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Are wormholes evidence for traversal of a higher dimension? Warning, pop science coming.. please correct what I’m getting wrong. Einstein’s equations of relativity showed the potential for existence of wormholes that can connect different points in space time. I understand the mechanisms for their practical implementation are nothing near feasible. However, based on the equations of gravitational “tunneling”, I can traverse back and forth between times and locations. Wouldn’t this require a higher dimension than 4d space time?
That is, we’re moving from a point that we would think of as the present to another point we would think of as the present. If this were feasible, Would These “presents” need to be on a traversable continuum?
To my lay brain, This seems as though there are points along a higher dimension where what we would consider the future is currently present, and what we consider the past is also present. That the world we see is determined and laid out as slices in a higher dimension that would be traversed with a wormhole, and that we normally traverse in a single direction.
| I guess so. At least according to the illustrations/analogies of folding paper. However there is nothing in Einstein's equations that require an existence of a higher dimension unlike in string theory. But if wormholes are proven to exist, then yes this could prove the possibility of higher dimensions since there is no other way for wormholes to work.
| {
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Why is pressure different at different points in the same level when the water is flowing? In fig.a below, water is to flow out from the yellow tank. But the flow is stopped because of the mercury in the green manometer. So the water is stationary. In this situation, the pressure at both the points A and B will be the same, which is ${\rho gh}$. Where ${\rho}$ is the density of water.
In fig.b, the mercury is removed from the manometer. So water flows out. In this situation, pressure at A is not equal to pressure at B. Even though A and B are at the same level. Can we give a simple explanation for such a pressure difference? I saw the basics of Bernoulli’s equation. But it does not give the reason. Thanks.
| The Bernoulli equation (which ignores the effect of friction) expresses conservation of energy density. Changes in pressure are included to account for the work done by pressure between the two points under consideration. If the cross section decreases, the velocity must increase. The pressure does work in increasing the kinetic energy (and can exert less force on water moving away). (By the way, the decreased pressure on the right side of the U-tube will not support a water column of the same height when the water is flowing.)
| {
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Permissible Electrostatic Potential Let us consider a $1D$ real function $V(x)$. When is this a classical electrostatic potential?
My take on the problem:
*
*$V(x)$ must be differentiable everywhere. In fact, we should be able to differentiate it $n$ times.
*$V(x)$ should vanish at $\pm \infty$.
I think these are necessary and sufficient conditions. Is this right? How do I deal with discrete charge distributions, where the potential is not differentiable at the points where the charges are present?
| I think that most conditions can be seen from the Poisson equation:
$$
\frac{d^2 V(x)}{dx^2} = -\rho(x).
$$
Thus, it should be differentiable everywhere except a few singular points. One often bypasses this issue by using generalized functions (delta-function and Heaviside step function).
However, there is no requirement that the potential vanishes in infinity. For example, the potential corresponding to a constant electric field of magnitude $E$ is
$$
V(x) = - Ex
$$
| {
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Generalized Pauli matrices I wanna know the generalized form of Pauli matrices, for example for $3\times 3$. And do they satisfy all of the properties of Pauli $2\times 2$ matrices?
I wrote $3\times 3$ but I couldn’t write all Hermitian $3\times 3$ matrices with those.
| As @Charlie asks, the Pauli matrices have several properties, which generalize in different manners. They certainly, together with the identity, provide a complete basis for 2×2 matrices, but they are also hermitian. If Hermiticity is important to you, you generalize them as in the link provided, that is along the Gell-Mann matrices' route for 3×3 matrices.
However, a far more tasteful and systematic basis is J J Sylvester's 1882 one of clock and shift matrices for d×d unitary matrices which you should know about, anyway. They are not hermitean in general, but they are more systematic (some would say "analytic in d").
For $ω= \exp(2iπ/d)$, a root of unity not equal to 1. The sum of all roots annuls, $1 + \omega + \cdots + \omega ^{d-1} = 0 $, so integer indices may be cyclically identified mod d.
The shift matrix is defined as
$$
\Sigma _1 =
\begin{bmatrix}
0 & 0 & 0 & \cdots & 0 & 1\\
1 & 0 & 0 & \cdots & 0 & 0\\
0 & 1 & 0 & \cdots & 0 & 0\\
0 & 0 & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots &\vdots &\vdots\\
0 & 0 & 0 & \cdots & 1 & 0\\
\end{bmatrix}
$$
and the clock matrix as
$$
\Sigma _3 =
\begin{bmatrix}
1 & 0 & 0 & \cdots & 0\\
0 & \omega & 0 & \cdots & 0\\
0 & 0 & \omega^2 & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & \omega^{d-1}
\end{bmatrix}.
$$
These matrices generalize $σ_1$ and the diagonal $σ_3$, respectively.
Since Pauli matrices describe quaternions, Sylvester dubbed the higher-dimensional analogs "nonions", "sedenions", etc.
These two matrices are also the cornerstone of Weyl's celebrated quantum mechanical dynamics in finite-dimensional vector spaces The clock matrix amounts to the exponential of position in a "clock" of d hours, and the shift matrix is just the translation operator in that cyclic vector space, so the exponential of the momentum. They are (finite-dimensional) representations of the corresponding elements of the Heisenberg group on a d -dimensional Hilbert space.
The following relations echo and generalize those of the Pauli matrices:
$\Sigma_1^d = \Sigma_3^d = I$,
and the braiding relation,
$\Sigma_3 \Sigma_1 = \omega \Sigma_1 \Sigma_3 = e^{2\pi i / d} \Sigma_1 \Sigma_3$, and can be rewritten as
$\Sigma_3 \Sigma_1 \Sigma_3^{d-1} \Sigma_1^{d-1} = \omega ~$.
The complete family of $d^2$ unitary (but non-Hermitian) independent matrices
$$
\left(\Sigma_1\right)^k \left(\Sigma_3\right)^j = \sum_{m=0}^{d-1} |m+k\rangle \omega^{jm} \langle m|,
$$
then provides Sylvester's well-known trace-orthogonal basis for $\mathfrak{gl} (d,ℂ)$, known as "nonions" $\mathfrak{gl} (3,ℂ)$, "sedenions" $\mathfrak{gl} (4,ℂ)$, etc... Since all indices are defined cyclically mod d, $\mathrm{tr}\Sigma_1^j \Sigma_3^k \Sigma_1^m \Sigma_3^n = \omega^{km} d ~\delta_{j+m,0} \delta_{k+n,0}$
| {
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Is $U^\dagger(R)\hat{H}U(R)=\hat{H}$ always true? Consider a Rotation transformation on momentum state,
$$U^\dagger(R)\hat{\mathbf{p}}U(R)=R\hat{\mathbf{p}}$$
Now the question is whether,
$$U^\dagger(R)\hat{H}U(R)=\hat{H}\,?$$
Here, $\hat{H}$ is the Hamiltonian of a free particle. Is it always true? Is there any counter examples?
My attempt:
\begin{align}
U^\dagger(R)\hat{H}U(R)&=\frac{1}{2m}U^\dagger(R)\hat{\mathbf{p}}^2U(R)\\
&=\frac{1}{2m}U^\dagger(R)\hat{\mathbf{p}}U(R)U^\dagger(R)\hat{\mathbf{p}}U(R)\\
&=\frac{1}{2m}(R\hat{\mathbf{p}})(R\hat{\mathbf{p}})
\end{align}
Is this always true that $$\frac{1}{2m}(R\hat{\mathbf{p}})(R\hat{\mathbf{p}})=\frac{1}{2m}\hat{\mathbf{p}}^2\, ?$$
If it is why? If not when it is not?
Note: This is an exercise from Coleman's course 253a (https://arxiv.org/abs/1110.5013). See equation (1.8) there. It would be better if the answer is provided using his notations.
| There is a dot product in fact $\mathbf p^2 = \mathbf p^{\dagger} \mathbf p$. So one may do the following:
$$
U^{\dagger} \mathbf p^{\dagger} \mathbf p \ U =
U^{\dagger} \mathbf p^{\dagger} U U^{\dagger} \mathbf p \ U = (U^{\dagger} \mathbf p \ U)^{\dagger} U^{\dagger} \mathbf p \ U = (R \mathbf p)^{\dagger} R \mathbf p = \mathbf p R^{\dagger} R \mathbf p = \mathbf p^{\dagger} \mathbf p
$$
Here I've denoted the transposed momentum $p^{\dagger}$ to unify notation, despite the fact, it is real, And in the last step, I've used, that the $R$-matrix is unitary.
| {
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Curl of electric field is not zero in the case of a steady current in a loop Say we got conducting circular loop connected to a battery . The electric field inside the loop obeys equation $\vec{J}=\sigma \vec{E}$.
Since the current flows in a circumferential way around the loop the electric field will be circumferential as well which implies that the curl of the electric field will be non zero.
Which is a contradiction ,what's wrong in the above reasoning.
Thank you
| Note that the curl of the electric field is not necessary to be zero in all cases. It's only valid for electrostatics.
Maxwell's equation here is valid through out the electrodynamics which given by :
$$\nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}$$
$$\nabla\times\mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$$
$$\nabla\cdot\mathbf{B}=0$$
$$\nabla\times\mathbf{B}=\mu_0\left(\mathbf{J}+\epsilon_0\frac{\partial \mathbf{E}}{\partial t}\right)$$
| {
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Interpretation of normal modes from the mathematical formula In the topic of small oscillations, the system below has a normal mode described by:
$$n_{1} = \frac{x1+x2}{2}.$$
This normal mode is represented as the symmetric mode:
In that case, the center of mass moves as a simple harmonic oscillator. However, the picture also shows that both of them start in the same initial conditions and move in phase. My question is where that information is on the normal coordinate $n_{1}$ since I can not relate the normal mode with the picture representing it. Where does it say the blocks must be strechted the same distance in the same direction in the formula of the normal coordinate?
| You should notice that the other normal coordinate is implied to be fixed at zero while you consider the motion along the normal coordinate $n_1$.
The normal coordinates of two particles (or blocks in this case) can generally be written as
\begin{align}
n_1 =& a_{11} x_1 + a_{12} x_2, \\
n_2 =& a_{21} x_1 + a_{22} x_2.\label{eq: n1n2}\tag{1}
\end{align}
In your specific case, $a_{11}=1/2$ and $a_{12} =1/2$. I didn't calculate $a_{21}$ and $a_{22}$, but you should be able to do so according to the definition of the normal modes.
The above set of equations can be solved for $x_1$ and $x_2$ in the form of
\begin{align}
x_1 =& b_{11} n_1 + b_{12} n_2, \\
x_2 =& b_{21} n_1 + b_{22} n_2, \label{eq: x1x2}\tag{2}
\end{align}
where $b_{ij}$ are determined by $a_{ij}$. In fact, by writing the above sets of equations by matrices and vectors, you can confirm that
\begin{equation}
\begin{pmatrix}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{pmatrix}
=
\begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{pmatrix}
^{-1},
\end{equation}
where $A^{-1}$ means the inverse matrix of a matrix $A$.
The first set of equations (\ref{eq: n1n2}) give the coordinate transformation from the coordinates $(x_1,x_2)$ that was convenient for your measurement and other operations to the normal coordinates $(n_1,n_2)$ that is convenient for calculation and some type of interpretation associated with the calculation. The second set (\ref{eq: x1x2}) gives the inverse transformation. That is, once you calculated the time-evolution $n_1(t)$ and $n_2(t)$ individually by solving the differential equations, you can predict the motion $x_1(t)$ and $x_2(t)$ of the respective particles by (\ref{eq: x1x2}).
The differential equations for the normal coordinates are
\begin{equation}
\frac{d^2 n_i}{dt^2} = -\omega_i^2 n_i(t),
\end{equation}
for $i=1,2$, where $\omega_i^2$ are constants, as long as the force acting on the block $l$ is of the form $F_l = - \sum_j c_{lj} x_j$ with some constants $c_{lj}$ as is the case for your problem. The function, $n_2(t) = 0$ for all $t$, is a valid solution for this equation for the initial condition, $n_2(0) =0$ and $[dn_2/dt](0) =0$. Suppose that this condition is satisfied through (\ref{eq: n1n2}) by particular values of $x_j(0)$ and $[dx_j/dt](0)$ ($j=1,2$) which are prepared by your putting hands on the spring-and-mass system at $t=0$. These
$x_j(0)$ and $[dx_j/dt](0)$ ($j=1,2$) also determine the initial values of $n_1(0)$ and $[dn_1/dt](0)$, and hence give a particular solution $n_1(t)$ of the differential equation above. With this $n_1(t)$ and $n_2(t) =0$, through (\ref{eq: x1x2}), the motion of the blocks is seen as
\begin{align}
x_1(t) =& b_{11} n_1(t), \\
x_2(t) =& b_{21} n_1(t).
\end{align}
If $b_{11} =b_{21}$, then $x_1(t) =x_2(t)$, i.e., the motion of the two blocks are the same. You should be able to see that actually $b_{11} = b_{21}$ for your system.
| {
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How we feel (perceive) exact size of object through our eyes? Light after getting reflected from objects gets focused on retina by our lens. The images formed on retina is small, which is then sensed by our brain and depending on distance we feel size of that object.
If an object is at particular distance from us, the image on retina is not going to be the exact size which we feel by our eyes after sensed by our brain. Our brain predicts size depending on the distance that brain sensed. But why the size we feel by our eyes is same as the size we feel by touch. Like if we see an object we can feel its boundary, and if we touch it and sense the boundary it does not extend or change.
Why the boundary of object we feel by eyes doesn't contradict reality?
| One eye perceives the angular size of an object. With two eyes we can get an estimate of the distance to the object. In combination these can let us estimate the actual size. Our estimate of size will often be dependent on the context in which the object is observed.
| {
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Heating cup in microwave? I heated my milk cup in the microwave today and noticed that the cup was hot but not the handle. Even if I heat it too much , cups handle temperature remains the same. How is that possible?
| Like Bob has mentioned, the container doesn’t absorb the microwave. The reason for this lies in quantum mechanics according to which the energy levels of atoms are quantised. This means the atom can’t have any arbitrary energy and can only absorb energy corresponding to the difference of the allowed energy levels (see figure below).
Microwave ovens radiate light having discrete energy chunks of $10^{-5}$ eV. This frequency is particularly chosen because water has energy level separation such that it strongly absorbs this light. And since most food we consume contains water, this is an efficient way of heating them provided that the container doesn’t absorb/reflect any of it. Glass and ceramics don’t have energy level separation corresponding to this frequency so they don’t absorb any.
So the only process by which it is getting heated up is by conduction through the water it is containing.
Even if I heat it too much , cups handle temperature remains the same.
Strictly speaking, this isn’t true. There will be some temperature difference in the two cases where you heat the water to different extents. However, depending on the material properties (thermal conductivity, shape) and the timescale at which the container is picked up, the temperature difference in the two scenarios might not be resolved by our hand as they are terrible at quantifying temperature.
So I recommend you to measure the temperature for different heating conditions and see if the temperature difference is resolvable by say an IR thermometer.
| {
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What causes pressure in a liquid? Say there is a mass inside a liquid. What causes the force on its upside downwards and the force on its downside upwards? It seems logical that the downward force is due to the weight of the fluid above it, is it similar with the upward force? Just collision of liquid particles?
| Consider a vessel of any shape.
At any 2 or more points is the vessel that are coplanar,
no matter what the shape of the vessel, the pressure will be equal.
Just collision of liquid particles?
If an object is immersed in the liquid, pressure will act on all directions on it from the liquid's particles.
The net pressure (the pressure acting on both sides of the object cancels each other) acting on the on an object at a depth $d$ in the liquid is given by $$P=d\rho g+P_0$$ where $P_0$ is the atmospheric pressure acting on the free surface and $\rho$ is the density of the liquid.
is it similar with the upward force?
There will be an upwards pressure from the liquid's particles due to the buoyant force too.
| {
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What conditions are necessary to guarantee uniform circular motion? Suppose an object is subjected to a force of constant magnitude, which is always directed to the origin. And suppose we know the initial position of the object relative to the origin, and the initial velocity of the object, can we determine if the object will perform uniform circular motion? If so, what conditions are necessary? Can we determine its position as a function of time from these givens?
I know that if we know that an object performs uniform circular motion, and we have the equations which describe its motion, for example $$ \mathbf r= \begin{bmatrix} \cos(t)\\ \sin(t)\\ \end{bmatrix} $$
we can find the velocity, and acceleration simply by taking derivitives. But can we go the other way around and deduce the equation of motion as I described above? Perhaps by solving the differential equation $$ m \ddot{\mathbf r} = - \lVert \mathbf F \rVert \frac{\mathbf{r}}{\lVert \mathbf r \rVert}$$
where $\lVert \mathbf F \rVert$ is constant?
|
But can we go the other way around and deduce the equation of motion as I described above? Perhaps by solving the differential equation
$$m \ddot{\mathbf r} = - \lVert \mathbf F \rVert \frac{\mathbf{r}}{\lVert \mathbf r \rVert}$$
where $||\mathbf F||$ is constant
This would not be the correct equation of motion.
You can just apply Newton's second law in Polar Coordinates:
$$\mathbf F=m\mathbf a=m(\ddot r-r\dot\theta^2)\,\hat r+m(r\ddot\theta+2\dot r\dot\theta)\,\hat\theta$$
For a force of constant magnitude always pointing towards the origin we have $\mathbf F=-F\,\hat r$, and so the equations of motion become
$$m\ddot r-mr\dot\theta^2+F=0$$
$$r\ddot\theta+2\dot r\dot\theta=0$$
which hold for any initial conditions.
