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Memory fading “striyers” in viscoelastics I was talking to a guy who does polymer moulding, and we were discussing a few industrial issues in melting, mixing and forming shapes, and with a few of my suggestions he rebutted with, "no that would create striyers". So for example, one mixing technique I suggested had the fluid melt pass through a grate at the end, and he didn't like that idea because it would create striyers.
My understanding at the time, from the context, was that he was referring to the viscoelastic nature of the fluid, that has a (fading) memory component. So in my example the product might still have lines far downstream left over from passing through the grate. But I've not heard this word "striyer", and wondered if anybody knows what it is?!
We were speaking in English, but if it helps, his mother tongue is German.
| What he is getting at is that, after the grate, the polymer molecules that passed on either side of the grate take time to "knit back together" again (i.e., form entanglements). So the structure of the polymer is temporarily weakened.
| {
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"timestamp": "2023-03-29T00:00:00",
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What new features does the Heisenberg Model have compared to the Ising Model? Both the Ising and the Heisenberg Models describe spin lattices with interaction on first neighbors. The Hamiltonian in each case is quite similar, despite the fact of treating de spins as Ising variables (1 or -1) or as quantum operators. In the Ising case it looks like
$$H_\textrm{Ising} = -~J \sum_{\langle i\ j\rangle} s^z_{i}\ s^z_{j}$$
where J is the coupling constant ($J>0$ for ferromagnet and $J<0$ for anti-ferromagnet), $\langle i\ j\rangle$ represents sum over first neighbors and $s^z$ is the spin in z direction. On the other hand, the Heisenberg model is
$$H_\textrm{Heisenberg} = -~J \sum_{\langle i\ j\rangle} \hat{S}_{i} \cdot \hat{S}_{j}$$
where the only difference lies in the spins being operators. (In both cases I took away the interaction with an external field for simplicity)
My Question is: What new phenomena does treating the spins as operators brings? I can see that $\hat{S}_{i}\ .\ \hat{S}_{j}$ takes account of the spin in every direction and not just z, but I can't see the physical implication of that.
| One of the main differences is that the Ising model lies on a discrete symmetry (the $Z_2$ symmetry) while the Heisenberg model lies on a continuous one (rotational symmetry). It will affect the phase transitions that these models undergo.
In particular, because of the Mermin-Wagner theorem, there can be no finite-temperature phase transition of the Heisenberg model in $d=2$ (leaving apart the very special case of the BKT transition).
This is not the case of the Ising model that undergoes a phase transition at finite temperature from a high-$T$ disordered state to a low-$T$ ordered state in $d=2$ (the exact solution has even been calculated by Onsager and later by others).
There's probably much more to it than this particular case, feel free to edit my answer if you feel like adding something.
| {
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Is there a way to estimate fan capacity at altitude? Provided you know the capacity of a fan (flow rate) at constant speed and at sea level, is there an analytical way to predict what the flow rate would be at altitude? Or is this specific to the fan's design?
| For a fixed speed, a fan, blower or any turbo-machine in general will deliver the same volumetric flow regardless of the ambient pressure since the machine essentially scoops out a volume of air as each blade of the machine passes the machine's inlet.
$$Q_{SL}=Q_{alt}$$
where $SL$ designates 'Sea Level' as reference and $alt$ as some higher altitude
But at higher altitudes there are fewer molecules per unit volume (lower gas density) and so the mass flow rate is lower with increasing altitude and barometric pressure.
$$\dot{m}_{alt} < \dot{m}_{SL}$$
So since the volumetric flow rates are the same then
$${\dot{m}_{alt}\over{\rho}_{alt}} = {\dot{m}_{SL}\over{\rho}_{SL}}$$
and
$$\dot{m}_{alt}={{\rho}_{alt}\over {\rho}_{SL}}\dot{m}_{SL}$$
But if we were to measure these mass flows as volumetric flows relative to sea level then
$${{\dot{m}_{alt}}\over {{\rho}_{SL}}}={{{\rho}_{alt}}\over {{\rho}_{SL}}}{{\dot{m}_{SL}}\over {{\rho}_{SL}}}$$
which becomes
$$Q_{Malt}={{{\rho}_{alt}}\over {{\rho}_{SL}}}Q_{MSL}$$
And the $Q_M$'s are the measured volumetric flows at altitude and sea level respectively.
This result shows the the measured volumetric flow is reduced as altitude increases by the ratio of air density as it decreases. And this is consistent with zero volumetric flow as one moves out of the atmosphere and no more scooping is possible.
| {
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Are electric fields produced by static electric charges different from those produced by time-varying magnetic fields? I came across an interesting yet perplexing statement in my physics textbook:
However, we hasten to add that electric fields produced by static electric charges have properties different from those produced by time-varying magnetic fields.
My questions are, simply put:
What exactly are these properties? And why do they differ?
| The only difference is in the structure of the fields:
*
*The field contribution from static charges are non-rotational (curl-free) but divergent.
*The field contributions from time-varying magnetic fields are divergence-less but have non-zero curl.
You can read this right out of the Gauss and Faraday's Laws in their differential form:
\begin{align*}
\mathbf{\nabla \cdot E} &= \frac{\rho}{\epsilon_0} \\
\\
\mathbf{\nabla \times E} &= - \frac{\partial \mathbf{B}}{\partial t} \;.
\end{align*}
The operators on the LHSs are called "divergence" and "curl" (read top to bottom).
I hasten to add that these are exactly the same kind of stuff, which is why I've written "contributions from" rather than "fields produced by" as if the fields might be of different kinds. In the presence of both static charges and varying magnetic field the total electric field will show both divergence and curl.
| {
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Lagrangian and finding equations of motion I am given the following lagrangian:
$L=-\frac{1}{2}\phi\Box\phi\color{red}{ +} \frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4$
and the questions asks:
*
*How many constants c can you find for which $\phi(x)=c$ is a solution to the equations of motion? Which solution has the lowest energy (ground state)?
*My attempt:
since lagrangian is second order we have the following for the equations of motion:
$$\frac{\partial L}{\partial \phi}-\frac{\partial}{\partial x_\mu}\frac{\partial L}{\partial(\partial^\mu \phi)}+\frac{\partial^2}{\partial x_\mu \partial x_\nu}\frac{\partial^2 L}{\partial(\partial^\mu \phi)\partial(\partial^\nu \phi)}=0 $$
then the second term is zero since lagrangian is independent of the fist order derivative. so we will end up with:
$$\frac{\partial L}{\partial \phi}=-\frac{1}{2} \Box \phi+m^2\phi-\frac{\lambda}{3!}\phi^3$$
and:$$\frac{\partial^2}{\partial x_\mu \partial x_\nu}\frac{\partial^2 L}{\partial(\partial^\mu \phi)\partial(\partial^\nu \phi)}=-\frac{1}{2}\Box\phi$$
so altogether we have for the equations of motion:
$$-\frac{1}{2}\Box\phi+m^2\phi-\frac{\lambda}{6}\phi^3-\frac{1}{2}\Box\phi=0$$
and if $\phi=c$ where "c" is a constant then $\Box\phi=0$ and then the equation reduces to $$m^2\phi-\frac{\lambda}{6}\phi^3=0$$ which for $\phi=c$ gives us 3 solutions:$$c=-m\sqrt{\frac{6}{\lambda}}\\c=0\\c=m\sqrt{\frac{6}{\lambda}}$$
My question is is my method and calculations right and how do I see which one has the lowest energy (ground state)? so I find the Hamiltonin for that?
| Looks good so far. To find the Hamiltonian you just use that if $L = T - U$ then $H = T + U$ (technically there are some extra assumptions there, but if your case it works out fine). Since $T = 0$ if $\phi$ is constant, you just need to find out which of those values $c$ minimize(s) the potential energy $-1/2 m^2 \phi^2 + \lambda/4! \phi^4$.
| {
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Neutral points in a system of charges on the vertices of a square There are four positive charges of equal magnitude placed at the four vertices of a square. Is there any point where the electric field vanishes (neutral point) within the square and in its plane, other than its center?
| I believe there is such a point.
Let us assume that unit charges are located in points $[\pm 1,\pm 1]$ of a plane with coordinates $x,y$. Let us consider the field on the abscissa ($y=0$). The $y$-component of the field is zero on the abscissa due to a symmetry. It is obvious that the $x$-component of the field at the "left infinity" ($x\rightarrow-\infty$) is directed to the left ($E_x<0$)) (provided the charges are positive and repel a positive charge). According to my calculation, the field at point,say, $[-\frac{1}{2},0]$ is directed to the right. Therefore, there is an intermediate point on the abscissa between $-\infty$ and $-\frac{1}{2}$ (therefore, not in the center) where the field vanishes. Symmetry will give three more such points.
By the way, it looks like the expression for the field given by @Haru Fujimura is not correct.
EDIT (9/18/2016): So let us calculate the $x$-component of the field at point $[-\frac{1}{2},0]$, assuming each charge equals +1. The distance from this point to the left bottom charge is $\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2}$, the relevant cosine is $\frac{1}{\sqrt{5}}$, so the contribution of this charge to the $x$-component of the electric field is $\frac{1}{\sqrt{5}}\frac{4}{5}\approx 0.36$.
The distance from point $[-\frac{1}{2},0]$ to the right bottom charge is $\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$, the relevant cosine is $\frac{3}{\sqrt{13}}$, so the contribution of this charge to the $x$-component of the electric field is -$\frac{3}{\sqrt{13}}\frac{4}{13}\approx -0.26$ (the minus sign arises because the contributions of the left and right charges have opposite signs). The calculation of the contributions of the top charges yields the same results. Thus, the field at this point is directed to the right.
EDIT [9/19/2016]: It is clear from the edited question that the point is required to be not just in the plane of the square, but also within the square. To adjust the proof, one can just consider point [-1,0] instead of [-infinity,0]: it is obvious that the field is directed to the left in this point. Therefore, there is an intermediate point between [-1,0] and [-1/2,0], i.e., within the square, where the field vanishes.
| {
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Space bends relative to what? We all are aware of gravitational waves, as it bends space and time, black hole squeeze space, but the squeezing, bending, expanding happens reference to what? Since the observable universe is the universe existing within itself, so it bends in reference to whose perspective?
| The easiest way to understand this is to understand the notion of extrinsic versus intrinsic curvature.
Extrinsically curved things follow non-straight lines relative to the space they are contained in.
Intrinsically curved things, however, obey non-Euclidean laws on their own surfaces.
Now, if you fit an intrinsically curved space into an enveloping Euclidean space, it will also be extrinsically curved, but the opposite is not true -- you can turn a sheet of paper into a cylinder and back and not distort it, but you cannot do this with an orange peel.
Now, let's get back to general relativity -- it explains gravity as curvature of spacetime, but this curvature is an intrinsic curvature -- it makes sense in spacetime, without having to refer to some larger space it is contained in. You can tell you're in a curved spacetime just by observing that Euclid's parallel postulate is violated. Most dramatically, this is true, because if you start out with two objects high above earth, initially at rest (and therefore, initially travelling in parallel paths through spacetime), their paths will intersect at the center of the Earth (at least, if the earth somehow didn't stop them).
| {
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How to find pseudo-force on a body in a non-inertial reference frame with respect to another non-inertial reference frame? If there is a body in a non-inertial reference frame (say accelerating with acceleration $A_1$) and an observer in another non-inertial reference frame (say accelerating with acceleration $A_2$) what would be the pseudo-force acting on the body (say it has mass $m$) as seen by the observer? Both $A_1$ and $A_2$ are in the same direction and along a straight line.
Do I simply need to find the relative acceleration of the first frame with respect to the second frame and multiply the relative acceleration with the mass of the body?
| The pseudo force depends only on the acceleration of the observer frame (which is why if the acceleration of the observer is zero (in the inertial reference frame case) there is no pseudo force.)
Therefore, given that the acceleration of the observer is $A_2$ the pseudo force acting on any body (which has mass $m$) observed will be $-m \cdot A_2$.
No, we do not take the relative acceleration of the observer and the body observed but the acceleration of the observer as measured by any inertial frame.
| {
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Force between proton in a conducting shell and electron outside of shell? There's a proton inside a conducting shell and an electron outside of it. Inside the shell, there is no field due to the electron, but the electron feels the field due to the proton. Therefore the electron should move towards the immobile proton, but what happened to Newton's third law? Does the whole shell move?
| The error that you have made is to apply Newton's third law incorrectly.
On the conducting sphere there are two sets of induced charge.
One set of positive and negative induced charges is due the field of the electron having to be negated so that there is no electric field inside the conducting sphere.
These positive and negative induced charges reside on the outside of the conducting sphere.
So you are correct in your assertion that the proton does not feel the effect of the electron directly.
Put another way - the proton does not know that there is an electron outside the conducting shell.
The other set of induced charges are produced by the proton and they are present to ensure that the electric field inside the conductor is zero.
These charges reside on both the inside (negative) and the outside (positive) of the conducting sphere.
So on the outside of the conducting sphere you have induced charges produced by the presence of the electron and the proton.
There is a force on the electron due to the induced charges on the outside of the conducting sphere
and the Newton's third law pair to this force is that
there is force on the induced charges on the outside of the conducting sphere due to the electron.
So indirectly, via the induced charges on the outside of the conducting sphere produced by the proton, the electron feels the presence of the proton.
Inside the sphere
there is a force on the proton due to the induced negative charges that it has induced on the inside of the sphere
and the Newton's third law pair to this force is that
there is a force on the induced negative charges on the inside of the sphere due to the presence of the proton.
| {
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Do we need a bounded domain for the Laplace equation to have a non-zero solution $u$? If we have the Laplace equation for electrostatics in free space, that is
$$\Delta u(x) = 0 \quad \quad x \in \mathbb{R}^3,$$
is the only solution $u = 0$? And also, we only get non-zero solutions for $u$ if we instead consider the Laplace equation on some bounded domain? How can I mathematically show this if this is in fact the case?
| I'm not sure that I understand what you mean by "bounded domain". This is an answer based on what I understand from the question.
To solve Laplace's equation uniquely, you have to specify either the Dirichlet or Neumann type boundary condition.
EDIT: The boundary of the region of interest can be at finite distances as well as at infinity. For example, to solve the potential due to a point change in free space, you have to solve Poisson's equation. Here, we take the potential going to zero at infinity. So if you have some finite boundary in mind that is not necessary. As you know the solution in this case is the familiar $u(r)\sim \frac{1}{r}$ Coulomb potential (which is not identically zero everywhere).
Now for Laplace's equation in absolutely free space (no charge anywhere), if the boundary condition is such that the potential vanishes everywhere on the boundary, then the potential will remain zero everywhere simply because Laplace's equation doesn't support local maxima or minima.
| {
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Time dilation for a clock in orbit Suppose that we want to compute the total time dilation for a clock located in an orbiting satellite relative to the clock in our cell phone on the ground.
Consider two different approaches below.
*
*Use special relativity and compute time contraction due to the relative velocity. Use approximation of General Relativity in the Newtonian limit and compute time dilation due to lower gravity and then find the total time dilation.
*Don't use Special Relativity. Stick to the approximation of General Relativity based on the symmetry and find Schwarzschild metric and the geodesic for the Earth limits. Find the time dilation assuming a relative velocity in the metric.
The question is:
Which of them are more justified and provide a better approximation? Are they equivalent? What happens when the relative velocity of the satellite is zero?
How good is the approximation in either of the two approaches above.
When we pick the second approach and use the Schwarzschild metric we get this equation:
$$ dt' = \sqrt{1-\frac{3GM}{c^2r}}dt = \sqrt{1-\frac{3r_s}{2r}}dt $$
where $r_s$ is the Schwarzschild radius: $r_s = 2GM/c^2$.
Here we not only assume the asymptotic flat metric to measure $r$ but also switch to Newtonian gravity when we want to cancel $v$:
$$ v = \sqrt{\frac{GM}{r}} $$
So it appears that in the second approach there are many more approximation assumptions.
| They are different effects. Special relativity (SR) will do the first part, just calculate from Newton the velocity and you get a slowdown at the clocks in the satellite. With GR you get a faster clock because of the gravitational time dilation. Both apply, subtract one from the other
In GPS orbits the GR effect is about 45 usec per day, and the SR effect is 7 usec per day. The net effect is 45-7= 38 usec per day. The GPS clocks go that much faster than earthbound clocks. If you were measuring frequencies at the GPS satellites you'd measure a redshift at the GPS satellites (faster clock, lower freq)
See https://en.wikipedia.org/wiki/Error_analysis_for_the_Global_Positioning_System
The clock reported times are adjusted for that.
