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Why does rotation occur? So it sounds like a dumb question, as it is very intuitive why rotation occurs. However, can you give me a scientific explanation as to why whenever I exert a force on a body it tends to move, but when it is tied to something or I am pushing it through the edge of an object, it rotates? What is there in a body which lets it rotate in place of simply moving?
| It's implicit in Newton's second and third laws together with one crucial assumption.
Consider the simplest system - two point masses linked by a massless infinitely stiff tether. Now impart a force on one and write down the equations of motion for the point masses given the tether. You will find that the system's rotation falls naturally out of the equations.
Indeed, for an arbitrary system of point masses, you can define their total angular momentum about their center of mass and show that the time rate of change of this quantity is indeed given by the nett torque about the center of mass imparted to the system if you assume that the force on particle A from B is equal and opposite to that on B from A and that the forces between the particles always act along the line joining them. This is a specialized case of Euler's second law, and Newton was aware of its holding when the these assumptions hold further to his second law. Indeed, it was thought experimentation along these lines that helped led him to his third law.
So Newton's second and third law together with the assumption of forces along lines joining particles is equivalent to conservation of angular momentum for systems of particles.
Rigid body rotation is simply the special case of the above when the forces between constituent atoms become very large for very small strains, i.e. the body is stiff and can thus only possibly move by Euclidean isometries, i.e. by translation and rotation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding atmospheric pressure using temperature of boiling water I was checking the temperature inside my oven using a multimeter with a temperature probe. I had set the oven's thermostat to 190o C, and it had reached equilibrium, but the measured temperature was well below what I had set the thermostat for. Having not previously used this temperature probe (new meter!), I was wondering if my probe and multimeter were accurate. So I put a pot of water on to boil, to see how close it would be to 100o C. The boiling water turned out to be 98o C. This made me reasonably sure that my oven's thermostat was either wildly out of kilter, or incapable of heating to the desired temperature. So I baked the item longer than the recipe called for. Problem solved, except for wondering why the water was boiling at 98oC instead of 100oC.
I figured that it must have something to do with atmospheric pressure, and is probably also wrapped up with the elevation of my house above sea level. The question I have, then, is this: is there a way to calculate the atmospheric pressure from the temperature I measured, or alternatively, determine what the boiling point of water should be at this moment -- to determine how accurate my meter is? My house is 25 m above mean sea level, and the current atmospheric pressure is 1040 hPa.
| Steam Table says that the saturation temperature for Water @1040 hPa is 100.7°C. Conversely, the saturation pressure at 98°C is 943.9 hPa.
Saturated Steam Table
Check the accuracy of your temperature probe as well as that of your barometer.
For the connection of Temperature, Pressure and Elevation, see Barometric formula
| {
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Heisenberg equation of motion In the Heisenberg picture (using natural dimensions):
$$
O_H = e^{iHt}O_se^{-iHt}. \tag{1}
$$
If the Hamiltonian is independent of time then we can take a partial derivative of both sides with respect to time:
$$
\partial_t{O_H} = iHe^{iHt}O_se^{-iHt}+e^{iHt}\partial_tO_se^{-iHt}-e^{iHt}O_siHe^{-iHt}. \tag{2}
$$
Therefore,
$$
\partial_t{O_H} = i[H,O_H]+(\partial_tO_s)_H \, , \tag{3}
$$
but this is not equivalent to what many textbooks list as the Heisenberg equation of motion. Instead they state that
$$
\frac{d}{dt}{O_H} = i[H,O_H]+(\partial_tO_s)_H. \tag{4}
$$
Why, in general, is this true and not the former statement? Am I just being pedantic with my use of partial and total derivatives?
| With some definitions to make time dependences explicit, your equation (4) can be made sense of. Let's take the following:
Let $O_s$ be an operator depending on time and other parameters $O_s:\mathbb{R}\times S\rightarrow \mathrm{Op}$, where $S$ is the space of the other parameters and $\mathrm{Op}$ is the space of operators on the Hilbert space. Let $\phi:\mathbb{R}\times\mathrm{Op}\rightarrow\mathrm{Op}$ denote time evolution of operators in the Heisenberg picture, given by $\phi_t(O)=e^{iHt}Oe^{-iHt}$.
Note that $(\partial_t \phi)_t(O)=i[H,\phi_t(O)]$ and $\partial_O\phi=\phi$ (because $\phi$ is linear in $O$). Now, given a parameter $p\in S$ we can define the function of time: $O_H:\mathbb{R}\rightarrow \mathrm{Op}$ with $O_H(t)=\phi_t(O_s(t,p))$. Our function $O_H$ is a one-parameter one, so it only makes sense to take its total derivative:
\begin{align}
\frac{dO_H}{dt}(t)=&(\partial_t\phi)_t(O_s(t,p))+(\partial_O\phi)_t\left[(\partial_tO_s)(t,p)\right]\\ =&
i[H,\phi_t(O_s(t,p))]+\phi_t\left[(\partial_tO_s)(t,p)\right]\\=&
i[H,O_H(t)]+e^{iHt}(\partial_tO_s)(t,p)e^{-iHt},
\end{align}
where in the first step I have applied the chain rule and in the others, the equalities we already had.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evolution of neutrinos flavor states What do we mean by saying that neutrino flavor states do not satisfy the schrodinger equation? How does the time evolution of states look like?
| Because of the way neutrinos are (I'm going to try to keep this simple) the flavor eigenstates are related to the mass eigenstates via:
$$
\vec{\nu}_f = \bf{U} \vec{\nu}_m,
$$
where the vector $\vec{\nu_f}$ has flavor eigenstate components ($\nu_{\mu}$, $\nu_{e}$, $\nu_{\tau}$), the vector $\vec{\nu}_m$ has mass eigenstate components ($\nu_{1}$, $\nu_{2}$, $\nu_{3}$), and $\bf{U}$ is a unitary transformation matrix. This expression is telling us that flavor eigenstates can be expressed as a linear combination of mass eigenstates. These mass eigenstates are the ones that satisfy the schrodinger equation:
$$
\hat{H}\vec{\nu_m}=\vec{E}\vec{\nu_m},
$$
where the vector $\vec{E}$ has entries of the energy eigenvalues associated with $\vec{\nu_m}$.
Now if we have a neutrino that begins in a flavor eigenstate ($|{\nu_e}\rangle$ for example) then we can apply the time evolution operator, $\hat{U}(t)$,to find how the flavor eigenstate evolves in time, i.e.
$$
|{\nu(t)}\rangle = \hat{U}(t)|{\nu_e}
\rangle \;.
$$
For simplicity let this be a free theory (no potential). Now because $\hat{U}(t) = e^{iHt}$ we will expand $|{\nu_e}\rangle$ into the linear superposition of mass eigenstates and find that:
$$
|{\nu(t)}\rangle = Ae^{iE_1t}|\nu_1
\rangle+Be^{iE_2t}|\nu_2
\rangle+Ce^{iE_3t}|\nu_3\rangle \;,
$$
where the (1,2,3) subscript denotes the mass eigenstates and the constants $(A,B,C)$ are given by the unitary transformation matrix $\bf{U}$ entries. This is clearly not an eigenstate of the Hamiltonian! Let me know if you want me to expand upon any of these ideas
| {
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Why do we assume weight acts through the center of mass? The weight of a body acts through the center of mass of the body. If every particle of the body is attracted by earth, then why do we assume that the weight acts through the center of mass? I know that this is true but I can't understand it. Does it mean that the earth does not attract other the other particles of the body ? Wouldn't it mean that girders would not need any support at the periphery if we erect a pillar at the center?
| Force that acts on the center of mass does not exert any torque on an
extended body. So, gravity 'acting on the center of mass' means a force
that accelerates, but does not rotate, its target.
There is tidal torque on
the Earth by the Moon, but this is because the Earth is not a rigid body, and
changes shape (and weight distribution) with tides. That means that the
Earth is effectively polarized by a gravity field, and that polarization
(tidal lobes) slows the Earth while adding to the Moon's orbital angular
momentum. This is not due to gravity directly, but due to the time-dependence of
Earth's shape change (it isn't a time-reversible effect, though a
conservative force field, like gravity, is).
There is clearly torque, too, in a Cavendish apparatus Cavendish torsion balance where the two objects are designed to act against a torsion spring by gravity force. So, the claim of action on the center of mass is sometimes false.
One can argue, of course, that a point mass exerts no
torque on a rigid object, because no equal-and-opposite torque can
meaningfully be exerted on the point object by gravity. It is difficult,
though, to generalize that argument.
| {
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How do Faddeev-Popov (FP) ghosts help path integrals? How does the inclusion of Faddeev-Popov ghosts in a path integral help to fix the problem of over counting due to gauge symmetries?
So, after exponentiating the determinant for the inclusion of either anti-commuting or bosonic variables and the corresponding extension to a superspace theory... why does that solve the problem exactly?
| In a nutshell, the Faddeev-Popov (FP) determinant (and its integral formulation via FP ghost variables) can be viewed as a compensating factor in the path integral $Z$ to ensure that the path integral $Z$ does not depend on the choice of gauge-fixing condition. See also this Phys.SE post.
For a simple gauge theory (like e.g. QED), it is not necessary to introduce FP ghosts. However, for more complicated gauge theories, it becomes convenient to introduce FP ghosts explicitly, and possibly to encode the gauge symmetry in a BRST formulation (such as, e.g., the Batalin-Vilkovisky (BV) formulation).
In fact, the action $S$ may for a non-trivial gauge theory in principle depend non-quadratically on the FP ghosts so that FP action terms have no simple interpretation via a determinant.
The BV formulation may in general be used to provide a formal proof that the gauge-fixed path integral $Z$ does not depend on gauge choice.
| {
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Independence of coordinates in the Hamiltonian formulation I was told the following statement
It is often true that the Lagrangian method leads more readily to the EoMs than does the Hamiltonian method. But because we have greater freedom in choosing the variable in the Hamiltonian formulation of a problem (the $q$ and the $p$ are independent, whereas the $q$ and the derivative of $q$ are not).
My question is the $p$ is dependent on the derivative of $q$ and the derivative of $q$ is depend on $q$, then why do we say that the $q$ and the $p$ are independent and we have greater freedom in choosing the variable in the Hamiltonian formulation of a problem?
| Coordinates $q$ and momenta $p$ are independent variables in Hamiltonian formalism. Dependence of $p$ on $\dot{q}$ appears after solving the Hamiltonian equations of motion
\begin{equation} \dot{q}^i=\frac{\partial H}{\partial p_i}.
\end{equation}
Doubling of independent variables in comparison with Lagrangian formulation is compensated by doubling of equations of motion.
We have greater freedom in choosing the variables because in Hamiltonian formalism we can perform change of variables of the form
\begin{equation} q \rightarrow q^{\prime} (q,p), \quad p \rightarrow p^{\prime}(q,p),
\end{equation}
if such a transformation preserves the canonical 1-form
\begin{equation} \sum\limits_i p_i \mathrm{d}q^i,
\end{equation}
while in Lagrangian formalism only $q \rightarrow q^{\prime} (q)$ transformations are allowed.
| {
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How matter waves travel faster than light? I read this in my physics textbook. It says that matter waves travel faster than light. Why is it so?
Also are matter waves G-Waves?
| In the original non-relativistic formulation of wave mechanics by Schrödinger, his waves did indeed travel faster than light, and in fact their velocity was infinite. The propagator which showed the influence of one space time point on another clearly allowed superliminal influences.
His theory had other problems as well: his waves were unstable under the motion of the observer. Even at a slow walk, a zero-momentum wave (flat) suddenly morphed into a real wave. These problems were probably among the reasons he initially decided not to publish the theory.
| {
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Can gravitons travel faster than the speed of light? It is common knowledge that not even light can escape black holes, but since the black hole is emitting gravitons, and the gravitons travel in the direction away from the black hole, and it is escaping successfully, are they faster than light?
| "Nothing can propagate faster than light" is a condition in GR.
Gravitons are not postulated by GR.
Hence the two (speed limit, and gravitons) are defined by two different theories and it would not be possible to compare things from two different theories unless we are able to combine the two theories, which has not happened yet.
Gravitons, or otherwise - gravity is the one that stops light from escaping the black hole. Same would not apply to gravity. Something like "gravity does not allow gravity to escape black hole", does not make any sense. If gravity stopped gravity to escape black hole, then nothing would experience gravity of a black hole from outside. Which is not true.
It is true that we can not get the information from inside event horizon in the form of light. However, in the form of gravity (or even in form of outer light), we do get the information about motion. A moving black hole tells us that whatever lies behind EH has changed its position in space with respect to other bodies.
Whatever lies behind EH, does influence things outside the EH, via gravity. It does cause change in the curvature of space continuously as it moves along. Something must be making out of the black hole - Something that bars everything else from doing so, and that must be gravity.
Even if gravity crosses the event horizon, it can not be said to travel faster than light.
| {
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How does a Galilean telescope form an enlarged image even though it has a diverging lens? I have been reading about Galilean telescope and the picture in the book is something like this:
After rays pass through the converging lens, there is a real image formed which is intercepted by the diverging lens but as I learnt before, diverging lens cannot form an enlarged image. So, is the ray diagram inaccurate?
| There is your answer, the Galilean telescope - the assembled telescope as a whole with EVERY component part of it in its correct place - magnifies.
But you did raise a concern in your question that a diverging (or concave) lens cannot form an enlarged image. That is also correct. Such a lens, standing alone, a single component not in a telescope and not in conjunction with another lens -indeed cannot magnify.
This is probably a poor simile, but a car wheel by itself cannot move, it will just sit where you happen to leave it. But put it in its correct place with other component parts, in this case called a "car", and that wheel suddenly has the capability of movement.
Sometimes, things can work only when they work together.
| {
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Why doesn't hydrogen gas exist in Earth's atmosphere? The root mean square velocity of hydrogen gas at room temperature is:
Gas constant: $R=8.31\ \mathrm{J\ K^{-1}\ mol^{-1}}$
Molar mass of hydrogen gas: $M=2.02\times10^{-3}\ \mathrm{kg/mol}$
$$\begin{align}
v &= \left(\frac{3\times8.31\ \mathrm{J\ K^{-1}\ mol^{-1}}\times300}{2.02\times10^{-3}\ \mathrm{kg/mol}}\right)^{\frac12}\\
&= 3356.8377\ \mathrm{m/s}\\
&= 3.356\ \mathrm{km/s}
\end{align}$$
The escape speed of Earth is $11.2\ \mathrm{km/s}$,
which is larger than the root mean square velocity of hydrogen gas.
But still, hydrogen gas doesn't exist in Earth's atmosphere. Why?
Have I made any mistakes in my calculations?
| Brief explanation in human words:
Hydrogen is a small atom so its mean velocity (3.4 km/s) would be much higher than air molecules (0.5km/s); some hydrogen atoms would be travelling faster than the escape velocity.
The escape velocity on the Earth is 11.2 km/s so you might expect the Earth to have hydrogen gas. However, 3.4 km/s is the average speed; a lot of the molecules would be travelling faster than this leading to a significant number escaping, and over time all would escape.
| {
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Why does stacking polarizers of the same angle still block more and more light? I have some sheets of polarization film. They came in a big box, all stacked at the same angle. I noticed that the entire stack of them lets almost no light through, even though they're all at the same angle.
I pulled out two, and those two also block more light than just one.
Why?
Is this because I have low-grade polarizers? Or because lining them up at EXACTLY the same angle is impossible? Or because the light that gets through the first one is not really polarized exactly to its angle — it's just that less of it is polarized away from its angle than before?
