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Einstein Field Equations and Electromagnetic Stress-Energy Tensor My question is: if we write Einstein field equations in this form:
$$R_{\mu\nu} - \dfrac{1}{2}g_{\mu\nu}R=8\pi \dfrac{G}{c^4}T_{\mu\nu}$$
Then the left hand side is one statement about the geometry of space-time and the right hand side is one statment about the distribution of momentun and energy right? My point is: what if we use the electromagnetic stress-energy tensor as the energy-momentum tensor?
My thought was: if I understood correctly, does this says that electromagnetic fields can also change the geometry of space-time making it bend as does the presence of matter?
Sorry if it makes no sense, or if it's completely nonsense, it's just a thought that came out, I'm just starting to study those things.
|
Yes. It does in fact mean that electromagnetic fields can also change the geometry of spacetime. Anything with energy and/or momentum affects the geometry of spacetime because, as you point out, the gravitational field equations exhibit a coupling of spacetime geometry to energy-momentum.
For more info in the case of electromagnetism coupling to gravity, see THIS.
|
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|
Schrodinger's cat experiment What's wrong in taking the cat as an observer in Schrodinger's experiment? Plz kindly elaborate! And if possible also describe about possible logics if the question bears the answer No.
|
The experiment is intended to highlight the problem of quantum superposition applied to macroscopic objects.
It's not inconceivable that a radioactive nucleus can be in a superposition of states. When you interact the nucleus with the detector, hammer and glass of poison, the wavefunction that describes these items become entangled with the wavefunction of the nucleus, and they must also enter a superposition of states. The wavefunction that describes the cat must also become entangled, hence the cat must also exist in a superposition of states (dead and alive). Finally when the experimenter looks at the cat the wavefunction that describes the experimenter also becomes entangled and the experimenter must also be in a superposition of states i.e. the states of having observed the cat to be alive and having observed it to be dead.
The point of the experiment is that common sense tells us we don't see cats that are both alive and dead, so somewhere along the line the argument presented above must break down. The Copenhagen interpretation is that the act of observation collapses the superposition, and insofar as the term observer has a meaning it is the point at which this happens. Since humans attach a great deal of importance to being conscious beings they tend to regard themselves as the observer, and hence the cat would not be considered an observer. The cat would presumably disagree.
The current interpretation (well, "a" current interpretation) of the experiment is described by the theory of decoherence and the Many-Worlds Interpretation and does not involve the concept of an observer. The act of observation does not have any effect, so the point is moot. Neither the cat nor the experimenter are observers in the Copenhagen sense.
The idea behind decoherence is that every object interacts with its environment, and the more complex the object the faster it interacts. The interaction breaks the superposition and the system divides into separate histories that cease to interact. While a nucleus is small and simple enough to remain as a superposition for a long time, the detector, hammer and galss vial are already too complex to remain in a superposition for a measurable time, let alone the cat and the experimenter.
|
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Why can't we store light in the form of light? We can store cold (ice), heat (i.e. hot water bag) and electrical charge (batteries). We can even "store" a magnetic field in a magnet. We can convert light into energy and then, if we want, back to light. But we can't store light in form of light in significant amounts. What is the explanation of that in physics terms?
|
The answer by John Rennie and subsequent comments reminded me of this TEDtalk about energy storage from light.
I don't know the details, but this is what I understand they did: they've studied the electronic and absorption properties of foils made of nanotubes, in particular when combined with the result of some impressive research on infrared imagery. The combined product gains the extraordinary property that it can absorb light and store the energy for longer periods of time and in a cleaner way than batteries (our main and perhaps only real method for energy storage). This energy could be free (because you could just attach these flexible foils to your window for example) and it could even be shared through the coherent re-emission of light (from your window to your neighbours window for example).
It's not storing light in the form of light - what the question asked for - but I think it's as close as we're able to get to storing sunlight for a semi-long period of time efficiently and conveniently, something photovoltaic cells are still struggling with. I imagine Justin Hall-Tipping is making it sound more advanced than it is at this time, but nonetheless it has some great potential and I think it's definitely useful to mention it here. I reiterate that I'm unaware of the details and am not an expert in this field though.
|
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What causes insects to cast large shadows from where their feet are? I recently stumbled upon this interesting image of a wasp, floating on water:
Assuming this isn't photoshopped, I have a couple of questions:
*
*Why do you see its image like that (what's the physical explanation; I'm sure there is an interesting one)?
*Why are the spots surrounding the wasp's legs circle shaped? Would they be square shaped if the 'feet' of the wasp were square shaped?
|
This is a great example of how nice it can be to reason about refraction stuff using Fermat's principle.
Let's reduce all this to 2 dimensions. The surface tension produces something like this:
Now if we want to know where a light “ray” needs to go to get from some light source, we just need to find the way that takes it the least time. Light is slower in water, so it wants to go as far as possible in air – of course, only if that's not too much longer to go. So far from the insect, a light ray would just enter the water straight perpendicular, since that minimises both the total waylength and the path in water.
However, right below the insect foot, that won't work – the foot itself is not translucent† – and, more importantly, a little way left or right from right below the foot the quickest path will still be right through the foot, since any other path will require the light to travel substantially more through water, while the total path length is only a little shorter.
so all these rays are “invisible”. Whether that works out this way depends on how far we are away from right-below-the-foot, so that makes a circular shadow, even when the foot itself has another shape.
†Actually, it is kind of translucent I suppose, but we know the small foot will only get hit by a tiny amount of light. So if that bit of light has to be spread over a whole lot of ground, there won't be much intensity down there.
|
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Newton's third law of motion: An apple resting on the surface of earth My friend asked me a question which was:
For an apple resting on the floor, what is the action and reaction force pair?
I said that the Earth exerts a gravitational force on the apple and the apple also pulls on the earth and both these forces are equal in magnitude and opposite in direction this is action and reaction pair.
He said:
Okay, Earth is exerting a downward force on the apple, so the apple
can not accelerate downward because of the normal force of the floor,
but what about the Earth? Since the apple is exerting a force on the
earth in an upward direction.
I said the Earth will not accelerate in an upward direction because the apple is in contact with the Earth we will treat the apple and the Earth as a single object and the force that the apple exerts on the Earth is internal to the object and the internal force can not cause the acceleration.
This is my own logic; I have not read it anywhere. So tell me, am I right?
|
You are basically correct. The apple and the earth exert equal but opposite forces on each other but, as they are in contact, neither of them can move. The forces are balanced by the internal pressure in the apple and in the floor.
Consider the case where the apple is not resting on the floor. In that case the apple will, of course, fall towards the earth. However, at the same time the earth falls upward towards the apple. As the earth has so much more mass than the apple, its acceleration is correspondingly smaller, due to $$a = f/m$$
Hence we never talk about the movement of the earth, but it is there. For the same reason, whenever a spacecraft uses a planet for a "slingshot" approach to increase its speed, the speed of the planet around the sun also changes, by a minute amount.
When the moon travels around the earth, what really happens is that both travel around their common centre of mass, which is still inside the earth, but not far from the surface.
|
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Do the proton PDFs change much with Q? Specifically, the second moments, do they change much from say 100 to 1000 GeV? Why or why not?
|
You can check directly yourself using the Durham HEPData page, which lets you plot PDFs as a function of $x$ or $Q^2$.
At large $x = 0.1$, they're fairly constant over a range of $Q^2$.
But at small $x = 10^{-4}$, they rise quickly with $Q^2$:
This is an illustration of Bjorken scaling at large $x$ and its breaking at small $x$.