If the relative velocity of the satellite was zero there would be no SR effect, it'd be 45 usec faster. And it would fall.
| {
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Special Relativity - Non-conservation of Newtonian momentum I'm studying special relativity and I don't quite understand why Newtonian momentum isn't conserved, because everytime I think of the math involved or imagine myself observing some system from various reference frames, it never makes sense to me...
I'll be very grateful to anyone who can explain this non-conservation of Newtonian momentum to me.
| Newtonian linear momentum is not conserved because your idea of how to calculate how different reference frames see the same thing is flawed. Usually when people get stuck on this point it's because they have trouble grasping the concept of how the passage of time is reference frame dependent in special relativity (SR). The Minutephysics channel did a great pair of videos with visualizations of how relative time works when applying the Lorentz transformations between reference frames:
$$\begin{align} t' & = \frac{t - \frac{vx}{c^2}}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \\
x' & = \frac{x - vt}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \\
y' &= y \\
z' & = z,\end{align}$$ and how to apply them to resolve the "twins paradox".
If I'm right about what's confusing you, you're stuck on what is known as Galilean addition of velocities:$$\mathbf{v}' = \mathbf{v} - \mathbf{u}_o,$$ where if an object is moving with velocity $\mathbf{v}$ and an observer is moving with velocity $\mathbf{u}_o$ then that observer will measure the objects velocity to be $\mathbf{v}'$. The derivation of that formula assumed an absolute rate of time's passage. The relativistic addition of velocities formula is somewhat more complicated, and can be derived from the Lorentz transformations (note: Wikipedia uses a different sign convention). If we break the vectors down into components parallel to (eg $\mathbf{v}_{||}$) and perbendicular to ($\mathbf{v}_\perp$) $\mathbf{u}_o$ then we get the following transformations: $$\begin{align}\mathbf{v}'_{||} & = \frac{\mathbf{v}_{||} - \mathbf{u}_o}{1 - \frac{\mathbf{v}\cdot \mathbf{u}_o}{c^2}} \\
\mathbf{v}'_\perp &= \left(\frac{\sqrt{1 - \frac{u_o^2}{c^2}}}{1 - \frac{\mathbf{v}\cdot \mathbf{u}_o}{c^2}}\right)\mathbf{v}_\perp.\end{align}$$
If you apply the rules correctly, you'll find that what is conserved is called the 4-momentum. It has four components, 0 through 3, that correspond to timelike (component zero) and space-like (components 1 through 3, often denoted as the ordinary vector part) parts. It is: $$\begin{align}p_0 &= \frac{m c}{\sqrt{1 - \frac{v^2}{c^2}}} \\
\mathbf{p} & = \frac{m \mathbf{v}}{\sqrt{1 - \frac{v^2}{c^2}}}.
\end{align}$$ You should recognize the space-like part of the 4-momentum as the Newtonian momentum times $\gamma = 1 \left/ \sqrt{1 - \frac{v^2}{c^2}}\right.$, and the time like part as the relativistic energy, $E = \gamma mc^2$, divided by $c$.
| {
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Is measurement of coordinates possible near strong gravity? We know that Schwarzchild metric describes an asymptotically flat spacetime. This means that far away from the event horizon we can safely interpret the $r$ coordinate as distance from the center.
But when close enough to the event horizon the curvature becomes significant and our common sense of $r$ breaks.
The question is that what is understood as measurement of the coordinates near very strong gravity?
| You seem to have some serious conceptual misunderstandings about the subject. Co-ordinates are not "measured" they are defined on a space time manifold. Proper time on the other hand can be measured. So please think about the concepts correctly. I will recommend reading Einsteins works on the subject.
| {
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Equations of motion for a free particle on a sphere I derived the equations of motion for a particle constrained on the surface of a sphere Parametrizing the trajectory as a function of time through the usual $\theta$ and $\phi$ angles, these equations read:
$$ \ddot{\theta} = \dot{\phi}^2 \sin \theta \cos \theta $$
$$ \ddot{\phi} = - 2 \dot{\phi} \dot{\theta} \frac{1}{\tan \theta} $$
I've obtained them starting from the Lagrangian of the system and using the Euler-Lagrange equations.
My question is simple: is there a way (a clever substitution, maybe), to go on and solve the differential equations? I would be interested even in a simpler, partially integrated solution. Or is a numerical solution the only way?
| Note that you can rewrite your second equation as $$ \frac{\ddot{\phi}}{\dot{\phi}} = -2\cot{(\theta)}\dot{\theta} $$ Each side is an exact differential in one variable, so we can integrate, and Wolfram|Alpha gives $$ \ln{(\dot{\phi})}=-2\ln{(\sin{(\theta)})}+C $$ for some integration constant $C$. We can exponentiate to get $$ \dot{\phi}=\frac{B}{\sin{(\theta)}^2} $$
Substituting this into the first equation yields $$ \ddot{\theta}=B^2\frac{\cos{(\theta)}}{\sin{(\theta)}^3} $$ This, too, can be integrated via the "energy trick": multiply by $ \theta $, then integrate. The LHS integrates by parts to $\dot{\theta}^2$ but the RHS looks sufficiently complicated I don't want to type it out on my phone.
| {
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Concentric Circular Loops We have two concentric circular wire loops. The inner loop has a stable, clockwise current. The outer loop does not have current. If the current of the inner loop increases, what is the direction of the current of the outer loop?
My first thought was "counter-clockwise" using the RHR, increasing flux, and so on. But after thinking a little more, I was wondering if the answer is "cannot be determined". I imagined the inner loop being very very small compared to the outer loop. In that case, the space between the outer and the inner loop would see an increase in B-field out of the page, and since that part has a much bigger area than the inner loop, it would have a bigger flux. The resulting current would be "clockwise" to counter that change. Is my logic in this case correct?
| According to Lenz's Law, the current induced in the outer loop by an increase of the clockwise current in the inner loop has to flow counterclockwise. https://en.wikipedia.org/wiki/Lenz%27s_law.
| {
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Existence of a Hamiltonian system It is well known that a two dimensional system to first order is locally Hamiltonian from Darboux' theorem. For example,
\begin{equation}
\dot x = f(x,y), \qquad \dot y = g (x,y)
\end{equation}
Admits the following Poisson structure,
\begin{equation}
\{x,x\}=\{y,y\} =0 , \qquad \{x, y\}= -\{y,x\} = F(x,y)
\end{equation}
Where $F\neq 0$ and Hamilton's equations being,
\begin{equation}
\dot x = F(x,y) \frac {\partial H}{\partial y}, \qquad \dot y =-F(x,y) \frac{\partial H}{\partial x}
\end{equation}
If now we have an $n$-dimensional system $\dot x_i=f(x_1,\dots , x_n)$ where $i=1,\dots , n$, can we in general give conditions for the admission of a Hamiltonian system?
If I had a system and wanted to solve it's dynamics, is there a way I could test to see if it is Hamiltonian? By this I mean, let us assume I have a collection of variables and can monitor their time evolution in a computational experiment. Is there a way I can use the very powerful theory of Hamiltonian mechanics to someway solve my own system? i.e how can I take it beyond the use of $q$s and $p$s to solve my own problems!
| There is the well-known condition for a system
$$\dot{x}_i = f_i(x_1,\ldots, x_{2n})\:,\quad i =1,\ldots, 2n \tag{1}$$
to admit a local Hamiltonian re-formulation:
$$\dot{x}_i = \frac{\partial H}{\partial x_i}(x_1,\ldots, x_{2n})\:,\quad \dot{x}_k = -\frac{\partial H}{\partial x_k}(x_1,\ldots, x_{2n}) \quad i =1,\ldots, n\quad k=n+1,\ldots, 2n\tag{2}$$
Assuming the $f_i$ of class $C^1$ on the open set $\Omega \subset \mathbb R^{2n}$, define the new functions
$$ F_j := \sum_{k=1}^{2n}S_{jk}f_k\quad j=1,\ldots, 2n $$
where
$$S=\left[\begin{matrix}0 & -I\\ I & 0\end{matrix}\right]$$
and $I$ and $0$ are viewed as $n \times n$ submatrices (the identity and the zero matrix respectively).
Then (1) admits a local Hamiltonian reformulation of the form (2) if and only if, everywhere on $\Omega$,
$$\frac{\partial F_i}{\partial x_k} = \frac{\partial F_k}{\partial x_i}\quad i,k = 1,\ldots , 2n\:.$$
Obviously this is a very particular case where we also suppose that the coordinates $x_1,\ldots, x_{2n}$ are canonical. It is however possible that an Hamiltonian re-formulation of the initial system arises after having also changed the initial coordinates.
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Possible Error in Assumption - Griffiths Quantum Mechanics In "Introduction to Quantum Mechanics" by Griffiths, right at the beginning of section 9.1.1 (Time-Dependent Perturbation Theory, The Perturbed System), Griffiths states:
Now suppose we turn on a time-dependent perturbation, $H'(t)$. Since $\psi_a$ and $\psi_b$ constitute a complete set [of the two-level system], the wave function $\Psi (t)$ can still be expressed as a linear combination of them. The only difference is that $c_a$ and $c_b$ are now functions of t:
I don't understand. You modify the Hamiltonian, you modify the solution basis - easy as that. Why on earth does he assume that if you add a time-dependent perturbation to the Hamiltonian the basis (for the two-level system that he considered in the section right before) will remain the same? And if this is indeed a mistake, then how valid is the assumption that the true wave function $\Psi (t)$ is merely a time-dependent linear combination of the two states $\psi_a$ and $\psi_b$?
| A basis is a set of wave functions such that a any wave function can be formed as a linear combination of basis wave functions. Often you choose them to be eigenfunctions of the Hamiltonian. But you don't have to.
If you change the Hamiltonian, you change the egienfunctions, so you change the most common choice for a basis.
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Dimension of representations of Lorentz Group In my Quantum Field Theory class we were discussing the finite dimensional representations of the Lorentz group. We discussed the vector representation which acts on 4 vectors, and then also the spinor representation that acts on spinors. My issue is that both of these representations are 4 dimensional.
I'm used to dealing with SO(3) where the representations are labeled by half integers and there is a unique mapping from each label to the dimension of the representation. Spin (1/2) gives 2 dimensional matrices, Spin (1) gives 3 dimensional matrices, and so on. But it seems we have lost that uniqueness in the Lorentz group. The vector representation and the spinor representation have the same dimension. Does this have to do with the non compactness of the lorentz group?
Thanks.
| No, the loss of uniqueness does not have to do with the non-compactness of the Lorentz group. The fact that there is only one irreducible representation of any given dimension is special to the group $SU(2)$ (and by extension, to $SO(3)$). It is not true even for other compact semisimple Lie groups. For example, $SU(3)$ has two distinct irreducible representations of dimension 3, which also happen to be conjugates to each other. It also has four representations of dimension 15, two pairs of conjugate representations.
You hopefully will soon learn that the finite-dimensional irreducible representations of the Lorentz group can be labelled by two numbers, which happen to correspond (in a roundabout sense) to representations of $SU(2)$.
In general, an irreducible representation for a semisimple Lie algebra can be labelled by the values of its independent Casimirs. $SU(2)$ has one independent Casimir, so you only need one number to specify a representation. $SU(3)$ and $SO(1,3)$ have two Casimirs, so you need two numbers.
And, as ACuriousMind pointed out, the spinor representation you're talking about isn't actually irreducible. It is the direct sum of two irreducible representations, $(\tfrac{1}{2},0)$ and $(0,\tfrac{1}{2})$ in one common notation. The vector representation is irreducible, and is labelled $(\tfrac{1}{2},\tfrac{1}{2})$.
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Why is it that the change in internal energy always uses the formula with Cv in regards to pressure/volume/temperature changes on a gas? Normally I would associate the use of $C_v$ with finding the energy taken into or leaving a system when the volume is kept constant. However, the formula to find $\triangle E_i$ (change in internal energy) is $nC_v \triangle T$. Why $C_v$? Also, does this apply to pretty much anything? Or are there limitations?
| If you keep the volume constant then the gas can do no work as $\delta W = P \Delta V = 0$ and so from the first law of thermodynamics the change in internal energy
$\Delta U = \delta Q - \delta W \Rightarrow \Delta U = \delta Q = n c_v \Delta T$
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One straw drinking from many containers of liquid One of my friends brought up a photo:
Which sparked a debate about whether the containers closest to the end of the straw would empty first. I was just wondering if someone could explain if the closest two containers would be empty before the furthest.
| Assuming the straw(s) are full of liquid and you are sucking very slowly (so we are just considering pressure, not fluid dynamics) the fluid levels in all the containers must be equal, otherwise the liquid would flow to equalise them.
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Consequence of approaching infinite mass for near-light speed particles Considering that the inertial mass of an object approaches infinity as the speed of the object approaches $c$, and that inertial mass equals gravitational mass, does this not imply that particles nearing $c$ would have gravitational mass approaching infinity?
Postulating that in fact gravitational mass nears infinity, would an object moving at near $c$, due to approaching infinite gravitational mass, cause profound gravitational acceleration on very distant objects? Infinity is a big value, even after decaying at $1/d^2$.
| Some time before the mass of the particle approaches infinity, surely its gravitational field must make it collapse into a micro-black hole. Or not, the whole thing sounds basically flawed the more you extrapolate its implications.
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Is Dirac's notation really necessary? One thing that's always bothered me in Dirac's notation is that it assumes that the Hilbert space contains a "continuum basis" of vectors $|x\rangle$, which happen to be eigenvectors of an operator $X$ (which has no eigenvalues, only a continuous spectrum that spans the whole space). Their inner product is distribution-valued, with $\langle x'|x\rangle = \delta(x'-x)$. There's also the cryptic normalization property: $\int |x\rangle\langle x| dx = Id$. According to this question, some of these can't be made rigorous even with something like the concept of "Rigged Hilbert Spaces".
So, is there another approach to quantum mechanics in general that sidesteps this issue entirely, without loss of descriptive power?
| The problem, in my view does not lie with the notation as such, but rather with the physical scenario that is being studied. Space-time is continuous infinite. Therefore, one requires a basis that is also continuous and infinite. So, regardless of how one would represent such a basis in terms of a notation, its orthogonality condition would necessarily have to include a Dirac delta function. In the end, one can take one's hat off to people such as Dirac who came up with some mathematical system that makes it possible to do calculations that can lead to predictions, which in turn can be tested in experiments. The mere fact that such predictions often agree with these experimental results, seems to indicate that this way to calculate these quantities using this mathematical formalism must be correct to some extent. It then becomes a challenge to the mathematitians to try and come up with an axiomatic system that can lead to this formalism in a consistent manner. This often implies that one would need to stretch the notions of integrals, vector spaces and such so that the formalism can work in a strict mathematical sense. Whether or not that happens to be the case usually does not stop physicists from using the formalism.
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Intensity of unpolarized light acted upon by a quarter wave plate Why does the intensity of unpolarized light remain unchanged when it pass through a quarter wave plate? A quarter wave plate produces a phase difference between e- ray and o-ray. But the intensity is changed for plane polarized light.
| An ideal quarter wave plate does not absorb any light.
The speed of light through the quarter wave depends on the orientation of the electric field vector relative to the plate. So one orientation is slowed down more than the other and the thickness of the plate is designed so that there is a change of phase of $90^\circ$ for one plane relative to another.
This represents a change of phase equivalent to a quarter of a wavelength.
An ordinary Polaroid actually absorbs light.
The light which emerges has the electric field oscillating in one plane the plane at right angles having been absorbed.
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Why wave theory cannot explain photoelectric effect and provides evidence for particle nature of light? I am able to understand how light can be modeled to have wave characteristics from Young's double slit experiment.
But I am unable to comprehend how we can understand light to have particle characteristics from the photoelectric experiment. How is it wave character not able to explain the phenomena observed in this experiment? And how is it that particle nature defeats the wave theory?
| Why don't you look a this video an A levels tutorial for the photoelectric effect.
Summary of the observations of the puzzling photoelectric effect:
*
*The electrons were emitted immediately - no time lag!
