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Density of states of 3D harmonic oscillator Consider the following passage, via this image:
5.3.1 Density of states
Almost all of the spin-polarized fermionic atoms that have been cooled to ultralow temperatures have been trapped by magnetic fields or focused laser beams. The confining potentials are generally 3D harmonic traps. So let's consider this case in more detail. You might be interested to note that Fermi's original paper on fermionic particles considered this case, not the 3D box case above. As we saw previously, ignoring the zero-point energy in each dimension the eigenvalues (accessible energy states) are given by $\epsilon(n_x, n_y, n_z)=n_x\hbar\omega_x + n_y\hbar\omega_y + n_z\hbar\omega_z$. In order to evaluate the various integrals, we first need to obtain the density of states per unit energy. A rough way to do this is to simply set $k_i=n_i$, so that
$$\epsilon^2 = k_x^2(\hbar\omega_x)^2 + k_y^2(\hbar\omega_y)^2 + k_z(\hbar\omega_z)^2 \equiv k^2(\hbar \overline\omega)^2,$$
where $\overline \omega = (\omega_x\omega_y\omega_z)^{1/3}$ is the mean frequency, and $dk_i/\epsilon_i=1/\hbar\overline \omega$. Because $k_i=n_i$ now rather than $k_i=\pi n_i/L$, th 3D density of states is given by
$$g(\epsilon) = \frac{k^2}{2} \frac{dk}{d\epsilon} = \frac{\epsilon^2}{2(\hbar\overline\omega)^3}.$$
for the first displayed equation,
shouldn't be $\epsilon^2 =\epsilon_{n_x}^2 +\epsilon_{n_y}^2 + \epsilon_{n_z}^2 + 2\epsilon_{n_x}\epsilon_{n_y} + 2 \epsilon_{n_x}\epsilon_{n_z} + 2\epsilon_{n_y}\epsilon_{n_z}$..?
if I assume $\omega_i=\omega$ for $i=x,y,z$
by
$\epsilon_{n_x}=\hbar \omega n_x $
$\epsilon_{n_y}=\hbar \omega n_y $
$\epsilon_{n_z}=\hbar \omega n_z $
$\epsilon_{n_x,n_y,n_z}=\hbar \omega(n_x +n_y +n_z)$
let $\vec{k}=(k_x,k_y,k_z)$ where $k_i=n_i$
$$\epsilon_{n_x}^2 +\epsilon_{n_y}^2 + \epsilon_{n_z}^2 = \hbar^2 \omega^2 (k_x^2 + k_y^2 +k_z^2 ) = \hbar^2 \omega^2 k^2 \not=\epsilon^2~?$$
And for second displayed equation, why it's not $$\frac{\pi k^2}{2} = \frac{1}{8}4\pi k^2~?$$
| The author is working by analogy with the 3D box case. The 3D box worked out easily because $\epsilon\propto k^2$, so that the number of states up to each energy could use the volume of an ellipsoid, which is well known. He/she makes the volume of an ellipsoid work in a "rough way" for the harmonic oscillator. He/she puts a state where each 3D box state is (with $k_i=n_i$), and calls this $\epsilon^2$ to get the energies right along each axis.
Your suggestion of $4\pi k^2/8$ was correct for the 3D box, but here he/she is using the 3D box result and correcting using $k_i=\pi n_i/L \to k_i=n_i$.
One way of calculating the 3D density of states directly is to use integral 4.634 of Gradshteyn and Ryzhik (7th). I can supply more details if interested. Also see http://arxiv.org/abs/cond-mat/9608032.
| {
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Why does the water bottle not rotate when it is half full? Consider this water bottle:
When it is full and thrown up in the air, it rotates at a constant velocity.
When it is less than 1/8th full, the water bottle rotates even faster than when it was full.
When it is half full, however, the water bottle rotates for one half-spin, and then it stops rotating.
Why is this?
By the way, I tested it with a 16.9oz bottle, but the bottles are mathematically similar.
| I guess, The most important question is wheter you will accelerate the fluid when throwing the bottle. This and conservation of angular momentum will be much more important than dissipation of energy in the fluid.
If it is full, the water, due to the constraints on the system, will be accelerated with the bottle when throwing. So it will just continue to spin.
If it is half full, the fluid will, while you throw the bottle, just flow around not taking angular momentum, once you let go of the bottle the total angular moment will be conserved, but the fluid will be accelerated trying to match the movement of the bottle, as the fluid is much heavier than the bottle, it will seem as if the rotation was stopped.
To confirm this hypothesis, spin up the bottle properly before throwing it when it is half full, then it will continue to spin in the air.
So on a last note, why does it spin faster, when it is filled 1/8? Because then the water will basically just act like a weight at the bottom of the bottle (because at the bottom the centrifugal force will be strong enought to keep the water in place), not leaving the end of the bottle while it is accelerated, thus allowing you to more efficiently transfer angular momentum (compare throwing a straw and a straw with a weight on the end).
| {
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Soliton solution for a diffusive system With a simple model for bacterial diffusion, I get this partial derivative equation :
$$\frac{\partial n}{\partial t} = D\frac{\partial^2 n}{\partial x^2} + d_1 n -d_2 n^2$$
where $n(x,t)$ is the population of bacteria, $d_1$ the rate of proliferation and $d_2 n $ is the rate of death for a bacteria.
I know that there should exist some soliton solution going from the solution $n=0$ to the solution $n=\frac{d_1}{d_2}$ but I have no clue how to treat such a problem. What are some methods to get about those soliton solutions?
| At least for an equilibrium situation, where $\dfrac{\partial n}{\partial t}=0$ and $\dfrac{\partial n}{\partial x}=0$, you would easily see that
$$d_1 n - d_2 n^2=0,$$
with the two solutions you anticipated.
I am not sure what you mean with a soliton solution for an equation like these, as the basis of your equation is not a wave equation. Probably you mean solutions that also satisfy the equation:
$$\frac{\partial n}{\partial t} = D\frac{\partial^2 n}{\partial x^2}$$
But it would lead to the same result. I hope this is at least a partially satisfying answer.
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Determining the three Euler angles from the acceleration I want to know, given the measurement of an accelerometer at rest (so not really an acceleration but a force per unit of mass) the inclination of this accelerometer, along the three axis.
For this, I find this PDF document. However, there's something I can't definitely understand; page 10, to results are found: the first is
$$\phi_{xyz} = atan\left( \frac{a_{y}}{a_{z}} \right) $$$$ \theta_{xyz} = atan\left( \frac{-a_{x}}{\sqrt{a_{y}^{2} + a_{z}^{2}}}\right) $$
While the second is
$$\phi_{yxz} = atan\left(\frac{a_y}{\sqrt{a_x^2 + a_z^2}}\right)$$
$$\theta_{yxz} = atan\left(\frac{- a_x}{a_z}\right)$$
Where $\phi_{abc}$ (resp. $\theta_{abc}$) is the roll (resp. Pitch) obtained after rotating the gravity vector around axis a, then b and finally c, $a$ the measured acceleration. What I can't understand is why we find two different equation for same angle. I know, it comes from matrixes multiplication, but that's truly unintuitive. Moreover, if we take $a_x = a_y = a_z = \frac{\sqrt{3}}{3}$, so $\theta = \phi = \frac{\pi}{4}$, the first equations give us
$$ \phi = \frac{\pi}{4} $$ $$\theta \approx -0.62$$
While the second equations lead to
$$ \phi \approx 0.62 $$ $$ \theta = \frac{-\pi}{4}$$
So, why do we find to different angles and equations for the same angle ?
| The solutions these equations give for the angles are different because they are for different rotation sequences. The first set of equations has the subscript xyz and the second has the subscript yxz, so these angles depend on the rotation sequence.
| {
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Meniscus attached to an inclined plate To be more specific, suppose a hydrophilic infinite plate is stuck into a semi-infinite region of water, above the water is a semi-infinite region of air, when the plate is stuck into the water vertically, the contact angle is $\alpha$, as shown in the figure below:
Needless to say, the menisci on both sides are symmetric, but what will happen when the plate is inclined for an angle $\beta$? Will the contact angle $\alpha$ remain unchanged? The meniscus on which side will be higher?
| the angle will change.
and it will be smaller at the inclination beta.
and greater at the other side
| {
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Number of Nodes in energy eigenstates I have a question from the very basics of Quantum Mechanics. Given this theorem:
For the discrete bound-state spectrum of a one-dimensional potential let the allowed energies be $E_1<E_2< E_3< ...$ with $E_1$ the ground state energy. Let the associated energy eigenstates be $ψ_1,ψ_2,ψ_3,...$. The wavefunction $ψ_1$ has no nodes,$ψ_2$ has one node, and each consecutive wavefunction has one additional node. In conclusion $ψ_n$ has $n−1$ nodes.
What is the physical interpretation for the number of nodes in the concrete energy eigenstate? I understand that the probability of finding the particle in the node point is $0$ for the given energy. However, why does the ground state never have a node? or why does every higher energy level increments number of nodes precisely by 1?
| The physical interpretation behind the increase of energy with the number of nodes can be understood in a very crude manner as follows:
Nodes are points of zero probability densities. Since the wavefunction is continuous, the probability density is also a continuous function. So the regions in the neighbourhood of nodes will have small probability densities. Physically, this means that the particle has less space to move around. That is, the particle is more confined and uncertainty in position $(\Delta x)$ decreases. This increases $\Delta p$ (due to the uncertainty principle), causing an increase in energy. Hence, the energy increases as the number of nodes increase. So the ground state should not have any node.
| {
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Torque in a non-inertial frame How can we calculate torque in a non-inertial frame? Take for instance a bar in free fall with two masses, one on either end, $M_1$ and $M_2$. Taking the point of rotation to not be the center of mass, i.e. $M_1\neq M_2$ and take the point of rotation to be the center, what is the proper way of analyzing the situation to come to the conclusion that there is no rotation?
| Follow the rules of motion:
*
*Sum of forces equals mass times acceleration of the center of mass: $$ \sum_i \vec{F}_i = m \vec{a}_{cm} $$
*Sum of torques about the center of mass equals change in angular momentum: $$ \sum_i (M_i + \vec{r}_i \times \vec{F}_i) = I_{cm} \dot{\vec{\omega}} + \vec{\omega} \times I_{cm} \vec{\omega}$$ where $\vec{r}_i$ is the relative location of force $\vec{F}_i$ to the center of mass.
So for an accelerating rigid body that is not rotating $\dot{\vec{\omega}} = \vec{\omega} = 0$ the right hand side of the last equation must be zero.
See https://physics.stackexchange.com/a/80449/392 for a complete treatment of how you go from linear/angular momentum to the equations of motion.
Also see https://physics.stackexchange.com/a/82494/392 for a similar situation where a force is applied away from the center of mass.
The rule that comes out of the above equations of motion are:
*
*If the net torque about the center of mass is zero then the body will purely translate
*If the sum of the forces on a body are zero (but not the net torque) then the body will purely rotate about its center of mass.
| {
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Coulomb barrier in nuclear fusion Suppose there exists an alpha particle in the nucleus.
Within a radius of 2 femtometer, the dominating force is the nuclear force but beyond this radius, the Coulomb force becomes the effective dominating force due to it's longer range force. Outside this radius is a daughter nucleus.
So let's suppose the alpha particle moves beyond the effective range of the nuclear force,
the electric potential energy due to the force between the alpha particle and the daughter nucleus is given by as
$$U_{B}=\frac{1}{4\pi \varepsilon _{0}}(\frac{2(Z-2)e^{2}}{R}).$$
Why is there the term $$2(Z-2)~?$$
| The electric potential between two point charges depends on the product of the charge of each.
The charge in the nucleus is due to protons, and the count of protons is given as $Z$ for a total charge of $Ze$.
The alpha decay removes two protons (and two uncharged neutrons). The charge of the alpha particle is $2e$, leaving a charge of $Ze - 2e$ in the nucleus after decay. Multiplying the two and rearranging, that becomes $2(Z-2)e^2$.
| {
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True randomness? I am a physics high-school student so my knowledge is not very deep on the subject.
We started learnning about quantum mechanics and on some processes that my teacher described as random. I began to think on the concept of randomness and question it, thinking how can a process or an outcome be determined without any cause, how an outcome be determined at all in complete randomness?
I searched the internet and figured the scientific community does not agree with me. I'd really like to understand how can true randomness exist? Why the scientific community rejects the idea that ''random'' events may just have a cause we are not aware of?
What am I missing?
Thanks for answers and sorry for bad English.
| Quantum mechanics is a theory where randonmness is inherently present. According to it, it is possible to desing experiments where we can only predict the probabilities of the different outcomes.
It has been argued, in the beggining of the theory, that this must happen becasue QM is an incomplete theory of nature. So, there must be hidden variables than the theory ignores, and that it is in this ignorance that the randonmness originates. These critics where done in the 30's.
Years later, a physicist called Bell, discovered that any theory that would include these hidden variables will inevitably be incompatible with QM. Since then, every experiment done has been in agreement with QM, so today it is quite accepted that QM implies the presence of pure randomness.
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Elements of particle mass From what I can tell, it seems that particles have two kinds of mass, the mass inherent in a fundamental particle itself, or for composite particles, additional mass associated with the Higgs field. Is that distinction correct?
If so,is the mass of associated with with the Higgs field equal to the mass of the energy required to hold the particles together?
| The Higgs field is a fundamental field in the standard model of particles. The particles which acquire mass due to the Higgs field are shown in the table.
The Standard Model of elementary particles (more schematic depiction), with the three generations of matter, gauge bosons in the fourth column, and the Higgs boson in the fifth.
Each particle in the table is described by a special relativity four vector, whose "length" is the mass in the table. All other particles, protons ,neutrons, atoms, molecules are a hierarchical addition of four vectors which will have an invariant mass, according to the rules of special relativity.
The masses induced by the Higgs field are very small as seen in the table ( except for the Higgs Boson itself, the Z and the W). The proton and neutron acquire their much larger mass by the addition of the innumerable four vectors of the quarks, antiquarks and gluons that it contains. If one added just the mass of the constituents the mass is a small fraction of the measured nucleon mass.
Thus most of the mass we measure for the proton is not from the Higgs field, but from the special relativity dynamics.
| {
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What about a surface determines its color? Light falls on a surface. Some wavelengths get absorbed. The other are reflected. The reflected ones are the colors that we perceive to be of the surface.
What is the property that determines, what wavelengths are reflected and what are absorbed? Is it electronic configuration of the object on which the light falls?
If yes, then if we know the electronic configuration of a surface can we make a model, which will predict the color it will show?
| Great question -- it opens up a lot of physics. My favorite example is "why do some conducting metals look gold/copper/etc. rather than grey?" Turns out it's due to relativistic effects acting on the various electrons' orbitals.
Then there's the blue jay, whose blue color is entirely interferometric rather than absorptive/reflective.
and more :-)
Edited to provide info about quantum mechanisms for color in metals. Quoting from wikipedia pages,
The characteristic color of copper results from the electronic
transitions between the filled 3d and half-empty 4s atomic shells –
the energy difference between these shells is such that it corresponds
to orange light.
(references available at the "Copper" page)
Whereas most other pure metals are gray or silvery white, gold is
slightly reddish yellow. This color is determined by the density
of loosely bound (valence) electrons; those electrons oscillate as a
collective "plasma" medium described in terms of a quasiparticle
called a plasmon. The frequency of these oscillations lies in the
ultraviolet range for most metals, but it falls into the visible range
for gold due to subtle relativistic effects that affect the orbitals
around gold atoms. Similar effects impart a golden hue to
metallic caesium.
References include this page as well as various books.
| {
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Two people are holding either end of a couch, is one person exerting more force than the other? I was carrying a couch with my flatmate yesterday, and I started thinking about this.
Often when carrying objects like this, one person will be taller and has thus lifted his end of the couch higher than the other person's. Additionally, one can intentionally lift his end of the couch higher, for comfort.
My question is, in situation is one person carrying more weight than the other?
| Although Floris made a clear picture involving only vertical forces, this picture is mostly useful when carrying washing machines or large chairs, where the 'height' of the object is more pronounced.