If it's because these are low-grade polarizers, can anyone recommend a linear polarizer that I can stack several of in a row at the same angle and still have a 100% probability of the light getting through?
I feel like I'm probably just misunderstanding polarization theory so please correct me.
| An ideal (theoretical) polarizer will only let in light along a certain axis.
This is impossible in real life. Any polarizer that you can purchase will let in light in a range of possible polarization angles. Additionally, manufacturing processes cannot guarantee perfect alignment with the apparatus holding the polarizer.
Therefore, when you are aligning them all up, you are not actually accomplishing that. They are still out of alignment, even if in small amounts.
| {
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Why is it so much easier to fall when walking down a slope? When I walk uphill on an icey road, I almost never slip/fall. But if I walk downhill or even on a relatively flat road, I slip/fall frequently.
What is the physics reason behind this, and are there any tips to reduce the chance of falling (other than shoe modifications and "being careful")?
| I think it is better to say that concept of torque is working here.
When you walk on floor, there is no force which can provide a torque to your body. But when you walk on a slope, a component of gravitational force provide you a torque which results in falling.
| {
"language": "en",
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How does the sun really produce light in terms of waves? Electromagnetic waves are caused by changing electric and magnetic fields, and these are caused by a charge possible oscillating like an antenna or a varying current etc.
My question is, with the sun, where is this source that causes the electric and magnetic fields to oscillate.
Everywhere I've read stated that it was due to the energy released from nuclear fusion, but when looking at the process of nuclear fusion there are no charges produced. How is this 'energy' supposedly producing the same effect as an electron oscillating?
| The Sun is made of plasma, which is a gas of bare ions and electrons. The energy released from nuclear fusion heats this plasma, and the moving charges emit electromagnetic waves.
Note that in other, colder radiating bodies (like a tungsten filament or a human body), the radiation is emitted by other means; in this case, electron transitions in atoms emit photons of various energy levels, and since each atom's energy level is affected by its environment, the discrete levels smooth out into what we call the blackbody curve.
Interestingly enough, despite the clear difference in mechanism at the molecular level, all of the systems I just described (including the Sun) follow the same approximate law: the blackbody curve! This is because the approximation in question is derived from statistical mechanics, which tends to ignore the exact mechanisms and motions at the microscopic level in favor of a macroscopic description.
| {
"language": "en",
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The position of an element on a periodic sound wave So I am looking at the equation used to locate of a small element relative to its equilibrium position on a periodic sound wave. The equation is defined as below:
$s(x, t) = A\cos(kx - wt)$
Now I understand why the use of a sinusoidal function, but the equation is expressed in terms of a Cosine and why not simply use a sin function? Is it just a convention or there is more?
| The general equation for the displacement could be
$$s(x, t) = A\cos(kx - wt)+B\sin(kx-wt)$$
It appears that the author has made a choice that $s(0,0) =A$ and $B=0$ which results in $s(x, t) = A\cos(kx - wt)$.
Others might have chosen $s(0,0) =0$ and $A=0$ which results in $s(x, t) = B\sin(kx - wt)$.
Another possibility is that a sound wave can be described in terms of a variation of pressure relative to atmospheric pressure (pressure wave) or a variation of position relative to an equilibrium position (displacement wave).
There is a $90^\circ$ phase difference between the pressure wave and the displacement wave and so the author might favour the pressure wave as a sine function which would result in the displacement wave being a cosine function.
| {
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Example of compact operators in quantum mechanics Can anyone give an non-trivial example of compact operators in quantum mechanics? Of course, any operator on a finite-dimensional Hilbert space is compact.
| All the normal quantum states of a given W* algebra of quantum observables are represented as compact (actually trace class) operators on a given Hilbert space (where the algebra of observables is represented).
In other words all the usual states of quantum mechanics in the Schroedinger representation, i.e. density matrices (both pure and mixed) are compact operators.
The Hamiltonians describing confined particles are not compact, but have compact resolvent. However one might argue that the resolvent of an observable is not strictly a physical quantity.
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Newton's Law of Cooling as boundary condition to Heat Equation I am reading a book on Partial Differential Equations, and they derived the 1 spatial dimension heat equation from scratch (using a rod as an exmaple with end points at $x=0$ and $x=L$) and then gave a couple possible boundary conditions with appropriate physical explanations of each. One boundary condition they used was Newtons Law of Cooling, expressed with this notation:
$$-u_x (0;t) = h(T_e - u(0;t))$$
$$u_x (L;t) = h(T_e - u(L;t)) $$
Where $u(x;t)$ is the temperature as a function of space and time, $u_x (x;t)$ denotes the partial derivative with respect to $x$, $T_e$ is the external temperature, and $h$ is a positive constant of proportionality.
I'm okay with the rate of change needing to be opposite signs for the two ends, as positive Flux is pointing in different directions on the two faces, but I feel as though the signs should be switched as they presented them. They stated:
"if $T_e$ is less than the endpoint temperature, then $u(0;t)>T_e$ and the rate of change will be negative, so that we may expect the heat to flow into the he rod from the hehe exterior$.
Mathematically, by the given equation, if $u(0;t) > T_e$, then $u_x (0;t)>0$, which would seem to imply the heat is flowing in ( I would think ), but physically, I dont see how this is possible. Wouldn't there want to be an equilibrium between the rod's end and the external temperature? This, to me, would suggest that heat would need to flow out of the rod, as the temperature is lower outside. I may be missing something vital here, and I would love an explanation!!
| In the situation where the temperature of the left end of the rod is less than the surrounding temperature then you got that the graph of the temperature should slant upwards in the right direction kind of like the in this graph
Remember that heat is always transferred from points with higher temperature to points with lower and this slant condition essentially guarantees that you will have a local from right to left transfer of heat at this point and since whatever heat moves left at an end point leaves the system this should indeed correspond to cooling.
| {
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Grashof number as a ratio of buoyant and viscous forces The Grashof number is supposed to be a ratio of buoyant forces to viscous forces.
I find this hard to believe, since if
$$F_b=\beta g \rho \Delta T$$
is the buoyancy force, the definition of the Grashof number,
$$\text{Gr}=\frac{\beta g\Delta T L^3}{\nu^2},$$
implies that the viscous force is something like $\frac{\rho}{L^3}\nu^2$, instead of something linear in $\nu$. How is this supposed to be the viscous force?
| you both are right. I think the closure point is that the velocity $U$ which is normally the velocity outside the boundary layer (BL) in the forced convection problem, here is defined in terms of the buoyancy force. This comes in when going from the dimensional form of the BL equations for natural convection to the non-dimensional form. Here,
$$u_o=[gβ(T_s-T_∞)L]^{1/2}$$
In order to clean the first term of the RHS of the momentum equation, see Page 565 of
Incropera F.P., Dewitt D. P., Bergman T. L., Lavine A. S., Fundamentals of Heat and Mass Transfer, John Wiley & Sons, 6th Ed, 2007.
So at the end of the day, although $U$ defines $Re$, true, this is a $Re$ number based on a $U$ velocity promoted by the buoyancy force.
Lastly, thank you guys. You help me to explain in detail this to my students.
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Optimal laser wavelength for heating air Lets say I want to heat air with laser,what wavelenght should I chose,ultraviolet,infrared or something in visible spectrum? To clarify,I want the laser beam to lose power and get converted to heat in shortest amount of distance possible. I am looking for maximum absorbtion, to convert the laser into heat.
When I think about it,two different wavelenght may produce equal heating of air,but the thing is one that have short range will produce heating that is more concentrated in space while other will heat air over longer distance so the energy will be spread over greater amount of air,I want that short range concetrated type heating so minimum quantity is heated to high as possible peak temperature.
| As you need to heat air to high temperature anyway, maybe you should consider a very high-power (pulse) laser that induces breakdown in air (https://www.rp-photonics.com/laser_induced_breakdown.html ); after the breakdown, air will be ionized and will absorb more power. Ideally, you should choose shorter-wavelength laser to be able to better focus the beam, but it is often harder to get high power at shorter wavelength. There are several processes that need to be taken into account: when the air ionization is so high that the plasma frequency exceeds the radiation frequency, the plasma becomes opaque for the radiation; heated air expands, which causes decrease of temperature; some other nonlinear effects, such as self-focusing, can be significant as well.
| {
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Deriving relativistic momentum (wikibooks) I asked a question about a derivation of relativistic momentum here, but I didn't really get an answer that helped me. So I looked up a different but similar derivation on Wikibooks (see here), and I have a different question now about this other proof.
Somewhere near the end they say:
I don't get this. What exactly is the principle of relativity here? I thought it was that the laws of physics are the same in each reference frame, so how did they come up with this equation? I see they're equating the change in classical momentum for R to the change in classical momentum for B, but wasn't the whole point of this thought experiment that we're deriving a relativistic momentum?
If someone could help me with this, that'd be great, because it's the only step I don't get!
| First a little clarification: this derivation is, although effective, a bit old-fashioned since it uses a concept like the relativistic increase of mass, which is now considered an outdated and confusing interpretation of processes in special relativity.
Nevertheless, the answer to your question comes from a few lines above the ones you cited: at a certain point, after having found that $u_{yR} \neq u_{yB}$, the derivation states:
If the mass were constant between collisions and between frames then although $2mu'_{yR} = 2mu'_{yB}$ it is found that $2mu_{yR} \neq 2mu_{yB}$
After that, the (deprecated) relativistic mass $m_A$ and $m_B$ are introduced in order for the inequality above to become an equality. The principle of relativity here is to be understood as "the equality which holds in a frame must hold in every other frame too"; in this case the equality is the conservation of linear momentum: if the equality didn't hold, you would have a frame in which it is conserved and a frame in which is not. But conservation of momentum is a basic law, therefore a new definition of momentum has to be introduced in order to restore conservation.
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Why does torque produce a force on the axis of rotation? If a door is rotated about its fixed axis in (outer) space, a force parallel to the door on the hinges will arise due to centripetal force on the centre of mass and conservation of momentum (Newton's third law).
But any torque on the door will create a force on the hinges which is equal to $t/r$ or torque divided by radius. I'm looking both an intuitive and mathematically based explanation for this fact. I can sort of 'see' why, but my understanding is vague and uncertain.
| If you view the torque as Force * Radius * Sin(angle between force and door) then you can see that t/r is the component of force perpendicular to the door if that force were applied at radius r from the hinge. Now imagine the forces on the hinge if you wanted to generate the same torque at various radii. The closer I push to the hinge, the more the hinge has to push back to get the door to rotate and the further out I push, the smaller the force, thus the 1/r relationship between the force on the hinge and the radius of the applied force.
| {
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Rolling in smooth inclined plane Can a body posses pure rolling in smooth inclined plane.In smooth
inclined plane the centre of mass of body is accelerating.So to maintain the pure rolling there should be angular acceleration.But no one provides the torque because the only forces Gravity and Normal reaction passes through centre.
| Whenever a simple roll starts rolling on earth, it is the earth itself that provides the angular momentum.
In your case, the roll on the say, for example, 30 degrees inclined plane always exerts its force perpendicular to the plane, i.e.pointing at (90-30)= 60 degrees into the earth. That means your roll kicks the earth a little sideways. So if your roll rolls to the east, it pushes the earth a bit to the west. If you would install your inclined plane say on a large cartwheel at rest, then indeed the cartwheel would behave like the earth and to rotate in the opposite direction as your roll.
You asked also how does the angular momentum get into the roll: The fact is that the different parts of the roll feel different resulting forces. The part closer to the plane experiences gravity, but also the restoring force from the resting point on the plane, while the part further away from the plane just experiences gravity and "drops" therefore. so you have a similar distribution of force as with a crank, which acts only on one side of a roll.
| {
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Why are all my separable solutions orthogonal polynomials in $L^2$? I spent a few hours today solving the Laplace and Schrodinger equation on a variety of domains, and kept finding solutions to the separated equations that were orthogonal (polynomials) in $L^2$, e.g. the quantum harmonic oscillator
$$-i u_t = u_{xx}-x^2 u$$
which yield the eigenvalue problem for the separable solutions
$$X''(x)+(\lambda-x^2)X(x)=0~~~~~~~~\text{or equally}~~~~~~~w''(x)-2xw'(x)+(\lambda-1)w=0$$
where $X(x)=w(x)e^{-x^2/2}$. The solutions to this equation are the Hermite polynomials, which are orthogonal in $L^2$ on $[-1,1]$. The Schrodinger equation for the hydrogen atom
$$i u_t = -\frac{1}{2}\nabla^2 u -\frac{u}{r}$$
has separable solutions in terms of the Laguerre and Legendre polynomials, again orthogonal, and Chebuchev polynomials appear in other circumstances. I'm wondering what it is about these physical problems that produces solutions with these properties, and how if at all these properties impact the physical phenomena they describe. Does this have physical significance or a physical explanation related to the symmetry of the problem?
The second answer to this question is quite relevant, but it's quite "hand-wavy", and I don't fully understand his argument.
| You have discovered the spectral theorem - (generalized) eigenvectors of self-adjoint operators like the Hamiltonian are orthogonal to each other and if the spectrum is discrete, they form an orthonormal basis.
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Can the Mikheyev-Smirnov-Wolfenstein (MSW) effect be modified by non-standard neutrino-neutrino interactions? The MSW effect describes how propagation of neutrinos through matter can resonantly enhance the neutrino mixing. The reason for this enhancement is that the presence of electrons in matter changes the energy levels of the neutrino propagation eigenstates, due to charged-current coherent forward scattering of the electron neutrinos (via weak interactions). This effect is especially important for the propagation of solar neutrinos.
Non-standard (i.e., beyond the Standard Model) neutrino interactions with electrons can modify the MSW effect (see e.g. p. 7 of this paper).
Now my question is: is it also possible that non-standard neutrino-neutrino interactions (such as the ones mediated by Majorons) modify the MSW effect?
| Indeed non-stardard interactions can modify neutrino mixing. Here's one paper that discusses a few types of NSI and their effect on the solar survival probability.
*
*R. Bonventre, A. LaTorre, J.R. Klein, G.D. Orebi Gann, S. Seibert, and O. Wasalski, "Nonstandard models, solar neutrinos, and large $θ_{13}$", Phys. Rev. D 88, 053010, arXiv:1305.5835.
I'm not personally aware of any paper that discusses the affect of Majorons in particular on neutrino mixing though.
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Momentum of an electron acting as a wave Was working on a problem with electrons acting as waves in diffraction. Part of the question asked me to calculate the momentum of the electron. Since I was dealing with waves I used the following equation:
$h=pλ \implies p = h/λ$
Since $λ = v/f$ we can substitute that in, resulting in $p = hf/v$.
Substituting in the de Broglie $h = E/f$ into the above equation we get $p = E/v$. Since we're talking about electrons the only energy that the electron has is kinetic so we can substitude $E = 0.5mv^2$ into the equation giving us $p = 0.5mv^2/v = 0.5mv$.
I've repeat that, $p = 0.5mv$. Any 4th-grade physicist knows that momentum is $mv$ so on one hand, I have mv and on the other I have a derivation saying the momentum is $0.5mv$. Is there a mistake in my derivation I'm not seeing?
P.S: I noticed something a bit later. $p = E/v \implies E = pv = mv^2$. See any similarities between this and another infamous equation in the realm of relativity?
|
Since I was dealing with waves I used the following equation:
$$h=pλ \implies p = h/λ$$
This is fine.
Since $λ = v/f$ ...
True for all waves, yes.
... we can substitute that in, resulting in $p = hf/v$.
This is okay, but we don't usually find people talking about the "frequency" of an electron matter wave.
