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What are the main algorithms the LHC particle detectors use to reconstruct decay pathways? I am just starting to look into how we understand the data from particle collisions.
My question is, what are the algorithms or ways that these detectors interpret the data? Are there standard approaches? Or if not, what are some good papers or places to look to get started in learning more about the implementation and/or details of how this works?
So far I haven't dug into any textbooks, but many articles on the web and this was somewhat helpful in pointing to where to look:
http://arstechnica.com/science/2010/03/all-about-particle-smashers-part-ii/
So from my understanding so far, there are a few different LCH "experiments", which are physical structures that are optimized to focus on specific aspects of data from a collision event. The detector measures all kinds of particle emissions and changes in electrical fields, and then seems to try to backtrack and figure out all the emission/decay events that might have taken place in that split second.
From my understanding so far, basically the computer programs used to compute these possible "decay pathways" must be using some standard algorithms or something, and must have built into them all possible particle emission pathways (like all possible Feynman diagrams if there is such a thing).
Are there any good resources or standard algorithms/approaches to understanding how particle detectors analyze their data?
| Well, if you have the time... CERN has all the technical design reports for its detectors online at http://cds.cern.ch/. They are excellent reading material.
Start with a search for "ATLAS technical design report" and "CMS technical design report" and work your way trough the references in those documents. Once you understand the geometry of the detectors (not a small feat), you can start reading about "trigger algorithms" and "reconstruction algorithms". You may have to pick up a thing or two about particle matter interactions and the GEANT simulation software.
Little warning... it took me almost two years to read trough just the parts that were important to my work...
| {
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The particle content of a given state In Carroll's we read
...The Unruh effect teaches us the most important lesson of Quantum Field Theory (QFT) in curved spacetime, the idea that "vacuum" and "particles" are observer-dependent notions rather than fundamental concepts.
I wonder are we talking about the same observer or two different observers using two different frames?
Put it another way, can the same observer who observed nothing using a frame detect a thermal spectrum if switched to another frame that is uniformly accelerating with respect to his previous frame?
Does it apply both ways? I mean if an observer detects particles, does this observant-dependent notion of particle content of a given state, implies that he can switch to another frame in the blink of an eye, and detect nothing?
| That is exactly the point: if the field is in a vacuum with respect to observer A, and observer B accelerates uniformly with respect to A, then B will observe a field state with nonzero particle content.
It doesn't really matter whether you talk about different observers on different frames of reference, or of a single observer who 'switches' their frame of reference. The notion of 'observer' is not meant to imply a specific person, nor does the experimenter himself have to be at rest or in inertial motion with respect to those coordinates. The terms 'observer' and 'frame of reference' refer to ways of making sense of observations more than actual people.
Finally, note that the Unruh effect can sometimes work the other way - i.e. if observer A sees a field with particle content, there may exist a second observer B in uniformly-accelerating motion w.r.t. A for whom the field is in vacuum - but this is not guaranteed. What the effect teaches us is that such observers B will observe a different particle content than A, but in general this need not be the vacuum.
| {
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How is photon time of flight/range over sub-millimeter distances measured? I was reading a paper that described how the force a low-thrust torsion pendulum was measured. In it, the paper states a laser is bounced off a mirror and the displacement is "...based upon the beam reflection time." The paper states that the device can measure sub-micrometer displacements.
Conceptually as I understand it, this measurement device would have 4 major components. A laser emitter, a mirror, a detector, and a controller. The controller would power on the laser, then note how long it takes for the detector to respond to the signal.
However, the time of flight difference for ranges this size are vanishingly small. For instance, it'd take a photon roughly $3.336×10^{-12}$ seconds to travel an additional 1mm along the beam path.
If you flip that over to cycles per second, that would suggest the controller has to operate at around ~300 GHz. Only then could it check the sensor often enough to have the temporal resolution to resolve a 1mm change in the beam path length.
This seems like an absurd clock speed for any sort of computer controller. Is there another component, or concept that I'm missing?
| Can you cite the paper, please?
Assuming that it's a modern version of an old experiment, my first guess would be, that the observation uses the fact that the light will induce a constant moment on the torsion pendulum. The response of the pendulum will be an oscillatory motion, for short amounts of time (seconds to minutes), that motion around the original equilibrium point is proportional to the time the light is turned on. The timescales on which such an experiment runs will be between seconds and many minutes, and not at the inverse of the laser frequency.
| {
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How can gravity affect light? I understand that a black hole bends the fabric of space time to a point that no object can escape.
I understand that light travels in a straight line along spacetime unless distorted by gravity. If spacetime is being curved by gravity then light should follow that bend in spacetime.
In Newton's Law of Universal Gravitation, the mass of both objects must be entered, but photon has no mass, why should a massless photon be affected by gravity in by Newton's equations?
What am I missing?
| It's just a simple concept according to Einstein's relativity when light travels through high gravation field or high mass containing object (i.e. same when a object has high mass means it can attract other objects of lesser mass) the photons present in the light gets attract towards the other object and we see light bending in the universe, but there is one thing is to note that photons are massless matters but in this case the object with higher mass attracts the object with lesser mass weather it is 0.
| {
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Finding the cosmological redshift of a galaxy in the expanding Universe Firstly, I understand what the Doppler effect is when it comes to sound or light waves.
From everything that I've read, we are told that the universe is constantly expanding since the all the radiation we observed is red-shifted.
Assuming we are observing a distant galaxy/start that is moving away from us, the EMR waves incident on us from that galaxy is red shifted.
My question is:
How do we know that the light is red shifted? When you measure its wavelength you are given just ONE value $\lambda$, right? How does one know what its original wavelength was to begin with? Its only after we know both the values (the actual wavelength when it was emitted and the wavelength that we measure hear on earth) that we can claim that the galaxy/star is moving away.
I always assumed that one would observe both the wavelength of the photons and the energy and there is some sort of disparity there that tells us the light is red-shifted. But that does not make sense since one usually determines the wavelength based on the energy of the photon......I am confused here.
| The answer to this question is that if you can only see one line or feature in the spectrum then the redshift cannot be measured unless you have some other information that leads you to guess what the line or feature in the spectrum is due to (e.g. the 21cm line of hydrogen at radio wavelengths is so strong and ubiquitous it can usually be identified immediately).
The more usual situation, especially in the optical and infrared parts of the spectrum, is that you have two, or often several more, features or lines in the observed spectrum.
If we have two lines with wavelengths in the rest (laboratory) frame $\lambda_1$ and $\lambda_2$, and say these are redshifted by an amount $(1+z)$, where $z \simeq v/c$ (approximately true when $v \ll c$). We label the observed redshifted wavelengths from the distant galaxy as $\lambda_1^{\prime}$ and $\lambda_2^{\prime}$, such that
$$ \lambda_1^{\prime} = (1+z)\lambda_1\ \ \ \ \ \ \ \lambda_2^{\prime} = (1+z)\lambda_2$$
The point of the algebra is that all line wavelengths are shifted by exactly the same factor $(1+z)$. Hence a pattern of lines in the spectrum (e.g. the Balmer series of Hydrogen, or a close pair of calcium H & K lines, or sodium D lines) is replicated and can be recognised easily as such in the redshifted spectrum. Then, with the lines identified, the redshift can readily be calculated from the observed wavelengths.
| {
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Glashow-Weinberg-Salam mass terms At the end of spontaneous symmetry breaking I get these mass terms:
$$W_{\mu}^{\pm}=\frac{1}{\sqrt{2}}\bigl(W_{\mu}^{1} \mp i W_{\mu}^{2} \bigr )$$
$$\mathcal{L}_{mass}=\frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{+}{W^{\mu}}^{-} + \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+}$$
So I have $$M_{W^+}=g \frac{v}{2} \quad M_{W^-}=g \frac{v}{2} $$
Is it right? Or there are too many terms and it is enough:
$$\mathcal{L}_{mass}= \frac{1}{2} g^2 \frac{v^2}{4} W_{\mu}^{-}{W^{\mu}}^{+} $$
| Notation $W^{-}, W^{+}$ may confuse in a sense that it may seem that here are two different particles which aren't connected by charge conjugation. But of course, $W^{+}$ is only $(W^{-})^{\dagger}$, so it is an antiparticle to $W^{-}$. So term $( W^{-} \cdot W^{+} )$ is simple $|W|^{2}$ (which is standard for the mass-term), and, of course, both of particle and antiparticle have the equal masses.
Also before making substitution
$$
\tag 1 W^{\pm}_{\mu} = \frac{1}{\sqrt{2}}(W_{\mu}^{1} \mp iW_{\mu}^{2})
$$
you can see that both of fields $W^{1}, W^{2}$ have equal masses. So of course that their linear combinations $(1)$ also have equal masses.
| {
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Wouldn't the presence of dark matter slow the expansion of the universe? If there is a huge element of dark matter in the universe, wouldn't this extra gravity prevent the accelerated expansion of the universe?
| The acceleration of the expansion is currently observed to be happening. This observation is one of the pieces of data we use to infer the amount of dark matter. It tells us that there can't be more than a certain amount of dark matter, because that would be incompatible with the observed acceleration.
| {
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Physical reason for Lorentz Transformation Seeing the mathematical derivation of the Lorentz Transformation for time coordinates of an event for two observers we get the term
$$t'=-\frac{v/c^2}{\sqrt{1-\frac{v^2}{c^2}}}x+\frac1{\sqrt{1-\frac{v^2}{c^2}}}t$$
Now how to make sense physically of the $t-\frac{vx}{c^2}$ factor.
I am looking for an argument along the lines of the following. When relating spatial coordinates, one observer measures the length separation between an event and the second observer in his frame and tells the other observer that this should be your length, which the second observer denies due to relativity of simultaneity and multiplies by the $\gamma$ factor to get the correct length.
| The physical reason IS the constancy of the velocity of light... since I'm writing in a tablet the answer won't be complete, but expect to get you to the mathematical cross-road.
Constancy of velocity of light implies that
\begin{equation}
\frac{d|\vec{x}|}{dt} = c, \quad\Rightarrow\quad d|\vec{x}| = c\,dt.
\end{equation}
Since $d|\vec{x}| = \sqrt{dx^2 + dy^2 + dz^2}$, it follows that
\begin{equation}
dx^2 + dy^2 + dz^2 = c^2\,dt^2 \quad\Rightarrow\quad dx^2 + dy^2 + dz^2 - c^2\,dt^2 = 0.
\end{equation}
From here it is straightforward to see that the set (or group) of transformations preserving this quantity are those known as Lorentz transformations.
Now I leave you to analize the "generalization" to nonvanishing intervals, for massive particles. Hint: define a four-dimensional metric!
(continuation... after a few days)
The interval
As exposed previously, the physical condition of constancy of the speed of light leads to the conclusion that
All equivalent observers are connected through a transformation which
keep the quantity $$dx^2 + dy^2 + dz^2 - c^2\,dt^2 = 0.$$
This can be generalized to the preservation of the quantity $$ I = dx^2 + dy^2 + dz^2 - c^2\,dt^2, $$ called interval.
Notice that the interval can be written as
$$
I = X^t\, \eta\, X =
\begin{pmatrix}
ct & x & y & z
\end{pmatrix}
\begin{pmatrix}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
ct \\
x \\
y \\
z
\end{pmatrix}
$$
Invariance of the interval
In order for the interval to be invariant under a transformation $X' = M\, X$, one needs to
\begin{align}
I &= I' \notag \\
X^t\, \eta\, X &= (M\, X)^t\, \eta\, M\,X \notag \\
&= X^t\, M^t\, \eta\, M\,X \notag \\
\Rightarrow\quad \eta &= M^t\, \eta\, M. \tag{*}
\end{align}
Therefore the problem is to find a set of transformations $M$ satisfying Eq. (*).
Two-dimensional case
Finding a general 4 by 4 matrix $M$ preserving the Minkowskii metric ($\eta$) requires a lot of algebra, but one can easily find the transformation preserving the 2 by 2 restriction to the $(ct,x)$-plane.
Propose a matrix $$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix},$$, and solve the equation
$$
\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} =
\begin{pmatrix} a & c \\ b & d \end{pmatrix}
\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} a & b \\ c & d \end{pmatrix} =
\begin{pmatrix} -a^2+c^2 & -ab+cd \\ -ab+cd & -b^2+d^2 \end{pmatrix},
$$
which is simple if one uses the identities $\cosh^2\theta - \sinh^2\theta = 1$, and the condition $M(\theta\to 0) = \mathbf{1}$.
Thus, $$M = \begin{pmatrix} \cosh\theta & -\sinh\theta \\ -\sinh\theta & \cosh\theta \end{pmatrix}.$$
Relation with the velocity
In a similar fashion of Euclidean geometry, in which
$$
\frac{y}{x} = \tan\theta,
$$
one uses the transformation $M$ above to relate the $ct$ coordinate with the $x$ coordinate
$$
\frac{v}{c} \equiv \frac{x}{ct} = \mathop{\mathrm{tanh}}\theta.
$$
Now,
\begin{align}
\mathop{\mathrm{tanh}^2}\theta &= 1 - \mathop{\mathrm{sech}^2}\theta \notag \\
&= 1 - \tfrac{1}{\cosh^2\theta} \notag \\
\Rightarrow\quad \cosh\theta &= \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \notag \\
\sinh\theta &= \frac{\frac{v}{c}}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}. \notag
\end{align}
Finally, from the relation $X' = M\, X$, one obtain the usual relations
\begin{align}
x' &= -\sinh\theta\cdot t +\cosh\theta\cdot x \notag\\
&= \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}\left( x - vt \right) \notag \\
t' &= \cosh\theta \cdot t -\sinh\theta\cdot \tfrac{x}{c} \notag\\
&= \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}\left( t - \tfrac{v}{c^2}t \right). \notag
\end{align}
| {
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What is entropy really? On this site, change in entropy is defined as the amount of energy dispersed divided by the absolute temperature. But I want to know: What is the definition of entropy? Here, entropy is defined as average heat capacity averaged over the specific temperature. But I couldn't understand that definition of entropy: $\Delta S$ = $S_\textrm{final} - S_\textrm{initial}$. What is entropy initially (is there any dispersal of energy initially)? Please give the definition of entropy and not its change.
To clarify, I'm interested in the definition of entropy in terms of temperature, not in terms of microstates, but would appreciate explanation from both perspectives.
| A higher entropy equilibrium state can be reached from the lower entropy state by an irreversible but purely adiabatic process. The reverse is not true, a lower entropy state can never be reached adiabatically from a higher entropy state. On a purely phenomenological level the entropy difference between two equilibrium states, therefore, tells you how "far" away they are from being reachable the lower entropy state from the higher entropy one by purely adiabatic means. Just as temperature is a scale describing the possibility of heat flow between interacting different temperature bodies, entropy is a scale describing the states of a body as to how close or far apart those states are in the sense of an adiabatic process.
| {
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Getting first and second time derivative of mean motion of spacecraft - without TLE? The Celestrak website provides information on reading the TLE ( Two-Line Element Set ) format. In Line 1, Column 34-43 & 45-52 give information about First Time Derivative of the Mean Motion and Second Time Derivative of Mean Motion.
Is there any way to calculate or estimate these two parameters for a spacecraft, if you don't have its TLE at disposal to simply read them from there ? And what do they actually mean ?
| The mean motion $n$ of a satellite is its angular velocity, averaged over one period. In other words, if the satellite rotates around the Earth with period $P$, its mean motion $n$ is
$$
n = \frac{2\pi}{P}
$$
If the Earth were a perfectly spherical symmetric object, and there were no other perturbing agents in the Universe (in other words, if Earth + satellite were a closed system), then $n$ would be an exact constant of motion.
