M,0}a^{-3} + \Omega_{K,0}a^{-2} + \Omega_{\Lambda,0}}.
$$
The age of the universe can then be derived as follows: from
$$
\frac{\text{d}a}{\text{d}t} = \dot{a},
$$
we get
$$
\begin{align}
\text{d}t &= \frac{\text{d}a}{\dot{a}} = \frac{\text{d}a}{aH(a)} = \frac{a\,\text{d}a}{a^2H(a)}\\
&= \frac{1}{H_0}\frac{a\,\text{d}a}{a^2\sqrt{\Omega_{R,0}a^{-4} + \Omega_{M,0}a^{-3} + \Omega_{K,0}a^{-2} + \Omega_{\Lambda,0}}}\\
&= \frac{1}{H_0}\frac{a\,\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}a + \Omega_{K,0}a^2 + \Omega_{\Lambda,0}a^4}}.
\end{align}
$$
Integrating yields the difference between the time that a signal is emitted and the time it is observed:
$$
t_{\text{ob}} - t_{\text{em}} = \frac{1}{H_0}\int_{a_{\text{em}}}^{a_{\text{ob}}}
\frac{a\,\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}a + \Omega_{K,0}a^2 + \Omega_{\Lambda,0}a^4}},
$$
and the age of the universe is
$$t_0 = \frac{1}{H_0}\int_0^1
\frac{a\,\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}a + \Omega_{K,0}a^2 + \Omega_{\Lambda,0}a^4}}.$$
This should give you the correct age.
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Calculating the mechanical power of a water pump Say I want to pump water from one container to another. The water levels are 3 meters apart, and I want to pump 10 litres per hour. I figure the mechanical power necessary, assuming no losses, is:
$$
\require{cancel}
\dfrac{10\cancel{l}}{\cancel{h}}
\dfrac{kg}{\cancel{l}}
\dfrac{\cancel{h}}{3600s}
= \dfrac{0.0028kg}{s}
$$
$$
\dfrac{0.0028kg}{s}
\dfrac{3m}{1}
\dfrac{9.8m}{s^2}
= \dfrac{0.082kg\cdot m^2}{s\cdot s^2}
$$
$$
\dfrac{0.082\cancel{kg}\cdot \cancel{m^2}}{\cancel{s}\cdot \cancel{s^2}}
\dfrac{\cancel{J} \cdot \cancel{s^2}}{\cancel{kg} \cdot \cancel{m^2}}
\dfrac{W\cdot \cancel{s}}{\cancel{J}}
= 0.082W
$$
But, I know from practical experience that real centrifugal pumps that can work at a 3m head are big and certainly require orders of magnitude more electrical power. What explains the difference?
Intuitively, I figure this must be because the pump must exert some force to balance the force of gravity from pushing water backwards through the pump, siphoning the water back to the lower container, then exert yet more force to accomplish what was desired, pumping to the higher container.
*
*How is this force calculated mathematically?
*Assuming an ideal electric centrifugal pump, can we establish the electrical power required by the pump, given the difference in heights between the containers?
*Does this apply to all pumps, or just centrifugal pumps?
| The number you calculate is proportional to the flow rate you put in, so a ten times faster flow rate will require ten times more electrical power, so if you do the same calculation with a flow rate closer to what you'd get from a fountain, you'll probably get a lot closer to the right order of magnitude.
This calculation is actually the right way to calculate the power used by the theoretically most efficient fountain pump. Of course a fountain pump doesn't quasi-statically move water from one reservoir to another, but instead gives the water kinetic energy. But pretty most of that kinetic energy is then converted into potential energy by the water's inertia. (Some will be lost to heat instead, but this can be minimised by making the flow laminar, e.g. by using a flow straightener, especially if the flow rate is quite high.) So this calculation tells you the power you have to put into the water in the form of kinetic energy, in order for it to reach 3 metres in height.
However, there will be losses in the pump itself. Mechanical pumps can be made very efficient if they can operate relatively slowly, pushing water from place to place without much inertia involved. However, for a fountain you have to start with non-moving water and accelerate it to a high velocity, and I suspect it's a lot harder to do that efficiently.
| {
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If particles can find themselves spontaneously arranged, isn't entropy actually decreasing? Take a box of gas particles. At $t = 0$, the distribution of particles is homogeneous. There is a small probability that at $t = 1$, all particles go to the left side of the box. In this case, entropy is decreasing. However, it is a general principle is that entropy always increases. So, where is the problem please?
| Right, there is a small probability that the entropy will decrease. But for the decrease by $-|\Delta S|$, the probability is of the order $\exp(-|\Delta S| / k)$, exponentially small, where $k$ is (in the SI units) the tiny Boltzmann constant. So whenever $|\Delta S|$ is macroscopically large, something like one joule per kelvin, the probability of the decrease is de facto zero.
If you have $10^{20}$ molecules of gas (which is still just a small fraction of a gram), the probability that all of them will be in the same half of a box is something like $2^{-10^{20}}$. That's so small that even if you try to repeat the experiment everywhere in the Universe for its whole lifetime, you have no chance to succeed.
Statistical physics talks about probabilities and quantities with noise, as the previous paragraph exemplifies. But there is a limit of statistical physics that was known earlier, thermodynamics. Effectively, we can say that thermodynamics is the $k\to 0$ limit of statistical physics. We just neglect that $k$ is nonzero – it is tiny, anyway. In this limit, the noise of different quantities disappears and the exponential $\exp(-|\Delta S| / k)$ is strictly zero and the decreasing-entropy processes (by any finite amount, in everyday SI-like units) become strictly prohibited.
| {
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North and south of magnetic field
The current I is flowing upward in the wire in this figure. The direction of the magnetic filed due to the current can be determined by the right hand rule.
Can we determine the north and the south of the magnetic field produced by the current I by using a hand rule?
| The concept of magnetic poles is only defined for localized magnetic systems, which include permanent magnets (or equivalently their surface currents) and induction coils.