*Increasing the intensity of the light increased the number of photoelectrons, but not their maximum kinetic energy!
*Red light will not cause the ejection of electrons, no matter what the intensity!
*A weak violet light will eject only a few electrons, but their maximum kinetic energies are greater than those for intense light of longer wavelengths!
The basic point is that the effect disappears at a threshold frequency and is not dependent on the intensity of the light impinging on the metal.
A wave formalism cannot explain all of these, because the energy in waves is additive, the more intense the beam, the more energy it delivers, but the photoelectric experiments show that this is not true for "light wave" + "electrons in metal" scattering.
No matter how strong the incoming beam of light, if it is below a threshold in frequency (depending on the metal), the electrons will not budge.It shows a one to one correlation that only a particle model can explain.
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Theoretically, could there be different types of protons and electrons? Me and my friend were arguing. I think there could theoretically be different types of protons, but he says not. He says that if you have a different type of proton, it isn't a proton, it's something else. That doesn't make sense to me! There are different types of apples, but they're still called apples!
He says that's how protons work, but can we really know that?
| you are merely arguing semantics. your friend is right because there is no way to distinguish one proton from another. you are right because we might some day find a way to distinguish one proton from another. you two seem to be arguing about what words we would use for the two different types of proton we might someday distinguish.
we might call them both protons (e.g. "this is a type 1 proton and that is a type 2 proton"). or we might come up with a new name for one of them (e.g. "this is a proton and that is an experton, which is exactly like a proton except..."). which way the language evolves probably has to do with how common the two different kinds of protons are. if all the protons on earth are type 1 protons, then we will probably give the type 2 protons a new name, but if earth has both type 1 and type 2 protons in equal quantities, then they will probably both still be called protons. you might say that when we discovered antiprotons, we had this exact choice to make. it seems we chose to give the particles a new name because all the protons on earth are type 1 (protons), not type 2 (antiprotons).
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Working principle of inverter I got project on the working of inverter from school. I know this that DC inverter has an alternator switch which constantly changes its direction so that magnetic field is produced in primary coil due to which current is induced in secondary coil and we get output AC. So according to all this Electromagnetic induction should be the working principle behind the working of DC inverter. But DC can't take part in EMI, I know alternator is being used but it doesn't feels right. I hope I am right till this point I am taking following image as reference:
Please don't get mad at me if I got everything wrong.
| Low-cost inverters like you see in a car work in this fashion. Ones used in power production, and higher-quality ones used in cars, improve the output by not switching 120 times a second (or 100 if you're in Europe), but many thousands of times. The output is sent into circuits that boost the voltage, sometimes without a transformer at all, and filter the output with capacitors. The result is a very accurate sine wave that is good enough to put on the grid.
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Motion of one body with reference to another I studied that Galileo was punished by the church for teaching that the sun is stationary and the earth moves around it. His opponents held the view that earth is stationary and sun moves around it.
The question i want to ask is if the absolute motion has no meaning,are the two viewpoints not equally correct or equally wrong?
Thanks in advance for any help.
| Because the sun has more mass and a higher inertia, so the gravitational attraction between the two will impact the Earth's momentum more than the Sun's.
If we look at the two bodies in isolation and have no prior knowledge of the system the only thing that breaks the symmetry is mass. If gravity exerts an equal force of attraction between the two objects, the earth will experience a greater change in momentum because it has less mass and less inertia, in other words the earth changes it's movement to accommodate the sun more so than the other way around.
That being said, when the two bodies move by each other (motion can be relative in this case) if they're at that right distance/speed/trajectory, where gravity overcomes escape velocity and the orbital momentum between the objects is enough to keep them from falling into each other, it will always be the earth that changes it's momentum and orbits the sun, and since the sun is so much more massive we see the extreme that we do in that earth does nearly all the orbiting.
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Measuring very small temperature differences Can one use a thermometer with $\pm$5 mK accuracy to measure a temperature difference of 2 mK (the measurement is near 100 mK temperature on a sample on an ADR)? Using the same thermometer, I am thinking to measure temperature of the sample, heat the sample slightly, measure temperature again, and take the difference. Does the $\pm$5 mK uncertainty cancel out when I take the difference? My thermometer is sensitive enough, my AC resistance bridge is capable of resolving such small temperature differences, but I want to know if the $\pm$5 mK is really an issue here.
| Yes, of course you can do it. The 'accuracy' is from a calibration, after all, and your temperature-difference determination establishes a short-term new calibration. If the apparatus were to have hysteresis (the meter pointer is sticky), or if there were interfering signals (the power supply ripple dominating
an electrical measurement over a short time), those could cause difficulties
that would interfere with your intention, but most measurement apparatus is
well controlled for those kinds of errors.
The key here, is that you are resolving differences below the 'it holds this
calibration for a year at a time' accuracy. If you can resolve them, with
a repeatable measurement, those
differences ARE measured, in every sense of the word.
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Why is velocity inversely related to pressure in a flow? I've seen the equations that give this relationship, and I understand the math and have seen it worked out in problems. But I don't have a qualitative, conceptual grasp on the relationship.
Is the pressure that which is exerted by a small element of the flow, or exerted on a small element? Maybe it is the pressure exerted on or by the boundary?
And once pressure is defined, why is it related to velocity? Why does something push less if you speed it up? I would have guessed a faster flow pushes harder.
| The "pressure is inversely related to velocity" formula is applied when fluid is in dynamic motion and $P =f/a$ as area is inversely to velocity. Instead, in $P = fv$ pressure is directly proportional to velocity and it is applied only on static fluid
$P〆v$ in static fluid
$P〆1/v$ in dynamic fluid
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Why is direct current needed to produce an electromagnetic field using a solenoid? I am performing an experiment for school investigating the magnetic force of a solenoid. While doing this experiment I realized that I needed to connect the solenoid to the DC output of the power supply instead of the AC.
I am perplexed since for current induction, fluxtuations in the magnetic field are needed. I thought this would be the same "the other way round", i.e. using a fluxtuating current to create a magnetic field.
| To be fair, you don't actually need continuous DC current, but you do need a waveform that is always in the positive (or always in the negative) voltage range. An AC source with a simple half-wave or full-wave rectifier (4 diode) will work fine without the need for a transformer or DC converter. It is still DC, it's just unregulated. Note that a half wave rectifier will only let you use half of the energy, with the same voltage but a 50% duty cycle pulse train. It might cause issues, depending what you are doing.
The important part is that the waveform does not go say from 0V to 12V to 0v to -12v etc... With AC, every time the voltage is positive the field will go in one direction, and in reverse when negative, so overall it is approximately cancelled.
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Models of gravitational waves: Do we need to modify our understanding? So, now we got the first observation of gravitational waves, and that's indeed an amazing revolution in the field of astronomy and cosmology. So one can start discussing whether the previous model(s) were completely to the point or not, isn't it ? I mean, for decades gravitational waves were kind of a myth in science, but now one can confront models and experiments, as in any experimental science.
So I was wondering whether the theory of gravitation needs to be adapted to some unexpected behaviours of the gravitational waves already detected, if they are alternative models describing the observations, and so on. In short, where are we in the scientific debate about modelling gravitational waves ?
I guess it's well too soon for discussing this now, but still I can ask the question ;-)
| I think the answer is clearly no, we don't. The detections have agreed extremely well with numerical models based on GR so far. So this is another test that GR has passed with flying colours.
It may well be that we will learn new things -- and with luck surprising things -- about the statistics of objects and this might influence cosmological models. Indeed the initial detection was of black holes with masses which were thought to be uncommon I think. However this will take time as there need to be many detections.
(It's clear from the comments below that 'myth' in the question is not meant in the sense it now often has of something presumed not to be true).
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Moment of Inertia of Annular Quadrant I am measuring the moments of inertia for various numbers of annular quadrants placed on a torsional oscillator. I know $\displaystyle{I=\frac{1}{2}M(R^2+r^2)}$ for a whole annulus. If I want the moment of inertia of only an annular quadrant, would I divide the formula above by four?
| It sounds like you are doing the following TeachSpin experiment.
If I understand correctly, brass quadrants are placed symmetrically on a rotation table, at the same distance from the axis as for the annulus.
When the axis is the same, the distribution of mass about the axis is the same for the quadrant as for the whole annulus. Because of this the same formula can be used, but using the mass of the quadrant instead of the whole annulus.
The quadrant could be stretched out to 360 degrees, every part of it remaining at the same distance from the axis, without changing the moment of inertia. This is known as the Stretch Rule. After doing that, you would be left with an annulus with the same dimensions (and volume) but $\frac14$ the mass (and density) of the whole annulus.
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What is the prevailing opinion in scientific community about Hans C. Ohanian's description of spin? In the paper What is spin?, Am. J. Phys. 54 (1986) 500, by Hans C. Ohanian, spin is described as a circulating flow of energy in the wave-field of a particle. Is this the generally agreed upon explanation of intrinsic angular momentum or just a fringe theory?
(A similar thread exists on Reddit, but I couldn't find a satisfactory discussion there.)
| I read the abstract.
The basic reason it is not referred to or used is that main stream physics has elementary particles as point particles in the standard model and any wave nature attributed to the particle is on the probability distribution of its location in space and time.
it can be shown that the spin may be regarded as an angular momentum generated by a circulating flow of energy in the wave field of the electron.
A point particle can have no circulating flow of energy. So it may be a correct mathematically description but not within the language/model of mainstream particle physics at present.
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Why aren't there simulateous spin z and x eigenstates? Consider the spin $1/2$ case and suppose I have an eigenstate of $S_z$. This has $\sigma_z=0$. The uncertainty principle states that $\sigma_z \sigma_x \geq \frac{\hbar}{2} |\langle S_y\rangle|$. In this case $|\langle S_y\rangle|=0$, so the uncertainty principle is obeyed. However, why can't $\sigma_x=0$ too then? I know it doesn't have to be the case just because the uncertainty principle wouldn't prevent it, but I'm wondering if there's some reason related to the principle?
| The simple answer why $\S_x$ and $\S_z$ can not be simultaneously specified is of course since they do not commute, i.e. $[S_x,S_z]=S_y$. This leads to your uncertainly relation $$\sigma_x\sigma_y\geq\frac\hbar2|\left<S_y\right>|$$
Since in your special case $\left<S_y\right>=0$ you can choose $\sigma_z=0$ without violating the bound. That $\sigma_x\neq0$ is already build in from the start.
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What actually oscillates in quantum harmonic oscillator with given frequency? Quantum harmonic oscillator is said to be describing motion of microscopic stuff (like atoms in molecules). But unless one keeps on measuring the position on the atom, it doesn't exist at all. It's in superposition of several possible positions (position eigenstates). How can we even assume oscillation and specifically with given frequency of oscillation?
| Hamiltonian eigenstates viz. $\hat{H}\left|n\right\rangle =E_n\left|n\right\rangle$ span the Hilbert space. At time $t$ the state vector may be written as $\left|\psi\left( t\right)\right\rangle =\sum_n \psi_n\left( 0\right)e^{-i\omega_n t}\left|n\right\rangle$ with $\omega_n=\frac{E_n}{\hbar},\,\psi_n\left( 0\right)=\left\langle n|\psi\left( 0\right)\right\rangle$. The $e^{-i\omega_n t}$ factors are complex phases. It is the relative phases of energy eigenstates of different eigenenergies that oscillates (though as other answers have noted, certain other things do as well).
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Clebsch-Gordan coefficients for all the normalized $S_\textrm{tot}^2$ eigenstates of a three spin system with spin $s_\textrm{tot} = 1/2$? The total spin operator is defined as $\vec{S}_\textrm{tot} = \vec{S}_1 + \vec{S}_2 + \vec{S}_3$ with $\vec{S}_1 = S\otimes\mathbb{I}\otimes \mathbb{I}$, $\vec{S}_2 = \mathbb{I}\otimes S\otimes \mathbb{I}$ and $\vec{S}_3 = \mathbb{I}\otimes\mathbb{I}\otimes S$. Furthermore, the square of the total spin is $S_\textrm{tot}^2 = \vec{S}_\textrm{tot}\cdot \vec{S}_\textrm{tot}$. My question is how to determine the normalized eigenstates of $S_\textrm{tot}^2$ with total spin quantum number $s_\textrm{tot} = 1/2$, that is, with eigenvalue $\frac{3\hbar^2}{4}$. I think I have to combine the two first spins, and then combine those with the third, but I'm not sure how!
| Yes. Combine spins 1 and 2 to obtain $j_{12}=0$ and $j_{12}=1$. Then combine each with the remaining spin 3 to get the total spin -$1/2$ states.
Note that there are two sets of spin-$1/2$ states, "coming from" $j_{12}=0$ and $1$ respectively. The states are distinct (as you will see) even if the total spin are the same. Indeed one could, for "local" reasons of symmetry or whatever, use any orthogonal combinations of states in those sets (presumably having identical total projections $M_s$.)
Moreover, you get two mathematically equivalent but physically different sets if you couple first spins 2 and 3, i.e. if you start by building $j_{23}=0$ or $1$, and then couple spin 1. These latter two sets can be written as linear combos of the former two sets.
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What is the difference between conventional current and electronic current? what is the difference between conventional current and electronic current?
How are they linked to one another?
| • Electric current can be either positive or negative, but conventional current is always positive.
• The conventional current for an electron flow is positive, whereas the electrical current is .
• For a flow of positive charges, both the electric current and the conventional current are the same.
• Since almost every electrical circuit uses an electron flow, it can be safely stated that the conventional current = – electrical current.
• In conventional current, the flow of electrons is assumed as a flow of protons on the opposite direction.
| {
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How to detect femtosecond pulses with slow PMT's? In pump-probe measurements, we can use femtosecond lasers to study ultrafast processes, such as molecular vibrations. PMT's are very slow compared to the pulses, so how can we use them to detect such short pulses?
I know that the pulses themselves are used for pulse characterization techniques such as FROG and SPIDER, but these are not used for pump-probe spectroscopy measurements, right?
| You can still do pump probe spectroscopy with a femtosecond laser. As long as the detector's gate is wide enough to capture the pulse, you should be fine. As for getting the vibrational information, the pulse needs to be passed through a spectrometer to get your vibrational information. (We did a similar thing for femtosecond CARS)
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Gravity between two Photons (I searched for an answer online already but I couldn't quite find what I was looking for...)
I thought about this for a long time now. If two Photons fly in the same direction, one behind the other one, for my understanding the one behind the other one should be pulled towards the photon in front of it due to it's gravity, and because it cant get faster it should increase it's frequency and therefore gain energy. The one in front cant be pulled backwards though because gravity travels with the speed of light itself(?) and therefore the gravity of the rear photon cant reach the one infront of it, which would therefore not lose energy.
But that would break the law of conservation of energy, wouldn't it? So I'm confused...
Am I thinking something wrong? Or how does it work/what would actually happen in this scenario?
Thanks for answers in advance!
| They do not interact because their rest masses are zero.
You can not attach to the photon neither an inertial nor a gravitational "mass" as
$$ m \neq\frac{\hbar \omega}{c^2}$$
See for example on this topic https://arxiv.org/abs/physics/9907017.
This part is pure kinetic part and is related to the momentum of a photon in the energy–momentum relation (with $m_{rest}=0$ )
$$E=p c $$
As example, for a moving particle with non-zero rest mass, the momentum part $E=p c $ does not influence its gravitational interaction (i.e. the gravitational potential that it creates). Because you can always consider the case in a comoving reference frame where $p=0$.
| {
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Flow rate of a liquid Can one use the following formula for calculating the flow energy rate of a fluid?
$Q = Cp_{fluid} T \dfrac{dm}{dt}$
where dm/dt is the mass flow, T the temperature and Cp the specific heat.
And how would it be related to the flow energy rate $\dfrac{dm}{dt}Pv$?
Thanks!
| only if the flow is isothermal. You derive this from the state equation (ideal gas law) but when you take the derivative wrt time each side you have to use the chain rule. And if temperature can change, then you need to include its derivative.
| {
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How attitude indicator (gyro horizon) adjusts to the Earth's curvature? Image a plane is flying from North to South and is crossing equator. How gyro horizon would maintain correct pitch angle? (Or East-West?)
I assume that pitch angle is correct at takeoff, so the further plane flights, the more difference would be between pitch angle relative to current g-force and pitch angle shown by attitude indicator.
