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Why the exponential part of Electromagnetic waves on both side of interface must be equal? Why the exponential part of Electromagnetic waves on both side of interface must be equal which implies equality of phases at the boundary at all the time to satisfy the boundary conditions. My text book and all the textbooks says the same thing without explaining the reason I am very confused. I am attaching a photo of my text book
| For simplicity, assume normal incidence at the point $r=0$, so that all the electric fields are parallel to the interface.
The E-field parallel to the interface is continuous (the same either side of the boundary).
$$ E_i \exp(i\omega_I t) + E_r \exp(i\omega_R t) = E_t\exp(i\omega_T t).$$
Now suppose that $\omega_I$, $\omega_R$ and $\omega_T$ were different.
Let's say that the equality works at $t=0$, so that $E_i + E_r = E_i$, which it has to be, since the sum of the E-field on each side of the boundary must be the same. We can then ask, at what other time is this true? And the answer is that it wouldn't be. There would always be some non-zero phase difference between the components at other values of $t$ (bar some special values if the frequencies were multiples of each other). The only way the relationship $E_i + E_r = E_t$ can be true for all $t$ is if the frequencies were the same.
The argument (and that's a self-contradictory pun!) doesn't change if you add the $\vec{k}\cdot \vec{r}$ parts, because you can always choose some point on the boundary where $\vec{k}\cdot \vec{r}=0$ and demand that the equality is true for all $t$ at that point. In a similar way (probably on the next page of your book) you can set $t=0$ and demand that the equality still holds for all $\vec{r}$ on the boundary - this leads to the law of reflection and Snell's law of refraction.
| {
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Accessible States in the Ergodic Hypothesis According to Wikipedia, the ergodic hypothesis is the assumption that
all accessible microstates are equiprobable over a long period of time.
My question is about the precise meaning of "accessible" here.
Consider the microcanonical ensemble for a thermally insulated system (say an ideal gas of $N$ particles and $E$ units of energy, confined to a box of volume $V$). As far as I can tell, the only thing that "accessible" means here is that the energy is constant, so when computing spatial averages we only average over the states of a given energy.
But why do we only worry about the energy here? Why not worry about only averaging over the states of constant momentum, or angular momentum, or any other quantity that could potentially be conserved?
(I realize, by the way, that momentum is not conserved in an ideal gas confined to a box, but surely there are thermodynamic systems where there is some conserved quantity other than energy, right?)
What exactly is so special about energy here? And are there situations where "accessible states" refers to more than just all the states of a certain energy?
| Generically, you could absolutely fix any number of conserved quantities you like. The energy need not even be among them. The energy is just the most common thing to fix because the second most common conserved quantity, momentum, is often not conserved due to boundary conditions in the systems that are common in statistical mechanics. There's no reason why this need always be the case though.
| {
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Why is it that we ignore height difference when applying Bernoulli in an airfoil I learn physics myself and sorry if this is a very simple question
*
*Why is it that we can apply Bernoulli on above and below the plane even if the are not in the same streamline?
*Why do we ignore height difference when doing so?
Any help whatsoever is highly appreciated.
| Bernoulli shouldn't be used to try and explain the workings of an airfoil, as the NASA page I linked to clearly explains.
It's basically an 'urban myth' that the Bernoulli principle can explain airfoil lift; it isn't true.
| {
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What does Art Hobson mean/explain in his article "There are no particles, there are only fields" regarding the double slit experiment? So I have been reading about fields in physics. I am reading Art Hobson's "There are no particles, there are only fields" published in The American Journal of Physics in 2013, and I am confused by something he states.
On page 14 he writes
"each quantum must carry information about the entire pattern that appears on the screen (in order, e.g., to avoid all the nodes). In this sense, each quantum can be said to be spread out over the pattern. If we close one slit, the pattern shifts to the single-slit pattern behind the open slit, showing no interference. Thus each quantum carries different
information depending on whether two or one slits are open" (Hobson 14).
Can someone please explain what it means that the quanta carry information. And what exactly are nodes and what role do they play?
| The double slit experiment with presumed material particles, such as electrons, demonstrates that the inner reality of nature is represented by waves. In quantum field theory (QFT) the fields are regarded as most fundamental than particles, which are described as oscillations (quanta) of the fields.
Note: The nodes refer to the interference pattern.
| {
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Mathematical Definition of Power I am a high school student who was playing around with some equations, and I derived a formula for which cannot physically imagine.
\begin{align}
W & = \vec F \cdot \vec r
\\
\frac{dW}{dt} & = \frac{d}{dt}[\vec F \cdot \vec r] = \frac{d\vec F}{dt} \cdot \vec r + \vec F \cdot \frac{d\vec r}{dt}
\\
\implies & \boxed{P = \frac{d\vec F}{dt} \cdot \vec r + \vec F \cdot \frac{d\vec r}{dt}}
\end{align}
I differentiated Work using its vector form formula $\vec F \cdot \vec r$
So I got this formula by applying the product rule. If in this formula $\frac{d\vec F}{dt}=0$ (Force is constant), than formula just becomes $P = \vec F \cdot \frac{d\vec r}{dt}$ which makes total sense, but this formula also suggests that if $\frac{d\vec r}{dt}=0$ then the formula for power becomes $P =\frac{d\vec F}{dt} \cdot \vec r$, which implies that if the velocity is zero that doesn't necessarily mean that Power of the object will also be zero!
But I don't find this in my high school textbook and I can't think of an example on that top of my head where this situation is true.
From what I have heard and read, if the velocity of the object is zero then power is also zero.
Can someone please clear my supposed misconception or give me an example of the situation where this happens?
| As others have already answered, $W = \mathbf F \cdot \Delta \mathbf r$ is a simplification and works only in a special case of constant $\mathbf F$. And so does your formulae.
One way to look at it physically is to recognize that work is not a function of position. Mathematically we usually describe it using the concept of inexact differential:
$$\delta W = \mathbf F \cdot d \mathbf r$$
This notation is used to underline the fact that you can integrate both sides and get the same number, but you may not rearrange this formula and in fact you can not (in general case) express $\mathbf F$ using $W$.
An example of an exact differential and what it allows you to do:
$$d \mathbf r = \mathbf v \, dt \implies \mathbf v = \frac {d \mathbf r} {dt}$$
P.S. There are some special cases where you can write $\mathbf F = \nabla \, W$, in those cases it is said that $\mathbf F$ is a potential force.
| {
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Why is are induced electric field's non-conservative while static electric fields are conservative? I have learned that the $E$-field induced by changing magnetic flux, such as in 'motional emf', is non-conservative in nature.
I am also aware that static $E$-fields are conservative in nature.
What is the reason for this difference in the nature of the $E$-fields?
| A force feild is called conservative when it can be expressed in terms of a potential energy (or potential), as
$$
\mathbf{F}(\mathbf{x}) = -\nabla U(\mathbf{x}).
$$
Helmholtz decomposition means that any field can be expressed as a sum of a potential and a solenoidal components, i.e., as
$$
\mathbf{E}(\mathbf{x}) = -\nabla \phi(\mathbf{x}) + \nabla\times\mathbf{A}(\mathbf{x}).
$$
Note that $\nabla\cdot(\nabla\times\mathbf{A})\equiv 0$, that is $\nabla\cdot\mathbf{E}=\nabla^2\phi$. Since the Maxwell equation for the electrostatic field is
$$
\nabla\cdot \mathbf{E} = 0,
$$
it is fully described by the scalar potential - adding a solenoidal component would not change anything!
On the other hand, since $\nabla\times\nabla\phi\equiv 0$, we have $\nabla\times\mathbf{E}=\nabla\times(\nabla\times\mathbf{A})$. For the magnetic field this means that this field is fully described by a vector potential, since we have a Maxwell equation
$$\nabla\times\mathbf{B}=0.$$
For the electric field it means that its solenoidal component is fully determined by the derivative of the magnetic field, since
$$
\nabla\times\mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}.
$$
In other words the solenoidal (i.e., non-conservative) component of the electric field is solely determined by the magnetic field changing in time.
To summarize, the distinction between the conservative and non-conservative components of the electric field is due to the form of the Maxwell equations, which are the experimentally determined laws of nature.
| {
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Is $(L^2, L_z)$ a complete set of commuting observables? According to the main definition we define a (C.S.C.O.) complete set of commuting observables $(A,B,C, \dots)$ if:
*
*Every commutator between the operators of the list is $0$
*If we fix the eigenvalues of the operators there exists a unique eigenvector with these eigenvalues.
(Anyway, there is a reference for the exact formal definition of this concept ? In every textbook I have this concept is introduced with just a brief discussion on the subject.)
If I follow blindly this definition I conclude that ($L^2$, $L_z$) is a CSCO, because if a fix a value of $l$ and a value of $m$ there exists a unique eigenvector (namely a unique spherical harmonic for every fixed value of $l$ and $m$).
But if this set is complete, why in the study of the Hydrogen atom I can add to the set the Hamiltonian $H$ ?
For myself the set must not be complete, because if I fix just one value of $l$ or either $m$, I can clearly notice the degeneracy.
I even think that the latter reasoning may serve as a method to find that the set of observables is not complete, but I haven't found any reference in the literature.
So, what parts of my reasoning are wrong ?
| What you’re missing is to account for “complete”. In practice this means: do you have enough observables to uniquely label quantum states? In the case of $L^2$ and $L_z$, it will not be enough to uniquely label hydrogen states, or the states of a 3d harmonic oscillator, or for that matter the states in any 3d central potential.
| {
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Dimension of matrix elements in scattering cross section This question is probably going to be somewhat trivial, but I am a little confused about the dimension of the matrix element that appears in the formula for the cross section of a scattering process.
I had always assumed that this kind of matrix elements should be dimensionless since they represent a probability amplitude. Anyway, this seems not to be the case looking at Schwartz's formula [5.22]:
$$
d \sigma=\frac{1}{\left(2 E_{1}\right)\left(2 E_{2}\right)\left|\vec{v}_{1}-\vec{v}_{2}\right|}|\mathcal{M}|^{2} \prod_{\text {final states } j} \frac{d^{3} p_{j}}{(2 \pi)^{3}} \frac{1}{2 E_{p_{j}}}(2 \pi)^{4} \delta^{4}(\Sigma p)
$$
On the LHS the dimension is $E^{-2}$, while on the RHS it is $E^{-2}E^{2n_f-4}[\mathcal{M}]^2$, where $n_f$ is the number of final particles. So for 2 particles in the final state the matrix element looks dimensionless, but for $n_f \geq 3$ it should have a negative mass dimension, e.g. for 3 particles $E^{-2}$. Is this correct?
| We wondered the same thing with a group of friends and came to the conclusion that you are right.
As a confirmation, if you have a copy of Peskin and Schroeder, you can deduce from equation (4.74) that $|\mathbf{p_1} \dots \mathbf{p_n}\rangle$ has mass dimension $-n$, and going back to the definition of the matrix elements $\mathcal{M}$ in equation (4.73), with the fact that the momentum delta function has mass dimension 4, you can deduce that the mass dimension of the matrix elements is, in general
$$[\mathcal{M}] = 4 - N$$
where $N$ is the total number of particles involved in the process. In the $3 \to 2$ case you mentioned, this is consistent with your conclusion that $[\mathcal{M}^2] = -2$.
| {
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Why does thermal conductivity of an alloy becomes nearly flat from $0.2I am trying to understand the effect of alloying on the thermal conductivity of an crystalline alloy. I have found a great many papers where I see thermal conductivity sharply decreases from 0 to 0.2 X of the alloy composition, then becomes nearly flat, then they start increasing again after 0.8X composition. But not one paper properly explains what thermal conductivity becomes flat from 0.2<x<0.8. Can anyone please help me with this?
| bufferlab gives a good answer, but I'll dumb it down a bit. Since you tag phonons (and not electrons), I'm assuming that most of your thermal transport is due to phonons.
When $x$ is near 0 or 1, then you basically have a crystal with some impurities. E.g. if you're dealing with $Si_xGe_{1-x}$, then if $x$ is close to 0, then you more or less have Ge, and if $x$ is close to 1, then you more or less have Si.
However, if $x$ is in the middle, then you just have a disordered mess where your phonons are scattered all over the place. In fact, it's a little difficult to even define phonons at this point, because phonons as we normally think of them exist only in regular lattice --- not alloys. Arguably, $x=0.4$ is "more" disordered than $x=0.2$, but in either case, it's so disordered that it doesn't make a huge difference. Hence, thermal conductivity is basically the same for a fairly wide range of $x$.
| {
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Pressure, entropy and enthalpy For an ideal compressible flow, this relation holds:
$$P = P(s, h)$$
where $s$ is the specific entropy and $h$ is the specific enthalpy.
I don't know why: I know that $s = s(e, v)$, but even using Maxwell relations, the density or the temperature is still inside the definition of $P$, so $P=P(s, h, T)$. Could anyone help?
| In Thermodynamics, it is quite common to have some perplexity on the possibility of expressing state functions as a function of other state functions. It is a safe attitude since it is not true that everything can be considered a function of everything else. A counterexample is the case of $P=P(T,v)$ (here, I am using the usual Thermodynamic convention of expressing with the same symbol the value of the pressure and its functional dependence on its independent variables). In general, it is impossible to obtain $v=v(P,T)$ for temperatures below the critical temperature due to flat isotherms in the coexistence region.
The case of $P(s,h)$ is different from the previous example. To show that it is possible to express pressure as a function of entropy and enthalpy, it is enough to start from enthalpy as a function of its natural variables (entropy and pressure) and notice that the relation
$$
h=h(s,P)
$$
can always be inverted to provide $P=P(s,h)$, based on the implicit function theorem, since
$$
\left.\frac{\partial{h}}{\partial{P}}\right|_s=v>0.
$$
| {
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Derivation of curl of magnetic field in Griffiths Can someone please derive how $$\frac{d}{dx} f(x-x') = -\frac{d}{dx'} f(x-x')~?$$
In Griffiths electrodynamics, this is directly mentioned. I'm really confused, can someone elaborate!
| I think that this can come from this argument:
You can prove that $$\frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}=-\nabla\left(\frac{1}{|\vec{x}-\vec{x}'|}\right)$$
If you take this change $$\vec{x}\rightarrow\vec{x}'$$ and $$\vec{x}'\rightarrow\vec{x}$$ the previous equality become $$\frac{\vec{x}'-\vec{x}}{|\vec{x}'-\vec{x}|^3}=-\nabla'\left(\frac{1}{|\vec{x}'-\vec{x}|}\right)$$
From the two equations you can get that $$\nabla\left(\frac{1}{|\vec{x}-\vec{x}'|}\right)=-\frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}=\frac{\vec{x}'-\vec{x}}{|\vec{x}'-\vec{x}|^3}=-\nabla'\left(\frac{1}{|\vec{x}'-\vec{x}|}\right)$$
I hope it is useful, by the way, i check the book Classical electrodinamycs second edition of Jackson in the page 33. Bye!
| {
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What is gravity's relationship with atmospheric pressure? I'm asking for clarification here. If Earth had the same atmospheric mass per square unit of ground but the Earth had suddenly gained mass so it had twice the gravity at the surface, would the Earth now have twice the atmospheric pressure just because of the doubling of gravity? I know pressure is defined as force per area, but I'm not sure if air pressure works the same way. When I look up gravity and air pressure on the Internet it just has information on air pressure with height, but I'm not looking for that.
| Air pressure has everything to do with height because it is caused by the force exerted from the volume of air above weighing down.
That said, if Earth had twice the gravitational field strength at the surface, by the inverse square law this would mean that the Earth doubled in mass. This would double the gravitational force on all objects if they remain at the same distance r from the Earth's centre as before:
$$F_{g,earth} = G \frac{M_{earth}m}{r^2}, \space \therefore F_{g,double} = 2F_{g,earth} = G \frac{2M_{earth}m}{r^2} $$
Therefore the force exerted by layers of air above would double. $P \propto F$, therefore atmospheric pressure $P_{atm}$ too would double.
| {
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Why every system tends to be more stable? An entire topic of inorganic chemistry, i.e. chemical bonding, which is also one of the most important topics, is based on the idea of stability. But whenever I ask why every system tends to get stable, I never get a satisfactory answer. People say that's how nature "wants" to be. How does a system know that it wants to get stable?
| Consider how many somethings there are in a mole. A whole lot, right?
Imagine you have a mole of something. And imagine that something has 3 states it can get into easily, and a lot of rare states. After awhile, statisticly it will be mostly in the common states. Because it's easy to get into them, and hard to get into the others. How many of the individual "somethings" are in each state depends on how easy they are to get into, and how easy they are to get out of. But once they've had time, they will be "stable". You'll have new ones getting into each state at about the same rate they leave that state.