However, several effects are not included in a two-body, idealized problem, and they make $n$ time-variable. One such effect is the Earth's oblateness, which makes the gravitational potential differ from a simple $1/r$ law. Another effect is the presence of the Moon, which also perturbs the satellite motion. If, instead of a human-built satellite, we were discussing the motion of moons around their planets, other perturbing agents would the gravitational influences of nearby planets and of the Sun.
The fact that $n$ is not constant in time is clearly seen in accurate positional data extending over many orbits. Fitting the ensemble of these data yields both a first and a second derivative, to allow more precise predictions of the satellite position in the nearby future. Since however the variation of $n$ is not subject to any simple law, it needs to be recomputed all the time.
So, in short, the answer is no, you cannot derive those parameters from data in your possession. In order to be able to do that, you should own the whole set of position measurements over very many orbits which is surely in the hands of both civilian and military space agencies.
If you wanted instead to compute from first principles what those parameters ought to be, you will have to study the importance of the Earth's oblateness, and of the Moon, in a restricted three-body problem.
| {
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Accelerating masses lose energy? If I understand this correctly, accelerating charges lose energy in the form of EM waves because they change the electric and magnetic fields, which "costs" energy. Does that mean that accelerating masses lose energy too, because they change the gravitational field (i.e. curve spacetime)?
| Yes, it an extremely small effect but it exists in Einsteins general relativity. There is one case of a double star where there rotation around each other seems to lose energy at rate that this phenomena should give according to general relativity
| {
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Given the spring constant & maximum kinetic energy; length of spring extension? I need to understand the following question before i right my exam tomorrow.
A body attached to a spring with spring constant 100 N/m executes simple harmonic motion. The maximum kinetiv energy of the body is 2J. Calculate the spring extension (in m) when its potential energy is equal to the kinetic energy.
I understand the concept of total mechanical energy, i just think im missing something small thats got to do with springs in order to calculate this.
| Write down the potential and kinetic energy as a function of position. When the spring is in the middle of the motion, all is kinetic. When it is at the extreme of the range, all is potential. Somewhere between these two extremes, the potential and kinetic energies will be the same; their sum should always be constant (when there is no loss).
Recall that
$$KE = \frac12 m v^2\\
PE = \frac12 k x^2$$
where $k$ is the spring constant, $x$ is the displacement, $m$ is the mass and $v$ is the velocity.
In this case - if max kinetic energy = 2J then potential energy = 1J when it is equal to the kinetic energy. Then you use the PE equation above to find the extension.
| {
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What happens when a slow wave reaches lower hybrid resonance? Lower hybrid resonance occurs when $n_{\perp}^2$ goes to infinity, and it occurs only for the slow wave solution, not the fast wave. Since $n_{\perp}$ is proportional to $k_{\perp}$, and $k = \frac{2 \pi}{\lambda}$, it means that the wavelength of the wave goes to zero. But what physically happens when the slow wave reaches the lower hybrid resonance?
I should mention that I'm talking about in the cold plasma model, where the fast and slow wave modes have meaning.
| Well, I am not sure if your statements are entirely accurate because the fast mode can approach the lower hybrid resonance. In fact, in this regime, it becomes effectively indistinguishable from an electrostatic whistler mode. At low frequency and oblique angles, the fast (or magnetosonic) modes are right-hand polarized (with respect to $\mathbf{B}_{o}$) electromagnetic waves, which happen to be on the same branch of the dispersion relation as whistler mode waves.
In any case, in the limit as $\mathbf{k}_{\parallel}$ $\rightarrow$ 0 the wave will become electrostatic and, for all intents and purposes, be a form of ion-acoustic wave (not to be confused with the much higher frequency electrostatic version that has $\mathbf{k}_{\perp}$ ~ 0) or lower-hybrid wave. At such oblique angles, the only important things are the wave frequency and that it is electrostatic. The name given to the mode is really just semantics.
Edits/Additions
I realized after the fact that I had forgotten to say that $\mathbf{k}$, in addition to knowledge of $\omega$ and polarization (e.g., electrostatic), are the only important things for these modes. The reason is that these properties let you know how and with what particles these modes interact.
At the lower hybrid resonance, an electrostatic wave can simultaneously couple and exchange energy/momentum with both electrons and ions. This is why the waves are so popular in current dissipation theories, since $\mathbf{j}$ = $\sum_{s} n_{s} \ q_{s} \ \mathbf{V}_{s}$ . If the waves can transfer energy/moment from(to) electrons to(from) ions, then have the capacity to limit $\mathbf{j}$. If the interaction is stochastic in nature, then the result can be an irreversible form of energy transformation (i.e., energy dissipation).
| {
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Amplification of magnetic field can we by any means amplify magnetic signal as we can with electric signal. As both electric and magnetic field can be represented in the form of a wave the analogy seems to be natural.
I want the input and output as magnetic signal.
| Electrical amplification is about using an input signal to modulate a larger amount of power that comes from a separate power supply of some sort.
And yes, there is such a thing as a magnetic amplifier that works on a very similar principle (even though the inputs and outputs are usually electrical). But you can't get an output value that's greater than the sum of the input values (control signal plus power source).
| {
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Magnetic Force on a Ferromagenetic Material I am currently working on a project involving solenoids, and I needed a force(Newtons, not a measure of magnetic field strength) equation. What I came up with after some digging around on the internet, is the equation:
$$F = (NI)\mu_0\frac{\text{Area}}{2g^2}$$
Where $F$ is force (in Newtons), $N$ is the number of turns in the coil, $I$ is the current being passed through the coil, $μ_0$ is the magnetic permeability of vacuum, and $g$ is the gap between the coil and the ferromagnetic material. (Area $A$ and $g$ can be any units, as long as you're consistent with the usage)
I don't know in which plane exactly the area $A$ is taken.
Assuming I have a rod, moving lengthwise into a solenoid, which plane would $A$ represent?
Plane a, plane b, or another plane that I did not consider relevant to this problem?
Rod:
Edit:
I was looking for the force an electromagnet would exert on a ferromagnetic material moving into the coil. something like this.
Edit:
If the equation I was using before does not work, I don't suppose anyone has the correct one?
Edit:
After looking at the equation some more, I realized I had written it wrong. It should be:
$$F = (NI)^2\mu_0\frac{\text{Area}}{2g^2}$$
| The atomic magnetic dipoles in a ferromagnetic material experience a torque that tends to line them up with the crystal axis and another which tends to line them up with any magnetic field. At normal temperatures they can often maintain such an alignment. If the field has a gradient they can also experience a net force. If a ferromagnetic bar is lined up with a current carrying solenoid, the field from the solenoid magnetizes the bar and then sucks it in due to the divergence (or convergence) of the field which occurs just inside and outside of the end of the solenoid. This principle is used a lot in electric current relays and fluid flow valves. (As I recall, the doors on the Boston subway trains in the 50's were opened and closed by rods being sucked into long solenoids.) The actual force is going to depend a lot on the properties of the ferromagnetic material. I would suggest an experimental approach (or buy a preexisting device). You might start with a premagnetized bar, but then there is the risk that the solenoid may modify the magnetization.
| {
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How long does it take plasma to radiate its heat? Lets say we have 1 gram of plasma (Argon) at 1 million kelvin confined in a vacuum with electromagnets. If we keep the magnets on but shut down the device that heated the plasma, how long will it take for it to cool off?
| Since your plasma is in a vacuum environment, the only way for it to loose energy is by radiation (conduction transfer through the magnets are neglected). You have thus to consider which bodies are surrounding your plasma and which radiative model is the best ton consider for them. I guess you can consider a black body with the simple Stefan equation.
The most common model is the NEC (net emission coefficient). Take a look at this article chich describes quite precisely the Argon plasma radiation exchange. With the NEC and P1 model you can take in account the emitted power and the absorbed energy by the plasma itself.
http://documents.irevues.inist.fr/bitstream/handle/2042/16649/CFM2007-0901.pdf?sequence=1
Once you get the power emitted by the plasma, I think you just have to integrate the energy balance equation to get the amount of energy lost over the time. Then you have to consider the temperature evolution of this plasma, and figure out when it actually stop existing (its temperature equals surrounding temperature? I don't know...). This can be quite tough because the temperature field isn't uniform (but maybe you can start with a uniform one to get first results).
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Electric Potential of Conductors I understand that charges in a conductor reside on the surface of a conductor. So why is it that the neutral inside of a conductor and the charged surface are at the same potential?
| Do you understand that the electric field within a conductor is zero? The charge is mobile, so the internal charge rearranges itself until there is no longer any force to move them: there is no field in the interior. If you understand that, then you will realize that a test particle within the conductor will feel no force, so no work will be done in moving the charge anywhere inside the conductor. So every point in the interior is at the same potential.
The same argument applies to the surface. Surface charges will move until there is no longer a force to move them. There is no electric field parallel to the surface, so the surface is an equipotential.
Finally, why is the potential at the surface the same as the potential in the bulk? Same argument, plus the realization that charges can migrate between the surface and the bulk, and will do so until there is no longer any force to move them: the surface and the bulk will then be at the same potential.
I hope I interpreted your question correctly ... it is a little vaguely worded.
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Photons to Represent a Wave I fear that I have a fundamental misconception about the "wave particle duality" of light, but in a related question, the answerer said, in some sense, that a light wave propagates until it hits something, at which point in time it (can) act(s) like a photon. Which is fine to me, but there are a finite number of photons in a wave front, so what happens if you "run out" of photons in a wave front? Certainly the wave needs to interact with everything it touches, so if you have a wave that only effectively has one photon, and it "hits" two electrons, how does it interact with both? Say you have two electrons both a distance $R$ from a photon emitter, emitting circular waves. Or something like that.
| The classical electromagnetic field given mathematically by Maxwell's equations can be proven to emerge from a confluence of individual photons, which photons are described by the Quantum Mechanical form of Maxwell's equations. Thus the classical wave is made up by zillions of photons with energy $h\nu$, where $\nu$ is the frequency of the classical wave. Have a look at this blog description of how this happens mathematically. The interference pattern of individual photons at a time is the same as the classical interference pattern because of this $h\nu$.
One photon does not a wave make in space. One photon can be described by a probability wave, which means the probability of being at an $(x,y,z,t)$, which manifests in the single photon at a time double slit experiments. It is an ensemble of photons that make up a classical wave. I like to think of it as analogous to a "stadium wave". One person does not a wave make.
| {
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How can I calculate the force that is applied on a tube by another tube? Let's say there is two tubes(cylinders with no tops or bottoms) with charges $q_1$ and $q_2$, radii $b_1$ and $b_2$, lengths $l_1$ and $l_2$. These tubes are located along the axis of each other's surfaces like in this figure:
If the electric field that the first tube creates on a point is;
$$
E = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{\sqrt{b^2 + (c-a)^2}} - \frac{1}{\sqrt{b^2 + (c+a)^2}}\right)
$$
where $b$ is the radius of the tube, $c-a$ is the distance between the centre of the furthest part of the tube and the point, $c+a$ is the distance between the centre of the closest part of the tube and the point, $q$ is the total cahrge on the tube and
$\epsilon_0$ is the electric constant. Here is the figure of the tube and the point for those who didn't understand from my description:
The question is how can I calculate the force between these two tubes?
Update:The electric field formula I found is not true since it is valid for a point on axis of the cylinder. Thus I would be pleased if you could show me how to solve the problem from the beginning.
| Without actually providing the mathematical details (which is left for the reader) the basic outline is this:
1.Select a differential segment (a segment of infinitesimal lateral dimensions) on the second cylinder.
2.Write the electric field expression for an infinitesimal charge on the segment.
3.Write the force equation and integrate over the entire surface.
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Sound difference between musical instruments I know that the difference between two musical notes is given by the sound frequency, and the difference in volume is given by the amplitude.
What I am wondering is why does the same note sound different on different musical instruments?
What in the wave makes the difference between the sound of a harmonica and the sound of a violin singing the same note?
| The different tonality of a note in different instruments stems from the different mixes of amplitudes in the harmonic frequencies that the instrument provides.
To be more concrete (and keeping to a slightly simplified view), you play the A note (440 Hz) and then you have the harmonic frequencies 880, 1320, 1760, ... ($440n$ where $n$ is the number of the harmonic). Each of the frequencies will have an amplitude or "volume" contributing to the sound produced by an instrument. Thus a particular set of amplitudes gives the instrument its tonality.
This is highly related to the concept of Fourier analysis used in many areas of physics.
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Why should any physicist know, to some degree, experimental physics? I've been trying to design a list with reasons why a proper theoretical physicist should understand the methods and the difficulty of doing experimental physics. So far I've only thought of two points:
*
*Know how a theory can or cannot be verified;
*Be able to read papers based on experimental data;
But that's pretty much what I can think of. Don't get me wrong, I think experimental physics is very hard to work on and I'm not trying to diminish it with my ridiculously short list. I truly can't think of any other reason. Can somebody help me?
| Here's a reason that hasn't been touched yet (but is alluded to by your question): to be able to form new theories.
A lot of the most interesting theories in physics comes from someone reading about an experiment and trying to explain the results. We wouldn't have relativity if Einstein didn't read about the Michelson–Morley experiment and going "hmm.. let's assume there are no errors, something funny is going on here".
There's still a lot of experiments published with unexpected results with incomplete or not so convincing explanations. Yes, a lot of them are in less glamorous fields such as fluid-mechanics or acoustics or crowd-dynamics. But once in a while we get interesting theories out of them and once in a while two seemingly unrelated fields yield a single unifying theory.
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Can a third type of electrical charge exist? Upon reading my book on physics, it mentions that there are only two discovered types of electric charges. I wonder if there could be a third type of elusive charge, and what type of effects could it have upon matter or similarly?
| In the Standard Model, electric charge $Q$ is actually part weak hypercharge $Y_W$ and part weak isospin $T_3$
$$Q = T_3 + \frac{Y_W}{2}$$
which can be either positive, zero (electrically neutral), or negative.
In this framework, that's it.
If, in fact, there is another type of electric charge (and its associated anti-charge), I believe it would be the case that the there would need to be three types of photons which would themselves be electrically charged and, thus, interact with each other.
This is in analogy with weak isospin where the three weak 'photons' ($W^+, W^0, W^-$) are isospin charged.
This would, of course change everything. But, we see only one type of photon and it is electrically neutral.
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Poynting vector plane wave I'm calculating the poynting vector for a plain wave and I have some doubt.
$$ \bar S = \frac 1 2 \bar E \times \bar H^* = ... = \frac {| \bar E|^2} {2 \zeta} \hat i_k $$
Now if I consider a cylindrical volume and apply the divergence theorem I get
$$ \int_{s_1} Re \,\, \bar S \,\,\hat i_n dS = - \frac {| \bar E|^2} {2 \zeta} A$$
$$ \int_{s_2} Re \,\, \bar S \,\,\hat i_n dS = \frac {| \bar E|^2} {2 \zeta} A$$
$$ \int_{s_l} Re \,\, \bar S \,\,\hat i_n dS = 0$$
where $s_1$ is the left face, $s_2$ is the right face and $s_l$ is the lateral face of the cylinder.
So I should have
$$ \int_{S} Re \,\, \bar S \,\,\hat i_n dS = - \frac {| \bar E|^2} {2 \zeta}A +\frac {| \bar E|^2} {2 \zeta} A = 0$$
Is it possible?
| Your calculation is right: it is telling you "what goes in, comes out again"!
