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How does one exert greater force on the ground by jumping? When one jumps, how does he/she manage to exert greater force on their ground than their weight?
Also, what is normal force and the reaction force (are they the same thing?) and by newton's third law, shouldn't the reaction(weight) when we are standing on the ground that the ground exerts on us send us flying above the ground- why doesn't the law apply here?
Finally, when we drop a hard stone on the ground why doesn't it bounce? Plus, why is the force exerted by the stone on the ground greater than its weight?
| In a standing state, your legs exert a force on the ground that correlates to your weight. Any motion or acceleration caused by the movement of your legs is added to this.
Think of jumping as generating upward momentum in your body. Your legs are causing you to accelerate upwards and gravity is causing you to accelerate downward. If you legs don't generate more acceleration than gravity, you will never leave the ground. Otherwise, your body builds up momentum until the acceleration caused by you legs stops. At this point, you leave the ground and travel upwards until gravity has reduced you upward momentum to 0. Then you start to fall.
Don't forget that momentum is stored energy and must be accounted for.
Sometimes a stone will bounce. This has to do with elasticity.
| {
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Derivation of the general Lorentz transformation The standard Lorentz transformation or boost with velocity $u$ is given by
$$\left(\begin{matrix} ct \\ x \\ y \\ z \end{matrix}\right) = \left(\begin{matrix}
\gamma & \gamma u/c & 0 & 0 \\
\gamma u/c & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{matrix}\right) \, \left(\begin{matrix} ct^\prime \\ x^\prime \\ y^\prime \\ z^\prime \end{matrix}\right) = L_u \,\left(\begin{matrix} ct^\prime \\ x^\prime \\ y^\prime \\ z^\prime \end{matrix}\right)$$
where $\gamma = \gamma(u) = 1/\sqrt{1-u^2/c^2}$. In the standard Lorentz transformation, it is assumed that the $x$ and $x^\prime$ axes coincide, and that $O^\prime$ is moving directly away from $O$.
If we drop the first condition, allowing the inertial frames to have arbitrary orientations, then "we must combine [the standard Lorentz transformation] with an orthogonal transformation of the $x$, $y$, $z$ coordinates and an orthogonal transformation of the $x^\prime$, $y^\prime$, $z^\prime$ coordinates. The result is
$$\left(\begin{matrix} ct \\ x \\ y \\ z \end{matrix}\right) = L \,\left(\begin{matrix} ct^\prime \\ x^\prime \\ y^\prime \\ z^\prime \end{matrix}\right)$$
with
$$L = \left(\begin{matrix} 1 & 0 \\ 0 & H \end{matrix}\right)\, L_u \,\left(\begin{matrix} 1 & 0 \\ 0 & K^\textrm{t} \end{matrix}\right)$$
where $H$ and $K$ are $3 \times 3$ proper orthogonal matrices, $L_u$ is the standard Lorentz transformation matrix with velocity $u$, for some $u < c$, [and 't' denotes matrix transpose]."
I have two questions:
*
*Why are two orthogonal transformations, for both the unprimed and primed spatial coordinates, necessary? That is, why isn't one orthogonal transformation sufficient to align the axes of the inertial frames?
*Why does the first orthogonal transformation use the transposed orthogonal matrix $K^\textrm{t}$?
| It's actually very simple. The general Lorentz transformation can be rewritten as
$$\left(\begin{matrix} 1 & 0 \\ 0 & H^\textrm{t} \end{matrix}\right)\,\left(\begin{matrix} ct \\ x \\ y \\ z \end{matrix}\right) = L_u \,\left(\begin{matrix} 1 & 0 \\ 0 & K^\textrm{t} \end{matrix}\right) \,\left(\begin{matrix} ct^\prime \\ x^\prime \\ y^\prime \\ z^\prime \end{matrix}\right)\,.$$
This corresponds to aligning the $x$ and $x^\prime$ axes with the direction of the relative velocity, and then applying the standard Lorentz transformation.
| {
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Quantum Excitations In the context of quantum mechanics, is the Schrödinger equation actually describing some sort of an actual wave in some field like light in EM field ? So all particles are actually waves in their respective fields ?
| In QM the Schrödinger equation, is the equivalent of Newton's law in Classical Mechanics. The Schrödinger equation describes the state of a quantum system (i.e. atoms, subatomic particles etc.), and how the quantum system changes over time. I think you are getting confused because there are two main places where the term wave appears.
(1) The Double Slit experiment
(2) The Wavefunction (Heisenberg Uncertainty Principle)
Waves appear in (1) because the Double Slit experiment, proves that particles can have the properties of both particles and waves.
The wavefunction (2) is the answer to Schrödinger's equations and describes the state of one or more partices in the system. Quantities, such as average momentum, average position etc.can be derived from the wavefunction - again the analogue of Newton's equations.
BUT most importantly, the wave of the wavefunction is not an physical wave. The wave described by the wavefunction is an abstract wave that describes the probability of the quantities.
Therefore, no, the Schrödinger equation is not describing different particles in their respective fields, it is describing particles as (1) particles and waves (because that's what they are), and (2) functions of probability.
| {
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Aufbau principle in modern quantum theory What is the rigorous definition of the Aufbau principle and the mathematical model used for its description?
From Wikipedia, we have that the principle postulates a hypothetical process in which an atom is "built up" by progressively adding electrons. For modeling the postulate, we use the Madelung rule, Pauli exclusion and "if two same energy orbitals are available, choose the less occupied one)". So, $n+\ell$ ordering with constraints. And I also know that the Aufbau principle has its exceptions.
| The Aufbau principle isn't rigorous because it's based upon the approximation that the electron-electron interaction can be averaged into a mean field. This is called the Hartree–Fock or self-consistent field method. The centrally-symmetric mean field results in a set of atomic orbitals that you can populate 2 electrons at a time.
The trouble is that the electron correlations mix up the atomic orbitals so that distinct atomic orbitals no longer exist. Instead you have a single wavefunction that describes all the electrons and does not factor into parts for each electron. For example this is explicitly done in the configuration interaction technique for improving the accuracy of Hartree–Fock calculations.
| {
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Should theory be the appropriate term? Should theory be the appropriate term? I mean, for example, because of the quantum field theory we have been able to find the subatomic particles that it theorized and make the Standard Model. Why then is is labeled as a theory? Also wave-particle duality is widely accepted fact yet labeled as a theory. What is up with that, why call it a theory. Maybe because it promotes the fact of idealism?
| Terminology doesn't matter much as long as people understand the ideas involved. You should call and idea whatever you want to call it as long as you are clear about the substance of your ideas.
Some commentators have stated that the Standard Model or other scientific results are well-grounded. So far, these ideas have not been refuted. If somebody invented a competing theory, then that theory would have to solve the same problems the current theory solves. Since the current theory solves a lot of problems that is difficult. However, any given theory is either right or wrong and no number of experimental results that agree with a theory prove it is true or that it is probably true or anything like that.
See "Realism and the Aim of Science" by Karl Popper, especially the first chapter, and on the definition issue see "Conjectures and Refutations" by Popper Chapter 3, Section 3.
| {
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Infinitely many points of spacetime? Please can someone provided me with academic literature (Journals/Books, titles & links) which discuss the current view on spacetime i.e. that there is not Infinitely many points of spacetime?
| This is a nice introduction to Loop Quantum Theory in which both space and time are the result of weaving together fundamental quantum of space itself. http://arxiv.org/pdf/1001.1330v1.pdf. I am sending you too a page which contains several links and options for the study of this theory. http://cosmicposts.wordpress.com/loop-quantum-gravity/
Enjoy it
| {
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Creating electricity from mains water pressure. Could someone cleverer than me help me out?
I had a crazy thought going through my head the other day and I can't lay my mind to rest until I get an answer.
Q. How much energy could be produced by using mains water pressure to turn a generator? And would it be feasible to install a system to feed whatever is produced back to the grid? Assuming that the system would be installed in a building where a constant water supply is needed so the generator would be turning continuously, and a rough water pressure of around 3-4 bar.
Thanks in advance for any help
| A generator converts mechanical power to electrical power; pressure alone is insufficient.
Assuming the flow in = flow out and a constant flow, the power output of the generator would then be proportional to the pressure difference between the inlet and outlet.
Thus, subtract the minimum pressure required by the building from the mains pressure and multiply that by the flow to find the potential power available for conversion.
Since generators aren't 100% efficient, the actual electrical power generated will be less.
| {
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BICEP2 and e-foldings during inflation After the BICEP2 results, we now know that $n_s = 0.96$ and $r = 0.2$.
From what I understand, this fits extremely well with the basic chaotic inflation model given by $V(\Phi) = \lambda \Phi^4$.
We also know that amplitude of density fluctuations is $\approx 10^{-5}$ and
Energy scale of inflation is around $10^{16}$ GeV.
My question: Given also this information, can we now make an educated guess for how many e-foldings happened during inflation ? Or at least a theoretical lower (upper ?) bound ?
PS: I am only referring to inflation of the patch of space that now contains our observable universe, not inflation of Universe as a whole.
| $r = 8(1-n_s)-\frac{8}{N_*}$ for monomial potential inflation models, equation 206 of this reference: http://lesgourg.web.cern.ch/lesgourg/Inflation_EPFL.pdf
where $N_*$ is "the number of e-folds between horizon crossing for observables scales and the end of inflation" (basically the observable number of e-folds).
so for $r=0.16$ (the dust-corrected value of the BICEP2 paper) and $n_s = 0.96$
$N_* = 50$
| {
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Why do the planets orbit the Sun? Why do the planets orbit the Sun? If it's because the gravity of Sun then why don't they just fall in and burn up?
| The planets are attracted towards the sun, as you would expect from the gravitational force. The planets don't fall into the sun, though, because their velocities are at right angles to that force. The planets end up being pulled by the sun into a circle. A planet's speed is constant, but its direction changes. I think it's easiest with a picture:
Gravity pulls the planet in a circular orbit.
| {
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Size of Universe after inflation I read in some website that during the period of inflation, the expansion of the universe underwent incredibly fast, and its size increased by a factor of $10 ^{50}$, see this link
In this field, I think, there is nothing for sure, but if there was really inflation, what does it mean that the size increased by a factor of $10 ^{50}$, but from what extent? From 1 mm? Or from the Planck length? Or something else?
Has it been established that (whatever was the original measure)
the factor is increased by $10 ^{50}$, and not for example of 40 or $10 ^{50^{60}}$?
| The best way to think about inflation (according to my University's cosmology expert) is to think as if you were measuring the universe with an ever-shrinking ruler.
So if you were measuring 1 meter, then your two reference points would now be separated by $10^{50}$ meters, because your meter is shrinking. If you were measuring things 1 mm apart, now they would be $10^{47}$ meters apart.
To directly answer your question, there is no definite size of the Universe, because it is infinite.
The definition of a factor implies that there doesn't need to be an original measure.
If there was an original measure (and the universe was not infinite), I bet you scientists wouldn't use a factor but would give the actual size!
| {
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How does particles gain electrical charges and repel each others? (electrostatic stabilization) When I study electrostatic stabilization, I understand that the particles have same charge and thus repel others, this is how colloid is stabilize. But how does particles gain electrical charges and repel each other in the first place? I think that there are weak electrostatic force on the particle and when electrolytes were added, it gain more charges. Thank you.
| John Rennie will probably have more details on the matter, but in general colloids (such as oil dispersed into soap water) are not so much stabilized by a net total charge of the mixture, but rather are stabilized by repulsions from separated charges. For example:
This is a cartoon representation of what an oil droplet in soap water looks like; note that the surface of the oil droplet is net negatively charged due to the carboxylate (or sulfonate) heads at the oil surface.
In essence, you have negatively charged oil droplets floating around in a positively-charged fluid. The net charge, however, is zero.
If two of these oil droplets attempt to collide, the negatively-charged heads on the surface will repel each other, preventing aggregation. This is what stabilizes the colloid.
You'd have to provide more details on what kind of colloid you're examining in order to tell if this is actually the mechanism stabilizing the colloid you're looking at.
| {
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Which is the smallest known particle that scientists have actually *seen with their eyes*? Which is the smallest particle that has been actually seen by the scientists? When I say "actually seen", (may be using some ultra advanced microscope or any other man made eye, using any wavelength or phenomena) I really mean it; just like we have seen the red blood cells.
Davidmh's answer is pretty much in line with what I am asking
| Obviously, the smallest particle that scientists have ever seen directly is a photon.
The question is a bit silly because it tries to eliminate any simple device like a photographic plate. But the human eye, its nerves and the visual cortex together are far more complicated.
| {
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In the B mode power spectrum, what is the relationship between the multipole number and the wavelength of the seed gravitational waves? One of the key datasets of the recent BICEP2 results is the B mode power spectrum shown below. The existence of these B modes implies the existence of gravitational waves prior to inflation.
My question is: what is the relationship between the multipole number of the B mode and the wavelength of the gravitational waves which generated it? My naive expectation is that it is one to one with some sort of scaling factor which includes information about the expansion of the universe and the conversion between spherical and linear coordinates.
| Wayne Hu (professor, Univ. Chicago) has a CMB website
that seems to answer this question.
It is stated that "The one-to-one mapping between wavenumber and multipole moment described in [a previous section] is only approximately true and comes from the fact that the spherical Bessel function is strongly peaked at $kD \approx l$"
where $l$ is the multipole moment (x-axis of the graph in the question); $D$ is the distance light can travel between recombination and present, and $k$ is the wavenumber.
| {
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Help understanding proof in simultaneous diagonalization The proof is from Principles of Quantum Mechanics by Shankar. The theorem is:
If $\Omega$ and $\Lambda$ are two commuting Hermitian operators, there exists (at least) a basis of common eigenvectors that diagonalizes them both.
The proof is:
Consider first the case where at least one of the operators is nondegenerate, i.e. to a given eigenvalue, there is just one eigenvector, up to a scale. Let us assume $\Omega$ is nondegenerate. Consider any one of its eigenvectors:
$$\Omega\left|\omega_i\right\rangle=\omega_i\left|\omega_i\right\rangle$$
$$\Lambda\Omega\left|\omega_i\right\rangle=\omega_i\Lambda\left|\omega_i\right\rangle$$
Since $[\Omega,\Lambda]=0$
$$\Omega\Lambda\left|\omega_i\right\rangle=\omega_i\Lambda\left|\omega_i\right\rangle$$
i.e., $\Lambda\left|\omega_i\right\rangle$ is an eigenvector with eigenvalue $\omega_i$. Since this vector is unique up to a scale,
$$\Lambda\left|\omega_i\right\rangle=\lambda_i\left|\omega_i\right\rangle$$
Thus $\left|\omega_i\right\rangle$ is also an eigenvector of $\Lambda$ with eigenvalue $\lambda_i$...
