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Separation between slip rings of DC generator
This is a DC generator which converts mechanical energy to Direct Current. The commutators in a DC generator are separated (as you can see in the image). It is explained in our book that it prevents the change in direction of current. But I don't understand.
Even if we don't keep them separated, how will the current change direction?
In short, what is the function of the gap between the two split rings?
| The direction of the current with respect to the magnets stays the same, but not with respect to the wire. Imagine that the positive terminus is on the left and the negative terminus on the right. The current starts out going $a\rightarrow b\rightarrow c\rightarrow d$, which is clockwise with respect to the magnets, as shown in the illustration. After the commutator turns 180 degrees, the positions of $b$ and $c$ are flipped, as are $a$ and $d$. If current still flowed from $a\rightarrow b\rightarrow c\rightarrow d$, the current would now flow counter-clockwise with respect to the magnets, which would reverse the torque. The commutator causes the current to flip, so that it now flows from $d\rightarrow c\rightarrow b\rightarrow a$, so that the current loop remains clockwise even when the wire loop has turned over.
(If there is no gap, the circuit will short, and no current will flow through the loop.)
| {
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Can entropy be equal to zero? I've searched for it but I only found contradicting answers from "scientists":
Dr. David Balson, Ph.D. states: "entropy in a system can never be equal to zero".
Sam Bowen does not refutes the following affirmation: "It is know[n] that entropy is zero when a pure crystalline substance is at absolute zero". He says there's residual motion in particles, but he doesn't say clearly that 0 entropy isn't possible.
I this link is a long discussion about that, it isn't really conclusive, but most people seem to agree that entropy can be zero.
I'm not very receptive to the idea of zero entropy.
| If we use the definition of entropy $$ S=-k_B \sum_i P_i ln P_i$$ where $P_i$ is the probability of the i'th microstate, then at 0K, we know the system is certainly in the ground state, $P_0=1$ so the definition returns zero.
The only "residual motion" in the system at this state is that due to the uncertainty principle in the ground state, but this does not change the fact that we know for certain which state the system is in, and therefore the entropy is zero.
| {
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What is the current radius of cosmological event horizon? Doing some crude calculations (using the value of $H_0$ at this point of time only, since it is time dependent but not distance dependent thanks to Johannes answer) what is the radius of cosmological event horizon at this point of time? (not looking for the changes of CEH through time)
From here we have for $H_0 $:
$$H_0 = 73.8 \pm 2.4 (\frac{km}{s})\frac{1}{Mpc}\tag{I}$$
We are seeking the distance $L$ s.t. $H_0L = c = 3\times 10^6 \frac{km}{s}$
$$L=\frac {c}{H_0} = \frac {3\times 10^6}{73.8 \pm 2.4} Mpc \tag{II}$$
Where 1 pc = 3.26 light years ($ly$),
$$ L=\frac {c}{H_0} = \frac {3\times 10^6\times10^6\times3.26}{73.8 \pm 2.4} ly \tag{III}$$
$$ L= \frac {9.78\times 10^{12}}{73.8 \pm 2.4} ly \tag{IV}$$
$$L=1.3\pm0.1\times 10^{11} ly \tag{V}$$
Is this calculation correct?
Would the correct calculation make sense? (By making sense I mean it would seem in accordance with some observation and not in contradiction to some other observations? Or results like this are unconfirmable, just mere flights of fancy were they do not relate to anything physical?
The only thing I could use to see it is not invalid (yes double negative, I cannot say it was valid) is the fact that observable universe is $45.7×10^9 ly$ but then again by that account $L=10^{123}ly$ would seem just as valid.
| The answer by Johannes is correct - the proper horizon distance in the concordance cosmology is ~46 billion light years. The reason that the answer in (1) was three times larger than that, when it should have been three times smaller, is that the value of c used was incorrect: The speed of light is $3 \times 10^5\ \mathrm{km}/\mathrm{s}$.
| {
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What is the simplest possible topological Bloch function? Kohmoto (1985) pointed out in Topological Invariant and the Quantization of the Hall Conductance how TKNN's calcuation of Hall conducance is related to topology, in which topologically nontriviality is said to be equivalent to impossiblility choosing a global phase of Bloch function $u_k (r)$ in Brillouin zone. As shown in the Figure, we can choose two distinct gauges in sector I and II, and the curvature is the loop integral of phase mismatch on boundary $\partial H$.
What is the simplest possible Bloch function that is
*
*topologically nontrivial, and
*an eigenstate of Bloch Hamiltonian?
Bloch Hamiltonian: $H(k_x,k_y) = \frac{1}{2m}(-i\partial + {\bf k}+e{\bf A}(x,y))^2 + U(x,y)$ where $U$ is lattice periodic.
| Surprisingly, according to Immanuel Bloch's group (no relation to F. Bloch!), the simplest topological Bloch function is the 1D staggered lattice. The topological invariant is the Zak phase, the Barry phase accrued by walking across the edge of the Brillouin zone. The article will explain it better than I can: Direct Measurement of the Zak phase in Topological Bloch Bands
| {
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Don't understand the integral over the square of the Dirac delta function In Griffiths' Intro to QM [1] he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being
$$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right)$$
(cf. last formula on p. 101). He then says that these eigenfunctions are not square integrable because
$$\int_{-\infty}^{\infty}g_{\lambda}\left(x\right)^{*}g_{\lambda}\left(x\right)dx
~=~\left|B_{\lambda}\right|^{2}\int_{-\infty}^{\infty}\delta\left(x-\lambda\right)\delta\left(x-\lambda\right)dx
~=~\left|B_{\lambda}\right|^{2}\delta\left(\lambda-\lambda\right)
~\rightarrow~\infty$$
(cf. second formula on p. 102). My question is, how does he arrive at the final term, more specifically, where does the $\delta\left(\lambda-\lambda\right)$ bit come from?
My total knowledge of the Dirac delta function was gleaned earlier on in Griffiths and extends to just about understanding
$$\tag{2.95}\int_{-\infty}^{\infty}f\left(x\right)\delta\left(x-a\right)dx~=~f\left(a\right)$$
(cf. second formula on p. 53).
References:
*
*D.J. Griffiths, Introduction to Quantum Mechanics, (1995) p. 101-102.
| Suppose I want to show $$\int \delta(x-a)\delta(x-b)\; dx = \delta(a-b) $$ To do that , I need to show $$\int g(a)\int \delta(x-a)\delta(x-b) \;dx \;da = \int g(a)\delta(a-b)\; da$$ for any function $g(a)$.
\begin{align}\textrm{LHS}& = \int \int g(a) \delta(x-a)\;da \ \delta(x-b) \;dx\\ &=\int g(x)\delta(x-b)\;dx \\&=g(b) \end{align} But $\textrm{RHS}$ clearly $=g(b)$ too.
The result follows putting $a=b=\lambda$
| {
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Do particle pairs avoid each other? Please end my musings Can you explain what happens when a particle and its antiparticle are created. Do they whiz away from each other at the speed of light or what? I suppose that they don't because otherwise they would never meet and annihilate each other, but then, if I had just been created with an antiparticle I would do all I could to stay away from him/her. On the other hand, for the sadistic/suicidal type, they might actually be attracted to one another.
[the question is serious, even though it's written light-heartedly. Please explain.]
|
Do they whiz away from each other at the speed of light or what? I suppose that they don't because otherwise they would never meet and annihilate each other,
Do you mean that you can have particles moving at speed of light? The difficulty of deaccelerating those particles what makes you thinking that renders the annihilation impossible?
It looks like you think that the only way to annihilate is to meet the original antiparticle. But, all elementary particles of the same type are identical. So, there is no need to wait for the recombination of the pair to annihilate. Why don't you consider that your particle may dispart and marry another different one?
Take some force that pulls the particles apart initially. Once they reach some distance, they do not annihilate and engage in interactions with other particles, where they also can annihilate. There is no need for speed (of light) nor need to beware the original counterparticle, once particles are sufficiently apart.
Might be physicists have another opinion
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What causes the structure visible in first few milliseconds of a nuclear explosion? Following on from this question, here is a high-speed photograph of a nuclear explosion, taken about 1 ms after detonation: (source) As anna v pointed out, several similar images can be seen in a Google image search for "high speed nuclear explosion photos".
The spikes at the bottom are know as "rope tricks" - they're caused by cables being heated by the radiation. But what causes that weird-looking structure in the fireball itself?
My suspicion is that the explosion at this stage is an expanding ball of air that's being ionised by the radiation coming from the nuclear chain reaction. If some regions of the surrounding air are ionised more easily due to differences in their temperature (or density or moisture content) prior to the explosion then perhaps this would explain the structure. The explosion is probably taking place in a desert, meaning that there could be quite a bit of turbulence in the ambient air due to thermal convection. However, this is just a wild guess, and I'd be interested to know if anyone has any knowledge or insight about this structure.
| There is a description on Wikipedia accompanying this picture that seems to indicate the mottled surface is the leading edge of a compression shock propagating through the air. In particular, it carries the imprint of the irregularities in the casing surrounding the bomb.
I am inclined to agree. While I admit there is no scale to indicate what the different levels of brightness mean, it seems there is considerable variation, and this seems difficult to attribute to the percent-level differences in air temperature and composition you might get at those scales (20 meters across). Furthermore, there seems to be far too little time for the wavefront to develop large-scale turbulence of its own given the densities involved - the time since detonation is quoted as "less than one millisecond," which is about a percent of the sound crossing time of something that large in the ambient air.
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Conformal Quantum Mechanics I heard the term Conformal Quantum Mechanics used today.
*
*What exactly does this mean?
*Why would one want to study this?
| Conformal groups are basically groups that preserve the angle between two points locally. SO(2,1) group is a prototype of conformal groups. Conformal field theory (theory developed from conformal groups) in zero spatial dimension is known as conformal quantum mechanics. It has applications in string theory, AdS_2-CFT correspondence. SO(2,1) group is a prototype of conformal groups. For more details refer to the following papers.
https://arxiv.org/abs/1506.05596;
http://cds.cern.ch/record/693633/files/197603099.pdf
| {
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Will two trains running along the equator in opposite direction experience same wear out? Two identical trains, at the equator start travelling round the world in opposite directions. They start together, run at the same speed and are on different tracks. Which train will wear out its wheel treads first?
Will their weight change?
Answer is given as : the train travelling against the spin of the earth. This train will wear out its wheels more quickly because the centrifugal force is less on this train.
How do the forces change when the frame of reference is same for both trains?
| As the question is probably more about reference frames, than actual wearing out of train wheels, i'll try to answer it.
The reference frame is not intertial, but rotating reference frame. There is an centrifugal force associated with movement of the reference frame: $F=\frac{mv^2}{r}$.
Assuming non rotating reference frame, there is an centrifugal force acting on both trains, but when one train moves in one direction and the other in other direction, their speeds sum with earths rotation speed with different signs: $\tilde{v}_1=v_e+v_t$ and $\tilde{v}_2=v_e-v_t$.
($v_e$ - trains speed in non-moving reference frame because of earths rotation, $v_t$ - trains speed against rotating reference frame.)
This means the force acting on trains in the direction of earths center will be:
$F_1 = mg - \frac{m(v_e+v_t)^2}{r}$ and $F_2 = mg - \frac{m(v_e-v_t)^2}{r}$. And as $F_1 \neq F_2$, wheels will wear out differently.
| {
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Phase difference of driving frequency and oscillating frequency Suppose a mass is attached to a spring and is oscillating (SHM). If a driving force is applied, it must be at the same frequency as the mass' oscillation frequency. However I'm told that the phase difference between the driving frequency and the mass's frequency must be $\frac{\pi}{2}$.
Why is that? I would have thought they should have to be in phase to be in resonance?
| The oscillator frequency $\omega$ says nothing about the actual oscillator phase. Let us suppose that your oscillator oscillates freely like this: $$x(t) = A_0\cdot\cos(\omega t + \phi_0),\; t<0.$$ At $t=0$ it has a phase $\phi_0$. Depending on its value the oscillator can be moving forward or backward with some velocity. If you switch your external force on at $t=0$ and onwards, say, to push your particle in a positive direction, then, depending on the particle phase, the force will accelerate or decelerate the particle.
Generally you write down the external force in the same way: $$F_\text{ext}(t)=F_0\cdot \cos(\omega t +\Phi_0).$$ This expression stays in the driven oscillator equation, namely, in the right-hand side. The resonance happens always, i.e., the external force will supply energy to the particle, but this supplying can start immediately if the force direction and the particle velocity direction coincide. Otherwise the external force first slows down the particle and only then starts pumping its amplitude.
The particle velocity phase is shifted with respect to the particle coordinate $$v(t) = -A \sin(\omega t + \phi_0)=A\cos(\omega t + \phi_0 +\pi/2),\; t<0.\;$$ So, when $\Phi_0 =\phi_0 +\pi/2\;$ the force is in phase with velocity (not with coordinate) - you have not only the same direction for the velocity and for the force, but also coincidence of instants when both the velocity and the force become zero (no time intervals with their opposite signs).
EDIT: The permanent phase shift of $\pi/2$ in a resonant case with friction (as described in user17581 answer) is a self-established thing and its meaning is simple - the external force in the end compensates exactly the friction force; the latter being proportional to velocity which is shifted by $\pi/2$ with respect to the coordinate time-dependence (so the oscillator oscillates as if it were free, without losses).
| {
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Rigorous approaches to quantum field theory I have been reading Quantum Mechanics: A Modern Development by L. Ballentine. I like the way everything is deduced starting from symmetry principles.
I was wondering if anyone familiar with the book knows any equally elegant presentation for quantum field theory. Weinberg's books start off nice with the irreducible representations of the Poincare group/algebra but the later chapters lose me with the notation. Also, most books I've read on QFT (Srednicki, Peskin and Schroeder, Mandl and Shaw) make a valiant initial attempt at a nice consistent framework but end up being a big collection of mathematical recipes and intuitive insights that seem to work but the overall structure of the theory seems to be sewn up. The relativistic equations crop out of rather flimsy arguments, the canonical anti/commutation relations are imposed out of density indeterminate air, functional methods are developed cause we know no better, infinities come about with renormalization theory to the rescue but it seems very alien from the initial context.
Is there any approach that ties all these things together in an elegant mathematical framework which accounts for all the patch up work that is needed? I am not talking about axiomatization just a global point of view that encompasses all the issues.
| I enjoyed Brian Hatfield's "Quantum Field Theory of Point Particles and Strings." Part 1 of the book is great IMO, Part 2 is useless. Also make sure to get the newest edition, there are many errors in the older editions, but the work is still good.
| {
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Why do we weigh less when falling?
I don't want to go to science world to find out because it would be a long round-trip.