The plane wave does indeed bear energy. The two nonzero parts of your calculation:
$$ P_1=\int_{s_1} Re \,\, \bar S \,\,\hat i_n dS = - \frac {| \bar E|^2} {2 \zeta} A$$
$$ P_2=\int_{s_2} Re \,\, \bar S \,\,\hat i_n dS = \frac {| \bar E|^2} {2 \zeta} A$$
are opposite in sign. What this is telling you is that whatever energy from the plane wave that goes into one end of the cylinder comes out again at the other.
It's analogous to a cylindrical hose with water going through it, with none being lost along the way (otherwise the hose would swell).
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How was the formula for kinetic energy found, and who found it? My questions mostly concern the history of physics. Who found the formula for kinetic energy
$E_k =\frac{1}{2}mv^{2}$
and how was this formula actually discovered? I've recently watched Leonard Susskind's lecture where he proves that if you define kinetic and potential energy in this way, then you can show that the total energy is conserved. But that makes me wonder how anyone came to define kinetic energy in that way.
My guess is that someone thought along the following lines:
Energy is conserved, in the sense that when you lift something up
you've done work, but when you let it go back down you're basically
back where you started. So it seems that my work and the work of
gravity just traded off.
But how do I make the concept mathematically rigorous? I suppose I need functions $U$ and $V$, so that the total energy is their sum $E=U+V$, and the time derivative is always zero, $\frac{dE}{dt}=0$.
But where do I go from here? How do I leap to either
*
*a) $U=\frac{1}{2}mv^{2}$
*b) $F=-\frac{dV}{dt}$?
It seems to me that if you could get to either (a) or (b), then the rest is just algebra, but I do not see how to get to either of these without being told by a physics professor.
| The author of the law of energy conservation was Hermann von Helmholtz (1821-94). See his classic 1847 paper "Über die Erhaltung der Kraft," translated into English as "On the Conservation of Force." (He called energy force.)
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How can metal objects become electrically charged, if current flow means that an equal number of electrons enter/leave the object? I am trying to answer to the question in the title. I am restricting my question to metal objects only.
Here is my logical reasoning:
*
*Current is the flow of charge over time.
*In a circuit (simple series circuit, let's say), the flow of current
is the same at every point in the circuit.
*Therefore, the same # of coulombs of charge is flowing at every
point in the circuit.
*Electrons are the "material" of the charge that is flowing.
*Therefore, equal flow of charge at every point in the circuit must
mean equal flow of electrons at every point in the circuit.
*Therefore, current can never cause a metal object to become
positively or negatively charged, because the net number of
electrons in the metal object will never change due to current flow. (!)
Of course, objects CAN become electrically charged, gaining or losing electrons. So something is wrong with my reasoning or my premises. I just don't know what it is. Where am I going wrong?
| If you have a complete circuit, every piece of metal will gain and lose the same number of electrons and will not have a net charge. If you connect two plates, one to each end of a battery, the battery will take charges from the plate connected to the positive terminal and send charges to the plate connected to the negative terminal until the voltage between the plates equals the voltage of the battery. At this point, no more current will flow, but each plate will have been given a net charge.
| {
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Direction of current in concentric cylinders Example 7.2 in David Griffiths E & M book (3rd edition) has a side view of 2 concentric cylinders, with smaller radius $a$ and larger radius $b$. The region in between $a$ and $b$ has conductivity $\sigma$. "If they are maintained at a potential difference $\textit{V}$, what current flows from one cylinder to the other for a given length $L$?"
The E field is pointing radially outward along $\textit{s}$. My question is: what direction is the current? Do electrons flow in the opposite direction of an E field? If so, does that mean the current is flowing radially inward, along $\textit{-s}$, from $b$ to $a$?
| The Lorentz force on a charge in an electromagnetic field is
$$F=q(E+v \times B) \ \ .$$
For an electron between the cylinders, $q$ is negative, and $E$ is defined as pointing outward, so the electron will experience a force radially inward. But due to the unfortunate sign convention used for currents, electrons flowing inward means that the conventional current is flowing radially outward.
| {
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Who does work while walking? While walking, the work done by friction is zero.
But who does the work, actually? How someone is getting displaced?
This situation also arises when someone climbs without slipping or is climbing a ladder.
| When you are walking you are doing work against gravity and friction.
Consider this - when you walk on a flat surface, you shift your body weight on to say right leg. Lift the left leg and move it by a step. For the next step, you shift the weight on the left leg, lift the right leg and move forward. Thus you move.
What is the work done?
The leg consists of foot and leg. This makes up approximately 6% of the total body weight. You have 2 legs. So one leg + foot weighs 3% of the total body weight.
Thus if a person weighs 50 kg, than the weight he shifts for each step will be 1.5 kg = 15 N
Thus when a person walks a straight distance of 1 km, he will be working approximately 15,000 J
Do not try to convert into calories. As the medical data of 'calories burnt' represent the heat generated by the body. It has nothing to do with the physical work done
I hope, I have provided an explanation...
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Deviation from 2D trajectory I need to find out how far a ball is from the predicted trajectory in 2 dimensional space and I know the start and end position of the ball in both dimensions. Along with that I know the initial velocity and the angle at which the ball was launched.
Every single variable is known in this picture.
The problem is that I can predict where the ball is going to land based on the angle α and the initial velocity v0, but the actual landing is a bit off due to variables assumed to be zero (wind, friction, etc.) and I need a mathematical way of calculating this deviation.
NOTE: The mathematical model may not include calculus as we haven't covered that part of our curriculum yet.
Any suggestions on how to do that?
| It is worth noting that the force due to air resistance is usually modeled as
$$\vec{F} \propto -\vec{v} \ \ \ \text{or} \ \ \ \vec{F} \propto -|v|^2 \hat{v} $$
Intuitively this makes sense: the faster you go, the more drag you should experience. The minus sign means that the force acts in opposition to the direction you are travelling.
Since the force is proportional to velocity, the force will change at each point in time. Therefore, it is impossible to model drag forces without using calculus. You would need to find the equations of motion and integrate.
That said, as others have mentioned you can just make an estimation of how much error will be introduced by:
*
*not knowing the initial conditions exactly
*ignoring non-conservative forces (friction, air resistance)
The best way to do this is to do the calculation and run the experiment and see how far you are off for a few different sets of initial conditions. That should give you a pretty good estimate of how much your model deviates from reality.
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Potential difference between point on surface and point on axis of uniformly charged cylinder Question:
Charge is uniformly distributed with charge density $ρ$ inside a very long cylinder of radius $R$.
Find the potential difference between the surface and the axis of the cylinder.
Express your answer in terms of the variables $ρ$, $R$, and appropriate constants.
$Attempt:$
I am struggling with determining which Gaussian surface to use. If I use a cylinder, then the cylinder would have an infinite area, right? How can I deal with that? If I use a sphere (since I am trying to find the potential difference between only two points, one on the surface and one on the axis), what will be the charge inside the sphere?
If I use a sphere as my Gaussian surface, I get:
$$\int \overrightarrow{E}.d\overrightarrow{A}=\frac{Q }{\epsilon _{0}}$$
$$\Delta V = -\int_{i}^{f}\overrightarrow{E}.d\overrightarrow{s}$$
$$E = \frac{\rho }{4\pi R^{2}\epsilon _{0}}$$
$$\Delta V = \frac{\rho }{4\pi R^{2}\epsilon _{0}} \int_{0}^{R}dR=\frac{\rho }{4\pi R\epsilon _{0}}$$
But this is wrong.
| Like Eternal Code said, using a cylinder inside the original problem cylinder is the right approach. If you use Gauss' Law, you should find that the electric field inside the infinitely long, uniformly charged cylinder is
$$E=\frac{ρr}{(2ε_0)} $$
Now, to calculate the potential difference between the surface and axis of the cylinder,
$${\Delta V}=-\int_0^R \frac{ρr}{(2ε_0)}dr$$
This gives the potential difference between the surface and axis of the cylinder as being
$${\Delta V}=\frac{-ρ(R^2)}{4ε_0}$$
| {
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Mass of small fluctuation around vacuum For a potential $V$, how do we define the mass of a small fluctuation around its vacuum? For example I have the potential
$$ V_\mathrm{eff}(\phi) = \frac{1}{2} \left(\frac{\rho}{M^2} - \mu^2\right) \phi^2 + \frac{1}{4} \lambda \phi^4. $$
What is the definition of the mass of small fluctuations around the symmetry-breaking vacuum and what is the physical meaning of it?
| The mass of a small fluctuation is usually defined as $$ \pm m^2= \frac{d^2V}{d\phi^2}\biggr|_\text{VEV}$$
The sign depends on your conventions. This makes sense in analogy with the canonical free field potential $$V_\text{free}=\pm \frac{1}{2}m^2\phi^2$$
for which the above formula is clearly right. More generally, we can expect any (reasonably smooth, [insert other obscure mathematical assumptions here]) potential to be well-approximated by a quadratic potential when it is close to an extremum, so we can define a mass in analogy with the harmonic oscillator - the free field is of course just that!
In your case, it yields, $$\pm m^2=\frac{\rho}{M^2} -\mu^2 +3\lambda \phi_\text{VEV}$$
| {
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Question about Metropolis Monte Carlo in the case of equal energies If configuration A is equal to configuration B in a Metropolis Monte Carlo method, do you still do the attempted update?
| Was initially posted as a comment. (Comment removed now)
The micro-states are changing nonetheless, a different point in the phase space of your system, so the system is evolving, even though the two states are part of the same macro-state. Finally remember that the Metropolis probability criterion for accepting moves is (in one of the simplest schemes of it): $$\frac{N(n)}{N(o)}\propto exp[−β[U(n)−U(o)]]$$
Where $N(o)$ and $N(n)$ are the probability densities of the old and new states respectively. Finally in the case of equal energies we have $\frac{N(n)}{N(o)}\propto 1$ and the attempted move is accepted.
Bear in mind that, another way to go about this, is by considering the fact that even when your system has been equilibrated, you still need it to evolve and explore further micro-states, as there may still be fluctuating elements in your system, like an interface between two coexisting phases.
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Is the Higgs Boson a Force Carrier? I am told there are four fundamental forces, and each of these forces has a boson that acts as its carrier.
Reading this http://www.fnal.gov/pub/science/inquiring/questions/higgs_boson.html I find that the Higgs Field is not a force field
The Higgs field is not considered a force. It cannot accelerate particles, it doesn't transfer energy. However, it interacts universally with all particles (except the massless ones), providing their masses.
So I can get this argument, but then the article goes on to claim
The Higgs particle is considered to be a carrier of a force. It is a boson, like the other force-transferring particles: photons, gluons, electroweak bosons. One may call the force mediated by the Higgs boson to be universal as the Higgs boson interacts with all kinds of massive particles, no matter whether they are quarks, leptons, or even massive bosons (the electroweak bosons). Only photons and gluons do not interact with the Higgs boson. Neutrinos, the lightest particles with almost zero mass, barely interact with a Higgs boson. Top quarks, which have about the mass of a Gold atom, have the strongest interaction with a Higgs boson.
So ...
*
*Is the Higgs Field a Force Field?
*Is the Higgs Boson a carrier of force?
*Does this mean there are actually 5 fundamental forces?
*Why is that elephant over there orange?
| The difference between the Higgs boson and the bosons of the three/four fundamental (depending whether you include gravity as a quantized theory or not) actions is that the latter are associated with gauge symmetries, while the Higgs plays a role in spontaneous symmetry breaking. Photons, W- and Z-bosons, gluons and gravitons arise from the requirement that the theory should be gauge invariant, while the Higgs field/boson does not.
In this sense, the Higgs field and the associated particle are not considered to form another fundamental force.
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Advantages/Disadvantages of "hanging off" a motorcycle when leaning The closest question I could find with regards to this subject was this one:
Countersteering a motorcycle
However, it does not address the specific physics of what I would like to know.
There are 3 ways to lean when turning a motorcycle:
*
*Upper body remains upright while the bike leans.
*Whole body remains aligned with bike.
*Most of the body "hangs off" the side leaning in.
I'm trying not to make any assumptions to allow for detailed and proper answers addressing issues I may not have considered; hopefully, without being too generic.
So to summarize, I would like to know whether the first 2 items are sufficient for all conditions or whether the 3rd has some physical properties necessary in certain conditions.
| All the force to accelerate or turn a motorcycle come from the wheel/ground interface. So to turn a bike quickly means a larger force on the wheel (from friction on the ground).
If the bike were upright when a strong sideways force were applied, it would quickly rotate the bike around its center of mass and dump the rider. Instead, the rider leans into the turn so that the sum of the friction and normal forces go through the center of mass and there is no torque to flip the bike.
The harder the turn, the more the center of mass must be displaced from the vertical. So the hardest turn possible would be limited either by the static friction of the tires, or by the amount the center of mass could be displaced. Hanging off allows for greater displacement.
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How can there be a voltage when there is no current? I'm told at school that the Electromotive Force (e.m.f) of a battery equals the potential difference between the terminals of the battery when there is no current.
How is that possible? How can there be a potential difference with no charge flowing?
| If you insert a dielectric in a circuit, you will not see any current but obviously there is a potential difference across the dielectric. To have a potential difference, you just need an electric field inside the material. This electric field might drive a current if the charges are mobile.
| {
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A Conceptual Problem With the Field Equations of General Relativity I have two questions:
*
*Suppose that we have an amount of energy in the form of a perfect fluid in the right hand side of Einstein field equations (energy momentum tensor), this will lead to a gravitational field, the gravitational field itself has energy, and this self energy also produces gravitational field ... in other words the gravity beget gravity !! ... due to this scenario we will have an infinite gravitational field!! ... what's wrong here?! is my reasoning wrong or is it the field equations that are not correct?
*Has this non-linear behavior of gravity(or maybe graviton!) anything to do with the fact that when we try to quantize gravity we encounter infinities?
| This is the famous back-reaction problem in perturbative gravity. To avoid it, we typically only work to a few orders in a perturbative series (though the PPN people will go farther than seems sane when doing numerical work, but you can't blame them considering that radiation only shows up at 2.5 PPN). It is unclear whether perturbative methods in general relativity converge.
What is clear, however, is that you can safely have exact solutions to general relativity where you solve this back-reaction problem nonperturbatively. In particular, there is an existant proof that the classical self-energy of a charged ball is finite, due to a cancellation of the infinite electromagnetic self-energy against the infinite gravitational self-energy.
| {
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A light so strong it has a shadow I have recently taken an interest in shadows. I know that in order for a shadow to exist that you must have a solid in the way of the light. My hypothesis is that there can be a light so strong, like a laser beam, that it acts like a solid in the sense where it doesn't let light pass through... is this plausible?
| Your hypothesis is basically correct, in theory, even in a vacuum.
Light consists of electromagnetic radiation. According to classical electrodynamics, electromagnetic fields in a vacuum are linear, which means that one light beam will pass right through another, completely unaffected. But according to quantum electrodynamics, at electric field intensities greater than the Schwinger limit, electromagnetic fields are expected to become nonlinear even in a vacuum, resulting in inelastic photon-photon scattering. That more or less means that a beam of light could in principle "block" another beam of light that crosses it.
However, the Schwinger limit at this point is purely a theoretical limit. Even the Extreme Light Infrastructure currently being developed, which will be at the frontier of producing super-intense laser light, will still produce light that's a couple orders of magnitude below the Schwinger limit.
| {
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If everything is relative to each other in this universe, why do we keep the Sun to be the reference point? and study the solar system and universe relative to it and why not relative to the Earth?
| A reference frame at rest with the Sun is, with a good approximation, an inertial system (much better than one at rest with our planet or other bodies in the Solar system, essentially in view of the hugely larger mass of the Sun). Physics in inertial reference frames has the simplest form. For instance the motion of planets around the Sun is described along ellipses with the Sun as one of the center, with a good approximation. The ultimate reason of this fact (assuming the Newtonian form of the gravitational law) is that I pointed out: If referring to another reference frame one has to include so-called, in a sense unphysical, inertial forces in addition to the gravitational one to explain the complicated motion of planets.
