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Magnet spinning between two other magnets Suppose, we have two magnets, MA, MB, and we have a third magnet MC in between the two magnets.
Each magnets' north pole faces the other magnets south pole, and the magnets are placed horizontally side by side. We spin the magnet MC at a speed between super fast and slow. What will happen then?
I mean, once MC's north pole faces MA's south pole and MC's south pole faces MB's north pole, then the next moment MC's south pole faces MA's south pole and MC's north pole faces MB's north pole. So what will happen then?
| The middle magnet is spinning, so it attracts and repulses the other two magnets once per rotation.
It is spinning "super fast" - that is so fast that the attraction and repulsion phases are super short. The other magnets are just too heavy to even start moving visibly in one or the other direction, before the direction of the force changes again.
We could say "nothing happens" - except the outer magnets oscilate slightly with each rotation.
If the middle magnet would spin "super slow", the others would just jump to the middle one, stick to it, and rotate with it as if it's all one magnet.
What happens if the middle magnet would spin with a frequency in betwen?
That's difficult, because much depends on how the rotation starts, and we only know it has started...
If the rotation starts slowly and gets faster then, up the middle speed, different things can happen during the first rotation. The magnets could stick to the rotating one, or move away a little bit; That whould have a big influence on what happens later.
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Statistical mechanics: What is a "microscopic realization" of a system? What is a "microscopic realization" of a system?
The context is statistical mechanics. The microscopic system consists of many atoms (too many to track individually) with an assigned probability density function f(x,y,z,Vx,Vy,Vz,t).
The macroscopic system consists of the atoms taken together, with macroscopic quantities computed as expectations of microscopic quantities.
| Statistical mechanics relies on a probabilistic understanding of the world and as such one needs to define a probability space. In classical statistical mechanics the probability space consists of a domain which is the set of all possible microstates (that is the position and velocity vectors of all the particles in the system) and a probability measure associated to this space. This probability measure ensures that all microstates incompatible with the constraints on your system (e.g. fixed number of particles, fixed volume etc...) have probability zero. The set of constraints on your system define what is called a statistical ensemble.
In equilibrium classical statistical mechanics, the probability measure is invariant under the law of classical mechanics and is therefore time independent. Traditionally, we tend to equate the operational notion of statistical ensembles and the probability measures they correspond to in equilibrium statistical mechanics and we use essentially four measures/ensembles (microcanonical ,canonical, grand canonical and canonical-isobaric).
Now, the fact that we use probabilities is in essence no different than our use of probabilities when we cast a dice, except that here the dice has many many faces (which correspond to all the microstates one can imagine).
Now, in the same way that when you cast a dice, a realization would be any number between 1 and 6 (for instance 3), then for a thermodynamic system with say fixed $(E,N,V)$, a microscopic realization can be any microstate compatible with these constraints (for instance all the particles in the corner of the box with one particle having all the energy $E$ of the system and the rest of the particles having no motion). You just have to imagine that you have in your hands a god like device that can tell you the instantaneous positions and velocities of all the particles in your system. Each time you perform a measurement with this device, you will be observing a microstate compatible with the statistical ensemble and hence observing a microscopic realization of this ensemble.
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Problem with derivation of phonons in crystal
In this derivation of phonon solutions, everywhere, we are forcefully assuming the wavelike characteristics along the length of the chain. While all we can deduce for finding out the fundamental frequencies is that the solution will be periodic in time,
and solution should be of the form
$\exp(i\omega t)$,
I am not getting how the derivers arrive directly at
$\exp(ikx - i\omega t)$. That one is in Kittel.
In the figure below,
Here also, somehow, $q$ has been deliberately has been involved linking it to distance.
PS: $n$ is a measure of distance along the chain.
| There is nothing wrong with looking for plane-wave like solutions of the form $A \exp (i (\omega t - k x) )$. Given the linearity of the equations, and as @ignacio pointed out the fact that the $\exp (i k x_n)$ form a basis of solutions, you can write a more general solution as a combination of these plane waves. This solution isn't necessarily periodic (think of a propagating wavepacket peaked at a certain position in space, for example).
| {
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Gravity and electromagnetism If light bends due to the curved spacetime,would not the act of bending light warp space? How does one describe curved light?
| Light does curve spacetime. This is discussed in the question Does a photon exert a gravitational pull?.
In many cases we are describing the interaction of very massive bodies with much lighter ones, for example a satellite orbiting a planet. In these cases it's a good approximation to ignore the curvature of spacetime caused by the lighter body. The sort of situation you describe would be the lensing of light by a massive object, and when calculating the bending of the light we ignore its affect on the spacetime curvature. We do this because we expect the effect of the light to be unmeasurably small, and because trying to include the effect of the light makes the calculation vastly more complicated.
| {
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If everything is relative to each other in this universe, why do we keep the Sun to be the reference point? and study the solar system and universe relative to it and why not relative to the Earth?
| It's all about the context in which you want to analyze particular issue.
If you are studying the solar system, the most suitable, would be to consider the sun as the center of the system.
If you are studying the Milky Way, the sun is not a good reference point, you should take the center of the galaxy.
Similarly, to locate the stars from an observer on earth, the "celestial sphere" is used, and that does not mean returning to the model of Ptolemy.
It's all about the scope that is intended in a particular analysis.
| {
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What is a flat rotation curve? Was reading about dark matter and the distribution of it throughout the galaxy. it said "For example, if rotation curves are flat this means-" what exactly does this mean?
| When you look at an image of a galaxy like our own, you can see that most of the visible mass is concentrated in the core and that the density of stars in this core region is approximately constant:
So let's compute the expected rotation curve.
The orbital velocity of a star at a distance $r$ from the centre of the galaxy is
$$ v=\sqrt{M(r)/r} $$
where $M$ is the mass of the galaxy inside the orbit $r$, and I've taken $G=1$.
Now let's pick a radius $r$ that is somewhere within the core region. Since the density $\rho$ (mass per unit area) is roughly constant in this region, the mass $M$ inside $r$ will be
approximately
$$ M(r)=\pi r^2\rho\qquad\mathrm{(core)} $$
And so the orbital velocity in the core region will scale as:
$$ v\sim\sqrt{r^2/r}=\sqrt{r}\qquad\mathrm{(core)} $$
Now pick a radius $r$ far outside the core. Since the vast majority of the inner mass $M$ is contained within the core, then as we increase $r$ the mass $M$ will remain roughly constant. The velocity will therefore drop off as
$$ v\sim 1/\sqrt{r}\qquad\mathrm{(outer)} $$
Plotting these two limiting cases ($\sqrt{r}$ in blue, $1/\sqrt{r}$ in red, and a 'hybrid' dashed):
we can see the sort of shape the rotation curve should have. However, the actual observed curve is roughly flat as $r$ increases outside of the core region:
This tells us that there is more mass than we can see (i.e. dark matter). Also, from the equation $v=\sqrt{M/r}$, we can see that this is only possible if the mass $M$ increases roughly linearly with $r$.
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How is electric field strength related to potential difference? I was looking at the derivation of drift velocity of free electrons in a conductor in terms of relaxation time of electrons.
Here a metallic conductor XY of length $\ell$ is considered and having cross sectional area A. A potential difference $V$ is applied across the conductor XY. Due to this potential difference an electric field $E$ is produced. The magnitude of electric field strength is $E = V/\ell$.
I don't understand this part,
How is electric field strength = potential difference divided by length? Where am I lagging in my concepts?
| The assumption is that the electric flux lines are going to go through the conductor parallel to the conductor. Under those conditions, the electric field within the conductor is going to have a constant magnitude, and point parallel to the conductor. I assume you're familiar with
$$E=-\nabla \phi\ \ ,$$
where $\phi$ is the electric potential. We then integrate along a line through the conductor
$$V=\int_{0}^{\ell} \nabla \phi \ dx = \int_{0}^{\ell} – E \ dx = - E \ell\ \ ,$$
which is the same as $E=V/\ell$ if you consider a positive $E$ to mean a vector pointing in the $-x$ direction instead of the $+x$ direction.
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Difference between heat capacity and entropy? Heat capacity $C$ of an object is the proportionality constant between the heat $Q$ that the object absorbs or loses & the resulting temperature change $\delta T$ of the object. Entropy change is the amount of energy dispersed reversibly at a specific temperature. But they have the same unit joule/kelvin like work & energy. My conscience is saying these two are different as one concerns with temperature change and other only at a specific temperature. I cannot figure out any differences. What are the differences between heat capacity and entropy?
| All comments miss a point, or maybe I didn't get it right. I see it this way,
lets start with other forms of energy (kinetic and potential, for example) of an harmonic oscillator. We know for a non damped HO the relation between position "$s$" and velocity "$v$" is given by total energy $E_o = \frac{1}{2}kx^2 + \frac{1}{2}mv^2,$ where $k$ is spring constant and $m$ is mass. This relationship (since energy is conserved here) can be turned in many ways. Differentiating (focusing on small values) both sides yields:
$\mathrm{d}E_o = kx \cdot \mathrm{d}x + mv \cdot \mathrm{d}v$ assuming conservative forces $\mathrm{d}E_o =$ and we can express
$\dfrac{\mathrm{d}v}{\mathrm{d}x} = - k \dfrac{x}{mv}$. Does the velocity depend on position, NO! the fact that the conservation law allows to relate two (measurable) quantities does not mean those two are affecting each other. The same is true for $C$ and $T$. Although heat capacity is found by known relationships does not mean they are based on each other. If we for the same amount of heat quantity get one sample to have more temp than another sample it means that for the first one more energy is stored in the system than second. Is $T$ is the kinetic energy the $C$ relates to potential energy. I hope I make sense.
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Demostrating possible equivalence of two tensors Is there anyway to see by inspection that a form like
$$a(x^2 )^{-3} (g _{μσ} x_{\rho} x_{ ν} + g_{μρ} x_{σ} x_{ ν} +g_{νσ} x_{ρ} x_{ μ} + g_{ νρ} x_{ σ} x_{ μ} ) $$
may be equivalent to (i.e reduced down to or reexpressed) $$
b(g _{μν} x_{ ρ} x_{ σ} + g_{ ρσ} x_{ μ} x_{ ν} )(x^2 )^{ −3} ?$$ where $g_{\mu \nu}$ is the metric tensor (diagonal).
I have tried to put in various permutations of $\mu \nu \rho \sigma$ and from $1111$ and $2222$ for example, I obtained the constraint that $a/b = 2$ but I am not really sure what this means. If I try the combination $1221$ e.g then it implies $b=0$, which seems to contradict my first result.
Does this mean that the two forms are not equivalent?
| Indeed, if no values of $a$ and $b$ work for across different sets of indices, then the forms are not equivalent.
In fact, these two forms are not equivalent even under the restriction of the metric being diagonal (and thus are not equivalent under a general metric). The diagonal case is easy to analyze, and you gave a good set of indices to do it: $\mu\nu\rho\sigma = 1221$. Then the large parenthesized part of the first expression becomes $g_{11} x_2 x_2 + g_{22} x_1 x_1$. The middle terms drop out because they involve off-diagonal parts of the metric. However, both terms in the second expression also involve off-diagonal metric coefficients, so the second expression is identically $0$.
Given tensor components $T_{\mu\nu\rho\sigma}$ and $S_{\mu\nu\rho\sigma}$, we have $T_{1221} = g_{11} x_2 x_2 + g_{22} x_1 x_1$ while $S_{1221} = 0$, so clearly $T \not\propto S$.
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Is there any evidence that matter and antimatter continuously appear and disappear on the edge of a black hole? I heard Stephen Hawking got a Nobel prize for this, someone said there was no evidence for it which I find quite strange since he got an award for it.
| There is plenty of evidence for the underlying quantum field theoretical description of the vacuum.
*
*The (complete) quark content of the nucleons has been measured, and includes both flavors not in the valence content (strange, charm (?)) and lots of anti-quarks. Everything that isn't valence content ($uud$ for a proton or $udd$ for a neutron) is called the "sea", and it comes from the same place as the pairs that Hawking is working with. The experiments I'm familiar with are NuSea and SeaQuest, but there are others.
*Pair-production (of electrons and positrons) happens all the time in high-energy photon interactions with matter (it's one of the main energy loss mechanisms) for photons with energy much higher than 1 MeV. In principle we should be able to do this in free space in the field of a very strong magnet, but I'm not aware of a actual realization of this; lots of theory papers in the google search, but no experiments. These pairs also come from the vacuum.
*Other kinds of pair knock-on reactions. Nuclear pion production in very forward kinematics is dominated by quark pair-production and so on.
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Projectiles Launched at an Angle with unspecified Initial Velocity I'm attempting to do my Physics homework, and I did the first one right, but that problem gave me initial velocity. This problem gives me only the angle relative to horizontal and the distance it covers. Can anyone help me figure out where to start? I've tried but I can't find any formula that I can find initial velocity without having time, or vice versa. Any help would be much appreciated. Here's the problem in full:
A golfer hits a golf ball at an angle of 25.0° to the ground. If the golf ball covers a horizontal distance of 301.5 meters, what is the ball's maximum height? (Hint: At the top of its flight, the ball's vertical velocity component will be zero.)
I realize that the vertical velocity component has something to do with it, but I can't figure out where that would fit in.
| First, you have to find the maximum height of the ball. In order to do so, we use the condition that $v_{y}=0$.
$$0=v_{0}\sin(\alpha)-gt_{H}$$
From this, the time to reach the maximum height $t_{H}$ is
$$t_{H}=\frac{v_{0}\sin(\alpha)}{g}$$
From the vertical displacement, we obtain the maximum height using $t_{H}$
$$H=\frac{v_{0}^{2}\sin^{2}(\alpha)}{2g}$$
But you dont know the initial velocity. Only the angle and the range. You find the range by setting $y=0$
$$0=v_{0}\sin(\alpha)t_{D}-\frac{gt_{D}^{2}}{2}$$
Following the same steps as above, you get the time of travel $t_{D}$ and from this you find the range to be
$$D=\frac{v_{0}^{2}\sin(2\alpha)}{g}$$
From all this you get
$$\frac{H}{D}=\frac{\sin^{2}(\alpha)}{2\sin(2\alpha)}\Longrightarrow H=\frac{D\tan(\alpha)}{4}$$
| {
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What is going on in front of and behind a fan? Why is it that when you drop paper behind a fan, it drops, and is not blown/sucked into the fan, whereas if you drop paper in front of a fan, it is blown away?
| Think about the air around the fan at any given time:
The amount of air flowing into the fan must equate to the air flowing out of the fan.
The amount of air that passes through an area in a given time is related to the velocity of the air i.e. the faster the air is moving, the more air that can flow through a fixed area/hole/slot.
The fan blades apply a force in order to boost the velocity of the air in order to 'blow'. Hence, the flux of air through the disc located at the fan blades must be equivalent behind and in front of the fan (continuity condition).
Your observation is that the air behind the fan is moving slower than the air in front of the fan. If we were confined to pulling air from directly behind the fan enclosed in a cylinder with cross-sectional area equal to the fan blade coverage, then the air moving behind the fan would need to move as fast as the air in front of the fan to satisfy our continuity condition.