However, you will find that, even when the object is mostly flat, the bottom person will carry most of the weight. The key here is the direction of the forces involved.
You will find that it's very difficult to lift a couch underneath you, 'tugging' it upwards. Instead, the top person will apply force in the direction that does not require any tugging (especially when you can't get a grip): perpendicular to the object. All you do, is offset the torque gravity incurs on the couch, where we take the bottom person as the pivot point. In the image above, the torque is provided by the component of gravity perpendicular to the couch (green arrow).
Since the top person is twice as far away from the pivot point as the center of gravity, he will only need to apply half of that component which was already less than the total gravity - i.e., much less than half of the total gravitational force. The top person has the easy job, although usually in a very awkward position.
The bottom person needs to account for the rest of the upward force required, along with a small horizontal component to make sure there is no horizontal acceleration (blue arrows) - i.e., a lot more, which is why being the bottom person is the heavier job.
| {
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Heating suprecritical water in a closed water tank How does the pressure of water in a closed tank evolve in the following setting:
- closed tank of 2 liters (filled up with water)
- water initially at 25°C and pressurized to 3 bars
The water is now heated up to 130°C, thus remaining a fluid (based on water property tables). The specific volume of water increases roughly by 6% due to the temperature increase. Since the water is in a closed tank, it cannot expand, thus the pressure needs to increase.
Which equation will allow me to calculate the resulting pressure in the water tank?
Thanks in advance for any hints and help
| For water, the equation of state is semi-empirical and so the choice depends on how accurate you want to be and the range of temperature/pressure/etc that you're interested in.
Equation 19 in Jeffery et al [1], where the parameters are given in equation 32, seems appropriate.
[1] Jeffery, C. A., and P. H. Austin. "A new analytic equation of state for liquid water." The Journal of chemical physics 110.1 (1999): 484-496. Direct PDF link here.
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Why do higher modes propagate more in the cladding of an optical fiber than lower modes? I am trying to understand the theory of inter-modal dispersion in optical fibers. It seems quite obvious that if higher modes have a greater angle of incidence in the fiber than lower modes, the path length of higher modes through the fiber is larger. This is because higher modes undergo more reflections, but they also have a greater part of the light wave traveling is the cladding. Here the speed of light is a little higher than in the core and therefore the higher modes are moving faster than lower modes.
The theory of a part of the light wave traveling in the cladding has to do with the evanescent field I think, but why do higher modes have a greater part of the wave traveling in the cladding in comparison with lower modes?
| At a smaller angle of incidence on the boundary (higher mode), the field of the evanescent wave penetrates more deeply into the optically rarer medium (cladding). See the derivation here, in which the characteristic depth of penetration of the evanescent electric field is given by
$$
\frac{c}{\omega}\left((n_1\sin(\theta_I))^2-n_{2}^{2}\right)^{-1/2}.
$$
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Units of vector differential operator del ($\nabla$) My book says that $\left[\nabla \cdot (\vec E \times \vec H)\right] = \mathrm{W/m^3}$. I see that $\vec E$ is in $\mathrm{V/m}$ and $\vec H$ is $\mathrm{A/m}$, so these multiplied is $\mathrm{W/m^2}$, but how does dotting with $\nabla$ give another $\mathrm{m^{-1}}$?
| Note on notation: I use $[\cdot]$ do denote the units of the quantity in brackets.
Derivatives always have units of $1/[\text{differentiation variable}]$. This can be clearly seen from the definition of the derivative in terms of difference qutionts:
$$ \partial_x f(x) := \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.$$
So if $x$ has some unit $[x]$ then $\partial_x f$ will have units $[f]/[x]$. (As the limit does obviously not change the units.)
As
$$\nabla = \begin{pmatrix} \partial_x \\ \partial_y \\ \partial_z \end{pmatrix}$$
and the coordinates in space carry the unit $\mathrm{m}$, you have that
$$[\nabla \cdot \vec V] = [\partial_x V_x + \partial_y V_y + \partial_z V_z] = [\partial_x V_x] = [V_x]/[x] = [V_x]/\mathrm{m}.$$
(Where I arbitrarily chose the $x$-component of the vector, as all components have the same units.)
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How to determine whether a nuclear transition would be electric octupole, or hexadecapole? The transition from one nuclear state to another is classified as quadrupole/octupole, etc, depending on the units on angular momentum transferred. But depending on the angular momentum of the two states involved, the net J can take different values. So what decides whether a nuclear transition would be electric octupole, or hexadecapole?
| A way to decide whether a gamma photon emition is electrical or magnetic is from the photon's parity.
Parity conservation requires: $π_i = π_f π_{ph} $ where the parity at the left side is the initial parity of the nuclei, and at the right the final parity of the nuclei and of the photon.
Thus, an electrical transitions happens when: $π_{ph}=(-1)^l $ and a magnetic when $π_{ph}=-(-1)^l $, where l is the the angular momentum of the photon.
From the conservation of momentum demands in general: $\bar j_i -\bar j_f = \bar j_{ph} $ where the subscripts $i$ and $f$ mean the initial and final state of the nuclei. Thus, it is $$\left|\bar j_i - \bar j_f\right| \leq j_{ph} \leq \bar j_i + \bar j_f $$.
Thus, to your question, by defining the order of the polypole by the angular momentum of the photon we have:
Radiation |Symbol| l photon momentum | photon parity |
*
*E dipole, E1, $\ell=1$, $\pi_\gamma=-1$
*B dipole, M1, $\ell=1$, $\pi_\gamma=+1$
*E quadrupole, E2, $\ell=2$, $\pi_\gamma=+1$
*B quadrupole, M2, $\ell =2$, $\pi_\gamma=-1$
*E octapole ,E3,$\ell=3$, $\pi_\gamma=-1$
*B octapole ,B3,$\ell=3$, $\pi_\gamma=+1$
*E hexadecapole, E4 $\ell=4$, $\pi_\gamma=+1$
*B hexadecapole, B4, $\ell=4$, $\pi_\gamma=-1$
and so on...
I hope this helps with your question.
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Why do our ears pop? Have you ever been on a train going through a tunnel or plane and your ears pop?I was wondering why this happens and I know it relates to pressure but don't know exactly the reason
| Measure first, think later. The best tunnel for experiencing ear discomfort around here is a train tunnel under a canal, the Drontermeertunnel. The graph shows the air pressure recordings I made in the train. A and B are the gates of the tunnel. Orange indicates when I experienced ear discomfort (although it wasn't easy to decide when discomfort started and stopped exactly). The graph shows that the ear discomfort corresponded to peaks of the oscillation.
The pressure increases before A because the train descends on a slope before the canal; it briefly increases when entering the gate; the average pressure decreases inside the tunnel due to the Venturi effect. The damped oscillation is the fundamental frequency of the tunnel. The tunnel is an open-end air column.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Derivation of an ordinary, Lagrangian/Hamiltonian and action formulation I am confused as to how the different formulations in physics are derived.
In many fields of physics, we usually begin with an ordinary formulation (e.g Newton's Laws in classical mechanics), and then move on to the Lagrangian, then Hamiltonian, and finally the action formulation. However, I don't understand how this chain of formulations are derived, one step at a time.
This physics.SE post deals with the derivation of Lagrangian from Newton's laws, and then I know that the Hamiltonian is obtained by changing the variable from $\dot{q}$ to $p$. What about action, then? How do we obtain the action?
And what about other fields of physics? Are there any ways to derive the Lagrangian and action, without just guessing or being given a specific Lagrangian?
| Regarding your last question, I'd like to add another important point. The Lagrangian is a real scalar function of space and time. Keeping that in mind, one tries to construct scalar objects with the physical fields of the theory, which could be scalars or vector (for instance, in electomagnetism) or tensors, etc. Also, one has to be careful about dimensions.
| {
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Why the speed of light in vacuum is same in all inertial reference frames?
If Cathy’s velocity
toward Bill and away from Amy is v = 0.9c, Cathy finds, by making measurements
in her reference frame, that the light from Bill approaches her at speed c, not at
c + v = 1.9c. And the light from Amy, which left Amy at speed c, catches up from
behind at speed c relative to Cathy, not the c - v = 0.1c you would have expected.
My question is why speed of light is same in all inertial frames (as in this example)? I know this is true from observations, but HOW ?
| It is that the speed of light is the same for all inertial frames that causes Special Relativity, not the other way around. As to why it is the same, nobody knows. Or why it is a finite value. There is some speculation about quantum gravity and a magical æther called spacetime, but nothing has been proven yet.
| {
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How can there be entropy change in this system? How can there be an entropy change in this system?
Suppose if I have a system consisting of liquid water, $1\, \mathrm{kg}$ at $290\,\mathrm{K}$, I stir it, and do say, $10\, \mathrm{J}$ of work on it, I can calculate the temperature change of the system given that:
$$U = cT \quad\mbox{ and }\quad S = c \ln \Omega$$
for $c$ constant.
From the fundamental equation of thermodynamics:
$$dU = dQ + dW = 0 + dW = 0 + 10 = 10\,\mathrm{J}$$
Hence:
$$dT = \frac{dU}{cM} = \frac{1}{410}\,\mathrm{K}$$
But how can there be a change of entropy in the universe when $dQ = 0$. I understand that we can calculate it using the formula for $S$ given, but I don't understanding how the fundamental equation allows this?
$$dQ = 0$$
and
$$dS = T^{-1}\,dQ$$
Hence, it may be concluded that:
$$dS = 0$$
Can someone tell me where my understanding is lacking, because obviously the entropy change is not zero in this case?
| The integral of dQ/T is equal to the entropy change of a system between two thermodynamic equilibrium states only if the path between the two states is reversible. If you want to determine the entropy change between two thermodynamic equilibrium states for a system that has undergone an irreversible process, you need to dream up a reversible path for the system between the same two thermodynamic equilibrium states, and calculate the integral of dQ/T for that path.
Answer #2 is pretty much completely incorrect.
| {
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Explanation of Michelson Interferometer Fringe Shift I have been working on an experiment where 2 glass microscope slides are pinched together at one end (so that there is a "wedge" of air between them) and placed in the path of a laser in one leg of a Michelson interferometer. When I move the glass slides (fractions of a mm at a time) so that the path of the laser is closer or further from the place where the slides are pinched, a fringe shift occurs. I cannot seem to explain why this is happening! Any help with explaining this phenomenon would be greatly appreciated! If any more specifics about the setup or dimensions of the slides are needed, please let me know.Also, a full "light to dark" fringe shift occured roughly every 4mm of moving the slides.
| This answer is based on Floris' insight that the slides might be bent.
Let's say the laser hits the slide at an angle of $\theta$ and travels through the panel at an angle of $\theta'=\sin^{-1}({n_a\over n_s}\sin(\theta))$. Let's assume the curvature is light enough that the laser essentially exits parallel to how it entered. I am also going to assume you are using a $\lambda=633\mathrm{nm}$ laser. We can express the change in phase from having no slides as $\phi={4\pi D\over\lambda}\sec\theta'\left({n_s\over n_a}-\sec(\theta'-\theta)\right)$ using a little trigonometry, where $D$ is the thickness of a slide. We want the curvature which is ${d\theta\over ds}$, where $ds=\sec\theta\,dx$. Let's express ${d\theta\over ds} = \left({d\phi\over d\theta}\right)^{-1} \left({d\phi\over dx}\right) \cos\theta$. You measured $d\phi\over dx$ to be ${2\pi\over 4\mathrm{mm}}$, and specified $n_s=1.52={n_s\over n_a}$. Let's say our angle of incidence is 10 degrees. We will now give this mess to Wolfram.
So at this point, we have to have a curvature of .06 degrees per mm to observe this fringe effect.
| {
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Physical Meaning of Cone used in Conic Section for Orbital Mechanics Does the polar angle (complement of $\theta$ below) of a cone which intersects a plane to yield a conic section have a physical meaning in orbital mechanics?
Note that the angle of incident planes, which form parabolas, ellipses, etc., is not the polar angle of the cone.
Here's an example. Consider the initial case of an elliptical orbit illustrated below:
Now, consider the mass of the central body greatly increasing. To conserve momentum, the elliptical orbit would shrink in area. In fact, the polar angle of the cone would decrease to reflect this proportional reduction in area:
[Note the smaller orbit and more narrow cone, despite keeping the same angle of the slicing plane as before.]
Is, therefore, the polar angle all and only a function of the mass of the orbiting bodies?
| The polar angle of the plane (the angle of the plane with respect to the symmetry axis) should relate to the eccentricity of the orbit. For $90^o$, the section is a circle and the eccentricity is zero. For an angle between the $90^o$ and the angle of the cone, you will get an ellipse with an eccentricity $0<\epsilon<1$. In both of these cases, the total mechanical energy of the system is negative.
If the angle is equal to the angle of the cone, you will have a parabola, and the total mechanical energy is zero.
At an angle less than the angle of the cone, you have a hyperbola, and the total mechanical energy is positive.
The kinetic energy and potential energy and angular momentum will all affect the related angles.
| {
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Is Gravity a Problem? I was watching the movie "Interstellar" recently and in that a character "Romly" mentions that if he could peep into the black hole "Gargantua" he could solve gravity.
I have a questions, is that a factual statement or is it for the movie's plot.
If yes then can someone explain what he means by solving gravity.
I am though interested in physics am not that knowledgeable in it.
| There is no problem with gravity. We have general relativity to describe it in full, and it works as far as we can tell.
The "problem" in the movie, as I understand it, was that it was infeasible to evacuate everyone from Earth. It takes an enormous amount of energy to get even a small amount of mass out of Earth's gravity well. The movie called for some futuristic ability to manipulate gravity so as to make this evacuation possible. Note such an ability is pure science fiction with no basis in reality.
Now as soon as you mention "problem" and "gravity" in the same sentence, everyone will immediately start thinking about how general relativity and quantum mechanics don't play well together. This is a real "problem" that many physicists work on, but note that in the movie "solving gravity" could be done even without "solving gravity + quantum mechanics."
| {
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Is this definition of orthohelium and parahelium incorrect?
"One electron is presumed to be in the ground state, the 1s state. An electron in an upper state can have spin antiparallel to the ground state electron ($S=0$, singlet state, parahelium) or parallel to the ground state electron ($S=1$, triplet state, orthohelium)."
From HyperPhysics
When they say "parallel to the ground state electron" then is it assuming that they are both spin up, or both spin down? If so, isn't it then ignoring the $S=1$ state with spin up and spin down:
$$|1\rangle |0\rangle = \frac{|+\rangle|-\rangle + |-\rangle|+\rangle}{\sqrt 2}$$
Therefore, if one electron is presumed to be in the ground state, 1s, state, if the spins can be opposite, a second electron can also occupy the ground state in $S=1$ orthohelium. Is this correct?
| The definition of orthohelium as having two parallel electron spins is correct. In the state $$\frac1{\sqrt2}\big(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\big)$$ the two spins are also parallel. It may seem that they are not, but note that this state has $m_z=0$, so both spins are perpendicular to the quantization direction. Indeed, the operator $S_+$ will turn this state into $$\big(|\uparrow\uparrow\rangle\big)$$.
Also, this is an eigenstate of $S^2$ with eigenvalue 2$\hbar^2$, just like the other members of the triplet, while $S_1 \cdot S_2$ has eigenvalue $+\frac{\hbar^2}{4}$. The positive sign indicates that the spins are parallel. Indeed for the singlet this eigenvalue is $-\frac{3\hbar^2}{4}$, which indicates antiparallel alignment.
| {
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What would put a harddisk drive (HDD) under 350G's of force? I always see the label and it says 350G's withstandable. What would put this over 350G's? Is it even possible to hit 350Gs of force to a hard drive?
|
Is it even possible to hit 350Gs of force to a hard drive?
Sure is. Drop it on the floor.
You are thinking about sustained forces. 350g sustained won't happen even in rocket launches. But momentary forces can easily peak at this level.
Note that the G limit on the drive is for when it's not running. No spinning drive will like 350g, except maybe in particular directions that will never happen in reality.