All this reasoning makes sense if disregarding cosmological issues where general relativity plays a crucial role, and instead adopting the Newtonian paradigm.
| {
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How to prove the Levi-Civita contraction? I want to prove the following relation
\begin{align}
\epsilon_{ijk}\epsilon^{pqk}
=
\delta_{i}^{p}\delta_{j}^{q}-\delta_{i}^{q}\delta_{j}^{p}
\end{align}
I tried expanding the sum
\begin{align}
\epsilon_{ijk}\epsilon^{pqk}
&=&
\epsilon_{ij1}\epsilon^{pq1}
+
\epsilon_{ij2}\epsilon^{pq2}
+
\epsilon_{ij3}\epsilon^{pq3}
\end{align}
I made the assumption that $\epsilon_{ij1} = \delta_{2i}\delta_{3j} - \delta_{3i}\delta_{2j}$, then I tried to argue the following using cyclical permutations
\begin{align}
\epsilon_{ijk}\epsilon^{pqk}
&=&
(\delta_{2i}\delta_{3j}-\delta_{3i}\delta_{2j})(\delta^{2p}\delta^{3q}-\delta^{3p}\delta^{2q})
\\&+&
(\delta_{3i}\delta_{1j}-\delta_{1i}\delta_{3j})(\delta^{1p}\delta^{3q}-\delta^{1p}\delta^{3q})
\\&+&
(\delta_{1i}\delta_{2j}-\delta_{2i}\delta_{1j})(\delta^{1p}\delta^{2q}-\delta^{2p}\delta^{1q})
\end{align}
and then I realized that this was getting long and messy and I lost my way.
How does one prove the Levi-Civita contraction?
| You need to use the fact that the only isotropic (invariant under rotations) tensors are $\epsilon_{ijk}$ and $\delta_{ij}$. Any other isotropic tensor must be expressible in terms of these two tensors. Clearly, $T_{ijlm}=\epsilon_{kij}\epsilon_{klm}$ is isotropic. It is a tensor that is (i) antisymmetric under the exchanges: $i\leftrightarrow j$, $l\leftrightarrow m$ and (ii) symmetric under the simultaneous exchange $(i,j)\leftrightarrow (l,m)$. This implies that
$$
T_{ijlm}= A\ (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})\ ,
$$
where $A$ is a constant. A simple computation, say $T_{1212}=1$, shows that $A=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the charge of an ion evenly distributed? This question relates to:
Gauss' law and ions?
Is the charge distribution in an ion spherically symmetric due to quantum mechanical effects or do we assume it when using Gauss's law, as in the linked question, to make the calculation easier?
I think it is the former but I am not sure.
| It will help you understand the quantum mechanical picture if you read up on atomic orbitals. These are the loci around the nucleus where the electrons have a probability to be found. You will see that the orbitals have a shape, which depends on the angular momentum of the state. The electrons carry the charge and thus you can interpret the plots as probability of the charge being there. They are not uniformly spherically symmetric except l=0. l=1 has two lobes and is elongated, and higher spins more lobes .
Now in the case of an ion, where an electron is missing: When close to an ion, as another atom will be, the shape of the positive charge will be the complement of the hole left by the missing electron, and it will not be uniformly spherically symmetric either, there will be lobes.
To assume that the single excess positive charge of the singly charged ion is at the center and be able to apply Gauss' law you have to be far enough that the details of the orbital shapes can be ignored. For milimmeter and micron distances this certainly will be true. For nanometer and smaller it becomes problematic and will depend on the specific problem one wants to address.
| {
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Riemann curvature tensor notation in Wald This question is entirely on tensorial notation in Wald's General Relativity. When specifying the properties of the Riemann tensor on pg39, he states:
$R_{[abc]}^{\quad \ \ \ d} = 0$
and
For the derivative operator $\nabla_a$ naturally associated with the metric, $\nabla_a g_{bc}=0$, we have $R_{abcd} = -R_{abdc}$.
and
The Bianchi identity holds: $\nabla_{[a}R_{bc]d}^{\quad \ \ e} = 0$
Questions:
*
*What do the square brackets around "abc" mean?
*Why does $R_{abc}^{\quad d}$ become $R_{abcd}$? What is the relation between the two?
*What does having $R$ in the square brackets mean?
Thank you.
| *
*The square brackets mean antisymmetrization. That is:
$$ X_{[a_1a_2\dots a_n]} = \frac{1}{n!}\sum_{P\in S(n)} \text{Sign}(P) X_{a_{P(1)}a_{P(2)}\dots a_{P(n}} $$
where $S(n)$ is the set of permutations of $n$ elements, and $\text{Sign}(P)$ is the sign of the permutation $P$, that is, $\text{Sign}(P)=-1$ if you need an odd number of element exchanges, and $\text{Sign}(P)=+1$ if you need an even number of element exchanges.
In particular,
*
*$R_{[abc]}{}^d = \frac{1}{6}\left(R_{abc}{}^d+R_{bca}{}^d+R_{cab}{}^d-R_{bac}{}^d-R_{acb}{}^d-R_{cba}{}^d\right)$
*$\nabla_{[a}R_{bc]d}{}^{e} = \frac{1}{6}\left(\nabla_{a}R_{bcd}{}^{e}+\nabla_{b}R_{cad}{}^{e}+\nabla_{c}R_{abd}{}^{e}-\nabla_{b}R_{ac
d}{}^{e}-\nabla_{a}R_{cbd}{}^{e}-\nabla_{c}R_{bad}{}^{e}\right)$
*You can "move" indices up and down using the metric tensor. That is,
$$R_{abcd} = g_{de}R_{abc}{}^e,\quad R_{abc}{}^d = g^{de}R_{abce}.$$
*The square brackets just affects the indices; the $R$ is inside because the antisymmetrization affects indices both from $\nabla$ and from $R$.
| {
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Converting two component product to four component notation Consider the product of two left Weyl spinors in the notation commonly found in supersymmetry,
\begin{equation}
\chi ^\alpha\eta_\alpha = \chi ^\alpha \epsilon _{ \alpha \beta } \eta ^\beta
\end{equation}
This is equal to,
\begin{equation}
\left( \begin{array}{c}
\chi ^\alpha \\
0
\end{array} \right) ^T\left( \begin{array}{cc}
\epsilon _{ \alpha \beta } & 0 \\
0 & \epsilon ^{ \dot{\alpha} \dot{\beta} }
\end{array} \right) \left( \begin{array}{c}
\eta ^\beta \\
0
\end{array} \right) = \bar{\eta} _L ^\ast \gamma _0 C \chi _L
\end{equation}
where I have used some common spinor identites and defined, $ \eta _L \equiv P _L \eta, \chi _L \equiv P _L \chi $ ($\eta $ and $ \chi$ are now four component spinors). I also use the defintion, $C \equiv i \gamma_0 \gamma _2 $. While I don't think anything is particularly wrong with this derivation, I have never seen a term like this in normal quantum field theory. It there a simpler way to reformulate this to correspond to common expression for such mass terms or is my uncomfort with this term due to my ignorance?
| Following this ref, one sees that, in some basis where the current is diagonal ($3.2.16$), then a term like $\chi \eta$ is just a part of the mass term ($3.2.17$).
| {
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Difference between heat capacity and entropy? Heat capacity $C$ of an object is the proportionality constant between the heat $Q$ that the object absorbs or loses & the resulting temperature change $\delta T$ of the object. Entropy change is the amount of energy dispersed reversibly at a specific temperature. But they have the same unit joule/kelvin like work & energy. My conscience is saying these two are different as one concerns with temperature change and other only at a specific temperature. I cannot figure out any differences. What are the differences between heat capacity and entropy?
| https://en.wikipedia.org/wiki/Standard_molar_entropy?wprov=sfti1
$$dQ = T \ dS \tag1$$
$$dQ = C \ dT \tag2$$
Interesting, right? In $(1)$, the whole $T$ multiplies the infinitesimal $\frac{\text{J}}{\text{K}}$. In $(2)$ it's the opposite: the whole $\frac{\text{J}}{\text{K}}$ multiplies the infinitesimal $T$.
But you hinted that you knew that yourself already. Let's cut to the chase: both are different beasts entirely, just like heat and torque are not related just because they carry the same unit (joules are newton-meters, right?).
However, if you still want a defining difference between them, other than "they're just different", I'd give you this:
Entropy by itself is not useful and cannot even be measured. What is useful are changes in entropy, or how it differs from one state to the other. In this sense, it's akin to internal energy and enthalpy, for which only relative values matter. Heat capacity, on the other hand, can have its absolute value determined experimentally, and it won't depend on a reference value like entropy does. Its absolute value is immediately useful, if you will. In this sense, it's akin to pressure and specific volume, for which absolute values make sense.
| {
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Is there any evidence that matter and antimatter continuously appear and disappear on the edge of a black hole? I heard Stephen Hawking got a Nobel prize for this, someone said there was no evidence for it which I find quite strange since he got an award for it.
| Direct experimental evidence of Hawking radiation is going to be exceedingly difficult to obtain. The radiation from stellar mass black holes is so small as to be undetectable, and we haven't (yet) worked out how to small black holes in the lab. At the moment there is no direct experimental evidence, and we have to accept we may not see any in our lifetimes.
However we can examine systems that are described by similar mathematics to see if they show analogous effects. A few years ago a team did an experiment of this type, and they claim to have seen the equivalent of Hawking radiation. There have also been experiments looking at acoustic analogues of black holes, though I don't think they have yet seen the analogue of Hawking radiation. These experiments don't prove a black hole radiates, but they show that the type of mathematics used to predict Hawking radiation from black holes does work when applied to other systems.
| {
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Difference between Veneziano amplitude and Virasoro shapiro amplitude I have been study about Veneziano amplitude and Virasoro Shapiro amplitude.
I want to summarize this two amplitude in the following way, please check that i am understand them properly.
Veneziano amplitude : they are open string amplitude for disk.
Usually we called four-open string tachyon on disk as Veneziano amplitude
Virasoro-Shapiro amplitude : they are closed string amplitude for sphere.
Usually we called four-closed string tachyons on shpere as Virasoro-Shapiro amplitude.
The similarity between Veneziano and Viraosoro-Shaprio amplitude are they both have Regge and hard scattering limit.
Is there are more difference and similarity for Veneziano and Virasoro-Shaprio amplitude?
| Well, there is the Kawai-Lewellen-Tye (KLT) relations, which says that a closed string amplitude is roughly speaking a product of two open string amplitudes. See e.g. Ref. 1.
References:
*
*Z. Bern, Perturbative Quantum Gravity and its Relation to Gauge Theory, Living Rev. Relativity 5 (2002) 5; Section 3.1.
| {
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How is force exerted on a wall equal to derivative of hamiltonian with respect to wall position? I'm trying to understand a solution of a problem in Landau, Lifshitz "Quantum mechanis. Non-relativistic theory" in $\S22$ "The potential well":
Determine the pressure exerted on the walls of a rectangular "potential box" by a particle inside it.
The first sentence in the solution makes me wonder how what it says is true:
The force on the wall perpendicular to the $x$-axis is the mean value of the derivative $-\partial H/\partial a$ of the Hamilton's function of the particle with respect to the length of the box in the direction of the $x$-axis.
Here $a$ is length of the box in $x$ direction. As I understand, this is a result from classical mechanics. But having re-read the chapters on Hamilton's function in "Mechanics" by Landau and Lifshitz, I still don't quite get how the force on the wall appears to be the derivative above.
So, the question is: how to derive this result?
| I Assume that the Hamiltonian only depends on $a$ through a potential term $V(a)$. This is the case more often than not. Then
$$\frac{\partial H}{\partial a}=\frac{\partial V}{\partial a}$$
Now, invoke Newton's second law:
$$ \vec{F}=-\frac{\partial V}{\partial \vec{x} }$$
and the result follows quite simply.
Alternatively, we can use the canonical (Hamiltonian) formalism. There, we of course have
$$\dot \pi_a=-\frac{\partial H}{\partial a}$$
where $\pi_a$ is the generalized momentum associated with the coordinate $a$. Of course, $\dot \pi_a$ is also a generalized force, so this is another way to find your result.
| {
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Hydrogen Bomb Mass to Energy? How much mass is converted to energy when a hydrogen bomb explodes? I remember an eighth grade chemistry class where, by going through the nuclear processes, my teacher estimated that roughly 2g of matter was converted in a fission bomb.This is a surprisingly small amount of mass! I have never seen the process involved in a fusion device.
| Of course all other replies do not consider the fact that in a hydrogen bomb only 30% of yield comes from fusion, ie. hydrogen. The rest comes from fission as in the known sequence:
fission (plutonium) -> fusion (hydrogen) -> fission (depleted uranium in the case)
So we have to correct all of the calculations by multiplying them by, say 30% and then we'll get the correct numbers. We have to remember that even though less hydrogen is used even more depleted uranium will be used.
| {
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Work done against gravity The work done against gravity is $mgh$, well at least that's what my textbook says.
I have a question:
I can apply a force say 50N, so total work done = $mgh + mah$. Where $ma$ = Force.
But the truth is irrespective of the force applied, the work done against gravity is always $mgh$. Why?
For example, when I move an object with a force, the work done is more, so work depends on the Force. But in case of gravity it always depends upon the weight
| You might want to change your question title to "Work done by gravity," because that is what is implied by the variables mgh. Of course, you can add a greater force than that of gravity, which would cause whatever object to which the force is applied to accelerate (since the forces are not in balance). No matter how much force you apply, the force of gravity (ignoring general relativity or changes in g by location or elevation) does not change, thus the work done BY gravity does not care about how much force you exert. Does that help?
| {
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Is there a classical analog to quantum mechanical tunneling? In comments to a Phys.SE question, it has been written:
'Tunneling' is perfectly real, even in classical physics. [...] For sufficiently large temperatures this can put the system above a hump in its potential energy.
and
the only difference between the classical case and the quantum mechanical one is that classical physics is a random walk in real time, while QM is a random walk in imaginary time.
I understand that in a system of particles with finite temperature some particles can overcome a potential barrier. That's how I interpret the first statement. I don't understand the business of "random walk in imaginary time". Can someone explain?
Update
What I was originally looking for was 1.) classical system that can transport mass through a forbidden region and 2.) explanation of "random walk in imaginary time". So far, I don't see anything for question 1.), but I think I'll grok 2.) if I invest some time and energy.
| Frustrated total internal reflection is an optical phenomenon. It's such a close analogue to quantum tunneling that I sometimes even explain it to people as "quantum tunneling for photons". But you can calculate everything about it using classical Maxwell's equations.
| {
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The purpose of auxiliary lens in microscope Can someone explain me the purpose of auxiliary lens in microscope.
The specification says:
Eyepiece: Extra wide field 10x Eyepiece w Spectacle Correction 30mm Ocular
Objective 0.7-45x , Auxiliary 2x
And the manufacturer claims that it is having magnification of 90X. Is it true ?