Substituting in the de Broglie $h = E/f$ ...
Ah, here's your problem. The de Broglie equation has nothing to do with energy or frequency --- the de Broglie relation is $p = h/\lambda$. The relation $E=hf$ is a result for massless particles, see below.
... Since we're talking about electrons the only energy that the electron has is kinetic ...
This is a reasonable statement for a free particle.
... so we can substitude $E = 0.5mv^2$
Nope, that's a nonrelativistic approximation muddled by a confusing notation. The relativistically correct statement is
$$
E^2 = (pc)^2 + (mc^2)^2
\tag 1
$$
where $E$ is the total energy, including the rest energy, and the kinetic energy is $K = E-mc^2$.
For massless particles, the Einstein equation (1), the de Broglie equation $p=h/\lambda$, and the wave equation result $\lambda f = v$ combine to give the result $E=hf$.
However for massive particles you find instead
\begin{align}
E^2 &= \left(\frac{hfc}{v}\right)^2 + (mc^2)^2
\\
K &= -mc^2 + \sqrt{\left({hfc}/{v}\right)^2 + (mc^2)^2}
\end{align}
which is much less helpful. Hence people don't talk much about the frequncies of matter waves.
| {
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Why don't humans burn up while parachuting, whereas rockets do on reentry? I guess it has something to do with their being both a high horizontal and a vertical velocity components during re-entry. But again, wouldn that mean there is a better reentry maneuver that the one in use?
| As others explained, the maximum speed of a parachutist is much smaller than that of orbital vehicles. But this is because parachutists jump from a relatively small height: the record jumps by Eustace (https://en.wikipedia.org/wiki/Alan_Eustace) and Baumgartner (https://en.wikipedia.org/wiki/Felix_Baumgartner) were performed from the height of about 40 km, and the maximum speed was about 1.3 km/s, whereas the orbital speed is about 7.9 km/s. @Cort Ammon said that "A parachutist dropping from orbit would have the same issues with burning up." That is correct assuming the parachutist dropping from orbit has an orbital speed. However, the low orbit height (say, 150-200 km) is comparable to the record height of parachute jumps. If a parachutist were dropped from such orbital height with zero initial speed, his/her maximum speed would be much smaller than the orbital speed. To achieve speed comparable to the orbital speed, a parachutist would need to be dropped from a height comparable to the Earth radius (about 6000 km).
| {
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Where does all the heat go during winter? I do not understand where actually the heat in our surroundings go during the winter season. Is it radiated out into space? I know it cannot coz global warming would not be a issue then. It might get absorbed but where? I tried figuring it myself but couldn't please help.
| Energy does get dissipated to space in Summer and in Winter. Think of CO2 like the difference between covering yourself in a very thin sheet or no sheet. In Winter, you'd still feel very cold if you had just a thin sheet covering you, and in Summer just a bit warmer. As CO2 increases, the sheet gets a little thicker each year.
In Winter, your part of the planet is tilted slightly more away from the sun so receives less solar energy, and hence is colder.
Global temperatures can also go down some years due to fluctuations in received solar energy. It's not actually the case that the earth is constantly getting warmer. It's that the earth's ability to trap received solar energy is improving, so that if solar energy is assumed to be constant, the earth would be warmer.
Locally, temperatures vary from day to day due to differences in air pressure causing warm air either to move to your locality or away from it, as well as factors such as wind and cloud cover.
| {
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How force exerted by spring is always opposite to the direction of displacement in Hooke's law Suppose a spring lying on a horizontal table, displaced from its equilibrium length by an external agent. The external agent is removed, the spring will head back to its equilibrium length. Here, the direction of spring force and displacement will be same.
But according to Hooke's law,
$$\mathbf{F}=-k\Delta\mathbf{x}$$
The minus sign tells us that the force exerted by spring is always opposite to the direction of displacement.
How is this? Please explain the reason for the minus sign.
Thanks.
| Occasioned by your comment @Fracher's answer :'But when moving from extreme to mean this does not seems true', i think that you mixed up the vectors 'displacement' and 'velocity' which can have opposite directions.
So i will try to explain the situation as simple as i can (and it is):
Hooke's law for a spring is often stated under the convention that $F$ is the restoring (reaction) force exerted by the spring on whatever is pulling its free end.
In that case, the equation becomes: $ F = − k X $
since the direction of the restoring force is opposite to that of the displacement.'
(as stated in wikipedia).
This is illustrated in the following diagram :(from karantonibg.blogspot.gr)
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Can a (micro) black hole be used to make a microscope? We have seen that black-holes can be used as a telescope. Is there a chance that light bending from a micro black-hole can be used to create a microscope?
| the difference between a microscope and a telescope:
microscopes are used to magnify small objects that are at a short
distance from the viewer whereas telescopes are used to magnify large
objects that are at a large distance from the viewer.
In gravitational lensing a black hole is used as a lens to see objects father away, so yes, in theory a small lens (a small black hole) can be used to view things that are close.
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Why is a nonzero VEV for a spinor field said to break Lorentz invariance? Consider a spinor field $\psi(x)$. Its vacuum expectation value is given by $$v=\langle 0|\psi(x)|0\rangle.$$ Using the fact that the vaccum is invariant under Lorentz transformation, we get, $$v=\langle 0|\psi(0)|0\rangle.$$ Why is it that, if $v\neq 0$, the Lorentz invariance is broken?
| The $v$ you write is itself a spinor, not a scalar. A non-zero spinor is obviously not invariant under Lorentz transformations, so a non-zero spinorial VEV breaks Lorentz invariance of the 1-point function.
| {
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Did inflation stop because of energy density drop or some other reason? If I am understanding big bang correctly... During big bang
Approximately $10^{−37}$ seconds into the expansion, a phase transition caused a cosmic inflation,
Which was 'free lunch' of energy and it continued for some time but then suddenly 'free lunch' was over and energy conservation started...
The universe continued to decrease in density and fall in temperature, hence the typical energy of each particle was decreasing.
and quarks and other particles formed....
Did inflation stop because of energy density drop or some other reason?
| There is no single answer, since there are many models for inflation and no one knows if any of them are correct.
Broadly speaking, what all the models contain is a scalar field at early times with a non-zero value, which is often termed a false-vacuum. The energy density of this field completely dominates the dynamics of the universe, like a super version of dark energy today. In such circumstances, the universe expands exponentially with time.
Whilst the field is present, the universe inflates, but at some point in time and for some reason (e.g. symmetry breaking), the scalar field decays to a true vacuum, converting it's energy into mass.
As I said, nobody knows why inflation stopped, just that the process of field decay must take long enough for the universe to inflate sufficiently to solve the various problems of the original big bang model.
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Can an accelerating frame of reference be inertial? In physics problems, the earth is usually considered to be an inertial frame. The earth has a gravitational field and the second postulate of the general theory of relativity says:
In the vicinity of any point, a gravitational field is equivalent to
an accelerated frame of reference in gravity-free space (the principle
of equivalence).
Does this mean that accelerating frames of reference can be inertial?
| I think I understand what you mean. In an inertial frame, an object on which no net forces act, moves with a constant speed along a straight line. On earth this doesn't even approximately hold.
If you still want to consider this as approximately inertial, the conclusion must be that the earth itself is present (as mass, matter or a potential) inside your frame. In other words, the frame is not equivalent to one that is accelerating, but there is a source of acceleration present within the approximately inertial frame.
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Is the magnetic field of a moving electron caused by length contraction in the direction of motion? Consider an electron moving relative to us. Because the space in the electron's rest frame is contracted relative to us in the direction of the electron's velocity, the electric field lines are squeezed in the same direction, so the electric field "density" is bigger perpendicular to the electron's motion (but smaller (zero?) in the direction parallel to its motion). Is this the qualitative source of the magnetic field?
| There is the effect called magnetism, and then there is the effect that you described.
Some basic things about magnetism: There's no magnetism between a moving charge and a still standing charge. There's no magnetism between charges that move along a line.
And now some basic things about the effect you described: There is an effect between a moving charge and a still standing charge. There is an effect between charges that move along a line.
It's reasonable to say that those are two different effects.
Let's consider a current carrying wire that contains an equal amount of plus and minus charges. There is no electric field next to that wire, although every electron's field lines are squeezed in the direction perpendicular to the motion. All those electrons' field lines added together results in identical field lines as in the case when there is no current.
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What is the correct way to estimate the work done by a climber? My teacher gave us a worksheet with word problems and their solutions. It is in German, so I have tried my best to translate it to English:
A 26 year old man climbs Mount Everest (8848 m) in only 8 hours 10 mins from the base camp at 5300 m. Estimate the "lifting work" (Hubarbeit) that the man exerted in the climb.
I thought to use the formula: $W = F \cdot \Delta S$, but I don't know what $F$ would be. I think the $\Delta S$ is $8848 - 5300 = 3548$. But then I looked at the solutions and my teacher used this formula: $W = m g h$. He guessed the man's weight and also used $g = 10\ \mathrm{N}/\mathrm{kg}$. I don't understand this.
Could someone maybe help me? Anyone have an idea?
| "Estimate the 'lifting work'" is the key part. Well, what is the lifting work? What are you lifting against? Gravity. What's the force gravity exerts? $F_{gravity} = mg$ (neglecting variations in $g$ as you go up the mountain). Therefore, the force in your formula $W=F\cdot \Delta S$ is the force of gravity.
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How do scientists know Iron-60 is created during supernovae? I know that the meteoroids contain Ni-60, which is formed after decaying Fe-60, and as per my study, I got to know that Fe-60 is formed during the time of a supernova. But I wonder how scientists know/find that these elements were created during that event?
| As always, it depends on what you mean by know/find. As aptly illustrated by Kyle Kanos, theoretical arguments show that $^{60}\mathrm{Fe}$ is naturally produced through stellar nucleosynthesis in the last stages of the life of massive stars, and then injected into the InterStellar Medium (ISM) by SN explosions. This has been known, from a purely theoretical point of view, for a long time, but what is the actual observational evidence?
Also known for a long time is that some radioactive elements are detected in situ, meaning from emission originated within some SN remnants. These include Nickel-56, Cobalt-56, Iron-56,... Recently, even Titanium has been detected in situ, strengthening our belief in the overall correctness of the theoretical picture mentioned by Kyle Kanos.
Iron-60 is trickier, because it is both rare (of order of $10^{-7}$ of the abundance of $^{56}\mathrm{Fe}$) and weak in its emission, so that so far no in situ detection has ever been made. Yet the INTEGRAL telescope has managed to detect the $\gamma$-ray line from its decay. This is an integrated measure over most of the sky, because like I said every source in the Galaxy is likely to be too faint to be detected individually, but their sum produces a cumulative effect that can be observed. This is the best that can be done nowadays.
$^{60}\mathrm{Fe}$ is somewhat similar in importance to the more abundant $\gamma$-ray brilliant isotope $^{26}\mathrm{Al}$. Wikipedia has a good article on it, which may complement what you know about $^{56}\mathrm{Fe}$.
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Why does each plate receive a charge exactly the same magnitude in parallel plate capacitor? When a parallel plate capacitor is connected through a cell, each plate of the capacitor receives a charge with the same magnitude, but with the opposite sign. Is it because of the battery or the area of the plates?
| Suppose you have a simple circuit with a capacitor and a power supply:
You want to create a charge on the capacitor, so you turn on the PSU to add some extra electrons to the upper plate:
But the number of electrons in your circuit is constant. The power supply can't create or destroy electrons. All it can do is act like a pump to move electrons round the circuit. Specifically it can pump electrons from the bottom plate round the circuit to the top plate:
So for every electron that you add to the top plate you have to remove an electron from the bottom plate, and that means the negative charge on the top plate is necessarily always equal to the positive charge on the bottom plate.
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Interpretation of geodesic constant of motion The Schwartzschild metric in standard coordinates with signature $(1,-1,-1,-1)$ is given by
$$ds^2=(1-\frac{r_s}{r})\ dt^2 - (1-\frac{r_s}{r})^{-1}\ dr^2 - r^2(d\theta^2+\sin^2\theta\ d\phi^2).$$
As the Schwartzschild metric is independent of time then it has a time-displacement symmetry described by a Killing vector $\xi^\mu$ given by
$$\xi^\mu = (1,0,0,0).$$
This implies that a particle free-falling on a geodesic path with four-velocity $P^\mu$ has a constant of motion $\epsilon$ given by
$$\epsilon=\xi_\mu P^\mu.$$
I understand that $\epsilon$ can be interpreted as the particle energy measured by a stationary observer far from the origin, where the metric is flat, with four-velocity $U^\mu=\xi^\mu$.
Can one also interpret $\epsilon$ as the particle energy measured by a local observer who is free-falling with the particle?
I presume one must somehow transform $\epsilon=\xi_\mu P^\mu$ to the local coordinates of the free-falling observer.
| The Killing vector is
$$
\xi_t~=~\sqrt{1~-~r_s/r}\partial_t
$$
and reduces to you case in the asymptotic region, or on the rest frame of any observer. For $\xi_\mu U^\mu~=~\epsilon$ what this tell us is the metric may be expressed as
$$
1~=~\epsilon^2~-~\frac{1}{{1~-~r_s/r}}(U^r)^2~-~r^2\left((U^\theta)^2~+~sin\theta(U^\phi)^2\right),
$$
which defines a Hamiltonian.
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Double-slit experiment with two different mediums How the interference pattern will look like in a double-slit experiment done with two different mediums? Air before the slit and glass after the slit.
| There would be no difference qualitatively. The pattern is due to path difference after the slit, which would change by a factor of refractive index $n$ and therefore would change the fringe width compared the case of air..
However, if the upper half is filled with water and bottom half is air, the pattern will change. A little analysis will tell the details of that too.
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Direction of normal force on stick on box
What is the direction of the normal force on the stick in this case, assuming gravity? Is it right angled with the stick? Or is it upwards? Or is it impossible to determine?
| Normal is a synonym for perpendicular.
The normal force is as you show it, perpendicular to the slanted object. Gravity is a separate force having a different agent (the earth) and plays no role in determining the direction of the normal force. Friction is parallel to the surface, and is not a normal force.
| {
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Wave packet expression Speaking in general about plane waves propagating along $z$ (electro-magnetic waves, for example; not necessarily particles represented as waves), a wave packet can be defined as
$$A(z,t) = \int_{\omega_1}^{\omega_2} A ( \omega ) e^{j (\omega t - kz) } d\omega$$
In particular, this expression is used when dealing with group velocity.
But a single plane wave is usually expressed as
$$B(z, t) = B_0 \cos ( \omega t - kz )$$
So, why is the complex exponential used above? Or is the actual $A(z,t)$ of the wave packet just the real part of the first expression?
Observation: I did not consider the form $B_0 \cos ( \omega t - kz )$ because I necessarily want real functions, but because this is the standard form a plane wave is presented and written.
| The complex form
$$A(z,t) = \int_{\omega_1}^{\omega_2} A ( \omega ) e^{j (\omega t - kz) } d\omega$$
is the most common because it is compact (easy to write and read). As usual in physics, one writes complex expressions and when one wants the "real" thing, one takes the real part of the expression.
However, if you really want to have an expression without complex numbers, you can write your wavepacket under the form
$$B(z,t) = \int_{\omega_1}^{\omega_2} \alpha( \omega ) \cos(\omega t - kz) + \beta( \omega ) \sin(\omega t - kz)\,d\omega,$$
both $\alpha( \omega ) $ and $\beta( \omega )$ being real numbers (and possibly $0 \leq \omega_1 \leq \omega_2$ if you like to consider only positive frequencies, which is all you need in the real world). This is for a packet travelling towards positive $z$. Change $\omega t - kz$ into $\omega t + kz$ for a packet travelling towards negative $z$.