Or do I understand gyros wrong?
| I've been crushing flat earth nonsense with solid facts and mathematics for about two years now. Their understanding of gyro physics (much like any of their other scientific understanding) is flawed, and they base most of their claims on these flawed understandings.
Gravity is an acting force on the attitude indicator gyro of a plane. This means that the force of gravity is applied to the Y axis of the gyro which keeps it continually vertical relative to earth's surface, or perpendicular to the center of gravity (a.k.a. level).
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Where am I going wrong in finding the center of mass of this sphere? A solid sphere of density $ρ$ and radius $R$ is centered at the origin. It has a spherical cavity in it that is of radius $R/4$ and which is centered at $(R/2, 0, 0)$, i.e. a small sphere of material has been removed from the large sphere. What is the the center of mass $R_{cm} = (x_{cm}, y_{cm}, z_{cm})$ of the large sphere, including the cavity?
My attempt:
$$\vec{R}=\frac{1}{M}\int \vec{r}dm$$
$M=\rho V=\rho (V_{total}-V_{cavity})=$total mass
$\vec{r}=x^2\hat x + y^2 \hat y+z^2 \hat z$
$dm= \rho dV$
$$\vec{R}=\frac{1}{\rho (V_{total}-V_{cavity})}\int_V \rho (x^2\hat x + y^2 \hat y+z^2 \hat z)dV$$
$$=\frac{1}{ (V_{total}-V_{cavity})}\int_0^x \int_0^y \int_0^{\sqrt{R-x^2-y^2}} (x^2\hat x + y^2 \hat y+z^2 \hat z)dxdydz$$
...but here I'm confused - how do I set up the integral to account for the cavity?
| You dont have to complicate using such integrals. Just do this. Assume the larger sphere to be complete and find its centre of mass (which is its geometrical centre of course). Now for the cavity, assume it is made of negative mass(no such thing of course, just eases the calculation).The mass of the smaller sphere would then be $-M/64$ (where $M$ is the mass of the larger sphere) Find its centre of mass co-ordinates. Then use the formula for the centre of mass for two point masses located at two points(which are the co-ordinates of the centre of mass of larger sphere and that of the smaller sphere/cavity) giving $R_{COM}$ = $(-\frac{R}{126},0,0)$
| {
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Should zero be followed by units? Today at a teachers' seminar, one of the teachers asked for fun whether zero should be followed by units (e.g. 0 metres/second or 0 metre or 0 moles). This question became a hot topic, and some teachers were saying that, yes, it should be while others were saying that it shouldn't be under certain conditions. When I came home I tried to find the answer on the Internet, but I got nothing.
Should zero be followed by units?
EDIT For Reopening: My question is not just about whether there is a dimensional analysis justification for dropping the unit after a zero (as a positive answer to Is 0m dimensionless would imply), but whether and in which cases it is a good idea to do so. That you can in principle replace $0\:\mathrm{m}$ with $0$ doesn't mean that you should do so in all circumstances.
| First, as long as the quantity under consideration possesses a unit, yes, because of the importance of the Consistency of Units or Dimensional analysis. Second, in a minor key: in experimental physics, a pure zero is not likely to happen in practice. Therefore, whenever a zero is very close 0.0000000000257, it is important to know the unit: is it in micro or gigaUnit? The factor makes a huge difference. Third: a measurement is often sounder with some notion of uncertainty. Hence, $m=0 \pm 0.001 \;\mathrm{kg}$ tells a lot about the precision one can expect.
To equate, add or multiply quantities, they should be consistent. When one writes $y=ax+b$, the quantities $y$, $ax$ and $b$ should have the same dimension, i.e. the unit of $a$ times the unit of $x$ should be the same as the unit of $b$.
I believe that one should not add a unitless $0$ to a distance, while adding $0$ meter to that distance makes sense. Even if the quantity is $0$ "unit", I believe it still does matter with products, see xkcd: dimensional analysis.
When it comes to applying a more complicated function (a logarithm, an exponential) to a dimensionful number, the discussion is more involved, see for instance Exponential or logarithm of a dimensionful quantity?. Some advocate for instance that a logarithm is dimensionless (from What is the logarithm of a kilometer? Is it a dimensionless number?).
[EDIT] For real-world fans of dimension analysis, Why dimensional analysis matters by UnitFact:
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Electromagnetic induction with light bulbs
I chose C as the voltage, which was initially split between the two lamps, will only be "used" by lamp 1, thus increasing the brightness. However, the answer is D.
| Assume that the copper wire a very much smaller resistance than that if the bulbs and that the resistance of the bulbs $R$ is independent of the current passing through them.
Let the induced emf in the loop be $2\mathcal E$.
The current in the circuit which produces a certain equal brightness for the two bulbs is $I = \frac{2\mathcal E}{2R}=\frac{\mathcal E}{R}$.
Suppose the whole area of the loop was $4A$.
With the connecting wire as shown in the right hand diagram let the area of the top loop be $3A$ and that of the bottom loop be $A$.
The emf induced in the top loop is now $1.5\mathcal E$ because the rate of change of magnetic flux in the top loop has increased. And the current $\frac{1.5\mathcal E}{R}$ which is larger than before so the top bulb is brighter.
For the bottom loop the current is $\frac{0.5\mathcal E}{R}$ which is smaller than before so the bottom bulb is less bright.
So the answer is C.
An alternative (quicker?) method.
Now put the wire exactly down the middle so that the induced emf in each loop, top and bottom, is $\mathcal E$.
So the current through each loop is exactly the same as before $I =\frac{\mathcal E}{R}$ and so the bulbs are as bright as before.
Now place the central wire so that it is very, very close to the wire and the bulb at the bottom.
The emf induced in the top loop is slightly less than $2\mathcal E$ and so the current through the top bulb is just less than $\frac{2\mathcal E}{R}$ which is slightly less than twice the current in the original circuit and so the top bulb is brighter than before.
For the bottom loop the induced emf is very small, so the current in that loop is very small and so the bottom bulb is dimmer than before.
So the answer is still C.
| {
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What happens to a photon as it experiences an increasing amount of force opposite to its momentum? Could it be frozen? Image that a photon is emitted from a sphere, moving orthogonally away from the sphere's surface. As the weight of this sphere increased, how would the increasing gravitational force affect the motion of the photon? Imagine that the mass of the sphere was sufficiently high to form a black hole - but the photon is emitted outside the black hole's event horizon - what happens to the photon then?
| Photon momentum does not correspond to its velocity, but its frequency, so if you applied a force opposite to its momentum, it would not slow down, but redshift.
As photons carry no charge, the only force applicable is gravity, and its general relativistic description is more kinematical than dynamical (the photon gets redshifted because the relative orientation of the velocity vectors of photon and observer changed while moving along a straight line through curved spacetime - no forces anywhere in sight).
As to where the energy goes, the classical answer is into the gravitational potential. The relativistic answer is a bit more complicated as it's not possible to associate an energy density with the gravitational field. Morally speaking, it's arguably nevertheless the right idea.
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Operator functions: why is $f(A)$ uniquely defined? In Nielsen and Chuang, they write: Let $A = \sum_a a|a\rangle \langle a|$ be the spectral decomposition of $A$. Define $f(A) = \sum_a f(a) |a \rangle \langle a|$. Apparently this is uniquely defined. I'm having trouble seeing why this is.
If we used some other orthonormal basis of eigenvectors of $A$, say $A = \sum_b b|b\rangle \langle b|$, then why is $\sum_a f(a)|a\rangle \langle a| = \sum_b f(b)|b\rangle \langle b|$? I think there must be some property about eigenvectors sharing the same eigenvalues, but I'm unsure about what I'm missing.
| This stems from a misformulation of the spectral theorem. In a proper mathematical text one never sees it stated using a basis precisely because that is not unique. Physicists often assume "for the sake of simplicity" a nondegenerate spectrum, where every eigenvalue has an algebraic multiplicity of 1, and consider nontrivial multiplicities a degenerate case that can be covered if needed.
The spectral theorem gives a unique decomposition using projectors instead of $|n〉〈n|$, which for finite-dimensional cases looks like
$$A = \sum_{α ∈ σ(A)} α P_α,$$
where $\{P_α\}$ are mutually orthogonal projectors. The application of an analytic function $f$ is then defined as
$$f(A) = \sum_{α ∈ σ(A)} f(α) P_α$$
and the uniqueness naturally follows: indeed, there's no point where this could become non-unique.
If you went one step further and decomposed each of the projectors
$$P_α = \sum_{k=1}^{ν_α} |v_α^{(k)}〉〈v_α^{(k)}|$$
in some orthonormal basis of its range, you'd obtain the decomposition used by Nielsen and Chuang, but as you noted, it would no longer be unique. (The eigenvalues still are, and their ordering does not matter.) The trick why this also works is then to note that if you group the summands putting ones using the same eigenvalue in an inner sum,
$$A = \sum_a a|a〉〈a| = \sum_{α ∈ σ(A)} α \sum_{k=1}^{ν_α} |a_α^{(k)}〉〈a_α^{(k)}|,$$
where $\{|a_α^{(k)}〉\}$ are $ν_α$ of the original eigenvectors $|a〉$ all corresponding to the eigenvalue $α$, then also
$$f(A) = \sum_a f(a)|a〉〈a| = \sum_{α ∈ σ(A)} f(α) \sum_{k=1}^{ν_α} |a_α^{(k)}〉〈a_α^{(k)}|.$$
By obtaining the same $f(a)$ for all $a$ which corresponded to the same eigenvalue $α$, the grouping is not violated, and once we agree that
$$\sum_{k=1}^{ν_α} |a_α^{(k)}〉〈a_α^{(k)}|$$
is the projector $P_α$, then
$$f(A) = \sum_{α ∈ σ(A)} f(α) P(α),$$
which as we know from above is the unambiguous definition.
| {
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How many photons can exist in a cubic box of unit volume, simultaneously? Suppose that we have a cubic box of unit volume. Simultaneously, how many photons can exist in such box? Is there any limit?
| You have to keep in mind that if we are not at absolute zero the atoms in the walls of the container will have thermal energy and therefore they will vibrate and emit/absorb photons. In an equilibrium situation, the electromegnetic radiation in the box will be blackbody radiation (assuming we are dealing with an isolated system). The expected number of photons will be in this case $N \propto V T^3$. Therefore, since volume is fixed, the answer will depend on the temperature of the box.
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Momentum and acceleration of an atom after emitting of a photon? I'm a second year physics student and we've been talking about light and the fact that it carries momentum. I've been thinking about a situation where there is an excited atom that has an electron at a high energy level. When the electron jumps back down to a lower energy level, it will release a photon. This photon has momentum and according to the conservation of momentum, the atom must gain the same amount of momentum in the opposite direction. I understand that the light is released and it immediately is moving at c, with momentum p = h/λ. This means that the momentum of the atom must be mv=-h/λ.
What's bugging me, is that the photon isn't accelerated, it has this momentum immediately when it is emitted, which means the atom must also have this momentum (but negative) when the photo is emitted. Does this mean that the atom goes straight from 0 velocity to a non-zero velocity without accelerating? Or is something else going on that I don't know about?
| The momentum depends on the wave length (energy). The time from the wave starts until it reaches its top phase will be the acceleration time. The acceleration amount depends on the mass of the atom.
| {
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Why are planets not crushed by gravity? Stars can be crushed by gravity and create black holes or neutron stars. Why doesn't the same happen with any planet if it is in the same space time?
Please explain it in simple way. Note: I am not a physicist but have some interest in physics.
| The particles which make up atoms are electrically charged, and they repel each other when they get too close to each other. Gravitational forces only attract one particle to another, and never repel, but they're extremely weak compared to the electrical force. To create a black hole, the gravitational force needs to overcome these repulsive forces between particles. For objects like the earth and the sun, the repulsive forces are much greater than the gravitational force.
| {
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Charge leakage from two suspended charged spheres Recently I was going through "Problems in General physics" by I E Irodov. In Electromagnetics chapter, there is a question how much is the charge leakage from two spheres suspended by a silk thread (3.3).
Two small equally charged spheres, each of mass $m$, are suspended from the same point by silk threads of length $l$. The distance between the spheres $x \ll l$. Find the rate $\frac{dq}{dt}$ with which the charge leaks off each sphere if their approach velocity varies as $v = \frac{a}{\sqrt{x}}$, where $a$ is a constant.
$$
\frac{1}{4 \pi \epsilon_0} \frac{q^2}{x^2} = \frac{mgx}{2l} .
$$
Take derivative then,
$$
{2q} \frac{dq}{dt} = \frac{4\pi \epsilon_0 mg}{2l} 3 x^2 \frac{dx}{dt} .
$$
One may simplify further by introducing expression for q from the electrostatic force equation and may do more arithmetic... For my point here, this much is enough, it is clear that dq/dt depends on dx/dt (it is vivid in the question too).
Here I did not understand the concept fully. I have two questions:
*
*What charge leakage means? Where the charge is leaking to?
*The expression for the charge leakage dq/dt depends on the relative velocity between the two charged spheres. If the spheres are held stationary or if the relative velocity is zero, according to the equation no charge leakage happens, right? Can any one explain physically why no charge leakage happens when relative velocity is zero?
| To your questions:
(1) Charge leakage $dq/dt$ is the charge loss current of the spheres due to the conductivity of the air.
(2) If the spheres are losing charge, the repulsive Coulomb force decreases and thus also the distance x decreases. If the spheres maintain a constant distance x, while they are kept apart by the repulsive Coulomb force, then there is no charge loss. That "no charge leakage happens when the relative velocity is zero" is not the causal sequence. The correct cause-effect relation is converse: the relative velocity is zero because no charge leakage happens.
| {
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Two Cases Of Harmonic Motion Caused By Gravity I'm a highschool student and we learned not so long ago Simple Harmonic Motion, and I'm trying to analyse "similar" cases which I thought of.
Here we have a body (with mass $m$) being affected by the gravity of a body whose mass is $M$, yet it doesn't collide with it (it just goes through its center depiste not being so realistic). I want to mathematically describle this motion.
Just as doing with SHM, we get the differential equation
$m\frac{d^2x}{dt^2}=\frac{GMm}{x^2}\Rightarrow x^2 \frac{d^2x}{dt^2}=GM:=k$
Obviously I don't have the tools to solve such equation (I can only solve easy separables and using Laplace Transform), But Wolfram gave me the following solution
Looking back, I realized that at $x=0$ the force is "infinite" and I kind of stopped there since I'm clueless (plus I couldn't find the constants)
Another case I thought of which might be more realistic is the following
Here the movment on the perpendicular bisector of course.
We have $\Sigma F=2(\frac{GMm}{d^2+x^2})\cos\alpha=\frac{2GMmx}{(d^2+x^2)\sqrt{d^2+x^2}}$ So
$\frac{d^2x}{dt^2}(d^2+x^2)^\frac{3}{2}=2GMx:=kx$, which Wolfram couldn't solve.
I'm pretty sure this might not be far fetched, I came here to see if anyone has any contributions to my understanding? Perhaps some good way to approximate one of the motions?
Thanks for reading this mess!
| Harmonic motion $(x=A\sin(\omega t))$ depends on having a restoring force which is proportional to the displacement from the equilibrium position.
This is not true for your 1st case. Although there will be an oscillation, it is not harmonic. You can avoid the infinite force at $x=0$ by giving mass $M$ a finite radius and constant density, and creating a narrow tunnel through one diameter. The force inside $M$ is then proportional to displacement from the centre of $M$, so if $m$ remains inside $M$ the motion will be harmonic.
In the 2nd case there is approximate SHM for small values of displacement $x$.
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The Schrodinger equation for quantisation of light? Why not with the Dirac equation? Light is a born relativistic. However, when we quantise the electromagnetic wave, we start with the time-independent Schrodinger equation, which is a non-relativistic equation. Why is this fine?
Wouldn't it more make sense if we use the Dirac equation for the quantisation of light? or do I misunderstand the physics of light with respect to the quantisation?
| Light obeys Maxwell's equations. So when we quantize light, the typical starting point is the Maxwell Lagrangian:
$$\mathscr{L} = \frac{1}{4}F^{ab}F_{ab}$$
where
$$F_{ab} = \partial_{a}A_{b} - \partial_{b}A_{a}$$
where $A_{a}$ is the 4-dimensional vector potential $(\phi, {\vec A})$ such that ${\vec E} = -{\vec \nabla}\phi$ and ${\vec B} = {\vec \nabla} \times {\vec A}$.
