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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
| I agree with answer of ACuriousMind, but I think it might also help to think about it like this....
$E^2=m_0^2c^4+p^2c^2 =m^2c^4$
where $m_0$ is the rest mass and $m$ is the relativistic mass (or inertial mass), defined as $m = \gamma m_0 = m_0 / \sqrt{1 - v^2/c^2}$.
The relatavistic mass increases as the momentum of the mass increases. At rest the two are equal to each other. As the speed of an object and its momentum increase the mass of the object increases.
so I think about it as
$E^2=m_0^2c^4+p^2c^2$
and
$E=mc^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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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!
|
For a ring:
$Ke = \frac{1}{2}m ( \int_0^{2\pi} (Rw_1-rw_1cos(\theta)+rw_2cos(\theta))² +(rw_2abs(sin(\theta))² d\theta) $
$Ke = \frac{1}{2}m \int_0^{2\pi} R²w_1²+r²w_1²cos²(\theta)+r²w_2²cos²(\theta)-2Rw_1²rcos(\theta)+2Rw_1rw_2cos(\theta)) -2rw1cos(\theta)rw_2cos(\theta)+r²w_2²sin²(\theta) d\theta$
$Ke = \frac{1}{2}m (R²w_1²+\frac{1}{2}r²w_2²+\frac{1}{2}r²w_2²-r²w_1w_2) $
$Ke = \frac{1}{2}mR²w_1² + \frac{1}{2}mr²(w1-w2)² $
With:
$abs(W_1) > abs(W_2)$
$w_2<0$ and $w_1>0$
The result take in account the sign of $w_2$, kinetics energy increase if $abs(w_2)$ decrease.
| {
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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?
| In a system, the total sum of forces when added together equals mass times acceleration:
$$
\sum F = \frac{\mathrm{d}p}{\mathrm{d}t} = \frac{\mathrm{d}mv}{\mathrm{d}t} = m\frac{\mathrm{d}v}{\mathrm{d}t} = ma
$$
Since the sum of the forces on the robots is zero, there is no acceleration. However, the tension of the string is not contingent on the movement. I will assume that there is an arbitrary force of 500 newtons being pulled by each robot. It is simple to calculate this. Since force is a vector quantity (because acceleration is a vector quantity), the direction is important.
$$
Robot \space 1:\\
\sum F = ma\\
500 - T = ma
$$
where $T$ is Tension. Since acceleration is $0$, we know that $ma$ must be $0$. Therefore, $T = 500$. The same can be done for the other robot, which is pulling the string the opposite way:
$$
Robot \space 2:\\
\sum F = ma\\
T - 500 = ma
$$
Again, if they are not moving, acceleration is zero and $ma$ is $0$. Therefore $T = -500$ (relative to Robot 1). The tension in the string would be $500 \mathrm{N}$.
There is acceleration if, and only if, the sum of the forces is not zero. In this case it is. The sensors are not measuring force, per se, but rather tension. They are still measuring force, but this force is offset by the other robot, so you must think of the system as a whole.
| {
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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?
| If the number of electrons and ions is exactly equal, it is still plasma. You are misunderstanding the quasineutrality requirement.
The term "plasma" was coined by Irving Langmuir with the phrase "We shall use the name plasma to describe this region containing balanced charges of ions and electrons", Oscillations in Ionized Gases Proc. Nat. Acad. Sci. U.S., vol. 14, p. 628, (1928)
| {
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Why are permanent magnets permanent? Let me see if I get it right. When an iron bar is attracted by a permanent magnet it becomes a magnet itself because all of its magnetic domains start to point in the same direction. When the iron bar is no longer attracted by the permanent magnet, it is no longer a magnet itself because its magnetic domains point in different directions again.
When iron is heated up to curie temperature and cooled down all of its magnetic domains also start to point in the same direction. ( If I am not wrong the atomic structure does not change)
So why is it permanent in the second case and not in the first ? (Correct me if I messed up something here)
| When you cool it the domains so formed align to get minimum energy and that is not the energy in which all are pointing in the same direction.
| {
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Kitchen floor dries faster with lights on? My mother used to leave the lights on in the kitchen after washing the floor, saying that it would dry faster.
Does this really happen, or is it just a superstition? If true, how substantial is the effect?
| In addition to the points mentioned by Andre Holtzner, there is also a non-thermal effect due to photons hitting the water molecules and kicking them out of the water directly. This process plays an important role in the vaporation of water due to irradiation by sunlight. You can think of the photons being at a temperature that is much higher than the ambient temperature. The energy of these photons will eventually thermalize to the much lower ambient temperature, raising the ambient temperature a bit. But when a photon hits a molecule, the kick it gets is similar to the kick it would get if it were in thermal contact with a heat source at an extremely high temperature, the difference is that the rate at which molecules are hit by photons is quite low. However, one collision can be enough to launch a molecule out of the water.
| {
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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?
| Shankar is being a little bit sloppy with the term "state," although not that sloppy since he does use the qualifiers "quantum" and "classical" to achieve clarity.
In classical mechanics, the position of a particle is represented by a point $\mathbf r$ in three dimensions $\mathbb R^3$. To rotate the configuration of the classical particle, then, one would act on the position with a $3\times 3$ rotation matrix $R$.
In quantum mechanics, the state of the particle is represented by a vector in a Hilbert space $\mathcal H$. To "rotate" its state (which is what happens to the system when you, for example, rotate your measurement apparatus according to the spatial rotation $R$), one would act on it with a unitary operator $U(R)$. Why unitary you might ask? This is a consequence of a deep theorem on symmetries in quantum mechanics called Wigner's theorem.
Note that both $R$ and $U$ are operators, but they operate on different vector spaces, $\mathbb R^3$ and $\mathcal H$ respectively, and they have distinct physical interpretations as a result.
| {
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Is there charge build up before a resistor? I understand that Kirchhoff's current law says that the current, $I$, is constant throughout a resistor, i.e. there is no build up of charge in a resistor. All charge going in to the resistor is the same as all charge coming out. In other words, Coulombs/sec going in = Coulombs/sec coming out.
However, suppose we have an ideal wire, i.e. a wire with no resistance (or a physical wire with very little resistance; but lets use an ideal one), if there is a battery providing a voltage, the current through the wire is infinite. Once we get to the resistor the current is a finite amount $I = dq/dt = V/R,$ in other words, the current has decreased. So, from what I understand there should be a little charge pileup at the entrance to the resistor, and in fact, this is what gives the resistor voltage to drive the current through. Is this true? If not, please explain.
| Crudely, electrons repel each other and even out the charge. While the influence of the electrons travels at a good fraction of the speed of light the electrons themselves do not move much. From this link
" The electricity that is conducted through copper wires in your home consists of moving electrons. The protons and neutrons of the copper atoms do not move. The actual progression of the individual electrons in a given direction through the wire is quite slow. The electrons have to work their way through the billions of atoms in the wire and this takes considerable time. In the case of a 12 gauge copper wire carrying 10 amperes of current (typical of home wiring), the individual electrons only move about 0.02 cm per sec or 1.2 inches per minute (in science this is called the drift velocity of the electrons.). "
| {
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Spectral irradiance units conversion I have a table of data containing irradiance of light at different wavelengths. This is how it looks like for 300.5 nm:
*
*Wavelength, nm: 300.5
*Wavelength, $\mu$m: 0.3005
*W/m$^2$/$\mu$m: 403
*W/m$^2$/nm: 0.403
My question is, how can I convert W/m$^2$/nm or W/m$^2$/$\mu$m to W/m$^2$?
And what exactly do those other units mean (W/m$^2$/nm)?
| All sources of light have a spread of wavelengths. There is no such thing as a light source that produces light of exactly one wavelength. Let's assume that the power emitted by your light source looks like this:
I just made up this curve, but the shape of the curve doesn't matter for this discussion. The $y$ axis shows the spectral irradiance and as you say this has units of $\text{W}/\text{m}^2/\text{nm}$. The power emitted at exactly $300.5$ nm is zero, but the power emitted over all wavelengths between $\lambda = 300.3$ nm and $\lambda = (300.5 + \delta\lambda)$ nm is the area under the curve between the dashed lines, that is:
$$ W(300.5\text{ to }300.5+\delta\lambda) \approx I(300.5)\delta\lambda $$
And the units of $I(\lambda)\delta\lambda$ are indeed $\text{W}/\text{m}^2$ as we'd expect for power.
More generally, to get the power emitted over a range of wavelengths from $\lambda_1$ to $\lambda_2$ you have to integrate the spectral radiance:
$$ W = \int_{\lambda_1}^{\lambda_2} I(\lambda) d\lambda $$
| {
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Choice of the $z$-axis in the Schrödinger equation for the hydrogen atom I am reading about the solution of the Schrödinger equation for the hydrogen atom and have a question about the choice of the $z$-axis. Most websites say that the $z$-axis is arbitrarily chosen. If so, why is the choice of the $z$-axis not part of the quantum numbers? Wouldn't two electrons with the same quantum numbers but different $z$-axes have different wavefunctions?
Say, I have a wavefunction of an electron with $n$=2, $\ell$=1, $m_\ell$=0, $m_s$ = 1/2 and the z-axis pointing in some particular direction. I can create a new wavefunction with the same quantum numbers by, say, rotating the z-axis around the x-axis by 1 degree. In fact, I can create an infinite number of different wavefunctions with the same quantum numbers by just rotating the z-axis. Since they are different wavefunctions, they should be able to fit into the same atom without violating the Pauli exclusion principle. Therefore, there should have been many more electrons in the hydrogen atom at $n$=2. But this is not the case. Could you please tell me what I am missing?
| Say we have two different coordinate systems $F=(x,y,z)$ and $F'=(x',y',z')$.
Consider the basis spanned by the eigenvectors of the $L_z$ operator for a given $l$, $\{|l \; m\rangle\; ; \;m=-l,\dots,l\}$. Now, one can find the eigenvectors of the $L_{z'}$ on this basis. In general, the eigenvectors of $L_{z'}$ will be linear combinations of the eigenvectors of $L_z$.
This means that if $|\psi'\rangle$ is an eigenvector of $L_{z'}$, then, there are values of $m$ for which
$$
\langle l \; m | \psi'\rangle \neq 0.
$$
So stating that you can create different wavefunctions is not true.
| {
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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.
`
| Choose a point P on the periphery of your gyroscope and paint it with some color. Then, as the gyroscope rotates take photos and if possible, record the time of each picture. Let's name O the origin around which rotates your gyroscope. Now, using the pictures and the times recorded, calculate the change in time of the angle between the line OP and some fix direction in space that passes through the origin O.
I hope that it helps,
Sofia
| {
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Tranverse electromagnetic waves TEM waves do not exist in waveguide. Is this the correct explanation- Both curl and divergence of TEM are zero inside the waveguide and because of the boundary conditions (electric field zero at every point of a hollow conductor), the only solution possible is zero i.e. trivial?
| TEM waves do exist in multi-conductor waveguides such as coaxial guides, in fact they exist in any homogeneous waveguide with more than one conductor. There are no propagating TEM, TE or TM modes if the cross section is inhomogeneous but at cutoff frequency the hybrid modes degenerate to the respective transversal modes.
| {
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Why is the unitary matrix relating the gamma matrices and their complex conjugates antisymmetical? In Messiah's Quantum Mechanics Vol. II, properties of the Dirac matrices are derived. There is so-called fundamental theorem, which states that,
Let $\gamma^\mu$ and $\gamma^{'\mu}$ be two systems of 4 fourth-order unitary matrices satisfying the relations $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu} I_{4\times 4}$ (similarly for primed gamma matrices). There exists a unitary matrix $U$, defined to within a phase, such that $\gamma^{'\mu}=U\gamma^\mu U^\dagger (\mu=0,1,2,3)$.
The unitarity of gamma matrices is defined as $\gamma^{\mu\dagger}=\gamma^0\gamma^\mu\gamma^0$. Then, it is easy to check that $\gamma^{\mu *}$ are also unitary and obey the relations $\{\gamma^{\mu *},\gamma^{\nu *}\}=2g^{\mu\nu} I_{4\times 4}$. Therefore, there is a unitary matrix, called $B$, such that
$\gamma^\mu=B\gamma^{\mu *}B^\dagger,\; \gamma^{\mu *}=B^*\gamma^\mu B^T$ ----- Eq.(1)
I can show that $BB^*=cI_{4\times 4}$ with $c$ a constant. I cannot show that $c=-1$. Can you guys help me out? Thank you very much!
The way to show $BB^*=cI_{4\times 4}$ is that by the first expression in Eq.(1), solve for $\gamma^{\mu *}=B^\dagger\gamma^\mu B$. Comparing this with the second expression in Eq.(1), we have
$B^\dagger\gamma^\mu B=B^*\gamma^\mu B^T \Rightarrow \gamma^\mu BB^*=BB^*\gamma^\mu$
So $BB^*$ commutes with $\gamma^\mu$ and by the basic properties of gamma matrices, $BB^*=cI_{4\times 4}$.
There is a hint in Messiah's book. It is that $BB^*$ is the same for whatever the system of 4 unitary matrices $\gamma^\mu$ used to define $B$.
So I choose a different system of $\gamma^{'\mu}$ and so $\gamma^{'\mu}=U\gamma^\mu U^\dagger$. Taking complex conjugates of both sides and applying Eq.(1), finally I get
$B'=c'UBU^T$ where $B'$ is for the system of $\gamma^{'\mu}$ and $c'$ is a second constant.
Simple calculation shows that $B'B'^*=UBU^TU^*B^*U^\dagger=UBB^*U^\dagger$, assuming $c'=1$ (normalization). But I cannot prove that $B'B'^*=BB^*$.
| Notice that both $B$ and $B^\star$ are unitary matrices and, therefore, $B B^\star$ must be unitary as well, which implies that its determinant must be either $1$ or $-1$. Since you have already determined that $B B^\star = c\,I$, this narrows the possible values of $c$ down to $c = \pm 1$.
About the expression $B^\prime B^{\prime\star} = U B B^\star U^\dagger$, having also proved that $B B^\star = c\,I$ implies that $B^\prime B^{\prime\star} = c\, U U^\dagger = c\,I$. Therefore $B^\prime B^{\prime\star} = B B^\star$.
I still miss a proof to exclude the $c=1$ solution. Sorry.
| {
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Newtonian tidal forces and curvature Today in my physics class, my lecturer said something which confused me. He said:
"Newtonian tidal forces are reinterpreted as a manifestation of curvature in General Relativity".
Now I know what tidal forces are (an effect of the force of gravity), a good example is the cause of the waves on the ocean because of the tidal forces with the moon. However I do not see how this shows curvature in the GR sense.
| curvature produces relative acceleration of geodesics because of equation of geodesic deviation (relation between riemann tensor and relative accleleration of geodesics) and also newtonian theory of gravity predicts tide-producing acceleration which explain two particle with separation x parallel to r ,cause evaluating relative acceleration and x prependecular to r ,have a stated amount of relative acceleration .so relative acceleration caused by curvature ..
But why we use these thing instead of force , you can find the answer in weightlessness , actually there isnt any real force in a free fall particle
See page 29 and 30
gravitation mtw
for tide-producing acceleration
| {
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Transverse doppler effect in light In most books to explain transverse Doppler effect the following example is given:
Consider a source that emits flashes at frequency f0 (in its own frame), while moving across your field of vision at speed v. There are two reasonable questions we may ask about the frequency you observe:
• Case 1:
At the instant the source is at its closest approach to you, with what frequency
do the flashes hit your eye?
• Case 2:
When you see the source at its closest approach to you, with what frequency
do the flashes hit your eye?
In the first case we observe from the trains frame, while in the second we do not.
The explanation for doing this is given as follows. If we observe from the ground frame the following error is supposed occur:
The error can be stated as follows. The time dilation result, ∆t = γ·∆t0, rests on the assumption that the ∆x0 between the two events is zero. This applies fine to two emissions of light from the source. However, the two events in question are the absorption of two light pulses by your eye (which is moving in the source frame), so ∆t = γ·∆t0 is not applicable. Instead, ∆t0 = γ·∆t is the relevant result, valid when ∆x = 0.
Here x0, t0 is the observation in the moving frame, and γ is the dilation factor.
My question is, for what events and why is ∆x0 not equal to 0. And why when we observe from the moving frame ∆x is supposedly 0.
| Transverse Doppler effect
On a plane, parallel lines are drawn. On each,lines, light sources (frequency is the same time) are moving in the opposite direction. Imagine light sources form Japanese letter エ. Transverse Doppler effect will not be.
| {
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How to know if a vehicle is moving without any external source of information? The situation is the following:
I'm inside a vehicle (plane or a car, it doesn't matter) and I need to know if the vehicle is moving at a constant speed BUT I cannot perceive any external change like visual changes, vibration, etc.