If you drop your hard drive from $1~\text{m}$ it will hit the floor at around:
$$
\sqrt{2\times 1~\text{m}\times 1~g}\approx 4.4~\text{m}\cdot\text{s}^{-1}
$$
At exactly $350~g$ it would come to a stop in:
$$
\frac{\left(4.4~\text{m}\cdot\text{s}^{-1}\right)^2}{2\times 350~g}=2.8~\text{mm}
$$
(Note that due to the way the math works out, the stopping distance when dropping from a height $h$ is just $hg/a$).
Since the actual impact will probably be a varying acceleration and the rigid case of the hard drive will probably deform less than $3~\text{mm}$, the actual peak acceleration can easily exceed $350~g$.
| {
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Creation and annhilation operator in the Heisenberg picture I am trying to calculate the time evolution of the creation/anni. operator in the Heisenber picture.
On this webpage http://quantummechanics.ucsd.edu/ph130a/130_notes/node191.html, they used the Heisenber equation of motion, but instead of using the operators in the Heisenberg picture, they used the operators in the Schrödinger picture.
Why does this work?
Best
| Because (assuming a time independent Hamiltonian, operators without subscript referring to Schrödinger operators, those with the subscript $H$ to Heisenberg operators):
\begin{align*}
[a_H(t), H_H(t)] &= e^{-iHt/\hbar} a e^{iHt/\hbar} e^{-iHt/\hbar} H e^{iHt/\hbar} - e^{-iHt/\hbar} H e^{iHt/\hbar} e^{-iHt/\hbar} a e^{iHt/\hbar} \\
&= e^{-iHt/\hbar} (aH - Ha) e^{iHt/\hbar} = e^{-iHt/\hbar} [a, H] e^{iHt/\hbar} \\
&= e^{-iHt/\hbar} \hbar \omega a e^{iHt/\hbar} = \hbar \omega e^{-iHt/\hbar} a e^{iHt/\hbar} = \hbar \omega a_H(t).
\end{align*}
Note that $e^{A}e^{-A} = 1$.
When looking carefully at the proof, one can see, that this calculation can easily be generalized to prove that $[A_H, B_H] = [A, B]_H$ (we only used the concrete properties of $H$ and $a$ in the last steps).
| {
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Sound source not in a straight line with the sound receiver - does that make a difference?
Hope the graphics will help me explaining my question better.
Let's say the box would be a room in the fourth floor, and the sound source would come from cars in the street.
It is clear that in case A the sound would enter the room and would be clearly heard from the human ear.
What I am interested to inspect is case B. The red line there represents a sound-proof shield. The shield would block the direct sound geometrically as you see. In front of the flat there would be no nearby building that could bounce the sound back to the room from the other direction. So, there is just air.
I know sound is not a bullet that travels in straight line, so my question is: Would there be a significant decrease in the sound heard inside the room if a sound-proof shield would be placed as in the picture in case B? Can we calculate a rough percentage?
| You will benefit by finding some tutorials on wave theory. In brief, assuming a spherical wavefront from the emitter, you are correct there's no direct path to the receiver. However, the edge of yourabsorber there causes diffraction (Huygen's principle), so thatsome of the sound wave (energy) will make its way to the receiver. You can see a demo of this, e.g., at mike-willis tutorial .
| {
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Flaws of Broglie–Bohm pilot wave theory? I recently learned about an oil drop experiment that showed how a classical object can produce quantum like behavior because its assisted by a pilot wave. How has this not gained more attention? What flaws does Broglie–Bohm pilot wave theory have in explaining particle behavior?
| I am not sure if the OP shared the link or not.
I happened to watch this very recently -
https://www.youtube.com/watch?v=WIyTZDHuarQ&t=199s
This is amazing explanation in terms of real visual.
Between 2:35 and 3:15 the video shows how the pattern is built over a period of time, while the jumps appear to be random at any one time.
Therefore things may not be random, as claimed by some parts/interpretations of QM.
I think the entanglement correlations also build over time, not due to randomness, but due to conservation/balancing. I have scrutinized recent experiment data closely and it gives some indication of such possibility. http://vixra.org/pdf/1609.0237v7.pdf
| {
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Are wavelength and the distance same thing? Can you clarify for me the following question: are wavelength and distance same?
I know wavelength is measured in terms of distance but when we have a look at the two equations:
$$
c=f\,\lambda\\
v=d/t
$$
it actually explains the same thing where $v=c$=velocity and $1/t$ is frequency. So $\lambda$ should be equal to $d$. So if $\lambda = d$, then why do we have two equations existing instead of one. Can we use any equation to calculate velocity?
| The relationship between wavelength and distance is similar to the relationship between frequency and duration, and no: neither pair is the same. You can see by using dimensional analysis.
Wavelength is distance divided by cycles. Frequency is cycles divided by time. Multiply the two, the cycles cancel out, and you get distance divided by time, or velocity.
For instance, if you look at a 90MHz FM radio wave (that's 9 x 10^7 cycles per second), the wavelength is about 3 1/3 meters (that's 3.333 meters per cycle). Multiply them together, and you get 3 x 10^8 meters per second. Bingo: speed of light.
| {
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Could some astronomical objects have superconducting properties? The colder it is, the more efficient the superconductivity process works. And as we know, if there is no star nearby, space gets pretty cold.
I do appreciate that many condensed, burnt out, stars may take a long time to cool off, but are there any other types of known astronomical objects that may feature superconductivity to create and/or maintain a very strong magnetic field?
| Doubtful you'll find anything within the Solar System, but there are neutron stars, which are thought to have regions which are both superconducting and superfluid (that link is one of the original references from almost 50 years ago - there is a ton of literature on the topic since, you could start with some of these).
| {
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Antenna direction I have a router with a wifi antenna that can be turned in any angle. I wonder what difference does the direction of the antenna make to the electromagnetic signals propagation? Where is the signal strength the biggest?
| I had the same question once, and scoured the Internet for advice. All I got was conflicting information, much of it from "experts". I ended up getting a signal strength app for my smart phone and one for my laptop and experimenting. In my house, with my router, I found no detectable difference between horizontal and vertical.
| {
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The speed of light/EM waves in vacuum; as if there was another one in non-vacuum? Q1: is there a speed of a photon other than in "vacuum"?
Q2: isn't "speed of light in vacuum" misleading?
If I understand, that light moves with speed of light until there is "something in between" (no matter what) (1)
What I ask for, is not a deeply explanation; it's just:
Children ask me:
*
*"But how can be light slower" if it is a constant?
My explaination is:
*
*"Till it collides" (not the deepest answer, I know)
The question (since I prob. are not at the pulls of new physics): Is there another speed of light than in "vacuum" / nothing crossing ?
(1) no discusss, what "what" is
| Speed of light is constant.
But in some substances , still transparent , light is absorbed and retransmitted ( with the same properties ) , spending some time. With not well transparent material, things are more complex. Anyway, between 2 obstacles, it's the vacuum and the speed remains constant and maximum.
How many are retransmitted and the specific speed depend on the light frequency ( dispersion , see the prism image ) , the transparency and translucency of the substance and its refractive index.
| {
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Difference between a reversible change and a reversible process? Question
In thermodynamics what is the difference between a reversible change and a reversible process?
Additional information
I am new to the topic of thermodynamics and getting confused about the difference, if any, between a reversible change and a reversible process. It seems to me that the difference is to do with the equilibrium of the system with the surroundings.
| A process is something that goes on, has duration. A change has no necessary relation to time. It is a statement of difference between the initial and the final state of the process.
| {
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Why do sea waves seem to be standing still when you look from the window of an airplane? Looking from the window of a passenger plane even at moderate altitude such that one can still recognize individual waves and even something like white foam, and small boats close to the cost line, it already looks like the water is not moving.
To make more clear what i mean, here is an examplein this video at around 10:40. In HD eye resolution it is much more intriguing, but the video shows the idea, that even when the plane is quite low, waves close at the beach appear to be "frozen".
Why is that? and does that effect have a name?
| I imagine this effect has to do with the fact that velocity is relative. When you're on the shore, you gauge the velocity of the waves with respect to the shore. When you're in a plane, you're likely gauging the velocity with respect to the other wave crests, which are moving at the same velocity and so there is no apparent movement.
| {
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How and when are the relativistic corrections applied to GPS satellites? It is known that there is a need to correct the onboard clocks to reduce the time difference from 38μs to 50ns. Where is relativity playing its role here? Why cant the clocks be simply synchronised with the ground clocks through telecommunication? If these are not possible then how are the clocks corrected?
| Due to relativity, the clocks on the GPS satellites move fast by about 38 µs per day. Which would be a problem, but not that big a problem because they all move fast by the same amount. Still you'd need to synchronize the clocks from time to time, because the satellite's position in space also depends on the clock.
HOWEVER, if you do that once a week, compensating for about 270 µs, you'd have to be able to do that absolutely at the same time for all satellites. If I'm driving happily through a town, directed by my TomTom, and the first satellite changes by 270 µs = 80 km, and then the second satellite changes ten seconds later, and so on, my TomTom will have a major problem with that, and so will I.
Much better to make sure that the clocks compensate for the known 38 µs per day by themselves all the time, so the compensation is only for a few nanoseconds.
| {
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Quantum entanglement and the big bang Prior to the Big Bang all matter was compressed into a point of high density. Why isn't all matter already entangled?
| Let $|\Omega\rangle$ be the quantum state that describes the whole universe. Certainly it doesn't make sense to talk about the entanglement of $|\Omega\rangle$ with something else, since $|\Omega\rangle$ describes everything. However, we can meaningfully discuss the entanglement of the marginals of $|\Omega\rangle$:
\begin{equation}
\rho_a=\text{Tr}_{a'}\left(|\Omega\rangle\langle\Omega| \right)
\end{equation}
and
\begin{equation}
\rho_b=\text{Tr}_{b'}\left(|\Omega\rangle\langle\Omega| \right)
\end{equation}
Here I have defined these two marginals by tracing out a part of the state-space (i.e. ignoring a part of the universe). Therefore,$\rho_a$ and $\rho_b$ are sub-systems of $|\Omega\rangle$: i.e. they may describe a "smaller" part of the universe. Now the notation of bi-partite entanglement can used to discuss the entangelment and/or separability of $\rho_a$ and $\rho_b$.
So to answer your question succinctly: some matter is indeed entangled with other matter, but everything in general is not entangled. This can be experimentally verified.
| {
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I don't understand black body radiation graphs
Let's look at the above graph.
*
*This black body graph is for the temperature of 5000K. Each temperature has a different black body graph?
*How am I supposed to read this graph? Do I start from the left, the right, or the peak?
*As wavelength is approaching zero, intensity is approaching zero. This is what my textbook says (experimentally). This is explained by Planck's hypothesis regaridng quantised energy. However, if energy is quantised, and thus the all or none principle applies, should it then simply just, with the shape of a straight line, collapse? i.e. the graph should approach its peak, then instantly drop down in a straight line because it can either absorb ALL or none. The graph does not depict this. Why?
*A black body of ANY temperature will emit all types of radiations (eg UV, X-rays, visible light etc). BUT, at higher temperatures eg 60000K, certain transitions in quantum states are favoured more, thus we would expect more intense radiation to prevail eg gamma.
| *
*The dependence of spectra radiance (y-axis) on wavelength (x-axis) is as follows (Planck's law):
$$I(\lambda,T)=\frac{2hc^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda k_BT}}-1}$$
As you can see, $I$ also depends on temperature T. For each T, we have corresponding curve.
*You should pay attention to the peak first. The peak will tell you at what wavelength (frequency), the spectra radiance is maximum. Then you should look at higher wavelength (same as classical) and lower wavelength.
*Will be answered later.
*You should read the Wien's displacement law.
| {
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Can we find actual rest mass of things on Earth Earth moves around the Sun and the Sun moves around the galaxy and the galaxy moves with unknown speed and direction. We have speed so the mass of us all altered.
Can we know the real rest mass? If so, can we deduce our speed in the universe?
| Earth moves around the Sun and the Sun moves around the galaxy and the galaxy moves with unknown speed and direction. We have speed so the mass of us all altered.
The relativistic mass is altered, but this is a somewhat archaic term these days, and is said to be a measure of energy. Nowadays when we say mass without qualification, we tend to mean rest mass. Like Rc and Jazz said, this doesn't change with speed. Instead it changes with gravitational potential, see mass in general relativity and the mass deficit. Unfortunately rest mass is also called invariant mass, which is rather confusing.
Can we know the real rest mass?
Yes, because the mass of a body is a measure of its energy content, and energy is conserved. But we have no accepted theories for that at present. For example the first free parameter of the Standard Model is electron mass.
If so, can we deduce our speed in the universe?
We don't need to deduce it. We can measure it. From the CMB dipole anistropy. We're moving at 627±22 km/s relative to the reference frame of the CMB. Or relative to the universe as a whole.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the scalar magnetic potential continuous? If we have two current-free spaces and separated by a surface current, we can solve the magnetic problem by solving two magnetic scalar potentials and then using matching conditions. My question is, is the general scalar magnetic potential continuous? Why?
| The potential for a static magnetic problem is defined through
$$
{\bf B} = - \nabla \phi
$$
(or you can define another potential for $\bf H$). Then since $\nabla \cdot {\bf B} = 0$ we have Laplace's equation for $\phi$ and that is why it is useful: we have lots of good methods for solving Laplace's equation. (Of course it won't work if $\nabla \times {\bf B} \ne 0$; in that case one should adopt another approach.)
The above equation implies that the answer to your question is that $\phi$ is continuous if and only if $\bf B$ is finite. At a boundary such as a surface you expect to find finite $\bf B$. At a current-carrying wire of arbitarily small radius, on the other hand, $\bf B$ can diverge.
| {
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What is known about Renyi entropy of a probability density function? I see most discussions about Renyi entropy to be using either of these two kinds of definitions, for $\alpha > 0, \alpha \neq 1$
*
*$H_{\alpha}(p_i)=\frac{1}{1-\alpha}\log \sum p_i^{\alpha}$ for a probability measure over a discrete set of events indexed by $i$.
or,
*
*$H_\alpha(\rho)=\frac{\ln\mathrm{tr}(\rho^\alpha)}{1-\alpha}$ when a physical system is described by a density matrix $\rho$.
Q1 : In the language of probability measure spaces is it clear as to what is that underlying structure in the second kind of definition?
Q2 : For a continuous probability density function has one thought of quantities like $H_\alpha (f) = \frac{1}{1-\alpha}\log [ \int f(\vec{x})^\alpha d\vec{x} ] $ ? (If yes, then can someone point to some reference?)
| Here is a reference for the continuous versions:
http://arxiv.org/abs/1402.2966
| {
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Why can microphone be much smaller than wavelength of sound? For sound from 20Hz to 20kHz, wavelength is 17m to 17mm, for sound at 2kHz, wavelength is 17cm.
And I saw tiny microphone which is much smaller than that. In electromagnetic, there is a smallest size for antenna of each wavelength (half wavelength???). And there is a law (IIRC) that if sampling frequency is smaller than half of the signal, than it is not possible to reconstruct the signal.
How can microphone size is much smaller than the wavelength? E.g. the head of the mic of singer, I think it is smaller than 1 cm
| A microphone is a transducer that converts variations in air pressure from sound waves into electrical signals.
Air pressure varies as the wavefront passes into the diaphragm (or the ribbon, or the condenser) of the microphone. The diaphragm needn't be as long as the wavelength, as it senses the wave from a "head-on" perspective rather than "looking at it" from the side.
| {
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Is it possible to build a thermoelectric nuclear power plant? Current nuclear power plants are essentially an enhanced version of a kettle, which seems like a stupidity caused by a lack of other options. We heat the water which turns to steam which rotates the turbine, which is total waste of energy due to the several conversions.
I googled a bit and found that actually there exists the thermoelectric effect which allows for converting heat to electricity directly. Yes, I didn't know about it until today. ;)
Is it possible to turn the heat from the nuclear reactor directly to electricity? Have there been any attempts to do it?
I am not asking why we do not use it currently, my question is about whether it's possible in principle and whether anyone has tried it.
| The efficiency of a thermoelectric generator is around 5 - 8%.
The efficiency of a large steam turbine power plant aproaches 40%.