Because a mole is a whole lot. A hundredth of a mole is still a whole lot.
We can't use that approach for human things, not as well. There are only 8 billion of us. We're all different genetically and we have different brains. The circumstances we react to are different. But sometimes it kind of applies to us too.
| {
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Why is stress defined in the way as it is? Stress is like pressure and it doesn't matter in which direction the force acts (given it is perpendicular to the surface).
I read in my book that if we have a rope which is being pulled on both sides by a force $F$ , then the stress at any cross section of the rope is defined as
$\sigma = \frac{F}{area}$
.
But my question is that since the rope is pulled from both the ends, the molecules of the considered cross section is being pulled by other molecules from both sides of the cross sections.
So the stress is due to both the forces.
So shouldn't stress be defined as
$$\sigma =\frac{2F}{area}$$
Edit : The answer of Bob D forces me to add an edit. By $\sigma = \frac{2F}{A}$ , I meant to say
$\sigma =\frac{|F_{left}|}{A}+\frac{|F_{right}|}{A} = \frac{2F}{A}$ .
Here $|F_{left}|$ and $|F_{right}|$ are the forces applied by molecules on the left and right of the considered cross section on the cross section.
Hope it is clear now.
| The standard definition is the correct one. You can see that by investigating some different configurations where stress is not constant.
First, consider the rope being pulled with only one force, let’s say only the force to the right. In that case the rope will accelerate to the right. The stress on the left side of the rope will be zero, and the stress on the right side of the rope will be $F/A$. It would not make sense to call the stress on the right $2F/A$ in this case, because there is only one force, but it turns out that this is the same stress as you get when you apply both forces. Adding that second force does not increase the stress at the right, it just makes it uniform through the rest of the material.
Another way to consider it is a cube sitting on the floor stationary in gravity. The bottom surface of the cube will have a pressure $mg/A$. Pressure and stress are closely related (pressure is the isotropic part of stress) so it makes sense that the stress at the bottom would also be $F/A=mg/A$. Similarly, at the top the pressure is zero as is the stress.
| {
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Nature of Work done by friction I have always been told that work done by friction can, at most, be zero, but never positive.
But consider two blocks placed one on top of the other, such that their surfaces in contact are rough. If we give the block on the top a certain horizontal velocity, then in crude words, we can say that friction will try to slow down the block on the top and speed up the other one, thus opposing relative motion. Then in this case, wouldn't the work done by the friction on the block at the bottom, be positive? Please correct me if I have gone wrong.
| The energy conservation principle tells us that energy is never consumed or produced, just converted. And the very nature of friction is that it only converts from kinetic energy to heat, never the other way round.
If we ignore the heating part (as the sentence "work done by friction can, at most, be zero, but never positive" does), the total energy in the system decreases by friction.
The top block loses energy by slowing down, and the bottom block gains energy by starting to move. The stable state will be determined by conservation of momentum, and result in both blocks moving together with the same velocity. If you do the math, then you'll find that the top block loses more energy than the bottom one gains, and that loss of kinetic energy is what the sentence tries to express.
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Formula for centripetal acceleration: simple proof that does not use calculus? I teach physics to 16-year-old students who do not know calculus and the derivates. When I explain the formula for centripetal acceleration in circular uniform motion, I use this picture:
Here,
$$\vec{a}_{\text{av}}=\frac{\Delta \vec{v}}{\Delta t}=\frac{\vec{v}_2-\vec{v}_1}{\Delta t}$$
and
$$\vec{v}_1=(v\cos\phi){\bf \hat x}+(v\sin\phi){\bf \hat y}, \quad \vec{v}_2=(v\cos\phi){\bf \hat x}+(-v\sin\phi){\bf \hat y}.$$
Combining these equations gives
$$\vec{a}_{\text{av}}=\frac{\Delta \vec{v}}{\Delta t}=\frac{-2v\sin\phi}{\Delta t}{\bf \hat y}, \tag 1$$
which shows that the average acceleration is towards the center of the circle.
Using $\Delta t=d/v=2r\phi/v$, where $d$ is the distance along the curve between points $1$ and $2$, gives
$$\vec{a}_{\text{av}}=-\frac{v^2}{r}\left(\frac{\sin \phi}{\phi}\right){\bf \hat y}.$$
As $\phi\to 0$, $\sin \phi/\phi\to 1$, so
$$\vec{a}_{\text{cp}}=-\frac{v^2}{r}{\bf \hat y}, \tag 2$$
which shows that the centripetal acceleration is towards the center of the circle.
Does there exist another simple proof of Equation $(2)$, in particular, that the centripetal acceleration is towards the center of the circle?
| Your students might like this approach. It uses Newton's technique of modelling a smooth change as a series of sharp, sudden changes.
Consider a particle bouncing around the inside of a cylinder, following a path that is a regular polygon.
The change of velocity at each bounce, for example at C, is clearly
$$\Delta v=2v\ \cos \theta\ \ \ \text{towards the circle centre}$$
The time, $\Delta t$, between bounces is
$$\Delta t = \frac{\text{CD}}{v}=\frac{2r\ \cos \theta}{v}$$
If we make the side length of the polygon smaller and smaller (by making $\theta$ approach $\pi /2$) then any small arc contains a large number $n$ of bounces and the same number of intervals $\Delta t$, so the mean rate of change of velocity is
$$a=\frac{n\Delta v}{n\Delta t}=\frac{v^2}{r}.$$ In the limit the polygon approaches a circle, and the mean acceleration becomes the instantaneous acceleration.
This method is essentially a variation on the classic derivation in the question. It avoids explicit use of $\lim \limits_ {\theta \to 0} \frac{\sin \theta}{\theta}=1$. Instead, the taking of this limit is hidden in the demand to consider smaller and smaller lengths of side of the polygon, so the polygon becomes a circle.
| {
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Why is it easier to handle a cup upside down on the finger tip? If I try to handle a tumbler or cup on my fingertip (as shown in fig), it is quite hard to do so (and the cup falls most often).
And when I did the same experiment but this time the cup is upside down (as shown in fig), it was quite stable and I could handle it easily.
In both the cases, the normal force as well as the weight of that cup is the same but in first case it falls down and in the other it is stable.
I guess that it is falling because of some torque but why is there no torque when it is upside down.
What is the reason behind this?
| Think of it like this: The center of gravity of the cup wants to be as low as possible, right? Consider your fingertip as a fulcrum point. So when the cup is above your fingertip, so is the center of mass, so the slightest tilt allows the cup to rotate so that the center of mass drops. When the cup is the other way, the center of mass is below your fingertip. Even a large tilt will go back to equilibrium without falling off, because it rotates back to normal.
| {
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On Poincare group’s Casimir operators We’ve defined Casimir operator for a group as an operator which commutes with all generators of that group. For the Poincare group we’ve found two Casimir operators: $p_\mu p^\mu$ and $W_\mu W^\mu$ where $W_\mu$ is the Pauli-Lubanski vector.
In checking that they are indeed Casimir operators, can I say that, since $p_\mu p^\mu$ is a scalar, it automatically commutes with all the generators? And same for the second Casimir operator.
| Unfortunately, Lorentz invariant operators are not automatically Casimir operators - you can see this since there are essentially infinite independent Lorentz scalars you can construct from $M_{\mu\nu}$ and $P_\mu$, whereas the dimension of the Cartan subalgebra of the Poincaré group can be shown to be finite. An example is $\frac12 M_{\mu\nu} M^{\mu\nu}$, which is actually a Casimir operator of the Lorentz subgroup - but in the full Poincaré group, this operator fails to commute with $P_\mu$, so it falls short of being a Casimir operator for the full group.
The essence of this lies in the fact that the commutator $[AB, C]$ equals $A[B, C] + [A, C]B$, which is not identically zero (perhaps you have gotten caught up in the terminology - it is identically zero for scalars as in numbers, not Lorentz scalars)
Thus the most straightforward method to prove their Casimir-ness is to simply crank through the commutation relations (a few tricks may be employed in the case of $W_\mu W^\mu$, but that is beyond the scope of this answer).
The converse, proving that these are the only 2 Casimir operators for the Poincaré group, is much trickier - see this excellent answer by David Bar Moshe for an exposition.
| {
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Galaxy Superclusters Okay quick question... why is it that when galaxies group together in superclusters they form a sort of branch-like shape? I personally would think they’d all just group together in some sphere or disk just like regular galaxies.. So what’s the physics behind the branching?
| galaxy clusters are formed where there was higher density of dark matter after the inflation epoch. The dark matter creates some gravitational wells in where the galaxy clusters are formed. These dark matter clusters are connected by lower density dark mater regions which may look like a branch. Some galaxies may have been formed on these branches.
| {
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Engine rotating a generator at its rated speed, but engine power exceeds power required? What would occur if a gasoline engine would be driving a generator (specifically, a permanent-magnet synchronous AC 3-phase sinusoidal generator) as its only load at the output shaft and if that engine is rated at i.e. 35kW @ 8000 RPM but at the same speed the generator is rated to produce 30 kW of electricity, at 120V / 250A?
Is a “rated power” of an internal gasoline engine a rating for its maximum possible power output at a certain RPM? Meaning that if a load (in this case, the generator) requires less power input from the engine then the engine throttle would be positioned as such to allow the required fuel amount to be consumed, but not the maximum safe possible (because it is unnecessary)?
If the load at that speed would be larger (i.e. 35kW, and not 30) then the throttle would need to be opened more to increase the fuel flow to produce the required power?
Or does the engine produce 35kW at 8000 RPM regardless of generator load and the 5kW is unused i.e. dissipated through heat?
Excuse me if this is too simple but I would like to find clarification regarding this.
Thanks
| Most gasoline engine powered small AC electric generators have a governor on the throttle to keep them running at a fairly consistant rate of revolutions per minute. This keeps the AC phase steady. When the generator has a load increase or decrease the governor will open or close the engine throttle accordingly to keep the revolutions per minute constant under different loads. When there is a heavy load the throttle will be open more, burning more fuel than when there is less load on the generator.
| {
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Hamiltonian formalism of General Relativity Textbook I've been reading Wald's book on General Relativity and in appendix $E_{2}$ it discusses the Hamiltonian formalism of General Relativity.I would like to understand it more, can you recommend me a textbook about ADM formalism?
P.S.
I have already read Eric Poisson book too.
| I can only point you to the summary by R. Arnowitt, S. Deser, C.W. Misner "The Dynamics of General Relativity", p. 227 from "Gravitation: An Introduction to Current Research", Editor Louis Witten, John Wiley & Sons, 1962.
from which you can see the original list of articles by the trio:
Arnowitt, R., and S. Deser, 1959, Phys. Rev., 113, 745 (I).
--- S. Deser, and C. W. Misner:
1959, Phys. Rev., 116, 1322 (II).
1960, Phys. Rev., 117, 1595 (III).
1960, Nuov. Cim., 15, 487.
1960, Phys. Rev. Letters, 4, 375.
1960, Phys. Rev., 118, 1100 (IV).
1960, J. Math. Phys., 1,434 (IIIA).
1960, Phys. Rev., 120, 313 (V).
1960, Phys. Rev., 120, 321 (VA).
1960, Ann. Phys., 11, 116 (VB).
1961, Nuov. Cim., 19, 668 OVA).
1961, Phys. Rev., 121, 1556 (IVB).
1961, Phys. Rev., 122, 997 (IVG).
| {
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What quantum gates are needed to get the state $|01\rangle+|10\rangle$ from $|00\rangle$? I was wondering if I start with two qubits in the state
$$|00\rangle$$
If it's possible to apply gates to get it to the state
$$\frac{|01\rangle + |10\rangle}{\sqrt{2}}$$
I have tried applying the Hadamard Gate, Controlled X etc, But I couldn't make this state. So I'm curious if it's possible and I am just missing something very obvious.
| Applying Hadamard + CNOT takes you from $|00\rangle$ to $\dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$. Now, just apply the single-qubit $X$ Pauli operator (which swaps $|0\rangle$ with $|1\rangle$ and vice-versa) to either one of the two qubits, and you get the target state $\dfrac{|01\rangle + |10\rangle}{\sqrt{2}}$.
| {
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How does light look like when it is 180° out of phase? When two lights are 180° out of phase, would it look like this?
In the photo below, the left side is flipped and it is a mirror image.
| When people talk about phases of a light field, they usually refer to the electric field $E$. To understand why this is of interest, consider two light sources in 1D. Their (skalar) electric fields at the position $z_0$ are given by $E_1(z_0)=|E_1| e^{i\phi_1(z_0)}$ and $E_2(z_0)=|E_2| e^{i\phi_2(z_0)}$. The relative phase $\phi_2(z_0) - \phi_1(z_0)$ of these fields determines, whether they are constructively or destructively interfering. Since photographic plates and ccd chips do not measure the electric field, but measures the intensity, $I\propto |E_{total}|^2 = |E_1(z_0) + E_2(z_0)|^2$, the relative phase is important.
What you are doing is fundamentally different. You took the intensity of an image $I(x,y)$ for all negative $x$ values and just reused it for all positive $x$. Not only is it rather meaningless to ask about a relative phase of two intensities in general, but you do not have two intensities at the same point in the image.
======
I'm not sure whether or not you know how two electric fields, which are 180deg out of phase look like. Here are two simulations. The light sources are located at $x=\pm 12$ and $y=0$. In the left image the electric field of the sources are out-of-phase by 180deg, and in the right image they are in-phase
In both I plot the intensity.
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Will the velocity be constant in a vertically oriented tube?
The tube here is placed into a water stream moving with a velocity $v$
My query is will the velocity be constant throughout the pipe or only in some places?
I have conflicting opinions on this, the equation of continuity would say the velocity is constant through the pipe but Bernoulli's equation would say it slows down the higher it went.
Where have I flawed conceptually here?
The person who solved the question said the velocity at the blue points will be the same but in the portion above the level of water, it will not. This just confused me even more.
Link to the question although not necessary.
|
My query is will the velocity be constant throughout the pipe
No. Flow will slowdown going up due to the stopping gravity force of water column. Check this scheme :
Water will raise-up to the height $h$ only due to dynamic pressure. Neglecting atmosphere pressure, maximum height until water can go-up can be calculated from hydro-static equilibrium which will be formed when water column of height $h$ gravitational pressure will be fully in balance with a flow dynamic pressure :
$$ P_d = P_g $$
Substituting dynamic pressure and gravitational pressure expressions gives :
$$ 1/2~ \rho u^2 = \rho gh $$
Solving for $h$, gives maximum possible water column height given flow velocity $u$ :
$$ h_{\text{max}} = \frac {u^2}{2g} $$
So, when flow velocity is zero,- $h=0$ too. If there exist maximum height to where water can raise (and hence flow velocity at that point $h_{\text{max}}$ also falling to zero too) - this indicates that water flow inside tube must decrease going-up.
| {
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Why don't opaque objects reflect light? My sister was doing a quiz and I tried to point her in the right direction by giving her scenarios to imagine. One of the questions in the quiz was:
Which of the following objects do not reflect light:
*
*Polished metal
*Mirror
*Undisturbed water
*Book
She suggested that the answer was "undisturbed water" and that made sense to me too.
But the answer given was "book", which didn't make sense to me. How can you even see the book if it didn't reflect light in the first place?
Is this terrible framing by her teacher or am I having a conceptual misunderstanding?
| In my opinion, one more thing which is possible here is that may be the question is
Which among the following causes diffuse reflection ?
Undisturbed water can also give specular reflections but the surface of a book have many microscopic as well as macroscopic irregularities which causes diffusion of the reflected rays and thus you can't see your face on your book's surface like you do in mirrors.
Hope it helps .
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General plane motion and freely floating rigid body
Consider a rigid rectangular plate of length $l$, width $w$ and thickness $t$ which is at rest and is floating freely in space (no gravity). The center of the plate is at $O_L$ with respect to global coordinate frame $O_G$. The initial pose (position and orientation) $\mathbf{T}$ of the rigid body is assumed to be known and is given by a $3\times 3$ Rotationmatrix and a $3\times 1$ translation vector. Also as shown in the figure, there are $n$ points on the rigid body whose position is known. On each of these points forces are applied which is also known. After time interval $\Delta t$ the pose of the rigid body is given by $\mathbf{T'}$.
Is the information provided above sufficient to find the new pose $\mathbf{T'}$? If not, what information is missing and how do I proceed to find the new value of $\mathbf{T'}$?.