How can I know if the vehicle is moving? Do I really can know?
Additional question Can I know my speed?
| Acceleration
If your exact need is, as you say, to determine "if the vehicle is moving at a constant speed" then there is a wide range of accelometers available on the market with various degrees of accuracy.
If the accelometer shows 9.8 m/s2 downwards (assuming you install it in a fixed orientation relative to the vehicle), then it is moving at a constant speed (possibly with a constant speed of 0 relative to the ground), if it shows anything else, then it's speed of movement is not constant.
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Why won't a block less dense than water fully submerge? Suppose we have an object of volume $1\: \mathrm{m^3}$. Mass of that object is $500\: \mathrm{kg}$, which means that the density of the object is $500\: \mathrm{kg/m^3}$.
If the object is in water it will float and half of it's volume ($0.5\: \mathrm{m^3}$) will be submerged in water (assuming that the density of water is $1000\: \mathrm{kg/m^3}$; as the object's density is half of water so half of it's will be submerged).
From the Archimedes principal we know that the object will displace the water of same mass as it. So the object will displace $500\: \mathrm{kg}$ water and $500\: \mathrm{kg}$ water = $0.5\: \mathrm{m^3}$ water.
We also know that the lost weight of an object = weight of water displaced by that object.
It means that the object will lose all of it's weight in water and as buoyant force is same as the weight of that object, the object should be submerged totally in water. But, that it is not possible, it will be submerged only half of it's volume. But how?
If the weight of displaced water is equal to weight of that object, shouldn't it be totally submerged?
| In this case you show that the net force acting on the object is zero in this situation, that is, half of the object is within the water and another half on the water. This position of the object is thus equilibrium position. If it would come totally on the water surface or within the water then the equilibrium will be lost, i.e. it comes to in-equilibrium position. Then object will try to return to its earlier equilibrium position.
That's why it will not sink.
| {
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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.
| Leverage
Often, a limiting factor in tasks required from a tractor is the amount of pulling force a tractor can apply without tipping over (the front rising up) - the engine is strong enough to do so.
Having the driving axle be high from the ground helps by simple lever action - see the illustration; twice the height means twice the maximum pulling force allowed before the tractor tips over. The other thing for the same problem that's commonly done is to attach a
large heavy object far to the front - https://www.google.lv/search?q=tractor+front+weight&tbm=isch
[edit] - a more accurate but more complex illustration of the actual leverages.
| {
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Atmosphere model Im working on project where I should simulate glider soaring. The goal is to create gliders that will look for regions with hot upwinds using evolution algorithms. That shouldn't be problem.
What I have problem with, is how to simulate the atmosphere with wind and temperature? I've read that meteorological simulators divide space into 3D matrix and compute temperature, pressure and wind speed for every cell.
What would be the simpliest atmosphere model I could use using 3D matrix? Please provide equations and example on how to compute.
Something like this http://hint.fm/wind/ but in 3D would be perfect, but it could be more simpler. I thought about matrix holding temperatures and differences in adjacent cells would give me vector with wind direction and speed but I'm not sure if that would work and if it isn't too easy for my simulation.
| NASA has created an atmospheric model that should do exactly what you want. It's called EarthGRAM.
It allows you to input a time history of coordinates (latitude, longitude, altitude) and will generate a 3D wind velocity, density, and temperature for each coordinate. The exact values of each output are randomly generated, but with statistical distributions based on measured historical data.
Unfortunately, it's export controlled, so you'll have to be a US citizen or permanent resident to get a copy.
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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?
|
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?
'speed' (or velocity) is the measure of 'motion' and is related to KE, a unitary mass acquires 'speed' equal to the square root of twice its KE, as you know, if v = 10, in one second the body travels the distance of 10 metres
By 'mass' whe mean the total mass of a body, the sum of all unitary masses
.
The relation of mass and speed is like the relation between sweets and jars: if 1 jar contains 10 sweets each, 2 jars contain 20 sweets (2 is the total mass of the jars).
Why whould more matter in movement would imply more motion? – Chirac
2 jars * 10 sweets is 20 = the 'quantity of sweets'. Has a jar anything to do with sweets? jars by sweets = total quantity of sweets, p = m * v, masses by velocity is the total 'quantity of motion'.
Momentum is the quantity of motion that a body possesses p = m * v, and depends on the value of speed and the number of unitary masses of a body. The greater its momentum, the greater the force it takes to stop it, the greater the force of impact on another body.
| {
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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?
$5n$, $10n$, $50n$?
| By definition, anything outside of the observable universe is unobservable. This has the annoying effect (eye twitch) of making it so we have practically no idea what the universe is actually like outside of what we can observe. We can assume that it is homogeneous and isotropic and that there are other large galaxies out there, but there is a non-zero probability that we live in a privileged area and outside of what we can observe there is nothing. There is also a chance that the universe is infinite and homogeneous. This is why Dirk said between zero and infinity, we can't observe anything about it and therefore we cannot know how many galaxies there are.
I do desperately wish there was an accurate number I could give you, but there are no estimations for the number of galaxies outside of our observable limits.
| {
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Frequency of an open air column Given only the length of an organ pipe to be $2.14 m$, is it possible to find what frequency it vibrates at? If I use the equation $f=\frac{v}{\lambda}$, does the $v$ apply to the speed of sound in the organ pipe or in air?
| $v$ applies to the speed of sound in the equation $f=\frac{v}{\lambda}$. Assuming air to be an ideal gas we can use the following equation to calculate the speed of sound in air:
\begin{equation}
v=331.3\sqrt{1+\frac{T}{273.15}}
\end{equation}
where $T$ is the air temperature in degrees Celsius.
The wavelength should be twice the length of the organ pipe, thus, the frequency is:
\begin{equation}
f=\frac{v}{\lambda}=\frac{331.3\sqrt{1+\frac{T}{273.15}}}{2L}
\end{equation}
Assuming that the air temperature is about standard room temperature (~20°C), the frequency is equal to
\begin{equation}
f=\frac{331.3\sqrt{1+\frac{20}{273.15}}}{2*2.14m}=80.190332Hz \approx 80Hz
\end{equation}
| {
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Complex Conjugate of Wave Function I've been reading through Griffiths QM book, and the only thing bugging me is they never fully described what $\Psi^* $ should be for any given function. I know it's the complex conjugate at the same time I think I just need concrete examples to solidify it in my head.
What is the corresponding $\Psi^*$ for
\begin{align}
\Psi_n(x,t) =& \sqrt{2\over a} \sin{n\pi x\over a} e^{-iE_nt} \qquad \text{(Infinite square well)} \\
\Psi_0(x,t) = &{m\omega\over{\pi \hbar}}^{1/4} e^{-{m\omega\over{2\hbar}}x^2-iE_0t} \qquad \text{(Simple Harmonic Oscilator)}\\
\Psi_k(x,t) =& Ae^{i(kx-{hk^2\over{2m}}t)} \qquad \text{(Free Particle)}
\end{align}
I think the part that is bugging me is that for the two prior cases the conjugate only alters the time term, but in the last equation, we are also altering the position term. How exactly should I rationalize this and come up with a good generalized concept of what $\Psi^*$ is?
| Take the time dependent Schrodinger equation
$$iħ \frac{∂Ψ}{∂t} = HΨ$$
and take the complex conjugate on both sides. The Hamiltonian is real, s.t. we get
$$-iħ \frac{∂Ψ^*}{∂t} = HΨ^*$$
But we can write the last equation otherwise, by inversing the direction of the time,
$$iħ \frac{∂Ψ^*}{∂t'} = HΨ^*$$
where $t' = -t$. It is a strange idea, isn't it?
Now, take the complex conjugate of the last wave-function you wrote - I assume for simplicity A = real,
$$Ψ^*_k(x,t) = A \exp\bigl(i\bigl[-kx - ħk^2\frac{-t}{2m}\bigr]\bigr)$$
You see what we got? The the time goes toward the past and the particle moves in opposite direction (back to the source).
This is $Ψ^*$ : the movie going backwards.
I am pretty sure that my answer can open more questions than you had before, but this is what I can say for the moment.
Good luck,
Sofia
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Can length contraction really be derived from time dilation? Does speed equal speed? I am referring to Wikipedia: "Length contraction can also be derived from time dilation." with the following proof which seems to be the result of a circular reasoning.
The proof uses only one and the same velocity v for the point of view of the observer and for the point of view of the reference frame of the observed object:
$\frac{L'}{L} = \frac{T'v}{Tv} = \frac{1}{\gamma}$
It seems that this is a circular reasoning, because we must first proof that both v of the formula are equal, and for this we need the same formula with v' and v:
$\frac{L'}{L} = \frac{T'v'}{Tv} = \frac{1}{\gamma}$
Example: A spaceship is travelling a distance of 8 Lmin (Earth reference frame) within 10 minutes (Earth reference frame), yielding a velocity of 0,8 c (Earth reference frame).
The point of view of the space ship is following from the proper time formula and the Lorentz contraction formula, both are multiplying with the reciprocal Lorentz factor 1/γ. The reciprocal Lorentz factor for v= 0,8 c is 0,6. We get 10 min. x 0,6 = 6 minutes and 8 Lmin x 0,6 = 4,8 Lmin. From the point of view of the spaceship it is travelling 4,8 Lmin in 6 minutes, yielding a velocity of 0,8 c (q.e.d.), so we proved that
v' = v.
Or am I wrong, and it is evident and without need of further proof that v' = v?
For the answer see the answer of Ben Crowell:
"v" is the relative velocity taken into account by the Lorentz factor between inertial reference frames which is the same for both frames. Thus v = v' can be derived directly from the SR postulates. And thus it seems that effectively length contraction can be derived from time dilation, as Wikipedia says.
|
Or am I wrong, and it is evident and without need of further proof that v' = v?
It may not be obvious, and it does require proof, but it is true, and it's not anything terribly deep or mysterious.
Consider two observers, Alice and Bob, moving away from each other. Alice says she's at rest and Bob is moving. Bob says the opposite. If they want to find out how fast the motion is, they can do it by sending signals back and forth and measuring how the time lag grows. The details of how they analyze the data are not important. What matters is that due to rotational symmetry (or reflection symmetry) they will both get the same data, and therefore arrive at the same result for $v$.
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How is the scattering length in 2d defined? Scattering length is 3d is well-defined. In the literature, one can also see scattering length in 2d. How is it defined? Can we even generalize it to 1d?
| One has to be careful when extending the concept of scattering length to lower dimensions. A straightforward extension of the 3D methodology to 2D is prone to lead to logarithmic divergences. Reason being that in 2D the radial Schrödinger's equation for the s-wave includes a negative centrifugal potential.
When carefully defining the scattering length as the hard-cylinder potential that gives the same low-wave number scattering solution phase shifts, one arrives at scattering lengths as defined by Verhaar et al:
Scattering length and effective range in two dimensions: application to adsorbed hydrogen atoms,
B J Verhaar, J P H W van den Eijnde, M A J Voermans and M M J Schaffrath
J. Physics A, vol. 17 (1984).
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What's Optimal About Six Legs According to Physical Laws? In many respects the insects can be regarded as the most successful class of animals in evolutionary terms. And one of the most common features of insects is that they (mostly) all have six legs.
Not discounting other traits, is there something about six legs that has helped insects achieve this success?
Can we use physical laws to analyze and determine an optimality of having six legs - perhaps such as stability?
|
Not discounting other traits, is there something about six legs that has helped insects achieve this success?
Spiders oftentimes have eight legs, mammals oftentimes have four. But centipedes have lots of legs, and millipedes have lots of legs. The reason mammals have four legs, and millipedes have lots and lots of legs isn't so much about optimality so much as that's how many legs their ancestors had.
The story of why people cut off the ends of the roast comes to mind. Great grandmother cut off the ends of the roast because a large roast wouldn't fit in the oven. The great grandchild follows this grand condition. Great grandmother did it because her oven was too small. The great grandchild's oven? It could easily accommodate a full-sized roast. Cutting off the ends of the roast is suboptimal.
Learning to cut off the ends of a roast because thats one's great grandmother did is a Lamarckian trait. Darwinian traits are even more strongly constrained. Spiders have eight legs because that's what ancestral spiders from hundreds of millions of years ago had. Insects have six legs because that's what ancestral insects from hundreds of millions of years ago had. Amphibians, reptiles, mammals, and birds have four appendages because that's what ancestral amphibians had hundreds of millions of years ago. Just as there's nothing optimal about cutting off the end of the roast now that the oven is bigger, there's nothing optimal about the number of limbs an animal has.
Oftentimes in evolution, the best answer for why some trait arose is that stuff happens, or rather, stuff happened.
| {
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Does zero free current entail zero $\vec H$? There are two kinds of magnetic fields (different authors give them different names), $\vec B $ and $\vec H$ which are related by the equation $$ \vec B = \mu_o (\vec H + \vec M)$$ where $\vec M$ is the magnetization.
Ampere's law for free currents states $$\oint_C \vec H \cdot d\vec l = I_{free} $$
This is my question: does zero free current entail zero $\vec H$?
My argument for that is this: since the contour integral of $\vec H$ is zero for all arbitrary curves C in a region of zero free current, $\vec H$ is necessarily zero. However this may not be a mathematically correct argument...
| I think your argument is completely flawed.
Consider a uniform H-field. The closed line integral of this field around any loop is zero - and there must be no free current through the loop. Hence there is no free current, yet the H-field is non-zero.
You might be better off thinking about this in terms of free current density.
In this case we can say
$$\nabla \times {\bf H} = {\bf J},$$
so that a zero free current density just means that the H-field has zero curl at that point, nothing more than that.
| {
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Who plays the role of centrifugal force in an inertial frame of reference? It is noteworthy to quote a sentence from my book,
It is a misconception among the beginners that centrifugal force acts on a particle in order to make the particle go on a circle. Centrifugal force acts only because we describe the particle from a rotating frame which is non-inertial.
Yes, the statement is undoubtedly right. But one thing that is annoying me is that If the particle moves on a circle due to centrifugal force from a rotating frame, what is the cause for motion of the particle on a circle from an inertial frame ? If there were only centripetal force , the particle would go towards the centre and never it would move in a circle. So, in order to move the particle on a circle , who will play the role of centrifugal force when viewed from inertial frame?? Please help.
| Let's consider a simple experiment in which a stone tied to a string is moving in a uniform circular motion in a horizontal plane. We can analyze this experiment from inertial and non-inertial frames.
An observer in an inertial frame sees the stone having a radial acceleration and concludes that there must be a radial force causing it. He observes the taut string and writes Newton's second law as $T = mv^2/r$, $T$ being the tension in the string, directed toward the center.
Now consider an observer in a frame of reference in which the stone is at rest. (Imagine him riding the stone.) He too observes the taut string. But he is at rest. To save Newton's second law, he imagines a force $F$ that exactly balance $T$ so that $F + T = 0$. This fictitious force $F$, is the centrifugal force.
As you see the observer in the inertial frame does not need $F$. It is contrived by the observer in the non-inertial frame so that he can still use Newton's second law.
Although Newton's second law is saved, Newton's third law is not. For in the non-inertial frame, the string pulls the stone with a force $T$ and as a reaction the stone also pulls the string by an equal and opposite force. While the stone experiences a centrifugal force $F$, no one experiences $-F$.
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If angular velocity & angular acceleration are vectors, why not angular displacement?
Are angular quantities vector?
... It is not easy to get used to representing angular quantities as vectors. We instinctively expect that something should be moving along the direction of a vector. That is not the case here. Instead, something(the rigid body) is rotating around the direction of the vector. In the world of pure rotation,a vector defines an axis of rotation, not a direction in which something moves.
This is what my book (authored by Halliday, Walker, Resnick) says. But they mentioned with a caution that
Angular displacements cannot be treated as vectors. ....to be represented as a vector, a quantity must also obey the rules of vector addition..... Angular displacements fail this test.
Why do they fail in this test? What is the reason?
| The wikipedia link does an adequate job of explaining this.
You can find a very mathematical discussion in this partial duplicate question, but here is my simple take on it.
The angular displacement in three dimensions does have a vector nature in the sense of having both a magnitude (the angle through which you turn something) and a direction (the axis about which it is turned).
However, for a true vector quantity ${\bf A} + {\bf B} = {\bf B} + {\bf A}$, and this is not true for angular displacement.
Hold a pencil with its point vertically upwards. Now rotate it through 90 degrees so the point is directly away from you (i.e. about a horiontal axis), followed by a 90 degree rotation about the vertical axis defined by the pencil originally. Note the position of the pencil point.
Now go back to the original position. Rotate the pencil through 90 degrees around its own vertical axis, followed by rotating it 90 degrees about a horizontal axis. The point ends up in a different place right?