I'd also mention that the $Q^2$-dependence of the PDFs is predicted from QCD by the DGLAP equations, so you can look at those to further understand the behavior.
|
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distance of electron from proton An electron is projected, with an initial speed of $1.10 \times 10^5 \text{m/s}$, directly towards a proton that is essentially at rest. If the electron is initially a great distance from the proton, at what distance from the proton is its speed instantaneously equal to twice its initial value?
i know it can be solved by equating the total energy as
K.E+P.E(electron)=K.E+P.E(Proton)
kinetic energy is $ \frac{1}{2}mv^2$. how can i find out potential energy of each particle?
|
The potential energy due to the electrostatic interaction between two particles of charges $q_1$ and $q_2$ is
$$
U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r}
$$
where $r$ is the distance between them.
|
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Influence of charged particle's own electric field on itself I read this in my textbook: A charged particle or object is not affected by its own electric field.
Since I find this completely unintuitive and my mind is yelling "wrong! wrong! how could a particle even distinguish between its own field and the external fields?" I would really like to hear an explanation about why this is true, if it is.
(A corollary question that comes to my mind after thinking about this is... What about gravity and mass considering the analogous situation? And the other two forces?)
|
The self force is the result of the radiated field created because of the force caused by another particle but it's not a force caused only by the particle to itself.
If we consider a particle with an extension and a charge density, there is no Coulomb force caused by a point of the particle over the rest of the point. The proof is easy: in the hydrogen atom, taking the first energy level the potential is caused only by the proton at the nucleous, if the electron would cause force over itself, the inner shells of the wave-function would change the potential in the outter shells.
Since the hydrogen atom energy levels have been evaluated empirically many time, this is a proof that a particles does not exert force over itself directly.
Best regards,
Sergio
|
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Why is electrical energy so difficult to store? Does anyone know a general answer to these questions? (I've asked them together because they're all pretty related, it seems.)
*
*Why is it that we find electrical energy so difficult to store? Do we just find energy difficult to store generally? (...surely not, we can store energy in a block by sending it to the top of a hill.)
*is there something in particular about charge/electricity that makes effective batteries difficult to produce, and, if so, what?
*Is the problem that we're having with storing the energy just an artefact of our use of the energy, or is it difficult to store electrical energy per se?
|
A general answer which is not of any particular use is that electrical energy, and the forms in which we store it, are typically very low entropy systems. The lower the entropy the more they "want" to dissipate and the harder it is to stop that tendency to turn into (ultimately) heat. Same way that it is a lot easier to store water that is 10 degC above ambient than 100 degC.
|
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How to determine the direction of a wave propagation? In the textbook, it said a wave in the form $y(x, t) = A\cos(\omega t + \beta x + \varphi)$ propagates along negative $x$ direction and $y(x, t) = A\cos(\omega t - \beta x + \varphi)$ propagates along positive $x$ direction. This statement looks really confusing because when it says the wave is propagating along $\pm$ x direction, to my understanding, we can drop the time term and ignore the initial phase $\varphi$ while analyzing the direction, i.e. $y(x, 0) = A\cos(\pm\beta x)$, however, because of the symmetry of the cosine function, $\cos(\beta x)\equiv \cos(-\beta x)$, so how can we determine the direction of propagation from that?
I know my reasoning must be incorrect but I don't know how to determine the direction. So if we don't go over the math, how to figure out the direction of propagation from the physical point of view? Why $-\beta x$ corresponding to the propagation on positive x direction but not the opposite?
|
First the assumption/definition is that $\omega$ and $\beta$ are positive constants.
Next you are asking about the phase velocity ie the velocity of a crest, a trough, any fixed point on wave profile.
This means that $\omega t \pm \beta x + \phi$, which can be called the phase of the wave, is a constant.
If you differentiate $\omega t \pm \beta x + \phi=\rm constant $ With respect to time you get the component of the velocity of the wave in the x-direction $\dfrac{dx}{dt} = \mp \dfrac{\omega}{\beta}$.
So with the bracket $\omega t + \beta x + \phi$ the wave is travelling in the negative x direction (negative component) and with the bracket $\omega t - \beta x + \phi$ the wave is travelling in the positive x-direction (positive component).
|
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Hamiltonian of a simple graph I have a spin system:
As shown in the picture, there are two spins S1 and S2, and a pair of interactions between them. One is a ferromagnetic interaction and the other is anti ferromagnetic interaction. I am trying to calculate the Hamiltonian of this system.
The Hamiltonian of the system is:
$$ H = -J_F S1_z S2_z +J_{AF} S1_z S2_z $$
$S1_z$ is the spin matrix for Z direction for spin 1 and $S2_z$ is the spin matrix for Z direction for spin 2. If we allow two random values for $J_F$ and $J_{AF}$, -0.5 and 0.5 respectively the Hamiltonian of the system is as follows.
$$ H = 0.5 S1_z S2_z + 0.5 S1_z S2_z $$
$$ = S1_z S2_z $$
$$ =
\begin{pmatrix}
1 & 0 \\
0 & -1 \\
\end{pmatrix}
\times
\begin{pmatrix}
1 & 0 \\
0 & -1 \\
\end{pmatrix}
$$
$$ =
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
$$
Am I able to calculate the Hamiltonian correctly?
|
The Hamiltonian of this system lives in a 4-dimensional Hilbert space since you have two spin $1/2$. Therefore, you should represent the spin matrix in this four dimensional space like this:
$S_1^z=\begin{pmatrix}
-0.5 & 0 &0 &0 \\
0&-0.5 &0 &0 \\
0 &0 &0.5 &0 \\
0 &0 &0 &0.5
\end{pmatrix}$ , $S_2^z=\begin{pmatrix}
-0.5 & 0 &0 &0 \\
0&0.5 &0 &0 \\
0 &0 &-0.5 &0 \\
0 &0 &0 &0.5
\end{pmatrix}$
The order of the four states along the rows and columns is $|DD\rangle,|DU\rangle, |UD\rangle, |UU\rangle$ where $U$ stands for spin up and $D$ stands for spin down.
In this case $S_1^z.S_2^z=\begin{pmatrix}
0.25 & 0 &0 &0 \\
0&-0.25 &0 &0 \\
0 &0 &-0.25 &0 \\
0 &0 &0 &0.25
\end{pmatrix}$
|
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Does the collapse of the wave function increase entropy of the atomic system itself? Does wave-function collapse cause the entropy of the atom (ie. the sub-atomic particle system that makes up the atom) to increase?
|
Disclaimer. I'm not sure it even makes sense to talk about changes in entropy of systems that undergo wavefunction collapse unless one also includes the measuring apparatus as part of the system. Having said this, here are my two cents that I hope are informative:
The (von-Neumann) entropy of a quantum system prepared in a state (density operator) is defined as
$$
S(\rho) = -k_B\mathrm {tr}(\rho\ln\rho)
$$
Where $k_B$ is Boltzmann's constant. In particular, If a quantum system is described by a pure state (the notion of state as an element of a Hilbert space that you learn when you start out in QM), then its entropy is zero. As a result, if you prepare an atom such that its quantum state is pure, then its entropy will not depend on which pure state it is prepared in. For example, whether it's prepared in an energy eigenstate, or a linear combination of energy eigenstates, its entropy after being prepared in any pure state will be zero.
In this sense, perhaps it can be said that the entropy of an atom that collapses from one pure state to another via projective measuement does not change.
In order for the entropy of a quantum system consisting of an atom to be greater than zero, you would have to prepare the atom in a statistical mixture of pure states. For example, you could envision coupling a large sample of atoms to a heat bath.
|
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Are two waves being in phase the same as saying that the two waves are coherent? If two waves are coherent, is it the same as them being in phase? Please correct if I'm wrong.
|
I believe that "in phase" means that the phase difference of the two waves is zero (and out-of-phase means that this difference is equal to pi). But more general if the phase difference of two waves remains the same during a time that is called "Coherence time", the fields are coherent during that time.