I understand that acceleration/deceleration would effect the weight and I can also imagine that someone at terminal velocity would weigh nothing but I can't get my understanding in terms of forces and how that would effect the weight. For example, what would a formula relating mass, speed (in direction of gravity), gravitational force and weight look like?
|
"I understand that acceleration/deceleration would effect the weight and I can also imagine that someone at terminal velocity would weigh nothing"
It looks like you think that only in terminal velocity we do not weight, and that's wrong. We measure weight by the normal force applied by the ground on us. That means that there must be a normal force. In the elevator, it first accelerates down with an acceleration $a<g$.
In terms of forces, you must accelerate at the same rate as the elevator so $ma_T=ma-mg$ the normal force must be smaller than just $mg$, and so you weigh less. In the terminal velocity of the elevator the total forces must be 0 as your speed is constant, so the normal must be again $mg$.
In free fall, the only force that acts on you if the gravity, and so you weigh nothing, in space for example, in the ISS you see them floating no because there's no gravity (gravity at the altitude of the ISS is almost the same as in the surface), but because gravity is the only force acting on the whole system, so they weigh 0.
| {
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Cyclic Coordinates in Hamiltonian Mechanics I was reading up on Hamiltonian Mechanics and came across the following:
If a generalized coordinate $q_j$ doesn't explicitly occur in the
Hamiltonian, then $p_j$ is a constant of motion (meaning, a constant, independent
of time for a true dynamical motion). $q_j$ then becomes a linear function of time.
Such a coordinate $q_j$ is called a cyclic coordinate.
The above quote is taken from p. 4 in Ref. 1.
What I don't understand is why $q_j$ is a linear function of time if $p_j$ is constant in time. In other words, why does $p_j$ constant in time imply partial $\frac{\partial H}{\partial p_j}$ is a constant? (In particular, $\frac{\partial H}{\partial p_j}$ could depend on any of the other coordinates or momenta.)
Reference:
*
*Patrick Van Esch, Hamilton-Jacobi Theory in Classical Mechanics, Lecture notes. The pdf file is available from the author's homepage here.
| OP is right. The text has an error. A cyclic coordinate $q_j$ does not have to be an linear function of $t$.
Example: Consider two canonical pairs $(q,p)$ and $(Q,P)$ with Hamiltonian $H= p Q +P$.
Then $q$ is cyclic, and therefore $p$ is a constant of motion.
$\dot{Q} =\frac{\partial H}{\partial P}=1$, so $Q$ is a linear function of time.
$\dot{q}= \frac{\partial H}{\partial p} = Q$, and hence $q$ is a quadratic function of time.
| {
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Do perfect spheres exist in nature? Often in physics, Objects are approximated as spherical. However do any perfectly spherical objects actually exist in nature?
| I suppose it depends on how perfect your perfect sphere is.
A drop of water in space with no other gravity effects would look spherical, but if you zoomed in enough to see edges of molecules, then it wouldn't be.
From there you can just keep getting smaller (atoms, protons and neutrons, quarks, etc.) until you get into things like string theorem and quantum foam.
I would say no, like straight lines, there will not be a perfect spherical physical object in the natural world.
| {
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How to get "real-time" temperature from sensor? The following is LM35 Thermal response time in air
The following is temperature reading from LM35 sensor. Horizontal axis is time in sec.
So this is not "real-time" temperature graph. The question is having thermal response graph, how to best adjust real readings from sensor to get "real-time" values?
| You would need to know the transfer function of your sensor, but as a first approximation the graph you have from the manufacturer seems compatible with something like:
$$\frac{dT_s}{dt}=k(T-T_s)$$
where $T$ is the outside temperature and $T_s$ the sensor reading. You could then estimate $T$ as
$$T=T_s+\frac{1}{k}\frac{dT_s}{dt}$$
To figure out $k$, notice that under a stationary $T$, i.e. what the manufacturer graph is showing, the response curve of $T_s$ would be
$$T_s-T_{s0} = T(1-e^{-kt})$$.
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What happens to heat waste produced by energy generation? What happens to heat waste produced by energy generation on earth that must be there according to the laws of thermodynamics? So, it never dissipates and remains on earth?
| The energy does not necessarily remain on earth. Earth maintains an energy balance with its surroundings. The sum of the energy reaching earth through radiation and the energy produced on earth must match the energy leaving earth through radiation. If the latter is too low, earth will warm up, thereby increasing its blackbody radiation and restoring the balance.
However, the energy production on earth (including the production in the core) is generally an extremely small term in this equation.
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Construction of the supersymmetric Faraday tensor When I first learned gauge theories in my introductory quantum field theory course, I was taught that the Faraday (field-strength) tensor can be constructed by computing the commutator of the gauge-covariant derivative:
$$[D_\mu,D_\nu]=-ieF_{\mu\nu}$$
Now, I am studying supersymmetry following Martin's SUSY primer, and in chapter 4.8, the author immediately writes down the super-symmetric field strength chiral superfield out of the vector superfield $V$:
$$\mathcal{W}_\alpha=-\frac{1}{4}D^\dagger D^\dagger D_\alpha V.$$
I would have liked a more gentle introduction to this in terms of something I am already familiar with: is there a way for me to have constructed this using the commutator of some 'gauge super-covariant derivative'?
| As far as I know, it is defined in that form in order to satisfy chirality
\begin{equation}D^{\dagger}_{\dot{\alpha}}W_{\alpha}=0\end{equation}
and gauge invariance
\begin{equation}\delta W_\alpha=0.\end{equation}
I have not seen a definition by a commutator anywhere.
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Force and Torque Question on an isolated system If there's a rigid rod in space, and you give some external force perpendicular to the rod at one of the ends for a short time, what happens?
Specifically:
What dependence does the moment of inertia have?
If it rotates, what is the center of rotation?
Does it matter that the rod is rigid? What happens if it's "springy", say a rubber rod instead.
Is there a difference between exerting a force for a short period of time, and having an inelastic collision (say a ball hits the end of the rod instead of you pressing).
| The moment of inertia is the rotational mass of the object and it is solely a function of mass distribution - the shape. Usually it does not matter if it is rotating, but if the rate of rotation is high enough the body could deform. This is a concern in the connecting rods of high speed engines.
The MI does depend on shape, so it will change if the material is not rigid. However, angular momentum will preserved so the object will rotate faster when the MI is smaller.
The center of mass does not change unless acted on by an external force. But that force is being applied from another mass, so an "external" force does not really exist.
Inelastic collisions should be analyzed using the conservation of momentum, and not energy. Then it does not matter the duration of the force, or the strength, just the integrated amount.
Energy is also conserved in the collision, but in an inelastic collision energy is transformed into heat preserving the conservation rule, while providing no help for our solution.
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Why does smoke go out the window of the car - and what if there's wind blowing instead of the car moving? When driving a car while smoking with the window open (safety and legal issues aside), I've noticed that the smoke tends to go outside the window.
*
*Why does the smoke go outside?
*If the car is standing still and there is wind blowing at the same velocity the car was going - will the smoke behave the same?
| The air flowing around the outside of your car near the windows will be at a reduced pressure due to the Bernoulli effect. There will also be some sort of vent letting air into the cabin, the vent intake must be located at a point in the airflow where you have higher stagnation pressure, so you get a net flow going in the vent and out the window, I'm guessing.
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Is there a Lagrangian formulation of statistical mechanics? In statistical mechanics, we usually think in terms of the Hamiltonian formalism. At a particular time $t$, the system is in a particular state, where "state" means the generalised coordinates and momenta for a potentially very large number of particles. (I'm interested primarily in classical systems for the sake of this question.) Since this state cannot be known precisely, we consider an ensemble of systems. By integrating each point in this ensemble forward in time (or, more often, by considering what would happen if we were able to perform such an integral), we deduce results about the ensemble's macroscopic behaviour. Using the Hamiltonian formalism is useful in particular because it gives us the concept of phase space volume, which is conserved under time evolution for an isolated system.
It seems to me that we could also consider ensembles within the Lagrangian formalism. In this case we would have a probability distribution over initial values of the coordinates (but not their velocities), and another distribution over the final values of the coordinates (but not their velocities). (Actually I guess these would need to be two jointly distributed random variables, since there could easily be correlations between the two.) This would then lead to a probability distribution over the paths the system takes to get from one to the other. I have never seen this Lagrangian approach mentioned in statistical mechanics. I'm curious about whether the idea has been pursued, and whether it leads to any useful results. In particular, I'm interested in whether the idea of phase space volume has any direct meaning in terms of such a Lagrangian ensemble.
| There is a field theory version of statistical physics. The temperature is like the imaginary time. In this way we can formulate theory by path integral with action determined by Lagrangian.
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Spinning spheres colliding In an ideal environment with no friction, in a vacuum, what happens to the velocity of the spin of two spheres spinning in perfect parity at two different velocities when they come into contact?
| The rotating surfaces of the spheres would just slide over each other at the instant of contact: no forces perpendicular to the line connecting the centers of the two spheres would exist (i.e. no torque would exist). They would undergo a perfectly elastic collision (no loss of energy, thus no friction), thus conserving angular (they just keep spinning) and linear (elastic collision) momentum.
The angular velocity of each sphere remains constant: the only 'usual' things which can change the angular velocity is exerting torque or changing the mass of the sphere. At contact, colliding spinning spheres which have no friction just don't care about the spinning.
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How to guarantee that a kilogram of antimatter will quickly annihilate another kilogram of matter? What I mean is, suppose we could somehow get a kilogram of antimatter and contain it safely. Now lets say we want to make a bomb using this kilogram, now, we have two ways, either store another kilogram of matter inside the bomb itself and let the matter and the antimatter touch each others when we want to bomb to detonate, or just expose the kilogram to the air and it will explode. But, my question here is simple, either of the previously mentioned ways will just allow the first particles touching each others to annihilate and sending the rest matter and antimatter in opposite ways making the reaction harder and slower to continue. I know eventually the whole kilogram will be annihilated, but it's all about reaction speed in explosives and that's the main difference between nuclear reactors and nuclear weapons.
So now, is there a way to ensure that the matter and antimatter will completely annihilate each others with a high rate of reaction ?
| The obvious answer, to my mind, would be to build an antimatter bomb like a fission bomb. You have a shell of matter (probably the containment unit) around the antimatter which is crushed by some kind of explosive, perhaps a nuclear explosive, creating a rapidly contracting dense shell that can partially overcome the centrifugal tendencies introduced by the antimatter detonation.
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How do transformers work? A transformer is basically a primary inductor connected to a voltage $U_P$ which you want to transform. You also have an iron rod and a secondary inductor. So when there is a current $I_P$ the iron rods becomes magnetic. When you connect the primary inductor to AC, that means that you'll have a changing current, which causes a change in flux which causes induction. My question is, is $U_S$ just the induction voltage created by the iron rod?
| According to ATL Transformers Ltd:
The transformer is based on two principles: first, that an electric current can produce a magnetic field (electromagnetism), and, second that a changing magnetic field within a coil of wire induces a voltage across the ends of the coil (electromagnetic induction). Changing the current in the primary coil changes the magnetic flux that is developed. The changing magnetic flux induces a voltage in the secondary coil.
The referenced article discusses Faraday's law
The Wikipedia article on magnetic core says
A magnetic core is a piece of magnetic material with a high permeability used to confine and guide magnetic fields in electrical, electromechanical and magnetic devices such as electromagnets, transformers, electric motors, inductors and magnetic assemblies. It is made of ferromagnetic metal such as iron, or ferrimagnetic compounds such as ferrites. The high permeability, relative to the surrounding air, causes the magnetic field lines to be concentrated in the core material. The magnetic field is often created by a coil of wire around the core that carries a current. The presence of the core can increase the magnetic field of a coil by a factor of several thousand over what it would be without the core
(my emphasis)
There's a useful looking article on this subject by a Dr A.M. Etamaly
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What is $v \, dp$ work and when do I use it? I am a little confused, from the first law of thermodynamics (energy conservation)
$$\Delta E = \delta Q - \delta W $$
If the amount of work done is a volume expansion of a gas in, say a piston cylinder instrument at constant pressure,
$$\Delta E = \delta Q - p \, dv$$
Here $p$ is the constant pressure and $dv$ is the change in (specific) volume.
So, when do I take into account
$$\delta W = d(pv) = p \, dv + v \, dp$$
I am assuming that for cases of boundary work, at constant pressure, the $v \, dp$ term is zero.
So under what conditions should I consider the $v \, dp$ term?
| Generally we call p dV as the displacemnt work done by a piston expanding in a cylinder containing some amount of gas. This is applicable to only NON FLOW PROCESSES. But the work v dp is applied to only flow processes or control volume processes. you can simply use the steady flow energy equation in place of this to calculate the work done in a flow process.
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Is it possible for a physical object to have an irrational length? Suppose I have a caliper that is infinitely precise. Also suppose that this caliper returns not a number, but rather whether the precise length is rational or irrational.
If I were to use this caliper to measure any small object, would the caliper ever return an irrational number, or would the true dimensions of physical objects be constrained to rational numbers?
| First of all, it doesn't make sense to assign an absolute number to a physical quantity like length and volume.
The number can be different with respect to different "units" of measurement.
But one can still question the ratio of two lengths, in this case:
According to the Bekenstein entropy limit(information), I guess there should be some maximum level of precision.
If a rod has an irrational length ratio, then it demands an infinite amount of information(write the length in a binary format like this sequence 11010100001000....).
Since irrational numbers don't have any pattern( repetition of a finite sequence) I guess it should be impossible to retain all that information in a Quantum Mechanical world.
Moreover one can be even more critical and accept the existence of length ratios only with a finite code(rather than a finite repetitive pattern)! so that one can deny even a rod with a non-integer rational number length ratio.
This sounds like the emergence of integers(quantas) in Quantum Mechanics!
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Exact diagonalization of graphene's tight binding Hamiltonian While directly diagonalize graphene's tight binding Hamiltonian, which is numerical. We have to use a finite-sized graphene.
So how to deal with boundary conditions? The usual solutions are zigzag or armchair condition, but to make our model more realistic to real infinite graphene plane, how about using periodic condition at boundaries while exactly diagonalizing the tight binding Hamiltonian?
| TL;DR version: you want periodic boundary conditions with an extra twist.
You don't want straightforward periodic boundary conditions as this would be solving for a structure that is periodic. You can solve for an infinite sheet of graphene using Bloch's theorem.
I'll give a couple of details:
Find a basic finite unit that you can tile to make the infinite sheet so that you cover the plane by placing copies of the tile at positions $ma_1+na_2$ with one copy for each integer pair $(m,n)$.
Using Bloch's theorem, a basis for the space of wavefunctions can be found as follows:
Pick phase angles $k_1$ and $k_2$. Now imagine we're working with periodic boundary conditions so the hopping matrix has elements corresponding to neighbouring pairs of atoms where the elements of the pair are on opposite sides of the tile. When you walk off one side and come back on the other, this must correspond to a step along a vector of the form $ma_1+na_2$ for some $m$ and $n$. You modify the hopping matrix so that it has an extra phase factor $\exp(mk_1+nk_2)$ (and the complex conjugate for the step going the other way). You now solve the tight-binding model numerically using this new "twisted" matrix. As you're dealing with a finite tile, if it has $n$ atoms then you expect $n$ eigenfunctions.