Taking the variation of the action with respect to $A_{a}$ gives the equation of motion:
$$0 = \nabla_{b}F^{ba}$$
which can be shown to be equivalent to Maxwell's equations in vacuum.
Doing a proper job of quantizing the above lagrangian requires several non-trivial tricks in quantum field theory to handle the fact that the Maxwell theory is a gauge theory and its evolution contains a constraint (this means that the naïve Hamiltonian is zero, and a "normal" Legendre transform won't just simply work). These technical issues CAN be resolved, but typically, students are expected to have the quantization of the Klein-Gordon and Dirac fields under their belt before they do so.
Neither the Klein-Gordon nor the Dirac equations are appropriate, because photons have spin 1.
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Why is nuclear force spin dependent? Why nucleons with parallel spins have greater nuclear force than the ones with anti-parallel spins?
I just want a clear and easy explanation. Thank you!
| Spin dependence of nuclear force is highly non trivial. As an example, exceptional stability of some particular nucleus with nucleon number (also known as magic number) 2, 8, 20, 28, 50, 82, and 126
can be explained by considering the spin-orbit interactions between nucleons.
To see the role of spin dependence, the deuteron nucleus is a good laboratory. The total angular momentum for the deuteron (or in general for a nucleus) is usually denoted by $I$, is given by
$$\hat{\vec{I}} = \hat{\vec{L}}+\hat{\vec{S_{p}}}+\hat{\vec{S_{n}}} \tag{1}$$
For the bound deuteron state $L = 0$ and $I=\hat{\vec{S_{p}}}+\hat{\vec{S_{n}}}=\hat{\vec{S}}$. A priori we can have $\hat{\vec{S}}=0$ or $1$ (following the rules for addition of angular momentum, with $\hat{\vec{S_{p,n}}}=1/2$).
The simplest form that a spin-dependent scalar potential could assume is $$V_{\text{spin}} = \frac{V(r)}{\hbar^{2}}\left(\hat{\vec{S_{p}}}.\hat{\vec{S_{n}}}\right)\tag{2}$$
There we assumed that, how $V_{\text{spin}}$ has spatial dependence is encoded in $V(r)$. Now, in terms of Casimir's $V_{\text{spin}}$ can we written as,
$$V_{\text{spin}}=\frac{V(r)}{2\hbar^{2}}\left(\hat{\vec{S^{2}}}-\hat{\vec{S_{p}^{2}}}-\hat{\vec{S_{n}^{2}}}\right)\tag{3}$$
For generic eigen functions $|S,S_{p},S_{n},S_{z}\rangle$, the expentation value of $\hat{\vec{S_{p}}}.\hat{\vec{S_{n}}}$ is given by
$$\langle \hat{\vec{S_{p}}}.\hat{\vec{S_{n}}}\rangle=\frac{\hbar^{2}}{2}\left(S(S+1)-S_{p}(S_{p}+1))-S_{n}(S_{n}+1)\right) $$
Now, since $S_{p,n}=1/2$, we obtain
$$\langle \hat{\vec{S_{p}}}.\hat{\vec{S_{n}}}\rangle = \frac{\hbar^{2}}{2}\left(S(S+1)-\frac{3}{2}\right)=\begin{cases}
+\frac{\hbar^{2}}{4}\quad\text{triplet state},\quad|S=1,\frac{1}{2},\frac{1}{2}\rangle \\
-\frac{3\hbar^{2}}{4}\quad\text{singlet state},\quad|S=0,\frac{1}{2},\frac{1}{2}\rangle \end{cases}$$
Now if $V(r)$ is attractive potential (i.e. $V(r)<0$) then
*
*strength of $V_{\text{spin}}$ increases for $S=1$,
*while it is reduced for $S=0$ configuration.
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Simple Electric Field Problem Solve the Electric Field distance z above a circular loop of radius r. The charge/length = $\lambda$
The arc-length is 2$\pi$r. So the smallest portion of the circle is 2$\pi r \delta \theta$ and charge is therefore
\begin{align}
q&=2\pi r \delta \theta*\lambda
\\
R&=\sqrt{r^2+z^2}= \text{constant}
\\
E&= \frac {1}{4 \pi \epsilon _o}\frac {q}{R^2}=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \delta \theta*\lambda}{\sqrt{r^2+z^2}}\end{align}
And we only need the z component.
$$E =\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \delta \theta*\lambda}{\sqrt{r^2+z^2}}\sin(\theta)=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \delta \theta*\lambda}{\sqrt{r^2+z^2}} \frac {z}{\sqrt{z^2+r^2}}$$
and everything is constant except for $\delta \theta$
$$E=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \lambda}{\sqrt{r^2+z^2}} \frac {z}{\sqrt{z^2+r^2}}\int_0^{2\pi}{\delta \theta}$$
So I thought the correct answer must be:
$$E=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \lambda z}{[r^2+z^2]^{3/2}} \cdot 2\pi$$
But the correct answer does not multiply by 2$\pi$
Correct: $$E=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \lambda z}{[r^2+z^2]^{3/2}}$$
Why was I wrong? where did I slip up?
Thanks!
| The length element should be $r d\theta$ not $2\pi r d\theta$. So the charge element is
$$dq=\lambda r d\theta$$
but not
$$dq=\lambda 2\pi r d\theta.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What does "point of application of force" mean in the given context? I faced a particular conceptual doubt while solving a textbook problem. I will initially write the complete question in my textbook and then try to boil it down to a single conceptual doubt so that it complies with the rules of Physics Stack Exchange.
Original Question:
A finite conductor $CD$ carrying current $i$ is placed near a fixed
very long wire $AB$ current carrying $i_0$ as shown in the figure.
Find the point of application and magnitude of the net ampere force on
the conductor $CD$. What happens to the conductor $CD$ if it is free
to move? (Neglect gravitational field)
My conceptual doubt is : What does "point of application of force" mean ? How to find it?
According to me the force doesn't act on a single point but on the whole wire $CD$, the force being maximum at end $C$ and minimum at end $D$.
P.S: I hope the question complies with the rules of the site. If not, please inform me, so that I can try to improve/re-frame the question. Thank you.
| Think about gravity for example. It acts on every point of an object, and one finds the center of gravity by doing a "weighted" sum of the point locations.
$$ x_{\rm CM} = \frac{ \sum x \Delta m}{\sum \Delta m} $$
If gravity varied by location as $g(x)$ then the above would be
$$ x_{\rm CM} = \frac{ \sum x \Delta m\, g(x)}{\sum \Delta m\, g(x)}= \frac{ \int x \,g(x)\,{\rm d}m}{\int g(x)\,{\rm d}m} $$
Now instead of gravity, substitute in the EM force $F(x)$
$$ x_{\rm CM} = \frac{ \int x \,F(x)\,{\rm d}m}{\int F(x)\,{\rm d}m} = \frac{ \int x \,F(x)\,{\rm d}x}{\int F(x)\,{\rm d}x} $$
NOTE: ${\rm d}m = \rho A {\rm d}x $ and the constant cross-section $A$ and density cancel each other between the top and the bottom of the fraction.
| {
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acceleration and potential energy due to conservative force how can we find acceleration due to a conservative force as a function of time when potential energy due to the conservative force is given as function of position(e.g. U(x)=x^2)
| For a conservative field, the force $F$ can be calculated from the potential energy gradient $dU/dx$
$$
F = -\frac{dU}{dx} = -2x
$$
where $U(x) = x^2$ in your example.
If the mass, $m$, is known, Newton's second law gives
$$
\ddot{x} = \frac{F}{m} = -\frac{2}{m} x
$$
which is the differential equation for an undamped oscillator. This linear differential equation can easily be solved for $x(t)$:
$$
x(t) = a\cos(\omega t-\phi)
$$
where
\begin{eqnarray*}
a &=& \sqrt{x_0^2+(\dot{x}_0/\omega)^2} \\
\omega &=& \sqrt{2/m} \\
\phi &=& \tan^{-1}\left(\frac{\dot{x}_0}{\omega x_0}\right)
\end{eqnarray*}
$x_0$ and $\dot{x}_0$ are the initial conditions, $x_0=x(t_0)$ and $\dot{x}_0=\dot{x}(t_0)$.
| {
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Does the top plate of a capacitor hold half of the capacitor's charge or all of it? I am a little confused conceptually about the charge of a capacitor held by the top plate. Is it equal in magnitude to $q$, or is it half of $q$? It makes more sense to me for it to be half of $q$, with the other half existent on the other plate, but I am thinking this is incorrect.
| When we say that the charge of a capacitor is $Q$, we mean that one plate has charge $Q$ and the other plate has charge $-Q$.
That is to say, we don't mean that the capacitor has a net charge of $Q$ (as I suspect you might be thinking) since, in fact, the net charge of a charged capacitor is zero.
By charged, we mean the capacitor has stored energy in more or less the same sense of a charged battery.
| {
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Example of a transformation that is not canonical Can someone please give me an example of a transformation that is not a canonical transformation?
| Consider the transformation $x \mapsto x, p \mapsto v = \frac 1 m [p - qA(x)]$ where $A$ is a vector potential. Then $\{v,v\} = \frac q {m^2} v\times B$ where $\{\cdot,\cdot\}$ is Poisson bracket and $B = dA$ is the magnetic field, so this transformation is not canonical.
More concretely:
With the Hamiltonian $H = (p-qA)^2/2m$, you find $\dot v_i = \frac q m (v\times B)_i$. If the form of Hamilton's equations is preserved, then $\dot v_i = -\partial K/\partial x_i$ where $K$ is the new Hamiltonian, the "Kamiltonian". But then $\partial_j \dot v_i$ must be a symmetric tensor, but this is not the case for arbitrary $B$.
| {
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Discrete Langevin Equation We have the Langevin equation, that describes the motion of a particle in a viscous medium, given by
\begin{equation}\label{Langevin}
\frac{dv}{dt} = -\gamma v + \zeta(t)
\end{equation}
With the conditions that
\begin{equation}
\langle \zeta(t) \rangle = 0
\end{equation}
\begin{equation}
\langle \zeta(t)\zeta(t') \rangle = \Gamma \delta(t-t')
\end{equation}
And, if we make the time discrete, by putting $t = n\tau$ we can obtain the relation
\begin{equation}
v_{n+1} = av_n + \sqrt{\tau \Gamma}\xi_n
\end{equation}
where $a = (1 - \tau \gamma)$ with the conditions
\begin{equation}
\langle \xi_i \rangle = 0
\end{equation}
\begin{equation}
\langle \xi_i \xi_j \rangle = \delta_{ij}
\end{equation}
My question is that I didn't know how I obtain the discrete equation from the continuous equation. I understand the $a$ but why the square root appears? What transformation between $\xi$ and $\zeta$ I should do?
| The short answer to your question is $\zeta(t) \rightarrow \sqrt{\frac{\Gamma}\tau}\xi_n$. The easiest reason to give for the square root is dimensional analysis. $\zeta$ is dimensionful, but $\xi$ is dimensionless, so using dimensional analysis in the variance equations will give you the square root.
In order to deduce this, you should think about how one discretizes differential equations. Working out this procedure, one gets $v(t) \rightarrow v_n$, $\zeta(t) \rightarrow \zeta_n$, $\frac{dv}{dt} \rightarrow \frac{v_{n+1} - v_n}{\tau}$, and $\delta(t-t') \rightarrow \frac{1}{\tau}\delta_{ij}$. The $\frac{1}{\tau}$ can be remembered based on dimensional considerations, or you can integrate/sum both sides in order to properly derive it. Now, simply plug these in and find the relationship between $\zeta_n$ and $\xi_n$.
| {
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Where does gravitational energy come from? We've all heard mass tells space how to curve and curved space tells matter how to move. But where does the energy to curve space come from? Likewise where does the energy that curved space uses to push planets around come from? I mean if I tell my son to clean his room, and he does, then I did not provide him the energy to do so.
| Mass is the same as energy by $E = mc^2$ So the energy that curves space-time is the mass that curves space-time. Then you could ask why does mass curve space? As far as I know that is equivalent to asking where do Einstein's field equations come from. For that one needs a complete and consistent theory quantum gravity which has Einstein's equation in the low energy limit. No one has that.
"We've all heard mass tells space how to curve and curved space tells matter how to move." This poetic but be a bit misleading. Einstein's equations simply describe in detail how space-time changes as a result of changes in matter. There isn't some process or mechanism happening between matter and space-time i.e when space-time changes there is a process that occurs that then can be used to describe what matter will then do.
Lastly, you are using gravitational energy in the newtonian way of looking at gravity. The gravitational potential which ultimately gives gravitational energy in newton's theory is buried in the metric.
| {
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Frozen lake ice formation Yesterday I went walking on the Lagorai mountain, in northern Italy. Passing near the Lake Erdemolo, I noticed a small part of frozen water from the lake, separated from the rest (the lake was not frozen at all). I found a quite strange ice formation, which I cannot fully comprehend.
At the borders of the pool there was a very thin layer of ice with no water underneath, and progressively going towards the center, more layers added up. A picture explains it better:
The layer at the center of the pool was very thick (a 5 kg rock hardly managed to brake it) and was covered with a thin layer of non-frozen water. Under this thick layer there was water.
Breaking the ice in various points I noticed that the various layers were supported by a lot of small columns and walls.
In the image the column in upside down. Those formations were collocated between the layers and the rocky ground.
I thought that the layers might be formed by an alternate solidification-evaporation process during which, once the first ice layer was formed, the water lowered its level (evaporating through the escape routes of the rocky ground). Maybe this can be caused by a higher outside temperature with respect to the ground one. I'm absolutely not sure about this explanation and, even so, I cannot figure out the cause of the columns between the layers. I guess it can depends of the velocity of the freezing process. If this is not too off-topic, can someone explain the cause of such formations?
| You might be seeing evidence of 'frost heave'. Moisture in the soil will,
in cold conditions, both freeze at the surface (adjacent to cold air), and
wick up the existing ice crystal from its (relatively warm) base near the
soil, to the cooler upper tip (where the newly arrived water freezes).
Thus, ice crystals can be seen to grow upward... if such a field of crystal
pillars is then covered with snow, you can (after a few freeze/thaw
temperature changes) get a large mass of ice in the air, supported by a few
pillars of ice that crystallized days ago from soil moisture.
Usually, the top layer of such a 'frost heave' patch is dirty.
| {
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Understanding rocket problem intuitively
A rocket is trying to land on a planet. The mass of the rocket is $1\,\rm kg$, and the gravitational acceleration of the planet is $1\,\rm m/s^2$. The rocket starts the free fall at $20\,\rm m$ above the surface of the planet (initial velocity is $0$), and can use the thrust for $2\,\rm s$ (the force of thrust is $1\,\rm N$).
When is the most reasonable height at which the rocket uses its thrust for two seconds? (By the way, we ignore the loss of mass due to the use of the thrust.)
I solved the question, but I'm not satisfied. I don't quite understand it intuitively.
Someone said $W = Fs$, and since $F$ (thrust) is the same, when $s$ (distance moved) is the greatest, the work done by the thrust, to counteract gravity, would be the greatest. Therefore, the most reasonable height to start using the thrust is when the height at which the rocket would end using its thrust is when it reaches the ground (the calculation to find the actual value of the height is very complicated, so I'll skip (it's not the main point of my question).
When I first tried to solve this, I thought that the chemical energy of the thrust would be used to counteract gravity, and since the chemical energy of the the thrust does not change by the velocity at which it moves, I thought that the height at which the thrust is used does not matter, as the total energy (potential energy due to the gravity + the kinetic energy of the rocket - the chemical energy of the thrust) stays the same, the final velocity would be the same, but this is not the answer.
Can anyone please help me why I may be wrong?
| To minimise the impact speed, there are two ways to think about this:
*
*You want to maximise the work done on the rocket:
$$\mathrm{Work = Force \times Distance}$$
*You need to minimise the time spent falling.
| {
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How much information can you obtain from a pulsar-black hole system? Imagine that we have detected an interesting source in the sky that we believe is generated by a pulsar orbiting a black hole.
The challenge here is the following:
What physically relevant information could you extract from the observation of this system?