The reason for this is that the north/south pole description of a magnet is, mathematically, a description of the magnetic (dipole) moment of the system, where the dipole approximation to the field is only valid "away" from the system. If the wire is infinite, you can't be "far" from it.
For loop, surface, or volume currents, respectively, the magnetic moment is defined as
$$
\mathbf{m}
=\frac{1}{2}\int_C\mathbf{r}\times I\,d\mathbf{l}
=\frac{1}{2}\int_S\mathbf{r}\times \mathbf{K}\,dS
=\frac{1}{2}\int_V\mathbf{r}\times\mathbf{J}\,dV.
$$
If your loop is an infinite wire, the magnetic moment is infinite and its direction depends on where you place your origin. Both of these tell you that this is an incorrect description of your system.
For a simple solenoid, there is a simple rule to get the north and south poles, which is best explained graphically:
Magnetic field lines come out of the north pole, loop around, and go into the south pole (and you can see that there is no analogue of this for a single long wire!).
Also, as far as this is concerned, solenoids and permanent magnets are much the same. This is because the multiple current-carrying coils of the solenoid look very much like the surface spin currents on a permanent magnet, which are the ones that create its magnetic field. (Where they do differ is in the values of the $\mathbf{H}$ field inside the magnet.)
| {
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Does relativistic mass have weight? If an object was sliding on an infinitely long friction-less floor on Earth with relativistic speeds (ignoring air resistance), would it exert more vertical weight force on the floor than when it's at rest?
| "Does gravity depend on relativistic mass or rest mass?" is a rather interesting question -- Einstein's initial approach was to say "the relativistic mass", and this was the pre-general relativistic answer, but this is not satisfactory, since the relativistic mass is only one component of the energy-momentum vector (and it would actually sound more reasonable to say it depends on the rest mass).
This is the theoretical motivation for general relativity, in which gravity depends on the stress-energy-momentum tensor, and only the time-time component of gravity, $\Gamma^i{}_{00}$, depends on the relativistic mass, whose density is also the time-time component of the stress-energy tensor, and the other components depend on on the other components of the stress-energy tensor by the Einstein field equations.
| {
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Error in Sear's and Zemansky's University Physics with Modern Physics 13th Edition (Young and Freeman)? I was reading up on the Ideal Gas Equation in University Physics with Modern Physics by Young and Freeman when I chanced upon a seemingly illogical mathematical equation.
Can anyone rectify this error? Or is it misunderstanding on my part?
Here is the portion (Page 600, Chapter 18, Equation 18.12):
$$pV = \frac{1}{2}Nm(v^2)_{av} = \frac{1}{3}N\biggl[\frac{1}{2}m(v^2)_{av}\biggr]$$
It should be clear that $\frac{1}{3} \neq (\frac{1}{3} \times \frac{1}{2})$.
| Nice catch!
For reference here is the book page.
:
See , though it may error in printing or anything else.The final equation they get $$pV=\dfrac23K_{tr}$$ is very correct.
The correct form of $eq.(18.12)$ must be $$pV=\dfrac13Nm(v^2)_{av}=\dfrac
{\color{red}{\huge{2}}}3N\bigg[\dfrac12 m (v^2)_{av}\bigg]$$
| {
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$\langle B|A \rangle$ expressed in terms of the Partition Function Say you have an electron departing from point A and reaching poing B after a time t.
According to some helping friend, the Partition Function for that electron going from point A to B can be written as
$$Z = \int_{A \to B} [\mathcal{D}x]~ e^{iS[x]}$$
where $\mathcal{D}x$ is the measure that sums up over all paths going from $A$ to $B$, and $e^{iS[x]}$ is the weight of each path, $S[x]$ is the action.
That friend states then "From this partition function all desirable quantities can be obtained."
Not having much idea about Feynman's Path Integral Formulation of Quantum Mechanics, I have looked around a bit, and I would like someone to confirm the following statement I make:
The amplitude for the electron to go from A to B in the time t can be found in terms of that Partition Function $Z$ given above, as
$$\langle B|e^{-iHt}|A\rangle = Z$$
Did I catch it right?
| What your friend actually meant is that you can obtain all desirable correlation functions. Assuming you're talking about a non-relativistic electron, consider a source term added to your action
$$
S'[x] = S[x] + \int dt \, J(t) x(t)\,.
$$
Now you can write any correlation function as a derivative of $\ln Z$ calculated at $J = 0$, i.e.
$$
\langle B| \hat{x}(t') | A \rangle = \left.\frac{\partial}{\partial J(t')} \ln Z\right|_{J=0}\,,
$$
$$
\langle B|\hat{T}\left\{ \hat{x}(t') \hat{x}(t'') \right\}| A \rangle = \left.\frac{\partial^2}{\partial J(t')\partial J(t'')} \ln Z\right|_{J=0}\,,
$$
and so on, where $\hat{T}$ is the time ordering operator.
Hope this helps you.
| {
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Mercury's Orbital Precession in Special Relativity I am researching Mercury's orbital precession. I have considered most perturbations and general relativity. I am still not satisfied. I need your help.
I need a solution to Exercise 13, Chapter 6, in Ref. 1 (which is Exercise 26, Chapter 7, in both Ref. 2 and Ref. 3).
The exercise is copied below:
Show that the relativistic motion of a particle in an attractive inverse square law of force is a precessing motion. Compute the precession of the perihelion of Mercury resulting from this effect. (The answer, about 7" per century, is much smaller than the actual precession of 43" per century which can be accounted for correctly only by general relativity.)
I have the solution to Exercise 7, Chapter 3.
References:
*
*H. Goldstein, Classical Mechanics, 1st edition, 1959.
*H. Goldstein, Classical Mechanics, 2nd edition, 1980.
*H. Goldstein, Classical Mechanics, 3rd edition, 2000.
| I think this is the Thomas precession, which is a kinematical effect that depends on the shape of the worldline and is independent of the nature of the force.
Wikipedia gives a low-speed approximation for the Thomas precession of $ω_T=av/2c^2$. For a circular orbit with radius $r$ and speed $v$, the precession per orbit is $$Δθ = (2πr/v)ω_T = πra/c^2 = πv^2/c^2$$
which agrees with the low-speed, low-eccentricity formula in Fausto Vezzaro's answer (using $v^2/r=GM/r^2$).