Any comments and suggestions are welcome : )
EDIT
In simple words what I wish to find is a solution (if possible) that says something like: shift the plate by so and so amount in $x$, $y$ and $z$ direction and then rotate by so and so amount about $x,y$ and $z$ axis respectively so that the plate lands at $\mathbf{T'}$.
Please note that the Forces remain constant during the short time interval $\Delta t$.
| The torque calculated from a point of an inertial frame (for example the origin $O_G$) is the time derivative of the total angular momentum: $$\tau = \frac{d\mathbf L}{dt}$$
And the angular momentum of the plate at a given time is:
$$\mathbf L = \int_v \mathbf r_G \times d\mathbf p = \int_v \mathbf r_G \times \frac{d\mathbf r_G}{dt} \rho dv$$
Where $\mathbf r_G$ is the position vector of the points of the plate from the origin $O_G$. But at the same time, by knowing the forces and their locations in the plate, the torque is known:
$$\tau = \sum_{i=1}^n\mathbf r_{Gi} \times \mathbf F_i$$
Equating this torque to the time derivative of the intergral of the angular momentum we have a diferential vector equation in $\mathbf r_G$ and $\frac{d\mathbf r_G}{dt}$, that should be solved with the boundary conditions that $\frac{d\mathbf r_G}{dt} = 0$ when $t = 0$.
This procedure is valid even if the body is not rigid. But that additional constraint means that for any point of the body, the distances to any other point don't change with time. Choosing axis parallel to global coordinate frame $O_G$, but with origin at an arbitrary point of the body, after a small time $\Delta t$ the position of all the other points move according to the infinitesimal rotation matrix $R$.
$$\Delta \mathbf r_b = R\mathbf r_b - \mathbf r_b = (R - I)\mathbf r_b \implies \frac{d \mathbf r_b}{dt} = \Omega \mathbf r_b$$
Where $\mathbf r_b$ are the position vectors relative to the selected origin in the body, and $\Omega$ is the matrix:
\begin{Bmatrix}
0 & -\omega_3 & \omega_2\\
\omega_3 & 0 & -\omega_1 \\
-\omega_2 & \omega_1 & 0
\end{Bmatrix}
The $\omega$'s are the instantaneous angular velocities relative to the coordinates axis. The cross product in the integral of the angular momentum becomes:
$$\mathbf r_b \times \frac{d\mathbf r_b}{dt} = \mathbf r_b \times \Omega \mathbf r_b$$
Expanding the cross product, the angular momentum at any given time, relative to the point in the body, can be expressed as: $\mathbf L = (\int_v \rho M dv) \omega$
where $M$ is the square matrix:
\begin{Bmatrix}
(y^2 + z^2) & -xy & -xz \\
–yx & (z^2 + x^2) & -yz \\
-zx & –zy & (x^2 + y^2)
\end{Bmatrix}
and $\omega$ is the column matrix:
\begin{Bmatrix}
\omega_1 \\
\omega_2 \\
\omega_3
\end{Bmatrix}
In particular, if the selected point in the body is the COM, we can use the second Newton's law for its movement:
$$\sum_{i=1}^n\mathbf F_i = m \frac{d\mathbf v_{COM}}{dt}$$
And equate torque relative to the COM to the time derivative of angular momentum also relative to the COM:
$$\tau = \sum_{i=1}^n\mathbf r_{COMi} \times \mathbf F_i = \frac{d(\int_v \rho M dv) \omega}{dt}$$
Of course the integral simplifies a lot if the density is constant, and if by coincidence the forces happens to rotate the body around one of the 3 main axis of inertia.
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Is (net) force invariant in special relativity? I am aware that acceleration is not invariant under lorenrz transformations, but I was sure that the first postulate of special relativity implied that newton’s second law in its original form, F=dp/dt, where p is the relativistic momentum, was invariant. However, the following 2 questions imply otherwise:
Relativistic electromagnetism and electromagnetic forces on 2 protons
Magnetic force between 2 moving charges
It seems that the magnetic force between 2 moving charges depends on the frame of reference, and even goes to 0 in the relativistic limit. Doesn’t this violate the first postulate of special relativity?
| The fact that Newton’s Second Law in the form
$$\mathbf F=\frac{d\mathbf p}{dt}$$
is relativistically form-invariant, meaning that in another inertial frame
$$\mathbf F’=\frac{d\mathbf p’}{dt’},$$
does not mean that force doesn’t transform under a Lorentz boost. It does transform, in the same way as the time derivative of relativistic momentum does. $\mathbf F’\ne\mathbf F$.
See this paper for a discussion of the Lorentz transformation of a three-force $\mathbf F$.
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Generalization of first law of thermodynamics What is the mathematical statement for the first law of thermodynamics, accounting for kinetic energy, potential energy, internal energy, work, heat and most importantly taking into consideration the work-energy theorem?
Also, is $∆U=∆Q-∆W$ only valid for systems whose center of mass is at rest in an inertial frame, or is it also valid for other systems?If so, please specify. Thanks
| The relationship you state is for a closed system with no change in center of mass kinetic or potential energy, viewed from an inertial frame. You also have to consider whether the system is open or closed. An open system can have mass flow in/out but a closed system cannot.
For the detailed relationships, see any good text on thermodynamics, such as one by Sonntag and Van Wylen, and search for first law on the web.
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Why can't dark matter lose energy by gravitational waves and collapse into itself? Because of lack of electromagnetic induction, dark matter can't lose its gravitational potential energy. That is preventing it from collapsing like an ordinary matter cloud in space.
But why can't dark matter lose energy by gravitational waves and collapse into itself?
| Dark matter was originally postulated to explain galaxy rotation curves, which needed an extended in space matter envelope to explain them. So it does collapse gravitationaly, but by the fact that its postulated extent is so large its effective density has to be low.
Evidently, if it is composed of particles , in order to be around a galaxy it has lost energy by gravitational interaction, but total collapse into the galaxy must take much longer, as the curves, by their existence, show.
After all look at the earth atmosphere, it does not collapse on the ground although it is gravitationally interacting with the ground, due to other forces ,mainly explained by thermodynamics, that balance the very small gravitational attraction for the atmosphere molecules.
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Why does air pressure decrease with altitude? I am looking to find the reason: why air pressure decreases with altitude? Has it to do with the fact that gravitational force is less at higher altitude due to the greater distance between the masses? Does earth’s spin cause a centrifugal force? Are the molecules at higher altitude pushing onto the molecules of air at lower altitudes thus increasing their pressure? Is the earths air pressure higher at the poles than at the equator?
| As you go higher, there are less air molecules (less weight) on a given area this is basically one reason why it decreases.
From the barometric formula, one can get the relation between the pressure and altitude. It's defined as
$$P = P_{0}e^{-\frac{mgh}{kT}}$$
so the relation between pressure and altitude is $P\propto e^{-h}$. Thus, as we go to higher altitudes pressure will exponentially decrease.
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Strong and Weak Interactions How do I determine whether an interactions is strong or weak if what i've been given is just the particles and the type of reaction:
Determine whether the following are strong or weak interactions?
*
*a muon decays
*a kaon decays
*pions are produced
| There are different arguments to use depending on the reaction. For instance :
*
*muon decay is due to weak interaction, leptons are not concerned by strong
*kaon decay, is also due to weak interaction, because kaons have a strange number equal to $+1$ while its decay products have no strangeness at all
*if pions are produced, it is likely to be a strong interaction process (pions being the lightest mesons, they are massively produced in hadrons collisions for instance). But once again you can think of processes involving weak interactions with pions in the final state; for instance when kaon decays
Your best tool to discriminate between strong and weak processes is to look at quantum numbers which are conserved by the former but not by the latter (strangeness, charmness, etc). If information is given, you can also use the fact that strong processes usually happen much faster than weak ones
| {
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Direction of dipole moment? Why is the direction of dipole moment taken from negative to positive charge when the field is directed in the opposite direction ?
| It actually does not matter as long as you keep track of the signs. It was probably defined this way in order to give a similar formulation for torque
$$\vec{\tau} = \vec{r}\times\vec{F}$$
dipole moment is defined as $\vec{p} = q\vec{d}$ and torque on a dipole
$$\vec{\tau} = \vec{p}\times\vec{E}$$
If it was defined otherwise you had to write it down like $\vec{\tau} = \vec{E}\times\vec{p}$
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Geometrical linearization in continuum mechanics In continuum mechanics, we often make use of "physical and geometrical linearization", e. g. during derivation the Navier-Cauchy equations (c. f. https://en.wikipedia.org/wiki/Linear_elasticity). I understand that physical linearization refers to the assumption that the constituive matrix $\mathbf{C}$ in our constitutive equation $$\mathbf{\sigma}=\mathbf{C}\cdot\mathbf{\epsilon}$$ is constant and does not depend on $\epsilon$ to facilitate calculations. But what is the point of "geometrical linearization"? To my understanding, it means using only the "symmetrcial part" of the displacement gradient $\mathbf{H}$, which defines the "engineering strains $\epsilon_i$ we use in our constitutive equation: $$\epsilon:=\epsilon_i=\mathbf{H^S}=\frac{1}{2}\cdot(\mathbf{H}+\mathbf{H^T})$$If we look at $\mathbf{H}$ and $\epsilon_i$ in a 2D case (merely for the sake of simplicity, 3D is no different), there seems to be little difference between them: $$\mathbf{H}=\begin{pmatrix}
\frac{du}{dx} & \frac{du}{dy}\\
\frac{dv}{dx} & \frac{dv}{dy}\\
\end{pmatrix}$$
$$\mathbf{\epsilon_i}=\begin{pmatrix}
\frac{du}{dx} & \frac{1}{2}\cdot(\frac{du}{dy}+\frac{dv}{dx})\\
\frac{1}{2}\cdot(\frac{du}{dy}+\frac{dv}{dx}) & \frac{dv}{dy}\\
\end{pmatrix}$$
In comparison, they both consist of the same derivatives and the only thing we achieve is a symmetrical strain tensor (for which we pay a price in the form of erroneous strains for rotations). So how exactly does that make things easier?
| Imagine two infinitesimally close points $a$ and $b$, the difference between their squared distances before and after the deformation can be measured by $((I+H)(b-a))^T (I+H)(b-a)-(b-a)^T(b-a)$, where $I$ is the identity matrix. So it is possible to think of the nonlinear strain tensor as $(I+H)^T(I+H)-I$. The accurate measurement can be replaced through the geometric linearization you described by $H^T+H$, up to a scaling factor. So there is some loss in accuracy in $H^T H$, which is of order $O(|H|^2)$, and a gain in efficiency in calculation.
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Is it the gravitational field created by an object which exerts force on another object or the interactions between their fields that exerts force? We are taught that Gravitational force exerted by an object is a two-step process:
*
*The object creates a field around it.
*The field exerts a force on bodies present in the field.
Now, since we know that an object any object of certain mass creates its own field, is it correct or incorrect to say that 'Field-field interactions exerts force on the bodies'. Can a similar doubt be raised in case of Electric fields created by charges?
| Here is the wikipedia article on classical fields that will help clarify:
A classical field theory is a physical theory that predicts how one or more physical fields interact with matter through field equations. The term 'classical field theory' is commonly reserved for describing those physical theories that describe electromagnetism and gravitation, two of the fundamental forces of nature. Theories that incorporate quantum mechanics are called quantum field theories.
Bold mine, to stress that a field in classical field theory interacts with matter not with another field.
The word interaction occurs once here:
The first field theory of gravity was Newton's theory of gravitation in which the mutual interaction between two masses obeys an inverse square law. This was very useful for predicting the motion of planets around the Sun.
and a second time:
Maxwell's theory of electromagnetism describes the interaction of charged matter with the electromagnetic field.
So fields are additive, with the algebra on which they are defined, vector or tensor afaik in classical physics. They do not interact with each other.
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Why is the net force acting on a massless body zero? I know that massless bodies can accelerate (in theory) even with the net force equaling to zero. But, why cannot there be a net force on a massless object? Why does it always have to be zero as a resultant in the end?
I'm talking about object whose mass is assumed to be zero, i.e. $m\to0$
| In truth, the answer to your question is that no massless objects exist.
Sometimes an object's mass is negligible compared to something else - such as strings carrying weights in a pully system, or such as a human being on the surface of the earth - and then we can safely ignore it in our calculation with no significant change in our results. Other times that same mass is not negligible when compared to other more similar-scaled objects - such as a heavy chain carrying those weights, or such as a human being compared to another human being.
In short: Massless is another way of saying so comparatively small that it can be ignored.
Such an object can still accelerate, though, and that similarly takes comparatively negligible force. We often indicate that as $F\approx 0$ force. But in reality, there is a non-zero force, albite small, when a massless object accelerates because that object actually does have a mass, albite small.
| {
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Did I make an electric circuit with my cat? This is definitely the dumbest question I've asked, but I'm curious. My cat was resting her head on my left hand while I pet her with my right. Suddenly, my left hand began receiving regular static electric shocks. Was I pulling electrons from her fur with my right hand, thus causing them to flow back through my left?
If it really is an example of a circuit, I was thinking it's a pretty hilarious way to explain circuits to students.
| Yes, you made a circuit, although, due to the bad conductivity of the contact between your left hand and the fur of your cat, the current was not continuous but concentrated in the sparks.
It was a circuit because your right hand was acting as a generator, where the electromotive force was due to the triboelectric effect. The remaining part of the circuit was your and your cat's bodies (conductors, although characterized by quite a high resistance). The circuit was closed at the contact between your right hand and her fur. This contact is not good and allows for the accumulation of charge until the threshold for a spark is reached. In the case of air, one needs a few kV per centimeter in order to have a spark. Distance between your skin and your cat's fur is less than a millimeter, therefore one could estimate the build-up of the order of tens or hundreds of volts before the spark.
Notice that the same kind of circuit could be obtained by using any static generator. Moreover, in recent years, there has been a strong interest in triboelectric generators based on nano-technologies, as efficient power generators for nanocircuits. See, for instance, this paper.
| {
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Parity transformations and Dirac Spinor I'm reading "No-Nonsense quantum field theory" and I have some doubts about the transformation law for Dirac Spinors as explained by the author. In the book the left chiral spinors $\chi$ and right chiral spinors $\xi$ are introduced as objects that have two components and behave under rotations $R$ around $x$-axis and boosts along $z$-axis $B$ as follows:
$$\chi_a \rightarrow R_{ab}^{(\chi x)} \chi_b
\\ \chi_a \rightarrow B_{ab}^{(\chi z)} \chi_b
$$
where $$R_{ab}^{\chi z} = \begin{pmatrix} \cos(\theta/2) & i\sin(\theta/2)\\\ i\sin(\theta/2) & \cos(\theta/2)\end{pmatrix} \\ \\
B_{ab}^{(\chi z)} = \begin{pmatrix} e^{\phi/2} & 0\\\ 0 & e^{-\phi/2}\end{pmatrix}$$
and
$$\xi_a \rightarrow R_{ab}^{(\xi x)} \xi_b
\\ \xi_a \rightarrow B_{ab}^{(\xi z)} \xi_b
$$
where $$R_{ab}^{\xi z} = \begin{pmatrix} \cos(\theta/2) & i\sin(\theta/2)\\\ i\sin(\theta/2) & \cos(\theta/2)\end{pmatrix} \\ \\
B_{ab}^{(\xi z)} = \begin{pmatrix} e^{-\phi/2} & 0\\\ 0 & e^{\phi/2}\end{pmatrix}$$
Then the author introduces the Dirac spinor:
$$\Psi = (\chi, \xi)^T$$
which tranforms under boosts as
$$(\chi, \xi)^T \rightarrow \begin{pmatrix} B^{(\chi z)} (\phi) & 0\\\ 0 & B^{(\xi z)} (\phi)\end{pmatrix} (\chi, \xi)^T$$.
So far I'm following the argument, but then the author claims the equation just above becomes:
$$(\chi, \xi)^T \rightarrow \begin{pmatrix} B^{(\xi z)} (\phi) & 0\\\ 0 & B^{(\chi z)} (\phi)\end{pmatrix} (\xi, \chi)^T$$ because under parity transformation we have $B^{(\xi z)} (\phi) \rightarrow B^{(\xi z)} (-\phi) = B^{(\chi z)} (\phi)$ and $B^{(\chi z)} (\phi) \rightarrow B^{(\chi z)} (-\phi) = B^{(\xi z)} (\phi)$. And then asserts that this implies that the Dirac Spinor $\Psi$ transforms under parity transformations as $$ \Psi = (\chi, \xi)^T \rightarrow (\xi, \chi)^T$$
I'm confused about why the last statement follows from the discussion above.
