Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Can electrical forces be reflected by non charged objects? If there is a charge in the center of a closed room, with just walls that are not charged: the electrical forces can traverse the walls, are they absorbed by the non charged walls, or are reflected back?
| Objects that are overall electrically neutral, but are made of positive and negative charges, can still interact electromagnetically. This is how wires work; even though the wire itself is not charged, it still carries current because the electrons in the wire are moving.
In the case of a wall, electromagnetic radiation impacting it interacts with the electrons on the surface of the wall. How much of the radiation is absorbed or reflected depends largely on how the electrons in the wall are allowed to move. If the electrons can move freely (as in most conductors), the radiation is mostly reflected. If the electrons' motion is restricted (as in most insulators), the radiation is mostly absorbed (if it roughly corresponds to an allowed transition between electron energy bands) or transmitted (if it does not).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How does an object in space travelling at constant velocity have a net force of zero acting upon it? If the definition of balanced forces is "two opposing forces that are equal" and an object with a net force of zero acting upon it means that the forces are balanced on the object, then it should follow that an object travelling through space at constant velocity should have a net force of zero acting upon it. But there was an initial force acting upon it when it was thrown into space. What is the force that is balancing that initial force to make the net force zero?
| You are right. If we have an object at rest and then we want it to start moving, we apply a force to the object. While the force is being applied, the object accelerates according to $F=ma$.
Now let's say we stop applying this force. Then there is no longer a net force acting on the object. Therefore, the acceleration is $0$ and the object now moves at a constant velocity.
I'm not sure if I fully understand where you are having difficulty, but it seems to me that you are thinking of objects that can "remember" the "history of forces". So that if we apply a force and then take the force away, we still need to apply an opposite force to undo what the first force did to cause the acceleration to be $0$. This is not the case. Once the first force is gone, the acceleration is then $0$.
However, if we wanted to stop the object and bring it back to rest, then we would need to apply a force opposite to the first force to produce an acceleration in the opposite direction.
Side note, for this answer I am assuming we are in an inertial reference frame and just applying a force to an object. Any discussed motion or velocity is assumed to be viewed from this reference frame. I know I need to specify this or else I might get attacked by the relative motion police (of course, what constitutes an attack is also relative, so maybe let's just all agree to boost to a frame where we don't have to worry about all of this). :)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why are RG flow fixed points associated with different phases? Why are RG flow fixed points associated with different phases? I thought the RG makes only statements about behavior near to critical points... a definite phase is far away from the critical point, right?
| RG fixed points themselves are not associated with phases. Rather, RG fixed points (and their basins of attraction) describe systems at a phase transition. You are correct that taking the limit as a system approaches the critical surface only allows us to make statements about the system near the transition, but the RG is also used more broadly. The different relevant perturbations of a fixed point describe the possible phases (low-energy behaviors). By tuning the relevant perturbations to zero, we reach criticality, e.g. phase transitions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Can we enhance evaporation rate with a vacuum pump? First, consider having water at 100C and 1atm and a heat source. If more heat added to water, we have more evaporation rate according to the following formula:
$$Q_{in}=m_{vapor}*h_{fg}$$
THEN
Consider having water at 100C and 1atm and a vacuum pump.
Can we change the water evaporation rate with the vacuum pump as we did change the evaporation rate with the heat source?
What are the formulas and calculations?
Thx
| As you drop the pressure, the boiling temperature drops. As the water boils the heat of vaporization comes from the water and the temperature drops. This should continue until the remaining water is frozen. You may need to find out if the heat of vaporization is a function of temperature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Use of generating function in canonical transformation In the theory of Canonical transformations, initially we use the fact that the new and the old system of $(q_i, p_i)$ with the Hamiltonian $H$ satisfy the modified Hamilton's principle. Now here, the use of the theory of generating functions makes sense. But after the development of Poisson bracket formalism of canonical transformations, what exactly is the use of generating function formalism? As such even, if it was tough to guess the transformation equation $$P_i=P_i(q_i,p_i,t) ; Q_i=Q_i(q_i,p_i,t)$$ as such, how would it be any easy to guess the generating function?
| *
*For various notions of canonical transformations (CTs), see this Phys.SE post.
*Presumably what OP calls "Poisson bracket formalism of CTs" refers to symplectomorphisms (at least if there is no explicit time dependence).
If we furthermore assume that the $2n$ new coordinates $(Q^1,\ldots, Q^n, P_1, \ldots, P_n)$ are canonical/Darboux coordinates, then we get $n (2n-1)$ differential conditions$^1$
$$ \{Q^i,P_j\}~=~\delta^i_j , \qquad\{Q^i,Q^j\} ~=~0~=~\{P_i,P_j\} ,\qquad i,j~\in~\{1,\dots,n\},\tag{A}$$
to satisfy in terms of the old canonical/Darboux coordinates $(q^1,\ldots, q^n, p_1, \ldots, p_n)$.
*On the other hand, if we instead use a generating function $F$ to define a CT, we are automatically guaranteed that the new coordinates $(Q^1,\ldots, Q^n, P_1, \ldots, P_n)$ are canonical/Darboux coordinates. Also it is often much simpler to solve for a generating function $F$ than to solve the full equation system (A). This is one possible answer to OP's question.
--
$^1$ The number of differential conditions corresponds to the number of independent entries in an antisymmetric $2n\times 2n$ matrix.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Negativity for a diagonal reduced density matrix Suppose one has a tripartite system A,B,C with density matrix $\rho$ , and with reduced density matrix $\rho_{BC}=\text{Tr}_A\ \rho$. Suppose $\rho_{BC}$ is a diagonal matrix.
As the partial transpose of a diagonal matrix corresponds to the same matrix, and since all the matrix elements are positive, the negativity
$\mathcal{N}(\rho_{BC})=\sum_i \frac{\vert \lambda_i\vert-\lambda_i}{2}$, where $\lambda_i$ are the eigenvalues of the partial transpose
$\rho_{BC}^{T_B}$
is zero.
How it is possible to say whether the state is entangled or not?
| A state with a diagonal density matrix is always separable (=not entangled): If
$$
\rho= \sum p_{ij} |ij\rangle\langle ij|\ ,
$$
then a separable decomposition is given by
$$
\rho = \sum p_{ij} \sigma_{ij}^A\otimes\sigma_{ij}^B\ ,
$$
with $\sigma_{ij}^A=|i\rangle\langle i|$ and $\sigma_{ij}^B=|j\rangle\langle j|$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is the randomized motion of particles produce heat? In the show on factomania from gregg foot I've heard that when sun light comes then some gas particles of air get more energy and velocity and as the average kinetic energy of molecules is Temperature the temperature increases for the same region . Now if thats true empty space should have no heat as it is and earth atmosphere should have. So is this explanation true. Or light carries radiation energy and heat energy separately ?
| Heat is energy transfer from one thing to another due solely to there being a temperature difference between the two. Things do not "contain" heat. The earth's atmosphere does not contain heat. Temperature is a measure of average translational kinetic energy component of a substances internal energy.
The three basic types of heat transfer are conduction, convection, and radiation. The first two require some type of substance to enable the transfer, such as a liquid, solid, or gas. The third (radiation) does not as the transfer is in the form of electromagnetic radiation. No medium is needed for electromagnetic radiation.
Heat transferred from the sun to the earth and its atmosphere is by electromagnetic radiation. This energy transfer occurs through the vacuum of space.
Hope this helps.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Earth bulge and axis precession We know that earth is not spherical because it spins. Does every object that spins bulge? for example a rotating wheel bulges? Also how moon and sun create the torque that shifts the axis of rotation (precession) if gravitational force acts on center of mass therefore no torque?
| Every object that spins must be accelerating. This means different forces (and usually stresses) from the same object at rest. Whether it bulges or not depends on the material and its response to that stress.
If you have a steel marble spinning at 1 revolution per minute, the rotational forces on it are tiny. Given the strength of the material, you probably would be unable to measure the deformation (or bulge) from the spin. As the object becomes larger, weaker, and faster, the deformation would increase.
...gravitational force acts on center of mass therefore no torque ?
Gravitational force acts across an entire mass, not just at the center. However, in some cases, the sum of these forces is identical to it acting only at the center. This is true when you have a uniform sphere or uniform spherical shell of material. So for things that are mostly spherical, the simplification that gravity acts only on the center of mass is reasonable, but perhaps not exact.
Because the earth deviates from a spherical shape, the sun and moon don't tug on it uniformly. The equatorial bulge gets pulled a bit differently and allows for a small torque that can cause precession. If the earth were exactly spherical, this torque would not exist.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why gas molecules move with different speed at a given tempreture? As per my understanding we know that molecules of an ideal gas are identical in all aspects (size, shape, mass). Since collisions are elastic in nature, they don't lose their kinetic energy. That means that kinetic energy of each molecule doesn't change over time. Then how do the molecules move with different velocity regardless of possessing same mass and kinetic energy ?
| That's a good question. Elastic collisions between isolated particles will indeed conserve energy and momentum. But, consider this: Suppose the particles' momenta before the collision are uncertain: they are only known within some range. Think about it a while and you'll realize that the uncertainty grows with each collision. The Boltzmann distribution is the situation where those changes in uncertainty reach equilibrium for a large number of particles with a given total energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why does the warm air rises up? Warm air has more energy than cold air. This means that according to the Einstein equation $E = mc^2$ the warmer air has a greater mass than the cold one. Why is the warm air rising, if it has a greater mass, which means that the attraction of gravity between the Earth and the warm air is greater?
| As other answers point out, the reason is buoyancy. This post is to show just how small the opposing (relativistic) effect of increased gravitational force is.
The increase in the gravitational attraction associated with kinetic energy is proportional to the Lorentz factor, $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$.
We could look up molecular masses, and get into the Maxwell-Boltzmann distribution which describes the range of speeds of gas molecules, but for guesstimation purposes, let's use the rule of thumb that typical molecular speeds in a gas are on the same order as the speed of sound through the gas--for air near the surface of the Earth, roughly 300 m/s.
Let's say we increase the temperature of our parcel of air by 10%, from 300 K (a warm day) to 330 K (roughly the hottest day in the hottest desert). That means we increase the average molecular velocity by 5%, from our fudged 300 to 315 m/s. This takes $\gamma$ from $1+5 \times 10^{-13}$ to $1+6 \times 10^{-13}$, an increase of 1 part in $10^{13}$.
That is how much the gravitational force increases. You could cancel that increase by moving half a part in $10^{13}$ further from the center of Earth. That's a third of a micron.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
What is the most efficient way to pour a liquid through a funnel? Be it in cooking or an experiment, when transferring liquid from one container to another using a funnel should you:
(a) pour it all straight into the funnel, filling it up and waiting for the liquid to drain through
(b) pour it such that the end of the funnel is still always filled but the volume of 'blocking' liquid in the funnel is small - so you pour it slowly
Does one of the above options achieve a faster rate flow of liquid through the funnel? I'm thinking bernoulli's principle is going to come into this? Intuitively the blocking liquid in (a) is going to increase the pressure at the bottom of the funnel and hence 'force' more liquid through...
| The Bernoulli equation for a non-turbulent fluid with density $\rho$ is
$$
P_\text{fluid} + \frac12 \rho v^2 + \rho g h = \text{constant}.
$$
At the top and the bottom of the funnel, where the fluid is exposed to the atmosphere, the fluid surface will move until the fluid pressure and the atmospheric pressure are the same. For an ordinary funnel, that effectively means that the inlet and outlet pressures are identical. If you are continuously filling the funnel, so that the fluid speed $v$ at the top is zero --- or alternatively, if the top of the funnel is so much larger than the outlet that the flow speed at the top is negligible --- then the outlet speed is
$$
v = \sqrt{2g\Delta h}.
$$
So doubling the distance $\Delta h$ between the exit of your funnel and the surface of your fluid will make your exit velocity about 40% larger.
An engineer friend of mine once solved a problem with a water reservoir that was draining too slowly by adding a longer hose to dangle under the outlet. It's the same trick: increase the flow rate by increasing $\Delta h$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to obtain $Y$ rotation with only $X$ and $Z$ rotation gates on the Bloch sphere? Let's say you have a system with which you can perform arbitrary rotations around the $X$ and $Z$ axis. How would you then be able to use these rotations to obtain an arbitrary rotation around the $Y$ axis?
I have seen somewhere that rotation around an arbitrary axis can be achieved by doing three rotations around two fixed axis.
$\hat{R}_\vec{n}(\theta)=R_Z(\gamma)R_X(\beta)R_Z(\alpha)$ for some angles $\gamma, \alpha, \beta$. But how do you actually use this? What if I want to rotate around the Y axis with an angle of $\theta$ , $\hat{R}_Y(\theta)$, then how do I find the values of $\gamma,\alpha,\beta$.
| Here it is in words, more details below.
*
*Rotate by $\alpha=-90^\circ$ about $z$ so as to bring your $y$ axis around to point along the $x$ axis.
*Rotate by desired angle $\beta=\theta$ about $x$.
*Rotate by $\gamma=90^\circ$ about $z$ to put the $y$ axis back where it was.
The information you need
for the more general case is on the Rotation formalisms Wikipedia page, and on the Euler angles Wikipedia page.
Your initial specification of the rotation is in the "angle-axis" form (rotating by angle $\theta$ about the axis $(0,1,0)$ in your case). This section describes how to convert from that to a $3\times 3$ rotation matrix. In your case
$$
\begin{pmatrix}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{pmatrix}
=
\begin{pmatrix}
\cos\theta & 0 & \sin\theta \\
0 & 1 & 0 \\
-\sin\theta & 0 & \cos\theta
\end{pmatrix}
$$
Then you need to convert the rotation matrix to Euler angles, and I'm guessing you want the $z$-$x$-$z$ extrinsic version in which the axes remain fixed,
but you should check this carefully. This section describes how to do that.
There are some caveats in that section about multiple solutions etc, which you should check.
Anyway, in your case
\begin{align*}
\alpha &= \text{atan2}(-\sin\theta,0) = -90^\circ \\
\beta &= \arccos(\cos\theta) = \theta
\\
\gamma &= -\text{atan2}(\sin\theta,0) = 90^\circ
\end{align*}
Check also the sign conventions, it's a notorious minefield.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If the molecular collisions are elastic will there be any dissipation in a fluid? Viscosity arises due to collisions of the molecules of one layer of a fluid with another in contact. But viscosity is a dissipative element leading to heating and dissipation. Where does it heat come from? Does it come from the molecular collisions being inelastic? If the collisions were elastic, would there be no viscosity or dissipation in a fluid?
| What is happening is that molecules from one layer collide with those of adjacent layers, transferring both momentum and kinetic energy. If there is an organized motion of the molecules (e.g., mean velocity gradient), the kinetic energy of the organized motion is continually converted to random kinetic energy as a result of the collisions. This happens even if the collisions between molecules are elastic. The increase in random kinetic energy is equivalent to an increase in internal energy of the fluid (i.e., temperature). So, what the viscous behavior of the fluid does is convert more useable mechanical energy to internal energy. An example of this is steady shear of a fluid between parallel plates, where mechanical energy (shear work) is continually applied at the boundary, but is continually being converted to random kinetic energy within the fluid.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Why are there so many different "colors" in the line spectra of simple elements? I understand that atoms emit photons of wavelength $\lambda$ when electrons transition to lower orbitals according to the equation $E = \frac{hc}{\lambda}$. Based on my understanding, those orbitals have fixed energies, and simple elements like oxygen should have a small number of orbitals. The number of transitions possible, and thus the number of wavelengths, should be quite small. How then is the large numbers of lines on the oxygen line spectrum explained?
| There are an infinite number of orbitals. Even for simple one-electron Hydrogen, there are an infinite number of energy levels, $E=-E_0/n^2$ for $n=1,2,3,...$ all the way to infinity.
So there are many possible transitions between energy levels, each corresponding to a spectral frequency. Some transitions have a much higher probability than others, so you see the most likely transitions, not all of them.
Addition: Some transitions are actually forbidden, based on “selection rules”. For example, in the dipole approximation for Hydrogen, the $\ell$ quantum number must increase or decrease by 1 in a transition.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there is any way to prove the Faraday's law of induction? I want to know if there is any way to prove the Faraday's law or is it just an experimental observed phenomena?
More specifically, is there any reason why the proportionality constant is 1? How did Faraday discover it? I also heard that we can prove Faraday's law using the principle of least action, but is that true?
| Every proof starts with axioms — things that are merely assumed rather than proven. The axioms should be motivated by their applicability to a broad range of phenomena, but utlimately they are just axioms. We test them by comparing their predictions to experiment. Axioms that apply to a broader variety of experiments are better.