Note that for $$\alpha( \omega ) = \Re(A( \omega )),\ \beta( \omega )=-\Im(A( \omega )),$$
you have exactly
$$B(z,t)=\Re(A(z,t)).$$
Indeed, using this definition of $\alpha$ and $\beta$, that is: $$A(\omega)=\alpha(\omega)-j\beta(\omega),$$ and the identity $\Re((\alpha-j\beta)(C+jS))=\Re(\alpha C+\beta S+j(\alpha S-\beta C))=\alpha C+\beta S$ where $C$ and $S$ stand for $\cos$ and $\sin$, you have:
$$\Re(A(z,t))=\Re \int_{\omega_1}^{\omega_2} (\alpha( \omega ) - j \beta( \omega ))\; e^{j (\omega t - kz) } d\omega=\\
\int_{\omega_1}^{\omega_2} \Re ((\alpha( \omega ) - j \beta( \omega )) (\cos (\omega t - kz) +j\sin (\omega t - kz))) \;d\omega=\\
\int_{\omega_1}^{\omega_2} \alpha( \omega ) \cos (\omega t - kz) + \beta( \omega )\sin (\omega t - kz) \;d\omega=B(z,t).\\$$
| {
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Charged plasma and ion grid in interaction in ion thrusters I was just wondering .....
In this Image of an ion thruster, when the positively charged particles pass through the grids, wouldn't they just bombard the negatively charged grid(a fraction of them). This means that there must be a constant adjustment to maintain the potential difference between the grids. Is this the reason for the high energy consumption of these engines (along with ionization..)?
| You say "constant adjustment" as if you expect a motor with a PID controller to be spinning cat's fur on the grid to keep its charge ;P
The grid keeps it charge with 1.) a battery and 2.) by connecting both electron guns' filament to the positive grid. An electron gun is a special filament that when heated (by passing current through it) expels electrons.
As the gas ions are passing by the negative grid, the electrons are jumping off and neutralizing the gas as it exits. This is just part of the circuit under normal operating conditions - the Thevenin equivalent voltage of the circuit would be such that it's adequately charged at all times.
Energy consumed by the system is 1.) work done to accelerate the ions and 2.) the work done to heat the electron gun filament.
Here's the circuit:
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What is an induced electric field? I have read in many books about induced current in a coil (Faraday's law), and also the motional emf across a moving conductor in a magnetic field. But somewhere I read about induced electric field due to a time varying magnetic field.
And I think that Induction of electric field is the fundamental phenomenon, and induced emf and current are the results of it
I am just a novice in physics.
Could someone explain me how these phenomenon (Induction of emf and Induction of electric field) are related to each other?
|
Current I through the solenoid sets up a magnetic field B along its axis and a magnetic flux Φ passes through the surface bounded by the loop.
Since charges are at rest (v=0) so magnetic forces Fm=q(v X B) cannot set the charges to motion.Hence induced current in the loop appears because of the presence of an electric field E in the loop.It is this electric field E which is responsible for the induced emf and hence for the current flowing in a fixed loop placed in a magnetic field varying with time.
using faraday's law
From equation (10) we see that line integral of electric field induced by varying magnetic field differs from zero.This means we can not define a electrostatic potential corresponding to this field.
Hence this electric field produced by changing electric field is non-electrostatic and non-conservative in nature.We call such a field as induced electric field
| {
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Why the electrons below the Fermi level do not conduct electricity? Physically, why is it that the electrons need to excited above the Fermi level to conduct electricity? In other words, why is the current zero when the electrons lie below the Fermi level? Does Pauli exclusion principle play any role here?
| Here is my way of thinking of it intuitively:
Each electron in an atom resides at a specific energy level (or state). Pauli's exclusion principle prevents any of them to occupy the same exact energy state.
When putting two equal atoms close together, they have identical electrons at identical energy states. Again Pauli's exclusion principle prevents two electrons to occupy the same energy state. So they must "fit" their energy states to be slightly lower or slightly higher, fitting in underneath and above one another to make it work.
When infinitely many equal atoms are packed together to form a solid material, infinitely many energy states must be crammed together and fitted underneath and above each other. This stack of many, many energy states is what we call a band, so closely packed that the band is as good as continuous.
All energy states are occupied, meaning that all spots in the band are taken. No electron can move anywhere. A current is impossible.
If an election would manage to reach the next band of tightly packed but this time empty energy states, then it would be alone. Any negligible amount of energy would make it move from spot to spot within this empty band. A current is now possible.
This is the idea. The Pauli exclusion principle is key. That is the reason why no electrons can move around in filled bands.
| {
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Does alpha radiation penetration depth decrease exponentionaly with distance? imagine there is 1mm² square area with thin film of Americium 241,kind of like the button in smoke alarms,lets theoreticaly imagine it gives 1000 becquerel of 5 MeV alpha radiation,that is 1000 particles per second.Now I know Americium gives different energy alpha and other kinds of radiation and the becquerel rate would be higher but for the sake of simplicity lets imagine it like I described.
Now lets imagine square target with 1mm² size that is ahead of the Americium,there is air between them,if the target is away set distance,like 2cm or 4cm,on average how many alpha particles will hit it per second?
What I want to know is the graph showing how many particles hit the target with increasing distance,is it exponentional or is it more complex?
| Alpha particles interact with matter via multiple small interactions.
The word "small" is used to indicate that the energy lost by an alpha during each interaction is small compared with the kinetic energy of the alpha.
So the alpha progresses through matter slowly losing kinetic energy.
So the graph of count rate which is proportional to the flux rate of alpha particles against absorber thickness looks something like this.
.
So all the alphas travel roughly the same distance through matter.
Here is a graph of range of alpha particles in air against energy of alpha particles.
So you 5 MeV alphas will hardly be absorbed by 4 cm of air.
At the other extreme is the all or nothing interaction where one interaction stops the passage of a particle.
This would result in an exponential drop of the flux rate with thickness of material.
| {
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Will a contiguous, low-Re, low-Ca, liquid body always become a sphere at zero gravity? Let's assume zero gravity, zero initial speed everywhere, $Re \ll 1$ and $Ca \ll 1$
Will such a liquid body always become a sphere or will it sometimes split?
Formally speaking, I'm talking about
$$
\lim_{viscosity -> \infty} \lim_{t -> \infty} ShapeAtTime(t)
$$
(Sufficiently high viscosity will also limit $Ca$, even though it is not directly in the expression)
I think it helps to think about this kind of experiment, but with an hourglass-like shape: Will its neck widen or expand at zero gravity?
| If the initial fluid blob had symmetric dumb-bell shape, then fluid pressure will be higher at its waist, and there will be flow from waist region to the two bulging regions, resulting in breakup into (at least) two smaller droplets (read up Rayleigh-Plateau instability). In other words, even if velocity is zero everywhere initially at $t=0$, you can always set up a situation where pressure gradient is not zero everywhere inside the fluid, resulting in a flow for $t>0$ and thus possible breakup. You can always have a flow so far as viscosity is finite, no matter how high, and this alone cannot prevent breakup.
| {
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Why doesn't Young's modulus change with length and diameter? In this question:
The Young modulus of steel is determined using a length of steel wire and is found to have the value $E$.
Another experiment is carried out using a wire of the same steel, but of half the length and half the diameter.
What value is obtained for the Young modulus in the 2nd experiment?
I know that the Young's modulus is an intrinsic property of a object. But what I found confusing is that, when I calculated the Young's modulus for the 2nd experiment, I got $2E$. But the answer was $E$, instead of $2E$.
However, my thought kept lying with the equation:
$$\text{Young's modulus} = \frac{\text{force}\times\text{length}}{\text{extension}\times\text{area}}$$
Doesn't the change in length and diameter affect the Young's modulus value? How can it be an intrinsic value for a object?
| From the given information how did you "calculate the Young's modulus for the 2nd experiment"?
You cannot assume that in the second experiment the force and extension are the same as in the first experiment because they are not.
For a given force the extension would be twice in the second experiment as that in the first experiment.
Update in response to a comment
$\text{extension}_1 = \dfrac{\text{force}\times\text{length}}{\text{Young's modulus}\times\text{area}}$
$\text{extension}_2 = \dfrac{\text{force}\times\frac{\text{length}}{2}}{\text{Young's modulus}\times\frac{\text{area}}{4}}=2\times \text{extension}_1$
| {
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How does $I = \mathrm{d}q/\mathrm{d}t$ work for a capacitor? When the capacitor is charging in a circuit consisting of a resistor, a capacitor and an alternating sinusoidal generator at $t=0$, the charge across the capacitor is 0 and the current is $I =\mathrm{d}q/\mathrm{d}t$. Does this make the current zero too? While it is max across the resistor in the same circuit and they are connected in series which means that the current should be the same in all the components of the circuit.
| If you apply a sinusoid generator to an RC series circuit, at t=0
the sine(2pi * f * t) is indeed zero. But the voltage on the capacitor
is NOT zero, rather the sum of I*R and the capacitor voltage is zero.
That's because 'a sine wave' means a sine wave that started
long ago, and at time t=0 it just came off a half-cycle of negative
voltage excursion. The capacitor has a negative charge at that time.
For the charge to be zero at t=0, perhaps one does not supply
a sine wave, but rather a MODULATED sine wave, amplitude zero
at all negative times? Capacitors have stored energy, which
means memory of the past excitation voltage.
The way to find the charge on the capacitor at t=0 is to
do an integral over all past times using sinewave drive, or
to use a different function than 'sinewave', which means your
analysis will include a turnon transient.
| {
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What happens to temperature as volume increases (charles law) Let us take Boyles law to start.
Assumptions:
*
*Gas is perfect.
*In a massless piston that can be expanded with no friction
*Adiabatic
If we were to decrease the volume of the piston, the pressure inside would go up because the gas molecules would be hitting the sides more often.
If we were to increase the outside pressure on the piston, the volume of the piston would go down until the internal pressure matched the external pressure.
We see here that Boyle's law is perfectly explained by the kinetic model of gases (T held constant).
Let's look at Charles law (P held constant)
If we were to increase the temperature of the molecules, their kinetic energy would increase and would therefore hit the piston with greater force and increase the volume of the piston until the internal and external pressures are equal.
But if we were to spontaneously increase the volume of the piston, the temperature would NOT increase as a result (to maintain the increased volume against the constant external pressure) because heat doesn't spontaneously arise.
So it seems that Charles law only works one way, but not the other. And that T and V are not intrinsically linked like P and V are in Boyles law. Is this true?
| Compressing a gas that is not at absolute zero will increase its temperature because a finite volume of gas at a finite temperature has a set amount of heat energy. when that set amount of heat energy occupies a smaller space its temperature rises. Conversely when it occupies a larger space its temperature goes down. so both laws are reversible.
| {
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Doppler effect and apparent frequency What is meant by "apparent frequency"?
I mean the answer we get by applying the formula; what does it signify?
If it is the frequency received by the observer, does it mean that the observer receives the same frequency no matter what the distance of the source?
Shouldn't distance of play a role?
| Doppler effect is the change in frequency (pitch) of a source when there is a relative motion between the source of the observer. It occurs in both sound and light waves.
| {
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Proton spin/flavor wavefunction I am currently working through Griffiths' Introductory to Elementary Particle Physics and I'm a little confused about a particle's spin/flavor wavefunctions. As a specific example, I've attached Griffths solution to the proton's wavefunction, and the formula he used to get it.
I understand the solution, but what confuses me is the ordering for the antisymmetric spins/flavors. As an example, looking at the first term, wouldn't the flavors still be antisymmetric in particles 1 and 2 if we just switched the udu and duu terms. We would get a different final solution for the wavefunction now due to terms cancelling out after expanding.
| Note that Griffiths is very careful to match each of these terms,
$$udu ~~\Leftrightarrow~ \uparrow \downarrow \uparrow,$$and if you match both terms at once you get two sign flips: $$(\downarrow \uparrow \uparrow - \uparrow \downarrow \uparrow)\otimes (duu - udu) = (-1)^2 (\uparrow \downarrow \uparrow - \downarrow \uparrow \uparrow)\otimes(udu - duu) $$and since $(-1)^2 = 1$ this is a non-issue.
So the real question you're asking is, why do we have to match these terms? And that's a good question and it has to do with how the 3 terms all play together (a sign flip on any individual term does nothing for consistency or inconsistency).
So the expression takes the form of "we're going to insert some $u_\uparrow$ state into the twice-antisymmetrized 2-quark state $$d_\downarrow u_\uparrow - d_\uparrow u_\downarrow - u_\downarrow d_\uparrow + u_\uparrow d_\downarrow,$$because we know we have two $\uparrow$ spins and two $u$ quarks and so one of these up-quarks has to be in the spin-up state." (Note that under $1^\text{st}\leftrightarrow2^\text{nd}$ interchange the above is in fact symmetric, that last term being exactly the first term with the two particles switching places.) Now the expression chooses to symmetrically insert this $u_\uparrow$ quark in the first position, the second position, and the third position, so that the result will still be symmetric here and will become antisymmetric after correcting for color charge.
What you're proposing by flipping the sign of the first term, therefore, is not symmetrically inserting this $u_\uparrow$ quark in each of the three spots, but inserting it in the first place with a 180-degree phase shift. And that naturally will not be properly symmetric here or antisymmetric afterwards.
| {
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Why does a system expand isothermally? Considering the first step of the Carnot process, heat is transferred from a bath to the system with both at the same temperature. But how does this process start? Why should the system spontaneously absorb heat and expand as a result. And even if I pull on the piston an infinitesimal bit, why should the process continue and do not stay at the new equilibrium state?
| It is as you say. On the isothermes the system is at the same temperature as the bath. Therefore no energy is exchanged at first. Now reduce the pressure on the system ("pull the piston"). The system reacts by increasing its volume to go back to mechanical equilibrium. It does work on you! It does work because a pressure difference implies a net force along which the piston moves.
If it were not for the bath providing energy in the form of heat, the temperature would decrease.
Since you are pulling very slowly, it will also stay in thermal equilibrium with the bath. So temperature keeps constant over the process. Should the internal energy $U$ depend on temperature only, as is the case for an ideal gas, we have $dU=0$.
But because energy leaves the system as work, it must be compensated for by a heat flow into the system, since $dU = \delta Q + \delta W$. That heat is absorbed from the bath, although they are at the same temperature.
The trick is to successively disturb the system infinitesimally out of mechanical equilibrium, giving it the chance to effectively stay in thermal equilibrium with the heat bath.
| {
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A helpful proof in contracting the Christoffel symbol?
Out of all of my time learning General relativity, this is the one identity that I cannot get around.
$$ \Gamma_{\alpha \beta}^{\alpha} = \partial_{\beta}\ln\sqrt{-g} \tag{1}$$
where $g$ is the determinant of the metric tensor $g_{\alpha \beta}$.
With the Christoffel symbol, we start by contracting
$$ \begin{align}
\Gamma_{\alpha \beta}^{\alpha} &= \frac{1}{2} g^{\alpha\gamma} (\partial_{\alpha} g_{\beta\gamma} + \partial_{\beta} g_{\alpha\gamma} - \partial_{\gamma} g_{\alpha\beta} ) \\
&= \frac{1}{2} g^{\alpha\alpha} ( \partial_{\beta} g_{\alpha\alpha}) \\
&= \frac{1}{2g_{\alpha\alpha}} ( \partial_{\beta} g_{\alpha\alpha})
\end{align}\tag{2}$$
where I took $\gamma \rightarrow \alpha$ and $g^{\alpha\alpha} = 1/g_{\alpha\alpha}$.