I've also attached a picture of the section of the book where I got this from:
| Under parity in spherical coordinate we have,
$$\textbf{P} \theta = \pi - \theta \\ \textbf{P} \phi = \pi + \phi $$
This explains why ,
$$ B^{(\xi z)} (\phi) \rightarrow B^{(\chi z)} (\phi)$$
Now we need to know what do you mean by a left-chiral spinor. Left-chiral spinor is an object which transforms like this boost,
$$ \chi' \rightarrow B^{(\chi z)} \chi \;\;\;\;\;\; \;\;\;\;\;\; eq.1$$
similar thing is true for the right-handed spinor.
Let's start with ,
$$ \xi' \rightarrow B^{(\xi z)} \xi \;\;\;\;\;\; \;\;\;\;\;\;$$
Now we apply parity to both sides.
$$ \textbf{P} \xi' \rightarrow \textbf{P} ( B^{(\xi z)} \xi ) \;\;\;\;\;\; \;\;\;\;\;\; \\ \textbf{P} \xi' \rightarrow \textbf{P} B^{(\xi z)} \textbf{P} \xi \;\;\;\;\;\; \;\;\;\;\;\; \\ \textbf{P} \xi' \rightarrow B^{(\chi z)} \textbf{P} \xi \;\;\;\;\;\; \;\;\;\;\;\; eq. 2$$
Now we need to ask what is $\textbf{P} \xi$. In order to answer this question, you should compare eq.1 with eq. 2. $\textbf{P} \xi$ is an object which transform like a left-handed spinor so it must be a let-handed spinor.
| {
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If i wrote something in the sky what would the radius of visibility on the ground? If i wrote something in the sky 8500-10,000 ft high 40 ft tall letters what would be the visibility radius on the ground? Or what formula would i use to come up with the answer?
| The size of the letters and their distance will be proportional to size and distance where they are to be calculated. This is arrived by calculating angle those letters subtend at eye.
Let say if their visibility radius is to be calculated at 1 ft distance then:
$$\frac{sz}{1 ft} = \frac{40 ft}{10000 ft}$$
=> $$sz = .0004 ft$$
| {
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Does the speed of light in our definitions take vacuum energy into consideration? We know that the speed of light decreases as it goes through a medium, and we also know that there is a certain vacuum energy that creates a sea of particles coming in and out of existence, which, naively, one could assume has an effect on the speed of light.
The speed of light is taken to be $299 792 458 \;m/s$ in a vacuum, but is this vacuum taken to be an exact vacuum, or does it take into account the vacuum energy?
If the latter, is the "actual" speed of light then faster (and thus, unknown)?
| The constant $c$ is commonly referred to as the speed of light, but really it's better interpreted as a conversion factor between distance and time units, or a maximum speed of cause and effect. The speed $c$ is the only one that observers in all frames agree on. Therefore if we found that the speed of light was not $c$, then observers in different frames of reference would not measure it to be the same. In particular, there would be some frame in which is was zero. This would be a preferred frame of reference. Having such a preferred frame is one way of breaking Lorentz invariance.
Theoretically, adding a vacuum energy term in general relativity does not break Lorentz invariance, and therefore we can be sure that it doesn't change the speed of light. This makes sense because the role played by vacuum energy in general relativity only becomes apparent on very large scales. This is why it was originally called a cosmological constant.
Experimentally, Lorentz invariance is one of the most precisely tested physical claims in all of history, and all tests have come up negative.
| {
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Can the gravity of one entire galaxy slow down an astrophysical jet emitted from its central black hole? Let say we are talking about our Milky way and Sagittarius A* emitted a astrophysical jet. If the galaxy was 150 000 ly wide would its gravity with dark matter included eventually slow down that jet keeping in mind that gravity of a disk loses strength slower than a sphere as a black hole is?
| No.
These jets are jets because they are faster than the escape speed of the various relevant objects (central black hole / Active Galactic Nucleus / host galaxy). Thus, gravity is not relevant in slowing it down. For order of magnitude, the escape speed of the Milky Way galaxy is about 650km/s, but these jets are relativistic, i.e. one or two orders of magnitude faster.
Instead, jets are slowed down by Ram pressure as they encounter the intergalactic gas. This is why they "puff up" (rather than fall back on a parabolic orbit). Classic example is the jet from M87:
| {
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Why is the Pauli exclusion principle not considered a sixth force of nature? Why is the Pauli exclusion principle not considered a sixth force of nature, given it produces such things as repelling of atoms and molecules in solids?
| The Pauli exclusion principle as understood nowadays, is a consequence of the spin-statistics behavior of fermions. We know that Spin representations are related to wave-functions with specific commutation relations which enforce anti-symmetry (all of this encoded in the Dirac-equation, which already accounts for spin) of the wave-function, thus not allowing for two fermionic particles to have the same quantum states in a system. As you can see, it is a statement related to statistics and spin. Both are very well understood and are a consequence of the commutation relations, not the fundamental interactions between single particles (I would prefer fields) which is what we call a force.
Forces on the other hand, are of a different nature. They are not a statistical effect but concern the point-like interactions among fields. Except for gravity, we describe all of them by a gauge (mediator) boson. We don't need to include such a bosonic field to account for the Pauli exclusion principle since the fermionic commutation relations already encode this effect.
| {
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What would happen if a 10-kg cube of iron, at a temperature close to 0 kelvin, suddenly appeared in your living room? What would be the effect of placing an object that cold in an environment that warm? Would the room just get a little colder? Would it kill everyone in the room like some kind of cold bomb? What would happen?
Don't think about how the cube got there, or the air which it would displace.
| The heat capacity of iron at room temperature is 0.444 J/K per gram (it changes with temperature, but let us leave that aside). That means it will want to absorb about 1,332,000 J of heat. That is a lot, but a bathtub with 300 litres of 40 °C water (about 10 degrees above a 300 K) will have about 12,900,000 J of internal energy - if you dumped the iron into the warm water it would not become cold (ignoring losses to the environment).
The real issue is how fast it would cool. Convective heat transfer from air to the metal gives a heat flow $\approx hA(T_{hot}-T_{cool})$ where $h\approx 10$ to 100 W/(m$^2$K) and $A$ is the area (about 0.0726 square meters for the iron). So the heat flow would in theory be on the order of 217.8-2178 J/s initially: sounds fairly impressive, but you would get the same flow in the other direction from a 600 °C cube. That flow will also soon slow, since the cube would be surrounded by cold air and whatever it is sitting on.
So the sad news is that the cube would not do anything super impressive. It would sit there, making air condense like around liquid nitrogen and soon be covered with rime frost. Probably some interesting crackling and perhaps cracking as it changed volume while heating. But no explosions, just a room with cold air along the floor.
| {
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Understanding dependent/independent variables in physics How does one determine the independent and dependent variables?
What do the terms mean?
Can they be derived from a formula?
For example I saw in a textbook $F = k\Delta l$, Hooke's Law, that $F$ is the independent variable. Is this because $\mathbf {F} $ is the subject, therefore it is independent?
| In an equation there is no inherent distinction between dependent and independent variables. There are to my knowledge only two contexts where the distinction makes sense.
Experimental: In an experimental context the independent variable is the one that the experimenter is controlling in the experiment. It is the treatment. For example, if the experimenter is determining the resistance of a resistor using a series of a few different voltages, then the independent variable would be the voltage and the dependent variable would be the current.
Statistical: In a statistical context the independent variable is the one that is known with no error. That is usually an approximation, so instead the independent variable is the one with negligible error. If none of the variables of interest have negligible error then unusual statistical methods must be used.
| {
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Will this spaceship collide with the star? (time dilation) I thought of the above thought experiment and arrive on 2 conflicting conclusions. I can't seem to identify the flaw in my reasoning.
Suppose there is a star 4 light years from earth that has will explode and turn into a white dwarf in 3 years (as measured in the earth frame). A spaceship travels to the star at 86% the speed of light.
According to earth's frame of reference, the journey to the star will take 4.5 years so the star will have turned into a white dwarf.
according to the ship's frame of reference, however, the journey will only take 2.25 years. Moreover, since the star is travelling relative to the ship in its own frame, the event of the star exploding will actually take 6 years. So the spaceship will collide with a white dwarf instead of a star.
I thought this had something to do with simultaneity, but I know that the events must be same in all frames of reference. My collusions imply that collisions occur between different bodies depending on the frame of reference, which can't possibly be true, can it? Where am I going wrong?
| It is a synchronization problem. When the two observers synchronize their clocks at earth (Let us assume the traveler accelerates almost instantaneously so it reaches full speed still at earth), the star that is current for the traveler is in the future for the person on earth. So for the traveler, the time left to explode is smaller than otherwise. Imagine the star becomes a red giant before exploding. This means that at synchronization time, it is possible that for the earth observer the star is still in normal stage, but for the traveler it will be already in red giant stage. When they synchronize their clock at earth, far away objects will still be out of synchrony. So the star will explode before the spaceship will collide, for both observers, of course. Both observers will agree on a single event in spacetime, such a the collision between two objects.
| {
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Does Work become state function in an Isothermal Process and what are other processes in which it happens? In a reversible isothermal process and for an ideal gas we know from the definition of Helmholtz free Energy
$dF= -SdT -PdV$.
And as temperature doesn't change for an isothermal process, $dT$ must be zero. So dF can be written negative of change in Helmholtz free Energy. Since $F$ is a state function and $dF$ a perfect differential, work also should be.
Also, does work become state function for adiabatic processes also? Please throw light on it.
| The equation dF=TdS-PdV applies only to two closely neighboring (i.e., differentially separated) thermodynamic equilibrium states, where P is the pressure calculated from the (equilibrium) equation of state (e.g., the ideal gas law) for the fluid. In an irreversible process, even if the boundary of the system is held at a constant temperature, this does not mean that the temperature interior to the system is uniform spatially. This spatial non-uniformity will also apply to adiabatic irreversible processes. So the entire fluid is isothermal only for a reversible path. In addition, in an irreversible expansion or compression, the force per unit area at the interface where work is being done (e.g., the inside face of a piston) is not equal to the pressure calculated from the equation of state. This force also includes viscous stresses resulting from rapid deformation of the fluid. Therefore, the equation for dF cannot be applied to this, and it is not a perfect differential all along an irreversible path. In addition, from this it follows that, for the irreversible path, the work is not equal to the change in F.
| {
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Fugacity of Bose-Einstein Condensation I'm studying Bose Einstein Condensation.
In the book "Huang K Statistical Mechanics 2 edition", page 288, the author gets the following result for the fugacity ($z$) as a function of temperature and specific volume (lambda is the thermal wavelength and small $v$ the specific volume):
I understand how $z$ is equal to 1. I don't understand how can one obtain the last result for the value of $z$ above the critical temperature.
| I am assuming you don't know how to get from 12.41 to the second equation of 12.52?
If you start from 12.41:
$$ \frac{1}{v} = \frac{1}{\lambda^3}g_{3/2}(z) + \frac{1}{V}\frac{z}{1-z}, $$
and take the infinite volume limit $V\rightarrow \infty$, so that the $1/V$ term above goes to $0$.
Then, you are left with:
$$ \frac{1}{v} = \frac{1}{\lambda^3}g_{3/2}(z), \\
g_{3/2}(z) = \frac{\lambda^3}{v}. $$
Then you solve it graphically by plotting each side of the equation and find their intersection (the "root"):
For finite volumes, the error goes as $\mathcal{O}(1/V)$.
| {
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What does Gibbs Energy represent physically? I was reading about Thermodynamics in Chemistry.
I know that Enthalpy change at a constant pressure by a system is actually released as heat by the system.
My question:
What exactly does Gibbs Free Energy and its change represent physically (on a similar note to what Enthalpy represents)?
To me Gibbs Energy seems very non-intuitive and more like a magical number which can predict the feasibility of a process.
| The Gibbs Free Energy represents the energy that is free to do useful work for a spontaneous process. In other words, it is the max work done by a process (at constant $T$ and $P$). The Gibbs free energy can tell us whether a process will proceed spontaneously or not.
The Gibbs free energy is not magical. Instead it is derived from the inequality
$$dS \ge \frac{dQ}{T}$$
which states that in a reaction, the change in entropy must always be greater than or equal to the heat transferred/$T$. If we consider the enthalpy for constant pressure
$$\Delta H_p = \Delta U + P\Delta V$$
and since
$$\Delta U = Q + W$$
we get
$$\Delta H_p = Q + W + p\Delta V$$
with $W = - P\Delta V$ then $\Delta H = Q$, and using the inequality above, we get
$$TdS \ge dH$$
and if we integrate and rearrange we get
$$\Delta G = \Delta H - T\Delta S$$
which is the Gibbs free energy and therefore has to be negative for a reaction to proceed spontaneously.
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Are we seen or not by an observer from a galaxy farther away than the age of our solar system? I looked at this question. There's one answer that alludes to what I'm asking, but I don't find it satisfactory, because to me the following is still a paradox:
The most distant galaxies we have ever observed appear to be around 13 billion light years away. Here's my question: Suppose the universe were not expanding, and suppose an observer in a galaxy 10 billion light-years away from us RIGHT NOW were to look directly at our planet (or rather at least the section of space our planet is occupying) with some magical telescope powerful enough to do such a thing. In this hypothetical scenario, both our planet and this galaxy would be static. The physical distance between the two doesn't change throughout all time. If we can observe such a far away galaxy (albeit as it was 10 billion years ago), could an observer in this supposed galaxy see earth? You'd say no, because light from our earth hasn't had enough time to reach the galaxy, with Earth being only 4.5 billion years old. But how is it that we could see such a galaxy from our viewpoint? Agreed that when light left that galaxy the earth was still a twinkle in the Milky Way's eye. But the light made the distance. At the same instant the distant observer looks at us and finds nothing?
Why is it that we can see them and they can't see us? Again, static environment. Light works one-way only?
What are the conditions for simultaneous mutual observation??
| Because, not accounting for the expansion of space, we are seeing what was in their galaxy 10 billion years ago if they are 10 billion light years away, we do not know what it looks like right now. If observers there are looking towards our part of space right now they will see what was here 10 billion light years ago, most likely the molecular cloud that we were formed from since our Sun is only about 4.6 billion years old. So, at this moment, we would both see only each others past.
| {
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Interpretation of Hooke's Law I often see people interpreting Hooke's Law $σ=Eε$ as,
"The deformation $ε$ that occurs when you subject a material to a stress $σ$."
This makes it sound like stress is an external stimulus that causes the material to deform. But from what I know, stress is an internal phenomenon, not an external one.
So technically, isn't it more correct to interpret Hooke's law as
"The stress σ that develops in a material given a deformation of $ε$"?
I would greatly appreciate it if someone could clear this up for me.
| Hooke's Law—in the standard form as you've written it—says that the normal stress $\sigma$ and normal strain $\varepsilon$ are linearly coupled by a constant of proportionality $E$ (termed Young's modulus). This is a good approximation for a long rod of a stable, constant-temperature solid for small axial deformations over moderate time scales. (All these qualifiers are needed to eliminate the effects of stresses in other directions, temperature dependence, creep, etc.) Note that no specific cause or effect regarding the stress or strain is implied.
| {
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Why do we need instantaneous speed? I am new to this topic and was just wondering about the use of instantaneous speed. I mean, we use to calculate the speed of car let us say at 5 sec. So we take the distance travelled in 4.9 to 5.0 seconds and divide it by time. We get instantaneous speed. We could simply as well have had taken distance travelled from 0 to 5 seconds and then divide it by time. So what is the use of instantaneous speed then?
| One way of looking at it is that instantaneous speed gives you more details about your journey, especially when your journey consists of variable speeds. The smaller the time interval in which you measure, the more information you have about your journey. But if your speed were constant, instantaneous speed would make no difference as it would be the same as the average speed at all times.
You can think about it like pixels on a computer screen. The smaller the pixels and the more pixels you have, the clearer and more detailed the picture will be. But if your picture is just the color red, it makes no difference how many pixels there are, even if the entire screen were just one big pixel, which is analogous to a car having constant speed.
Another thing to note is that even when you say instantaneous speed, it is not truly instantaneous but rather the average speed within a much smaller time interval
| {
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What causes water droplets to drop in periodic, but not uniform time intervals? A little while ago I noticed water droplets forming from a slightly overflowing reserve in my sink. They dropped in a special periodic time pattern, which was not uniform. Instead two droplets would be created and fall right after each other followed by a longer pause than the distance in time between the two droplets. Then the pattern would repeat, resembling something like "oo…oo…oo…" and so on, if "o" represents a single droplet and "…" a pause longer than the distance in time between two droplets "oo".