What I do not understand is the statement/argument "Since this vector is unique up to a scale." I do not see how the argument allows to state the equation following it. What axiom or what other theorem is he using when he states "since this vector is unique up to a scale"?
| I don't know exsactly what it means with scalar but remember that:
A vector $|\Psi\rangle$ is invariant up to a phase because a global phase is always ruled out when you calculate, for example, with the state $e^{i\theta}|\Psi\rangle$, the expected value of and observable $A$ using this state is
$$\langle A \rangle = \langle\Psi | e^{-i\theta} A e^{i\theta} |\Psi \rangle = \langle\Psi | A | \Psi\rangle,$$
which is the same result as if you calculate the expected value by just using the vector $|\Psi\rangle$. So for all practical purposes, you can use just a state $|\Psi\rangle$ in your punch line of your proof.
| {
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Proving a step in this field-theoretic derivation of the Bogoliubov de Gennes (BdG) equations In derivation of the BdG mean field Hamiltonian as follows, I have a confusion here in the second step:
$H_{MF-eff} = \int d^{3}r\psi_{\uparrow}^{\dagger}(\mathbf{r})H_{E}(\mathbf{r})\psi_{\uparrow}(\mathbf{r})+\int d^{3}r\psi_{\downarrow}^{\dagger}(\mathbf{r})H_{E}(\mathbf{r})\psi_{\downarrow}(\mathbf{r})
+\int d^{3}r\triangle^{\star}(\mathbf{r})\psi_{\downarrow}(\mathbf{r})\psi_{\uparrow}(\mathbf{r})+\int d^{3}r\psi_{\uparrow}^{\dagger}(\mathbf{r})\psi_{\downarrow}^{\dagger}(\mathbf{r})\triangle(\mathbf{r})-\int d^{3}r\frac{|\triangle(\mathbf{r})|^{2}}{U}$
$ = \int d^{3}r\psi_{\uparrow}^{\dagger}(\mathbf{r})H_{E}(\mathbf{r})\psi_{\uparrow}(\mathbf{r})-\int d^{3}r\psi_{\downarrow}(\mathbf{r})H_{E}^{\star}(\mathbf{r})\psi_{\downarrow}^{\dagger}(\mathbf{r})
+\int d^{3}r\triangle^{\star}(\mathbf{r})\psi_{\downarrow}(\mathbf{r})\psi_{\uparrow}(\mathbf{r})+\int d^{3}r\psi_{\uparrow}^{\dagger}(\mathbf{r})\psi_{\downarrow}^{\dagger}(\mathbf{r})\triangle(\mathbf{r})-\int d^{3}r\frac{|\triangle(\mathbf{r})|^{2}}{U}$
$= \int d^{3}r\left(\begin{array}{cc}
\psi_{\uparrow}^{\dagger}(\mathbf{r}) & \psi_{\downarrow}(\mathbf{r})\end{array}\right)\left(\begin{array}{cc}
H_{E}(\mathbf{r}) & \triangle(\mathbf{r})\\
\triangle^{\star}(\mathbf{r}) & -H_{E}^{\star}(\mathbf{r})
\end{array}\right)\left(\begin{array}{c}
\psi_{\uparrow}(\mathbf{r})\\
\psi_{\downarrow}^{\dagger}(\mathbf{r})
\end{array}\right)+const.
$
with
$H_{E}(\mathbf{r})=\frac{-\hbar^{2}}{2m}\nabla^{2}$
In the second step, we have taken
$\int d^{3}r\psi_{\downarrow}^{\dagger}(\mathbf{r})\nabla^{2}\psi_{\downarrow}(\mathbf{r}) = -\int d^{3}r\psi_{\downarrow}(\mathbf{r})\nabla^{2}\psi_{\downarrow}^{\dagger}(\mathbf{r})$............(1).
I can prove (by integration by parts and putting the surface terms to 0) that $\int d^{3}r\psi_{\downarrow}^{\dagger}(\mathbf{r})\nabla^{2}\psi_{\downarrow}(\mathbf{r}) = \int d^{3}r \nabla^{2}\psi_{\downarrow}^{\dagger}(\mathbf{r})\psi_{\downarrow}(\mathbf{r})$
but how is it justified to now take
$\int d^{3}r \nabla^{2}\psi_{\downarrow}^{\dagger}(\mathbf{r})\psi_{\downarrow}(\mathbf{r}) = - \int d^{3}r\psi_{\downarrow}(\mathbf{r})\nabla^{2}\psi_{\downarrow}^{\dagger}(\mathbf{r})$
in order to prove (1) ?
| The negative sign comes from the anti-commutation of the fermion operators $\psi$ and $\psi^\dagger$. That is $\psi^\dagger(r)\psi(r') = - \psi(r')\psi^\dagger(r) + \delta(r-r')$. The delta function just gives the some constant that will be absorbed into the $const$ term. Maybe it is bothering you that this term is formally divergent but this is removed by any regularization, such as putting the system on a lattice.
| {
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How to show the invariant nature of some value by the group theory representations? Let's have Dirac spinor $\Psi (x)$. It transforms as $\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right)$ representation of the Lorentz group:
$$
\Psi = \begin{pmatrix} \psi_{a} \\ \kappa^{\dot {a}}\end{pmatrix}, \quad \Psi {'} = \hat {S}\Psi .
$$
Let's have spinor $\bar {\Psi} (x)$, which transforms also as $\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right)$, but as cospinor:
$$
\bar {\Psi} = \begin{pmatrix} \kappa^{a} & \psi_{\dot {a}}\end{pmatrix}, \quad \bar {\Psi}{'} = \bar {\Psi} \hat {S}^{-1}.
$$
How to show formally that
$$
\bar {\Psi}\Psi = inv?
$$
I mean that if $\Psi \bar {\Psi}$ refers to the direct product (correct it please, if I have done the mistake)
$$
\left[\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right) \right]\otimes \left[\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right) \right],
$$
what group operation corresponds to $\bar {\Psi} \Psi$?
This question is strongly connected with this one.
| If we assume that
$$\Psi {'} = \hat {S}\Psi$$
and
$${\bar{\Psi}}{'} = \bar {\Psi} \hat {S}^{-1},$$
it follows that the product of the two transforms as
$$(\bar{\Psi}\Psi)'={\bar{\Psi}}{'}\Psi {'}=\bar {\Psi} \hat {S}^{-1}\hat {S}\Psi=\bar{\Psi}\Psi,$$
which is a consequence of
$$\hat {S}^{-1}\hat {S}=\mathbb{1}.$$
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Renormalizing with external momenta set to zero I've often seen in textbooks that authors renormalize diagrams by setting external momentum to zero. Under what conditions is this justified?
An example of this is done in Manohar and Wise's book on Heavy Quark Physics after they renormalize QED and then calculate the operator renormalization, $ Z _S $, of
\begin{equation}
S = \frac{1}{ Z _S }\bar{\psi} _b \psi _b = \frac{ Z _\psi }{Z _S } \bar{\psi} \psi
\end{equation}
where $ \psi_b $ is the bare field. They calulate this through the diagram,
$\hspace{5cm}$
where the cross indicates an operator insertion. The authors then say ``The operator, $ S $, contains no derivatives (and $Z _S$ is mass independent in the $\overline{MS}$ scheme), so $Z _S$ can be determined by evaluating (the diagram) at zero external momentum (and neglecting the (fermion) mass).'' Are these the two conditions necessary,
*
*The operator doesn't have derivatives
*The quantity of interest has no mass dependence
and if so how do we know that the quantity you want to calculate (in this case, $ Z _S $) is mass independent, ahead of time?
| In general, derivative couplings lead to momentum-dependencies in scattering amplitudes. This can be seen from the fact that the Fourier transform of a derivative operator corresponds to a multiplication by the relevant momentum. A mass dependence is implicit through by having a momentum, since the momentum of a fermion depends on its mass. In this case, setting momenta to zero would remove information about the coupling.
However, when the coupling does not contain derivatives, there are no momenta resulting from a Fourier transform. Hence, one can simplify the problem by setting them to zero.
| {
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Does the transmission axis matter for sending polarized light through polarized glass? If I have polarized light and I send through only one polarized glass plane, does the transmission axis matter, or will the intensity be halved no matter what.
| The axes definitely matter. If you put light through a linearly polarized glass pane, the output light will be entirely polarized along the polarization axis of that pane. The intensity of the output light will be
$$I_{\textrm{out}} = I_{\textrm{in}}\cos(\theta)^2$$
where $\theta$ is the angle between the polarizations of the ingoing light and the pane, and $I_{\textrm{in}}$ is the intensity of the input light.
This all assumes the ingoing light is linearly polarized. If it's not, you can always decompose it into a sum of linearly polarized parts, one parallel to the polarization axis of the pane and one orthogonal. Once you've made that decomposition the output intensity is simply the intensity of the input component that's aligned with the axis of the pane.
Note that if $\theta=\pi/4$ we get $I_{\textrm{out}}=I_{\textrm{in}}/2$.
| {
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Free particle propagator - Evaluating Integral In path integral formalism, when evaluating the free particle propagator, we obtain the functional integral of the form,
$$ K_0 = \lim_{n\rightarrow\infty} \bigg( \frac{m}{2\pi i\tau}\bigg)^\frac{n}{2} \int \prod_{i=1}^{n-1}dx_i \; \exp\bigg(\frac{im\tau}{2}\sum_{j=0}^{n-1}(x_{j+1} - x_j)^2\bigg). $$
To begin with, for I have first solve for some finite $n$ case and then generalise by induction. But for even the finite case, I am confused how to do the integral. For instance, with $n=2$ case, I will get an integral of the form,
$$ I = \bigg( \frac{m}{2\pi i\tau}\bigg) \int dx_1e^{\frac{im\tau}{2} \bigg((x_{2} - x_1)^2+(x_{1} - x_0)^2\bigg)}. $$
How am I supposed to solve this integral? I know it just involves some trivial trick of substitution, but am just confused which way to go.
EDIT :
After a bit of evaluation of the integral, I end up with an integral of the form
$$ \int_{0}^{\infty} e^{iax^2}\;dx $$ and making a substitution therewith $ s= -iax^2 $ and $ dx = \frac{ds}{\sqrt{-ias} } $, but how do evaluate this kind of integral (i mean what kind of contour can we choose)
$$ \int_{0}^{i\infty} \frac{ds}{\sqrt{-ias}} e^{-s} $$
PS : A diagram of the contour will be really helpful :)
| Note that all your integrals are Gaussians in differences of positions at successive instants $(x_k - x_{k-1})$ so implement a change of integration variable from $x_k \longrightarrow (x_k - x_{k-1})$. You will have $N-1$ (straightforward) integrations to perform with $x_0$ and $x_N$ held fixed.
| {
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How big was the first transistor? The first working point-contact transistor made in 1947 by Bell Labs. I'm looking for specific dimensions, all I've been able to find is "Fits in the palm of your hand".
| From "The Transistor, A Semi-Conductor Triode", by J. Bardeen and W. H. Brattain, Phys Rev. 74(2), 230-231 (1948):
"The device consists of three electrodes placed on a block of germanium as shown schematically in Fig. 1. Two, called the emitter and collector, are of the point-contact rectifier type and are placed in close proximity (separation ~0.005 to 0.025 cm) on the upper surface. The third is a large area low resistance contact on the base."
So, the actual device was much smaller than your (or my) palm. Now, the support for the device was probably bigger. Sze's book had a picture of the device on the cover long ago - not sure about new editions.
(Replica of the first transistor)
| {
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Deriving $F = ma$ - Newton's Second Law of Motion Context:
In my textbook it is given: 'momentum' short for 'linear momentum':
Mass = $m$, momentum is $p=mv$. In time $\Delta t$, momentum changes by $\Delta p$, the rate of change of momentum is:
$$\frac{\Delta p}{\Delta t} = \frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$
My Doubts:
*
*Isn't a $\Delta$ sign missing beside the $t$ in the second fraction, and thus it should be $\frac{\Delta(mv)}{\Delta t}$
*How did they derive the third fraction from the second. I tried a lot but can't seem to get that.
My Work:
I have looked at this question - How does $F = \frac{ \Delta (mv)}{ \Delta t}$ equal $( m \frac { \Delta v}{ \Delta t} ) + ( v \frac { \Delta m}{ \Delta t} )$?, but it's a totally different equation.
My Final Question:
Can someone please clear my doubts about this equation and help me understand how does:
$$\frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$
Thanks a lot !
| *
*Yes. It should be:
$$\frac{dp}{dt}=\frac{d(mv)}{dt}$$
I'm using $d$ instead of $\Delta$ because I am thinking about the limit where the changes in $p$ and $t$ are very small. Then these are called infinitessimal changes, and denoted by a $d$.
*Usually, when one considers simple problems in Newtonian mechanics, what one does is study a given object with a fixed and constant mass, $m$. This means that $dm=0$ at all times by definition.
We have, using the product rule (which only holds for infinitessimal changes):
$$\frac{d(mv)}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$$
Can you guess what happens to the first term on the right hand side of the equal sign?
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Magnetism due to relativity? So I have been reading in some books that magnetism does not have to be assumed a priori, but can be obtained from the electric field + special relativity.
And I have seen how this leads to the common formula for the magnetic field of a current carrying wire.
Fine.
What about materials that are inherently magnetic? Such as iron, or magnetite?
Surely their magnetic field is not a consequence of relativity? (if yes, who's moving and with respect to whom?)
| Oliver Heaviside showed that magnetic fields are a result of moving charges, and that relativity is not really needed. He showed that if gravity moves at a finite speed, then a magnetic like effect arises.
Bohr's atom is non-relativistic, but still has the correct magnetic properties.
Relativity itself is due to the failure to observe the etherfer, and that the speed of electric and gravitational field, might be more to the nature of the medium than the fields themselves.
| {
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Viewing glass from an oblique angle When I view most glass from the side it's green which I've found out is due to impurities in the glass specifically from iron oxide.
Why is it when I view the larger face from an oblique angle, it isn't nearly as green? I cannot personally notice any different on the piece I have next to me even when I hold it at an angle that would be almost looking at the edge of the glass. It is pretty small (about 2.5" x 5" x .0625" or about 61mm x 127mm x 2mm, l x w x h) but I feel like it's big enough that I'd be looking through enough glass to get the green.
| Even when held at a glancing angle, the light within the glass follows a relatively short path. As light crosses the boundary between the air an the glass its angle changes (refraction). Light that leaves the glass at a glancing angle and enters your eye was traveling much closer to perpendicular to the surface when it was within the glass. It hasn't passed through a thick layer, hasn't encountered many impurities, and hence hasn't had its color changed much.
| {
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The shape of the graph of the equation: $V= -r I + E$ I have recently collected data (for a school experiment) in order to measure the EMF and the internal resistance of a solar cell. The data complied with the equation: $V = -rI + E$, i.e. the voltage decreased as the current increased and I have a negative gradient. However, I am having some trouble as to explaining why the graph is the way it is, that is, why is there a negative gradient?
Here's my rationale: Using Ohm's law, we can see that decreasing the (external) resistance increases the current flowing throughout the whole circuit (it is a simple series circuit with just a solar cell, ammeter, variable resistor and a voltmeter (in parallel, obviously)). Now, the voltage drop across the solar cell should increase as well (assuming its internal resistance in constant) because $v = V_r = Ir$, where $v$ is the 'lost volts'. For this reason the voltage drop across the load must decrease(which is what we see from the graph) because $E = V + v$. Thus, the voltage drop across the resistor decreases as current increases.
I don't know to what extent this is correct or if it is correct at all but I would appreciate any and all help given. Also, as you can probably tell, the level of physics used here is very basic, so I would also appreciate it if the answers explained it in this way as well, although, where possible, do not sacrifice accuracy for the sake of simplicity.
| Your reasoning is correct.
If the solar cell is modelled as a voltage source $E$ in series with an internal resistance $r$ and the cell is connected to a load resistance $R_L$, the series current is given by Ohm's law:
$$I = \frac{E}{r + R_L}$$
or
$$E = (r + R_L)\cdot I $$
The output voltage $V$ of the solar cell is the voltage across the load resistance which is, by Ohm's law, $V = R_L\cdot I$.
Thus
$$V = R_L\cdot I = E - r\cdot I $$
| {
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Why do particles in high pressure air always flow to lower pressure? The title really says it all: Why is this case? A "Feynman type" answer would be really appreciated as I'm more of a layman that a physicist.
| You mention particles, so I'll provide a micro-scale analogy. In a gas or liquid, particles are always interchanging energy and momentum due to collisions with other particles. So we can treat them as if their velocity and direction is always random and changing.
Imagine a line in the middle of a field, with 10 people (say: $10^{22}$ particles in a typical amount of gas) on the left side, and 1000 people ($100 *10^{22}$ particles) on the right side. Every minute, every person rolls a dice and if it lands on one, moves to the other side of the field.