In reality the fields do not remain coherent for infinite interval and after the coherence time the phase difference drifts.
|
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|
What fraction of peak horsepower do typical 4 door passenger vehicles use? I was surprised when I looked at the power rating of the engine used on a Humvee. It's only ~190 horsepower, which is exceeded by many sedan engines.
So an obvious question is why doesn't my Camry SE burn more gas than a Humvee and I think it's just because during regular use, it doesn't use anywhere near 250 HP.
Does anyone know about how much mechanical power a typical sedan engine puts out during highway driving?
Also, can someone speculate what the torque vs RPM graphs of a sedan engine and a Humvee engine look like and what the typical operating points are?
|
At normal highway speeds of 75, or less, about 20 percent based how auto makers are gearing the transmissions in most cars.
|
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What Pauli matrices should I use for this two spin system? Consider the Hamiltonian $H = -J_\text{F}S^{(1)}_zS^{(2)}_z + J_{AF}S^{(1)}_zS^{(2)}_z$, describing the graph
Here, F means ferromagnetic and AF means antiferromagnetic interactions. I am having problem with the value of $S^{(1)}_zS^{(2)}_z$.Someone suggested to me that
$$S^{(1)}_z=\frac{1}{2}\begin{pmatrix}
-1 & 0 &0 &0 \\
0&-1 &0 &0 \\
0 &0 &1 &0 \\
0 &0 &0 &1
\end{pmatrix},\quad S^{(2)}_z=\frac{1}{2}\begin{pmatrix}
-1 & 0 &0 &0 \\
0&1 &0 &0 \\
0 &0 &-1 &0 \\
0 &0 &0 &1
\end{pmatrix},$$ and therefore $$S^{(1)}_z\cdot S^{(1)}_z=\frac{1}{4}\begin{pmatrix}
1 & 0 &0 &0 \\
0&-1 &0 &0 \\
0 &0 &-1 &0 \\
0 &0 &0 &1
\end{pmatrix}.$$
On the other hand, from page 7 of these notes on Pauli spin matrices, I know that for two spin systems $$\Sigma_z = \begin{pmatrix}
2 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & -2 \\
\end{pmatrix}.$$ I asked the person but never got a reply. I don't see what is the difference between $S^{(1)}_z\cdot S^{(1)}_z$ and $\Sigma_z$; when I should use what?
|
Essentialy $S^{(1)}_{z}.S^{(2)}_{z}=\sigma_{z}^{1}\otimes\sigma_{z}^{2}$ and you can easily check that $\sigma_{z}^{1}\otimes\sigma_{z}^{2}=\sigma_{z}^{1}\otimes I_{2\times 2}. I_{2\times 2}\otimes\sigma_{z}^{2}$. And $\Sigma_{z}=\sigma_{z}\otimes I_{2\times2}+I_{2\times2}\otimes\sigma_{z}$.
|
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|
Why planets are orbiting only in one plane? Since gravity is three dimensional why planets are orbiting only in one plane around sun.
|
The planets generally are traveling in a plane. The Sun's axis is tilted about 30 degrees to this plane. Pluto's orbit is roughly 17 degrees off of the Earth's orbit. Eris is 44 degrees. There is a distribution of smaller bodies at all inclinations.
http://en.wikipedia.org/wiki/Invariable_plane
http://en.wikipedia.org/wiki/List_of_notable_asteroids#Retrograde_and_highly_inclined
|
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|
How to describe heat transfer between two solid materials? A general equation for dealing with heat transfer between one material and a region of insulating material. I've seen basic heat transfer equations for one material, but I'd love to see an explanation of how to do two.
|
Basically you can use Fourier's Law
$$
q = -k\frac{dT}{dx}
$$
with the appropriate boundary conditions between the two materials. The basic issue is that at the interface between the two materials, there is a jump discontinuity in the value of the thermal conductivity, and you have to take this into account in solving the equation.
I did a very detailed calculation in a related post Heat transfer between two surfaces that you might find useful in this regard.
Also, if you want to actually calculate something about something in the real world, you may find this list of thermal conductivities useful.
Addendum. In response to comments below Lorenzo's response.
Consider two bars of length $L$ and of uniform (but unequal) thermal conductivities $k_a$ and $k_b$. Let the heat flow $q_0>0$ be constant along the bars, then Fourier's Law shows that the temperatures $T_a$ and $T_b$ of bars $a$ and $b$ satisfy
$$
T_a(x) = -\frac{q_0}{k_a} x + C_a, \qquad T_b(x) = - \frac{q_0}{k_b}x+C_b
$$
for some constants $C_a$ and $C_b$. Now suppose that the left end of bar $a$ is located at $x=-L$ and the right end of bar $b$ is at $x=L$ so that they are joined at $x=0$. Suppose additionally that the left end of bar $a$ is at temperature $T_L$ and the right end of bar $b$ is at $T_R$, then we have the boundary conditions
$$
T_a(-L) = T_L, \qquad T_b(L) = T_R
$$
which tells us that
$$
C_a = T_L -\frac{q_0}{k_a} L, \qquad C_b = T_R+\frac{q_0}{k_b}L
$$
So that
$$
T_a(x) = T_L - \frac{q_0}{k_a}(x+L), \qquad T_b(x) = T_R -\frac{q_0}{k_b}(x-L)
$$
In particular, at $x=0$ we find
$$
T_b(0)-T_a(0) = (T_R-T_L)+q_0\left(\frac{L}{k_a}-\frac{L}{k_b}\right)
$$
In particular, there is, in general a jump discontinuity in the temperature at the interface between the two materials unless the temperatures at which the ends of the bars are being kept are related by
$$
T_R - T_L = q_0\left(\frac{L}{k_b}-\frac{L}{k_a}\right)
$$
|
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How is parity relevant to determining angular momentum? Question:
Particle A, whose spin $\mathbf{J}$ is less than 2, decays into two identical spin-1/2 particles of type B.
What are the allowed values of the orbital angular momentum $\mathbf{L}$, the combined spin $\mathbf{S} = \mathbf{s}_1+\mathbf{s}_2$ (where $\mathbf{s}_1$ and $\mathbf{s}_2$ are the spin vectors of the B particles), and the total angular momentum $\mathbf{J}$ of the decay products? Give the results from two cases: first if particle A has odd parity and then if A has even parity.
My thoughts:
Particle A can be spin-1/2, spin-1, or spin-3/2. Since $\mathbf{J}<2$, we see that there are four possibilities for A:
$$
\begin{align*}
&(1): \;\;\mathbf{S}_A = 1/2 \quad\quad \mathbf{L}_A = 0 \quad\Rightarrow\quad \mathbf{J} = 1/2 \\
&(2):\;\;\mathbf{S}_A = 1/2 \quad\quad \mathbf{L}_A = 1 \quad\Rightarrow\quad \mathbf{J} = 3/2 \\
&(3):\;\;\mathbf{S}_A = 1 \quad\quad\quad \mathbf{L}_A = 0 \quad\Rightarrow\quad \mathbf{J} = 1 \\
&(4):\;\;\mathbf{S}_A = 3/2 \quad\quad \mathbf{L}_A = 0 \quad\Rightarrow\quad \mathbf{J} = 3/2 \\
\end{align*}
$$
The total spin of the B particles can be either $1$ or $0$, and each particle can individually have an orbital angular momentum, along with the angular momentum of the particles as a system. With this thought, cases 1,2, and 4 are impossible because the orbital angular momentum of the B particles is an integer, as is their total spin (and therefore their total angular momentum too). Thus we find that only case 3 is allowed, so the total angular momentum of the B particles is $1$ and their orbital angular momentum is $0$ (so $\mathbf{J}=1$).
I have a strong feeling that this is incorrect, because the question asks for the cases when A has odd parity and even parity (what does that even mean?!) so I suspect there should be more than one possible answer. Where did I go wrong?
|
Hmmm, an old question without a satisfactory answer. I'll have a go.