Perhaps they are multiplying the power of Objective, Eyepiece and Auxiliary to get 90X. But is correct to say the magnification is 90 X
BTW this is the microscope I am talking about and the specification is given at the end.
| Auxiliary lens normally serve to cover different zoom ranges.
for example here: http://www.2spi.com/catalog/ltmic/ZTX-3E-Microscope.shtml they say:
Objective lens:
Zoom type objective lens that has a range of 1x to 4x comes "standard"
with the Series 3 "package". We also have available optional auxiliary
lens that literally "screw on" to the existing standard lens to make
possible the zoom function covering different zoom ranges. Such
auxiliary lenses are available as 0.5x, 0.75x, 1.5x, and 2x. The
installation of a 2x auxiliary lens permits a zoom function of 2x to
8x. A 0.5x auxiliary lens, reduces the zoom function range to 0.5x to
2x.
| {
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Acceleration: A ball sliding down the inside of a semicircular bowl Edit: To clarify, this is a homework question. But I couldn't care less about getting the answer to this specific question as it's not even assigned yet. It looks interesting and I want to understand the concepts behind it. Not understanding these things bothers me :p
Here is the given question:
The face of block M in the figure below is shaped like a semicircular bowl of radius R. A mass m is placed at the top-left corner of the bowl and then let go. Find the acceleration of block M relative to the surface it is sitting on when m is a distance of 0.8R from the bottom of the bowl. There is no friction between M and m, or between M and the surface on which it sits.
There is a diagram provided:
I define $cos(\theta) = \sqrt{1-0.2^2}$ , as if the ball is 0.8R from the bottom it is 0.2R from the top, then using the Pythagorean Identity to solve for $cos(\theta)$.
It states the answer is $a_{block}=\frac{-mgcos(\theta)}{M}$
My answer was that $a_{block}=\frac{-mgsin(\theta)cos(\theta)}{M}$
I split the $F_g$ vector into its parallel and perpendicular components such that $F_{g\perp}=m_1gsin(\theta)$
The parallel component is irrelevant for this question.
$F_N=F_{g\perp}=m_1gsin(\theta)$
Then finding the horizontal component of $F_N$ is simple. Drawing a triangle it is apparent that:
$F_{Nx}=F_Ncos(\theta)$
$F_{Nx}=m_1gsin(\theta)cos(\theta)$
Since $a=\frac{F}{m}$, I get my equation: $a_{block}=\frac{m_1gsin(\theta)cos(\theta)}{M}$.
Could anyone enlighten me on what my mistake is? I think I'm running into a similar issue in another problem involving a block sliding down a wedge that is free to move as well (literally this exact problem, but instead of a semicircular surface it's simply a wedge).
| do it with conservation of energy
first of all find out the potential energyat the top of the semi circular bowl
keep it equal with the kinetic energy at lowest point
| {
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Reflection - reaction force direction Let's say an object hits a wall. When the object is reflected does the direction of the reaction force caused on the wall look like the red arrow? Does that direction depend on how "strong" object is reflected?
| Since you are saying an object i am considering it to be a rigid particle.Now,since the particle strikes the surface as in your figure.it gives a downward force on the surface and hence the reaction is obviously upward.
As you must be knowing normal reaction is perpendicular to the surface.So break the black arrow in componentsi.e you can imagine the ball to be coming rightwards along Y and towards the surface by X.For Y no reaction force and the reaction N is due to the component X and hence directed upwards.
Regarding your second question-magnitude of N depends on a number of factors-not only on the speed but also the rigidity of the wall,ball.What you can tell by the change of speed is the impulse given $\int \vec f dt=m(\vec v_2-\vec v_1)$.As you can see by the cannot in velocity,you cannot predict the strength ofthe reaction as time comes as a product which in turn depend on their material.
| {
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$\rm Lux$ and $W/m^2$ relationship? I am reading a bit about solar energy, and for my own curiosity, I would really like to know the insolation on my balcony. That could tell me how much a solar panel could produce.
Now, I don't have any equipment, but I do have a smartphone, and an app called Light Meter, which tells me the luminious flux per area in the unit lux.
Can I in some way calculate W/m2 from lux? E.g. the current value of 6000lux.
| Lux is a unit of "illuminance" and is based on the eye's response to light and each wavelength is weighted based on the percentage the eye is capable of perceiving. The curve is loosely a bell curve so at deep purple the eye may only "see" 5% of the W/m2 available while in the fat green part of the visible curve it may "see" 90%. So the ability to quantify lux, properly, means having a full spectral power distribution table of the incident light in W/m2/nm from about 390 to 810 nm, multiplying the value at each wavelength by the average human's eye response at that wavelength, and summing the results.
So lux as a measure of radiant energy only has value if you want to know how bright it will appear to a human, not any meaningful engineering measurement of spectral quality or heat energy.
| {
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Why does this condition ensure that the residue of the propagator is 1? The corrected propagator is given by $$\Delta'(q)=\frac{1}{q^2+m^2-\Pi^*(q^2)-i\epsilon}$$
($\Pi^*$ is the sum of all irreducible one-particle amplitudes) I get that the residue of the original propagator around the pole $q^2=-m^2$ is $$\frac{1}{2\pi i}\oint_{\text{around }q^2=-m^2} \frac{dq^2}{q^2+m^2-i\epsilon}=\lim_{q^2\rightarrow -m^2}\frac{q^2+m^2}{q^2+m^2}=1$$ and that the corrected propagator must have the same residue $$\frac{1}{2\pi i}\oint \Delta'(q)dq^2=1$$ So how does the condition $$\left[\frac{d\Pi^*(q^2)}{dq^2}\right]_{q^2=-m^2}=0$$ ensure the second integral above?
EDIT: Devouring complex analysis literature. Have already edited some things that weren't quite right. For anyone interested, I'm using Weinberg Vol 1 and this in section 10.3, ~p. 430.
| The original propagator has a pole at $q^2=-m^2$, the mass shell. For $m=\sqrt{-q^2}$ be the true mass of the particle, we have $\Pi^*(-m^2)=0$ and require that the residue of the modified propagator is unity around $q^2=-m^2$. Recall that for a meromorphic function $f(z)$ we have
$$\oint f(z)dz=2\pi i\sum_k\operatorname{Res}_{z_k}(f)$$
Thus $$\oint \Delta'(q^2)dq^2=2\pi i\operatorname{Res}_{-m^2}(\Delta')$$
Note that the pole in $\Delta'$ is simple. Thus we have
$$\operatorname{Res}_{-m^2}(\Delta')=\lim_{q^2\rightarrow-m^2}(q^2+m^2)\Delta'(q^2)=1$$
Inserting the definition of $\Delta'$, we get
$$\lim_{q^2\rightarrow-m^2}\frac{(q^2+m^2)}{q^2+m^2-\Pi^*-i\epsilon}=1$$
which is $0/0$ on the left, i.e. indeterminate. Using L'Hopital's rule, we find
$$\left.\frac{d\Pi^*}{dq^2}\right|_{-m^2}=0$$
as was to be shown.
| {
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What technology can result from such expensive experiment as undertaken in CERN? I wonder what technology can be obtained from such very expensive experiments/institutes as e.g. undertaken in CERN?
I understand that e.g. the discovery of the Higgs Boson confirms our understanding matter. However, what can result form this effort? Are there examples in history where such experiments directly or indirectly lead to corresponding(!) important new technology? Or is the progress that comes from developing and building such machines greater than those from the actual experimental results?
| In practice very little new technology results from experiments like those at CERN. While they are pushing the envelope on some things like the design of resonators, power klystrons and particle detector technology, the immediate technological return on those things is relatively small, even though one can argue that modern x-ray imaging (tomography) has profited highly from folks who did high energy and nuclear physics as students.
More importantly, though, facilities like CERN and the hundreds of associated institutions are teaching tens of thousands of students to think far outside the box. Very few of these students will stick with high energy physics in the long run (there are not that many paid jobs there). Most of them will go on to do other things, and they will use what they have learned about technology and management under difficult conditions at an extreme science/engineering project like LHC to push the envelope in whatever they will be doing in their lives. That is of enormous economic value to participating nations. If all of us would be happy with plain vanilla jobs, then all of us would still be stuck on the farm.
Apart from that, having been part of something like that is a constant source of pride in anybody's life who has been there. It's not something you forget the day you walk out the door. It gives you an idea of what humans can do, if they set their minds to something!
| {
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Tensor product notation convention? For two particle state, the Dirac ket is writren as
$$\lvert\textbf{r}_1\rangle \otimes \lvert\textbf{r}_2 \rangle. $$
Then how do we write its bra vector,
$$\langle\textbf{r}_1\rvert \otimes \langle\textbf{r}_2\rvert ~~\text{or}~~\langle\textbf{r}_2\rvert \otimes \langle\textbf{r}_1\rvert ~~\text{?} $$
Is there any rule or convention? I'm just asking the order of bra vector.
| It is a matter of definition of whether you want to revert the order of vector spaces on tensor products or not when going to the complex conjugate vector space, i.e. in physics jargon: from ket-spaces to bra spaces. Different authors use different conventions.
In particular, in the case of super vector spaces with Grassmann-odd elements, in order to minimize sign factors, different conventions are useful for different tasks.
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How many wavefunctions are in a minimal basis set for benzene? I am reading Modern Quantum Chemistry by Szabo and Ostlund and on page 62 he says "A minimal basis set for benzene consists of 72 spin orbitals." I tried to understand this number but failed.
Previously he illustrates his concept of a minimal basis set with the Hydrogen molecule $\text{H}_2$, where he starts off with one spacial orbital per atom. Then he combines both orbitals to obtain two properly symmetrized spacial wavefunctions and then multiplies with the two spin states (spin up, spin down) to arrive at a total of four spin orbitals.
In the case of benzene however things get more complicated. We have now 12 atoms that we need to consider and we need to construct properly symmetrized wavefunctions. It is clear that we have more options now, but how do we get to 72?
| You have one 1s atomic orbital for each H atom, and one 1s, 2s and three 2p for each C. This makes a total of 36 atomic orbital in the whole molecule, and so you have 72 spin states.
The 2s and 2p's orbitals are going to hibridate giving three 2sp$^2$ orbitals and one 2p orbital, so it has its characteristic $\pi$-delocalized electronic estructure on the ring.
| {
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Specular or Diffuse Reflection - Overhead Projection Question:
What type of reflection is exhibited as light waves from an overhead projector reflect off a white projection screen and are scattered throughout a room?
I said specular reflection, but my friend says diffuse because the light rays are expanding outward. What's correct?
| Your friend is correct.
It's not specular, because that would mean it's a mirror, and you'd only see the image if you were at the angle of reflection. Further, a mirror does not act as an image plane, so you might have difficulty ( :-( ) perceiving the image. Basically you'd need another lens to re-image the source.
The diffuse surface acts as an image plane. Each point in the plane radiates the incoming light in a Lambertian distribution. As such, it's essentially the same as a printed image, which as you know you can view from any angle by focussing on the image plane itself.
| {
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Why is the phase space a symplectic manifold rather than a manifold with a metric? Why does phase space require a symplectic geometry rather than a metric? Is there some scenario where a metric is unable to describe the notion of length in phase space, specifically in relation to the uncertainty principle?
| *
*A metric structure $g$ and
*a symplectic structure $\omega$
are two very different structures, although sometimes they can co-exist in a compatible way.
Unlike a symplectic structure, there are no Jacobi-like identity and no Darboux-like theorem for a metric structure.
There exists a unique torsionfree metric connection $\nabla$ on a pseudo-Riemannian manifold $(M,g)$, known as the Levi-Civita connection, while there exist infinitely many torsionfree symplectic connections $\nabla$ on a symplectic manifold $(M,\omega)$.
Unlike a pseudo-Riemannian manifold $(M,g)$, there are no scalar curvature, or local invariants on a symplectic manifold $(M,\omega)$.
The group of local symplectomorphisms on a symplectic manifold $(M,\omega)$ is infinite dimensional, while the group of local Killing symmetries on a pseudo-Riemannian manifold $(M,g)$ is typically finite dimensional.
A metric structure on spacetime is important for general relativity (GR), but irrelevant for the Hamiltonian formalism.$^1$ If we artificially assign a metric on phase space, it would typically not be preserved under symplectomorphisms.
On the other hand, note that in the Hamiltonian formalism, the cotangent bundle of the configuration space is born with a canonical symplectic structure, cf. e.g. this Phys.SE post.
--
$^1$ Note in particular that in the Hamiltonian formulation of GR, known as the ADM formulation, the pertinent phase space is different from spacetime itself, and also different from the configuration space of all possible spacetime metrics. The configuration space of all possible spacetime metrics is endowed with the DeWitt metric.
| {
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DC Generator with magnet as rotor DC generators convert the AC current in them by split ring commutators right and the graph of the current will be like this
but the question is how would be the graph if the magnet is the rotor and not the armature? Me and a part of my friends are on the opinion that the the graph will be like this and a few of my friends say that it will be like the usual
I would like to know how it would be
| When the magnet is rotating and the coils are stationary, you have an AC generator unless you play tricks with brushes. This is because the net flux change in the coil after a complete revolution is necessarily zero - and thus the net voltage must be zero too.
The sinusoidal waveform in your second sketch is therefore what you expect to see in a brushless generator with rotating magnet. You are right.
| {
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Why can I see my breath after I pop my ears? I have this weird but reproducible thing that I can see my breath for less than a second even in hot/humid weather. The key to do that is to pop my ears - I have to do that often, it's like I'm on a flight, not sure why my ear is that way... Anyway even if I just try to force more air into my mouth, keeping it flexed (not blowing it up like a balloon)... At the moment I open mouth and let the air out I can see it. I noticed it on a hot day in the car, against the darkness of the dashboard I could see the vapor. Also just about anywhere, anytime. It's barely visible though. My friend can see his breath during or after rain (!)... I think it has something to do with pressure and relative humidity - but then why does it work in very humid weather too, also in the bathroom? If I breath out normally it does not happen at all! I'd like to know the scientific explanation (and if it's normal or there is something wrong with me).
| The relative humidity of air is pressure dependent. Your method of popping your ears involves increasing the pressure of the air in your mouth. And if you sufficiently compress a volume of air that has a high relative humidity, you can increase the air's relative humidity beyond it's saturation limit, which causes the water vapor in the air to start to condense, producing a visible mist.
You can find videos and descriptions of this phenomenon by googling "mouth mist trick".
| {
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Why does Energy-Momentum have a special case? I was reading Energy-momentum, and I came across this simplified equation:
$$E^2 = (mc^2)^2 + (pc)^2$$
where $m$ is the mass and $p$ is momentum of the object. That said, the equation is pretty fundamental and nothing is wrong when looked upon, I similarly also believed this but I came across a "special" cases where this does not apply:
*
*If the body's speed $v$ is much less than $c$, then the equation reduces to $E = (mv^2/2) + mc^2$.
I find this really crazy, because first Einstein, always wanted to create a theory\equation that applied to every aspect of physics and has no "fudge" factors, that said irony is present from Einstein. Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given.
Most importantly, why does this not work, if velocity is "much" slower than light? What do they mean by "much slower", what is the boundary for "much slower"?
Regards,
| Special equations of the kind you mention are useful as they elucidate limiting behaviors.
Your example is no different from a statement like: the hypotenuse $c$ of a right triangle with legs $a$ and $b$ with $b \ll a$ is given by $c = a + \frac12 b^2/a$. The Pythagorean relation $c^2 = a^2 + b^2$ is valid for any right triangle, but the special case $b \ll a$ is useful to elucidate how much the hypotenuse extends beyond the larger leg in the limit of ultra-sharp right triangles.