In order to have uniform circular motion, we need
*
*$\dot r$, $\ddot r$, and $\ddot\theta$ to all be $0$ and,
*$r$ and $\dot \theta$ to be non-zero (they also need to be constant, but that follows from point 1).
This occurs when
$$F=mr\dot\theta^2$$
$$\dot r=0$$
So this shows that in order to have uniform circular motion we need for our initial conditions
*
*The force magnitude is equal to $mr(0)\cdot(\dot\theta(0))^2$
*$\dot r(0)=0$
If these two properties are not true of the initial conditions then you will not get uniform circular motion. You can determine what the motion will be from the general equations of motion we obtained above.
| {
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How does an observer outside an accelerating body rationalize the effects of pseudo force? Consider this following thought experiment: There are two tracks, one with a line of cameras and another on which a car with transparent sides moves (*). Now, imagine attaching a pendulum to the roof of the car, as the car accelerates to gain speed, we see that the pendulum makes an angle with the vertical in a direction opposite to which the car is accelerating in from the photographs which the line of cameras will take.
Using the photographs, we can construct a position time curve and we would find that the acceleration of bob in the car has some sudden shifts. A person inside the car can understand this phenomenon as the effect of pseudo force(**), that is the pendulum bob experiences a force opposite to the direction in which the car is accelerating. However, how does a person outside the car rationalize this phenomenon occurring?
*: We can think of it as ripping off the car door so that we can see the pendulum's motion inside the car at all times.
**: If we attach a frame of reference to the car, then that frame is accelerating, and hence we have to consider the -Ma pseudo force on it. Where M is the mass of the pendulum and a is the acceleration of the car.
| I think the best answer here is that, from the perspective of the inertial observer, there just isn't anything to explain, as long as you've accepted Newton's laws. The pendulum bob is doing exactly what Newton's first law predicts it will do: maintaining its momentum in the absence of any forces on it.
The need for explanation arises only for the non-inertial observer, who sees a violation of Newton's 2nd law insofar as he sees the bob accelerate but with no force to explain it. So he has two options: either invent a force to explain the acceleration or postulate that his reference frame itself is accelerating. This is analogous to how Einstein revolutionized our understanding of gravity.
| {
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In the entropy formula, how do we know which one is a valid microstate for a particular macrostate? Consider a very long vertical cylinder containing air in thermodynamic equilibrium. Observe that the air column is necessarily bottom heavy. The macrostate is described in part by a pressure gradient that is due to gravity. All corresponding microstates for this particular macrostate will have a matching density distribution. It seems that a top heavy distribution of the air molecules is not a valid microstate of that macrostate. And neither is a uniform density distribution along the cylinder length. When gravity is reversed, the density distribution also reverses.
When the gravity force is removed or equalized along the cylinder's length by turning it horizontal, it seems the density distribution necessarily becomes uniform in equilibrium. A microstate where all the air molecules are concentrated in one area appears to not be a valid microstate of that macrostate because the density gradient causes a pressure gradient (all else equal) that is different than the uniform pressure of the uniform density system, hence the two macrostates are not identical.
Can we therefore say that entropy forbids density fluctuations in a gas-filled system in thermodynamic equilibrium? In other words, can we say that the only valid microstates of a particular macrostate are those that always match the macrostate parameters?
| Remember that macro state parameters are derivable from the distribution of micro states. Concretely, macro state parameters like pressure, density etc are averages over micro states of like velocity, number density etc.
It is not true that for a gas in a gravitational field, the uniform density micro state does not contribute because it does not reproduce the correct bulk parameter. Instead the statistical interpretation would be that this micro state occurs with a very low probability in comparison to the micro state that is bottom heavy, which occurs with a very high probability. Therefore when one averages over all possible micro states, the average is dominated by the bottom heavy configurations thereby reproducing the correct macrostate parameter.
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Is a reflected wave on a string of the same form as of the incident Griffiths says that if a incident sinusoidal wave on a string gets reflected, it's form will be sinusoidal as well.
Why is it so? Does it hold for all wave forms in any medium?
| You already understand how a string can sustain and propagate sinusoidal waves in both directions. Having a reflector at one end is a boundary condition, one that is compatible with the string's motions. Since the string allows propagation both ways the boundary condition representing the model of the reflecting wall constrains the ratio of the reflected and incident sinusoidal waves.
The boundary condition is a model, if the wall is non-linear then the boundary condition is also non-linear and an incident wave may generate harmonics by that model as function of the incident wave amplitude and you get a reflection that is not a pure sinusoidal.
Even the wave generator is represented by a boundary condition that has usually two parts (1) an outgoing wave of given amplitude and a (2) partially absorbing/reflecting wall. The latter is important if there are waves going towards the generator, e.g., usually from reflections from the other end. (In dc electronics the battery is represented by a perfect voltage source plus a resistor, or in the case of an amplifier connected to an antenna the amplifier is represented by an ideal wave source of fixed amplitude independently prescribed and imposed on the rest of the circuit and its output impedance is that partially absorbing/reflecting "wall" to the waves incoming towards the amplifier.)
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Why don't we use the concept of axis of mass in place of center of mass? Being a high school student, I read the concept of center of mass and it was written in my book that
When a spinning ball is projected with some velocity , then all the points on the ball have complicated paths except the center of that ball which follows the well known parabolic trajectory. And hence we define that point as center of mass.
However, I think that all the points on the any axis about which the ball is spinning follow the parabolic trajectory and are not influenced by spin .
Edit : Most of the answers argued that the rotation axis may change because of torque but the main point to note here is that we can't differentiate between two axis in case of a sphere since it is symmetric from all sides and also that a sphere can't rotate about more than one axis at a time . So saying that it will rotate about different axis is I think meaningless.
So is it okay to define axis of mass in place of center of mass for sphere or other symmetric bodies or am I wrong somewhere ? If not, give a proper reason.
| If the sphere is fully symmetric, then you can't define "the" axis of mass, because ANY axis of the sphere would be equally good, so it's not a well-defined quantity. Only the center is well defined.
In general, in 3D Newtonian physics, an object actually has three "axis of mass," expressed by the Inertia Tensor. This is used to good effect in-game physics simulations, where a long, thin object tumbles differently from a short, fat object. This is also why objects typically end up spinning around one "main" axis of rotation (the "biggest" axis in the tensor) but can use off-center angular momentum to "flip" around one or both other axes. Google the "spinning corkscrew in zero gravity" YouTube videos for some illustrations.
| {
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Can quantum measurements be the origin of thermodynamic arrow of time? We can practically consider that the microscopic interactions are symmetric with respect to time(as we can neglect weak force for many cases which is the only interaction that can violate $T$ symmetry). So I thought that the asymmetry might be due to the irreversibility of quantum measurements. But this is only applicable for interpretations where wave function collapses like Copenhagen etc. What is the answer to this question in Many-worlds interpretation, Consistent histories, etc? Also in this page, they gave that the initial conditions of the universe are the reason for $T$ asymmetry in the 2nd law of thermodynamics. But I am not sure what they mean. Do they mean that the universe had a very low entropy at the beginning?
| No, because "measurements" and "wave function collapse" are not part of the system modeled by QM; they're interpretations. As such, no physical phenomena can be effects of them.
| {
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How is a free theory defined? In field theory, I've seen a free theory described as
*
*A field with the specific Lagrangian density ${\cal L}=|\partial\phi|^2+m^2\phi^2$
*A field whose equation of motion yields a linear set of solutions
*A field with non-interacting i.e. free normal modes
The first seems too specific, the second seems too general, and the third seems ill-defined. I was hoping that these three could be extended to solve any of those problems or if there is some way to unify, say, the first and the second then maybe that final description would strike right.
| By defining the action of a free moving object.
By free I mean it is not confined in a potential but I'm sure someone will argue self action. I mean we need a point mass or something which carries charge which implies a field. Moving charge implies work done. Which implies force which implies accelration which implies radiation which implies self action but don't think about it too much.
At least in string theory we start by defining the action of a free string first.
For me the action came before the lagrangian and from it one can create a lagrangian.
Why? One can have an action without defining a lagrangian but the opposite is not true.
Edit: I guess people disagree.
| {
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How is torque transmitted between inclined surfaces? In the picture below, in a), a body K1 is pivotably attached to a bearing. My question is about the torque that results from a force exerted onto a surface of the body K1.
A first force F1 applied orthogonally onto the surface should result in a torque M1 in clockwise direction.
Is it correct that a second force, F2, applied almost parallel to the surface will result in a torque M2 in counterclockwise direction?
My thoughts are, F2 is split into F2t and F2o (transversal and orthogonal components) by the surface of the body K1. To get a torque, F2o is multiplied by the lever b and F2t is multiplied by the lever a (M2 = F2t * a - F2o * b > 0). As a>b and F2t>F2o, the torque from the force F2 results in counterclockwise direction.
Applying these thought to the two bodies K1, K2 in b), a torque of M3 applied to the body K2 will result in a torque M4 in the body K1. (The bodies won't move because they are in each others movement path)
Is this correct or am I forgetting something? What is the job of friction in this case? From looking at b), K2 should push K1 away by applying a clockwise torque, but that is wrong then, right?
Suppose there is enough friction so that no slippage occurs.
| Don't wrap yourself around in pretzels. Even for planar cases, assume they are defined in 3D (with z-axis out of plane) and use the cross product to define torque
$$ \vec{\tau} = \vec{r} \times \vec{F} $$
which expands to
$$ \pmatrix{ 0 \\ 0 \\ \tau_z} = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix} \pmatrix{F_x \\ F_y \\ 0} $$
and projects into 2D as
$$ \tau_z = -y F_x + x F_y $$
The nature of force does not matter. Use the combined normal and frictional forces to find the net torque, or just an individual component to gauge the effect of on the body.
The same goes for moment of impulse in case you have contacts.
$$ \vec{\gamma} = \vec{r} \times \vec{J} $$
| {
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A simplified version of the Fizeau experiment for measuring the speed of light? Long ago in high school I saw a short film in which as I recall the apparatus was just a spinning paper plate (with holes along its edge) with a light source -- I don't recall a spinning mirror or a half-silvered mirror -- and the entire experiment could be done in a classroom. The basic idea that even though the speed of light was very fast you could still spin the plate fast enough to affect whether the light was visible or not struck me as very clever and almost something that could have been done by Galileo except that getting a steady speed of rotation without a motor would have been hard.
So my question is, is there a simplified version of the Fizeau experiment that could be performed within a normal-sized room with just an electric motor, pie plate and a laser pen that will give a reasonable (if less accurate than in Fizeau's actual setup) value for the speed of light or am I remembering wrongly?
| Fizeau used a folded light path that was very long, and mirrors to bounce the beam back and forth, and a very high-speed motor running the slotted disc. There is no way to spin a paper plate fast enough in a classroom to perform the fizeau experiment with a classroom-sized baseline for the light beam.
| {
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Why does a capacitor act as a frequency filter? What is it about a capacitor which allows it to filter frequencies?
I understand the construction of a high-pass RC filter, and the mathematics behind it, but I'm struggling to find an explanation of the physics behind the phenomenon.
In my mind I can picture the broad spectrum signal hitting the capacitor, but I feel like the "output" behaviour would be mush, not a controlled and predictable behaviour. I'm not a physicist, but I'd like to understand this problem better.
What is the physical behaviour which allows a capacitor to act as a high or low pass filter?
| FWIW, I think of it as the capacitor having lower impedance at higher frequencies. As the frequency component of a signal gets higher, the capacitor in the RC filter diagram above looks more and more like a piece of wire, thus allowing more of the signal amplitude to be developed across the resistor. At low frequencies, the cap impedance is high, compared to the resistance of the resistor, so more signal appears across the capacitor, and less across the resistor. (We're taking our output across the resistor.)
| {
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Why is the electrostatic force felt in straight lines? When two positive charges are kept close, they get repelled in the direction of a line joining both the charges. Why is it so?
Also, why is the repulsion in a straight path?
In both the cases, the potential energy of the charge which gets repelled decreases. What makes it repel in a straight line such that the line passes through both charges?
| Just to elaborate on the symmetry argument - Let's suppose in your first diagram you are observing the two charges from the side and we assume, as you have done, that the repulsion direction is vertical and to the right. If we now observe the two charges from the top looking down, we are presented with the exact same situation as before and we would say the charge should now be repulsed horizontally and to the right. But that contradicts the direction we initially assumed - the repulsion cannot depend on how you look at the two charges. Continuing with this type of argument one can only conclude that the repulsion must be along the line connecting the two charges.
| {
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What are the best resources for Crystallography? I am undertaking a module in nanosurfaces and I was unaware that I would require some knowledge in Crystallography. The information that I must know regarding this area are Miller indices, Symmetries, Bracket conventions, so that I can answer some example questions such as:
*
*sketch for a face-centred cubic structure the (100), (110) and (111) planes and identify their in-plane symmetries
*draw the (0001) plane of a hexagonal structure with the principal in-plane directions of type <1-100> and <-2110>;
*sketch the (210) plane of a simple cubic structure and calculate its angle to the (100) plane? Remind yourself of the Ewald sphere construction for reciprocal space and diffraction.
What are the best sources of information for Crystallography? What books would you recommend?
I have come across books such as :
*
*The basics of crystallography and diffraction by C. Hammond ;
*Fundamentals of Crystallography by Carmelo Giacovazzo;
*Crystallography made crystal clear by Gale Rhodes.
Would any of these be good for the information I need?
| I personally used The basics of crystallography and diffraction by C. Hammond in conjunction with my material science class and the book did a more than outstanding job in explaining the basic concepts that you would like to know.
| {
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Confusion about the dimension of a Hilbert Space in Quantum Mechanics In Quantum Mechanics, the quantum state of the physical system lives in an infinite-dimensional Hilbert space and can be written in terms of two different bases, the position basis (uncountably infinite) and the energy basis (countably infinite). Apparently, the two bases are of different cardinalities, which violates a theorem in Linear Algebra that all bases of a vector space must be of the same cardinality. How to explain this confusion?
| I am a mathematician and not a physicist, so I don't know the physical context and other physicists here are very welcome to correct my answer if it is wrong.
I think your confusion comes from the unfourtantate terminology of "basis" in the context of Hillbert spaces. There are two different concepts both happen to be called basis:
*
*Hamel Basis: That's just a basis in the sense of linear algebra. ie A linearly independent subset such that any vector can be written as a finite linear combination of the members of your basis.
*Hillbert Orthonormal basis: That's a linearly independent orthonormal subset of your Hilbert space such that any vector in your vector space can be approximated with arbitrary precision using finite linear combinations of your basis. It follows mathematically that any vector in your hillbert space can be expressed as an infinite sum of members of your basis.
I don't know the physical context, but I suspect that what you call an "energy basis" is the second type of basis I talked about above and not the first one(ie Hillbert Orthonormal basis and not Hamel Basis). However, the theorem of linear algebra which you quote refers to Hamel Basis and not Hillbert orthonormal basis. Hence, you are applying the linear algebra theorem in a wrong way/invalid context and that should solve the confusion.
See this link of Wikipedia:
https://en.wikipedia.org/wiki/Hilbert_space#Orthonormal_bases
| {
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Is Penrose's CCC consistent with Penrose's singularity theorem? According to Penrose's Conformal Cyclic Cosmology (CCC), there were universes prior to ours, prior to the singularity of our universe.
But how is this claim compatible with his famous singularity theorem, according to which spacetime geodesics cannot be extended beyond a singularity?
I believe Penrose doesn't deny the big bang singularity. Then how does he make sense of 'spacetime prior to the big bang singularity' in CCC?
| My reading of the paper is that there is a singularity, in the sense of the singularity theorems, at the beginning of each cycle, but it's physically irrelevant because physics is precisely scale invariant there. The FLRW metric in terms of conformal time is $ds^2 = a(η)^2 (dη^2 - d\mathbf Σ^2)$, where the "coordinate" $\mathbf Σ$ ranges over a sphere*. If the scale factor is physically irrelevant then the space is effectively a sphere at the big bang even though $a(η)\to 0$. (In particular see the top half of the left column of page 2761.)
* I think space has to be positively curved in this model because the future infinity of de Sitter space is spherical, but he doesn't seem to say that in the paper unless I missed it, so I'm worried that I'm missing something.
| {
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Argument on why spin correlation functions in Ising model decay exponentially with a correlation length? I'm reading Quantum Field Theory in Strongly Correlated Electronic Systems, Nagaosa.
Consider 1D Ising model,
$$H=J_z\sum_i S^z_iS^z_{i+1}.$$
on page 3, it says
The groud stae is 2-fold degenerate because the Hamiltonian is invariant
under the transformation $S^i_z \rightarrow -S^i_z$, performed at all sites $i$.
Calling these two ground states $A$ and $B$ and assuming that the system at the right-hand side is in state $A$, and at the left-hand side in state $B$, then somewhere there
must exist a boundary between region $A$ and region $B$. This boundary is called a kink or soliton. Because at finite temperature this excitation occurs
with a finite density, the spin correlation function $F(r) =\langle S^z_iS^z_{i+r}\rangle$ will decay exponentially with a correlation length $\xi$.
I know how to directly calculate the correlation function, but I wonder how the argument for exponential decay of correlation function is made here and how to understand it.
Any help would be highly appriciated!!
| Let me write the Hamiltonian
$$
H = -J \sum_i S_i^z S_{i+1}^z.
$$
This choice will avoid some annoying (and irrelevant) signs.
One way to formulate the statement in the OP precisely is as follows.