You get a basis for the whole plane by considering all possible wavefunctions as you vary $k_1$ and $k_2$, ie. you get $n$ bands with each band parameterised by $k_1$ and $k_2$.
For graphene you can use a repeating two atom unit: one atom has a bond going straight out to the left and the other has one going straight out to the right.
Have a look at section 2.1.2 in this course.
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Could an ultra-relativistic particle tunnel directly through a stellar mass black hole? It occurred to me in passing that the Lorentz contraction of a black hole from the perspective of an ultra-relativistic (Lorentz factor larger than about 10^16) particle could reduce the thickness of a black hole to less than the DeBroglie wavelength of the particle.
It would seem to me that under those conditions the particle would have a non-insignificant probability of tunnelling right through the black hole rather than being adsorbed by it.
Is this so? If not, why?
| The important thing is the cross-sectional area of the horizon, and this is independent of Lorentz transformation, since the $y$ and $z$ coordinates are not changed.
Additionally, you can calculate that light will be captured by the horizon with non-zero cross section, and the geodesics ultra-relativistic particles will asymptote to the geodesics of null particles.
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Can low-gravity planets sustain a breathable atmosphere? If astronauts could deliver a large quantity of breathable air to somewhere with lower gravity, such as Earth's moon, would the air form an atmosphere, or would it float away and disappear? Is there a minimum amount of gravity necessary to trap a breathable atmosphere on a planet?
| The speed of oxygen at room temperature (293k) is 1720km per hour so if the escape velocity of the moon or planet is greater than that then at least you will have oxygen. If you want some nitrogen in the mix then you will have to google it's speed like I did for oxygen;-)
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Equivalence principle question I understand the equivalence principle as "The physics in a freely-falling small laboratory is that of special relativity (SR)." But I'm not quite sure why this is equivalent to "One cannot tell whether a laboratory on Earth is not actually in a rocket accelerating at 1 g".
| I'm not sure if the O.P. was satisfied with the previous answers or not but I feel that there is something that wasn't answered, at least in the way that the question asked for it. The reason why the equivalence between a free falling system and a Special Relativity Rest Frame implies the equivalence between a laboratory on Earth and a rocket accelerating at 1 g is this:
1) First, notice that a laboratory in earth is not the same system as a free falling system, altough both of them feel gravity. The laboratory (or just elevator) on earth is meant to represent a system that feels gravity but holds its place do to some other force, namely, the normal in the ground or the tension in the elevator.
2) Suppose that the first equivalence is right: A free falling system is equivalent to a rest system in Special Relativity. Great, how we get to the second equivalence?
3) Add an upward force to the free falling system, namely, add a tension going upwards to the free falling elevator and you will get an elevator hunging up in earth.
4) According to the first equivalence, if a free falling system was the same as a rest frame, then a free falling system plus a force acting against gravity should be the same as a rest frame with the same force applied.
5) So, an elevator hunging up in Earth is the same as a elevator feeling a force upwards with no gravity.
This is the logical way to go from the first equivalence to the second one, just add the tension in the elevator to go from "free falling system" to "Laboratory on earth" so that the second part of the equivalence goes from "rest frame of Special Relativity" to "rocket going upwards at 1g".
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Question on Radiance equation The radiance equation is
$$
L = \frac{d}{dA} \frac{2(\phi)}{dW cos(\theta)} (watt/srm^2)
$$
where $\phi$ is the flux.
I am thinking, should not be the cosine term on the numerator instead of the denominator? Having the cosine in the denominator will make L goes to infinity if $\theta = 90$ , which does not make sense to me. My understanding is that if $\theta = 90$ (i.e. flux direction is perpendicular to the surface normal), then $L$ should equal to zero (and not infinity).
| Generally, the radiative transfer equation is given by
$$
\frac{1}{c}\frac{\partial I_{\nu}(\vec{r},\vec{n},t)}{\partial t} +
\vec{n}\cdot\frac{I_{\nu}(\vec{r},\vec{n},t)}{\partial \vec{r}} =
\rho(\vec{r},t) \kappa_{\nu} (\vec{r},t)\left\{-I_{\nu}(\vec{r},\vec{n},t)+S_{\nu}(\vec{r},\vec{n},t)\right\}.
$$
To monochromatic intensity is defined by noting that $I_{\nu}(\vec{r},\vec{n},t)\cos\!\Theta \,d\nu\,df d\Omega\, dt$ at the position $\vec{r}$ gives the radiative energy
in the frequency interval $\nu,\nu+d\nu$ transported through the surface element $df$ into the solid
angle element $d\Omega$ enclosing the direction $\vec{n}$ during a time increment $dt$,
see the appended figure.
Your definition looks a bit strange, there is probably a time differential $dt$ missing in the denominator and the $2\phi$ should be the energy differential $dE$. However, $\cos\theta$ in your definition of the radiance correctly appears in the denominator. Looking at my figure, if $\Theta = 90^{\circ}$ the energy propagates along the area element $df$ instead of though it, so the radiance is indeed ill defined in this case.
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How should I simulate the electric potential field from a wavefunction? I was interested in making what I thought would be a simple simulation of an electron encountering a positron by numerically solving the Schrodinger equation over several time steps, but I've run into 1 problem: the potential used in most textbooks (to solve the hydrogen atom for example) are still of a classical point particle, rather than a wavefunction. How do I generate the potential field from two real, moving and interacting wavefunctions?
| If you look at hwlau answer, you see that you can change coordinates from x1,x2 to the center of mass and relative coordinate. Just like you do for the Hydrogen atom.
If you work in the system of coordinates where the COM is stationary, the solution will be simple - identical to the Hydrogen atom, in fact.
My guess is, however, that this equation will not contain all the interesting physics, because there is no way for the two particles to annihilate and emit photons. To see that you will need to work in 2nd quantization, I guess.
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Infinite square well in momentum space As we know the eigenfunctions for a particle of mass $m$ in an infinite square well defined by $V(x) = 0$ if $0 \leq x \leq a$ and $V(x) = \infty$ otherwise are:
$$\psi_n (x) = \sqrt{\frac{2}{a}} \sin \left(\frac{n\pi x}{a} \right).$$
How does the ground state wave function look like in momentum space? As far as I recall I have to integrate $\psi_n(x)$ over the whole of space with the extra factor $\frac{e^{(-i p x / \hbar)}} {\sqrt {2 \pi \hbar}}$ (everything for $n = 1$).
In the solutions to this problem they integrated over $-a \leq x \leq a$ while I would've integrated from $0$ to $a$.
Am I somehow missing something or is this solution just plain wrong?
A further question: How would I check whether or not my resulting $\psi(p)$ is an eigenstate of the momentum operator? Just slap the momentum operator in front of my function and see if I get something of the form $c \psi(p)$, where $c$ is some constant? Or how does this work?
| This question is not well-posed from scratch. There is no Momentum Operator for the problem you are considering. Your geometric space is a bounded region of the real axis, so no translation group can be defined and no self-adjoint generator of translation (the momentum observable) exists. The symmetric operator $-i\frac{d}{dx}$ is not essentially self-adjoint on a natural domain like the one of $C^1$ functions vanishing at $0$ and $a$. It admits infinite self-adjoint extensions and there is no preferred physical choice.
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When do I apply Significant figures in physics calculations? I'm a little confused as to when to use significant figures for my physics class. For example, I'm asked to find the average speed of a race car that travels around a circular track with a radius of $500~\mathrm{m}$ in $50~\mathrm{s}$.
Would I need to apply the rules of significant figures to this step of the problem?
$$ C = 2\pi (1000~\mathrm{m}) = 6283.19 $$
Or do I just need to apply significant figures at this step?
$$ \text{Average speed} = \frac{6283.19~\mathrm{m}}{50~\mathrm{s}} = 125.664~\mathrm{m}/\mathrm{s} $$
Should I round $125.664~\mathrm{m}/\mathrm{s}$ to $130~\mathrm{m}/\mathrm{s}$ since the number with the least amount of significant figures is two?
| https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/02%3A_Measurement_and_Problem_Solving/2.04%3A_Significant_Figures_in_Calculations
Just revisited this in chemistry this year. You actually have to apply the correct number of significant figures based on the rules of operations (addition/subtraction or multiplication/division) as you perform each calculation in a problem with multiple calculations. The purpose of using significant figures is not to get accurate results or results closest to what a calculator would get. The point of using significant figures is not to mislead the reader or person following your work in thinking that you used exact numbers when in fact you did not. So the more steps you have in a problem, the further your answer will be from the answer a calculator would derive. If you apply the correct number of significant figures in every step, you are accurately accounting for what you actually MEASURED (this is why they are applied to measured numbers, not exact numbers). And if you’re sitting somewhere computing the problem, you did not measure anything. We also do not just chose to use an extra significant figure or two but have to stick to the rules of operation. ie- least amount of significant figures for multiplication and division and lowest decimal place for addition and subtraction.
Hope the link above with consistent explanations and videos helps.
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What are the "loopholes" in past Bell's theorem experiments? I am intrigued by the following Phys.org article:
Researchers began using photons in 1980s to test Bell's theory and determine if Einstein's reasoning is right or wrong. Since then, researchers have used various quantum states to test the theory but continued to have loopholes in their methods, therefore falling short of a definitive result. Luo said the new collaboration would, for the first time, be using several different quantum systems—including photons, ions, quantum dots and solid-state ensembles—to test the theory across large distances and hopefully eliminate all possible loopholes, he said.
—Physics researchers join effort to finally complete quantum theory
I am familiar with Bell's inequalities, and I would like to know: how did the experimental results fall short? Why is this time going to be different?
| The main loopholes were the detection (efficiency) loophole and the locality (or communication) loophole (http://en.wikipedia.org/wiki/Loopholes_in_Bell_test_experiments ). I don't know why or if this time it's going to be different.
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Andrew's experiment In Thomas Andrew's experiment, consider the dome shaped saturation region. If we increase the pressure at constant volume until we reach the critical point, why does the density of vapours rise and the density of liquid fall? Moreover, why will the meniscus or the separation boundary of two phase slowly become less distinct and finally at critical point disappear?
Will vapour pressure increase or decrease as we move vertically up in the P-V plane in the saturation region?
| Regarding your second question, it is intimately linked to the feature of scale invariance of critical phase transitions. Essentially what goes on is that a condensing vapour, within the coexistence region, forms droplets of some particular size (more precisely, a distribution of sizes with some particular scale), which depends on the temperature. As you approach the critical point within the coexistence region, this characteristic size increases without bound, and the result is that you will have droplets of all possible sizes from the very smallest to the largest allowed by the container. This causes the blurring of the liquid/vapour meniscus.
This phenomenon is of a particularly universal character, and is ubiquitous in nature. For an illustration in a weird place look up for instance the theory of percolation.
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Electrostatic charge leakage What are the ways electrostatic charged objects leak charge in humid conditions?
Can airborne particles pick up charge by contact, then be repelled hence removing charge? If so would it be a significant factor?
| There are two ways charge can leak in humid air. I'll give a basic description here, but I encourage you to Google around the area as you'll find lots of papers reporting measurements of conductivity of humid air and charge leakage in humid air.
The first leakage mechanism is that in the presence of water vapour virtually all materials have an adsorbed layer of water on their surface, and if this water is thick enough charge will conduct through it just as charge conducts through bulk water. See for example this paper, though note it's behind a paywall. Hygroscopic and polar materials adsorb water most easily, while low energy and non-polar surfaces like PTFE do not significantly adsorb water.
The second mechanism is that the conductivity of air increases with humidity. See for example this paper: there's no special significance to this paper other than it was the first Google found for me. The conductivity must be due to charged particles in the air, but I'm not sure it's understood what these particles are. My guess would be dust or pollen. It would be interesting to measure the conductivity of filtered air compared to unfiltered air, but I couldn't find any papers describing this. I'm also not sure why the conductivity increases with increasing humidity. I would guess that water molecules adsorbed onto the charged particles stabilise them in some way.
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Partition function of a gas of $N$ identical classical particles Partition function of a gas of $N$ identical classical particles is given by
$$ Z~=~\frac {1}{N! h^{3N}} \int \exp[-\beta H(p_1.......p_n, x_1....x_n)]d^3p_1...d^3p_n,d^3x_1...d^3x_n $$
in this above equation we use $N!$ as the total number of sub-systems of a system of identical particles. and $ h^{3N} $ to make the partition function dimensionless. I am not clear how $ h^{3N} $ is used to make it dimensionless.
| Notice that the $e^{-\beta H}$ is dimensionless, while each factor of $dp$ contributes one factor with the dimensions of momentum while each $dx$ contributes one factor with the dimensions of length. Therefore each factor $dp dx$ contributes a factor with dimensions of angular momentum. Since there are $3N$ of these factors (N particles and 3 dimensions) in the integration measure, the integral has a total dimension of angular momentum to the power of $3N$. On the other hand, $h$ has dimensions of angular momentum, so dividing by $h^{3N}$ makes the full expression dimensionless.
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How accurate are our calculations about distant stars keeping in mind their enormous distances? Since many stars are hundreds of light years away from the Earth and therefore, what we observe of them today is really their distant past, how can we say anything with certainty about their composition, size and nature? Betelguese, for example, is said to be in the last stage of stellar evolution, but taking into consideration its approximately 640 light years of distance from the Earth, aren't we actually making assertions about the star that are 640 years old? If that is the case, how can we claim to know the actual present status of the star?
| What we infer about stars from the light we see at the Earth is indeed "old news". However, for almost all practical purposes in stellar astrophysics this doesn't matter. The phases of a star's life last millions or billions of years and most of the individual stars that are studied are within say 30 thousand light years of the Earth -so the light travel time is a negligible fraction of the evolutionary timescales.
The example you picked is an interesting one. Betelegeuse is probably in the very last phase of its life before exploding as a supernova. This final phase is about the shortest in a stellar lifecycle and it is possible that Betelgeuse has already exploded, though more likely we will have to wait a few tens of thousands of years more.
Another interesting example is that when it comes to other galaxies, it is possible to study individual luminous (very massive) stars. So, if we study very massive stars in the Andromeda galaxy (2.5 million light years away) it is likely that most of these stars (the most massive stars have the shortest lifetimes) have already exploded as supernovae, since their lifetimes are less than a few million years.
If we want to say what a star is like now, then we have to rely on an extrapolation forward in time of our stellar evolution models. This is not of great concern because (a) the extrapolation is not that great for most stars (see above) and (b) even where the extrapolation is a large fraction of the stellar lifetime, the models are tested using different stars at all phases of their lives so we have reasonable confidence in them.