The spins of the two $B$ may combine as
\begin{align}
\text{singlet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> - \left|\uparrow\downarrow\right\rangle}{\sqrt2},
&
\text{or triplet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> + \left|\uparrow\downarrow\right\rangle}{\sqrt2}.
\end{align}
Since the two $B$ have spin-1/2, they obey Fermi-Dirac statistics and their total wavefunction must be antisymmetric under exchange. Therefore the antisymmetric spin singlet can only be paired with $L=0,2,\cdots$, which have even parity; the symmetric spin triplet must be paired with $L=1,3,\cdots$, so that the parity of the orbital angular momentum wavefunction makes the entire wavefunction change sign if the two $B$ are interchanged.
The intrinsic parity of $B$ doesn't contribute to the overall parity of the final state, since there are two of them; if the $B$ have negative parity, the overall intrinsic parity of the pair is still positive. Thus the allowed final states are
\begin{align}
\text{positive parity}&: & \text{spin singlet (antisymmetric)} && L&=0 \\
\text{negative parity}&: & \text{spin triplet (symmetric)} && L&=1
\end{align}
I've omitted the spin singlet combined with $L=2$, since we do not have that much angular momentum. Likewise there is no combination of the spin triplet and $L=3$ which can be produced from a $J=2$ initial state.
If $A$ has definite parity and parity is conserved in the decay, only one of these possibilities is allowed.
In fact, the way that angular momentum combines (with the possibility of the spin triplet plus the $L=1$ wavefunction adding to a $J^P=0^-$ final state) means that the parity has more to do with the allowed final spin state than $A$'s spin does:
$$
\begin{array}{r|cc}
& \text{spin 0} & \text{spin 1} \\
\hline \text{parity}+ & \text{decay to singlet},0^+ & \text{decay forbidden} \\
\text{parity}- & \text{decay to triplet},0^- & \text{decay to triplet},1^-
\end{array}
$$
|
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Place each foot on a scale: can you add the two to find your weight? I frequent a blog from a British psychologist, and every Friday he likes to pose an interesting puzzle or riddle. The Monday after that he posts the answer. They're good fun, and IANAP but this week's answer made my it-might-not-be-quite-as-simple-as-that detector go off.
My question boils down to this: let's say I have two identical scales, and I stand on the scales with one foot on each scale. The scales read W1 and W2. Does my weight equal W1 + W2?
|
Well, It would work, but only with the condition that your weight doesn't shift around between the time you look at the first reading and the next. But either way, the sum of the readings on the scales always equals "more than twenty stone" however much that specific weight may be, as long as the body in question is in equilibrium, that is to say, not moving. Most scales operate by spring, and Hooke's law states that the force applied to a spring is directly proportional with the extension or compression of the spring by this force. The force used need not be gravity alone, but could be momentum or great amounts of heat, but the less of these there are, the more accurate your weight reading is. In a real life scenario, I would stand still, then take a snapshot of both scales together so that my weight doesn't shift.
|
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Where are we : On level ground or on a ramp - moving in a train? Let's say we are traveling in a train. The path has two parts: one at ground-level and the other moving up on the ramp. The ramp has an inclination of $\arctan\frac{a}{g}$ with the horizontal, where $a$ is the acceleration of the train on level ground and $g$ is the acceleration due to gravity.
The train does not accelerate on the ramp, but moves with a constant velocity.
Can we comment where we are (sitting inside the train of course!) when we have only a pendulum hanging on the roof to observe. (windows are blackened)
|
Although the gravitational/inertial force causing the pendulum to tilt in the same way in both cases (see Equivalence principle link of @zhermes) thus not allowing you to see whether you are on accelerating or on the slope, you will probably be able to feel the difference.
The reason is that although the force parallel to the train is equal (which causes the pendulum to behave in the same way), the normal force acting on you is not so depending on the value of $\arctan\left(\frac{a}{g}\right)$ (whether it is sufficiently larger than 0) you will probably be able to feel whether you are on the incline or not.
[ADDITION after discussion with @markovchain]
Thinking along the line of parallel and normal forces I have to expand my answer a bit: a RIGID pendulum will look exactly the same on the flat and sloped part. A FLEXIBLE pendulum however, will be straight when moving up the slope, but curved when accelerated on the horizontal surface because it will have both a component pulling it to the left and a component (gravity) pulling it down.
|
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How does the solar sailing concept work? Wikipedia describes solar sailing as
a form of spacecraft propulsion using a combination of light and high speed ejected gasses from a star to push large ultra-thin mirrors to high speeds.
I understand the part where ejected gasses bump into the sail pushing the spacecraft. On the other hand, I don't understand how light can do this, since light has no mass.
How does that work? Does this mean that if I have a mirror balancing on a needle I would be able to push it over with my flashlight?
|
I think you are really asking "how can light deliver an impulse to the sail". The answer is that although light has no mass it does carry momentum. When light is reflected off the sail, conservation of momentum requires that the sail changes momentum by twice the momentum of the light. The extra kinetic energy of the sail comes from the red shift of the reflected light.
This question has several answers that discuss the momentum of light in some detail.
|
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Capacitance of a capacitor Why does the capacitance of a capacitor increase if the distance between the two plates of a parallel plate capacitor is decreased? I think, with decreasing distance between the two plates, the force of attraction between the charges on the two plates will increase, and as a result more charge will be stored.
|
A capacitor has a capacitance depending on the maximum voltage sustained across the plates $C_{\text{max}}=\frac Q{\Delta V_{\text{max}}}$.
The voltage sustained between the plates depends on field strength between the plates. The strength of the electric field between the plates is equal to the voltage sustained between the plates divided by the distance between the plates $E=\frac {\Delta V}{\Delta x}$.
The strength of the electric field is limited by Electric Breakdown.
Electric breakdown occurs when the strength of the field between the plates exceeds the strength of the dielectric between the plates.
Under a high enough electric field the dielectric partially ionises. The ions in the field accelerate and collide with other molecules in the dielectric creating more ions.
The ions form conducting channels between the plates which allows current to flow in spontaneously discharge i.e. sparks/lightning between the plates .
Combining these concepts we see;$$C_{\text{max}}=\dfrac {Q\Delta x}{E_{\text{max}}}$$
the capacitance is proportional to the distance between the plates $C_{\text{max}}\propto \Delta x$.
|
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Can we change frame of reference twice in a single problem? My question has an inclined plane of mass $M$ and simple block kept on it, of mass $m$ (Both on a table). All surfaces are friction-less. Both of the objects would move, block down the incline and inclined plane parallel to the table, somewhat opposite to the block. Can the two equations I make be from the Free Body Diagram (FBD) of incline, in GROUND frame, and FBD of block in the INCLINED-PLANE frame?
Or do I need to solve in a single frame(either GROUND or INCLINED-PLANE)?
|
I'd recommend that you stick to one frame, otherwise you'll have to make coordinate transformations to make your equations consistent and that's significantly more work than it's worth.
Also, be careful that if the inclined plane accelerates (which it will if I understand your setup correctly) then you have to introduce so-called fictitious forces to solve the problem correctly in the incline's (non-intertial) frame, and that's also, in this case, probably more trouble than it's worth in my opinion.
|
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What does it mean that an inspiral is 'adiabatic'? Binary systems emit gravitational radiation. This causes the system to lose energy, which results in a shrinking of the semi-major axis. I have read on countless occasions that this 'inspiral' is adiabatic (here for example). What does this mean that the shrinking is adiabatic?
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The $Q=0$ is just a convenient definition for thermodynamics. The key to adiabacity is that the process is approximately in an equilibrium state at every step. In thermodynamics, this means that you want the process to be reversible, which means that you want the entropy change during the process to be zero, and since $dS = \frac{dQ}{T}$, this means that the heat added is necessarily zero.