Consider the variables $\delta_i=S_i^zS_{i+1}^z$. Since $\delta_i=1$ when the spins at $i$ and $i+1$ agree and $\delta_i=-1$ when the spins at $i$ and $i+1$ disagree, you can identify them with the kinks in your question (that is, there is a kink between $i$ and $i+1$ when $\delta_i=-1$).
Introducing the variables $\delta_i=S_i^zS_{i+1}^z$, the Hamiltonian becomes
$$
H = J^z \sum_i \delta_i.
$$
It follows that the random variables $\delta_i$ are independent and identically distributed. One can easily compute their expectation: since
$$
P(\delta_i = 1) = \frac{e^{\beta J^z}}{e^{\beta J^z} + e^{-\beta J^z}},
$$
one has
$$
\langle \delta_i \rangle = \frac{e^{\beta J^z} - e^{-\beta J^z}}{e^{\beta J^z} + e^{-\beta J^z}} = \tanh(\beta J^z).
$$
Finally, noting that $S_i^zS_{i+r}^z = \delta_i\delta_{i+1}\cdots\delta_{i+r-1}$, we obtain
$$
\langle{S_i^zS_{i+r}^z}\rangle = \langle\delta_i\delta_{i+1}\cdots\delta_{i+r-1}\rangle = \langle \delta_i \rangle^r = (\tanh(\beta J^z))^r.
$$
In words, the fact that kinks proliferate in the system (at each $i$, there is a positive probability that a kink is present, so there will be a positive density of them in the system) prevents the ordering of the spins.
| {
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Meaning of complex conjugate in $T$-symmetry I have a question about the meaning of complex conjugation in time reversal symmetry in quantum mechanics.
$T$-symmetry in classical mechanics is defined simply by the substitution $t \to -t$. If I have an external magnetic field it is not enough and I have to substitute $ \textbf{B} \to - \textbf{B} $. This makes sense because reversing time makes the "external current" generating the magnetic field run backwards, therefore a time symmetry that acts on the system as a whole is indeed supposed to reverse the signs of the magnetic fields as well.
In quantum mechanics T symmetry is given by an operator that acts on a generic wave function as $ T \psi\left(\textbf{x}, t \right) = \psi^{*} \left(\textbf{x}, -t \right) $. The meaning of $t \to -t$ is clear but what about complex conjugation? I know it makes Schroedinger equation invariant if the Hamilton is invariant but what does it have to do with time reversal? Is there a way to justify it like we justify the correspondence $ \textbf{B} \to - \textbf{B} $ in electromagnetism? Because if not it appears to me that $T$ kind of has to do with reversing time but it isn't really a time reversal of the system.
| Time reversal operator $T$, when acting on $i$, must result in $TiT^{-1} = -i$. This comes from the fact that $TxT^{-1} = x, TpT^{-1} = -p$, and commutation relation $[x, p] = i\hbar$.
Combine this fact with linearity of time reversal operator, we conclude that $T$ is antiunitary operator, and can be decomposed in the form $KU$, where $U$ is unitary and $K$ is complex conjugation operator. (For reference, it is proved by Wigner that all symmetries of quantum mechanics must be unitary or anti-unitary)
$U$ can of course vary depending on the system you are working on.
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What is the equivalent to $\Box A^\alpha =- \mu_0 J^\alpha$ using differential forms? The set of equations $$\Box A^\alpha = -\mu_0 J^\alpha$$
can be found in section 12.3.5 of Griffiths's book. From what I understand, the real-valued functions on both side of the equations are the coefficients of some $1$-forms with respect to a chart. Thus, I am wondering how the equivalent index-free equation involving differential forms looks like.
| Maxwell's equations can expressed in the language of differential forms.
In terms of tensors, Maxwell's equations can be written as
$$
\nabla_{\nu}F^{\mu\nu} = \mu_0J^{\mu}\,, \nabla_{[\alpha}F_{\mu\nu]} = 0
$$
where $F_{\mu\nu}$ is the Faraday tensor.
In terms of differential forms, these equations are written as
$$
d{\bf F} = 0, d{\star}{\bf F} = \mu_0{\bf J}\,,
$$
where ${\bf F} = \frac{1}{2}F_{\mu\nu}dx^{\mu} \wedge dx^{\nu}\,$, $\star$ is the Hodge star operator and ${\bf J} = \frac{1}{3!}{\cal J}_{\alpha\beta\sigma}dx^{\alpha}\wedge dx^{\beta}\wedge dx^{\sigma}$ is the 3-form associated with the current density four-vector. The components of ${\cal J}_{\alpha\beta\sigma}$ are related to those of the current density 4-vector.
The Faraday 2-form, ${\bf F}$, satisfies, ${\bf F} = d{\bf A}$, where ${\bf A} = A_{\mu}dx^{\mu}\,.$
See https://en.wikipedia.org/wiki/Maxwell%27s_equations
and
https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field
Note that there are different sign conventions so be careful when consulting different websites/books on this subject.
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Understanding of electric potential's integration form I already known that the potential difference when a charge moves from A to B is
But I still have confusions about what does the infinitesimal of vector $s$ refers. I mean when you change the movement of the charge from B to A, the $\Delta V$ should be opposite number of it. But if the E and S's direction is opposite, the dot product of E and S should be negative. Since the range of integration is reversed, the out come of the delta Vab is same to delta Vba.
Help is really appreciated.
| you appear to have accounted for the change in direction twice. You need to first recall why reversing the limits change the signs in good old integration of functions in the cartesian plane. Although we simply put the minus sign, the reasoning behind it is that the new or infinitesimal change in x is negative of what it was before (since you are varying x in the opposite direction) and you directly account for the new with a minus sign outside the integral and the old dx inside
Here, since you are already accounting for the new displacement vector in the proper way, you no longer need to use the "reversing limits reverses signs rule" as that would be repetitive
| {
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Generalised Lorentz force expression from Classical Mechanics by Goldstein I am reading chapter 7 in the 3rd edition of Goldstein's Classical mechanics textbook and the expression for the Lorentz force is confusing me. I cannot scan it so I am just going to write it out verbatim and formulate my question afterwards. Here is the extract of page 298 from the text:
In terms of $\phi$ and $\mathbf{A}$, the Lorentz force is $$\mathbf{F} = q\{-\nabla\phi+\frac{1}{c}\frac{\partial \mathbf{A}}{\partial t} + 1[v \times(\nabla\times \mathbf{A})]\}\tag{7.67c}.$$
This suggests that we should generalize the force law to
$$\frac{dp_{\mu}}{d\tau} = q\left(\frac{\partial (u^\nu A_\nu)}{\partial x^\mu}-\frac{dA_\mu}{d\tau}\right).\tag{7.68}$$
The first equation is the three three dimensional Lorentz force express using the vector and scalar potentials (As a note I think the second term should be $-\frac{\partial A}{\partial t}$ but the above is as written.)
I am unsure howhow you reach the second equation from the first expression, I would appreciate any help in understanding this problem.
| TL;DR: The total derivative term
$$\frac{dA_\mu}{d\tau}~=~\gamma\frac{dA_\mu}{dt}
~=~\gamma\left(\vec{v}\cdot\vec{\nabla} A_\mu+ \frac{\partial A_\mu}{\partial t}\right)$$
in eq. (7.68) is correct. It should not be a partial derivative.
Before trying to read the relativistic formulation in section 7.6, I would strongly recommend you to fully understand the non-relativistic derivation in section 1.5, which essentially features the same issue.
| {
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Shouldn't We modify the field in force equation $\mathbf{F}=q\mathbf{E}$? Consider charge particle $q$ in electric field $\mathbf{E}$. The force on the charge is given by
$$\mathbf{F}=q\mathbf{E}$$
Now we know that charge $q$ will also produce an electric field. Due to this field, the field already present in the space should be modify. And thus we should use the modified version of the field. But we don't? (atleast I didn't see).
So the question is if the above reasoning is correct what should be the correct expression? If it's wrong why?
| In the absence of particle acceleration the only way I know to alter the force on the particle is by proximity to a (neutral) conductor which would induce asymmetric surface charge on the conductor of the opposite polarity to the particle's.
The quantitative calculations are beyond intro physics. I'll let others delve into Legendre polynomials etc.
| {
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Effect of motion on ball in a moving cart I know that in case of cart moving with an acceleration we are supposed to apply the concept of pseudo acceleration to judge the motion of ball in frame of the cart.
Now , consider the case where cart is moving with constant velocity. What will be the trajectory of the ball? Is there something called pseudo velocity?
situation
| I think your terminology is confusing you - the correct term is pseudo-force, not pseudo-acceleration.
Introducing a horizontal pseudo-force in the reference frame of an accelerating cart allows us to pretend that objects that are stationary in the reference frame of the cart are in equilibrium, even though we know they are really not.
If the cart is moving with constant velocity (i.e. zero acceleration) then objects that are stationary in the cart’s reference frame really are truly in equilibrium, so we do not need to introduce a pseudo-force.
Another way to look at this is to note that the pseudo-force on an object of mass $m$ will be $ma$ where $a$ is the cart’s acceleration. But if $a$ is zero then the pseudo-force is also zero.
| {
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Is Loop quantum cosmology replacing the Big Bang singularity with a big bounce? Am I correct in understanding that in the cosmological version of Loop quantum gravity, namely Loop quantum cosmology, the universe in all models starts with a big bounce? Are there other models, for example, the pre-big bang condition with unconnected loops?
| Loop Quantum Cosmology is a finite, Symmetry reduced model of LQG, which for a layman means it arose from Loop Quantum Gravity. It predicts a Quantum bridge between the Expansion and the Contraction of the Cosmological branches. So, instead of having Big Bang, you have a Big Bounce. Loop Quantum Cosmology aims to describe a lot more than Bounce, it mathematically describes the Inflation in the Early Universe. For more, you can read the articles of Francessca Vidotto and Aurelien Barrau. The first Link is given here. https://tel.archives-ouvertes.fr/tel-01737503. They talk about some basic aspects of Loop Quantum Cosmology and the Loop Quantum Cosmology aspects of the Dark Universe
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Confusion on repeated index for Einstein Summation The rule for Einstein notation is that the same dummy index cannot be repeated twice. However suppose I want to compute Christoeffel symbols:
$$
\Gamma^{\alpha}_{\beta\gamma} = \frac{1}{2}g^{\alpha\sigma}(\partial_\beta g_{\gamma\sigma}+\partial_{\gamma}g_{\sigma\beta}-\partial_{\sigma}g_{\beta\gamma})
$$
Now if my metric is diagonal, then only the terms $\alpha = \sigma$ survive, hence we have:
$$
\Gamma^{\alpha}_{\beta\gamma} = \frac{1}{2}g^{\alpha\alpha}(\partial_\beta g_{\gamma\alpha}+\partial_{\gamma}g_{\alpha\beta}-\partial_{\alpha}g_{\beta\gamma})
$$
Of course now the problem is that the index $\alpha$ is repeated three times. However, it makes perfect sense to me when I do the computation. Is there some exception to the "not repeated twice" rule?
| The Einstein summation rule is true for tensor-equations. Once you assume a form for the metric (diagonality), the equation you get is no longer a true tensor-equation (it is only true in some coordinate-systems). This is the reason why you need to write the summation by-hand from this point on.
| {
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A tilted disk rolling on the floor First of all, an image to describe the situation we have:
Background
A uniform disc is rolling without slipping on a flat surface. The disc itself is also in circular motion about the point $O$. I have tried this with a roll of cellotape so the situation itself seems plausible. What I want to find is the radius of the disk in terms of $g$, $\omega$ and $\theta$.
My analysis
(apologies for the lack of any diagrams from here onwards)
Now, something must be keeping the disc from just falling over. If I take the angular momentum about the point of contact, I get $\overrightarrow{L}=\frac{3}{2}mr^2 \omega$. The vector itself is angled at $\frac{\pi}{2} - \theta$ to the floor, and is swinging around as the disk moves in space.
Some force is providing the torque for this to be possible. This force is the gravitational acceleration. There is no torque due to normal, frictional or centrifugal forces (?) about the point of contact. If we compute the torque due to gravity about the point of contact, we get $\overrightarrow{\tau} = mgr \cos{\theta}$.
The angular momentum vector is rotating about an axis perpendicular to it and inclined at $\frac{\pi}{2} - \theta$ to $+z$. Let us call the angular velocity about this axis $\Omega$. We can find $\Omega$ by picking the center of mass of the disc to study. The distance from $O$ to the COM is $r\tan{\theta}$. So we have $\Omega r\tan{\theta} = \omega r$, therefore $\Omega = \omega \cot{\theta}$.
Therefore, we can find the required torque to be $\overrightarrow{\tau}_{req} = L \Omega$. Equating $\overrightarrow{\tau}_{req}$ and $\overrightarrow{\tau}$, we get $$r = \frac{2g \sin{\theta}}{3 \omega^2}$$
Can someone please check my analysis? Since I came up with this myself, I don't have anything to refer to. (I also feel a little unsure whether centrifugal force has any role here.)
| This problem is in Introduction to Classical Mechanics by David Morin as problem 9.23.
Let the rate of precession of the coin be $\Omega$. Let the moments of inertia be $I = \frac 14 mr^2 $ and $I_3 = \frac 12 mr^2$ respectively. In this situation, it is most convenient to find $\mathbf{L}$ about the center of the coin.
The important thing here is to (temporarily) forget about the motion of the center of the coin in space (as it does not contribute to the changing part of $\mathbf L$). The angular velocity is then $\mathbf{ω}-\Omega \hat{\mathbf{z}}$. The minus sign occurs because they point in opposite directions.
Your mistake is that you failed to include the $-\Omega\hat{\mathbf{z}}$ part of the coin's rotation. The coin, on top of spinning about $I_3$ with angular velocity $\omega$, is also rotating about the $z$-axis with angular velocity $\Omega$. This can be most easily visualized by imagining yourself sitting some height above the center of the coin, always facing in the direction of the $x$-axis.
The next few lines is the heart of the problem.
Now, we are interested in finding the non-vertical component of $\mathbf{L}$, which we shall denote as $L_{\parallel}$. The $\mathbf{ω}-\Omega \hat{\mathbf{z}}$ can be reexpressed as $\omega - \Omega \cos\theta$ perpendicular to the coin and $\Omega \sin\theta$ downwards along the coin.
$\omega - \Omega \cos\theta$ perpendicular to the coin translates into a contribution $I_3(\omega - \Omega \cos\theta)\sin\theta$ to $L_{\parallel}$.
$\Omega \sin\theta$ downwards along the coin translates into a contribution $I\Omega \sin\theta \cos\theta$ to $L_{\parallel}$.
Putting the above two together gives us a total of $$L_{\parallel} = mr^2 \left(\frac 12 \omega \sin\theta -\frac 14 \Omega \sin\theta \cos\theta\right)$$ Since $L_{\parallel}$ is precessing with frequency $\Omega$, we must also have $$\left| \frac{\text d \mathbf{L}}{\text dt}\right| = L_{\parallel} \Omega$$
The other equations are $$R \Omega = r\omega$$ $$F_f = m(R-r \cos\theta)\Omega^2$$ $$\left| \frac{\text d \mathbf{L}}{\text dt}\right|= r(mg\cos\theta - F_f \sin\theta)$$
where $R$ is the radius of the point of contact and $F_f$ is the friction.
Solving all the equations will obtain $$\Omega = \sqrt{\frac{g}{\tan\theta \left( \frac 32 R - \frac 54 r \cos\theta\right)} }$$ Therefore, precession is only possible when $R > \frac 56 r \cos\theta$.
| {
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Matsubara sum with log term How do I compute the Matsubara sum
$$\sum_n \log\left(-i\omega_n +\frac{k^2}{2m}+\mu\right)?$$
If I have sums like
$\sum_n \frac{1}{i\omega_n -m}$, I can sum it up by calculating the sum of residues of the function $\frac{1}{z-m}g(z)$ at the poles where $g(z)=\begin{cases}
\frac{\beta}{\exp (\beta z)+1} \text{ for Fermions}\\
\frac{\beta}{\exp (\beta z)-1} \text{ for Bosons}
\end{cases}$
But how do I do I compute in this case where there is a $\log$ term and there are no poles.
| One possible approach is writing the sum of logs as a log of product and using the formulas for infinite products in Gradshtein and Ryzhik.
| {
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Why are we allowed to make algebraic calculation on units Is there math proof that we can cancel out units in Physics? For example:
$\require{cancel}distance = \frac{meters}{\cancel{second}} * \cancel{second}$.
So we see that seconds cancel out and we left with meters which is correct but how is it possible if it is not actually division (meters per second) it is only our interpretation of speed. Why are we allowed to make algebraic calculations on units? Also, we can use dimensional analysis to check that we do everything correctly but why? Why does it always work?
| Good question. I don't have any elegant, deep mathematical answer for you. But let's consider an object moving at 3 m/sec and consider a time interval of 5 sec. We can lay out the object's progress along a line consisting of 5 sections, in each of which the object has moved 3 m. So the total distance moved is 3m + 3m +3m +3m +3m = 15m. Knowing multiplication, we can also find the total distance as 3m x 5 = 15m. Notice that in this last calculation time doesn't play a role. I'm simply multiplying a distance by a pure number. So (3 m/sec) x (5 sec) is really 3m x 5. The seconds cancel. That to me would justify treating the sec unit like a number to be canceled.
| {
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Why doesn't a mercury thermometer follow the rules of volume dilatation? let's consider a classic mercury thermometer.
I do not understand why it does not behave like a "normal" thermometer which exploits volume dilatation. In a normal thermometer, I'd say that the mercury length would be proportional to its temperature.