Therefore, the fan must be pulling air from a larger area behind the fan than the output air. If we think about this as a volume, it would be some sort of expanding cone as shown in diagrams you see of the air flow.
| {
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What happens to a photon after it is absorbed by an antenna? I recently have read about interception of wireless information, however this mentions that people can intercept the information, and then somehow the recipient also gets the information. Regardless of this context, what happens to the actual photon if it is absorbed by one antenna how can another person receive the same signal? Is it that when the photon is absorbed exciting the electron, the electron will then leap back to the lower energy state causing it to emit another photon? so the antenna acts as a receiver & transmitter? to be honest I'm confused overall in how antennas work.
| The radio waves or microwaves that are used for communication don't contain just one photon. They contain a bunch. (Maybe someone will do the math for how many photons a standard radio broadcast antenna is producing each second; it'll blow your knee-high off even if you're wearing sandals over them.)
Consider for example a frequency-modulated signal. The information is contained within the frequencies/energies of the outgoing photons. You might remove one photon, but there are many others with the same signal/frequency information also traveling in different directions that you don't intercept.
Thus, intercepting just one photon doesn't destroy these communications. But it does destroy the photon; it is absorbed by the material of the antenna and is gone.
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How to determine your position underground? Crossed my mind after random rant on wikipedia that lead me to articles about chronometers and measuring position.
Let's assume I were trapped in the underground laboratory with lots of equipment but without any access to the surface. Would I be able to properly determine my position (latitude, longitude and altitude), and if so, what instruments are needed? (and mny what's the coolest way to do it :)
I thought about measuring Coriolis effect, which could lead to latitude measurement, and earth's gravity map could give more hints, but it's still far too imprecise.
| John's answer gives some ideas for latitude and longitude. You could measure your altitude (read depth) by measuring the weight of a known mass. In a perfectly uniform, spherical Earth, the weight is proportional to your distance from the center.
| {
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Can you choose the variables of a state function? I'm confused. I was first introduced to entropy as a state function of internal energy and volume
$$S(U,V) \Rightarrow dS = C_v\frac{\mathrm{d}T}{T} - p\frac{\mathrm{d}V}{T} $$
wich is the thermodynamic identity, but now I see that it can also be written as
$$S(T,p) \Rightarrow dS = C_p\frac{\mathrm{d}T}{T} + k\mathrm{d}p $$
I don't understand, if I can choose the parameters I could just decide that entropy is a function of the number of molecules of my gas so there would be no variation in entropy as long as it's in a closed container...
Can someone explain to me the logic behind this ?
| Note that I am working in unit-mass bases (with lower-case symbols).
*
*Since,
*
*We have,
*
*So that,
Notes:
*
*The internal energy is a function of the constant volume specific heat, and the latter is a function of temperature.
*The enthalpy is a function of the constant pressure specific heat, and the latter is a function of temperature.
*Remember the state postulate: the state of a simple compressible system is completely specified by two independent, intensive properties (intensive: independent of mass. e.g., P, T, or v "or p the density, the reciprocal or v"). This is true as long as there is no electrical, magnetic, gravitational, motion or surface tension effects.
*Notice the left-hand side (ds) and the right-hand ones (du, dh, dT, dv, dP), and think about integration. What would be the variable of integration in each of these equations? That is where the dependency (functionality) came from.
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A James Clerk Maxwell Disproof One of my favorite physicists to learn about was James Clerk Maxwell, for the fact that he unified the study of E&M in physics and he would often disprove theories that did not work as a Mathematician.
I remember my physics professor mentioning a disproof Maxwell had done with what was then the current knowledge of how currents flowed or were held through capacitors... This is probably way off from what he actually did disprove since it has been over 5 years since I had been to this particular lecture but could anyone tell me or demonstrate the exact disproof by counterexample that Maxwell had done with capacitance?
| Are you talking about the famous derivation of the displacement current, where Ampère's law is both true and false depending on what surface you choose to integrate through, despite the same boundary, as below:
(Image from WikiMedia commons http://commons.wikimedia.org/wiki/File:Displacement_current_in_capacitor.svg)
The solution of this was to add a term to Ampère's equation that depends on the time derivative of the electric field. And then, there is no paradox anymore.
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Eigenfunctions for $1s$ hydrogen Schrodinger equation I am a computer scientist and started my Phd in material science. The second course os my Phd is material simulation by computer. One the task is show the verification of the eigenfunction $1s$ from time-independent Schrodinger equation.
I dont want any answer, I just need a few tips.
I already found one example in Eisberg book (Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles, pg ~ 242 Example 7-2).
In the Eisberg book I can't find the points where the $\Psi_{2 1 1}$ have influence in the demonstration.
| The time-independent Schrödinger equation for the hydrogen atom is
$$-\frac{\hbar^2}{2m}\vec \nabla^2\psi-\frac{e^2}{4\pi \epsilon_0r}\psi=E\psi $$
If your aim is just to verify that the $1s$-wave function
$$\psi_{100}=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}\hspace{2cm} a\equiv \frac{4\pi\epsilon_0\hbar^2}{me^2} $$
is indeed an eigenfunction, then your task isn't all that hard. Simply plug take the necessary derivatives, add the second term, and check that the outcome is indeed of the form $C\psi_{100}$ where $C$ is a constant.
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Blowing your own sail?
How it this possible? Even if the gif is fake, the Mythbusters did it and with a large sail it really moves forward. What is the explanation?
| The question is whether this is a form of "pulling yourself up by your own bootstraps". Clearly the driving force here is the leafblower the skater is wearing, which takes ambient air with zero average momentum and sends it out a vent with large average momentum. One would expect, absent some deviousness having to do with turbulence, that the skater would accelerate faster if he ditched the umbrella and pointed the leafblower's exhaust to the rear.
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What makes a material adhesive (or sticky)? I currently have a bandage on my arm that isn't sticking to me but definitely sticks to itself. It brought up the question:
What makes a material sticky? What's happening on the molecular level? Why do some materials stick to some things better than to others?
| There are several different types of adhesion. As pointed out by Ryan S. :
Electrical, like when a balloon sticks to your head after a good rubbing. Chemical, like plastic cement will "melt" the plastic pieces to be bonded together. And the most common, physical, which is extremely hard to make an example of without a picture. Therefore, I give you a picture.
What we have here, is paper. Normal paper. (Thanks for the picture Arstechnica!)
Glue, or other adhesive (like gum) can squeeze itself into the spaces between the fibers, and harden. The hardening while in an irregularly shaped form is what makes glue sticky. (That, and some amount of chemical bonding.)
I would think that in a case of a bandage, the electrical adhesion would be negligible. Probably most of the "stickiness" would come from the glue pressing into a crack/ around a hair etc. I would also guess that it would have a fair amount of chemical adhesion, but I would think that that would occur after the bandage had been worn for some time.
Then again, the manufacturers have probably found a formula that gives an extremely "bond happy" substance.
http://en.wikipedia.org/wiki/Adhesive_bandage
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Forces while squeezing a toothpaste tube When I squeeze a tube of toothpaste, I am working with 2 squeeze forces toward tube. What causes the toothpaste to go out of the tube? (instead of remaining stationary)
| From a fluid dynamics perspective, your applied force induces motion via the Navier-Stokes equations:
$$
\rho\left(\frac{\partial}{\partial t}+\mathbf v\cdot\nabla\right)\mathbf v\propto\mathbf f_{body}
$$
where $\rho$ is the density of the toothpaste, $\mathbf v$ the velocity of the toothpaste, and $\mathbf f_{body}$ your applied force. Since the density is non-zero, the only way for the equation to balance is if $\mathbf v\neq 0$; it will then flow out the only orifice it has.
Alternatively, you can view this as your squeezing is shrinking the volume of the container, which must displace the material filling that volume.
| {
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What technology can result from such expensive experiment as undertaken in CERN? I wonder what technology can be obtained from such very expensive experiments/institutes as e.g. undertaken in CERN?
I understand that e.g. the discovery of the Higgs Boson confirms our understanding matter. However, what can result form this effort? Are there examples in history where such experiments directly or indirectly lead to corresponding(!) important new technology? Or is the progress that comes from developing and building such machines greater than those from the actual experimental results?
| Just to mention the latest developments brought up from neutrino science, it seems that the useless, wimpy, weakly interacting neutrinos (and their anti-neutrino sisters) can be harnessed to detect and monitor nuclear reactors anywhere in the world that violate non-proliferation agreements
I remember a discussion about a really big neutrino beam that could be in principle be directed to a rogue nuclear reactor and disrupt their isotopes, but I'm not sure if that is really feasible. But the fact is that 20 years ago this kind of technology wasn't even on the map, and now is being seriously considered for development and implementation
regarding detection: http://www.tandfonline.com/doi/abs/10.1080/01440389008403936#.VBi4Ry4qvdE
regarding bomb deactivation: http://arxiv.org/abs/hep-ph/0305062
| {
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Differentiating the Hamiltonian Operator, $\hat{H}$ Firstly let $\hat{H}$ denote the full energy of the electromagnetic wave. I'm trying to differentiate the Hamiltonian operator with respect to the components of momentum, i.e. $$\frac{d}{dp_x} \frac{d}{dp_y} \frac{d}{dp_z} \hat{H}$$
To do so, I need to write the Hamiltonian as the components of momentum. Using the Dirac equation, I think it would be correct to say
$$\hat{H}=\beta mc^2+c(\alpha_x p_x +\alpha_y p_y +\alpha_z p_z)$$
As
$$\left(\beta mc^2 + c(\alpha_1 p_1 + \alpha_2 p_2 + \alpha_3 p_3)\right) \psi (x,t) = i \hbar \frac{\partial\psi(x,t) }{\partial t}$$ and $$\hat{p}=p_x \mathbf e_x+p_y \mathbf e_y+p_z \mathbf e_z$$according to Wikipedia.
I'm not familiar with what I think is matrix mechanics so I would like someone to explain the purpose of $\beta$ and $\alpha$ and how to differentiate the aforementioned equation, provided it is correct.
| The momentum operators $\hat{p}_x,\hat{p}_y,\hat{p}_z$ are operators. They can be represented as infinite dimensional matrices acting on an infinite dimensional Hilbert space. The Hamiltonian $\hat{H}$ can also be represented as a infinite dimensional matrix.
One can generalize the notion of a differentiation to differentiation by matrices:
http://en.wikipedia.org/wiki/Matrix_calculus#Derivatives_with_matrices
However the momentum operators are constant matrices. You can't differentiate by a constant.
Why you would even try to do this is another matter. Differentiating with operators is not a useful in any physics I'm familiar with because the complete set of commuting observables whose eigenstates span the Hilbert space are taken to be constant due to them usually being some generators of some Lie algebras.
| {
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Did Newton argue that particles speed up when entering a more dense medium? Statement:
Newton argued that particles speed up as they travel from air into a dense, transparent object, such as glass.
From this source, I gather that he did argue that the light particles sped up when entering a more dense medium. However, it just doesn't make sense. I thought they seemed to slowed down.
| "From this source, I gather that he did argue that the light particles sped up when entering a more dense medium. However, it just doesn't make sense."
It does make sense. They enter the medium at a higher speed, then slow down. If Newton had said they continue to move at a higher speed within the medium, he would have been wrong. But I don't think he said so - "higher speed in a more dense medium" seems to be a mythology serving the opponents of Newton's emission theory of light.
| {
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Can a bullet really fly through space forever? Some people says that if it would be possible to shoot a bullet so high that it would get over the Earth gravitational pull, the bullet could fly through space forever, because of no deceleration of friction (and if there wouldn't be any more particles in the space, of course).
But according to quantum physics, every single particle in the universe has a non-zero probability of existence anywhere in the universe. So gravity (produced by gravitons) attract on infinite distance - so there is no place in the universe (even in universe without the particles except Earth) without gravity.
So is it true that bullet would have to stop because of the gravity (even minimal we don't consider the expansion of the universe and curve of the universe)?
| I think your confusion is about whether a bullet could truly escape the earth's gravity if gravity extends forever. You are worried that the bullet will always feel a non-zero gravitational force, and therefore it will stop at some point.
You are correct that the bullet will feel a non-zero gravitational force. This can be seen from newtonian gravitation; it is not necessary to discuss quantum mechanics. However, as the bullet gets farther and farther from the earth the force gets smaller and smaller. The faster the bullet goes, the faster the force gets smaller. If the bullet is going fast enough, it can happen that the force gets too small too fast, and it cannot stop the bullet, and the bullet will continue forever. The minimum speed the bullet needs to go in order to beat out gravity is called the "escape velocity".
So the moral of the story is that even though the bullet always feels the gravity of earth, if it is going fast enough, the gravitational force will quickly become very weak and the bullet will keep moving away from the earth forever.
| {
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Standard Usage of the word "per" Math guy here. What is the usual meaning of "x per y per z?" Is this (ignoring details) (x/y)/z or x/(y/z)?
Sorry to be mundane.
| Two examples that come to my mind are acceleration and intensity.
Acceleration is measured (in SI units) in "meters per second squared" = $\text{m}/\text{s}^2$, but it is also commonly said as "meters per second per second," which matches your first option.
Likewise, intensity is measured (in SI units) in "joules per square meter per second" = $\text{J}/(\text{m}^2\text{s})$, which also matches your first option.
I can't claim this is always the case, but I would guess most of the time it is.
| {
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Difference between high-level and low-levels of electromagnetic radiation can someone please explain me what we mean by 'high-level' or 'low-level' in electromagnetic radiation? for example, it is believed that high-level microwave radiation is harmful to human but not the low-level one.
what is this level here we are talking about?
if the frequency and wavelength is same, how high-level and low-level radiation differs?
| An important factor in determining how much energy electromagnetic radiation carries is its intensity, which is just the power per area. In fact, it can be found in the wikipedia article that intensity is sometimes taken to be synonymous with 'level' (even though it is not really correct), so surely this is the right variable to look at.
| {
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Why does Energy-Momentum have a special case? I was reading Energy-momentum, and I came across this simplified equation:
$$E^2 = (mc^2)^2 + (pc)^2$$
where $m$ is the mass and $p$ is momentum of the object. That said, the equation is pretty fundamental and nothing is wrong when looked upon, I similarly also believed this but I came across a "special" cases where this does not apply:
*
*If the body's speed $v$ is much less than $c$, then the equation reduces to $E = (mv^2/2) + mc^2$.
I find this really crazy, because first Einstein, always wanted to create a theory\equation that applied to every aspect of physics and has no "fudge" factors, that said irony is present from Einstein. Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given.
Most importantly, why does this not work, if velocity is "much" slower than light? What do they mean by "much slower", what is the boundary for "much slower"?
Regards,
| The above answers are all good, but I want to add something else. You can derive the energy-momentum relation from at least principle, the action is $A=-m\int \sqrt{1-v^2}{\rm d}t$ . Lagrangian is $\cal L$$=-m\sqrt{1-v^2}$, then you can get momentum $p$ from derivative with respect to $v$ (this is the real momentum, not $mv$). Energy $H=E=pv-\cal L$ , just as $E^2=p^2+m^2$ you get.
| {
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Derivation of the average Velocity formula with constant acceleration (using calculus) I have been looking at
$$
v_{avg} = \frac{v_{i} + v_{f}}{2},
$$
when the acceleration is constant, where $v_i$ is equal to the initial velocity and $v_f$ is equal to the final velocity.