Note: I am posting an answer with some possible information that we could obtain, but I will NOT mark my answer as the correct one.
| The system you describe could be considered analagous to the famous Hulse-Taylor binary system, which consists of a pulsar and a "quiet" neutron star companion in a close orbit.
A paper by Weisberg et al. (2010) describes how a detailed timing analysis of the pulsar alone will yield the masses of both components (to 5 decimal places in the case of the Hulse-Taylor system).
In addition you can of course measure the spin properties of the pulsar - period, spin-down rate, braking index, glitches etc.
You also get the basic orbital parameters - the period, period derivative(s), precession of periastron etc. that allow you to test whether the orbital decay predicted by GR through the emission of gravitational waves is accurate.
Additional tests of GR might come through measuring the Shapiro delay if the system inclination was reasonably high. There are also predicted changes in the pulse shape of the pulsar expected to be caused by coupling between the spin and orbital angular momentum a.k.a the Geodetic effect. There may also be effects caused by the spin of the black hole if the components were close enough together - a.k.a. the Lense-Thirring effect - which might tell you about the spin of the black hole.
| {
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Change in enthalpy equal to heat transferred
*
*Is the relation $ \Delta H = Q _P $ valid for both reversible and irreversible processes or only for reversible processes?
$Q_P$ is the heat exchanged at constant pressure.
*Specifically, is it valid for this case: saturated water (mass fraction = 1) is enclosed in one partition a box having two partitions of equal volume, the partition being evacuated. The partition is broken and the evacuated region gets filled with a mixture of water and steam. The temperature of the apparatus is maintained at $100^\circ\, \mathrm{C} $ (therefore the process will also be isobaric, from the phase diagram of water).
| I don't know if it is not true for every irreversible process, but it is certainly not true for the process you described in item 2. And it is not true for the case of a so-called constant pressure irreversible expansion or compression of a gas, where, during the deformation, the external pressure is held constant at a value different for the initial pressure of the gas.
In item 2, from the first law of thermodynamics, the heat added is equal to the change in internal energy of the water, not the change in enthalpy. The change in enthalpy of the water is greater than the change in internal energy (and thus greater than the amount of heat added).
| {
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Thermal equilibrium of universe I am just an interested layman in the field of cosmology. In the usual account of the development of the universe, it is stated that about 380.000 years after the big bang the electromagnetic radiation was in thermal equilibrium with matter.
Does this mean that the universe was (and maybe is) an isolated thermodynamic system that was (and is) in internal thermodynamic equilibrium? Further, it is stated that this radiation is seen in todays cosmic microwave background radiation corresponding to a perfect blackbody radiation with a temperature of 2.7K after the expansion of the universe. Could this expansion be considered as an adiabatic expansion of the radiation system?
Then the radiation supposedly decoupled from matter due to recombination of electrons with atomic nuclei. Does this mean that after this (and now) radiation and matter (on the average) are no longer in thermal (thermodynamic) equilibrium in the universe?
| Yes. You are completely right on all accounts. At one point you suggest the universe is still in thermodynamic equilibrium --- this is not the case. As you later point out, after recombination, photons are baryons do decouple. Once average densities become low enough that the diffusion time (the time it takes for things to transfer heat to each-other) is longer than the dynamical time (the time it takes things to move and interact gravitationally), sub-systems also separate (e.g. galaxies and galaxy clusters). Most things are now moving towards gravitational equilibrium (via collapse into filaments, clusters, galaxies, clouds, stars, planets, etc).
| {
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Is an electron attracted to one of the magnetic poles ...? I noted the question above had been posted. And I wanted to comment, but nay, it was locked out. However, what of the old 'CRT' tubes, in which magnetic fields are used to steer the electron stream ? Now I know the 'Electron's' do not get attracted, or repulsed directly to the source of the magnetism, but the magnetic field certainly has an effect. So if one were to extend the CRT tube to ... well, rather long, would the electron eventually move to, or away from the 'North' or 'South' pole ?
| Thanks for all the input. I think I just answered my own question. (The site suggested refining question, but can't get there easily). Aurora Borealis. I hope that answers the question I answered. (Grin).
Charged particles hit the earths magnetic fields and follow those lines to the poles. I want to thank one of the users who had queried, "How long a CRT" and made me think of a infinite magnetic field, which of course, the only example I could think of is the earth.
So, thanks again for all the input!
Cheers!
| {
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Calculating height of a ball formula based on bounce I'm trying to figure out a formula for the new height of a ball, up until it stops bouncing.
The time iteration between each could be for example one second; and gravity of course would be 9.8
I've tried doing research at different websites such as physics @ illinois.edu and whatnot;
I want it to be regardless of properties of the ball and ground (such as material of the ball and ground).
So assume height starts at 10 meters, after one second it would be x height, then after another second it would bounce and go up to x height, then go down to x height, back up to x height, etc.
Assume it's bouncing straight up and down, so the only plane it moves on is the height plane.
The reasoning behind this is for a programming application; however once I have the formula for the ball bounce I can write the program itself, I just need to figure out the formula.
| R.H=h/4 [t2÷t1] ²
t1=Time of free fall
t2=Time of flight
h=Free falling height
We can calculate the first Rebouce by this equation.
| {
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How do one show that the Pauli Matrices together with the Unit matrix form a basis in the space of complex 2 x 2 matrices? In other words, show that a complex 2 x 2 Matrix can in a unique way be written as
$$
M = \lambda _ 0 I+\lambda _1 \sigma _ x + \lambda _2 \sigma _y + \lambda _ 3 \sigma_z
$$
If$$M = \Big(\begin{matrix}
m_{11} & m_{12} \\
m_{21} & m_{22}
\end{matrix}\Big)= \lambda _ 0 I+\lambda _1 \sigma _ x + \lambda _2 \sigma _y + \lambda _ 3 \sigma_z $$
I get the following equations
$$
m_{11}=\lambda_0+\lambda_3 \\ m_{12}=\lambda_1-i\lambda_2 \\ m_{21}=\lambda_1+i\lambda_2 \\ m_{22}=\lambda_0-\lambda_3
$$
| Let $M_2(\mathbb{C})$ denote the set of all $2\times2$ complex matrices. We also note that dim$(M_2(\mathbb{C}))=4$, because if $M\in M_2(\mathbb{C})$ and
$M=\left(
\begin{array}{cc}
z_{11} & z_{12}\\
z_{21} & z_{22} \\
\end{array}
\right)$, where $z_{ij}\in \mathbb{C}$,
then
$M=\left(
\begin{array}{cc}
z_{11} & z_{12}\\
z_{21} & z_{22} \\
\end{array}
\right)=z_{11}\left(
\begin{array}{cc}
1 & 0\\
0 & 0 \\
\end{array}
\right)+z_{12}\left(
\begin{array}{cc}
0 & 1\\
0 & 0 \\
\end{array}
\right)+z_{21}\left(
\begin{array}{cc}
0 & 0\\
1 & 0 \\
\end{array}
\right)+z_{22}\left(
\begin{array}{cc}
0 & 0\\
0 & 1 \\
\end{array}
\right)$.
The standard four Pauli matrices are:
$I=\left(
\begin{array}{cc}
1 & 0\\
0 & 1 \\
\end{array}
\right),~~
\sigma_1=\left(
\begin{array}{cc}
0 & 1\\
1 & 0 \\
\end{array}
\right),~~
\sigma_2=\left(
\begin{array}{cc}
0 & -i\\
i & 0 \\
\end{array}
\right),~~
\sigma_3=\left(
\begin{array}{cc}
1 & 0\\
0 & -1 \\
\end{array}
\right)$.
It is straightforward to show that these four matrices are linearly independent. This can be done as follows.
Let $c_\mu\in \mathbb{C}$ such that
$c_0I+c_1\sigma_1+c_2\sigma_2+c_3\sigma_3=$ O (zero matrix).
This gives
$\left(
\begin{array}{cc}
c_0+c_3 & c_1-ic_2\\
c_1+ic_2 & c_0-c_3 \\
\end{array}
\right)=\left(
\begin{array}{cc}
0 & 0\\
0 & 0 \\
\end{array}
\right)$
which further gives the following solution:
$c_0=c_1=c_1=c_3=0$.
It is left to show that $\{I,\sigma_i\}$ where $i = 1,2,3$ spans $M_2(\mathbb{C})$. And this can accomplished in the following way:
$M=c_0I+c_1\sigma_1+c_2\sigma_2+c_3\sigma_3$ gives
$\left(
\begin{array}{cc}
c_0+c_3 & c_1-ic_2\\
c_1+ic_2 & c_0-c_3 \\
\end{array}
\right)=\left(
\begin{array}{cc}
z_{11} & z_{12}\\
z_{21} & z_{22} \\
\end{array}
\right)$
which further gives the following equations:
$c_0+c_3=z_{11},~c_0-c_3=z_{22},~c_1-ic_2=z_{12},~c_1+ic_2=z_{21}$.
Solving these equations, one obtains
$c_0=\frac{1}{2}(z_{11}+z_{22}),~c_1=\frac{1}{2}(z_{12}+z_{21}),~c_2=\frac{1}{2}i(z_{12}-z_{21}),~c_3=\frac{1}{2}(z_{11}-z_{22})$.
| {
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General method of deriving the mean field theory of a microscopic theory What's the most general way of obtaining the mean field theory of a microscopic Hamiltonian/action ? Is the Hubbard-Stratonovich transformation the only systematic method? If the answer is yes then what does necessitate our mean field parameter to be a Bosonic quantity ? Is the reason that all of directly physical observable quantities should commute?
| In second quantization, the mean field approximation consists in approximating some combination of operators $A$ by a $c$-number $\langle A\rangle$. For example,
for Bose-Einstein condenstate $A=b_0$, for Cooper pairing $A= a_{\mathbf{p}\uparrow}a_{-\mathbf{p}\downarrow}$, in the Hartree-Fock approximation $A=a_{\mathbf{p}}^+a_{\mathbf{p}}$, in the charge density wave state $A=a_{\mathbf{p}+\mathbf{q}}^+a_{\mathbf{p}}$. Here $a_i$ and $b_i$ are fermionic and bosonic annihilation operators.
The same in the Hubbard-Stratonovich transformation: we can couple any desired combination $A$ of field operators to auxiliary field.
In all mentioned cases, $A$ is indeed bosonic. Since bosonic operators can have large occupation numbers $\langle A\rangle\gg1$, deviations of $A$ from $\langle A\rangle$ can be relatively small, and also noncommutativity of $A$ and $A^+$ can be neglected (if we have the right choice of $A$):
$$
\frac{\sqrt{\langle (A-\langle A\rangle)^2\rangle}}{\langle A\rangle}\rightarrow0,\qquad\frac{\langle[A,A^+]\rangle}{\langle A\rangle}\rightarrow0.
$$
This makes the approximation accurate.
For fermionic quantities $A$, the mean field approximation does not have much sense because the occupation number is limited, $\langle A\rangle\sim 1$. Therefore, $A$ will be strongly fluctuating with respect to $\langle A\rangle$, and the approximation will be inaccurate.
Update number 2 following the discussion in comments
Average value of the fermionic opearator can be nonzero, $\langle A\rangle\neq0$ only at coherent mixtires of different numbers of fermions in the system, e.g. $|\psi\rangle=\alpha|N\rangle+\beta|N+1\rangle$. Generally, such superpositions of integer- and half-integer angular momentum states are forbidden by superselection rules.
In any case, the average $\langle A\rangle$ of a fermionic quantity $A$ is either zero or remains small in the macroscopic limit, so it is a bad candidate for a mean field order parameter.
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Why do some chemicals take electrons from other chemicals? How can some chemicals, if they have an equilimbrium of electrons, take away electrons from other chemicals?
One example I believe is placing a small amount of gallium on top some alluminum and watching the alluminum melt.
Why does the gallium, if it is at equilibrium state, need more electrons?
|
One example I believe is placing a small amount of gallium on top some alluminum and watching the alluminum melt. Why does the gallium, if it is at equilibrium state, need more electrons?
No, gallium dissolving aluminium is not an example of electron exchange but of formation of an alloy.
Exchange of electrons between atoms in chemical reactions can roughly be explained by the Octet Rule. Atoms with less than 8 electrons in their outer electron shell (so called valence electrons) tend to shed or acquire electrons to make up a total of 8 (an octet), which happens to be a very stable configuration.
An example is the reaction involving aluminium foil and iodine crystals, which starts vigorously even at room temperature.
In it, $\mathrm{Al}$ loses three valence electrons, becoming an $\mathrm{Al^{3+}}$ cation, with the electron configuration of $\mathrm{Ne}$ which has a full octet.
The iodine atom $\mathrm{I}$ on the other hand acquires one electron, becoming an iodide $\mathrm{I^{-}}$ anion, with the electron configuration of $\mathrm{Xe}$ which also has a full octet.
The overall reaction is thus:
$$2\mathrm{Al}(s)+3 \mathrm{I_2}(s)\to 2\mathrm{AlI_3}(s)$$
Elements that tend to acquire electrons are called electronegative and are mostly found on the right hand side of the Periodic Table. Elements that tend to lose electrons are called electropostive and are mostly found on the left hand side of the Periodic Table.
| {
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Is there a low-bugdet method to measure micronewtons? I will have to measure very small EM forces (on the order of 10^-6...10^-5 Newtons) acting on a small copper sphere or plate (of about 0.05 kg).
I could hang the copper object with a long thread and use a micro-Newton sensor but I can't find any vendor that gives prices without contact so i guess these are very expensive equipment for me.
Then I thought maybe I could tie a lightweight rod in the center to the thread with 2 identical copper objects on the ends with balance.
It would be very similar to the Coulomb's Torsion Balance set up.
But I'm confused about what to do with this setup.
For example, there could be 10^-6 Newtons from 0.5 meters distance (half the rods length) to center. Would such a small moment produce any measurable angle of twist in a common silk thread? Even if it did, I wouldn't be able to calibrate the scale and know how much force there is.
Or should I try to add a second but much shorter rod and try to measure the "amplified" force with less sensitive force sensor?
Or is there a more common method I'm not aware of?
| Rewritten in response to comments.
The torsion balance is often used to measure gravitational attraction of small objects.
You could make a torsion balance with your copper sphere. Add a metal sphere B as shown in the diagram. Calculate the gravitational attraction. Measure the deflection. Use this to calibration your balance.
If gravity is weaker than the forces you are interested in, charge the spheres and calculate the electrostatic attraction/repulsion.
| {
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Does a trumpet operate using an open air column or a closed air column Just as the title states. I could not find a coherent answer online.
Many thanks in advance
| The trumpet is considered a closed-ended pipe, but this is not a straight pipe. The flared end (at the bell) modifies the overtone structure so that all the integer multiples (rather than just the odd multiples) of the fundamental are present. The fundamental is also modified by the flare. See Thomas Rossing's The Science of Sound.
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Young's Double Slit Experiment-Problem Can it be proved that there would be no change in the "fringe width" when the main illuminated slit(s) is shifted to a position, which makes an angle of $\Theta$ with the original position of the source slit?
My try - I first found out the fringe width in a normal double slit where the position of the source slit is unchanged. However, after that, As I tried for the new position of the slit, I got stuck and couldn't equate the two values of fringe width.
|
Hope you understand by the above explanation . I had a sample question also.
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What do the commutators of the Hamiltonian with the spin operators mean precession-wise? I proved that $[H, S_z] = 0$, while $H$ and $S_x,S_y$ do not commute.
I showed this using matrix representations, now I am to comment on my results with respect to spin precession and I need help for that - how exactly do commutators represent something about precession and rotation along the 3 axes?
| The commutator of an observable like $S_x$ or $P_y$ (the spin in the x direction and the y component of the momentum respectively) with the hamiltonian will tell you about the time evolution of that observable. It tells you how that observable changes with time.
This is given through the heisenberg equation:
$\frac{dA}{dt} = \frac{i}{\hbar}[H, A]$
where A is an operator (like $S_x$ or $P_y$) and I have assumed that A does not depend explicitly on time, i.e $A = A(x(t),p(t))$ and not $A = A(x(t), p(t), t)$.
Now from the equation above, we can see that if A commutes with the hamiltonian, $[H,A] = 0$, then $\frac{dA}{dt} = 0$ and thus A is constant in time, it is conserved.
So when you showed that $S_z$ commutes with H while $S_x$ and $S_y$ do not, you showed that the z component of a particles spin is constant in time, while the x and y components are not constant, they precess.
| {
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Why are only some portions of the EM spectrum able to pass through a metal dog bowl?