If we start with quantum electrodynamics (which has very broad applicability), then one of the axioms is that the components of the electric and magnetic fields are all encoded in a single gauge field $A_a$, where the index $a$ takes values in $\{0,1,2,3\}$. These four index-values correspond to the four dimensions of spacetime (0 for the time dimension, and 1,2,3 for the space dimesions).
To describe how the electric and magnetic fields are encoded in the gauge field, first define
$$
F_{ab}\equiv \partial_a A_b - \partial_b A_a
\tag{1}
$$
where $\partial_a$ denotes the partial derivative with respect to the $a$-th coordinate in spacetime. (The gauge field is a function of all four of these coordinates.) This quantity is antisymmetric, $F_{ab}=-F_{ba}$, so it has six independent components. The three components $F_{ab}$ with $b=0$ are the components of the electric field $E_a$, and the three components $F_{ab}$ with $a,b\neq 0$ are the components of the magnetic field $B_{ab}=-B_{ba}$, which are usually written with a single index like this:
$$
B_1 \equiv B_{23}
\hskip2cm
B_2 \equiv B_{31}
\hskip2cm
B_3 \equiv B_{12}.
\tag{2}
$$
(Notice the cyclic pattern.)
With these identifications, equation (1) implies Faraday's law. To see this, first notice that equation (1) implies
$$
\partial_a F_{bc} + \partial_b F_{ca} + \partial_c F_{ab} = 0.
\tag{3}
$$
(Again, notice the cyclic pattern.) The components of equation (3) with all indices non-zero say that the divergence of the magnetic field is zero. The components of equation (3) with one index equal to zero, say $a=0$, give Faraday's law:
$$
\partial_0 B_{bc} + \partial_b E_{c} - \partial_c E_{b} = 0.
\tag{4}
$$
(Disclaimer: I didn't check to see if my sign-conventions for $E$ and $B$ are standard, but if they're not, then just reverse all of the signs in equation (2). It's just a convention.)
I also heard that we can prove it using the principle of least action is that true?
Formulating things in terms of the gauge field allows the equations of electrodynamics (classical or quantum) to be expressed using the action principle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why is $E=mc^2$ and not $E=m\frac{c^2}{2}$? Kinetic energy for a moving object is the integral of force with respect to distance, often given as:
$$E=m\frac{v^2}{2}.$$
This would imply that for mass moving at the speed of light, the kinetic energy would be:
$$E=m\frac{c^2}{2}.$$
This puts it off from the Einstein result by a factor of two. Why the discrepancy?
| You cannot make this 1:1 correspondence between the classic kinetic energy of a particle and the rest energy of a relativistic particle.
$E = mc^2$ does not apply only to objects moving at the speed of light, but instead to all objects, moving or not. As such, it isn't a relativistic correction of the kinetic energy, but describes the rest energy of a particle.
The relativistic energy of a moving particle is given (from special relativity) by $$E^2 = m^2 c^4 + p^2 c^2$$
This can further be simplified to $$E = \gamma m c^2,$$ where $\gamma = \frac{1}{\sqrt{1-\beta^2}}$ (the Lorentz factor) and $\beta = \frac{v}{c}$.
$\gamma$ is $1$ if the object is at rest, which reduces this to the famous $$E = mc^2$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How should I imagine a multi-particle state in a free QFT? It is reasonable to think of single-particle Focks states as of plane waves. Indeed, since $|p\rangle=a^\dagger_p|0\rangle$ and $\langle x|p\rangle\sim \operatorname{e}^{ipx}$, we conclude that the state $|p\rangle$ can be thought of as a plane wave in the position representation.
What about multi-particle states, such as $|p_1,p_2\rangle=a^\dagger_{p_1}a^\dagger_{p_2}|0\rangle$? Naturally, I would imagine those to be superpositions of plane waves (since in the free theory we are dealing with equations obeying the superposition principle). But is it really like that? Is the matrix element $\langle x|p_1,p_2\rangle$ really equal to the sum of plane waves?
| Previous answers look right, but I hope I may be able to add clarity.
In a free-field theory, you can read $a^\dagger_{p_1}a^\dagger_{p_2}|0\rangle$ as 'the occupation number of mode $p_1$ is 1 and the occupation number of mode $p_2$ is 1'. If you want to go to the language of spinor-valued (i.e. not operator-valued) wavefunctions in space, then the form of the wavefunction for such a state is $\frac{1}{\sqrt{2}}(\phi_A({\bf r}_1, \chi_1) \otimes \phi_B({\bf r}_2,\chi_2) + \phi_A({\bf r}_2, \chi_2) \otimes \phi_B({\bf r}_1, \chi_1))$ where A and B are the two modes, 1 and 2 are labels attached to the particles, and I assumed they are Bosons (so I wrote a symmetrised state). $\bf r$ refers to spatial position, $\chi$ refers to spin state. Now, depending on what calculations you are doing, this second notation may or may not be useful. It is basically an attempt to handle a field by using the methods of single-particle quantum theory, which can be useful sometimes, and maybe gives some physical intuition, but ultimately will not allow you to complete the kinds of calculations that particle physics needs for scattering problems etc. However it can sometimes be useful for looking at photons propagating through an optical system, or things like that.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Do Maxwell's equations predict the speed of light exactly? I know that $\frac{1}{\sqrt{\mu_0\varepsilon_0}}$ is equal to the speed of light but is this prediction accurate? I mean is it 100 percent accurate?
| In the current SI system $\epsilon_0$ is defined as $\epsilon_0=1/c^2 \mu_0$ so in the current SI system $c=1/\sqrt{\epsilon_0 \mu_0}$ is clearly exact. Furthermore, $c$, $\epsilon_0$, and $\mu_0$ are themselves all exact defined quantities with no uncertainty individually.
In the new SI system starting next year we will still have $\epsilon_0=1/c^2 \mu_0$ so in the new system $c=1/\sqrt{\epsilon_0 \mu_0}$ will still be exactly true. However, $\epsilon_0$ and $\mu_0$ will now themselves each be uncertain quantities. Under the new system, $\epsilon_0=e^2/2hc\alpha$ and $\mu_0=2h\alpha/ce^2$. All of these quantities are exact with the exception of the fine structure constant, $\alpha$, which has an experimental uncertainty of 0.23 parts per billion. Note that the contribution of the uncertainty in $\alpha$ is such that the uncertainties cancel out and while $\mu_0$ and $\epsilon_0$ are each individually uncertain their product is exact.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 4
} |
Can we say that bosons attract each other? We know that bosons donot follow Pauli exclusion principle, thus they can occupy the same state. But is it equivalent to say that bosons attract each other?
| No. When we say that two particles "attract," we usually mean that there is some intermediary field causing the attraction, i.e. a "force carrier." You can easily make theoretical models of bosons which do not attract. (These would be called "free particles.") However, sometimes people say that bosons "like to be together." This is because it takes less energy to put bosons "near" each other than fermions. However, that doesn't mean they attract. You could make fermions attractive or bosons repulsive by tweaking the intermediary force carrier. It's just that, all else being equal, bosons take less energy to put next to each other than fermions. So while there might be a fuzzy sense in which bosons "attract," I would say that this is an abuse of language that mostly just causes extra confusion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Why can't we see images reflected on a piece of paper? Why can't you see a reflected image on a piece of paper? Say you put a pen in front of the paper, even when light rays are coming from other sources, hitting the pen, reflecting back, and hitting the paper, there is no reflection.
What's wrong with the following "ray diagram" and why such even don't happen and the image of the pen don't form on the paper (right side is a paper)?
When then can you see the image of a torch when you shine it on the paper?
When you put a convex lens in front of the pen, why you can now see the image of the pen on the paper?
| A reflective or polished surface respects and preserves the angular proximity of incoming beams on its points. The less the angular proximity of incoming intensity is preserved in the outgoing intensity profile, the more diffuse and "matte" looks the surface
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "72",
"answer_count": 9,
"answer_id": 6
} |
Negative Miller indices and parallel planes The integers are usually written in lowest terms, i.e. their greatest common divisor should be 1.
https://en.m.wikipedia.org/wiki/Miller_index
Does this mean, that parallel planes are generally equivalent, particularly does $(200)$ for example even exist or is this a wrong notation of $(100)$?
Also what is the correct notation for $(\bar{1}00)$? Since $(100)$ is parallel to $(\bar{1}00)$ I think $(100)$ is correct. Is there kind of a rule or a convention to invert the signs to obtain the maximum positive indices?
| I like Pister's fold-up crystal for this: http://www-bsac.eecs.berkeley.edu/~pister/crystal.pdf
One sees then that $(\bar{1}00)$ is on the other side of $(100)$ and that those are perpendicular to $(010)$. In a cubic crystal like silicon all six cubic planes $\{100\}$ are equivalent.
In non-centrosymmetric crystals opposite faces can be different. For example, in GaAs, one could say that the $(111)$ surface is gallium-terminated and $(\bar{1} \bar{1} \bar{1})$ is arsenic-terminated.
(200) etc is a notation for x-ray diffraction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Could quantum fluctuations spawn real matter? Would it be plausible for fluctuations in the QED vacuum to spawn actual matter (such as quarks, electrons the constituents of a hydrgen atom) given enough time and space?
| The answers above are correct. Kramer and Steane make reference to gravitation. This is where things get strange, for a quantum vacuum is only defined in a local region. A black hole for instance has local internal frames patched together by transition functions that under derivatives give connections terms that further with covariant derivatives give curvatures. Because of this you have not a vacuum but a set of vacua. For this reason a transformation on one vacuum can result in a vacuum plus particles or bosons.
Coleman showed how a potential function that with a hump, say a quartic function, that is also asymmetric may generate a bubble of spacetime. Quantum states in the right can tunnel into the left, and where there is an energy gap. This means there might be a transition that produces bosons or particles. This radiation may in general be bubble cosmologies.
So in general to generate particles you either need a set of vacua, such as those in local regions of spacetime containing an event horizon, or some potential function with different vacuum states. You can't get particles generated out of a global vacuum state. That is unless there is some sort of instability that causes that vacuum to transition into another vacuum at lower energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 0
} |
Does a rock use up energy to maintain its shape? A rock sitting on land, the ocean floor, or floating in space maintains its shape somehow. Gravity isn't keeping it together because it is too small, so I'm assuming it is chemical or nuclear bonds keeping it together as a solid. If not it would simply crumble apart. So, what type of energy maintains the shape of a rock, where did this energy come from, and is it slowly dissipating?
As a corollary, if a large rock is placed on top of a small rock, is the energy required to maintain the shape of the small rock 'used' at a greater rate?
| There are various mechanisms that keep solid things together, they all have one thing in common: They reduce energy to a minimum! When you want to break it apart, it costs you energy to do so!
Examples of bonds are:
Hydrogen-Bonds, which are very weak and come from an asymmetry of the electron around the proton, in such a way that it is energetically favourable to form bonds instead of repel each other.
Ion-bonds, which can be quite strong, but the materials are often recalcitrant (brittle). Materials having ion-bonds are not pure, they are a mixture of two different elements, one positively charged, another negatively charged and they form molecules together, mainly due to the Coulomb force.
There are many more!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 0
} |
Two People Pushing Off of Eachother, Newton's Third Law, and Unbalanced Force Different versions of this question have come up all over the internet. Usually it deals with tension in a rope or two people pushing on each other with the same force. I am trying to understand 2 people pushing each other with different forces.
Say two people of equal mass are standing on a frictionless surface, touching palm to palm, and they push off of each other at the same time with different amounts of force. Say the person on the left pushes with 100 N and the person on the right pushes with 70N. What happens to this system? It seems like there should be a net force of 30 N to the right, but at the same time, each person pushing should feel the same force pushing back as per Newton's 3rd Law.
If I push on you with 100 N then I also feel the 100 N from the force pair and we slide apart each having been accelerated by 100 N. If you push on me with 70 N then I feel the 70 N force and I exert 70 N back on you, we slide apart each having felt 70 N this time. But what happens when we both push with a 100 N and 70 N respectively?
Also, I have seen elsewhere on this site that if 2 people were to push on each other with the same force under conditions like those in the scenario above, they each feel the same force but fly apart at double the final velocity because the same force has been applied for twice the distance. Is this actually the case?
|
Say the person on the left pushes with 100 N and the person on the right pushes with 70N. What happens to this system?
This is not possible. They can only push on each other with the same force.
Suppose the person on the left is a weightlifter and capable of pushing very forcefully, and suppose that the person on the right is unconscious and not capable of exerting any force voluntarily. Even in such an asymmetrical circumstance the force will be the same. If the person on the right is rigid then the person on the left can generate their full force and the natural rigidity of the person on the right will return the equal and opposite force without effort. If the person on the right is limp then the person on the left cannot generate much force at all since the limp person will deform under minimal force. Either way the force is automatically guaranteed to be equal and opposite.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Explanation of formation of Newton's rings I was asked to do a lab report in my University about Newton's rings experiment that we made in lab. I understand that the aim of the experiment is to measure the wavelength of a light after the formation of Newton's rings and I understand the mathematical derivation of formulae used for that (relation between radius and wavelength due to constructive or destructive interferences).
I am confused a little about the behaviour of the light inside the setup. When light enters the plano-convex lens what are the cases that lead the light to reflect and refract in a specific way that when interference occures bright fringes are formed due to a constructive one or dark fringes are formed due to a destructive one?
I hope somebody can help me to understand how the rings are formed exactly so that can help me understand what I am working on.
| From what you've written, I deduce that you know that the path difference between light reflected back up from a point P on the lower surface of the lens and reflected back up from the point Q immediately below it on the plane surface is $\frac{r^2}{R}.$ Here, $R$ is the radius of curvature of the lower face of the lens and $r$ is the 'ring radius', that is the distance from the point of contact between lens and surface to point Q. This path difference can be expressed as $\frac{r^2}{R\lambda}$ wavelengths, $\lambda$ being the wavelength of the light. So the path difference between the light reflected from P and Q gives rise to a phase difference of $2\pi\frac{r^2}{R\lambda}.$
There is another phase difference, independent of any path difference. Light travelling through air and reflected from glass (as at Q in our case) experiences a phase reversal, that is a phase change of $\pi$. There is no phase change for light travelling through glass and reflected from the glass-air interface (at P in our case).
So the light reflected from P and Q will differ in phase by $2\pi\frac{r^2}{R\lambda}-\pi.$
Light reflected at P and Q will superpose and interfere constructively if the phase difference is $2\pi n,$ with $n = 0, 1, 2…,$ that is a whole number of cycles. The condition for a bright ring is therefore
$$2\pi\frac{r^2}{R\lambda}-\pi=2\pi n\ \ \ \ \ \text{in other words}\ \ \ \ \ \frac{r^2}{R\lambda}=n+\frac{1}{2}.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What is the difference between enthalpy and internal energy? What is pv in enthalpy's equation? if we transfer energy to a system in the form of work does it reflect in its internal energy fully and if it does why do we need a term called enthalpy which seperates pv term in its equation?if pv is not included in internal energy then why on doing work internal energy of the system changes?
| Enthalpy is a state property and is defined as
$$H=U+pV$$
Where $H$ is the enthalpy, $U$ internal energy, and $pV$ is the product of pressure and volume. So enthalpy is a state property derived from internal energy and other properties, $p$ and $V$.
Enthalpy is simply a useful derived property for analyzing certain type of thermodynamic problems. It is not a fundamental property such as internal energy, entropy, pressure, volume and temperature.
Hope this helps.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the function of this complicated tensioning system? I saw this arrangement for tensioning overhead cables from my train window (schematic below). Why not just have one pulley wheel leading directly to the weights? What function do the additional pulleys serve? For that matter, what are the cables for? They're clearly not power lines.
| You already got an answer to your main question - why are there pulleys? I will attempt to answer the second question - what is this wire?
There are two possible answers.
First, I found the following image at http://www.rail.co.uk/rail-news/ecml-suffers-another-failure/
The label tells us that this is a "along-track conductor" pair. Such conductors may be used to distribute power over longer distances (without the power line rubbing against the pantograph - think of it as a parallel circuit). The cable expands and contracts with temperature which is why you need a tensioning mechanism with a lot of "play" (and which maintains constant tension, which a spring won't do quite as well - at least not as easily).
There is a lot of detail about overhead electrification in this document if you want to read more.
But then, I found another picture on this site - one that is almost identical to yours:
The caption says
Figure 11: Overhead line suspension system. The weights and pulley system is designed to maintain contact wire tension. Photo: Author.