How can you derive this using calculus?
$$
v_{avg} = \frac{\Delta x}{\Delta t} = \frac{1}{t_{f} - t_{i}} \int_{t_{i}}^{t_{f}} v(t) dt
$$
We know that
$$
v(t) = at + v_0,
$$
thus by substituting this into the previous integral yields
$$
v_{avg} = \frac{1}{t_{f} - t_{i}} \left[a\frac{t_f^2}{2} + v_{0} t_f - a\frac{t_i^2}{2} - v_{0} t_i\right].
$$
But there's nothing else, which I can think of. Idea?
| There is one error in the derivation, if you want to have $v(t_i)=v_0$, you must have
$$v(t) = v_0 + a(t-t_i)$$
You also have to use the fact that $v_f = v(t_f)$. Once you use all this, you should be able to divide out $t_f-t_i$ in the corrected version of your last line and get the result you seek.
| {
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Handedness of Reference Frames? I am developing a new derivation of the Lorentz transformation which I think and hope is more attractive to students than those I have seen in currently available texts. I am carefully defining and discussing the important concepts of homogeneity and isotropy of space.
My question is this: am I justified in assuming that a transformation between inertial reference frames cannot reverse handedness? I understand that parity is not conserved in all particle interactions, but my question concerns two relatively moving observers. Can one of them distinguish that the other has adopted the oppositely handed coordinate frame? References would be appreciated.
|
[...] my question concerns two relatively moving observers. Can one of them distinguish that the other has adopted the oppositely handed coordinate frame?
What information does observer A have about observer B? If all A knows about B is B's state of motion, and if he assumes that B is going to choose coordinates in which he is at rest at the origin, then there is a multiparameter family of possible coordinate systems that A could impute to B. These could differ by rotation, parity, and time-reversal. If A assumes that the psychological arrow of time is universal, then there's still rotation and parity. What you're asking about is basically the distinction between the Poincaré group and the Lorentz group.
Note that rotation and parity are similar in that we can't fix either without reference to some object. See The Ozma Problem .
I am carefully defining and discussing the important concepts of homogeneity and isotropy of space.
This is tough to do totally rigorously at the freshman level. Einstein's 1905 attempt to formalize it wasn't right. IIRC he says that the transformation has to be linear by the homogeneity of space. But really that's not quite right since you could, e.g., transform into an accelerated frame. That doesn't require GR according to modern definitions.
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Why is the shape of lightning or an electric spark a zig-zag line? Why is the shape of the lightning (or an electric spark) always of a zig-zag nature? Why is it never just a straight line?
Image source.
| The lightning is just electricity, a huge burst of electrons that try to find the path of least resistance through the molecules of the atmosphere to the ground.
The electrons come from many different places in the clouds
and the atmosphere is not homogeneous: there are differences in humidity, temperature, density, particle count, velocity, etc., and this is particularly true in thunderstorms. Charges build up all over the place and electric fields exist all over as well, sooner or later one or more of these electric fields reach critical strength, or perhaps a cosmic ray ionizes a path through the middle of it where things start to conduct and there it goes: lightning and thunder.
While moving electrons are subject to change of direction from magnetic fields. The energy is coming from differences in statics voltage levels or electric fields
The electric arc forms where the air is most easily ionized (which is what makes it conduct electricity) and generally in the direction of shortest distance between the electrical charge and the ground (the electrically lower potential).
| {
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Spiral galaxies and gravity lenses Spiral Galaxies must have a great deal more mass than elliptical galaxies of the same size in order to account for the flat velocity curve. I've seen references of eight to ten times the visible mass. So then Spiral Galaxies should act as much better gravity lenses than an elliptical (of the same size). Do we have any evidence either for or against the Dark Matter in Spiral Galaxies acting as stronger gravity lenses than their elliptical counterparts?
| The problem is that galaxies come in all sorts of sizes and therefore with different lensing strengths. The experiment would be to measure the lensing of elliptical galaxies then compare this with their mass and see if the lensing looks bigger than the observed mass would suggest.
The trouble is that while lensing measurements give us the total mass it's difficult to get a good measure for the mass of the normal matter alone in elliptical galaxies. Various attempts to do this have been reported. A quick Google found me The stellar and dark matter distributions in elliptical galaxies from the ensemble of strong gravitational lenses (nice concise title :-).
It does seem that elliptical galaxies do have less dark matter than spiral galaxies, though the experimental constraints are still rather poor. In some special cases, such as M87, the presence of dark matter has been inferred from the dynamics, but again the errors are still large.
There is quite a nice article on this subject on the Astrobites website.
| {
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Is the ground state closest to the uncertainty relation? For simplicity, suppose we are only talking about discrete energy levels, i.e., bound state case. The energy levels are $E_1, E_2\cdots$, and the corresponding wave functions are $\psi_1, \psi_2 \cdots$.
My question is, is it true that $\sigma_x \sigma_p$ is minimum when $n=1$ for the eigenstates?
I came across this question because I found harmonic oscillator and infinite potential well problems satisfy this statement, so I want to know if this is a general case.
I think this may be true because for the ground state, there is no node ground state wave function. Thus the $\sigma_p$ may be small compared to other eigenstates.
| Take it as you want, but this is the way I interpret the necessary existence of a ground state (at finite energy) for any bound system in quantum mechanics. The idea, in my view, consists in finding the minimum energy value of a Hamiltonian of the form $H = \frac{\vec{p}^2}{2m} + V(\vec{x})$ under the statistical constraint that, say, $\Delta x \Delta p_x \geq \frac{\hbar}{2}$.
A rapid account of this strategy (I have seen a more rigorous reference somewhere else but I can't retrieve it for now) can be found here.
| {
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Trouble understanding the concept of true and apparent weight I need help understanding the concept of true weight vs apparent weight. I understand this much: if someone is standing in an elevator on a scale, the further up they go the less the reading on the scale becomes. But why is this? Is it that distance affects the force of gravity? The further away the object goes [from the Earth's surface] the less the attractive force? Also, if on some other planet with radius $r$ an object is some distance $d$ away from the surface and is 1% less than its true weight on surface, what is the ratio $d/r$?
| While it is true that the gravitational force dissipates with respect to distance squared, that is not the reason a scale would output a "different weight". A scale does not actually measure weight, only it's response to it. That is, the scale reads the normal force. If the elevator was motionless, the normal force would be equivalent in magnitude to your weight. Therefore, the scale reading would also happen to be your weight.
Now consider an accelerating elevator. We can easily analyze what's going on mathematically.
(Normal Force) - (Your Weight) = (Your mass) * (acceleration).
Normal Force = (Your mass * acceleration) + (Your weight).
That is the normal force is the sum of your weight and the relative force associated with the accelerating elevator.
Physically, you can think about the electrons (which account for the normal force) in the scale. When the elevator is motionless (or constant speed), and you are standing on the scale, those electrons will push back with equal force. However, when the elevator is accelerating upward, those electrons are forced to be closer to your feet. The response then is simply a larger normal force.
| {
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The Earth is spinning, so why don't we jump and land on a different location? I know there are similar questions on StackExchange but I think it is different and detailed.
The earth is spinning 465 meters/second so why don't we jump and land on a different location?
I googled about this question and I got some answers:
*
*When we jump, we (and the atmosphere itself) also spin along with the earth so we don't land at a different location. (But why we are also spinning along with earth ?)
*The earth is so big and we are very small relative to earth so the tiny jump won't make any difference.
But according to answer 1, We are also spinning with the earth. But the question is, "why are we also spinning along with earth" ??
Regarding the answer 2, it won't make any difference but logically it doesn't make sense or I just don't understand it.
Here is the followup question,
"When we jump why don't we thrown out of earth due to centrifugal force of the spinning earth ??"
Please provide some detailed answer to these questions. With logic if possible.
This phenomenon is not possible I know, but even if it exists, then traveling will be so easy, we just have to hang on space (with help of helicopter etc.. ) and we can travel the whole world in 24 hours :D
| When planes fly in the west direction they are basically doing what you suggest: the position of the Sun stays almost fixed and the Earth rotates below them. Still to do that they burn a lot of fuel: first they need to come at a stop with respect to the Sun, which means getting some speed with respect to the Earth, then they have to keep this speed winning the air resistance and preserving the altitude.
As when you jump you do nothing of the above, that's why you land in the very same position.
| {
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Will 5 pizzas in the same Hot Bag stay warmer than 5 pizzas in 5 separate Hot Bags? For example, say I am delivering 5 pepperoni pizzas to 5 different addresses. In one scenario, I Keep all 5 in the same insulated Hot Bag, I carry that bag to the door, and I quickly remove one of the pizzas from the bag to give to the customer. In the other scenario, I use a separate Hot Bag for each pizza. This would mean that only one bag would need to be opened while the other 4 bags could stay closed.
Which method would keep the pizzas warmer?
| With the two options, the temperature difference and bag insulation R-factor are the same.
The five-bag option has five times the surface area (if truly "separate" and not stacked). So the one big bag should be the better choice.
A less snicker-inducing version of the "two people - two sleeping bags - how to keep warmer" problem...
| {
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How do we estimate $10^{23}$ stars in the observable universe? Now, I read somewhere, that there are $10^{23}$ stars in the observable universe. How did scientists estimate this?
| An alternative method to John's answer is to look at the total number of atoms in the observable universe. Thanks to measurements of the cosmic microwave background, we have a fairly precise estimate of this number. Indeed, we know that ordinary matter makes up about 4.9% of the energy content of the universe. In this previous post, I calculated that this corresponds to about
$$
N_A = 7\times 10^{79}
$$
atoms in the observable universe. 75% of these atoms is hydrogen, and nearly 25% is helium, so the average mass of an atom is
$$
m_A \approx 0.75\,m_\text{H} + 0.25\,m_\text{He}\approx 2.9\times 10^{-27}\;\text{kg}.
$$
Next, we need an estimate of the average mass of a star. If our own solar neighbourhood is representative, we find according to this article an average stellar mass of about 1/4 the mass of the Sun:
$$
M_\star \approx 0.25M_\odot\approx 0.5\times 10^{30}\;\text{kg}.
$$
So an average star contains about
$$
N_{AS} = M_\star/m_A\approx 1.7\times 10^{56}
$$
atoms. Combining this with the total number of atoms in the observable universe, we arrive at an estimated number of
$$
N = N_A/N_{AS} \approx 4\times 10^{23}
$$
stars. Of course, we assumed here that all matter is locked up in stars, which is not true: in fact, according to this article about 75% of baryonic matter consists of diffuse intergalactic gas. And according to this post, only 6% of baryonic matter is within stars. In that case, our estimated number of stars lowers to $\approx 2\times 10^{22}$.
(Thanks to Ben Crowell for the comments)
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Ball Bearing Inside a Hollow, Spinning Rod: where is the logical flaw? As described in the title, suppose we have a frictionless, hollow rod that is rotating in the $xy$-plane with some fixed angular velocity $\omega$. The rod is pivoting around its midpoint. Suppose we place a ball bearing inside of this rod such that the diameter of the ball bearing is equal to the diameter of the hollow opening of the rod. Additionally, assume plane polar coordinates: $(r,\phi)$.
So, the constraint equation is simply $$G(\phi,t) = \phi - \omega t = 0.$$ Now, suppose that we want to find the $\phi$ component of the constraint force on the ball bearing. I know that blindly applying the generalized Lagrange equations yields
$$ \lambda = 2mr\dot{r}\dot{\phi},$$
which then implies
$$ F_{\phi,constraint} = \lambda \frac{\partial G}{\partial \phi} = 2mr\dot{r}\dot{\phi},$$
but I am not certain that time-dependent constraints can be treated in the same way. What exactly is incorrect with the following argument?
The bead is constrained to rotate in the $xy$-plane with constant
angular velocity $\omega$, hence $\ddot{\phi} = 0$. Thus, since there
is no angular acceleration we know that the net force on the bead in
the $\phi$ direction must be zero. However, the only force on the bead
in the $\phi$ direction is the constraint force from the rod on the
bead. Therefore, the $\phi$ component of the constraint force is zero.
I want to justify this by saying that the bead feels equal and opposite $\phi$ forces from the rod, as it is surrounded on all sides and thus the net force would be zero, but I think there is something fundamentally wrong with this argument. Any insight is greatly appreciated.
| You seem to realize that the rotational analog for Newton's law is important here. This law states that the net torque $\tau$ on an object and its angular momentum $L$ are related by $\tau = \dot{L}$.
If I read you question correctly you seem to think that because $\ddot{\phi}=0$, that the angular momentum $L$ must be constant. However, this isn't true. Recall that the angular momentum $L$ is given by $L=mr^2\dot{\phi}$, where $m$ is the mass of the bead and $r$ is its radius. Now in this case the radius of the bead $r$ is changing while $\dot{\phi}$ and $m$ stay constant. Therefore the angular momentum $L$ is in fact changing. So the flaw in the argument is saying that $\ddot{\phi}=0$ implies $\dot{L}=0$.
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Time energy uncertainty principle $ \sigma _{H}\sigma _{Q}\geqslant \frac{h}{4\pi }\frac{d\left \langle Q \right \rangle}{dt}$
$\Delta E = \sigma _{H}$
$\Delta t = \frac{\sigma _{Q}}{d\left \langle Q \right \rangle / dt}$
$\Delta E \Delta t \geq \frac{h}{4\pi }$
Q is any observable
I know that $\Delta E$ represents the standard deviation of energy distribution, but what does $\Delta t$ represent precisely?
I read an answer saying "It is the average time of the expectation value to change by one standard deviation, but I don't understand this sentence, I need some clarification.
| As an example, http://en.wikipedia.org/wiki/Particle_decay, you can regard $\Delta t$ as the lifetime of the particle decayed.
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Penning Trap Simulation I'm currently working on a particle tracker and I would like to implement a Penning trap. I think I might have a problem with the field of the electrical quadrupole. My idea was to place 2 dipoles and overlay their electrical fields, so that the tracked particle would be in the middle of the resulting rectangle(similar to the wikipedia article). But the resulting trajectories were not stable and didn't have the right shape. I didn't calculate if the initial conditions would work out so this could be a reason, but I wanted to know if designing it this way would be working out at all?
| The underlying problem of your trap design is that two dipoles can't make a quadrupole!
You are probably aware of the multipole expansion of electric (or magnetic) fields. This is where the terms "dipole" and "quadrupole" come from. As a physics student, one often encounters the exterior multipole expansion. This is what you use when there are charges inside some volume, and you want to describe the electric potential (or field) on the outside of (and far away from) that volume with as few and simple terms as possible.
Closeley related, but less well known, is the interior multipole expansion. This is relevant when the charges are placed on the outside of some volume, and you want to describe the potential inside that volume. The interior multipole expansion has surprisingly simple terms when you state it in Cartesian coordinates. (Except for the normalization constants, which are not really standardized and often simply dropped.)