I have recorded this with my camera and made it available here: https://nextcloud03.webo.hosting/s/eY5cAsb3XsxDy57
Is there any theory capable of predicting such a behaviour (ideally qualitatively and quantitatively)?
(I am grateful for any suggestion for improving the question and its associated tags.)
| The droplet formation process depends on gravity, surface tension, the nozzle diameter, and any velocity that the water has before it begins forming up into a droplet at the nozzle tip. When one droplet breaks off the nozzle tip, it tends to leave behind a velocity field in the water right next to the nozzle tip which perturbs the formation of the next droplet. This makes the behavior of droplet #2 dependent on the behavior of droplet #1, since droplet #1 establishes the initial conditions for the formation of the next drop.
Note also that because the growing droplet has mass and the surface of the droplet acts like a springy membrane, there will be a natural frequency at which the system likes to oscillate. Note further that the mass goes like ~(cube root of radius) and the compliance of the surface goes like ~(some godawful nonlinear function of the radius). This means that the natural frequency is a strong function of the displacement of the droplet, meaning in turn that the system is strongly nonlinear.
The presence of strong nonlinearity, feedback effects between consecutive droplets, and sensitive dependence on initial conditions add up to something called quasiperiodic behavior, which is related to (but different from) chaotic dynamics. Water droplets breaking off a nozzle are a good example of quasiperiodicity.
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Trapped Radiation inside a Faraday Cage Say you have a Faraday cage mesh with holes small enough to reflect the radio waves. What happens to radiation when it is emitted from inside the Faraday cage?
Does it keep reflecting off of the inside of the cage forever?
| No. Unless the cage is a perfect conductor (which it is impossible in real life to make or find), a portion of the energy of the waves will get absorbed by the walls of the cage on each reflection, eventually causing the wave to attenuate or "die out". Given the incredible speed of light, these reflections happen close to a billion times each second in a cage around $1\, \mathrm{m}^3$ in volume, so the attenuation happens really quickly.
The absorbed radiation (electromagnetic energy) gets transformed into another form of energy, usually heat. The walls of the conductor heat up slightly as a result.
Typically, the radiation penetrates a characteristic distance called the "skin depth" inside the material of the conductor, before falling to around $1/3$ of its initial strength. For example, copper has a skin depth of around $8.5 \, \mathrm{mm}$ for a radiation of frequency $60\, \mathrm{Hz}$. The conductor will get heated up as the wave is attenuating; it is a continuous process.
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Proper terminology for comparing decibels Decibels are a log-scale rather than linear unit, so for example 10 dB + 10 dB is about 13 dB, not 20dB. What then would be the proper terminology for comparing decibels?
For example, how would we describe the relationship between 13 dB and 10 dB? Would we say that "13 dB is 3 dB louder than 10 dB", or that "13 dB is 10 dB louder than 10 dB?" Is going from 10 dB to 13 dB "an increase of 3 dB" or "an increase of 10 dB"?
Or, to avoid confusion, is it better to completely avoid comparing decibels in terms normally associated with addition and subtraction, and always describe in terms of scale ("twice as loud", "half as loud", "300% louder", etc)?
| As mentioned by The Photon in the comments, you add decibels according to normal arithmetic, so 10 dB + 10 dB = 20 dB. However, you need to be careful with what that means. Saying that 13 dB is 3 dB louder than 10 dB signifies that the pressure amplitude of the 13 dB signal is roughly twice that of the 10 dB signal.
There is a reason for this usage, confusing as it may seem. Human perception of loudness is roughly logarithmic, so increasing the signal by 3 dB sounds like (again, only roughly) the same increase in loudness regardless of what the initial level was. (More precise measures of loudness exist, such as the phon, described here.) From a human perception standpoint, saying that the signal increased in amplitude by, say, 50 mPa really doesn't tell you very much. On the other hand, if you want to work on the physics of sound, human perception is somewhat irrelevant, and it is usually best to just work in units of pressure directly.
| {
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Help proving bound on POVM measurement probabilities I am trying to follow Nielsen and Chuang's 1 proof that the difference in measurement probabilities is bounded by the difference between two unitary operators applied to a given state.
Can someone show me how to get from Equation 4.66 to 4.67 in the proof below (see page 195 in the 10th anniversary edition):
1 Nielsen and Chuang, "Quantum Computation and Quantum Information"
| $$|\langle \psi| AB \phi\rangle | \leq ||\psi|| \: ||AB \phi|| \leq ||\psi||\: ||AB|| \: ||\phi|| \leq ||\psi|| \: ||A||\:||B||\: ||\phi||$$
In our case $||\psi||=1$ and $||A||, ||B|| \leq 1$.
because one of $A$ and $B$ is unitary and the other is part of a POVM.
| {
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Where does the law of conservation of momentum apply? Take the scenario of a snowball hitting a tree and stopping. Initially, the snowball had momentum but now neither the snowball nor tree have momentum, so momentum is lost (thus the law of conservation of momentum is violated?). Or since the tree has such a large mass, is the velocity of the tree is so small that it's hardly noticeable?
If the explanation is the latter, this wouldn't hold for a fixed object of smaller mass. So in that case, how would the law of conservation of momentum hold?
| The ground applies a frictional/constraint force , so there is a net external force on the system and hence, Conservation of momentum can't be used on the snowball and tree as a system.
| {
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Can thermodynamics be considered logical? One of the laws says that heat won't flow from cool to warm and at the same time this same theory claims that there is a finite (albeit tiny) chance that it will, because there is always such a microstate.
We can also have a situation where all air molecules in the room can be found in the left side of the room and none in the right side, because it is one of the microstates therefore it can happen and the entropy will drop. So how can we say that the entropy always increases when it can decrease too sometimes?
| You need to distinguish between the net transfer of energy in the form of heat between two bodies and the energy transfer that can occur between the individual molecules of the two bodies.
Take the simple example of heat transfer by conduction. Body A is placed in contact with body B where $T_{A}>T_{B}$ prior to contact. We know that net heat transfer only occurs spontaneously (naturally) from body A to body B until, as some point, they reach thermal equilibrium.
Now consider what is going on between the individual molecules of body A and body B. The fact that the temperature of A is higher means that the average translational kinetic energy of the molecules in A is greater than B. But the kinetic energy of the individual molecules in A and B vary about the average value of all the molecules of the bodies. This mean that the kinetic energy of some of the molecules in B may be greater than some of those in A. If those molecules happen to collide, energy is transferred from body B to A, i.e., from the lower temperature body to the higher temperature body.
But those collisions are outnumbered by the collisions between the higher kinetic energy molecules of A with the lower kinetic energy of B so that, on average, net energy is transferred from the higher temperature body A to the lower temperature body B. Or to put it another way, the probability of higher energy molecules of A colliding with lower energy molecules of B is greater then the reverse.
Hope this helps.
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How does a deadcat work? This is a deadcat. It's a fluffy cover you put on top of a microphone to prevent wind noise on the recording. Lavalier and head-mounted microphones often use pieces of foam over the mic for the same reason. But how? I can't come up with any reasonable physics justification for why it would eliminate wind noise while leaving voices basically untouched.
| Wind noise is generated when wind hits a surface. The rigid structure of the surface resonates at particular frequencies, generating the noise. A fluffy cover or foam has no resonant frequencies, so does not generate noise, and at the same time shields the microphone itself from the wind. Likewise the fluffy cover minimally impedes sound waves (vibration in the air).
| {
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What experiment confirms $\mathbf{J}^2 = \hbar^2 j(j+1)$? I learned that if we measure the spin angular momentum of an electron in
one direction $J_z$, we get $\pm \frac{1}{2} \hbar$. But if we measure
the magnitude of the angular momentum $\mathbf{J}^2$, we should get
$\frac{3}{4} \hbar^2$. What experiment gives the latter result?
As @user1585635 notes, measuring $J_x$, $J_y$, and $J_z$ separately and
summing their squares gives $\frac{3}{4} \hbar^2$. This is not what I'm
looking for. First, if I measure $J_z$, measure angular momenta in
three directions separately, and measure $J_z$ again, the two
measurements of $J_z$ aren't guaranteed to be the same, since $J_z$,
$J_x$, and $J_y$ don't commute. But $\mathbf{J}^2$ commutes with $J_z$.
Second, when the spin is not $\frac{1}{2}$, say it's $j$, summing the
squares of components of $\mathbf{J}$ gives 3ℏ²², where ² should
be ℏ²(+1) doesn't always give $\hbar^2 j(j+1)$. (Thanks to
@MichaelSeifert for pointing that out)
This question is not a duplicate of
Why is orbital angular momentum quantized according to $I= \hbar \sqrt{\ell(\ell+1)}$?.
That question is about how $\mathbf{J}^2$ is derived mathematically.
Mine is about how it is confirmed experimentally.
| It is true that $J_x$, $J_y$ and $J_z$ does not commute.
But $J_z$ does commute with $\textbf{J}^2=J_x^2+J_y^2+J_z^2$. So there can be an eigenstate of both $J_z$ and $\textbf{J}^2$, where you can simultaneously measure both the z component of the angular momentum and the magnitude of the angular momentum.
If the possible eigenstates of $J_z$ each has an eigenvalue of from $-l\hbar$ to $+l\hbar$, each eigenstate will also be an eigenstate of $\textbf{J}^2$ with eigenvalue of $l(l+1)\hbar^2$.
For example, if the possible eigenvalues of $J_z$ are $-\hbar,0,+\hbar$, each of these three eigenstates of $J_z$ will also be an eigenstate of $\textbf{J}^2$ with eigenvalues $2\hbar^2$.
Note that the way we measure the magnitude of angular momentum is not by measuring the individual $J_x$, $J_y$ and $J_z$ components and then summing them up. That would be the classical way of measuring things.
In QM, we need to find the eigenvalues of the operator $\textbf{J}^2=J_x^2+J_y^2+J_z^2$ and then take the square root of its eigenvalues to find the magnitude of angular momentum we are able to measure physically.
| {
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Does a planar object balance on a unique point? Consider a horizontal planar convex 2D object (say lying on x-y plane) with uniform density. Under constant gravitational force (say in -z direction), does it always balance on a unique point lying on the object (i.e. sum of the torques vanishes with respect to a unique point)?
I guess the answer is yes and if so then I want to conclude that the point must be the object's center of mass (that is the object will balance in any orientation w.r.t. that point), assuming the uniqueness of center of mass.
Possibly the question is trivial but I have been confused over this for some time. Any comments or answer will be appreciated.
EDIT: Added the assumption of convexity, as otherwise the point of balance may not lie on the object.
Let me add that, one may assume existance of the point. I am more interested in showing that there can not be two or more points of balance.
| Saying that the object balances on a point, suggests to me that the point is on the perimeter of the object (lke a corner). Given that, then the object can be in an unstable equilibrium on any extruding point as long as the point is below the center of gravity. If you are going to drill a hole and insert a axle, then the equilibrium will be stable if the axle is at or above the center of gravity.
| {
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$P=\epsilon_o \chi E$ or $\epsilon_o \chi E_o$ Suppose the polarisation inside a dielectric is given by $P$, then is it related to the electric field as $\vec{P}=\epsilon_o \chi \vec{E}$ where $E$ is the field inside the dielectric or is is $E$ the original field that would have been present in that region in absence of the dielectric?
| Good question! $\vec{E}$ is the total field. i.e. the external field in the absence of the dielectric plus the field due to polarisation of charges in the dieletric.
Source: Griffiths, Introduction to Electrodynamics 4Ed., p181. (Griffiths actually says the field due to anything except the polarisation, which could include free charges inside the dielectric, plus the field due to polarisation).
| {
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What does it mean that a neutron has a 'negative' magnetic moment? Most questions about this ask why, or how, a neutron has a magnetic moment at all, or why it is negative....
But I am curious as to what it means, physically or experimentally, for a magnetic moment to be 'negative'....
I am reading that a neutron's 'angular-momentum' spin is pointing in the opposite direction of its 'magnetic-moment' spin, but I thought that the quantum spin of a particle WAS its magnetic moment spin....
At any rate, what experiment(s) showed that neutrons have a 'negative' magnetic moment spin? Maybe reading about that will help...
Edit: P.S.: I still don't understand how a neutron's negative magnetic moment is actually manifested... Perhaps an antiproton or electron, being negatively charged, is antiparallel, or whatever, but a neutron is neutral....
| It means that a neutron’s magnetic moment vector is in the opposite direction from its spin angular momentum vector.
| {
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Will speed of light in any denser medium will be same for all the inertial frame in that medium? As we know the speed of light in vacuum is constant for everyone (at least for all inertial frames), i.e., if we run away or toward a beam of light (in vacuum) the speed would be $c$. It doesn't change.
But imagine a beam of light traveling in a denser medium (where the speed of light decreases), say, water. Let the speed of light in water be $s$. Suppose a fish is swimming away from that beam of light at velocity $V$. For the fish will the speed of light will be $s-V$ or $s$?
Thank you for spending your precious time.
| If we assume linear media, then it is easy enough to transform the E- and D-fields and show that permittivity is not a relativistic invariant.
Thus the refractive index of a medium is not relativistically invariant and observers in different inertial frames measure a different speed of light in a medium.
For example, if the wave motion is parallel to the velocity difference $v$ between the two frames S and S', then
$$n' = \frac{n + v/c}{1 + nv/c} .$$
(Shen 2004).
However, note that you can no longer define a single refractive index in S' and it will depend on the direction in which the waves are travelling.
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How can we discern so many different simultaneous sounds, when we can only hear one frequency at a time? As I understand it, the eardrum works like any other kind of speaker in that it has a diaphragm which vibrates to encode incoming motion into something the inner ear translate to sound. It's just a drum that moves back and forth, so it can only move at one rate or frequency at any given time.
But humans have very discerning ears and can simultaneously tell what instruments are playing at the same time in a song, what the notes in the chord of one of those instruments is, even the background noise from the radiator. All of this we can pick apart at the same time despite that all of these things are making different frequencies.
I know that all of these vibrations in the air get added up in a Fourier Series and that is what the ear receives, one wave that is a combination of all of these different waves. But that still means the ear is only moving at one frequency at any given time and, in my mind, that suggests that we should only be able to hear one sound at any given time, and most of the time it would sound like some garbled square wave of 30 different frequencies.
How can we hear all these different frequencies when we can only sense one frequency?
|
So too much frequency and you do lose the ability to decipher it and it starts to just sound like nosie?
If that happens, it's because of how your brain interprets the signals that it receives from your ears, and not because of the physics of how your ears work. If the total sound pressure level is not so great as to damage your hearing, then your ears will faithfully report all of the frequency components to your brain, but there seem to be limits to how many distinct "signals" your brain is able to derive from that information at the same time.
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Do photons "lose energy" when they are absorbed? Recently in my biology class I learned about an experiment in which isolated and illuminated chlorophyll pigments fluoresce in the red part of the spectrum, but also, the solution of the pigments gets hotter. Are the photons that are reflected as the electrons fall back to their ground states lower in energy than the photons absorbed?
| You use "absorbed" in the title, and "reflected" in the text, so I'm not clear on what your conception of the process is. In fluorescence the incoming photon is completely destroyed, and a new photon of lower energy is generated. This is evident because the incoming light is typically blue or ultraviolet, and the emitted light is green, or yellow, or red, ... some color that is associated with photons of lower energy than blue. The energy deficit between the incident and emitted light has to go somewhere. In the situation you described, it eventually ends up in the thermal energy of the system.
| {
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Can sound be used for propulsion? I'm no physicist so this might seem absurd.
I Remember watching a cartoon as a kid where the character uses a powerful speaker to propel his cart and I was wondering if this was actually possible.
Being a highschooler I am aware to propel forward you shoot something backward.
So maybe in the case of a speaker it could "shoot out" sound waves?
| No, you can't. Here is why:
When a loudspeaker is producing sound, it is pushing forward to produce the compression part of the wave in air, then it pulls backwards to produce the rarefaction part of the wave, then forward, then back, etc. This means there is no net force applied to the air from the cone and no net reaction thrust applied by the air to the cone. Your vehicle would just vibrate back and forth as the cone works back and forth and no propulsion would result.
| {
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Polarity in a magnetized Möbius strip When a flat iron or Alnico washer is magnetized one of the faces develops a north polarity and the other, south. The geometric shape here is simple.
However, when a standard Möbius strip (or one of given thickness, radii of curvature and torsion of edges) is magnetized, which regions develop a north polarity and which regions south and according to which geometrical or other mass distribution criterion/law?