The insulator is a bit more clearly visible in this picture, but it seems plausible that this is a picture of the same mechanism that you saw. That means this is indeed providing tension for the contact wire, which is the wire that the pantograph rubs against.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 2
} |
Why are green screens green? How/why do green screens work? What's so special about the color green that lets us seamlessly replace the background with another image and keep the human intact?
Are there other colors that work similarly?
| I may misunderstand the question, but the method of selecting the background based on colour you are asking for is called chroma keying.
In digital post-processing, all pixels which are sufficiently green are considered background and hence treated as transparent. What is "green" is configurable, often in HSV colorspace.
J.G.'s answer elaborates why green usually works best. Blue screens are common, too.
From Wikipedia:
Chroma key compositing, or chroma keying, is a visual effects/post-production technique for compositing (layering) two images or video streams together based on color hues (chroma range)... to remove a background from the subject of a photo or video... A color range in the foreground footage is made transparent, allowing separately filmed background footage or a static image to be inserted into the scene. [...] This technique is also referred to as color keying, colour-separation overlay (CSO; primarily by the BBC), or by various terms for specific color-related variants such as green screen, and blue screen – chroma keying can be done with backgrounds of any color that are uniform and distinct, but green and blue backgrounds are more commonly used because they differ most distinctly in hue from most human skin colors. No part of the subject being filmed or photographed may duplicate the color used as the backing.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 0
} |
How do we know that light cannot travel faster than it does? We assume the speed of light in vacuum is its maximum speed but can we not assume that it could be faster, or slower?
| We used to think it could. Experiments proved that wrong.
In the 19th century, it was widely believed that light, being wave-like, must be propagating through a medium that permeates the universe, just like sound waves propagate through air. This hypothetical medium was called "luminiferous aether".
Many experiments were conducted to prove the existence of aether, and in particular to answer an important question: is aether stationary, or is it dragged along by moving matter? The most famous of these experiments was the Michelson-Morley experiment, which attempted to detect the Earth's motion through stationary aether, by comparing the speed of light in beams parallel and perpendicular to this motion.
However, the experiment did not produce the expected result, and in fact all the experiments on aether led to an odd conclusion: the speed of light seemed to always be the same, regardless of the motion of the light source, the observer, and the path taken by the light. In time, this led to Einstein formulating his theory of relativity, and the aether theory being abandoned.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why is torque defined as $r × F$ and not $F × r$? Is it merely due to popular convention or does it supply any special clarification regarding other physical quantities?
| I don't know the correct explanation but I think that $\boldsymbol{r}\times \boldsymbol{F}$ signifies that the $\boldsymbol{r}$ vector rotates because of the $\boldsymbol{F}$ vector. To produce the same result of a screw coming out when we apply a force- This is the direction of torque, whereas $\boldsymbol{F}\times \boldsymbol{r}$ will mean that the $\boldsymbol{F}$ vector rotates because of the $\boldsymbol{r}$ vector, which would be the opposite direction to which the screw comes out (downward).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
What determines the color of the light emitted in a Tokamak? We see images of Tokamak plasma with all sorts of colours from red to purple. Why do we see any light at all, since the plasma should be so hot to have dissociated all its electrons? It is all from contamination or unwanted cooling?
| As you correctly state, a plasma is composed by a certain density of charged particles (ions, electrons,...). Due to many different reasons, such as the presence of external and internal (self consistent) electromagnetic fields, these charged particles are moving under the action of various forces. It is possible to show that a charged particle that is accelerated will be emitting electromagnetic radiation (not only in the visible spectrum).
Emission of electromagnetic radiation due to the accelerated motion of charged particles in a plasma is often an important part in the energy balance of a plasma system (as the Tokamak configuration you are talking about).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
In what context is enthalpy a convenient concept? Internal energy $U$ is clearly an important concept; the first law of thermodynamics states that for an isolated system internal energy is constant $(\Delta U=0)$ and that for a closed system the change in internal energy is the heat absorbed by the system $Q$ and work done on the system $W$ $(\Delta U=Q+W)$.
Enthalpy $H$ is the sum of the internal energy and the product of the pressure and volume of the system $(H=U+PV)$. I was taught that enthalpy is a preferred quantity to internal energy for constant pressure systems where $\Delta H=Q$, as opposed to constant volume systems where $\Delta U=Q$. But why would anyone care about what quantity is equal to $Q$ under certain conditions instead of simply reporting $Q$? Enthalpy seems redundant in this context.
Is enthalpy a convenient concept in other contexts, such as systems with varying pressure? Is it described by any fundamental laws as internal energy is by the first law?
| In addition to what @Bob D and and others said about the use of enthalpy (primarily) in performing energy balances on continuous flow systems, enthalpy is also important in quantifying the temperature dependence of the equilibrium constant for chemical reactions (via the van't Hopf equation) and the temperature dependence of vapor-liquid equilibrium of single- and multicomponent chemical systems. And, of course, such equilibrium constants are important in designing and operating distillation equipment and chemical reactors.
So, if you are ever going to be working with industrial scale continuous flow systems, you are going to be working with enthalpy rather than internal energy. And, if you are ever going to understand phase equilibrium and chemical equilibrium, you better have more than a nodding acquaintance with enthalpy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Periods of non-circular Schwarzschild orbits I have been thinking about non-circular orbits in the Schwarzschild spacetime. How would you define a period of one orbit? I was thinking, in terms of proper time, for $r$, how long it takes to go from one apogee to another. For $\phi$, again in terms of $\tau$, how long it takes to cover $2\pi$. What about $t$, though? Is my reasoning wrong?
| $\let\a=\alpha \let\b=\beta \let\phi=\varphi \let\De=\Delta \def\D#1#2{{d#1\over d#2}} \def\dr{\dot r} \def\dt{\dot t}
\def\dx{\dot x} \def\dphi{\dot\phi} \def\half{{\textstyle {1 \over 2}}}$
If I could assume you can read Italian I would have an easy life - had
only to give a link. But since I find it unlikely, I'll write a
synthesis of essential points.
Consider a general metric
$$d\tau^2 = g_{\a\b}\,dx^\a dx^\b.\tag1$$
We are interested in timelike geodesics, which can be parametrized with
proper time $\tau$. Then coordinates are functions of $\tau$:
$$x^\a = x^\a(\tau) \qquad dx^\a = \D {x^\a}\tau\,d\tau$$
and from (1) we have
$$g_{\a\b}\,\dx^\a \dx^\b = 1.\tag2$$
It can be shown that geodesics obey a variational principle with
lagrangian
$$W = \half\,g_{\a\b}\,\dx^\a \dx^\b.$$
Eq. (2) shows that $W$ is a constant of the motion, with $2W=1$ on
a timelike geodesics.
Schwarzschild's metric, restricted to the plane $\theta=\pi/2$, is
$$d\tau^2 = \left(\!1 - {1\over r}\!\right) dt^2 -
{dr^2\over 1 - 1/r} - r^2 d\phi^2$$
where units were so chosen $G=1$, $c=1$, $2M=1$ ($M$ Sun's mass). Then
$$2W = {r - 1 \over r}\,\dt^2 - {r \over r - 1}\,\dr^2 -
r^2 \dphi^2 = 1$$
Since $W$ doesn't depend on $t$ and on $\phi$, we have the constants
of the motion
$${r - 1 \over r}\,\dt = E \qquad r^2 \dphi = J.\tag3$$
Substituting (3) into (2) we have
$$\dr^2 = E^2 - \left(\!1 - {1 \over r}\!\right)\!
\left(\!1 + {J^2 \over r^2}\right)\!.\tag4$$
Integrating eq. (4) by separation of variables we get $\tau(r)$ and
the radial period. Unfortunately an elliptic integral is involved.
As to $\phi$, from the second of (3) and (4) we have
$${J^2 \over r^4} \left(\!\D r\phi\!\right)^{\!\!2} =
E^2 - \left(\!1 - {1 \over r}\!\right)\!
\left(\!1 + {J^2 \over r^2}\right)$$
which gives $\phi(r)$, again as an elliptic integral. An approximation
is possible to deduce perihelion precession (I'll not show how to do it).
Instead I find a problem if the azimuthal period is of interest. The
reason is the following. Let's start form perihelion: when $\phi$
increases by $2\pi$ we are not yet arrived at another perihelion,
because of precession. This shows qualitatively that azimuthal period
is less than radial one. But a further increment by $2\pi$ brings us
still farther from perihelion, and I expect that the latter $2\pi$
variation of $\phi$ takes a different time from the former (larger or
smaller?) So it seems that a well definite azimuthal period doesn't
exist.
Edit. But an average period can be defined. let $T_r$ be the radial period, $\De\phi$ the perihelion advance in time $T_r$. Then in the average $\phi$ advances by $2\pi$ in time
$$T_\phi = {2 \pi\,T_r \over 2\pi + \De\phi}.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/446084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Can we ever "measure" a quantum field at a given point? In quantum field theory, all particles are "excitations" of their corresponding fields. Is it possible to somehow "measure" the "value" of such quantum fields at any point in the space (like what is possible for an electrical field), or the only thing we can observe is the excitations of the fields (which are particles)?
| In QFT, it's not possible to measure the value of quantum fields at any point in space. This is because quantum fields are not in spacetime (per the Copenhagen Interpretation, Transactional Interpretation, and others which include the concept of wave function collapse). They are calculated entities which we infer from the behavior of particles (which are in spacetime). When a measurement is made, a particle appears. For example, when a photon hits a photographic plate, it is a real thing and we can measure it's position. But the underlying electromagnetic quantum field can only be calculated.
The wave function calculates the quantum field at any point. But the calculation does not tell us the strength of the field. It tells us the probability of detecting (measuring) a particle.
For example, let's say that we're talking about the electromagnetic field and the calculation is for predicting the likelihood of detecting a photon in a particular position. Here's an image of the relationship of the quantum field to the particle:
The red grid graphs the wave function's calculation of the probabilities that a particle will be detected. The green film shows where a photon has actually been measured in spacetime. In this sequence, its path has been measured in 4 positions.
This image is a still from an excellent 5-minute film by Fermilab on QFT 3. It addresses your question. Also, see this article on measurement in quantum mechanics in an encyclopedia for laypeople which addresses this issue in straightforward terms.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/446492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Please help identify this physics apparatus!
This was my grandfather’s and have no idea what it is only that it is some piece of physics equipment!
The main black cylinder doesn’t seem like it wants to rotate but not sure if it should?
| It is a spark radio transmitter.
The first working radios.
Video: https://www.youtube.com/watch?v=YSf93g0heUA
Pics: https://www.google.com/search?q=spark+radio+transmitter&source=lnms&tbm=isch&sa=X&ved=0ahUKEwi-68m5vJjfAhXMx1kKHVuUASQQ_AUIDygC&biw=1920&bih=930
This one looks awfully similar and might give you some help finding out model and such:
http://www.samhallas.co.uk/bt_museum/radio.htm
Remember 300 baud modems that you put the handset into? This was the top of the line once upon a time too. It's why we have "SOS" in our language rather than relying upon a simple "Oh God, we need help!"
And if Tesla'd ever realized be was using Morgan's money to succcessfully invent radio transmission rather than failing at wireless power transmission, we'd've never called these "Marconi Spark Gap Transmitters"... but he didn't.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/446618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 2,
"answer_id": 1
} |
Difference in spectrum of green laser and green LED In an experiment I conducted I used a spectrometer to find the spectrum of green laser and green led.
this is what I found:
LED spectrum:
Laser spectrum:
why is the spectral width of the LED is wide compared with the laser?
| Also there are green lasers that produce light directly from semi-conductor materials (InGaN -see wiki). The green LED uses the same material but probably less pure. The laser is a tiny chip ( very pure, very few crystal defects, very carefully grown) with mirrors at either end. Due the exact dimensions of the crystal and distance between the mirrors a narrow range of optical frequencies is produced. An LED is a bigger device with no mirrors and light scatters out.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/446858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Falling object and the side wind I don't have a big knowledge on Physics so I'm sorry in advance if it doesn't make sense..
If a brick is falling from 100m tall-building when there's the side wind of 30m/s, is there any way I can find how far the brick would have traveled from the origin?
If there's any equation I can utilize please let me know. I'm from architecture and trying to find out how to protect pedestrians from the falling objects. Thank you very much!
| The usual, and only simple, way of calculating the sideways displacement $d$ of a projectile in a cross wind $W$ is
$$
d= W(T_{\rm air}-T_{\rm vacuum})
$$
Here $T_{\rm air} $ is time to target, (or ground in your case) taking into account air resistence, and $T_{\rm vacuum}$ is the time the projectile or brick would take to reach the target (or ground) if there were no air resistance.
You can easily derive this by considering the motion in the frame of the wind. You can also give a more complicated derivation by considering the sideways component of the drag force and relating the backwards component of the drag to change in time to reach the target. The hard thing, of course, is computing the drag.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/446915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the "lowest energy"? In many textbooks I come across the term lowest energy. For example in atomic structures, electrons are placed in orbitals in order for the atom to have the lowest energy. But what is this energy? Potential- or kinetic energy or the sum of the two?
| The lowest energy of a quantum system is the minimum eigenvalue of the Hamiltonian of the system. The Hamiltonian is the operator which corresponds to the total energy of the system, so it is the sum of the kinetic and potential energy. This is often also referred to as the ground state energy of the system.
but what is this energy?
Heuristically, the ground state energy is the energy which the system has simply by existing. Normally, this doesn't make any sense: If I just make the system do nothing, just sit there, then it would have zero energy. We wouldn't need the fancy name.
But it turns out that this is impossible. We can never "pin down" a system and make it completely still, due to the famous Heisenberg uncertainty principle,
$$\Delta x \Delta p \geq \hbar/2.$$
The system must have some momentum, to satisfy the inequality. In turn, it will also have some energy. This is why we can never reach absolute zero! No matter how hard we try, there always remains the jiggling of the ground state.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/448569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Can a battleship float in a tiny amount of water? Given a battleship, suppose we construct a tub with exactly the same shape as the hull of the battleship, but 3 cm larger. We fill the tub with just enough water to equal the volume of space between the hull and the tub. Now, we very carefully lower the battleship into the tub.
Does the battleship float in the tub?
I tried it with two large glass bowls, and the inner bowl seemed to float. But if the battleship floats, doesn't this contradict what we learned in school? Archimedes' principle says "Any floating object displaces its own weight of fluid." Surely the battleship weighs far more than the small amount of water that it would displace in the tub.
Note: I originally specified the tub to be just 1 mm larger in every direction, but I figured you would probably tell me when a layer of fluid becomes so thin, buoyancy is overtaken by surface tension, cohesion, adhesion, hydroplaning, or whatever. I wanted this to be a question about buoyancy alone.
| Note that the water does not need to have been present - this calculation gives just the way to calculate the non water volume occupied by the floating object (that hence is unavailable for water).
So in your example, assume you put the ship into the container. If you filled it up to the same water level without the ship being present - that is the displacement caused by the ship.
BBB
wBBBw wwwww w w
wBBBw vs. wwwww NOT w w
wwwww wwwww wwwww
In this sketch, 6 w have been "displaced".
Hence your example works even with a tiny gap. You can argue that hydraulic machines work with 0 gaps. The piston is the boat in the cylinder.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/448673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 8,
"answer_id": 3
} |
The water analogy seems to imply that power = current. Why is this incorrect? Many people think of the water analogy to try to explain how electromagnetic energy is delivered to a device in a circuit. Using that analogy, in a DC circuit, one could imagine the power-consuming device is like a water wheel being pushed by the current.
In the case of an actual water wheel, the more water that flows per unit of time, the more energy gets delivered to the wheel per unit of time: power = current, but in electric circuits power = voltage x current.
Why is this?
| Power is defined as work done per unit time. So a mass of water moving from one potential to some other lower potetial can do work when it hits the wheel. How much work per unit time? It depends how much mass falls times height times gravitational constant g and all of that divided by time. Water current, on the other hand is just total volume or mass that flows per unit time. Sure, it is connected to the work done or power but is not the same thing. If you define gravitational potential difference as gH and water flow as dm/dt then to have power you have to have curent times this potential difference: gHdm/dt...
In a conductor, work is done also and energy per unit charge or potential difference is given by U (voltage) and current by dQ/dt so it seems to me that everything is same...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/448724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 9,
"answer_id": 8
} |
Future pointing light cones in Black Hole in Schwarzschild Coordinates In examining black holes in Schwarzchild Coords (ie without resorting to other coords) the r coord becomes timeline within the event horizon and the t coord spacelike.