This preprint gets a Thomas precession of 7.163″/hyr from a more careful calculation that takes the eccentricity into account. It also says that this is a problem for general relativity, which isn't true (calculations in GR automatically include SR "effects"), but I suppose the special-relativistic calculation is correct in spite of that.
This preprint, which was mentioned in a comment by Pulsar, derives a similar result with no mention of Thomas precession, and then a result twice as large (14.3″/hyr, one third of the GR prediction) from what is allegedly a more careful treatment.
| {
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What's the physical significance of using fourier transform for diffraction? I am studying some basic idea of diffraction and there mention in far field, the diffraction pattern could be understood by Fourier transform. But I just don't understand what's the physical fact for that. And why cannot use Fourier transform for the the near-field case?
Also, when I am trying to understand the theory of diffraction, it ends up with some complicate math (integrals). I want to learn that but the books I am reading are not easy to understand. Anyone recommends some good books or video lectures (more theoretical but to explain most of the math in plain way)?
| To give an unmathematical catchy answer, let's look at Fraunhofer diffraction in double slit experiment.
Interference at the observation plane depends on slit parameter $d$. What is the frequency of slits? E.g. $1\,\text{mm}\frac{1}{d}$: number of slits per length. Concluding frequiency in the setup. The following argumentation links this frequency to the fourier transform. The physical significance is in the real optics setup. The setup is easier described, when transformed in fourier space.
The double slit link on high school level above gives all the math without integrals. After visualizing the following concept, you see that the integrals are just math to convert the diffraction pattern in fourier space via fourier transformation.
Using trigonometry first compute phase difference $\Delta\phi(\theta,d)$.
Go deeper in this concept using a sketch to visualize phase difference of $n\cdot\lambda$, $n\in\mathbf{N}$ as bright maxima in diffraction pattern. There is no magic in the next step. It's just a another point of view: Try to grasp $\frac{1}{d}$ as a parameter on its own: $\Delta\phi(\theta,\frac{1}{d})$.
Fourier space is a synonym for frequency domain. Acoustics examples are given in Eichenlaubs SE answer and ptomatoes optics explanation.
No literature: Calculate and understand yourself with the links above.
| {
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Physical interpretation of Poisson bracket properties In classical Hamiltonian mechanics evolution of any observable (scalar function on a manifold in hand) is given as
$$\frac{dA}{dt} = \{A,H\}+\frac{\partial A}{\partial t}$$
So Poisson bracket is a binary, skew-symmetric operation
$$\{f,g\} = - \{f,g\}$$
which is bilinear
$$\{\alpha f+ \beta g,h\} = \alpha \{f, g\}+ \beta \{g,h\}$$
satisfies Leibniz rule:
$$\{fg,h\} = f\{g,h\} + g\{f,h\}$$
and Jacobi identity:
$$\{f,\{g,h\}\} + \{g,\{h,f\}\} + \{h,\{f,g\}\} = 0$$
How to physically interpret these properties in classical mechanics? What physical characteristic each of them is connected to? For example I suppose the anticommutativity has to do something with energy conservation since because of it $\{H,H\} = 0$.
| If we consider for simplicity a 2d phase space (q,p), then we can interpretate the poisson bracket between two functions f(q,p) and g(q,p) as the vector product of their gradients, which are vector fields in this plane:
$[f,g]=(\nabla f\times \nabla g)\cdot \mathbf{e}_z$
where $e_z$ is a unit vector perpendicular to the plane.
From that definition all the properties are obvious.
We can imagine the following physical analogy for the equation of motion, the gradient of the hamiltonian act like magnetic field $B$ and the gradient of the function is the velocity $v$, in formulas:
$\partial_tf\, \mathbf{e}_z= \nabla f\times \nabla H = \mathbf{v}
\times \mathbf{B}$
which is the expression of the Lorentz force.
| {
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Bleaching groundstate I'm reading an article about two-dimensional infrared spectroscopy and I don't understand the following sentence.
Bleach or stimulated emission contributions yield negative signals.
What are "bleach contributions"? I have never heard of it and cannot find a suitable explanation on the Internet. Can somebody explain that to me?
The article I'm referring to is J. Chem. Phys. 121, 5935 (2004)
| I feel obliged to post another answer since the first one contains a mistake.
As you deplete the ground state, you also populate the excited state, and when the photon interacts with the excited molecule, then instead of being absorbed it generates another photon through stimulated emission. The paper you cite treats this contribution as a negative signal.
Eventually, at 50/50 population you reach an equilibrium, where the amount of absorbed light is equal to the amount of light produced by stimulated emission, and it looks like your medium is not absorbing light at all, which is why it is called bleaching (i.e. loosing its color).
| {
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What is our estimated running speed on Moon's surface? I was wondering if we have the chance to run on the Moon's surface, how would you expect it look like?
I expect our velocity will increase for the same work we do on Earth, but not sure if this will be multiples in term of gravity variations.
How do you think our maximum speed would reach?
| Maximum speed can be ~ (because friction etc. are slowing you down and the moon isn't a perfect sphere)
$\frac{v^2}{r} = \frac{GM}{r^2}$ $-$ $(1)$
where $r$ is radius of moon & $M$ is mass of the moon.
EDIT : Since then in this case your normal reaction is $0$ , means you are almost flying , speed up a little bit more and you will fly off the moon along a tangent line onto a bigger radius and now you are flying actually along the moon like a satellite , and not running .
if you are running on the surface of moon provided you have that kind of energy in you . Gravity won't effect directly as gravity is already perpendicular to your direction of motion always ,but friction will be less as friction is $\mu_k N$ and N will be less since gravity force is less , otherwise no effect , I am assuming you aren't trying to jump , if you try to jump then your projectile range will be more than on earth , as gravity is less , you'll be having more time in air and as there's no atmosphere , no viscous force by air .