Unfortunately, there is no consensus on how to name or order the terms that come up in the interior multipole expansion. But here is one way to state them, up to the octupolar terms:
$$\Phi(x,y,z) = \sum_{i,|j|\leq i}c_{i,j}\, P_{i,j}(x,y,z) $$
$$ \begin{align}P_{0,0} &= 1 && \nonumber \\
P_{1,0} &= z
&P_{1,+1} &= x
&P_{1,-1} &= y \\
P_{2,0} &= -\frac{1}{2}(x^2+y^2)+z^2
&P_{2,+1} &= xz
&P_{2,-1} &= yz \\
&&P_{2,+2} &= x^2-y^2
&P_{2,-2} &= xy \\
P_{3,0} &= -\frac{3}{2}(x^2+y^2)z+z^3
&P_{3,+1} &= x^3+xy^2-4xz^2
&P_{3,-1} &= x^2y+y^3-4yz^2 \\
&&P_{3,+2} &= zx^2-zy^2
&P_{3,-2} &= xyz \\
&&P_{3,+3} &= x^3-3xy^2
&P_{3,-3} &= y^3-3x^2y \quad .\end{align}$$
The $P_{0,0}$ term is the monopole term. The corresponding field, given by $\vec{E}_{0,0} = -\vec\nabla P_{0,0} $), evaluates to $0$.
The $P_{1,j}$ terms are the dipole terms. The corresponding fields are
$$\begin{align}
\vec{E}_{1,0} &= -c_{1,0} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \\
\vec{E}_{1,1} &= -c_{1,1} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \\
\vec{E}_{1,-1} &= -c_{1,-1} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\quad,
\end{align}$$
which are homogenous fields in the $x$, $y$, and $z$ direction.
The $P_{2,j}$ terms are the quadropolar terms. If the magnetic field of your trap points along the $z$-axis, then $P_{2,0}$ gives the electric field that you need for the Penning trap:
$$ \vec{E}_{2,0} = c_{2,0} \begin{pmatrix} x \\ y \\ -2z \end{pmatrix}\quad . $$
The $P_{3,j}$-terms are the octupolar terms. You hopefully won't need them for your Penning trap (although some traps use them for coupling different modes).
As you can see from the polynomials, or from the corresponding vector fields, there is no way in which you can add two dipoles to make a quadrupole. These terms are all orthogonal to each other.
Exterior and interior multipole expansion are sometimes called "irregular solid harmonics" and "regular solid harmonics", respectively. See this post for a better overview, and for Mathematica code on how to generate the polynomials.
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Estimating the force needed to increase height of a mountain? How would you estimate the force required by a tectonic plate to make the height of a mountain increase when it pushes against another?
I've used a method to try and do it for Mt Everest and have ended up with 8x10^(-7)N required to increase its height which I don't think sounds reasonable
By the way, I was doing this for fun (it doesn't have any real life value, I don't think)
| In really simple terms the equation f=mg can be utilised. It calculates the gravitational potential energy of the mountain and therefore the energy needed to lift a mountain.
| {
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What is the difference between diffraction and interference of light? I know these two phenomena but I want to know a little deep explanation. What type of fringes are obtained in these phenomena?
| Both interference and diffraction result from superposition of the EM waves. Inteference result from the superposition of two different coherent sources whereas in diffraction superposition result from different parts of the same source. So we speak about diffraction resulting from a wide slit or circular aperture and interference resulting from two slits or a number of slits. Though light in these slits may originate from the same source we consider them as different sources.
| {
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Can time be interacted with? Astronauts come back to Earth younger than they would have been had they stayed on Earth for that same period of time. They are traveling so fast relative to the Earth that time slows down for them. Does that mean that the astronaut interacted with time? Does time interact with speed? And if so, does that mean time is made up of some kind of fundamental particles, like gravity has the graviton? What does time have?
| As for "why", that is either simple or complex! The simple answer is that it is a consequence of the speed of light being constant for all observers. That means when travelling relative to each other they both measure time differently. This site provides a simple math explanation
OTOH, time is measured differently in different gravitational fields, and that is a lot more complex - General Relativity
| {
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Derivation of Maxwell stress tensor from EM Lagrangian From Noether's theorem applied to fields we can get the general expression for the stress-energy-momentum tensor for some fields:
$$T^{\mu}_{\;\nu} = \sum_{i} \left(\frac{\partial \mathcal{L}}{\partial \partial_{\mu}\phi_{i}}\partial_{\nu}\phi_{i}\right)-\delta^{\mu}_{\;\nu}\mathcal{L}$$
The EM Lagrangian, in the Weyl gauge, is:
$$\mathcal{L} = \frac{1}{2}\epsilon_{0}\left(\frac{\partial \vec{A}}{\partial t}\right)^{2}-\frac{1}{2\mu_{0}}\left(\vec{\nabla}\times \vec{A}\right)^{2}$$
Applying the above, all I manage to get for the pressure along x, which I believe corresponds to the first diagonal element of the Maxwell stress tensor, is:
$$p_{x} = \sigma_{xx} = -T^{xx} = \frac{-1}{\mu_{0}}\left(\left(\partial_{x}A_{z}\right)^{2}-\partial_{x}A_{z}\partial_{z}A_{x}-\left(\partial_{x}A_{y}\right)^{2}+\partial_{x}A_{y}\partial_{y}A_{x}\right)+\mathcal{L}$$
But I can't see how this can be equal to what is given in Wikipedia.Why is this?
| Hint: The canonical stress-energy tensor from Noether's theorem is not necessarily symmetric, and often needs to be improved with appropriate improvements terms. This is e.g. the case for EM. See also e.g. this Phys.SE post and links therein.
References:
*
*Landau and Lifshitz, Vol.2, The Classical Theory of Fields, $\S$33.
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What would happen if an accelerated particle collided with a person? What would happen if an accelerated particle (like they create in the LHC) hit a person standing in its path?
Would the person die? Would the particle rip a hole? Would the particle leave such a tiny wound that it would heal right away? Something else?
| A charged particle will create charge separation (ionization) along its path. This will cause harmful chemical reactions to occur in the body, including DNA damage. The effects of these chemical reactions depend on their amount. The body can heal from a low amount on its own, while a high amount will cause radiation sickness and probably death.
This can be calculated, but also deduced by comparison.
A single proton from LHC has 4 TeV of energy. This is much less than the probable energy of cosmic rays' protons.
According this plot, approximately four TeV protons hit each square meter each month. Once per year, they are protons of 10E16 eV, i.e. 10000 times harder, than those from LHC.
Cosmic ray protons hardly ever reach Earth's surface but astronauts can be exposed to them. No damage to astronauts from individual protons has been registered. Some astronauts report rare unusual light flashes in their eyes which may be caused by particles penetrating their brains or retinas.
P.S. Also see the answer about BEAMS of particles, which can definitely damage the body.
| {
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What is an order parameter? I've seen order parameter used in two different ways. One is to distinguish between an ordered and an unordered phase, like whether the net magnetization is stable or not. The second way is to distinguish what the magnetization is, up or down.
More broadly, does it just mean a macroscopic observable? For us to see it at large scales, doesn't it have to be relatively stable?
EDIT: by macroscopic observable I mean like the net magnetization, which is an average of a local operator.
| An order parameter distinguishes two different phases (or orders). In one phase the order parameter is zero and in another phase it is non-zero. It does not have to be macroscopic.
For example, in the BCS theory of superconductivity the order parameter is called the gap $\Delta$. It can be interpreted as the binding energy of a Cooper pair, namely two electrons that become correlated over long ranges due to an attractive interaction. The order parameter, or gap, shows up as a minimum energy that would be needed to excite a single-particle in the system.
| {
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Two soft questions about spin and the particle nature of electrons How can we define spin as the spin of an electron around it's own axis if an electron is described by a probability cloud of finding an electron in a point in space? How does that probability cloud spin around it's own "axis"(I find this ill-defined too) and create magnetic field? Also, when is the elctron in an atom described as a particle and follows particle principles?
| Spin is not defined as the spin of electron around its own axis. Spin is the intrinsic angular momentum of the electron - intrinsic meaning it does not arise from the electron's motion, but is a property of electron itself.
The electron in the atom "can" be described as a particle if you are using the Bohr model of the atom. Quantum mechanical picture of the electron is a more accurate description and should be used when dealing with atom-sized objects if you can deal with the mathematics.
| {
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Finding 3-Sphere Christoffel connection coefficients using variational calculus, Sean Carrol problem I have A 3-Sphere with coordinates $x^{\mu} = (\psi,\theta,\phi)$ and the following metric:
\begin{equation}
ds^2 = d\psi^2 + \text{sin}^2\psi(d\theta^2 + \text{sin}^2\theta d\phi^2)
\end{equation}
I know how to get the connection coefficients using the metric derivatives etc, but I'm looking for a way to do this through calculus of variations. A problem in Sean Carroll (Exercises 3.11 question 8 a) Introduction to General Relativity suggested varying the following integral to find the connection coefficients:
\begin{equation}
I = \frac{1}{2}\int g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{v}}{d\tau} d\tau
\end{equation}
So I have a lagrangian:
\begin{equation}
\mathcal{L} = \dot{\psi}^2 + (\text{sin}^2\psi) \dot{\theta}^2 + (\text{sin}^2\psi)(\text{sin}^2\theta)\dot{\phi}^2
\end{equation}
Which I put into the Euler-Lagrange equation:
\begin{equation}
\frac{\partial}{\partial \tau}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}^\mu}\right) - \frac{\partial \mathcal{L}}{\partial x^\mu} = 0
\end{equation}
Am I on the right track here? What is the strategy for relating this back to the connection symbols? The literature isn't too clear and I'm struggling to make the connection.
| The strategy is to recall the geodesic equation,
$$
\frac{d^2x^\lambda}{dt^2}+\Gamma^\lambda_{\,\mu\nu}\frac{dx^\mu}{dt}\frac{dx^\nu}{dt}=0\tag{1}
$$
From your Lagrangian, you'll end up with equations of the form
\begin{align}
\ddot{\psi}&=f(\psi,\,\theta,\,\phi,\,\dot{\psi},\,\dot{\theta}\,\dot{\phi})\\
\ddot{\theta}&=g(\psi,\,\theta,\,\phi,\,\dot{\psi},\,\dot{\theta}\,\dot{\phi})\\
\ddot{\phi}&=h(\psi,\,\theta,\,\phi,\,\dot{\psi},\,\dot{\theta}\,\dot{\phi})\\
\end{align}
to which you relate to (1) index-by-index.
| {
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Heisenberg's Unified Field Theory While searching in the Internet, I came to know about Werner Heisenberg's attempt to obtain an Unified Field Theory (see the book Introduction to Unified Field Theory of Elementary Particles). But also it has been mentioned that since this theory is non-renormalizable, it wasn't a successful candidate for the UFT. Was non-renormalizability the only reason? If not then what were the other reasons? Specifically, my question is that was the theory abandoned because it was wrong or because it wasn't practically efficient? If it was wrong then can some instances be provided?
| Good references were provided by Conifold here.
To sum up, there were several problems with this initial approach: spinors where observable physical objects due to their dimension of $-\frac{3}{2}$, the introduction of a dimensioned coupling constant lead to non-renormalizability, etc.
You may want to check out this video proposing changes to this initial approach, leading to some quite interesting results:
A non-linear spinor theory of elementary particles
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Nature of tension in a massless rope I'm faced with a conceptual problem, where I am supposed to describe the motions of the following system (two masses on each end of a totally inelastic rope hanged on a pulley):
I understand that there is the gravitational force acting, and also a second, opposing force in the rope. My problem is that I don't understand the nature of this second force. Where does it come from? How do I summarize it for each of the masses? From my solutions manual I know that it must be equal $|T_1|=|T_2|$ for each masses, but that sounds plain counterintuitive for me. Can you please shed some light on this?
UPDATE: the problem is formulated so that the pulley and the rope are both massless, and the bearing has no friction.
| The nature of the rope force is such that the accelerations of the two masses are connected. If the two heights are $y_1$ and $y_2$, then the their sum is constant, and their derivatives are equal to zero
$$ y_1 + y_2 = \ell \\
\dot{y}_1 + \dot{y}_2 = 0 \\
\ddot{y}_1 + \ddot{y}_2 = 0 $$
So the nature of the force is such to enforce $\ddot{y}_1 = - \ddot{y}_2$ such that when you do a free body diagram you will get
$$ m_1 \ddot{y}_1 = T - m_1 g = - m_1 \ddot{y}_2 \\ m_2 \ddot{y}_2 = T - m_2 g $$ which is 2 linear equations solved for 2 variables $T$ and $\ddot{y}_2$
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Lorentz Transformation at t=0 Suppose I have two reference frames $S$ and $S'$, where $S'$ is moving with velocity $v$ with respect to $S$.
The Lorentz transformation equation for time in reference frame $S$ is given by:
$$t'=\gamma\left(t-x\frac{v}{c^2}\right)$$
where $\gamma$ is the Lorentz factor.
Now for an event happening at $t=0$ and at some position $x$ in $S$ frame, then $t'<0$ for $S'$ frame. But if $x=0$ then $t'=0$.
How should I interpret this result?
| The interpretation is that two events being simultaneous as measured in frame $S$ doesn't imply that the events are simultaneous in frame $S'$. Which events count as being "simultaneous" depends on the frame of reference. This is known as the relativity of simultaneity.
Added clarification due to comment:
The coinciding of the origins is an event, call it event $A$. The coordinates of $A$ as measured in $S$ are $t_A=0$, $x_A=0$, and the coordinates of $A$ as measured in $S'$ are $t'_A=0$, $x'_A=0$.
You're also considering another event, call it $B$, whose coordinates as measured in $S$ are $t_B=0$, $x_B=x$. As measured in $S'$, the $t'$ coordinate of $B$ is $t'_{B}=-\gamma x v/c^2$.
$t'_B$ being negative (if $x>0$) means that according to $S'$, event $B$ occurs before event $A$. However, if $x=0$, then $B$ is the same event as $A$, so $t'_B=0$ and $x'_B=0$.
There is no physical discontinuity between the situations where $x=0$ and $x>0$, i.e., between the situations where $t'_B=0$ and $t'_B<0$. It's just that the further away event $B$ is from event $A$ spatially, the bigger the difference will be between $t'_B$ and $t_B$.
| {
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Why is the Moon not redder at moonrise/moonset? Okay we all know about Raleigh Scattering, which makes the sky blue. And by the same token, sunsets appear red because sunlight traveling through more atmosphere will "lose more blueness" as it's scattered away.
But what about the Moon? The Moon is just reflected sunlight, so when the Moon is setting on the horizon, it should appear reddish right? But I've never seen that happen.
Now I know lunar eclipses are red, so I'm not discounting the principle of Raleigh Scattering or anything. But there seems to be something else at play causing the normal rising/setting Moon to not turn red.
| Is this not a question about human physiology rather than physics?
We are discussing the variations in observed response of the human eye to light with different qualities: overall intensity, and distributions over the visible spectrum.
This question seems to ascribe all of the observed differences (or lack thereof) to the quality of the light; "the Moon [is] not redder" is the basis of the question.