Therefore the light cone is tilted by 90 degrees. However, how do we say which direction in r is future and which is past? (textbooks jump to the forward light cone being radially inward with minimal explanation).
This question is further complicated by the theoretical existence of white holes (even though they are not thought to be a physical reality) where the future light cone is pointing radially outwards. Is there a less hand wavy approach to explaining all of this?
| You can't expect an explanation if you stick to singular coordinates,
where there is no smooth way to go from without to within.
Shift to well-behaved coordinates (e.g. Kruskal-Szekeres). Then
light-cones behaviour is quite simple and you may easily see which
their relationship must be wrt to Schwarzschild coordinates.
In K-S coordinates light-like geodetics are all at 45°, $t=$ const
lines are straight lines through the origin, $r=$ const lines are
equilateral hyperbolas with 45° asymptotes (see e.g. wikipedia article
https://en.wikipedia.org/wiki/Kruskal-Szekeres_coordinates for
help).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/448884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Physical Model for a Black Body I am learning about black body radiation and it was stated in a textbook that
The best model for a black body is a large cavity with a very small hole in it which absorbs all incident radiation. The radiation hits the walls and rattles around inside and comes to thermal equilibrium with the walls before emerging from the hole.
However, the definition for a black body is an object that absorbs all incident radiation. In this case, for incoming radiation that did not hit the hole but the external walls of the cavity, they could either be reflected or absorbed. How then is the cavity a black body?
| I think in this case the author is referring to the small hole as being a good model for a black body. As mentioned any light which hits the hole will travel inside the cavity and reflect on the walls of the cavity until it is absorbed. The larger the cavity the more likely it is to absorb since it's less likely to be scattered back outside of the hole.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How is pressure an intensive property? I've seen this question asked before but I can't find an answer to the specific point I'm troubled with. From the kinetic theory of gases, pressure results from molecules colliding with the walls of a container enclosing a gas, imparting a force upon the wall. Now, if we split the container into two halves, I am told that pressure remains the same on either side of the partition assuming the gas has uniform density throughout the container. But If we split the container into two, isn't there effectively half the number of molecules striking the wall on each side so the pressure should also be halved? Shouldnt pressure be dependent on the number of molecules?
| Yet another way of looking at it: if the container is in equilibrium, then in aggregate the particles on one side the partition are the same as the ones on the other side. Every side a particle hits the partition on side A, that particle stays on side A, but would have moved to side B if it weren't for the partition. But there's another particle on side B that would have moved to side A. So those particles "cancel out" (again, in the aggregate). The "missing" particles that aren't are on side B because of the partitions are replaced by particles that stay on side B. If the container is equilibrium, then by definition, all the regions of the container contain essentially the same particles, so it doesn't matter whether it exchanges its particles with the neighboring regions (which is what happens without a partition), or it keeps its own particles (which is what happens with a partition). Putting in a partition doesn't affect the macrostate (other than the partition itself).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 4
} |
Intuitive explanation for the free field Lagrangian? The free field Lagrangian is
$$\mathcal{L}=\frac 1 2 \partial^\mu\phi\partial_\mu\phi-\frac 1 2m^2\phi^2$$
with sign convention $(+,-,-,-)$.
Plugging this into the Euler-Lagrange equations gives the KG equation.
In class we were given the semi-intuitive explanation that this resembles the classical Newtonian Lagrangian of $L=T-V$ since the first term resembles $p^2/2m$ and the second part resembles the potential.
Is there a more satisfying explanation to this Lagrangian?
| The Lagrangian for a real scalar field $\phi\left(\mathbf x,t\right)$ given by OP with the sign convention $(+,-,-,-)$ is expressed alternatively as
\begin{equation}
\mathcal{L}\left(\phi,\boldsymbol{\nabla}\phi,\overset{\;\centerdot}{\phi}\right)\boldsymbol{=}\frac12\overset{\;\centerdot}{\phi}{}^{\,2}\boldsymbol{-}\frac12\left(\left\Vert\boldsymbol{\nabla}\phi\vphantom{\tfrac12}\right\Vert^2\boldsymbol{+}m^2\phi^2\right)
\tag{01}\label{01}
\end{equation}
Comparing \eqref{01} to the usual expression for the Lagrangian $\;L\boldsymbol{=}T\boldsymbol{-}V$, we identify the kinetic energy of the field as
\begin{equation}
T\boldsymbol{=}\frac12\!\int \overset{\;\centerdot}{\phi}{}^{\,2}\mathrm d^3\mathbf x
\tag{02}\label{02}
\end{equation}
and the potential energy of the field as
\begin{equation}
V\boldsymbol{=}\frac12\!\int \left\Vert\boldsymbol{\nabla}\phi\vphantom{\tfrac12}\right\Vert^2\mathrm d^3\mathbf x\boldsymbol{+}\frac12\!\int m^2\phi^2\mathrm d^3\mathbf x
\tag{03}\label{03}
\end{equation}
The first term in this expression is called the gradient energy, while the phrase $^{\prime\prime}$potential
energy$^{\prime\prime}$, or just $^{\prime\prime}$potential$^{\prime\prime}$, is usually reserved for the last term.
Reference : $^{\prime\prime}$Quantum Field Theory$^{\prime\prime}$ by David Tong, $\S$ 1.1.1 An Example: The Klein-Gordon Equation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can two electrons repel if it's impossible for free electrons to absorb or emit energy? There is no acceptable/viable mechanism for a free electron to absorb or emit energy, without violating energy or momentum conservation. So its wavefunction cannot collapse into becoming a particle, right? How do 2 free electrons repel each other then?
| Basically, because the process doesn't simultaneously conserve energy and momentum; this is why we say that it's mediated by a virtual photon.
In more technical language, this means that the photons that are exchanged between two interacting electrons are allowed to be "off shell", where the "shell" is the relationship $E^2 = p^2c^2$ (for a massless particle). Real particles are required to be on-shell, but virtual particles are allowed to stray from that condition, at least by some amount.
In the Feynman-diagram description of the scattering between two electrons you often find diagrams where one electron emits a photon which is then absorbed by the other; this is a virtual photon and as such both the 'emission' and 'absorption' processes are exempt from these considerations.
As always, though, it bears repeating that Feynman diagrams are calculational tools, and none of the virtual particles that appear in those diagrams actually physically exist in any definable sense. The fact that the 'emission' and 'absorption' processes, as well as the virtual photon itself, seem to defy energy or momentum conservation is purely a quirk of the way in which we've chosen to interpret limited chunks of our calculation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Violating Newtons First Law! Suppose you are inside a very large empty box in deep space , floating ( i.e not touching the box from anywhere initially).The box is at complete rest.
Now you push the box forward from inside.
Now you would go backwards but the box will move forward to conserve momentum.
However since you were inside the box your force is an internal force but the box would have moved forward.
So doesnt this violate newtons 1st law
as an internal force made a body move from state of rest?
| Internal forces can make things move. Nothing says they can't. It's just that any internal forces have an "equal and opposite force" that is also internal, so momentum is conserved, at least when considering internal forces. This has everything to do with Newton's third law and nothing to do with Newton's first law. Newton's first law says nothing about the distinction between internal and external forces.
If you want to apply Newton's first law, all you say is "I was at rest, the box pushed on me, I experience a net force, so I start to move."
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
How can we define energy other than the definition that it's a capability to do work? It is actually a property of energy that it can do some work not an actual mean to define it because we cannot define a thing on the basis of what it is doing or what it can do.
| "[…] we cannot define a thing on the basis of what it is doing or what it can do."
Why not? Try defining a progressive wave other than in terms of what it's doing!
In my opinion "The energy of a system is the amount of work it can do" is an excellent starting definition of energy. It enables one to derive the Newtonian formula ($\frac{1}{2}mv^2$) for kinetic energy, and formulae for potential energy in uniform and inverse square law fields, escape velocities, closest distances of approach and so on.
Later on one discovers various difficulties with the "amount of work it can do" definition. The Second law of Thermodynamics is, on the face of it, a glaring example. If heat is a form of energy, but you can't turn it all into work, that seems to generate a contradiction. [This argument is (deliberately) rather sloppily stated.] Another example: according to quantum mechanics an oscillatory system (e.g. a diatomic molecule) has a zero point energy, that is a minimum energy that cannot be removed from the system, so the system's energy is greater than the amount of work it can do!
There are ways of getting round these difficulties and preserving the "amount of work" definition of energy, but they come at the cost of caveats and interpretative devices (arguably sophistry) that spoil the original simplicity of the idea. In fact one may wish to abandon "the amount of work it can do" as a definition of energy, and to look upon energy as a conserved quantity which can be calculated for various systems by specific formulae or equations. An admission of defeat?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why is lattice QCD called non-perturbative? Like, if you are approximating a smooth structure with a discrete lattice, isn't this like a perturbation from smooth space-time?
If Feynman diagrams are a perturbative method, why are Feynamn diagrams on a lattice/grid called non-perturbative?
| In general, by a perturbative approach, we mean an approximation of the form,
$$f = f_0 + \epsilon f_1 + \epsilon^2 f_2 + \dots$$
where $\epsilon$ is the perturbation parameter, for some solution $f$. That is to say, one can approximate the behaviour of the solution by this series.
However, summing all the terms does not mean you will recover the exact solution; in most cases perturbative series are asymptotic series.
On the other hand, lattice QCD is an approach which is not described as perturbation theory because it does not follow this scheme, and in principle one recovers the exact solution in the appropriate limit.
To convince yourself of the distinction, consider the differential equation,
$$\frac{\mathrm d f}{\mathrm dx} = g(x).$$
If we choose to discretize it (very naively), we obtain a linear system,
$$\frac{f_{i+1}-f_{i}}{\Delta x} = g_i$$
which is a totally different approach to plugging in a series expansion like the one above. That being said, Feynman diagrams are always a perturbative approach, so Feynman diagrams on a lattice are a perturbative approach, but putting the theory on a lattice is not what makes it perturbative.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
I have a question regarding the Painlevé-Gullstrand (PG) metric with factor 2 I have a question regarding the Painlevé-Gullstrand (PG) metric.
If we have the line element in a radial fall we get:
$$d\theta = d\phi = 0$$
$$ds^2 = -dT^2 + \left(dr+\sqrt{\frac{r_s}{r}}dT\right)^2.$$
Writing out the binomial formula we obtain:
$$ds^2 = -dT^2 + dr^2 + 2 \sqrt{\frac{r_s}{r}} dr dT + \frac{r_s}{r} dT^2.$$
If we now want to write down the metric tensor, we should obtain:
$$g_{\mu\nu} = \begin{pmatrix}
1 & 2\sqrt{\frac{r_s}{r}} \\
2\sqrt{\frac{r_s}{r}} & \frac{r_s}{r} -1\\
\end{pmatrix}.$$
So am I right, that the factor 2 also comes into the metric?
| No, the off-diagonal metric components are $g_{rT}=g_{Tr}=\sqrt{\frac{r_s}{r}}$ without a factor of 2.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/449959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why are ropeway pillars tilted? While skiing I have noticed that ropeways pillars are usually tilted to be perpendicular to the slope (fig.1). If the gravity is pulling straight down, why aren't they vertical as they are supposed to support ropeway's weight? Is there something more they "do"?
Also there are pillars which function is not to support weight but to "push" the rope down in order to keep it tense, why is it so crucial? Why the rope needs to be tense?
fig.1
| The cable can only put a force on the mast that is perpendicular to itself: It runs over rollers which do not allow it to transfer any force in the direction of the cable.
As such, the force on the mast is exactly given by the angle between the oncoming and the outgoing cable. This force is roughly perpendicular to the slope of the cable (assuming a small deflection of the cable by the mast), especially if the cable has several supports in a roughly straight line.
The mast is setup in such a way, that the sum of this perpendicular force and its own weight is roughly in the direction of the mast itself. And since the weight of the mast is by far the smaller force of the two, you see the strong diagonal setup.
Of course, the cabins pull the cable straight down, even though the masts only take a force perpendicular to the cable. The difference is taken by the cable itself, which has a much higher tension at the upper end of the ropeway than at the lower end due to this.
As to why you need so much tension on the cable: The lower the tension, the greater the slack of the cable between the masts, and the more masts you need. It's generally just much cheaper to build two very well-anchored stations and use a high tension, than to build more masts. And that assumes that the additional masts can actually be built at sensible costs, which may not even be the case.
Also, the higher the tension, the higher the designed, directional forces of the cable relative to the forces introduced by wind, and seat-rocking passengers. And thus the safer the seat of the cable on the rollers.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/450078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Potential Difference of a battery - What does it mean? I have studied current electricity for a while now. When I look back at basic concepts, I am quite clear about what current, electron, resistance is. But I cannot imagine about the potential difference or voltage of a battery. Or in a circuit, it is said that potential drops across a resistance, why is that so?
What does it mean to have a potential difference? I asked my friends too, but none of them have quite understood the concept too. So, can you clarify the concept of p.d? using analogies or any way that might easy.
| I think it is a measure of how much energy it takes to move a charge over a certain path.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/450445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Propagator for W boson I've found in different literature that some write the propagator for the W boson as $\frac{g_{\mu\nu}-\frac{k_\mu k_\nu}{M^2_W}}{k^2-M^2_W+iM_W\Gamma_W}$ and others like $\frac{g_{\mu\nu}-\frac{k_\mu k_\nu}{M^2_W}}{k^2-M^2_W}$, without the $iM_W\Gamma_W$ term in the denominator.
Is there a reason for that or am I missing something?
| Putting an $iM\Gamma$ in the denominator of a propagator for bosonic particle with mass $M$ makes the particle unstable, decaying with rate $\Gamma$. However, it also makes the propagator non-unitary, since the decaying particle doesn't turn into anything else; it just disappears, and probability is lost. So the propagator with the $\Gamma_{W}$ cannot be the true propagator. Nevertheless, if you want to do a tree-level calculation for a certain process that contains a $W^{\pm}$ intermediate, including the instability of the boson will improve the accuracy of your calculation.
In fact, neither of the propagators you cite is actually the "true" $W$ propagator. I put "true" in quotation marks, because the kind of propagator you need to use to calculate radiative corrections is actually dependent on the gauge. Moreover, in addition to the vector bosons, you typically also need to include scalar (Fadeev-Popov) ghosts in the representation of the gauge field. These complications arise in both non-Abelian gauge theories and spontaneously broken gauge theories; and the electro-weak theory describing the $W^{\pm}$ is both!
So the propagators listed in the question will not work if they are used for anything beyond tree-level calculations. If you try to use such expressions in radiative corrections, you will get gauge-dependent and nonunitary results. Thus, if you are already limited to tree level, you might as well include the boson decay term, which is why many people do.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/450544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why did Einstein pick his definition of synchronous clocks? In his paper "On the Electrodynamics of Moving Bodies", Einstein defines synchronous clocks as clocks which satisfy $t_B-t_A=t'_A-t_B$. $A$ and $B$ are two points separated by empty space, and identical clocks are located at $A$ and $B$. A light signal is emitted from $A$ at time $t_A$ as measured by the clock at $A$. The signal arrives at $B$ at time $t_B$ as measured by the clock at $B$. The light signal is immediately reflected at $B$, and it returns to $A$ at time $t'_A$ as measured by the clock at $A$.
Why does he choose this definition for synchronous clocks? Why is this a sensible definition for synchronous clocks?
He does say this which might provide some insight into why he chose this definition: "We have not defined a common “time” for
A and B, for the latter cannot be defined at all unless we establish by definition
that the “time” required by light to travel from A to B equals the “time” it
requires to travel from B to A."
I can't seem to understand this quotation, because I'm not sure what he means by "time" in this sense. So far time has only be defined at the locations of the two clocks, so you can't use one clock to measure the time for the light signal to move between the points, you would have to use a measurement from both clocks, but there is not necessarily any relation between the time displayed by the two clocks as far as I understand the situation.
| This method of synchronization works because both clocks can agree on the speed of light.
If we want to determine whether or not two distant events are simultaneous, or more generally the order in which events occur, then we need to factor in the time it takes for the signals to reach us. We can easily find this time if we know the distance of the event, and the speed of the signal.