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Is a photon really massless? If a photon travels at a speed of light and its massless then it must have no energy but this is not the case as we see in photo electric effect. Also help me to know what are photons made of, how are they created?.
| No, a photon does not need to have mass to be able to interact with matter. In fact it is its energy which is important in interactions. For the photoelectric effect the incoming photon must contain enough energy to displace the electrons on the metal of the photodiode.
The question of what are photons made of is a pretty deep and difficult question, one that can only be answered with particle physics. But this is what I do know. You can create photons from pair annihilations (electron+positron, etc). And you can cause a photon to be ejected if you let an excited electron go down an energy level in an atom/molecule.
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Velocity of car down an Inclined Plane I was learning to make a car game and came across this situation where my car is on an inclined plane. It's initial velocity is 0. Now, the problem is that my memory serves me this formula to calculate the final velocity
$$v_f^2 - v_i^2 = 2as$$
Since, $v_i = 0$, the equation would be
$$v_f^2 = 2as$$
For an inclined plane $a = g \sin \theta$,
$$ \Rightarrow v_f^2 = 2 g s \sin \theta $$
However, my memory doesn't serve me what to do when you do not know how far the car would have to travel on this inclined plane.
My problem in short:
$\bullet$ I need to add Velocity to the car which is on the inclined plane
$\bullet$ What is the formula I should use? does the formula mentioned above of any good in this case?
$\bullet$ If yes, how would I come over this displacement drawback (drawback in the sense, I would not know how far the car would travel)
$\bullet$ If no, what would you use to add the velocity to the car?
| You can cast your problem in terms of what is the velocity of the car at any instant of time after it started. The answer is $v = a_{\rm eff} t$ from the first kinematical equation. In the same time, the car would've traveled a distance $s = (1/2)a_{\rm eff} t^2$ from the second kinematical equation.
The other option is - if your inclined plane is of any fixed length $L$ and the time $t$ of concern to you is the time in which it has completely traversed the plane, you can cast everything in terms of this distance, by using the second of these equations to write $t_f = (2L/a_{\rm eff})^{1/2}$, and $v_f = a_{\rm eff} t_f$.
Thus, in order to model everything perfectly, you need to have one of these as inputs - either $L$, or the instant of time $t$.
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Is it possible to create a parachute large enough to stop all velocity? This idea came to me while playing Kerbal Space Program. I noticed that the larger my parachute was, the slower my rocket would fall back down to Kerbin. I would like to know if it is possible to create a parachute so large in the real world that it might stop all velocity, essentially making whatever is attached to it float in mid-air. Common sense is telling me "no," but I could always be wrong, and I would love some explanation behind whether or not it is possible.
| Yes it is possible. The trick is to have a parachute which is large enough that it's Schwarzschild radius extends down to the object it is lifting.
Under such a circumstance, the parachute would stop ALL motion of the object it is lifting.
PS I just watched Interstellar :D
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Where on Earth does the mass of 1 kg actually produce a 1 kg reading on a digital scale? Gravity on Earth varies by about 0.1% between poles and equator. If someone was buying/selling something mass critical e.g. gold, where is the standard place on Earth where a 1 kg mass produces a 1 kg reading as measured by a device like the following:
Based on some responses, I should give a very specific example. We take the International Prototype Kilogram mass from its repository in Paris. We then take it to one of the poles and measure it using this type of device (also shown above). We then take it to the equator and do the measurement again. The numbers are different.
Is there a place on Earth specified where the scales will read exactly 1kg?
| In addition to the good answers already given a couple of points.
1) if you used scales where one mass balances another mass like this one
Then you would not have problems with any variation in $g$.
2) I did a google search to check what people use to measure the mass of gold (and also diamond) and everything came back as digital scales like the one shown in the question.... seems like noone uses the old type shown in point 1) above any more.
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What do physicists mean by "information"? On the question why certain velocities (i.e. phase velocity) can be greater than the speed of light, people will say something like:
since no matter or "information" is transferred, therefore the law of relativity is not violated.
What does information mean exactly in this context?
It may help to consider the following scenarios:
If a laser is swept across a distant object, the spot of laser light can easily be made to move across the object at a speed greater than $c$. Similarly, a shadow projected onto a distant object can be made to move across the object faster than $c$. In neither case does the light travel from the source to the object faster than $c$, nor does any information travel faster than light.
Read more: https://www.physicsforums.com/threads/phase-velocity-and-group-velocity.693782/
| In the context of relativity and nonproagation of information at greater than lightspeed between two separated points $A$ and $B$, "information" simply means any particle, feature of in a field (EM, quantum field, curvature in spacetime ...), message or so forth that could allow a causal link between $A$ and $B$, i.e. could make $B$'s physics depend on $A$'s presence (and contrariwise).
Therefore we know, for example, that, in relativistic limits, the wonted heat diffusion equation $(\partial_t-k\,\nabla^2)T=0$ cannot be correct, for its solution in 1D is a superposition of heat kernels $\frac{1}{\sqrt{4\,\pi\,k\,t}}\exp\left(-\frac{x^2}{4\,\pi\,k\,t}\right)$. Suppose $A$ sits at $x=0$ and imparts an impulse of heat at $x=0$ (i.e. heats a tiny region near $x=0$ intensely and quickly) and $B$ at $x=L$ has agreed to raise a flag as soon as $B$ senses a rise in the temperature at $x=L$. The time dependence of the heat kernel, we see that the temperature begins to rise at $L$ at $t=0$, so the signalling speed in this case between $A$ and $B$ would be arbitrarily fast and limited only by the signal to noise ratio of $B$'s measurement.
Sometimes it is stated that the signal, or "information propagation" speed in a dispersive medium is the group velocity, since this is approximately the speed at which any narrowband modulation is propagated on a carrier light wave and indeed this does seem to impose speed limits of $<c$ when applied to regions of anomalous dispersion in optical mediums. But this is only an approximation which breaks down for very wideband signals. Ultimately one needs to return to basic causality limits, e.g. as found by the Paley-Wiener criterion, to work out what limitations there must be on things like optical dispersion.
| {
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Statistical physics and momentum conservation In statistical physics one usually looks at energy as a conserved quantity and e.g. in the canonical ensemble assumes a constant average energy of the ensemble. Now why don't we usually do this for other conserved quantities like momentum? Why not do a 'canonical' ensemble with momentum exchange? Is it more complicated or simply never useful?
| As the previous answers have noted, momentum conservation often isn't useful. However, there are exceptions -- especially in transport problems. Although you're not necessarily directly using the canonical ensemble for these problems, they are decidedly statistical mechanics problems. For example, (quasi)momentum conservation in phonon normal scattering processes leads to a modified distribution function. (See equation 6.8 in this paper -- often regarded as a classic.)
| {
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What does the exponential decay constant depend on? We know the law of radioactivity:
$$N=N_0e^{-\lambda t}$$
where $\lambda$ is the exponential decay constant. My question is: This constant depends of what?
| The transition probability per unit time of a nucleus from an initial state i to a final state f, representing the decayed system, is modeled by Fermi's Golden Rule:
$$\lambda=T_{i\rightarrow f} = \frac{2\pi}{\hbar}\left|\left\langle i\left|H'\right|f\right\rangle\right|^2\rho$$
Where $T_{i\rightarrow f}$ is the transition probability from state $i$ to state $f$ per unit time, $H'$ is the matrix element of the the transition operator, and $\rho$ is the state density about the final nuclear energy.
The experimental measurement of a decay constant provides a benchmark for validation of theoretical models of the physics of nucleon-nucleon interactions and nuclear energy structure. In some rare cases, the decay probabilities are so minute that the Golden Rule provides a useful a priori estimate of the decay likelihood, which can guide the design of experimental measurements of such rare decays.
| {
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What is the point of this type of graph? (Projectile) What is the point of graphing vs range, during kinematic motion?
| Generally speaking, when we graph experimental data it's nice for the graph to be a straight line. This is because it's easy to tell if a graph is a straight line just by putting a ruler on it. If a graph is a curve it's hard to tell whether it's a parabola, ellipse, sine wave or some other curve without diving into some maths.
In this experiment you are varying the launch angle $\theta$ and measuring the range $R$. Let's have a look at what a graph of range against $\theta$ looks like:
This obviously isn't a straight line. So can we change the graph a bit to make it a straight line? Well, if we have a look at the Hyperphysics article on trajectories we find the range is given by the equation:
$$ R(\theta) = \frac{2v_0^2}{g}\sin\theta\cos\theta \tag{1} $$
and this obviously isn't a straight line because it's quite a complicated function. But suppose define a new variable $x$ as:
$$ x = \sin\theta\cos\theta $$
Using this new variable the equation (1) for the range becomes:
$$ R(x) = \frac{2v_0^2}{g}\, x $$
The parameters $v_0$ (the launch velocity) and $g$ (the acceleration due to gravity) are just constants, so the range is just proportional to $x$. This means a graph of the range against $x$, i.e. against $\sin\theta\cos\theta$, should be a straight line. And here's what that graph looks like:
And it is a straight line!
So that's why when you do the experiment you graph $R$ against $\sin\theta\cos\theta$. It's because the graph should be a straight line. Any deviations away from a straight line will mean either you messed up the experiment or there is some other factor that isn't included in equation (1).
In fact you may well find your graph isn't quite straight. That's because equation (1) doesn't include the effects of air resistance.
| {
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What does it mean to say "a paramagnetic material is attracted to an external magnetic field?" I'm just having a hard time wrapping my head around what actually goes on when a paramagnetic material is exposed to an external magnetic field. I understand that the individual dipoles line up so that they point in the direction of the field, but why does that need to happen for there to be magnetic attraction? And what exactly is being attracted in the first place? If I imagine the dipoles are caused by little rings of current, are the rings themselves pulled in the direction of the field? Also, what happens if the dipole starts out pointed out in exactly the opposite direction of the external field?
| A paramagnetic material is attracted not by the magnetic field but the force on it is towards the direction in which the magnetic field is increasing. In a constant field it would stay in place. This follows from the simple fact that all systems try to obtain a potential energy minimum.
| {
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Are there eight or four independent solutions of the Dirac equation? I edited the question as a result of the discussion in the comments. Originally my question was how to interpret the four discarded solutions. Now I'm making a step back and hope that someone can clarify in what sense it is sensible to discard four of the eight original solutions of the Dirac equation.
From making the ansatz ${\mathrm{e}}^{+ipx}$ and ${\mathrm{e}}^{-ipx}$, with $E=\pm \sqrt{ (\vec p)^2 +m^2} $ we get eight solutions of the Dirac equation. $u_1, u_2, u_3 , u_4$ and $v_1,v_2,v_3,v_4$.
Conventionally the four solutions ($u_3 , u_4,v_3,v_4$.) following from $E=- \sqrt{ (\vec p)^2 +m^2}$ are said to be linearly dependent of the remaining four solutions with $E=+\sqrt{ (\vec p)^2 +m^2}$ two ($u_1,u_2$) are commonly interpreted as particle and two ($v_1,v_2$) as antiparticle solutions.
Nevertheless, in order to be able to construct chirality eigenstates we need the other four solutions and I'm unsure in how far we can then say that four of the eight solutions are really linearly dependent. A chiral eigenstate must always be of the form $ \psi_L= \begin{pmatrix} f \\ -f \end{pmatrix} $ for some two component object $f$. In order to construct such an object we need all eight solutions. For example $\psi_L= u_1 - u_3$, as can be seen from the explicit form of the solutions recited below.
In addition, I'm unable to see that the eight solutions are really linearly dependent, because for me this means that we can find numbers $a,b,c,d,e,f,g,h \neq 0$, such that $a u_1 + b u_2 + c u_3 +d u_4 + e v_1 + f v_2 + g v_3 + h v_4=0$. As pointed out in the comments, this can be done, but only for one point in time. Is this really enough? In what sense is then for example the basis used in the Fourier expansion $\sum_n (a_n e^{in x} + b_n e^{-in x}) $ linearly independent? With the same reasoning we could find numbers for one $x$ to show that all these $e^{in x}$ and $e^{-in x}$ are linearly dependent...
The explicit solutions
This is derived for example here
Two solutions follow from the ansatz ${\mathrm{e}}^{-ipx}$ with $E=+ \sqrt{ (\vec p)^2 +m^2}$ and two with $E=- \sqrt{ (\vec p)^2 +m^2}$ .
In the rest frame the solutions are
$$ E=+ \sqrt{ (\vec p)^2 +m^2} \rightarrow u_1 = \begin{pmatrix} 1 \\ 0 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} \qquad u_2 = \begin{pmatrix} 0 \\ 1 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} $$
$$ E=- \sqrt{ (\vec p)^2 +m^2} \rightarrow u_3 = \begin{pmatrix} 0 \\ 0 \\1 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} \qquad u_4 = \begin{pmatrix} 0 \\ 0 \\0 \\ 1 \end{pmatrix} {\mathrm{e}}^{-imt} $$
Analogous four solutions from the ansatz ${\mathrm{e}}^{+ipx}$, we get four solutions.
$$ E=+ \sqrt{ (\vec p)^2 +m^2} \rightarrow v_1 = \begin{pmatrix} 0 \\ 0 \\1 \\ 0 \end{pmatrix} {\mathrm{e}}^{imt} \qquad v_2 = \begin{pmatrix} 0 \\0 \\0 \\ 1 \end{pmatrix} {\mathrm{e}}^{imt} $$
$$ E=- \sqrt{ (\vec p)^2 +m^2} \rightarrow v_3 = \begin{pmatrix} 1 \\ 0 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{imt} \qquad v_4 = \begin{pmatrix} 0 \\ 1 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{imt} $$
Examples for chiral eigenstate are, with some two component object $f$
$$\psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} \hat = u_1 - u_3 = \begin{pmatrix} 1 \\ 0 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} - \begin{pmatrix} 0 \\ 0 \\1 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} \qquad \text{ or } \qquad \psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} \hat = u_2 -u_4 $$
$$\psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} \hat = v_1 - v_3 \qquad \text{ or } \qquad \psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} \hat = u_2 - u_4 $$
And similar for $\Psi_R = \begin{pmatrix} f \\ f \end{pmatrix}$.
Are four of the eight solutions really dependent? If yes, how can this be shown explicitly ? Any source, book, pdf would be awesome. Is it possible to interpret the solutions $(u_3,u_4,v_3,v_4)$ that can be discarded for many applications, but that are needed in order to create chirality eigenstates?
| There are only 4 independent solutions because:
$$v_{1,2}(+E, +\vec{p}) e^{+i(Et-\vec{p}.\vec{x})} = u_{3,4}(-E, -\vec{p}) e^{-i((-E)t-(-\vec{p}).\vec{x})}$$
(where $u$ and $v$ here don't include the propagating part contrary to your notation). You could choose to work with $u_1,u_2, u_3, u_4$ but it is more convenient to use $u_1,u_2, v_1, v_2$ where only "positive" quantities are defined.
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Habitable zones around other stars I have a question about measuring the boundaries of habitable zones on other planets.
Is it okay to assume that, if Sun's habitable zone starts at a distance $R_0$ and its luminosity is $L_0$, we can calculate any other star's with luminosity $L$ habitable zone's inner boundary as $$R= R_0 \sqrt{\frac{L}{L_0}}$$?
Formula was derived from $$F=\frac{L}{4\pi R^2}$$ where $F$ is the Flux, $R$ is the distance to the star. I assume that the Flux at the boundary should remain the same as it was in the solar system, thus $F_1=F_2$;
If thpse were incorrect assumptions I would like to know what am I missing.
Thank you!
| This looks fine, BUT, indeed have a good read of the wikipedia pages on the topic of habitable zones to see all the complications there are in deducing where around a star the conditions may be "habitable". In addition I recommend, if you are serious about this calculation, reading the short article by Kane & Gelino (2012) and visiting the accompanying Habitable Zone Gallery.
The formula you have gives a reasonable baseline and estimate, but should probably not be extrapolated too far away from solar-like stars.
Complications include:
Do you want your habitable zone to be habitable for a particular length of time? Massive stars change their luminosity fast.
You are assuming a circular orbit, but planet orbits may be elliptical.
Your calculation assumes that your planet has a similar atmosphere (and therefore gravity etc.) as the Earth. Different atmospheres lead to different surface temperatures. Low metallicity stars will have planets without any carbon dioxide! There are also external factors to do with stellar magnetic activity that can drastically influence atmospheres (and have done so in our solar system); that means stars of different type and age could have altered habitable zones.
Very small stars would have very close-in habitable zones, but then tidal effects could be very important.
Both very massive stars and low mass stars (especially when younger) can have strong ultraviolet radiation fields that may preclude life (as we know it).
The calculation assumes that all the heat required comes from the star. But it could also come from the radioactive decay of rocks or by tidal heating in the case of a moon orbiting a larger planet (think Io, Europa).