The appearance (to a human observer!) of a color wheel under sunlight and moonlight changes greatly. Yet we do not theorize about the changes in the pigments under various light levels. We work out the sensitivity of the rods and comes in the retina, and conclude that moonlight is not intense enough to trigger color vision in the human eye.
Perhaps the question should be deferred, pending measurement of the spectrum of moonlight for the moon at different elevations.
| {
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The pole and the barn paradox (stretching the pole) The basic variant of this paradox makes sense to me but I have problems in proving mathematically variation of this non-paradox.
So lets imagine that we have barn and pole which is stationary to the barn and they both have rest length $L_0$. Moreover, we are given team of people who are rushing towards the barn with speed $v$. When they reach the barn, they grab the pole and continue. We are asked to find what will be the length of the pole according to Team Pole if Team Barn says that the members of Team Pole grab the pole simultaneously.
So as Team Barn says that the pole is grabbed simultaneously, there is no stretching or compression from their perspective-only Lorentz contraction. Solving the equations from their perspective I get that they will claim that the pole's length is $L_0/\gamma$ and from here according to Team Pole the length must be $\gamma(L_0/\gamma)=L_0$. So far so good but lets now solve first for Team pole.
Team Barn says that they have grabbed the pole simultaneously so from team pole's perspective-they have grabbed the pole with $\Delta t=(v L_0/c^2)\gamma$. this leads us to conclusion that there is stretching of the pole from Team Pole's perspective. Solving all the equations I get: $$L_{new length}=L_0/\gamma+v(v L_0/c^2)\gamma=L_0 \gamma$$
which is in contradiction with the first answer. My question is where is my mistake in calculating the new length first from Team Pole's perspective. I suppose that there must be something wrong with the calculations in the second way because the first way makes sense in all the perspectives.
| There's going to be a distortion of the pole during the acceleration period that goes away. The pole bearers should observe that they are holding a pole with length $L_{0}$, assuming that it's rigid. Note that it will start out length contracted from their potential, though, so during your $\Delta t$, they will stretch the pole out to its rest length.
| {
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Weight versus gravitational force What is the difference between weight and gravitational force? I am a beginner who want to study physics properly.
| Newton law of gravitation is given by:
$$F = G \frac{m_1 m_2}{r^2}=\left(G \frac{m_1}{r^2}\right) m_2$$
The gravitational constant, $G$, the weight of Earth, $m_1$, and the radius are constants, so:
$$G \frac {m_1}{r^2}=(6.6742 \times 10^{-11}) \frac{5.9736 \times 10^{24}}{(6.37101 \times 10^6)^2}=9.822$$
Hence, the equation simplifies to
$$F =(9.822) m_2$$
where 9.822 is the gravitational acceleration, $g$. Therefore you obtain the equation $W = mg$.
| {
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A hypothetical question on mechanics Being located in a tropical region, I am quite acquainted with the Ceiling fan. I have a question about it.
If the top, that is, the axle (I'm not sure of the terminology: I mean the part which is thin, rodlike, attached to ceiling)...is rigidly fixed, then when the fan is turned on, the blades spin.
On the other hand, when the local mechanic brings the fan down and puts it on the floor, then when it is powered on the blades are fixed and the axle spins.
My question is this:
If, neglecting friction, I keep both top and bottom of the fan free to move (like maybe in outer space), and I turn on the fan, what will happen?
|
If, neglecting friction, I keep both top and bottom of the fan free to move (like maybe in outer space), and I turn on the fan, what will happen?
Some satellites use something very much like this to keep the satellite pointing in the right direction, and do so without using rockets. It's called a reaction wheel. The motor is rigidly connected to the satellite body. What about the fan blades? There's no air to move in space, and moving air isn't the point of a reaction wheel. The point of a reaction wheel is to make the satellite rotate a bit so it points in the right direction. Reaction wheels use a solid flywheel instead of fan blades.
When your fan breaks down you just call your electrician or mechanic. That's a bit tougher in space. Astronauts replaced one of the reaction wheels on the Hubble Space Telescope during one of the Hubble repair missions. The Kepler satellite on the other hand is out of the reach of any repair mission. The breakdown of two of its reaction wheels pretty much put Kepler out of the planet finding business.
| {
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Storing a Planet-sized Chunk of Metal Inside a Star Would it be physically possible to "store" a planet-size or larger sum of metal, say gold or platinum, inside a star by letting it fall to the core?
Would it be possible to detect which stars had these treasures inside them?
(This is for a Sci-Fi project, but I'd like to root it in reality).
| "Storing" something implies the purpose is to put is somewhere safe so that it can eventually be retrieved. Heavy metals should eventually sink to the center of a star, but how are you going to retrieve it? Even in a science fiction context, it's hard to imagine a plausible means or retreiving a large pile of heavy metal from the center of a star. Therefore, I'll say no, it's not possible.
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Moment of inertia of rods Ok so I'm extremely comfortable with calculating moment of inertia of continuous bodies but how do we do it for a system not continuous.
For example if 3 rods of mass $m$ and length $l$ are joined together to form an equilateral triangle what will be the moment of inertia about an axis passing through its centre of mass perpendicular to the plane.
i know that moment of inertia of each rod is $ml^2/12$ and c.o.m is at centroid?
also if 2 rods form a cross then to calculate the moment of inertia about its point of intersection would it be correct to sum up the individual moment of inertia of the rods form??
| @DrChuck's answer is correct. Generally speaking the total moment of inertia is the sum of the moments inertia calculated individually. You have to be careful about the the axis of rotation thought: if you wanted to calculate the moment of inertia (with respect to any axis) of a T shape created from 2 identical rods, you would calculate the moment of inertia of each rod differently because the axis of rotation as seen from the perspective of the rod would be different in each case.
Also if it's homework we can neglect the contact point of the different rods. In engineering we would have to alter the rods to make them stay together in a shape.
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What does the $c$ in $eV/c^2$ stand for? I have been wondering(also searching) for what does the $c$ in eV/$c^2$ stand for?
(For example, mass of the electron is $0.511 \, \text{MeV}/c^2$.)
I have read that this unit has been derived from Einstein's equation, $E=mc^2$, but how it is possible, We use the symbol $c$ for Coulomb.
Also, tell how to convert this to our normal units of mass ($\mathrm{eV}/c^2 \to \mathrm{kg}$).
| For physicists it can be very annoying that our historically evolved units of measurement cause the speed of light $c$ to differ from unity. So physicists often apply a trick to avoid distracting conversion factors corresponding to the numerical value of (powers of) the speed of light popping up in their equations. That trick is simply to define your own units such that length and time units are treated consistently from a relativistic perspective.
So one can, for instance, select an energy unit like the mega-electronvolt ($\text{MeV}$) and use this along with a relativistically consistent unit for mass written as $\text{MeV}/c^2$. Such an expression denotes nothing more than the amount of mass corresponding with one MeV of rest energy.
Similarly, one can use an expression like $\mathrm{s}c$ ($\mathrm{second} \times c$) to denote a unit of distance. This notation, however, is less common, as one usually refers to such a distance unit as light second.
In particle physics it is common to start from the giga-electronvolt ($\text{GeV}$) as energy unit, and and fix all other units such that both the speed of light $c$ and the reduced Planck constant $\hbar$ both have value unity. This leads to units $\text{GeV}/c^2$ for mass, $\hbar / \text{GeV}$ for time, and $\hbar c/ \text{GeV}$ for distance.
| {
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Is there a difference between the speed of light and that of a photon? As in the title I am curious whether there is a difference between the speed of photon and the speed of light, and if there is what is the cause of such a difference?
| I haven't done this in a long time, but my understanding from Feynman's QED is that the speed of a photon is unknown (Heisenberg) - photons traveling in a vacuum are around the speed of light +-, but at any instant the speed differs due to uncertainty. The photons going faster and slower than light speed cancel in the same manner that photons bouncing off a mirror cancel all but the equal angle reflection.
| {
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What happen to a spoon which is detached from the satellite? Suppose a spoon is a part of satellite but after detachment from the satellite Does it fall to the ground straight or does is take a parabolic path or any other path before coming to the surface of Earth
| The spoon will keep moving until orbital decay takes control and it reaches the earth over several orbital periods.
To explain why, imagine a spoon hung to a satellite traveling at a constant velocity (ignoring any accelerations, for ease of understand) of $x$ and therefore all its components must be travelling at the exact same velocity to make sure their relative velocity is $0$ m/s, that said the spoon must be also travelling at $x$ velocity but as orbital decay slowly pulls the object by slowing it down it falls to earth gradually.
Furthermore, space rocks and other objects could disturb the object by slowing it down or pushing it towards earth and such or destroying it in process.
The key effect here you might want to see is orbital decay: http://en.wikipedia.org/wiki/Orbital_decay
| {
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Is this an entangled state? Is the following state entangled?
$\left| \psi \right> = \alpha_0 \beta_0 \left| 00 \right> + \alpha_0 \beta_1 \left| 01 \right> + 0 \left| 10 \right>+ \alpha_1 \beta_1 \left| 11 \right>$
I 'know' it is an NOT an entangled state, but I here's where I am a bit confused. It says $ \alpha_1 \beta_0 = 0$ which implies either $ \alpha_1 = 0$ or $\beta_0 =0$ , but that would be a contradiction. (since we assume $\left| 00 \right>$ and $\left| 11 \right>$ have non-zero probability amplitudes)
Also, if this is NOT an entangled state, how can I factor it to $\left| \phi_1 \right> \otimes \left| \phi_2 \right> $?
| It is an entangled state, as you concluded yourself it cannot be written as a direct tensor product. To recap again:
$$
\left(
\begin{array}{c}
\alpha_0\\
\alpha_1\\
\end{array}
\right)
\otimes
\left(
\begin{array}{c}
\beta_0\\
\beta_1\\
\end{array}
\right) =
\left(
\begin{array}{c}
\alpha_0 \beta_0\\
\alpha_0 \beta_1\\
\alpha_1 \beta_0\\
\alpha_1 \beta_1\\
\end{array}
\right)
$$
If $\alpha_1 \beta_0$ is set to 0, and the rest kept non-zero, writing $\left| \psi \right>$ as a direct tensor product fails, as there no $\alpha$ or $\beta$ satisfying these conditions at the same times.
More generally, when it comes to questions of type "is this state entangled or separable", it is much simpler to tackle it at the level of density matrices. For instance, given any pure 2-qubit state, such as yours, a simple criterion indicating a separable state would be to calculate the density matrix $\rho$, trace out (integrating out undesired degrees of freedom) $\rho$ w.r.t. the basis of one of the qubits to obtain the reduced density matrix of the other qubit, then calculate $\operatorname{Tr}(\rho_{\rm reduced}^2).$ The state is separable iff the latter is equal to $1,$ otherwise it is entangled. In other words, if the reduced density matrix has rank $1,$ then it is a pure state (as opposed to mixed), and thus the composite 2-qubit state must be a separable one.
Let's take a separable state and verify the above scheme. Suppose we have $|\psi\rangle=|\phi_1\rangle|\phi_2\rangle,$ then $\rho=|\psi\rangle \langle \psi|.$ The reduced density matrix of the first qubit is then obtained by tracing out in the basis of the second: $\rho_1=\operatorname{Tr_2}(\rho)=|\phi_1\rangle \langle \phi_1|.$ Squaring and tracing $\operatorname{Tr}(\rho_1^2)=1,$ thus, confirming that $\psi$ is indeed a separable state.
Specifically for your example, the normalization condition gives you a relation between the coefficients, namely, $|\alpha_0\beta_0|^2+|\alpha_0\beta_1|^2+|\alpha_1\beta_1|^2=1$ $:(1).$ Then by performing the above calculation (well-encouraged exercise), you should get to an equation of type $Tr(\rho_1^2)=|\alpha_0|^2|\beta_0|^2+|\alpha_1|^2|\beta_2|^2 <1,$ clearly smaller than $1,$ which directly follows from the normalization criterion $(1).$ Hence, $|\psi\rangle$ describes indeed an entangled state of the 2-qubit system.
| {
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Final velocity of a spring I need to calculate the velocity of an object when it is thrown by a spring; we have to calculate the velocity from $U=\frac 12 kx^2$.
Now I know that $U=L=F\cdot S$ and $S=\frac{v_f^2-v_0^2}{2a}$. Here's what I did:
$$U=L=F\cdot S=ma\cdot S\rightarrow L=ma\cdot \frac{v_f^2}{2a}=\frac{m\cdot v_f^2}{2}$$
$$v_f=\sqrt{\frac{2L}{m}}\rightarrow \sqrt{\frac{2\frac 12 kx^2}{m}}=x\cdot\sqrt{\frac km} $$
Is this correct?
| Yes, alternatively you can use that the potential energy of the spring is transformed into kinetic energy of the object. This is simpler than considering the work $L$ done by the spring. The result is however the same:
$$
U=E_{kin}\\
\Rightarrow \frac{1}{2} k x^2=\frac{1}{2}m v^2\\
\Rightarrow v=x\sqrt\frac{k}{m}
$$
| {
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What does the "size of the universe" mean if the Universe is infinite? There are questions that may seem to be similar to this one, but I've yet to find an answer.
I have come to understand that a flat universe, that is to say a curverature of $k=0$ which means that $S_k(r) = r$. The FLRW metric polar coordinates:
$$ds^2 = -dt^2 + a^2(t) \left[ \frac{dr^2}{1 - kr^2} + r^2d\Omega^2 \right]$$
Now, since only $r$ is being altered, $dt = 0$ and $d\Omega$ = $0$.
$$ds^2 = a^2(t) \frac{dr^2}{1 - kr^2}$$
This can be integrated to:
$$s(r) = \frac{\sin^{-1}(\sqrt{k}r)}{\sqrt{k}}$$
So, by definition, the maximum value is when $\sin^{-1} = 90^\circ = \frac{\pi}{2}$ which occurs when $\sqrt kr=1$ To find the highest value, we replace $\sin^{-1}(\sqrt kr)$ with $\pi \over 2$ and get:
$$s(r)_{\text max} = \frac{\pi}{2\sqrt k}$$
Therefore, as $k \to 0,\space s(r)\to \infty$. Given $k=0$, there is an infinite possible distance.
Now that we have that out of the way, when physicists talk about the size of the universe, by which I mean "when the universe was the size of a grapefruit" r a similar comparisson, space must have still been infinite, so what is this a description of?
| All statements like "when the universe was the size of a grapefruit" refer to the currently observable universe. As the universe has a finite age and light travels at a finite speed (and there is nothing infinite going on with expansion), the observable universe is a finite patch.
I discussed some of the different notions of horizons in answering another question. The "observable universe" is taken to extend out to the particle horizon. That is, it includes precisely the points in our current time slice whose past worldlines (assuming they simply go with the expansion of space and have no peculiar velocity with respect to our reference frame) intersect the interior of our past light cone.
If you think of galaxies as marking these points, these are precisely the galaxies that we can see assuming arbitrarily good telescopes, since the light reaching us today was emitted as the galaxy crossed our past light cone.