The constant speed of light is convenient, since everyone can agree on its value, it's $c$. The method of synchronization also ensures distance is fixed, as the signal is returned immediately with no delay. So if both clocks can agree on the speed and distance, then it means they must agree on time (otherwise, they are out of sync).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/450705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
It seems that the Euler equation in thermodynamics and The first law of thermodynamics are in contradiction The Euler equation in thermodynamics are as followed:
$U=TS-PV+\mu N$
But The first law of thermodynamics states that
$dU=TdS-PdV+\mu dN$
But I think that The Euler equation can be written by
$dU=TdS+SdT-PdV-VdP+\mu dN+Nd\mu$
Then, $SdT-VdP+Nd\mu=0$
But I don't think that this is always correct.
Edit
I see that it's only correct if the system's homogeneous. Can you give me an example of a homogeneous system and nonhomogeneous system?
| This is the Gibbs–Duhem equation. It's related to extensiveness of energy, entropy, volume and number of particles.
The assumption is that the edge effects are much smaller than the volume effects, in particular, when combining two systems the total energy is the sum of energies of the smaller systems, without "interaction terms".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/450978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does the inverse-square law apply to linearly polarized light? It's a stupid question but: We did and experiment using linearly polarized microwave radiation generators and receivers. Our teacher asked to check experimentally if the receiver measurements are proportional to the intensity of the radiation I or to the intensity of the electric field $E$. To check that we took the measurements M of the receiver to different distances R between the receiver and the generator. She said that if the receiver goes with I the diagram of $M=f(\frac{1}{R^2})$ will fit better to linear fitting than the $M=f(\frac{1}{R})$ diagram. If the reverse happens the receiver goes with $E$. So that's why I am confused. Shouldn't the $\frac{1}{R^2}$ relationship indicate that the receiver's measurements are proportional to $E$?
| Nope. Point sources of waves have fields falling off as $1/r$ while intensities fall off as $1/r^2$. To see why, think about conservation of energy.
A point source is spherically symmetric, and so the intensity should only be a function of the distance $I(r)$. Intensity is defined as the area flux density of energy due to the wave. Consider a sphere of radius $r$ around the point source. The total power going through this sphere should be the product of the area and intensity:
$$P=4\pi r^2 I(r).$$
And since the power leaving the sphere should be the same at any distance (all the energy has to leave the sphere in order to be conserved) then the intensity is given by
$$I(r)=\frac{P}{4\pi}\frac{1}{r^2}.$$
Which clearly falls off as $1/r^2$. To see why the field (electric field, in your case) goes as $1/r$, remember that intensity and field $\phi$ are related as
$$I=\kappa|\phi|^2,$$
With $\kappa$ being a constant. As such, we can then say that
$$|\phi|=\sqrt\frac{P}{4\pi\kappa}\frac{1}{r},$$
Which falls off as $1/r$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/451100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What is a good site to use for finding physical constants at times when NIST web resources are affected by government shutdowns? There are multiple resources for finding physical constants (say, particle data, atomic spectra, or even special-function identities) which are generally hosted by NIST, and which form essential parts of the scientific infrastructure of multiple fields.
However, on occasion, these essential NIST-hosted websites can become unavailable for extended periods of time, particularly when NIST's funding is affected by government shutdowns in the US.
What are suitable alternatives which can be used in this situation?
| The Particle Data Group is the only way to go.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/451460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The reasoning behind doing series expansions and approximating functions in physics It is usual in physics, that when we have a variable that is very small or very large we do a power series expansion of the function of that variable, and eliminate the high order terms, but my question is, why do we usually make the expansion and then approximate, why don't we just do the limit in that function, when that value is very small (tends to zero) or is very large (tends to infinity).
| The key reason is that we want to understand the behavior of the system in the neighborhood of the state rather than at the state itself.
Take the equation of motion for a simple pendulum, for example:
$$\ddot{\theta} = -\frac{g}{\ell}\sin(\theta)$$
If we take the limit where $\theta \rightarrow 0$, we find $\ddot{\theta}= 0$, and we would conclude that the pendulum angle increases or decreases linearly with respect to time.
If we however take a Taylor expansion and truncate at the linear term, we find $\ddot{\theta} = -\frac{g}{\ell}\theta$, which is a simple harmonic oscillator! This expansion shows us that in the neighborhood of $0$, the system returns back to $0$ as if it was a simple harmonic oscillator: completely unlike what we could state in the limit approximation above.
In fact, you could consider the limiting behavior around a state to be the zeroth-order component of a local expansion, which holds true straightforwardly for the example above since the limit term contributes no terms to the dynamics of the pendulum (but correctly notes that the angle increases/decreases linearly very close to $0$).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/451588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 1
} |
$Q$-factor for damped oscillator (not driven)? How would this be defined?
Some of the Q-factor definitions I have encountered include:
$$Q=2\pi\frac{\text{Energy stored}}{\text{Mean power per cycle}}\\Q=2\pi\frac{\text{Energy stored}}{\text{Energy lost per period of
oscillation}}\\Q=2\pi\frac{1}{\text{Fractional power lost per cycle}}$$
However, none of these seem to work for a non-driven, damped oscillator. The first two won't work because energy stored is not a constant, and unless fractional power lost per cycle is a constant (is it, and if it's then how do you show that?) the third won't work either.
| The Q-factor tells you something about the frequency response of a driven system to a constant amplitude driver when a steady state (constant amplitude of driven system) has been reached.
The driver supplies energy to the driven system which at steady state results in the energy stored in the driven system staying constant (constant amplitude) and there is also a constant rate of energy (power) dissipation from the driven system..
With a non-driven system there is no input of energy into the system and so the energy (amplitude) of the oscillating system just decreases with time.
The Q-value is ratio of the total energy stored in the oscillating system (at some time) divided by the energy lost in the following single cycle.
For small amounts of damping (large values of Q) the Q-value is the number of oscillation such that the amplitude drops off to approximately $\frac {1}{25}^{\rm th}$ of its original value.
The Q-value is also equal to $\frac{\pi f_0}{\alpha}$ where $f_0$ is the natural frequency of the undamped oscillator and $\alpha$ appears in the term ${\rm e}^{-\alpha t}$, where $t$ is the time, which controls the rate at which the amplitude of the oscillations decay.
I have just found out that this is a duplicate question - Definition of the Q-factor?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/451690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Projective measurement using two mode squeezed state? Let me define two mode squeezed states as $ \left | \xi \right>_n=\exp\left(\xi \hat{a}^\dagger \hat{b}^\dagger-\hat{a} \hat{b} \xi^\star\right)\left | n,0 \right>$ where $\left|n,0\right>$ is the number state with $n$ photons in mode 1 and none in mode 2. Now, let me define a projective
measurement $\{\left | \xi \right>_n\left<\xi\right|,\mathbb{I}-\left | \xi \right>_n\left<\xi\right|\}$. Is it possible to implement this projective measurement in the lab? have you seen any paper which has been discussed such a measurement?
| If you are happy with destructive measurements, this can be done in principle in the lab. Let $\rho$ be the input state, and write $|\xi\rangle = S|n,0\rangle$.
*
*Undo the squeezing transformation $S$, i.e., transform $\rho'=S^\dagger\rho S$. Note that $S^\dagger$ is just another squeezing transformation which can in principle be implemented in the lab. Correspondingly, this means that now you need to perform the measurement $\{|n,0\rangle\langle 0,n|,I-|n,0\rangle\langle 0,n|\}$.
*Use photon number resolving detectors to measure both modes. Again, this is something which can be done in principle (to a certain accuracy). Count the outcome $(n,0)$ towards the first outcome, and all others towards the second.
Formally speaking, what you are looking for is the probability $p_\xi$ for the first outcome, which we can rewrite as
$$
p_\xi = \langle \xi|\rho|\xi\rangle
=\langle n,0|S^\dagger \rho S|n,0\rangle
= \langle n,0| \rho'|n,0\rangle\ ,
$$
where the last term is exactly what the proposed measurement determines.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/452101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
are there changing magnetic and electric fields that are not EM radiation? Let us consider these two Maxwell equations:
$$\frac{\partial \vec{B}}{\partial t}=-\vec{\nabla}\times \vec{E}$$ and
$$\frac{\partial \vec{E}}{\partial t}=\frac{1}{\epsilon_0}\left(-\vec{J}+\frac{1}{\mu_0}\vec{\nabla}\times \vec{B}\right).$$
When we consider faraday's law of induction, we usually assume that the changes are slow, and thus we can neglect radiation by assuming that the left hand side of the second equation is zero. That is, a changing current creates a changing magnetic field, which in turn creates a changing electric field, per the first equation.
I cannot understand this. First, if we can neglect the change in E from the second equation, should not we also neglect the change in B in the first equation? Second, this imply that we can have changing electric and magnetic fields that are not electromagnetic waves. But are not all changing magnetic or electric fields EM waves? or is this approximation equivalent to a charge moving at constant speed, in which the change in E and B are not due to radiation but just to the translation motion of the static field lines
| Regular induction in a coil satisfies Maxwell’s equation (usually the integral form is used) and it’s usually not considered radiation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/452560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Where does it getting wrong , when using $v^2 - u^2 = 2as $ down the incline, for different object having different moment of inertia? Well, Consider a situation there is a sphere and a ring, of same mass $M$ and radius $R$. They both starts rolling down the inclined plane. We know moments of them as well, $$I_\text{sphere}=\frac{2}{5}MR^2$$ and $$I_\text{ring}=MR^2$$ respectively. So, We know that sphere will have more transitional kinetic energy, so more velocity, so it will take less time to reach at bottom.
The question is while using equation for both, $$v^2 - u^2 = 2as, $$ initial velocity is $0$ for both, final velocity are different for both, but acceleration and distance traveled same. So, where is the blunder happening?
And also the equation $$v=u+at,$$ if velocity for sphere is greater, then what about the time? Why is the time taken less? Where are the equations getting wrong or is it me getting it wrong?
| What you have left in calculations is acceleration,, a≠g nor a=gsinα.
$$a= gsinα-F/M$$
where α is angle of incline, F is force of friction and M remains mass..
Since friction acting on both are different, their acceleration are different for same distance s.
Same goes for your second equation
v=u+at, here a is different. Same goes for t, dont you think one with faster translation kinetic energy reach sooner?
Hope this helps..
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/452670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Photons exerting force Since gravity affects photons, and forces always work in pairs.
Does this mean that photons have a resultant force.?
And would we be able to harness this resultant force to move objects using light?
|
Does this mean that photons have a resultant force.? And would we be able to harness this resultant force to move objects using light?
Photons are elementary particles in the standard model of particle physics . They have zero mass, energy=$h*ν$, and spin $1$.
They are described by a four vector of special relativity, i.e., they have energy and momentum, and the length of this four vector, the invariant mass, is zero.
The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.
As photons have mass zero, they have no rest system.
As they have momentum, when they interact according to the rules of particle interactions, they transfer a dp/dt, i.e. a force, to the particle they interact with.
Here is a video giving an explanation of radiation pressure .
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/452880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Would someone be able to see where they are headed once they crossed the event horizon? Let us say an explorer was studying a supermassive black hole and ended up blundering past the event horizon. From my understanding, which may be wrong, the only paths allowed after crossing the event horizon are paths that bring the object closer to the singularity. Therefore, if the hapless explorer faced the direction of travel, would none of the photons ahead of the explorer be able to reach the explorer's eyes?
| If there is other infalling matter, then certainly they may be able to see it. A simple example is that if they hold their own hand in front of them, they will still be able to see their hand. This is an example of the equivalence principle, one form of which states that spacetime is always locally flat, so that on small enough scales, gravitational effects are not observable.
The infalling observer may also be able to see things at fairly large distances, if those things fell in at a late enough time. The easy way to tell that this is true is to look at a Penrose diagram.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/452990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
number density of photons My lecturer put up as a solution to the number density of photons to be $n_{o\gamma}=\tfrac{8\pi}{c^3}\int^{\infty}_0(\tfrac{kT}{h})^3\tfrac{x^2dx}{e^x-1}$. The integral on the right hand side being the Riemann zeta function with value which is approximately 1.2. But then he said this gives the value of the number density as $n_{0,\gamma}=400cm^{-3}$ for T=2.7. But if we input all the value in the first equation I gave this doesn't give anywhere close to that value.
I think perhaps what he is done is made a small blunder and the first equation gives the total number of photons in the universe and then we must divide by the volume of the universe, but even then $\tfrac{8\pi}{c^3}\int^{\infty}_0(\tfrac{kT}{h})^3\tfrac{x^2dx}{e^x-1}=1.98\times10^8$ and the volume of the universe is about $4\times10^{80}$ but dividing the first of the by the second returns a value of order $10^{-73}$which is still way off from what he said it should be.
Could anyone please explain this to me ?
| Putting this exact formula into Wolfram Alpha actually gives me $400\text{ cm}^{-3}$, assuming that your evaluation of the integral is correct. We can check to see that the units here make sense by doing some dimensional analysis. The equation is
$$\frac{8\pi}{c^3}\left(\frac{kT}{h}\right)^3\int^{\infty}_0\frac{x^2dx}{e^x-1}$$
Let's break down the units here:
$$\frac{8\pi}{c^3}\text{ is in }\frac{\text{seconds}^3}{\text{meters}^3}$$
$$\left(\frac{kT}{h}\right)^3\text{ is in }\left(\frac{\text{Joules}\cdot\text{Kelvin}^{-1}\cdot{\text{Kelvin}}}{\text{Joules}\cdot\text{seconds}}\right)^3=\text{seconds}^{-3}$$The product of these two quantities is in $\text{meters}^{-3}$ - the units of number density. Since the integral as I've written it is dimensionless, the result of this equation is indeed a number density, with a value of $4\times10^8{\text{ m}}^{-3}=400\text{ cm}^{-3}$ in cgs units - closer to the value your professor gave you. The difference in scale is simply that your professor wrote their answer in $\text{cm}^{-3}$ rather than $\text{m}^{-3}$, as you calculated it.
The remaining discrepancy with your result of a factor of $2$ is an incorrect evaluation on your part of the integral. The Riemann zeta functions is
$$\zeta(s)=\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{x^{s-1}}{e^{x}-1}$$
You want to calculate $\zeta(3)\Gamma(3)\approx2.4$.
Note, finally, that experiments tell us that the number density of photons is indeed about $410\text{ cm}^{-3}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/453123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What gives mass to black hole? I like to know when a dying star collapsed into a black hole, is there anything inside or on the event horizon that is interacting with higgs field?
| It is important to understand that energy can have mass as well. For example 99% of the mass of the proton is due to the binding energy of the quarks inside it. Only about 1% comes from the interaction of the quarks with the Higgs field. So only the masses of the elementary particles are due entirely to the Higgs field. Since a black hole is a bound state not an elementary particle we don't need to invoke the Higgs field to explain all its mass.
General Relativity treats mass and energy as equivalent and related by Einstein's famous equation $E=mc^2$. Both mass and energy contribute to the curvature of spacetime. So while we talk about a black hole having a mass of $M$, we could equally well talk about it having an energy of $Mc^2$. Both are the same in GR. This means we do not need to invoke the Higgs field to explain the gravitational field of a black hole, or indeed any massive object. We could have constructed our black hole from massless photons and it would still have a mass $M$ related to the energy $E$ we put into it by $E=Mc^2$.
So the bottom line is that we don't need to worry about the Higgs field when working with calculations in GR. This is just as well since we do not have a theory of quantum gravity, so the problem of reconciling gravity with quantum field theory is currently unsolved. We live in hope that as and when we do understand quantum gravity it will become clear how the Higgs field is involved.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/453497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Conceptual understanding of quantum harmonic oscillators The way I understand it is that we have the time-independent Schrödinger equation for a particle described by a wave function $\psi$ in a potential V(x)
$$-\frac{\hbar}{2m}\frac{d^2}{dx^2} \psi + V(x)\psi(x) = E\psi(x)\tag{1} \, .$$
If I were to approximate the potential $V(x)$ to be quadratic in $x$ (for example $V(x) = \frac{1}{2} m \omega x^2$) we know that this describes a harmonic oscillator.
If we then solve equation (1) for $\psi$ and $E(x)$ we will find that
$$E_n = \hbar\omega \left(n+\frac{1}{2} \right)$$ and the corresponding $\psi_n$.
Now it is mostly the sentences I can't really formulate.
Like for example which of these statements is correct:
*
*Our particle described by $\psi$ is a harmonic oscillator.
*Our particle described by $\psi$ is modeled by a harmonic oscillator.
*Let us consider a particle described as a harmonic oscillator...