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Amplitude of Probability amplitude. Which one is it? QM begins with a Born's rule which states that probability $P$ is equal to a modulus square of probability amplitude $\psi$:
$$P = \left|\psi\right|^2.$$
If I write down a wave function like this $\psi = \psi_0 e^{i(kx - \omega t)}$, I find $\psi_0$ inside.
If $\psi$ is called probability amplitude then what is $\psi_0$ called? Is it perhaps called an amplitude of a probability amplitude?
| $\psi_0$ is the initial amplitude and $\psi$ is the amplitude after some amount of time (or more appropriately, the amplitude after some change in spacetime coordinates). The function you have provided: $\psi=\psi_0e^{i(kx-\omega t)}$; is a function of space and time. So it tells me that given some initial function $\psi_0$ at some time $t$ or position $x$ it will have evolved into some function $\psi$. Both $\psi$ and $\psi_0$ are probability amplitudes, its just that $\psi_0$ is the probability amplitude at some initial point.
So my answer is that you call it the initial amplitude, or amplitude naught.
| {
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What is the proof that a force applied on a rigid body will cause it to rotate around its center of mass? Say I have a rigid body in space. I've read that if I during some short time interval apply a force on the body at some point which is not in line with the center of mass, it would start rotating about an axis which is perpendicular to the force and which goes through the center of mass.
What is the proof of this?
|
What is the proof of this?
The proof of Chasles’s Theorem relating to the rotational and translational displacements of a rigid body is done in very many texts including Appendix 20A of this document produced by MIT.
In essence this shows that, if a force whose line of action does not pass through the centre of mass of a rigid body, the applied force is equivalent to the same magnitude and direction force which passes through the centre of mass of the body which produces only translation acceleration of the centre of mass of the body and a couple which produces only rotational acceleration of the body.
The actual motion of the body is dictated by the sum of these two accelerations.
Suppose a force $\vec F$ is applied to a body whose line of action does not pass through the centre of mass $C$ of the body.
Adding two forces $\vec F_1$ and $\vec F_2$ at the centre of mass of the body such that $\vec F = \vec F_2$ and $\vec F_1 + \vec F_2 =0$ as shown in the diagram below.
There is now a force $\vec F_2 (= \vec F)$ acting along a line through the centre of mass of the body, $C$, which will only produce a translational acceleration of the body and a couple consisting of the two forces $\vec F$ and $\vec F_1$ and of magnitude $Fd$ which will only produce a rotational acceleration of the body.
| {
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Lorentz transformation of classical Klein–Gordon field I'm trying to see that the invariance of the Klein–Gordon field implies that the Fourier coefficients $a(\mathbf{k})$ transform like scalars:
$a'(\Lambda\mathbf{k})=a(\mathbf{k}).$
Starting from the mode expansion of the field
$$\phi'(x)=\phi(\Lambda^{-1}x)=
\int \frac{d^{3}k}{(2\pi)^{3}2E_{k}}
\left( e^{-ik\cdot \Lambda^{-1}x}a(\mathbf{k})
+e^{ik\cdot \Lambda^{-1}x}b^{*}(\mathbf{k})
\right),$$
it's easy to see that it equals
$$\int \frac{d^{3}k}{(2\pi)^{3}2E_{k}}
\left( e^{-i(\Lambda k)\cdot x}a(\mathbf{k})
+e^{i(\Lambda k)\cdot x}b^{*}(\mathbf{k})
\right).$$
when using the property $\Lambda^{-1}=\eta\Lambda^{T}\eta$. Then doing a change of variable $\tilde{k}=\Lambda k$ and considering orthochronous transformations so that the Jacobian is 1, then I get the wanted result ($a'(\Lambda\mathbf{k})=a(\mathbf{k})$) when comparing with the original mode expansion. However, this is not quite right as I would have to justify that $E_\tilde{k}=E_k$ but I can't see how.
| The important insight is that it's actually the whole combination
$$
\frac{d^3 k}{2(2\pi)^3 E_\mathbf k}, \qquad E_\mathbf{k} = \sqrt{\mathbf k^2 + m^2}
$$
that forms a Lorentz-invariant measure. To see this, note that if we define $k= (k^0, \mathbf k)$ and use the identity
$$
\delta(f(x)) = \sum_{\{x_i:f(x_i) = 0\}} \frac{\delta(x-x_i)}{|f'(x_i)|}
$$
then we get
$$
\delta(k^2 - m^2)=\frac{\delta(k^0 - \sqrt{\mathbf k^2+m^2})}{2\sqrt{\mathbf k^2+m^2}} + \frac{\delta(k^0 + \sqrt{\mathbf k^2+m^2})}{2\sqrt{\mathbf k^2+m^2}}
$$
so the original measure can be rewritten as
$$
\frac{d^3 k}{2(2\pi)^3 E_\mathbf k}=\frac{d^3k\,d k^0}{2(2\pi)^3 k^0}\delta(k^0 - \sqrt{\mathbf k^2+m^2}) = \frac{d^4k}{(2\pi)^3}\delta(k^2-m^2)\theta(k^0)
$$
which is manifestly Lorentz invariant for proper, orthochronous Lorentz transformations. The rest of your manipulations go through unscathed, and you get the result you want!
Hope that helps!
Cheers!
| {
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Confused over complex representation of the wave My quantum mechanics textbook says that the following is a representation of a wave traveling in the +$x$ direction:$$\Psi(x,t)=Ae^{i\left(kx-\omega t\right)}\tag1$$
I'm having trouble visualizing this because of the imaginary part. I can see that (1) can be written as:$$\Psi(x,t)=A \left[\cos(kx-\omega t)+i\sin(kx-\omega t)\right]\tag2$$
Therefore, it looks like the real part is indeed a wave traveling in the +$x$ direction. But what about the imaginary part? The way I think of it, a wave is a physical "thing" but equation (2) doesn't map neatly into my conception of the wave, due to the imaginary part. If anyone could shed some light on this kind of representation, I would appreciate it.
| In short, “a wave traveling in the $+x$ direction” has nothing to do with actual motion of a wave packet. In spite of some mathematical similarities, wave function isn’t anything physically like a gravity wave on water surface. In quantum “waves”, there is no water (or gas, other 3D continuum, string, or anything else that can convey a physical meaning to values of $Ψ(x,t)$).
| {
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How do we make symmetry assumptions rigorous? I have, for instance, a problem with a spherically symmetric charge distribution. I deduce here, in order to solve the problem easily, that the corresponding electric field must be symmetric. How is this type of deduction rigorously justified?
| Let me first answer this question in a particular class of electrostatics problems.
In the case of a localized charge distribution in electrostatics (one in which the charge density vanishes outside of some ball around the origin), the general solution for a potential that vanishes at $\infty$ is
$$
\mathbf \Phi(\mathbf x) = k\int_{\mathbb R^3} d^3x' \frac{\rho(\mathbf x')}{|\mathbf x - \mathbf x'|}
$$
In this case, we often want to show that if $\rho$ has certain symmetries, then $\Phi$ shares those symmetries. Let's first derive a little result:
Suppose that we make an invertible transformation $T$ on the charge distribution (say a translation for a rotation for example), then the new (transformed) charge distribution $\rho_T$ will be
$$
\rho_T(\mathbf x) = \rho(T^{-1}\mathbf x).
$$
What will the resulting potential $\Phi$ be? Well, let's compute
$$
\Phi_T(\mathbf x)
= k\int_{\mathbb R^3} d^3 x' \frac{\rho_T(\mathbf x')}{|\mathbf x - \mathbf x'|}
= k\int_{\mathbb R^3} d^3 x' \frac{\rho(T^{-1}(\mathbf x'))}{|\mathbf x - \mathbf x'|}
$$
we can perform the resulting integral via a change of variables
$$
\mathbf u = T^{-1}(\mathbf x') \implies \mathbf x' = T(\mathbf u)
$$
and the formula for changing the variable of integration for volume integrals gives
$$
d^3 x' = J_T(\mathbf u) d^3 u
$$
where $J_T$ is the jacobian of the transformation, so that the transformed potential becomes
$$
\Phi_T(\mathbf x) = k\int_{\mathbb R^3} d^3u \,J_T(\mathbf u)\frac{\rho(\mathbf u)}{|\mathbf x - T(\mathbf u)|}
$$
Now back to the show.
To see how this formula for the transformed potential is used to answer your question about symmetries, let's consider a translationally symmetric charge density;
$$
\rho(\mathbf x - \mathbf x_0) = \rho(\mathbf x), \qquad \text{for all $\mathbf x_0$}
$$
In this case, the transformation $T$ is $T(\mathbf x) = \mathbf x + \mathbf x_0$. The Jacobian is just 1, and our the formula we derived above for the transformed potential gives
$$
\Phi_T(\mathbf x) = k\int_{\mathbb R^3} d^3u \frac{\rho(\mathbf u)}{|\mathbf x - (\mathbf u-\mathbf x_0)|} = k\int_{\mathbb R^3} d^3u \frac{\rho(\mathbf u)}{|(\mathbf x - \mathbf x_0) - \mathbf u|} = \Phi(\mathbf x - \mathbf x_0)
$$
On the other hand, the translational invariance of the charge density tells us that
$$
\Phi_T(\mathbf x)
= k\int_{\mathbb R^3} d^3 x' \frac{\rho_T(\mathbf x')}{|\mathbf x - \mathbf x'|}
= k\int_{\mathbb R^3} d^3 x' \frac{\rho(\mathbf x')}{|\mathbf x - \mathbf x'|}
= \Phi(\mathbf x)
$$
so combining these results gives
$$
\Phi(\mathbf x - \mathbf x_0) = \Phi(\mathbf x)
$$
Namely, the potential is also translationally symmetric. A similar procedure can be used for other symmetries. Try rotational invariance for example on your own!
Hope that helps!
Physics Rocks.
Addendum. I think you can show similar things for generic Neumann or Dirichlet boundary value problems in electrostatics in which, for example, you don't just want to solve Poisson's equation for a localized charge distribution with vanishing potential at infinity, but I haven't worked out the details.
| {
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Changes in Water Bonding Angle I heard something recently in a casual discussion, but have yet to be able to confirm it: is there any evidence that the bonding angle for a water molecule, currently defined as 104.5, has been either steadily growing or shrinking since it was first observed and measured? Is there any possibility that water samples from different locales would produce different results? I am speaking strictly of water in as pure a form as possible, not water affected by impurities or physical manipulation.
| No, there has been no change to the bond angle of pure water,1 nor does anyone expect to measure one. The reason is this angle is set by fundamental physical constants - things like the mass of the proton and the speed of light - albeit in a complicated way. Such things are called constants for a reason. Now there are some theories out there, a number of them a bit far-fetched, that speculate that maybe some of our constants have been evolving extremely slowly over billions of years, but certainly there has been no detected variation over the course of human scientific inquiry. Such a discovery would be huge, causing much agitation in all of physics and chemistry.
It should also be noted that the bond angle can both be calculated from theory and measured from experiment, and neither method yields exactly $104.5^\circ$. Now your theoretical calculations might improve if you make fewer simplifying assumptions than your predecessors, and your measurements might improve if you develop better apparatus and technique, but the true underlying value of the bond angle is not changing.
1 Defined here as oxygen-16 bonded to two atoms of hydrogen-1. Now the angle could vary with isotopic abundances, so perhaps someone was referring to changing isotopic abundances of a source of water. You might see this, for instance, in extreme climate change, since as temperatures rise the fraction of "heavy" water ($\mathrm{D}_2\mathrm{O}$, etc.) evaporated from oceans increases relative to "normal" water.
| {
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Usage of Schrödinger equation vs Madelung equations It is well known that Madelung formulation is alternative to the Schrödinger Formulation, cf. this previous Madelung transformation Phys.SE post. I wanted to know what makes Schrödinger's formulation superior to that of Madelung, and in particular, when does the Madelung formulation fail?
| This is a translation from the German Wikipedia entry about the Madelung equations:
Due to their non-linearity, the Madelung-equations are difficult to use in practise. However, they show that there exist non-linear equations that are based on linear equations.
Hope this helps.
| {
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Internal/Rotational angular momentum I have some difficulties to understand the relation between the internal and the rotational angular momentum of a rigid body which is also known as König's theorem, so what physical intuition lies behind this equation? (I googled it but I can't find anything, so I will also appreciate any references if possible.)
edit:
$$H(O/Rg) = OG \times p + H^\star$$ (as vector quantities)
$p$ is the moment of momentum , $G$ is the center of mass , $Rg$ is a gallilean reference of frame , $H$ is the angular momentum with respect to a point $O$ fixed in that reference , and $H^\star$ is the angular momentum in the CM reference .
| König's theorem is essentially a statement of conservation of angular momentum. If you consider the angular momenta of a system of particles, they had better add up to the same value no matter what frame you consider, provided that you are computing the angular momentum about the same fixed point in space. In a fixed frame, a system of particles has some angular momentum, which you have denoted $H(O/Rg)$. In a frame of reference co-moving with the center of mass of that system of particles, the total angular momentum appears to be $H^*$. König's theorem states that the connection between these two frames that resolves the apparent discrepancy in observed angular momentum is to adjust for the angular momentum of the center of mass itself, as viewed from the Gallilean refrence frame. This angular momentum is $\mathbf{r}_{CM} \times \mathbf{p}_{CM}$ where $\mathbf{r}_{CM}$ is the position of the center of mass relative to the origin $O$ and $\mathbf{p}_{CM}$ is the momentum of the center of mass of that system of particles, again relative to $O$.
| {
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Water-cooled fast neutron reactors Can anyone explain why fast neutron reactor designs use sodium/lead/salt cooling, instead of water (heavy/light)?
Is that because neutron absorption by water would not allow to break even in fuel cycle?
Will heavy water help here?
Or water slows neutrons so efficienly so that even if we reduce amount of water inside the reactor (by increasing flow speed) - it still will significantly lower neutron energy, while sodium does not slowdown neutrons at all?
| Please see a discussion of advantages and drawbacks of sodium cooling in fast neutron reactors at http://en.wikipedia.org/wiki/Sodium-cooled_fast_reactor . You mentioned one of the advantages: neutrons slow down much less in collisions with sodium nuclei than in collisions with hydrogen nuclei of water.
| {
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What would the collision of two photons look like? Could someone explain to me what the collision of two photons would look like? Will they behave like,
*
*Electromagnetic waves: they will interfere with each other and keep their wave nature
*Particles: they will bounce like classical balls
I assume that energy of that system is too small to make creation of pairs possible.
| By time symmetry, two photons with sufficient mutual energy necessarily are capable of annihilating each other to produce an electron-positron pair, since one of the decay modes for positron-electron annihilation is the production of two gamma photons. It's just harder to arrange experimentally, since unlike the electron and positron the uncharged photons have no attraction for each other.
Here's an experimental exploration of two-photon positron production:
D.L Burke et al, Positron Production in Multiphoton Light-by-Light Scattering, SLAC, June 1997.