In this photo you can see part of the bowl, which I turned upside down and placed on top of various electromagnetic sources to determine whether the waves could pass through it. The graph in the background illustrates my results.
For the shorter wavelengths, I had to guess. I knew that ultraviolet would be blocked, but I assumed that x-rays and gamma rays could go through the bowl.
To me, is seems that wavelengths in the range of, say, 100nm to 900nm are more easily stopped by the bowl. Even though I used a metal bowl, I think a plastic or paper bowl would produce similar results.
So my question is: why are the waves in the middle of the spectrum stopped by the bowl and not those at the ends?
Here are some of the items I used in my experiment:
a TV Remote, a Bluetooth device, an IPhone, our Home WiFi, a Cellular LTE, a Sprinkler Remote, and a Radio.
All of the items on my list were able to pass through the bowl except the TV remote, which I assume was infrared.
| Long-wavelength and short-wavelength radiation passes for different reasons: long-wavelength does not actually pass, but bends around the bowl due to diffraction (or is reflected from the walls), as the wavelength is greater or of the order of the dimensions of the bowl, and short-wavelength radiation passes because its frequency is higher than the plasma frequency of the metal of the bowl. Plastic and paper can give very different results (for example, in the visible range, if the plastic is transparent).
| {
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Why can the $1$-point correlation function be made to vanish? The $1$-point correlation function in any theory, free or interacting, can be made to vanish by a suitable rescaling of the field $\phi$.
I would like to understand this statement.
With the above goal in mind, consider the following theory:
$$\mathcal{L} = \frac{1}{2}\left((\partial\phi)^{2}-m^{2}\phi^{2}\right)+\frac{g}{2}\phi\partial^{\mu}\phi\partial_{\mu}\phi.$$
What criteria (on the Lagrangian $\mathcal{L}$) is used to determine the value of the field $\phi_{0}$ such that the transformation $\phi \rightarrow \phi + \phi_{0}$ leads to a vanishing $1$-point correlation function $\langle \Omega | T\{\phi(x_{1}\phi(x_{2})\}| \Omega \rangle$?
| The 1-point function is constant in spacetime because of translation invariance, i.e. $\langle \phi(x)\rangle = \phi_0\in\mathbb{R}$ for all $x\in\mathbb{R}^4$. Obviously, the 1-point function of $\phi'(x) := \phi(x) - \phi_0$ is zero since the expectation value is linear. So $\phi\mapsto \phi' = \phi + \phi_0$ gets rid of the non-zero 1-point function. This works for all Poincaré-invariant Lagrangians.
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What is the role of pillars in bridges?
As I can see in the picture, there are so many pillars which are holding the bridge. This picture gave a question to me that what are these pillars doing below the bridge?? An appripriate answer could be "these are providing support to bridge".
I tried to get the answer as follows:
In the first image there are two pillars holding a bridge of mass $M$, since gravitaional force is acting downwards thus pillars are bearing a force of $\frac{1}{2}Mg$.
In the second image there are four pillars bearing a force of $\frac{1}{4}Mg$. I'm assuming that mass of bridge is uniformly distributed and each pillar is bearing an equal amount of the load.
Now the question is that since the pillars are bearing the force, so if we make strong enough pillars to bear a large force then there will be no need of so many pillars.
But that is not the case, we see a large number of pillars holding a bridge. What is wrong with the work I did? Shouldn't the number of pillars depend upon the strength of the pillars we make rather than the length of the bridge ??
I shall be thankful if you can provide more information about this topic.
| In addition to the other answers, I think it's important to understand that you can't just make an arbitrarily strong pillar. Or rather, you can't make the ground the pillar stands on arbitrarily strong. So, depending on the geology that the bridge is standing on, you may need more or less pillars.
Of course you can get around this by making the pillars very big to spread the load, or sinking them very deep into the ground, but it may well be cheaper and more practical to have lots of pillars. Remember that this is engineering: everything is about complicated tradeoffs involving money, appearance, function & safety.
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If an electron is in ground state, why can't it lose any more energy? As far as I know, an electron can't go below what is known as the ground state, which has an energy of -13.6 eV, but why can't it lose any more energy? is there a deeper explanation or is this supposed to be accepted the way it is?
| You might as well have asked: "Why can't the string on my guitar produce a lower sound?" The short answer is: the lowest sound depends on the quality of the string, on the tension of the string and on the restricted length of the string. So it's just a matter of configuration! It's the same with an electron in a hydrogen atom. The mass of the electon, the Coulomb-force between it and the nucleus, the distance between these two and the restricted space determine it's lowest energylevel. There is nothing deeper than that. It's because of the way (we think) the atom is built. And if you want to know what the exact energy is for that level, then you will need Bohr's model of the atom or quantummechanics to calculate it.
| {
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What does this symbol $\odot$ mean? While reading through a physics textbook, I came across the use of sub-scripted ☉s.
Here's the context:
Stars between 0.5M☉ and 10M☉ will evolve into red giants...
I'm assuming it's to do with the life-span of a star; however, I don't know exactly how.
I searched google/wikipedia however it simply stated it represented the sun
| The symbol in question, $\odot$, usually denotes the Sun. The solar mass, $M_\odot$, is often used as a unit of mass in astronomical/astrophysical texts. Another example is the solar luminosity, $L_\odot$.
| {
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Phonons and heat conduction What is the physical picture of heat conduction in a poor conductor? In particular, I'm curious about the role of phonons in conduction in poor conductors. I know that phonons (within the harmonic approximation) move without being scattered and would lead to infinite thermal conductivity. This problem is resolved by including anharmonic terms in the Hamiltonian so that there are phonon-phonon scatterings.
*
*But how do the phonon-phonon scatterings reduce the thermal conductivity? I wish to understand this both physically and mathematically. The expression for conductivity can depend upon various quantities and scatterings must be affecting one of those.
*How does this phonon picture explain the fact that when we heat a poor conductor the heat propagates gradually from the hotter to the cooler end? If they are delocalized collective excitations, shouldn't they heat up all parts of the substance at the same time?
I don't have a condensed matter background and therefore, a detailed but not-too-technical answer will be helpful.
| Thermal conductivity is defined as a ratio of energy flux and temperature gradient (up to a factor). If phonons move freely, arbitrary energy flux can exist without temperature gradient (so the finite phonon velocity does not make the thermal conductivity final). See details on the role of scattering, say, at http://www.physics.iisc.ernet.in/~aveek_bid/PH208/Lecture%208%20phonons-thermal%20properties.pdf
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Small nucleus emission from a larger nucleus Like alpha decay, is there the possibility of a small (n,z) nucleus coming out of a large (N,Z) nucleus? Why lithium and beryllium don't decay out of big nuclei as helium does ?
| The decays you are considering are determined completely by the energetics. The alpha decay occurs so readily because the alpha nucleus is so tightly bound. Lithium and Beryllium are much less tightly bound so the energetics are less favorable. You can explore the possibility of such decays by looking at the masses of the nuclei involved. If the mass of the initial nucleus exceeds the sum of the masses of the product nuclei, the decay is possible. The extra energy goes into the kinetic energy of the products.
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What's happening when a force is applied but there's no displacement? I've thought of an example that can explain my doubt properly.
Let's take an electric motor which is connected with an object through a rope so that when the motor starts to rotate, it brings the object towards itself. However, the mass of the object is such that the motor can't move the object.
So, the moment I switch on the motor, Electric energy is given to it but does not produce any displacement of the object, and therefore, it doesn't do any work. Since energy is given to the motor, in what form is it transformed? What's happening inside the motor considering that it's not doing work?
I hope I explained properly the problem and I'm grateful to whoever will be able to clear up my doubt!
| I do apologize for not simply commenting, as I am not able to do so, but I found this answer to a different question that explains why a motor releases heat energy when held stationary. Again, apologise for not really giving any original answer, but thought this would help.
| {
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Partition function for classical indistinguishable particles and Bose particles We have two particles that can be in either level $E_0 = 0$ or in level $E_1$.
If we treat them as Bose particles, then the partition function will be:
$$ Z = 1 + e^{-\beta E_1} + e^{-2\beta E_1}, $$
whereas if we treat them as classical indistinguishable particles we'd get:
$$ Z = \frac{(1+e^{-\beta E_1})^2}{2!} = \frac{1}{2} + e^{-\beta E_1} + \frac{e^{-2\beta E_1}}{2}. $$
Why the discrepancy?
| In quantum statistics there there is never any over-counting. In this system containing two particles (= a quantum many-body system) the only three possible many-body states are
$$\vert \psi_A \rangle = \vert E_0 \rangle \otimes \vert E_0 \rangle \qquad \text{with energy} \quad E = 0$$ (with which I denote particle one being in state $E_0$ and particle two also being in state $E_0$) or $$\vert \psi_B \rangle = \frac{1}{\sqrt{2}}\left(\vert E_0 \rangle \otimes \vert E_1 \rangle + \vert E_1 \rangle \otimes \vert E_0 \rangle\right) \quad \text{(bosonic symmetry) with energy} \quad E = E_1$$ or $$\vert \psi_C \rangle = \vert E_1 \rangle \otimes \vert E_1 \rangle \qquad \text{with energy} \quad E = 2E_1$$
(The notion of particles in a quantum many-body system is dubious.) Hence the correct solution is $$ Z = 1 + e^{-\beta E_1} + e^{-2\beta E_1}.$$
Alternatively, you could think about having two isolated quantum systems (far away from each other), in which case they are in effect distinguishable (because they are far away from each other). In this case the partition function is $$Z = \left(1 + e^{-\beta E_1}\right)^2 = 1 + 2e^{-\beta E_1} + e^{-2\beta E_1}.$$
However, in classical statistical mechanics this question is a tricky one. As far as I know, we always decide to 'forget' all information of individual particles and just treat them as an ensemble with a given temperature, pressure, etc (macroscopic observables). Writing down a partition function and doing classical statistical mechanics with distinguishable particles doesn't really make sense - hence the discrepancy.
The Gibbs factor of $N!$ is just a way of recovering the correct answers in classical thermodynamics, it does not arise from considerations of real distinguishability.
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Strange quark decay into two down quarks and an anti-down quark I saw that a $\Sigma^+$ can decay into $n+\pi^+$, which means that the $s$-quark must decay into $dd\bar{d}$. However, is there a Feynman diagram to represent this? I cannot find one for either the $\Sigma^+$ decay or the $s$-quark decay. I have only just started learning so I am not sure. Thank you.
| This is a leading order Feynman diagram (which may be referred to as a 'gluonic penguin') for a $s \to dd\bar{d}$ transition. It also holds for any other kind of $q \to q'q'\bar{q'}$ flavour-changing neutral current transition (i.e. when $q$ and $q'$ have the same charge). FCNC decays are pretty interesting because the leading order diagrams contain a loop, which makes them more sensitive to contributions from BSM particles entering at loop-level.
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Confusion between two different definitions of work? I'm doing physics at high school for the first time this year. My teacher asked us this question: if a box is slowly raised from the ground to 1m, how much work was done? (the system is only the box)
Using the standard definition, $W = Fd\cos(\theta)$, the work should be 0, because the sum of the forces, the force due to gravity and the force of the person, is 0.
However, using the other definition he gave us, $W = \Delta E$, work is nonzero. $\Delta E = E_f - E_i$ , so that would be the box's gravitational potential energy minus zero.
My teacher might have figured it out but class ended. Does anyone have any insight?
| F is not the sum of the forces on the block, it is the force which is doing the work. It is either the force provided by the person (if you want to find the work done on the block by the person) or the force of gravity (if you want to find the work done on the block by gravity). You choose.
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Entanglement and wormholes: Are they the same? Some "recent" studies (e.g., http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.111.211603) popularized in the news (e.g., http://news.mit.edu/2013/you-cant-get-entangled-without-a-wormhole-1205 or https://www.ias.edu/ideas/2013/maldacena-entanglement) argue that quantum entanglement results from the creation of a wormhole. I am not a quantum whisperer so I was not able to determine the validity of these reports.
Is there any merit to this idea or is this a cute mathematical idea without any testable hypotheses? It is a cute idea and could get around some pesky issues with our understanding of entanglement but are there any serious flaws/holes in this idea (pun was not intended but can be for humor's sake)? To be clear, I am not looking for a philosophical discussion of this idea. I am curious if the arguments hold water.
If they do, then does that mean we can estimate wormhole lifetimes based upon the distribution of entangled vs. non-entangled particles after time $\tau$?
| Non-locality of entanglement is not experimentally yet fully proven beyond doubts/loopholes. It must have been proven mathematically though. Non-locality of Entanglement correlations has not been sufficiently scrutinized yet. All efforts are geared towards proving non-locality via Bell's inequality. Bell's inequality is not sufficient to prove non-locality. All it can prove is that static hidden variables (within entangled pair) are not capable of producing different kind of correlations.
Taking example of perfect anti correlated Bell's state, the correlations are -
1) 100 % anti correlation
2) 50/50 outcome when measured a particle in any one angle over and over.
3) Same spin (both up or both down) SQ(sin(A/2)) times, when measured at relative angle of A over and over.
So comparison (even if the worm hole do exist), can not give any real outcome.
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Free and bound current- and charge density in Maxwell equations The first and fourth Maxwell equations are often denotet in vaccum:
$$
\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}
$$
$$
\nabla \times \mathbf{B} = \mu_0\left( \mathbf{j}+\epsilon_0 \frac{\partial{\mathbf{E}}}{\partial{t}}\right)
$$
and in matter:
$$
\nabla \cdot \mathbf{D} = \rho_f
$$
$$
\nabla \times \mathbf{H} = \mathbf{j}_f +\frac{\partial{\mathbf{D}}}{\partial{t}}
$$
It is also often written, that the whole charge density $\rho = \rho_f+\rho_b$ is the sum of free and bound one. Similiarly $\mathbf{j} = \mathbf{j}_f+\mathbf{j}_b$, meaning the whole current density is the sum of the free and the bound one.
I ask then:
Why do the equations look like they do?
In vaccum there are no bound quantities, so using the "whole" quantities seems counterintuitiv. At the same time, I would expect bound quantities considered in matter. Is it just a problem of denotation? Something feels terribly off here.
| The electric field $E$ is the field we apply, what we express with the first Maxwell equation is that its sources must come from the total density charge $\rho$. In a material, there will be some fixed charges, so the presence of $E$ will induce some dipoles, and this will make Polarization $P$ appear. Then polarization is related with the bound charge density $\rho_b$, while the rest of charges that are free to move we associate it with free density charge $\rho_f$ In vacuum, there is no bound density charge, so we have $\rho = \rho_f$.
In a presence of a material, we define displacement field $D$, or response field as
$$
D= \epsilon_0 E + P
$$
We normally consider the ideal situation for linear, homogeneous and isotropic material such that $P=\chi \epsilon_0 E$, where $\chi$ is electric susceptibility. In that way we could write $D=\epsilon_r \epsilon_0 E$, where $\epsilon_r = 1+ \chi$ is relative electric permitivity. $D$ is then taking into account the presence of free charges and bounded charges, althoug its sources are only the free charges. This can be seen easily, because the contribution of the Polarization is actually negative. It can be shown that $\nabla \cdot P =-\rho_b$, and as we know $\nabla \cdot E = \frac{\rho}{\epsilon_0}$. So if we take the divergence of the $D$ defined above and we use this results we have:
$$
\nabla \cdot D = \epsilon_0 \nabla \cdot E + \nabla \cdot P = \rho - \rho_b =\rho_f
$$
since by definition $\rho = \rho_f + \rho_b$
| {
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"url": "https://physics.stackexchange.com/questions/295588",
"timestamp": "2023-03-29T00:00:00",
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Mock equation of state I am student of astrophysic i need to meaning of "mock equation of state".
And is there any specific way to generate mock equation of state.
| I guess that what is meant is a relationship between the pressure of the gas and it's density, temperature and composition.
No, there is no specific way to make something that you have defined in a completely general way, without describing any context - is this a star, a compact object, the interior of a star forming cloud, an ionised nebula?
A mock equation of state for a neutron star might be
$$P = A\rho^{\alpha},$$
where $P$ is pressure, $\rho$ is density and $A$ and $\alpha$ are constants of your choosing to reflect you chosen physics. You might even have several different $A,\rho$ value in different density intervals to cope with changes in the equation of state and composition that are made such that $P$ vs $\rho$ forms a continuous relationship.