As you might see, I have the basic mathematical framework, but I can't really grasp the words/concepts surrounding it.
| The same problem arises classically. If you have a mass on a spring we have a harmonic oscillator. But do we say the mass is a harmonic oscillator? Perhaps. If so then we would say for your question that the particle is a harmonic oscillator
But the spring is an important part of the system too. If we really want to be picky I would say that that the spring mass system constitutes a harmonic oscillator system. In that case we would say the particle in the given potential can be treated as a harmonic oscillator system.
In that sense anything which has a Hamiltonian that looks like
$$
H = ax^2+bp^2
$$
can be called a harmonic oscillator.
In any case, all of the sentences you wrote down are reasonable and would be understood by anyone.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/453638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Mass hanging from spring in free fall
Q: A mass $m$ hangs from a massless spring connected to the roof of a box of mass $M$. When the box is held stationary, the mass-spring system oscillates vertically with angular frequency $\omega$. If the box is dropped falls freely under gravity, why does the angular frequency increase?
So initially, we have $\omega=\sqrt{\frac{k}{m}}$, where $k$ is the spring constant. I read that the new angular frequency when in free-fall is $\omega'=\sqrt{\frac{k}{\mu}}$, where $\mu=\frac{Mm}{M+m}$ is the reduced mass, i.e. $\omega'=\sqrt{\frac{k(M+m)}{Mm}}$ which is clearly greater than $\omega$.
I'm not sure why this is true - I suspect it has something to do with the box oscillating (as it's no longer held stationary), but I'm not too familiar with the concept of reduced mass so I'd really appreciate a good explanation.
| Despite the weird geometry of the problem, this is simply two masses connected by a spring, and we can solve that problem using Lagrangian Mechanics. Let's assume that the masses have mass $M$ and m, and position coordinates $x_M$ and $x_m$ respectivly. They are also separated by a distance $d$, and the spring constant is $k$.
We will use $\alpha$ as the generalized coordinate representing how far the masses have stretched from equilibrium:
$$
\alpha = x_M - x_m - d
$$
Note also that
$$
\dot{\alpha} = \dot{x}_M - \dot{x}_m,
$$
because $d$ is a constant, and we can choose a frame in which the total momentum of the system is zero, which gives us
$$
m \dot{x}_m = -M \dot{x}_M
$$
We can write the kinetic energy as
$$
T = \frac{1}{2} m \dot{x}_m^2 + \frac{1}{2} M \dot{x}_M^2,
$$
and then transform to our generalized coordinate $\alpha$ using the above two equations:
$$
T = \frac{1}{2} m \dot{x}_m^2 + \frac{1}{2} M \Big(\dot{x}_m \frac{m}{M}\Big)^2
$$
$$
T = \frac{1}{2} m \Big( -\dot{\alpha} \frac{M}{m+M}\Big)^2 + \frac{1}{2} M \Big( \Big( -\dot{\alpha} \frac{M}{m+M}\Big) \frac{m}{M}\Big)^2
$$
$$
T= \frac{1}{2} \frac{m M^2}{(m+M)^2} \dot{\alpha}^2+ \frac{1}{2} \frac{m^2 M}{(m+M)^2}\dot{\alpha}^2
$$
$$
T= \frac{1}{2} \frac{m M}{m+M} \dot{\alpha}^2
$$
$$
T= \frac{1}{2} \mu \dot{\alpha}^2,
$$
where $\mu$ is the reduced mass.
The potential energy is much easier to find:
$$
U = \frac{1}{2} k \alpha^2.
$$
Our Lagrangian then is
$$
L = T-U
$$
$$
L = \frac{1}{2}\mu \dot{\alpha}^2 - \frac{1}{2} k \alpha^2,
$$
which can be plugged into the Euler-Lagrange equation
$$
\frac{d}{dt} \frac{\partial L}{\partial \dot{\alpha}} - \frac{\partial L}{\partial \alpha}=0.
$$
This equation is easy to solve by hand:
$$
\mu \ddot{\alpha}+k \alpha = 0.
$$
This is clearly the equation for a simple harmonic oscillator with angular frequency $\sqrt{k/\mu}$, which is the answer you were looking for!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/454020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Explanation of Lenz's Law phenomena If we drop a magnet through a copper pipe (without it touching any of the sides), it would fall slower than it would if there were no pipe.
Having the pipe otherwise accelerate the magnet would be in violation of Lenz's law and conservation of energy. I agree that, if there is any interaction between the copper pipe and the magnet, then it should be to decelerate the magnet.
But, why is there even any interaction between the pipe and magnet in the first place? Why can't there just be zero interaction? This way, the magnet is only affected by gravity, just how it would be if it were just a normal wooden or plastic pipe!
Would zero interaction be a violation of some other kind of laws? Such as, the Lorentz force law, or Biot-Savart, or something else? Would zero interaction violate special relativity? Or, some rules of quantum mechanics?
| Yes it has interction.
Beacuse the top side of the pipe will act as a opposite polarity and repel
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/454276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Will any charge oscillating in space create an EM wave? Would it be correct to say that any charge oscillating in space (regardless of the spacial amplitude) at a given frequency will emit an EM wave of the same frequency?
related: What change in an EM field is required to create an EM wave?
|
Will any charge oscillating in space create an EM wave?
Almost always yes, but a counterexample was given by G.A. Schott, "The Electromagnetic Field of a Moving Uniformly and Rigidly Electrified Sphere and its Radiationless Orbits," Phil Mag Suppl 15 (1933) 752. For a uniformly charged spherical shell of radius $b$, considering only retarded solutions, the radiation reaction force at time $t$ is proportional to $v(t-2b)-v(t)$. You get no radiation if there's periodic motion with period $2b/n$.
It doesn't seem to be possible to access Schott's paper online, but there is a 1964 paper by Goedecke that has a discussion of this type of thing. Schott was opposed to quantum mechanics, and his work is often cited by kooks, such as Randell Mills, who claims to have disproved quantum mechanics and discovered an unlimited source of energy. However, there appears to be nothing wrong with Schott's actual work in the 1933 paper.
Reading between the lines, Schott's hope seems to have been that he could salvage classical physics by showing that a hydrogen atom would not have to collapse by radiation, since it might be possible for the electron's orbit to satisfy this condition for not radiating. But this would only seem to work if the electron was a rigid, extended body orbiting at some speed that happened to be exactly matched to its own dimensions, and the electron also can't satisfy the condition unless the diameter of the orbit is less than the radius of the sphere.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/454375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Water mixture at different temperatures Let's say that we mix homegeneously and instantly cold water at $t^\circ $ C and hot water at $T^\circ$ C (like in a water tap) in ratio $p:1$. My question is the following: What is the instant temperature of this mixture? Is there a law in that sense?
I am a mathematician, not a physicist;all I know about that is Newton's cooling law. However I can't see how to apply it here.
| If the mixture is homogeneous and instantaneous, the temperature is simply determined by the ratio p. The law is the law of mixtures.
Newton's law of cooling is not applicable even if the instantaneous homogeneity is not assumed. That assumes heat loss form a body to the surrounding.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/454605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
What does the 'cosmological constant' represent? Newtons theory of gravity involves a gravitational constant $G$, however one does not refer to it directly, we speak instead of gravity or the force of gravity.
Now Einsteins introduced a cosmological constant in his equation for GR, and ever since it's been referred to as such; however, this usage seems strange to me, given the above parallel; it seems to me we ought to be referring to something physical, which can then be quantified and whose quantification involves the cosmological constant.
Does the cosmological constant then refer to a cosmological force? A force that is, contra gravity, repulsive? And which is even weaker than gravity, kicking in on the scale of galaxies rather than solar systems?
|
Does the cosmological constant then refer to a cosmological force?
The Cosmological Constant acts as “repelling” gravity. This can be seen from the Friedmann equations, here it works opposite to matter and radiation density due to the negative pressure it exerts.
This just means that the universe expands accelerated in case the CC dominates matter and radiation density (which is negligible today) in accordance to current observational data. But we still don’t talk about forces because General Relativity concerns geodesics and hence objects in free fall.
Forces come into play if we regard extended bodies and are called tidal forces then, known as “spaghettification” in the context of black holes. In our accelerated expanding universe (due to the CC) an extremely long rubber band would be stretched due to tidal forces which can be seen from second Friedmann equation also called acceleration equation.
In the ideal case, the perfect fluid assumption on which the FRL modell is based tidal acceleration is scale independent.
I’m quite late, not sure if you read that.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/454729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Why does angular momentum being constant prove Kepler's first law? So I was watching this video and this video on Kepler's first law in order to understand the proof of Kepler's first law. He started off by saying that for an ellipse, the distance from a focus point to a point on the ellipse equals to $$r = \frac{a(1-\epsilon^2)}{1+\epsilon \cos(\theta)}$$ (where $\epsilon$ is the eccentricity, $a$ is the semi-major axis and $r$ is the distance). He then proceeds to take the derivatives of the polar unit vectors and derives acceleration in polar coordinates in order to use in Newton's second law.
He then equates Newton's second law to the gravitational force and substitutes $mr^2\frac{d\theta}{dt}$ for angular momentum. After some calculation and substitution, he gets to the point where $$L^2=GMm^2a(1-\epsilon^2)$$ ($a$ is the semi-major axis and $\epsilon$ is the eccentricity) and he says it is done because the angular momentum is constant. I don't really understand the reasoning behind this, could someone please explain?
Also, am I correct in saying that the orbit must be elliptical because if $\epsilon$ was greater than $1$, $L^2$ wouldn't exist. Or is it because of something else that I'm missing?
Source videos:
Part 1: Lecture 15.6 - Deriving Kepler's 1st Law, Pt. 1; HMC Mooc on YouTube
Part 2: Lecture 15.7 - Deriving Kepler's 1st Law, Pt. 2; HMC Mooc on YouTube
| The planets equations of motion can be represented in polar coordinate $r$ and $\phi$
$$r\,\ddot{\phi}+2\,\dot{r}\dot{\phi}=0\tag 1$$
$$\ddot{r}-r\,\dot{\phi}^2=-\frac{\mu}{r^2}\tag 2 $$
where
$\mu=G\,M$
if we multiply equations (1) from the left with $r$ we get:
$\frac{d\left(r^2\,\dot{\phi}\right)}{dt}=0\quad \Rightarrow\quad r^2\dot{\phi}=h$
where ${h}/2$ is the area velocity of the radius of the planets. ($dA/dt=1/2\,r^2\,\dot{\phi}$)
Area velocity
$$\frac{d\,\vec{A}}{dt}=\frac{\vec{r}\times \vec{v}}{2}$$
and angular momentum
$$\vec{L}=\vec{r}\times\,m\, \vec{v}$$
$\Rightarrow\quad |L|=2\,m\,|h/2|=m\,|h|$
the angular momentum is equals $2m$ times the area velocity.
The solution of equation (2) is:
$$r=\frac{h^2\,/\mu}{1+e\cos(\phi-\omega)}\tag 3$$
comparing equation (3) with a standard polar equation of ellipse:
$$r=\frac{a\,(1-e^2)}{1-e\,\cos(\phi)}\quad \tag 4\Rightarrow $$
$$\frac{h^2}{\mu}=a\,(1-e^2)\tag 5$$
with $|h|=\frac{|L|}{m}$ we obtain:
$$\boxed{L^2=G\,M\,m^2\,a\,(1-e^2)}$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/455066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
3D perfectly elastic collision between two points There is a high probability, I think, that this question is a duplicate of some other question ... but to may knowledge, it hasn't been posed in this exact manner:
Assume we have 2 points, $P_1$ and $P_2$, of mass $m_1$ and $m_2$ in a world coordinate system $(O, \vec{i}_0, \vec{j}_0, \vec{k}_0)$. The point $P_1$ is moving with the constant velocity $\begin{bmatrix} v_{x1i}\\ v_{y1i}\\ v_{z1i}\end{bmatrix}$ while the point $P_2$ is stationary. The point $P_1$ undergoes a perfectly elastic collision with $P_2$. How will these two points move after the collision?
My attempt
This problem is about the conservation of linear momentum: therefore the momentum of the system formed by these two points remains constant. Before the collision the momentum of the system is:
$$\vec{P}_{init} = m_1\cdot \begin{bmatrix} v_{x1i}\\ v_{y1i}\\ v_{z1i}\end{bmatrix} + m_2 \cdot \begin{bmatrix} 0\\0\\0\end{bmatrix}$$
After the collision, the linear momentum of the system is:
$$ \vec{P}_{fin} = m_1\cdot \begin{bmatrix} v_{x1f}\\ v_{y1f}\\ v_{z1f}\end{bmatrix} + m_2 \cdot \begin{bmatrix} v_{x2f}\\v_{y2f}\\v_{z2f}\end{bmatrix}$$ The unknows are $v_{x1f}, v_{y1f}, v_{z1f}, v_{x2f}, v_{y2f}, v_{z2f}$. But we have only three equations $\vec{P}_{init} = \vec{P}_{fin}$ and six unknowns ... One can also use the law of conservation of energy to obtain another equation:
$$ \frac{m_1}{2} \cdot \left(v_{x1i}^2 + v_{y1i}^2 + v_{z1i}^2\right) = \frac{m_1}{2}\cdot \left( v_{x1f}^2+v_{y1f}^2+v_{z1f}^2\right) + \frac{m_2}{2}\cdot \left( v_{x2f}^2+v_{y2f}^2+v_{z2f}^2\right)$$ but there are still only four equations and six variables ...
| This is an indeterminate problem, there is infinity of solutions. To make it determinate, one has to add more assumptions to the model.
For example, one can add the assumption that the particles are not points, but perfectly solid spheres. Then we get two more equations (due to the fact that change of momentum of both spheres must be along the line joining the spheres in the instant of collision), so we have 6 unknowns and 6 equations, so collision of two spheres is a determinate problem. Adding a third sphere to the collision would make the problem indeterminate again, though.
Or one can assume that one of the particle is constrained to move along some prescribed axis. Then we have only 4 unknowns and 4 equations should be enough to determine them.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/455657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Why we neglect the $\hbar ω/2$ in the Hamiltonian of the the Electromagnetic Field? After the quantization of the electric and the magnetic field, we get the Hamiltonian of the electromagnetic field:
$$H= \hbar ω(a^{\dagger}a +1/2) .$$
with $\hbar$ the planck constant and $a^{\dagger}$ the creation operator.
Why can we neglect the term $\hbar ω/2$ in many cases, e.g. when we want to describe the Rabi Hamiltonian, where we just take $H= \hbar ωa^{\dagger}a$ .
| In general the term cannot be neglected as it causes the Casimir-Polder effect. In most cases it is just a constant and then it can be ignored, except for the fact that its present theoretical value is something like $10^{123}$. This value is ludicrously large. ZPE is an open problem of theoretical physics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/455835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Apparent frequency as function of distance So the Doppler effect says that the frequency of sound changes due to relative motion of source and observer. My question is if there any expression that tells how the apparent frequency changes in terms of the distance between the observer and source.
I know we have
$$f'=\frac{(v\pm v_O)}{(v\pm v_s)},$$
but there is no explanation for the rate of change in frequency.
| To a pretty good approximation, the frequency of a sound wave does not change as it travels. In a plane wave solution, all points along the wave are oscillating up and down with the same period; and so the number of cycles per second is the same at the source as at the observer, no matter where the observer is located.
There are some minor effects (due to non-linearities in the medium) which can cause frequencies to change. However, these are usually negligible, and are usually glossed over in introductory physics classes.
EDIT: You commented that there should be a relation between distance and frequency because the pitch changes when it comes closer. But that's not quite true; the pitch changes while it is moving, which is not quite the same thing.
Suppose you have a speaker that is 10 m away and at rest, and it is emitting a 440 Hz tone. While the speaker is at rest, you hear a 440 Hz tone. If the speaker then moves towards you, you will hear a higher frequency while the speaker is moving. But if it stops 5 m away from you, the frequency you hear will return to 440 Hz.
In principle, if you know what the "shifted frequency" was, you could figure out how fast the speaker was moving; and if you measured the amount of time that you heard the shifted frequency, you could multiply the velocity by the amount of time to find the distance the speaker travelled. But that only tells you the displacement of the speaker during its motion, not its initial or final distance from you.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/456140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does the heat capacity of an object come into play in thermal radiation? So say there's a cube in space acting as a blackbody.
Each side is 2 metres. Initial cube temperature is 400 Kelvin. Mass is 15 kg.