So let's see how the distribution changes, purely driven by randomness:
Seconds; # on left side, # on right side
0 10 1000
1 175 835
2 285 725
3 358 652
4 407 603
5 440 570
6 548 652
$\infty$ 505 505
Of course if you conduct this experiment, there will be some noise because of the small number of participants; with trillions and trillions of particles it will be a perfect equilibrium.
| {
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How can point-like particles in an ideal gas reach thermodynamical equilibrium? Having learned that the particles of an ideal gas must be point-like (for the gas to be ideal) I wonder how they can reach thermodynamical equilibrium (by "partially" exchanging momentum and energy). First the probability of two point-like particles to collide is literally zero, and second, they can only collide head-on which implies that they can only "swap" their momenta and energies.
How is this puzzle to be solved?
| Ideal gasses do not exist. In real gasses, molecules do take up volume. This does affect the pressure-volume curve.
Still, an ideal gas model is close enough to reality to be useful. It is more useful if not taken so literally as to include its most unrealistic features.
| {
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Is speed of light ( Sun-to-Earth ) related to the movement speed of solar system? The speed of light has been measured to be 299 792 458 m/s.
Now, the Solar System is traveling at an average speed of 828,000 km/h (230 000 m/s). Summing up the numbers we get close to 300 000 000 m/s
Does it mean, that the speed of light can be actually a relative number based on the movement of the source?
Does it mean that it IS actually 300 000 000 m/s but is calculated incorrectly due to sun moving round the galaxy?
| No the speed of light in vaccuum is an absolute constant $c$ = 299 792 458 m/s
The way to add up relativistic speeds is:
$u' = \frac{u-v}{1-\frac{uv}{c^2}}$ to account for the constancy of the speed of light
You cannot simply add them up.
Edit: This also applies to normal everyday speeds. The reason we don't use this formula is because the speeds we are dealing with are too small and it simply doesn't matter. The difference is very very small to be noticable.
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Requirements for hydrogen fusion What are the requirements for hydrogen atoms to go through fusion? Is it a ratio of heat to pressure or are there specific heat and pressure values that must be met (per atom or per mole?) Are there other requirements?
| The Wikipedia article answers most of your questions.
What are the requirements for hydrogen atoms to go through fusion?
Two atoms must overcome the coulomb barrier, which can be done by forcing two atoms very close together, or by leaving them moderately close for long periods of time, which allows them to tunnel through the barrier.
Is it a ratio of heat to pressure or are there specific heat and pressure values that must be met (per atom or per mole?)
It's fairly complicated but it does require very high temperatures and very high pressures.
For reference fusion happens at the centre of the sun, but does not produce very much energy per unit volume (about the same as a living mammal) so for a viable reactor we need much higher energies than the centre of the sun.
Wikipedia quote:
The core produces almost all of the Sun's heat via fusion ...
The energy production per unit time (power) of fusion in the core varies with distance from the solar center. At the center of the Sun, fusion power is estimated by models to be about 276.5 watts/m3. Source
Are there other requirements?
There are many different ways of producing nuclear fusion and each has its own set of requirements see Muon catalysed fusion as a way of causing nuclear fusion without the requirement of heat or pressure.
Really nuclear fusion is not difficult, there are many reactors that fuse elements, the difficulty is the fact that they require so much energy as an input that they are uneconomical as a power source.
| {
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State of constant motion Why does an object remains in its state of constant motion if there are no forces acting on that object?
My understanding is that all the energy of the motion will be kept inside and a change in the speed needs a change in the amount of the energy stored by that object. But how is that energy stored in a form of speed?
Does anything changes inside the object's structure if it is moving?
| the main reason is that the space is uniform, and that there is nor absolute reference point in the universe.
basically, what appears to be moving at constant speed to you, will be moving at a different constant speed or even not moving at all to another observer who is moving at constant speed in reference to you. since your point of view is not any better than his, then who said that the object is moving at all? it might as well be staying in the same place.
| {
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Does mass affect velocity when travelling through frictionless medium? I found the following question on an standardized test, and was debating with some friends what the answer would be:
A car of mass M is travelling with a constant velocity through a plane in which friction is non-existent. An object of mass m (m = M/3) that is falling perpendicularly to the car lands inside of it. How will the velocity of the car be affected?
This illustration can help explain the problem.
My initial thought was that the velocity would be the same, given that friction is non-existent and that the momentum of the falling object is perpendicular to that of the car. However, some friends suggested that, since the mass of the car increases, the velocity should decrease.
| Your friends are correct.
If there is no force in the left-right direction, then linear momentum will be conserved in that direction. Because the new composite object has more mass than the original object, it will have a lower speed to the right.
What about energy?
Kinetic energy is not conserved in this case, because the collision is inelastic. The kinetic energy lost in the collision (and the downward momentum lost, for that matter) is absorbed by whatever wall is preventing the composite object from continuing to move downward.
| {
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How can we detect cosmic background radiation? From what I understand, CMB is the left over radiation from the Big Bang. As all matter, including the Earth, was made during the Big Bang and then as the universe expanded that matter/energy got further and further apart, but as that matter can't move faster than the speed of light then we should be behind the CMB which travels at the speed of light. So from my understanding the CMB would have passed us and most of the matter of the universe a long time ago so we shouldn't be able to detect it. Also on a side note how is the CMB constant; shouldn't it be a flash? Sorry if the question doesn't make much sense. I am a student so please use layman's terms.
| You are assuming the Big Bang happened at a point, so the CMB is a shell of radiation expanding outwards from that point. However the Big Bang happened everywhere so every point in the universe is a source of the CMB. The CMB radiation we are detecting today comes from regions of the universe that were about 13.8 billion light years away at the moment the CMB was emitted (those points are a lot farther away now).
The fact that the Big Bang happened everywhere is a difficult conceptual issue for non-physicists. See my answer to the question Was the singularity at Big Bang perfectly uniform and if so, why did the universe lose its uniformity? for a non-physicist friendly discussion of this.
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Orthogonality of summed wave functions Problem. I know that the two wave functions $\Psi_1$ and $\Psi_2$ are all normalized and orthogonal. I now want to prove that this implies that $\Psi_3=\Psi_1+\Psi_2$ is orthogonal to $\Psi_4=\Psi_1-\Psi_2$.
My naive solution. From the premises, we know that
$$\int_{-\infty}^\infty \Psi_1^*\Psi_1 dx=\int_{-\infty}^\infty \Psi_2^*\Psi_2 dx=1$$
and
$$\int_{-\infty}^\infty \Psi_1^*\Psi_2 dx=\int_{-\infty}^\infty \Psi_2^*\Psi_1 dx=0$$
We also have $(z_1+z_2)^*=z_1^*+z_2^*$
$$\int_{-\infty}^\infty \Psi_3^*\Psi_4 dx = \int_{-\infty}^\infty (\Psi_1+\Psi_2)^*(\Psi_1-\Psi_2)dx \\
=\int_{-\infty}^\infty(\Psi_1^*+\Psi_2^*)(\Psi_1-\Psi_2)dx\\
=\int_{-\infty}^\infty(\Psi_1^*\Psi_1-\Psi_1^*\Psi_2+\Psi_2^*\Psi_1-\Psi_2^*\Psi_2)dx\\
=1-0+0-1=0\,,$$
which is equivalent with what we wanted to prove. Is this a legitimate proof? Is there any simpler way to do this? I am afraid I still haven't grasped how wave functions behave mathematically, so I may have missed somethings very obvious here.
Edit: The solution manual somehow uses normalization factors for $\Psi_3$ and $\Psi_4$. How are these factors when you don't actually know the exact functions? And how does this relate to the concept of orthogonality?
| This problem could be done more simply through the application of linear algebra. You want to prove that
$$\langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle = 0$$
The inner product is analogous to the dot product of linear algebra, and it is distributive. Distributing, we find that
$$\begin{aligned}
\langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle &= \langle \psi_1 - \psi_2 | \psi_1 \rangle + \langle \psi_1 - \psi_2 | \psi_2 \rangle \\
&= \langle \psi_1 | \psi_1 \rangle - \langle \psi_2 | \psi_1 \rangle + \langle \psi_1 | \psi_2 \rangle - \langle \psi_2 | \psi_2 \rangle
\end{aligned}
$$
Because $\psi_1$ and $\psi_2$ are orthogonal and normalized, you know $\langle \psi_i | \psi_j \rangle = \delta_{i j}$. Substituting, the above expression evaluates to $1 - 0 + 0 - 1 = 0$, demonstrating that the two vectors are indeed orthogonal.
Your approach - using the integrals - was also valid, and fundamentally similar to mine here. However, by noting that the relation you used ($\langle \psi_1 | \psi_2 \rangle = \int_{-\infty}^{\infty} \! \psi_1^* \psi_2 \, \mathrm{d}x$) satisfied the definition of an inner product, the integrals can be omitted.
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Electron distribution around atom when moving I do not have much experience on this but if an atom has some electrons around nucleus and the atom itself it is moving at some speed does that affect the distribution of electrons around?
I am presuming that the interaction between the nucleus and electrons has a constant speed $c$. Anything I found so far is a calculation for interactions that presume an infinite speed.
As an argument I am thinking of relativistic Doppler effect that does not change proportional with $v/c$. So I am thinking that maybe the speed does affect the distribution and so that is why the difference in the emission energy.
| The speed only affects if there is an acceleration. If the atom is moving at a constant speed, you can do a galilean transformation, move with the atom, and it will be at rest.
The two simplest reasons for an atom to accelerate are collision with other atoms and bombardment.
The first one happens everytime you have a gas or a liquid. The atoms wiggle around, hitting each other. In this case, the speeds are relatively low, and the biggest effect is due to the interaction between both electron clouds. You can increase the speed by raising the temperature, but then you stop having a gas and it becomes a plasma.
The other possibility is that a particle hits the nucleus, setting it in movement. That is a classical (and very academic) example of the sudden approximation. In this case, the wavefunction at the beginning at the end are the same, with the only difference that the end ones are moving.
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The other side of the lever If I have a lever, but I can see only up to the hinge and not the other half, can I know whether the other half is 1 m long with a weight of 3 kg on it, or 3 m long with a weight of 1 kg on it?
| Yes, you can. The moment of inertia of the lever will be different in each of the two situations. Let us assume that the lever is massless and the weight is a point mass. In the first situation, $I = mr^2 = (3)(1)^2 = 3 \text{ N} \cdot \text{m}$. In the second situation, $I = mr^2 = (1)(3)^2 = 9 \text{ N} \cdot \text{m}$.
Because the moment of inertia is greater in the second setup, the lever arm will be slower to respond to applied torques (that is, since $\tau = I \alpha$, a larger moment of inertia corresponds to a smaller angular acceleration for a given torque). To practically determine the difference, you can place a heavy weight on one end of the lever and use a stopwatch to record the amount of time the lever takes to reach a certain angular displacement. The one with a smaller time is setup with a 3 kg mass 1 m away, and the one with a higher time is a 1 kg mass 3 m away.
| {
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Can a laser be designed to ionize muonic atoms so as to prevent a-sticking? Muon catalyzed fusion is currently little more than a lab curiosity today in part because of how many hydrogen nuclei can be fused before the muon is carried away by an alpha particle. Deuterium+deuterium reactions are ten times more likely than deuterium tritium reactions to result in a muon sticking to a helium ion. I am wondering if some one can calculate the ionization energy needed to prevent that from happening and to speculate if a laser can be built to do it.
If it is possible, it may help pave the way to clean low-temperature fusion energy that produces more power than is used to make it.
| Muon mean lifetime is 2.2 µs. There's your problem. Muons mass 105.7 MeV/c2, about 200 times that of the electron. If you wanted to ionize a hydrogen atom, you would need 13.6 eV. If you wanted to ionize a muonic hydrogen atom, you would need about 2813 eV or about a 0.441 nm photon. Start building your laser.
| {
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Can magnetic flux be negative I am studying magnetic flux linkage in an ac generator and it appears to be that magnetic flux linkage is negative half the time, how can this be?? Also with lenz's law why is emf defined as negative when magnetic flux is increasing and how does this relate to the direction of the current?
| Magnetic flux is a scalar quantity and its positive/negative sign indicates the direction of the magnetic field.
And the Faraday's law of induction is a quantitative version of Lenz's law, which may help your understanding:
$\oint_{\partial \Sigma} \mathbf{E} \cdot \mathrm{d}\boldsymbol{\ell} = - \frac{d}{dt} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{S}$
| {
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The degeneracy of the ground state I wonder how can i know the degeneracy of the ground state of certain elements?
I'm doing Boltzmann distribution problems, and I'm not sure what to do. I have to calculate ratio of ions in 3p excited state, regarding ground state for Na$^+$ and Mg$^{+2}$ ions. I think that 3p degeneration is 6, but I don't know what else to put in this fraction.
Also, I've noticed that sometimes we just ignore that part of equation (count it as one) but I'm not sure when I'm allowed to do that.
| Each s sublevel (l=0) can be occupied by 2 electrons.
Each p sublevel (l=1) can be occupied by 6 electrons.
Each d sublevel (l=2) can be occupied by 10 electrons.
Each f sublevel (l=3) can be occupied by 14 electrons.
Each sublevel can be occupied by 4l + 2 electrons.
But are you sure that it is Na+ and Mg2+ (not Na and Mg+)? Na+ and Mg2+ don't have any valence electrons.
| {
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Why do rocket engines have a throat? Diagrams of rocket engines like this one,
(source)
always seem to show a combustion chamber with a throat, followed by a nozzle.
Why is there a throat? Wouldn't the thrust be the same if the whole engine was a U-shaped combustion chamber with a nozzle?
| As addition to @tpg2114 answer, I suggest also to read about de Laval nozzle and Rocket engine nozzle on wikipedia.
Some typical values of the exhaust gas velocity
for rocket engines burning various propellants are:
1.7 to 2.9 km/s (3800 to 6500 mi/h) for liquid monopropellants
2.9 to 4.5 km/s (6500 to 10100 mi/h) for liquid bipropellants
2.1 to 3.2 km/s (4700 to 7200 mi/h) for solid propellants
so that definitely makes sense to have nozzle)
| {
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What happens with a tunneling particle when its momentum is imaginary in QM? In classical mechanics the motion of a particle is bounded if it is trapped in a potential well. In quantum mechanics this is no longer the case and there is a non zero probability of the particle to escape the potential through a process call quantum tunneling.
This seems extraordinary from the point of classical mechanics because it implies the particle must cross a zone where it has imaginary momentum. I understand that from the point of view of quantum mechanics there is a non zero probability for the particle to be in such zones.
What is it know about the behaviour of the particle in this zone?
Links to research experiments or papers would be appreciated.
| One of the strangenesses concerning the behaviour of particles crossing a tunnel barrier is that one : crossing the tunnel barrier at superluminal group velocity is theoretically possible (see Wigner time and Hartman effect).
This has been experimentally confirmed by experiments conducted by R. Chiao with the participation of P. Kwait and A. Steinberg at Berkeley University (1993-1995). These experiments were devoted to the measurement of the tunnelling time of photons at a wavelength of 702 nm.
These measurement were achieved by comparison with the travelling time of "EPR twins" photons. These EPR twins were travelling the same optical path length, but in vacuum. They were created by Parametric Down Conversion, hence at the same time than their "EPR twins" crossing the tunnel barrier. The comparison of arrival times rested on the use of a Hong, Ou and Mandel interferometer.
The tunnel barrier consisted of a dielectric mirror with eleven quarter-wavelength layers of alternately high index material (titanium oxide with n = 2.22) and low index material (fused silica with n = 1.45). The total thickness of the eleven layers was 1.1 μm. This implied an in vacuo traversal time across the structure of 3.6 fs.