I am curious to know because such a Möbius strip (of rectangular section) has only one surface and only one edge.
Is magnetic polarity and strength distribution after magnetization influenced by changed geometry ( by homeomorphism ) ?
It may be easy to make a flattened thin Möbius strip looking like a recycling symbol to apply a magnetizing current. Thanks in advance for references.
|
When a flat iron or Alnico washer is magnetized one of the faces develops a north polarity and the other, south.
That's one way that a flat washer could be magnetized. But, it also could be magnetized in other directions (E.g., with one edge north and the opposite edge south. Or, it could be magnetized with several alternating pairs of north and south poles.
Magnetization doesn't know about "faces" or "surfaces" or "topology." Magnetization happens in the bulk of the material. Imagine starting with a solid sphere of material, magnetizing the sphere, and then carving material away until a Möbius strip-shaped piece remains. The carving would not change the way that the material was magnetized.*
* assuming that the process doesn't generate too much heat!
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How can magnet field go through non-transparent medium? We know that the change in the magnetic field is carried on electromagnetic waves so when magnets move relative to each other they radiate. But if the medium is not letting em cross in some length then the change in the magnetic field is blocked which means they won't be able to feel a force on each other if there is a medium that blocks the EM wave so do magnets produce radiation in length that cannot be blocked?
| Normally, electromagnetic response of media is frequency-dependent, e.g. gamma-particles and x-rays can go through most things.
You are interested in magnetic field due to moving magnets - that would be very low frequencies: hertz, milli-hertz etc. I guess a lot of media are transparent at those frequencies.
One needs to be particularly careful when working with phenomena where you move magnet or charge from one place to another and leave it there. Such analysis should really be done in time-domain since there is no lower frequency cut-off there. Dispersive (frequency-dependent) response would then become a response function of the material.
As far as your question regarding media that block magnets - Meissner effect in superconductors comes to mind. Superconductors will block any (static) magnetic field as long as it is below the critical value.
| {
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If a jet engine is bolted to the equator, does the Earth speed up? If a jet engine is bolted to the equator near ground level and run with the exhaust pointing west, does the earth speed up, albeit imperceptibly? Or does the Earth's atmosphere absorb the energy of the exhaust, and transfer it back to the ground, canceling any effect?
| Total angular momentum with respect to the center of mass, in practice the center of Earth, is conserved. When the airplane takes off, it acquires an angular momentum antiparallel to the Earths rotation axis. The total angular momentum of Earth plus atmosphere increases by the same amount. This angular momentum will eventually distribute over the entire planet. Long before this proces is completed the plane lands or crashes. Assuming it makes no turns its landing will decrease the earth + atmosphere AM back to its original value. If it turns while being up it again imparts angular momentum to the atmosphere. In the end it all adds up to zero but there are transient atmospheric effects.
| {
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When we use Lorenz gauge or Coulomb gauge, the result formula for electric $E$ and magnetic field $B$ is same or different? Gauge condition can be chosen as you like or not?
is the Lorenz gauge is the only one correct? If Coulomb gauge can obtained exactly same results as Lorenz gauge for the electromagnetic fields E and B, we can choose what we like, either Coulomb gauge or Lorenz gauge. If use different gauge we get different formula of electromagnetic fields $E$ and $B$, then we have to choose a corrected gauge to get the correct electromagnetic field formula. I am not clear that when we use Lorenz gauge or Coulomb gauge, the result formula for electric $E$ and magnetic field $B$ same or different?
| Gauge condition is any human imposed restriction on the functions $\varphi(\mathbf x,t), \mathbf A(\mathbf x,t)$ that does not change electric and magnetic field implied by those potential functions.
For any single physical situation, one can use either potential functions obeying the Coulomb gauge condition or those obeying the Lorenz gauge condition. Both choices give the same electric and magnetic field.
| {
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Principal quantum number of the classical particle The example 7.9 in this page shows the principal quantum number of the classical particle.
A small 0.40-kg cart is moving back and forth along an air track
between two bumpers located 2.0 m apart. We assume no friction;
collisions with the bumpers are perfectly elastic so that between the
bumpers, the car maintains a constant speed of 0.50 m/s. Treating the
cart as a quantum particle, estimate the value of the principal
quantum number that corresponds to its classical energy.
In the result, the principal quantum number of the cart is huge due to the high kinetic energy. If the cart moved very very slow, can we find the cart at other place? Why not?
| In the link you give it says "as though the cart were a quantum particle", so to ask:
If the cart moved very very slow, can we find the cart at other place
is to ask if the kinetic energy is very very small: "can it behave as a true quantum particle".
In the link they answer using the "bohr correspondence principle".
Considering that the energy is found to be 0.05J , our best timing is nanoseconds, and $h=6.62607015×10^{-34}$ Js the HUP always holds as if h=0, so there is no envelope in which a probable location can be measured.
The problem makes it clear that the relationships used are for the large energies, where the quantum formalism and the classical one give the same result, for low energy levels there is no connection between classical and quantum."However we cannot apply classical formalism to a quantum system in a low number quantum state". Your "very low energy" falls in this category. In classical physics there are no "probable states" for simple kinematic problems.
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Finding an exact value for energy in perturbation theory
Supose a particle of mass $m$ and electric charge $q$, subject to harmonic potential in 1D, is placed in an area with electric field $\vec E = E \hat u_x$. Determine the exact change in its energy spectrum caused by interacting with this field.
I started by writing down the Hamiltonean operator as:
$\hat H = \frac{1}{2m}\hat p_x^2 + \frac{1}{2}m\omega^2 \hat x^2 + qE \hat x$, where the last term represents the perturbation caused by interacting with the electric field. (We can treat it as a perturbation since usually $q \ll 1$.)
Using this, for any state $|n \rangle$, the first order correction to the energy is always going to be zero, since, using the ladder operators: $\hat a_{\pm} = \sqrt{\frac{m\omega}{2\hbar}}(\hat x \pm \frac{i}{m\omega} \hat p)$,
$$\epsilon_1 = \langle n | \hat W |n\rangle = qE\cdot 2\sqrt{\frac{2\hbar}{m\omega}} \langle n | \hat a_+ + \hat a_-|n\rangle = C (\sqrt{n+1}\langle n |n+1 \rangle + \sqrt{n}\langle n |n-1 \rangle) =0,$$
because the states are orthogonal to each other. Therefore, the first order correction is zero. However, this still leads to an approximate answer:
$$E(n) \approx \epsilon_0 (n) + q\epsilon_1 + O(q^2) = \epsilon_0 (n) + O(q^2), $$
so I don't understand how to get to an exact value for the change in energy, specially since in my Quantum Mechanics class we didn't cover higher order corrections.
Is there another way to approach the problem that I'm not seeing?
| This isn't intended as a perturbation theory problem. (It actually can be solved, to all orders in perturbation theory, but that would be unthinkably arduous.) The actual point is the notice that the potential, including the linear potential due to the electric field, is still a quadratic function of the position,
$$V(x)=\frac{1}{2}m\omega^{2}x^{2}+qEx=\frac{1}{2}m\omega^{2}\left(x-\frac{qE}{m\omega^{2}}
\right)^{2}-\frac{q^{2}E^{2}}{2m\omega^{2}}.$$
The full Hamiltonian therefore represents another harmonic oscillator (centered around a different location), with exactly the same frequency, but a different ground state energy. Thus
$$E_{n}=\left(n+\frac{1}{2}\hbar\omega\right)-\frac{q^{2}E^{2}}{2m\omega^{2}},$$
which does agree with the first-order perturbative expression in the question.
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Finding contradiction in equations Question:- In the figure shown, Coefficient of friction between the blocks C and B is $0.4$. There is no friction between block C and ground.The system of blocks is released from rest in the shown situtation.Find the accelerations of masses.(Given $m_{B}=5kg , m_{C}=10kg , m_{A}=3kg$ )
I made two cases
$1)$ Block B and C move together.
$2)$ Block B and C will not move together.
When I solve equations for Case $1$, I find that block will move together
$$30-T=3a_{3}$$
$$T-f=5a_{5}$$
$$f=10a_{10}$$
$$f \leq 20$$
Note that if blocks B and C will move together then $a_{3}=a_{5}=a_{10}=a$
On solving we get $a=\frac {5}{3}m/s^2$ and $f=16.67N$
I got the following equations for case $2$.These equations should somehow contradict each other because case $1$ is correct.Despite my efforts I was unable to find how they will contradict.
\begin{align}
30-T&=3a_{3}\\
T-f&=5a_{5}\\
f&=10a_{10}\\
f&\leq {20}\\
a_{3}&>a_{10}\\
\end{align}
Note that $a_{3}=a_{5}$
Adding equation $1$ and $2$, we get
$$30-f=8a_{3}$$
So, $30-8a_{3}\leq 20$ and $10a_{10}\leq 20$
This gives $a_{3}\geq \frac54$ and $a_{10}\leq 2$
As $a_{3}>a_{10}$ but above inequalities of $a_{3}$ and $a_{10}$ satisfies it for certain interval.
Can anybody help to find contradiction.
I have also asked same question on maths stackexchange since it involves inequalities
| There should be different between static frition coefficient $\mu_s$ and the kinetic friction ceofficient $\mu_k$, and $\mu_s$ must be greater than $\mu_k$. If $\mu_s \lt \mu_k$, there will induce non-physical phenomena.
Another concept is that the kenetic friction (as long as two badies have relative motion) is a constant $f_k = \mu_k N$ within the simple kenetic friction model. It is not less or equal than condition. Only in the static firction (when two bodies have no relative motion) $f_s \le \mu_s N$.
Therfore, in your problem, in setting $\mu_k = 0.4$, meaning $\mu_s \gt 0.4$.
In your case 1, the friction between Block B and Bolck C are static friction (there is no relative motion), $f=17 N$, $f_k = \mu_k 10 M_B=20 N$, these values satisfy $f \lt \mu_k 10 M_B \le \mu_s 10 M_B$. It is ok to apply static friction for this case.
But in the case 2, there is relatice motion, the kinetic friction between Block B and Block C:
$$
f_k = \mu_k 10 M_B = 20 N
$$
Once there is relative motion between B & C, the kenetic friction is a fixed number $20 N$.
The acceleration for Block C:
$$
a_C = 20 / 10 = 2 m/s^2.
$$
and the acceleration for Block A and Block B system:
$$
a_B = a_A = (30 - 20) / (3 + 5) = 1.25 m/s^2
$$
This results in block C an acceleration $2 m/s^2$, while the the system A & B has an acceleration $1.25 m/s^2$. It is not Physical.
Since the kenetic friction force is a fixed number as long as B and C having relative motion. There will be no flexible interval for the OP's condition to hold true.
| {
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Newton's 3rd Law and a bouncing ball just want to clear something up.
Take a ball that's been dropped to the ground. Gravity acts and this ball as it has mass and then the ball now moves to the ground with a constant force of say ($X$).
Now when the ball makes contact with the ground, Newtons 3'rd Law takes effect (no air resistance):
If object A (the ball) exerts a force on object B (the floor), then object B will exert an equal force on object A in the opposite direction. (Action has equal opposite reaction).
Now here's where I get confused.
If the ball (which has a constant force when it hits the ground ($X$)) experiences the same constant force in the opposite direction ($-X$, minus indicating opposite direction), then the total force acting on the ball should be net ZERO ( $X + (-X) = 0$).
So there are no net forces acting on the ball, so why does it BOUNCE BACK? What am I missing?
Shouldn't the ball just stay on the ground? Bouncing back means a force greater than (-X) was applied to the ball giving it upward motion. Where did it come from?
|
Now here's where I get confused.
If the ball (which has a constant force when it hits the ground ($X$))
experiences the same constant force in the opposite direction ($-X$,
minus indicating opposite direction), then the total force acting on
the ball should be net ZERO ( $X + (-X) = 0$).
Yes the ball and the floor (Earth) exert equal and opposite forces on one another per Newton’s third law but they don't "cancel" each other. To determine the effect of the equal and opposite force on each object you need to apply Newton’s second law $F=ma$ to each object individually.
$$a_{ball}=\frac{F}{m_{ball}}$$
$$a_{Earth}=\frac{F}{m_{earth}}$$
The acceleration of the earth is so small due to its large mass that only the acceleration of the ball is observed.
The actual force experienced by the ball and earth on impact depends on the nature of the impact (elastic or inelastic} as well as the stopping distance/time of the ball. In this regard one can apply the work energy theorem which states that the net work done on an object equals its change in kinetic energy.
Hope this helps.
| {
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What is the meaning of $F=ma$? Does it mean that an object with mass $m$ will have an acceleration $a$ if an external force $F$ is applied to it? I know this is a very simple question, but I am just learning physics. I am seeing the basics of how a block on a horizontal frictionless surface gets accelerated by a force F in any direction.
However, I ask myself if F=ma is only used on the object the force is being applied to, not the source of the force (which could be a finger, another block, or anything that pushes)
Could one use F=ma in order to calculate the force an object of mass M would produce on its own if it already had an acceleration.
Thank you for you helpful responses in advance
|
However, I ask myself if F=ma is only used on the object the force is
being applied to, not the source of the force (which could be a
finger, another block, or anything that pushes)
$F_{net}=ma$ is Newton's second law. Newton's third law says that the object the force $F$ is applied to exerts an equal and opposite force $F$ on the object applying the force (finger, another block, or anything that pushes). The effect on the object applying the force will depend on the net force it experiences, per Newton's second law applied to it.
Could one use F=ma in order to calculate the force an object of mass M
would produce on its own if it already had an acceleration.
"...had an acceleration" implies that the object of mass M is no longer accelerating, that is, it is no longer being subjected to a net force, i.e., $F_{net}=0$. It is therefore continuing at the velocity it had when the force was removed and therefore neither producing or experiencing a force (assuming no friction or air resistance). However such an object could produce a force on another object if collided with another object.
Hope this helps.
| {
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Hypothetically, why can't we wrap copper wires around car axles and turn them into electromagnets to help charge the batteries? We already have a magnetic core, why can't we use it to recharge the batteries? The only problems I see with it are potentially wiping magnetic data, but doesn't the electromagnet have to be revolving around the damageable device?
| Yes. This is how regenerative braking works.
There is a style of brakes that is used in many electric or hybrid cars that, rather than converting the motion of the wheels into heat energy the way conventional brakes do, instead convert the motion of the wheels into electrical energy that recharges the car's battery. By doing this, they can improve the performance of the car and extend how far it is capable of traveling when driven in environments with frequent braking, like most cities.
There's no technical reason why such a braking system couldn't be installed on a car with a purely internal combustion engine, though I expect that the energy that they produce might be wasted once the battery has been fully recharged, and the extra cost probably isn't worth the marginally increased battery performance.
| {
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Centrifugal Force Dilemma While learning Classical Mechanics, I am confused on nature and application of centrifugal force.
In my textbook, it is written that centrifugal force is a pseudo force that, depends on reference frame, but I can't understand that if it is pseudo force then why we feel something pushing us outwards during a tight turn in vehicle.
Also, I am very confused that when we will apply centrifugal force, since centrifugal will cancel centripetal force so how will the object move in circle in absence of radial acceleration? (My understanding)
Please clarify my confusion and tell any flaws in my understanding if any,
| It is not you who is moving outwards and squeezing into the car door. It is the car door which is moving inwards into you.
Your inertia makes your body tend to keep moving straight.
Just like when standing in a bus that brakes - you feel pushed forwards, but in fact it is the bus that is being pulled backwards underneath your feet.
When turning, the car door won't allow that your body continues straight. That would break the door. So it pushes you inwards along with it. That is then the centripetal force - it is inwards and there is no other force involved here.