So in net you'll reach a destination in fewer steps and quickly because dissipative forces are less and will require less energy consumption to reach a particular destination .
But a slight decrement will be caused in your projectile range since Buoyant forces will also not be present because of almost $0$ atmosphere .
However , beware of all the craters :)
EDIT : the formula $(1)$ comes from the fact that if you are running on the surface means your distance from centre isn't varying that much . And the moon is pulling you towards the centre providing you the necessary centripetal acceleration to allow you to turn along the moon surface .
| {
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Why does increasing the temperature of a thermistor decrease it's resistance? Surely, upon an increase in temperature, the atoms within the thermistor would vibrate with more energy and therefore more vigorously, hence making the electrons flowing through the electric circuit more likely to collide with one of the atoms, so increasing resistance.
However, the effect of temperature on a thermistor is contrary to this. I can't understand how it can be.
It's analogous to running across a playground: if everyone is still you're less likely to collide with someone, however if everyone is constantly moving from left to right then a collision is more likely.
So why does an increase in temperature decrease the resistance of a thermistor?
| Thermistor with this particular temperature behavior are commonly semiconductors. In a semi-conductor, there is an energy gap between the (filled) valence and the (empty) conduction band. At zero temperature, no charges are in the conduction band and the resistance should be infinite as the system behaves basically like an insulator.
If you turn on the temperature, some electrons will start to occupy the conduction band and thus contribute to conduction, lowering the resistivity.
| {
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Does inertia increase with speed? I have heard that when the speed of the object increase, the mass of the object also increase. (Why does an object with higher speed gain more (relativistic) mass?)
So inertia which is related to mass, increase with speed?
So, if I accelerate on a bus, my mass will increase and my inertia will increase for a while on the bus, until the bus stops?
| I think It has more to do with acceleration than speed. What do you compare a constant speed to without knowing the inertial frame for the universe. For instance if the earth is traveling toward the constellation Leo at 390 km/s what would happen if you blasted off in a rocket in the opposite? After accelerating would you be moving 390 km/s or would you be sitting still compared to what you were doing?
| {
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What is the derivation for the exponential energy relation and where does it apply? Very often when people state a relaxation time $\tau_\text{kin-kin}, \tau_\text{rot-kin}$,, etc. they think of a context where the energy relaxation goes as $\propto\text e^{-t/\tau}$. Related is an approach to compute it via
$$\tau=E(0)/\left(\tfrac{\text d E}{\text d t}\right)_{E=0}.$$
Both are justified from considering dynamics for which
$$\frac{\text d E}{\text d t}=-\frac{1}{\tau}(E-E(0)).$$
My question is: What fundamentally leads to this relation?
I conjecture it relates to a Master equation, which mirrors the form "$\dot x=Ax+b$". But I'm not sure how the degrees of freedom in the Master equation translate to the time dependence of the macroscopic energy value. There will also be a derivation from the Boltzmann equation somehow, for some conditions, but what is the general argument and where does it work?
| This kind of exponential decay toward "equilibrium" can be derived when one looks at a Markov process.
In this case, if we call $S_t$ the state of the system at time $t$ and $S_{t+1}$ the state at time $t+1$, one has for the evolution:
$S_{t+1} = T S_t $
where $T$ is called the transition matrix. This implies that $S_t = T^t S_0$. The idea is then to introduce the set of eigenstates $E_i$ such that $T E_i = \lambda_i E_i$. The set of $\{E_i\}_{i=1...}$ is a mathematical set of vectors and they need not always corresponds to a probability state. In fact, since the solution is unique for any given $S_0$ it implies that there can only be one probability state such that $T E_p = E_p$ i.e. such that $\lambda_p = 1$. Now, starting from any state $S_0 = \sum_i S_i^0 E_i$, one has then $S_t = \sum_i S_i^0 \lambda_i^t E_i = S_{eq} + \sum_{i \neq p} S_i^0 \lambda_i^t E_i$. $T$ is a positive definite matrix and in spectral theory, one can show that $\lambda_p = 1$ is the heighest eigenvalue, it therefore means all other eigenvalues are smaller than $1$. Let us call $\lambda_2$ the second highest eigenvalue of $T$, we then have:
$S_t-S_{eq} \approx S_2^0 \lambda_2^t \approx S_2^0 e^{t \ln \lambda_2} \sim e^{-t/\tau_2}$
where $\tau_2 = -1/\ln \lambda_2$. $\hspace{0cm}$
At the end of the day the idea is that the initial state can be projected on eigenstates among which only one is physical and happens to have the highest eignevalue of value 1, this one corresponds to the equilibrium state.
The major assumption here is that the dynamics is markovian.
| {
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Comparing Static Frictions In this figure, which of the static frictional forces will be more?
My aim isn't to solve this particular problem but to learn how is static friction distributed . Since each of the rough-surfaces are perfectly capable of providing the $-1N$ horizontal frictional force but why don't they ? This is kind of ambiguity that who will provide a bigger share in total static friction. And as the surface have different $\mu$, so we can't even invoke symmetry.
| The ground will provide all of the static friction. Imagine what would happen if the upper block contributed even a tiny amount to the static friction: It would have to move forward due to the reaction force. Having M2 inch along you pull M1 (which stays stationary) would be very strange indeed.
Static friction always acts to prevent relative motion. It acts from the ground because the ground is the first place where relative motion would kick in if you accelerated it (friction from the top would be able to prevent at least some of the relative motion)
| {
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Effective mass in Spring-with-mass/mass system Suppose you have a particle of mass $m$ fixed to a spring of mass $m_0$ that, in turn, is fixed to some wall. I'm trying to calculate the effective mass $m'$ that appears in the law of motion of the particle (suppose the system is isolated):$$m'\ddot x=-k(x-x_0).$$
I've read somewhere that this should be $m'=m+m_0/3$, but I'm getting a different result.