Therefore, I should be able to measure, for instance, 37 of body temperature, also starting with the thermometer at 38: there would be a contraction, but the measure would be correct! Why does this not happen? And why if I measure for instance, 38, and I try to cool the thermometer by putting it inside cold water, it does not become cooler? Why should I cool it by shaking it?
It seems a very not ideal thermometer... but what are the causes of these non-idealities?
| I think you are speaking of a clinical thermometer which records the maximum temperature it reaches. The thermometer has a narrow kink in the bore near the bulb that causes the mercury thread to break at that point when the volume of mercury in the bulb shrinks (the image you've posted actually shows that). As a consequence the top of the thread does not retract from the high-point reading. (One might worry about the mercury above that break-point shrinking, but there is very little mercury in the thread, most of it is in the bulb. Consequently there is little effect from the volume of the thin thread getting smaller.)
The reason that the thermometer is designed this way is so that the doctor or nurse can take their time in reading the thermometer --- which would otherwise begin to read lower temperatures as soon at it is removed from the patients mouth, or wherever.
Shaking the thermometer after it has coooled to room temperature causes the mercury in the broken thread to reconnect with the mercury in the bulb, and allow it to be used again.
| {
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Simplification using Newton's second law I am not sure if my simplification works in this problem:
Problem:
I have a beam which is strap around with cargo straps. First picture presents section through second picture. So applying Newtons second law:
$$F=G=\frac{m_{beam} g}{2} $$
so the F is the force in the cargo straps. Therefore I can calculate which king of cargo straps I need to use. Is this simplification legit or is it not approved method?
Black box is the beam and red lines are the straps. Green arrow is $G$
and red arrows are the $F$.
| The forces you have calculated are acceptable. However there are other things you will need to consider from an engineering point of view.
Not so important:
Because the straps are not directly connected to beams, rather the beam is resting on it. The beam will also deflect due to self weight. To be safe I would also check the slope of the beam at the ends given they are free to rotate.
$$\theta=\frac{MgL^{2}}{24EI}$$
where $E$ is the young's modulus of elasticity and $I$ is the moment of inertia.
You don't want the angle to be too large otherwise the strap may slide up the beam. If the angle is large, you will need to check the friction between the strap and the beam given the load Mg/2.
Important:
You will also need to check the internal capacity of the beam itself to make sure it can support it's own weight. The maximum shear force in the beam is Mg/2, you will need to check if the material can handle this. In addition, the beam bends when it is held from the straps.
You will get a maximum bending moment in the middle of the beam of:
$$M_{b}=\frac{MgL}{8}$$
The maximum stress that arises from carrying the beam is $f$:
$$f=\frac{M_{b}D}{2I}$$
where $D$ is the depth of your beam and $I$ is the moment of inertia.
If the stress you calculate is larger than the tensile or compressive strength (e.g. for steel $f=300MPa$) of your beam then your beam may get damaged during transport.
Also initial loading may be uneven, use straps that can take the whole load individually.
| {
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What does it mean to treat space and time on equal footing? I often read from textbooks that in relativity, space and time are treated on an equal
footing. What do authors mean when they say this?
Are there any examples that show space and time are treated on an equal footing? Conversely, what examples show that space and time are not treated on an equal footing?
| Putting space and time on the same footing means to treat time as another dimension in addition to the other three physical dimensions. In the context of relativity, time is treated as another dimension (but within this idea of Spacetime space and time are not the same).
In classical Newtonian physics, space is treated within the ideas of three dimensional space. In this approach, time is absolute, as oppose to relativity.
| {
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Meson as hadron and boson In wikipedia page about hadrons the following image appears:
I can understand why the intersection between hadrons and fermions are baryons, as a way to say a baryon is a kind of hadron composed of several quark fermions.
However, what is the meaning of the intersection between hadron and bosons labeled in the picture as mesons? If I understand correctly, a meson consists of one quark and one antiquark, nothing related to any boson.
| (Anti)quarks are spin 1/2, so two of them is either spin-0 or spin 1: bosons.
But it's more than that. The pions are (one of) the force carrying boson in quantum hadrodynamics, which is an effective field theory of nuclear interactions. Moreover, reactions like:
$$ p \rightarrow n + \pi^+ $$
look a lot like:
$$ \nu_e \rightarrow e^- + W^+ $$
in terms of coupling isospin and weak-isospin.
There is also something called "Vector Meson Dominance" used to describe higher energy photons' interaction with hadronic matter. Basically, the photon can be dressed as vector (spin-1) meson: $\rho$, $\omega$, and $\phi$. At the quark level, this is:
$$ \gamma \rightarrow q\bar q $$
while it the effective field theory level, the vector mesons are mixed into the photon state.
So they are very much bosons.
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How could I see the distance light traveled from an airplane? It was nighttime. I was flying on an airplane. As we were landing we passed over a highway. I saw cars below with their headlights on. I could see that the light from their headlights only lit up a certain distance in front of them. Any object that fell past that distance would have probably looked dark to the driver.
I'm confused as to how I was able to see this thousands of feet up in an airplane. The light from the cars' headlights only traveled a few feet in front of the cars but also traveled thousands of feet to the airplane I was on?
How is this possible?
Also, how could I see the beam-light structure of the light, if the light is able to be seen by an observer in any direction? I feel like none of this makes any sense and we were never taught anything to clarify this in E&M. Maybe I just missed something truly important.
| I just want to add one thing to BowlOfRed's answer.
Any object that fell past that distance would have probably looked dark to the driver.
Let $x_0$ be the distance from the driver to an object. Typically a few meters. Let $x_1$ be the distance from the object to the plane. 1000's of meters.
For an object lit up by the headlight to be visible in the plane, the intensity I must be bright enough to see. Light from the headlights spreads out and hits the object. Reflected light spreads out too
$$I_{plane} \propto \frac{1}{x_0^2}\frac{1}{x_1^2}$$
To be visible to the driver,
$$I_{driver} \propto \frac{1}{x_0^2}\frac{1}{x_0^2}$$
So $I_{driver} >> I_{plane}$
Objects will only be visible to the plane if $x_0$ is very small.
| {
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Why does particle measurement cause quantum wavefunctions to collapse When we attempt to measure a certain property of a particle, how and why does its wave function collapse? I've tried to find answers on my own, but they've been far too complicated for me to comprehend. Would appreciate any answer with limited complex jargon, and more simplistic explanation, if possible.
| It postulates III that says,
If the particle is in a state $|\psi\rangle$, measurement of the variable (corresponding to ) $\Omega$ will yield one of the eigenvalues $\omega$ with probability $P(\omega)\propto |\langle\omega|\psi\rangle|^2$. The state of the system will change from $|\psi\rangle$ to $|\omega\rangle$ as a result of the measurement.
The postulates doesn't have proofs.
Another aspect of this postulate says the measurement of the variable $\Omega$ changes the state vector, which is, in general, some superposition of the form
$$|\psi\rangle=\sum_{\omega}|\omega\rangle \langle\omega|\psi\rangle$$
into the eigenstate $|\omega\rangle$ corresponding to the eigenvalue $\omega$ obtained in the measurement. Which is called the collapse or reduction of the state vector.
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Glass state of water There is a lot of works on hard-sphere glasses where the spheres or other particles are squeezed and fail to find the global energy minimum being jammed in amorphous state. Is it possible to form a water glass by (perhaps, quickly) compressing water to very high density?
| Yes, water can form a "glassy" solid states, called amorphous ice.
Depending on conditions and formation process it can have quite different properties. In particular it is not necessary for it to be compressed to very high density: it is actually possible to obtain amorphous ice that is less dense than liquid water, by vapour deposition (see Wiki). High-density forms also exist.
All known forms of amorphous ice are found at relatively low temperature (~100-160K). Therefore it is not found on earth outside laboratories, but it is quite common in space, in particular in the interstellar medium.
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Do 2d CFTs define healthy 4d QFTs? When doing 2d CFTs we typically complexify coordinates and formally consider $\mathbb C^2$, with the understanding that, in the end, we are to restrict to the real slice $\bar z=z^*$. If we do not impose this, but regard $z,\bar z$ as truly independent, does the resulting object define a healthy four-dimensional QFT? If so, is such theory Poincaré invariant? (i.e., does $\mathfrak{vir}\otimes\overline{\mathfrak{vir}}$ contain a subgroup isomorphic to $\mathfrak{so}(4)\subset\!\!\!\!\!\!\small+\ \mathbb R^4$?) Is it even conformal? (i.e., same as before but with $\mathfrak{so}(1,5)$.)
| Correlation functions in 2d CFT are single-valued on $\mathbb{C}$ but not in $\mathbb{C}^2$.
For example, in minimal models, four-point functions are of the type
$$ Z(z) = \sum_{i=1}^n c_i F_i(z)F_i(\bar z) $$
where $F_i$ is a hypergeometric function or generalization thereof, and $z$ is the cross-ratio of the four positions. In a free boson CFT, they look like
$$ Z(z) = z^a \bar z^a $$
where the exponent $a$ is generally not an integer.
If $z$ and $\bar z$ are no longer complex conjugates, these expressions become ambiguous. An interpretation in terms of QFT on $\mathbb{C}^2$ seems difficult.
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Quantum mechanics and rigorous math I was reviewing a little of quantum mechanics in a rigorous way, so i realized there is a lot of concepts similar in words but different in its meanings, i would appreciate any help to understand it:
Kets -> It is the "physics entity" you are measuring: Momentum, energy, etc...
Operators -> What you actually use to get the results of Momentum, energy, etc...
Eigenkets -> The basis, the possible states the operator can have: exp: Spin up or spin down
Eigenvalues -> The possible values you can get.
Eigenfunctions -> Reserved to wave functions
Eigenvectors -> ?
State vector -> ?
| Very short and brief summary of quantum mechanics, please comment if there are mistakes
Ket vs. State
Each equivalence class of kets (vectors) which differ from each other by a (multiplicative) complex number is called a ray in Hilbert space (which is a complete inner product space). Each ray represents a state uniquely.
The state has many representations, the most famous one is the wave function, the so-called position space representation. Another one is the momentum wave function, in momentum space. It is possible to switch between them.
Hermitian operators (linear maps on the Hilbert space) represent observables, the eigenvalues of such an operator are the possible outcomes of a measurement of the corresponding observable, the quantum system takes on a corresponding eigenstate (eigenket) immediately after measurement.
A complete set of compatible observables (C.S.C.O) is a set of observables $A_1,A_2,...,A_n$ whose eigekets can constitute an orthonormal basis of the Hilbert space, such that every eigenket (basis vector) corresponds to a unique tuple $(a_{i_1}^{(1)},a_{i_n}^{(2)},...,a_{i_n}^{(n)})$ of eigenvalues of the C.S.C.O. Each basis vector is denoted $|a_{i_1}^{(1)},a_{i_n}^{(2)},...,a_{i_n}^{(n)}\rangle$. The general state is an "infinite (complex) linear combination" of this basis, The evolution of the state is determined by Schrödinger equation.
The probability of getting some eigenvalue (as an outcome of a measurement of some observable) is (in case of degeneracy) the (sum of) the squares of the magnitude of the coefficients (of the previous linear combination) that correspond to eigenkets of the considered eigenvalue.
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Net Work Done When Lifting an Object at a constant speed I am confused about the amount of work done when lifting an object at a constant speed. If you find the work done by you on the object and the work done by gravity on the object and add them the net work would be 0. How is there an increase in Potential Energy if the net work done on the object is 0? I was told that 0 was the incorrect answer and the net work should be equal to the potential energy. Which answer is correct?
| If you lift a heavy box off of the floor and place it on a high shelf you have done work to increase the box's gravitational potential energy. The net work you have done against gravity is not zero. If the box later falls off of the shelf this potential energy is converted to kinetic energy as gravity accelerates it downwards.
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Can coldness be converted to heat energy? We know that the heat can be converted into heat energy with the help of thermoelectric generators, but why can't we generate energy from coldness?
Like the temperature of the universe in 1 K, can this be used in the near future to be used as an energy resource for probes or satellites?
Here is the link to the article that made me think about this. Somewhere in the middle it is written that scientists can harness the cold energy using some active input method.
I think this article is poorly written.
| Heat is not “converted to heat energy”. Heat is the transfer of energy due solely to a temperature difference. Without a temperature difference there can be no heat. The consequences of that transfer can, but doesn’t necessarily, result in work.
In the case of the thermoelectric effect, heat can generate a voltage which in turn can produce electrical work. So in order for the thermoelectric effect to generate a voltage using a 1 K temperature source, you would need a sink with a temperature between 0 K and 1 K.
I hope this helps.
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Shouldn't the projection on the y-axis, of a radial vector in a circle making an angle theta with the x axis, also be the radius?
I don't understand this diagram... the projection is meant to be a shadow, right? So shouldn't the shadow of the radial vector as shown in the circle on the y-axis, be the entire y-axis (since the radius is the same, everywhere)? Why is the projection shown to be smaller than the radius?
For instance, let the origin be O. So shouldn't the projection of the phasor, on y-axis, be OA and not Asin(theta)?
| If think that shadow is not going to be the proper term here, since if you want to talk about shadows you have to say from where the light rays are coming. What we're using here is the idea of orthogonal projection in this, if you want still to think about shadows, the light rays come orthogonal to one of the axis.
For example, if you want to find the orthogonal projection of the radius on the $y$ axis, your light rays will come at $90^\circ$ on the $y$ axis. In this case is clear that the whole $y$ axis will be covered by the shadow of the radius only when the radius is exactly aligned with the $y$ axis. Same goes for the projection on the $x$ axis where now the light rays come at $90^\circ$ to the $x$ axis. One of this "light rays" is even shown in the circle by the dashed lines.
It should be clear with this discussion that different angles that the radius makes with the $x$ axis, will have different length for the orthogonal projection on the two axis. The idea of using orthogonal projection boils down to the possibility of defining, geometrically and analytically, the $\sin\theta$ and $\cos\theta$ functions which are defined on right triangles. The right triangle in the figure is the one which has the origin, the tip of the radius and the point o the projection on the $x$ axis as vertices.
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Why does ponytail-style hair oscillate horizontally, but not vertically when jogging? Many people with long hair tie their hair to ponytail-style:
Closely observing the movement of their hair when they are running, I have noticed that the ponytail oscillates only horizontally, that is, in "left-right direction". Never I have seen movement in vertical "up-down" direction or the third direction (away-and-back from the jogger's back). Why is the horizontal direction the only oscillation?
| I think the longitudinal oscillations of a ponytail are quickly damped (more precisely overdamped), since they involve layers of hair sliding along each other, as well as inelastic collisions of the back of the neck. On the other hand, the transversal oscillations require merely twisting the ponytail near the elastic band holding it together.
I think it would be harder to make an argument based on the forces causing the oscillations, since stepping from one foot onto another involves displacement in all directions: forward/backward, up/down, left/right.
Update
Those interested in exploring this subject deeper may be interested in article The shape of a ponytail and the statistical physics of hair fiber bundles, as well as in a more extended abstract discussion in this thesis (open access).
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Can spacetime be curved even in absence of any source? Einstein's equation in absense of any source (i.e., $T_{ab}=0$) $$R_{ab}-\frac{1}{2}g_{ab}R=0$$ has the solution $$R_{ab}=0.$$
But I think $R_{ab}=0$ does not imply that all components of the Riemann-Christoffel curvature tensor $R^c_{dab}$ be zero (or does it?). From this can I conclude that spacetime can be curved even in absence of any source?
| You are right. $R_{ab}=0$ does not imply $R^{a}_{bcd}=0$. For one thing, $R_{ab}$ has 10 components (in $n=4$ dimensions), whereas $R^{a}_{bcd}$ has $20$ components. The simplest example I can think of is Schwarzschild solution, which has $R_{ab}=0$ everywhere but $R^{a}_{bcd}\neq0$. If you allow the inclusion of a cosmological constant, then the de Sitter metric is an example of an empty solution with non-trivial spacetime curvature. As pointed out here
https://physics.stackexchange.com/a/105336/96768
A spacetime containing gravitational waves is empty but with non-trivial Riemann tensor.
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Can liquids exert/be affected by normal force? I am doing a problem in which I am drawing free-body diagrams for different objects in a system, and two of the objects which are included in my scenario are Earth and a body of water which is resting directly on earth (the tank/container is being ignored).
What is the force(s) that are present in between the Earth and the water, or in other words, what force prevents a liquid from passing through a solid object? I am looking for a force similar to the normal force, but as far as I can tell, the normal force can only exist between two SOLID objects, not a solid and a liquid.
Thank you for helping me out!
|
I am looking for a force similar to the normal force, but as far as I can tell, the normal force can only exist between two SOLID objects, not a solid and a liquid.
Not quite; liquids can also apply/receive a normal force. The catch is that additional normal forces must act in the other two directions as well. (For example, a normal force in the x-direction requires normal forces in the y- and z-directions.) The reason is that the ideal liquid can't sustain a shear stress, and all other stress states than equitriaxial stress (the compressive version is called hydrostatic stress, also known as pressure) would result in a shear component. Thus, only equitriaxial loading is possible at equilibrium.
If we consider gravity, then the normal forces on the bulk liquid aren't all the same (because the body force of the weight represents an additional load that breaks the cubic symmetry), but each infinitesimal element of the liquid is still under equitriaxial stress.
With that constraint met, there's no problem in assigning normal forces; one example would be the force that pushes back when you try to compress a liquid in a container.
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Transformations in classical field theory and configuration space When transforming a field in classical field theory the transformation of the four-gradient of this field follows automatically. At least this is what i have learned in my lectures.