The next steps to take now, I have no clue. MTW gives a hint by saying to use the results from some exercise, which are,
$$\det A = \det||A^{\lambda}_{\ \ \rho}|| = \tilde{\epsilon}^{\alpha\beta\gamma\delta}A^{0}_{\ \ \alpha}A^{1}_{\ \ \beta}A^{2}_{\ \ \gamma}A^{3}_{\ \ \delta} $$
$$(A^{-1})^{\mu}_{\ \ \alpha}(\det A) = \frac{1}{3!}\delta_{\alpha\beta\gamma\delta}^{\mu\nu\rho\sigma} A^{\beta}_{\ \ \nu} A^{\gamma}_{\ \ \rho}A^{\delta}_{\ \ \sigma} $$
$$ \mathbf{d}\ln|\det A| = \mathrm{trace}(A^{-1}\mathbf{d}A) ,\tag{3}$$
where $\mathbf{d}A$ is the matrix $||\mathbf{d}A^{\alpha}_{\ \ \mu}||$ whose entries are one-forms.
I fail to reason why the metric turns into the determinant from what I have done and then becomes the result at the top.
| Recall the matrix identity
$$\tag{1}\log\det M=\operatorname{tr}\log M.$$
If $M=M(\lambda)$ is differentiable in $\lambda$, then
$$\tag{2}\frac{d}{d\lambda}\log\det M=\operatorname{tr}\left(M^{-1}\frac{d}{d\lambda} M\right).$$
The proof of $(1)$ for symmetric matrices follows from the usual formulae for the trace and determinant in terms of eigenvalues$^{1}$.
As for the Christoffels, we have
$$\Gamma^i{}_{ij}=\frac{1}{2}g^{ik}(\partial_i g_{jk}+\partial_j g_{ik}-\partial_k g_{ij})=\frac{1}{2}g^{ik}\partial_j g_{ik}=\frac{1}{2}\operatorname{tr}(g^{-1} \partial_j g).$$
The last equality is just what the contraction of indices means for the (symmetric!) matrix $g=(g_{ij})$, and there is an error in the indices in OP's post. Now, using $(2)$ we have
$$\Gamma^i{}_{ij}=\frac{1}{2}\partial_j\log \det g.$$
This can be brought into the form
$$\Gamma^i{}_{ij}=\partial_j \log\sqrt{|\det g|}$$
by the usual rules of calculus.
$^{1}$ For symmetric matrices, such as $g$, it is easy because $g$ can be diagonalized. For other matrices you might need a Jordan normal form to compute $\log M$.
| {
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Why viscosity depends on the scale of things? Water feels like honey for bacteria and air is very viscous for small insects.
My question is why viscosity depends on the scale of things?
| As @Tropilio indicated, viscosity does not depend on the scale of things. But in bacteria and small insects, the flow passages are very small (i.e., the surface to volume ratio is very large), and the viscous drag on the flowing fluid occurs at the flow surfaces. So viscosity has a bigger relative effect when the fluid is flowing through small flow passages than through large flow passages. From the "Hagen_Poiseuille" pressure-drop/flow-rate relationship, for a given flow rate, the pressure drop varies inversely with the 4th power of the tube diameter.
| {
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Why doesn't increasing resistance increase brightness if $P=I^2\cdot R$ Light bulb brightness increases with power, $P$.
So why doesn't increasing $R$ increase $P$ and hence increase brightness as $P=I^2\cdot R$ due to $P=I\cdot V$ and $V=I\cdot R$?
I read increasing $R$ decreases brightness.
| You can use any of these formulas to calculate $P$:
$$P = I^2 \cdot R$$
$$P = \frac{V^2}{R}$$
They are both correct and will give same result. You can not tell which one is "dominant".
But to use these formulas you need to know not only $R$ but also $I$ or $V$. And to analyze these formulas you need to know how $I$ or $V$ change when you change $R$.
In case you connected the bulb to a power supply with produces constant voltage $V$ it's easier to use the second formula. You can use the first one either, but you should remember that when $R$ increases the $I$ changes as well. The result would be the same: $P$ decreases.
If you connect the bulb to a power supply which produces constant current $I$ then both formulas would tell you that $P$ increases when $R$ increases.
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Speed of block sliding on frictionless ramps Here's the question:
My book says the answer is C. How is it not A?
I know that all the potential energy is transferred to kinetic energy. With algebra, knowing Kinetic energy is (1/2) * m * v^2 and gravitational potential is mgh, I solve for h which results in (v^2)/2g
Ok so since this is a proportional reasoning problem my focus is that h = v^2 meaning height is directly proportional to the square of velocity. That being said, if we half the velocity, that means that some value (let's use p for the variable) multiplied by height gets me to (1/2v)^2.
That being said, if we refer back to the equation h = (v)^2 and v is being halved, it should look like this:
h * p (<- growth factor) = (0.5v)^2. Now it looks like the right side of the equation grew by a factor of (1/4). Think about it. If your velocity is 10, the right side of the equation becomes 100. If you half that velocity, the right side of that equation becomes 25. You can see that the right side of the equation grew by a factor of 1/4. That means that $p$ (my growth factor variable) should also be (1/4). $h * 1/4 = h/4$. NOT $3h/4$. Where did I go wrong?
| I have another solution:
Ok basically we can think of 2 phases. One where it goes from the first ramp down. The other phase is from down to another ramp, however this time at some height where velocity is equal to half the velocity at the bottom.
Considering the first phase, we know that all potential get's transformed into kinetic. Thus: 1/2mv^2 = mgh. Solving for v (but leaving it squared), we get v^2 = 2gh. I'm going to leave that for now, and as a matter of fact, I will call that v1 so (v1)^2 = 2gh.
Now for the second phase, at the bottom it has the same velocity as the final velocity at the first phase. All this energy gets transferred to a mix of some kinetic and some potential (and this is because we aren't going all the way back up to the top where velocity is 0 again).
Therfore, (1/2)m(v1)^2 = (1/2)m(v2)^2 + mgH.
Now, I used capital H because this is the actual height I am looking for.
v2 is the velocity that we must reference to find capital H. Before I move on to what that equals, let me divide out mass.
(1/2)(v1)^2 = (1/2)(v2)^2 + gH.
Ok so the question is saying to find H when we are at half the velocity at the bottom. so v2 = (1/2v1). Therefore:
(1/2)(v1)^2 = (1/2)((1/2)(v1))^2 + gH
Now for (1/2)((1/2)(v1))^2, I distribute out the ^2. so (1/2)(1/4)(v1)^2 which simplifies to (1/8)(v1)^2. Now back to the equation:
(1/2)(v1)^2 = (1/8)(v1)^2 + gH
Combining like terms I get this now:
(3/8)(v1)^2 = gH.
Looks like we are getting somewhere. Remember that (v1)^2 = 2gh? I'm going to substitute that for (v1)^2.
(3/8)(2gh) = gH
divide g out
(3/8)(2h) = H
Multiply the 2
(3/4)h = H or H = (3h/4).
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What is the significance of the phase constant in the Simple Harmonic Motion equation? The displacement of a particle performing simple harmonic motion is given by $x = A \sin(\omega t + \phi)$ , where $A$ is the amplitude, $\omega$ is the frequency, $t$ is the time, and $\phi$ is the phase constant. What is the significance of $\phi$. How is it used? Please explain the meaning of the phase constant
|
What is the significance of $\phi$?
The phase angle $\phi$ represents the relation between the displacement and velocity of the simple harmonic oscillator at the point in time arbitrarily designated as $t=0$. In particular,$$\tan\phi = \omega \frac{x(0)}{v(0)}$$
The point in time at which $t$ is zero is completely arbitrary. With a different time axis given by $t' = t-t_0$, the state of the SHO can be expressed as $x(t) = A \sin(\omega t' + \phi')$, where $\phi' = \phi + \omega t_0$.
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Acceleration of car. One dimensional motion easy problem A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.
My attempt at solving the problem:
$$a(x) = \frac{v - u}{t}$$
where
$v =$ final velocity
$u =$ initial velocity
$$$$
I get the answer as $4.05 \space ms^{-2}$
But the correct answer given to the problem is $8.10 \space ms^{-2}$.
They used a different equation to reach that answer.
Did I use the wrong equation? I have the average velocity and not the instantaneous veolcity?
| The displacement is equal to the area under a velocity $v$ against time $t$ graph as shown below.
If the body starts from rest and its final velocity is $v_{\rm f}$ then the average velocity is $\dfrac{v_{\rm f}}{2}$ and that is were your missing "$2$" comes from.
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What causes change in planet's angular velocity? A satellite moving in an elliptical orbit will increase in angular velocity as it nears a planet. I understand that this is consistent with angular momentum. But what causes the increase in angular velocity if there is no torque acting on the satellite?
| The gravitational force is a central force so there is no change in the angular momentum of the planet about the Sun but that does not mean that the angular velocity cannot change.
You can think of it as the moment of inertia of the planet about the Sun getting smaller as the planet gets closer to the Sun thus increasing its angular velocity.
As the planet is moving faster the planet has gained kinetic energy at the expense of a loss of gravitational potential energy due to the planet being closer to the Sun.
The force of attraction on the planet due to the Sun is doing work to increase the kinetic energy of the planet because as the planet is getting closer to the Sun there must be a component of the planet's displacement in the direction of the gravitational force.
| {
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An appropriate way to store neodymium magnets Okay so I've bought a few small neodymium magnets to play around with, they're very powerful and I really like them, but I was wondering what's the actual best way of storing those magnets in a way that doesn't affect their magnetic fields or degrades them in any way.
I'm currently storing them stuck to one another, is it a good practice? Thanks a lot!
| Before modern rare earth permanent magnets, magnets required a 'keeper', metal bar that would shunt the flux between poles. This would prevent a loss in magnetization that could occur over time for materials like AlNiCo.
But with rare earth magnets like NdFeB keepers are not required. They will hold their strength, even when stacked.
Perhaps the most important thing regarding storage is to keep them stored in a secure place where small children cannot get to them. Swallowing these magnets can lead to pinching and internal bleeding of the gut. That's for small magnets.
For bigger rare earth magnets there is the danger of the magnet accelerating to high velocity, or metal objects around the magnet accelerating. Pinching forces can cut off circulation in fingers and bones can be broken! So these magnets require extreme care in handling to constantly make sure they are outside the range of other ferromagnetic objects. These magnets should be stored by themselves in sturdy, thick walled wooden boxes.
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The equilibrium concentration of vacancies In the derivation of the equilibrium concentration of vacancies by statistical mechanics method, I was stumped by this procedure (marked by "?").
$\textbf{Physical Model}:$
1.Solid viewed as a collection of $N$ atomic sites;
2.Each site may or may not be occupied, and assume now that $N_o$ sites are occupied and $N_v$ sites are vacant;
3.If a site is not occupied then the system has an additional energy, namely the formal energy $E_v$;
$\textbf{Solution}:$
1.multiplicity function
$$\Omega = C_N^{N_o} = \dfrac{N!}{N_o!N_v!}$$
2.entropy:
$$S=k_B\ln \Omega = -N k_B (c\ln c+(1-c)\ln(1-c)) \qquad (c=\dfrac{N_v}{N} \quad ;\quad (1-c) = \dfrac{N_o}{N})$$
3.the internal energy ($\textbf{?}$)
$$U = N c E_v$$
(Why we don't consider the internal energy of the whole system rather than the vacancies ? )
4.The Helmholtz free energy
$$F = U-TS = N(cE_v + k_B T (c \ln c + (1-c) \ln (1-c)))$$
and taking $c \ll 1 $
$$\dfrac{F}{N} = c E_v + k_B T c \ln c$$
5.equilibrium concentration (by minimizing the Helmholtz free energy.)
$$ c \rightarrow e^{-\dfrac{E_v}{k_B T}}$$
So what's the missing points to understand the marked procedure above?
|
Why we don't consider the internal energy of the whole system rather than the vacancies?
Every atom being exactly at a crystalline site represents the minimum energy configuration. (Aside: This configuration only happens at absolute zero temperature.) What this energy is is irrelevant; it's some constant. You might as well treat it as zero.
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What is a "Standard value"? Temperature:a measure of the warmth or coldness of an object or substance with reference to some standard value.
I really tried searching lots AND lots for what is "Standard value" is... But I still don't understand what does it mean in that sentence.
Can someone please tell me what do they mean by that?
| At present the standard value is defined as follows:
The kelvin, unit of thermodynamic temperature, is the fraction $\dfrac{1}{273.16}$ of the thermodynamic temperature of the triple point of water.
It follows that the thermodynamic temperature of the triple point of water is exactly $273.16 $kelvins, $T_{\rm tpw} = 273.16\, \rm K$.
The symbol, $T_{\rm tpw}$, is used to denote the thermodynamic temperature of the triple point of water.
At its $2005$ meeting the CIPM affirmed that:
This definition refers to water having the isotopic composition defined exactly by the following amount of substance ratios: 0.000 155 76 mole of $^2\rm H$ per mole of $^1\rm H$, 0.000 379 9 mole of 17O per mole of $^{16}\rm O$, and 0.002 005 2 mole of $^{18}\rm O$ per mole of $^{16}\rm O$.
The scale of temperature which is in general use is called the International Temperature Scale of 1990 (ITS-90) and thermometers are calibrated with the use of a number of fixed points whose temperature has been defined. and agreed internationally.
The reason for having this second scale is that direct measurement of thermodynamic temperature is extremely difficult and very time consuming ie not practical for most applications.
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Is it correct to say that matter and antimatter are opposite in all quantum properties that are conserved? Matter and Antimatter are always created in pairs, with the exception of CP Symmetry Violation. Thus, in order for some quantum properties to be conserved, these properties must be opposite in the particle and antiparticle created.
An example of that is electrical charge, which is a conserved quantum property. When a particle-antiparticle pair is created, they must have opposite charges for charge to be conserved.
Therefore, is it right to conclude and define an antiparticle as a particle with opposite conserved quantum properties?
|
Therefore, is it right to conclude and define an antiparticle as a particle with opposite conserved quantum properties?
You have the right idea.
Note that antiparticles are required to ensure that a theory is causal. In other words, a measurement at $x$ should not affect a measurement at $y$ if the separation between the two coordinates is space-like [i.e., $(x-y)^2<0$]. One finds that for correlations between observables to vanish like this, each particle $\chi$ must have a corresponding antiparticle $\overline{\chi}$ with the same mass, but opposite internal quantum numbers.
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Accretion neutron star; mass gain 0.8m? My textbook on Astrophysics says the following about accretion (translation):
Assume we have a particle with mass $m$ that falls on a neutron star; $R\approx 10$ km, $M\approx 1.4M_{\text{sun}}$, so $v_{ff}=2GM/R\approx 0.64c$, and $E_{kin}=1/2mv^2\approx0.2mc^2$. The particle falls with a considerable fraction of the speed of light on the surface of the neutron star, and has a kinetic energy that is a considerable fraction of its rest-mass energy ($mc^2$). The source of this kinetic energy is the gravitational potential energy of the particle. When the particle gets slowed down, this energy is converted to heat and radiation. If the gravitational potential energy is emitted like this, then de mass os the neutron star only increased by $0.8m$.
I don't understand why the mass isn't just $1 m$. Because at the beginning (at $\infty$) the particle has no gravitational potential energy, so its only energy is its rest-mass energy $mc^2$. But then it gains energy, so it's total energy would be $mc^2 + 0.2mc^2$. When it is slowed down at the surface, it loses this acquired energy, so its total energy should again be $mc^2$, and therefore the neutron star did gain $mc^2$ instead of $0.8mc^2$.