Galaxies that started out too far away from us in an infinite universe haven't been able to get their photons to us. And indeed expansion will prevent most of them from ever getting to us.
The scale factor $a$ when the universe was the size of a grapefruit is simply the radius of a grapefruit divided by the radius of the current observable universe (about $46\ \mathrm{Gly}$), or something like $10^{-28}$ (corresponding to a redshift of about $z = 10^{28}$). The idea is that the galaxies (or rather their precursor quantum fluctuations) inside this grapefruit-sized volume are exactly the galaxies inside our observable universe today. In comoving coordinates the grapefruit is the same $46\ \mathrm{Gly}$ in radius then as our observable universe is now.
| {
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Planets and Pluto? Neptune? If one of the rules to be a planet is that it needs to clear ALL objects from their orbit, does this also make Neptune a non-planet? Since it has thus far failed to clear Pluto from it's orbit. Or does this rule not really apply to planets and we should welcome Pluto and a few other dwarf planets into our family of planets?
| Neptune actually is the dominant gravitational force in the region of the Kuiper belt in which Pluto resides. In fact, if you look at the image below, the belt is being cleared out by Neptune:
In fact, there is a class of objects, suitably named the plutinos, that have been captured by Neptune. Solar system models have actually shown that Neptune was formed much closer to the Sun than it is presently, and in drifting outwards, it has pushed these plutinos out.
The key trait about the plutinos is that they have an orbital resonance with Neptune. Here, orbital resonance is the effect where two objects exert periodic gravitational force on one another. In the case of the plutinos, they all have an average resonance of 2:3 (they orbit the Sun twice for every three orbits of Neptune)--in fact, Pluto was the first object that this resonance was first noticed. This resonance developed because Neptune pushed the plutino objects outwards as it migrated further from the Sun (i.e., as the dominant gravitational body, clearing the path).
| {
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How stable are Benard Cells when the thermal gradient begins decreasing? I've found lots of information on the formation of Benard Cells and convection currents but very little information about what happens to the self-organized structures when the energy gradient being applied begins to get smaller.
Do the complex structures persist as the gradient is diminished?
Do they break and then form slightly less complex structures?
| The dimensionless Rayleigh number characterizes buoyancy driven convection. When the Rayleigh number is below a critical value, heat transfer is primarily by conduction (e.g. no Benard convection cells). When the Rayleigh number is above the critical value, heat transfer is primarily by convection (e.g. Benard convection cells spontaneously form and persist). The Rayleigh number is linearly proportional to the temperature difference, and can be written in terms of the temperature gradient.
| {
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Forces Create Angular Acceleration And "Straight" Acceleration - But How Much Of Each? Let me set up the following problem for a rectangle floating in space:
*
*We know its dimensions.
*We know its mass.
*There's a force pushing it for a known amount of time - we know the angle & magnitude of the force.
*We know the point on the rectangle the force is being applied.
Here's a picture I made of the situation:
I can use the torque & moment of inertia equations to determine the angular acceleration this rectangle will experience.
But I would also imagine this rectangle will experience some "straight" acceleration.
For example in this picture, I can see the rectangle rotating counterclockwise, but also moving in an up-left direction.
My Question: I would use my torque & moment of inertia equations to determine the angular acceleration of the rectangle. But that's only part of the motion it would experience. How would I calculate the "straight" acceleration it has?
My best guess is that the "straight" acceleration is just going to be f/m, but since the rectangle is also rotating while the force is being applied, then the force vector keeps changing and this will make for some difficult computations?
| If you look at the planar motion, with the force $\vec{F}=(F_x,F_y)$ going through a point $\vec{r}=(x,y)$ applied to a rigid body, then the equations of motion are:
$$ \begin{align} F_x & = m \ddot{x}_C \\ F_y & = m \ddot{y}_C \\ (x-x_C) F_y - (y-y_C) F_x & = I_C \ddot{\theta} \end{align} $$
where $(x_C,y_C)$ is the location of the center of mass and $I_C$ is the mass moment of inertia about the center of mass.
Note that $\ddot{x}_C$, $\ddot{y}_C$ and $\ddot{\theta}$ are the linear and angular accelerations of the center of mass.
| {
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Defining creation and annihilation operators Creation and annihilation operators can be defined in several different ways, some more general than others. We usually choose to denote by $a$ the annihilation operator and by $a^\dagger$ the creation operator.
It can be seen that the two operators must be Hermitian adjoint of each other. But is there a reason why we make this particular choice (i.e., which operator doesn't have the dagger)?
| It boils down to a matter of convention. Nothing stops you from choosing the annihilation operator to be $a^\dagger$.
Still, in quantum field theory, you decompose e.g. a scalar quantity in plane waves
$$ \Phi(x) = \int \frac{d^4 k}{4 \pi} \left( a(k) e^{ik\cdot x} + a^*(k) e^{-ik\cdot x} \right)$$
where obviousely the second part is the c.c. of the first part. Here, it is usual to combine $a$ with $e^{ikx}$ which annihilates a particle from your Fock state.
But again, this is just convention and you are free to choose which operator is the "real thing" and which is "just the adjoint" any way you like.
| {
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Kinetic energy vs. momentum? As simple as this question might seem, I failed to intuitively answer it.
Let's assume there is a $10,000$ $kg$ truck moving at $1$ $m/s$, so its momentum and KE are: $p=10,000$ $kg.m/s$ and $KE=5,000$ $J$.
Now if we want to stop this truck, we can simply send another truck with the same mass and velocity in the opposite direction to collide with the first one, and both of them will stop because of the conservation of momentum.
But what if we want to stop the truck by doing work in the opposite direction of motion ?
Let's assume there is a rope with an end tied to the back of the the truck, and the other end is tied to a $400$ $kg$ motorcycle moving at $5$ $m/s$, then its $p=2,000$ $kg.m/s$ and $KE=5,000$ $J$.
Now we have a truck moving in one direction with the same kinetic energy as the motorcycle which is moving in the opposite direction, but the truck has more momentum. So will the truck stop by the energy (or work) of the motorcycle ?
If yes, then how is the momentum conserved, and if no, then where does the energy of the motorcycle go?
Ignore any friction forces.
| The truck keeps going. Assuming the rope does not break, then the kinetic energy ends up in elastic potential energy in the rope. The rope will stretch. If you ask what happens if you assume a rope that cannot stretch, bzzt. No such thing. The rope will either stretch or break.
| {
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What is the Weak force? In this Particle Physics books I'm reading it explains Weak force with Beta Decay, a Neutron turns into a Proton after emitting an electron, so after it emits an electron one of the neutrons down-quarks turn in to an up-quark, and for this there needs to be a force, which is the Weak force. This interaction is carried by the W and Z bosons. Now, I asked someone what Weak Force is and they said that it's the interaction between Protons and Electrons(Electronegativity). I would like to know what exactly the relation between the Weak Force and Electronegativity is, and if my initial definition of the Weak Force is wrong, I would appreciate it if I were corrected.
| The weak force acts on particle that have weak hypercharge, just as electromagnetism acts on objects with electrical charge and gravity acts on objects with mass.
All the quarks and leptons have weak hypercharge, so the weak force act on them.
The description you have been given of beta decay is incomplete because it does not conserve the overall lepton number (one of the few symmetries that the weak force respects). That is, you've been told
$$ n \to p^+ + e^- \,,$$
but the correct reaction includes an anti-neutrino
$$ n \to p^+ + e^- + \bar{\nu}_e \,.$$
The weak force also manifests in flavor changing hadron decays such as
$$ K^+ \to \pi^0 + \mu^+ + \nu_\mu \,.$$
and in parity violating scattering.
| {
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Explanation for negative specific heat capacities in stars? I've just found out that a negative specific heat capacity is possible. But I have been trying to find an explanation for this with no success.
Negative heat capacity would mean that when a system loses energy, its temperature increases. How is that possible in the case of a star?
Musn't there be a source of energy to increase the temperature of any system?
| Consider a satellite in orbit about the Earth and moving at some velocity $v$. The orbital velocity is related to the distance from the centre of the Earth, $r$, by:
$$ v = \sqrt{\frac{GM}{r}} $$
If we take energy away from the satellite then it descends into a lower orbit, so $r$ decreases and therefore it's orbital velocity $v$ increases. Likewise if we add energy to the satellite it ascends into a higher orbit and $v$ decreases.
This is the principle behind the negative heat capacity of stars. Replace the satellite by a hydrogen atom, and replace the Earth by a large ball of hydrogen atoms. If you take energy out then the hydrogen atoms descend into lower orbits and their velocity increases. Since we can relate velocity to temperature using the Maxwell-Boltzmann distribution this means that as we take energy out the temperature rises, and therefore the specific heat must be negative.
This is all a bit of a cheat of course, because you are ignoring the potential energy. The total energy of the system decreases as you take energy out, but the decrease is accomplished by decreasing the potential energy and increasing the kinetic energy. The virial theorem tells us that the decrease of the potential energy is twice as big as the increase in the kinetic energy, so the net change is negative.
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For a massless pulley moving upwards with acceleration, is the upward force equal to the downward force? Imagine a massless and frictionless pulley with two weights hanging either side of the pulley by a massless string.
Like this except not attached to a ceiling
Rather than being fixed to a ceiling, the pulley is being pulled upward by an external force F, with the weights and string still attached.
Due to Newton's 2nd Law,
$\Sigma F_y=F-2T=ma$,
where $T$ is the tension in the string on either side of the pulley and $a$ is the vertical acceleration of the pulley.
Clearly, since there is a net upward force, the pulley itself will accelerate upwards.
But because the $m=0$,
$F-2T=0$.
Does this not then suggest that the pulley has a constant velocity?
| In the equation $F_{net}=ma$, normally we would assume that $F_{net}=0$ implies $a=0$ on the right-hand side. However, for a massless object, we can satisfy the equation by having $F_{net}=0$, $m=0$, and $a\ne0$. In reality, of course, the pulley is not massless, so $m$ is small, $a$ is some nonzero number, and $F_{net}$ is small.
The above reasoning is the justification for the usual assumption that low-mass objects transmit forces unchanged, e.g., that the tension in a rope is the same value throughout the length of the rope.
| {
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Why is $\pi$ used when calculating the value of $g$ in pendulum motion? I am trying to intuitively understand why $\pi$ is used when calculating the value of $g$ using the harmonic motion of a pendulum:
$$g ~=~\frac{4\pi^2L}{T^2}.$$
Does it have something to do with the curvature? I am thinking something along the lines of that, aswell as the fact that the oscillation of a pendulum would follow a circular path. The squared value of it would be because it was performed in 3d space.
I am just looking for a mathematically intuitive understanding of this.
| First notice that simply by considering the dimension of the parameters involved, one can deduce that the time period of oscillations should go like
$$T\propto\sqrt{\frac{\ell}{g}}. $$
This is because $g$ is acceleration hence has the dimensions of Length over Time squared and so the only way the quotient can have the dimension of time is to have the quotient under a square root sign.
The proportionality constant of $2\pi$ cannot be deduced in this manner. For that one has to solve the involved differential equation for the motion. Which happens to have the "circular" functions $\sin$ and $\cos$ as solutions. So I guess one could say that the $\pi$ comes from those function.
| {
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Is a suit that hides a soldier's heat signature fundamentally possible? I recently played "Crysis", a game where the protagonist wears a suit that allows the player to hide both himself and his heat signature. Then I watched Iron Man 3, where a kid suggests that Tony Stark should have implemented retro reflection panels in his suit.
So I'm thinking, well, as is the nature of things, people are going to be pursuing this sort of thing in real life too, pretty soon. But I'm trying to figure out whether a suit can contain a person's heat signature without emitting the heat somewhere. Is such a thing fundamentally possible to do, without over-heating the person within?
| Such technology is in its infancy, but it definitely exists. The images below are produced by several companies promoting their thermal/IR camouflage clothes. Obviously the applications are well-suited for the military, so who knows what more the military has developed.
This last image is made by a company called Blucher Systems. The link provides much more detail about the how and includes a neat little video of the suit in action. They claim a maximum of a 4 degree C difference between the "ghost" suit and ambient.
| {
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How can I estimate the density of fog? I'm working on investigating the effect fog has on drag. I have assumed an air density of $1.225 \frac{\text{kg}}{\text{m}^3}$ for dry air, but I don't know what value for density I could assume that would be typical of fog.
I can't even reason out whether or not fog is more dense or less dense than dry air: I know that air density is lower at higher humidity since water vapor is less dense than air, but it seems to me that fog should have a higher density than air, since you can observe fog being more dense closer to the ground, and collecting in valleys.
What air density is typical of fog?
| Anyone who has ever seen fog pour over a mountain range can tell you that it is significantly more dense than either moist or dry air. It does, after all, settle to the bottom of valleys....
See https://www.dailymail.co.uk/video/news/video-1031660/Incredible-wave-fog-rolls-Canadian-mountain-range.html for a nice example.
I have no clue where to get a quantitative answer, but qualitatively it seem denser.
I came here looking for for an explanation for a significant loss of range in my EV when driving through dense fog. It is again, an empirical qualitative result, but it does seem like I needed to charge a lot more than when I do this same reasonably routine drive in dry air.
| {
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Unification of the electroweak theory Can the electroweak theory be described by the spontaneous symmetry breaking of $SU(3)$ to $SU(2)\times U(1)$?
| In fact it is possible, see the paper Spontaneous Breaking of Symmetries by Li Fong Li. In general for the Adjoint Representation of $SU(n)$(the octet for $SU(3)$) you can have the following breaking (when $\lambda_2>0$, a parameter in the general potential):
$SU(n) \rightarrow SU(l)\times SU(n-l)\times U(1), \;\; l= [\frac{1}{2}n] $
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"timestamp": "2023-03-29T00:00:00",
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What is the four-dimensional representation of the $SU(2)$ generators? Recently, I have been learning about non-Abelian gauge field theory by myself. Thanks @ACuriousMind very much, as with his help, I have made some progress.
I am trying to extend the Dirac field equation with a coupling to a $SU(2)$ gauge field:
$$(i{\gamma}^{\mu }{D}_{\mu}-m)\psi =0$$
where
$${ D }_{ \mu }=\partial _{ \mu }+ig{ A }_{ a }^{ \mu }{ T }_{ a }$$
the ${ T }_{ a }$ is the $SU(2)$ Lie group generator, with $[{ T }_{ a },{ T }_{ b }]=i{ f }^{ abc }{ T }_{ c }$, and the ${\gamma}^{\mu }$ are the Dirac matrices. When I write explicitly the first part of the Dirac equation, with spinor form $\psi=(\phi,\chi)^T$, I get (spatial part):
$$\begin{pmatrix} 0 & { \sigma }^{ i } \\ -{ \sigma }^{ i } & 0 \end{pmatrix}\partial _{ i }\begin{pmatrix} \begin{matrix} \phi \\ \chi \end{matrix} \end{pmatrix}+ig\begin{pmatrix} 0 & { \sigma }^{ i } \\ -{ \sigma }^{ i } & 0 \end{pmatrix}{ A }_{ a }^{ i }{ T }_{ a }\begin{pmatrix} \begin{matrix} \phi \\ \chi \end{matrix} \end{pmatrix}$$
My problem is: I only known the linear representation of ${ T }_{ a }$ is Pauli spin matrix from text book, but they are the set of 2-dimension matrixes, In above expression, I need to know the 4-dimension matrix of ${ T }_{ a }$ because of the spinor is 4-dimension, I
check some test book, but didn't find the explicitly statement of the 4-D matrix.