This circumstance kind of contradicts my understanding of the Lagrange formalism in classical mechanics. In classical mechanics the generalized coordinates and their derivatives are understood as coordinates in configuration space. Therefore i can perform a transformation in configuration space only concerning the generalized coordinates and not their derivatives.
So there is probably a strong difference between the Lagrange formalism of fields and the Lagrange formalism of classical mechanics which I do not understand yet.
Question:
Why is it possible to transform $q(t)$ in classical mechanics without transforming $\dot{q}(t)$, but it is not possible to transform $\phi(x)$ without transforming $\partial_\mu \phi(x)$?
| That seems to be a misunderstanding. The field-theoretic case behaves in essentially the same way as the point mechanical situation.
*
*On one hand, the arguments of the Lagrangian density $${\cal L}(\phi,\partial_t\phi,\partial_x\phi,\partial_y\phi,\partial_z\phi,t,x,y,z)$$ are independent variables in the same way as the arguments of the Lagrangian $$L(q,\dot{q},t)$$ are independent variables.
*On the other hand, inside the actions $S[\phi]$ (and $S[q]$), the derivatives $\partial_{\mu}\phi$ (and $\dot{q}$) depend on $\phi$ (and $q$), respectively.
The proof is a straightforward generalization of my Phys.SE answer here.
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Anti-symmetrized total tensor of two anti-symmetric tensors Suppose we would like to anti-symmetrize a tensor
$$T^{\mu_1, \mu_2,\ldots, \mu_n} = G^{[\mu_1, \mu_2,\ldots, \mu_r]} H^{[\mu_{r+1},\ldots, \mu_n]},$$
where $G$ and $H$ are anti-symmetric. One could do this iteratively by applying the anti-symmetrizer
$$\Lambda_{1,2,\cdots, n} = \frac{1}{n}\left(1-\sum_{i=1}^{n-1} P_{i,n}\right)\Lambda_{1,2,\cdots, n-1}, \quad \Lambda_{1,2} = \frac{1 - P_{1,2}}{2}$$
where $P_{i,n}$ swaps the indices $i$ and $n$, for instance
$$P_{1,2} T^{\mu_{1} \mu_2} = T^{\mu_2 \mu_1}$$
however this method makes no use of the fact that $G$ and $H$ are already anti-symmetrized. My question is whether there exists a simplified anti-symmtrizer which anti-symmetrizes $T$ with a minimal amount of operations.
One immediate simplification is that we know that $\Lambda_{1,\ldots,r}$ will have no effect on the tensor $G$ and thus no effect on $T$ either. And something similar for $H$.
One idea is that this problem is somehow related to finding all partitions like $\{1,\ldots, r\}$ and $\{r+1,\ldots, n\}$ via swapping elements between them, but I cannot figure out what the correct algorithm should be. I do however know that there are a total of $\binom{n}{r}$ such partitions meaning that we should have a sum of that many products of operators in the end. In the end, this problem is related to a few problems in both relativity and fermionic statistics, but I cannot seem to find a solution anywhere.
| One can think of this problem as all different swaps with the components of the two tensors. With this in mind, we start by defining the index ordered tensors
$$G^{\{\mu_1,\ldots, \mu_r\}} = G^{[\mu_{j_1},\ldots, \mu_{j_r}]},$$
where $(j_1,\ldots, j_r)$ is a permutation of $(1,\ldots,r)$ such that $$\mu_{j_1}< \mu_{j_2} <\ldots < \mu_{j_r}.$$
This definition is important since we then not longer have to worry about the ordering in a particular anti-symmetrized tensor, for instance
$$G^{\{\mu_1, \mu_2, \ldots\}} = G^{\{\mu_2, \mu_1, \ldots\}}.$$
Then the anti-symmetrizing process is done with
$$T^{[\mu_1\ldots \mu_n]} = \sum_{i_r = r}^{n} \sum_{i_{r-1}}^{i_r-1} \cdots \sum_{i_1 = 1}^{i_2 - 1} \text{sgn}\binom{1\cdots n}{i_1 \cdots i_n} P_{r,i_r} P_{r-1,i_{r-1}}\cdots P_{2,i_2} P_{1,i_1} G^{\{\mu_1,\ldots\mu_r\}} T^{\{\mu_{r+1},\ldots\mu_n\}}$$
where $i_{r+1},\ldots,i_n$ is the ordered list of all indices occurring in $T$. SO for instance if we had the case
$$G^{\{\mu_3,\mu_4, \mu_6, \mu_7\}} T^{\{\mu_2,\mu_5,\mu_1\}},$$
then $(i_1,i_2,i_3,i_4,i_5,i_6, i_7) = (3,4,6,7, 1,2,5)$, since we first order the sublists. It turns out that under this swapping process the first $(i_1,\ldots, i_r)$ will in fact remain ordered. Ideally I would still like to simplify this somewhat into a recursive form such as that of the anti-symmetrizer.
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The thin target approximation in particle physics In the first paragraph of chapter 2 in this paper, the authors say that an effective volume is reduced to an effective area, when the thin target approximation is valid.
What is the thin target approximation, and why this approximation is valid when the interaction length of the particle is much longer than the thickness of the medium? When the target is thick, do we expect different types of interactions?
It would be appreciated, if you could explain qualitatively with formulas.
| When a beam of $N$ particles hits a target of density $n/A$ (e.g. per square cm) with a reaction cross section of $\sigma$, the number of scattered particles is:
$$N_S = N\frac n A \sigma $$
It is standard to use an areal density for the target because that is what the beam sees: stuff in its way. The depth isn't relevant. That is the thin target approximation.
Note that if the material has a volumetric density of $\rho$, then the mean free path of the beam is:
$$ \lambda = \frac 1 {\rho\sigma} $$
If the target is of length of order $\lambda$, then then a significant fraction of the beam is going to be scattered by the time it reaches the back for the target, and that part of the target won't see $N$ incident particles, it will see $N/e$ particles, as the number of particles at depth $z$ will be:
$$ N(z) = Ne^{-z/\lambda} $$
Your linked article is dealing with neutrinos, where $\lambda$ is on the order of light years, meaning the approximation:
$$ N(z) = Ne^{-z/\lambda} \approx N $$
is valid.
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How does radiation (heat) take away momentum? In another post, I was taught that when we are moving (running, for example), radiation (in the form of heat, both from our muscles and friction with the ground) takes away from our momentum. That makes perfect sense and I understand.
I was also taught that heat/radiation is directional.
My question is: since radiation is directional, is that why it takes away from momentum (which is also directional)? Is it correct to say that? In other words, does radiation take away from momentum because they are both directional concepts?
Also, does radiation, since it is directional, dissipate in the same direction of momentum, in the opposite direction, or no particular direction?
I am not an expert, hoping just for simple clarification, if possible.
| Radiation will not "know" that it is being emitted by a uniformly moving body, so from the prespective of the body it will be emitted uniformly in all directions. Momentum will be lost to the body to a very small degree, though, because the mass of the body will be reduced by a very small degree due to energy/mass equivalence. However, the body will not change its velocity- only its mass and therefore its momentum. In practice, the change is too small to observe.
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How to transform velocity 4-vectors to Zero Momentum Frame I have a particle $p$ with speed $u$ in lab frame approaching a stationary particle $q$.
The $p^{\mu}$ and $q^{\mu}$ velocity 4-vectors are:
$$p_{LAB}^{\mu}=\gamma_u(c, u, 0, 0)$$
$$q_{LAB}^{\mu}=(c, 0, 0, 0)$$
To get to ZMF, I need a standard lorentz boost with speed $v=u/2$:
$$p_{ZMF}^{\mu}=
\begin{pmatrix}
\gamma_{v} & -\gamma_v \beta_v & 0 & 0\\
-\gamma_{v} \beta_v & \gamma_v & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{pmatrix}\begin{pmatrix}
\gamma_{u}c\\
\gamma_{u}u\\
0\\
0\\
\end{pmatrix}=\gamma_u \gamma_{\frac{u}{2}}
\begin{pmatrix}
c-\frac{u^2}{2c}\\
\frac{u}{2}\\
0\\
0\\
\end{pmatrix}
$$
and
$$q_{ZMF}^{\mu}=
\begin{pmatrix}
\gamma_{v} & -\gamma_v \beta_v & 0 & 0\\
-\gamma_{v} \beta_v & \gamma_v & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{pmatrix}\begin{pmatrix}
c\\
0\\
0\\
0\\
\end{pmatrix}=\gamma_{\frac{u}{2}}
\begin{pmatrix}
c\\
-\frac{u}{2}\\
0\\
0\\
\end{pmatrix}
$$
The magnitude of the first spatial component of $p_{ZMF}^{\mu}$ is a $gamma_u$ times more thanthe first spatial component of $q_{ZMF}^{\mu}$. I would expect that in the ZMF, they are opposite sign but otherwise equal. Is this expectation wrong, and if not, what am I doing wrong?
| Here are alternative approaches to @Philip's answer (where the masses of your particles are assumed equal).
*
*Although Galilean-velocities add linearly, [real-world Minkowskian-]velocities don't...as you found. In fact, Euclidean-slopes also don't add linearly. (Linearity of Galilean-velocities is more the exception, rather than the rule.)
However, angles (arclengths of "unit circles", or twice-the-areas of "unit-circular"-sectors) are additive.
If you were given these two rays in ordinary euclidean geometry,
how would you find the ray in the "center" between these two rays?
...so that if you rotated to that center-ray, the other rays would be on opposite sides of it, oriented symmetrically. We would call that center-ray the "angle bisector".
In the diagram below, let's measure slopes with respect to the vertical.
If the slope of the angled-ray is $u$,
then the slope of that center-ray ("angle bisector") would be $$m=\tan\left(\frac{\arctan(u)}{2}\right)$$
(This should agree with the result gotten using the relative-slope formula $\tan\theta=\displaystyle\frac{m_2-m_1}{1+m_2m_1}$ with @Philip's method.)
What would the analogous quantity be for special relativity?
*The following approach also works when the masses are unequal (and is generalizable to multiple particles).
The "center-of-momentum" frame is the frame where the "spatial-component of the total 4-momentum" is zero, which means that this frame has a 4-velocity along the direction of the total-4-momentum.
So, one can determine this 4-velocity as the unit-vector along the total-4-momentum:
$$\hat p_{COM}
=\frac{\tilde p_{1}+ \tilde p_{2}}{ |\tilde p_{1}+ \tilde p_{2}| }
=\frac{\tilde p_{1}+ \tilde p_{2}}{ M_{COM}c },
$$
which can be written in terms of the invariant-mass of the system $M_{COM}$.
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Isn't $\epsilon_{ij}$ an isotropic, rank-2 tensor? Definition of isotropic tensor: components are unchanged after rotation: $T_{ij}\rightarrow T_{ij}' \equiv R_{ia}R_{jb}T_{ij} = T_{ij}$
MathWorld says there is only one rank-2 isotropic tensor, $\delta_{ij}$.
But with
$$\epsilon_{ij}=\left(\begin{matrix}0&1\\-1&0\end{matrix}\right)$$
there is no change either:
$$\epsilon_{ij}\rightarrow\epsilon_{ij}'=R_{ia}R_{jb}\epsilon_{ab}=\epsilon_{ij}$$
So it seems to me that $\epsilon_{ij}$ is also a rank-2 isotropic tensor, in addition to $\delta_{ij}$.
What am I getting wrong?
Notes:
*
*$R_{ij}=\left(\begin{matrix}\cos a&-\sin a\\ \sin a&\cos a\end{matrix}\right)$
*I asked at math.stackexchange, but got no answer. Maybe this is more Phys Math Met as in Boas, which I was reading when this question came up.
| I think @mikestone's comment is correct. The MathWorld site is talking about tensors in 3 dimensions, not 2D. In 3 dimensions (actually, in all dimensions $\geq3$), it can be shown that any isotropic rank-2 tensor is proportional to the identity ($\delta_{ij}$), see Richard Fitzpatrick's notes here, for example.
In two dimensions, there are two rank-2 isotropic tensors, $\delta_{ij}$, and what you have called $\epsilon_{ij}$. Here's a quick way to list all the isotropic tensors in 2D that I have drawn heavily from the fantastic analyses here and here. There are two equivalent ways to define an isotropic tensor: one is the way you have defined it, saying that $A$ is an isotropic tensor iff $$A = R\cdot A\cdot R^T,$$
but an equivalent way is to say that $A$ is isotropic iff it commutes with the generators of rotations, $L_i$, for example $[A,L_z] = 0$. Now, let's look at this condition in 2D. It turns out -- if you do the calculations -- that $$L_z = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix},$$ precisely (the negative of) what you called $\epsilon_{ij}$!
Now consider an arbitrary 2D tensor $A$, and demand that it satisfy the commutation relation given above:
$$\Bigg[ \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}, \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\Bigg] = 0,$$
and you should be able to see that this just means such an arbitrary isotropic tensor in 2D should look like: $$A = \begin{pmatrix}a & -b \\ b & a\end{pmatrix},$$ and you should be able to see two things:
*
*An arbitrary isotropic tensor in 2D is itself a rotation, since $$A = \frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix}, \quad \quad \alpha=\arctan{\left(\frac{b}{a}\right)}$$
*An arbitrary isotropic tensor in 2D can be written as: $$A_{ij} = a \delta_{ij} -b \epsilon_{ij}.$$
Thus, there are precisely two independent isotropic rank-2 tensors in 2D, $\delta_{ij}$ and $\epsilon_{ij}$.
Note: I've not spoken too much about generators and how to derive $L_z$ in 2D since the answer would get too long, but I'd be happy to explain it if necessary.
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How is most of the forces occurring around us related to electromagnetism? Here is an extract from my textbook-
Most of the phenomena occurring around us can be described under electromagnetism. Virtually every force that we can think of like friction, chemical force between atoms holding matter together, and even the forces describing processes occuring in cells of living organisms, have its own origin in electromagnetic force.
So, how are these forces related to that electromagnetic force and how do they have origin in it?
| In physics, There are four fundamental forces of nature which give a rise to other phenomenological force, for example, friction. Such forces are commonly encountered in everyday physics and are approximately described by empirical equations because the concept of force is meaningful only if one knows how to solve problems involving forces.
The
phenomenological forces, such as friction, the contact
force, and the viscous force, can be described by relatively simple
empirical mathematical expressions, but when examined in detail they
can be explained as the macroscopic manifestation of complicated interatomic
forces.
The four fundamental forces are the gravitational and electromagnetic interactions, which produce significant long-range forces whose effects can be seen directly in everyday life, and the strong and weak interactions, which produce forces at minuscule, subatomic distances and govern nuclear interactions. Some scientists hypothesize that a fifth force might exist, but these hypotheses remain speculative.
how do these forces are related to that electromagnetic force and how do they have origin in it?
Let's talk of normal force as friction is much harder to examine at the microscopic level (see here for more):
The origin of the normal force is similar to the origin of tension in a
string. When we put a book on a table, the molecules of the book exert
downward forces on the molecules of the table. The tabletop moves
downward until the repulsion of the molecules in the table balances
the force applied by the book. The more rigid the surface, the smaller
the deflection. Because no surface is perfectly rigid, compression always
occurs. However, the compression is usually too slight to notice
and for most purposes, we can assume ideal surfaces that are perfectly
rigid.
In equilibrium, the normal force $N$ exerted on a body by a surface is
equal and opposite to the resultant of all other forces that act on the body
in the perpendicular direction.
Edit: Intermolecular forces are the forces that mediate the interaction between atoms, including forces of attraction or repulsion (electromagnetic forces) which act between atoms and other types of neighboring particles, e.g. atoms or ions.
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Unruh effect: should Minkowski vacuum transform according to different observers? It's known that the Minkowski vacuum is observed as a thermal bath for Rindler observers, in paticular:
$\langle0_{M}|N_{M}|0_{M}\rangle=0 \space\space\space\space\space\space\space $ (1)
$\langle0_{M}|N_{R}|0_{M}\rangle \neq 0\space\space\space\space\space\space\space\space$ (2).
However, from eq(1) to eq(2) we change from inertial observers to Rindler observers, so why the $|0_{M}\rangle$ remains the same in these two equations, since we know that quantum state vectors transform according to different observers?
| The fact that in your two equations there's $|0_M \rangle$ comes from what you want to measure. The first one is trivial to interpret: It is the number of particles in the Minkowski vacuum for the Minkowski observer. But the second one is a bit tricky: It is the number of particles in the Minkowski vacuum from the point of view of the Rindler observer. You could have for example $\langle 0_R | N_R | 0_R \rangle$ but it is identically null. The Unruh effect is precisely the measure of the number of particles in the Minkowski vacuum from the point of view of the Rindler observer.
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If $Q = mC_p\Delta T$, shouldn't $C_p$ be updated every increment? Considering that the specific heat capacity of a material is a function of temperature, is it wrong to assume that $C_p$ is constant for notable differences in temperature ($>50\mathrm{K}$)?
It is something that I just noticed but can not find a conclusive answer to and it bothers me that I just do not know this already!
| To reflect the dependence of $c_p$ on temperature, the equation should be $Q = m\int_{T_0}^{T_f}c_p(T)\enspace dT$. You perform the integration to obtain Q. Yes, $c_p(T)$ is a function of temperature, in general, but for solids over a small temperature range it is relatively constant.
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How come the formula $W=Fd$ doesn't apply for energy stored in springs? I always thought that work is like the energy transferred and it is given by $W=Fd$, but this concept gets problematic for springs.
If the force $F$ is applied to a spring which compresses it by a length $d$, then apparently the energy stored in the spring is $$E_{p}=\frac{1}{2}kd^2=\frac{Fd}{2}$$
Why is the energy transferred to the spring not $Fd$?
| I think that this question shows a misunderstanding of calculus rather than that of the
physics.