The data in Fig. 3(a) implies that after traversing the tunnel barrier, the peak of a photon wave packet arrived 1.47 ± 0.21 fs earlier than it would had it traversed only vacuum.
Tunneling Times and Superluminality: a Tutorial, Raymond Y. Chiao
http://arxiv.org/abs/quant-ph/9811019
| {
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How can one reasonably theoretically model polycrystalline materials? Many techniques are taught in advanced solid state courses but they are almost all derived for perfectly crystalline materials. For example, band structure really only appears theoretically when you look at periodic potentials that are pretty big in at least one direction.
But in experiment you often end up using materials that are polycrystalline (e.g., an evaporated film). Then, it is only periodic inasmuch the individual grains making up the sample are periodic, but they are both small (if your grains are ~1um wide, then you really only have 1000-10000 atoms in one direction, per grain) and randomly oriented, which messes up any theory I possibly know.
How can one analyze these? Is it possible? Which theories can still be used reasonably and which have to be thrown out?
Thank you!
edit: Sorry, I realized this could be a little ambiguous. I know you can probably model them computationally, but I mean more analytically, not just brute force.
| I can't give a general answer, but let me give an example. In graphene, grain boundaries have been modeled as dislocations. Their effect on transport properties can be estimated from an (analytic) Dirac equation model, and these estimates can be verified computationally, in this case using nonequilibrium Green's functions methods.
Reference: http://www.nature.com/nmat/journal/v9/n10/abs/nmat2830.html (or http://arxiv.org/ftp/arxiv/papers/1007/1007.1703.pdf)
| {
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Sum of acceleration vectors If a point mass has some accelerations $\mathbf{a_1} $ and $\mathbf{a_2} $, why is mathematically true that the "total" acceleration is $\mathbf{a}= \mathbf {a_1}+\mathbf {a_2}$?
| The expression "Total accelration" does not fit if the accelrations have different directions. The vector resultant is actually the "net accelration", or the combined effect of these two accelrations, or equivalently, forces. The vector resultant makes sure that only the effective components are added, and the opposing effects cancel out.
Maybe an example can help. Consider the following system, where a mass m is acted upon by two accelrations.
The vector resultant makes sure that the $a\sin \theta$ components are cancelled and the $a\cos \theta$ components are added up. The resultant gives the physically perceived view of motion of the object. A simpler answer would be that accelration is a physical quantity with a direction(i.e. a vector), and when you want to combine two accelrations, you calculate their vector resultant.
| {
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What's the difference between Raman scattering and second harmonic generation in crystal? As far as I know, the Raman scattering is from the stokes and anti-stokes scattering that the laser light interact with molecular vibrations. So we know that ""laser light interact with molecular vibrations"" in crystal can cause Raman scattering.
As for second harmonic generation, the pump waves with frequency w go through the crystal creating waves with 2w. I've searched information from internet and there are all explained this phenomenon by math methods, but I want to know exactly the physical model that what's happening in the crystal which create second harmonic. Is it the same as Raman scattering (incident waves interact with molecular vibrations)?
|
I want to know exactly the physical model that what's happening in the crystal which create second harmonic
One physically intuitive model for thinking about light-matter interactions is in terms of an energy level picture. In this picture, light propagating through a material can be thought of as series of absorptions and emissions. In one of these cycles the material absorbs a photon, which will excite the material into a virtual state (i.e. not a true energy state of the material). Since this state is virtual it will almost immediately decay back down into a real state emitting a photon in the process.
Below is an image which shows 3 different situations. The Rayleigh scattering does not change the energy of the light because the decay is back into the original state. Raman scattering though will decay into a different state (the different energy levels due vibrational modes of the molecules) with the difference in the energy resulting in a change in color/frequency of the scattered light.
In second harmonic generation a photon is absorbed and before the material decays into a real energy level, a second photon is absorbed. When the material re-radiates the energy it only emits a single photon, so effectively 2 photons were combined to create a single photon with twice the energy/frequency, and the final state of the material is in the same level it was to begin with (unlike the Raman scattering cases).
| {
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Can doppler shift be used to find the MH370 black boxes? The Australian ship Ocean Shield has detected multiple pings from the black boxes onboard the missing Malaysia Airline Flight 370, specifically on 4 lines of bearing according to this article. The same article also states that they need a few more lines of bearing in order to further narrow down the search area.
So my question is pretty much as in the title - can they use doppler shift to help figure out where the pings are coming from? IIRC, they used the doppler shift from the satellite data to figure out that the aircraft was most likely on the southern of the two arcs that were proposed several weeks ago.
| Doppler shift occurs only when the sender, the receiver or both are moving relatively to each other. As the black boxes rest at the bottom of the ocean and the search ships move relatively slow, there won't be any significant Doppler shift. However, if the Ocean Shield receives several signals at different locations (the location of the Ocean Shield), the position of the black box can be triangulated.
The Doppler shift from the signals the satellites received was most probably used to determine the speed of MH370 to extrapolate the most likely area in which the plane might have gone down.
| {
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Evolution of minimization of surface tension What are governing equations (or/and variational principles) for evolution of a simply connected body of water in vacuum?
Initial state - for time $t=0$ we have a bounded simply connected set $\Omega$ containing water, for simplicity - macroscopic velocity is zero and macroscopic temperature is constant.
| I'll assume that you have a well-defined interfacial energy $\gamma$ between water and "vacuum".
If there's only mechanics involved, Navier-Stokes equations with a boundary condition that the total stress along the normal $\vec n$ of the boundary (left hand side below) is normal and equal to the product of mean curvature $H$ and interfacial energy, along the normal:
$$
-p\vec{n} + 2\eta(\nabla v + \nabla v^T)\vec{n} = \gamma H \vec{n}
$$
where $p$ is the pressure, $\eta$ the dynamic viscosity and $v$ velocity.
| {
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In calculating work done by a constant force over a constant distance, why doesn't the subject's initial velocity matter? Assume a point-mass $m$ is travelling in a straight line, and a force $F$ will act on $m$ (in the same direction as $m$'s velocity) over a constant distance $d$; why doesn't $m$'s velocity matter to the calculation of work done on $m$ by $F$? Work is defined such that, in this example, the work done by $F$ on $m$ is equal to $Fd$, but it seems that if $m$ were moving slower, it would spend more time in the field, allowing $F$ more time to act on $m$, thereby doing more work. In fact, if $m$'s velocity were very great, it would hardly spend time in $F$'s field at all (so very little work done). Maybe I misunderstand work; can someone address this confusion of mine?
| Well, the reason it doesn't matter is that work is defined as
$$W = \int\vec{F}\cdot\mathrm{d}\vec{s}$$
so if you keep the force the same and the distance the same, this remains the same, regardless of what you do with the initial velocity.
Of course, that definition probably isn't particularly satisfying. So consider this: when an object is subject to a force, the rate at which that force changes its energy is given by the power,
$$P = \vec{F}\cdot\vec{v}$$
And since power is the rate at which the force transfers energy, the total work done will be the integral of power over time,
$$W = \int P\,\mathrm{d}t$$
or $W = P\Delta t$ for constant power.
When an object moves fast, then yes, it spends less time in the region with the force ($t = d/v$), but also that force produces more power $P = Fv$. These two effects cancel out exactly:
$$W = P\Delta t = (Fv)\biggl(\frac{d}{v}\biggr) = Fd$$
So the work done is the same either way.
| {
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If the MH370 black box did sink to 15000 ft, how long would it have taken? I have been following MH370 on the news just as everyone and latest reports seem to indicate that the black-box could be found. A recent info-graphic http://t.co/lyBBE9C2hF shows the insurmountable depth of the oceans and how the black-box could have sunk 15,000 ft! I wonder how long it would have taken for it to sunk to the bottom of the sea-bed? What is the equation of motion for a sinking object at sea, ignoring under-water currents?
| The relevant equation is the kinematics with linear drag. In this case, there is a resistant force that acts opposite gravity (i.e., upwards) and is linear to the velocity at which it travels:
$$
\mathbf F_D=-b\mathbf v
$$
where $b$ is some fluid- and object-dependent constant.
Using Newton's 2nd law,
$$
m\ddot{\mathbf x}=m\mathbf g - b\dot{\mathbf x}
$$
If we assume a one dimensional case,
$$x(t) = \frac{c_1 m e^{-b t/m}}{b}+\frac{g m t}{b}+c_2$$
If you know what the constants are (depends on assumptions at the boundaries, e.g. was it stationary or moving), then you can figure out the time it took.
| {
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Projection operators and their subspaces (of Hilbert space) I've been watching Susskind's lectures on Quantum Entanglement, and something he said regarding (non-)commuting projection operators confused me.
Consider two subspaces {$|a\rangle$} and {$|b\rangle$} of Hilbert space, with operators $K$ and $L$ for which:
*
*$K |a\rangle = \lambda |a\rangle (1)$
*$L |b\rangle = \mu |b\rangle (2)$
Now considers operators $P_K $ and $P_L$ that project any vector in Hilbert space onto their respective subspaces, that is:
*
*$K (P_K |\psi\rangle) = \lambda (P_K |\psi\rangle) $
*$L (P_L |\psi\rangle) = \mu (P_L |\psi\rangle) $
We want to find simultaneous eigenstates of both $K$ and $L$. If $P_K$ and $P_L$ commute: $P_K (P_L |\psi\rangle) = P_L (P_K |\psi\rangle)$. Now the left-hand satiesfies $(1)$, and the right-hand side satisfies $(2)$, so these are the required states.
In fact, if $P_K$ and $P_L$ operators commute, they share a complete set of eigenstates. The eigenstates of projection operators are those that span the subspace they project onto, so apparently $P_K$ and $P_L$ project onto the same subspace, which means they're the same operator? Then, is the statement: "projection operators commute $\rightarrow$ they're the same" correct, or do they somehow project states onto the same subspace in a different way?
Furthermore, we can imagine the subspaces geometrically as 'planes', and where these planes intersect we can find states that satisfy both $(1)$ and $(2)$. Now, according to Susskind, if $P_K$ and $P_L$ do not commute, finding such states is impossible. If the previous paragraph holds (does it?), then them commuting implies the intersection of their subspaces is the entire subspace. I don't know what non-commuting means geometrically, but shouldn't there be a case where the intersection of their subspaces isn't the entire subspace (for example, imagine two 2D perpendicular planes intersecting each other on a 1D line)? Susskind's comment seems to contradict that, and can't see exactly where I'm going wrongly.
| "projection operators commute → they're the same"
Are you sure he said this predicate ? or it is your conclusion?
Anyway, it is not true ! Consider two dimensional X-Projector And Y-Projector , they commute but they are not the same!
| {
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Why is the divergence of a magnetic field equal to zero? We know due to Maxwell's equations that:
$$\vec{\nabla} \cdot \vec{B}=0$$
But if we get far from the magnetic field, shouldn't it be weaker?
Shouldn't the divergence of the field be positive?
If we define the vector field as a function of distance, then if the distance increases then the magnitude of the vector applied to a distant point of the "source" should be weaker.
Is my reasoning correct?
| Divergence means the field is either converging to a point/source or diverging from it.
Divergence of magnetic field is zero everywhere because if it is not it would mean that a monopole is there since field can converge to or diverge from monopole. But magnetic monopole doesn't exist in space. So its divergence is zero everywhere.
Mathematically, we get divergence of electric field also zero without the delta function correction. In this case electric monopole exists in space i.e., positive charge or negative charge, and divergence is not zero wherever there is point charge or source, because field is converging to or diverging from that point/source. so after the delta function correction we get the correct result for divergence of electric field.
| {
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Is there a known equation for evolution of classical particle probability density? Suppose we have some very imprecise knowledge of classical particle's coordinates and momentum: what we can only tell is the probability density to find it in some point of phase space. This is (almost?) all what is usually known by quantum state function.
For quantum particle, there's an equation, which governs such initial state — it's Schrödinger equation.
Is there any known equation, which would similarly govern evolution of classical particle in some external potential, given initial probability density in phase space?
| It seems to me that you're looking for the Boltzmann transport equation:
$$
\frac{\partial f}{\partial t}+\frac{\mathbf p}{m}\cdot\nabla f+\mathbf F\cdot\frac{\partial f}{\partial\mathbf p}=Q+\left(\frac{df}{dt}\right)_{\rm coll}
$$
with $f$ the distribution in phase-space, $\mathbf p$ the particle momentum, $Q$ some source term, and the RHS an interaction term based on collisions. Here we can use $\mathbf F=-\nabla\Phi$ for some potential $\Phi$.
There is also a Fokker-Planck equation that uses the Lorentz force for $\mathbf F$ but is otherwise the same. The Fokker-Planck equation without the collision term is called the Vlasov equation.
| {
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How does the frequency of a particle manifest itself? In terms of wave-particle duality for, let's say a photon; how would the frequency practically manifest/demonstrate itself? Like, i understand that the frequency is related to the energy a particle has, but frequency in my mind suggests oscillation about a point. Is the photon physically oscillating through space as it travels? I wouldnt imagine so. Which periodic occurrence is referred to when one talks about the frequency of a particle?
| The easiest way to see frequencies is in interference. Imagine you have waves coming towards a wall. Imagine too that the frequency of the waves is way higher that what you can see. You cannot directly observe the waves, but you will see that the wall is wet a few centimetres over the surface. Now, instead of one wave, you have two coming from different points. In this case, you will see that your wall has wet areas and dry areas.
The distance between the dry and wet areas depends on the size of the waves, and assuming they go at a uniform speed, the size (or wavelength) and frequency are inverses.
How do you measure this experimentally? Pretty much this very same way, take light beams and make them interfere, the size of the features will be proportional to the size of the light, or inversely proportional to the frequency.
| {
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2D Gauss law vs residue theorem I used to have a vague feeling that the residue theorem is a close analogy to 2D electrostatics in which the residues themselves play a role of point charges. However, the equations don't seem to add up. If we start from 2D electrostatics given by $$\frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} = \frac{\rho}{\epsilon_0},$$ where the charge density $\rho = \sum_i q_i \delta(\vec{r}-\vec{r}_i)$ consists of point charges $q_i$ located at positions $\vec{r}_i$, and integrate over the area bounded by some curve $\mathcal{C}$, we find (using Green's theorem) $$\int_\mathcal{C} (E_x\, dy - E_y\, dx) = \frac{1}{\epsilon_0} \sum_i q_i.$$ Now, I would like to interpret the RHS as a sum of residua $2\pi i\sum_i \text{Res}\, f(z_i)$ of some analytic function $f(z_i)$ so that I would have the correspondence $$q_i = 2\pi i\epsilon_0 \text{Res}\, f(z_i).$$ For this to hold, the LHS would have to satisfy $$\int_\mathcal{C} (E_x\, dy - E_y\, dx) = \int_\mathcal{C} f(z)\, dz,$$ however, it is painfully obvious that the differential form $$E_x\, dy - E_y\, dx = -\frac{1}{2}(E_y+iE_x)dz + \frac{1}{2}(-E_y + iE_x)dz^*$$ can never be brought to the form $f(z)dz$ for an analytic $f(z)$.
So, it would appear that there really isn't any direct analogy between 2D Gauss law and the residue theorem? Or am I missing something?
| The analogy follows with the right definitions. The "flux" of the "vector" $E(z)$ through a contour $\Gamma$ is $\mathrm{Re}\left(\int_\Gamma E(z)^*\,\mathrm{d} z\right)$.
I think you may have forgotten the conjugate in the relationship between the "Electric field" and the complex potential $\Omega$: $E(z) = -(\mathrm{d}_z \Omega(z))^*$.
So it is the conjugate of $E(z)$, not $E(z)$ itself, that is holomorphic, being given by $E(z)^* = -\mathrm{d}_z\Omega(z)$. Work out the direction of the vectors from the Cauchy-Riemann relations and you'll see that you need a conjugate to make $-\nabla \phi$ equal to the real part of the derivative $-(\mathrm{d}_z\Omega(z))^*$.