For example, an ideal degenerate gas of electrons (accompanied by ions, say in a white dwarf) obeys $P \propto \rho^{5/3}$ at low densities, but become $P \propto \rho^{4/3}$ at higher densites. Then when neutronisation takes place the gas pressure again becomes dominated by non-relativistic neutrons with $P \propto \rho^{5/3}$, but if a neutron gas become relativistic then $P \propto \rho$. Alternatively you might adjust the value of $\alpha$ to reflect a (non-ideal) hardening of the equation of state due to nucleon repulsion at high densities by allowing $\alpha \rightarrow 2$.
| {
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What is the origin for the name "partition function" given to $Z = \sum e^{-\beta E_i}$? Does anyone know why is the function $Z = \sum e^{-\beta E_i}$ called "partition function"?
For example, does it have a connection to the mathematical term "partition of $A$" which is a representation of the set $A$ as a disjoint union of it's subsets (and defines an equivalence relation over $A$)?
EDIT: The explanation below and the explanation here indeed almost give me a full answer. I just want to be sure:
We are divding the whole quantity be it's energy states and not by it's particles. This means that a class in a partition can have lots of particles and can be related to one energy state exactly. And we have to know the distribution function of the particles in the system.
Am I right?
or mayby every particle has it's own distribution?
| It is appropriate to call $Z$,
$$Z=\sum_{i \in \, \mathrm{states}}\exp \left( -\beta E_i\right)$$
the partition function as it describes how probabilities are distributed amongst all the states with energies $E_0, E_1$, and so forth. To see this, note that the expected value of a property $Q$ is,
$$\langle Q \rangle = \frac{1}{Z}\sum_{i \in \, \mathrm{states}}Q_i\exp \left( -\beta E_i\right).$$
This is analogous to the fact that in probability, for a discrete variable $X$ which can take values $\{x_i\}$ the expected value is,
$$\langle X \rangle = \sum_{i} x_i p_i$$
for a normalised distribution $\sum_i p_i = 1$, with probabilities $p_i$ for each value $x_i$. Thus the partition function provides an appropriate weight for each state.
It is denoted by $Z$ after the German word, Zustandssumme which roughly translates to a sum over states, which is what we are instructed to do,$\sum_{i \in \, \mathrm{states}}$, to obtain it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/296156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can one derive Schwarzian derivative action as low energy effective field theory invariant under global $SL(2,\mathbb{R})$? In a recent paper (page 47, below eq (4.173)) they make a passing claim that the Schwarzian derivative action can be derived using effective low energy field theory reasoning. I imagine they mean that if I want to construct a least derivative action which is invariant under global $SL(2,\mathbb{R})$ transformations of the coordinates, then I will end up with Schwarzian derivative. I was wondering if this has been worked out anywhere. Also, using the same approach, what are the higher derivative invariants that I can possibly construct as 'less relevant' terms.
| The answer to this question is rather simple. Let us assume that our field is $f(x)$ and we want to find an effective low energy action invariant under the $SL(2,\mathbb R)$ transformations,$$f(x)\to\frac{a\, f(x)+b}{c\, f(x)+d}~,\tag{1}$$with $a,b,c,d\in\mathbb R$ and $ad-bc=1$. Let us look at these transformations one by one:
*
*Translations: $f(x)\to f(x)+b$ imply that effective low energy action can only be a function of derivatives of the $f(x)$ field, $f'(x),f''(x),f^{(3)}(x),\ldots$.
*Scaling: $f(x)\to a\, f(x)$ implies that the effective action is a function of ratios of the fields, $\dfrac{f^{(n)}(x)}{f^{(m)}(x)}$.
Since we are working with effective action we want an action with the least number of derivatives. Since $\dfrac{f''(x)}{f'(x)}$ is a total derivative our lowest derivative candidates are $\dfrac{f'''(x)}{f'(x)}$ and $\left(\dfrac{f''(x)}{f'(x)}\right)^2$. Let's assume our action is,
$$\dfrac{f'''(x)}{f'(x)}-\mu \left(\dfrac{f''(x)}{f'(x)}\right)^2~.\tag{2}$$
*Finally imposing invariance of the action under inversion: $f(x)\to\dfrac1{f(x)}$ one can easily fix the value of $\mu=\frac32$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/296382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Osmotic Pressure: Comparing Solutes and Ideal Gases The equation describing osmotic pressure is
$\Pi=\frac{n}{V} RT $
which is just like the ideal gas equation law
$PV=nRT$
So how much of an analogy is there between ideal gases and solutes? Is there a extended version of the equation for osmotic pressure that matches the van der Waals equation of state?
If solutes can be in a state like a gas, can they also be in a state like a liquid? Is this distinct from a precipitate which would be the analog of a solid?
| The van t'Hoff equation for the osmotic pressure can be expanded by a virial expansion similarly to the extension of ideal gas equation. In the gas case, the coefficients of the virial expansion are related to the van der Waals constants a and b. Thus you can probably also obtain a similar equation of state for the osmotic pressure $\Pi$ of a solute. However, their interpretation will probably be different.
| {
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Why is bench pressing your bodyweight harder than doing a pushup? Why does bench pressing your own bodyweight feel so much harder than doing a push-up?
I have my own theories about the weight being distributed over multiple points (like in a push-up) but would just like to get a definite answer.
| When you're doing press-ups, your arms have a mechanical advantage over the mass of your body. The force that they exert does not pass through your COM and, instead, exerts a torque on your mass with your feet being the fulcrum of the lever.
When you benchpress, your arms are exerting a force that passes straight through the mass of the weights. As such, there is no torque and no mechanical advantage.
Put another way, with each exercise, your arms move the same distance x. With a benchpress, the weights move this full distance, but with pushups, your COM does not move the full distance; thus, the work done during pushups is lower.
| {
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Electric field associated with moving charge I have recently started to learn about the electric field generated by a moving charge. I know that the electric field has two components; a velocity term and an acceeleration term. The following image is of the electric field generated by a charge that was moving at a constant velocity, and then suddenly stopped at x=0:
I don't understand what exactly is going on here. In other words, what is happening really close to the charge, in the region before the transition, and after the transition. How does this image relate to the velocity and acceleration compnents of the electric field?
| When the charge is moving at constant velocity, it emits a non-uniform electric field that can be calculated from the Lienard-Wiechert potentials. This field has a velocity component but no acceleration component, as the charge is not accelerating.
When the charge is not moving, it emits a spherically symmetric electric field that can be calculated from Coulomb's Law. The field, once again, has a velocity component and no acceleration component, as the charge is not accelerating (and the velocity component reduces to the Coulomb field when the velocity is zero).
In order to transition from moving to not moving, the charge must accelerate. Here, again, the charge's fields may be calculated from the Lienard-Wiechert potentials, but now there is a nonzero acceleration component to the field, which corresponds to radiation. The shape of this field can be reasonably approximated, for short accelerations, by requiring that the electric field lines be continuous through the transition, and this approximation appears to be used in your diagram.
When the charge suddenly stops, its field does not change instantaneously across all space. Rather, the change from non-uniform velocity field to Coulomb field propagates outward at the speed of light. If the charge suddenly stopped at $x=0,t=0$, and we examined the field at time $t$, we would find that the Coulomb field was present in the region $r < ct $, while the non-uniform velocity field was present in the region $r>ct $. This is exactly what your diagram depicts.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Using a Ring Imaging Cherenkov (RICH) detector I'm not clear on how RICH (Ring Imaging Cherenkov) detectors identify particles, as much as I've tried to read up on it. So they measure the angle Cherenkov light is emitted at, which is related to the refractive index of the material, speed of light and the speed of the particle. How can that be used for particle ID? And why are these detectors particularly good at differentiating between pions and kaons?
I realise this seems like a question that should be easily solved by googling, but I'm not having any luck there!
| A RICH alone doesn't ID a particle. The only thing it measures is velocity.
But by combining that information with either momentum (from a spectrometer) or energy (from a calorimeter) you can often deduce the species or at least constrain it to some range of possibilities.
As for why they are used to separate pions and kaons (I assuming that you have a particular detector system in mind for this, probably something at the LHC), these particles are both (relatively) long-lived, light hadrons. They are hard to reliably distinguish from one another based on decay products or shower parameters in a calorimeter, but their masses are sufficiently different that there will often be a measurable difference of velocity at interesting momenta.
In other contexts it may be more difficulty to distinguish pions from electrons and Cerenkov devices can be tuned for that as well.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/297037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Lorentz transformation, problem with derivation I have a question about the Lorentz transformation:
In the derivation it's said that two systems S and S' should at $t=0$ and $x=0$, overlap. We get the following transformation rules:
$t'=\gamma_0(t-v_0x/c^2)$
$x'=\gamma_0(x-v_0t)$
$y'=y$
$z'=z$
My question is: What happens if at $t=0$ they dont overlap? Can I just add a constant to both the time and coordinate $x$ transformation equation, to account for the misalignment at $t=0 $?
$t'=\gamma_0(t-v_0x/c^2) \color{Red}{+ T}$
$x'=\gamma_0(x-v_0t) \color{Red}{+ X}$
| There are three types of transformations that preserve the spacetime interval.
*
*Boosts: these are transformations like the one you gave. They transform between systems moving relative to each other, which have the same origin. If we write spacetime coordinates as column vectors like
$$ X = \left( \begin{matrix}
t \\
x \\
y \\
z \end{matrix} \right), $$
then Lorentz boosts are of the form $X \rightarrow X' = \Lambda X$ where $\Lambda$ is a $4\times4$ matrix obeying $\eta = \Lambda^T \eta \Lambda$ where $\eta$ is the metric. For example, the Lorentz boost in your question is given by the matrix
$$ \Lambda = \left( \begin{matrix}
\gamma & -\gamma v_0 & 0 & 0 \\
-\gamma v_0 & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{matrix} \right). $$
(Here we have set $c=1$.)
*Rotations: these are transformations between systems that are oriented differently, which also have overlapping origins but no relative motion. There is no difference between rotations in special relativity and rotations in ordinary mechanics. They are of the form $X' = RX$ where
$$ R = \left( \begin{matrix}
1 & 0 \\
0 & O \\
\end{matrix} \right) $$
and $O$ is a $3\times3$ matrix satisfying $O^TO = I$ and $\det(O) = 1.$ $O$ is just a normal spatial rotation matrix. Notice that $R$ only acts on the spatial coordinates and leaves $t$ unchanged.
*Translations: these are transformations between systems with different origins, but no relative motion or differing orientation. They are of the form $X' = X + C$ where $C$ is just a constant column vector.
The most general type of transformation that preserves the spacetime interval can combine boosts, rotations, and translations. For example, if you want to transform between two systems with different origins that are moving with relative velocity, then you just combine a boost with a translation:
$$ X \rightarrow X' = \Lambda X + C. $$
This is indeed exactly what you conjectured, and your final equations are correct.
As for terminology, the first two transformations (boosts and rotations) are called Lorentz transformations. The set of all Lorentz transformations is called the Lorentz group and is denoted by $SO(1,3)$. The set containing both Lorentz transformations and translations is called the Poincare group.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/297123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why faraday rotation is not effective in case of circular polariation? As in case of linear polarization the plane of polarization gets rotated. Why it's not happening with circular polarization?
| The Faraday effect is the rotation of the polarization plane (proportional to the length $l$ and magnetic field $B$ of linearly polarized light when propagating along a magnetic field $B$ applied to a (transparent) material. Linearly polarized light can be decomposed into left and right circularly polarized light. The rotation of the linear polarization direction is due to a difference in the refractive index for right and left circularly polarized light induced by the magnetic field leading to a phase difference proportional to $l$ and $B$, which explains the rotation of the linear polarization. Thus the influence of $B$ on the phase velocity of left and right circularly polarized light is essential to explain the Faraday effect .
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/297230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to distinguish "system" and "environment" in quantum decoherence Quantum decoherence distinguishes the whole big system into "system" and environment, and shows how system, when density matrix is traced over environment, comes to be decoupled from environment.
But this requires distinguishing environment from system, and I do not get how clear separation is possible. Doesn't the fact the whole big system is quantum should bring caution to separating systems arbitrarily, especially considering special relativity effects?
| The fact that the whole system is a quantum system is not relevant to this issue and neither does special relativity prohibits this separation as it is not physical.
The separation is made on the basis of what is the system that is the subject of an experiment. The environment is then everything else, including you.
Decoherence then comes from the practical impossibility of totally decoupling the system from this environment. As such the environment interacts with the system potentially causing it to collapse in an eigenstate of this interaction. We say this interaction causes collapse (or however you prefer to interpret measurement) when the information about the state is widely repeated throughout the environment, thus being classical as measuring the environment is now equivalent to having measured the state in the same basis. This is einselection.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/297329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sensation of atmospheric pressure Pressure is force divided by area, and force is mass times acceleration.
Now, the newton [N] is the force needed to accelerate 1 kg by 1 m/s, and the kilogram-force [kgf] is the force needed to accelerate 1 kg by g m/s, where g is the standard gravity.
The standard atmospheric pressure is set at 101325 Pa, which would translate to ~10333 kgf/m^2.
So why don't we all implode? What is missing from the picture?
| Solids and liquids are what we call incompressable. They do not change their volume a great deal when a pressure is applied to them. A comparable metaphor for you is that you can park the wheel of a car on top of a phone book. That phone book is under thousands of pounds of force, but it doesn't implode because solid materials tend to not change volume when the force is applied.
Meanwhile, the gasses in our body are all at roughly atmospheric pressure, so they already press outwards with a pressure equal to the pressure going in. However, if you rapidly change pressures (such as a rapid decompression in the event of an airline cabin losing pressurization), you very quickly feel just how much force was being applied by all of those gasses for a few moments while they all equalize.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/297498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is the Earth gaining or losing mass over time? The earth presumably loses mass because molecules of the atmosphere disassociate and fly off into space where the solar wind carries them away.
On the other hand the earth gains mass because particles of dust and meteorites strike the earth and accumulate onto it.
Which force is dominant? Does the earth gain mass over time or lose mass?
| We really don't know the figures of mass loss, or mass gain through cosmic dust accumulation.
Mass loss through Hydrogen and Helium
According to some calculations, the Earth is losing 50,000 tonnes of mass every single year, even though an extra 40,000 tonnes of space dust converge onto the Earth’s gravity well, it’s still losing weight.
If you take the lower end of the mass accumulation estimation from cosmic dust, that is 5 metric tonnes daily, this results in a figure of 1825 metric tonnes a year, resulting (if you take 50,000 tonnes a year of $\mathrm H_2$ and $\mathrm{He}$ as accurate), in a definite answer that the Earth is losing mass.
A break even figure for mass equilibrium is 137 tonnes of comic dust daily, which is almost midway between the (widely differing) estimates of 5 to 300 tonnes of cosmic dust thought to fall on Earth daily.
Cosmic Dust Estimates
Even though we consider space to be empty, if all the material (cosmic dust mainly) between the Sun and Jupiter were compressed together it would form a moon 25 km across.
Satellite observations suggest that 100-300 metric tons of cosmic dust enter the atmosphere each day. This figure comes from the rate of accumulation in polar ice cores and deep-sea sediments of rare elements linked to cosmic dust, such as iridium and osmium.
But other measurements – which includes meteor radar observations, laser observations and measurements by high altitude aircraft — indicate that the input could be as low as 5 metric ton per day.
The answer is we really don't know, but as we seem to have more accurate figures on mass loss than mass gain, it seems we can have more confidence that Earth has a net mass loss.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/297622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Is the Photovoltaic Effect observed only in P-N junctions? So far, all the sources I've found explain the photovoltaic effect in relation to p-n junctions when talking about solar cells (since that's what they're made out of), but I was wondering if one could explain the photovoltaic effect in general without having to explain what p-n junctions are, as if it is only observed in them.
| Another example of a system that exhibits photovoltaic action is a metal-semiconductor junction, or a Schottky junction.
In general photovoltaic action will occur at an interface involving a semiconductor. At this interface, say the interface between an n and p doped semiconudctor, there is an inherent asymmetry that causes electrons to want to go one way and holes to go the other.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/297706",
"timestamp": "2023-03-29T00:00:00",
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