For constant forces, the work is defined as:
$$ W = F \cdot s$$
For applying this to calculate work of variable forces
Consider an interval of extension $(x_o, x_o+h)$, if we were to shrink the size of $h$ to become very small, then we can say that the force is more or less constant over this interval under consideration, and hence we can apply the definition of work for constant force:
$$ \Delta W = F \Delta h$$
Shrinking $h$ to zero and summing this quantity over all such intervals possible from $x_o$ to $x_{f}$, where $x_f$ is the final displacement, we get the work by an integral::
$$ W = \int_{x_o}^{x_f} F dh$$
| {
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Relativistic energy of harmonic oscillator What is the relativistic energy of an harmonic oscillator:
$$\frac{m_0 c^2}{\sqrt{(1-\frac{v^2}{c^2})}}+\frac{1}{2}kx^2$$
Or
$$\frac{{m_0 c^2}+\frac{1}{2}kx^2 }{\sqrt{(1-\frac{v^2}{c^2})}}$$
I think the first one is true but I need an exact logic or derivation.
| What you need here is the special relativity version of the work-energy theorem.
The proof is given in many places, including that Wikipedia page, but you start out from the idea that the relativistic force is given by :
$$\vec{F}=\frac{d}{dt}(\gamma m_0 \vec{v})$$
and you will (after some math) get :
$$m_0c^2(\gamma_2-\gamma_1)=\int_{\vec{x_1}}^{\vec{x_2}}\vec{F}\cdot d\vec{x}$$
Where the subscripts refer to initial and final positions. The expression on the left is the change in kinetic energy, and the expression on the right is simply the work done.
For the simple harmonic oscillator in your case that reduces to :
$$W_{21}=\frac{1}{\gamma_2-\gamma_1}\frac k 2 (x_2^2-x_1^2)k$$
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Mathematical Definition of Power I am a high school student who was playing around with some equations, and I derived a formula for which cannot physically imagine.
\begin{align}
W & = \vec F \cdot \vec r
\\
\frac{dW}{dt} & = \frac{d}{dt}[\vec F \cdot \vec r] = \frac{d\vec F}{dt} \cdot \vec r + \vec F \cdot \frac{d\vec r}{dt}
\\
\implies & \boxed{P = \frac{d\vec F}{dt} \cdot \vec r + \vec F \cdot \frac{d\vec r}{dt}}
\end{align}
I differentiated Work using its vector form formula $\vec F \cdot \vec r$
So I got this formula by applying the product rule. If in this formula $\frac{d\vec F}{dt}=0$ (Force is constant), than formula just becomes $P = \vec F \cdot \frac{d\vec r}{dt}$ which makes total sense, but this formula also suggests that if $\frac{d\vec r}{dt}=0$ then the formula for power becomes $P =\frac{d\vec F}{dt} \cdot \vec r$, which implies that if the velocity is zero that doesn't necessarily mean that Power of the object will also be zero!
But I don't find this in my high school textbook and I can't think of an example on that top of my head where this situation is true.
From what I have heard and read, if the velocity of the object is zero then power is also zero.
Can someone please clear my supposed misconception or give me an example of the situation where this happens?
| The work done by a force is not defined by $W=\mathbf F\cdot\mathbf r$. Work is instead defined in terms of a line integral over a path (your equation just assigns a work for a force and position, which does not match what we mean by the work done by a force). We have
$$W\equiv\int\mathbf F\cdot\text d\mathbf r\to\text dW=\mathbf F\cdot\text d\mathbf r$$
So when we have $P=\text dW/\text dt$ we just have
$$P=\frac{\text dW}{\text dt}=\frac{\mathbf F\cdot\text d\mathbf r}{\text dt}=\mathbf F\cdot\frac{\text d\mathbf r}{\text dt}=\mathbf F\cdot\mathbf v$$
So there is no $\mathbf r\cdot \text d\mathbf F/\text dt$ term in the expression for power. This works out conceptually as well: the power output of a force should not directly depend on the position of the particle (i.e. the location of the origin) in question.
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Independent variables of the inner energy in thermodynamics the free energy is a function of the volume V and the temperature T. These two variables are called independent. On the other hand on can find the volume and the temperature as a part of the ideal gas equation, which means that they are related to each other.
Therefore I dont understand, why they are named independent.
I'd appreciate it if someone could help me.
| The ideal gas equation links the temperature volume and pressure, so it allows you to determine one of these parameters in terms of the other two. So we can consider the ideal gas equation as determining the pressure in terms of volume and temperature, leaving $T$ and $V$ independent.
| {
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Why do low-energy waves produce heat, but high-energy waves do not? Radio waves, microwaves and infrared are known to produce heat and even cause burns, while visible light and ultraviolet are not. This seems counterintuitive to me, as the latter contain the highest amount of energy.
Why is this? Does it have to do with the quantity of waves rather than the energy?
| I'd like to add to the other answers by mentioning that absorbance is a function of photon frequency (thus also the photon energy, through $E = h \nu$). For instance, here's a Wikipedia article on the absorbance of water.
| {
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Meaning/origin of the boltzmann curve I have physical data and I wanted to use a logistic function to fit them. In this context, I came across the so-called Boltzmann-fit which fits nicely to the data but since I wonder where does it come from:
I struggle to separate (if even possible) that Boltzmann-fit from the Maxwell-Boltzmann distribution curve and the Boltzmann equation.
I'm also not able to find that much about this fit which makes me wonder as Boltzmann contributed probably among few others some of the most crucial findings.
In short: Where does this "fit" come from? Is it the Boltzmann equation? Is it cconnected to the Maxwell-Boltzmann distribution? Is it something else?
edit: Some "official" information:
| Most physicists would call this fitting by a Fermi function, since Fermi function
$$
f(E)=\frac{1}{1+e^{\beta(E-\mu)}}\approx_{\beta\rightarrow 0} e^{\beta(\mu-E)},
$$
which gives you Boltzmann distribution only in a high temperature limit (second equality). While it is not clear where the term Boltzmann-fit comes from, fitting by this curve is the essence of logistic regression (although sometimes hidden in mathematical details), as it approximates a probability distribution where the data can take one of two values.
Another place where it frequently arises is when solving ordinary differential equations with two stable states (again, often well beyond physics).
| {
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Why every system tends to be more stable? An entire topic of inorganic chemistry, i.e. chemical bonding, which is also one of the most important topics, is based on the idea of stability. But whenever I ask why every system tends to get stable, I never get a satisfactory answer. People say that's how nature "wants" to be. How does a system know that it wants to get stable?
| A system state is called stable if the system returns to that state when is disturbed by a small amount. As well as that, if the system gets similar enough (which might not actually be very similar) to a stable state the system will transform to the stable state.
In this sense a system tends to become stable because stable states are the only states a system can remain in.
In energy terms a stable state is usually a state of lowest energy. The reason is that usually it is easier to lose energy than to gain it. Atoms rearrange themselves to lower-energy configurations and can lose the extra energy all at once or in stages in a variety of ways. It is usually harder to acquire the energy to change to a higher-energy configuration, and so the lower energy configuration is more stable. There are exceptions - ice is in a lower-energy state than water, but there is often enough available energy that ice is more likely to gain energy than water is to lose it.
| {
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Why is it impossible to measure position and momentum at the same time with arbitrary precision? I'm aware of the uncertainty principle that doesn't allow $\Delta x$ and $\Delta p$ to be both arbitrarily close to zero. I understand this by looking at the wave function and seeing that if one is sharply peeked its fourier transform will be wide.
But how does this stop one from measuring both position and momentum at the same time?
I've googled this question, but all I found were explantions using the 'Observer effect'. I'm not sure, but I think this effect is very different from the intrinsic uncertainty principle.
So what stops us from measuring both position and momentum with arbitrairy precision?
Does a quantum system always have to change when observerd? Or does it have to do with the uncertainty principle?
Thank you in advance.
| The OP wrote:
I understand that for a given wavefunction that if $\Delta x$ is small, $\Delta p$ will be big and how this arises from fourier transformations. But I fail to see how this prevents anyone from doing a simultaneous measurement of both $x$ and $p$ with infinite precision.
This seems to boil down to the issue of what "simultaneous measurement" means. What it means to simultaneously measure two observables $A, B$ is to perform a single measurement on the system, obtaining the values $a$ and $b$, such that, immediately after the measurement, the system is in a state where the value of $A$ is certainly $a$ and the value of $B$ is certainly $b$.
In other words, the result of the measurement is that the system is in a simultaneous eigenstate of $A$ and $B$. Since there are no simultaneous eigenstates of $x$ and $p$ (as the OP already understands), this isn't possible for this particular pair of observables.
| {
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Why is it easier to handle a cup upside down on the finger tip? If I try to handle a tumbler or cup on my fingertip (as shown in fig), it is quite hard to do so (and the cup falls most often).
And when I did the same experiment but this time the cup is upside down (as shown in fig), it was quite stable and I could handle it easily.
In both the cases, the normal force as well as the weight of that cup is the same but in first case it falls down and in the other it is stable.
I guess that it is falling because of some torque but why is there no torque when it is upside down.
What is the reason behind this?
| Think about the position of the center of mass of the cup.
Assuming it's not one with an oddly heavy bottom or flimsy walls, the wallwill place that center of mass around the middle of the hollow interior of the cup.
Your finger acts now as a fulcrum point. Thus, the center of mass and your finger tip, in effect, form a very weird pendulum, with the center of mass being the bob, and your finger tip the anchor.
Now, is a pendulum more stable and easy to control when it's pointing down, or when it's pointing up? Which way is this "pendulum" pointing (draw an arrow from your fingertip to the center of mass)?
| {
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Why don't we have just one type of charge? Why do we need two? Why do we need to admit two types of charges (positive and negative)? Can't there be a third type?
| If you take a collection of materials (glass, amber, polythene, perspex, pvc, polystyrene....) and rub them with a soft insulating material, they acquire charges. You discover that all these rubbed materials can be put into just two categories. All those in one category repel each other and attract any of those in the other category. Two categories; two sorts of charge.
This is the gist of the original argument, dating back some three hundred years. Subsequent discoveries have given us no reason to challenge it. For example, electrons are repelled by suitably rubbed amber or polythene but attracted by glass or perspex and therefore fall into the negative charge category.
What is more, equal quantities of the two sorts of charge can be represented by $+R$ and $-R$ in which $R$ is a positive real number with unit. For example the force on charge $Q_2$ due to charge $Q_1$ is
$$\mathbf F=\frac{Q_1 Q_2}{4 \pi \epsilon_0 r^2} \mathbf {\hat r} $$
and if $Q_1$ is negative and $Q_2$ is positive, $\mathbf{F}$ is in the opposite direction to $\mathbf {\hat r}$, the unit vector in the $Q_1Q_2$ direction.
| {
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Please explain the meaning of below statement Newtons second law is a local law.
(In the book,it says that it means that it applies to a particle at a particular instant without taking into consideration any history of the particle or its motion.)
Um, I couldn't understand what do they mean by " taking into consideration any history of the particle or its motion ".
If possible ,please explain it with an example.
| When the word local is used here in Newtonian physics to describe an object or interaction, it means “happening at a particular point in space at a particular time”. This means that you must ignore it’s previous history/trajectory.
For example, if we consider a local force $F(x)$ acting on a particle, this means that this force acts at a point in space (the location of the particle $x$) at an instant of time at $x$.
I think the need to emphasise the concept of locality in Newton’s laws of motion, may have arisen from the idea at the time that Newton’s Law of Universal Gravitation given by
$$F = \frac{GMm}{r^2}$$
seemed to imply that this force acted instantly at a distance $r$, and was therefore termed nonlocal.
We know now that gravity is an emergent property resulting from the curvature of spacetime as described by the Einstein field equations and does not act instantaneously, rather at the speed of light $c$.
| {
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Completeness relation of spherical harmonics In spherical coordinates, the resolution of the identity can be written as $$ 1=\int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta\, d\theta\, |\theta,\phi\rangle\langle\theta,\phi| \equiv \int d\Omega |\Omega\rangle\langle \Omega|,$$ where $|\Omega\rangle = |\theta,\phi\rangle$. For spherical harmonics $Y_{lm}(\Omega)$, we then have $$ \delta_{l'l}\delta_{m'm} = \int d\Omega\, Y_{l'm'}^\ast(\Omega)\, Y_{lm}(\Omega).$$ The resolution of the identity in the angular momentum basis is given by $$ 1=\sum_{l=0}^{\infty}\sum_{m=-l}^l |l,m\rangle\langle l,m|,$$ so that $$ \langle \Omega \mid\Omega'\rangle = \sum_{l=0}^{\infty}\sum_{m=-l}^l \langle\Omega\mid l,m\rangle\langle l,m\mid \Omega'\rangle\iff \delta(\Omega-\Omega')=\sum_{l=0}^{\infty}\sum_{m=-l}^l Y_{lm}(\Omega) Y_{lm}(\Omega').$$
Now, the term $\delta(\Omega-\Omega')$ is often rewritten as $\frac1{\sin\theta}\delta(\theta-\theta')\delta(\phi-\phi')$. How does one find this expression?
| For $\delta^{(2)}(\Omega-\Omega')$ to behave like a delta function, we should get $1$ when we integrate it over the surface of the unit sphere. In other words, we should have that
\begin{equation}
1=\int {\rm} d^2 \Omega \delta^{(2)}(\Omega-\Omega') = \int {\rm d}\theta {\rm} d \phi \sin \theta\delta^{(2)}(\Omega-\Omega')
\end{equation}
You can see this will work out if we take
\begin{equation}
\delta^{(2)}(\Omega-\Omega')=\frac{1}{\sin\theta} \delta(\theta-\theta')\delta(\phi-\phi')
\end{equation}
but will not work out if we do not include the $(\sin\theta)^{-1}$ factor. Indeed if we do not include this factor, then we will get $\sin\theta'$ instead of $1$.
More generally, in $D$ spacetime dimensions, one should write the $D$-dimensional Dirac delta function as
\begin{equation}
\frac{1}{\sqrt{|g|}}\delta^{(D)}(x)
\end{equation}
In spherical coordinates on the unit sphere, $\sqrt{|g|}=\sin \theta$. This is another argument to explain the factor of $(\sin\theta)^{-1}$.
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Engine rotating a generator at its rated speed, but engine power exceeds power required? What would occur if a gasoline engine would be driving a generator (specifically, a permanent-magnet synchronous AC 3-phase sinusoidal generator) as its only load at the output shaft and if that engine is rated at i.e. 35kW @ 8000 RPM but at the same speed the generator is rated to produce 30 kW of electricity, at 120V / 250A?
Is a “rated power” of an internal gasoline engine a rating for its maximum possible power output at a certain RPM? Meaning that if a load (in this case, the generator) requires less power input from the engine then the engine throttle would be positioned as such to allow the required fuel amount to be consumed, but not the maximum safe possible (because it is unnecessary)?
If the load at that speed would be larger (i.e. 35kW, and not 30) then the throttle would need to be opened more to increase the fuel flow to produce the required power?
Or does the engine produce 35kW at 8000 RPM regardless of generator load and the 5kW is unused i.e. dissipated through heat?
Excuse me if this is too simple but I would like to find clarification regarding this.
Thanks
| For a real application, the engine would have some type of feedback control that attempted to hold the engine at a constant rpm. If the load increased, the rpm would decrease, and more fuel would be sent to the engine to get back to the rpm setpoint. The opposite would happen if the load decreased. For the "extreme" case where the load exceeded the engine's capability, the engine would go to full throttle, but it wouldn't be able to meet the rpm setpoint, and it wouldn't be able to produce the required electrical power output.
| {
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Work done when you bring an bringing an object down from a height I am going to explain this question through an example.
Suppose I lift an object I apply a force $mg $ then I apply additional force, that would be $ma $ so total force would be $m(g+a)$. My doubt is that the work done by a person in lifting a box by applying more force than the weight that should be $m(g+a) *h $ right?
I would love to know if my answer is right or if someone could kindly correct me
| Normally in lifting, you apply a momentary extra force to get the object moving, and then proceed at a constant speed. As Dale points out, the extra work goes into kinetic energy.
| {
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Some questions regarding graviational potential energy and the concept of potential energy in general Starting off, I first want to know the relation between work and potential energy.
$\Delta\mathbf U = - W $
How was this expression formulated?
Moving on,
My second doubt was in the derivation of the expression:
$\mathbf U = \frac{-GMm}{r}$
*
*Why are we bringing the point mass from infinity to $\mathbf (x = r)$
*What's with the negative sign?
*Who's work done are we talking about here?
Another thing,
Why are we supposed to move the object without an acceleration? Is that even possible?!
Last but not the least,
$$\mathbf V_B - V_A = \frac{U_B - U_A }{m}$$
Where,
A mass $\mathbf m$ has been brought from point A to B under a gravitational field
and,
$\mathbf V_B - V_A$ is the change in potential
$\mathbf U_B$ and $\mathbf U_A$ denote the gravitational potential energy when the mass m is at point B and A respectively.
Can you explain this equation to me? and what $\mathbf V_B - V_A$ mean?
I know these questions are pretty obvious and dumb but I am a fresh mind to physics and the deep concept of potential energy. Also, it's my first time posting here. Forgive me if my questions weren't concise enough. My physics teacher just skimmed through this topic!!!
| Let's start by the definition of work. For simplicity I use,
$W = F \cdot \Delta x$
$W = m a_x \Delta x$
recall, $v^2 -v_o^2 = 2 a_x \Delta x $, so
$W = \frac{1}{2} m (v^2 -v_o^2 )$
$W = \frac{1}{2} m v^2 - \frac{1}{2} m v_o^2$
$W = K_f - K_i$
where $K$ is the kinetic energy.