Say the heat capacity is 500 J/kgK. How would that affect thermal radiation? Or is it not a factor? How does it affect the rate of cooling? And how would you calculate this?
| We can answer your question by considering the first law of thermodynamics on a body with uniform temperature $T$, which is only losing thermal energy via radiation:
$$ \frac{dU}{dt}=\dot{Q} - \dot{W}$$
where
$$ U = \rho c_p T = \frac{m}{V}c_pT$$
$$ \dot {Q} = -\epsilon \sigma T ^4$$
$$ \dot{W} =0 $$
where $ \dot {Q} = -\epsilon \sigma T ^4$ is the emitted radiation which depends on the emissivity $\epsilon$.
Substituting into the first law,
$$ \frac{m}{V}c_p \frac{dT}{dt} = -\epsilon \sigma T ^4$$
Integrate from $t=0$ to $t$, noting that $T(0) = T_0$, and solve for $t$:
$$ t = \frac{3 m c_p}{V \epsilon \sigma}\left(T^{-1/3} - T_0^{-1/3}\right)$$
You can see that heat capacity definitely affects how fast the body cools. Specifically, larger heat capacity means longer cooling times since more energy is stored in the body at a given temperature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/456738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Young tableaus for $SO(n)$ I know how to use young tableaus to find irreducible representations and their dimensions of $SU(n)$. Are there similar rules for $SO(n)$?
| This is not a complete answer to this question
and is adapted from my answer to Irrep decompositions for $SO(N)$ tensors for $N>3$ as it also seems to apply here. Indeed you can do decompositions and find the dimension of irreps using young tableau also for SO(n). The young tableaux describe permutation of indices and thus are relevant for all lie algebra's coming from $GL(N)$. The application to $SO(N)$ is described well in Group theory: Birdtracks, Lie's, and exceptional groups (paywalled).
The Mathematica application lieART can also do the decompositions for and find the dimensions for you for all classical and exceptional Lie algebras including $SO(N)$. The documentation in https://arxiv.org/pdf/1206.6379.pdf also explains the algorithm used to do the decomposition. The only downside is that it does not work with Young Tableaus for SO(N). This means that you will have to convert your irrep in terms of young tableau (or number of boxes per row, i.e. {4} for four index symmetric, or {1,1} for the anti-symmetric) into their Dynkin label description of the same irrep (I did not find this easy. If you know an easy way please tell).
Alternatively you can work with the description in terms of the dimensionality of the irrep but this has the downside that it is confusing when there are degeneracies in this dimension. I still found the latter easier so I converted all the young tableau to the dimension of the irrep in order to be able to feed it as input to lieART.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/456932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How does the negative energy solution to the Dirac equation predict the antielectron? Please, can someone explain how the negative energy solution can be used to predict the existence of the antielectron?
| If you have a complex field, you can write it as a sum of two real fields. Now if you construct the observables for both fields, you'll find that are all identical except the charge operator, that gives a negative charge to one field and positive for the other. Then you have identical particles except for the charge. You can check Quantum Field Theory 2010 from Franz Mandl chapter 2.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/457064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Derivation of the work-energy theorem We state the following version of work-energy theorem :
$$ K_2-K_1=Fd=W $$
Where acceleration is assumed to be constant, so is the force $F$.
Then the physicists proceed by writing
$K_2-K_1=F[x(t_2)-x(t_1)]$
$$=F(x_2-x_1)$$
Notice, $x(t)$ was a polynomial of time $t$ with highest degree of $2$. Now $x_2-x_1$ is just a quantity.
Then they write $K_2-K_1=F \left. x \right|_{x_1}^{x_2}=
F \displaystyle \int_{x_1}^{x_2} \, dx $
$$K_2-K_1=\displaystyle \int_{x_1}^{x_2} F \, dx $$
If $G(x)$ is a function such that $\dfrac{\mathrm dG(x)}{\mathrm dx}=F$
then one finds, $$K_2-K_1=G_2-G_1$$ Substituting $G(x)=-U(x)$ yields, $K_2-K_1=-U_2+U_1$
Or this $$E \equiv K_1+U_1=K_2+U_2$$
My question is, isn’t $F$ constant here too? If it is constant then how they apply these equations to non-constant forces like Spring force $F(x)=-kx$?
| The force is constant in the result because you start out by assuming it constant from the very beginning. The expression $$Fd=W \qquad \text{(or }\; F(x_2-x_1)=W\text{)}$$ only holds true for constant force. The general expression is $$\int_{x_1}^{x_2} F \;\mathrm dx=W$$ If you start out with that and redo the whole derivation, you'll end up with the same result (and a few of your steps will be redundant) but with no assumption of constant force from the beginning.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/457227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Rotational work done So the formula for rotation work is
$$\mathrm dW=T\cdot\mathrm d\theta$$
where T is torque.
While solving a question like in case of a body rolling down an incline (pure rolling), we usully equate the change in kinetic energy equal to work done by gravity (just the translational work). Why do we not count the rotational work done when gravity does provide a torque?
In almost all question of mixed motions (translational and rotational) I have come across, none of them have the application of rotational work done, so I would also like to know when do we apply this concept.
| First, for something like a ball rolling down an incline, gravity has no torque about the center of the ball. The force that causes the ball to start rolling is friction. On a frictionless incline the ball would just slide down the incline without rolling.
Now, with that out of the way, it turns out that we do take into account "rotational work" due to friction. Let's assume a constant friction force $f$ and say that the ball with radius $r$ is released from rest on the incline. We know that the net torque about the center of the ball is given by
$$\tau = fr$$
using Newton's second law we also have
$$\tau=I\alpha$$
where $I$ is the moment of inertia and $\alpha$ is the angular acceleration.
Now, since our torque is constant (since $f$ and $r$ are both constant) we know two other things. First, the work done by friction is given by
$$W=\tau\Delta\theta=I\alpha\Delta\theta$$
and second, we can apply our constant acceleration equations. This means that
$$\omega^2=2\alpha\Delta\theta$$
where $\Delta\theta$ is the angle the ball rolls through some time after release, and $\omega$ is the angular velocity at that same point in time.
Combining everything we end up with
$$W=\frac12I\omega^2$$
and this result might look familiar. It is what we usually associate "rotational kinetic energy" with. So we do take into account the "rotational work", we just do it implicitly with $\frac12I\omega^2$ rather than explicitly (this is similar to how we use potential energy to implicitly take into account the work done by conservative forces rather than explicitly calculating the work done by said forces).
For example, if the ball starts at a height $h$ above the ground on the incline, using energy conservation at the bottom of the incline we have
$$mgh=\frac12mv^2+\frac12I\omega^2$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/457330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why does sound behave differently in water than in air? I noticed in some experiments at home that sound does not behave the same in water than in air. Is there a good scientific explanation to this?
I noticed that the sound sounded distorted in water but not in air.
I also used a software that I could use to hear the sound as if I had ears that are meant for underwater. I do not have the files because they are self wiped after I am done
| When any wave enters a different medium, the wavelength and direction can change (from the refractive index of the medium). For example, there is a 33% change in refraction between water and air (for light).
Sound waves are less affected than waves at light speeds, but there is still an effect.
http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/refrac.html
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/457644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Torque due to friction on a series of disks Recently I was looking at two situations involving friction and torque.
The first situation seemed pretty straightforward at first. A disk of mass $m$ and radius $r$, with a coefficient of static friction $\mu _s$ with the ground, is given a force $F$ originating at its center. I have drawn a free body diagram below.
$\hskip2in$
For this situation, the applied force is exactly equal to the frictional force, that is to the coefficient of static friction is high enough, and that the applied force is low enough, such that $\mu_sN = F$, we can pretty trivially show that $\Sigma_x F=0$ and $\Sigma_y F=0$. However , when solving for the net torque, we arrive at $\Sigma \tau = \mu _srF$, which means a non-zero net torque is applied to the disk.
This paradoxically, at least to me, means that the ball is spinning in place without actually moving. How could this be?
The second situation is very similar to the first, except that a second disk, with the same mass and radius, has been placed directly next to the first disk. The coefficient of static friction between the two disks is $\mu_{sb}$. I have drawn another free body diagram below.
$\hskip2in$
The situation is very similar to the first, however I hypothesize that there is a torque ($f$) due to friction between the first and second disks. I think that the direction of the force points downward, as when the the first disk tries to rotate due to the friction with the table, the second disk resists this change, thus causing a force opposite the motion of the spin.
However, I am at a complete loss as to how to calculate this force new force.
Any help at all would be appreciated, thank you!
|
Lets write the equations for a free body diagram and see if we can determined the force $F_r$
Left disk
$$-m\,a_1+F-F_c-\mu\,N_1=0\tag 1$$
$$-m\,g-F_r-N_1=0\tag 2$$
$$-m\,r^2\,\alpha_1+\mu\,N_1\,r+F_r\,r=0\tag3$$
right disk
$$-m\,a_2+F_c-\mu\,N_2=0\tag 4$$
$$m\,g+F_r-N_2=0\tag 5$$
$$-m\,r^2\,\alpha_2+\mu\,N_2\,r-F_r\,r=0\tag 6$$
Constraint Equation
$x1-x2=0\quad \Rightarrow\quad $
$$a_1=a_2\tag 7$$
and
$F_r-\mu_r\,Fc=0\tag 8$
Where:
$F_c\quad N_1\quad N_2$ are constrain forces
$\mu$ and $\mu_r$ are the friction coefficients
$a_1$ and $a_2$ are the translation accelerations
$\alpha_1$ and $\alpha_2$ are the rotational accelerations
We have 8 equations for the 8 unknowns:
$a_1,\alpha_1,a_2,\alpha_2\,N_1,N_2,F_c,F_r$
Result
$$\boxed{F_r=\frac{1}{2}\,\frac{\mu_r\,F}{1-\mu_r\,\mu}}$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/457716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Charge $Q$ in cavity inside conductor A point charge Q is inside a cavity in an uncharged conductor. Is the force on Q necessarily zero?
The explanation in the answer goes as
No. For example, if it is very close to the wall, it will induce a charge of the opposite sign on the wall, and it will be attracted.
This sounds weird to me because:-
*
*First of all, it says if Q is very close to the wall, it will induce a charge of the opposite sign on the wall but opposite sign shall develop every time on the inner surface of cavity irrespective of the particle's position inside the cavity.
*Isn't it that getting attracted would signify experiencing a force because of charge's own field since the field inside the cavity is just because of the charge placed [BY GAUSS LAW].
|
First of all, it says if Q is very close to the wall, it will induce a charge of the opposite sign on the wall but opposite sign shall develop every time on the inner surface of cavity irrespective of the particle's position inside the cavity.
Yes, the presence of the charge Q will always induce charges of the opposite sign on the inner wall (field inside the conductor should be zero and thus, Gauss law tells us that the inner surface must have a total of -Q charge). However, if the charge is placed exactly at the centre, the charge distribution over the inner surface is symmetric, so there is no force on the charge. But in all other cases, the charge will experience a force due to the induced charges on the inner surface.
Isn't it that getting attracted would signify experiencing a force because of charge's own field since the field inside the cavity is just because of the charge placed [BY GAUSS LAW].
*
*Firstly, field as calculated by Gauss law is due to all the charges, not just the ones enclosed by the closed surface over which we apply Gauss law.
*The induced charges are exerting a force on Q. As far as i know, a charge does not exert a force on itself.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/457798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Conceptual understanding of operators in QM Do operators in QM represent in some fashion the action of the measurement apparatus on a state being measured? Usually operators in QM are introduced as abstract transformations whose eigenvectors/eigenvalues are axiomatically the possible results of measurement, with an explanation along the lines of "because it works". However it seems like a coincidence that the operators that determine possible measurement results are, well, operators that transform states on which they act, as though the dynamical act of measurement itself were being modeled by a coarsely-grained apparatus-state interaction during the process of measurement, and the possible results of measurement are those fixed-point states for which the operator isn't "scrambling things up" during measurement (i.e the eigenfunctions). For example the momentum operator is associated with infinitesimal spatial translations, which makes sense because an apparatus that measures momentum has to in some fashion probe how a state translates in space without changing it. Has a view like this been fleshed out? It seems like it could shed some light on the measurement problem; it would make sense for the dynamical evolution of states being acted on by operators to eventually settle down (collapse) to the fixed points of the operator.
| This is the essence of the decoherence-based understanding of measurement. The idea is that the operators in the Hamiltonian act upon the system's state, causing it to decohere into a probability distribution at a certain basis, called a "pointer basis". For example, because physical interactions are generally based on the position operator, they often cause (approximate) decoherence in the position eigenbasis, which is why things appear to be localized - particles have trajectories and positions.
The operators that represent measurement in QM can hence be considered to "work" because they are diagonal in the same basis as the pointer-basis that the measurement-process induces decoherence in.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/457908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
Dot product in E&M
I'm learning graduate level E&M. Textbook is a famous Jackson book. What I would talk now is about pp.295-298 in 3rd ed. I attached the photo of p.298.
It says (paragraph above eq.(7.15) and footnote in the photo) that $\vec{n}\cdot \vec{n}=1$ doesn't mean n is unit vector if n is complex vector. And it discusses about the form of n satisfying above relation.
But it looks weird to me.
When I learned linear algebra/mathematical physics, I learned that in complex domain it is more natural to define inner product as
$\vec{a}\cdot\vec{b}=\Sigma a_i^\ast b_i$. If we use this definition there would be no problem of being not unit vector. Why did Jackson stick to definition of dot product in real domain?
| The vector $\hat{n}$ is meant to have real components, so the definitions are equivalent.
The conjugation is applied to calculate the Poynting's vector only because we like to work with complex exponentials, but you should only care about the real part. Consequently, you can either use $S\propto\Re e(\vec{E})\times\Re e(\vec{H})$ or use the cross product normally but conjugating the second one.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/458121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Friction acting as an internal force I was solving this problem in my assignment:
Assuming a frictional force F acts on the block of mass m, a force -F will act on plank of mass M. Hence, the net work done by frictional force should be zero, as friction is an internal force , but option D is given incorrect. What's the error in my reasoning?
Thanks in advance.
| Assume the unit vector $\hat i$ points to the right in your diagram, the bottom block is on a frictionless surface, block $M$ starts from rest and block $m$ has an initial velocity $v_{\rm i} \,\hat i$.
The magnitude of the frictional force on both blocks is $F$ and their directions are opposite (Newton's third law).
The acceleration of block $m$ is $-\frac Fm \,\hat i$ and that of block $M$ is $+\frac FM \,\hat i$.
As there are no horizontal external forces on the two block system then the momentum of the two block system is conserved.
The final velocity of the two blocks is $\frac{m}{M+m}v_{\rm i}\, \hat i$.
The velocity-time graphs for the two blocks are shown below.
The area under a velocity-time graph is the displacement and from the graph it is clear that the displacement of mass $m$ is greater than that of mass $M$.
So although both frictional forces have the same magnitude the frictional force acting on mass $m$ does more work, $\int \vec F \cdot d\vec x$, than the frictional force acting on mass $M$.
The work done on mass $m$ is negative because the frictional force and the displacement are in opposite directions and the work done on mass $M$ is positive because the frictional force and the displacement are in the same direction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/458433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Different expressions for distance & displacement : $\int$$d$$|\vec r|$, $\int$$|$$d$$\vec r$|, and $|$$\int$$d$$\vec r|$ I came across these expressions in my book. And the book says that all these are different from each other.
The expressions are : $\int$$d$$|\vec r|$, $\int$$|$$d$$\vec r$|, and $|$$\int$$d$$\vec r|$
Are they different from each other?
I know that
*
*$|$$\int$$d$$\vec r|$ means magnitude of displacement,
*$\int$$|$$d$$\vec r$| means total distance,
*But what about $\int$$d$$|\vec r|$?
I think it should mean total distance too, but I’m not sure if $\int$$d$$|\vec r|$ and $\int$$|$$d$$\vec r$| have the same meaning, do they?
Edit : Some of the answers say that the question is not very clear, and that a little more explanation would help. I’m not sure what else to add, so I’m attaching a picture of that page
| It is the radial part of the total distance along some path. For example, for a circle there is an angular part of the total distance ($2\pi r$), but there is no any radial part contribution: $dr=0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/458586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Why is the far point of human eye infinite? In my exams, the presence of this question, which unfortunately I couldn't answer, made me wonder why is the far point of an eye infinite?
First thing that came into my mind was that how come we can see till infinity?
Far point of eye is sometimes described as the farthest point from the eye at which images are clear. As stated here
There's obviously a limit to a distance where the eye can see. If there is, then why isn't that taken in consideration for accurate measurements?
| Seeing an object is equivalent to seeing the light reflecting off this object. Without outside interference, light traveling from a far away object is identical to light traveling from a close object.