In fact the thermodynamic efficiency of a large steam turbine power plant is over 90%, so it's about as efficient as anything could be. The maximum possible efficiency of a steam driven engine is given by the idealised model called a Carnot engine. The efficiency is ultimately limited by the difference in temperature of the hot and cold ends of the engine, and modern power plants get pretty close to this theoretical maximum.
Thermoelectric generators tend to be used only where other restrictions force their use. For example the Curiousity rover uses a thermoelectric generator with an efficiency of about 6%. The lower efficiency is balanced out by a lack of moving parts, and of course the non-availability of water on Mars from which to make steam.
| {
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Does isotropy imply homogeneity? This question comes from exercise 27.1 in Gravitation by Misner, Thorne and Wheeler. They required the following:
Use elementary thought experiments to show that isotropy of the universe implies homogeneity.
I know homogeneity as the universe is the same everywhere at a given time, and isotropy is related to direction.
I wonder how the isotropy of the universe implies homogeneity.
| When MTW say the universe is isotropic, they mean it is isotropic everywhere i.e. at all points in the universe.
It's easy to construct universes that are isotropic at a single point and not homogeneous, for example CuriousOne's suggestion of a ball with density that is a function of distance from the centre. However this ball is only isotropic if you are at the centre of the ball. If you require the ball to be isotropic everywhere you necessarily require it to be homogeneous.
MTW actually give you the answer (in a technical form) to exercise 27.1 in the paragraph just above the exercise next to the side note:
Isotropy implies fluid world lines orthogonal to homogeneous hypersurfaces
| {
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How can a material conduct heat but not electricity Mica is a good conductor of heat but an electrical insulator.
According to free electron theory (which applies only to metals) free electrons carry heat and electricity. Therefore, thermal conductivity is directly proportional electrical conductivity.
What about dielectric materials? In the case of mica, it conducts heat but not electricity, so what are the carriers responsible for this behavior. If the are electrons as they are in metals then why they don't carry electricity too?
| Electricity needs charged particles (or quasi-particles) to conduct it. Heat can be conducted with almost any quasi-particle. Diamond is one of the best conductors of heat in existence, and it's because of phonons, ie quasi-particles of lattice vibrations, which are strong because the diamond lattice is strong.
| {
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Is air infiltration a type of convective heat transfer (convection) I have a building / physics question...
A major source of heat loss in homes and buildings is infiltration through cracks (warm air from inside seeping out). Wondering if this falls in the category of convection as a mode of heat transfer?
|
Wondering if this falls in the category of convection as a mode of heat transfer?
Yes.
You are discussing heat transfer due to air traveling through a crack in a wall. Any heat transfer due to a moving fluid is convective heat transfer.
If there is wind, then it is further categorized as forced convection. If there is no wind, just bouyancy effects due to different gas densities, then it is called natural convection.
For home heating/cooling, conductive and radiative heat transfer will also contribute. The relative importance of convective heat transfer through cracks in the walls depends on the construction. Houses with double-walled construction will have less air leakage than house with single-wall construction.
| {
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Do electrostatic fields really obey "action at a distance"? In an electromagnetic theory class, my professor introduced the concept of "action at a distance in physics".
He said that:
If two charges are at some very large distance, and if any one of the charge moves, then the force associated with the charges changes instantaneously.
But according to Einstein, no information can travel faster than the speed of light.
So photons (the information carriers in electromagnetic force) cannot instantaneously deliver information.
So that we associate a field with the two charges and if any charge moves, there is a deformation in that field and this deformation travels with the speed of light and conveys the information.
If the field deformation information cannot travel more than the speed of light, how does the force instantaneously change at very large distances?
| To reach the Lienard-Wiechert potentials or to prove Feynman's equation (exposed in his lectures without proof), it's necessary to begin with the so-called retarded potentials expressed here conveniently by the following.
\begin{equation}
\phi\left(\mathbf{r},t\right)=\dfrac{1}{4\pi\varepsilon_{o}}\iiint d^{3}\mathbf{r}^{\prime}\dfrac{\rho\left(\mathbf{r}^{\prime},t-\dfrac{\|\mathbf{r}^{\prime}-\mathbf{r}\|}{c}\right)}{\|\mathbf{r}^{\prime}-\mathbf{r}\|}\:, \quad \text{scalar potential}
\tag{01a}
\end{equation}
\begin{equation}
\mathbf{A}\left(\mathbf{r},t\right)=\dfrac{\mu_{o}}{4\pi}\iiint d^{3}\mathbf{r}^{\prime}\dfrac{\mathbf{j}\left(\mathbf{r}^{\prime},t-\dfrac{\|\mathbf{r}^{\prime}-\mathbf{r}\|}{c}\right)}{\|\mathbf{r}^{\prime}-\mathbf{r}\|}\:, \quad \text{vector potential}
\tag{01b}
\end{equation}
Now, what explains the non-instantaneous action are the terms in parentheses. For example, if the action was instantaneous then the scalar potential $\:\phi\:$ at point $\:\mathbf{r}\:$ in time $\:t\:$ would be that produced electro-statically by the charge density $\:\rho\:$ from various points $\:\mathbf{r}^{\prime}\:$ in that SAME MOMENT t:
\begin{equation}
\dfrac{1}{4\pi\varepsilon_{o}}\iiint d^{3}\mathbf{r}^{\prime}\dfrac{\rho\left(\mathbf{r}^{\prime},t\right)}{\|\mathbf{r}^{\prime}-\mathbf{r}\|}\ne\phi\left(\mathbf{r},t\right)
\tag{02}
\end{equation}
But the term
\begin{equation}
t-\dfrac{\|\mathbf{r}^{\prime}-\mathbf{r}\|}{c}= t^{\prime}
\tag{03}
\end{equation}
"repairs" exactly this, taking into account that the electromagnetic disturbance from charges at $\:\mathbf{r}^{\prime} \:$ needs a time interval
\begin{equation}
\dfrac{\|\mathbf{r}^{\prime}-\mathbf{r}\|}{c}
\tag{04}
\end{equation}
to arrive at $\:\mathbf{r}\:$ "running" with velocity $\:c\:$.
| {
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Water well fluid dynamics In my home village everyone has their own water well at the foot of the mountain (limestone) . The wells are over 100 feet deep but they are not cemented or have the area between the hole and the casing (PVC) sealed with bentonite or anything like it. Even though the well is 100+ feet deep, inside the well deep the water comes up to within 15 feet of the surface. I suppose the stone is fractured even though during the drilling they have reached meters and meters of solid limestone.
I assume that deeper wells are safer to drink /use and cleaner since the bad things (remains of cow manure, plain old dirt or septic waste) rarely reach say 100+ feet. However, since the casing is not sealed and the water inside the well reaches within 15-20 feet of the surface, is it safe to assume that that the water sometimes is as clean as it would be in a 15-20 feet well? The casing on top is solid, only the last few meters at the bottom of the well have it with holes for the water to come in.
Now the question is this: Suppose "dirt" leaches down from the surface, moves through cracks in stone and at 10 feet deep touches the PVC casing and....does it go down 100+ feet or does pressure stop it from doing so. What happens what the submersible pump (100+ feet down) turns on, does the water mix up and down?
Thanks guys,
| As posed there is no definite answer to your question.
It is certainly possible for water to enter the well at a depth of 100 feet, and have sufficient pressure that it eventually rises to within 15 feet of the top. That just means that the "water table" in that area is 15 feet below the surface.
If the sides of your well are reasonably well sealed so only water that had to seep through many feet of rock/soil enter the well, this will improve filtration - but it does not guarantee there is no mixing.
The physical principle here is that if you connect two vessels, and a steady state is allowed to be reached, then the water level in them will equalize:
| {
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Why, when and where is Gauss's law applicable?
Why is it said that Gauss's Law is mainly applicable for symmetric
surfaces/bodies? Why not for asymmetric surfaces?
I want a logical explanation!
BTW my teacher said that Gauss's law is applicable for any surface/body but in the case where symmetry does not exist, the calculation becomes a bit tedious. I did not get what he meant by that. Can someone please help me get a clear cut concept about when and where Gauss's law can be applied? Please note I'm not asking for the rigorous proof of Gauss's law.
| As your teacher says, it holds for every surface, but a look at the law itself, should clear out why some form of symmetrie is desirable:
$$
\iint_S \vec{E} .\mathrm{d}\vec{A}=\iint_S E . \cos\theta . \mathrm{d}A = \frac{Q}{\epsilon_0}
$$
Here, $E$ and $\theta$ are position-dependent, so to calculate the integral, you need to take care of a position dependent magnitude and a position dependent angle of the electric field, not an easy task if your aim is to find the electric field.
How much easier if you can use some reasoning before starting the calculation. E.g. for the field of a point charge, you can take a sphere with the charge in the center as surface, just for convenience. Since for symmetrie reasons, the field should be the same for every point on the sphere and the direction of the field should be perpendicular on the surface, it becomes much easier:
$$
\iint_S \vec{E} .\mathrm{d}\vec{A}=\iint_S E . \cos\theta . \mathrm{d}A = E\iint_S 1 . \mathrm{d}A =\frac{Q}{\epsilon_0}
$$
| {
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Is gravity's acceleration rate - squares of all natural numbers? I've seen some science\history documentary in which they made a replica of Galileo's inclined plane experiment.
They rolled a ball down the plane; and it's progressed in length units each unit of time was - 1, 4, 9, 16 etc. I realized it was squares of natural numbers; but I didn't really understand why this happens. I expected it to be exponential; as in - 1, 2, 4, 8, 16 etc.
I've checked a few more videos since; and now I understand that this progression is "time squared". That makes sense; but I still don't exactly understand why it works this way.
Also in some other video, which is a segment from some science show, they said the ball suppose to roll - 1, 3, 5, 7 etc. Did they just got it wrong?
So why does it work this way - time squared; and what other basic things in physics progress this way; can you give some examples?
What about energy needed to accelerate mass?
| The distance between successive lines is 1,3,5... making their absolute position 1,4 (1+3), 9 (1+3+5)...
This follows from double integration of the equation of motion with a constant force:
$$v = \int \frac{F}{m} dt = a\; t\\
x = \int v\; dt = \int a\; t\; dt = \frac12 a t^2$$
So the position increases quadratically with time for constant acceleration (constant force)
As for your question about "other basic things in physics": this applies to any situation where there is a constant force. The energy needed to go faster increases (since kinetic energy goes as velocity squared) but the power is velocity times force, so as the speed goes up the power goes up as well.
| {
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Free fall into circular motion If I'm on a roller coaster free falling from height $h$ and then suddenly start going into horizontal motion with a radius $r$ of turn what is the $g$-force I experience?
I worked out the equation like this but am not sure if it is correct:
*
*(1) instant velocity of free-fall $v=\sqrt{2 g h}$
*(2) uniform circular motion acceleration $a = \frac{v^2}{r}$
*(3) $g$-force $ gf = \frac{a}{g} = \frac{v^2}{g r}$
My doubts are:
*
*I don't know if I can use uniform circular motion equation since $v$ is not constant
*Where is the g-force directed towards? The center of the turn?
|
I don't know if I can use uniform circular motion equation since v is not constant
The equation for centripetal force is independent of whether the motion is uniformly circular or not.
However, irrespective of the radius of the track, the velocity at that point, and the weight of the roller-coaster, or whether the equation for centripetal force is still valid, etc; since it becomes horizontal at that particular point, any force that must cause your 'g-force' must be acting in the horizontal direction. As no such force acts in the horizontal, the g-force on you as soon as you go horizontal is zero. As simple as that.
| {
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What happens if the load on the electrical generator exceeds its generation power? And why? What happens if the load on the electrical generator exceeds its power generation? and why?
To be more precise, suppose we have a standard induction generator operating at frequency $\nu=50\:\mathrm{Hz}$ and voltage $V_0$, and rated to produce a maximum power $P_0$, and that we connect this to a load $R<V_0^2/P_0$, which will try to draw more power than the generator's capacity. Obviously the details will depend on the type of generator, but, generally speaking: what will be the generator's response, and what physical processes are involved?
| The generator circuit breaker should trip to avoid damage. But if that doesn't work, or doesn't exist, the generator will not be able to maintain its speed. On a typical AC system, this will cause the frequency to drop from the standard 50 or 60 Hz, as well as drop the output voltage.
| {
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Bounds on dimension of a purification? Let $\rho \in H_A$ be a density operator, $H_A$ is finite dimensioal, it is well known that $\rho$ has a purification in some larger hilbert space. Let $b$ be the minimum dimension for $H_B$ such that there exists a purification of $\rho$ in $H_A \otimes H_B$.
Are there special classes of $\rho$ for which there is a known upper bound on $b$?
| Consider any purification of $\rho$: $|\psi\rangle \in \mathbb{H}_A \otimes \mathbb{H}_B$. This purification necessarily admits a Schmidt decomposition of the form:
$$|\psi \rangle=\sum_i \sqrt{\lambda_i} |\alpha_i\rangle_A |\beta_i \rangle_B $$
with the state $\rho$ being of the form:
$$\rho=\sum_i \lambda_i |\alpha_i\rangle \langle \alpha_i |$$
This implies that the minimum dimension of $\mathbb{H}_B$, $b$, must be the rank of $\rho$. It follows that the upper bound on $b$ is the dimension of $\mathbb{H}_A$, and the class of states which saturate this bound are the states of maximal rank.
Is this what you're looking for?
| {
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A simple derivation of the Centripetal Acceleration Formula? Could someone show me a simple and intuitive derivation of the Centripetal Acceleration Formula $a=v^2/r$, preferably one that does not involve calculus or advanced trigonometry?
| Imagine a object steadily traversing a circle of radius $r$ centered on the origin. It's position can be represented by a vector of constant length that changes angle. The total distance covered in one cycle is $2\pi r$. This is also the accumulated amount by which position has changed..
Now consider the velocity vector of this object: it can also be represented by a vector of constant length that steadily changes direction. This vector has length $v$, so the accumulated change in velocity is $2 \pi v$.
The magnitude of acceleration is then $\frac{\text{change in velocity}}{\text{elapsed time}}$, which we can write as:
$$a = \frac{2 \pi v}{\left(\frac{2\pi r}{v} \right)} = \frac{v^2}{r} \,.$$
Q.E.D.
Aside: that derivation is used in a lot of algebra/trig based textbooks.
| {
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Does body weight affect the speed when going downhill on a mountain bike? We know heavier objects fall faster when dropped at certain height. I was wondering if I am going downhill on my mountain bike without any peddling, will I travel faster or slower because I am fat?
| No and yes. At first, your assumption is not quite correct. In vacuum, all masses fall at the same speed. The reason is the that the mass cancels in the equations of motion:
$ma=mg$
$a=\ddot{x}=g$
To be more precise: the inertial mass and the gravitational mass are the same. Therefore, they cancel.
However, things change when you take air resistance into account. Of course, a feather has much more air resistance than a stone, consequently the feather falls slower.
In the equations of motion, the air resistance is proportional to the square of the velocity and a geometric factor that describes the shape of your object/body only, but not to the mass.
More specific to your question: Although you have a higher mass, the air resistance of your body compared to the bodies of other persons may be quite similar. Especially if you try to make you as small as possible. This means that you will have an advantage compared to less heavier people, as long as you manage to have not much more air resistance than them.
| {
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Do black holes violate the uncertainty principle? If black holes have mass but no size, does that imply zero uncertainty in position? If so, what does that imply for uncertainty in momentum?
|
If black holes have mass but no size, does that imply zero uncertainty
in position? If so, what does that imply for uncertainty in momentum?
I mean to say that the particles which were originally separate have
theoretically come to occupy the same point in space. Does the
uncertainty principle apply to this phenomenon?
Zero size doesn't violate the uncertainty principal. It's knowing the exact location that is impossible with the uncertainty principal because you can't measure a position without moving it. Granted, black holes have enormous mass so any change in position is much smaller, but uncertainty principal still applies.
What you're describing sounds like the Pauli Exclusion Principle and that's true to an extent but there are ways that can be explained. One is that we don't really know what happens at a singularity. The math breaks down and we've never seen one, so it's a model that we admit isn't complete.