Photon-photon interactions (scattering) via virtual particle pairs gives you two-photon physics, which looks at the probabilities of photon-photon production of particle pairs much heavier than electrons.
| {
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Difference between torque and moment What is the difference between torque and moment? I would like to see mathematical definitions for both quantities.
I also do not prefer definitions like "It is the tendancy..../It is a measure of ...."
To make my question clearer:
Let $D\subseteq\mathbb{R}^3$ be the volume occupied by a certain rigid body. If there are forces $F_1,F_2,....,F_n$ acting at position vectors $r_1,r_2,...,r_n$. Can you use these to define torque and moment ?
| moment is turning effect produced by a force . while torque is due to rotation of body.
| {
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Can we make images of single atoms? I was wondering how far in imaging physics had gotten. Do we hold the technology to actually take a picture of, say, an alpha particle, or even a single atom?
I realise we aren't talking about camera pictures, so what kind of imaging techniques have taken images/is most likely to be the technique to take an image of a single atom?
| We can, and have, taken "pictures" of individual atoms in materials.
One example is this image taken at IBM :
Each dot is a single atom that was placed on the substrate. You can find similar images taken by IBM here.
This image was made using a technique called "Scanning Tunneling Microscopy" (STM) which relies on quantum tunneling. It isn't an optical technique, like the microscopes we use in school. But it uses a probe (which is just a sharp tip) which is at some voltage with respect to the sample being imaged, and because of the voltage difference between the probe and the sample, electrons tunnel through the space in between and a "tunneling current" is observed. This tunneling current can be a function of several different things, like the applied voltage difference, and the height of the sample from the tip etc.
The resolution of this technique is ~ $0.1 \text{ nm}$, and can resolve individual atoms.
A similar technique is Atomic Force Microscopy (AFM), which relies on measuring the force between the tip and the sample. The elaborate (and extensive!) details are in the wikipedia page that I linked to, but the principle of operation is still very similar. If the force between the sample and the tip is kept constant, then the height between them will change. And if you can keep track of the height, you get an "image" of the sample.
This technique is also capable of resolving individual atoms, but I think not quite as well as STM, but I could be wrong about that.
| {
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What happens to string theory if spacetime is doomed? What is expected to happen with string theory, if physics is reformulated according the lines hinted at by the twistor-uprising business discussed in this question and its answers for example and spacetime is doomed, as Nima likes to say?
Will it be incorporated into this new picture with "emergent" spacetime and surface again in certain limits? Or more generally, what effects will this twistor-revolution have on string theory, if any?
| My earlier impression that Nima's slogan "spacetime is doomed" could potentially lead to a reformulation of string theory, is wrong and an overinterpretation of Nima's very enthusiastic comments. As Lumo says in his nice clarifying comments, it is rather the other way round and "spacetime is doomed" is in fact a result that came out of string theory quit some time ago already:
The fact that spacetime is doomed is one of the main results of string theory, primarily; twistor variables and other things are just individual manifestations of this stringy conclusion and it would be silly for Nima or anyone else to claim the priority in the "spacetime is doomed" paradigm shift which was derived and said a decade before him by string theorists such as Seiberg and Witten. So your causality is upside down. Your question is like asking what would happen to Einstein's research if the world respected relativity. Well, its key epoch would victoriously end.
Concerning the particular wording "spacetime is doomed", you may see that it was used by David Gross in 2004, right after he won the Nobel Prize:"Everyone in string theory is convinced that spacetime is doomed." The statement "Space and time may be doomed" (Witten) and "I am almost certain that space and time are illusions" (Seiberg) come from the 1990s, see many google results.
| {
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Proper notation for normalized scalar? I have not been able to find a resource to tell me the standard notation for a normalized scalar value. Normalized vectors (i.e. unit vectors) are typically denoted by placing a hat over the variable, something like:
$${\bf \hat{e} = \dfrac{e}{||e||} }$$
However, does the same apply to normalizing (and nondimensionalizing) a scalar? Would it be correct to write:
$$ \hat{L} = \dfrac{L}{L_0} $$
This is assuming that $L$ and $L_0$ are just scalar lengths. If I am defining my own notation, is it verboten to call this something like $\bar{L}$ (with a bar)?
If it makes any difference, I am a mechanical engineer and this will be going in my thesis.
| I know in electrical engineering, particularly in power transmission fields, sometimes people use the so-called per-unit system, where quantities are normalized to the corresponding base value. Sometimes, subscript like $U_{\text{p.u.}}=\frac{U}{U_0}$ ($U_0=U_{\text{base}}$) is used. I don't think there are any proper notations for normalized scalars. You can follow the conventional notations in the particular area as long as there is no ambiguity.
| {
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Why is electric potential scalar? I can't conceptually visualize why it would be so. Say you have two point charges of equal charge and a point right in the middle of them. The potential of that charge, mathematically, is proportional to the sum of their charges over distance from the point ($q/r$). But intuitively, my thought process keeps going back to the concept of direction and how the electric field at that point would be zero. So why would the electric fields cancel while the electric potentials just add up algebraically?
| Suppose in a equilibrium system where two negative charges is accompanied by one positive test charge in the middle. Here the resultant electric field at the centre is zero. Due to opposite and equal electric line of force from two negative charges but to bring that test charge from zero potential to the middle some work is done which is nothing but the potential energy value you get by adding algebraically. Though I am commenting in 2016 feel free to give me positive rating. :)
| {
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Whistle Physics I'm looking for a simple explanation of how a whistle operates. I know that forcing air over a sharp lip can set up a wave in a resonating cavity, but how? "Most whistles operate due to a feedback mechanism between flow instability and acoustics"--yes, but what does that feedback mechanism look like?
I was surprised to be unable to find a basic diagram online demonstrating how a whistle operates. I did find lots of images like this:
. . . but such images are unhelpful since they don't show exactly what's producing the oscillation!
| Air in a chamber resonates like a spring: It has mass and springs back when compressed or decompressed. When air is blown across the opening, and slightly down, it will push the air inside down & cause a disturbance which will cause the air inside to start to resonate. As it resonates down, it deflects the air stream down into the opening, pushing the air down more. As the air inside springs back, it deflects the air stream upward, so it's no longer pushing down. It also drags air away from inside by the Bernoulli effect, helping to pull the air inside up. Any vortexes formed are incidental.
| {
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How is possible for current to flow so fast when charge flows so slow? How is it possible for current to flow so fast when charge flows so slowly?
We know electrons travel very slowly while charge travels at ~the speed of light.
| It seems you are contrasting the speed of propagation of current with the speed of the individual charge carriers.
These two things are clearly separate. There are many examples. Consider sound.
A fire cracker goes off at the other end of a football field from you. You hear the sound a few 100 ms later. The air molecules that were by the firecracker didn't end up by you. They didn't travel far at all. However, they pushed on their neighbors, which pushed on their neighbors, etc, all the way to your ears. This pushing can propagate a lot faster than individual molecules can move.
Think of a long hollow cardboard tube filled with small balls just a little smaller than the inside diameter of the tube. All the balls are touching each other. You push on one ball on one end and move it 1 mm. The ball at the other end then moves 1 mm. However, none of the balls themselves moved more than 1 mm and they did that as slowly as you pushed, yet the propagation of the push was instantaneous on your human scale.
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How does positive charge spread out in conductors? I know that when there are excess positive charges in a conductor, for example, a metal sphere, the positive charges will spread out over its surface. However, I am confused about how this excess charge spreads out over the surface, if protons cannot move and only electrons can move.
Can someone please inform me on how the excess positive charge spreads out over the surfaces of conductors?
| For a conductor to display positive charge it means that some of the electrons have been removed from it. The positive charge displayed homogeneously on the conductor's surface is the result of charge balance, between the positive stationary charges of the surface atoms and the negative charge of the electrons.
Because it is a conductor,the electrons are freely moving in a conduction band. and the charge difference will be uniformly distributed .
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Can a black hole bounce? Is there a limit to the amount of matter that a black hole can accrete per second and if so could a certain sized black hole bounce off a dense enough surface?
| Look, there's no limit to the accretion because the blackhole simply starts growing as it accretes more mass. I don't think it can "bounce off" any surface.
Yes, if the gravitational forces are comparable, for example, if it encounters another black-hole, they can get in equilibrium, bounce off each other or simply merge.
This happens in the universe.
Every galaxy is found to have a supermassive black-hole at the center. We have evidence for galaxies merging along with their black-holes. Sometimes the galaxies just tear each other apart from tidal influence and the black-holes simply don't interact.
Although, I have never heard of anything like bouncing or merging happening on garden-variety black-holes. The truth is, we haven't detected any significant sum of black-holes. All we have and are sure of are the black-holes at center of galaxies.
| {
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Cylindrical wave I know that a wave dependent of the radius (cylindrical symmetry), has a good a approximations as $$u(r,t)=\frac{a}{\sqrt{r}}[f(x-vt)+f(x+vt)]$$ when $r$ is big. I would like to know how to deduce that approximation from the wave equation, which is this (after making symmetry simplifications):
$$u_{tt}-v^2\left(u_{rr}+\frac{1}{r}u_r\right)=0$$
Proving that's a good approximation is easy (just plug it in the the equation), I want to know how to deduce that from the above equation.
I've been searching and I found this: http://vixra.org/abs/0908.0045, which actually solved me a couple of problems, but the way they do it looks a bit clumsy to me, saying for example that "assuming the function $g$ depends on $r$ so some terms just go away..."
Thanks in advance.
| Use the following identity:
$$ r^{-\alpha} \partial^2_{rr} \left( r^\alpha f(r) \right) = f_{rr} + \frac{2\alpha}{r} f_r + \frac{\alpha(\alpha - 1)}{r^2} f $$
Now, by inspection and comparing the above equation to the cylindrical wave equation you have that
$$ u_{tt} - \nu^2(u_{rr} + \frac1r u_r) = u_{tt} - \nu^2 \left( \frac{1}{\sqrt{r}}\partial^2_{rr} \left[ \sqrt{r} u\right] + \frac{1}{4r^{5/2}} \sqrt{r} u\right) $$
So writing $U = \sqrt{r} u$ we have that
$$ U_{tt} - \nu^2 U_{rr}+ \frac{\nu^2}{4r^2} U = 0 $$
So that $U = \sqrt{r} u$ solves the 1 dimensional wave equation up to a term that decays quickly (as inverse square). Hence $u$ is approximated by $1/\sqrt{r}$ times a solution of the 1 dimensional wave equation when $r$ is large.
| {
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Maxwell Stress Tensor in the absence of a magnetic field I'm having some trouble calculating the stress tensor in the case of a static electric field without a magnetic field. Following the derivation on Wikipedia,
*
*Start with Lorentz force:
$$\mathbf{F} = q(\mathbf{E} + \mathbf{v}\times\mathbf{B})$$
*Get force density
$$\mathbf{f} = \rho\mathbf{E} + \mathbf{J}\times\mathbf{B}$$
*Substitute using Maxwell's laws
$$\mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}$$
*Replace some curls and combine
$$\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right] + \frac{1}{\mu_0} \left[(\boldsymbol{\nabla}\cdot \mathbf{B} )\mathbf{B} + (\mathbf{B}\cdot\boldsymbol{\nabla}) \mathbf{B} \right] - \frac{1}{2} \boldsymbol{\nabla}\left(\epsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right)
- \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)$$
*Get the tensor
$$\sigma_{i j} = \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0} \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)$$
*Assuming B=0:
$$\sigma_{i j} = \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right)$$
*Assume flat surface with perpendicular field (z-direction)
$$\sigma_{z z} = \epsilon_0 \left(E^2 - \frac{1}{2} E^2\right)=\frac{\epsilon_0}{2} E^2$$
This is the formula given in e.g. The Feynman Lectures in Physics Vol. 2 (Page 31-14), and some other text books.
However, this derivation seems to assume a magnetic field until the final steps. Since most terms in eq. 4 result from the initial v x B term (even those that depend only on E, $(\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E}$ and $\frac{1}{2} \boldsymbol{\nabla}\epsilon_0 E^2$ ), these should not be present in my case, and in fact eq 4 should be as simple as
$$
\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E}\right]
$$
Tensor calculus is not my strong point. To me it is not clear how to get from eq 4 to eq 5, and how modifying eq 4 alters the resulting stress tensor. Will it really still be the same as eq 6? To me it seems strange that removing terms would not affect the result, yet this seems to be what many text books claim. Or is there some reason why the initial v x B term can not be removed, even when there is no magnetic field?
| The stress tensor is defined in such a way that taking its divergence brings you back to the force density (apart from the cross product term which corresponds to the time derivative of the Poynting vector). In the case of $\mathbf{B}=\mathbf{0}$, we take the divergence in index notation, i.e.
$$f_j=\partial_i\sigma_{ij}=\epsilon_0\left(\partial_i(E_iE_j)-\frac12\delta_{ij}\partial_i(E_kE_k)\right)=\epsilon_0\left(\partial_iE_iE_j+E_i\partial_iE_j-\frac12\partial_j(E_kE_k)\right).$$
In order to see that this expression corresponds to the first term in your expression 3., we recast this in index-free notation:
$$\mathbf{f}=\epsilon_0\left((\boldsymbol{\nabla}\cdot\mathbf{E})\mathbf{E}+(\mathbf{E}\cdot\boldsymbol{\nabla})\mathbf{E}-\frac12\boldsymbol{\nabla}(\mathbf{E}\cdot\mathbf{E})\right).$$
Making use of the identity
$$\frac{1}{2} \boldsymbol{\nabla} \left( \mathbf{A}\cdot\mathbf{A} \right) = \mathbf{A} \times (\boldsymbol{\nabla} \times \mathbf{A}) + (\mathbf{A} \cdot \boldsymbol{\nabla}) \mathbf{A}$$
and the Maxwell equation (for $\mathbf{B}=\mathbf{0}$)
$$\boldsymbol{\nabla}\times\mathbf{E}=\mathbf{0},$$
we find that
$$\mathbf{f}=\epsilon_0(\boldsymbol{\nabla}\cdot\mathbf{E})\mathbf{E}.$$
From this we can see that it does not matter at which stage of the derivation the magnetic field is omitted.
| {
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Why is there no time dilation for frequency of a wave? Since frequency of a wave is a function of time, then for a particular ray of the wave, why would the frequency remain the same when observed from a moving reference frame? The frequency should change according to time dilation. No?
| Both the the frequency and wavelength change. Obviously so because $\nu\lambda = c$ and $c$ is a constant. If you have a specific example of where you think the frequency doesn't change update your question with the details and I'll update my answer to address it.
| {
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How does my car acquire satellite signal? I just got XM radio again after a brief period without it. A customer service rep said that the satellite needs to beam my signal to me in a six minute time window. How exactly does this happen? How does the satellite know where to find me, and how exactly does it "talk" to my car? Why does it have a six minute time window?
I tagged this as radiometry, but I'm not quite sure what topic this is.
| It doesn't find you - XM uses three (IIRC) satellites in geostationary orbit as long as you are on the right continent you should get a signal.