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What do we see while watching light? Waves or particles? I'm trying to understand quantum physics. I'm pretty familiar with it but I can't decide what counts as observing to cause particle behave (at least when it's about lights). So the question is what do we see with our eye-balls?
*
*We point a laser (or any kind of light source) to the wall. We see its way from point A to B. Do I "see" a particle or a wave?
*Let's see an average object. It pretty much looks like their "pieces" a observing that influences their behave. Does this means while we're watching a light it acts like particle in quantum level?
| You are seeing particles. However there's more to this than meets the eye so I need to explain exactly what I mean by this.
Light is neither a particle nor a wave. Instead it is a quantum field. As a general rule while light is travelling it appears as a wave, but when the light quantum field is exchanging energy with anything it does so in quanta that appear as particles i.e. photons.
You see because light excites electrons in rhodopsin molecules in the cells in your retina. Since this is an energy exchange (from the quantum field to the rhodopsin molecule) the interaction looks like absorption of a photon.
So you are seeing particles.
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Capacitance of two non parallel plates What is the formula for capacitance of two non parallel plates at an angle with each other?If the plates were parallel then the value can be calculated as
(PermittivityX area of one plate)/distance between them.But what happens in case the plates are tilted at an angle?The question came to mind while trying to understand electrostatic separator.What would be the derivation of the formula for capacitance of two non parallel plates placed at an angle?
I did get a method from https://web.archive.org/web/20160417130540/http://www.davidpublishing.com/davidpublishing/upfile/12/15/2011/2011121573197833.pdf
Equation 6 from the above link above helps but it is independent of the length of the plates which doesn't seem likely.
| Let's do some calculus.
Suppose you have two plates, almost parallel (off by an angle $\alpha$). The plates lie in the XY plane, from $(0, 0)$ to $(x_1, y_1)$. At $x = 0$, the plates are separated by a distance $z_0$, and at $x = x_1$, the plates are separated by a distance $z_1$.
We'll now consider an infinitesimally small element of both plates. (Since parallel capacitances add, and all the infinitesimal pairs are in a parallel configuration, we can use integration)
\begin{align}
\tan \alpha &= \frac{z_1 - z_0}{x_1} \\
\mathrm{d}C &= \varepsilon \frac{\mathrm{d}A}{\delta z} \\
\mathrm{d}A &= y_1 ~\mathrm{d}x \\
\delta z &= z_0 + x \tan \alpha \\
\therefore C &= \int\mathrm{d}C \\
&= \int\limits_A \varepsilon \frac{\mathrm{d}A}{\delta z} \\
&= \int\limits_0^{x_1} \varepsilon \frac{y_1 ~\mathrm{d}x}{z_0 + x \tan \alpha} \\
&= \varepsilon ~ y \left[ \cot \alpha \ln(z_0 \cos \alpha + x \sin \alpha) \right]_0^{x_1} \\
&= \varepsilon ~ y_1 \left(\frac{\ln(z_0 \cos \alpha + x_1 \sin \alpha)}{\tan \alpha} - \frac{\ln(z_0 \cos \alpha)}{\tan \alpha} \right) \\
&= \varepsilon ~ y_1 \left( \frac{\ln(1 + (x_1 / z_0) \tan \alpha)}{\tan \alpha} \right) \\
&= \frac{\varepsilon ~ y_1}{\tan \alpha} \ln \left( 1 + \frac{x_1}{z_0} \frac{z_1 - z_0}{x_1} \right) \\
&= \frac{\varepsilon ~ y_1}{\tan \alpha} \ln \left( \frac{z_1}{z_0}\right)
\end{align}
If you assume $\alpha$ is small, then $\tan \alpha \approx \alpha$, which gives
\begin{align}
C &= \frac{\varepsilon ~ y_1}{\alpha} \ln \left(1 + \frac{x_1}{z_0} \right)
\end{align}
This conclusion is the same as the Eq. 6 in the paper you linked.
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Has a body angular momentum and torque only in a circular path? In different contexts, my book(Principles of Physics by Resnick, Halliday ,Walker) , they wrote
For torque, the path need no longer be a circle and we must write the torque as a vector $\vec{\tau}$ that may have any direction. .... Note carefully that to have angular momentum about $O$ , the particle does not itself have to rotate around $O$ .
That's what they wrote. But I am really confused why they wrote so. In fact I can't imagine torque & angular momentum without circular motion. Why did they tell so? What is the cause?? Please explain.
| Actually, your book is correct. Even if the most usual uses of angular momentum involve circular or rotating motion, this is not the general case. An object moving in a straight line has angular momentum in a reference frame in which the origin does not fall on the the line.
To see this simply remember the definition of angular momentum $\overrightarrow{L}=\overrightarrow{r}\times m\overrightarrow{v}$, and torque $\overrightarrow{\tau}=\overrightarrow{r}\times \overrightarrow{F}$. Both can defined for any object regardless of their motion. One example where you use it for linear motion is when solving problems involving conservation of angular momentum and a mass gets ejected from a rotating one. The ejected mass will move in a straight line, but total angular momentum is conserved when you include that of straight moving mass too.
If you compute the angular momentum of a free moving mass (no applied forces), you will obtain that it is constant across the entire path, as you would expect.
In the picture below it's shown both: that the angular momentum of a stright moving objects exists, and that it is a constant (if the velocity is constant)
This is because $r \sin{\alpha}$ is constant along the trajectory (is the distance between the two parallel lines in the picture.
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Is it possible to make laser beams visible midair without smoke? My question is: Is it possible to make laser beams viable midair without smoke? I thought it would be a great idea to have a (second) smartphone or pc screen without having a solid screen. The reason why it has to be without smoke is that smoke would be too dependent on the environment.
If it is possible to make laser beams appear in some way midair (without smoke, just air), would this be possible? I am not sure if its possible with two lasers "collide" or some other way.
Wouter
| Not really; you need to have the laser light pass through particles in a medium. Laser light is made of photons; in order to see the laser, photons must be reflected off of a something to your eyes. You cannot otherwise "see" a photon because photons don't interact via the electromagnetic force - in other words, photons don't emit photons. To see the laser beam, you have to see photons reflected off of something, and so you have to have it passing through some medium for this to work. Smoke works well, as does anything that's thick and fills the air. Normal air does not work well. So technically you could use something besides smoke, but you need to have the beam pass through some medium.
| {
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Restrictions on Bell-type inequalities While deriving and proving Bell-type inequalities of the form
$|E(a,b)-E(a,b')|+|E(a',b)+E(a',b')|\leq 2$
I know that the conditions on the operators $O_a$ and $O_b$ are that they must be bounded by $\pm 1$.
Joint operator $O_{ab}\equiv O_a O_b$ is consequently bounded by $\pm 1$.
However, is there any such bound on the correlation $E(a,b)$ given by operating by $O_{ab}$ on whatever state you're studying? Does $E(a,b)$ necessarily NEED to be bounded by $\pm 1$ as per the definition?
(I know that it sometimes is as a result of, say, operating on the singlet state, but is this a consequence or a condition?)
Thanks in advance!
| The CHSH inequality was not proved for any state, but for the spin singlet or for the photon singlet, s.t. to your question
"is there any such bound on the correlation E(a,b) given by operating by O_{ab} on whatever state you're studying? Does E(a,b) necessarily NEED to be bounded by ±1 as per the definition?
As I say, you cannot work with whatever state.
On other states there are other inequalities, you can follow in the arXiv quant-ph the works of Adan Cabello.
But WHY are you interested to break the limit +- 1 on E(a,b)? What is exactly the problem?
Good luck,
Sofia
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How can the Andromeda Galaxy alone have more mass than the Local Group? I was reading about the Local Group, and it is stated that the total mass of the group is $1.29±0.14 × 10^{12}$ Solar masses.
The number actually felt a bit low to me because I know the Milky Way galaxy alone is close to a trillion solar masses, so I checked the Andromeda Galaxy's mass to find it $1.5 × 10^{12}$ Solar masses.
The Local Group also contains more than 54 galaxies, with the Milky Way and Andromeda being the most massive, so by only summing the masses of the Milky Way and Andromeda, we already exceed the mass estimated for the group. Not to mention the other 52 less massive galaxies.
So how could this be correct ?
Does it have something to do with the calculation of the dark matter mass ?
| It is difficult to estimate the masses of either galaxies or clusters of galaxies, and it depends on what source you consult. You can get different values from different sources, for instance in this reference: the local group is estimated to have $5.27\times 10^{12} M_\odot$ (which does include dark matter). I didn't have access to the wikipedia source.
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Disturbing a line of infinite alternating charges Consider a line of infinite number of alternating charges. All are point charges having charge of same magnitude and are placed in a line. Neglect the effect of gravity here. Consider one of the positively charged particle. I disturb it a little upwards.
What will happen then? Will they buzz off from each other at a point of time?
What I actually think is that considering a slightly disturbed positive particle,
I have tried to explain my reasoning via the image above.
The length of the lines represent the force acting on the disturbed positive charge. This will continue till infinity, but the forces will get more faint. As you can see, the net force is downwards, and since the negative charges will attract it too, the entire line will move with it forming a bump or a wave. This will come back to its starting point back, where we have no force acting on it but a velocity in the downward direction. So it will continue to move and therefore I think the line will move in a wave with increasing amplitude.
But, if it is so, at one point, the amplitude should be large enough for the wave that the particles in a side would face repulsion from the other side and therefore buzz off.
But I don't think so it will happen because the nearby ones will still keep it in its place and won't let it go. So according to me they shouldn't buzz off.
Am I right? What actually will happen? Will the system disintegrate?
| The original system as described in inherently unstable (or at best metastable).
According to Earnshaw's Theorem, "a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges"
So the introduction of the slightest disturbance must cause a massive, total "buzzing off".
The system will disintegrate...
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Best way to heat something in aluminum foil? Let's say we have a wet piece of paper, wrapped in aluminum foil, that we need to heat up in the fastest and most energy efficient way possible (no flamethrower).
What would that be?
Details regarding the methods would be highly appreciated.
| Put it in a microwave at the right power, and it will quickly heat up withouth burning (and heat whatever is inside). But do not try this at home. Too much power and you will get the microwave oven on flames.
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Proper way to write the units to indicate that they include an offset? Past a certain point of complexity, I get rather confused with physical units, so I am asking a physicist for help.
I have a code that represents temperature, with a resolution of [0.5 °C], whose value ranges from 0 to 255 as the temperature goes from -41 °C to the max.
The equation is:
Temperature code [units?] = 82 + 2 * Actual temperature [°C]
i.e.:
0 = -41.0 °C
1 - -40.5 °C
2 - -40.0 °C
...
81 = -0.5 °C
82 = 0 °C
83 - 0.5 °C
...
Are the units of the "Temperature Code" in the included graph one the following?
[0.5 - 82 °C]
[0.5 - 41 °C]
[-82 + 0.5 °C]
[-41 + 0.5 °C]
[0.5 °C - 82]
[0.5 °C - 41]
[0.5 + 82 °C]
[0.5 + 41 °C]
[82 + 0.5 °C]
[41 + 0.5 °C]
[0.5 °C + 82]
[0.5 °C + 41]
If not, what are the units?
EDIT: CLARIFICATION.
I know what the conversion is.
I know that the internal units of the code are "counts".
I know that, if there were no offset, the units of the code would be [0.5 °C].
What I don't know is how to include the effect of the offset in the units.
Why do I need to know?
*
*To document the code correctly
*For the sake of doing a strict units analysis in data conversions, to confirm that the code converts correctly; for that, each variable must be documented with the correct units
*To help others understand the code
| While it is not a general answer, there is an engineering standard that holds that thermometer readings in non-absolute scales and temperature differences in the same scales have both different written notation and different spoken reading.
The prescribed convention is:
*
*Thermometer readings get the degree symbol before the unit as in $$0 ^\circ\mathrm{C} = 32^\circ\mathrm{F}$$ which is read
"zero degrees Celsius equal thrity-two degrees Fahrenheit".
*Temperature differences reverse the order of the degree marker and the unit, so that discussing the relative size of the scale increments one writes $$1 \,\mathrm{C}^\circ = 9/5 \,\mathrm{F}^\circ$$ and reads
"one Celsius degree equals nine-fifths of a Fahrenheit degree".
I've only ever seen this in textbooks in the physics world, but I'm told there are engineering disciplines where it is expected.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What would Earth have been without the Moon? Would Earth rotation have been more slowed down because of the tidal effect from the Sun, as seems to be the case with Mercury and Venus? Due to the giant impact hypothesis the angular momentum from the impact was increased and split.
If Earth not would have been a two part angular momentum system, but a single planet with the same rotation at that time as it have today, what would the tidal effect from sun have done to earth?
Are there any theories meant to explain the distribution of angular momentum in the Solar system? Roughly!
The reason for my interest is of course the question how important the Earth-Moon system's creation was to the development of which we are a part. It's a pretty well-shaped planet we live on.
| Tidal forces drop rapidly with distance - the derivative of $1/t^2$ is $-2/r^3$. Further, the difference in radius of the orbits of Earth and Mercury is a little more than a factor 3x and radius of mercury is about 2.5x smaller than that of earth.
From the orbits we gather the tidal effect is 27x smaller - from the radius we gather that moment of inertia is about 100x larger. The tidal torque probably scales with radius, so the final effect of tidal drag (rate of deceleration) on a moonless earth is about 1000x smaller than for Mercury.
Given that Venus is also not tidally locked it is reasonable to assume that Earth, even without the moon (and the impact that caused it) would probably still be rotating.
Of course without the lunar tides, there might have been no tidal pools and no evolution as we know it... But I think there would still be days and nights. Weekends might last a little bit longer, is all.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/149439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where do ultra-high-energy cosmic rays come from? Physicists have detected an amazing variety of energetic phenomena in the universe, including beams of particles of unexpectedly high energy but of unknown origin. In laboratory accelerators, we can produce beams of energetic particles, but the energy of these cosmic rays far exceeds any energies produced on Earth. So my question is, from where do these ultra-high-energy cosmic rays come from?
| Here is a topical proposal (from arxiv.org/abs/1602.06961):
The recent detection of the gravitational wave source GW150914 by the
LIGO collaboration motivates a speculative source for the origin of
ultrahigh energy cosmic rays as a possible byproduct of the immense
energies achieved in black hole mergers, provided that the black holes
have spin as seems inevitable and there are relic magnetic fields [B ≥ 10^11 Gauss] and
disk debris remaining from the formation of the black holes or from
their accretion history. We argue that given the modest efficiency <
0.01 required per event per unit of gravitational wave energy release, merging black holes potentially provide an environment for
accelerating cosmic rays to ultrahigh energies.
The authors cannot provide any precise composition and spectral features of the
ultra-high energy cosmic rays but they claim that
The only direct evidence of an association between UHECRs and BH
mergers can be obtained by the observation of gravitational waves in
coincidence with high-energy neutrinos.
Let us hope that future observations of black-hole mergers tell us more!
An update (with more data on compact astronomical objects mergers):
New computations based on the multimessenger astronomy data from the neutron star-black hole merger GW170817 seem to validate the relevance of mergers of compact astronomical objects like neutron stars as a prominent source of high energy cosmic rays in the 20-1000 PeV range.
Remark : 1000 PeV is the lower energy limit of ultra-high-heavy-cosmic-rays.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can the relative position of two masses connected by a spring become negative?
Consider the following diagram. Two masses of 1 kilogram each are attached by a spring of 1 N/m. The $x$-axis is chosen such that $x_1(0)=0$ and $x_2(0)=L$ where $L$ is the length of the spring in relaxation (no restoring force acting on the masses).
At $t=0$ the speed of the first mass on the left is $v$ directed to the right. The objective is to find the relative position of the masses, $d(t)=x_2(t)-x_1(t)$ for $t \ge0$.