But for other processes, the notion of heat added doesn't necessarily make sense, but an equliibrium process does. Typicallly, this is when you change some parameter on a process slowly enough that you don't change the basic physical situation. In addition to weak gravitational radiation causing inspiral where the orbiting objects are still orbiting in closed orbits, you get "adiabatic approximations" in quantum mechanics--for instance, when you consider a infinite square well where the walls are expanding at a small enough rate that the particle remains in what is approximately a ground state the entire time. Even more simply than that, you can consider a ball on a string problem, where the ball is rotating in two dimensions on a plane perpendicular to the string. If you change the length of the string at a rate slow enough that the ball stays rotating in what is approximately a circle and the string remains taut, you are adiabatically changing the system.
|
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Relating generalized momentum, generalized velocity, and kinetic energy: $2T~=~\sum_i p_{i}\dot{q}^{i}$ According to equation (6) on the first page of some lecture notes online, the above equation is used to prove the virial theorem. For rectangular coordinates, the relation
$$
2T~=~\sum_i p_{i}\dot{q}^{i}
$$
is obvious. How would I show it holds for generalized coordinates $q^{i}$?
|
Your answer looks complicated and I feel it misses the point. The fact that $2T=\sum_l p_l \dot{q}_l$ is fundamentally due to Euler's homogeneous function theorem. This states that
$$
\text{if }f(\alpha \mathbf{x})=\alpha^k f( \mathbf{x})\text{, then } \mathbf{x}\cdot\nabla f(\mathbf{x})=kf(\mathbf{x}).
$$
When stated like that, I would even hesitate to add a proof - simply differentiate with respect to $\alpha$ and set it to $1$.
The theorem holds for Lagrangians that depend quadratically and homogeneously in the velocities. This means specifically that $L(q_l,\dot q_l)=T(\dot q_l;q_l)-V(q_l)$, and $T(\alpha \dot q_l;q_l)=\alpha^2 T(\dot q_l;q_l)$. (I use the notation $T(\dot q_l;q_l)$ to emphasize that the $q_l$ dependence is as parameters, with $T$ being fundamentally a homogeneous quadratic function of the $\dot q_l$.) The homogeneous function theorem then states that
$$2T(\dot q_l;q_l)=\sum_l\dot q_l \frac{\partial T}{\partial \dot q_l}.$$
Since $V$ is independent of the $\dot q_l$, each partial derivative equals the corresponding momentum $p_l=\frac{\partial L}{\partial \dot q_l}$.
Note in particular that your first assumption (that the functional dependence of $T$ on the $\dot q_l$ be time-independent) is not necessary.
|
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Alternative methods to derive the static potential in the NR limit of QED In QED, one can relate the two-particle scattering amplitude to a static potential in the non-relativistic limit using the Born approximation. E.g. in Peskin and Schroeder pg. 125, the tree-level scattering amplitude for electron-electron scattering is computed, and in the non-relativistic limit one finds the Coulomb potential. If one allows for 1/c^2 effects in the non-relativistic expansion, one also finds spin-dependent interactions (e.g. spin-orbit, see Berestetskii, Lifshitz, Pitaevskii pg. 337).
Are there any alternative methods for calculating a two-particle non-relativistic potential?
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If you set $c\to \infty$ in the QED Hamiltonian you obtain a non-relativistic Hamiltonian whose potential only includes a pseudo-Coulomb term $(1/r)$, because this is the only term of order $c^0$ in the QED interaction. No further calculation is needed to obtain the potential.
|
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Path integral with boundary and bulk terms I was wondering if their is a general strategy for computing path integrals with a mix of boundary and bulk integral actions. Do people use divergence theorem to convert the action into bulk integrals, or is there some other trick?
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The usual Feynman rules may still be derived. If there are both boundary fields and bulk fields, one needs to include the bulk-to-boundary propagators, too. The same problem has to be solved in the AdS/CFT correspodence. Search for "Witten diagrams" to see how the Feynman diagrams are ultimately done in this context.
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Hamiltonians, density of state, BECs When working with Bose-Einstein condensates trapped in potentials, how can one tell what the density of state of a system of identical bosons given the Hamiltonian, $H$? (I have been told that it is possible.)
Suppose the Hamiltonian is some 2D harmonic oscillator -- so $$H=p^2/2m+(1/2)(a^2x^2+b^2y^2) \quad ?$$
I think there is some general formula, something like $$\rho(E)=[gV_dS_dp^{d-1}/(2\pi\hbar)^{d}] (dp/dE) \quad ,$$ where $d$ is the dimension of the space we are working in, $g=2s+1$ where $s$ is the spin of the particles and $V_d$ is the $d-$dimensional "volume", so for a fixed volume box, this is the volume of the box.
But what is $V_d$ in this case? And is $p$ simply $$p^2=2m[E-(1/2)(a^2x^2+b^2y^2)] \quad ?$$
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This is definitely possible. A good resource is this statistical physics script. The density of states for an electron gas is calculated explicitly there, check out eqn. (5.21) and following. Then, some BEC calculations are made explicitly, starting on page 119. V is the (physical) volume of the system you look at.
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Eye sensitivity & Danger signal Why are danger signal in red, when the eye is most sensitive to yellow-green?
You can check luminosity function for more details...
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Yes, It's true...
We know that our eyes have three types of cone cells - S (short), M (medium) & L (large). The naming is done in order to differentiate the cells from "which cell absorbs which color". S to Blue, M to Green and L to red. The peak wavelength of L is 564 nm, yellowish-green. The peak of M is 534 nm, bluish-green. The peak of S is 420 nm, bluish-violet.
This can also be predicted from the responsivity curves. At some comparatively higher magnitude, both M & L cells are sensitive to this 564 nm. So, our eye is most sensitive to yellowish-green. You're right. There's one more thing to notice the the Luminosity function is based on two types of vision. Photopic and scotopic. Photopic vision is when you have brighter conditions (sufficient enough) for eye vision. Hence, the sensitivity towards green is only on well-lit conditions (cone-cells). While on the other hand, scotopic vision is during the night-time when the rod-cells are highly sensitive to even dim lights. Both cone & rod cells are sensitive to green (but, different wavelengths).
Have a look at the Purkinje effect for more stories on greenish...
Why Red?
Okay, enough with all those sensitivities. Now, to Physics. The reason behind the red-theme for traffic lights is nothing more than just Rayleigh scattering. The amount of scattering is inversely proportional to the fourth power of the light's wavelength. $S\sim1/\lambda^4$
Red, having the higher wavelength is the least scattered. So, it can travel longer distances without appreciable loss by scattering than any other colored lights. Our eyes may seem to be sensitive when the green light is nearer to us (probably, a km). But, red can reach us more effectively. And so, we can spot it very easily at farther distances. (All circle the same thing)
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What happens when a compact fluorescent lamp implodes? When a incandescent lamp implodes, the filament burns up and the current is interrupted.
Yesterday, a compact fluorescent lamp imploded and blew all the switches / fuses. I'm not sure if the same would happen with ordinary fluorescent lamps.
Can it be that the charge stored in the capacitor is released backwards? Is something else going on?
|
It very much depends on the control gear, there are a variety of designs for circuits for starting and running fluorescent tubes. I don't think any of them involve capacitors but there is likely to be an inductor that might cause a voltage spike when the circuit is broken. Normally, an implosion of the tube would not by itself cause a short circuit and so would not trip circuit breakers (surely you don't have quaint 20th C fuses)
|
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What is the speed of acceleration of the inflation of the universe? Is the inflation speed of the universe accelerating or is it a constant speed of expansion proportional to distance between objects.
|
*
*When we talk about universe inflation we should have in mind that the four cosmic forces did not appear yet, because at inflation no elements were formed yet.