I'm guessing that the six minutes is something to do with ensuring your receiver is authorized . Presumably it broadcasts some sort of key that unlocks your receiver when you pay - so you pay and within 6mins it will have sent the key to turn on your unit.
| {
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How to evaluate commutator with angular momentum? I need to evaluate the commutator $[\hat{x},\hat{L}_z]$. I believe the $L_z$ is referring to the angular momentum operator which is:
$L_z = xp_y - yp_x$
using this relationship i end up with:
$[x,L_z] = x(xp_y - yp_x)-(xp_y - yp_x)x$
my next step is substituting in for the p operator but i still dont get anywhere. any suggestions???
| I'll help you with a general solution. First, you can write the angular momentum as $L_{i}=\epsilon_{ijk}x_{j}p_{k}$. Where $e_{ijk}$ is the Levi-Civita symbol. Now write a general commutator as $[x_{l},L_{i}]$ where $l,i$ run from 1 to 3. With this you have
$[x_{l},L_{i}]=[x_{l},\epsilon_{ijk}x_{j}p_{k}]$
Now, you use the following identity for commutator $[A,BC]=[A,B]C+B[A,C]$. With this in mind, you get
$[x_{l},\epsilon_{ijk}x_{j}p_{k}]=\epsilon_{ijk}[x_{l},x_{j}]p_{k}+\epsilon_{ijk}x_{j}[x_{l},p_{k}]$
Now, the commutator $[x_{l},x_{j}]$ is zero and $[x_{l},p_{k}]$ is equal to $i\hbar\delta_{lk}$. So, you are left with
$[x_{l},\epsilon_{ijk}x_{j}p_{k}]=[x_{l},L_{i}]=i\hbar\epsilon_{ijk}x_{j}\delta_{lk}=i\hbar\epsilon_{ijl}x_{j}$
From this result you can find your particular commutator.
| {
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What is potential energy in special relativity? I know what is rest energy $E_0=m_0 c^2$, total energy $E=\gamma E_0$, kinetic energy $E- E_0=(\gamma-1) E_0$, and momentum $p=\gamma m_0 c$. But what is potential energy in special relativity?
| Special relativity doesn't alter the fact that interactions between particles "store energy" in the form of "potential energy," alhtough special relativity does alter the terms you listed, all of which have to do with the energies possessed by particles either by virtue of their motion, or their mass.
For example, in special relativity, electromagnetic interactions can be said to "store potential energy." When two charged particles interact via the Coulomb force for example, there is an interaction energy between them that deserves to be called potential energy just as much as in pre-relativity classical physics. In fact, classical electrodynamics exhibits Lorentz invariance and is in this sense a fully relativistic theory without alteration.
Any other form of energy that is called "potential energy" in a non-relativistic context probably also deserves this designation in a relativistic concept (I, at least, can't think of any counterexamples).
| {
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Relation between the determinants of metric tensors Recently I have started to study the classical theory of gravity. In Landau, Classical Theory of Field, paragraph 84 ("Distances and time intervals") , it is written
We also state that the determinanats $g$ and $\gamma$, formed respectively from the quantities $g_{ik}$ and $\gamma_{\alpha\beta}$ are related to one another by $$-g=g_{00}\gamma $$
I don't understand how this relation between the determinants of the metric tensors can be obtained. Could someone explain, or make some hint, or give a direction?
In this formulas $g_{ik}$ is the metric tensor of the four-dimensional space-time and $\gamma_{\alpha\beta}$ is the corresponding three-dimensional metric tensor of the space. These tensors are related to one another by the following formulas $$\gamma_{\alpha\beta}=(-g_{\alpha\beta}+\frac{g_{0\alpha}g_{0\beta}}{g_{00}})$$ $$\gamma^{\alpha\beta}=-g^{\alpha\beta}$$
Thanks a lot.
| *
*Consider the $4\times 4$ matrix $g_{\mu\nu}$ with zeroth row $g_{0\nu}$.
*Now for $i=1,2,3$, add to the $i$'th row the zeroth row times $-g_{i0}/g_{00}$.
*This produces the following matrix
$$\begin{bmatrix}
g_{00} & g_{01} & g_{02}& g_{03} \\
0 & -\gamma_{11} & -\gamma_{12}& -\gamma_{13} \\
0 & -\gamma_{21} & -\gamma_{22}& -\gamma_{23} \\
0 & -\gamma_{31} & -\gamma_{32}& -\gamma_{33}
\end{bmatrix}.$$
*Such row manipulations do not change the determinant. So it is still $g=\det(g_{\mu\nu})$.
*On the other hand the determinant can be expanded in the zeroth column to yield $g_{00}\times \det(-\gamma_{ij})$.
*Hence we obtain the result $g=-g_{00}\det(\gamma_{ij})$.
| {
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Parallel circuits - Overall resistance decreases with additional resistor Let's say that there is a parallel circuit with two identical resistors in parallel with each other. If a third resistor, identical to the other two, is added in parallel with the first two, the overall resistance decreases.
Why does this overall resistance decrease?
|
Why does this overall resistance decrease?
A more elegant, sophisticated way to see why is through the notion of duality.
In electric circuit theory, conductance (the reciprocal of resistance) is dual to resistance. Other dual pairs are:
voltage - current
series - parallel
inductance - capacitance
Thevenin - Norton
and so on ...
For example, consider Ohm's Law: $v = iR$. The dual is: $i = vG$
You probably intuitively understand that adding a resistor in series increases the total resistance.
The dual of this is adding a conductance in parallel increases the total conductance.
But, if the conductance increases, the reciprocal, i.e., the resistance, decreases.
Mathematically:
Conductances in parallel add:
$G_{total}=G_1 + G_2 + G_3 = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} = \dfrac{1}{R_{total}}$
or
$R_{total} = \dfrac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}$
Now, it's clear that adding another resistor in parallel, increases the denominator thus decreasing the total resistance.
| {
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spectral functions Please, I would like to understand why you call the function $A(k,\omega)$ (here :The Spectral Function in Many-Body Physics and its Relation to Quasiparticles ) a spectral function? For me, as a mathematician, a spectral function is a function which writes : $F(S)=f( \lambda(S))$ where for example $S$ is a symmetric matrix and $\lambda(S)$ is the vector of eigenvalues of $S$. Thank you in advance, Sincerely.
| If you perform spectroscopy on a material (be it angular resolved photoemission spectroscopy (ARPES) or scanning tunneling spectroscopy (STS) or whatever method you fancy), the quantity you measure is roughly related to $A(k,\omega)$ (with additional prefactors and matrix elements depending on your method of choice.
Thus, performing spectroscopy on a sample provides you with information on $A(k,\omega)$, and hence we call it the spectral function.
| {
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What is the effect of refractive index of an object for imaging? My Question is as follows.
What is the effect of refractive index of an object for imaging (Photographs by high speed camera) on its size and shape information incurred from image?
Lets say ,
I keep the camera focal length, aperture, distance between camera & object, light intensity of the background constant.
Put two objects of same diameter 'd' with different refractive index n1 and n2 in front of the camera one at a time.
Size incurred from image of those object is of diameter d1 & d2.
My question is will d1 & d2 be the same ? or it will differ? if it differs how can I co-relate analytically/ theoretically or by ray tracing projection?
| A while back I did an experiment imaging oil droplets in water, where we used oils of differing refractive index. Is this the sort of thing you're interested in?
If so, assuming your camera is in focus it will accurately record the size of the oil droplet so varying the refractive index will not cause the apparent size of the drop to change. However the difference between the refractive index of the oil and the refractive index of the water will affect the contrast.
To see this imagine using an oil with the same refractive index as water. Because there is no refractive index at the oil/water boundary there will be no reflection of light and therefore the oil drop will be invisible. As you increase the refractive index of the oil you increase the amount of light reflected at the edges of the drop and it becomes easier to see.
| {
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Absolute zero and Heisenberg uncertainty principle I got to read Volume I of Feynmann's lectures. It said that at absolute zero, molecular motion doesn't cease at all, because if that happens, we will be able to make precise determination of position and momentum of the atom. We do know that Heisenberg uncertainty principle holds for microscopic particles in motion. But what then is wrong to consider that all molecular motion ceases at absolute zero? In other words, does the uncertainty principle not hold when there is no motion?
Need some help!
| Motion does not cease at absolute zero if the system you are looking at has a zero point energy.
In many systems, e.g. crystals, at low temperatures the atoms/molecules behave as harmonic oscillators, and the energy of a harmonic oscillator cannot be reduced to zero: there is always some minimum energy called the zero point energy. This means that at absolute zero the atoms in a crystal will not be stationary. There will be a small vibration corresponding to the zero point energy. This is most obvious for light atoms like Helium where the zero point energy is enough to keep the system liquid, so even at absolute zero Helium will not solidify unless it's put under pressure.
The situation is different for a free particle. In that case, at absolute zero the momentum is zero but then we have no knowledge about where the particle is (i.e. $\Delta x = \infty$). If we want to measure where the particle is we have to put some energy in, but then of course the system is no longer at absolute zero and the momentum is now non-zero.
| {
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A thought on definition of momentum Well, this is a simple, basic and I think even silly doubt. The first time I saw the definition of momentum as $p = mv$ I started to think why this is a good definition. So I've read the beginning of Newton's Principia where he said that momentum is a measure of quantity of motion.
Well, this started to make sense: if there's more mass, there's more matter and so there's more movement going on. If also there's more velocity, the movement is greater. So it makes real sense that quantity of motion should be proportional both to mass and velocity.
The only one thing I've failed to grasp is: why the proportionality constant should be $1$? What's the reasoning behind setting $p = mv$ instead of $p = kmv$ for some constant $k$?
Thanks in advance. And really sorry if this doubt is to silly and basic to be posted here.
| It's a good question! Physics is all about linking your intuition to science, so it's good that you're thinking about this. The statement that momentum should be proportional to mass and velocity is intuition. Like elfmotat says, you can choose your constant as long as it's consistent with units, I guess.
If you want another reason, consider the time-derivative of the equation $p=mv$. It's Newton's Second Law! $F=ma$. In other words, $F=\frac{dp}{dt}$, which is a great reason for k to be 1.
| {
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Why is electrical energy so difficult to store? Does anyone know a general answer to these questions? (I've asked them together because they're all pretty related, it seems.)
*
*Why is it that we find electrical energy so difficult to store? Do we just find energy difficult to store generally? (...surely not, we can store energy in a block by sending it to the top of a hill.)
*is there something in particular about charge/electricity that makes effective batteries difficult to produce, and, if so, what?
*Is the problem that we're having with storing the energy just an artefact of our use of the energy, or is it difficult to store electrical energy per se?
| We need to create a battery that would instantly store a large amount of electricity at one time. Ex. When a bolt of lighting strikes it gives off a very large amount of power. However a battery needs time to take that energy and change it over to a chemical for storage.
Lets say a bolt of lightning is 500 gallons of water and the battery that we presently use is a 5 gallons bucket. If we take that 500 gallons of water and pour it fast into a 5 gallon bucket, it would instantly over pour over to the floor. Therefore we need a large enough bucket or a bolt of lightning in this case that battery that can hold all that engergy at one time
| {
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How to determine the direction of a wave propagation? In the textbook, it said a wave in the form $y(x, t) = A\cos(\omega t + \beta x + \varphi)$ propagates along negative $x$ direction and $y(x, t) = A\cos(\omega t - \beta x + \varphi)$ propagates along positive $x$ direction. This statement looks really confusing because when it says the wave is propagating along $\pm$ x direction, to my understanding, we can drop the time term and ignore the initial phase $\varphi$ while analyzing the direction, i.e. $y(x, 0) = A\cos(\pm\beta x)$, however, because of the symmetry of the cosine function, $\cos(\beta x)\equiv \cos(-\beta x)$, so how can we determine the direction of propagation from that?
I know my reasoning must be incorrect but I don't know how to determine the direction. So if we don't go over the math, how to figure out the direction of propagation from the physical point of view? Why $-\beta x$ corresponding to the propagation on positive x direction but not the opposite?
| Here is (in my opinion) an easier answer. I'm sure you know that the line y = x crosses the x-axis in the origin (0, 0). Now we shift the entire line to the right in such a way that it crosses the x-axis in the point (1, 0). The consequence of this is that the function of the line changes. It becomes y = x - 1. Here is that minus-sign again. The entire line is shifted to the right because you have subtracted $1$ from all the input values. It's the same thing for a sine function, although the input is in radians then. The -1 in y = x - 1 has the same effect as - w.t in the sine function. In wave functions we call it a phase-shift. This can be a time-dependent phase shift (+/- $\omega t$) and/or a constant phase shift ($\phi$). Easy as that ;-)
| {
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Proof that flux through a surface is independent of the inner objects' arrangement $$\Phi=\iint_{\partial V}\mathbf{g} \cdot d \mathbf{A}=-4 \pi G M$$
Essentially, why is $\Phi$ independent of the distribution of mass inside the surface $\partial V$, and the shape of surface $\partial V$? That is, I'm looking for a mathematical justification for the characterization of flux as a kind of "flow" of gravitational field through a surface.
The reason I worry about that is this. I have encountered the following proof of Gauss' Law for gravity:
$$\iint_{\partial V}\frac{-MG}{r^2}\mathbf{e_r} \cdot \mathbf{e_r}d A=\frac{-MG}{r^2}\iint_{\partial V}d A$$ $$=-MG4 \pi$$
However, two assumptions are made in the above proof that I've not convinced myself hold true for the general case:
*
*The Gaussian surface is a sphere.
*It is assumed that all mass is concentrated at the centre of the sphere.
Returning to my original question:
*
*Why is it true that when the Gaussian surface is deformed (not adding or removing any mass to inside the surface, of course), the amount of flux, $\Phi$, through the surface is equal to the flux in the spherical case?
*Why is it true that the positions of the masses inside the surface doesn't affect $\Phi$?
Edit: removed reference to Poisson's equation, changed much of question to make it clearer.
Further edit: I am happy that $$\iint_{\partial V}\mathbf{F}\cdot d \mathbf{A}=\iiint_{ V}( \nabla \cdot \mathbf{F})dV$$
However, my problem is with the following equation:
$$\nabla \cdot \mathbf{g}=-\rho G 4 \pi $$
Which can be used to get
$$\iint_{\partial V}\mathbf{g}\cdot d \mathbf{A}=\iiint_{ V}(-\rho G 4 \pi)dV=-M G 4 \pi$$
| Its because the divergence is undefined at the origin and 0 everywhere else. Thus as long as the volume you integrate over contains the origin, it doesn't matter what region you integrate over.
| {
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Calculating the mass equivalency of a song? I've recently become fascinated with the idea of sound energy having a theoretical equivalent mass. I've read over this thread:
Do light and sound waves have mass
I understand this part: $m_{eq}=E/c^2$ and $E=A\rho \xi^2\omega^2$
Where I am getting tripped up is the way to measure $E$.