Similarly, the equation $E = mc^2 + \frac12 mv^2$ describes how much larger is the total energy $E$ of a moving object compared to its rest energy $mc^2$, in the limit of the ratio of the object speed over the speed of light dropping to zero. This energy difference is known as the kinetic energy, and the limiting equation makes it clear that in the limit of slow movements, kinetic energies in Einstein's theory are consistent with those in Newton's theory.
| {
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What is zero impedance in AC circuit? If a capacitor is connected with an inductor, then because $$Z=\frac{1}{j\omega C}+j\omega L,$$ the Z may be zero. Does that mean when I apply a voltage, the current will be infinite large?
What's more, in transmission line theory, the characteristic impedance could be $\sqrt{L/C}$ when $R=0 \ and~ G = 0$. Why capacitors and inductors could generate real impedance?
|
Does that mean when I apply a voltage, the current will be infinite
large?
No, not even in the context of ideal circuit theory. It's a bit subtle since we're using phasor voltages and currents and that requires a couple of assumptions to hold in order to be valid.
When those assumptions don't hold, we have to see what the 'infinity' (division by zero) is trying to tell us.
It means that if a sinusoidal voltage of frequency $\omega = \frac{1}{\sqrt{LC}}$ is across the series LC combination, the amplitude of the (sinusoidal) current increases with time at a constant rate.
Which is to say that there is no AC steady state solution for the current. This is the correct interpretation of the 'infinity' when solving for the phasor current.
Recall that, in order to use the notion of impedance in this context, we assume the circuit
*
*is driven by sinusoidal sources with the same frequency $\omega$
*is in AC steady state which means the amplitudes of the voltage and
current sinusoids are constant with time
But, in this case, the 2nd assumption doesn't hold. If one solves the differential equation for the series LC circuit driven by a sinusoidal voltage source, one finds that when the source frequency is $\omega = \frac{1}{\sqrt{LC}}$, the current is of the form
$$i(t) \propto t\cos(\frac{t}{\sqrt{LC}} + \phi)$$
so the amplitude of the current goes to infinity as $t \rightarrow \infty$. But note that at no time is the current actually infinite even in this ideal circuit theory context.
Also note that there is no phasor representation for a current of this form.
Now, for a physical LC circuit, the amplitude will increase until some physical limit is reached, e.g., the capacitor breaks down, or the source internal impedance limits the current.
Why capacitors and inductors could generate real impedance?
For a transmission line, the characteristic impedance is
$$Z_O = \sqrt{\frac{R + j\omega L}{G + j\omega C}} $$
In the case that $R = G = 0$, we have the (square root of the) ratio of two imaginary numbers which produces a real number.
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How do photons "decide"? I was reading that when horizontally polarized light hits a vertical Polaroid all the light is blocked out. But when the Polaroid is off the vertical, some but not all photons "decide" to jump into the new plane of polarization. Could this be a "road less traveled" kind of effect?
If a run of two or three photons make the jump then conditions are affected in such a way, that the next photon is less likely to make the jump. Then as one or more photons get blocked, conditions cool down a bit increasing the likelihood that another run of jumps will occur: a mechanism of so called "deciding".
| There is no such mechanism. The probability for a photon to pass through a polarizer at an angle $\theta$ is $\cos^2(\theta)$, regardless of what has happened before, and regardless of how many photons "at once" try to pass through it. As Bell's theorem tells us, the quantum world is really random (or non-local).
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Complex Quantum Wave Can the complex nature of quantum wave arise from the fact that particle is represented as wave packet in spatial frequency and particle's total energy is represented as wave packet in time frequency?
Those wave packets are connected since $E=p^2/2m+V$, but since wave-like particle posses definite location and definite energy, they make sense. Now we have $\cos (\text{phase})+i\centerdot \sin (\text{phase})=e^{i \centerdot \text{phase}}$ and with proper summation two dimensional information could be emerged into one scalar valued function $\psi (x,t)$.
| No, this is not a valid explanation. Pardon my possible simplifications, but I understand the reasoning in the following way:
*
*$k$ bears independent information
*$\omega$ bears independent information
*We can "store" only one piece of information in the real line, so we need "two folders", which is provided by the two parts of a complex number.
This is simply not how the whole thing works. The argument is completely detached from what the wavefunction actually is and what it is used for. The "information content" of $\omega$ and $k$ is not generally decomposable into separate $A_1(k), A_2(\omega)$, and it certainly isn't divided up the way you suggest. The full $A(k,\omega)$ is not imposed but determined by the Schrödinger equation.
But to give a simple counterexample to show the invalidity of the argument by it's own means: Actual wavefunctions occur in 3D space, so we have $\vec{k}=(k_x,k_y,k_z)$. With $\omega$ this means "four information folders". So complex numbers are not enough. You could pass to quaternions. But we do not do that in quantum mechanics. Why? Because it is not actually needed, the premises of the argument are incorrect.
Physics sometimes just postulates objects and they work - the postulation cannot be fully explained. Nevertheless, I like the argument for the complex wave-function by Jerry Schirmer: We "need" the formalism to produce actually observed wave phenomena such as interference, but also "flat" and "non-wavy" probability distributions, which however have the potential of interfering.
| {
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What allows us to distinguish between two different voices even if they're singing the same note? I love Freddie Mercury. I love Robert plant. I once heard Robert Plant sing a Freddie Mercury song at a tribute concert. Plant sang the song beautifully, but it was obvious that the song was different from the original -- it was obvious that Freddie Mercury was not singing the song.
From what I understand, sound can be described as a sinusoidal wave. Now, based on the assumption that both singers can hit the exact same pitch with the exact same intensity, what other parameter can we modify in a sinusoidal wave that would allow us to distinguish between the two voices?
| No voice sings in a "pure tone", i.e., while the voice is in tune, the sound signal is composed of various harmonic frequencies. This gives you the "color" of the voice, and that makes the two voices distinct.
| {
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Are there other places to look for the gravitational wave, other than space? The recent BICEP2 announcements about the existence of gravity waves made the news, and made a big pitch for cosmic inflation, in the process they also claimed to detect the existence of gravitational waves. Have the results of the experiment been validated? Have they been confirmed by the European counterparts? If not would there be other ways to look for the gravitational wave- are there any lab experiments that could prove/disprove its existence?
| Since the original BICEP2 publication there has been increasing suspicions that what they had seen was actually just a signal from interstellar dust. We've been waiting for data from the Planck experiment on the dust signal, and that data has just been released. Sadly it looks as if BICEP2 did indeed just see dust and not signals from gravitational waves.
All is not yet lost. The BICEP team plan to improve their experiment and evidence for gravitational waves may still emerge. However it's currently looking as if they've just added another entry to the list of failed GW experiments.
There's a good review of terrestrial GW detection experiments in the answers to Is there a good chance that gravitational waves will be detected in the next years?.
| {
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Alternative liquid for Galileo thermometer So a friend of mine broke my Galileo thermometer recently. The glass tube and the liquid inside were lost, but the bulbs survived. I've cleaned out an old tall glass candle, and tried filling it with water. Even when the water is steaming hot the bulbs still float, so the liquid in there was definitely not water.
It smelled somewhat like gear oil, so I'm guessing it might have been an oil, but I'm trying to think of low density (clear) liquids that I could acquire to fill the tube with. Preferably water soluble, so I can calibrate it by adding water until it's accurate.
Suggestions?
(No, I am not buying a new one. I am an intelligent human being, I have been presented a challenge, and I will use science to overcome it—not mere money.)
| You would be wise to somehow determine the exact fluid used by the original manufacturer.
Consider that each of the floats has a fixed density, and has a temperature marked on its hanging tag. So you need a liquid which will have the correct, different density at each temperature marked on a tag. In short, the liquid you choose must match both the original liquid's density, and coefficient of volume expansion.
An alternate method would be to first settle on a liquid that has the correct density. You could adjust ethanol with water at $20^o$C until the appropriate float is suspended. Then slowly change the temperature, note when each of the remaining floats moves, and change the marking on the tag of that float...
Good luck...
| {
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Will 5 pizzas in the same Hot Bag stay warmer than 5 pizzas in 5 separate Hot Bags? For example, say I am delivering 5 pepperoni pizzas to 5 different addresses. In one scenario, I Keep all 5 in the same insulated Hot Bag, I carry that bag to the door, and I quickly remove one of the pizzas from the bag to give to the customer. In the other scenario, I use a separate Hot Bag for each pizza. This would mean that only one bag would need to be opened while the other 4 bags could stay closed.
Which method would keep the pizzas warmer?
| I don't think there will be any difference. Provided the bags are ideal insulators and the pizzas are at the same temperature, there will be no exchange of heat. Even with all of them in the same bag, there will be no exchange of heat as no temperature gradient exists!
| {
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What causes the random movement of particles inside a conductor? I'm reading about currents in electricity right now, and it was mentioned that even if there's no electric field inside a conductor, charged particles inside are still undergoing random movement.
I wanted to know what forces cause this random movement to occur? Or if it's not any force which causes this mysterious movement, then what is it?
Thanks.
| In the quantum mechanical description of a conductor all energy levels of the conductor are filled up to some specific energy level, called the Fermi level. This is because of the Pauli exclusion principle, which says that electrons with the same spin cannot occupy the same energy level and thus causes higher energy levels to be populated. Therefore, even at zero-temperature electrons have some amount of kinetic energy. This motion is, however, random and does not contribute to a net current. To get a current an electric field must be supplied.
Some more info:
http://hyperphysics.phy-astr.gsu.edu/hbase/solids/fermi.html
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Why Earth's atmosphere behaves like an ideal gas? In every book I found this sentence like an assumption, without explananions, somebody can help me understand it better?
| Look at the definition of ideal gas .
An ideal gas is a theoretical gas composed of many randomly moving point particles that do not interact except when they collide elastically. The ideal gas concept is useful because it obeys the ideal gas law, a simplified equation of state, and is amenable to analysis under statistical mechanics. One mole of an ideal gas has a volume of 22.7 L at STP (standard temperature and pressure).
At normal conditions such as standard temperature and pressure, most real gases behave qualitatively like an ideal gas. Many gases such as nitrogen, oxygen, hydrogen, noble gases, and some heavier gases like carbon dioxide can be treated like ideal gases within reasonable tolerances
From the list and knowing the composition of the atmosphere one can see that the simple classical model of non interacting elastically scattering molecules is good enough for most purposes. Even though there exist interactions and scatterings follow quantum mechanical laws, to first order the ideal gas is a good approximation.
It is an assumption, but a good approximation of what is going on really.
| {
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Where does the $\partial \vec{E}/\partial t$ term from Maxwell's equation go in Ampere's Law? One of Maxwell's Equations (ME) is:
$$\nabla\times\vec B = \mu_0\vec J+\epsilon_0\mu_0 \frac{\partial \vec E}{\partial t}.$$
While Ampere's Law (AL) is:
$$\nabla\times\vec B = \mu_0\vec J.$$
Griffiths E&M book derives that form of AL using the Biot-Savart Law and applying Stokes' theorem. Intuitively, it makes sense to me: a steady current is going to give rise to a magnetic field around it. But then I have trouble reconciling it with the ME I posted above -- while AL seems to say that for a given steady current $J$ you get a straightforward $B$, the above ME seems to say that for a given $J$ you could get many combinations of $B$ and $E$.
How is this reconciled?
| I believe that Ampere's Law is wrong in some situations. When Maxwell looked into it, he discovered that Ampere's Law is not always true, so he modified it to get the Ampere-Maxwell Law.
According to my understanding, this can be shown if you have a wire with a current flow that is charging a capacitor. If you calculate the magnetic field between the capacitor plates simply using Ampere's Law, you will get zero, however, using the Biot-Savart Law, there will be a magnetic field. The Ampere-Maxwell Law corrects this.
Note: this is my understanding, I've just been learning about electromagnetism, so I'm not an expert.
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What is the difference between diffraction and interference of light? I know these two phenomena but I want to know a little deep explanation. What type of fringes are obtained in these phenomena?
| Diffraction occurs when a wave encounters an obstacle or a slit these characteristic behaviors are exhibited when a wave encounters an obstacle or a slit that is comparable in size to its wavelength, whereas Interference is the phenomenon where waves meet each other and combine additively or substractively to form composite waves. In a sense there are similarities in the fact that both phenomena from a given wave produce other waves (with in general different frequency or phase and/or amplitude etc.). The main difference is the mechanism, diffraction involves a wave and some obstacle or object which deflects the wave or bends it and intereference involves a wave which combine with other waves. In physical experiments both these phenomena can happen and be part of the same overall phenomenon.
The (geometrical) type of fringes can be similar in some cases to both phenomena or different, i dont think there is a general pattern here (as one can check in wikipedia images).
| {
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Classical toy models of particles with intrinsic spin Related to my question here (spacetime torsion, the spin tensor, and intrinsic spin in einstein cartan theory), I'd like to be able to put test particles on a manifold with non-zero torsion and see how this affects the motion.
The action for a free particle is usually given as:
$$S_{free} = -m\int d\tau = - m\int \sqrt{\frac{\partial x^\mu(\lambda)}{\partial \lambda}\frac{\partial x^\nu(\lambda)}{\partial \lambda}g_{\mu\nu}(\lambda)}\ \ d\lambda$$
where $\tau$ is the world line length, and $\lambda$ is some parameter to describe the particle path $x^\mu(\lambda)$. I assume this is a scalar particle, since rotations will not affect its description.
*
*Is there a term I am leaving out if we consider non-zero torsion?
*What is the corresponding model for a free spinor particle? (I've seen classical spinor fields discussed, but never a particle)
*What about for higher spin?
*What about for arbitrary spin? (or even in classical models, are we limited to representations of the manifold tangent space?)
| Note that in classical systems, spin is not quantized but just a parameter, so the question of ''higher spin'' is not really meaningful. There remains only the question of statistics.
Lagrangian principles for classical Fermions were first discussed in:
J.L. Martin,
Generalized classical dynamics, and the ‘classical analogue’ of a Fermioscillator, Proc. R. Soc. Lond. A 251 (1959), 536-542.
| {
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$ε_0$ affects electric field intensity, but $μ_0$ doesn't affect magnetic field intensity? I'll be honest: this question is actually a homework problem. I've spent the past hour going through Google and several textbooks trying to answer the question "Why does $ϵ_0$ affect electric field intensity but $μ_0$ does not affect the magnetic field intensity?" I don't understand much about electromagnetism, but I haven't found an explanation for this. From what I've seen, $μ_0$ seems to be inherently related to the magnetic field strength in the same way that $ϵ_0$ is related to the electric field strength.
Can anyone help me by providing a very general answer or by pointing me to a good resource? I don't want or expect anyone to do my homework for me, but a nudge in the right direction would be much appreciated!
| The only thing I can see them going for is the fact that only two of $\epsilon_0$, $\mu_0$ and $c$ are independent, and typically, a modern view will fold $\epsilon_{0}$ into the definition of charge, and declare $c$ to be the fundamental constant used to transform space into time in special relativity, making $\mu_{0}$ a prediction of the theory.
I couldn't tell you whether this is what your book is going for, though.
| {
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What would happen if an accelerated particle collided with a person? What would happen if an accelerated particle (like they create in the LHC) hit a person standing in its path?
Would the person die? Would the particle rip a hole? Would the particle leave such a tiny wound that it would heal right away? Something else?
| While a single LHC particle wouldn't be doing much harm, being hit by the LHC beam would be certainly deadly and it would damage the machine badly. Any dense matter that comes into the LHC beam will instantly act as a beam dump. We have a very good idea about what happens in the LHC beam dump, see e.g. http://iopscience.iop.org/1367-2630/8/11/290/pdf/1367-2630_8_11_290.pdf for a few data points.