Now, let's look at the conservation of energy.
$E_i = E_f$
$K_i + U_i = K_f + U_f$
$U_i - U_f = K_f - K_i$
The RHS is work. To get a $\Delta U$ we must have final minus initial so,
$- (U_f - U_i) = W$
$W = -\Delta U$.
Note that work is proportional to the change in potential energy. that means that to calculate work you do not need the exact value of the potential you just need to know how much it changes.
Let's try to find ( and define) gravitational potential.
$W = F \cdot \Delta x$
$W = F \cdot (x_f - x_i)$
$W = \frac{GMm}{r^2} \cdot (x_f - x_i)$
Let's replace ($W = -\Delta U= U_i - U_f$)
$U_i - U_f = \frac{GMm}{r^2} \cdot x_f - \frac{GMm}{r^2}\cdot x_i$
Let's look at the meaning of each term, $r$ is the distance between the two objects, $x_i$ is the initial distance between the objects, $x_f$ is the final distance between the objects. If you want to calculate the current gravitational potential between two planets, i.e. $r = x_f$, by comparing both sides of the equation we get (notice what is happening to the negative sign)
$U_i = - \frac{GMm}{x_i^2} x_i $
$U_f = - \frac{GMm}{x_f^2} x_f = - \frac{GMm}{x_f} $
We are only interested in the final position because it corresponds to the current positions of the planets. The initial one is not interesting to us because it does not tell us anything about what is happening between them now. Because it does not change our final answer we can assume that $x_i$ was very large and $U_i$ was practically zero. So in general it is correct to say.
$U = - \frac{GMm}{x} $
Who moved the planets together? gravity. The negative sign (which came out of the calculations) makes sure that you end up with a correct sign for your work.
Finally $\Delta V$, There are two masses in this equation $M$ and $m$. $\Delta V$ tells us what is the gravitational potential energy of M on 1 kg object at different locations ($r$).
| {
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Why can ropes pull but not push? If I have object that is heavy, I can pull it with rope but cannot push it. Why? What breaks the symmetry of the system? I can push or pull anything if I choose, so why is the push possibility not possible?
| I think this is the most complex answer possible to OP's question; I don't know if I should be proud or ashamed of myself.
I can pull it with rope but cannot push it. Why? What breaks the symmetry of the system?
The system has symmetries, but not the one you are thinking about, you are confusing yourself. The true symmetries arise from the following facts:
*
*The robe has fixed lenght $l$
*The rope is bendable but not stretchable
Think about it: you have an object and a rope attached to it, this divides 3D space in two regions: the first one is a sphere of radius $l$ and center in the position of the object; if you have the other end of the rope, the one not attached onto the object, in this region then everything is fine! In fact the rope is bendable!
But you cannot have the other end of the rope outside the sphere, because this would mean that the rope has broken since it has a fixed lenght $l$! This implies that if you try to get the other end of the rope outside the sphere then the center of the sphere must move to prevent you from getting into that impossible configuration! (Impossible without breakage or deformation)
| {
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Why don't opaque objects reflect light? My sister was doing a quiz and I tried to point her in the right direction by giving her scenarios to imagine. One of the questions in the quiz was:
Which of the following objects do not reflect light:
*
*Polished metal
*Mirror
*Undisturbed water
*Book
She suggested that the answer was "undisturbed water" and that made sense to me too.
But the answer given was "book", which didn't make sense to me. How can you even see the book if it didn't reflect light in the first place?
Is this terrible framing by her teacher or am I having a conceptual misunderstanding?
| This is a case where there's a word that technically refers to a general category, but is often used to refer to a particular subset of that category. The term "reflection" can refer to any redirection of light, but is often understood to refer to specular reflection. If the teacher had asked "Which of the following objects do not specularly reflect light?" it would have been more precise, but then there would be students wondering what "specularly" means.
| {
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Derivation operators as arguments or the Hamiltonian In a book I am reading about QFT (Quantum field theory by Mark Srednicki ,page 48), I see the following equation:
$$
\int \mathcal D p\mathcal D q \exp\left[i\int_{\mathbb R} dt (p\dot q - H_0(p,q)-H_1(p,q) +fq+hp)\right] $$
$$=\exp\left[i\int_{\mathbb R} dt (H_1(i\delta/\delta h(t),i\delta/\delta f(t)))\right]$$
$$\times \int \mathcal D p\mathcal D q \exp\left[i\int_{\mathbb R} dt (p\dot q - H_0(p,q)+fq+hp)\right] \tag{6.22}
$$
where
*
*$H=H_0+H_1$ is the Hamiltonian of the system, and $H_1$ is a small perturbation.
*$\mathcal D q$ denotes path integral.
*$\delta/\delta f(t)$ is the functional derivative operator; see here for the definition.
I do not understand how this equation works. Why could we insert differential operators in the place of $p,q$. And if I take the part $\exp\left[i\int_{\mathbb R} dt (H_1(i\delta/\delta h(t),i\delta/\delta f(t)))\right]$ under the integral sign, I find that it the operator has nothing to act on.
How to understand this equation?
| Let's look at an ordinary integral
\begin{equation}
\mathcal Z(h) = \int dx e^{-f(x) + xh},
\end{equation}
with $f(x)$ such that the integral is convergent. It should be obvious that
\begin{equation}
\int dx \, x \, e^{-f(x)+xh} = \frac{\partial}{\partial h} \int dx \, e^{-f(x)+xh}
\end{equation}
or
\begin{equation}
\int dx \, x^n \, e^{-f(x)+xh} = \frac{\partial^n}{\partial h^n} \int dx \, e^{-f(x)+xh}.
\end{equation}
Now using those you can easily get
\begin{eqnarray}
I(h) & = & \int dx \, g(x) \, e^{-f(x)+xh} = \int dx \, \sum_{n=0}^\infty\frac{g^{(n)}(0)}{n!}x^n \, e^{-f(x)+xh}=\sum_{n=0}^\infty \frac{g^{(n)}(0)}{n!}\int dx \, x^n \, e^{-f(x)+xh} \\
& = &
\sum_{n=0}^\infty \frac{g^{(n)}(0)}{n!}\frac{\partial^n}{\partial h^n}\int dx \, e^{-f(x)+xh}=g\left (\frac{\partial}{\partial h} \right) \int dx \, e^{-f(x)+xh}
\end{eqnarray}
In your specific case $g(x)$ is an exponent
| {
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Analytic solution to Kepler's Problem, exegesis From 'Solving Kepler's Problem' by Colwell, the first analytic solution to Kepler's Problem used a theorem of Lagrange, and later Burmann, to invert Kepler's equation. When you look on the internet for a proof you find these lines that begin the section on Burmann's theorem (copied straight from Whitaker: A Course in Modern Analysis) Burmann's Theorem
Given $f(z)$ analytic on a region, $\phi(a)=b$, and $\phi'(a) \neq 0$, then Taylor's theorem gives:
$$\phi(z)-b = \phi'(a)(z-a)+\frac{\phi''(a)}{2!}(z-a)^2+...\tag{1}$$
If it is legitimate to revert this series the result is:
$$z-a=\frac{\phi(z)-b}{\phi'(a)}-\frac{1}{2}\frac{\phi''(a)}{\phi'(a)^3}[\phi(z)-b]^2+...\tag{2}$$
How do you get from the first equation to the second?
| first solve this equation for x:
$$-2\,{\frac {\varphi _{{b}}}{\varphi _{{{\it ss}}}}}+2\,{\frac {
\varphi _{{s}}}{\varphi _{{{\it ss}}}}}\,x+{x}^{2}
=0$$
where:
*
*$\varphi_b=\phi(z)-b$
*$\varphi_s=\phi'(a)$
*$\varphi_{ss}=\phi''(a)$
*$x=z-a$
you obtain :
$$x={\frac {-\varphi _{{s}}\pm\sqrt {{\varphi _{{s}}}^{2}+2\,\varphi _{{{
\it ss}}}\varphi _{{b}}}}{\varphi _{{{\it ss}}}}}
\tag 1$$
Taylor series for $$\sqrt{1+a\,\varphi_b}=1+\frac 12 a\,\varphi_b-
\frac 18 a^2\varphi_b^2$$
where $a=2\frac{\varphi_{ss}}{\varphi_s^2}$
substitute this result to equation (1) with $~\pm\mapsto +~$ you obtain:
$$z-a={\frac {\varphi _{{b}}}{\varphi _{{s}}}}-\frac 12\,{\frac {\varphi _{{{\it
ss}}}{\varphi _{{b}}}^{2}}{{\varphi _{{s}}}^{3}}}
$$
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Why can't dark matter lose energy by gravitational waves and collapse into itself? Because of lack of electromagnetic induction, dark matter can't lose its gravitational potential energy. That is preventing it from collapsing like an ordinary matter cloud in space.
But why can't dark matter lose energy by gravitational waves and collapse into itself?
| It is really hard to create gravitational waves and we only see a significant amount of energy going into gravitational waves under extreme circumstances. Any obvious example of this is the black hole mergers detected by LIGO, or the first indirect detection was from the loss of energy from a pair of neutron stars orbiting each other.
So a cloud of dark matter cannot lose energy and collapse by emitting gravitational waves because it cannot generate gravitational fields intense enough to create those gravitational waves. In principle a cloud of dark matter could create gravitational waves, but the energy is so ridiculously low that no significant amount of energy would be lost even over the whole age of the universe.
However dark matter, along with any type of matter, can collapse by a process called gravitational sorting. The particles of dark matter can exchange energy with each other so low energy ones sink inwards and higher energy ones move outwards. You'll find a discussion of this in the answers to If dark matter only interacts with gravity, why doesn't it all clump together in a single point? and How can dark matter collapse without collisions or radiation?
For more on creating gravitational waves see Is it possible to produce gravitational waves artificially?
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Why does air pressure decrease with altitude? I am looking to find the reason: why air pressure decreases with altitude? Has it to do with the fact that gravitational force is less at higher altitude due to the greater distance between the masses? Does earth’s spin cause a centrifugal force? Are the molecules at higher altitude pushing onto the molecules of air at lower altitudes thus increasing their pressure? Is the earths air pressure higher at the poles than at the equator?
| The air pressure at a given point is the weight of the column of air directly above that point, as explained here. As altitude increases, this column becomes smaller, so it has less weight. Thus, points at higher altitude have lower pressure.
While gravitational force does decrease with altitude, for everyday purposes (staying near the surface of the Earth), the difference is not very large. Likewise, the centrifugal force also does not have significant impact.
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Why doesn't decoherence spoil the double slit experiment? Imagine firing one electron at a time at a double slit. Clearly the wave function interacts with the atoms of the material, and presumably many electrons do not pass through. Why does decoherence from these interactions not spoil the experiment?
The question has been asked before, but there is no answer
Edit, to clarify the question: since the electron wave function interacts with the atoms of the material in which the double slit is cut, I naively expect that decoherence would make the system classical, no matter how carefully the experiment is set up. I must be misunderstanding the decoherence mechanism that prevents macroscopic systems being in quantum superpositions. The question is, why doesn't this decoherence spoil the double slit experiment? Can anyone explain why decoherence ensures Schrodinger's cat is alive or dead, but does not ensure the electron goes through one slit or the other?
| One should distinguish the idealized model discussed in textbooks and real interferometers. Decoherence is indeed an issue in many experiments, which is why realizing such interferometers in practice has been challenging.
What is more surprizing, is that some degree of decoherence is present even in the simplest discussions of the two-slit experiment, as not all the particles arrive at the screen - some of them escape in space, while others land on the non-transparent part of the wall with the slits. Thus, the first attempts of literally realizing an Aharonov-Bohm experiment in solid state devices, with the two particle beams confined within two waveguides (arms of a ring) resulted in phase rigidity - AB oscillations with phase either $0$ or $\pi$. To make the phase change continuously, one had to introduce artificially particle losses, as in this paper. Since then the decoherence in AB interferometers was studied extensively, both experimentally and theoretically. There have been even proposals of using controlled decoherence for measurements, as here.
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Why is the net force acting on a massless body zero? I know that massless bodies can accelerate (in theory) even with the net force equaling to zero. But, why cannot there be a net force on a massless object? Why does it always have to be zero as a resultant in the end?
I'm talking about object whose mass is assumed to be zero, i.e. $m\to0$
| Massless objects cannot be accelerated. You cannot apply force on them. Massless object ALWAYS move on the shortest path between two points. The only two things that you can do to a massless particle are 1) in presence of gravity the shortest path between two points is not a straight line (i.e. Euclidian straight line) so the path may look curved to an observer (like you and me)
2) it can be absorbed and emitted by massive particles. For example. Photons in an optic wire seems to be moving on a curved path but what actually happens is that the photon is absorbed and re emited by the atoms.
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How does a capacitor get charged instantly in AC whereas it takes infinite time in DC? I have seen when a capacitor is connected to a dc source it takes infinte time to charge, but when connected to ac it takes the potential of the source instantly,
probably the approach in the books is not adequate, please clarify,
Here is a link that mentions the time constant for DC https://www.electronics-tutorials.ws/accircuits/ac-capacitance.html
And here is one that describes the AC https://physicscatalyst.com/elecmagnetism/growth-and-delay-charge-R-C-circuit.php
| The site that you mention says
When a capacitor is connected across a DC supply voltage it charges up to the value of the applied voltage at a rate determined by its time constant.
However the time constant is $\tau = RC$ so it is not a property of the capacitor by itself, but rather the circuit.
Their example circuit for the AC case has a resistance of 0. So the time constant is $\tau=0$. Therefore it will instantaneously charge for both the AC and the DC cases.
| {
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If i wrote something in the sky what would the radius of visibility on the ground? If i wrote something in the sky 8500-10,000 ft high 40 ft tall letters what would be the visibility radius on the ground? Or what formula would i use to come up with the answer?
| Let's assume that you want to write your letters in a square large enough for a little phrase and spacing. Call $\theta$ the angle of the effective field of view (FOV) of the eye (most of our visible area is a little blurred, we have almost a 180° view but we can see clearly only a small angle). here's a little picture for a better explanation
that's why we can see little object near us but only big objects at greater distances.
Speaking of formulas "reading" a square with a side lenght L at an altitude H requires a minimum FOV of
\begin{equation}
\theta_{min} = 360°\frac{L}{2\pi H}
\end{equation}
where the differences between the square and the sperical sector are assumed negligible and the observer is "under" the centre of the square (if you want a more precise formula we can discuss later, I also don't know the precise value of $\theta_{min}$, I assume it's about $\approx$ 10° for be able to read a text, but more research/googling is needed).
this image should be helpful to visualize the geometry.
if $\theta_{min}$ is too low you must decrease the altitude H o make the text bigger (increase L).
If the text is closer or bigger enough to be read, we can start moving out from the area below the square (that is also the nearest). Given H, L and $\theta_{min}$ we can calculate the radius by first calculating the max distance $D_{max}$ from the square to have it readeable inverting the formula that i wrote before (where H is swapped with $D_{max}$ and the fact that we are viewing the square a little to the side is considered negligible)
\begin{equation}
D_{max}= 360°\frac{L}{2\pi \theta_{min}}
\end{equation}
and finally we can find the radius of visibility on the ground $R_{max}$ using the pythagoras theorem.
\begin{equation}
R_{max} = \sqrt{D_{max}^2 - H^2}
\end{equation}
Post Scriptum
The calculations in this explanation are super semplifed for a better and easier understanding, but you should still achieve a good result. If you have any question we can discuss in the comments.
The formulas should be correct, it's my first time here and I hope I didn't make any mistake.
Sorry for the bad english
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Flavor changing weak neutral current and $Z^0$ boson In Peskin QFT p.725, it says:
The $K^0$ meson could decay by $Z^0$ exchange if this boson coupled to a flavor changing weak neutral current.
However, $Z^0$ boson does not couple to a flavor changing weak neutral current. Just look at the $Z^0$ current at Peskin QFT (20.80):
The $Z^0$ boson preserves the weak flavor.
Question: So what is the point and the big deal behind Peskin QFT p.725 paragraph re-emphasizing the $Z^0$ boson (which does not have a violating CP-violating current )?
(The 1-loop diagrams of $W$ bosons couple to a flavor changing weak neutral current - but it is suppressed by GIM mechanism. If you want to discuss and explore this, please feel free, but be as comprehensive as possible -- do not assume we know GIM mechanism...)
| We know by now that the Standard Model (SM) is a pretty good description of nature, so the effects of Beyond the Standard Model (BSM) physics are going to be small, at least in the regimes that we are able to probe experimentally. For that reason, it is very difficult to see the effects of BSM physics on a process where the SM prediction is large: the fractional change will be very small. Phenomenologists therefore like to pay a lot of attention to things that the SM predicts are zero (maybe due to an exact symmetry) or very small (loop/CKM/Yukawa/etc. suppressed). Then the effect of BSM physics could be fractionally large, and thus easier to discover.
My reading of Peskin's intentions here is to point out that FCNCs are such a phenomenon which is very small in the SM, and thus represents an idea experimental target for probing the SM and searching for new physics.
| {
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How many different vacuums are there in the string theory landscape? Different sources give different estimates, from $10 ^{100}$, $10 ^{500}$, $10 ^ {20,000}$, while others write that there are infinitely many of them.
|
How many different vacuums are there in the string theory landscape ?
I never studied string theory, however this question should be bend in slightly different direction.