Essentially, I want to measure $E$ and eventually $m$ of a song (for it's entire duration). The only equipment I have is an iPhone app that measures intensity (dB) (apparently it's fairly accurate for levels below 100 dB). I plan to play the "music" at about 80dB over speakers in a room that is 50'x30'x15' and I will assume the temperature is at room temperature. If your wondering, it's for a conceptual audio art piece.
| You could also set an upper bound if you know the phone's battery capacity (usually written on the battery) and how long you get out of a full charge playing the song on repeat. You can estimate the energy required to play the song by
$$ \text{Energy per song} = \text{Battery capacity} \times \frac{\text{Song length}}{\text{Battery life}}$$
This is an upper limit because because, obviously, not all of the battery power goes into playing the song. In fact most will go into other things like wifi and heat. If you want a more accurate estimate you can turn off as many other things as possible. If waiting for the phone to die isn't doable then you could run down say 30% of the battery and correct for that, but discharge curves aren't necessarily linear.
| {
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Am I properly identifying the resistor terminal voltage?
The frequency applied to a circuit of voltage 120 V with a real coil and a resistor has a value of 50 Hz. The resistance of the resistor is 10 $\Omega$. The voltage at the resistor terminals $u_1=60V$. The voltage at the coil's terminals is $u_2=90V$.
$$\nu=50\text{ Hz}$$
$$R=10\ {\Omega }$$
$$U=120\text{ V}$$
$$u_1=60\text{ V - resistor terminal voltage}$$
$$u_2=90\text{ V - coil terminal voltage}$$
Find:
*
*the intensity of the current $I$.
*the parameters of the coil.
I think by "parameters of the coil", it is meant the resistance and the impedance of the coil.
$$I=?$$
$$L=?$$
$$R_L=?$$
I've been trying a bit, but I am quite poor at physics. This problem is suggested in a book, I want to prepare for a testpaper.
What I've been thinking of is to calculate $$\cos\phi=\frac{U_r }{U}=\frac{60}{120}=\frac{1}{2}$$ $$\implies \phi=\pi/3$$ But I am not sure if the resistor terminal voltage is the same thing with $U_r$. Is this right?
| Ok , it was a nice one.!
Let's consider the real inductor as a pure inductor plus resistance pair.
the value of I(rms) can be get using the Pure resistor outside this pair,
U=I(rms)*R2 ; r2=10;
I(rms)=6;
Now potential drop across real inductor Ulr (90) , across pure inductor Ul and across resistance Ur are pythagorean triplets.
Ulr^2=Ul^2+Ur^2;
We get
225=R^2+(50L)^2 ; I=6A;
Now , see circuit as a whole;
potential drop is ==> sqrt[(10+R)^2 + (50L)^2 ] 6 ; ie. ZI=120
we get => 100 + R^2 + 2R + (50L)^2 = 400;
use the two equations to get R and L
| {
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Weak isospin and types of weak charge My understanding is that QCD has three color charges that are conserved as a result of global SU(3) invariance. What about SU(2) weak? Does it have two types of charges? What I'm getting at is:
U(1) --> 1 type of charge
SU(2) --> ?
SU(3) --> 3 types of charge
Does SU(2) have two types? If not, what is the relation between SU(N) invariance and the number of charge types?
Idea: Maybe both I and I_3 (weak isospin and its third component) are conserved before electroweak symmetry breaking? Is that true? If so, then that would answer my question.
| @Michael Brown is right. The SM has 12 exactly conserved currents.
*
*All local invariances, a fortiori also imply global invariances, if you ignore (for the sake of argument) the spacetime variability of transformation parameters/angles. So SU(3) has 8, not 3 conserved charges, RG, BG, .... The group has 8 generators. Likewise, SU(2) has 3, not 2 conserved currents: you know this from spin, where each of the 3 projections of a rotationally invariant system is conserved. U(1) has one conserved charge.
*SSB does not affect the number of conserved currents, in sharp contrast to explicit symmetry breaking: The currents are still conserved, except they have a special nonlinear form (their leading term is linear, not bilinear in the fields, so the corresponding charges shift the fields "nonlinearly"). The symmetry is hidden, and much less apparent, but it is still there, which is why these symmetries are so powerful: they control systematically the divergences of the corresponding QFT. (Actually, though, the 3 charges corresponding to the 3 broken generators are ill-defined/divergent themselves, although their corresponding currents are conserved: the symmetry is still there.)
*There are further approximate symmetries in the SM, meaning that their charges are violated by a "small" amount (a technical characterization), or even quantum anomalies (collective quantum action of the Dirac sea of fermions coupling chirally to them).
Further see 149324
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Inertial Frames of Reference - Description of an Inertial Frame of Reference An inertial frame of reference is described as being a frame of reference in which the first law of Newton (the law of inertia) holds. This means that all events as described with respect to this frame of reference must have a zero net force acting on it and therefore traces a straight line with a uniform non-translatory motion. But, I have read in some books, especially "Introduction to Special Relativity" by the well-known Robert Resnick, wherein his definition of an inertial frame of reference also refers to such a frame of reference as being an unaccelerated system. This is where I am confused.
How can we describe a frame of reference as being unaccelerated if we occupy the frame of reference itself? No mechanical experiment conducted solely confined to a single frame of reference can determine the absolute motion of the frame of reference relative to another frame of reference. All that can be understood is that there is a certain uniform relative motion between frames of reference and no more. Is Robert Resnick saying that the inertial frame of reference is unaccelerated with reference to another frame of reference?
|
all events as described with respect to this frame of reference must have a zero net force acting on it
This is wrong: there may exist nonzero forces in an inertial frame of reference. These have to obey Newton's laws. If the forces do not obey Newton's laws in your frame of reference, then it is not an inertial frame.
Newton's laws are correct only in an inertial frame of reference. What is an inertial frame of reference? One in which Newton's laws are correct.
| {
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Number of configurations of the universe I have read that quantum mechanics says that the amount of possible particle configurations is $10^{10^{122}}$ to be exact in the universe. Do we know this figure to be exactly true to the exact figure? Wouldn't we need to know a true theory of quantum gravity to know the exact answer? Is the amount exactly that figure or just an estimate?
| This is clearly an estimate, not an exact figure. It comes from ideas about quantum gravity (not proven but very strong conjecture) that say that the maximum entropy of a region is
$$ S = \frac{A}{4 \ell_P^2}, $$
where $A$ is the area of a surface bounding the region and $\ell_P \approx 10^{-35}\ \mathrm{m}$ is the Planck length.
Now entropy is a measure of the number of configurations available to a system (units where $k_B=1$):
$$ S = \ln \Omega. $$
Putting this together with the observed size of the universe for $A$ gives roughly the figure you mention.
| {
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does a rotating moving body in "flat" space curve its path because of frame dragging? I am not a physicist.
let's say we have a space with an object in it, where all other gravitational bodies are so far away that their affect on the shape of the space is negligible.
let's say the object is moving, and rotating such that we can say (for simplicity's sake) that it is moving to the right, and rotating clockwise (so we are considering rotation around the z axis, and movement in the x axis). therefore the +y side of the object is moving faster than the -y side of the object relative to the underlying space.
it seems to me that this object should experience frame-dragging between the body and the underlying space and so curve its path such that it appears to curve very very very slightly upwards in y.
is this correct?
| The object doesn't move relative to fixed space--there is no fixed space per special relativity--so forget that. The question is "why is there frame dragging?": because there is normal Newtonian-like gravitational attraction, and you need the frame-dragging field to make the whole thing relativistically invariant--that is, consistent for all inertial observers. I mean this in the same sense that the electric field needs a magnetic field to build a fully Lorentz invariant theory.
| {
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Electronic filter Can you explain, please, step-by-step how an electronic filter does work?
For example, high pass filter. I know It's a trivial things, but I can't get it completely. Don't bring me formula and etc. Just explanation in three words.
| A SIMPLE HIGH PASS FILTER: T-CONFIGURATION - no maths
For a high pass filter you can have a resistor with resistance $R$, and two capacitors with capacitance $C$, with the capacitors connected to the resistor which is grounded at one end. When a signal is entering the filter from the left, the high harmonics in the signal will pass through the capacitor on the left and close circuit through the resistor, no problem. The high frequencies will also pass the capacitor on the right. However the low frequencies that are below the lower cut-off frequency, $1/2\pi RC$, will encounter a very large resistance from the two capacitors and the circuit will be as if it is an open circuit for these low frequencies. Similarly, one could use two resistors and one inductance instead. The principle is very similar. I hope this helps you a bit.
| {
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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?
| Electron and hole generation takes place when a bond breaks.
Average energy of these particles under thermal equilibrium is $KT=0.026\,eV$.So when a semiconductor block sits at room temperature its average energy is KT. But to break a Si-Si bond we need 1.1eV of energy. So if we want to break a silicon bond we have to take* n* particles to converge at Si=Si bond so that it generates $1.1\,eV$ and breaks the bond.
$$1.1\,eV=n \cdot KT \Rightarrow n=42 (Particles)$$
So we need 42 particles to break a bond. Obviously the probability($p^{42}$) of this happening is very small. This explains why why a very small fraction of silicon atom actually contribute to e-h pair at $300k$. As the temperature rise $KT$ increases and hence probability of particle converging to a single point increases.
In simpler terms we need electrons and holes(for semiconductors) for conductivity. EHP generation takes place when energy is equal to $1.1\,eV$(for Si). Probability changes as we raise the temperature. So temperature is the main factor for conductivity in an intrinsic semiconductor.
| {
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Weyl Ordering Rule While studying Path Integrals in Quantum Mechanics I have found that [Srednicki: Eqn. no. 6.6] the quantum Hamiltonian $\hat{H}(\hat{P},\hat{Q})$ can be given in terms of the classical Hamiltonian $H(p,q)$ by
$$\hat{H}(\hat{P},\hat{Q}) \equiv \int {dx\over2\pi}\,{dk\over2\pi}\,
e^{ix\hat{P} + ik\hat{Q}} \int dp\,dq\,e^{-ixp-ikq}\,H(p,q)\; \tag{6.6}$$
if we adopt the Weyl ordering.
How can I derive this equation?
| The basic Weyl ordering property generating all the Weyl ordering identities for polynomial functions is:
$$
((sq+tp)^n)_W = (sQ+tP)^n
$$
$(q, p)$ are the commuting phase space variables, $(Q, P)$ are the corresponding noncommuting operators (satisfying $[Q,P] = i\hbar $).
For example for n = 2, the identity coming from the coefficient for the
$st$ term is the known basic Weyl ordering identity:
$$
(qp)_W = \frac{1}{2}(QP+PQ)
$$
By choosing the classical Hamiltonian as $h(p,q) = (sq+tp)^n$ and carefully performing the Fourier and inverse Fourier transforms, we obtain the Weyl identity:
$$
\int {dx\over2\pi}{dk\over2\pi} e^{ixP + ikQ} \int dpdqe^{-ixp-ikq} (sq+tp)^n =(sQ+tP)^n
$$
The Fourier integral can be solved after the change of variables:
$$
l = sq+tp, m = tq-sp
$$
and using the identity
$$
\int dl e^{-iul} l^n =2 \pi \frac{\partial^n}{\partial v^n} \delta_D(v)|_{v=u}
$$
Where $\delta_D$ is the Dirac delta function.
| {
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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?
| Light is an electromagnetic wave. When light arrives at a solar sail the electric and magnetic fields that make up the light accelerate the electrically charged partciles in the atoms of the sail, after all, that's what electric and magnetic fields do. If these particles are accelerated their momentum is changed and so for momentum to be conserved, that momentum must come from somewhere. It must come from the light. Computing the details of which direction those forces act in requires some mathematics, but I hope this will convince you that in principle light can 'push' something.
So ultimately a solar sail works for exactly the same reason that a magnet can push another magnet, because charged particles in electric and magnetic fields experience a force.
| {
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Cantilever Beam - Maximum Shear of the Beam
A cantilever beam $3\ \text{m}$ long is subjected to a moment of $10\ \text{kNm}$ at the free end. Find the maximum shear of the beam.
The answer is "There is no vertical load, shear is zero"
How come? The shear should be sliding upward at the end of the wall and downward to the free end.
| The answer is "no vertical load, the shear is zero" because the moment on the end of the cantilever beam only creates a bending moment on the beam which results in only a normal force on the beam and no shear force.
| {
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How can a massless boson (Gluon) mediate the short range Strong Force? I thought massless particles were mediators for long range forces such as electromagnetism and gravitation. How can the massless gluon mediate the short range strong force?
| Unlike photons, gluons carry the "charge" of the strong force (confusingly, it's called "color"). This means that, unlike photons, gluons interact with each other. The effect is that, rather than spreading out in all directions (as photons do), gluons tend to stick together and form strings. For example, two quarks (which have color) are not connected by a spread-out field, but by a string. The effect is that the force between the quarks doesn't weaken with distance, but is approximately constant. For an imperfect analogy, think of two balls connected by a rubber band.
Now consider trying to pull two quarks apart. Because the force (the tension in the "string") is roughly constant, the energy required is proportional to the separation between the quarks. (Work equals force times distance.) The tension is huge, about 50 tons (!! yes !!), so it requires an enormous amount of energy to separate the two quarks even by the size of an atomic nucleus. Thus, every object we see that is significantly larger than a nucleus is color neutral (no strings attached ;). Therefore, no long range forces - all because gluons aren't color neutral.
| {
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kinetic energy of the stone Suppose we have a man traveling in an open car (roof open) with speed $v$ towards right (man faces right). He throws a stone (mass $m$) towards right, in his frame-forward with speed $V$.
In the car's frame, the total energy imparted to it was $$E=E_f-E_i=\frac{1}{2}mV^2$$
In the ground frame, the total energy given to it was $$E=E_f-E_i=\frac{1}{2}m((v+V)^2-v^2)=\frac{1}{2}m(V^2+2Vv)$$
What part am I getting wrong on? Is it okay to get such difference?
| You are not wrong. The only thing that may be wrong, is the interpretation. Energy is a conserved quantity, but only within an inertial reference frame. If you change to another reference frame, the energy will also be different. This is what you calculated.
Suppose you throw two rocks with the same velocity $V$. The first one, you throw at a target moving alongside the car. The second one you throw at a stagnant target by the side of the road. Which one will take more damage?
Brain teaser: Compare a car crashing at a parked car with velocity $V$ to a heads-on collision between to cars driving at velocity $\frac{1}{2}V$. Is this the same collision? (Extensive answers can be found elsewhere on this site)
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Out of phase voltage current source and resulting power Examining the following graph, I am trying to understand the power plot. The power appears to take on a negative value when the current changes direction or the voltage changes polarity. Negative power does not make sense to me. In a proper representation of the power curve shouldn't the power remain above the dotted line?