My attempt
Initial condition
$$
\begin{cases}
x_1(0)=0\\
\dot{x}_1(0)=v \\
x_2(0)=L\\
\dot{x}_2(0)=0 \\
\end{cases}
$$
Applying Newton's second law to the first mass:
\begin{align*}
\ddot{x}_1(t) &=-(x_1-x_2)\\
s^2 X_1 -s x_1(0) -\dot{x}_1(0) &= -(X_1-X_2) \\
s^2 X_1 -v &= -(X_1-X_2) \\
(s^2+1) X_1 -X_2 &= v
\end{align*}
Applying Newton's second law to the second mass:
\begin{align*}
\ddot{x}_2(t) &=-(x_2-x_1)\\
s^2 X_2 -s x_2(0) -\dot{x}_2(0) &= -(X_2-X_1) \\
s^2 X_2 -sL -0 &= -(X_2-X_1) \\
-X_1 + (s^2+1) X_2 &= sL
\end{align*}
Solving the simultaneous equations, I have
\begin{align*}
X_1 &= \frac{(s^2+1)v+sL}{s^2(s^2+2)} \\
X_2 &= \frac{(s^2+1)sL+v}{s^2(s^2+2)}
\end{align*}
Partial fraction expansion,
\begin{align*}
X_1 &= \frac{L/2}{s} +\frac{v/2}{s^2} +\frac{-sL/2+v/2}{s^2+2} \\
X_2 &= \frac{L/2}{s} +\frac{v/2}{s^2} +\frac{sL/2-v/2}{s^2+2}
\end{align*}
\begin{align*}
D &= X_2-X_1 \\
&=\frac{sL-v}{s^2+2}
\end{align*}
Inverse Laplace transform for $D$,
\begin{align*}
d(t) &= L \cos (t\sqrt{2}) -\frac{v}{\sqrt{2}}\sin(t\sqrt{2}) \\
&=\sqrt{L^2+\frac{v^2}{2}}\cos\bigg(t\sqrt{2}+\tan^{-1}\bigg(\frac{v}{L\sqrt{2}}\bigg)\bigg)
\end{align*}
Question
Why can $d(t)$ become negative? What does it mean?
| I haven't combed through your math, but I suspect that you made some mistakes along the way. For instance, in your first set of equations $\ddot{x}_1(t)=-(x_1-x_2)$ should actually be $\ddot{x}_1(t)=(x_2-x_1-L)$. You can check yourself on this fact because the force ought to be $0$ when $d=L$. Furthermore, you need to be careful about your minus signs because when the force is positive on mass-2 then it ought to be negative on mass-1 (remember Newton's Third Law?).
I'm going to give you some tips for working on this problem and others like it, but first, I'm going to give you a word of advice: use the Laplace Transform only as a last resort; it is rarely necessary and almost always cumbersome.
Here's the method for approaching this problem: instead of making more work for yourself by taking the Laplace Transform, make less work by recasting the problem to be in terms of the variable $d=(x_2-x_1)$ at the very beginning and then subtracting the two equations of motion from one another. This change of variables looks like this:
$$
\ddot x_1=((x_2-x_1)-L)\\
\ddot x_2=-((x_2-x_1)-L)\\
\ddot x_2-\ddot x_1=-2((x_2-x_1)-L)\\
\ddot d=-2(d-L)
$$
From here, if you make one more change of variables to $X=d-L$ the problem becomes almost trivial to solve. And once you've solved it you can change your variables back to be in terms of $d$. Then, by plugging in your initial conditions (written in terms of $d$, of course), you should have no problem getting the solution.
You'll find that $d$ might still sometimes be negative in this new solution if $v$ is sufficiently large. This is due to the artificiality of the problem: real springs don't perfectly obey Hooke's Law and when the masses get close enough they would actually run into each other, but our model here doesn't take any of that into account. For this reason, our model allows the masses to pass through each other. Thankfully, the only thing that a negative value of $d$ means is that mass-1 is to the right of mass-2 - all of the kinematical properties still behave correctly.
| {
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Kirchhoff's current law with a nonlinear resistor It is said that by Kirchhoff's current law
$$
\frac{e - v_c}{R_1} = c\frac{dv}{dt} + f(v_c) + i\tag{1}
$$
and from Kirchhoff's voltage law
$$
v_c(t) = iR_2 + L\frac{di}{dt}\tag{2}
$$
from the following diagram:
It is easy to see equation (2) but I dont see how equation (1) was obtained.
| Kirchhoff's current law says that the current entering any junction is equal to the current leaving that junction.
The current through $R_1$ must therefore add up to the current through the three legs of the circuit.
$$ I_{\mathrm{R}_1} = I_{\mathrm{cap}} + I_{\mathrm{nonlinear\ R}} + I_{\mathrm{R}_2} $$
Using the terms in the diagram, two of these already have an explicit label:
$$I_{\mathrm{nonlinear\ R}} = f(v_c)$$
$$I_{\mathrm{R}_2} = i $$
So we only have two other currents that need to be written in terms of the other defined variables in the diagram.
The voltage drop across $R_1$ is $e - v_c$ (the battery minus the capacitor voltage). So the current through that resistor is
$$ I_{\mathrm{R}_1} = \frac{e - v_c}{R_1} $$
The charge on a capacitor is related to the voltage across it by $Q = C V$. So the current through the capacitor is:
$$I_{\mathrm{cap}} = \frac{dQ}{dt} = C \frac{d}{dt} v_c $$
Putting that all together gives:
$$\frac{e - v_c}{R_1} = c\frac{dv_c}{dt} + f(v_c) + i$$
| {
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How do microwaves heat moisture-free items? Today I learnt that microwaves heat food by blasting electromagnetic waves through the water molecules found in the food.
Does that mean food with 0% moisture (if such a thing exists - dried spices?) will never receive heat from a microwave oven? And how in that case is a microwave able to melt plastics etc., which contain no obvious water?
| This Isn't About Water
Microwave heating actually has nothing to do with the moisture content of items. It has everything to do with the amount of electric dipoles (polar molecules) in the item of concern. Water molecules (with many other organic molecules) happen to be electric dipoles. (That is, one side of the molecule has a positive charge and the other side has a negative charge.)
When the oven uses a microwave to make an electric field, all electric dipoles move to align with that field. If the direction of that field quickly flips, you give these dipoles kinetic energy. As you increase a group of molecules' kinetic energy, you increase the temperature of that group.
Any material containing significant amounts of electric dipoles will heat in a microwave. Also, resonance of water molecules have nothing to do with heating food in microwave ovens. The oscillations of the waves in microwave ovens is too slow for resonance to play a role.
For further reading, check out wikipedia's articles on Microwave Ovens and Dielectric Heating, which should answer your questions more throughly.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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Width of the decay of Higgs boson into dimuon According to Standard model, the partial width of the decay of Higgs into dimuon (up to tree level) is:
$$\Gamma\approx\frac{m_H}{8\pi} \left(\frac{m_{\mu}}{\nu}\right)^2$$
with the Higgs mass $m_H=125 GeV$, muon mass $m_{\mu}=0.106 GeV$, and the vacuum expectation value of the Higgs field $\nu=246 GeV$, apparently the decay width is extremely small. Then why is the width of the resonance peak in the plot from ATLAS so wide? If it's due to experimental errors then is there any meaning in comparing it with the theoretical result? I'm having trouble understanding this. Could somebody please explain it for me?
| The decay width of a particle is antiproportional to its lifetime. Looking at the partial width of the $H \rightarrow \mu \mu$ decay, one could expect that the lifetime of the Higgs is large. This would be correct, if the Higgs could only decay to muons. In other words: The Higgs decaying to muons has a low probability (a low branching ratio). This comes from the fact that the muon masses are relatively small, therefore the coupling of the Higgs to the muons is small too.
We can also look at this from the other way: The Higgs is heavy and there are no conservation rules that would prevent its decay, hence its lifetime is very short. A short lifetime means a large decay width. The total decay width is given as the sum of the partial decay widths:
$$\Gamma_H^{tot} = \Gamma_H^{b\bar{b}} + \Gamma_H^{WW} + \dots + \Gamma_H^{\mu \mu} + \dots$$
Since $\Gamma_H^{tot}$ is large and $\Gamma_H^{\mu\mu}$ is small, we can conclude (again) that the branching ratio to muons is small.
In the nice plots from ATLAS that you show, the Higgs can decay to anything, hence we observe the large decay width, even when looking only in the muon spectrum.
Maybe this can be better understood by using Heisenbergs uncertainty principle. Since the lifetime of the Higgs is short, it's energy (mass) is not well known, it is smeared around its true mass. Thus when it decays, the invariant mass of the decay products is smeared as well, no matter what particles it decays to.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/150232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How do you define a "universe" (in the context of multiverse)? How do you define a universe (in the context of multiverse)?
The traditional definition of universe is something like
"The Universe is all of spacetime and everything that exists therein".
But in multiverse theory one talks about "universes" and this raises the question of how one can in principle distinguish between two different universes.
A similar question is asked here and the top answer says "In addition, as can be shown by some holographic arguments, it is not really meaningful to talk about things that could exist outside our own cosmic horizon (or universe)." But here is an article in the Quanta Magazine which says
"If the universe that we inhabit had long ago collided with another universe, the crash would have left an imprint on the cosmic microwave background (CMB)"
So it seems that at least in some multiverse theories, different universes can send and receive physical information (light or gravitational force) between each other. So the question is
If light or gravitational force can travel between two universes then how would you know both are not part of the same universe (other than by a priori assumption)?
| As far as I know, "a" Universe is caracterised by fundemental constants such as the speed of light, Newton's gravitationnal constant, the Planck constant and so on.
You could distinguish between two different universe from the variation of these values I think.
| {
"language": "en",
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How does An Electric Field Create a Dipole Moment of a Rydberg Atom? I know that an Rydberg Atom will not usually have a Dipole Moment - as the positive nucleus are surrounded by a negative electron cloud, so there is no uneven charge distribution.
However, I also know that a Rydberg Atom with experience an Electric Dipole Moment when in an Electric Field. So my question is, How does An Electric Field Create a Dipole Moment of a Rydberg Atom?
| The Rydberg electron - the electron in the high n level - is highly polarizable and very weakly held. The binding energy of the electron is very small.
A small electric field will distort the wavefunction of the Rydberg electron so that it spends more time on one side of the atom than the other. The result is the formation of dipole.
Rydberg electrons in high levels are very sensitive to electric fields and easily ionized in 'field ionization' where electric fields as small as 5V /cm may pull the electrons off the atoms. This is the basis of the experimental ZEKE spectroscopy technique
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/150514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the difference between electromagnet and solenoid? What is the difference between electromagnet and solenoid? Both these terms seem as the same thing to me. The only difference that I can find seems to be that an electromagnet contains a soft iron core. I'm sure there must be some other difference between the two and I hope someone can clear this matter up for me.
| Electromagnets can be of various types ranging from simple solenoids to superconducting electromagnets, bitter electromagnets to explosively pumped flux compression generator. Thus, electromagnet can be considered as a universal set and solenoid is one of its subsets.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Coulomb's Law in the presence of a strong gravitational field I was under the impression that the $1/r^2$ falloff of various forces were because of the way the area of a expanding sphere scales. But that strict $1/r^2$ falloff would only be globally true in a strictly Euclidean geometry, yes? So, if you had an electron and a positron in a warped space due to gravity, wouldn't that $1/r^2$ term in Coulomb's Law end up needing to be adjusted because of the warping of space leading to an increase or decrease in effect due to gravitational lensing or scattering?
I found one related question here: Does Gravity Warp Coulomb's force?
But it only has one down-voted and unclear answer.
| Yes. Strictly speaking you can't apply Coulomb's law, or in general any law about the falloff of something with distance, in curved space.
Instead you have to shift to a field-based formalism. You can calculate the way the electromagnetic field propagates through a curved background—basically you take Maxwell's equations in tensor form and replace ordinary derivatives with covariant derivatives wrt the spacetime metric (for gory details, see Wikipedia). That may result in the effective falloff being weaker or stronger than $1/r^2$ depending on specific conditions, as the EM field lines will be distorted by the curvature.
| {
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What limits the doping concentration in a semiconductor? Si and Ge can be blended in any ratio, $\mathrm{Si}_x\mathrm{Ge}_{1-x},\ 0\le x\le 1$. So do
InxGa1-x.
So what exactly causes doping impurities inside Si/Ge/etc. to saturate at $\sim 10^{-19}\ \mathrm{cm^{-3}}$?
| If you dope the semiconductor too much it becomes what is called degenerate.
At normal levels of doping, the dopant atoms generate localized states in the material that can donate electrons or holes by either thermal excitation or optical excitation (e.g., a photon hitting a solar panel). The more you increase the doping the more likely it becomes that individual defect atoms will be closer together and merge into a midband (impurity state). At that point it stops behaving like a semiconductor because you've destroyed the necessary energy gap.
This basically means it becomes a bad metal, which by definition means that the conductivity of the material increases with temperature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/150747",
"timestamp": "2023-03-29T00:00:00",
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Why is Graphene Transparent? Graphene is always in the news now a days and its key features are that it is; very strong, conductive and transparent. It is so transparent that each layer of graphene will only absorb 2% of Light passing through it.
But what is it about the structure of Graphene which makes it (almost) transparent?
| I assume the biggest factor is the thickness. Graphene is a layer of carbon one atom thick. Light is absorbed/reflected by the top layers of a material and if you make any material into a layer one atom thick you'll find it increases transparency a lot.
The thing that is special about graphene is that it forms bonds in a 2D layer where most materials don't.
| {
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What is the procedure (matrix) for change of basis to go from Cartesian to polar coordinates and vice versa? I'm following along with these notes, and at a certain point it talks about change of basis to go from polar to Cartesian coordinates and vice versa. It gives the following relations:
$$\begin{pmatrix}
A_r \\
A_\theta
\end{pmatrix}
=
\begin{pmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
A_x \\
A_y
\end{pmatrix}$$
and
$$\begin{pmatrix}
A_x \\
A_y
\end{pmatrix}
=
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
A_r \\
A_\theta
\end{pmatrix}$$
I was struggling to figure out how these were arrived at, and then I noticed what is possibly a mistake. In (1), shouldn't it read
$$A_r=A_x+A_y$$
Is this a mistake, or am I making a wrong assumption somewhere?
I'm kinda stuck here, and would appreciate some inputs on this. Thanks.
| This properly belongs on math.se, but to properly derive these you need to remember that we can write a vector in terms of basis vectors. The vector $\vec{A}$ is unchanged, but it is just expressed as a different linear combination:
$$\vec{A} = A_x \hat {x} + A_y \hat{y} = A_r \hat{r} + A_\theta\hat{\theta} $$.
Because you can write $\hat{r}$ as a linear combination of $\hat x$ and $\hat y$, i.e., $\hat {r} = \frac{x}{r} \hat {x} + \frac{y}{r} \hat y = \cos\theta \hat{x} + \sin\theta \hat{y}$, and similarly for $\hat{\theta} = -\sin\theta \hat{x} + \cos\theta \hat y$, you can solve for the $A_r$ and $A_\theta$ in terms of $A_x$ and $A_y$.
| {
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How introduce sound wave in molecular dynamics or dissipative dynamic particles? How could I introduce sound wave in molecular dynamics or dissipative dynamic particles?
What do specify which is applicable between molecular dynamics and dissipative particle dynamics? problem is simulating a 50 nm bubble in a MHz sound waves.
| DPD doesn't have attractive forces so I don't see how it's possible to simulate a bubble with it. And a 50 nm bubble with MD? I hope you have a lot of computing power!
Assuming you use MD: I think the most elegant solution would be to use NVT with periodic boundary conditions, and modify the MD code you're using so that the cell vectors in one of the directions are sinusoidally varied at the desired frequency.
| {
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Does constant velocity of center of mass imply linear momentum is conserved? I know that if momentum is conserved for a system, you can derive that the velocity of the center of mass of that system is constant. I was wondering if the second condition also implies the first: if I can demonstrate that the velocity of the center of mass of a system is constant, does that imply that linear momentum is conserved in the system?
| The linear momentum of a system is given by $\vec{p} = m \vec{v}$.
If you differentiate this with respect to time in an inertial frame, you have:
$$
{d\vec{p} \over dt} = m {d\vec{v} \over dt} + {dm\over dt} \vec{v}
$$
If $\vec{v}$ is constant with time, this becomes
$$
{d\vec{p} \over dt} = {dm\over dt} \vec{v}
$$
Which means, for $d\vec{p}/dt$ to be zero (or $\vec{p}$ to be constant), you must have at least one of $\vec{v}$ or $dm/dt$ equal to zero.
Which means, either your object doesn't move at all, or, if it is moving, doesn't lose or gain any mass.
| {
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Physical significance of temperature Some books say when heat flows into a monatomic gas at constant volume, all of the added
energy goes into an increase in random translational molecular kinetic energy. But when the temperature is increased by the same amount in a diatomic or polyatomic gas, additional heat is needed to supply the increased rotational and vibrational energies. Thus polyatomic gases have larger molar heat capacities
than monatomic gases.
Does the absolute temperature reflect translation kinetic energy of gases only? If all types of kinetic energy of gas particles are related to temperature, why polyatomic gases have larger molar heat capacities than monatomic gases?
| Absolute temperature relates only to translational degrees of freedom (connection to pressure via momentum exchange with a supposed exterior membrane).
Since energy is constantly being randomly reshuffled between translational and non-translational degrees of freedom, the molar heat capacity is greater.
| {
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What will happen to a permanent magnet if we keep the same magnetic poles of two magnets close together for a long time? What will happen to permanent magnet's magnetic field or magnetic ability if we keep same magnetic poles of two permanent magnet for long time?