So what mistake am I making?
| As you wrote yourself "this energy is converted to heat and radiation".
Especially the radiation will not be captured by the neutron star, I will simply radiate away as the particle falls towards the neutron star.
As for the heat, it will also radiate way, but over longer time-scales.
So if you subtract the radiation and heat, I suspect you are left with an energy of $0.8mc^2$.
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Do physicists use particle "energy" to refer to kinetic energy? In 1963, this paper was written about the effects of radiation on solar panels. The paper states that:
When electrons at energies greater than 145 KeV and protons at energies
greater than 98eV bombard a silicon crystal, they can displace an atom from the
crystal lattice, producing a lattice vacancy and a recoil atom which comes to rest
as an interstitial atom.
However, the resting energies of electrons and protons are far greater than this, at roughly 511 KeV and 938 MeV respectively. I concluded that the paper was referring to kinetic energy rather than total energy, and adjusted my calculations based on this conjecture.
So: Was I correct to assume that the paper referred to kinetic energy, or was it instead some other measure of the particles' energy?
More generally, is there a standard meaning for a particle's "energy" when referring to such particles moving at relativistic speeds?
| Yes. Bombardment implies kinetics. The (rest) mass of electrons and protons is fixed, it would make no sense to discuss it as a variable. More generally, a particle's energy can be considered to be composed of its rest mass, its kinetic energy, and it's potential energy. There's no single meaning to the term, in fact, energy is an abstraction meaning there is no substance or thing called "energy".
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Translation of Vectors I am a bit confused about translation of vectors. In the first class in physics itself we are told that we can translate vectors as we like to the desired position to do whatever that we are trying to do. For example, if someone draws two random vectors then to get the sum, we translate them, make a parallelogram and draw the diagonal as the resultant.
However I have some doubts on this. In the following example, clearly we cannot translate the vectors.
Consider this rigid body. We want to calculate the torque about origin of a force.
Now if we translate the force vector, then we would obtain the following.
Obviously the situation are very different and its not equivalent.
So are we really allowed to translate vectors?
| What you are describing is a property of vectors. Vectors are not defined by their location in space. They are only defined by their magnitude and direction.
Intuitively speaking, a vector describes displacement from its start point to its end point. The displacement between these two points is only defined by how much space there is between them, and the direction the start point "faces" across this space. It doesn't matter where in a plane or space the vector is because the displacement it describes is the same. If I have a three-dimensional force vector with a magnitude of 10 N facing north (I use "north" very loosely), regardless of whether this vector is positioned here on Earth or on the Sun it is identical in quality.
As you say, when working with vectors algebraically their position is of no importance because the displacement they describe is not affected by their position. However, this is simply not true when describing forces in the real world. A force, as a vector, is still defined by just its magnitude and direction. The way a force interacts with the real world does not change its properties. Nevertheless, where we direct that force affects the translational, rotational, and vibrational motion of the object. The position vector is more of a mathematical tool than anything else. The way moment is defined, taking the cross product of the position vector and the force vector will give us the moment about the start point of the position vector. The position vector, similarly, is still defined only by its magnitude and direction, regardless of its application in real life.
| {
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is spontaneous magnetization and magnetic susceptibility a thermodynamic properties? Are spontaneous magnetization and magnetic susceptibility a thermodynamic property? How do you determine whether a property is thermodynamic or not?
| A macroscopic magnetic system is a thermodynamic system like that of a hydrostatic system. The former is described by the thermodynamic variables $(\textbf{H},\textbf{M},T)$ and the latter by $(P,V,T)$. Therefore, magnetization $\textbf{M}$ (not necessarily spontaneous) is a thermodynamic property and it's the analog of volume $V$ for a hydrostatic system or a fluid. The magnetic field $\textbf{H}$ is the analog of fluid pressure $P$. The susceptibility is a response function for magnetic systems and is the analog of the compressibility of a fluid system.
If by thermodynamic property you mean gross property of the system then yes. Both the magnetization and susceptibility are properties belonging to the system as a whole.
But if by thermodynamic property you mean thermodynamic coordinates or state variables then those are the set of macroscopic quantities required to completely describe the state of the system at a particular instant of time in equilibrium. They are $(\textbf{H},\textbf{M}, T)$ for a magnetic system.
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Can you build a compass that is attracted to the South Pole? Was just curious, since all compasses point to the North Pole.
South is just the opposite polarity of of North, so it seems very likely, but I've never seen an example of this. Is there any videos demonstrating this?
Could a South attractor be added to a standard compass to help confirm the integrity of the North's signal? (For situations where the compass is being affected by another magnetic source).
| If possible do as @AccidentalFourierTransform explained in a comment, namely:
Get a standard compass. Clean off the paint in one end of the needle, and pain the other one. Congrats, now you have a compass that is attracted to the South Pole!
Be aware, that some compasses are embedded in an oil capsule, so disassembling will destroy them mostly.
Also remember that the needle uses always both poles, as the magnetic field influences the structure of the needle as a whole. Otherwise it wouldnt't work, so the easiest is to use your imagination.
Perhaps you meant the declination which must be adjusted on the compass, which can influence the way you've got to take, when hiking close to the north or south? Navigation 101
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Is the Moon in a "Freefall" Around the Earth?
The force of gravity keeps our Moon in orbit around Earth. Is it correct to say that the Moon is in “free fall” around Earth? Why or why not?
I think the answer is yes. The moon is falling towards the Earth due to gravity; but, it's also orbiting the Earth as fast as it's falling towards it. This balance between the 2 forces means the moon is essentially "freefalling" towards the Earth. Is my thinking correct? Thanks.
| The moon is falling towards but "missing" Earth. Had it no sideways motion, it would certainly just fall straight down and crash. So yes, you can say that.
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Can relative velocity be found? Can you find the relative velocity between two cars with constant velocity of U1 and U2=-U1. In other words is there any kind of experiment you can do to understand that you are moving also and not just seeing the other car with 2U1?
|
"...to understand that you are moving also and not just seeing the other car with 2U1"
That you are moving with respect to the ground? Yes. You can look at the ground and see that it is moving relative to you, so you must be moving relative to it.
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Brewster's Angle In the explanation of Brewster window, wherever in the literature I have looked up, it considers the unpolarized light with only two polarization components: the s and p components.
But, in case of unpolarized light, the electric field oscillates in all directions, so at Brewster angle the p polarized light will not be reflected, I understand. But what about the other electric field components, other than s one? I suppose they will be reflected too. If this is the case, how does it come that the reflected light is completely polarised?
Thanks!
| Every linear polarization can be decomposed along two directions, one perpendicular to the other.
You correctly know that one of them will be partially reflected and partially transmitted, while the other completely transmitted. For a linear polarization along one of these two directions, you have a correct understanding of what is happening. For every other direction, you can think at it as the superposition of two components, one in each direction; therefore part of it (corresponding to the component in the direction in which we have also reflection) will be partially reflected and partially transmitted, while the other part (corresponding to the component in the direction in which we have only transmission) will be transmitted.
This results in a linear polarization of the reflected wave.
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Experimentally measure velocity/momentum of a particle in quantum mechanics In the context of quantum mechanics one cannot measure the velocity of a particle by measuring its position at two quick instants of time and dividing by the time interval. That is,
$$ v = \frac{x_2 - x_1}{t_2 - t_1} $$ does not hold as just after the first measurement the wavefunction of the particle "collapses".
So, experimentally how exactly do we measure the veolcity (or say momentum) of a particle?
One way that occurs to me is to measure the particle's de Broglie wavelength $\lambda$ and use $$p = \frac{h}{\lambda}$$ and $$v = \frac{p}{m}$$
to determine the particle's velocity. Is this the way it is done? Is there any other way?
| Your method for measuring observables is perfectly good but there are many other ways to measure observable quantities.
Firstly, there is no perfect way to measure these observables, but the most commonly used one is to measure its deflection when it is passing through a magnetic field. In cloud chambers, charged particles are passed through a magnetic field of known strength $B$. Using the formula $R=\frac{p}{qB}$, where R is the radius of the circle that is formed when the charged particle moves into a magnetic field, the momentum and velocity can be calculated. This method is used in many places like CERN.
Even though this method works only for charged particles, most particles in the Standard model are charged and deflect when they are passed through a magnetic field.
EDIT 1: For specific observables however, there are certain experiments such as for Spin there is the Stern-Gerlach experiment.
Hope this helps
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Transforming a sum to an integral: why does it work? The problem at hand has been discussed in loads of previous questions (1, 2, 3)), and my version can be stated as follows.
Consider the sum $$\sum_{\mathbf k} \ln(1+e^{-(\alpha+\beta \varepsilon_k)}) \ .$$ We are summing over, say, two dimensional k-space lattice $\mathbf k = \frac{2\pi}{L}(n_x,n_y),$ where the $n_x,n_y$ run through the positive integers. I want to understand how we can write this as an integral. Our energy is given by $\varepsilon_k = \frac{\hbar^2 k^2}{2m}$.
| Why does it work? By assuming the separation distance between the points in the discrete space is negligibly small compared to the total volume, we can make use of the definition of the Riemann integral.
Say we are summing over discrete points $\mathbf k$ in k-space $$\sum_\mathbf{k}\ f(\mathbf k)\ .$$
If we want to rewrite this as an integral we have to assume that the
separation between the points in k-space $\Delta k$ is negligible in
comparison to the volume of the k-space $V$.
In the following we can then use the definition of the Riemann
integral
\begin{align}
\sum_{\mathbf k} f(\mathbf k) &= \frac{1}{\Delta k} \sum_{\mathbf k}
f(\mathbf k) \Delta k \\
&\equiv \frac{1}{\Delta k} \int_{\mbox{all space}} f(\mathbf k) d\mathbf k
\ ,
\end{align}
where, in the last step we used our assumption that our seperation
distance is negligibly small compared to the volume of the whole
space.
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Bounce of a ping pong ball vs bounce of other things (How does the material affect ?) Yesterday I saw that a ping pong ball bounced nicely on a hard surface but poorly on a mattress.On the contrary ,I saw that other things (a pen for example) bounced poorly on a hard surface but nicely on a mattress
Please I want to know the reason behind the motion and more importantly why the difference is there
EDIT:
I found that certain materials rapidly lose their Kinetic energy on striking hard surfaces but a mattress manages to store the kinetic energy of the object as elastic potential energy and returns it in accord with its elasticity.
What I find intriguing now is the properties of the material which contribute to loss of KE. Any help is highly appreciated.
NOTE: I ensured that the pen hit the surface with minimum area in contact (the side from which the nib comes out) and the nib was removed.
The pen material didn't look like it could be easily be deformed.
| See here the shape of the object matters.
As the ping pong ball has following two reasons -
*
*The spherical shape provides it a perfect impact while returning through the hard surface. While the mattress has the resistance towards the motion of the ball and lets it loose its impact very quickly.
*The pen has the unfurnished (as compared to the ball) which effects its bounce on the hard surface and even lets it loose the energy within the short time which is the reason it doesn't bounce or return back. While mattress provides a good amount of unbalanced forces on the pen.
$$Impact={1\over A}$$ (here $A$ is the area of the object)
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What will be the velocity component in $x$-direction?
In the figure, the particle is hitting the surface at an angle $\theta$ and velocity $V_2$ along the $y$ direction.
Is there a name for this velocity? Can it be called orthogonal velocity?
I have worked out the velocity to be as following:
$V_x = \frac{V_2}{\cot(\theta)}$.
Kindly guide me with this.
| I think you would describe the direction of $V_2$ as oblique, in contrast to normal (perpendicular to the surface) or glancing/grazing (almost parallel to the surface). The same terminology is used for light rays.
Well that is your conceptual question answered, so no more guidance is needed - ok?
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Relation between source slit aperture and it's distance from the plane of the secondary slits in Young's Double Slit Experiment In Young's Double Slit Experiment (YDSE), the relation between the source aperture (my book says source size... I'm assuming they mean aperture) s and distance from the plane of the two slits S must be such that s/S < λ/d (angular width) for the interference fringes to be obtained. Where λ is the wavelength of the light used and d is the distance between the two slits. Where'd this relation come from? And why must it be so?
| As one could see s/S < λ/d is an inequality and not an exact equation. It's a recommendation and advice based on experience.
To draw the full picture let me tell you some historical facts. At the beginning the scientists (Grimaldi in the sunny Italy describes this) used a pinhole in a darkend window (and a mirror in front if the window to direct the suns light into a horizontal direction through the window) and bird feathers as multi-slits. What they got were blurry fringes of different colors.
Image from Wikipedia.
Later it was used monochromatic light and double slits. And it was detected that fringes occur behind small openings too (Airy disc).
Image from Wikipedia.
Even later it was investigated that single photons over time produce such intensity distributions too. To bring it to the point, behind every edge light gets deflected into a intensity distribution. But we don't see this in our everyday life. Why?
Having an extended light source the light will strip edges from different directions and a shadow behind this edge not only not show fringes but wasn't sharp at all. The advise for slit experiments is to use a point-like source of light. The better advice would be to use light from a collimator like this
Image from Wikipedia.
So the advise of the inequality s/S < λ/d one has to follow to get nearly light of parallel rays to get well separated fringes.
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Why does a simple pendulum or a spring-mass system show simple harmonic motion only for small amplitudes? I've been taught that in a simple pendulum, for small $x$, $\sin x \approx x$. We then derive the formula for the time period of the pendulum. But I still don't understand the Physics behind it. Also, there's no angle $x$ involved in a spring-mass system, then why do we consider it an SHM only for small amplitudes?
| It just means that the pendulum will only execute SHM with small angles for which the pendulum subtends as it oscillates.
For larger angles the motion of the pendulum deviates from being simple harmonic; that's why the small angle approximation is required in the derivation.
EDIT:
Even for a horizontal mass-spring system the displacement of the mass from its equilibrium position cannot be made too large, otherwise the simple-harmonic motion ceases to be valid.
But whether considering a pendulum undergoing SHM or a mass-spring system the displacement from equilibrium position is $\propto$ to the negative of the acceleration towards the equilibrium point as these are in opposite directions which is the reason for the minus sign.
| {
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Current as the time derivative of the charge I have been told that the current $i$ can be defined as
$ i = \displaystyle\frac{dq}{dt} $,
where $q$ is the charge and $t$ is the time.
I do not understand this definition because, if the charges are moving so that the net charge remains constant in an infinitesimally thin cross-section of a wire, $q$ is constant with time and hence $dq/dt = 0$. That result would mean that no constant current can exist unless the charge change has a linear dependence with time (i.e. $q = q(t) \propto t$).
As I assume my reasoning is wrong, where is my mistake?
Thank you.
| Realise that, in case of electric current, it is the free electron that moves while the nucleus stays fixed. Hence, when we say charge, in this case, we actually refer to the free electrons that are moving inside the wire.
We do not consider the positive nuclear charges. Obviously the net charge is $0$ across any cross section if we consider the positive charges; else there will be a non zero electric field inside the wire and the current flow will be in haphazard directions.
And, to add more, current is the flow of charges through a conducting wire. Since only the free electrons are mobile while the nuclei are not (from the frame of the observer standing outside the wire on the ground), the flowing charges are the electrons only and that is what matters most, as I have mentioned earlier.
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Condensation of Water. Classroom Controversy In our test there was a question that went like so:
Question 4 You have a glass of iced water on an unshaded picnic table and went for a walk for 30 minutes. When you return you noticed
the glass has water on the outside of it.
a. In terms of heat transfer explain what has happened to the glass of water.