*scientists say that at 0.03 second the inflation reached four light-years, also said that if it had continued at the same rate, it would have been disappeared during the second have of the first second.
*before the Big Bang the inflation speed was zero. And time is zero.
*now we can say:
Light travels during four year = 300,000 km/s x 60 s x 60 min. x 24 hours x 365 days x 4 years = 37.843 x 10ᴧ12 km, this distance was within 0.03 of first second of Big Bang, means 37.843 x 10ᴧ12 km / 0.03 seconds = 12.6 x 10ᴧ14 km / Sec. Thus, inflation speed at the peak (before the effect of the four cosmic forces) in regard to speed of light is = 12.6 x 10ᴧ14 / (3 x 10ᴧ5) = 4.2 x 10ᴧ9 times the speed of light.
-s = v 1 * t + 1/2 a * tᴧ2:
s: distance at certain time = 37.843 x 10ᴧ12 km. v 1: speed at zero time = 0. a: acceleration. t: time = 0.03 second.
37.843 x 10ᴧ12 km = 0 + 1/2 a * 0.03ᴧ2
a = (2 * 37.843 x 10ᴧ12) km / 0.03ᴧ2 = (75.686 x 10ᴧ12) / (9 x 10ᴧ-4) = 8.41 x 10ᴧ16 km/sᴧ2.
*the acceleration at universe inflation stage is: 8.41 x 10ᴧ16 km/sᴧ2
*When the elements were formed starting with Hydrogen, the inflation speed came down due appearance of the four cosmic forces that started action, and the universe expanding started with much lower speed than the speed of light.
|
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Direction of the Area Vector (with regards to magnetic dipole) I'm learning about torque on a conductive coil in a magnetic field. I have been taught that $\vec\tau = \vec\mu \times \vec{B}$, where $\vec\mu$ is the magnetic dipole moment. Also, $\mu = I\vec{A}$, where $\vec A$ is the area vector of the loop.
To find the direction of the area vector, I am told to use the right hand rule with regards to the current in the loop (curl your fingers in the direction of current, and your thumb points in the direction of the area vector).
My question is: Why does this give the correct direction for the area vector? Is the area vector just defined to be this way to avoid nasty usage of minus signs, or is there some other reason for this?
My guess is that whoever formalized this law/equation (not sure what correct term is for this instance) started with the direction of torque, and worked backwards defining the direction of $\vec\mu$ and $\vec{A}$ to reduce or eliminate stray minus signs in the equations. However, this is, of course, just a guess; I want to know what the true reason is.
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As far as I know, the area vector is a purely mathematical object whose definition is related to the orientability of the surface (in this case, a disk). This is a property of surfaces embedded in an Euclidean space that allows to choose surface normal vector to the surface at every point. For an oriented surface, this normal is determined so that we can use the right-hand rule to define a clockwise direction of loops on the surface, which by the way, is needed if we want to apply Stokes' theorem.
|
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Proof that a spherical lens is stigmatic In geometric optics, we generally allow that, for example in the case of a convex lens, rays coming from a particular point get refracted towards another particular point on the opposite side of the lens.
How is this proven from Snell-Descartes' law? Do we need to use the paraxial approximation in order for this to be true, or is it actually exactly true?
|
Usual spherical lenses are only approximately stigmatics, that is to say the image of a source-point is itself a point. You need paraxial approximation to ensure approximate stigmatism and avoid spherical aberration.
You can see it by playing around with just a plane interface between two different transparent media like the surface of water. Using paraxial approximation, you can prove the relation $$\frac{n}{HA} = \frac{n'}{HA'}$$ where $n$ is the refraction index of the medium where the object $A$ lies (for example a fish, then $n=1.33$), $H$ is the orthogonal projection of $A$ on the interface, $A'$ is the image and $n'$ is the refraction index of the medium where the observer thinks $A'$ is (for example the air so $n'=1$). If the fish is $40$ cm under the water, you will see it's image only $30$ cm away from the interface (that's why it's difficult to catch a fish in an aquarium on first try).
As you need spherical interfaces to build your lens and it's not even working with planar interfaces (which are a particular case of spherical interfaces), it can not be true for any type of spherical lens.
Here are two animations you can try to see how stigmatism works:
*
*one with plane interface
*and another with spherical mirror
|
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How old is SUN ☉? How do we know/calculate the exact age of sun ☉ ? ie. 4.57 billion years. What is the way to calculate it?
|
We do know that the solar system has formed along with the sun (being a third generation star). There are some isotopes distributed (though rarely) throughout the solar system. In Earth, we can do a radio dating for such decaying elements, which can be found in rocks, meteorite locations, etc..
For confirmation, we can check our observations along other areas of our planet, or some other planet in our own solar system. Now, there are many other methods to determine the age, especially using the Standard model of astrophysicists.
Another way is to have a look at the luminosity and spectra of other stars (star clusters can provide more data, as there is a lump) using the standard parallax measurement, plot them in the H-R diagram and finally compare the observed main-sequence stars with sun, and thereby have a guess on its total age...
|
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Galaxies moving away at the speed of light As an arts student, I really find those cosmological questions hard to understand and hence come here to seek your kind help.
The Hubble constant $H_0$ is estimated to be about 65 km/s/Mpc, where 1 Mpc (megaparsec) is around 3.26 million light-years. At what distance would galaxies be moving away at exactly the
speed of light? (I found that there is something called Hubble Radius, but is this the same as Hubble Radius?) If there were galaxies farther than the Hubble radius, how would they appear to us?
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The Hubble sphere is the locus of points where the Hubble flow is equal to c. The Hubble sphere is not an event horizon.
All observed objects with a redshift greater than about 1.46 are outside the Hubble sphere. In other words, these objects are receding superluminally. Furthermore, these objects were outside of the Hubble sphere, and were receding superluminally at the time the light was emitted.
The diameter of the observable universe is greater than the diameter of the event horizon. We will never observe events that are currently happening outside of the event horizon, but we can still see events from galaxies that are currently outside of the event horizon via the light they emitted long ago.
|
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The rule breaker, emissivity + reflectivity = 1 If emissivity and reflectivity are inversely proportionate, why does glass have a high emissivity of around 0.95-0.97 as well as being very reflective for IR Radiation?
normally it works but not with glass!
Can anyone explain this?
|
To me the problem is not linked to the reflectance, but to the surface finish.
Picking up an image of a reflection in the visible part of the spectrum is possible even on a black carbon surface if that surface is polished well enough. To convince yourself of this, simply pick a dark material with a well polished surface (or a not so well polished surface but placed at an highly inclined angle). Then place the surface to reflect the sky day light. And you will see the surface reflecting the color of the sky even if the surface color is blue, red or any other dark color/tint.
Thus, since the glass window is a well polished surface, it will reflect any image of the surrounding in the thermal infrared. This is also true for the sky in the visible part of the spectrum as would tell you any bird that survives hitting a cleaned glass window at spring time... ;-)
|
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Is lattice enthalpy positive or negative? I've learnt that the lattice enthalpy (defined as the energy change from a solid ionic lattice to separate gaseous ions) is always positive, obviously. However, I've seen it explained as the opposite other places, so it's negative.
What is correct?
|
Lattice energy is defined as when 1 mole of cation and 1 mole of anion combined to form 1 mole of solid crystal, energy is released. The energy released is known as lattice energy because we know anion and cation has highewr energy than crystal lattice therefore when bond is formed between cation and anion energy is released.
That's why lattice energy is negative.
|
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How does the freezing temperature of water vary with respect to pressure? I know when the pressure is reduced, the boiling temperature of water is reduced as well. But how does the pressure affect the freezing point of water?
In a low-pressure environment, is water's freezing temperature higher or lower than $0\sideset{^{\circ}}{}{\mathrm{C}} \, ?$
|
What happens when water becomes ice - it expands.