Will any magnetic loss happen over the long period of exposure or does the magnetic strength remain the same?
Sorry if my logic is wrong. Please explain this.
| They will repel each other and slowly they will lose their magnetic property which is also called demagnetization
| {
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Why does room temperature water and metal feel almost as cool as each other? From what I've read about heat, temperature and conductivity, I understand that the reason water at room temperature feels colder than most other things at the same temperature (like wood, air, cotton) is because of its higher thermal conductivity. That is, it transfers heat quickly from my body to itself, as well as within itself.
(Assuming the thermal conductivity is the only reason why different materials feel colder or warmer), what I don't understand is why metals feel about as cold as water, while their thermal conductivities are 100-to-200 times higher than that of water (Water's is ~0.58 W/mK, the values for metals range from 50 to 400).
I suppose there is more to why materials at identical temperatures suck heat faster; what is it?
| Several things are happening here that may make the sensations of touching metal and touching water similar when they are at room temperature (~ 25 C), although the thermal conductivities are a couple of orders of magnitude different. The sensation of coldness comes from the loss of heat from the part of your body contacting the material. The rate of heat transfer is proportional to both the difference in temperature between materials and the surface area of the transfer interface. Since the delta-T is about the same, the surface area is the significant variable. As Nathan Reed stated, the conforming nature of a liquid significantly (a couple of orders of magnitude?) increases the surface area of the heat transfer interface. This may offset the difference in thermal conductivities.
| {
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Formula for the magnetic field due to a current loop I need expressions for the $\mathbf B$ field generated by a circular current loop at a point off-axis from the ring's axis of symmetry.
The ones I came across on the internet aren't very convincing. I verified them with Mathematica, and none seems to be correct ─ I'm checking whether $\nabla \times \mathbf B = I \hat{\mathbf e}_\theta$ and $\nabla \cdot\mathbf B =0$, but the examples here don't satisfy those (so e.g. the latter will have $\nabla \times \mathbf B=0$).
So, more generally: given a ring of current, what is the magnetic field it generates at an arbitrary point? Can this be calculated exactly?
| As mentioned in the accepted answer, the second link in the original question (now hijacked) does have the correct exact expression for the magnetic field, which is essentially given by direct line integration of the vector potential into
\begin{align}
\mathbf A
&=
\frac{\mu_0 I}{4\pi}
\oint_C \frac{\mathrm d\mathbf l}{r}
\\ & =
\frac{\mu_0 I}{2\pi}
\frac{\sqrt{z^2+(R+\rho)^2}}{2R\rho}
\Bigg[
\left(1-\frac{2R\rho}{z^2+(R+\rho)^2}\right)
K\mathopen{}\left(\sqrt{\frac{4R\rho}{z^2+(R+\rho)^2}}\right)\mathclose{}
\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
-
E\mathopen{}\left(\sqrt{\frac{4R\rho}{z^2+(R+\rho)^2}}\right)\mathclose{}
\Bigg]
\hat{\mathbf e}_\phi
,
\end{align}
where $K(k)$ and $E(k)$ are complete elliptic integrals of the first and second kind, evaluated at the position-dependent argument
$$
k=\sqrt{\frac{4R\rho}{z^2+(R+\rho)^2}},
$$
and where obviously $\rho^2=x^2+y^2$ and the current ring has radius $R$ in the $x,y$ plane. The magnetic field $\mathbf B$ can then be obtained by taking the curl of this vector potential, as usual. (Don't trust the exact detailed constants on my expressions, by the way ─ double-check everything if you're going to use it.)
The catch here, as mentioned in the accepted answer, is that the elliptic-integral functions have singularities at the argument $k=1$, which is reached when $z=0$ and $\rho=R$ (i.e. at the current ring), and those singularities need to be handled carefully when evaluating the derivatives in $\mathbf B=\nabla \times\mathbf A$: they almost certainly can be done rigorously, but to do so, you need to treat the derivatives of the singular functions as derivatives in the distributional sense, at which point they will give Dirac-delta contributions to the derivative that exactly match what's required for Ampère's law to hold with a Dirac-delta current density.
| {
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QFT and violation of Heisenberg uncertainty principle In some QFT books is said that a free electron can emit a virtual photon as long as it reabsorbs the photon and returns to its original state within a time:
$$\Delta t<\dfrac{\hbar}{2\Delta E}$$
That inequality DOES VIOLATE the Heisenberg Uncertainty Principle. Why is that POSSIBLE? If it were said in a time
$$\Delta t\geq \dfrac{\hbar}{2\Delta E}$$
I would not be so puzzled.
| I have been thinking about this question. Tell me what you think:
1) The relation
$$\Delta \tau \Delta \Gamma \geq \dfrac{\hbar}{2}$$
is RIGHT, applied to the timelife AND width of RESONANT states in QFT. A completely stable particule would have "zero width" and infinite timelife.
2) The equation
$$\Delta t\Delta E<\dfrac{\hbar}{2}$$
is ALSO right, but applied not to REAL particles, but to VIRTUAL particles. Taking into account that the definition of virtual particle is one that is off-shell mass according to relativity, the opposite inequality in the Heisenberg Uncertainty Principle becomes possible. The propagator in QFT can not be ZERO, so the HUP vindicates the existence of particles popping out from vacuum in any moment of time and VIOLATING the relativistic dispersion relationship
$$E^2-p^2c^2=m^2c^4$$
Suppose that the transferred momentum is VERYLARGE, so we can neglect the rest mass term in the last equation. Thus we have TWO solutions, E=qc and E=-qc, where q is the transferred momentum. In particular, the transferred momentum can be negative...So the inverted inequality is possible. Let me know if I am making some terrible mistake...
| {
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Physical examples where changing the order of limits yields wrong result In mathematics it is generally not allowed to change order of limits. For example it is not always true for a sequence of functions $f_n$, that $\int_a^b \left(\sum_{n=0}^\infty f_n(x) \right) dx = \sum_{n=0}^\infty \left(\int_a^b f_n(x) dx\right)$. (Note that series $\sum_{n=0}^\infty\ldots$ and the integral $\int_a^b \ldots dx$ are mathematically defined via limits of sequences).
In my experience it happens a lot in physics lectures, that limits are changed in their order without any additional comment (such as mentioning Fubini's theorem or uniform convergence). It also seems to me that there are not many examples relevant for physics where changing the order of limits yield wrong results.
I'm looking for good physical examples showing to students that one has to be careful when he changes the order of limits. So for which physical example the order of the limits is important and you get a wrong result, when you change it?
| The low-frequency ($\omega\rightarrow 0$), long-wavelength ($q\rightarrow 0$) conductivity of an electron gas in the random phase approximation depends on the order in which those two limits are taken.
Intuitively, if you take the $\omega\rightarrow 0$ limit first, you're talking about a static, long-wavelength potential to which the electrons adjust, so the conductivity is imaginary (non-dissapative). If you take the $q\rightarrow 0$ limit first, you're talking about a uniform, slowly varying field applied to the gas, so you get a current (real conductivity). See page 27 of these notes for a discussion.
Basically, you order the limits one way to learn about Thomas-Fermi screening, and the other way to learn about DC currents. Changing the limits here is not "wrong" per se, but it may get you an answer to a question you didn't ask!
| {
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Half-integer eigenvalues of orbital angular momentum Why do we exclude half-integer values of the orbital angular momentum?
It's clear for me that an angular momentum operator can only have integer values or half-integer values. However, it's not clear why the orbital angular momentum only has integer eigenvalues.
Of course, when we do the experiments we confirm that a scalar wavefunction and integer spherical harmonics are enough to describe everything. Some books, however, try to explain the exclusion of half integer values theoretically. Griffiths evokes the "single valuedness" argument, but he writes that the argument is not so good in a footnote. Shankar says that the $L_z$ operator only is Hermitian when the magnetic quantum number is an integer, but his argument isn't so compelling to me. Gasiorowicz argues that the ladder operators don't work properly with half-integer values. There are some low impact papers (most of them are old) that discuss these subjects, although they are a little bit confusing.
So, basically, my question is: Does anyone have a decisive argument on why do we exclude the half-integer values from the orbital operator spectrum?
| From $\mathbf{L}=\mathbf{Q}\times \mathbf{P}$ we have $L_z=Q_xP_y-Q_yP_x$. Then, introduce the following new operators (assuming units of $\hbar=1$):
\begin{align}
q_1=\frac{Q_x+P_y}{\sqrt{2}},\\
q_2=\frac{Q_x-P_y}{\sqrt{2}},\\
p_1=\frac{P_x-Q_y}{\sqrt{2}},\\
p_2=\frac{P_x+Q_y}{\sqrt{2}}.
\end{align}
It is immediate to check that
$$
[q_1,q_2]=[p_1,p_2]=0,\quad [q_j,p_k]=i\delta_{j,k},
$$
so $q_i$ and $p_i$ are formally position and momentum operators of two different systems. It terms of these new operators
$$
L_z=\frac12(p_1^2+q_1^2)-\frac12(p_2^2+q_2^2).
$$
Therefore $L_z$ is nothing but the difference of two independent harmonic oscillator Hamiltonians, each having mass $M=1$ and angular frequency $\omega=1$, $L_z=H_1-H_2$, $H_i=\frac12(p_i^2+q_i^2)$. The spectrum of the harmonic oscillator Hamiltonians $H_1$ and $H_2$ is $(n_1+1/2)$ and $(n_2+1/2)$, respectively, with $n_1$ and $n_2$ positive integers. Finally, since $[H_1,H_2]=0$, the spectrum of $L_z$ is
$$
(n_1+1/2)-(n_2+1/2)=n_1-n_2.
$$
This is the difference of two integer numbers, so it is an integer.
| {
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Conformal Killing fields on Schwarzschild I am trying to understand which are the conformal Killing Fields on the Schwarzschild spacetime.
I say that $X$ is a conformal Killing field on $S$ ($S$ is Schwarzschild) if there exists a function $f: S \to \mathbb{R}$ such that
\begin{equation}
\mathcal{L}_X g = fg,
\end{equation}
where $g$ is the Schwarzschild metric, and $\mathcal{L}$ is the Lie derivative.
I know that the time translation, $\partial_t$, and the rotations, $\Omega_{ij}$ are Killing fields, therefore conformal Killing fields, with $f$ constant and equal to $0$.
In the Minkowski spacetime, for example, parametrized by $(x^0, x^1, x^2, x^3)$, I know that the ``dilation'' field, i.e. the field
$$
\sum_{i=0}^3 x^\lambda \partial_{x^\lambda}
$$
is a conformal Killing field, with $f=2$. I would like to understand if there is an analogous in Schwarzschild.
| In the Schwarzschild geometry, the Schwarzschild radius breaks naive dilation symmetry. In the simple case of a radial dilation $r \to \lambda r$, the geometry is only preserved by $R_S \to \lambda R_S$. So, it naively seems like it would be difficult to find a working dilation, even just a radial dilation.
I went to some effort (as an exercise for myself) to find a working dilation, and failed. What I have found is that the vector field
$$
X = t \partial_t + r \sqrt{1-\frac{R_S}{r}} \, \partial_r
$$
which approaches $0$ as $r \to R_S$ and $r \partial_r$ as $r \to \infty$ is almost a conformal Killing field. However, the lie derivative of the metric is
$$
\mathcal{L}_X g = 2 \sqrt{1-\frac{R_S}{r}} \left( \begin{array}{cccc}
-\sqrt{1-\frac{R_S}{r}}- \frac{R_S}{2r} & & & \\
& \left( 1 - \frac{R_S}{r} \right)^{-1} & & \\
& & r^2 & \\
& & & r^2 \sin^2 \theta
\end{array} \right)
$$
So, the tt component of the metric spoils everything. I spend a small amount of effort trying to modify this vector field (adding it a timelike component, adding in explicit time dependence, etc.), but so far nothing has worked.
| {
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Electric field inside a diode When a voltage is applied to a diode (forward or reversed bias) the depletion zone is changed due to charges change in this region. My question is in both case (forward or reversed bias), how the electric field that is responsible of moving the charges in the P and N region is established ? Is it the same mechanism of electric field establishment inside a conductor i.e surface charges density making the electric field?
| I don't think any of the current answers address your question as I understand it.
My understanding is that your question is effectively: What spatial distribution of charge gives rise to the electric field in a diode?
In the case of a uniform dielectric, the internal field is created by surface charges on the two surfaces, such are usually very small in width. In a pn junction this is not the case.
https://ecee.colorado.edu/~bart/book/book/chapter4/gif/fig4_3_1.gif
Figure a of the above image shows the charge density profile of a pn junction. Notice that you have both nonzero charge density in a sizeable area, the depletion region. In the case of two dielectrics, instead of a pn junction, you would get charge accumulating only along the surface of the interface. This non uniform charge density also means that the electric field is not simply a constant in the material, but rather varies along the depletion region, something that ia very different than for a simple resistor.
This charge density comes the bare donors/acceptors that cannot move, and are neutralized at the ends of the depletion region by the electrons and holes which accumulate in very small regions due to being free to move.
| {
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For creating beats how small the difference should be between the two frequencies It is said that to create beats we need two "slightly" different frequencies, and subtract it.
1- My question is why do we need slightly different frequencies? Why not large difference?
2- Also how slightly different? what is the limit on maximum difference?
| The beat frequency is very simply:
$$f_{beat}=|f_1-f_2|$$
So there is no limit on how far apart they can be. In demonstrating the beat frequency one frequently uses frequencies that are slightly apart because it produces the typical "beating."
If for example you were using the frequencies $561.6$ Hz and $300$ Hz you would get a resulting frequency of $261.6$ Hz which is in fact middle C.
In this described case you wouldn't actually notice a beating, but rather you would just hear the pure note C. It wouldn't sound like two frequencies at all, it would just sound like a single crisp note.
This is why in a lab demonstration you use frequencies that are 10 - 15 Hz apart. It produces a beat noticeable to the human ear. Using large differences produces "musical" (for lack of a better word) notes.
| {
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Temperature of electroweak phase transition How does one estimate the temperature at which electroweak phase transition (EWPT) occurred? Somewhere I have read it is around 100GeV but the reason was not explained.
| higgs field causing electroweak transition. before higgs field w,z boson photon all are massless. but higgs field gives masses to them and symmetry spontaneously break. because if particles having masses then su(2)u(1) symmetry break. all gauge symmetry break if we give mass to the particles.so transition must have to have below 125.6 gev that is the mass of higgs boson.
peoples are thinking about there may be another background field which causes strong and electroweak seperation.so that field can break the underlying symmetry.
In the standard model, the Higgs field is an SU(2) doublet, a complex scalar with four real components (or equivalently with two complex components). Its (weak hypercharge) U(1) charge is 1. That means that it transforms as a spinor under SU(2). Under U(1) rotations, it is multiplied by a phase, which thus mixes the real and imaginary parts of the complex spinor into each other—so this is not the same as two complex spinors mixing under U(1) (which would have eight real components between them), but instead is the spinor representation of the group U(2).
The Higgs field, through the interactions specified (summarized, represented, or even simulated) by its potential, induces spontaneous breaking of three out of the four generators ("directions") of the gauge group SU(2) × U(1): three out of its four components would ordinarily amount to Goldstone bosons, if they were not coupled to gauge fields.
However, after symmetry breaking, these three of the four degrees of freedom in the Higgs field mix with the three W and Z bosons (W+, W− and Z), and are only observable as spin components of these weak bosons, which are now massive; while the one remaining degree of freedom becomes the Higgs boson—a new scalar particle.
The photon as the part that remains massless
| {
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Why do floating water drops form spheres? Consider a drop of water floating in an inertial frame in STP air (e.g., the ISS). Intuitively, the equilibrium shape of the drop is a sphere.
How would one prove that? Is it equivalent to showing that the minimal surface area for a simply connected volume in $\mathbb{R}^3$ with a sufficiently smooth boundary is that of a sphere, i.e., the result of the isoperimetric inequality?
| A Community Wiki answer to make some other people's comments permanent and tie some loose ends up.
To add to Mark Mitchison's Answer, the reason that the prevailing shape is the one that minimises surface energy as he states is that, in the case of water, the liquid's total energy is an (almost) constant offset (the potential and kinetic energy of the molecules within the body of the liquid) plus the surface energy, so that minimising the latter is almost equivalent to minimising the former.
Given the experimental fact that most liquids are nearly incompressible, the energy change wrought by the internal pressure field that changes with the body's shape is utterly negligible compared with the changes in the energy associated with the surface tension, so that the latter sets the shape.