All in all: The "centrifugal force" is merely an illusion. There is no such force. But the centrifugal effect, if we choose to call it that, is very real.
| {
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Prove $\exp (-i \phi(\hat{n} \cdot \vec{\sigma}))=\cos \phi-i(\hat{n} \cdot \vec{\sigma}) \sin \phi$ using certain property I'd like to prove $$e^{-i \phi(\hat{n} \cdot \vec{\sigma})}=\cos \phi-i(\hat{n} \cdot \vec{\sigma}) \sin \phi$$ using $$\sigma_{i} \sigma_{j}=\delta_{i j} I+i \varepsilon_{i j k} \sigma_{k},$$ where $\phi$ is a real constant, $\sigma_i$ are Pauli's matrixes and $\hat{n}$ is a unit vector. Every proof I've seen uses another procedures, such as Nakahara's Quantum Computing: From Linear Algebra to Physical Realizations (page 23). How could I do this?
| Your relation dictates $(\hat{n}\cdot \vec{\sigma})^2=I$, hence, for integer m,
$$
(-\hat{n}\cdot \vec{\sigma})^{2m}=I, \qquad (-\hat{n}\cdot \vec{\sigma})^{2m+1}= -\hat{n}\cdot \vec{\sigma}.
$$
So, what is
$$
\exp \Bigl ( i\phi (-\hat{n}\cdot \vec{\sigma}) \Bigr )= \cos \Bigl ( i\phi (-\hat{n}\cdot \vec{\sigma}) \Bigr ) +i\sin \Bigl ( i\phi (-\hat{n}\cdot \vec{\sigma}) \Bigr ) ~~?
$$
| {
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Why do we need to introduce thermodynamic potentials? Each thermodynamic potential (Enthalpy, Helmholtz free energy, Gibbs free energy) is the same as the 1st law of thermodynamics. Then, why do we need them? Why did people introduce them in the first place?
| Each potential describes a different physical situation, i.e. they differ by the controlled variable of the experiment being described by them.
Take for example the Helmholtz free energy. Note that
$$
dF=-PdV-SdT
$$
so F is a function of volume and temperature: $F(T,V)$, since it follows from the equation that small variations in V or T cause a variation in F. It therefore describes a situation in which these variables are controlled, while their conjugate variables depend on them. An experiment connected to a heat bath and confined to a constant volume can be described by this potential. A different experiment, e.g. when pressure is constant (and volume may vary), will be described by the Gibbs free energy, and so on.
| {
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Quantum Harmonic Oscillator Virial theorem is not holding I'm asked to calculate the average Kinetic and Potential Energies for a given state of a quantum harmonic oscillator. The state is:
$$
\psi(x,0) = \left(\dfrac{4m\omega}{\pi\hbar}\right)^\frac{1}{4}e^{\frac{-2m\omega}{\hbar}x^2}
$$
The thing is, calculating
$\langle T\rangle=\int_{-\infty}^{\infty}\psi(x)(-i\hbar)^2\frac{d^2}{dx}\psi dx=\left(\dfrac{4m\omega}{\pi\hbar}\right)^\frac{1}{2}\int_{-\infty}^{\infty}e^{\frac{-4m\omega}{h}x^2}dx-\left(\dfrac{4m\omega}{\pi\hbar}\right)^\frac{1}{2}\int_{-\infty}^{\infty}x^2e^{\frac{-4m\omega}{h}x^2}dx=\hbar\omega$
Where I used that the momentum operator is $p=-i\hbar\frac{d}{dx}$
$\langle V\rangle=\dfrac{m\omega^2}{2}\left(\dfrac{4m\omega}{\pi\hbar}\right)^\frac{1}{2}\int_{-\infty}^{\infty}x^2e^{\frac{-4m\omega}{h}x^2}dx=\dfrac{\hbar\omega}{16}$
But then the Virial Theorem is not satisfied. I've read the virial theorem holds for any bound state and all states in a Quantum Harmonic Oscillator are bound. Can someone point out where I am going wrong?
Thank you
| The ground state of the harmonic oscillator is (see Wikipedia for example): $$\psi_0(x) = \left(\frac{\alpha}{\pi}\right)^{1/4} e^{-\alpha x^2/2},\quad \quad \text{where }\quad \alpha =\frac{ m \omega}{\hbar}$$
Your math is correct, it's just that the state you have is not a bound state of the harmonic oscillator, the parameters are slightly off. If you use the state provided above, you can indeed show that: $$\langle T \rangle = \frac{\hbar \omega}{4} = \langle V \rangle.$$
| {
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Why does a wooden spoon creates bubbles when put in hot oil? This question might be a bit weird, but I just asked myself why a wooden spoon creates bubbles when put in oil at about 170°C. My idea is, that the water in the spoon reacts with the Oil, but why does this just starts to happen when the Oil reaches approx 170°C? Why does this not happen at 100 or 120°C ?
| Wooden spoons are porous, and have oil, water, and air embedded in their surfaces after being used for a while.
Putting a spoon like this into hot oil will boil the water in the surface, and expand any air there, thus creating bubbles of water vapor mixed with air.
If you then allow the oil and the spoon to cool off together, the remaining air in the (hot) spoon will contract and pull oil into the deep pores of the wood as it does. This will form a thick layer of oil-saturated wood on the spoon surface, and suppress bubble formation the next time you stick it into a container of hot oil.
| {
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How warm are radioactive metals? I read that radium is warm to the touch -- is that because of actual heat or is that because, for example, the radiation it emits creates the sensation of warmth? How high of a temperature can a radioactive element or isotope actually have?
| The interesting number for application in radioisotope thermoelectric generators is the power by weight. It is not very high for radium, as the half-life is 1600 years. The Pioneers use generators containing 13 kg of $^{238}$Pu with a halflife of 88 years. That has a power density of 0.54 W/g.
The linked wikipedia article says: "A half-gram sample of $^{210}$Po reaches temperatures of over 500 °C." (source at Argonne Laboratory).
| {
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Meaning of adding a term to the Hamiltonian in a quantum harmonic oscillator Let $H$ be the Hamiltonian in a harmonic oscillator,
$$
H = \sum_{n=0}^{\infty} \hbar \omega \left (n+\frac{1}{2} \right ) |n\rangle \langle n|.
$$
Suppose we introduce the interaction
$$V = \sqrt{2} \hbar \omega
(|0\rangle \langle 1|+|1\rangle \langle 0|
$$
and the new Hamiltonian is $H+V$. How can we understand the physics under this interaction? Is there any classical interpretation?
| Well the eigenstates would change, but not all of them; since if $H'=H+V$, $H|n \rangle= H'|n \rangle$ for all $n>1$, where $|n \rangle$ is eigenstate of the original QHO with energy: $E_n = \hbar \omega \big(n+\frac{1}{2} \big)$. Now in the case of $n\leq 1$:
$$H'|0 \rangle=E_0|0 \rangle+ \sqrt2 \hbar \omega |1 \rangle$$
$$H'|1 \rangle=E_1|1 \rangle+ \sqrt2 \hbar \omega |0 \rangle$$
Finding the new eigenstates is therefore equivalent to finding the eigenvectors of the following matrix:
$$
\begin{pmatrix}
E_0 & \sqrt2 \hbar \omega\\
\sqrt2 \hbar \omega & E_1
\end{pmatrix} = \begin{pmatrix}
\frac{\hbar \omega}{2} & \sqrt2 \hbar \omega\\
\sqrt2 \hbar \omega & \frac{ \hbar \omega}{2}
\end{pmatrix}= \frac{\hbar \omega}{2}\begin{pmatrix}
1 & \frac{1}{\sqrt2} \\
\frac{1}{\sqrt2} & 3
\end{pmatrix}
$$
You will find the eigenvalues and associated eigenvectors to be: $\frac{4-\sqrt6}{4}\hbar\omega \text{ , } \frac{4+\sqrt6}{4}\hbar\omega$ and: $ \frac{-(2+\sqrt6)|0 \rangle +\sqrt2 |1 \rangle}{6\sqrt2 + 2\sqrt{12}} \text{ , } \frac{(\sqrt3 - \sqrt2)|0 \rangle +|1 \rangle}{6-2\sqrt6}$ respectively.
So your new Hamiltonian $H'$ will have eigenbasis:
$$\Big\{ \frac{-(2+\sqrt6)|0 \rangle +\sqrt2 |1 \rangle}{6\sqrt2 + 2\sqrt{12}}
, \frac{(\sqrt3 - \sqrt2)|0 \rangle +|1 \rangle}{6-2\sqrt6} , |n \rangle\Big\} \text{for all n>1} $$
and the energies are the same for $n>1$ and for $n \leq1$ they are the eigenvalues stated above.
Using this you can solve the S.E. and find your dynamics!
| {
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Why is energy lost here? Let's say a $1 \ \text{kg}$ block is moving.
With a speed of $1 \ \text{m/s}$ so its kinetic energy is $\frac{1}{2} \ \text{J}$. Now let's gently place a block of mass $3 \ \text{kg}$. Now as linear momentum is conserved due to lack of external forces on the system the blocks move together with velocity $1/4 \ \text{m/s}$ but the energy is now $\frac{1}{8} \ \text{J}$ which is lesser than it used to be.
Where has the energy gone?
| "let's gently place a block of mass 3 kg"
Maybe you should rethink the system. You cannot "add stuff" just like this. Energy is conserved in an isolated system. Adding a block just like this is not what I would consider as isolated at all.
Put the two blocks (or one) and then tell us what you want to do.
| {
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Non-Analytic Equations and Chaos Could anyone please tell me an example of an equation with no analytic solution(s) that is not a chaotic one? And what is the physical meaning of having analytic solution? For instance, the three body problem does not have in general analytic solution and it leads to chaos. But I don't know if this is a general statement. I have absolutely no idea. Could anyone explain me, please?
| As a real everyday physical model, consider a non-damping pendulum: ($\ddot{\theta} + \frac{g}{l}\sin\theta = 0$). It does not have a general analytical solution (Wikipedia) and yet it is not chaotic.
| {
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How to calculate the heat that leaves the furnace through one opening? I recently read in the newspapers that one man died through the flame that left the iron furnace. I wasn't sure about this. Iron is melting at 1500 °C. The furnace they used is 8 cubic meters, he was standing one meter from the furnace. The opening is 1 squared meter. By the story, he was outdoors when malfunction door opened and made him burns from which he died. The temperature in our city that day was 35 °C. I wasn't sure about the story is this heat really enough to cause the burns that can kill you. I tried to calculate the heat that left with this formula (Q=m×c×ΔT) but I have no idea of the air heat after the accident, so I can't calculate ΔT.
I'm sure I'm missing something, could someone point me in the right direction?
| $Q=m×c×ΔT$ is not useful here.
What you can do is apply Stefan - Boltzmann (as an approximation), which tells us that:
$$P=A\varepsilon \sigma T^4$$
where:
*
*$P$ is the power emitted by the grey body radiator,
*$\varepsilon$ is the emissivity, which we can approximate as $1$,
*$\sigma$ equals $5.670373\times 10^{-8}\mathrm{Wm^{-2}K^{-4}}$,
*$A$ is the surface area of the radiator,
*$T$ is the temperature of the radiator.
This gives a number of $560\mathrm{kW}$ (kilowatt).
Note however that not all of this radiation reached the unfortunate victim, as it is emitted in all directions of a hemisphere. You can estimate the portion of this radiation that reaches the man from his frontal surface are and the surface area of that hemisphere.
| {
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Is projectile motion an approximation? Doesn't the acceleration vector points towards the center of the Earth and not just downwards along an axis vector. I know that the acceleration vector's essentially acting downwards for small vertical and horizontal displacements but if the parametrization of projectile motion doesn't trace out a parabola, what is the shape of projectile motion?
| Just as the motion of body around the earth is ellipse (1st Kepler law replacing sun by earth), so is the motion of a projectile. Notice that almost everything we deal is an approximation, the earth is not a massive perfectly rounded ball, and we are neglecting the air, so it is not a sin to consider the motion of the projectile as a parabola (at least in our everyday experience as, for example, throw a stone in the river)
| {
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Will liquid nitrogen evaporate if left in an unopened container? SOS! I left work today and got a horrible feeling that I forgot to put the lid back on a large container of liquid nitrogen which contains many racks of frozen cells in it. If this did happen, how long would it take liquid nitrogen to evaporate? Does it start to evaporate as soon as it is exposed to oxygen? Will all the liquid nitrogen be gone from the container when I go back tomorrow?
| Depending on the type of container, there’s another risk: condensation and ice build-up. All containers need to be vented, but there’s a reason why the vents are designed to be one-way (nitrogen gas goes out, air can’t freely get in). If left open to air, the air will condense on surfaces near the liquid, and ice will start to form. Whether or not this is a problem depends on the specifics. But for a narrow-neck dewar left open, it could be catastrophic (if an ice plug forms in the neck, the vent could be blocked, pressure would build, and it could explode).
| {
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Is it even theoretically possible for a perfect clock to exist? I have heard that even atomic clocks lose a second every billion years or so. That raises the question, is it even theoretically possible for a perfect clock to exist, one that never gains or loses time?
| Imagine a totally perfect clock. It measures the time in the place where it is situated. Its own proper time. But due to general relativity the speed of time is affected by nearby objects, gravitational fields, spacetime curvature, speed of motion and gravitational waves.
The ideally perfect clock would look not that perfect for anyone who is located in a different place, surrounded by different bodies that curve the spacetime, and moves around at different speeds.
So, even ideally perfect clock would be not that much useful because it would measure its own proper time, in its own reference frame and in its own place relative to other objects, which is surely different than the proper time of any other object.
| {
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Momentum of a relativistic atom I have been solving a problem where you should derive the formula for doppler effect when a source of light is an atom moving at a relativistic velocity v. I understood everything in the solution except for why was the formula for impulse of the atom used as if it's not relativistic i.e. p=m*v?
Is it maybe because the atom is so heavy that the lorentz factor wouldn't change the momentum noticeably?
| There is nothing wrong on writing the relativistic momentum as $p = mv$ as long as you are saying that $m = \gamma m_0$, where $m_0$ is the rest mass (inertial mass). In other words, the equation $p = mv$ is relativistic, the Lorentz factor is simply compacted onto the mass, meaning that the mass you are using to calculate isn't the rest mass but rather the relativistic mass.
| {
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Is conformal transformation a coordinate transformation or not? I have been looking through several questions and answers on conformal transformation in the stack exchange community. While they have helped me gain a better understanding of the basic issues, I am also somewhat confused. Here are couple of them:
*
*While reading the post, Conformal transformation/ Weyl scaling are they two different things? Confused!, I became convinced of the fact that Conformal transformation is an actual transformation of the co-ordinates, leaving the metric intact while Weyl transformation is an actual transformation of metric, leaving the co-ordinates intact. However in the answer, What is the significance of the conformal invariance of electrodynamics in a covariant formulation?, they convey a completely opposite message. Thus I end up being confused.
*I understand that, tracelessness of the stress-energy tensor is said to be implied by the conformal invariance, which is sustained if the theory is massless. However, when proving this statement, what is actually used, is the Weyl transformation. This makes me think that, whether the invariance under Weyl transformation automatically implies conformal invariance!
| Conformal transformation/mapping is a term from the complex analysis. In its original sense it is indeed a coordinate transformation:
The conformal property may be described in terms of the Jacobian derivative matrix of
a coordinate transformation. The transformation is conformal whenever the Jacobian at
each point is a positive scalar times a rotation matrix (orthogonal with determinant
one). Some authors define conformality to include orientation-reversing mappings whose
Jacobians can be written as any scalar times any orthogonal matrix.[1]
These transformations are widely used in physics, e.g., for solving electrostatic problems, elastic problems, in field theory, etc.
| {
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"timestamp": "2023-03-29T00:00:00",
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Least count of cesium clock and maximum possible significant figures for time We know that a second is defined as being equal to the time duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the fundamental unperturbed ground-state of the caesium-133 atom(form wikipedia).Hence least count of this clock will be $\frac{1}{9192 631770}$ seconds
From my understanding of least count and significant figures I conclude that it is not possible to measure time duration less than $\frac{1}{9192 631770}$ seconds accurately. Is my conclusion correct?
Is there any workaround for this?
| Your conclusion is not correct. Current atomic clocks based on optical transitions are much more accurate and precise than that. So time durations less than 1/9192631770 s can indeed be accurately measured.
The kernel of truth to your conclusion is that such time durations cannot be accurately known in SI units. However, SI units have no particular existential prominence. Just because we cannot accurately measure such durations in SI seconds does not mean we cannot measure them at all.