Hence $\mathrm{Re}\left(\int_\Gamma E(z)^*\,\mathrm{d} z\right)$ is the real part of an everyday contour integral and the flux is thus the real part of $2\,\pi\,i$ times the sum of residues at the poles of $E(z)^*$, whence the 2D Gauss law follows.
| {
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What is an intuitive explanation using forces for the equatorial bulge? The earth is not a sphere, because it bulges at the equator.
I tried fiddling with centripetal force equations and gravity, but I couldn't derive why this bulge occurs.
Is there
(a) a mathematical explanation using forces (not energies) and
(b) a simple intuitive explanation to explain to others why the bulge occurs?
| There is a wikipedia article which describes the effect http://en.wikipedia.org/wiki/Equatorial_bulge
Basically the bulge is caused by the rotation of the Earth. The centripetal force is given by $F=m\omega^2 r$. Therefore the poles feel a lesser force than the equator which wants to spin out into a disc. This is balanced by gravity which wants the Earth to be spherical.
Mathematically the flattening of the Earth is
$$ f = \frac{5}{4} \frac{\omega^2r_a^3}{GM}$$ where $r_a$ is the average radius and $f$ is the ratio of the major and minor radii.
| {
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In what sense do Goldstone bosons live in the coset? Goldstone's theorem says that if a group, $G$, is broken into its subgroup, $H$, then massless particles will appear. The number of massless particles are given by the dimension of the coset, $G/H$. It is then often said that the Goldstone boson's live in the coset. In what sense is this statement true? The Lagrangian is not invariant under transformations of the coset so what does this "living" explicitly mean?
To be explicit we can consider the linear sigma model:
\begin{equation}
{\cal L} = \frac{1}{2} \partial _\mu \phi ^i \partial^\mu \phi ^i - \frac{m ^2 }{2} \phi ^i \phi ^i - \frac{ \lambda }{ 4} ( \phi ^i \phi ^i ) ^2
\end{equation}
We define,
\begin{align}
& \phi _i \equiv \pi _i \quad \forall i \neq N\\
& \phi _N \equiv \sigma
\end{align}
and give $\sigma$ a VEV.
The spontaneously broken Lagrangian is,
\begin{equation}
{\cal L} = \frac{1}{2} \partial _\mu \pi _i \partial ^\mu \pi _i + \frac{1}{2} ( \partial _\mu \sigma ) ^2 - \frac{1}{2} ( 2 \mu ^2 ) \sigma ^2 - \lambda v \sigma ^3 - \frac{ \lambda }{ 4} \sigma ^4 - \frac{ \lambda }{ 2} \pi _i \pi _i \sigma ^2 - \lambda v \pi _i \pi _i \sigma - \frac{ \lambda }{ 4} ( \pi _i \pi _i ) ^2
\end{equation}
The Goldstone bosons, $\pi_i$, exibit a $O(N-1)$ symmetry, but this is not the coset group symmetry. So where in the Lagrangian do we see this symmetry?
| I understand the statement in the following way:
Pions, which are pseudo-goldstone bosons of chiral symmetry breaking, are described by the introduction of a unitary matrix $U(x)$, defined as
$$U(x)=\text{exp}\left(2i\pi^a(x)T^af_\pi^{-1}\right),$$
where $\pi^a$ is the pion field, $f_\pi$ is the pion decay constant and $T^a$ are the generators of the broken symmetry, i.e. the coset space. The pion Lagrangian can be written down in terms of $U(x)$
$$\mathcal{L}=-\frac14 f_\pi^2\text{Tr}\partial^\mu U^\dagger\partial_\mu U,$$
which by expanding the exponential form results in
$$\mathcal{L}=-\frac12\partial^\mu \pi^a\partial_\mu \pi^a+\dots,$$
where dots denote higher order terms. Thus, the statement that goldstone bosons live in the coset space can be related to the fact that the fields themselves are linked to the generators of the coset.
This can be understood in terms of Goldstone's theorem: if the original Lagrangian exhibits a continuous symmetry, the number of goldstone bosons is equal to the number of generators of the broken symmetry. Take for example the linear sigma model: if your original theory is $O(N)$-symmetric, it has $N(N-1)/2$ symmetries. If the symmetry is broken spontaneously, you end up with $O(N-1)$, leaving you with $(N-1)(N-2)/2$ symmetries. The amount of broken symmetries is the difference, i.e. $N-1$. But this is precisely the number of pions you have in your theory. We can conclude that the pions are linked directly to the broken symmetries, i.e. the coset space.
| {
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Why can colors be mixed? We can combine colored light, creating other colors, at least in terms of visual perception.
But how it the result physically "a different color" - if it is at all?
Or is all this not a physical question to begin with - but only about our eye and brain?
To have an example, we
*
*have an incandescent bulb, showing "white" light, and
*combine red, green and blue light in intensities such that it looks roughly the same.
It is not central to the question whether it is exactly matching the white light - but certainly interesting to understan whether it could perfectly match, and why.
| In my answer to this question: (What is the sun's spectral series?), I give a very detailed answer about why mixing colours of light produces other colours and how it is purely a result of biology and evolution. I also delve a bit into the structure of the human eye and why, in fact, only three colours are necessary to reproduce all of the colours we can see. It touches on much of what Floris has written, but it has more pictures and is written slightly more for a layman explanation.
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Finding the vacuum which breaks a symmetry I will start with an example. Consider a symmetry breaking pattern like $SU(4)\rightarrow Sp(4)$. We know that in $SU(4)$ there is the Standard Model (SM) symmetry $SU(2)_L\times U(1)_Y$ but depending on which vacuum we use to break this symmetry, in a case you can totally break the SM symmetry, with the vacuum :
$$\Sigma_1 = \begin{pmatrix}
0& I_2 \\
-I_2 & 0
\end{pmatrix}$$
and in another case, these symmetry is preserved, with the vacuum
$$\Sigma_2 = \begin{pmatrix}
i\sigma_2& 0 \\
0 & i\sigma_2
\end{pmatrix}$$
In the first case (with $\Sigma_1$), the generators corresponding to the SM symmetry are part of the broken generators so the SM symmetry is totally broken.
In the second case ($\Sigma_2$), the SM generators are part of the unbroken generators then the SM symmetry is preserved. As you can read, I know the answers but not how to find them !
So, my questions are :
*
*How is it possible in general (not only for the $SU(4)\rightarrow Sp(4)$ breaking pattern) to construct the vacuum that breaks the symmetry ?
*Is it possible, when constructing the vacuum, to ensure that the vacuum will (or not) break a sub-symmetry like the SM symmetry in the previous example ?
| *
*As your QFT theory text should tell you, for an action invariant under G, addition of a Higgs potential only invariant under its subgroup H will spontaneously break the generators in G/H. You ought to do due diligence and study and understand and reproduce all examples of elementary classics such as Ling-Fong Li, PhysRev D9 (1974) 1723. There are, of course, far too many such treatises in the groaning literature!
*Usually it is possible, but this is a question depending on the particular circumstances of G and H. If you are ambitious, you could run through the tables of Slansky's 1981 review article to reassure yourself. For your particular example above, the answer is "yes". SU(4) has 15 generators, Sp(4) has 10, and the SM has but 4. Preserving your symplectic metric $\Sigma_1$ preserves Sp(4), but you may check by writing down the generators that the alternative does not; nevertheless you may rewrite $\Sigma_2= 1\!\!\! 1 \otimes i\sigma_2$ with the 2x2 identity matrix on the left, preserving SU(2) of course; and the right group trivially preserves itself, a U(1), at the very least!
| {
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Studying Quantum Electrodynamics? As an electrical/computer engineer, I already have a relatively thorough understanding of classical electromagnetism.
From what I understand though, classical EM is only an approximation to quantum electrodynamics. I'm very curious about how it all really works though.
So as an ECE engineer, what would be the best way to approach quantum electrodynamics? (assuming taking a course at a community college is not an option)
| In order to gauge your level of previous knowledge, you could try Engineering quantum electrodynamics, by Dietrich Marcuse ( https://books.google.com.br/books?id=SGd5AAAAIAAJ&dq=editions:STANFORD36105030208412&hl=pt-BR&redir_esc=yEngineering quantum electrodynamics )
| {
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Can molten metal be suspended in air? I wondered if magnets could be used to hold a drop of molten liquid metal in air (not for any particular reason just because it could be done), but was disappointed when a quick Google search showed the metal would lose its magnetic traits before it melted.
Are there any other forces that could be used to suspend a drop of molten liquid metal in air such as sound waves, high pressure air, electric currents, or anything else?
| Yes, it is possible to magnetically levitate molten metal. This is not due to ferromagnetism however. As seen in the below references, the metal sample is placed within a tapered conducting coil, which carries alternating electric current in the ~400kilohertz range. This sets up a magnetic field gradient inside the coil and causes eddy currents in the sample. The applied magnetic field is excluded from the interior of the sample. The sample experiences a force in the direction of decreasing magnetic field strength.
See the following for more information:
http://www.youtube.com/watch?v=intDuSJ2_PA
http://www.youtube.com/watch?v=Q6Zrnv4OtbU
http://www.google.com/patents/US2686864
http://www.google.com/patents/US2686865
http://www.cs.duke.edu/~reif/temp/MagneticLevitation/LevitationSuveys/Levitation%20in%20Physics.pdf
http://www.modlab.lv/publications/mmp2010/pdfs/015-020.pdf
| {
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Is the molecule of hot water heavier than that of cold water? We know that the molecule of hot water($H_2O$) has more energy than that of cold water (temperature = energy)
and according to Einstein relation $E=mc^2$ ,this extra energy of the hot molecule has a mass.
Does that make the hot molecule heavier?
| I am going to take a different approach from DavePhD and Floris.
"Hot"ness or temperature more generally is a thermodynamic idea, and can't really be applied to an individual molecule. Dave and Floris have avoided the issue by simply comparing and energetic molecule with a less energetic one, and that is reasonable, but it makes their answers frame-of-reference dependent. Presumably they are working in the rest frame of the macroscopic samples from which they drew their test particles. All very reasonable.
I'm going to make my usual argument about scale of inspection.
A mole of hot water is more massive than a mole of cold water because when examined at the human scale you can't differentiate the kinetic energy of the molecules from any other form of internal energy (and energy is mass). The scale of this change is that figured by Floris.
Examined at the level of a single molecule than each particle has the same mass--properly defined as the Lorentz scalar formed by contracting it's four-momentum with itself: $m^2 = \bar{p}\cdot\bar{p}$--and the fast molecule has more kinetic energy--$T = E^2 - m^2 = (\gamma - 1)m$--than the slow one. This approach follows the conventions of particle physics where we don't use the term "relativistic mass".
The upshot is a "Yes and no" answer, or perhaps a "You're not quite asking the right question".
| {
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Why does the Fermi Surface cross the Brillouin zone boundary at right angles? I'm not sure why the fermi surface crosses the Brillouin zone boundary at right angles. I understand that this is normally the case, but not necessarily always.
I'm aware that the fermi surface is a constant energy surface up to the filling point. The Brillouin zone is in reciprocal space.
| This answer nothing to the OP's question, please don't vote up anymore; and anyone who know the answer to this question please share.
This is mainly due to the time reversal symmetry.
Consider the Bloch equation:
$$[-\frac{\hbar^2}{2m}\nabla^2+U(r)]\psi_{nk}=\epsilon_{nk}\psi_{nk}$$
Recall that $\psi_{nk}=e^{ik\cdot r}u_{nk}$, then we have:
$$[-\frac{\hbar^2}{2m}(-i\nabla+k)^2+U(r)]u_{nk}=\epsilon_{nk}u_{nk}$$
Now we want to prove $\epsilon_{nk}=\epsilon_{n-k}$, take the complex
conjugate of the above equation and change $k\to-k$:
$$[-\frac{\hbar^2}{2m}(-i\nabla+k)^2+U(r)]u_{n-k}^*=\epsilon_{n-k}u_{n-k}^*$$
We can see that $\epsilon_{n-k}$ is the same set of eigenvalues as
$\epsilon_{nk}$ of the same Hamiltonian $H_k$. Thus they must be
equal.
Now let's answer your question:
Consider one particular band $n_0$, its zone boundary are $-K/2$ and
$K/2$.
$\epsilon_{n_0,K/2+\Delta k}=\epsilon_{n_0,-K/2+\Delta
> k}=\epsilon_{n_0,K/2-\Delta k}$
Let's $\Delta k$ tends to infinity small, the above equation just
means that the first derivative of energy band near the zone boundary
is zero.
So that when the filling of electrons doesn't modify the band
structure, you will always see the fermi surface perpendicular to the
zone boundary if time reversal symmetry is respected.
| {
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Could an orbiting mass generate power? If an heavy object (e.g. 10 tons) orbiting around Earth at 370 miles high, is connected with a cable back to Earth, we assumed either Earth is going to pull the mass or vice versa (or it will fall back to Earth). Assuming correct pressure/release/pull can be applied from the ground to prevent the fall, or it naturally starts spinning around in a circular motion - what do you think about its power generation capability?
This was a fun chat between friends, so just wanted to run it by.
| This would be space elevator type of setup but it is too short to stay in orbit. It should extend beyond the geostationary orbit (~22,000 miles). International space station is a an altitude of ~200 miles and goes around the planet every 90 mins. Even if you build a proper space elevator there is no way to generate more energy than you already use to build it.
| {
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simple question about collimating lens Let me preface this by saying that I have a very limited knowledge of optics -- basically, I know enough to be dangerous. So I have a square Fresnel lens that measures about 10.5 inches on a side. The focal length is about 11.5 inches. I'm trying to find a presumably-smaller glass collimating lens that will take the light from the Fresnel lens and convert it to a beam with an indefinite focal length (or at least, a focal length of a few feet). I don't care about image preservation, I just mainly want to transfer some focused light from point A to point B with a minimum of loss. I basically need to know what kind of lens to buy and where to place it. I've tried single and double convex and concave lenses and seemed to have the most luck with single convex, but wasn't able to get much of a beam past a couple of inches out. The attached picture shows kind of what I am looking for (lens on left represents Fresnel lens, lens on right represents collimating lens).
Any help you could give me with this would be great.
| Since your diagram indicates a point source, you only need one lens to do this. If you place the Fresnel lens at a distance equal to its focal length from the light source you will get a parallel beam of light.
| {
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Interpretation of Conjugate Momentum in Field Theory The conjugate momentum density, following as a conserved quantity with Noethers Theorem, from invariance under displacement of the field itself, i.e. $\Phi \rightarrow \Phi'=\Phi + \epsilon$, is given by $\pi=\frac{\partial L}{\partial ( \dot{\Phi})}$.
On the other hand the physical momentum density, following as a conserved quantity with Noethers Theorem, from invariance under translations, i.e. $\Phi(x) \rightarrow \Phi(x')=\Phi(x+\epsilon)$, is given by $\Pi = \frac{\partial L}{\partial ( \dot{\Phi})} \frac{\partial \Phi}{\partial x}$.
Does anyone know a enlightening interpretation of the conjugate momentum? Furthermore why do we, in quantum field theory, impose commutation relations with the conjugate momentum instead of the physical momentum?
(For brevity all possible indices are supressed)
| The momentum you call $\Pi$ is the momentum in a certain spatial direction. The momentum you call $\pi$ is the momentum in a direction in field space.