Additionally, since the power is a product of voltage and current, I would think that the power plot should be a maximum value during where the plots of the voltage and current intersect, and should be greater in magnitude the the plots of the voltage and current.
| Now to answer your question, first I want to point out what $V$ and $I$ should be representing. The negative values of these functions represents the change in the direction of either the electric field $E$, or the current density $J$, which is a vector with a magnitude of current per unit volume.
Now if you think about these things in this way, then negative power starts making sense, as it just means, that the power flux (the propagation of energy) direction reverses at some point and starts travelling to the opposite direction than the one you arbitrarily define.
Hope this is clear enough.
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Why does tea rise in the pot but water doesn't? I was wondering when I boil water in a pot it only shakes too much while boiling. But I could not figure out why tea rises in the pot when we boil it. It is also a liquid but it starts rising up till it jumps out of the pot and tries to kill the flame.
I think this happens due to change in the adhesive properties of the water due to addition of milk, sugar and tea leaves to it. Am I right? if not what the reason behind it?
| The phenomenon at play here is foaming. Due to the heat the water will start to boil. In normal water this will result in small bubbles that pop quickly. When you add milk, you create a mixture which has the ability to foam due to proteins and surfactants present in the milk. This foam will rise up the pan and eventually boil over.
A common misconception is the explanation for example on this website (which is about pasta, but the story in terms of physics/chemistry is the same) where they say that the surface tension of water is increased due to starches. This is plain wrong. You cannot increase the surface tension of water by adding stuff, you can only decrease it (see e.g this book by Israelachvilli).
| {
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Smaller Airy disk with another lens? Is it possible to reduce the airy disk size produced by one lens with another lens placed after the previous one? For example, parallel ray incident on first lens L1 (without aberration), then there is Airy disk on the back focal plane of L1, if image this pattern by another lens, how does the resulting Airy disk size change?
| Airy disk intensity is described by
$$I(r) = \left(
\frac{ 2 \cdot J_1(v) }{v}
\right)^2
\qquad \text{ with } v= \frac{2\pi}{\lambda}r\cdot NA
$$
The Bessel function $J_1(v)$ of first kind has a first minimum at
$$J_1(3.832) = 0$$
$$ \Rightarrow r_{\text{min}}=0.61\cdot \frac{\lambda}{NA}$$
At a given wavelength, the first minimum of the airy disk $r_{\text{min}}$ only depends on the numerical aperture NA. Imagine this NA as a capability to collect light.
Your proposal is to use a second lens to change the size of the Airy disk. A well designed second lens (diameter, focal length and position) is able to reduce the size of the Airy disk pattern.
In short terms: You can shink the size of the Airy disk, if your two-lens-system has a higher NA.
| {
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Conservation of Angular momentum in the dipole selection rules If the total angular momentum J of an atom is not changing during a dipole transition, where does the angular momentum for the photon come from?
| One must distinguish two conditions: whether the eigenvalue of $|\vec J|^2$, the squared total angular momentum, is changing; and whether the whole vector $\vec J$ is changing.
The latter is guaranteed in a dipole transition: one can't keep the whole vector constant. At most, you may satisfy the former condition: the length of $\vec J$ may stay constant so that the eigenvalue of $|\vec J|^2$ remains $J(J+1)\hbar^2$. But the whole vector will change. In quantum mechanics, we may only choose one more component to a maximum set of commuting observables. Aside from $|\vec J|^2$, we usually choose $J_z$, and the value of $J_z$ must change during the dipole transition because the photon is indeed carrying away some angular momentum.
One may change a vector by a nonzero amount even though its length may stay constant.
This notation is confusing because we should talk simply about $\vec L$ when we discuss the electric dipole transitions: the spin $\vec S$, the other term in $\vec J$, isn't interacting with the electromagnetic field in this approximation. If we do talk about the relevant part of the angular momentum only, $\vec L$, we will see that the selection rules actually do require $\Delta L = \pm 1$. There are lots of refinements for the selection rules you could be interested in but because of the rough appearance of your question, I won't go into details. See this summary table and the article around it for various selection rules.
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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?
| If the metal heats up the magnet above around $500^\circ\mathrm F$ the magnet will lose it magnetic characteristics, and will not regain those properties when it cools down.
| {
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Derivation of $ E=h\nu$ Is it possible to derive the relation $ E=h\nu$ from Schrodinger equation or the basic principles of quantum mechanics or is it something which is considered to be an axiom with no explanation?
| No. The Schrodinger equation tells you how the state of a quantum system evolves, this is not specific to photons, and cannot be used to derive any facts about them. I'm assuming that by "basic principles" you mean the postulates of quantum mechanics that hold for every quantum system (like Born's rule etc.) in which case the same goes for these.
You need a specific quantum model for the photon to even start talking about photons, and this will involve more specific physical input than just the basic principles of quantum mechanics.
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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?
| As long as there's a non-zero net force acting on the object, it will have a non-zero acceleration and therefore it will continuously change its velocity: $$\vec{F} = m\vec{a}.$$
In the case of gravity, the force is inversely proportional to the distance between the objects squared: $$\vec{F}_G = G\frac{m_1m_2}{r^2}\frac{\vec{r}}{r},$$ where $\vec{r}$ is the vector connecting the two objects and $r=|\vec{r}|$ its length. So the closer the objects are, the stronger the force or -equivalently- the acceleration. Notice that the acceleration is only zero if the objects are infinitely far apart. (I'm assuming no drag, let's only consider the gravitational force for simplicity)
If you pull the earth in the same direction the object is falling so that you maintain the same distance $r$ at all times the object will just keep falling with a constant acceleration. If you pull it faster, the distance will increase and the acceleration will therefore decrease, meaning the velocity of the object will increase more slowly than before. But it will never decrease. Pulling the earth more slowly will only decrease how much the acceleration would increase if you hadn't pulled, so again the velocity keeps increasing.
So to summarize, the object's velocity will always increase, unless you can get the distance to infinity, which should only take you about - an infinite amount of time. And even then you can only get to a constant velocity, never a decreasing one. You need a repulsive force for that (or additional attractive forces on the other side of the object).
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Third-order phase transition in Landau theory $F=\frac{a}{2}m^2+\frac{u}{4}m^4+\frac{v}{6}m^6-hm$, where $F$ is the free energy, $m$ is the order parameter, $h$ is the external field, $a=a_0(T-T_c)$, and $a_0>0,u>0$ and $v>0$.We know this free energy expansion describes a second-order phase transition. How to write down a free energy such that the transition is a third-order phase transition?
| Although third-order transitions are rare to see, but it is not hard to write down Landau theory for third-order transitions. Let $m$ be the order parameter, the free energy simply reads
$$F=a m^4+ b m^6+\cdots$$
with $b>0$, and $a$ is the driving parameter, which triggers a third-order transition at $a_c=0$. To verify, just calculate the saddle point from $\partial_mF=0$ so that
$$m=\left\{\begin{array}{ll}\sqrt{-2a/3b} & a<0,\\
0 & a>0.\end{array}\right.,$$
and hence
$$F=\left\{\begin{array}{ll} 4a^3/9b^2 & a<0,\\
0 & a>0.\end{array}\right.,$$
which obviously becomes singular in the third-order derivative $\partial_a^3 F$ at $a=0$. Following this line of thought, one can play with Landau theory, and write down
$$F=a m^{2(n-1)}+b m^{2n}+\cdots$$
for $n$th-order ($n\geq2$) phase transitions. But I don't think such theoretical construction really meaningful, because the vanishing lower order terms require fine-tuning of the model, and can hardly be realized physically.
However beyond Landau paradigm, third-order transitions can happen in topological quantum phase transitions. One known example is transition between 2D Chern-insulators, described by the following Hamiltonian
$$ H =\sum_{k} c_k^\dagger (\sin k_x\sigma_x+\sin k_y\sigma_y+(\cos k_x+\cos k_y-2+m)\sigma_z)c_k,$$
where $c_k$ is the electron operator of the momentum $k$. The topological mass $m$ is the driving parameter: $m>0$ ($m<0$) corresponds to the topological (trivial) phase, while $m=0$ is critical. It is easy to show that the free energy (at zero temperature) of the fermion system reads $F\simeq\int\mathrm{d}^2k\sqrt{k^2+m^2}\sim -|m|^3$, so $\partial_m^3 F$ is singular at $m=0$, and hence a third-order transition. During this transition, the gap of the Dirac cone (at $k=0$) closes and reopens, leading to $\pm1$ change in the Chern number of the occupied band, which reflects the underlying change of the band topology. However there is neither symmetry broken nor local order parameters involved in this transition, so it is an example of third-order transition that can not be described by the Landau theory.
| {
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"timestamp": "2023-03-29T00:00:00",
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Indistinguishable particles in quantum mechanics If you have two particles of the same species , Quantum mechanics says that $\Phi_{m_{1},x_{1},p_{1},m_{2},x_{2},p_{2}}=\alpha\Phi_{m_{2},x_{2},p_{2},m_{1},x_{1},p_{1}}$
But I don't understand why $\alpha$ doesn't depend on $x$ , $p$ . If $\alpha$ depends on $SO(3)$ invariants as $x^2 , x.p , p^2$ etc then it will be the same on all reference frame why does one require that it doesn't depend on these variables? Even if it depends on $p ,x$ $\alpha$ is a phase factor so it doesn't affect anything why should this be important?
EDIT : I figured out the answer to the second question , for $\alpha$ is a complex number that carries no indices so it cannot change
| Let us step back for a moment to answer your question.
We consider a system of $n$ indistinguishable particles. What does that mean ?
Let $S_n$ be the set of all permutations of $n$ elements, and let $\sigma \in S_n$. Then if $P(\sigma)$ is the (unitary) operator representing $\sigma$ on the $n$-particles Hilbert space $\cal H$, the property of "indistinguishability" means that the two vectors $|\psi\rangle$ and $P(\sigma)|\psi\rangle$ represent the same physical state, and this should be true for any state $|\psi\rangle \in {\cal H}$. In other words, we must have
$$
P(\sigma) = e^{i \chi(\sigma)} {\mathbf 1}
$$
where $\chi(\sigma)$ is some real number.
If I understand your question correctly, what you ask is why couldn't we have the number $\chi(\sigma)$ to be an operator depending on the momentum operator $\hat {\vec P}$ or the position operator $\hat {\vec R}$ (for example). But obviously $P(\sigma)$ would not be proportional to the identity operator $\mathbf 1$, violating the above conclusion.
I hope this help !
| {
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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!
| To answer the question simply, $E=mc^2$.
Energy is a manifestation of mass, and mass is a manifestation of energy. In a fusion or fission process, the total "energy" of the system remains constant, it just changes shape.
By "energy" I mean the totality of the already present energy, and the bound energy of the mass that takes part in the reaction.
| {
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Elastic collision of rotating bodies How would you explain in detail elastic collision of two rotating bodies to someone with basic understanding of classical mechanics?
I'm writing simple physics engine, but now only simulating non-rotating spheres and would like to step it up a bit.
So what reading do you recommend so I could understand what exactly is happening when two spheres or boxes collide (perfectly in 2 dimensions)?
| I worked on a physics engine written in C# that does just this.
Here are my notes on this topic.
Objects have both translational and rotational momentum.
When two objects collide, the overall algorithm goes like this:
1> Find the total momentum of both objects. Calculate the translational and rotational momentum, the vector sum of this is the total momentum of the object.
2> Split the momentum using the usual momentum splitting equation you would ordinarily use. (As in here)
Each object now has their new momentum. The next step is to decide how much of that momentum is translational and rotational.
3> Imagine a vector A which goes from the point of collision to the center of mass of the object that was hit. The component of the incoming momentum vector which is parallel with A forms the new translational momentum vector, the rest of the vector represents rotational momentum.
The extra notes I have linked to show more details on my methematical working, and also a description of how to handle inelastic collisions.
You can find the physics engine here, and an implementation of the collision handling here
| {
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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?
| The law of energy conservation certainly also applies here. What is not reflected will be absorbed. R+A=1 has to hold. And according to Kirchoff's law E=A (absorptivity is emissivity) you could write R=1-E.
If you have to consider transmission through the glass as well, then R+A+T=1 would hold.
You do not specify what type of glass you are talking about, what wavelength (or wavelength band) and what incident angle you are talking.
E.g. for window glass you will have approx. 10% reflection and 90% emissivity at room-temperature radiation (brightness temperature) and normal incidence (=0° =perpend. to surface).
For incidence angles greater than 81° you will get 50% reflection and 50% emissivity.
If you state that glass has a high IR-reflectivity then you might think of lowE-coated glass as used in insulated glazing units. In this case you will have to consider all coated and uncoated layers of all windows pane. But again for each layer the law of energy conservation will apply.
You can find a lot read more about the emissivity and reflectivity of window glass and all relevant dependences in my open-access publication.
| {
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"timestamp": "2023-03-29T00:00:00",
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A water balloon in vacuum: does it boil? If I put water in a vacuum it will boil. But what if I put this water inside a balloon ?
I searched for answers and fount this video: http://www.youtube.com/watch?v=9q8F3ClUuV0
It appears that the water isn't boiling but I am still not satisfied. The vacuum on the video could be just not enough to make the water boil at its temperature. Or it might be because of the pressure of the balloon.
If I heat up a water balloon over 100°C at atmospheric pressure, it will boil and grow. So why not at 20°C in a vacuum ?
| Water boils when the pressure is less than its vapour pressure (there is a table of vapour pressure vs temperature here).
At 20ºC the vapour pressure is 2339Pa, so if your balloon exerts a pressure greater than this the water won't boil. If the pressure exerted by the ballon is less than this, the water will start to boil and the steam generated will inflate the balloon. This will increase the pressure until the pressure has risen to 2339Pa, and at this point the water will stop boiling and you're left with water and steam in equilibrium.
If you heat water to 100ºC the vapour pressure rises to 1 atm (101325Pa), which is of course why water boils at 100ºC at sea level. However if you have a really strong balloon capable of exerting a pressure of greater than 1 atm the water won't boil at 100ºC even in a vacuum, and you'd need to raise the temperature above 100ºC to make it boil.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/59943",
"timestamp": "2023-03-29T00:00:00",
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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...
| Yes, every observer who makes a local measurement of the speed of light will get the same result, i.e. $c$, regardless of where the light came from.
I've used the word local because in general relativity the speed of light can differ from $c$ if it's not measured at your location. The most famous example of this is probably the fact that light slows down to a halt as it approaches the event horizon of a black hole. However even in general relativity a local measurement always returns the value $c$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does a thermal temperature gun work? I once worked as a kitchen porter over a winter season.
We had fun with thermal temperature guns (like these) which I learned can be used for measuring the temperature of something a reasonable distance away (aside from the obvious use of laser tag), which to my mind is pretty impressive.
How do they work?
| They basically measure the intensity of the infrared blackbody radiation in some wavelength region and calculate the temperature needed to give that intensity according to Planck's law.
| {
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