Therefore the far point of the human eye isn't limited by distance, but by circumstances obscuring light beams from reaching us.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/458700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How can I recreate the greenhouse effect in my car? I am trying to preheat my car during winter mornings using the greenhouse effect. I understand the greenhouse effect in cars works by visible light passing through the glass, with most UV and infrared being blocked by the glass. The visible light is either absorbed by the interior and turned into heat or reflected largely as infrared and later turned into heat. Darker colors absorb more light.
I set a 500W halogen spotlight inches from my car windshield and turned it on. Unfortunately, after approximately 1 hour, I do not see any increase in interior temperature. What might I be doing wrong?
| Greenhouses are viable because the Sun's irradiation is (1) free and (2) strong (maximum of about 1 kW/m²). These advantages aren't available with your current configuration.
To heat your car, you'd want a mechanism that offers at least 100% efficiency (e.g., a resistance heater or heat pump) that transfers energy directly to the interior by dissipation (e.g., an electric heater) or convection (e.g., by convected heated air, sometimes used at truck stops, as shown below, although the efficiency here would be degraded from heat loss through the duct insulation).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/459062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Does the resistance of the voltmeter affect the behavior of this circuit? I have this setup.
It consists of a battery of no internal resistance with voltage $V$ and a resistor with resistance $R$. It also consists of a voltmeter of some (not so large) resistance as good ones should have.
Now my question is, will the resistance of the voltmeter matter for this particular circuit?
I think that the voltmeters need to be high resistance only when there are multiple resistances in the circuit... and as the internal resistance of the battery always contributes to the total resistance in real cases, a voltmeter is always expected to be very high resistance.
(Please correct me if I'm wrong)
I have seen other questions related to voltmeters on this site and googled up but couldn't find an answer.
| You have it backwards. The resistance of virtually every digital voltmeter is standardized at 10 megohms. The idea is the circuit should not "see" the meter. Almost no current flows through the meter. Conversely anmeters should have the lowest resistance possible so there is almost no voltage drop across them. The reason DMMs make lousey battery testers is because they do not load the battery. The resistor in your drawing would provide a load for a test. The problem with failing batteries is their internal resistance keeps going up as much as voltage going down..When the battery is fresh the internal resistance is very low compared to the load resistor. As it goes up, more of the battery's voltage drops across it's internal resistance and less apoears across the load resistor. A basic ohms law voltage divider.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/459142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Are the net microstates of the universe increasing? In physics and chemistry we learn that entropy is given by
$$S=k\ln\Omega$$
where $S$ is entropy, $k$ is Boltzmann's constant, and $\Omega$ is the number of microstates in the system. We also learn by the 2nd Law of Thermodynamics, that the entropy of the universe always increases. So if we apply both ideas, can we say that the number of microstates in the universe is always increasing? Or is this a naïve approach to the subject?
| In statistical mechanics, the word entropy is used for $k\ln\Omega$ where $\Omega$ is the number of microstates that are compatible with specified conditions, such as a given total energy and/or a given total volume. If no conditions are specified, then it's the total number of possible microstates. Therefore, whenever we talk about (or read about) entropy in statistical mechanics, it is important to specify what, if any, conditions are being imposed.
Here's an example. Suppose we start with an airtight box full of gas sitting inside a bigger airtight box with no gas. If the smaller box breaks open, so that the gas molecules are free to move into the larger space, then the number of possible microstates of the gas has suddenly increased. In other words, if we define $S$ in terms of the number of microstates compatible with the given constraints, then $S$ has suddenly increased, at least if we pretend that the change in the available volume is the only information we have.
(By the way, in quantum theory, we count mutually orthogonal microstates.)
Now: is the entropy of the universe always increasing?
Although there are many possible microstates, the universe is in only one of them. We have only partial information about which one we're in, so many (maybe infinitely many) microstates are consistent with the partial information we have. We can define $S=k\ln\Omega$ where $\Omega$ is the number of microstates that are consistent with the information we have, and this $S$ may change with time, if it is finite at all. I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post.
Some related thoughts can be found here:
Explain the second principle of thermodynamics without the notion of entropy
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/459496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can we know whether it’s a $1D$ or a $2D$ motion just by looking at the position-time relation? How do I know whether it is a $2D$ or a $1D$ motion, just by looking at position-time, or velocity-time, or acceleration-time equations?
Maybe the question is not very clear, I’m not sure I’m getting it across properly, so I’ll try to use some examples to make the question clearer.
We have a position time equation : $\vec r$ = $6t^2$$\hat i$ + $3t^2$$\hat j$
It’s easy to see that it is a $1D$ motion, because its locus equation is a straight line.
Likewise, $\vec r$ = $5t$$\hat i$ + $2t^2$$\hat j$ is a $2D$ motion, because its trajectory equation is a parabola.
Other examples of two-dimensional motions are :
$\vec r$ = $30t$$\hat i$ + ($20$ - $10t^2$)$\hat j$ (projectile motion)
$\vec r$ = $sin2t$$\hat i$ + $cos2t$$\hat j$ (Circular motion)
How did I know that these were $2$-Dimensional motions? I checked their trajectory equations.
My question is, is it possible to know just by looking at position-time equations, whether the body is moving in a straight line or changing its direction (i.e $2D$ motion), without checking its equation of trajectory?
| My solution:
2D case
given the position vector $\vec {R}$ with the parameter $t$
$$\vec{R}=\left[ \begin {array}{c} x\\y\end {array} \right]=\left[ \begin {array}{c} f \left( t \right) \\ g
\left( t \right) \end {array} \right]
$$
Ansatz $y(x)=a\,x+b$
,the slope $a$ must be const.!, $\quad$ with $a$
$$a=\frac{\partial y}{\partial x}=\frac{\frac{d g}{dt}}{\frac{df}{dt}}=\text{const}\tag 1\quad \frac{df}{dt}\ne 0 $$
3D case
$$\vec{R}= \left[ \begin {array}{c} x\\ y\\
z\end {array} \right]
=\left[ \begin {array}{c} f \left( t \right) \\ g
\left( t \right) \\ u \left( t \right) \end {array}
\right]
$$
Ansatz $z(x,y)=a\,x+b\,y+c$
,the slope $a$ and $b$ must be const.!,$\quad$ with $a$
$$a=\frac{\partial z}{\partial x}=\frac{\frac{d u}{dt}}{\frac{df}{dt}}=\text{const}\tag 2,\quad \frac{df}{dt}\ne 0$$
and
$$b=\frac{\partial z}{\partial y}=\frac{\frac{d u}{dt}}{\frac{d g}{dt}}=\text{const}\tag 3,\quad \frac{dg}{dt}\ne 0$$
Example
$$\vec{R}=\left[ \begin {array}{c} 6\,{t}^{2}\\ 3\,{t}^{2}
\end {array} \right]
$$
$a=\frac{1}{2}=\text{const}$ ,linear function
$$\vec{R}= \left[ \begin {array}{c} \sin \left( 2\,t \right)
\\ \cos \left( 2\,t \right) \end {array} \right]
$$
$a=\tan(2\,t)$ ,nonlinear function
$$\vec{R}= \left[ \begin {array}{c} {t}^{2}-10\\{t}^{2}+5
\\{t}^{2}\end {array} \right]
$$
$a=1\quad,b=1$ ,linear function
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/459702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why magnetic field starts from north pole and ends in south pole? In magnets such as bar magnets the magnetic field lines are starting from North pole and ends in south pole..but I don't know what is the reason for it..why this happens.
| Because it is a fundamental property of nature. Magnetic field lines forms closed loops. Two bar magnets held over unlike poles (a north end and a south end) would attract. You can imagine a field line connecting the two ends. That is a loop by itself.
In fact, all about magnetic field lines that physicists know is about as much as you know yourself, that the field lines trace out closed loops between ends of the bar magnet.
(Field lines of a single bar magnet)
If you spread out iron fillings, they arrange themselves along this pattern of field lines. Magnetic fields are not really predictions of anything else. They were discovered in this fashion. That they always seem to form loops.
I am aware that there are theories on the existence of a single pole existing all by itself (monopole). It still waits to be confirmed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/459778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Why my cloud chamber won't work? let me briefly introduce myself, I'm a highschool student from Indonesia, I'm currently making a cloud chamber for my science fair project, I did what the videos on youtube told me, but for some reason my cloud chamber won't work. Here are some pictures of my setup
So, I glued felt on top of my aquarium and soaked it with alcohol (95%), put about 4 kg of dry ice under my metal tray (baking tray) and I let the alcohol and dry ice do their job for about 15 minutes, it only showed some particles raining but I didn't see any tracks of muons. I used one bright flashlight ( about 8000 Lumens ) and the room was dark enough that I couldn't see my hand. I also covered the aquarium glass with black cloth to maximize the darkness. Can anyone help check what did i do wrong? ( and what kind of camera setup should I use to record the muons tracks? )
| The type of light one uses seems to make a difference.
When using a linear LED source, I saw nothing (not even the "snow"). Then I switched to an old-style (filament bulb) flashlight, I saw the "snow" AND a few tracks. Using a point-source LED (3rd attempt) saw nothing (again). All other parameters (chamber, temperature, source) were the same. When I changed back to the filament bulb, I saw the "snow" and a nice long track.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/459937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Length Contraction Scenario Suppose a space ship is traveling from star A to star B at some significant fraction of the speed of light. In the frame of the ship, the distance A to B is less than the distance in A and B's rest frame. Is it possible for the ship to quickly increase its speed so that in its frame the ship is then closer to A compared to when it was going slower? That is, in the time interval that it is accelerating it moves further away from A due to its speed but the effect of increased length contraction has it end up closer? That sounds like an odd situation to be in.
| Results are somewhat different.
Suppose the observer aligns his x-axis passing through stars and Star A has some negative x-coordinate (-x1) and B has some positive (x2). Clearly observer stands at origin (0). We clearly quantify our definitions:
*
*Distance between A and B: x1+x2
*Distance between observer and A: x1
Now we can apply length contraction concept to 1. and not to 2..
Thus the observer in spaceship will see star A and B come closer due to increased length contraction. But A will nonetheless move away from observer in his frame. then in his frame velocity of A was always in negative x-axis before acceleration and acceleration tends to increase its magnitude so distance of A from his origin will increase.
PS: If we where to apply Newtons laws to A and B in observer's frame , both would seem to accelerate back at sane time and undergo acceleration for sane duration. However, according to relativity of simultaniety, B would start to accelerate first than A and hence travel some distance towards A.
A similar result is Bell's spaceship paradox
https://en.m.wikipedia.org/wiki/Bell%27s_spaceship_paradox
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/460049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Hamiltonian of a quantum heat bath I have seen the Hamiltonian for a heat bath written as:
$$ H_B = \hbar \int_0^\infty \omega b(\omega)^\dagger b(\omega) d\omega $$
I was hoping to understand this equation better. This suggests that the heat bath is written as a sum of harmonic oscillators with raising/lowering operators $b^\dagger(\omega)$ and $b(\omega)$. I understand that is that this is just the continuous limit of the quantised electromagnetic field.
Please could someone explain the following:
1) Does the raising operator $b^\dagger(\omega)$ correspond directly to a mode in the heat bath with frequency $\omega$? ie does $b^\dagger(\omega)b(\omega)$ give us the number operator for the number of excitations of frequency $\omega$? or does the argument $\omega$ mean something else?
2) How does one alter this to change the temperature of the heat bath? My current interpretation is that this equation is for the vacuum/zero temperature case, so do we just add a Boltzmann factor inside the integral, to change the temperature or is there more subtlety required?
3) If you have any other intuitions/tips/tricks for thinking about this equation, I would appreciate them!
Many thanks!
| *
*Yes the raising operator $b(\omega)^\dagger$ corresponds to a mode of frequency omega and $n(\omega) = b(\omega)^\dagger b(\omega)$ count the number of excitations. You can see that the mode has a frequency of $\omega$ because that is the prefactor of the mode in the Hamiltonian, so one excitation of that mode increases the energy by $\omega$.
*The Hamiltonian, quite generally, does not depend on temperature. The Hamiltonian tells you the possible energy levels of the system (and their degeneracy). Temperature (through the Boltzmann distribution) is related to how likely you are to find the system in a given energy level. You need to have the Hamiltonian first before you start talking about distributions or temperatures. Consequently the Hamiltonian should always be deperature independant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/460317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is compressive strain equal to streching strain? The stress induced on a piece of material (like a rectangular beam) is defined as:
$$\sigma = E\epsilon$$
where $E$ is the elastic modulus and $\epsilon$ is the strain.
The strain on the other hand is defined as the ratio of the new length of the beam to the original length of the beam:
$$\epsilon = \frac{L}{L_{0}}$$
So, for example, if we were to stretch a beam to $1.5$ its original length, the strain would be $\epsilon = \frac{1.5L_{0}}{L_{0}}=1.5$ (let's imagine this would be a realistic situation).
But if we were to compress the beam to half its original length, the strain would now be $\epsilon = \frac{0.5L_{0}}{L_{0}}=0.5$.
I find this confusing: In both cases, the deformation of the beam is the same, but in the stretching case the stress would be much higher! It doesn't make sense to me. Why do we induce a much higher stress to the beam by stretching (rather than compressing it) even though the deformation of the beam is the same?
| This is not correct. Stretch $\lambda$ is the ratio between length increments so for an undeformed part it would be close to 1, while the strain for an undeformed part would be close to zero. The stretch in both cases is:
$$\lambda_1 = \frac{\text{d}L_f}{\text{d}L_0} = 1.50, \qquad \lambda_2 = \frac{\text{d}L_f}{\text{d}L_0} = 0.50$$
The true strain is given by the finite strain theory and is equal to $\varepsilon = (\lambda^2-1)/2$:
$$\varepsilon_1 = 0.625, \qquad \varepsilon_2 = -0.375$$
For small strains (stretch close to one), the approximation $\lambda \approx 1+\varepsilon$ can be used, although in this case we are not in small strains and this approach is bad:
$$\varepsilon_1 \approx 0.500, \qquad \varepsilon_2 \approx -0.500$$
In the first case the strain is underestimated by 25% and in the second case it is overestimated by the same amount.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/460587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Are all solutions of Maxwell's equation related by a gauge transformation? Consider Maxwell's equation (without source):
$$ \partial_\mu F^{\mu \nu} = 0 \implies \partial_\mu \partial^\mu A^\nu = \partial_\mu \partial^\nu A^\mu.$$
Can we find a pair of classical field configurations $A^\mu(x),A'^\mu(x)$ such that they both satisfy the equation above (assuming similar boundary conditions) but are not related to each other by a gauge transformation of the type:
$$A'^\mu(x) = A^\mu(x)+\partial^\mu \varphi(x) \quad ?$$
If it's impossible, how could we argue/show this?
Answer: Thanks to my2cts' answer, I've found 2 solutions $A$ and $A'$ not related by a gauge transformation : $A^\mu=(0,e^{-i(t-y)},0,0)$ and $A'^\mu=(0,0,e^{-i(t-x)},0)$. It makes sense since they both give rise to different EM fields, which are invariant under gauge transformations.
| Of course there are many solutions that are unrelated by a gauge transformation, namely plane waves of any frequency, propagation direction, polarisation and phase and superposition thereof.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/460728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why aren't particles constantly "measured" by the whole universe? Let's say we are doing the double slit experiment with electrons. We get an interference pattern, and if we put detectors at slits, then we get two piles pattern because we measure electrons' positions when going through slits. But an electron interacts with other particles in a lot of different ways, e.g. electric field, gravity. Seems like the whole universe is receiving information about the electron's position. Why is it not the case and the electron goes through slits "unmeasured"?
Bonus question: in real experiments do we face the problem of not "shielding" particles from "measurement" good enough and thus getting a mix of both patterns on the screen?
| As long as these interactions are weak and do not distinguish between the two slits, they can be disregarded.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/460855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "66",
"answer_count": 6,
"answer_id": 1
} |
Angular acceleration of a double compound pendulum How can I calculate the angular acceleration of a double compound pendulum? I'd like to know what the angular acceleration of each of the pendulum's center of mass will face at any point in time.
PS - Not a physics guy but love the subject would really like to learn. I'm programmer and I'm trying to simulate a double compound pendulum as an experiment.
Update - Got it. Found the equations here -> http://www.team.kdm.p.lodz.pl/master/Jankowski.pdf
| The deflection angles of both pendulums are the degrees of freedom of the system.
First pendulum will have $\ddot{\theta}_1$ and second one will have $\ddot{\theta}_1+\ddot{\theta}_2$.
This is very trivial, maybe you are trying to ask something else?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/460980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.