As a side note, we can detect phase differences less than $2\pi$, so the limit would not be 1/9192631770 s anyway. What is important is not the frequency, but how well that frequency stays in phase so that you can detect differences. Right now that is on the order of $10^{-16}$ for Cesium clocks.
| {
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Is there a difference between instantaneous speed and the magnitude of instantaneous velocity? Consider a particle that moves around the coordinate grid. After $t$ seconds, it has the position
$$
S(t)=(\cos t, \sin t) \quad 0 \leq t \leq \pi/2 \, .
$$
The particle traces a quarter arc of length $\pi/2$ around the unit circle. This means that the average speed of the particle is
$$
\frac{\text{distance travelled along the arc of the circle}}{\text{time}}=\frac{\pi/2}{\pi/2} = 1 \, .
$$
However, since the motion of the particle is circular, the distance travelled is not the same as the displacement. The displacement of the particle would be $\sqrt{2}$, and so the average velocity would be
$$
\frac{\text{straight line distance from initial position}}{\text{time}} = \frac{\sqrt{2}}{\pi/2} = \frac{2\sqrt{2}}{\pi} \text{ at angle of $\frac{3}{4}\pi$ with the positive $x$-axis} \, .
$$
Here is the part I don't quite understand: over an interval, the average speed of the particle is different from the magnitude of its velocity. In the above example, the former is $1$, whereas the latter is $\frac{2\sqrt{2}}{\pi}$. However, the magnitude of the instantaneous velocity of the particle is the same as the instantaneous speed: here, they are both equal to $1$. We can mathematically prove this by considering the following limit
$$
|S'(t)| = \lim_{h \to 0}\frac{|S(t+h)-S(t)|}{|h|}=\lim_{h \to 0}\frac{\sqrt{\left(\sin(t+h)-\sin t \right)^2+\left( \cos(t+h)-\cos t\right)^2}}{|h|} \, ,
$$
which turns out to be equal to $1$. Hence, the magnitude of the instantaneous velocity is $1$. And clearly, the instantaneous speed of the particle is
$$
\lim_{h \to 0}\frac{h}{h} = 1 \, ,
$$
since the distance travelled along the arc between $S(t+h)$ and $S(t)$ is simply $h$ units. However, will this always be the case? Is the magnitude of the instantaneous velocity of a particle always equal to its instantaneous speed?
| By definition, $$\left|\text{instantaneous velocity}\right| = \text{instantaneous speed}.$$
However,
\begin{aligned}
\left|\text{average velocity}\right| &= \left|\frac{\text{displacement (i.e., change in position)}}{\text{time elapsed}}\right|\\
&= \frac{\left|\text{displacement (i.e., change in position)}\right|}{\text{time elapsed}}\\
&\leq \frac{\text{distance travelled}}{\text{time elapsed}}\\
&= \text{average speed}.
\end{aligned}
| {
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"timestamp": "2023-03-29T00:00:00",
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Explosion of an asteroid in orbit I've learned from an answer here on this site that if a body were to split apart in orbit the center of mass will continue to be on the same orbit. (Couldn't find the post)
But let's say an asteroid is to blow up in two pieces such that the smaller piece reverses its direction and velocity and thus stays in the same orbit, but going backwards.
But now the bigger piece must be faster than the velocity of the center of mass, this higher velocity changes the orbit of the bigger mass.
Now the center of mass is not in line the smaller piece continues the same orbit while the bigger piece changes the orbit. This doesn't seem right. What went wrong?
For example this question https://www.toppr.com/ask/question/an-asteroid-orbiting-around-a-planet-in-circular-orbit-suddenly-explodesinto-two-fragments-in-mass/
An internal explosion causes the center of mass to move farther and farther away from the Earth.
I'm very certain I've made astronomical blunders but I would like to know my mistake.
| If a body splits in orbit, the center of mass of the body won't necessarily be in the same orbit, before and after, even though the sum of angular momentum of the pieces will be conserved. As an example, consider an asteroid broken into 2 pieces which both have escape velocity in opposite directions. The center of mass of that asteroid wouldn't even have a bound orbit with the main body anymore.
| {
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Why the formula of kinetic energy assumes that the object has started from an initial velocity of zero? According to my physics textbook, the formula of kinetic energy is:
$$
W = \frac{1}{2}mv^2
$$
Where $m$ is is mass of the object and $v$ is the velocity of the object. The equation is calculated from this (according to that book as well): ($a$ is acceleration, $s$ is displacement, $t$ is time and $F$ is force whose value is $ma$)
$$
\begin{align}
W &= Fs \\
&= mas \\
&= ma \cdot \frac{1}{2}at^2
\end{align}
$$
Here comes the problem. According to that book: ($u$ is initial velocity)
$$
s = ut+\frac{1}{2}at^2
$$
But in the equation of kinetic energy, $s$ is replaced by $\frac{1}{2}at^2$, which is only possible when the initial velocity ($u$) is zero ($s = ut+\frac{1}{2}at^2 = 0t+\frac{1}{2}at^2 = \frac{1}{2}at^2$). Why the initial velocity is assumed to be zero here? What if the object has a non-zero initial velocity?
| Formula $W=Fs$ is the formula for work, that is the energy transferred to the body by the force. It is the difference of kinetic energies between the final and initial state. If you're starting from $v=0$ and assume $E_{kin}(0)=0$, you have
$$
E_{kin}(v)
= E_{kin}(0) + ma \cdot \frac12at^2
= 0 + \frac12 mv^2
$$
If you start from the velocity $u$ and change it to $v=u+\Delta u=u+at$, you have
\begin{align}
E_{kin}(v) &= E_{kin}(u+\Delta u) = \\
&= E_{kin}(u) + ma \cdot (ut+ \frac12 at^2) \\
&= \frac12mu^2 + mu \cdot at + \frac12 m (at)^2 \\
&= \frac12mu^2 + mu\Delta u + \frac12 m (\Delta u)^2 \\
&= \frac12 m(u+\Delta u)^2 = \\
&= \frac12 mv^2
\end{align}
so the formula still holds.
| {
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"url": "https://physics.stackexchange.com/questions/613617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Superposition of eigenstates in statistical mechanics Consider the simplest case in quantum statistical mechanics, where we find the density of states in the case of a cuboidal 3 dimensional box. In the derivation we take only those states which are product seperable into wavefunctions along the three directions i.e. can be denoted by three quantum numbers $(n_1, n_2, n_3)$ henceforth written as $|n_1,n_2,n_3\rangle$ . However I feel that even states which are not product seperable should be considered. For example a particle in the system could be in the state $\frac{|1,0,0\rangle+|0,1,0\rangle}{\sqrt{2}}$. This will alter the counting of number of states. Why are such states excluded?
| The point is not that there's a priori something special about separable states compared to non-separable states, the point is that you'd like to sum over a complete set of states -- that is, a basis for the Hilbert space. The separable states you describe happen to be particularly convenient states, because they are energy eigenstates. This is especially important when you want to derive something like the density of states. But if you wanted to calculate the partition function, for instance, you're technically free to choose any basis you'd like (although if you know the energy eigenstates, then those will usually be the most convenient).
One frequently confusing point when comparing quantum mechanics to classical mechanics is that every quantum system, no matter how small, has an infinite number of possible states. An easy example is a qubit: whereas a classical bit only has the states 0 and 1, a qubit has any linear superposition $\alpha | 0 \rangle + \beta | 1 \rangle$, of which there is a continuous infinity. But when you're calculating (say) a partition function of a Hamiltonian for a single qubit, you only need to sum over a complete basis, which will contain just two states.
| {
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Is a wave function a ket? I just started with Dirac notation, and I am a bit clueless to say the least. I can see Schrödinger's equation is given in terms of kets. Would I be correct to assume if I were given a wavefunction, say $\Psi(x)=A\exp(-ikx)$, would I be able to just use the notation $\lvert \Psi\rangle =A\exp(-ikx)$?
| You are indeed confusing eigenket and eigenfunction. The eigenkets are the possible forms your system takes after a measurement, they are the eigen vectors (eigenkets) of the measurement operators (ex $\hat{H}$) you choose to apply on your eigenfunction (your system). The eigenfunction of a system can be expressed as a superposition of eigenkets (integral ->continuous or series ->discret).
They are both vectors in the Hilbert space you are working in thought
Don't hesitate if you need clarification
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Definition of Ensemble Im studying statistical mechanics and came across the ensembles.
*
*Now system of large number of particles can be defined by an ensemble which contains elements (infinite of them) where each element is the mental copy of system at a particular time and time average of any quantity of system can be assumed as same as ensemble average (avg over these mental copies)
*In another book I saw that an ensemble can also be defined as a collection of a very large number of assemblies which are essentially independent of one another but which have been made macroscopically as identical as possible.
Now my doubt is that the element in the 1 case is same as assembly in case 2 or different?
How they differ from one another?
| Ensemble is many systems, evolving from different initial conditions. Ergodicity, an important assumption in statistical physics reasoning, is that time averaging (for a single system) is equivalent to the ensemble averaging (over many systems.)
Note that the same definition of ensemble is used in quantum mechanics, where it cannot (generally) be replaced by averaging over a single system (since a measurement collapses wave function.)
Ensemble (mathematical physics)
In physics, specifically statistical mechanics, an ensemble (also statistical ensemble) is an idealization consisting of a large number of virtual copies (sometimes infinitely many) of a system, considered all at once, each of which represents a possible state that the real system might be in. In other words, a statistical ensemble is a set of systems of particles used in statistical mechanics to describe a single system.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Infinite acceleration without loss of energy in a vacuum, gravity-less void? Suppose we have a void that is free of any other objects or forces, even gravity (impossible I know but hypothetically). If one were to propel an object forward, would that object be able to accelerate infinitely without requiring further addition of energy?
My line of thinking may be totally off, but an object that had energy applied to it loses the energy after work is done right? And when it comes to movement that initial acceleration energy dissipates when it encounters resistance through friction or gravity in a vacuum.
But if one were to remove all external forces aside from the initial burst of energy, that energy would have nowhere else to go, other than to propel the object forward faster and faster?
It’s been awhile since I’ve done calculus and physics so forgive me if I’m missing something ha.
| A fixed amount of energy will not lead to an object moving faster and faster. A fixed amount of energy would give the object a fixed velocity and that's that.
A fixed force that kept being applied would make the object move faster and faster. But the longer the time interval, the greater the total energy needed to apply that force over that interval. You don't get anything for free.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is the magnetic field $B$ a pseudo-vector? Physically speaking, "pseudo-vectors" are vectors $v\in \mathbb{R}^3$ which transform as $ v'= (\det {R})v$ if the "system were to transform as $R\in O(3)$". However, what does this mean mathematically? And in particular, why is the magnetic field $B$ a pseudo-vector?
I would imagine that by "vectors", we actually mean smooth differential forms with the isomorphism $v\mapsto \sum v_i dx_i$, and by "transforming the system as $R\in O(3)$", I would imagine that it means we are applying a pullback on 1-forms corresponding to the map $x\mapsto Rx$. Assuming that $B =*dA$ where $*$ is the hodge-star operator, how would $\det R$ be factored into this transform?
| I don’t know the math jargon, but the magnetic field is a pseudo vector because it is determined for a given configuration of moving charges by the right hand rule. This means that if you look at a mirror reflection of a physical setup with the B vector drawn in, it will be reversed. Angular momentum is another example of this. This property of reversing direction relative to the rest of the scenery upon mirror reflection is how I recognize pseudo vectors.
| {
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"timestamp": "2023-03-29T00:00:00",
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D'Alembertian of a delta-function of a space-time interval (i.e. on the light-cone) How one differentiates a delta-function of a space-time interval? Namely,
$$[\partial_t^2 - \partial_x^2 - \partial_y^2 - \partial_z^2] \, \delta(t^2-x^2-y^2-z^2) \, .$$
Somewhere I saw that the result was $$4\pi\delta^{4}(x).$$ However, have no idea on how to obtain it.
| Using the theory of generalized functions, let us regularize the Dirac delta distribution$^1$
$$ \delta(x^2)~=~\lim_{\varepsilon\searrow 0^+}\delta_{\varepsilon}(x^2), \qquad x^2~:=~x^{\mu}\eta_{\mu\nu}x^{\nu}, \tag{1}$$
via a smooth $C^{\infty}$-function$^2$
$$ \delta_{\varepsilon}(x^2)
~=~ \frac{1}{\pi}\frac{\varepsilon}{(x^2)^2+\varepsilon^2}
~=~-\frac{1}{\pi}{\rm Im} \frac{1}{x^2+i\varepsilon}, \qquad \varepsilon~>~0. \tag{2}$$
Eqs. (1)-(2) are a well-known representation of the Dirac delta distribution.
It is now mathematically well-defined to apply the d'Alembert operator on the smooth function (2). We calculate:
$$ \partial_{\mu}\delta_{\varepsilon}(x^2)~=~\frac{1}{\pi}{\rm Im} \frac{2x_{\mu}}{(x^2+i\varepsilon)^2},\tag{3}$$
$$ \partial_{\mu}\partial_{\nu}\delta_{\varepsilon}(x^2)
~=~\frac{1}{\pi}{\rm Im} \left(\frac{2\eta_{\mu\nu}}{(x^2+i\varepsilon)^2}-\frac{8x_{\mu}x_{\nu}}{(x^2+i\varepsilon)^3}\right),\tag{4}$$
$$\begin{align} \Box \delta_{\varepsilon}(x^2)~=~&\frac{1}{\pi}{\rm Im} \left(\frac{8}{(x^2+i\varepsilon)^2}-\frac{8x^2}{(x^2+i\varepsilon)^3}\right) \cr
~=~&\frac{1}{\pi}{\rm Im}\frac{8i\varepsilon}{(x^2+i\varepsilon)^3}\cr
~\longrightarrow~& -4\pi \delta^4(x)
\qquad \mathrm{for} \qquad \varepsilon ~\searrow~ 0^+.
\end{align}\tag{5}$$
Eq. (5) implies OP's sought-for proposition.
Proposition.
$$ \Box \delta (x^2)~=~-4\pi \delta^4(x).\tag{6}$$
Sketched proof of the last line in eq. (5). Consider a test function $f\in C^{\infty}_c(\mathbb{R}^4)$, i.e., an infinitely often differentiable function $f$ with compact support. Then
$$\begin{align}
\int_{\mathbb{R}^4}\! \mathrm{d}^4x & ~ f(x)~ \Box \delta_{\varepsilon}(x^2) \cr
~\stackrel{x=\sqrt{\varepsilon} y}{=}~&
\int_{\mathbb{R}^4}\!\frac{\mathrm{d}^4y}{\pi} ~ f(\sqrt{\varepsilon} y)~{\rm Im}\frac{8i}{(y^2+i)^3}\cr
~\longrightarrow~& f(0) \int_{\mathbb{R}^4}\!\frac{\mathrm{d}^4y}{\pi} ~ {\rm Im}\frac{8i}{(y^2+i)^3}\cr
~\stackrel{y=(t,\vec{r})}{=}~&~ 4 f(0) {\rm Im} \int_{\mathbb{R}_+}\!r^2\mathrm{d}r \int_{\mathbb{R}}\!\mathrm{d}t \frac{8i}{(r^2-t^2+i)^3}\cr
~\stackrel{a:=\sqrt{r^2+i}}{=}&~ 4 f(0) {\rm Im} \int_{\mathbb{R}_+}\!r^2\mathrm{d}r \int_{\mathbb{R}}\!\mathrm{d}t \left(\frac{2i}{t^2-a^2}\right)^3\cr
~=~~~& 4 f(0) {\rm Im} \int_{\mathbb{R}_+}\!r^2\mathrm{d}r \frac{-i}{a^3}\int_{\mathbb{R}}\!\mathrm{d}t \left(\frac{1}{t-a}-\frac{1}{t+a}\right)^3\cr
~\stackrel{\text{residue thm.}}{=}&~ 4 f(0) {\rm Im} \int_{\mathbb{R}_+}\!r^2\mathrm{d}r \frac{-i}{a^3} 2\pi i \cr
&\qquad \left(0+\left.\frac{3}{(t+a)^2}\right|_{t=a}+\left.\frac{3}{(t-a)^2}\right|_{t=-a}+0\right)\cr
~=~~~& 12\pi f(0) {\rm Im} \int_{\mathbb{R}_+}\!r^2\mathrm{d}r \frac{1}{a^5} \cr
~=~~~& 12\pi f(0) {\rm Im} \int_{\mathbb{R}_+}\!\frac{r^2\mathrm{d}r}{(r^2+i)^{5/2}} \cr
~=~~~& 12\pi f(0) {\rm Im} \left[\frac{r^3}{3i(r^2+i)^{3/2}}\right]^{r=\infty}_{r=0} \cr
~=~~~& -4\pi f(0) \qquad \mathrm{for} \qquad \varepsilon ~\searrow~ 0^+,
\end{align}\tag{7}$$
because of, e.g., Lebesgue's dominated convergence theorem. Here we have implicitly assumed that $a:=\sqrt{r^2+i}$ is the square root branch in the upper complex plane. $\Box$
--
$^1$ In this answer we work in units where the speed-of-light $c=1$ is one, and we use the Minkowski sign convention $(−,+,+,+)$.
$^2$ The $i\varepsilon$-prescription has an interpretation in terms of Wick rotation, cf. e.g. my Math.SE answer here.
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