As QMechanic's excellent link (the "Isoperimetric Inequality" Wikipedia page) points out the (hyper)sphere is the shape that minimises the surface area of a given enclosed body.
| {
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How do we find the accuracy of atomic clocks? We say that atomic clocks are the most accurate clocks ever made, they may lose or gain $x$ seconds in $y$ years. How do we find this uncertainty because we do not have an ideal clock to compare with the clock that we made?
| The nearest we have to a standard is International Atomic Time. This is:
TAI as a time scale is a weighted average of the time kept by over 200 atomic clocks in over 50 national laboratories worldwide.
The errors in individual clocks can be assessed by comparing them to the weighted average.
Re the mention of pulsars in your comment, pulsars slow down over time and indeed atomic clocks are used to track their slowing. Also pulsars occasionally glitch. So atomic clocks are more accurate than pulsars are.
| {
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Force of a Train Imagine that there are two trains and the first train is twice as long as the second train. They have the same mass per unit length and they are traveling at exactly the same speed.
If the first train hit me, would it hit me with twice as much force as the second train? These are two distinct situations: 1) I am hit by the first train only, 2) I am hit by the second train only.
Force is mass times acceleration, so if the one train has twice the mass, then it seems likely that it would have twice the force. But I am not sure.
| The difference in force to stop the trains you are talking about here is the difference in force is needed to bring the train to a stop within a particular distance Let me tell you what I mean.
When you try to stop the train, you'll obviously be dragged in front of the train. Say the dragging causes an uniform force (due to the friction from the ground) $f_1$ to act on the heavier train, and $f_2$ to act on the other. $f_1$ and $f_2$ act such that both trains come to rest within the same distance $x$. You say that both trains travel with the same velocity $v$, but have different masses. So, let the heavier train have mass $M$ and the lighter one have mass $m$.
The trains thus have kinetic energies $\frac 12 M v^2$ and $\frac 12 m v^2$.
In stopping the trains, the work you'll have to do to is $W_1=f_1x=\frac 12 Mv^2$ for the heavier train and $W_2=f_2x=\frac 12 mv^2$ for the other one. Once we divide the two, $${f_1x \over f_2x}={\frac 12 Mv^2 \over \frac 12 mv^2}\implies \frac {f_1}{f_2}=\frac {M}{m}$$
Thus, $$f_1 > f_2$$
By Newton's third law, the force you exert on the train is equal what the train exerts on you. Thus $$\boxed {force \ exerted \ by \ the \ heavier \ train \ on \ you \ > \ force \ exerted \ by \ the \ lighter \ train \ on \ you}$$
| {
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Does lunar module need the same amount of fuel for landing and take off? Let's assume there is no atmosphere and let's assume there is no change in weight due to fuel consumption, will reactive rocket need the same amount of fuel for landing on a planet as for take off?
In theory - I think - you need the same escape velocity to get rocket to orbit as you need to break it to 0 speed after free fall from the orbit, but this changes if the descend is slower than free fall, is that right?
What is the real world(moon) situation in case of lunar module? (extrapolating for fact that the Apollo Lunar Module leaves the descend stage behind)
| Now it occurred to me that the escape velocity/breaking may be symmetrical - that means you need the same amount of energy to counter the gravity on the way up as you need on the way down - but what also matters is how long you stay "hoovering" in the gravitation filed. This is what consumes fuel no matter what way you go and in reality both descend and ascend are slower than free fall(/escape speed). So your total fuel consumption will consist of fuel for generating the escape/breaking velocity plus fuel to compensate for the fact that you are not breaking or accelerating instantly.
I think that in the case of a rocket we can assume that the take off is much slower than the descent, which means you need more fuel for the take off.
Am I right?
| {
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How would gravitons couple to the Stress-Energy tensor? How would gravitons couple to the Stress-Energy tensor $T^{\mu\nu}$? How did physicists arrive at this result? I've read that it follows from the analysis of irreducible representations of the 4-dimensional Poincaré group, but is this accurate?
| It is not known yet.
Gravitons are from quantum mechanics model, while stress-energy tensor is from General relativity (GR) model.
Two models are not connected until quantum gravity created.
Also, gravitons were never observed, so they are pretty hypothetical.
Simultaneously, it is known, that metric tensor is "generated" by stress-energy tensor. Metric tensor is from GR model. Also GR model contains gravitation waves.
Gravitation waves were never observed too. (Gravitation waves were never observed directly, i.e. so that they affect matter on Earth, although they were confirmed indirectly, by predicting of energy loss in rotating heavy (neutron) star systems).
If gravitons exist, they should be a quantum representation of gravitation waves. And, it is known from it, that gravitons should have spin of 2.
This is the sequence: GR -> gravitation waves -> spin of 2.
Two last parts are hypothetical.
| {
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How to reconstruct the dependence of the potential from a coordinate? An ion moves along the x-axis of a black box with a speed $V$ and returns in a time
$$T=a V^b$$
where $a$ and $b$ are some known constants. Having this, can we reproduce the dependence of a field potential $U(x)$ of this box?
So far I have managed to do this:
Adding to an initial velocity some $dV$ we get the increase in a time $dt$ and so $$T+dt=a (V+dV)^b \, .$$
We can derive $dt$ by subtracting the initial $T$ value
$$dt=a(V+dV)^b-aV^b \, .$$
Using the equivalence formula for $dV$ approaching zero we find
$$dt=aV^b(1+b\frac{dV}{V})-aV^b=abV^{b-1}dV \, .$$
We can find the acceleration
$$\frac{dV}{dt}=\frac{V^{1-b}}{ab} \, .$$
Similarly, as $V=(T/a)^{\frac{1}{b}}$ we have
\begin{align}
dV
&= \left(\frac{T+dt}{a} \right)^\frac{1}{b}
- \left(\frac{T}{a} \right)^\frac{1}{b} \\
&= \left(\frac{T}{a} \right)^\frac{1}{b} \left(1+\frac{dt}{bT} \right) - \left( \frac{T}{a} \right)^\frac{1}{b} \\
&=\frac{T^\frac{1-b}{b}dt}{ab} \, .
\end{align}
This yields again
$$\frac{dV}{dt}=\frac{T^\frac{1-b}{b}}{ab} \, .$$
Now I assume it is our acceleration which the field imparts to the particle, thus
$$\frac{dV}{dt}=\frac{f}{m}=\frac{dU}{dx m} \, .$$
From there I am not sure whether or not I can integrate the equation
$$dU=m\frac{T^{\frac{1-b}{b}}}{ab} \, dx \, .$$
So, are the time $T$ there is a constant in relation to $dx$ or not? The answer I get from the last equation is
$$U(x)=U(0)+m\frac{T^{\frac{1-b}{b}}}{ab}x \, .$$
I am confused by two things here:
*
*The difference between the answers derived from the second and the first approach:
$$\frac{dV}{dt} = \frac{V^{1-b}}{ab} \quad \text{and} \quad \frac{dV}{dt} = \frac{T^\frac{1-b}{b}}{ab}$$
*The possibility of integrating that way.
Are there other ways to get the definite answer for this task?
| As it follows from the Mechanics by Landau-Lifshitz (which can be found here — page 28), the desired dependence can be found in the form of
$$x(U)=\frac{1}{2\pi \sqrt{2m}}\int_{0}^{U} \frac{T(E)dE}{\sqrt{U-E}}$$
Where $T(E)$ from the law of conservation of energy and the initial statement $T_{½}=\alpha V^\beta$
$$T(E)=2\alpha (\frac{2E}{m})^\frac{\beta}{2}$$
The period $T$ given there is twice as big as a time of return $T_{½}$ due to the fact, that ion has done only a half of its full way by the time it returns to initial point.
From now I am not familiar with methods of integrating such equations, so I have used Mathematica database for this purpose (please, let me know if the answer is wrong, for I cannot make sure that the answer is correct):
$$U(x)=\frac{1}{2}m \left ( \frac{2x\pi}{\alpha}\right ) ^{\frac{2}{\beta+1}}\frac{\Gamma(\frac{\beta+3}{2})}{\Gamma(\frac{\beta+3}{2})}$$
$\Gamma$ here is a gamma function. Note that the answer $U(x)$ is dimensionally correct.
| {
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Electromagnetic radiation bending on Earth Most articles say that a radiowave is able to propagate itself beyond the horizon because it is reflected off by the ionosphere (and the Earth itself).
But do radio waves also get bent according to the Earth's curvature due to gravity thus transmitting beyond the horizon without need for the bounce effect?
| Calculating the path that a light ray takes in a gravitational field is a complicated business, but for the special case of a light ray coming from infinity and escaping to infinity there is a convenient approximate formula for the angle, $\theta$, the light ray is deflected:
$$ \theta \approx \frac{4GM}{r_0 c^2} $$
Where $M$ is the mass of the deflecting object and $r_0$ is the distance of closest approach. If we feed in the mass of the Earth and set $r_0$ to the smallest value it can have (the radius of the Earth) we get a deflection of $\theta \approx 2.8 \times 10^{-9}$ radians, or about $0.0000002$ degrees.
So the gravitational deflection of any electromagnetic wave by the Earth is entirely negligable.
| {
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Exploring beyond event horizons In the expanding universe the velocity of separation between galaxies depends upon how far they are. If they are much far away will they have relative velocity of separation greater than speed of light and if so how can we even detect such galaxies.
There are things like Quantum information that can travel faster than light using entanglement, is there any any possibility to detect such unseen horizons using such effects?
|
If they are much far away will they have relative velocity of separation greater than speed of light and if so how can we even detect such galaxies.
We can't detect such galaxies. Redshift goes to infinity at the cosmic horizon, and we cannot see beyond. Note that the cosmic horizon is different from the Hubble sphere: At the former, relative velocities according to parallel transport along the light ray reach $c$, whereas recession velocities reach $c$ at the latter. As far as observable effects go, the Hubble sphere is largely irrelevant.
is there any any possibility to detect such unseen horizons using such effects?
No, entanglement cannot be used that way - it is completely useless without a classical channel of information to 'compare notes'. If one of your particles vanishes behind an event horizon, such a channel is unavailable (Note that from your point of view, the particle will actually never cross the horizon, but gets frozen in time and become unobservable due to redshift).
The fact that the particles are quantum correlated instead of classically doesn't really matter. Instead of using entangled particles and the cosmological horizon, you could take a red ball and a blue ball and put them into two boxes without looking. Then, throw one of the boxes into a black hole and open the remaining one. You'll instantaneously know which ball ended up in the black hole - but what will such an experiment tell you about its interior?
| {
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Does wave interference happen only to same frequency waves? As the title says, from books and results from internet, I find that examples of wave interference always have the same frequency, only different in phase constant.
So, I'd like to know if wave interference happens only to same frequency waves
| No; wave interference takes place whenever two waves of any frequency, same, nearly the same or widely different interact. An air molecule next to your ear, for example, can only respond to the sum of all the different sound waves reaching it at any moment.
The results are simpler when the two waves are closely related, or some simple multiple of each other. One common effect caused by the interference closely related frequencies is the phenomenon of beats.
This video, MIT Physics Demo Tuning Forks Resonance & Beat Frequency 720 at about the $1:30$ mark, shows the result of adding two sound waves of almost the same frequency. The rider on the tine of one of the two tuning forks detunes it slightly. The two resulting waves can interferes constructively, then destructively, and then back again.
I've done this demonstration with a group of students around the apparatus. By asking the students to individually raise their hand when they heard the loudest sound, it becomes clear that the moment when the two waves arrive in step for any student depends on their position around the apparatus.
In the general case, with no assumptions about the frequency, shape, phase or amplitude of the two waves, the Superposition Principle applies.
Consider the first of the waves. What would the medium be doing at that instant, if this were the only wave acting? Say we're talking water waves, and the answer is that the water would be 20 cm above its normal position. Do the same for all the other waves present, keeping track of each answer, and noting in each case whether the water is above or below its undisturbed position. Finally, add up all the individual displacements. The result is the position of the water at that instant. Move forward a short period of time and repeat. The sequence of displacements versus time give nature of the resulting wave.
| {
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The subtle differences between angular momentum and centrifugal force I am a mathematician wanting to understand the differences between the concepts of angular momentum and centrifugal force.
The following two ideas are clear to me from a physical point of view, but I have a difficult time discerning the difference between them as people tell me they are different but do not give me any explicit reason as to why.
*
*Angular momentum is a vector quantity (taken in the physical sense) of a mass's rotational velocity about some axis.
*Centrifugal force is defined on the axis of a rotational reference frame, which depends on the inertia of the object.
My question: What is the subtle difference between this notion of "rotational reference frame" and the notion of the vector quantity of a mass's rotational velocity?. Are they not physically the same point (vector quantity) of rotation about an axis, thus making the meaning of centrifugal force a relative way to speak about the meaning of angular momentum?
I hope this question isn't too naive. I have been really hoping to understand the subtle difference between these physical concepts in a straightforward way.
| Angular momentum is around, centrifugal force is out (from the axis). Subtle in the sense that they are related, but distinct?
| {
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What makes Sun's light travel as parallel beams towards earth?
Sun's light appear to travel as parallel beams towards earth $_1$. Sun produces electromagnetic radiations through pp chain and other reactions in Photosphere $_2$.
I don't see whether these reactions send photons in that neatly arranged parallel rays, or else any other effects make the rays have such parallel beautiful motion.
So what makes Sun's light travel as parallel beams towards earth?
Links: A very good news for those who wants to desperately know the answer. I have replaced the word "ray" with "beam", as both have difference in their meaning. I have extracted a small passage from the webpage Light beam - Wikipedia :
A light beam or beam of light is a directional projection of light energy radiating from a light source. Sunlight forms a light beam (a sunbeam) when filtered through media such as clouds, foliage, or windows.
If the passage is saying truth, we are only left with the question, how clouds (the question is concerned with this) form beams?
Reference: $_1$ Crepuscular rays-Wikipedia $_2$ Sun-Wikipedia
| Light from the sun actually converges on the earth - the sun is about 220 Earths across, so light from one edge and light from the other edge must converge to reach the same point on Earth. The angle is small enough (about 0.5 degrees) that for most practical purposes we can consider the sun to be either a point source or a uniform flat source as needed.
| {
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Is there a commonly used unit of measure (other than temperature units) that is not absolute? I live in a country where we use Degree-Celsius(°C) to measure the temperature.
Sometimes from one day to the other, the temperature rises from 10°C to 20°C and I hear people say, "Wow! Today is twice as hot as yesterday!".
I try to explain that today is not two times hotter than yesterday because Celsius(°C) is not an absolute unit of measure, that is 0° does not mean the absence of temperature.
If I convert Celsius(°C) to Fahrenheit(°F) or Kelvin(K) it gets clear, but I wish I could provide another example of non-absolute unit of measure to clarify things.
In short, do you know any other unit of measurement (except for temperature ones) where 0 does not mean the absence of the physical phenomenon that is being measured?
| Best one I can think of is shoe (or clothing or ring) sizes. Another common measure might be wire gauge.
There are many others, but almost all are less common and I doubt they'd be effective.
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Is Boltzmann distribution contradicting with the fundamental assumption of statistical thermodynamics? In equilibrium statistical physics the fundamental assumption of statistical thermodynamics states that the occupation of any microstate is equally probable (i.e. $p_i=1/\Omega, S=-k_B\sum p_i\,{\rm ln}\,p_i=k_B{\rm ln}\,\Omega$). But for isolated system in equilibrium we also have Boltzmann distribution which states $p_i=e^{-\beta E_i}/Z$, where $E_i$ are the allowed energy levels. So the two $p_i$ matches if and only if there is one single allowed energy level. How can we resolve this conflict?
| The equal probabilities are meant for states of an isolated system with constant total energy. Each state with this energy is then equally probable.
The Boltzmann probabilities are meant for systems in thermal contact and equilibrium with reservoir of definite temperature - in that case the energy of the system may change due to interaction with the reservoir, so it makes sense that probability of state is correlated with its energy.
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Cosmic Background Radiation and redshift vs. temperature I get that the CMB has gone from high energy photons, to low microwave photons today. And that is probably due to the redshift from the expanding Universe. But, since CMB is a black body it is temperature dependent, i.e. it has a high peak at high temperatures, but then gradually gets less "peaky" and goes towards longer wavelengths.