My reasoning is as follows. Suppose the particle is at position $x$. The lenght of the spring is $x$ and we can suppose that its center of mass is at $x/2$. So the spring/particle center of mass is at: $$X= \dfrac {\frac{m_0}{2} + m}{m+m_0}x$$
Differentiating two times we get $$\ddot X = \dfrac {\frac{m_0}{2} + m}{m+m_0}\ddot x$$
Now, the only external force causing acceleration to the center of mass is the ceiling reaction to elastic force, that is exactly $-k(x-x_0)$. Thus:$$-k(x-x_0)=(m+m_0)\ddot X=(\frac{m_0}{2} + m)\ddot x $$and so I'm getting:$$m'=\frac{m_0}{2} + m.$$
Could you please point out where am I wrong (if I am) and possibly how is the result demonstrated?
| You can understand in a simple way the factor $1/3$ which gives you the approximate solution in the low frequency regime (more on this later) in the following way. Start by writing the kinetic energy of your system as:
$$K = \frac{1}{2} m \dot{\delta}(\ell)^2 + \int_0^\ell \frac{1}{2} \lambda \dot{\delta}(u)^2 du$$
where $\delta(u)$ is the displacement of the point of the spring which is in the $x=u$ position in the equilibrium configuration and $\lambda=m_0/\ell$ the linear mass density of the spring. The spring has length $\ell$ when unstretched.
If you suppose an harmonic motion for the mass at a very low frequency, the stretching of the spring will be approximately uniform, which means
$$\delta(u) = \frac{u}{\ell} \delta(\ell)$$
Accepting this approximation by substituting in the expression for kinetic energy one gets
$$K = \frac{1}{2} m \dot{\delta}(\ell)^2 + \frac{1}{2} \frac{m_0}{\ell} \dot{\delta}(\ell)^2 \int_0^\ell \frac{u^2}{\ell^2} du$$
and after an integration
$$K = \frac{1}{2} \left( m +\frac{1}{3} m_0 \right) \dot{\delta}(\ell)^2$$
which is the expected result.
The system has an infinite number of degrees of freedom, which means that it will have an infinite number of oscillation modes. If $m\gg m_0$ the lowest frequency mode will be approximately described as an oscillation of the mass with an uniform stretch of the spring. In the higher frequency modes the mass will be nearly fixed, and there will be a nearly stationary elastic wave on the spring.
The flaw in your reasoning consist in supposing that the external force applied to the system mass+spring is $-k(x-x_0)$. The applied force is really the tension of the spring at his fixed point, which is not $-k(x-x_0)$ for a spring with mass when there are accelerations.
| {
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How can a car's engine move the car? Newton's First Law of Motion states that an object at rest or uniform motion tends to stay in that state of motion unless an unbalanced, external force acts on it.
Say if I were in a car and I push it from the inside. It won't move. So how is the engine of a car capable of moving the car?
| Each force causes reaction (3rd law). If move a car from the inside the car moves you as well. That's because you are pushing or pulling. However the engine does not push but converts energy in other directions, usually a rotating one (the same as riding a bicycle). This rotating force has its counter-force which is reaction of ground.
Pushing a car you use a friction force. Pushing from the inside this force is caused bythe object itself, not solid ground.
| {
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Quantum Mechanical Operators in the argument of an exponential In Quantum Optics and Quantum Mechanics, the time evolution operator
$$U(t,t_i) = \exp\left[\frac{-i}{\hbar}H(t-t_i)\right]$$
is used quite a lot.
Suppose $t_i =0$ for simplicity, and say the eigenvalue and eigenvectors of the hamiltionian are $\lambda_i, \left|\lambda_i\right>$.
Now, nearly every book i have read and in my lecture courses the following result is given with very little or no explanation:
$$U(t,0) = \sum\limits_i \exp\left[-\frac{i}{\hbar}\lambda_it\right]\left|\lambda_i\right>\left<\lambda_i\right|$$
This is quite a logical jump and I can't see where it comes from, could anyone enlighten me?
| Without loss of generality, let's take the $|\lambda_i\rangle$ to be orthonormal. Notice that, by the spectral theorem, the hamiltonian can be written as follows:
$$
H = \sum_i \lambda_i P_i, \qquad P_i = |\lambda_i\rangle\langle \lambda_i|
$$
Each operator $P_i$ is a projectors onto the subspace spanned by $|\lambda_i\rangle$. Notice, in particular, that
$$
P_i^2 = P_i, \qquad P_iP_j = P_jP_i = 0
$$
and a mathematical induction argument gives
$$
P_i^n = P_i
$$
for all $n\geq 1$. Now, for notational simplicity let
$$
\mu = -\frac{i}{\hbar}t
$$
Then we have
$$
U(t,0) = e^{\mu H} = \sum_{n=0}^\infty \frac{1}{n!}\mu^nH^n
$$
but notice that using the properties of projection operators written above, we have
$$
H^n = \sum_{i_1, \dots, i_n}\lambda_{i_1}\cdots\lambda_{i_n}P_{i_1}\cdots P_{i_n} = \sum_i\lambda_i^nP_i
$$
and therefore
$$
U(t,0) = \sum_i\sum_n\frac{1}{n!}(\mu\lambda_i)^nP_i = \sum_ie^{\mu\lambda_i}P_i
$$
as desired.
| {
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Why the magnetic flux is not zero? If $\vec{\mathbf B}=B\vec{\mathbf a}_z$, compute the magnetic flux passing through a hemisphere of radius $R$ centered at the origin and bounded by the plane $z=0$.
Solution
The hemisphere and the circular disc of radius $R$ form a closed surface, as illustrated in the figure; therefore, the flux passing through the hemisphere must be exactly equal to the flux passing through the disc.
The flux passing through the disc is
$$\Phi=\int_S\vec{\mathbf B}\cdot\mathrm d\vec{\mathbf s}=
\int\limits_0^R\int\limits_0^{2\pi}B\rho\,\mathrm d\rho\,\mathrm d\phi
=\pi R^2B$$
The reader is encouraged to verify this result by integrating over the surface of the hemisphere.