Also, one could argue that a particle in a singularity is no longer a particle, but the singularity is the particle. This approach addresses the Pauli exclusion principal problem but it raises another question, Conservation of information.
at least, that's how I look at it.
| {
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7/2 versus 9/2 for diatomic heat capacity Question
I calculated the classical heat capacity of a diatomic gas as $C_V = (9/2)Nk_B$, however the accepted value is $C_V = (7/2)Nk_B$.
I assumed the classical Hamiltonian of two identical atoms bound together as
$$
H = \dfrac{1}{2m}( |\bar{p}_2|^2 + |\bar{p}_2|^2)+
\dfrac{\alpha}{2} |\bar{q}_1-\bar{q}_2|^2.
$$
I calculated the partition function of $N$ particles as
$$
Z
=
\left(
\iiint_{-\infty}^{\infty}
\iiint_{-\infty}^{\infty}
\iiint_{-\infty}^{\infty}
\iiint_{-\infty}^{\infty}
e^{-\beta H}
~d^3q_1~d^3p_1~d^3q_2~d^3p_2
\right)^N
\propto
V^N T^{(9/2)N}.
$$
I calcuated the heat capacity as
$$
C_V
=
\dfrac{\partial }{\partial T}
\left(
k_B
T^2
\dfrac{\partial \ln(Z)}{\partial T}
\right)
=
\dfrac{9}{2}k_BN.
$$
Why does the classical argument fail?
Classical Derivation
The partition function is
\begin{align}
Z
&=&
\left(
\frac{1}{h^6} \int \mathrm{e}^{- \beta H(\bar{q}_1,\bar{q}_2,\bar{p}_1,\bar{p}_2)} ~d^{3}q_1 ~d^{3}q_2 ~d^{3}p_1 ~d^{3}p_2
\right)^N
\\&=&
\left(
\frac{1}{h^6} \int \mathrm{e}^{- \beta ((|\bar{p}_1|^2+|\bar{p}_2|^2)/(2m)+\alpha |\bar{q}_1-\bar{q}_2|^2/2)} ~d^{3}q_1 ~d^{3}q_2 ~d^{3}p_1 ~d^{3}p_2
\right)^N
\end{align}
A useful gaussian integral
\begin{align}
\int_{-\infty}^{\infty} e^{-\gamma (x-x_0)^2}dx = \sqrt{\dfrac{\pi}{\gamma}}
\end{align}
The partition function can be evaluated using separated integrals
\begin{align}
\iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta |\bar{p}_1|^2} ~d^{3}p_1 = \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta |\bar{p}_2|^2} ~d^{3}p_2 = \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3
\end{align}
and
\begin{align}
\iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta \alpha |\bar{q}_1-\bar{q}_2|^2/2 } ~d^{3}q_1 ~d^{3}q_2
=
\left(
\sqrt{\dfrac{\pi}{\beta \alpha/2}}
\right)^3
\iiint_{-\infty}^{\infty} ~d^{3}q_1
=
\left(
\sqrt{\dfrac{\pi}{\beta \alpha/2}}
\right)^3
V
\end{align}
The last set of integrals are improper integrals. One has to take the limit as the space approaches infinite containment. In that limit, integrating one set of variables $d^3q_2$ approaches the limit of a finite Gaussian term, while the other $d^3q_1$ approaches the diverging value of the total volume of the gas.
The partition function is
\begin{align}
Z
&=&
\left(
h^{-6}
\left(\sqrt{\dfrac{\pi}{\beta}}\right)^3
\left(\sqrt{\dfrac{\pi}{\beta}}\right)^3
\left(
\sqrt{\dfrac{\pi}{\beta \alpha/2}}
\right)^3
V
\right)^N
\\&=&
\left(
h^{-6}
\left(k_B T \pi\right)^{9/2}
\left(
\dfrac{2}{\alpha}
\right)^{3/2}
V
\right)^N
\\&=&
\left(
h^{-6}
\left(k_B \pi\right)^{9/2}
\left(
\dfrac{2}{\alpha}
\right)^{3/2}
\right)^N
V^N
T^{9N/2}
\end{align}
| The potential energy for a diatomic molecule is not
$$
U(\vec{q}_1, \vec{q}_2) = \frac{\alpha}{2} |\vec{q}_1 - \vec{q}_2|^2
$$
but is instead
$$
U(\vec{q}_1, \vec{q}_2) = \frac{\alpha}{2} (|\vec{q}_1 - \vec{q}_2| - r_0)^2,
$$
where $r_0$ is the equilibrium bond distance. The important difference here is that in your version, any displacement of the vector $\vec{q}_1 - \vec{q}_2$ will result in a quadratic change in the potential energy; whereas in the correct version, there will be two directions in "configuration space" that correspond to no change in the potential energy. Remember that the equipartition theorem basically says that every degree of freedom that contributes quadratically to the energy will then contribute $\frac{1}{2} k$ to $C_V$. These two spurious energetic degrees of freedom are what are giving you $C_V = \frac{9}{2} k N$ instead of $C_V = \frac{7}{2} k N$.
Just to show that I'm not making this up, let's do the integral. Define $\vec{Q} = \frac{1}{2}(\vec{q}_1 + \vec{q}_2)$ and $\vec{r} = \vec{q}_1 - \vec{q}_2$.
$$
\begin{align}
I = \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta \alpha (|\bar{q}_1-\bar{q}_2|-r_0)^2/2 } ~d^{3}q_1 ~d^{3}q_2
&= \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta \alpha (r-r_0)^2/2 } ~d^{3}Q ~d^{3}r \\
&= \left[ \iiint_{-\infty}^{\infty}~d^{3}Q \right] \left[ \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta \alpha (r-r_0)^2/2 } ~d^{3}r \right]
\end{align}
$$
The first integral gives a factor of $V$ as before. The second one is a little more complicated. The angular contribution is obviously $4\pi$, leaving
$$
I = 4 \pi V \int_0^\infty r^2 \mathrm{e}^{- \beta \alpha (r-r_0)^2/2 } ~dr
$$
This last integral isn't of the standard "useful Gaussian integral" form, and will not give a result that is exactly proportional to $\beta^{-1/2}$. However, in the limit of low temperature, it does approach this limit. Define $\tilde{r} = \sqrt{\beta \alpha} (r - r_0)$; then the integral becomes
$$
I = \frac{4 \pi V}{\sqrt{\beta \alpha}} \int_{-\sqrt{\beta \alpha} r_0}^\infty \left( \frac{\tilde{r}}{\sqrt{\beta \alpha}} + r_0 \right)^2 e^{-\tilde{r}^2/2} \, d \tilde{r}.
$$
In the low-temperature limit, we have $\beta \to \infty$, meaning that the lower limit of integration becomes $- \infty$ and the first term in the parentheses vanishes; thus, in this limit,
$$
I \approx \frac{4 \sqrt{2} \pi^{3/2} V r_0^2 }{\sqrt{\beta \alpha}} \propto V T^{1/2}
$$
as desired.
EDIT: The exact integral above can't actually be evaluated in closed form, but it can be expressed in terms of the normalized error function erf(x):
$$
I = \frac{4 \pi^{3/2} V}{\sqrt{2}} \left[ \left(\frac{r_0^2}{\sqrt{\alpha \beta}} + \frac{1}{(\alpha \beta)^{3/2}} \right)\left( 1 + \text{erf} \left( \frac{r_0 \sqrt{\alpha \beta}}{\sqrt{2}} \right) \right) + \sqrt{\frac{2}{\pi}} \frac{r_0}{\alpha \beta} e^{-\alpha \beta r_0^2/2}\right].
$$
Note that if we set $r_0 \to 0$, we recover your result above (with $I \propto T^{3/2}$.) However, for non-zero $r_0$, we get a leading-order result proportional to $\sqrt{T}$, and a leading-order correction proportional to $T^{3/2}$ (as well as even smaller corrections proportional to $e^{-\alpha \beta r_0^2/2}$ times various powers of $T$, arising from the exponential term and the asymptotic expansion of the erf function.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
A way to determine if a body accelerates or loses speed at a certain time With given vectors for acceleration and velocity, is there a way to determine if a body accelerates or decelerates at a certain time-interval? Can this be determined, for instance, by simply observing the vectors?
|
With given vectors for acceleration and velocity, is there a way to determine if a body accelerates or decelerates at a certain time-interval?
Given the velocity vectors of a (specific) body (with respect to specific members of a suitable reference system) throughout a trial, in particular the velocity vectors $\mathbf v_{\text{initial}}$ at the beginning of the trial and $\mathbf v_{\text{final}}$ at the end of the trial, then
*
*the body is said to have "gained speed" (on average, over the course of the trial) if $\| \mathbf v_{\text{final}} \| \gt \| \mathbf v_{\text{initial}} \|$,
where the double bars denote the magnitude (norm) of the vector; and
*the body is said to have "lost speed" (on average, over the course of the trial) if $\| \mathbf v_{\text{final}} \| \lt \| \mathbf v_{\text{initial}} \|.$
Can this be determined, for instance, by simply observing the vectors?
No, not merely "by observing". Instead, the magnitudes of vectors, in comparison to each other, are to be measured; i.e. derived from observational data, to obtain unambiguous commensurate values.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Did the Sun form around a solid core? When Jupiter formed I assume like the other planets it started as tiny clumps of matter that eventually came together, became gravitationally bound and then eventually captured a lot of gas. I've also heard it was capable of collecting a lot of solid ice due to its distance from the Sun. Anyway, if Jupiter were larger we might be living in a binary star system. So, my question then becomes, did the Sun have a similar beginning to Jupiter and in what way was it different? Did the Sun form around a solid core?
| The Sun did not form around a solid core. Rather, it seems to have formed from a cloud of collapsing gas that may have been further enriched by matter from a nearby supernova. Gravitational force caused the collapsing cloud to start spinning, and the spinning compressed it into a disc with a bulge in the center that became the Sun. Here is a better explanation: http://image.gsfc.nasa.gov/poetry/ask/a11379.html.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
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Why are most antennas in cellular networks +/- 45° polarized? I've just been asked a strange question that I cannot find an answer to (even on the internet it seems I can't find any explanation for this) and I ended up wondering why most of the antennas which work as "base stations" are nowadays +/- 45° polarized. While I do understand the meaning of polarization diversity, it is still not clear to me why antennas' producers chose those 2 polarizations instead of horizontal / vertical or other 2 orthogonal angles whatsoever. Moreover, I am sure that the radiators inside an antenna cannot perfectly have a +/- 45° polarization. What happens then if those are "rotated" a little bit (let's say +55° / - 35°)? Is the orthogonality the only important thing, or is a correct orientation (+/- 45°) important too and why is that?
Bonus question: I think that the polarization of mobile phones has nothing to do with the polarization of base station antennas (because of reflections and multi-paths their transmissions' polarizations could be received "rotated" in comparison to how it was sent), is that a correct assumption?
EDIT: I can't really remember the source, but I read somewhere that we don't use H-Pol, because the ground greatly attenuates the field in that case.
| A motive is: "Seen from the front or rear of a handset user, the polarization will be dominated by the vertical component (the most handsets have a linear polarization because they use patch or monopole antennas), but in the lateral direction a handset is typically held at a large angle to the vertical, between the mouth and the ear of the standing or seated user, typically at least 45°."
Antennas for base station in wireless communications, Zhi Ning Chen and Kwai-Man Luk. pp. 40
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why must heat supplied in the definition of entropy be reversible? Can't it be irreversible after all it is a state function? The definition of entropy contains the term $Q_\text{rev}$ which means the heat supplied or taken out reversibly. I thought yes it can be after all only the initial & final states are important as entropy is a state function irrespective of the process heat is transferred. However I was baffled when I first read Clausius' theorem where it is written that $dS \geq \dfrac{Q}{T}$. If $Q$ is transferred irreversibly, then $dS$ is greater than $\dfrac{Q_\text{irrev}}{T}$; if the heat transfer is reversible, then only $dS$ equals $\dfrac{Q_\text{rev}}{T}$. So, does that mean entropy depends on the process heat energy is transferred?? Then, how can it be a state function? Where am I mistaking? Please explain.
| For an infinitesimal heat transfer $\delta Q$ the inequality of Clausius states that
$$\Delta S = S_1-S_0 = \int_0^1 {\dfrac {\delta Q_\text{rev}}{T}} > \int_0^1 {\dfrac {\delta Q_\text{irrev}}{T}}$$
Here $\delta Q_\text{rev}$ and $\delta Q_\text{irrev}$ denote reversible and irreversible heat transfers, respectively. Thus if the process is reversible and we know what $\delta Q = \delta Q_\text{rev}$ is at each step then we can calculate the entropy change from the integral $\Delta S = \int_0^1 {\dfrac {\delta Q}{T}} $. But if the process is irreversible the integral $\int_0^1 {\dfrac {\delta Q}{T}}$ only gives a lower bound for the entropy change not the actual change.
The difference between the entropy change and the integral is the internally generated entropy by the process $$\Delta S - \int_0^1 {\dfrac {\delta Q}{T}} = \sigma_\text{irrev} $$ and is characteristic to it. For example, a resistor $R$ with a dc current $I$ through it and kept at constant temperature in steady state generates $\dot \sigma = \dfrac {I^2R}{T}$ entropy per unit time and sheds the same to its environment along with ${\dot q = I^2R}$ heat flux.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is double-slit experiment dependent on rate at which electrons are fired at slit? I am a mathematician and I am studying string theory. For this purpose I studied quantum theory. After reading Feynman's book in which he described the double-slit experiment (Young's experiment) I was wondering if I send one electron per day or per month (even more), could I see the interference pattern?
| When a time interval between photons or other particles that bombard a foil with two slit is more than milliseconds, the phenomenon of the double slit diffraction/interference will be absent.
For example, in a paper:
V. Krasnoholovets, Sub microscopic description of the diffraction phenomenon, Nonlinear Optics, Quantum Optics, Vol. 41, No. 4, pp. 273 - 286 (2010); also http://arxiv.org/abs/1407.3224
the author describe a mechanism of the diffractionless of photons on one pinhole/slit in the case of a very low intensity of statistically single photons. NOTE: the author describes REAL EXPERIMENTS.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Where does the $(\ell + x)^2\dot\theta^2$ term come from in the Lagrangian of a spring pendulum? I am reading some notes about Lagrangian mechanics. I don't understand equation 6.9, which gives the Lagrangian for a spring pendulum (a massive particle on one end a spring).
$$T = \frac{1}{2}m\Bigl(\dot{x}^2 + (\ell + x)^2\dot{\theta}^2\Bigr)\tag{6.9}$$
I don't understand where the component $(\ell + x)^2\dot{\theta}^2$ is coming from. If we say the $x$-component is radial and $y$ is tangential, so we have according to this $\vec{v}^2 = v_{x}^2 + v_{y}^2$, then $y = (\ell + x)\sin\theta$ by small angle approximation we have $y = (\ell + x)\theta$, but then if we choose this coordinate system then $V(x,y)$ equation doesn't make sense specifically the potential from gravity! If someone could shed some light into this that would be nice.
| I'm not sure why you're talking about an $x$ and $y$ component of velocity when you're working in a polar coordinate system. Maybe you're confusing $x(t)$ (the extension of the spring as a function of time) with the Cartesian coordinate $x$. These are very different things. To understand what the radial component of the velocity is, assume the pendulum isn't swinging ($\dot{\theta}=0$). In this case you know the speed of the mass is $$\frac{d}{dt}(\ell+x)=\dot{x}$$ since $\ell$ is a constant. This gives the radial component of the kinetic energy $$T_{r}=\frac{1}{2}m\dot{x}^2$$
Now assume the pendulum is swinging but there is no radial motion ($\dot{x}=0$). We know that the kinetic energy of an object in circular motion is $$T_{\theta}=\frac{1}{2}mv^2$$ where $v$ is the tangential speed of the object and $m$ is its mass. In this case, we know the tangential speed is $$v=\dot{\theta}r$$ where $r$ is the radius of the circle traced out by the path of the object. Since we're considering the case with $\dot{x}=0$, we have a well defined constant radius given by $r=\ell+x$. This gives the radial kinetic energy $$T_{\theta}=\frac{1}{2}m\dot{\theta}^2(\ell+x)^2$$ The important thing to note is this kinetic energy relation holds even with a changing radius. The Lagrangian is a function of time anyway (although not explicitly in this case), so this isn't a problem. You can then take the kinetic component of the Lagrangian to be the sum of these kinetic energy relations to give $$L=\frac{1}{2}m\Bigl(\dot{x}^2 + (\ell + x)^2\dot{\theta}^2\Bigr)-V$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If gravity were to suddenly change, would the lift generated by a airfoil also change If gravity were to suddenly change, would the lift generated by a airfoil also change?