With a pressure change apply Le Châtelier's principle - "If a system at equilibrium is subjected to a change in pressure . . . . . . . then the equilibrium shifts in such a way so as to undo the effect of the change".
So in the case of the volume of ice being bigger than the volume of water when the pressure is increased the system tries to reduce the pressure by converting ice into water.
|
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Higher order covariant Lagrangian I'm in search of examples of Lagrangian, which are at least second order in the derivatives and are covariant, preferable for field theories. Up to now I could only find first-order (such at Klein-Gordon-Lagrangian) or non-covariant (e.g. KdV) ones. Also some pointers to the literature about general properties of such systems are welcome.
Thanks
|
I) As user Vibert mentions in a comment, the Euler-Lagrange equations are not modified$^1$ by adding total divergence terms to the Lagrangian density
$$ \tag{1} {\cal L} ~ \longrightarrow ~{\cal L} +d_{\mu}F^{\mu}. $$
Adding total divergence terms leads to an inexhaustible source of higher-order Lagrangians.
II) Generically, without some cancellation mechanism in place [such as, that part of the Lagrangian density is (secretly) a total divergence] an $n$-order action would lead to $2n$-order Euler-Lagrange equations.
III) Example. The Einstein-Hilbert (EH) Lagrangian density
$$\tag{2} {\cal L}_{EH}~\sim~\sqrt{-g} \left\{g^{\mu\nu} R_{\mu\nu}(\Gamma_{LC},\partial\Gamma_{LC})-2\Lambda\right\} $$
depends on both second-order temporal and spatial derivatives of the metric $g_{\mu\nu}$. This is of course an important example. Here $\Gamma_{LC}$ refer to the Levi-Civita (LC) Christoffel symbols, which in turn are first order derivatives of the metric $g_{\mu\nu}$. However, it is possible to add a total divergence term to render the Lagrangian density first order, as user drake mentions in a comment.
Thus the Euler-Lagrange equations for the Einstein-Hilbert action $S_{EH}[g_{\mu\nu}]$, i.e. the Einstein field equations (EFE), are not of fourth order, as one may naively have expected, but still of second order.
IV) Higher-order Lagrangians are also discussed in many Phys.SE posts, see e.g. here and here.
--
$^1$ Note that adding total divergence terms (1) may affect consistent choices of boundary conditions for the theory.
|
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How does the wavelength change in relativistic limit? In the text, it reads that the momentum of a particle will change if it is moving at speed close to light speed. In the general case, the wavelength is given as
$$
\lambda = \frac{h}{p}
$$
and
$$p = \frac{mv}{\sqrt{1-v^2/c^2}}$$
when $v \to c$, $p\to\infty$, so is it say that the wavelength is ZERO? I don't understand why the wavelength will change to zero if it is moving at speed very close to light speed?
|
Lorentz contraction! The measured de Broglie wavelength in the direction of propagation vanishes because that's what special relativity says happens. The wavelength has to go as $h/p$ as you wrote, so why does it surprise you that when $p$ gets large the wavelength gets small?
|
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$\beta^+$ decay We've been discussing radioactive decay at school, and I grasped everything except for $\beta +$ decay. When I googled radioactive decay, I immediately found out they dumbed down radioactive decay for us, which is probably why they didn't care to explain what they did, they just showed some calculations. We have never discussed neutrino' s and antineutrino's, they leave that out of the equation, which is no problem since they have negligible mass and no charge.
So we've been taught that a proton and electron form a neutron, which I have also discovered is not true (I'm discovering a lot of new things :P). I learned that this is caused by the spontaneous (?) change from up to down quarks and vice versa.
However, at school I must keep to the 'rules' and by those rules I don't really understand $\beta +$ decay. I see $\beta -$ decay as follows:
An electron flees a neutron and leaves a proton. That's why you get an atom with a higher atomic number.
However, how would this work with beta + decay? Can it even be dumbed down to this kind of high school thinking?
|
Neutron mass ~ mass of proton + 2 electrons.
Beta- decay, by Einstein's formula, the remaining mass is converted into kinetic energy, the speed of the proton and electron.
Beta- is an exothermic reaction; it gives off energy.
Beta+ is endothermic; it needs energy to happen, the kind of energy that is available inside a nucleus or the heart of a star or a reactor.
|
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How do you tell if a metric is curved? I was reading up on the Kerr metric (from Sean Carroll's book) and something that he said confused me.
To start with, the Kerr metric is pretty messy, but importantly, it contains two constants - $M$ and $a$. $M$ is identified as some mass, and $a$ is identified as angular momentum per unit mass. He says that this metric reduces to flat space in the limit $M \rightarrow 0$, and is given by
$$ds^2 = -dt^2 + \frac{r^2 + a^2 \cos^2\theta}{r^2 + a^2}dr^2 + \left(r^2 + a^2 \cos^2\theta \right)d\theta^2 + \left(r^2 + a^2 \right)\sin^2\theta d\phi^2 $$
and $r$, $\theta$ and $\phi$ are regular spherical polar co-ordinates.
But I don't understand why this space is obviously flat. The Schwarzschild metric also contains terms involving $dt^2$, $dr^2$, $d\theta^2$ and $d\phi^2$ but is curved. I always thought that a metric with off diagonal elements implied a curved space, but clearly I was very wrong.
Question: How do you tell if a metric is curved or not, from it's components?
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You tell if a space (or spacetime) is curved or not by calculating its curvature tensor. Or more unambiguously one of the curvature scalars (e.g. Ricci, or Kretschmann) since these don't depend on the coordinate system, but all of the information in the scalars is also contained in the Riemann tensor.
It is not necessarily obvious whether a given metric is curved or flat. You can take a perfectly flat spacetime and express it in some bizarre coordinate system, in which the metric has nonconstant off-diagonal terms. It's a simple exercise to take flat space and use the tensor transformation laws for the metric, with some arbitrary weird coordinate transformation that you just made up. You'll see what I mean.
|
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Why the chemical potential of massless boson is zero? In Bose-Einstein condensation, the chemical potential is less than the ground state energy of the system($\mu<\epsilon_g$). But why does the massless boson such as photon have zero chemichal potential($\mu=0$)?
|
The chemical potential is a complementary variable to $N$, the number of particles (of a certain kind), and they get combined in the same sense as $-\beta,H$ and similar pairs. The chemical potential "punishes" too high or too low number of particles in grand canonical and similar distributions such as
$$\exp(-\beta(H-\mu N))$$
In the derivation of similar terms in the exponential in the distribution, it's important that all the extensive quantities such as $H, N$ are conserved. You may view $\beta,\beta\mu$ and similar coefficients as Lagrange multipliers that impose the conservation of $H,N$ etc.
The distribution is maximizing the number of microscopic rearrangements given the fixed specified values of the conserved quantities such as $H,N$ etc.
However, for massless bosons, there doesn't exist any sense or approximation in which the number $N$ of these particles would be conserved. So the states with higher or lower numbers $N$ can't be punished by any $\exp(\beta\mu N)$ factor. It always takes "zero work" to change the number of these massless bosons by one. For example, it's trivial to create a photon; in fact, an accelerating charge is emitting an infinite number of photons (a source of infrared divergences in quantum field theories). The number $N$ of particles like photons isn't even finite so it's clear that the coefficient multiplying it has to be zero for the product to be well-defined.
|
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Is it possible to "add cold" or to "add heat" to systems?
Amanda just poured herself a cup of hot coffee to get her day started.
She took her first sip and nearly burned her tongue. Since she didn't
have much time to sit and wait for it to cool down, she put an ice
cube in her coffee and stirred it with a metal spoon. After a moment,
she felt the spoon warm up, but when she took another sip, the coffee
was cooler. She was pretty sure, the ice added cold to her coffee,
and the coffee added heat to her spoon.