So, as mentioned in title, What is the 4-dimension representation of the $SU(2)$ generators, or how can I calculate it?
| Comment to the question (v4): OP seems to effectively conflate spacetime symmetries and internal gauge symmetries. They act in different representations, or more precisely as a tensor product of representations.
For instance the fermion $\psi$ carries two types of indices, say $\psi^{\alpha i}$, $\alpha=1,2,3,4,$ and $i=1,2$. The fermion acts
*
*as a $4$-dimensional Dirac spinor representation under Lorentz transformations.
*as a $2$-dimensional fundamental representation of the gauge group $SU(2)$ under gauge transformations.
Similarly, the $4\times 4$ Dirac matrices $\gamma^{\mu}$ and the $2\times 2$ $SU(2)$ gauge group generator $T^a$ act on different representations. The product of $\gamma^{\mu}$ and $T^a$ is a tensor product. In particular, the term $\gamma^{\mu}T^a\psi$ in OP's formula again carries two types of indices, and is evaluated as
$$ (\gamma^{\mu}T^a\psi)^{\alpha i}~=~(\gamma^{\mu})^{\alpha}{}_{\beta}~ (T^a)^{i}{}_{j}~\psi^{\beta j}. $$
| {
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Is it a postulate or a well proven fact that speed of light remains constant w.r.t any observer? We usually heard that speed of light in vacuum $c$ remains same no matter how observer is moving?
I am wondering whether is it taken as a postulate or a proven phenomenon that $c$ is constant irrespective of observer's speed?
|
I am wondering whether is it taken as a postulate or a proven phenomenon that c is constant irrespective of observer's speed?
Either one. Both.
Einstein took it as a postulate in his 1905 paper on special relativity. From it, he proved various things about space and time.
The frame-independence of $c$ is also experimentally supported. This is what the Michelson-Morley experiment showed (although it was not interpreted correctly until much later).
You can also take other postulates for special relativity, describing the symmetry properties of space and time. In this case the constancy of $c$ becomes a theorem rather than an axiom. From a modern point of view, this approach makes more sense than Einstein's 1905 axiomatization, which puts light in a special role and defines $c$ as the speed of light. Nowadays we know that light is just one of several fields, and $c$ is not the speed of light but rather a conversion factor between space and time units. The symmetry approach goes back to W.v.Ignatowsky, Phys. Zeits. 11 (1911) 972, and can be found in various other modern presentations, such as this one or my own.
| {
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Is $E^2=(mc^2)^2+(pc)^2$ or is $E=mc^2$ the correct one? I have been having trouble distinguishing these two equations and figuring out which one is correct. I have watched a video that says that $E^2=(mc^2)^2+(pc)^2$ is correct, but I do not know why. It says that $E=mc^2$ is the equation for objects that are not moving and that$ E^2=(mc^2)^2+(pc)^2$is for objects that are moving. Here is the link to the video: http://www.youtube.com/watch?v=NnMIhxWRGNw
| The equation $$E^2=(mc^2)^2+(pc)^2$$ represents the correct energy-momentum relationship. It gives the total energy $E$ for an object of invariant mass (rest mass) $m$ that is observed to move with momentum $p$. This equation is applicable regardless whether the object is observed to be in motion ($p \ne 0$), or is observed to be at rest ($p = 0$). In the latter case, the energy-momentum equation simplifies into the well-known $E=mc^2$.
As an aside (some might call it nitpicking), when discussing the generics of the energy-momentum equation, it is good form to write the equation such that both sides of the equation are independent of the frame of observation chosen: $$ E^2 - (pc)^2 = (mc^2)^2$$
Same math, different physics. (Note that this relativistically invariant relationship is simply the expression for the square norm of the energy-momentum four-vector.)
| {
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Is this expression for the kinetic energy of a spinning disk revolving about a second axis correct? My question is motivated from a question from another user. You can see the configuration of the rotating system here: https://physics.stackexchange.com/q/143377/.
I am not interested in all the complicated arguments of his question, but only of the expression for the total kinetic energy. My answer was that the rotational KE can be expressed as the addition of the KE of the center of mass plus the KE relative to the center of mass, which results in this expression:
$$E_k=\frac{1}{4}mr^2\omega_2^2+\frac{1}{2}md^2\omega_1^2$$
(Notice that this result is independent of the sign of $\omega_2$).
But the original OP claims that the right expression is
$$E_k=\frac{1}{2}md^2\omega_1^2+\frac{1}{2}mr^2(\omega_1-\omega_2)^2,$$
based on answers obtained on other forums (which I checked) and even the moderators in those forums seem to agree with it. The OP itself does not know enough physics to come up with its own answer, but still does not believe mine for the reasons given above.
So, my question is: I am missing something pretty obvious here? which of the expressions is the correct one (if any?)
Thanks!
| This is simple, take the linear and angular velocity of the center of mass (point B) and combined them with the inertial properties
*
*Linear Velocity of B : $\vec{v}_B = (0,d \omega_1,0)$
*Angular Velocity of B : $\omega_B = (0,0,\omega_2-\omega_1)$
*Mass of disk $m$
*Mass Moment of Inertia of Disk $I_{zz} = \frac{m}{2} r^2$, where $r$ is the radius of the disk.
*Kinetic energy is $T = \frac{1}{2} m (\vec{v}_B \cdot \vec{v}_B) + \frac{1}{2} I_{zz} ( \vec{\omega}_C \cdot \vec{\omega}_C) = \frac{m}{2} d^2 \omega_1^2 + \frac{I_{zz}}{2} (\omega_2-\omega_1)^2 $
$$ \boxed{ T = \frac{m}{2} \left( d^2 \omega_1^2 + \frac{r^2}{2} (\omega_2-\omega_1)^2 \right) } $$
| {
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Can there be force without motion? I am confused. Can you have a force or tension without motion?
Take for instance two robots with jet packs connected via a cord, each is flying in opposite directions.
The tension of the cord is measured through a sensor of some kind. At some point, the net forces of the robots becomes zero and they no longer are moving, yet the sensor of the cord is still reading a force.
So what is causing this force if there is no relative motion?
| A net force acting on a body should do some work. For the simplest conservative case, the applied work will appear as the change in potential energy of the system. The object is not moving. But the potential energy creates a tension on the system. An example is a spring with its one end fixed. You apply a force that causes the spring to extend. The applied work appears as the increase in potential energy of the spring system which causes a tension on the spring.
Why tension is there even there is no net force?
The applied force causes the potential energy of the robots-cord system to increase. This increase in potential energy has to be accounted to some kind of force; otherwise, it will violate the energy conservation theorem. The net force is zero. However, each robot is accelerating in order to maintain that state.
... if there is no relative motion?
Obviously, there are no absolute motions. We can only speak about relative motions and relative measurements. There is no universal fixed reference for any motion or measurement.
| {
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Why is quasineutrality required for a gas to turn into a plasma? Why is quasineutrality a required condition for a plasma to exist?
Quasineutrality means that no density of electrons and ions should almost be equal but not exactly equal. Can anybody explain this this condition is required?
| Plasma potential.
Often electrons move faster than ions and leave the plasma at a higher rate. The plasma becomes positive until the positive 'plasmas potential' slows down the rate of electron loss from the plasma until it is the same as the rate of positive ion loss. The electrons are partially held in the plasma by the attractive positive plasma potential. The positive ions are pushed out a bit more.
So plasmas are mostly a bit positive and quasineutral.
| {
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Using the force law to obtain total energy of an electron as a function of its radius I am working on a problem which starts saying determine the total energy of a hydrogen atom with an electron moving with momentum $p$ at a radius $r$.
For that part I got:
$E = \frac{p^2}{2m_e} - \frac{e^2}{r}$
Which is just the kinetic energy plus the potential energy. If I didn't goof big time that should be good to go.
But now it asks:
"Use the force law to obtain the total energy as a function of radius. What radius corresponds to the lowest possible energy?" and I am completely lost.
What force law is it asking for here?
If it's worth noting, this is a problem trying to get you to understand the failure of classical mechanics at quantum levels.
| If the orbit is circular, then $p=\rm{const}$ and $r=\rm{const}$. $E$ is constant and negative (for a bound state) even though the orbit is not circular. So, one can determine $r$ from this equation: $$r=\rm{e}/\left(p^2/2m_e-E\right).$$
The minimum is zero (no kinetic energy, only the negative potential one), which is not supported experimentally.
| {
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How do you pronounce $\vec{A} \cdot \vec{B}$ and $\vec{A} \times \vec{B}$? I'm French.
I would like to know:
*
*How do you pronounce $\vec{A} \cdot \vec{B}$ : "A scalar B" or "A dot B" ?
*How do you pronounce $\vec{A} \times \vec{B}$ : "A vectorial B", "A vector B", "A cross B" or "A times B" ?
In French we say "A scalaire B" and "A vectoriel B".
| The first bullet would be read "$A$ dot $B$" or "The dot product of $A$ and $B$"
The second bullet would be read "$A$ cross $B$" or "The cross product of $A$ and $B$"
| {
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Adding a tracer to the surface of a water droplet I have a 2 mm water droplet generated by a syringe and falling down. I am using two perpendicular cameras to capture simultaneous frames from it. I need to track the droplet during the time and reconstruct it (through consecutive frames). However, only the border of the droplet (which is an ellipse) is distinguishable in each frame and I do not have any traceable point on the surface of the droplet to use it as a tracer (each point on the droplet surface will give me a point inside the corresponding ellipse on each frame). I want to add one or few very small particles/tracers in the order of my pixel dimension (e.g. 0.05 mm or even less which are still detectable through my cameras) such that while the droplet wobbling or revolving during falling down, always remain on the surface of the droplet and don't go inside it. Which material or technique do you think is more appropriate for this purpose?
Please let me know if you know any reference or paper used your suggestion.
for more information please see this Phys.SE question as well.
| I would consider using water with a dye like a deep blue to be nearly opaque. Illuminate with an array of LEDs. Each bright spot reflected from the droplet is an LED. You can add a few strategic red LEDs among all white as reference.
Working out the most convenient geometry will tell if you need the LEDs on some curved surface in space or if a flat panel will do. I think you can get pretty good data this way and quite high resolution as far as number of points on the drop.
| {
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How would behave theoretical matter with negative mass? I wonder if there is any possibility to evaluate theoretical characteristics of matter with negative mass? This is not thing of anti-mass but classical matter just with negated weight. I thing that if you would put it on surface of the Earth it would fly to heaven, wouldn't it? Could this be space-freedom redefined problem solution or am I just curious too much? Can't we simply use working functions/formulas with negate mark?
| Well, one of the basic facts from General Relativity (the commonly accepted theory of gravity) is the so-called Equivalence principle, which basically states that all kinematic behaviour of a particle in the gravitational field does not depend on the particle's internal properties (like mass). So it would not fly to heaven, no.
But the gravitational field generated by this particle could differ from the one generated by positive mass. Or it could not. Actually, physics (very much like history) does not allow us to ask "what would be if ...".
Let's take a quick look at QFT (the commonly accepted theory of matter and fundamental forces - everything except gravity). For bosons, the mass-defining parameter in the Lagrangian is $m^2$. One can either make $m^2$ less than zero (which would make your field tachyonic) or greater than zero, which does not fix the sign of $m$. So basically, there is no way of having negative-mass bosons in QFT. Same goes for fermions, where you can specify $m$ explicitly, but the observable mass is just the absolute value of this $m$.
We see that in QFT, there is no such thing as negative mass. So it is very much likely that it is in fact unphysical.
| {
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Confusion with rotation operator definition in Shankar In Shankar quantum mechanics on page 306-307 it has the following:
12.2. Rotations in Two Dimensions
Classically, the effect of a rotation $\phi_0\mathbf{k}$, i.e., by an angle $\phi_0$ about the $z$ axis (counterclockwise in the $x\ y$ plane) has the following effect on the state of a particle:
$$\begin{align}
\begin{bmatrix}x \\ y\end{bmatrix}
\to
\begin{bmatrix}\bar{x} \\ \bar{y}\end{bmatrix}
&=
\begin{bmatrix}
\cos\phi_0 & -\sin\phi_0 \\
\sin\phi_0 & \cos\phi_0
\end{bmatrix}
\begin{bmatrix}x \\ y\end{bmatrix} \tag{12.2.1}\\
\begin{bmatrix}p_x \\ p_y\end{bmatrix}
\to
\begin{bmatrix}\bar{p}_x \\ \bar{p}_y\end{bmatrix}
&=
\begin{bmatrix}
\cos\phi_0 & -\sin\phi_0 \\
\sin\phi_0 & \cos\phi_0
\end{bmatrix}
\begin{bmatrix}p_x \\ p_y\end{bmatrix} \tag{12.2.2}
\end{align}$$
Let us denote the operator that rotates these two-dimensional vectors by $R(\phi_0\mathbf{k})$. It is represented by the $2\times 2$ matrix in Eqs. (12.2.1) and (12.2.2). Just as $T(\mathbf{a})$ is the operator in Hilbert space associated with the translation $\mathbf{a}$, let $U[R(\phi_0\mathbf{k})]$ be the operator associated with the rotation $R(\phi_0\mathbf{k})$. In the active transformation picture
$$\lvert\psi\rangle \underset{U[R]}{\longrightarrow} \lvert\psi_R\rangle = U[R]\lvert\psi\rangle\tag{12.2.3}$$
The rotated state $\lvert\psi_R\rangle$ must be such that
$$\langle X\rangle_R = \langle X\rangle\cos\phi_0 - \langle Y\rangle\sin\phi_0\tag{12.2.4a}$$
Specifically I'm confused about how it describes the rotation operator $R(\phi_0\mathbf{k})$ and then the operator $U[R(\phi_0\mathbf{k})]$.
It says that $R(\phi_0\mathbf{k})$ is the operator associated with the rotation but the second to last sentence before equation 12.2.3 seems to imply that $R(\phi_0\mathbf{k})$ just denotes the rotation itself and $U[R(\phi_0\mathbf{k})]$ denotes the operator. Where am I going wrong?
| $R(\phi_0,k)$ is the operator that rotates your co-ordinate system. But it is not suitable to apply a 2x2 matrix, as in this case, to a vector in Hilbert space. Mind that Hilbert space is unlike an ordinary orthogonal position space.
Thus, using $U$, you map the operator into an equivalent operator which can operate on vectors in Hilbert space.
| {
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What is a rocket engine thrusting against in space? I know Newton's third law of motion might be the answer for this but still I am wondering how the rockets could thrust in the empty space and move in the opposite direction. I guess an astronaut wouldn't be able to push in the empty space with his hands or legs to move himself, but with a rocket engine it's possible. How? What might be the explanation for this in General Relativity?
| If I'm not wrong, it's basically the same principle in which an astronaut would throw something in empty space and, with so, move in the opposite direction.