So, is it the decreasing temperature of the Universe that is increasing the wavelength of the CMB radiation, or is it the expansion, and therefore redshift, or are they somewhat linked ?
| Red-shifted black-body radiation is just like black-body radiation at a lower temperature, that is why you might think the two effects are mixed. The CMB comes from the time when the universe was about 379,000 years old, the time at which (as the universe expanded) adiabatic cooling caused the energy density of the plasma to decrease until it became favorable for electrons to combine with protons, forming hydrogen atoms. This recombination event happened when the temperature was around 3000 K. At this point, the photons no longer interacted with the now electrically neutral atoms and began to travel freely through space, resulting in the decoupling of matter and radiation. The the original peak of the CMB was at visible light with a temperature of about 3000K, and due to the redshift the peak is observed at at a lower wavelength of about 160.2 GHz, in the microwave range of frequencies. You can give this observed distribution a color temperature of 2.72 K (which will continue to drop due to redshift as the universe expands).
| {
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General question about the potential barrier problem: Why does $\exp( kx)$ diverge when $x>0$ in the case when $E < V(x)$? For the two images below, the first potential barrier has particles approaching it where $E > V_o$ & the second has a particle that has $E < V_o$, where $E$ is the energy of the particles and $V_o$ is the potential of the barrier:
For the first case, when $x < 0$, the particles have a part that is reflected from the barrier (marked by coefficient $B$) and a part that is transmitted (has a coefficient $A$). When $x > 0$, the particles do not have any part that is reflected from the barrier and only a part that is transmitted (has a coefficient $C$).
For the second case where $E < V_o$, the particles again can be reflected or transmitted when $x < 0$, but for the solution when $x > 0$, the book says:
(Images from "Quantum Mechanics Concepts And Applications" by Nouredine Zettili)
My question is:
I am confused about why there is this difference here? What about $E$ being less than $V_o$ that we need to look if $\exp(kx)$ diverges or not? Why was this not taken into account in the first case when $E > Vo$, i.e doesn't $C \times \exp(ikx)$ also diverge in the first case when $E > V_o$?
| Complex and real exponentials are fundamentally different mathematical objects. Recall Euler's Formula:
$e^{i z} = \cos{z} + i \sin{z}.$
So, for a real value of $k$, we have
$e^{i k x} = \cos{k x} + i \sin{k x}.$
This is a sum of sinusoidal functions - and we know that sinusoidal functions have no limit as $x \rightarrow \infty$ -- certainly they don't go to infinity, since $|e^{i k x}| = 1$ for all $x$.
You can see here a graph of a real exponential (the one that diverges):
http://www.wolframalpha.com/input/?i=e^x+from+x+%3D+-20+to+x+%3D+20
And here you can see the graph of the real and imaginary parts of a complex exponential ($\exp(i k x)$, which does not go to infinity):
http://www.wolframalpha.com/input/?i=e^{ix}+from+x+%3D+-20+to+x+%3D+20
Real exponentials, on the other hand, diverge. We have $\lim\limits_{x \rightarrow \infty} e^{k x} = \infty.$
EDIT: For more info, you may wish to look at this question Boundary Conditions in a Step Potential
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Why does a pitcher with lemon juice have foam, while one with pure water does not? Whenever I pour water into lemon juice (pouring directly from the tap into the pitcher, not quietly along its edge) I get a foam on top:
The same pitcher with water (same water tap, pitcher, time between the water poured and the picture, temperature):
Another answer discusses the formation of foam when pouring water but I did not find any mentions of lemon juice being particularly surface active (for instance Kitchen Mysteries: Revealing the Science of Cooking only mentions that lemon juice adds water to various sauces and modifies the surface active elements like oil, but does not mention lemon juice's own surface active properties).
Which substances in lemon juice could help to form such a persistent foam (it lasts at least 10 minutes)?
| I too have seen this effect with pure, unsweetened lemon juice.
To form a foam, (1) a surfactant is needed to lower the host liquid's surface tension (2) one needs to do mechanical work on the liquid to swell the surface area of the bubbles/foam and (3) the foam needs to be made faster than it breaks down. The Foam Wikipedia Page has a good summary of this.
The pouring of water and the pouring of lemon juice both supply the required work. So the only difference must be the presence of surfactants in the lemon juice. Citrus fruits contain significant amounts of oils and lipids, particularly in the skin and the matrix that makes up the cells of liquid in the lemon's flesh. All kinds of weird things- phenols for example, are found in lemons as well. So some of these compounds are clearly lowering the water-air surface tension to make foam as described in the Wiki page. I'm speculating that most of them come from the skin/flesh matrix, but this hypothesis would be very hard to test (as you'd need to extract the lemon juice without crushing the fruit, thus contaminating the juice with the skin).
| {
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Opacity/transparency of conductive meshes to charged particles (electrons/ions) When using a conductive (metal) mesh, effectively a metallic woven fabric, in vacuum applications as a "grid" for charged particle optics, how does one calculate (or at least estimate) the opacity or transparency of this grid?
$$
\begin{array} { l } { \mathrm { d } = \text { Wire diameter } } \\ { w = \text { Aperture width } } \\ { \mathrm { p } = \text { Pitch } } \\ { \mathrm { A } _ { 0 } = \text { Open area } } \\ { \mathrm { A } _ { 0 } = \frac { \mathrm { w } ^ { 2 } } { ( \mathrm { w } + \mathrm { d } ) ^ { 2 } } \times 100 } \\ { \mathrm { Nr } = \frac { 25,4 } { \mathrm { w } + \mathrm { d } } } \end{array}
$$
If given the wire diameter $d$, the aperture width (from edge to edge) $w$ or the pitch (from center to center) $p$, one can determine mathematically the smallest (macroscopic) particle that could go through the aperture, as if using the mesh as a sieve. Only particles with a cross-sectional area smaller than the open area $A_0$ will make it through.
Electromagnetically, the question is more complex, because it then entails the wavelength of the photons passing through it, and the effects of diffraction and interference.
What about for charged particles? How would someone go about calculating this? If the grid/mesh is biased, it will attract particles of one charge, accelerating them on approach, decelerating them as the depart, and vice-versa for the opposite charge. What then, prevents 100% of the charged particles from being absorbed by the grid itself?
Examples:
*
*The amount of current generated by a cathode, passing through a given grid (either biased or grounded), and making it to the anode.
*The amount of detected current by a Faraday cup with a grid (biased or grounded) in front of it.
| Generally, people simply look at the fraction of area covered and assume that the transparency to ions and electrons will be the same as the transparency calculated by open area/total area.
For most applications for ions and electrons woven grids are not used - grids can be used which are are formed by electrochemistry (i think) from flat sheets of material.
A typical grid might be 88 % transparent Molybdenum grid with squares with sides of length 1 mm or 0.5 mm - it is quite expensive to purchase. If you are not worried about magnetic fields you can get some nice Nickel grid material.
The action of potentials on grids is complicated because of lensing effects; any aperture will act as a lens and I do not think it would ever be possible to use a grid to collect 100 % of ions or electrons because of this.
I have an experiment I worked on where we could see a 90% tranmission grid was 90% transmissive because the current drops by 90% when the grid is in front of the Faraday cup. To do this experiment the grid needs to be mounted on a moveable support inside the vacuum. In this experiment the ions had a few keV of energy. Hope this is hepful.
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What is going on in a rotating magnet in a quantum scale? If there is a rotating magnet in an empty space and there is no outer field acting on it. Rotating in such a way that after half rotation magnet's N pole will be in the place where magnet's S pole was. I imagine that spin of the electrons inside magnet are following this rotation. Axis of rotation of electrons is perpendicular to the line connecting S and N pole of the electron. Do electrons inside the matter change their spin in "steps" or is it continuous movement? Are electrons trying to counteract this movement (like when you are trying to change rotation axis of macroscopic object)?
| A rotating magnet can be handled as an "orbital" angular momentum problem, no spins needed, but nothing stops you from using the usual spin-orbital coupling to calculate the total magnetic moment of a realistic solid state system. What matters for the interaction of the magnet is its total magnetic moment. Since angular momentum is quantized, the magnet can only change its angular momentum under the influence of an external field in a quantized way. There is no heating involved when such changes occur, the magnet will simply rotate faster or slower. Yes, the Einstein-de Haas effect is quantized.
Now, if you are asking about spins in a magnetic field (whether it is generated by the spins themselves of external), a single spin can only be up or down. Since the energy difference between a spin in up vs. down state is small compared to $kT$ in magnets at room temperature, magnetization has to be treated as a thermodynamic problem, in which the average magnetization is a thermodynamic average over a macroscopic number of spin states. Unless we go to really small systems with only a handful of atoms at low temperature or we do magnetic field resonance with many spins in a correlated state, the quantum mechanical properties of a magnetic system won't show up as an obvious effect.
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Drift velocity of charges in current In an electric curcuit, charges (electrons e.g.) move randomly around very, very fast. When a current is set in a curcuit, the charges still move randomly, but have a drift velocity around the curcuit. This is only in the order of about 0.1 mm/s.
The question is short and simple, and maybe the answer is too.
Why are collisions happening so much more frequently while drifting than when there is no current?
Turning a flashlight or an electrical heater on must gives a large increase in collision frequency to produce that much more energy than in the electrostatic case. Or what.
| The collision frequency of electrons in a metal at room temperature is given by the thermal distribution of the electron velocities (please note that this is already a somewhat questionable approximation, metals really require a quantum mechanical treatment). I do not believe that this collision frequency increases much when a current flows trough the metal. What does happen, though, is that on average there is no energy transfer between the electrons and the lattice if there is no drift, because the electrons are in thermodynamic equilibrium with the lattice. When we add an electric field electrons accelerate a little between any two collisions and then they are not in thermal equilibrium with the metal ions any longer. As a result they will shed their additional kinetic energy to the ions in these collisions
which will heat the metal.
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Given a wave function $\psi(x)$, is there always a potential $V(x)$ such that $\psi(x)$ is an eigenstate? Given any unit norm wave function $\psi(x)$ which is in the Hilbert space, can we always find a $V(x)$ such that the $\psi(x)e^{-i\omega t}$ is a solution of the corresponding Schrödinger equation? (I mean the Hamiltonian which uses the potential $V(x)$.)
| I'm not quite sure where OP is going with this, but from a mathematical perspective, assuming that the wave function $\psi:\mathbb{R}\to \mathbb{C}$ is twice differentiable and different from zero, consider the (possibly complex) potential
$$ V(x)~:=~\hbar\omega+\frac{\hbar^2}{2m} \frac{\psi^{\prime\prime}(x)}{\psi(x)}. $$
This will trivially satisfy TDSE.
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Why do things here on Earth fall down? I want to have an answer with that question above for my physics lesson. I really don't have an idea about it, so, I ask help from you guys and hope that someone can help me with it.
| In 1600 Newton discovered a formula that could explain why things fall down on earth
things fall due to the gravitational force of the Earth. The formula for the intensity of the force is: $$F = G \dfrac{m_1 \cdot m_2}{r^2}$$
where $m_1$ and $m_2$ are the 2 masses of the object, $r$ is the distance between them and G is a constant called Universal Gravitational Constant (also discovered by Newton) and is $6,673 × 10^{−11}$
This is the GENERAL formula.
Now consider the gravitational force of the Earth; if we plug in the equation the mass of the Earth and the distance of an object on the surface of the Earth and the centre of it (which is just the radius of the Earth) we get:
$$F=G \frac {M_{earth} \cdot m_{object}}{r_{earth}^2}=6,673 × 10^{−11} \cdot \frac{(5,9736 \cdot 10^{36})\cdot m}{(6372,795 \cdot 10^3)^2}\approx m\cdot 9,81 $$
$9,81$ is the result of all the calculation and physicist call this $g$ which simplifies all the calculations. If you want to calculate the gravity force on an object by the Earth you just have to take the mass of the object and multiply it by $9.81$
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Can we test QFT on a curved spacetime? It is possible to extend a quantum field theory to a curved spacetime. But does this lead to predictions that can be tested and measured? Had it been confirmed?
The underlying reason I am asking this is: curved spacetime means emergence of gravity and therefore General Relativity regime. And we know that GR and QFT are incompatible. I realise that in order to include gravity, one should put its Lagrangian in from the very beginning and this, I guess, does not work. But does the current mathematical framework for extending the known field theories to a curved spacetime work?
| Cosmology and inflation provide a hugely important test of quantum field theory in curved spacetime. During inflation, there is a scalar field (the inflaton) that is providing the vacuum energy that drives inflation. This scalar field is subject to the rules of quantum field theory in curved spacetime. The quantum fluctuation of this scalar field lead to the temperature fluctuations we see in the Cosmic Microwave Background today, and as these fluctuations grow due to gravitational collapse, they produce the structure we see today in galaxies and clusters. Measuring properties of the CMB and large scale structure is thus a test of quantum field theory in curved spacetime from the inflationary epoch.
| {
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"url": "https://physics.stackexchange.com/questions/156709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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How is mass reduced in a normal chemical reaction which releases energy like heat, sound or light? I was reading this link.
Just above the paragraph titled "OTHER CONSERVATION LAWS", it says that
"This conversion of mass to energy happens well below the speed of light, in a very small way, when a stick of dynamite explodes. A portion of that stick becomes energy, and the fact that this portion is equal to just 6 parts out of 100 billion indicates the vast proportions of energy available from converted mass."
I think this is incorrect. The chemical energy in the dynamite is converted to heat, light and sound. There is no nuclear reaction taking place to convert mass into energy. Am I wrong?
| As a matter of fact, chemical reactions can reduce mass just like nuclear reactions. I find it hard to accept, but it's true.
When the molecules in the dynamite explode, bonds between atoms are broken and reformed in different configurations. The result of this is that the net electrical potential energy in the resulting molecules is less than the electrical potential energy of the original stick. Now here's the cool part – that means it has less mass. Like, literally, less mass. Like, if you let the heat, light, and sound dissipate, and ultracarefully collect and weigh all the reaction products (impossible in practice, of course), it would weigh a tiny bit less than the original stick.
I find it helpful to think of a simpler example. Take two hydrogen atoms, and an oxygen atom. Allow them to run into each other. Their electron orbitals merge and hybridize. As their electrons settle into their new, shared, lower-energy state, they release photons. These photons carry away energy, and therefore mass, from the atoms. The resulting $\mathrm{H_2O}$ molecule literally weighs less than the two hydrogen and one oxygen beforehand.
Weird!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/156861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Are there new physics scenarios that predict low lying hadrons? There is a significant ongoing experimental effort to search for new hadrons with masses in the GeV range. This is used to find the spectra of QCD bound states, with a particular emphasis on finding exotic resonances such as the tetraquark. To my knowledge, they have not found any state whose mass is in contradiction with the theoretical prediction using lattice QCD thus far (though e.g., there are a few tetraquark candidates, such objects are not in conflict with lattice predictions).
These searches are clearly very important as they confirm our understanding of QCD and in particular, they verify the validity of lattice QCD, which can subsequently be used to study new phenomena. But my question is, are there any mainstream new physics scenarios which predict a deviation in the spectra of QCD and could be found at for example, LHCb?
EDIT: I'm interested in changes to the low energy ($\sim$ GeV) bound state spectrum measured by these experiments
| If you mention about exotic state hadrons which are composite of two anti quarks and two quarks, yes there are some new observations. The recent and approved one is Z(4430) which observed by LHCb group from CERN http://home.web.cern.ch/about/updates/2014/04/lhcb-confirms-existence-exotic-hadrons It consists $c^-cdu^-$ quarks and anti quarks.
There is also an on-going research from CMS group which is not approved yet, but supported by a group from Fermilab as well, they expect to see resonance through the $B^±$ meson decays and called the observed particle Y(4140) http://news.discovery.com/space/lhc-spots-mysterious-y4140-particle-121120.htm
These new hadrons are mostly called by their own mass as Y(4140) $mass = 4.4140 GeV/c^2$ and Z(4430) $mass = 4.430 GeV/c^2$ and we would say they are in low energy scale
I am not theoretical physicist mostly works experimental but I know the importance of these kind of hadrons because they seem like two mesons compositions but theoretically they are not at all.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/156921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Two superimposed sounds traveling through low- and high-density matter: is separation possible? Two superimposed sounds (at source: $s(t) = s_1(t) + s_2(t)$; the two sound components overlap completely in time, partially in spectra) travel through low- and high-density matter and are recorded from two different positions (see figure). Knowing the two different densities $d_1$ and $d_2$ (I don't actually know them yet but it is just a matter of some research), would it be possible to separate the superimposed sounds from each other?
It is from an actual research problem but this far I have only considered signal processing techniques. I thought people with a different background might see the problem differently. I'd of course acknowledge any contributions that would further tackling the problem.
DSP, including filters and blind source separation has been tested already.
| I don't see how this can be done, given the problem as-stated. What defines it as being 2 superimposed sounds, rather than just 1 sound, other than just an arbitrary definition? What stops me coming along and saying: "No, it's actually 4 superimposed sounds, or 27!"? If both sounds are coming from the same source then any shift in speed or frequency of each superimposed sound as it travels to each microphone will be the same, so I don't see how you can use this to separate them. Why do you think this high/low density material thing would make a difference?
If the 2 sounds were coming from different sources then they would take different times to reach the 2 microphones, so in that case I'm sure you could make some process or algorithm to separate them.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/156999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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