According to Maxwell's equations the magnetic flux over a closed surface must be zero, why in this case does not happen?
| The flux through the closed hemisphere is zero,
$$\Phi_{\mathrm{hemi}}+\Phi_{\mathrm{disk}} = 0.$$
This allows us to find the flux through the hemisphere knowing the (more easily calculable) flux through the disk,
$$\Phi_{\mathrm{hemi}} = -\Phi_{\mathrm{disk}}.$$
| {
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Induced current using a reference system bound with a moving charge Suppose we have a charge moving at velocity $\mathbf{v}$ in the same plane of a square wire.
If I sit in a reference frame where the square wire is still, since the charge is moving with velocity $\textbf{v}$ in this coordinate system, I will see an induced current in the wire.
$$\textbf{B} = \frac{\textbf{v}}{c^2} \times \textbf{E} $$
$$ \frac{d\phi_B}{dt} \neq 0 $$
Now, what If I choose a reference system where the charged particle is at its origin?
According to this frame, since the the electric charge(and its electric field) is static, $\text{rot}\,\textbf{E}$ will be zero.
$$\nabla \times\mathbf{E} = 0$$
But this means that there is no induced current.
Are my assumptions right? If not, how should I estimate the induced current in a reference system bound with a moving charge at its origin?
| There's no contradiction (there never is in relativity problems...): a transient current flows in the square as it passes by the charge.
You're correct that the conservative electric field in the charge's rest frame means that the line integral around a square in that frame is zero, but that's not the appropriate integral to use for a moving wire.
The induced current in the square is determined by the line integral of E around the square at a fixed time in the square's rest frame (the "primed" frame). To calculate this quantity in the charge's rest frame (the "unprimed" frame) the integrand $\boldsymbol{E' \cdot} d\boldsymbol{l'}$ must be transformed:
*
*the electro-magnetic field transforms to the static coulomb field of the charge.
*the path infinitesimals transform according to $dx = \gamma \, dx'$ and $dy = dy'. (t'$ is fixed.)
It's that $\gamma$ that changes the character of the integral to be calculated and gives a non-zero result. This effect is a typically small relativistic correction, but even a small correction matters when the non-relativistic result is 0.
| {
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Convergence of Light on the Retina So, I've learned about lens ray diagrams-but the problem I'm having is that when ray diagrams are drawn for a point of an image, they converge to another point, but there are two problems that I see with regards to our eyes:
1) In the simplest drawn case it's at least three rays converging to the same point, but our eye would only have once receptive cell at a point, how does it take in all of them?
2) At any given time, isn't there more than one point being sent to the same point on the back on the retina? We draw 2D cases in high school physics, but I feel like there should be more than one point that gets sent to the same point. In any case, part one is the more pressing matter.
| You are a bit confused.
Rays are a geometrical representation of the classical electromagnetic waves and are very good for optics situations.
The cells in the retina are in the quantum mechanical regime, because to react to light they absorb photons at specific molecular energy levels of the molecules in the cells which send signals of detection to the brain .
Photons are individual particles which in their great plethora build up the classical electromagnetic light. They are discrete in space and time, and the cells in the retina have many molecules, of order 10^18 or so ( there are 10^23 molecules in a mole of matter) so a great number of photons can excite a great number of molecules in your optically drawn "ray".
Do not confuse geometric arguments with physics arguments.
| {
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How does relativity explain gravity, without assuming gravity I have seen the "objects pull down on space-time" explanations, but they assume a "pull down" force themselves. Could anyone explain the space-time explanation without assuming gravity in the first place?
| Massive objects distort spacetime, as described by the Einstein Field Equations. In turn, this causes particles to accelerate: the GR equivalent of $\mathbf{F}=m\mathbf{a}$ are the geodesic equations:
$$
\frac{\text{d}^2x^\alpha}{\text{d}\lambda^2} + \Gamma^{\alpha}_{\mu\nu}\frac{\text{d}x^\mu}{\text{d}\lambda}\frac{\text{d}x^\nu}{\text{d}\lambda} = 0,\qquad \alpha=0,\ldots 4,
$$
with
$$
\Gamma^{\alpha}_{\mu\nu} = \frac{1}{2}g^{\alpha\beta}
\left(\frac{\partial g_{\beta\mu}}{\partial x^\nu} +
\frac{\partial g_{\beta\nu}}{\partial x^\mu} -
\frac{\partial g_{\mu\nu}}{\partial x^\beta}\right),
$$
the so-called Christoffel symbols, and $g_{\mu\nu}(x^\alpha)$ the spacetime metric. In the absence of matter, $g_{\mu\nu}(x^\alpha)$ is constant, so that the geodesic equations reduce to
$$
\frac{\text{d}^2x^\alpha}{\text{d}\lambda^2} = 0,
$$
which describe a constant motion, as expected.
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Stable Nuclei - Deviation from equal protons and neutrons While studying the semi-empirical mass formula for nuclei, I came across an "asymmetry term" whose function, as far as I understand, is to build in the fact that nuclei "prefer" to have equal numbers of protons and neutrons. This is explained by the Pauli exclusion principle; the neutrons and protons are distinguishable, and hence have fill separate energy levels. Hence, if the number of neutrons and protons is close, the nuclei will have a smaller energy.
Yet, it is observed that large stable nuclei have more neutrons than protons. Why is this so? If nuclei prefer to have equal protons and neutrons, shouldn't the stable nuclei lie along the N=Z line? (in the image)
| Neutrons are subject to the strong force, holding the nucleus together, but not to the electric force, which pushes only the protons apart. In that sense, it's more stable for a nucleus to have a few spare neutrons to space the protons apart and keep them stuck together.
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How can the big bang occur mathematically? As we know time began with the big bang. Before that there was no time, no laws, nothing. Mathematically how can an event take place when no time passes by? How did the big bang took place when there was no time?
Note my question is not about weather big bang took place or not, my question is about is this a mathematical anomaly? Thanks
| You asked for a purely mathematical answer and received a correct physical answer from John Rennie. If you want a purely mathematical answer, that purely mathematical answer has to exist within a well-defined mathematical theory. As John has explained, we don't have a well-defined mathematical theory of the Big Bang that is also an acceptable, complete physical theory. We do, however, have one pretty serviceable, well-defined mathematical theory of the Big Bang, which is the one provided by general relativity (GR).