Unlike with previous generations of machines, where the beam was sometimes dumped into the air, the LHC beam is contained in a vacuum tube and nobody can cross the beam accidentally. There are, however, many areas around the machine which would cause serious radiation exposure, if people were admitted during operations. Safety is of enormous concern around these experiments.
| {
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Why does electron move in an elliptical path? According to Sommerfeld's atomic model, an electron moving around a central positively charged nucleus is influenced by the nuclear charge. As a result of which, the electron moves in an elliptical path with the nucleus situated at one of the foci. But how the path of electron becomes circular to elliptical? I've trouble understanding this.
| Considering the way matter waves are associated with all moving particles, it seems inconceivable to me that electrons cannot move in other than elliptical orbits.
Close examination of the harmonics & resonance effects of phase waves & matter waves, it becomes apparent that they need to move in elliptical orbits in order to harmonize the way they do. Group waves it should be remembered are made up of many thousands if not millions of phase waves, these too have to harmonize, unless electrons travel in elliptical orbits, they cannot do that.
| {
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Simple estimation of the critical temperature of water I'm trying to develop fermi estimation skills and I came up with a question for which I don't even know where to start from. Here goes:
Is it possible to estimate the critical temperature (say in Kelvin degrees) of water in a simple way using fermi estimation?
By critical temperature I mean the temperature of the point at the end of the coexistence line of water and vapour. See this plot.
| I guess you can use the Van der Waals equation and some estimates of the molecule volume and intermolecular attractive forces. The parameters of the critical point depend on these characteristics, so you need to assess them.
| {
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Why rubber is incompressible material? Why rubber is incompressible material? I know its Poisson's ratio is nearing to 0.5. So I don't understand physically, what it means by 0.5 Poisson's ratio and incompressibility. When I tried searching it, I found that rubber (or similar polymers) conserve volume after deformation and so they are incompressible. But same is the case with steel (Poisson's ratio around 0.3), it conserves volume after deformation.
So can someone explain this?
| Assuming a cube of side l with normal stress in one direction, its final volume after deformation would be:
$$V=(l+d1)*(l-d2)^2=l(1+d1/l)*l^2(1-d2/l)^2$$
$$ d1/l=\frac {\sigma}{E}=\epsilon $$
$$ d2/l=\frac {\nu \sigma}{E} $$
$$\nu =0.5$$ $$d2/l=\frac { \sigma}{2E}$$
$$V=l^3(1+\epsilon)(1-\epsilon /2)^2$$
$$V=l^3(1-\epsilon+\epsilon^2/4+\epsilon-\epsilon^2+\epsilon^3/4)$$
Ignoring the $\epsilon $ of higher order we obtain:
$$V=l^3(1-\epsilon+\epsilon)=l^3$$
$$\Delta V=l^3-l^3=0$$
So the volume does not change, thus it is incompressible.
| {
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Complex theory in physics I'm a physics graduate and usually encounter with complex numbers in physics.
For example, in electrical engineering,
Why do we express capacitive reactance as an imaginary number..?
Can't it be expressed as real, as we generally do?
Forgive me, if my question is logically incorrect, But I do want to know why "i" is involved..?
| For your engineering problems, the complex numbers are usually introduced to simplify the problem solving. Just take the complex reactance as an example. If you trace back to the place where you introduce the complex numbers, you will find that it is when you solve the "forced oscillation" equations$$\frac{d^{2}q}{dt^2}+2\beta\frac{dq}{dt}+\omega^2q=F_0cos\omega t$$Indeed, this equation can be solved purely using real numbers. However, it saves time by converting it into a complex equation, solving it in the complex field and then converting the solution back to the real number field (check any classical mechanics book for a detailed discussion. Don't read engineering physics textbooks if you want to understand the logic behind it.)
There are fields of physics in which complex numbers are widely "believed to be intrinsic" rather than just a problem-solving short-cut. Indeed, complex number field and real number field have different algebraic structures and sometimes phenomena are better described using the complex algebra. However, even this does not mean that complex numbers are the only choice. For example, one can work with a 2-dimensional real space equiped with a structure matrix $$\left( \begin{matrix} 0 & -1\\ 1 & 0 \end{matrix} \right)$$This space is, in some sense, equivalent to the complex field and both work well for describing the nature.
Physics can go even beyond complex numbers. For example, Grassmann numbers are used in describing spinor fields, accounting for the anti-commutativity. On the other hand, one is free to work with a set of matrices possessing the same anti-commutative property.
In my opinion, it is not the complex numbers themselves that are important; rather, it is the algebraic properties they possesses which makes them the right language to describe the nature. If you learn some serious algebra, you will have many substitutes to the complex numbers, but complex field is definitely easy to compute.
| {
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Origin of quark masses Does all the mass of the quarks in the standard model come from the Higgs sector or is there also a contribution to quark masses due to QCD chiral symmetry breaking?
| 99.9% of the mass of a hadron or a meson comes from confinement in QCD. Confinement is a special feature of QCD due to its non abelian symmetry which leads to a negative beta function. It is confinement that also leads to a breaking of the chiral symmetry at about 200 MeV or the radius of a hadron (about 1 femto meter).
| {
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Why is the Moon not redder at moonrise/moonset? Okay we all know about Raleigh Scattering, which makes the sky blue. And by the same token, sunsets appear red because sunlight traveling through more atmosphere will "lose more blueness" as it's scattered away.
But what about the Moon? The Moon is just reflected sunlight, so when the Moon is setting on the horizon, it should appear reddish right? But I've never seen that happen.
Now I know lunar eclipses are red, so I'm not discounting the principle of Raleigh Scattering or anything. But there seems to be something else at play causing the normal rising/setting Moon to not turn red.
| This is just an opinion, but the moon on the horizon is simply less visible than the sun is. I suspect that color changes it makes are more subtle and less easily noticed. However full moons are often noticeably orange. Here is a page with a wonderful time lapse view.
http://www.pikespeakphoto.com/moon-rising.html
| {
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Gibbs free energy and maximum work I'm getting confused between two important results.
The Gibbs free energy is $G = H-TS$
where $H$ is the enthalpy and $S$ is the entropy.
When the temperature and pressure are constant the change in the Gibbs energy represents maximum net work available from the given change in system .
But $dG = VdP-SdT$, so at constant temperature and pressure i'm getting $dG=0$. This is the criteria for phase equilibria.
I'm getting Gibbs free energy change at constant $T$ and $P$ as maximum work in one relation and zero in another. How are these compatible?
| More of a layman's answer because @higgss and @By_Symmetry answered with very good support.
I think a lot of the times that the 'work definition' of Gibbs Free Energy comes up during the discussion of hypothetical thermodynamic processes, "what if the volume decreased?", or "what if this chemical decomposed into these two chemicals?" so what happens to $G$?
Well, really what you're asking in those hypothetical scenarios is:
"What is $\Delta G$?" and yes, if $\Delta G < 0$ then the hypothetical process would happen. So the sign tells us whether it will happen, but the magnitude i.e. the actual value of $\Delta G$ is the maximum amount of work that can be extracted from the process if it is carried out reversibly.
Again, @higgss and @By_Symmetry gave pretty complementary and complete answers, but I thought that this might serve as a helpful, conceptual note in addition.
| {
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If light is electromagnetic then can light produce electricity or attract metals? what I mean to say is that if light is electromagnetic in nature then shouldn't it show electric or magnetic properties on matters? Like if light falls on a metal it should produce current due to its electric nature but it doesn't. Secondly due to its magnetic nature shouldn't it attract metal object or get deflected t words then or magnetite them due to its magnetic field?
| An example of a charged, magnetic particle that can be manipulated by light is the electron. With light it can be excited into a higher atomic or molecular orbital. When it falls back it emits light. Light or better the closely related microwaves can flip its magnetic moment in EPR.
| {
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What are the "other" Hadamard matrices? The Pauli matrices
$$ X = \begin{pmatrix}0&1\\1&0\end{pmatrix}, Y=\begin{pmatrix}0&-i\\i &0\end{pmatrix},\,\text{and}\, Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix} $$
can be used to construct the Hadamard gate
$$ H=\frac{1}{\sqrt{2}}(X+Z)=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}. $$
$H$ is Hermitian and two other Hermitian matrices arise when doing a similar computation with the $Y$ matrix:
$$ \frac{1}{\sqrt{2}}(X+Y)=\frac{1}{\sqrt{2}}\begin{pmatrix}0 & 1-i \\ 1+i & 0\end{pmatrix}\quad\text{and}\quad\frac{1}{\sqrt{2}}(Y+Z)\begin{pmatrix}1 & -i \\ i & -1\end{pmatrix} $$
Do these matrices have some names and if yes, do there exist known properties about them?
| Realize that an arbitrary rotation around the axis $\mathbf{n}$ is given by $R_\mathbf{n}=\cos(\alpha/2)I-i\sin(\alpha/2)\hat{\mathbf{n}}\cdot\mathbf{\sigma}$ and an arbitrary unitary operator can be written as $U=\exp{(i\gamma)}R_\mathbf{n}$ with $\gamma$ some phase factor. Thus, in general, any operations on the qubit can be seen as a rotation with some phase factor. Nevertheless, in the literature only the Hadamard, X,Y,Z and the phase shift gate are usually mentioned since these gates are conceptually the most practical to use on a single qubit.
| {
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How detectors in particle colliders can differentiate neutrons from antineutrons? Their mass is the same. None of them interacts with EM fields. And their decay (around 1000s) is far too slow to see their decay products yet in the detector.
How is it then possible to differentiate them?
| The experiment which has measured the most stringent limit on neutron to anti-neutron oscillations (i.e. produce neutrons, let them fly for some time and then look if you find anti-neutrons) has used a 130 micrometer thick and 110 cm diameter carbon foil. This target had a probability greater than 99% for anti-neutrons to interact (and thus produce secondary particles) and a high transparency for neutrons.
This target was surrounded by a tracking detector (for measuring the momentum of charged particles), scintillator counters and a calorimeter (for measuring the energy of charged and neutral particles). These are used to look for the products of an anti-neutron annihilating with the protons or neutrons of the carbon nuclei (while the neutrons would mostly only scatter elastically, i.e. leave the foil without breaking up).
Admittedly, the neutrons used in this experiment have relatively low energy, 0.002 electronvolt on average, corresponding to a velocity of 600 meters per second.
| {
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Moment of inertia of rods Ok so I'm extremely comfortable with calculating moment of inertia of continuous bodies but how do we do it for a system not continuous.
For example if 3 rods of mass $m$ and length $l$ are joined together to form an equilateral triangle what will be the moment of inertia about an axis passing through its centre of mass perpendicular to the plane.
i know that moment of inertia of each rod is $ml^2/12$ and c.o.m is at centroid?
also if 2 rods form a cross then to calculate the moment of inertia about its point of intersection would it be correct to sum up the individual moment of inertia of the rods form??
| The moment of inertia for a system of $n$ point masses, $m_i$, at distances $r_i$ from the pivot is simply:
$$ I = \sum m_i r_i^2 \tag{1} $$
We normally calculate $I$ by integration, i.e. we take each point mass to be an infinitesimal element of our continuous object and integrate to add up the moments of inertia of all those elements.
In your case let's call the three rods $A$, $B$ and $C$, then our initial equation (1) can be written as:
$$ I = \sum m_{Ai} r_{Ai}^2 + \sum m_{Bi} r_{Bi}^2 + \sum m_{Ci} r_{Ci}^2 $$
where all we've done is divide up our sum into the infinitesimal parts that belong to the three masses. But from equation (1) we know that $I_A = \sum m_{Ai} r_{Ai}^2$, and likewise for $B$ and $C$, so the total moment of inertia is just:
$$ I = I_A + I_B + I_C $$
So just calculate the separate moments of inertia for all the objects in your system then add them together. In your particular case the objects are identical so the total is just the moment of inetria of a single rod multiplied by three.
| {
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Are critical exponents below and above the critical point always same? The scaling relations don't distinguish the the critical exponents below and above the critical value. In the mean field level, I understand these critical exponents are same whatever one approaches the phase transition from the order phase or disorder phase. However, beyond the mean field treatment, are they always same?
Are there examples where they are different below and above the critical value? Or are there some theoretical arguments that the must be same? Can anyone give me some hints or good reference?
| For what it's worth, it is claimed that the critical exponents differ above and below the critical point for some exactly solvable 2-dimensional model: http://www.ujp.bitp.kiev.ua/files/journals/49/11/491114p.pdf (Ukr. J. Phys., v.49, #11, p.1122 (2004)).
| {
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Is not force required to start each and every motion,uniform or non-uniform however they might be?
A body which is at rest will remain at rest; and a body which is in motion will continue motion in a straight line as long as no unbalanced force acts on it.
This is Newton's 1st law of motion,everyone knows. But what Newton did not tell is that how the motion of the body starts. In the real world,in order to move a body,force must be given so as to overcome the frictional force. Suppose, between a body and surface,the limiting friction is $f_s$; so we have to apply a force $F$ such that $$F > f_s$$ . The extra force $ F - f_s$ does start the motion. In a frictionless world, how does a body start motion??? Obviously,to move it from rest, again some force must be given,however small it might be,right? If it is so, then there is practically no uniform motion because initially the body starts motion under the application of force,however small time it takes. Newton only told what happened after the motion starts but did not mention how the motion began. So , isn't force required to start each and every motion? And hence is there any uniform motion practically? Plz help.
[ I don't support any Aristotleian philosophy!]
| I think that a force will be definitely required to accelerate a body from rest to a velocity, however, once it has achieved that velocity you should be able to cease applying the force and it will continue that at that velocity in a frictionless ideal world. After you cease applying the force it should be undergoing uniform motion, but before that it will not due to the application of the force, as I understand it.
NB: I am a plebeian beginner when it comes to physics, so I may be entirely wrong.
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Do transmitters create magnetic fields and radiation? In the company I work for (I am a software engineer) we develop a system which uses a transmitter (antenna) that creates a magnetic field. I also know that radio station transmitters create radio waves, so I am somewhat confused.
Do the coils built into the transmitter create an EM radiation when a current passes through them or do they create a magnetic field? Maybe both?
I have very limited knowledge of physics, so I would appreciate an intuitive rather than a formal type of answer.
| Every change in a magnetic field automatically creates an electric field and vice versa. For technical purposes, however, magnetic antennas create a stronger magnetic near field (i.e the field that can be measured less than a wavelength away from the antenna), while "electrical" antennas create a stronger electric near field. So depending on application, engineers will favor one type over the other.
Independently of the type of antenna, magnetic or electric, the far field (i.e. more than a couple of wavelengths away from the antenna), the energy density in both the magnetic and the electric field components will be the same.
Having said that, some or many of the coils in your transmitter may not be meant to be antennas to begin with. In modern power transmitters they are more likely power combiners and matching elements.
| {
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Is there a difference between the speed of light and that of a photon? As in the title I am curious whether there is a difference between the speed of photon and the speed of light, and if there is what is the cause of such a difference?
| Comment to the question (v4):
It seems relevant to mention that there in principle could be a difference between the universal speed limit constant $c$ (which is usually casually referred to as the speed of light in vacuum), and the actual speed of light in vacuum if the photon has a rest mass, at least from an experimental point of view. Of course, no non-zero rest mass of the photon has been detected so far. See also this Phys.SE post and links therein.