The majority of the class understood this question and answered it correctly.
b. Would there have been more or less water on the outside of the glass if the picnic table was in the shade? Explain.
This question caused a lot of controversy with the majority of the students (including myself) believing that the shade would have caused more water on the outside. Whereas the teacher and a few students thought that the sun would have caused more water on the outside.
The reasoning that the teacher provided was not very convincing and so we have come to this forum to ask what is the correct answer to part b and most of all WHY?
We are 16 -17 years of age if you need to know the level for the explanation.
| I noticed that one commenter said that the question is inherently ambiguous concerning whether you count condensation that has rolled off the side of the glass.
Question: Would there have been more or less water on the outside of the glass if the picnic table was in the shade?
Indeed, if I were to interpret it in a particularly amusing way, in the shade less water would evaporate from inside the glass into the air, and hence there would be less water on the outside of the glass.
That aside, another commenter also mentioned that you can't expect the ice to remain unmelted if you leave it in the sun for half an hour, and it is very likely that after the ice has melted the sun will dry up the outer side of the glass by evaporation. In the shade it is easy to imagine the glass of iced water remaining iced when you come back, so there will still be water condensing on the side of the glass.
Therefore I don't see much need for complicated reasoning here.
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Is it possible to make a destructive interference generator? Is it possible to make a light source that shifts waves of light to cause destructive interference to cancel the other light source out to make it fully dark?
| in theory yes you can do it. But real word is harsh, the light you see every day is mix of almost all wave lenght so to interfire whit that you would need emiter that send all waves but moved by pi/2 (to make it simple i asume every wave is sinusoidal). It might work at still viev but at dinamic you wolud have to know the future to destroy incomig light. Beside that it will work as long as you stay behind you foton destroyer (sorry for broken eng)
| {
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Can an object falling in vacuum generate electricity by itself? When an object falls through vacuum, gravitational potential energy is converted to kinetic energy. Is there some way to get electrical energy out of the equation by itself (i.e. somehow convert the gravitational potential energy to electrical energy)? Is this physically possible? If so, what properties must this object have?
By by itself, I mean without using any external (possibly stationary) "reference object" (e.g. a stationary coil), so a magnet falling through a coil does not count, i.e. the electricity is generated solely by the object that is falling. Note that the object itself can be arbitrarily complex internally, just that whatever mechanism it has inside must also be falling along with the object.
| Floris posted an answer that assumes the object is large enough for different parts of it to experience noticeably different gravitational forces. This is one way to accomplish it and written rather well, so I won't discuss that case further.
If you require that the object is small enough that all parts of it would be at approximately the same gravitational potential as every other part at any given time, then the answer is "no". Free fall is a geodesic, which means in its frame, the internal components would experience no real difference from a situation where it is not falling, so there wouldn't be a change that would allow it to produce energy for itself.
Looking at it a slightly different way, converting the gravitational potential energy into electrical energy would mean not all is converted into kinetic energy, which means you'd effectively be slowing the fall of the object compared to something not producing electricity. You already said we can't have it interacting with the massive body through anything but gravity, so you can't have the fall slowed by anything and, thus, all energy must transform to kinetic.
| {
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Can Zener Breakdown be converted to Avalanche breakdown? Wikipedia says:
The Zener effect is distinct from avalanche breakdown. Avalanche breakdown involves minority carrier electrons in the transition region being accelerated, by the electric field, to energies sufficient for freeing electron-hole pairs via collisions with bound electrons. The Zener and the avalanche effect may occur simultaneously or independently of one another. In general, diode junction breakdowns occurring below 5 volts are caused by the Zener effect, whereas breakdowns occurring above 5 volts are caused by the avalanche effect. Breakdowns occurring at voltages close to 5V are usually caused by some combination of the two effects. Zener breakdown voltage is found to occur at electric field intensity of about 3×107 V/m.[1] Zener breakdown occurs in heavily doped junctions (p-type semiconductor moderately doped and n-type heavily doped), which produces a narrow depletion region.[2] The avalanche breakdown occurs in lightly doped junctions, which produce a wider depletion region. Temperature increase in the junction increases the contribution of the Zener effect to breakdown, and decreases the contribution of the avalanche effect.
My question is can in a diode, primarily under Zener breakdown, increasing the reverse-bias voltage cause avalanche breakdown ?
| Once you have any kind of breakdown, it dramatically reduces the differential resistance contributed by the depletion region, so if you double the external voltage, you're not doubling the voltage across the depletion region, but increasing it just a little bit, while an increasing fraction of the voltage drop occurs in the contacts and bulk semiconductor and elsewhere.
So, I'm going to say probably not in practice, or at least not much. But in principle, if you crank up the external voltage enough, you will eventually get enough voltage across the depletion region to induce avalanche current, which would supplementing the (presumably) much much larger amount of zener current. (I could be wrong, this is just a guess.)
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Looking for a good casual book on quantum physics I'm looking for something that is going to blow my mind without any scientistic ideas (e.g. something that sounds like science, but doesn't have anything in common with science), written by a professional physicist who spent a lot of time considering "what it all means". I'm reasonably proficient in math and stats, but I'd prefer something that I could spend time listening to in my free time.
Any recommendations on good and exciting books on quantum physics written by scientists?
| I very much enjoyed "The Quantum Universe: Everything That Can Happen Does Happen", by Brian Cox and Jeff Forshaw. I believe that it was written for a slightly-above-lay audience, so your knowledge of maths and stats should be more than sufficient to get through it. The book has a witty and conversational style, but is not in any way dumbed-down. Both authors are academics and public science communicators by trade: Jeff Forshaw is a Professor of Particle Physics at Manchester, where his co-author Brian Cox is an "Advanced Fellow" in the same field, and a well-known TV personality.
I should point out as an aside that the title "Professor" has more gravitas in the UK than in some other Western nations; specifically it is not equivalent to a lecturer. However, I was unable to produce the name of a specific Chair occupied by Jeff Forshaw.
| {
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What does one second after big bang mean? Consider the following statement:
Hadron Epoch, from $10^{-6}$ seconds to $1$ second: The temperature of the
universe cools to about a trillion degrees, cool enough to allow
quarks to combine to form hadrons (like protons and neutrons).
What does it mean to say "from $10^{-6}$ seconds to $1$ second"?
How is time being measured?
One particle might feel just $10^{-20}\ \mathrm s$ having passed and another could feel $10^{-10}\ \mathrm s$ having passed.
Is saying "1 second after the big bang" a meaningful statement?
| While not at all obvious, it turns out that our best models of cosmology suggest that there exists a special frame of reference in which the distribution of the entire universe's matter and energy appears extremely uniform on very large (i.e. cosmological) scales. When we talk about the age of the universe, we always mean the age as viewed in this special frame. You are completely correct that particles that are moving very quickly with respect to this special frame will measure a very different age of the universe.
(Note that the existence of this special frame is completely compatible with special and general relativity, which say that the laws of physics themselves look the same in any inertial reference frame, not that the distribution of matter does.)
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Method of image charges for a point charge and a non-grounded conducting plane I know how to solve Laplace's equation for a point charge in front of a grounded conducting infinite plane. But I want to know what happens (both physics and math) when the infinite conducting plane isn't grounded, or is connected to a potential $V$.
| If you would like to define V at infinity as zero and the conducting plane as not grounded, you can also think of the solution as a superposition of two different elctrostatic cases: Take the fields expression of a single charge and a grounded plane, and sum this with the fields given off by a plane of fixed potential.
| {
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Density of states of Bogoliubov quasiparticles For a simple fermionic system the formula for calculating the density of states (DOS) is $N(E) = \sum_{n}\delta(E-E_{n})$ where $\{E_{n}\}$ is the set of eigenvalues obtained after diagonalizing the hamiltonian. Now to diagonaloize a hamiltonian with pair correlation terms ($\sum_{k}c_{k\uparrow}^{\dagger}c_{-k\downarrow}^{\dagger}$) Bogoliubov transformation ($c_{k\uparrow}=u_{k}\gamma_{k\uparrow}-v_{k}^{\ast}\gamma_{-k\downarrow}^{\dagger}; c_{-k\downarrow}^{\dagger}=v_k\gamma_{k\uparrow}+u_{k}^{\ast}\gamma_{-k\downarrow}^{\dagger}$) is used. Now after diagonalizing we get a set of eigenvalues in the form:$\{E_n,-E_n\}\forall n$. Now to find the density of states I found a formula like this: $N(E)=\sum_{k}|u_k|^2\delta(E-E_k)+|v_k|^2\delta(E+E_k)$ where $\{E_k\}$ is the set of positive eigenvalues only. I don't understand this particular formula for density of states of bogoliubov quaisparticles. If anyone can explain it that would be very helpful.
| Some information is missing, but I think that maybe if you expand the terms ($\sum_{k}c_{k\uparrow}^{\dagger}c_{-k\downarrow}^{\dagger}$) with the Bogoliubov transformation, some ortogonal operators may cancell and so you can separate the hamiltionian in terms of each $\gamma$ operator and $|u_k|^2$ and $|v_k|^2$. After that maybe you can separate the whole system into positive and negative energy states.
| {
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Would a gas "weigh" less than a liquid if they have the same mass? Thought experiment: I acquired two boxes of the same dimensions and same weight. One box contains $1\ \mathrm{kg}$ of water at room temperature while the other box has $1\ \mathrm{kg}$ of water, but in steam form, because the temperature of the box is above $100^\circ\mathrm{C}$. The volume of the boxes is large relative to the amount of space the $1\ \mathrm{kg}$ of water would take (let's arbitrarily say $10\ \mathrm{L}$). Both boxes contain the same amount of air (at $1\ \mathrm{atm}$) which is why the second box has water in steam form at $100^\circ\mathrm{C}$.
I put each box on a simple electronic scale to measure their respective weights. Unsurprisingly, the box containing water comes out to be $1\ \mathrm{kg}$. But what about the box containing steam?
My guess: Electronic scales measure the amount of force being exerted on it, then divide that force by $g$, to get the mass of the object. I think the box with steam in it will be exerting less force on to the scale and therefore the scale will think its mass is less than $1\ \mathrm{kg}$.
| If both boxes are the same size and weight, contain the same mass of water, and the same mass of air, the weight of both boxes will be the same, and the buoyant force on the boxes from the air that they displace will be the same. With "all things being equal", the two boxes will weigh the same when put on the electronic scale. However, all things are not equal. The box with the steam in it is substantially hotter than the box with the water in it. Due to this, the hot box will heat up the pan of the electronic scale, causing the air under the pan to heat up. This will produce a small amount of "lift", because the hot air is less dense than the air that is under the pan of the scale when the "water" box is weighed. Assuming a high precision for the hypothetical electronic scale, this will lead to a lower weight reading for the box containing steam than the box containing water.
| {
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Schrödinger equation and non-Hermitian Hamiltonians Is the Schrödinger equation still valid if we use a non-Hermitian Hamiltonian with it? By this I mean does:
$$\hat{H}\psi(t) = i\hbar\frac{\partial}{\partial t}\psi(t)$$
if $\hat{H}$ is not Hermitian?
| There is nothing stopping you from writing $\hat H\Psi(x,t)=i\hbar \partial_t\Psi(x,t)$ for arbitrary $\hat H$. The physics is in $\hat H$, not in the differential equation.
Now, if you want to get there, why not write even more generally
$$
\hat{\cal O}\Psi(x,t)=\partial_t\Psi(x,t)
$$
and get rid of the $i\hbar$ factor?
| {
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Doppler effect differs when the entity moving is different
Why is the doppler shift different when whether the source is moving or the observer? Although they are traveling at the same speed. Noticing that the difference can't be neglected when the speed is a big fraction of the speed of sound.
| You would think that it wouldn't matter whether it's the source or the observer that's moving in the doppler effect. An argument might be that when the observer is moving, you can just choose a difference reference frame in which the observer is instead stationary and the source is moving, or vice versa.
The problem with this argument is that the medium, air, can also be travelling. In shifting reference frames, the velocity of the medium is also changing, so we can't change reference frames without also altering the physics of the problem. There's actually a version of the formula that takes the velocity of the medium into account, that resolves this apparent paradox.
The result is that the source and observer velocities aren't interchangeable.
| {
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What kind of wave motion is described by grass moving in the wind? You know that sort of 'rolling' illusion when wind blows across long grass, like in the "amber waves of grain" line from America the Beautiful
It's not the same motion as dropping a rock in water, which causes an up and down motion.
And if wind blows across gravel, or water, it just shears it. The water or rocks don't recover like the individual grass blades do.
It's nothing like plucking a string, but kind of like plucking a cantilever beam. Except, I'm curious about what the aggregate motion of hundreds/thousands of blades of grass is known as
| There are three basic kinds of mechanical waves: Transverse waves, longitudinal waves, and surface waves. Surface waves propagate along an interface between differing media.
One way to describe wind blowing across tall grass is as a surface wave. The grass is held in place by its roots. The tops and sides of the grass present a surface to the wind similar to water, which is held in place by gravity and surface tension. Wind blowing across the surface of water creates circular motion of water molecules that appears as surface waves. Likewise, wind blowing across grassland may create a modified circular motion of the grass stalks.
The grass rises and falls in sinusoidal waves subject to the restoring force of the grass stalks and the density of the grass.
In the case of waves through grass, most of the energy probably is carried by air, although some may be transferred from grass stalk to stalk.
Another possibility may be to describe this phenomenon as an unstable bedform. The shapes are wavelike, like ripples in sand, but grass doesn't hold its shape as long as sand because of the restoring force of the grass blade. The wave shapes are caused by variations in windspeed.
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Liquid vs. gas cooling I have an aluminum can that needs cooled. I put it in the refrigerator, where it is cooled by the cold air surrounding the can.
If I were to place the same can in water that had been cooled to the same temperature as the refrigerated air, would the can cool down faster?
I'm inclined to assume the liquid would cool the can faster due to the density of the molecules surrounding the can, but I am interested in what people who know physics have to say.
| Heat transfer can occur by radiation, convection (natural or forced), conduction, or through a phase change. For the case of an Al can whose temperature is changing by a few degrees, it may be the case that conduction dominates (and conduction also plays a part in convection).
The most important material property in conduction is the thermal conductivity, which characterizes the rate of heat transfer for a given temperature difference. The thermal conductivity of a material depends more on its bonding type than its density; conductive heat transfer is much more efficient in strongly bonded solids (in which the lattice vibrations that carry thermal energy can propagate quickly) than in liquids and much more efficient in liquids than in gases (in which molecular interactions are relatively infrequent). But it is true, as you noticed, that denser types of matter generally conduct heat better.
Air's thermal conductivity is about 0.02 W/m-K; water's is about 0.6 W/m-K.
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Ball inside an accelerating frame Why does a ball inside a moving bus at rest start moving backwards when the bus suddenly accelerates? Also does the moving ball have some acceleration?
This is my theory:
Initially the bus and the ball are at rest. When the bus starts accelerating, due to inertia of rest, the ball resists change in motion and tends to remain at rest. Since it is an accelerating frame, pseudo forces of unknown origin act on the ball in the direction opposite to the direction of motion of the bus. This pseudo force is responsible for the ball to move in the opposite direction with some acceleration. This explanation with respect to the accelerating frame.
But how do I explain this fact from the frame of reference of a person on the road? I understand that the acceleration of the ball is in the same direction as that of acceleration of bus. There is no other force to balance this force which is in the direction of motion of bus. So the ball should have moved in a direction along the direction of the bus. But the reverse happens. Please help me.
| Analyse it this way .