As we know time began with the big bang. Before that there was no time, no laws, nothing. Mathematically how can an event take place when no time passes by? How did the big bang took place when there was no time?
In the context of GR, the answer to this is extremely simple. An event in GR is a point in the manifold that represents spacetime. The Big Bang is a singularity, and singularities are not points on the manifold. Therefore the Big Bang is not defined mathematically as an event (or set of events).
| {
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Why does sound move faster in solids? I know that the molecules are closer together in solids, and I know thicker springs also respond carry waves faster than thinner springs, but for some reasons I can't understand why.
The molecules will have a larger distance to move before colliding with another molecule, but in a thicker medium wouldn't that time just be spent relaying the message between multiple atoms? Why is the relaying between a lot of tight knit atoms faster than one molecule moving a farther distance and colliding with another?
| Think of it this way. Elasticity is a property of material that allows it to store energy and release it without dissipating. Solids have high elasticity, therefore, they can store and release energy quite efficiently. Liquids and gases have low elasticity. They are also viscous and dissipate energy instead of transmitting it. Please note that I am not connecting speed of waves to viscosity but merely pointing out its dissipative nature.
| {
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Where does the extra equation come from to determine the forces from an object on a table? I have a question about basic statics but somehow I cannot manage to find the answer on my own (btw, this is not a homework. It's been so many years since school for me...).
The problem is very simple: we have an object with weight $D$ at a given location on table wit with four legs ($F_1$ to $F_4$). What is the force applied on each leg? (for simplicity, I'm just using the same labels $F$ and $D$ for both the location and the force)
$W$, $H$, $x$, $y$ and $D$ are given. To find the forces on each leg, as far as I remember, I have to consider two general equations:
$\sum F=0$ and $\sum M=0$. So I have:
$$
F_1 + F_2 + F_3 + F_4 - D = 0
$$
Also, considering the moments round the point $F_1$:
$$
W(F_2+F_3) - xD = 0
$$
$$
H(F_3+F_4) - yD = 0
$$
But this just give me 3 equations! I missing one more equation and cannot figure it out.
| As you have noticed yourself, your system is simply underdetermined. In order to find a unique solution you need to add some extra constraints in addition to Newton's equations. Imagine a table with more than four legs: the more legs you add, the more unknown forces you have. But the number of equations does not change. If we instead remove a leg we find a unique static solution.
See also the Wikipedia page about statically indeterminate systems.
| {
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Tensor equations in General Relativity In the context of general relativity it is often stated that one of the main purposes of tensors is that of making equations frame-independent.
Question: why is this true?
I'm looking for a mathematical argument/proof about this fact.
| What we want of a law of nature is that is has the same form for every equivalent observer.
Therefore, these laws should be construct with geometrical objects which transform into themselves up to multiplicative factors. This is also known as an homogeneous transformation under certain group (typically Lorentz or diffeomorphism).
The geometrical object which satisfy this homogeneous transformation rule are tensors (there are also spinors). Thus physical theories are described (so far) successfully by these objects (or fields).
| {
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Why are the magnetic moment and the angular moment related? Why are the magnetic moment and the angular moment related? I've always read everywhere that they are related but found nowhere a satisfactory explanation of the cause
| The way I understand it, a charged particle, which has an angular momentum, will have a magnetic moment associated with it. This could be because an angular momentum is associated with some kind of rotation. For a charged particle, this rotation could be thought to constitute a current loop. And a current loop can always be associated with a magnetic moment at its centre. Thus charge + rotation (angular momentum) --> magnetic moment.
The moment comes out to be proportional to the angular momentum.
For the quantum case, the angular momentum is probably linked to some rotation we don't understand yet, which could be giving rise to the magnetic moment...
In some sense, this is just a longer version of Amey Joshi's answer, but hope it helps. (Though I AM a year late! :P)
| {
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Solving a light ray worldline with the geodesic equation I'm having trouble solving the geodesic equation for a light ray.
$$ {d^2 x^\mu \over d\tau^2} + \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} = 0 $$
I apologise, but I'm a bit new to this, but I have the initial $x^\mu$ and initial $dx^\mu\over d\tau$. I'm just not sure how to use them to solve the equation for $x^\mu$.
I would logically start with
$$ {dx^\mu \over d\tau}_{initial} = v^\mu $$
and suppose that the initial acceleration would be
$$ {d^2 x^\mu \over d\tau^2}_{initial} = - \Gamma^\mu_{\alpha\beta} v^\alpha v^\beta $$
But that doesn't really help me integrate it, since I've only got constants for the initial condition. How would I solve this for $x^\mu(\tau)$?
Furthermore, I feel that this equation may not apply to light rays, as their proper time ought to be $0$, right?
| $\tau$ can be seen as proper time scaled by an arbitrary factor. For the limit of approaching a null geodesic, yes $\tau$ as proper time goes to zero, but use an arbitrary factor approaching infinity, and the combination can be kept finite. Or ignore all that; $\tau$ can be any arbitrary parameterization along the geodesic.
I notice someone has already asked about this: What is the physical meaning of the affine parameter for null geodesic?
| {
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"S-duality" between confinement and the Higgs mechanism? I feel picked by the second to last sentence in this answer to a question about what would happen if EM and QCD were spontaneously broken, which says
"In fact, there is a sense in theoretical physics in which confinement is complementary to Higgsing – it's the same thing in different ("opposite", "S-dual") variables"
to explain further why a broken $SU(3)$ gauge group could not be confining.
I'd like to see a further explanation of this.
For example which of the weak/strong regime would correspond to the higgsing/confinement regime? Is it as I naively expect from the cited comment, the strong coupling constant that is used to establish the S-duality, or does it work via a more unified higher energy scale coupling constant? Heck, I'd just like to know in some more detail than just this comment, how it works.
| Experimentally there are the running coupling constants, which are being used in models like GUTS etc which aim to eventually unify all four forces, strong, weak, electromagnetic, and hopefully gravity.