For example, consider vertical waves on a horizontal string. The "field" $y(x, t)$ represents the vertical displacement of the string at point $x$. Then $\Pi$ is the ordinary horizontal momentum, since waves on the string move horizontally, while $\pi$ is the ordinary vertical momentum, since the string itself only moves up and down.
| {
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What is the Physical Significance of Tr(A) w.r.t. Matrix Representations in Group Theory I've seen the post on mathoverflow.SE asking almost the same question, and I have indeed flipped through said answers, but most are in a more general context ie quantum mechanics and do not provide a conceptual answer with physical interpretation. Anyone able to offer any insight, or even an example in the aforementioned context? ( More particularly the theory of representations for symmetry groups. )
| In physics one tends to write (for a Yang-Mills field), $A_{\mu}^i$, where $\mu$ is the spacetime index and $i$ is the `group' index. To be more specific, it means that $A_{\mu}$ take values on (i.e., is contracted with the generators of) a Lie algebra, $$A_{\mu} = A_{\mu}^i T^i = A_{\mu}^i (T^i)_{mn}, $$ where in the las equality the explicit matrix indices have been written.
Thus, $Tr(A_\mu A_\nu)$ means $$Tr(A_\mu A_\nu) = A_{\mu}^iA_{\nu}^j \; Tr[(T^i) (T^j)] = A_{\mu}^iA_{\nu}^j \; (T^i)_{mn} (T^j)_{nm} .$$
As you might notice the trace acts on the matrix indices of the group generators.
| {
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Does electricity flow on the surface of a wire or in the interior? I was having a conversation with my father and father-in-law, both of whom are in electric related work, and we came to a point where none of us knew how to proceed. I was under the impression that electricity travels on the surface while they thought it traveled through the interior. I said that traveling over the surface would make the fact that they regularly use stranded wire instead of a single large wire to transport electricity make sense.
If anyone could please explain this for some non-physics but electricly incline people, I would be very appreciated.
| The short answer is the surface. Being in a car during a lightening strike or high voltage line drop would kill you. Also think of the Tesla videos where someone is wearing a suit of armor and doesn't die from the arcs of electricity hitting him in the head; the difference in potential from his head to his feet, although only for a moment, is enough to kill him otherwise.
| {
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Feynman diagram for annihilation What is the difference between these two Feynman diagrams?
They should both describe the same physical process, annihilation between an electron and a positron.
| The first process corresponds to $e^{-}e^{+}\to e^{-}e^{+}$ (Bhabha scattering), where the final and initial states are the same, consisting of an electron and positron. However, the second process is $e^{-}e^{+}\to \gamma \gamma$, where instead the final state is that of two photons. The scattering amplitudes will be different. Notice that the first diagram requires an insertion of the photon propagator,
$$-\frac{i\eta_{\mu\nu}}{q^2 +i\epsilon}$$
whereas the second diagram has a fermionic internal line, requiring a propagator,
$$\require{cancel} \frac{i\left(\cancel{q}+m_f\right)}{q^2-m_f^2 +i\epsilon}$$
In addition, the second diagram will contain polarization vectors, as the photons are not internal lines but rather external. For a comprehensive overview of QED, see Peskin and Schroeder's text.
| {
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Is refraction sharp or smooth? Refraction: light changes direction of propagation when entering a material with a different refractive index.
Does the direction of propagation of light change sharply and almost instantaneously (as shown in the diagram) or smoothly?
| Generally speaking, it's impossible to physically observe a strictly "sharp" phenomenon in physics (there still some exceptions of course). However one phenomemon can have some characteristic constants which vary much more faster compared to others phenomema.
Here, in the geometrical optics approximation, since you are only interested in what happens far from the interface, it is "ok" to describe a sharp direction change of your ray path.
But if you are zooming to see what is going on close to the interface, the ray path will always experience some gradient of the index of refraction $\nabla n$.
To calculate the ray path through a middle with a gradient of index $\nabla n$, one can use the Fermat's principle, i.e. a ray of light always takes a path with a stationary optical length $S$, so that :
$$\delta S=\delta\int_\mathcal{C}n(s)\,ds=0$$
where $s$ is the curvilinear coordinate along the ray path.
From this principle follows the eikonal equation which is an Euler-Lagrange-like equation for rays path:
$$\frac{d}{ds}\left(n(\mathbf{r})\frac{d\mathbf{r}}{ds}\right)=\frac{\partial n}{\partial\mathbf{r}}$$
where $\mathbf{r}\in\mathbb{R}^3$ are the ray path coordinates.
Solving this equation will give you the curved ray path $\mathbf{r}$ associated to the gradient $\nabla n=\frac{\partial n}{\partial\mathbf{r}}$ that the ray is experiencing for its propagation through the medium.
| {
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Higgs boson production via positron-electron collision One of the suggested diagrams for the Higgs production is the following:
so basically an electron-positron pair annihilates and forms an (excited?) Z boson, which then decays into another (less excited?) Z boson and a Higgs boson.
Why can't the electron-positron pair decay directly into a Higgs boson?
Charge and lepton number would be conserved anyway, and if the pair has enough energy to produce the $Z^*$ boson in the first place it should have enough energy to produce the Higgs boson... ?
| The electron-positron pair can produce directly a Higgs boson, but this process is very suppressed, because the coupling between the leptons and the Higgs is proportional to the tiny mass $m_e$:
$$g_{\rm Hee}=-i\frac{ m_e}{v},$$
where $v\approx 246 \,\rm{GeV}.$
On the other hand, the process $e^+ e^-\to H Z$ is more likely to happen, because the coupling between $H$ and $Z$ is proportional to the $W$ mass:
$$ g_{\rm hZZ}^{\mu\nu}=i g \frac{M_W}{\cos^2\theta_W} g^{\mu\nu}. $$
In the latter case we also have to take in account the propagator of the $Z$ boson, which introduces a suppression factor of order $1/m_Z^2$, but at the end we still have a larger cross section.
| {
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Deriving formula for time dilation Last week in class we derived the formula for time dilation using light clocks and got $$t=\gamma t_0\quad \gamma=\left(1+\left(\frac vc\right)^2\right)^{-1/2}$$
So far so good. However, after class I was thinking of an alternate proof, but for some reason I got another result and I can't tell why.
Basically you have a train with an observer A inside who emits a beam of light to the left which is reflected off a wall at distance $d$ from A. The time it takes for the beam to get back to the observer is $t_0=\frac{2d}{c}$ which is the proper time.
Now consider an observer B outside the train. The train is moving at a velocity $v$ to the right relative to B. Thus the time it take for the light to hit the wall is $\frac{d}{c+v}$ and the time it take for it to return to A is $\frac{d}{c-v}$. Thus the dilated time is
$$t=\frac{d}{c+v}+\frac{d}{c-v}=\frac{2dc}{c^2-v^2}=\frac{2d/c}{1-\left(\frac vc\right)^2}=\frac{t_0}{1-\left(\frac vc\right)^2}=\gamma^2t_0$$
Where did I go wrong?
| Light always travels with $c$. That is why Michelson-Morley's experiment is so famous and groundbreaking.
If you send a beam of light from point A to point B, and it gets reflected and comes back to B than it doesn't matter, if A is moving relative to B with velocity v or not. You still get the same total velocity for the light: $c$.
Another thing, in SR you cannot change perspectives in the middle of experiments and treat measurements made by different observers as equal. Not only $t$ is affected by the difference in velocity. Your $d$ is also perceived differently from different frames of reference.
Incidentally, if I understand your set-up correctly, it looks like the light is traveling perpendicular to the direction of the movement of your train, and yet you simply add the values of $c$ and $v$ (regardless of the constancy of $c$).
| {
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Is frequency quantized in the black body spectrum? I'm aware that there're some questions posted here with respect to this subject on this site, but I still want to make sure, is frequency quantized? Do very fine discontinuities exist in a continuous spectrum like the black body spectrum?
The quantization of photon energies
| Frequency is not quantized, and has a continuous spectrum. As such, a photon can have any energy, as $E=\hbar\omega$. However, quantum mechanically, if a particle is restricted by a potential, i.e.
$$\hat{H}=-\frac{\hbar^2}{2m}\nabla^2 + \hat{V}$$
for $V\neq 0$, the energy spectrum is discrete. For example, in the case of the harmonic oscillator,
$$E_n=\hbar \omega \left( n+\frac{1}{2}\right)\quad n=0,1,2,\dots$$
| {
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Why can't an excess of electrons or holes by themselves cause current flow? I am a beginner in electrical engineering. Often times (most cases actually), the underlying physics aren't really explained to us and we are just left to assume that it works "because it works." This is never enough for me in classes etc and I always end up doing a follow up of physics side of the spectrum.
My question is, you have a battery with excess electrons built up on the negative lead, and excess holes on the positive lead, why is it that in our universe then connecting a conductive compound (like.. copper) to just the negative lead does not produce current, or vice versa connecting the wire to just the positive lead.
After all, if we move a magnet passed a conductor, there is a tiny induced current. How can the magnet do this, but an excess of electrons or electron holes repelling each other cannot?
| If you connect a piece of wire to just one terminal of a battery, the whole wire tends to gain the same potential as that of the terminal it is connected to, this requires an instantaneous current and is achieved very quickly.
The need of connecting the wire to both terminal, is that the internal chemical mechanism of the battery maintains a potential difference across the wire, thus the wire never reaches a desired potential and continues to try to reach it by flowing current.
The moving magnet also provides a stable potential difference between 2 points on a conductor which just by transference of electrons isn't diminished or nullified, because it is controlled by the moving action of the magnet.
Whereas, the excess/lack of electrons, provide a potential differece only till redistribution takes place and whole conductor gets to the same potential.
| {
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Exact diagonalization to resolve ground state degeneracies I am studying a perturbed Toric Code model that is not analytically solvable.
On a torus the ground state degeneracy of the unperturbed model is 4.
Once we turn on the perturbation there is a change in the ground state degeneracy.
I would like to detect this change in ground state degeneracy numerically using exact diagonalization techniques.
On my computer I have stored the action of the Hamiltonian on a set of basis states. So if you give me some state $\left|\psi\right\rangle$ I can give you
$\hat{H}\left|\psi\right\rangle$ in terms of the basis states. Now I used this information to compute the spectrum using the Lanczos algorithm and the Jacobi-Davidson algorithm. While the spectrum itself is correctly reproduced neither of these algorithms reproduces the degeneracy of the ground state correctly - not even in the unperturbed case.
Hence the following question: What are common exact diagonalization algorithms for this type of many-body system that correctly resolve the degeneracy of the ground state?
I am looking forward to your responses!
| What implementations of the two algorithms are you using? In my understanding, the Jacobi-Davidson algorithm is supposed to be good at dealing with degeneracy.
In my personal experience, using the ARPACK package is usually good enough even for highly degenerate eigenvalue problems (~10 fold or even more), as long as a reasonably large workspace is provided.
ARPACK uses what's called the Implicitly Restarted Lanczos Method. All it requires is the application of the matrix on any given vector, which is exactly what you have.
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Why does squeezing a water bottle make the water come out? This seems natural, but I can't wrap my head around it when I think about it.
When I squeeze an open bottle filled with water, the water will spill out. When I squeeze a bottle, the material collapses where I squeeze it, but expands in other areas, resulting in a constant volume. If the volume is constant, then I would think that the water shouldn't spill out.
If I were to guess, there is something related to the pressure my hand is creating inside the bottle, but I'm not entirely sure.
| The surface area of the bottle is conserved, but the volume is not. Squeezing the bottle deforms it into a shape whose volume to surface area ratio is lower than it was previously.
As an example consider a bottle whose cross-section is initially a circle. The volume of the bottle will be $V_0=\pi r^2h$ where $h$ is the height of the bottle, and the surface area will be $A_0=2\pi rh$.
As we squeeze the bottle we deform it into some kind of ellipse. If we choose the major and minor axes of the ellipse to be $a=xr$ and $b=\sqrt{2-x^2}r$, then the new surface area of the bottle is given approximately by
$$
A_1\approx h 2\pi\sqrt{\frac12(a^2+b^2)}=2\pi rh\quad\Rightarrow\quad\frac{A_1}{A_0}=1,
$$
where we have neglected the end caps. So, the surface area of the bottle is conserved. The new volume is given by
$$
V_1=\pi abh=\pi r^2h x\sqrt{2-x^2}\quad\Rightarrow\quad\frac{V_1}{V_0}=x\sqrt{2-x^2},
$$
where we have again neglected the end caps. So you can see that even though the surface area is conserved, the volume is maximum only when $x=1$. A plot of the last term, $x\sqrt{2-x^2}$ is shown below to confirm this fact.
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Why electrons have less energy than photons with the same wavelength? I am studying quantum physics and I have a question: what is the physical explanation for electrons having less energy than photons with the same wavelength?
Energy of a photon : $E = h c/\lambda$.
Energy of an electron: $E = h^2/(2m\lambda^2)$
| A slightly different viewpoint is the following. First, note that the formulae given by the OP do not in themselves exclude the possibility that the electron has the same or greater energy than a photon of the same wavelength. Indeed, for small enough $\lambda$ the non-relativistic formula $E = h^2/(2m \lambda^2)$ predicts that the electron has greater energy than the photon. The "critical" wavelength $\lambda_c$ where the crossover happens is found by equating the two expressions, giving
$$ \frac{hc}{\lambda_c} = \frac{h^2}{2m\lambda_c^2} \quad \Longrightarrow \quad \lambda_c \approx \frac{h}{mc},$$
ignoring the factor of 2. But this quantity has another meaning: it is the Compton wavelength. This is the wavelength at which the electron kinetic energy is approximately equal to its rest mass:
$$ \frac{h^2}{2m\lambda_c^2} \approx m c^2. $$
Above this energy, the spontaneous production of electron-positron pairs starts to become important. Therefore, the non-relativistic concept of a "single electron" loses its meaning once the energy becomes comparable to a photon of the same wavelength.
| {
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Effect of wavelength on photon detection When some photon detector detects a photon, is it an instantaneous process (because a photon can be thought of as a point particle), or does the detection require a finite amount of time depending on the wavelength of the photon?
EDIT: I guess what I am wondering is if a photon has a wavelength and travels at a finite speed, then if a photon had a wavelength of 300,000,000m, would its interaction with the detector last 1s? Or does the uncertainty principle say that a photon with wavelength 300,000,000m (and therefore energy E), it cannot be known exactly when it hit the detector with an accuracy better than 1s. Or is it more like this: suppose there is a stream of photons moving towards the detector with wavelengths of 300,000,000m and they reach the detector at a rate of 10 photons/second and the detector has a shutter speed such that the shutter is open for 1s at a time, then it would record 10 photon hits (records all the photons). But if the shutter speed is only 0.5s, then it would record 2.5 hits on average?
EDIT2: I'm not interested in the practical functioning of the detector and amplification delays. I'm looking at and ideal case (suppose the photon is 'detected' the instant an electron is released from the first photomultiplier plate). It is a question regarding the theory of the measurement, not the practical implementation.
| Well I would argue a much simpler and shorter explanation:
That the measurement of a photon will collapse the wave function of the photon and therefore the system will no longer be a quantum mechanical one but classical. Before the measurement there will be a error due to the uncertainty principle, see it like the photon wave is hitting the detector but the photon is not yet detected. Once detected, the wave function is localized.
So to answer your question. It is an instantaneous process. But due to QM there will be a uncertainty in your measurement
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How are complex sound waves combined? Audio is often explained by single frequencies. Typically this is a sound wave:
plot sin(x) * 2 from 0 to 10
However we usually deal with more complex sounds, more specific various frequencies and amplitudes. Varying amplitudes within same frequency etc. in the same time frame.
How are such sound patterns visualized? Say we have the previous tone + this one:
plot sin(5x) * 2 from 0 to 10
Would this be the end-result? Would a wave like this travel trough the air?
plot (sin(5x) + sin(x)) * 2 from 0 to 10
Even a mono track can have a rather detailed sound picture of say a philharmonic playing some composition. Would such a recording (or live observation for that matter – but that would give a different picture I guess as one would have multiple sound sources) be a complexified × ten folds version of the two sine waves?