If the bus is infinitely long such that there is no force from the walls and there is no friction also. Then in the road frame the ball stays where it is ( because no force acts on it ) while the bus moves forward.
In the bus frame the bus is at rest while to the bus the ball accelerates back which you have explained by the pseudo force .
Friction enters then rolling enters the scene sort of complicating the scene .
If there is Friction
Let u be the coefficient of friction then
From ground frame the FBD of the ball is
f=uN=ma towards the right.
From accelerated bus frame
f=uN=m(a+A) (towards the left)
mA being the pseudo force
| {
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How do I find the time evolution of a ket? I have a question which reads:
Let \begin{bmatrix}
{E_0} & 0 & A \\
0 & E_1 & 0 \\
A & 0 & E_0
\end{bmatrix}
be the matrix representation of the Hamiltonian for a three-state system with basis states $|1>, |2> \mbox{and } |3>$.
a. If the state of the system at time $t$ = $0$ is $|\psi(0)>=|2>$ what is $|\psi(t)>$?
b. If the state of the system at time $t$ = $0$ is $|\psi(0)>=|3>$ what is $|\psi(t)>$?
$\textbf{My attempt at a solution:}$
a. For both problems we can use $|\psi(t)> = \hat{U}(t)|\psi(0)>$ where $\hat{U}= e^{\frac{-i\hat{H}t}{\hbar}}$. Since
$$|2> = \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}$$
is an eigenvector with eigenvalue $E_1$ we can simply replace the Hamiltonian in the time evolution operator by $E_1$, so $$|\psi(t)> = e^{\frac{-iE_1t}{\hbar}}|2> $$
Is this correct? I am finding other solutions online which have a different answer, although I can't see how this could possibly be wrong, unless my representation for $|2>$ is wrong.
Assuming this is the correct way of doing this, I am having a hard time doing b. I can find the eigenvalues and eigenvectors of the hamiltonian easily, and can represent |3> = $(0,0,1)^T$ as a linear combination of those vectors, thereby allowing me to operate on it. However, my final answer is in terms of |1> and |3>, which I feel is incorrect somehow.
| Your reasoning is perfectly correct. Here it is in a complete form.
Let us write the Hamiltonian in the following way to make things clearer
$$ \hat{H} = E_0(|1 \rangle \langle 1|+|3 \rangle \langle 3|) + E_1|2 \rangle \langle 2| + A(|1 \rangle \langle 3| + |3 \rangle \langle 1|) $$
It is then straightforward to see that :
*
*$|2 \rangle$ is an eigenstate with eigenvalue $E_1$ as you have already noticed. Hence if the initial state is $|2 \rangle$ then $$|\psi(t)\rangle = e^{-iE_1t/\hbar}|2\rangle$$
*$(|1\rangle+|3\rangle)$ and $(|1\rangle-|3\rangle)$ are eigenstates with respective eigenvalues of $E_0 + A$ and $E_0 - A$. Hence if the initial state is $|3\rangle = \frac{1}{2} [(|1\rangle+|3\rangle) - (|1\rangle-|3\rangle)]$, then $$|\psi(t)\rangle = \frac{1}{2} \left[ e^{-i(E_0+A)t/\hbar}(|1\rangle+|3\rangle) - e^{-i(E_0-A)t/\hbar}(|1\rangle-|3\rangle)\right]$$
I hope my explanation was clear !
Cheers
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What is the difference between mean free path and intermolecular distance? Why is the mean free path not be equal to the intermolecular distance?
A particle moving in a particular direction should strike the object in that direction after the traveling the same distance as the distance between them initially.
| The difference lies in the cross section of the particles.
Consider two equally large volumes containing an equal amount of particles, but the particles in volume A are twice the radius of the particles in volume B. In this case, the inter-particle distance is the same in both volumes, but the mean free path in volume B is four times the mean free path in volume A.
If the density is $n$, and the cross section of the particles is $\sigma$, then
$$\mathbf{mean \,interparticle\, distance\!\!:}\,\qquad \langle r \rangle \sim \frac{1}{n^{1/3}} \qquad\mathrm{(independent \,of \,cross\,section)},
$$
while
$$
\mathbf{mean \,free\, path\!\!:}\,\qquad \ell = \frac{1}{\sigma n} \qquad\mathrm{(inversely \,proportional \,to \,cross \,section)}.
$$
For a real world example, consider a Lyman $\alpha$ photon that first enters an HII cloud of ionized hydrogen atoms with a density of $1\,\mathrm{cm}^{-3}$, and subsequently an HI cloud of neutral hydrogen atoms with the same density. The clouds have the same inter-particle distance, but the while the HII cloud is transparent to the Lyman $\alpha$ photon (because the probability of interaction is virtually zero), the HI cloud will likely scatter the Lyman $\alpha$ photon multiple times.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minkowski spacetime with a twist? Minkowski spacetime the has a flat metric of Lorentzian signature (-1,1,1,1).
It is well known (c.f. Geroch 1967 & citation there) that whether a manifold admits a metric of Lorentzian signature is equivalent to the question of whether it admits a nowhere vanishing timelike vector field. By Geroch's theorem the spacelike hypersurfaces of Minkowski spacetime are diffeomorphic to each other, the diffeomorphism being produced by the integral curves of such a timelike vector field.
Since the metric on each spacelike slice of Minkowski spacetime is the (flat) Euclidean metric, the diffeomorphism must be an isometry, but among the possible isometries are rotations.
Question: is there any sense to the idea any each slice may be related another by a rotation? What physical or mathematical justification would there be for [disallowing] allowing such diffeomorphisms? (If the timelike vector field had curl, could one have a corkscrew hole in ~Minkowski space?)
The independent rotation of spatial hypersurfaces being something like this...
| Let me try answering this (assuming I understood the question).
In 4 dimensions you do not have a unique axis of rotation. Instead, there are two of them (so-called stationary plane which is fixed under a rotation). So in Minkowski space, when you rotate something around a time axis, you also rotate it around one of spatial axes (say $z$), and it will be the same as rotation in 3D around $z$. If you instead fix stationary plane to be purely spatial, then you get a Lorentz boost.
Feel free to correct me if anything.
| {
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When can one omit a total time derivative in the Lagrangian formulation? I am studying Lagrangian and Hamiltonian mechanics and i am using Landau & Lifshitz and Goldstein books. Both of them state that a modified lagrangian $$L'=L+\frac{df}{dt}$$ gives the same solutions than $L$ wich i kind of understand but its not the main problem.
In landau there is a problem in which a pendulum whose attachment point is oscillating. setting up the equations is not a problem for me, but when he gives the solutions he states that he omits total time derivatives as if it was the most obvious thing to do, i guess that this omission is related to the "invariance" of the Lagrangian but i fail to se the direct relation with this. How do you know from which function are you supposed to omit the total time derivarive?
How do you identify this function?... etc etc
The example that i'm talking about is in Landau & Lifshitz book page 11 exercise 3)b.
| Remember that the physics you get from a Lagrangian is due to a variational problem where you seek to extremize the action
$$S = \int_{t_1}^{t_2} L \, dt$$
So, the reason you can remove a total time derivative from your Lagrangian is because its contribution to the action is fixed:
$$\int_{t_1}^{t_2} \frac{df}{dt} \, dt = f(t_2) - f(t_1)$$
and thus has no impact on the variational problem. As for which total time derivatives to remove, that can be a bit of an art. With experience you can see in advance that shifting the Lagrangian by a total derivative might simplify things.
Also note that the boundary conditions play a role here, as you can clearly see above, where one gets $f$ evaluated at $t_1, t_2$. Usually this doesn't matter, especially in simple mechanics problems, but it's worth remembering for those odd cases where the boundary conditions become important.
| {
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Why does space have the topology of a three sphere? Suppose that $U(x)$ is an element of the gauge group say $SU(2)$ and suppose $U(x)=1$ as $|\vec{x}|\to\infty$. Then, why does space have the topology of $S^3$?
This is done in Srednicki page 571. Note that I'm not asking how to prove that $SU(2)\cong S^3$. What I'm asking is how to prove that when $U(x)=1$ as $|\vec{x}|\to\infty$ the space $\mathbb{R}^3$ is compactified to $S^3$ space.
| I think i got it,correct me if i am wrong.
We consider stereographic projection from the North pole $p$. Since Stereographic
projection is a one-to-one correspondence between {$S^n−p$} and $R^n$ and since,
$U(x)=1$ as $|\vec{x}|\to\infty$ we can regard $U(\infty)$ as the image of the
point $p$ then, instead of having a map between $R^n$ and $SU(2)$ we can consider
a map between $S^n$ and $SU(2)$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/317165",
"timestamp": "2023-03-29T00:00:00",
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Conservation of momentum in refraction Light, when passing through a boundary refracts. How is momentum conserved here? There can't be an impulse, the energy doesn't change.
| Momentum is conserved because the refracting medium/media at the boundary experience an equal and opposite change in momentum when the light changes speed and direction.
Light can push things - usually imperceptibly.
I'd stop there - but you did use the "how" word.
It's an electromagnetic interaction. The photon changes energy as the EM oscillations experience a change in local charge density.
Some energy is needed for an EM wave to approach a stationary charge, climbing the electric potential.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/317342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a $n$-dimensional system such that the minimal action from a path from $x$ to $y$ is the scalar product? Suppose we work (with a particle) in $\mathbb{R}^n$.
Is there a Euler-Lagrange equation associated to the particle in question such that the minimal action of all path going from a position $x\in \mathbb{R}^n$, to another position $y\in \mathbb{R}^n$ is precisely the scalar product $\langle x,y\rangle$ ?
| The non-relativistic free point particle with Lagrangian
$$L~=~\frac{m}{2}\dot{\bf q}^2 \tag{1}$$
and with Dirichlet boundary conditions
$$\tag{2} {\bf q}(t_i)~=~{\bf q}_i\quad\text{and}\quad {\bf q}(t_f)~=~{\bf q}_i,$$
has Dirichlet on-shell action
$$S({\bf q}_f,t_f;{\bf q}_i,t_i)~=~ \frac{m}{2} \frac{({\bf q}_f-{\bf q}_i)^2}{t_f-t_i}.\tag{3}$$
The minimum of the off-shell action functional, which OP asks about, is by definition the Dirichlet on-shell action. See e.g. my Phys.SE answer here and links therein.
| {
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If free quarks can't exist, how did the universe form? As I understand, the Big Bang started with a photon gas that then created the other particles. Thus obviously there would be some free quarks in the early Universe unless quarks are always created in pairs for some reason. How does physics resolve this?
| In the first stages of the Universe Quarks and Gluons were asymptotically free. This state of matter is called Quark-Gluon Plasma. Then, as the temperature of the Universe kept decreasing, the so-called hadronization (quarks combine to form hadrons) took place.
The coupling constant of the QCD (which, to make it simple, sort of represents the intensity of the strong interaction between quarks) is a $\textit{running coupling}$: it means it's not really a constant, but it varies with the energy scale. As you can see from the picture below, the $\alpha_{QCD}$ decreases at high transferred momentum. This means that quark tends to behave ALMOST as free particles when the energies are really high. You can also view them as a gas of fermions (quarks) and bosons (gluons).
In these conditions ($\alpha \ll 1$), a perturbative approach is possible: we use pQCD (perturbative QCD).
Quark-Gluon Plasma can be obtained nowadays by high energies collisions of heavy-nuclei. This is achieved at CERN, for example, by the ALICE experiment, by means of Pb-Pb collisions at $5.02$ TeV.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/317672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Gaussian integral formula for matrix product I am looking for a way to prove that
$$ \det (M \cdot N) = \det(M)\det(N) \tag{0}$$
Where $M$ and $N$ are matrices with continuous indices, so that $\det$ is a functional determinant. A way to show that $(0)$ is wrong would also be welcomed.
This question is about the following formula,
$$
\int\text{d}\vec{x} \exp(- \sum_{ij}x^i A_{ij}x^j) = \left (\det A_{ij}\right )^{-1/2}\left (2\pi\right )^{D/2}. \tag{1}
$$
Now, we would like this identity to be compatible with,
$$
\int\text{d}\vec{x} \exp(- \sum_{ijk}x^i A_{ik}B_{kj}x^j) = \left (\det A\cdot B\right )^{-1/2}\left (2\pi\right )^{D/2} = \left (\det A\right )^{-1/2}\left (\det B\right )^{-1/2}\left (2\pi\right )^{D/2}.\tag{2}
$$
Any idea how to prove this? I am interested, eventually, in the generalisation of this formula to path integrals, namely, given the path integral
$$
\int\mathcal{D}\phi \exp\left[- \int\text{d}x\text{d}y \phi(x)M(x,y)\phi(y)\right] =C \left (\det M\right )^{-1/2}, \tag{3}
$$
where now $\det M$ is a functional determinant, i ask the question whether it makes sense to write the generalised formula,
$$\begin{align}
\int\mathcal{D}\phi \exp\left[- \int\text{d}x\text{d}y \text{d}z\phi(x)M(x,y)N(y,z)\phi(z)\right] =& \left (\det M\cdot N\right )^{-1/2}\cr =& \left (\det M\right )^{-1/2} \left (\det N\right )^{-1/2}.\end{align} \tag{4}
$$
[UPDATE]: I might have an answer now: let us just consider,
$$\det M\cdot N = \prod_i \lambda_i[M\cdot N],\tag{5}$$
where $\lambda_i[M\cdot N]$ are the the eigenvalues of the matrix $M\cdot N$. This formula is valid even for continuous matrices, such as the laplacian operator $\partial^2 \delta(x-y)$.
If the commutator $[M,N] = 0$, then the two matrices can be diagonalised in the same basis, and $\lambda_i[M\cdot N] = \lambda_i[M]\lambda_i[N]$, with no sum over $i$. Then formula (4) can be proven at least in the simple case in which the commutator vanishes.
A trivial example of this is for $M = A$ and $N = A^{-1}$, for any invertible matrix $A$, which leads to $\det A\cdot A^{-1}=1$. Also, in case $M\cdot M^T = f(x) \delta(x-y)$, this would imply that
$$\det M\cdot M^T = (\det M)^2 = \det f(x) \delta(x-y) = \prod_x f(x)\tag{6}$$
and so on. These seem trivial cases, but since we are talking of functional determinants they constitute a powerful computational tool.
How much do you agree with this attempt of a solution? It is not very formal, but i don't see where it could go wrong.
| The statement seems to be wrong even for an infinite number of discrete indices.
Consider for example the vector space of square integrable functions on the positive integers, i.e. sequences $\{f_1,f_2,\cdots\}$ s.t. $\sum_{i>0} |f_i|^2 < \infty$, and consider the shift operator $S:f \mapsto Sf$, where
$Sf = \{ f_2,f_3,\cdots \} \ . $
Consider furthermore the operator $S^\dagger: f \mapsto S^\dagger f$ with
$S^\dagger f = \{ 0, f_1,f_2, \cdots \} \ . $
Now
$SS^\dagger f = \{f_1,f_2,\cdots \} = f \ \ \ \text{ but } \ \ \ S^\dagger S f = \{ 0, f_2,f_3,\dots \} \ . $
That is, $SS^\dagger$ has all eigenvalues $1$, while $S^\dagger S$ has one eigenvalue $0$ and all other eigenvalues $1$. Hence
$ \det( SS^\dagger) = 1 \ \ \ \ \ \text{while} \ \ \ \ \det (S^\dagger S ) = 0 \ .$
Now if it were true that $\det(M N) = \det(M) \det(N)$, than we would have proven that $1 = 0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/317771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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