I realise that if gravity were to increase, then weight would also increase, leading to a change in the resultant force on the object (say a aeroplane). However, would the lift itself change, or just the resultant force.
I'm assuming it would due to possible change in fluid density, but am not sure.
If it were to change, would the opposite affect happen if gravity suddenly decreased?
| If gravity changes, then so will the density of air. Air pressure at the surface is proportional to the weight of the column of air above it, so if gravity increases, the pressure would, too. That will in turn affect the lift from an airfoil. NASA says lift varies linearly with density. I suspect the two effects would pretty much balance each other.
| {
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"url": "https://physics.stackexchange.com/questions/194787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does a propeller being pushed through the air have more friction if free to spin? I just passed a boat being pulled on a trailer and noticed the propeller spinning in the wind. I began considering the possibility that a propeller free to spin might cause less friction against the forward motion of a force pushing the propellar forward.
If that were the case, we could create a race car where the passenger area is free to spin as a propeller, shaped as a screw sort of, with the inside spinning at a constantly opposite speed so the driver himself didnt spin, but the vehicle's body gained improved aerodynamic performance.
So, what's the science behind this? Does a spinning device like a propeller experience less friction than an identical shape unable to spin?
| The net drag force on the surface of the propeller is a function of the relative velocity of the fluid against the propeller's surface.
The simple model for drag force that can be (loosely) applied to the question regarding the propeller is
$$F_d=C_dA\frac{1}{2}\rho v^2$$
where $v$ is the relative velocity, $A$ is the projected area normal to the direction of $v$, $\rho$ is the fluid density, and $C_d$ is a linearized drag coefficient particular to the geometry considered.
In either case (propeller fixed or rotating under the wind's force) receives the same incoming velocity from the the trailered boat's speed. But at the surface of the propeller the relative velocity is reduced in the case of the spinning propeller which modes away from the direction of the incoming wind, thus reducing the relative velocity.
Therefore the moving propeller you observed spinning indeed reduces friction over what would occur if the propeller remained stationary.
NOTE: although drag forces were reduced, nothing was really gained in terms of energy expenditure. In other words by having the propeller spinning you don't take any less load of the engine that is pulling the boat trailer. Initially the higher wind frictional forces were required to accelerate the propeller into a spin, and at steady state there is momentum exchange to maintain the spin. The terminal (spinning) velocity is an equilibrium of forces - that which the wind supplies, and that what is consumed in the bearing friction of the propeller shaft. So the shaft frictional losses see some of the energy flow that was once all in the energy lost to drag over the still propeller. The total energy losses are the same. Sorry- no perpetual motion machine.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194897",
"timestamp": "2023-03-29T00:00:00",
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What is the viscosity difference between a solid and a liquid The pitch drop experiment, for example, shows bitumen as a liquid, even though it appears to be a solid, and then there is the "glass: solid or liquid" debate. Is there a numerical value in viscosity that defines the boundary between a solid and a liquid?
Or is there another numerical identifier that separates the two?
| No.
The elasticity of the solids is a liquid kind of character. The non newtonian fluids provide a solid kind of character.
As everything is just electromagnetism, you can split your thoughts down the an single atom level, and play it with magnets on real scale. You can push your hands between magnets, and you can make them flow.
In the wikipedia article about Cold welding Richard Feynman explains this well;
The reason for this unexpected behavior is that when the atoms in contact are all of the same kind, there is no way for the atoms to “know” that they are in different pieces of copper. When there are other atoms, in the oxides and greases and more complicated thin surface layers of contaminants in between, the atoms “know” when they are not on the same part.
— Richard Feynman, The Feynman Lectures, 12–2 Friction
If you bend a steel the atoms are changing their positions just like they do in a fluid. There is no way to define exactly which is the exact difference between solid and fluid, which would be exactly valid for all materials. You can't barely do it to even a single material; You think that water freezes at 0 °C? Think again, as you might have it liquid down to -48 °C - at Standard pressure!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/194948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Is it normal for radiation levels to be elevated near a medical imaging lab? I work in a general-purpose, commercial office building where the first floor is dedicated to an x-ray / imaging lab for medical diagnostics. The public lobby of this building is routinely experiencing radiation level increase from 0.10 uSv/h (Background) to 0.96 uSv/h in approximately 20 - 30 second intervals as measured by a geiger counter that calculates uSv/h in 10 second increments. To put it another way, my geiger counter typically reads 0.10 uSv/h, but when walking past the lobby I often see the reading increase to 0.96 uSv/h for several cycles. My assumption is that radiation is leaking from the imaging lab and into the lobby as x-rays are being taken, causing accidental exposure to members of the public (myself included).
Does anyone know if this is to be expected or otherwise have insight into whether or not such relatively small dosages are harmful to the public?
| I don't think 0.96 $\mu$Sv/h or 365.25$\times$24$\times$0.96 = 8.5 microsieverts/year is a problem. To put this in perspective,
*
*Every year, we receive natural radiation : 2,000 microsieverts
*CT scan : 7,000 microsieverts
*Additional radiation in a life time for those living around Fukusima : 10,000 microsieverts
*US radiation worker in a year : 50,000 microsieverts
*Astronaut in the International Space Station for 6 months : 80,000 microsieverts
*In a smoker's lung in a year : 160,000 microsieverts
Therefore, don't worry about the amount of radiation leaking from your lab. It is very very small.
P.S. I get this number from a Veritasium's video called "The Most Radioactive Places on Earth".
| {
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Force division of moving pulleys? I am a second grade at a middle school and I was reading a physics workbook to prepare for a test. And I was solving pulley problems and one problem made my brain stop.
The problem asked me what would the minimum force of F would be when the weight of the pulleys were 30N. I checked the answers and the way to solve it. The workbook told me that the force on each string holding the moving pulley equals to 1/n. (n = the numbers of string holding the pulley) Why is it true?
| This is a mechanical advantage problem. For fixed pulleys, only the direction of motion is changed, and there is no mechanical advantage. A 1 N force directed downward on one side of the fixed pulleys (the small ones) produces a 1 N force directed upward on the rope on the other side of the fixed pulleys.
For a movable pulley (the large one), there are two supporting strands on it, so the mechanical advantage is 2. Each strand supports 1/2 of the total weight, which is the weight of the load and the weight of the pulley. The mechanical advantage multiplied by the force F must equal the total weight. This means that F equals (450N + weight of the large pulley)/2. If the weight of all the pulleys taken together is 30 N, the question can't be answered. However, if the weight of just the large pulley is 30 N, the answer is F = 240 N.
Note that I don't have to include the weight of the small pulleys because their weight is supported by the beam that they are attached to.
Regarding why this is true:
Work = Force x distance.
For the movable pulley, F is 1/2 of the total weight, since there are two supporting strands (n=2). When you pull up on one of the ropes supporting the movable pulley, and you pull 2 m of rope through the pulley, the load only rises 1 m, because you remove 1 m of rope from each side of the movable pulley. Work = Force X distance for you when you pull up on the rope, and work = force X distance for the weight that is moving up. This means that work into the device equals work coming out of the device, which (assuming an efficiency of 100%) is a requirement if energy (work) is going to be conserved.
If you have the time to set this pulley arrangement up in a tutorial session, or your teacher gives a demo, the above explanation will make somewhat more sense. Good luck on your test.
| {
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What would happen if we tried to run a motor in space when it is not attached to anything to provide support to it? I know that a when a motor runs it generates torque and that torque can be used to do useful work. On the other hand, the motor needs strong support that absorbs the reaction torque. In our case let us assume that that support is provide by a workshop floor on which the motor is firmly attached ( The workshop floor is essentially the earth) The earth receives the reaction torque and being the massive object it is, it doesn't move.
Now let us imagine that we took our motor into a space where there is no gravitational field. What would happen if we tried to run the motor. Assuming the motor is powered by a battery pack. The battery pack and its control electronics are neatly packed around the stator.
Would the motor rotate at all? Would the rotor and the stator rotate in opposite directions?
Would there be a transfer of energy from the batteries to rotational mechanical energy
| Your intuition was correct - the shaft will rotate in one direction and the housing/stator will rotate in the other.
If you look up "moment of inertia" you will find that it is the rotational equivalent of mass. For almost any reasonable motor the moment of inertia of the shaft/rotor windings will be smaller than the moment of inertia of the housing/stator. Since the total angular momentum of the motor must be zero (it has nothing to react against), the rotational speed of the shaft will be greater than the rotational speed of the housing.
As an equation, let $\omega _s$ be the angular velocity of the shaft, $\omega _h$ be the angular velocity of the housing, and $I_s$ and $I_h$ the moments of inertia of the shaft and housing. Then$$I_s \omega _s = -I_h \omega _h$$ where the difference in signs reflects the opposite rotation directions of the two, and $$I_s \omega _s + I_h \omega _h = 0$$
Assuming $I_s < I_h$, then $\omega _s > \omega _h$
| {
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Is Gauss' law valid for time-dependent electric fields? The Maxwell's equation $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r})=\frac{\rho(\textbf{r})}{\epsilon_0}$ is derived from the Gauss law in electrostatics (which is in turn derived from Coulomb's law). Therefore, $\textbf{E}$ must be an electrostatic field i.e., time-independent. Then how is this equation valid for the electric field $\textbf{E}(\textbf{r},t)$ which is time-dependent (for example, the electric field of an electromagnetic wave)? Can we prove that $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r},t)=\frac{\rho(\textbf{r},t)}{\epsilon_0}$ ?
EDIT: I have changed $\boldsymbol{\nabla}\cdot \textbf{E}=0$ to $\boldsymbol{\nabla}\cdot \textbf{E}=\frac{\rho}{\epsilon_0}$ in the question.
| No we cannot prove it; Maxwell postulated that it would hold dynamically because it made the most sense for it to do so as he pondered the displacement current problem. As you likely know, Maxwell pondered the inconsistency between Ampère's law for magnetostatics and the charge continuity equation. Ampère's law for magnetostatics reads $\nabla\times \vec{H}=\vec{J}$; when we take the divergence of both sides of this equation we get $0=\nabla\cdot\vec{J}$ for any magnetic field with continuous second derivatives. This violates the charge continuity equation; we need $0=\nabla\cdot\vec{J} + \partial_t\,\rho$. So we need to add a term to the right of Ampère's law in the dynamic case whose divergence is the charge density $\rho$. The easiest solution is to assume that Gauss's electrostatics law holds in the dynamical case: then we add the electric displacement to the RHS of Ampère's and it has the right divergence to make everything properly in keeping with the continuity equation. Note that we can also add an arbitrary vector of the form $\nabla\times \vec{N}$ to the electric displacement for this to work, but this degree of freedom doesn't affect Gauss's law.
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Question about Planck's constant
*
*How did Dirac or whoever came up with it know that the momentum operator in quantum mechanics is $-i\hbar\frac{d}{dx}$?
*How did he know the $\hbar$ was in there?
*How did all these physicists know they had to put $\hbar$?
*Does it go back to Planck's paper on blackbody radiation?
*I know it is to make the units correct but why couldn't they use another constant with the same units as $\hbar$, like a constant which was a little bigger than $\hbar$?
*Also I know that $\hbar = \frac{h}{2\pi}$ but basically I am asking why $h$ and not some other constant?
| Note: This is a brief summary.
Wikipedia is helpful, if you can't look anywhere else at the moment. It notes that
$$\hat{p}=-i\hbar\frac{\partial}{\partial x}$$
because of the de Broglie relation,
$$p=\hbar k$$
where $p$ is momentum and $k$ is the wave vector.
de Broglie's equations, in turn, relate to the de Broglie wavelength,
$$\lambda = \frac{h}{p}$$
which follows from a variant of the Planck relation.
That's the very, very, very, very basic way to look at it. A proper treatment would, of course, be found in a proper textbook on the subject.
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Where does the equation $p=\frac{1}{c}\sqrt{T^2 +2mTc^2}$ come from? Where does the relativistic formula
$$p~=~\frac{1}{c}\sqrt{T^2 +2mTc^2}$$
come from? What is the derivation from Einstein's formula? $T$ is the kinetic energy $m$ is the mass $p$ is the momentum.
| We can write total energy $E$ two ways:
\begin{equation}
E^2=p^2c^2+m^2c^4 \\
E=T+mc^2,
\end{equation}
where $T$ is kinetic energy.
Eliminating $E$ from those two equations will give you the desired result.
| {
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Why does humidity cause a feeling of hotness? Imagine there are two rooms kept at the same temperature but with different humidity levels. A person is asked to stay in each room for 5 minutes. At the end of experiment if we ask them which room was hotter, they will point to the room with the higher humidity. Correct right?
How does humidity cause this feeling of hotness?
| For the thought experiment in your question, the answer is: it depends on the temperature of the room!
Cold Room
If the room is cold, humidity will actually make it feel colder! This is because water vapor has a much higher heat capacity than dry air, meaning that it takes more heat to raise or lower its temperature. So a volume of air with a lot of water vapor can transfer more heat to (or from) your skin than the same volume of dry air.
Warm Room
If the room is moderately warm (around 27–37 °C, I think), it's close enough to your own body temperature that radiative heat transfer is minimal, so the main difference in perceived heat is due to the water vapor's effect on your body's ability to use evaporation (of sweat) to cool itself. This is what the other answers are talking about.
Hot Room
If the room is hot, heat capacity again plays the dominant role. A given volume of high-humidity air contains much more heat than an equal volume of dry air, so it is able to transfer a lot more heat to your body. This is why dry saunas are frequently operated in excess of 100 °C, while a steam sauna at that temperature would be instantly scalding!
In a hot room there is also the issue of condensation. If your skin is at or below the dew point, water will condense out of the air onto you. Condensation is an exothermic phase change, so it will transfer additional heat to your skin.
| {
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In general, why do smaller guns have more felt recoil? Why is recoil easier to control on a more massive gun compared to a smaller gun with the same bullet. Presumably the bullet leaves both guns with the same momentum, but the larger gun seems easier to control. Since the momentum you have to control is the same in both cases, why do we perceive less recoil on a bigger gun?
| Felt recoil is partly a matter of momentum, partly a matter of force.
When a bullet with mass m leaves a gun with a velocity v, the gun must have an equal-but-opposed momentum MV, where M is the mass of the gun and V is the recoil velocity, or $$mv + MV = 0$$. If there are two possible gun sizes, $M_1$ and $M_2$, each will have a recoil velocity $V_1$ and $V_2$. If, for instance, $M_2 = 2M_1$, $$M_1 V_1 = M_2 V_2$$ and $$V_1 = 2V_2$$ Why does this matter? Consider kinetic energy. Let $K_1$ be the kinetic energy of $M_1$, and $K_2$ is that of $M_2$. Then $$\frac {K_1}{K_2} = \frac {\frac{M_1{V_1}^2}{2}}{\frac{M_2{V_2}^2}{2}} = \frac{M_1}{M_2} {(\frac{V_1}{V_2})}^2 = \frac{1}{2} 2^2 = 2$$
So the lighter gun has twice the kinetic energy of the heavier gun and this shows up in two ways. First, since both guns need to stop in about the same distance, the force applied to the lighter gun must be greater than that applied to the heavier. By Newton's First Law, this means that the lighter gun pushes harder on the hand or shoulder of the shooter. Second, the duration of acceleration must be smaller for the lighter gun, since $$S_1 = \frac{a_1{t_1}^2}{2} = \frac{{V_1}t_1}{2} = S_2 = \frac{a_2{t_2}^2}{2} = \frac{{V_2}t_2}{2} $$and $${V_1}{t_1} = {V_2}{t_2}$$ or $$ \frac {t_1}{t_2} = \frac{V_2}{V_1} = \frac{1}{2}$$ So not only is the recoil force greater for the lighter gun, it lasts a shorter time and is therefore "sharper".
| {
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How and why the phrase "quark force increases with distance"? I have seen that phrase "force between quarks increases with distance" at many resources, some even relatively credible (albeit written for general audience).