My end goal is to better understand PCM audio and how digital audio works. A starting point is to better understand the physics behind audio. Then again a most of what I find deals with one frequency samples and the like. (I'm likely missing some terminology.)
| Yes, you have the right idea. You will want to learn about Fourier analysis, which lets you take a complicated-looking waveform like your third figure and analyze it to say "this is two sine waves, frequencies 1 and 5, equal amplitudes, zero relative phase."
I like to think of a piano as an inverse Fourier transform machine: you push the keys to tell the piano "please generate frequencies C, E, and G, with the C having larger amplitude than the others" and the piano makes the air vibrate for you. Your auditory system then does the ordinary Fourier transformation: with ear training, you can take those vibrations and say "Oh, a major triad, with a strong root."
| {
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Why fermions have a first order (Dirac) equation and bosons a second order one? Is there a deep reason for a fermion to have a first order equation in the derivative while the bosons have a second order one? Does this imply deep theoretical differences (like space phase dimesion etc)?
I understand that for a fermion, with half integer spin, you can form another Lorentz invariant using the gamma matrices $\gamma^\nu\partial_\nu $, which contracted with a partial derivative are kind of the square root of the D'Alembertian $\partial^\nu\partial_\nu$. Why can't we do the same for a boson?
Finally, how is this treated in a Supersymmetric theory? Do a particle and its superpartner share a same order equation or not?
| Steven Weinberg starts with symmetries of relativity, and quantum framework, and arrives in chapter 5 of volume one at Dirac equation. He first gets the quantum field for spin 1/2 particles (without reference to any Lagrangian or wave equation), then he constructs vacuum expectation value of the field and its
adjoint, this leads to Feynman-Dyson propagator. Earlier in the analysis he gets the Dirac operator when he calculates spin sums. Acting the Dirac operator on the FD propagator then yields, up to a sign, the four dimensional Dirac delta function for x-y. From this follows the Lagrangian density.
Mathematically, the linearity depends on the fact that (\vec Pauli matrix . unit momentum vector)^2 = identity matrix.
If you need more details, please consult my monograph (Mass Dimension One Fermions, Cambridge Monographs on Mathematical Physics, Cambridge University Press, 2019). And one may evade the linearity you write about.
Wait for a preprint soon -- on spin 1/2 bosons with linear wave equation.
| {
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How to find the electric field due to a point charge in 3 dimensions? A point charge with charge $+q$ is situated at $(x,y,z)$. How do I find the electric field at $(p,q,r)$? $E=k\frac{q}{r^2}$, right? So why isn't the electric field, $E=k\frac{q}{(x-p)^2+(y-q)^2+(z-r)^2}$ ???
Question:
A charge $q=1\,\mu C$ is placed at a point (1m, 2m,4m). Find the electric field at point P (0m,-4m,3m).
I end up with $E$ = 2.38x10$^2$ N/C; the given answer is 1.46x10$^3$ N/C
| The difference between your (correct) answer and the incorrect given answer is a factor of 6.16, which is the length of $(x,y,z)-(p,q,r)$ in meters. Your textbook answer is the value of the potential at $(p,q,r)$, in volts.
| {
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Why doesn't this model plane fly? I have been designing a model plane for Design Technology for the past month or so, and today I laser cut my final design and assembled, it then tested it. Upon testing the plane does not get any lift, whereas the previous testing model which was virtually the same did.
The plane is built using Balsa Wood, and Assembled with hot glue (I used as little glue as possible to reduce weight :) )
Any Ideas?
Image:
| Most model planes don't have wings aerodynamic enough to lift it by those means. Honestly, put a rotor strong enough on one side of a plain board and it will fly also.
| {
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Nonzero ground state energy of the quantum harmonic oscillator Since $\frac{1}{2}\hbar \omega$ is the zero point energy of the ground state of the harmonic oscillator, then there is no way to extract this energy.
Therefore, in what way is this value different from zero? Is not it just about where we decide to choose the reference point?
| You are correct in that for any given harmonic oscillator we can define the zero of the energy so that the ground state has zero energy. However, there are two things to point out.
*
*Coming from a classical perspective, it's still a curiosity: The harmonic potential itself, $\frac{1}{2} m \omega^2 x^2$ has $0$ as its minimum, and the ground state energy lies exactly $\frac{1}{2}\hbar \omega$ above that minimum. So, even if we shift the energy around to have the ground-state energy be $0$, then the potential minimum is at $-\frac{1}{2}\hbar \omega$, i.e. there still is a finite difference between the energy of the ground state and the minimum of the potential.
*Different oscillators have different $\omega$ and thus different zero point energy, so we can't choose a reference point where they all have zero ground state energy. Or maybe in your system, $\omega$ isn't even a constant but depends on other parameters of your system. This is the handwaving explanation behind the Casimir effect: Bringing two metal plates closer together reduces the zero point energy and, as a consequence, those two metal plates experience an attractive force.
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Twin Paradox: Still a Paradox? Alright, so David Griffiths in his "Introduction to Electrodynamics" states that the Twin Paradox is not a paradox at all since the traveling twin returns to Earth. By returning to Earth, the twin had to reverse direction, thus undergoes acceleration, and therefore cannot claim to be a stationary observer.
However, what if the traveling twin simply Skypes the twin that is on Earth. The twin on earth will still appear older, which would make no sense since in that case the rocket can be seen as the stationary frame of reference while the Earth "travels" at a speed close to the speed of light. No acceleration is undergone, yet the paradox remains.
Is Griffiths just completely glossing over important nuance again?
| No absolute relativity as long as a constant reference exists this reference is speed of light (this reference will be responsible for defining the moving body
Earth twin will see that other twin's clock is slow because it is really slow
But space twin will not see that other twin's clock is slow because his time
Perception is also slow
Space twin will see earth moves far from him but by shorter distance than the distance seen by earth twin, so if we supposed that space twin is moving far from earth with the exact speed of light he will never see that earth is moving because he is moving with the light that is moving from earth so he will not receive image updates
| {
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A question about Feynman diagram and symmetry factor Consider a $\varphi^3$ theory:
$$
Z_1(J) \propto \exp\left[\frac{i}{6} Z_g g\int \mathrm{d}^4 x \left(\frac{1}{i}\frac{\delta}{\delta J}\right)^3\right] Z_0(J),
$$
where
$$
Z_0(J) = \exp\left[\frac{i}{2} \int \mathrm{d}^4 x \mathrm{d}^4 x' J(x)\Delta(x-x')J(x')\right].
$$
That is
$$
Z_1(J) \propto \sum_{V=0}^\infty \frac{1}{V!}\left[\frac{i}{6} Z_g g \int \mathrm{d}^4 x \, \left(\frac{1}{i}\frac{\delta}{\delta J}\right)^3\right]^V \times \sum_{P=0}^\infty \frac{1}{P!}\left[\frac{i}{2} \int \mathrm{d}^4 y \, \,\mathrm{d}^4 z\, J(y)\Delta(y-z)J(z)\right]^P.
$$
In particular, we can consider the term when $V=2, P=3$. Calculation shows that
\begin{equation}
- i \frac{1}{2!}\frac{1}{3!} \frac{(Z_g g)^2}{6^2*2^3} \left[\int \mathrm{d}^4\, x_1 \, \mathrm{d}^4 x_2 \, \left(\frac{\delta}{\delta J(x_1)}\right)^3 \left(\frac{\delta}{\delta J(x_2)}\right)^3\right] \left[\int \mathrm{d}^4 y \, \mathrm{d}^4 z \, J(y)\Delta(y-z)J(z)\right]^3\\=
- i \frac{1}{2!}\frac{1}{3!} \frac{(Z_g g)^2}{6^2*2^3} \int \mathrm{d}^4 x_1 \, \mathrm{d}^4 x_2 \, \left[3^3*2^4 \Delta(x_1-x_1)\Delta(x_1-x_2)\Delta(x_2-x_2) + \\ 3^2\times 2^5\Delta(x_1-x_2)\Delta(x_1-x_2)\Delta(x_1-x_2)\right]\\=
-i(Z_g g)^2 \int \mathrm{d}^4 x_1 \, \mathrm{d}^4 x_2\, \times \left[\frac{1}{2^3} \Delta(x_1-x_1)\Delta(x_1-x_2)\Delta(x_2-x_2) + \\ \frac{1}{2\times 3!}\Delta(x_1-x_2)\Delta(x_1-x_2)\Delta(x_1-x_2)\right],
\end{equation}
where $\Delta(x_1-x_1)\Delta(x_1-x_2)\Delta(x_2-x_2)$ and $\Delta(x_1-x_2)\Delta(x_1-x_2)\Delta(x_1-x_2)$ correspond to their Feynman diagram. Then the question is why $\frac{1}{2^3}$ and $\frac{1}{2\times 3!}$ are just the reciprocal of symmetry factors of the corresponding Feynman diagram respectively?
In the general case of $V, P$, why the coefficients of the terms in the result of calculation are just the reciprocal of symmetry factors of the corresponding Feynman diagram respectively?
| This is exactly the point of the symmetry factor.
Let's call the term in $Z$ that we're considering $T$.
Without considering the symmetric exchanges that produce the symmetry factor, the contribution of each diagram to $T$ is simply its associated term without any numerical factor in front (a factor of 1). This is because when we count every possible exchange of vertices, propagators, derivatives, etc. that leaves the Feynman diagram invariant, this number neatly cancels out the factorials in the Taylor expansion and our choice of 1/6 and 1/2 in the field Lagrangian. If the symmetry factor for a diagram is 1, each of these exchanges gives rise to an identical term in the $T$.
When a diagram has a symmetry factor that is not 1, some of these exchanges mentioned above no give rise to additional terms. Hence the contribution of that particular diagram must be divide by the symmetry factor $S$.
This is a confusing topic, I wrote a note specifically on this kind of counting here
| {
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Gauss’s Law inside the hollow of charged spherical shell Use Gauss’s Law to prove that the electric field anywhere inside the hollow of a charged
spherical shell must be zero.
My attempt:
$$\int \mathbf{E}\cdot \mathbf{dA} = \frac{q_{net}}{e}$$
$$\int E \ dAcos\theta = \frac{q_{net}}{e}$$
$$E \int dA = \frac{q_{net}}{e}$$
$E\ 4\pi r^2 = \frac{q_{net}}{e}$ and since it is a hollow of a charged spherical shell the $q_{net}$ or $q_{in}$ is $0$ so: $E = 0$.
Is my reasoning on this problem correct? Essentially $E$ is $0$ because there is no charge enclosed.
| You may have forgotten to consider the case where $\vec E \perp\vec A$. Then, also flux is zero. But, it is easy to tell using symmetry that then $\vec E$ would form closed loops which is not permissible. Hence, $E$ has to be zero units.
And yes, your reasoning is correct. You can show this for any (imaginary) shell inside your shell. Hence, Electric Field is zero everywhere.
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Noether's theorem and gauge symmetry I'm confused about Noether's theorem applied to gauge symmetry. Say we have
$$\mathcal L=-\frac14F_{ab}F^{ab}.$$
Then it's invariant under
$A_a\rightarrow A_a+\partial_a\Lambda.$
But can I say that the conserved current here is
$$J^a=\frac{\partial\mathcal L}{\partial(\partial_aA_b)}\delta A_b=-\frac12 F^{ab}\partial_b\Lambda~?$$
Why do I never see such a current written? If Noether's theorem doesn't apply here, then is space-time translation symmetry the only candidate to produce Noether currents for this Lagrangian?
| The current
$$J^a=\frac{\partial\mathcal L}{\partial(\partial_aA_b)}\delta A_b=-\frac12 F^{ab}\partial_b\Lambda$$
is a conserved current. It is indeed a direct consequence of the Noether theorem. However, this current does not represent any physical observables since it is not gauge invariant.
This is explained in great details in this paper.
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What is the symbol Å? I saw this symbol like:
$$\lambda=3000\overset{\circ}{\text{A}}$$
and I don't know what this means. Is it a frequency? (since $\lambda$ is usually used for frequency)
| The angstrom was intended to be a tenth-metre, but because the definition being used was more exact than the metre at the time, and based on a mis-estimation of the Stockholm prototype, a different name is used. In a similar vein, the X-Unit is defined as a thirteenth-metre, but also is out by some factor, so is not so called.
It was first used to measure spectral lines of the visible spectrum, but the length is convenient size for atoms and molecules. A hydrogen atom has a diameter of 1.058 A.
| {
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Will glass always break in the same way? This question has had me thinking for a while. If I have two large panes of glass and a rock or similar item is thrown in exactly the same place on the glass, would the two panes break in the same way. Does the shattering of glass follow any rules or is it always random and subject to other variables? Could you predict the shattering of glass down to the smallest shards or again, is it random?
| I believe the breaking itself will be fairly deterministic. Since I would expect that quantum mechanical uncertainty will play only a very small roll at those scales.
However I do suspect that the results will be different. But this would be due to the structure of the glass. By this I mean the imperfections within it. Because these will be the spots where the glass will fail first.
I think the location of these imperfections can be called random and are formed during fabrication. One source would be uneven cooling, which could add internal stresses. Other sources I can think of are impurities and air bubbles. All of these should be able to be modeled, however not knowing the exact initial and boundary conditions should give plenty of room for uncertainty.
| {
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Which clock is the fastest inside an accelerating body? The picture shows an accelerating spaceship with two clocks inside it. It is so far away from all other bodys that gravity is of no importance.
Will the bottommost clock be slower than the topmost one? Or will both clocks have the same speed?
| The bottom clock will run slower than the top clock.
The simple way to explain this is to appeal to the equivalence principle. This tells us that locally an acceleration is equivalent to a gravitational field. So if the rocket is accelerating at some acceleration $a$ this is equivalent to two clocks sitting stationary in a gravitational acceleration $a$.
We can use this to calculate the gravitational time dilation between the clocks because to a first approximation the relative time dilation is given by:
$$ \frac{\Delta t_{top}}{\Delta t_{bottom}} = \frac{1}{\sqrt{ 1 + \frac{2 \Delta\Phi}{c^2}}} $$
where $\Delta t_{top}$ is the time interval measured by the top clock, $\Delta t_{bottom}$ is the time interval measured by the bottom clock and $\Delta\Phi$ is the difference in the Newtonian gravitational potential. If the distance between the clocks is $h$, then the difference in the potential is simply:
$$ \Delta\Phi = ah $$
so:
$$ \frac{\Delta t_{top}}{\Delta t_{bottom}} = \frac{1}{\sqrt{ 1 + \frac{2ah}{c^2}}} $$
Let's do this calculation for an acceleration of $1g$ and a rocket length of $100$ m. We're taking the upward direction as positive, which means the acceleration is negative because it points down. The relative time is:
$$\begin{align}
\frac{\Delta t_{top}}{\Delta t_{bottom}} &= \frac{1}{\sqrt{ 1 + \frac{2 \times -9.81 \times 100}{c^2}}} \\
&= 1.00000000000001
\end{align}$$
The ratio is possibly better written as $1 + 10^{-14}$ i.e. there are thirteen zeros after the decimal point. This is an extraordinarily small effect, but it can be measured. Indeed it was measured by the Pound-Rebka experiment.
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How can one (formally) determine the particle content of a free field theory? Here's my question: Suppose I'm given a free field theory, where my fields are functions $\phi:\mathbb{R}^4 \rightarrow V$, and the equations of motion are a system of linear Lorentz-invariant differential equations (Given an action of Lorentz algebra on $V$). Is there a way of saying formally or abstractly what the 'particle content' of this field theory is? Ie, how many particles are there and what are there spins/masses?