Wiki on String-Landscape gives estimates in range $[10^{272,000}; \infty]$. But does it really matters ?
Question should sound like:
How many different vacuums are there in the string theory landscape compatible with the Standard Model of particle physics ? Not so much according to this paper. Add requirement that fermions must be well-defined and other particle physics known limits and you'll get only $10^{15}$ valid subspace of string theory solutions.
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Number of variables in the Hamilton-Jacobi equation In Goldstein's Classical Mechanics, while introducing the Hamilton-Jacobi equation, he argues that the equation $$H(q_1, ... , q_n; \frac{\partial S}{\partial q_1}, ..., \frac{\partial S}{\partial q_n}; t) + \frac{\partial S}{\partial t} = 0$$ is a partial differential equation in $(n + 1)$ variables $q_1, ... , q_n; t$.
He then proceeds to say that the solution (if it exists) will be of the form $$S(q_1, ... , q_n; \alpha_1, ... , \alpha_{n+1}; t)$$ where the quantities $\alpha_1, ... , \alpha_{n+1}$ are the $(n + 1)$ constants of integration.
How is time a variable? Isn't it the parameter we're integrating over?
Perhaps this warrants some context. He introduces the Hamilton-Jacobi equation with the motivation to find a canonical transformation that relates the canonical coordinates at a time $t$ -- $(q(t), p(t))$ -- and the initial coordinates $(q_o, p_o)$ at $t = 0$. I hence get that time must be a variable here. However, it is still the parameter we integrate the Hamilton-Jacobi over in order to get $S$, right? Where does the $(n+1)^{th}$ constant of integration come from?
| *
*The HJ equation is a non-linear 1st-order PDE in $n+1$ variables $(q^1,\ldots,q^n,t)$, namely the generalized positions and time, which in principle enter on equal footing.
*A complete$^1$ solution, known as Hamilton's principal function $S(q,\alpha,t)$, should not be conflated with the off-shell action functional $S[q;t_f,t_i]$ nor the (Dirichlet) on-shell action $S(q_f,t_f;q_i,t_i)$, cf. my Phys.SE answer here.
References:
*
*H. Goldstein, Classical Mechanics; Section 10.1 first footnote.
*L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1 (1976); $\S$47 footnote on p. 148.
--
$^1$ A complete solution to a 1st-order PDE is not a general solution [1,2], despite the name!
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Will this spaceship collide with the star? (time dilation) I thought of the above thought experiment and arrive on 2 conflicting conclusions. I can't seem to identify the flaw in my reasoning.
Suppose there is a star 4 light years from earth that has will explode and turn into a white dwarf in 3 years (as measured in the earth frame). A spaceship travels to the star at 86% the speed of light.
According to earth's frame of reference, the journey to the star will take 4.5 years so the star will have turned into a white dwarf.
according to the ship's frame of reference, however, the journey will only take 2.25 years. Moreover, since the star is travelling relative to the ship in its own frame, the event of the star exploding will actually take 6 years. So the spaceship will collide with a white dwarf instead of a star.
I thought this had something to do with simultaneity, but I know that the events must be same in all frames of reference. My collusions imply that collisions occur between different bodies depending on the frame of reference, which can't possibly be true, can it? Where am I going wrong?
| $t'=0$, $x'=0$
B: $t'=6-4\sqrt3\approx -.928$, $x'=8-3\sqrt3\approx 2.804$
C: $t'=4/\sqrt3\approx 2.309$, $x'=0$
As you can see, the order of events in the earth frame is first $A$, then $B$, then $C$ (the ship leaves earth, then the star explodes, then the ship reaches the star). The order of events in the ship frame is first $B$ (occurring .928 years before $A$), then $A$, then $C$ (the star explodes, then the ship leaves earth, then the ship reaches the star).
So everyone agrees that the star explodes before the ship gets there. We should have known this even before we calculated because $B$ and $C$ are spacelike separated.
This is essentially an elementary homework problem and I'm mildly ashamed of myself for answering it, but I want you to see that if you took half the time you've been spending on these questions and spent it learning relativity, you'd easily be able to do this for yourself.
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Does Work become state function in an Isothermal Process and what are other processes in which it happens? In a reversible isothermal process and for an ideal gas we know from the definition of Helmholtz free Energy
$dF= -SdT -PdV$.
And as temperature doesn't change for an isothermal process, $dT$ must be zero. So dF can be written negative of change in Helmholtz free Energy. Since $F$ is a state function and $dF$ a perfect differential, work also should be.
Also, does work become state function for adiabatic processes also? Please throw light on it.
| The fact that work can equal the change in a state function, as in the case of an adiabatic process where work equals the change in internal energy, does not mean that work is a state function.
A state function is a system property. Work (and heat) is never a state function because work is not a property of a system. Work is the transfer of energy to or from a system. It is not the energy of system itself, which is its internal energy.
Hope this helps.
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Why is there a specific negative sign in front of the $m_{12}$ term of the 2HDM Higgs potential? Why is there a specific negative sign in front of the $m_{12}$ term of the 2HDM Higgs potential?
(but not for the $m_{11}$ and $m_{22}$)
See for example: https://arxiv.org/abs/1106.0034
Eq. (2) Page 6:
$$
V = m_{11}^2\Phi_1^\dagger\Phi_1 + m_{22}^2\Phi_2^\dagger\Phi_2 -m_{12}^2(\Phi_1^\dagger\Phi_2+\Phi_2^\dagger\Phi_1) + \frac{\lambda_1}2(\Phi_1^\dagger\Phi_1)^2+\frac{\lambda_2}2(\Phi_2^\dagger\Phi_2)^2+\lambda_3\Phi_1^\dagger\Phi_1\Phi_2^\dagger\Phi_2+\lambda_4\Phi_1^\dagger\Phi_2\Phi_2^\dagger\Phi_1+\frac{\lambda_5}2\left[(\Phi_1^\dagger\Phi_2)^2+(\Phi_2^\dagger\Phi_1)^2\right].\tag{2}
$$
| Usually, arbitrary signs and phases are chosen to minimize the number of explicit signs and phases that appear later, for convenience. In the paper you linked, the sign chosen for $m_{12}^2$ ensures that it appears with a positive sign in the mass terms for the charged scalars in equation (5), and with a positive sign on the diagonal elements for the mass matrix for uncharged scalars in equation (7).
| {
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How to compute gauge potential $A$ from the field strength $F$? Let $F=dA+A \wedge A$ be the field strength that solves vaccum Yang-Mills equation.
The question is: how to recover the gauge potential $A$? Is there any standard way? or any theorem stating the solvability? Suppose the metric is $g=g_{\mu \nu}dx^{\mu}dx^{\nu}=\eta_{ab}e^ae^b$, $e^a$ is tetrad basis.
| To supplement the answers that were posted earlier, here's an explicit example from ref 1. This example proves that $A$ is not always uniquely determined by $F$, not even up to gauge transformations, not even locally.
Work in four-dimensional spacetime with coordinates $(w,x,y,z)$, and take the structure group to be $SU(2)$. Let $T_1$, $T_2$, $T_3$ be the generators of $SU(2)$, and remember that the only element of $SU(2)$ that commutes with all of the $T_k$ is the identity element. Consider the gauge potential
$$
A = T_1 w\,dx + T_2 y\,dz + \alpha T_1\,dz
$$
where $\alpha$ is an arbitrary nonzero real number. The field strength is $F=dA+A\wedge A$ with
\begin{align}
dA
&= T_1 dw\wedge dx + T_2 dy\wedge dz
\\
A\wedge A
&= [T_1,T_2]w y\,dx\wedge dz
\\
&\propto T_3 w y\,dx\wedge dz.
\end{align}
Observe:
*
*$F$ is independent of $\alpha$, so we have a family of gauge different potentials that all give the same $F$.
*The three two-forms appearing in these expressions for $dA$ and $A\wedge A$ are linearly independent, and their coefficients are the three different generators of $SU(2)$.
Now, consider whether or not these potentials can be gauge-equivalent to each other (equal to each other up to gauge transformations). When we apply a gauge transform $g$ to $A$, its effect on $F$ is $F\to g^{-1} Fg$. But we already know that all of these $A$s give the same $F$, so we must have $g^{-1} F g=F$. Thanks to the observations highlighted above, this implies $g^{-1} T_k g=T_k$ for all $k$, which in turn implies $g=1$. Therefore, the potentials $A$ with different values of the coefficient $\alpha$ cannot be gauge-equivalent to each other, even though they all have the same field strength $F$.
Reference:
*
*Mostow and Shnider (1983), "Counterexamples to Some Results on the Existence of Field Copies" (https://projecteuclid.org/euclid.cmp/1103940415)
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Interpretation of Hooke's Law I often see people interpreting Hooke's Law $σ=Eε$ as,
"The deformation $ε$ that occurs when you subject a material to a stress $σ$."
This makes it sound like stress is an external stimulus that causes the material to deform. But from what I know, stress is an internal phenomenon, not an external one.
So technically, isn't it more correct to interpret Hooke's law as
"The stress σ that develops in a material given a deformation of $ε$"?
I would greatly appreciate it if someone could clear this up for me.
| Unfortunately, both descriptions are misleading because both describe the relationship as a one-way cause-and-effect relationship. The correct way to describe it is
the stress is proportional to the strain
It doesn’t matter if the stress is given and the strain is obtained by Hooke’s law or if the strain is given and the stress is obtained by Hooke’s law. Either way they are proportional to each other.
Unfortunately, whether it is Newton’s laws, Ohm’s law, Hooke’s law, or Maxwell’s equations, the tendency to verbally express such equations in cause-and-effect language is fairly strong and common in many introductory physics courses. It is almost universally inappropriate. A real cause-and-effect relationship is given by an equation of the form $$f(t)=g(t-\Delta t)$$ In this equation $g$ is the cause and $f$ is the effect and $0<\Delta t$ so that the cause always precedes the effect. In an expression like Hooke’s law both stress and strain are happening at the same time and causes happen before effects. It is a simple proportionality, not a cause-and-effect relationship.
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How can $1/V$ be equal to $0$ in Boyle's Law? In relation to ideal gases, Boyle's Law states that pressure is inversely proportional to volume under constant temperature. In other words,
$$P \propto 1/V$$
Below is a graph that plots pressure, $P$, against inverse volume, $1/V$.
How can $1/V$ ever equal zero? How is this possible?
|
How can $1/V$ ever equal zero?
Physically, it can't. This is just a point on a line - it does not have to correspond to an actual physical situation. It is like saying the average family has $2.3$ children - this does not mean any actual family can have $2.3$ children - it just means if you take a large number of families and divide the number of children by the number of families, this ratio approaches a limit of $2.3$.
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Why do neutrons have magnetic moments with spin? Neutrons are neutrally charged, but apparently they have magnetic moments with spin.
What is the intuition behind this?
| A neutron is a composite particle consisting of three pointlike quarks. These carry fractional electric charge which adds up to zero, so the neutron has no net charge. But since the quarks do carry nonzero charges, if they are spinning around inside the neutron then it is possible for the neutron to possess a magnetic moment while not possessing a net charge.
The fact that the neutron has a nonzero magnetic moment despite having no net charge while also possessing a well-defined radius were clues that it had internal structure (the quarks), which was proven in the deep inelastic scattering experiments done at SLAC in the late 1960's.
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How can there be current without charge? This might be a stupid question, but I actually think that it is not so obvious:
When solving Maxwell equations, depending on the problem usually a charge and current density are assumed.
However, how can there be any current without charge density?
I understand that in most problems where this question arises there is no net charge. So for example in a conductor where a voltage is applied across it there is no net charge, however nevertheless a current flows.
That is quite clear to me, but formally should one not have to assume the existence of charges even if the net charge vanishes.
Is this problem solved in quantum mechanics where the charge and current are manifestations of a function satisfying the Dirac equation?
| In the absence of a changing magnetic flux, a current flow requires a separation of charges. If you connect a long looping uniform conductor to the terminals of a battery, a variable charge density is require to maintain a uniform field and current density in the conductor. There will be a positive charge density near the positive terminal of the battery and a negative charge density near the negative terminal. Within the conductor, the field driving the current will be proportional to the gradient of the charge density. (Toward the positive end, excess flux will leave through the sides of the conductor.)
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Where does the law of conservation of momentum apply? Take the scenario of a snowball hitting a tree and stopping. Initially, the snowball had momentum but now neither the snowball nor tree have momentum, so momentum is lost (thus the law of conservation of momentum is violated?). Or since the tree has such a large mass, is the velocity of the tree is so small that it's hardly noticeable?
If the explanation is the latter, this wouldn't hold for a fixed object of smaller mass. So in that case, how would the law of conservation of momentum hold?
| Momentum is conserved only if there is no net external force on the system.
Consider the snowball and the tree as the system. In your case, the earth provides an external force on the tree, so the momentum of the snowball/tree system is not conserved. If the tree is "suspended" (not attached to the ground) momentum would be conserved, but the final velocity of the tree would be very small and hardly noticeable due to its large mass.
If the system is taken to be the snowball, tree, and earth, momentum is conserved , but the final velocity of the tree and earth (assuming the tree stays attached to the earth) is infinitesimally small due to the very large mass of the earth.
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Can thermodynamics be considered logical? One of the laws says that heat won't flow from cool to warm and at the same time this same theory claims that there is a finite (albeit tiny) chance that it will, because there is always such a microstate.
We can also have a situation where all air molecules in the room can be found in the left side of the room and none in the right side, because it is one of the microstates therefore it can happen and the entropy will drop. So how can we say that the entropy always increases when it can decrease too sometimes?
| There are two different views on thermodynamics, two approaches.
We can build thermodynamics as a phenomenological theory based on some postulates. According to this theory, heat will not flow from a cold body to a hot one. Never ever. And in this theory there is no concept of a microstate at all. But this theory is phenomenological and approximate.
But instead of starting with the postulates of thermodynamics, we can build thermodynamics based on statistical mechanics. This theory is more precise and fundamental than phenomenological thermodynamics. And this theory predicts that sometimes heat can flow from a cold body to a hot one. That is, the second law of thermodynamics holds only statistically, on average. (However, there is a formulation of the second law of thermodynamics, which, as far as we know, is absolutely accurate: it is impossible to build a perpetual motion machine of the second kind.)
When it comes to statistical thermodynamics, I don’t know of a perfect textbook. But for a basic level I would recommend the following two books:
*
*Kittel Thermal physics
*Huang Statistical mechanics
Theirs content partially overlaps, partially complements each other. Better to start with Kittel.
| {
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"timestamp": "2023-03-29T00:00:00",
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Do atoms always move in phase within the unit cell for acoustic mode phonons? In my condensed matter book it says 'For the acoustic mode, all atoms in the unit cell move in-phase with each otehr (at $k=0$) whereas for optical modes they move out of pahse with each other (at $k=0$)'.
I saw that in the example given this is true, but is it always the case? Is it easy to show from the definition of an acoustic mode ($\omega \rightarrow0$ as $ k \rightarrow 0$) that the relative ampltiudes of atoms within the unit cell will always be in phase?
| At $k=0$, acoustic phonons correspond to macroscopic translation of the entire crystal, which naturally are completely in phase and cost zero energy. This behavior is essentially by definition and is always true as acoustic phonons are the Goldstone modes of translation. In other words, the energy of the crystal should be the same if it is located in London or Leiden.
If there was some out of phase component at $k=0$ that cost zero energy, it would mean the crystal would spontaneously deform and reorganize with zero energy cost. Such behavior would imply the material's structure is unstable, which can happen at a structural phase transition. However, structural transitions are limited to very specific temperatures and pressures - under general conditions you do not have any out of phase motion that costs zero energy
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/606201",
"timestamp": "2023-03-29T00:00:00",
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What experiment confirms $\mathbf{J}^2 = \hbar^2 j(j+1)$? I learned that if we measure the spin angular momentum of an electron in
one direction $J_z$, we get $\pm \frac{1}{2} \hbar$. But if we measure
the magnitude of the angular momentum $\mathbf{J}^2$, we should get
$\frac{3}{4} \hbar^2$. What experiment gives the latter result?
As @user1585635 notes, measuring $J_x$, $J_y$, and $J_z$ separately and
summing their squares gives $\frac{3}{4} \hbar^2$. This is not what I'm
looking for. First, if I measure $J_z$, measure angular momenta in
three directions separately, and measure $J_z$ again, the two
measurements of $J_z$ aren't guaranteed to be the same, since $J_z$,
$J_x$, and $J_y$ don't commute. But $\mathbf{J}^2$ commutes with $J_z$.
Second, when the spin is not $\frac{1}{2}$, say it's $j$, summing the
squares of components of $\mathbf{J}$ gives 3ℏ²², where ² should
be ℏ²(+1) doesn't always give $\hbar^2 j(j+1)$. (Thanks to
@MichaelSeifert for pointing that out)
This question is not a duplicate of
Why is orbital angular momentum quantized according to $I= \hbar \sqrt{\ell(\ell+1)}$?.
That question is about how $\mathbf{J}^2$ is derived mathematically.
Mine is about how it is confirmed experimentally.
| I suppose we measure spin component $S_z$ by measuring the shift in energy spectrum when we add a magnetic field, which gives an extra term $B_z S_z$ in the hamiltonian.
Similarly, to measure $S^2$ we probably need a term that relates it to some energy.
The one I thought of is Fine Structure correction of the hydrogen spectrum, in which the energy is different for different values of total angular momentum $j$.
But $\vec{J}=\vec{L}+\vec{S}$ so the eigenvalues j depends on $S^2$ being $\frac34 \hbar^2$.
| {
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