What is the reason behind that, when the area law for the confining phase clearly gives the potential energy as increasing linearly with distance and hence the force being a constant?
| Perhaps the resources you have seen are confusing "force" with "energy", which is a common misunderstanding that frequently leads to mistakes in terminology. Energy must be applied to overcome the color force and increase the distance between quarks. The formation of a new quark/anti-quark pair when the gluon field becomes too stretched may provide an opportunity to confuse the constant strong force field with the increasing potential energy necessary to overcome that field.
Another source of confusion may be a misunderstanding of the color field, thinking it extends throughout space like an electric or magnetic field, rather than being confined to a narrow string-like gluon field between quarks.
For what it's worth, the Wikipedia article on color confinement says that the strong force between a quark pair "acts constantly - regardless of their distance...", citing T. Muta, Foundations of Quantum Chronodynamics (2009), and A. Smilga, Lectures on Quantum Chromodynamics (2001), so they all appear to have stated it correctly.
| {
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Travelling near the speed of light, will radar give me sufficient warning of approaching debris? If I’m travelling in space near the speed of light, I would like to use radar to give me advance warning of approaching hazards. However, will I have enough time to be able to react?
From a perspective on the spaceship, the radar beam travels away from me at the speed of light, so I might imagine that I would have ample warning of an object positioned at rest one light-hour away. However, observed from a position of rest, the radar beam is just slowly pulling away ahead of the spaceship, and the radar pulse would arrive at the obstruction only shortly before the spaceship.
I suspect therefore that I would not have enough time to react, but can someone shed some light on this for me please?
| Per Bort's comment, this is easier to think about from the spaceship's frame rather than the debris's frame. A rock is traveling toward your spaceship at .99 times the speed of light. You send out a pulse of light that intercepts the rock when it's one light-minute away from the ship. The pulse bounces back and arrives at the ship 1 minute later. The rock arrives another $1/.99-1\approx .01$ minutes after that. That's how much warning you get.
| {
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Conical train wheels I've been reading about how the conical shape of train wheels helps trains round turns without a differential. For those who are unfamiliar with the idea, the conical shape allows the wheels to shift and slide across the tracks, thus effectively varying their radii and allowing them to cover different distances while rotating at the same angular velocity.
A cross-sectional view of the tracks and wheels generally looks something like:
But what about a configuration like the following?
I read in an online article that wheels in the second configuration may more easily slip and derail from the tracks (assuming there are no flanges to prevent them from doing so). But I can't convince myself using physics why that might be.
Is one of these two configurations actually more reliable than the other?
| Shift the upper configuration to the left a short distance at equilibrium. Result: the left wheel goes a little up, the right goes a little down, the train tilts clockwise, the center of mass is to the right of the centerline between the wheels, and therefore the center of mass provides a restorative force to push the train back to the right.
Shift the lower configuration to the left a short distance at equilibrium. The argument proceeds in reverse and the center of mass provides an anti-restorative force, pushing the train further to the left. Pain ensues.
You're trading $m \ddot x = - k x$ (harmonic oscillator) for $m \ddot x = k x$ (exponential diverger) and praying that the implicit drag forces keep the thing diverged only a small amount. That's a risky game, no doubt.
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Why is the index of refraction different for different wavelengths? The index of refraction can be written as
$$n=\frac{\lambda_v}{\lambda_m}$$
where $\lambda_v$ is the wavelength in a vacuum and $\lambda_m$ is the wavelength in the medium. I’ve been told that since wavelength appears in the definition of an index of refraction, an index of refraction varies with wavelength. However, why would that be the case? The index of refraction is a ratio; if a wavelength of one wave is different from that of another wave passing through the same medium, the index of refraction should not be different for each wave, since they would have had different wavelengths in a vacuum too. So why is the index of refraction dependent on the wavelength?
| You are mixing up two different things.
The refractive index is usually defined in terms of the velocity of light:
$$ n = \frac{c}{v} $$
where $v$ is the velocity in the medium. However the velocity is related to the frequency and wavelength by:
$$ v = \lambda f $$
so:
$$ n = \frac{\lambda_0 f_0}{\lambda f} $$
The frequency of the light, $f$, doesn't change as the light enters the medium, so $f_0 = f$ and the $f$s cancel to give:
$$ n = \frac{\lambda_0}{\lambda} $$
which is the equation you cite. The equation is not based upon any assumptions about the variation of $n$ with the frequency/wavelength of the light.
However the refractive index does change with frequency. This effect is called optical dispersion. The cause is the way the interaction of the light and the electrons in the medium change with frequency. See for example the questions Why do prisms work (why is refraction frequency dependent)? and Why does the refractive index depend on wavelength?.
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What determines the probability of creating a particular particle in a collision? When discussing events at the quantum level, we deal in probabilities and not absolutes. Articles I've read on particle physics state that a particle has a probability of being created in a collision. What determines this probability?
Assuming we have the energy and other criteria met, which would allow us to create range of particles (please feel free to expand what particles would make sense for an example), why do certain particles have a higher probability of being created than others?
| Once one specifies a quantum field theory, typically in the form of a Lagrangian density, one can calculate the probabilities of various outcomes in collisions.
A quantum field theory is a theory based on fields that obeys quantum mechanics and special relativity. The so-called Standard Model is perhaps the most famous quantum field theory, and certainly the most successful in reproducing experimental observations.
The probability of a particular outcome is related to the number of ways that that outcome could happen. Indeed, as in famous double-slit experiments, possible intermediate states could interfere constructively or destructively. Some ways that an outcome could happen are more probable than others. This typically happens because one outcome proceeds by an interaction with a bigger coupling constant $g$, or with particles that are lighter.
The full details of these calculations are rather involved. You can make rough guesses for which outcomes are most probable by considering the power of the coupling constant that the probability is proportional to, i.e. $g^n$. If an outcome requires lots of interactions, and so high powers of $n$, it will be improbable, because $g<1$. You must also consider "phase-space" - if there are few ways in which a final state could satisfy energy-momentum conservation, the total probability for that final-state will be small.
| {
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Can a magnet damage a compass? I've heard the claim before that a magnet can ruin a compass, and was about to repeat it to my son when I realized it sounds like complete nonsense. Googling turned up such unsubstantiated and illogical answers as this one and unanswered questions as this one but nothing that sounded reasonable to me and gave a convincing explanation. Perhaps my Google bubble is at work.
Anyway, since SE is generally very reliable, I thought this was the right place to ask, before I pass on untested nonsense to my son. Help me break the chain of untested pseudoscience via oral tradition: does a magnet actually do permanent damage to a compass, or just temporarily prevent it from detecting magnetic north?
If it actually does do this, please explain how that is so.
| If the field is big enough it can physically destroy the compass needle. But that might be the least of your problems at that point
| {
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Wavelength vs Wavenumber etiquette When am I supposed to use the terminology of EM "wavenumber", instead of "wavelength" (or frequency)?
The concepts of wavelength and frequency are no problem for me, but wavenumber (number of wavelengths per unit length) seems redundant to me as a student engineer and proto-physicist. And then there's use of energy levels at higher frequencies.
| Traditionally wavenumber is used in molecule spectrums such as infrared spectrums in organic chemistry where it is given in the incoherent SI-unit $\textrm{cm}^{-1}$. Mostly because one obtains convenient numbers on the axis. Also in most of the wave equations it is used, because again you can make the convenient substitution $k \equiv \frac{2\pi}{\lambda} = \frac{p}{\hbar}= \frac{\sqrt{2 m E }}{\hbar} $ which is commonly done in solving the Schrödinger equation for simple boundary constraints. Thus it depends on the exact context if you would use the first or latter.
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Pointwise and uniform convergence. Examples from physics I am a first-year mathematical student, and from a mathematical perspective I understand the difference between pointwise and uniform convergence of sequences and series of functions. However, I have been wondering about what phenomenons from physics (or other sciences using math) are described by sequences or series that converge pointwise, but not uniformly. Can you give any such examples? Thank you.
| Sciences use mathematics only as a tool. In almost all such applications, mathematical problems (such as pointwise vs uniform convergence) are not inherent to the scientific problem at hand, but arise from the mathematical model and are indicative of its limitations.
For example, when modelling a large collection of particles (be it in a solid (crystal) or fluid), one may describe their distribution via Dirac $\delta$-functions. Mathematically, this has certain issues (they are strictly not functions in the mathematical sense and not differentiable), which occur precisely because real particles are not point-like. But the point-like particle model may still be very useful in many applications.
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Is wave-particle duality not clear from the single-slit experiment? In experiments it is easy to discern between 2 and more-than-2 fringes on a screen, making the double-slit experiment the default one for wave-particle tests.
Let's say we shoot massive particles (e.g. electrons) towards a slit. Would the image behind it be the same no matter if we consider the electrons to be classical particles or wave-packets?
My interpretation, using an ideal (infinitely-narrow) slit, is that the (interpretation of the particles as) classical particles would produce an image with sharp boundaries, while a wave would imprint a gaussian-like distribution on the screen.
| There is still interference at a single slit resulting in a Fraunhofer pattern. Just consider both edges of the split as starting point of a new wave.
Generally you're right. But, in a single slit, the electrons could still be deflected by the atoms that make up the slit. This - I think - leaves more room for discussion than the double-slit. It is propably just a matter of what resonated with people first and would gives students the least amount of headaches..
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Are Hubble Telescope Images in true color? Like many others, I have marveled at the images made available from the Hubble Space Telescope over the years. But, I have always had a curiosity about the color shown in these images. An example is shown below. Are the colors we see, such as the yellows, blues, and so on the true colors or are they applied by some kind of colorization method to enhance the image quality for realism.
| The classic color mapping for Hubble is described in Flase-color astrophotography explained.
What you have is (in the Hubble palette):
Line Freq True False
Ha (656.3 nm) Red -- Green
S-II (672.4 nm) Red -- Red
O-III (500.7 nm) Green -- Blue
An example of this for true color from John Nassr at Stardust Observatory at Coming to Life Again…. Hubble’s 25th Anniversary Honored with New Hi Definition Photo of Pillars of Creation in the Eagle Nebula:
And false color from wikimedia commons (the pallet that of the Hubble)
You will note that the true color image is dominated by red - which is the Hydrogen alpha, and Sulfur-II lines (which show up as reds, greens and yellows in the false color).
The colors in astrophotography are specific remappings of frequencies into other frequencies so that our eyes are capable of perceiving the subtle differences between different emissions from elements. Otherwise, we have difficulty seeing the structure that is there and indicated by different elemental densities in the nebulae.
I will point out that the H/S/O palette given above isn't the only one. The image in the question has a link to the fastfacts tab which states:
The image is a composite of separate exposures made by the WFPC2 instrument on the Hubble Space Telescope. Three filters were used to sample narrow wavelength ranges. The color results from assigning different hues (colors) to each monochromatic image. In this case, the assigned colors are:
F658N ([N II]) red
F656N (H-alpha) green
F502N ([O III]) blue
In that image, the doubly ionized nitrogen is used as red rather than the doubly ionized sulfur in the Eagle Nebulae picture.
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How do I know what variable to use for the chain rule? In my textbook the tangential acceleration is given like this:
$$a_t=\frac{dv}{dt}=r\frac{dw}{dt}$$
$$a_t=rα$$
I understand that the chain rule is applied here like this:
$$a_t=\frac{dv}{dt}=\frac{dv}{dw}\frac{dw}{dt}=rα$$
What I don't understand is why we have to apply the rule in this specific way. Say I write like this:
$$a_t=\frac{dv}{dθ}\frac{dθ}{dt}$$
This way, I end up with entirely different result. How do I know how the chain rule must be applied?
| In general you are allowed to use any parameter $q$ to describe the motion of an object as $\vec{r}(q)$ where the parameter changes with time $q=q(t)$. The parameter can be an angle or a distance or any combination that best makes sense.
So now you have expressions for velocity and acceleration defined from the chain rule
$$ \vec{v} = \frac{{\rm d} \vec{r}}{{\rm d} t}
= \frac{{\rm d} \vec{r}}{{\rm d} q} \frac{{\rm d} q}{{\rm d}t} = \vec{r}\,' \dot{q} $$
$$ \vec{a} = \frac{{\rm d} \vec{v}}{{\rm d} t}
= \frac{{\rm d} \vec{v}}{{\rm d} q} \dot{q} + \frac{{\rm d} \vec{v}}{{\rm d} \dot{q}} \ddot{q} = \frac{{\rm d} (\vec{r}\,' \dot{q})}{{\rm d} q} \dot{q} + \frac{{\rm d} (\vec{r}\,' \dot{q})}{{\rm d} \dot{q}} \ddot{q} =\\
\vec{a} =\vec{r}\,'' \dot{q}^2 + \vec{r}\,' \ddot{q}$$
So what the book did is set $q=\theta$, $\dot{q}= \omega$ and $\ddot{q} = \alpha$. This results in the following for circular motion
$$\vec{r} = R\,\vec{n}(\theta)$$ where $\vec{n}$ is the normal direction away from the center (which varies by $\theta$). Now you have
$$ \vec{v} = \frac{{\rm d}R \vec{n}}{{\rm d}\theta} \omega = R \omega \frac{{\rm d}\vec{n}}{{\rm d}\theta} = R \omega \vec{e}$$
where $\vec{e}$ is the tangetial direction
$$\vec{a} = \frac{{\rm d}R \omega \vec{e}}{{\rm d}\theta} \omega +\frac{{\rm d}R \omega \vec{e}}{{\rm d}\omega} \alpha = R\omega^2\frac{{\rm d} \vec{e}}{{\rm d}\theta} +R \alpha \vec{e}=R \alpha \vec{e} - R\omega^2 \vec{n}$$
So the tangential component is $R \omega$ and the centripetal $R \omega^2$.
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What is the ratio of cosmic ray energy to dark matter energy in our solar system? In any cubic meter of space in our solar system there is predicted to be some amount of dark matter. Also in a cubic meter of space is a known average amount of cosmic ray energy. What is the ratio of cosmic ray energy to dark matter energy in our solar system? I'm curious if they are on the same order of magnitude or far from it.
| The integrated local interstellar cosmic ray energy density is claimed by Webber (1998) to be about 1.8 ev/cm$^3$. The energy density of dark matter (mostly rest mass energy) in the solar system is thought to be around $0.43\pm0.1$ GeV/cm$^{3}$ (Salucci et al. 2010).
Both numbers likely have quite big error bars, but as you can see, there is an 8-9 orders of magnitude difference.
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How to estimate if an image is in focus I am building a test measurement (optics) to look a rectangular slit opening (1mm x 15 microns). The slit opening is illuminated by a white LED and using a microscope objective to magnify it to 10X on an image sensor.
What would be the possible method to estimate if the image is in it's best focus?
Current method, using a line profile and look for peaking position.
Any other ideas ?
| Since you're imaging something that is basically a 2D rectangle function you would need an infinite amount of spatial frequencies reconstruct it in the Fourier domain.
You could take the FFT of the image and vary the image or object distance until you got the highest possible spatial frequencies.
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Why aren't calculation results in error propagation at the center of the range? We have two copper rods, with $L_1$ and $L_2$ as their lengths respectively, and we want to glue the two bars together, with glue that's infinitesimally thin.
$$\begin{align}
L_1 &= 20 ± 0.2\ \mathrm{cm} \\
L_2 &= 30 ± 0.5\ \mathrm{cm}
\end{align}$$
To calculate the length of the composite bar, $L$, as well as its uncertainty, we can do the following (which I admit is a rather crude method, but is done for completeness):
$$\begin{align}
L_\text{MAX} &= 20.2 + 30.5 = 50.7\ \mathrm{cm} \\
L_\text{MIN} &= 19.8 + 29.5 = 49.3\ \mathrm{cm}
\end{align}$$
Therefore, $L = 50 ± 0.7\ \mathrm{cm}$.
This, although is a long method, is a correct as the length $L$ is just a sum of the values of $L_1$ and $L_2$ with an uncertainty of the range of values possible divided by two.