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How the polarization of electromagnetic wave is determined? What help us determine the polarization of electromagnetic wave . Does perpendicular electric and magnetic field determine it or does the direction of propagation ?
| You can uniquely define the polarisation of a plane wave from any of the following:
*
*The electric field vector as a function of time $\vec{E}(t)$ and the magnetic field (or induction) $\vec{H}(t)$ (or $\vec{B}(t)$;
*The wavevector $\vec{k}$ and two scalar functions of time, the latter being the transverse components (in the plane at right angles to $\vec{k}$) of either the electric or magnetic field (or magnetic induction);
In the case of a nearly monochromatic wave, the vector functions of time in 1. and the two scalars in 2. can be reduced to complex scalars, which define the amplitude and phase of sinusoidally varying quantities. The alternative 2. together with an implicit knowledge of the wavevector is what we are using when we represent a write down a pure polarisation state as a $2\times 1$ vector $\psi$ of complex scalars, called the Jones vector. The scalars define magnitude and phase of the two transverse components of $\vec{E}$ (or $\vec{H}$, $\vec{B}$, as appropriate).
If only the relative phase of the two complex scalars is important, we can represent a pure polarisation state by an implicit definition of the wavevector and three real scalars: the Stokes parameters $s_j = \psi^\dagger \sigma_j \psi$, where the $\sigma_j$ are the Pauli spin matrices.
A partially polarised state is most readily thought of in quantum terms: we consider a general partially-polarised state to be a classical probabilistic mixture of pure polarisation states, defined by the $2\times2$, Hermitian complex density matrix (as well as an implicit definition of the wavevector direction). An equivalent characteristation is through the Mueller calculus, as discussed in my answer here. The classical description, in terms of random processes, is much fiddlier, messier and subtler than the quantum, and takes a full chapter in Born and Wolf, Principles of Optics" to describe (Emil Wolf was one of the pioneers in the rigorous description of partially polarised and partially coherent light).
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Derivation of temperature-volume relationship for a reversible adiabatic expansion of an ideal gas We start with $\delta q = 0$ and $dU = C_{V}(T)dT = \delta w$. Why can we take the heat capacity at constant volume, when this process is an expansion so volume increases?
| This is valid for ideal gas whose molar number is constant $n$. Why?
When a fluid changes volume, the equation
$$
dU =dQ - pd V
$$
is obeyed. Formally dividing by $dT$ we obtain
$$
\frac{dU}{d T} = \frac{d Q}{dT} - p\frac{d V}{d T}.
$$
If we now consider only processes where $V$ remains constant, the relation
$$
\frac{dU}{d T}\bigg|_{V=const.} = \frac{d Q}{d T}\bigg|_{V=const.}
$$
is obeyed.
Because of the right-hand side, we call this quantity heat capacity at constant volume and denote it $C_V(T,V)$. Generally it is a function of both $T$ and $V$ and is sufficient to express the change in the internal energy in the above way only when $V=const.$
However, for ideal gas internal energy is a function of $T$ only - let me denote it $U^{(id)}(T)$. This means that the condition on the left-hand side is superfluous - its presence does not matter. The value of the derivative can be calculated without any condition on the process:
$$
\frac{dU(T,V)}{d T}\bigg|_{V=const.} = \frac{dU^{(id)}(T)}{d T}.
$$
Thus the heat capacity at constant volume can be calculated as ratio of delta $U$ to delta $T$, process being immaterial. Because of the right-hand side, we know the result is a function of $T$ only - let me denote this function $C_V^{(id)}(T)$.
Now any change in the internal energy of ideal gas, irrespective of whether any other quantity is constant or not, can be written as
$$
dU^{(id)}(T) = \frac{dU^{(id)}}{dT}(T)d T.
$$
From the previous equation it follows that for any process,
$$
dU^{(id)}(T) = C_V^{(id)}(T)dT.
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/141130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Eigenvalues being physical observables I think I'm comfortable with the PDE solutions to the Schrodinger equation. But as soon as we start putting these values in a matrix (in dirac notation), I lose my understanding and everything becomes plug-and-chug math magic.
I'm wondering if anyone has an understanding as to why eigenvalues of the eigenstates of a matrix correspond to physical observables. That is, how can we show, using wave mechanics, that the eigenvalues to our eigenfunctions in our PDE can correspond to eigenvalues to our eigenvectors in a matrix? And can we use this understanding to get a better idea of what's going on when we look at the eigenvalues of eigenvectors of a rotation of our matrix?
| A measurement is not a primitive in physics. Rather, a measurement is a physical process that takes place according to the same laws of physics as any other physical process. Those same laws apply to the measurement apparatus, to the person doing the measurement and to the records he makes of the measurement. What distinguishes a measurement from any other kind of physical process? One property that is necessary for a particular process to count as a measurement is that it is possible to copy the result. That is, it has to be possible for the result to be present in one system before being copied and in more than one system afterward.
So what sort of operators represent results that can be copied in this way? According to quantum mechanics, systems evolve unitarily. Any unitary operator can be written in the form:
$$
U = \Sigma_a e^{i\phi_a}|a\rangle\langle a|,
$$
where the $|a\rangle$ form an orthonormal set. To copy a result this operator would have to leave the operator being copied unchanged and the only operators it leaves unchanged are normal operators. The normal operators that represent whether something happens or not are projectors, so the values attached to those projectors, eigenvalues, represent possible measurement outcomes.
For a more detailed discussion see
http://arxiv.org/abs/quant-ph/0703160.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to make a rotating linearly polarized (not circular polarization) beam from a single beam? One way to make a linearly polarized beam rotating at frequency $\Delta f\approx10\mbox{MHz}$ is by combining two circularly polarized beams, one left-handed and one right-handed, and where one beam is at a frequency $f$ and the other at $f\pm\Delta f$. Is there another way to do this using only a single beam passed through some active optical device (eg electro-optic-modulator)? I'm interested in the mid-infrared (10.6um).
| Faraday Effect. The plane of polarization in a medium is rotated when exposed to a magnetic field. Solenoid, glass rod.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/141322",
"timestamp": "2023-03-29T00:00:00",
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Normalization of a wavefunction that's superposition of two unknown energy eigenfunctions Question:$$\psi(x)=A(3u_1(x)+4u_2(x))$$where $u_1(x)$ and $u_2(x)$ are energy eigenfunctions.
How to normalize function $\psi(x)$?
My intuitive solution:
I got $$\int^{\infty}_{-\infty}|\psi(x)|^2dx=A^2\int^{\infty}_{-\infty}(9u_1^2(x)+16u_2^2(x)+12u_1(x)^{*}u_2(x)+12u_2(x)^{*}u_1(x))dx$$
Since $u_1(x)$ and $u_2(x)$ are eigenfunctions, the total area under each of them squared must be 1, thus: $$\int^{\infty}_{-\infty}|\psi(x)|^2dx=A^2(9+16+\int^{\infty}_{-\infty}12u_1(x)^{*}u_2(x)dx+\int^{\infty}_{-\infty}12u_2(x)^{*}u_1(x)dx)$$
Since in the remaining integral, $u_1(x)^{*}u_2(x)$ and $u_2(x)^{*}u_1(x)$ are the definition of the inner products of $u_1(x)$ and $u_2(x)$, which, in quantum mechanics, is the square-root of probabilities of getting $u_1(x)$ or $u_2(x)$ knowing that the initial state function is the other one, inside the integral it's just constants, which will make the integral become infinite. If so, the area under the absolute squared wavefunction will be infinite regardless of the value of A.
I am kind of stuck at the infinite integrals. Could someone please give me a hint?
| We are told that $\left|1\right>$ and $\left|2\right>$ are energy eigenstates, meaning they are eigenstates of the Hamiltonian $\hat{H}$, a Hermitian operator: ${\hat H}^\dagger = H$. Eigenstates of a Hermitian operator with different eigenvalues are orthogonal (see this): $\left<m|n\right> = \delta_{mn}$ (using Kronecker delta), assuming the eigenstates are normalized. So, normalizing the state $\left|\psi\right>$ gives
$$
\begin{eqnarray}
1 = \left<\psi | \psi \right> &=& \left[A(3 \left|1\right> + 4 \left|2\right>)\right]^\dagger A(3 \left|1\right> + 4 \left|2\right>) \\
&=& \left|A\right|^2 (3 \left<1\right| + 4 \left<2\right|) (3 \left|1\right> + 4 \left|2\right>) \\
&=& \left|A\right|^2 (3^2 \left<1|1\right> + 3*4 \left<1|2\right> + 4*3 \left<2|1\right> + 4^2 \left<2|2\right> ) \\
&=& 25 \left|A\right|^2
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/141413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What happens to waves when they hit smaller apertures than their wavelenghts? I was wondering this for quite a long time now. Let's say you have a water wave (like ripples, not the ones you see during tsunamis) with wavelength 10 m. Imagine you put a boundary with an opening of 1 m. Will diffraction be observed? According to my research, no. But, then, what would be seen?
| http://www.nature.com/nature/journal/v445/n7123/box/nature05350_BX1.html
You can find a complete answer in above link.The theory which is introduced can be valid for mechanical waves such acoustic and water waves.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/141556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
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Baryon in terms of quark fields – spinor index structure What is the most general way to write down a current describing a baryon made from quarks $\psi_i^\alpha$? Let's say we suppress flavour indices but want to write colour $(i,j)$ as well as spinor $(\alpha,\beta)$ indices. Then I suppose the colour structure would be something like
$$
\psi_i \psi_j \psi_k \varepsilon^{ijk}
$$
in order to be antisymmetric. How can one explicitly add spinor indices to this expression and how would one then decide if this describes a baryon of spin 1/2 or 3/2?
Edit To be more clear, I stumbled upon this reading Witten's Baryons in the 1/N expansion. There he investigates a 2d model of $SU(N)$ QCD in section 9. In equation (39), page 109, he introduces a current "with the quantum numbers to create a baryon":
$$
J(x) = \psi_1(x) \psi_2(x) \cdots \psi_N(x)
$$
He then goes on to say
"Since we are not keeping track of Dirac indices, the Dirac indices have not been written in (39). If one wishes to keep track of Dirac indices, one should choose in (39) the same Dirac component for each of the $N$ quark fields. For instance, one may consider the positive chirality component of each quark field."
I do not understand why this choice is useful and if it is only true due to peculiarities of spinor representation in two dimensions. I thought I would understand if I saw the explicit structure as far as SU(3) in 4d is concerned.
| I found an explicit form for a baryon current in a paper by B.L. Ioffe. I still have problems understanding a certain aspect, which I thought warranted a new question here: Charge conjugation matrix in baryon current
Anyways, to be complete, a current describing the isobar $\Delta^{++}$ can be found in equation (13) there and has the structure
$$\eta_\mu (x) = \left(u^i(x) C\gamma_\mu u^j(x)\right) u^k(x) \varepsilon^{ijk}.$$
A possible current for a proton can be seen in equation (46):
$$\eta(x) =\left(u^i(x) C \gamma_\mu u^j (x) \right) \gamma_5\gamma_\mu d^k (x) \varepsilon^{ijk}.$$
So whereas a spin 1/2 baryon only carries one spinor index, a spin 3/2 one has an additional Lorentz index.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/141647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Power loss due to dipole antenna position mismatch If we have two dipole antennas, it is well known that in order to transfer maximal power, two dipole antennas should be parallel and on the same height, which means that line that connects their middle points is perpendicular to both antennas. I wonder now, if two dipoles are parallel but not on the same height, how is received power calculated?
| You'd need to consider the angle between the two antennas and the distance between them. Dipole antennas do not radiate uniformly into $4\pi$. If you're looking up or down at the poles of the antenna, you will see no radiation (ideally). Looking at a direction transverse to this, the radiation is at a maximum. The angular dependence of the far (electric) field is given by:
$$E_\theta \propto \frac{sin~\theta}{r \lambda}$$
the power radiated per unit surface area is just:
$$P/A \propto E_\theta^2 \propto \frac{sin^2~\theta}{r^2 \lambda^2}$$
These equations just come from applying Maxwell's equations to an alternating current in the dipole antenna geometry. So you can see that the power per unit area at the receiver depends on the relative distance ($r$) and angle ($\theta$) between the transmitting and receiving antennas. Here, $\theta = 0$ and $\theta = \pi$ correspond to the poles of the antenna. This should be enough to get you a relative power loss given an offset in height if you already know the power radiated in the parallel case.
The diagram below illustrates the angles, height offset and the distance r:
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/141805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Kinetic energy vs. momentum? As simple as this question might seem, I failed to intuitively answer it.
Let's assume there is a $10,000$ $kg$ truck moving at $1$ $m/s$, so its momentum and KE are: $p=10,000$ $kg.m/s$ and $KE=5,000$ $J$.
Now if we want to stop this truck, we can simply send another truck with the same mass and velocity in the opposite direction to collide with the first one, and both of them will stop because of the conservation of momentum.
But what if we want to stop the truck by doing work in the opposite direction of motion ?
Let's assume there is a rope with an end tied to the back of the the truck, and the other end is tied to a $400$ $kg$ motorcycle moving at $5$ $m/s$, then its $p=2,000$ $kg.m/s$ and $KE=5,000$ $J$.
Now we have a truck moving in one direction with the same kinetic energy as the motorcycle which is moving in the opposite direction, but the truck has more momentum. So will the truck stop by the energy (or work) of the motorcycle ?
If yes, then how is the momentum conserved, and if no, then where does the energy of the motorcycle go?
Ignore any friction forces.
| Regarding scenario 1:
we can simply send another truck with the same mass and velocity in
the opposite direction to collide with the first one, and both of them
will stop because of the conservation of momentum.
If one assumes a totally inelastic collision then, by conservation of momentum, it is true that both trucks (objects) stop.
This requires that the initial kinetic energy of the objects is entirely converted to another form. The total change in kinetic energy equals the total work that is done during the collision process.
So, to be sure, work is done during the collision process since the trucks (objects) are, shall we say, permanently deformed.
I assume you know this already and have taken this into account.
Regarding scenario 2:
If we assume that the rope can stretch, dissipating energy in the process, then we have the following results:
(1) eventually, the truck and motorcycle have the same velocity which is equal to
$$v_f = \frac{(10,000 - 2,000)\mathrm{\frac{kg\cdot m}{s}}}{10,400 \mathrm{kg}} = 0.769\mathrm{\frac{m}{s}}$$
(2) the final kinetic energy is
$$KE_f = \frac{1}{2}10,400 \; \mathrm{kg}\cdot \left(0.769\mathrm{\frac{m}{s}}\right)^2 = 3,077 \mathrm{J}$$
so the work done (energy dissipated) by the rope is
$$W = \Delta KE = 10,000 - 3,077 = 6,923 J $$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/141891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
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Why the EMF of a battery doesn't depend on distance between the two electrodes? I have read that resistance of a conductor is directly proportional to its length. So, the EMF of a conductor (electrolyte of battery) should increase with increase in length/distance between the electrodes..is it not so?
| If you are talking about the open circuit voltage of the battery, then internal resistance and resistance of any conductors connected to the battery are irrelevant. The voltage across a resistor is proportional to the current thru it, see Ohm's law. When there is no current, there is no voltage drop across a resistor, so both ends are at the same voltage.
Battery voltage does go down with current because batteries have internal resistance. The resistance of the wire to the battery is usually very small compared to the internal resistance of the battery, so except for very high current applications (like starting a car), the resistance of the wires can be ignored.
Bigger batteries usually have less internal resistance (more stuff in parallel), and are therefore capable of higher current. Internal resistance does go up with distance between electrodes, you you can't confuse that with distance between the external contacts of the battery. Round single cells, like the common AA for example, a electrodes and electrolyte rolled up. A D cell has less internal resistance that a AA cell of the same type because the overall flattened cell area is larger. The distance between electrodes is still the same, there is just more area all rolled up.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/141925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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