Note the LEP line on the left, that is where experiments are up to now. I do not expect the LHC to improve on this, because of the complexity of the studied interactions .
What is the relevance to the question? It shows that the behavior of strong interactions has an opposite trend from the electroweak tendency to unification, and at best , the three may meet at a point. The unification is proposed with various models to be found in the literature and the breaking of the symmetry will depend on the model. As only the electroweak part is considered standard, and the models make sure that the breaking reproduces the standard model symmetries. Example here. As experimental verification seems very impossible at the moment it is not the symmetry breaking mechanism that will decide among models.
| {
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How is it that the Earth's atmosphere is not “blown away”? The Earth moves at a high rate of speed around the Sun, and the solar system is moving quickly around the Milky Way. How is it that the Earth's atmosphere is not “blown away”?
| I dont think @Brendon is right. Best exemple of an object moving trougth space and being blown away is a comet (blown away part is the tail).
Best answer I think si that the Earth's atmosphere IS blown away but to a lesser level due to its gravitationnal force (Keeping the Earth in one pieces).
| {
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Quantum Mechanics, Uncertainty Principle-- help understanding notes There is a section of my notes which I do not understand, hopefully someone here will be able to explain this to me. The notes read (after introducing the uncertainty operator):
If the state $\chi_A$ is an eigenstate of $\hat O_A$ then the uncertainty is zero and we measure it with probability 1. However, if $\hat O_B$ is another observable which does not commute with $\hat O_A$, then the uncertainty in any simultaneous measurement of the two observables will be infinite.
I understand the first sentence, but I can't see how to justify/prove the second one. Can someone tell me how the second sentence is justified, please?
| You can justify thus:
$\psi=\ Ae^{i\frac{(px-E t)}{h}}$ is an eigenstate of a free particle. The momentum $\ p$ is well defined and its in the eignestate of momentum operator (as $O_p\psi=\ -ih\frac{d\psi}{dt}=p\psi$ ) This means the probability of finding the particle with momentum $\ p$ is $\ 1$.
Operators $\ x$ and $\ p$ don't commute. Now look at $\ x$ .The particle has equal probability of being found anywhere on the x axis since $\psi^*\psi=constant$, all over the axis. This means you have no idea where the particle is. Uncertainty is as high as it could be.
Therefore uncertainty in position tends to infinity and in momentum it is zero for the given case.
| {
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Does potential energy in gravitationall field increase mass? I was just taught (comments) that any type of energy contributes to mass of the object. This must indeed include potential energy in gravitational field. But here, things cease to make sense, have a look:
*
*I have object at some distance $r$ from radial source of gravitational field.
*The potentional energy is calculated like this: $E_p = m*a*r$ where $a$ is gravitational acceleration and $m$ is mass of your object.
*But that means, that the object is a bit heavier - because of the potentional energy of itself - $m = m_0 + \frac{E_p}{c^2}$ ($m_0$ here is the mass without the potentional energy)
*That would mean that the gravitational force is a bit stronger at higher distances.
Now, I do understand that the rules of physics are not recursive and the mass and force will be finite. But what is the correct approach to this situation? What is the correct equation for potentional energy?
| Yes. All energy contributes to the overall mass. PBS Space Time has a really clear video on this subject: The Real Meaning of E=mc² In the video, Gabe points out that Einstein originally wrote the equation as m = E/c², and describes how all energy has the property of mass, including potential energy.
Edit to add: it's important to note (as is mentioned in the video) the potential energy of a particle within a system does not change the mass of the particle itself. Instead, it contributes to the mass of the system within which the particle has potential energy. Also note that a particle considered in isolation has no potential energy.
| {
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What happens when a photon hits a beamsplitter? Yesterday I read that we can affect the path and the 'form' (particle or wave) of a photon after the fact (Wheeler's delayed choice experiment). Part of what is puzzling me is the beam-splitter. Are the individual photons actually being split into two new photons of lesser energy?
This question implies that you cannot split a photon but it seems that beam splitters do exactly that.
| A single photon is a quantised packet of Electromagnetic energy, the smallest indivisible unit imposed by boundary conditions according to quantum mechanics. In this regime I find it easier to think of the photon as a particle with a 'polarization' degree of freedom which can be horizontal $\left|H\right>$,vertical $\left|V\right>$ or in any linear superposition of the two.
When the photon meets a beam splitter it acts like a quantum particle and takes both paths with some probabilities. Much like the electron taking both paths in Young's famous double slit experiment.
It is not the actual photon being split into two new ones, only the 'particle-like' position wavefunction holds the information about where the photon is. Quantum mechanically, we can treat photons exactly like any other bosonic particle.
| {
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Solving equation of motion of two massive particle exerting the gravitational force each other I'm trying to analyze the motion of the particles which exert the gravitational force each other. Let $M_1$, $M_2$ be the masses of the particles, and the equation of motion of particle $M_1$
$$
F=G\frac{M_1M_2}{r^2}=M_1\ddot{r} \\
\ddot{r}r^2-GM_2=0
$$
I find it doubtful that the simple motion has difficult equation of motion.
Should I just use some form of successive approximation?
| I suggest to see the scenario physically first, and proceed from there.
Here, the gravitational force does not change the distance from the center, but it provides the centripetal force for the circular motion (or any conic section for that matter, based on initial values).
Assuming it is a circular motion for now, we can just get right into the mathematical description of the motion.
For such systems, masses move in a circle with their COM at the center.
Because there is no other force in system, the COM does not accelerate. It is free to move at a constant velocity though, but this is (usually) irrelevant when analyzing the system.
Because they move in a circle about the COM, just get the distances from their COM, and equate the gravitational force to the centripetal force. This way you have fully described the motion.
Example for $M_2:$
$$
\dfrac{GM_1}{\underbrace{R}_{\text{distance between bodies}}} = \dfrac{{v_2^2}}{\underbrace{R_{2_{cm}}}_{\text{distance of 2 from COM}}}
$$
Similar for $M_1$.
| {
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