It's not thrusting against something but throwing energy and power by burning fuel according to the law of inertia... I could be wrong so I'd like someone more knowledgeable to double-check, please :)
| {
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Difficulties in understanding basic energy equation in quantum mechanics While reading a text book about basics of Quantum Mechanics, I came across a situation in which it is said that
$E=\hbar\omega$ and also
$E = \frac12mv^2=p^2/2m$
where
$h$ Planck's constant
$\hbar=\frac{h}{2\pi}$ Planck's reduced constant
$\omega=2\pi f$ angular frequency
$m$ mass
$v$ velocity
$p$ momentum
But if I take the first definition,E=(h/2pi)*w,then
E=(h/2pi)*2*pi*f (because w=2*pi*f)
= h*f
= h*(v/λ) (because v=fλ)
= p*v (de-Broglie's wave-particle duality p=h/λ )
= mv*v (because p=m*v, the momentum)
E = m*v^2
This is not same as definition $E=\frac12mv^2$.
What am I missing in the derivation above?
| The relationship
$$
v=\lambda\nu=\frac{\omega}{k}
$$
describes the phase velocity ($v_p\neq v$) and not the group velocity ($v_g=v$), so it should be
$$
v_p=\frac{\omega}{k}=\frac{\hbar\omega}{\hbar k}=\frac{E}{p}=\frac{p}{2m}=\frac{v}{2}\tag{1}
$$
which does follow from the de Broglie relation ($p=h/\lambda=\hbar k$). Inserting (1) into your 3rd line gives
$$
E=pv_p=mvv_p=mv\left(\frac12v\right)=\frac12mv^2
$$
| {
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Radio wave propagation in ionosphere Radio communication is based on the concept that a radio signal incident on the ionosphere is reflected if the frequency of the wave matches the plasma frequency.
But what exactly happens? Is it based on the electrons in the ionosphere absorbing and re-radiating the energy if the frequencies match?
| Exactly the same happens as when light reflects off a metal surface. In both cases you have an electron gas that interacts with the light. In the case of a metal it's a dense (almost) free electron gas, and in the ionosphere you have a very dilute electron gas formed by ionisation of air molecules.
The incoming electromagnetic wave causes the electrons to oscillate, and as the electrons oscillate they emit EM radiation. If the forward direction the induced radiation emitted by the electrons interferes destructively with the incoming wave, and in the reverse direction induced radiation emitted by the electrons interferes constructively with the incoming wave. The result is that the wave is reflected.
The dense electron gas in a metal interacts with the incoming EM radiation so strongly that even a micron thick layer of metal is effectively perfectly reflecting i.e. reflects 100% of the incoming light. Because the electrons are so dilute in the ionosphere the reflection is far less efficient, though of course the ionosphere is a lot think than a micron and anyway you don't need perfect relection for radio transmission.
| {
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Comparison between entropy and internal energy Why is entropy change a better way of determining a spontaneous process compared to the change in internal energy?
| The change in internal energy is not a relevant quantity for spontaneous evolution of a system. Consider an isolated system made of two blocks of the same material at two different temperatures such that $T_1>T_2$. Heat will flow from $1$ to $2$ but the total change in internal energy is $\Delta U_{1+2}=0$. This information is therefore not useful.
On the other hand, the change in total entropy is $\Delta S_{1+2}=Q/T_2-Q/T_1>0$ and can tell us about the direction of the process.
| {
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Why do we add the spin angular velocity and orbital anglar velocity when asked to calculate total angular velocity of Gyroscope? Normally when we talk of angular velocity we mean how the angle of a vector changes with time with respect to an origin.Thus the oribital angular velocity of gyroscope makes sense to me.However I find that we add another type of angular velocity -spin angular velocity- to find total angular velocity.This seems a bit ambiguious as this angular velocity is not due to change in angle about our origin about which we calculated the orbital angular velocioty.Thus adding both to get angular velocity seems confusing to me.
`
| It's as simple as adding the two vectors, the vector that determines the orbital rate relative to your reference frame and defined origin, and the vector that defines the spin angular velocity relative to the spin axis of your gyroscope. The two vectors are tip to tail connected and their sum is just the vector connecting the origin to the tip of the spin vector.
| {
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What exactly is color? What is color?
“The property possessed by an object of producing different sensations on the eye as a result of the way the object reflects or emits light”…. You might say.
So take a white object and paint it “red”. What property of the paint now makes the object appear red?
“Pigmentation!” … you might say.
Good. Now that I’ve got you thinking in line with what I’m thinking, please help me understand what pigmentation actually is. I’m looking for an explanation that goes down to the microscopic level. What is it about the elemental structure of different pigmentation that alters the way that pigmentation (color) is seen by our eyes when applied over an object?
| OK the answer is not microscopic level but beyond that. It is in atomic level.
*
*Metal complexes are often colored. These colors come from the d-orbitals because they are not involved in bonding. This is because they do not overlap with the s and p orbitals of the ligands. Most transitions related to colored metal complexes are either d–d transitions or charge band transfer.
*In centrosymmetric complexes, d-d transitions are forbidden by the Laporte rule. However, forbidden transitions are allowed if the center of symmetry is disrupted, resulting in a vibronic transition. The color of such complexes is much weaker than in complexes with spin-allowed transitions.
*In Metal-to-Ligand Charge Transfer, electrons can be promoted from a metal-based orbital into an empty ligand-based orbital. These are mostly likely when the metal is in a low oxidation state and the ligand is easily reduced. Ligands that are easily reduced include CO, CN- and SCN-.
*An electron may jump from a predominantly ligand orbital to a predominantly metal orbital (Ligand-to-Metal Charge Transfer or LMCT). These can most easily occur when the metal is in a high oxidation state.
*Coordination complex color results from the absorption of complimentary colors. A decrease in the complimentary color wavelength can be observed by UV-Vis spectroscopy. This decrease is correlated with the electric field of ligands, and indicates the energy gap is increasing.
These are the complex and detailed answers to the question
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Does drag produce heat? When a solid object moves through a fluid drag is produced. Does this drag produce heat?
I believe drag should produce heat as it is the friction between fluid and surface of object. Is this true or not?
| Actually drag is NOT completely the friction between the object and fluid. Sometimes there can be almost no friction, but still highly significant drag. In a non viscous fluid, there is no friction between the object and the fluid, but there IS still drag. The phenomenon of ram pressure transfers momentum losslessly between the object and the fluid without any friction. I discuss these ideas more fully in my answer to the question "Where does the energy from a parachute go?".
In a low viscosity fluid, the energised fluid will slowly heat owing to friction with itself, not with the dragged body.
So yes, ultimately all the energy transferred to the fluid from the dragged body will wind up as heat, as it must from the first law of thermodynamics, but this is often not through direct friction.
In blunt ended spacecraft atmosphere re-entry, the effect is even more dramatic. The blunt end squashes the air, which cannot get out of the way and the air's temperature rises adiabatically just as a bike pump gets hotter if you hold your finger over the output hole and squash the plunger in suddenly. Very little friction is involved and, as long as the spacecraft can withstand the adiabatic temperature rise, it rides on a cushion of air which thrusts the atmosphere further from the aircraft out of the way. The drag is ram pressure, and the frictional heat (which would be extremely dangerous) is only generate by the fluid dragging on itself through viscous shear, at a good safe distance from the spacecraft. See the links in my other answer for my answer to the question "Where does the energy from a parachute go?".
| {
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How fast does an electron travel in a circuit? How is it possible to calculate the speed of an electron in a circuit? What factors does it depend on?
| Individual electrons may have a range of speeds in a circuit (thermal motion, scattering, absorption, photon etc..)
However the current (or drift velocity) gives the average speed of the whole electron cloud (not a single electron).
Note again single electrons may have a range of speeds (from slow to very fast, near $c$). It is the electron cloud that makes the current and this has another velocity.
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What is the link between observation wavelength and spatial resolution of an instrument? It seems that when capturing and emitting EM waves matter proceeds differently depending on the wavelenght. According to an answer from another question, EM energy is captured following three modes: electronic transition (e.g. for visible light), rotational and vibrational absorption. To which we can at least add induction phenomena (e.g. radio wave) which works in an unrelated way.
Morover diffraction limit imposes restrictions on the finer details observable.
How the frequency and energy capture process used to observe an object affects the resolution at which we can observe the spatial features of an object?
A corollary question is: are there regions of the EM spectrum that cannot be observed because there are no possible instrument sensitive to these frequencies?
| The answer would have to depend on a specific scale. i dont think there is an answer to this question without determining a scale.
A scale will fix the dimensions and the relative wavelengths that are relevant to the spatial resolution of objects.
Lets say one wants to observe a cubical object which radiates (very high at ultra-violet).
At visible wavelengths the spatial features of the object are determined. At ultra-violet the spatial features (for the same scale and distance) will be blured due to the amount of radiation at that wavelength (the object will glow intensely there, thus boundaries are blurred).
Effectively a boundary is similar to a phase transition between areas of different wavelengths (thus the resolution wavelength can give different boundaries for same object).
Plus the spatial features even in visible wavelengths can be blurred due to atmospheric distortions or attenuation or shifts over distances (which are comparable to the range of wavelength used in the resolution).
| {
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Correlation between equations of elliptical orbits and pendulums The equation for the period of a pendulum is:
$$T=2π\sqrt{\frac{L}{g}}$$
Where 'g' is the acceleration due to the gravitational field and 'L' is the length.
The equation for the period in of a body travelling along an elliptical orbit is:
$$T = 2π\sqrt{\frac{a^3}{GM}}$$
Where 'a' is the semi-major axis. I can see that this is derived from Kepler's 3rd law.
Is there are similar equation to Kepler's 3rd law for pendulum periods?
If a pendulum string is of fixed length, does that essentially make the motion of a circular orbit? It seems like a very similar relationship between length, gravity and periods.
My question: "Is there a deeper, more fundamental relationship between these equations?
| The equation for the period of a pendulum $(T=2π\sqrt{\frac{L}{g}})$ is only an approximation. That equation assumes, among other things, that gravity doesn't change with height, and that $\sin(\theta) = \theta$.
Even if there were a connection between that approximation and elliptical orbits, that would not imply any connection between the true period of a pendulum and the period of an elliptical orbit.
| {
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Tractor wheels -- large vs small This question has stumped me for over a month now:
Why is it that a tractor has large wheels at the back and small wheels in the front?
Current ideas:
*
*small wheel in front --> lower center of mass--> less likely to tip over, moment.
*large wheels in back provides more torque, since friction is the driving force of the wheel.
Let's assume that there's a load behind the tractor and a cable is connected to the tractor and cart that carries the load. FBDs and moment equations for the wheel and load are highly appreciated.
How does the extra torque help pull the load? and how does it provide more torque? if the engine has a CC moment of 100, the torque from the friction opposes that, and why would I want more torque from the friction? Wouldn't that slow down how fast I'm able to pull the load?
Sum of Moment at the wheel's center=Applied moment- F_f*radius of wheel.
| One of the reasons is that if the wheel axle is above the attachment point it drives the wheels downwards when pulling increasing traction.
Think of the opposite, where the attachment point is really high it will force the front wheels off the ground limiting the pull force so the tractor does not flip.
So what happens with the tall wheels and the attachment point below the axle is that the stability of the tractor does not limit the pulling force.
| {
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What is the phase shift incurred by a sound wave as a result of reflection? While studying waves I read the fact that a sound wave gets shifted by $\pi$ as a result of reflection against a surface. But I am unable to prove that fact.
Assuming the interface to be a node I can prove that there is a phase shift of $\pi$ but speaking generally how do I deduce it? If we do not assume that the interface is a node I saw in some case it's not true.
Am I misunderstanding something here? Does the sound wave manipulate itself so as that the interface becomes node? How does it do that?
| A wave e.g
$$\sin (kx + \omega t + \phi)$$
when reflected runs in the opposite direction. In other words gets a rotation by $\pi$ or what amounts to the same thing gets a phase shift by $\pi$.
$$\sin (kx + \omega t + \phi + \pi)$$
Tentative proof:
Let's say a wave $\psi \sim e^{i(kx-\omega t)}$
on reflection the wave should propagate opposite (or rotated by $\pi$)
i.e $\psi \sim -e^{i(kx-\omega t)}=e^{i\pi}e^{i(kx-\omega t)}=e^{i(kx-\omega t + \pi)}$
| {
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How can the linear momentum can be understood physically? Currently reading Classical Mechanics by Herbert Goldstein, and I'm trying to understand every concept physically. Speed can be understood physically, as the distance traveled within a certain amount of time, it makes sense to me. By contrast, I can't attribute a physical explication to linear momentum. How can I understand it physically? Why do we multiply mass by speed?
| The idea of momentum is the idea of "quantity of motion". You want to be able to formulate in some sense the idea of "how much motion there's in this particle?" and you can think about the simplest model for that like that: the faster the particle moves intuitively more motion there is, so the quantity of motion should be proportional to the velocity. Also, if we agree that mass is a measure of quantity of matter, the more mass there is, more matter in movement you have and so more motion. In that case, quantity of motion should be proportional to mass as well. In that case, you can simply choose units such that this becomes equality, so that you define $\mathbf{p} = m\mathbf{v}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How many more galaxies are out there in the Universe (beyond the observable radius)? Let's say that the number of large galaxies in the observable universe is $n$ (approximated to 350 billion).
If the universe is homogenous and isotropic, what are the estimations for the total number of large galaxies in it?
$5n$, $10n$, $50n$?
| You assume homogeneous and isotropic structure. That limits the possibilities as follows: It is just as likely that there are observers at the 'edge' of our Observable Universe as there are here. Assume two exist opposite each other. They both see the same Universe we do. Repeat the argument with them. Now consider they could be in any direction. So, instead of the Observable Universe being ~93E9 ly in diameter, this new construct has a diameter proportional to the number of these observers/edges. In principle there are three possibilities: 1. The Universe is flat. 2. The Universe has spherical curvature 3. The Universe has hyperbolic curvature. All of these are consistent with a isotropic, homogeneous Universe. The Universe is flat to a very good approximation, so that if it is curved, then the curvature is not noticable with our current methods. This implies that at a minimum the Universe is several hundred times larger than the Observable Universe. So given your assumptions say that the diameter is between 1E+13 ly and ∞. My take, not that you asked, is that asking to quantify what is by definition unknowable makes as much sense as trying to figure out how many Angels can dance on the head of a pin: meaningless.
| {
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What is $Q_p/Q_e$ experimentally? What is the experimental value of the ratio between the proton and the electron charge? Or more generally, is there a table that lists the ratio of the different nuclei charges to that of the electron?
|
[PDG] quoted something like $(q_p+q_e)/e$ without defining these quantities,
That is exactly what you asked for. Recall that the charge on the electron $q_e$ is negative and that on the proton $q_p$ is positive, so the sum there is exactly the difference in their magnitudes. Taking it as a fraction of the defined base charge $e$ makes it a dimensionless value that does not depend on units, and
$$ \frac{\left| q_p + q_e\right|}{e} \le 1 \times 10^{-21}$$ is a pretty good measurement.
| {
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