Now, if we want to calculate the area with the following length and width, as well as the uncertainty, we could use a method similar to the one described above:
$$\begin{align}
W &= 20 ± 0.2\ \mathrm{cm} \\
L &= 10 ± 0.2\ \mathrm{cm} \\
A_\text{MAX} &= (20.2\ \mathrm{cm})(10.2\ \mathrm{cm}) = 206.04\ \mathrm{cm}^2 \\
A_\text{MIN} &= (19.8\ \mathrm{cm})(9.8\ \mathrm{cm}) = 194.04\ \mathrm{cm}^2
\end{align}$$
In this case, the answer without the uncertainty, $10\ \mathrm{cm} \times 20\ \mathrm{cm} = 200\ \mathrm{cm}^2$, is not the center of our range of values. Although 200 isn't the smack center, there does exist a center, which in this case is 200.04.
The actual uncertainty of the area is 6, which is in fact half of the range of the maximum and minimum, giving us a final answer of $200 ± 6\ \mathrm{cm}$.
The way we have defined the propagation of uncertainties in physics is such that the answer is not necessarily the smack center of the minimum and maximum value range, but is instead the product of the two measurements, the two lengths in this case. This approach to the first problem made a lot of intuitive sense, however, I cannot understand why the final answer is 200 (which is not the center of the range) ± 6, and why this answer gives a different range of values than the range calculated using the long, crude method.
I am a high school student who has not covered calculus yet, which is what prevented me from understanding the proof of adding fractional or percentage uncertainties when we multiply or divide quantities. Any help will be greatly appreciated, thanks in advance.
| I think you should have a look in this http://ipl.physics.harvard.edu/wp-uploads/2013/03/PS3_Error_Propagation_sp13.pdf it should clear up how to do error estimation in a more rigorous manner. I would only worry about the first two sections on addition and multiplication if you haven't covered calculus yet.
This method of taking the maximum and minimum value and halving the difference will only result in a symmetric error if the operations you are dealing with are linear. So for example if I was to work out the value and error of some quantity $y=x^2$ and I know the value of $x$ and its error $\delta x$ can you see why the error in $y$ isn't symmetric under your method? Put in some numbers and try it for yourself.
If you have any follow up post it as a comment and I will try and answer you.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is heat conduction impeded at interfaces between dissimilar materials? Sound in air essentially echoes off concrete walls, rather than penetrating them, because of the difference in the material properties of air and concrete.
By analogy, are there pairs of solid materials where their interface would be very inefficient at propagating heat? Perhaps one material has heavy atoms and soft bonds and the other has light atoms and stiff bonds, and neither has free electrons.
If this phenomenon exists could it be used to create super-insulators, by laminating together large numbers of very thin layers of the two materials?
| Sound in air is a percussion wave - way different. Heat via conduction does not have an interface barrier. It would just be the sum of the resistance.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/199357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
thin film interference of light
In a thin film interference (reflective system) I know that condition for maxima is
$$2\mu t\cos(r)=(2n\pm 1)\frac{\lambda}{2}$$ and for minima is $$2\mu t\cos(r)=n\lambda$$ and for transitive system it's just the opposite.
but what happens if then film is very small such that $$\lim_{t \to 0}$$ i.e. thin film is too thin? My teacher told my that condition for minima is satisfied because then $\delta x = \lambda /2$ and hence film appears dark. How is this possible? and similarly what happens of film in too thick ? I am guessing interference does't happen then , but what would be explanation for it ?
| At the reflection in an interface from low to high refractive index, a phase shift of $\pi/2$ occurs; no such phase shift occurs on the second interface (high to low). As a consequence, for sufficiently thin films, there is indeed destructive interference- so your very thin film looks dark. You can sometimes see this on soap bubbles just before they pop - the go from shiny and colorful to (patches of) dull and colorless, usually at the top (as liquid is pulled to the bottom of the bubble by gravity).
An explanation of the phase shift is given in this answer - note that the premise of that specific question is backwards but the answer is correct...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Eddy Currents – Tubes with slits When a magnet falls down a tube, eddy currents form and flow around the tube, perpendicular to the direction in which the magnet falls.
However, when there is a vertical slit in the tube, are either no eddy currents formed (since they cannot complete a rotation), or alternatively do much smaller eddy currents form, as suggested by the following figure?
Further to this, a recent HSC (Australian Physics Examination) tested this,
The official answer to this question C. If it is true that small eddy currents are formed, why then is this the case?
| Probably in the last case of slit direction of magnetic field is on the side walls upwards as shown and thus there is no interaction of this field with the magnet. Here it seems it is to be considered to neglect radial magnetic field (even though it do exist). In complete loop magnetic field is centralized so it interacts with the magnet making it go really slow.
| {
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"timestamp": "2023-03-29T00:00:00",
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Any fractal physical model that generates time series which demonstrate heavy-tailed (non-Gaussian) behavior in some form? I know that fractal structures have power-laws in various forms "hidden" in them. I am looking for the most simple fractal model that I can find that generates time series with, say, Pareto-distributed increments. Or any other heavier-than-Gaussian -tailed distribution.
By generating time series, I mean that the system can be described by some kind of observations, any type. Temperature, some integral measure, derivative of some property, etc. For example, consider a system of agents organized in a network model, and define a sort of interaction between them on a graph. The observed measure of biggest cluster, or sum of vertex weights, or number of vertices with weight bigger than some specific number, etc., iterative, could be a time series.
By fractal properties I mean property of self-similarity. For example, having structure (for an agent-based model that could be graph structure of connectivities) that is defined by the same rule on all scales. For the mentioned exmaple of graph structure, that would be structure among clusters that copies structure of nodes inside every single cluster.
I am trying to come up with the most simple system that has fractal properties, demonstrates non-normality of the whole while having normality or binary behavior at the lowest level, and has clear connection between the two.
Any help would be much appreciated!
| your answer is hidden in the self organized criticality (SOC) concept.
there is many model that have applied in many branch of science that obey SOC.
this concept for the first time was introduced by Bak et al with "Abelian Sand pile model".
this model is the simplest model can receive to critical state (showing power law and infinite temporal-spatial correlation) without any control parameter.
here you can find many things about this concept. even the simulation code of the model.
| {
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"timestamp": "2023-03-29T00:00:00",
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Probing beyond the black hole event horizon Black holes are interpreted to have a "break down" of general relativity at their point of singularity. The region near the singularity is expected to be described by some theory of quantum gravity.
Since quantum gravity is expected to describe the physics near the black hole center, black holes would make for great laboratory environments for observing the effects of quantum gravity.
The problem with using black holes to probe quantum gravity, other than them being very far away, is that black holes supposedly prevent anything from escaping beyond the event horizon.
I want to ask if there are any successful or serious theoretical attempts at overcoming the problem stated above.
(This question does not concern Hawking radiation.)
| There are a number of attempts at constructing theories of quantum gravity, of which string theory and loop quantum gravity are the most developed. However none of these theories have been developed to a point where they can make uncontroversial predictions about what happens near a black hole singularity.
The only even passably convincing attempt is using a streamlined version of loop quantum gravity called loop quantum cosmology. When applied to the Big Bang singularity this predicts that gravity becomes repulsive at distances around the Planck length, and this prevents the singularity forming. The Big Bang becomes a bounce, and geodesics can be continued through the Big Bang. However there is limited acceptance of this as a valid description, and of course it describes a different type of singularity. I'm not sure what work has been done to describe the black hole singularity with LQG, though apparently some progress has been made.
Although you ask about the central singularity rather than the even horizon, I should mention the recent ideas about black hole firewalls. However these ideas remain highly controversial.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/199938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is blocking the magnetic field of this neodymium magnet? I opened up my hard drive and pulled out (among other things) the top and bottom braces for the reading arm.
Each bracket contains a really powerful neodymium affixed to its underside.
When the two undersides come within a few inches of eachother, they accelerate toward each other and are seemingly unstoppable.
However, when they are placed close together but back to back (i.e. the magnets on the underside do not face eachother) there is no attraction whatsoever!
Somehow the material the bracket is made of is completely shutting down the magnetic field. This is strange because I tried placing both magnets on different sides of metal boards, wood, plastic etc and they are always attracted. But back to back, even up close, there is no attraction.
How does this work? Why is this happening?
| Those are nickel/steel brackets that have high magnetic permeability and saturation. Due to the magnets shallow thickness the magnetic fields are completely diverted into the bracket preventing any mag fields from penetrating through it. Notice how the magnet is strongly attracted to the bracket due to magnetic attraction, if you unglue the magnet from the bracket and attempt to block two magnets from eachother you will notice that one magnet will (strongly) stick to one side and if you attempt to attach a magnet on the other side it will push the other magnet off the other side. This happens due to oversaturation and the magnetic fields begin competing with eachother.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which power equation to use: $P = I^2 * R$ or $P = V^2 / R$? Given are ideal max voltage $V = 200\;\mathrm{V}$ and max current $I = 5\;\mathrm{A}$.
Therefore:
*
*ideal resistance is $$R = \frac VI = \frac{200 \;\mathrm{V}}{5\;\mathrm{A}} = 40 \;\mathrm{\Omega}$$
*ideal max power is $$P=IV = 5 \;\mathrm{A}* 200\;\mathrm{V} = 1000\;\mathrm{W}$$
*1st power equation: $$P = I^2 * R$$
*2nd power equation: $$P = \frac{V^2}R$$
Say the real resistance is $$R = 20 \;\mathrm{\Omega}.$$ I presume I am to use the first equation since the other one gives a power above the max power and can't be true.
$$P = I^2 * R = 25 * 20 \;\mathrm{W}= 500\;\mathrm{W}$$
or
$$P = \frac{V^2}R = \frac{40000}{20} \;\mathrm{W}= 2000\;\mathrm{W}$$
What if the real resistance was greater than the ideal, e.g. $R = 60\;\mathrm{\Omega}$. Then I presume I would use the second equation since the first one is above the max power.
$$P = I^2 * R = 5^2 * 60 \;\mathrm{W}= 25 * 60 \;\mathrm{W}= 1500\;\mathrm{W}\\
P = \frac{V^2}R = \frac{40000}{60} \;\mathrm{W} = 666\;\mathrm{W}$$
I think I have found out which equation to use, however I would like to know why this is the case.
| Both equations are valid. You just made an error in taking suddenly 20 $\Omega$ instead of 40 $\Omega$.
$$P = I^2 \cdot R = 25 A^2 \cdot 40 \Omega = 1 kW$$
$$P = U^2 / R = 40000 V^2 / 40\Omega = 1 kW$$
The power is always determind by the current that runs through the part of interest times the voltage drop over it. If there are other parts in a serial circuit that have a resistance you cannot take the voltage the battery supplies but the real voltage drop that occurs at the resistor.
When exchanging the resistor the voltage and/or the current through your circuits change. You cannot work with all the stats given initially.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
What makes neutron heavier than a proton? The mass of proton is 1.672 × 10−27 kg while it is 1.675 × 10−27 kg. Both are made up of 3 quarks each. Then what makes proton lighter than a neutron?
| Note that the combined rest masses of the quarks (~10 MeV/$c^2$) account for about 1% of the proton and neutron mass (~938 MeV/$c^2$), the main contribution to the mass are the gluons from the Strong Force. Since the composition of the proton and neutron are different, so is the force that binds them.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Why is heat flux sometimes assumed to be proportional to surface temperature? Suppose a fluid flows over a heated surface. In "Newtonian heating", it's assumed that the heat flux through the surface, $q$, is proportional to the surface temperature, $T_\text{surface}.$
Written as an equation, it's$$
q ~=~ h \, T_{\text{surface}}
\,,$$where $h$ is heat transfer coefficient.
Questions:
*
*What is the physical significance of assuming that heat flux is proportional to the surface temperature?
*Why might someone make this assumption?
| I guess you wanted to say
$$
q ~=~ h \, \left(T_\text{fluid} - T_\text{surface}\right)
\,,$$
right? So this means that the heat flux density $\left(\mathrm{W} \cdot {\mathrm{m}^{-2}}\right)$ is proportional to the temperature difference between the fluid and the wall. It seems reasonable since when $T_\text{fluid} = T_\text{surface},$
$$
\require{cancel}
q
~=~ h \, \cancelto{0}{\left(T_\text{fluid} - T_\text{surface}\right)}
~=~ 0
\,,$$you don't have any heat flux anymore.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Is the change in orbital of an electron the only way a photon is created I would like to know if there are any other ways in which photon's are being emitted other than in the case an electron's orbital around a nucleus changes.
| That's a give and take. The emission of photons is always based on an energetic level of particles higher than the surrounded world. To reach this higher level it needs an receive of photons.
A moving electron will be deflected in a non parallel to the movement magnetic field and emit photons. How the electron reach the kinetic energy for this propagation? By electromagnetic interaction with other particles. Photons are responsible for the increasing kinetic energy. And they give this energy back in braking situations. And also in the form of photons.
So the change in orbital of an electron is only one possibility to recive or emit photons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How to calculate tide times? How do I calculate the time of the tides at a given location?
I'm not interested in the amplitude of the tides, just the times when they occur.
| There is a lot more to it than just astronomy. For example, the tide times inside Boston Harbor are significantly different from those on the southeast coast of Cape Cod.
It is true that the primary force behind tides is the position of the Moon, but the macro tidal bulges take a long time to propagate around/across oceans, and then the shoreline shape makes a big difference at the end.
Further, the two-a-day tidal pattern in the Atlantic is pretty much symmetric, while the Pacific Coast tides follow a "low-low" and "high-high" pattern. That is, the two low tides are of repeatably different level, and same for the high tides.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can you huddle next to a fridge in sub-zero temperatures and keep warm? There's a saying I've heard in so many places.. "It was so cold that we used to huddle next to our refrigerator to keep warm..." I had heard this phrase uttered some 30 or so years ago, and it's stuck with me ever since...
Which gets me thinking...
Imagine it's -40 degrees (Fahrenheit or Celsius, it's the same number for both scales). Your fridge is by comparison capable of blasting chilled air at +4 degrees Celsius (39.2 degrees Fahrenheit)... give the temperature difference between environment and the refrigerator, could an average human of body temperature of ~37 deg C potentially warm themselves by an open fridge blasting chilled air at +4 deg C in a surrounding environment of -40 deg C and keep "warm"?
| Suppose you would actually go sitting inside your refrigerator and close the door, so that you are in an environment of +4 degrees Celsius. If you are literally warming up, then you was colder than 4 degrees Celsius in which case you probably had died from hypothermia.
If you are wearing protective clothing, the clothing itself can be much colder than +4 degrees Celsius after being exposed to -40 degrees Celsius for a while. If you then go to the refrigerator, your clothing will warm up. However your body will keep cooling down if it wouldn't preventing itself from cooling down by producing heat. This is because it is warmer inside the clothing, where your body is, than outside, in the refrigerator, so heat will float out.
The speed of the heat going out is proportional to the difference between the heat inside and outside the clothing, at the refrigerator we have that the difference is $\Delta T = 37^\circ C - 4^\circ C = 33^\circ C$, but outside we have $\Delta T = 37^\circ C - -40^\circ C = 77^\circ C$, so you will cool down twice as slow inside the refrigerator.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Car Crash Scenario Two vehicles travelling at 80mph in the same direction.
Vehicles are directly behind each other.
12 meter distance between them.
Front door of car in front rips of and hits the front window of the car behind.
At what speed did the door make contact with the car?
How would one go about calculating this?
====
My current guess is as follows: The car behind would take approximately 0.34 seconds to reach the door in front, the door in front would have a forward velocity starting at 80mph but would quickly be decelerating due to no longer being attached to the car in front. Considering the car behind reach the door in approximately 0.34 seconds, i would guess that the speed of the door would be still very close to 80mph.
But this is a pure guess, is there a formula i can use here to accurately calculate this?
| The door is not slowing down because of its mere detachment from the front car. It is slowing down because of air drag on the door. That drag would depend on the area of cross-section interaction, the mass of the door and whether the door is tumbling. One might be able to estimate a worst-case scenario, but there isn't really a way to calculate an accurate.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
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