I should mention that I understand how to determine this empirically- One can often just 'look at the Lagrangian' and figure out what the kinds of particles there are and what their masses are. I'm looking for a more formal way of saying this.
| Yes, the solutions to your linear equations of motion will furnish linear representations of the Poincare group. Particles will correspond to the irreducible representations present. The unitary representations of the Poincare group were classified by Wigner using the method of induced representations [see Weinberg Ch. 2]. They are labeled by a mass and a representation of the little group (SU(2) for massive particles, ISO(2) or SO(2) for massless particles). These quantum numbers arise as casimirs of the algebra (mass is $P^{\mu}P_{\mu}$ and the Pauli Lubanski vector $W_{\mu}\approx\epsilon_{\mu \nu \rho \sigma}J^{\nu \rho}P^{\sigma}$ and its square tells you the little group and corresponding casimir.) The Casimirs are constant on representations of ISO(3,1) so each irreducible representation gives you a particle, and the casimirs tell you the mass and spin. You can examine your equations of motion more closely. If there are additional symmetries you may further decompose the particles according to the irreducible representations of the internal symmetries.
| {
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If I'm floating in space and I turn on a flashlight, will I accelerate? Photons have no mass but they can push things, as evidenced by laser propulsion.
Can photons push the source which is emitting them? If yes, will a more intense flashlight accelerate me more? Does the wavelength of the light matter? Is this practical for space propulsion? Doesn't it defy the law of momentum conservation?
Note: As John Rennie mentioned, all in all the wavelength doesn't matter, but for a more accurate answer regarding that, see the comments in DavePhD's answer .
Related Wikipedia articles: Ion thruster, Space propulsion
| See Solar Sails http://en.wikipedia.org/wiki/Solar_sail.
As other people have pointed out, this is extremely inefficient energy-wise, but has the advantage of being purely passive - no need to carry an energy source, and few or no moving parts to fail.
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Which side of wave-particle duality to choose in a given situation How does one know whether, in treating a certain problem, one should consider particles as waves or as point-like objects? Are there certain guidelines regarding this?
| As a general rule the wave model is most useful when you're looking at the propagation of light and the particle model is most useful when you're looking at the light wave exchanging energy with something else.
If you take the good old Young's slits experiment as an example, the wave model well describes how the light diffracts at the slits, but you need the photon model to explain how the light interacts with the CCD or photographic plate recording the diffraction pattern.
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Feynman Diagram in $\phi^3$ theory I'm slightly befuddled by is what it means when I'm asked to
Draw the Feynman diagram in momentum space for the two point function of $\frac{\lambda}{3!}\phi^3$ theory for order $O(\lambda^2).$
I can draw Feynman diagrams, and I thought two-point function meant
$$\langle0\|\phi(x)\phi(y)\|0\rangle$$
and what I know about $ O(\lambda^2)$ is that it will have more diagrams than $ O(\lambda).$
Other than that, I'm a bit lost. I mean, I'm not even sure if this is a really simple calculation or quite a long one.
Apologies to myself if anything I've written above is embarrassing.
| Order $O(\lambda^2)$ means that your diagram includes two such $\lambda\phi^3/3!$ vertices. Since overall you would have 6 legs of which 2 are the external (you are calculating a two point function with just two external legs) you have to contract four of them. This gives you a loop diagram (Well, there is more than one loop diagram but only one type is 1PI)
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Does the Standard Model require neutrinos to be massless? I am an undergraduate student in Physics, I have a basic understanding of Particle Physics and Quantum Mechanics but none whatsoever of Quantum Field Theory.
I know that Neutrino mixing requires neutrinos to be massive (but why? Physically, couldn't neutrinos mix if they were massless?), and that their mass is usually estimated to be lower than an upper threshold.
But mathematically, does the Standard Model actually predict an upper limit on the neutrino mass, or does it just say that they are massless?
In the former case, what is it stopping it from predicting a lower limit?
In the latter case, so is it wrong?
| As could be quickly found from a wikipedia search, the 'classical' Standard Model of particle physics indeed predicts massless neutrino's. Therefore, the experimental evidence of neutrino oscillations is a strong indication that the Standard Model is missing some important physics.
This is not a huge problem (it's an exciting challenge, though!), since the Standard Model doesn't claim 'absolute truth'. It has been known for quite a while that the Standard Model is just an effective theory which it works very well in most situations that are relevant on the energy scales that are accessible to us right now, but does not claim to be a theory of everything (which it couldn't anyways, since it's missing gravity). Another wikipedia article lists some possible explanations for the observed neutrino mass and oscillations
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/113242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Query on an operator acting on a function I have a naive question about an operator acting on a well-behaved function. Let us say, we are talking about space translation operator acting
on a function $\psi(x)$:
$$\hat{T(a)}\psi(x)=\psi(\hat{T}(x))=\psi(x+a)$$
For a rotation operator $\hat{R}$, the equation becomes:
$$\hat{R}\psi({\bf{r}})=\psi(\hat{R}({\bf{r}}))=\psi(\hat{R}{\bf{r}})$$
I saw these in text books. My question is that acting the operator directly on the argument ($x$ in first case and $\bf{r}$ in second case) of the function, is it a property of these linear operators? I ask this because, in my studies (Demkov 1971) on inversion in a sphere of radius $a$, I see that
$$\hat{M}\psi({\bf{r}})=\frac{a}{r}\psi(\hat{M}({\bf{r}}))=\frac{a}{r}\psi(\frac{a^2}{r^2}{\bf{r}})$$
Here, the said inversion has been denoted by $\hat{M}$. Demkov shows that $\rm{Schr\ddot{o}}dinger$ equation can be inverted w.r.t. a sphere with a transformed Hamiltonian and the inverted wave function. Here,
$$\hat{M}\psi({\bf{r}})\neq\psi(\hat{M}({\bf{r}}))$$
Can anyone please tell me about the correct procedure?
| I believe that the reason is to make the operator $\hat M$ unitary. The $\frac{a}{r}$ comes from the Jacobian of thee transformation $$\mathbf r \mapsto \frac{a^2}{r^2} \mathbf r.$$ This transformation is $$(r, \varphi, \theta)\mapsto (\frac{a^2}{r}, \varphi,\theta)$$
so clearly $r^2 \, dr = -a^2dr'$. This minus sign will be canceled by the inversion changing the orientation. Now consider $$ \langle \hat M \psi| \hat M \psi\rangle =\int r^2 dr\, d\cos\theta\, d\varphi \, (\hat M \psi) (\hat M \psi^*) = \int r^2 dr\, d\cos\theta\, d\varphi \frac{a^2}{r^2} \psi(\frac{a^2}{r^2} \mathbf r) \psi^*(\frac{a^2}{r^2}\mathbf r).$$
$$ = \int dr'\, d\cos\theta\, d\varphi\, \frac{a^4}{r^2} \psi(\mathbf r')\psi(\mathbf r').$$
Since $a^4/r^2 = r'^2$ the operator $\hat M$ is unitary.
A similar prefactor does not appear for rotations and translations since their Jacobians are unity (that is, they preserve volume).
| {
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"timestamp": "2023-03-29T00:00:00",
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Is it possible to create nothing? Is it possible to create nothing?
Lets say you take a cube serving only as a a shell. Then expanded the cube.
Or a balloon, the size of really really small. Then expanded the balloon using outside forces to pull on the exterior. Could it be possible to create a shell that harnesses... well.. nothing but space? No atoms inside other than the exterior? I mean nothing.
Can you create 0? A vacuum isn't what I'm asking here.
Of course trying to do this, in my eyes, would simply rip the exterior apart because there's nothing getting inside to fill in the gap. Any thoughts?
| I have been thinking along the same lines.
Seems to me it could be the answer to a lot of questions.
If one could make a space of nothing (no gravity, no gas, no light etc )
One could then look at one partical or atom at a time to examine the properties without external force impacting on the results.
How one could make a space of nothing is beyond my pay level.
This idea of nothing could also be why the universe expansion is speeding up.
If the outer galaxies are travelling into "nothing" to my way of thinking there would be less or no resistance therefore there is "nothing" to slow it down (maybe gravitational forces from behind).
I know the Dark Matter idea being the reason for expànsion but we don't really know that is the correct answer.
Anyway merry Christmas to all.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/113502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An error in Gravitation by Misner Thorne and Wheeler? I was studying on Gravitation the PPN formalism.
Since in equation (39.41) pag. 1087, the term
$1 + \dfrac{v^2}{2}+(2+\gamma)U = 1 + \dfrac{v^2}{2}+3U$ (the second in GR)
looked odd, I tried (several times) to derive the formula, but I always find:
$1 + \dfrac{v^2}{2}+U$
Is this one of the two famous errors metioned by Wheeler?
I am really going crazy over this...
| Equation 39.41 is quite different in my edition:
$$ A^0_0 = 1 + \frac{v^2}{2} + U + O(\epsilon^4) $$
This is from the hardcover edition published 12/31/1973 (ISBN 0-7167-0334-3). Prior to this the paperback version was published 9/15/1973 (ISBN 0-7167-0334-1).
If your edition has an ISBN number ending in 1 or 2, then this is the revised and correct equation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/113593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Infinities in Newtons law of gravity (for point particles) Newtons law of gravity for two particles of mass $m_1$ and $m_2$ is:
$G\frac{m_1.m_2}{r^2}$.
Supposing that the particles are point particles then gravitional attraction will bring them closer together, and in fact infinitesimally closer together. Now in Newtons time there was no theory, as far as I am aware of inter-atomic forces that would have kept these two particles apart, so the gravitional attraction is asymptotically infinite. This is nonsensical, and either one can say that point particles cannot arbitrarily approach one another, or that particles can never be point particles and must have extension - this in fact includes the previous solution, as the notional point positions of the centre of mass of a particles with extension cannot obviously approach one another.
In Classical Mechanics, would this have counted as evidence of either particles cannot be point masses; or of some then unknown repulsive force that acts at very small distances.
What does the historical record show?
| Classical mechanics has only theoretical point masses. In the simplest case of taking the particle's dimensions to a point, the following argument would hold.
The mass of a particle would be given by its mass density times its volume.
Take the gravitational field:
$$
{\bf g}({\bf r}) = -G\frac{m_1}{|{\bf r}|^2}{\bf \hat{r}},
$$
$m_1$ will be proportional to volume times density of the material the particle is made of.
The volume is proportional to $r^3$, therefore the infinity at $r=0$ is avoided; there will be 0 mass there. The force between $m_1$ and $m$ goes like:
$$
{\bf F}({\bf r}) = m {\bf g}({\bf r}),
$$
so it will be zero there as well.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/113660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Apparent dimensional mismatch after taking derivative Suppose I have a variable $x$ and a constant $a$, each having the dimension of length. That is $[x]=[a]=[L]$ where square brackets denote the dimension of the physical quantity contained within them.
Now, we wish to take the derivative of $u = log (\frac{x^2}{a^2})-log (\frac{a^2}{x^2})$. Here, we have taken the natural logarithm. It is clear that $u$ is a dimensionless function.
$$\frac{du}{dx} = \frac{a^2}{x^2}.\frac{2x}{a^2} - \frac{x^2}{a^2}.(-2a^2).\frac{2x}{x^3} \\
= \frac{1}{x} - 4. $$
Here, the dimensions of the two terms on the right do not match. The dimension of the first term is what I expected. Where am I going wrong?
| I think the second half of your derivative is wrong:
$ \frac{d}{dx} \log\left( \frac{a^2}{x^2}\right) = \frac{x^2}{a^2} \cdot \frac{d}{dx} \left(a^2 x^{-2}\right) = \frac {x^2} {a^2} \left(-3a^2\right) x^{-3} = \frac{-3 a^4}{x} $
which has the correct dimension.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Is evaporative cooling more efficient with dry or moist air? I live in India, and in the summer season, the temperature can reach up to $45 \sideset{^\circ}{}{\mathrm{C}} .$ We use Split 1.5 Ton AC in our small office. The idea is to put an evaporative cooler on the inlet side of the heat exchanger of AC to give it more efficient cooling.
Will it help to increase efficiency? or COP? By how much?
| This is what i have seen work well in my home town. If your climate us dry evaporator cooler will help. 1st make syre your condensing unit outside gas some shade cover then blow the evaporated air onto to the condensing unit outside. If you blow tge evaporated air inside it will raise the humidity inside your office, and humid air is more difficult to cool (takes more energy to cool )
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/114156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Force between two current carrying parallel wires, multiple proof? Having two current carrying (currents $I'$ and $I$) wires of length $a$ parallel to the $z$-axis, one with end points $(0,0,0)$ and $(0,0,a)$ and one from $(a,0,0)$ to $(a,0,a)$, I'm looking for the force on the second one due to the first one. Here is my problem:
I know I'm suppose to get (which I get by finding the magnetic field,...) $$F=\frac{\mu_0II'}{2\pi a}a$$ but I also know that I can compute this with the formula
$$F=\frac{-\mu_0II'}{4\pi a}\int_0^a\int_0^a (dr'.dr) \frac{(r'-r)}{|r'-r|^3}$$ where $r$ and $r'$ are along the wires. When I tried, I got rid of the absolute value by separating the second integral in two parts : $\int_0^r$ and $\int_r^a$. But I don't get anywhere near the solution! Can you help me?
| Sorry for bad-diagram.
This is how I proceed.
Between two parallel current-carrying wires, the magnetic-force is given by
(mu*i1*i2*l)/(2*pi*r), where, mu=magnetic-permeability of medium, i1,i2: currents, l=length of wire on which force is to be found, r=perpendicular separation between the wires.
so, considering diagram, and given data, magnetic force on second wire, due to firstwire is:
(mu * I' * I * a)/(2*pi*a)
= (mu * I' * I )/(2*pi)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/114226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which Force would help find the mass? Two masses, A and B, are connected to a rope. A constant upward force 86.0N is applied to box A. Starting from rest, box B descends 12.1m in 4.70s . The tension in the rope connecting the two boxes is 32.0N.
What is the mass of B?
What is the mass of A?
My work and what I am struggling with:
I am trying to find the acceleration experienced by B, with that I will find the mass.
I find the acceleration using:
$$2\Delta_y/t^2=a$$
$$a=1.0955m/s^2$$
Then I have a problem for B do I use this equation:
$$\sum F_y =32-M_b*g=M_ba$$
$$Or$$
$$\sum F_y =86-M_b*g=M_ba$$
| The two possible equations you bring up only differ by the magnitude of the upper force. So you're asking: should you use $32\ \text{N}$ or $86\ \text{N}$ as the upward force on box B?
To help you arrive at your own answer: Ask yourself which object is actually exerting the force on box B. Is it whatever is exerting the force "F" on object A? Is it the rope connecting the two boxes? Is it box A? Something else?
Once you identify which object is responsible, your task of determining which force to use becomes easier.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/114354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Is darkness really light? According to this wikipedia article
"Consequently, most objects that absorb visible light reemit it as heat. So, although an object may appear dark, it is likely bright at a frequency that humans cannot perceive."
Can someone explain this to me in lay man's terms? Does this mean that darkness is really light?
| Mike Dunlavey's answer is accurate. But, I will try to add little more info on the topic.
Your cat or dog can see things that are invisible to you because of UV detecting ability, a python can see things that are invisible to you because of infrared detecting ability.
The cat, python, dog, most probably you will be thinking, why do we detect rays only of certain wavelength or frequency? why can't we see what others see?
The waves of different wavelength or frequency are detected by different ways. For example, infra-red waves are detected by Thermopiles, Bolometer, Infrared photograpic film. The pythons use relatively the same principle to detect these waves and form images using pits, which are heat sensitive channels.
(Python Image credits:Wikipedia)
UV or Visible light range is detected using photocells, photographic films. Relatively the same principle is used in detecting visible light using photochemical reactions in some of the animals. Similarly, different waves involve different waves of detection, the animals don't have all the privileges of detection. If you would had all the privilege, you can surely see what your animals can see.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/114503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 1
} |
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