Would you agree?
|
I definitely disagree.
Cold is a relative feeling, cant be added;
Coffee will release temperature to equalize the temp.
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Towing of asteroid I recently studied that NASA has planned to tow and place it in the orbit of the moon.
My doubt is when asteroid is placed in the orbit near moon.since the gravitational field of earth is very high.what will it revolve around the moon or the earth.
Can anyone clarify my doubt ??
|
The setup is just like this: Earth around Sun, Moon around Earth and finally asteroid around Moon. Here's a phrase I've got from Nat Geo (assuming this is the news you've studied)
NASA wants to identify an interesting asteroid flying around deep space, figure out a way to capture the spinning and hard-to-grab orb, nudge it into our planetary region and then set it into orbit around the moon, the agency announced Wednesday.
So, NASA really plans to make the orbit of asteroid around Moon. There's nothing complicated in it. If you're thinking that the orbits depend on gravitational field only, then you can also ask, "Why the moon orbits around Earth and not the sun which has high gravitational field?" And you can expect an answer that the Earth is nearer relative to Sun.
The same goes to the asteroid. It's much nearer to the moon. Gravitational field being an inverse-square law, says that the Earth's field is larger - but the effective flux on the asteroid is small when compared to that of the moon.
But, it confused me when you asked - "what will it revolve around - the moon or the earth?" while NASA already declared that it's gonna place the asteroid around the Moon..!
|
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Dimensional Regularization involving $\epsilon^{\mu\nu\alpha\beta}$ Is it possible to dimensionally regularize an amplitude which contains the totally antisymmetric Levi-Civita tensor $\epsilon^{\mu\nu\alpha\beta}$?
I don't know if it's possible to define
$\epsilon^{\mu\nu\alpha\beta}$ in e.g. $$d-\eta$$-dimensions where $\eta$ is considered small and which we set to zero in the end.
So what are your thoughts?
|
This problem was already mentioned in the original 't Hooft-Veltman article and solved by Breitenlohner and Maison. This solution is known by the name "HVBM scheme" (after 't Hooft, Veltman, Breitenlohner and Maison).
A clear description of this regularization procedure is given for example in the following dissertation by Barbara Jäger. It consists basically of splitting the metric into a $4$-dimnsional and $(d-4)-$ parts, assuming the Levi-Civita tensor to have non-vanishing components only in the 4-dimensional subspace. In addition $\gamma_5$ is assumed to anti-commute with the $\gamma$ matrices of the 4-dimnsional subspace and commutes with the others. This procedure leads to consistent Ward identities.
As mentioned in Jäger 's thesis, this procedure leads to a higher complexity in the computation of the Feynman diagrams, but there exist computer algebra programs implementing this scheme.
|
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How can I prevent my son building up static on his trampoline? Whenever my three year old son plays on his trampoline, it doesn't take very long for him to start building up a significant amount of static electricity. His hair stands on end (which is quite amusing), but when I help him down we both get a nasty static shock (which is not amusing).
He finds the shock upsetting, and I'm concerned that it will discourage him from enjoying his trampoline.
How can I prevent the build-up?
One suggestion I have seen on a parenting forum is to ground the trampoline frame with a cable and a metal stake. However my understanding is that this would enable him to discharge by touching the frame (rather than me), but it would not prevent the static occurring in the first place.
If it's relevant, it's this trampoline and is on a grass lawn. My son bounces in his socks, no shoes. In case of link rot, it is an 8ft trampoline with a net enclosure.
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Dryer sheets rubbed on you before you both get on there, works great for us!
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If photons can be absorbed by electrons, wouldn't that mean light has a charge? I am a biochemistry and molecular biology major. If photons can be absorbed by electrons, wouldn't that mean light has a charge? Electrons only attract positive charges. Isn't it?
|
No. Photons (of electromagnetic radiation) are produced by accelerating electric charges. This means that an excited electron (at a higher energy state) emits energy in the form of electromagnetic radiation proportional to the difference between the two energy levels (orbitals) between which the transition of electron take place.
If such an excited electron can emit energy as EM radiation and return to a lower energy state, then the symmetry which is it can absorb a photon an get to an excited state (depending on the energy of photon) is also possible. Sometimes, an energetic photon can also knock off electrons from metal surfaces (photoelectric effect).
This doesn't mean that the photons should have charge. Instead, they interact with charged particles and transfer their energy. In short, they're just the electromagnetic force carriers.
|
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Physical Interpretation of the Integrand of the Feynman Path Integral In quantum mechanics, we think of the Feynman Path Integral
$\int{D[x] e^{\frac{i}{\hbar}S}}$ (where $S$ is the classical action)
as a probability amplitude (propagator) for getting from $x_1$ to $x_2$ in some time $T$. We interpret the expression $\int{D[x] e^{\frac{i}{\hbar}S}}$ as a sum over histories, weighted by $e^{\frac{i}{\hbar}S}$.
Is there a physical interpretation for the weight $e^{\frac{i}{\hbar}S}$? It's certainly not a probability amplitude of any sort because it's modulus squared is one. My motivation for asking this question is that I'm trying to physically interpret the expression
$\langle T \{ \phi(x_1)...\phi(x_n) \} \rangle = \frac{\int{D[x] e^{\frac{i}{\hbar}S}\phi(x_1)...\phi(x_n)}}{\int{D[x] e^{\frac{i}{\hbar}S}}}$.
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One of the the numerical values of the weight $\exp{\frac{i S}{\hbar}}$ is going to have a maximum contribution to the Feynman path integral. You've probably seen a probability density plot in 2D or 1D. The classical path is going to be the one that minimizes the action. Think of it as a maximum probability density moving from one most probable position to another. The classical action Lagrangian density is going to contribute the most to the path integral.
|
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Does Earth produce metallic elements in its core? Does Earth produce metallic elements in its core?
|
The densest elements (metals) in the Earth fall to the center, due to the gravitational force. The densest elements are radioactive, Earths core is radioactive, Uranium-238 is one radioactive species at the center of the earth.
U-238 decays to Thorium via alpha emission, due to the electric force
$$^{238}_{92}\text U_{146}\to\quad^{234}_{90}\text {Th}^{2-}_{144}+_2^4\text{He}^{2+}_2$$
The Thorium produced is also radioactive and usually decays via beta minus decay, due to the weak force
$$^{234}_{90}\text {Th}_{144}\to\quad^{234m}_{91}\text{Pa}_{143}^++\,^{0}_{-}e^-_0+ \bar\nu_e $$
The Uranium metal has transmutated to become Thorium metal which further transmutated producing Protactinium. Many further transmutations occur (probabilistically) until the original nucleus reaches a stable Lead-206.
Nuclear transmutation effectively converts a radioactively unstable metal to a less massive metal by ejecting subatomic particles from the nucleus.
Yes many metals are being produced in nuclear decay cycles, heavier radioactive metals in Earths core are becoming lighter. The metals produced are chemical different, but numerically there is no more metal atoms than were originally.
These decays also produce a lot of heat. This heat keeps the planet hot in the center and also keeps the outer core and mantle liquid.
|
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Invariance, covariance and symmetry Though often heard, often read, often felt being overused, I wonder what are the precise definitions of invariance and covariance. Could you please give me an example from field theory?
|
The definitions of these terms are somewhat context-dependent. In general, however, invariance in physics refers to when a certain quantity remains the same under a transformation of things out of which it is built, while covariance refers to when equations "retain the same form" after the objects in the equations are transformed in some way.
In the context of field theory, one can make these notions precise as follows. Consider a theory of fields $\phi$. Let a transformation $T$
$$
\phi \to\phi_T
$$
on fields be given. Let a functional $F[\phi]$ of the fields be given (consider the action functional for example). The functional is said to be invariant under the transformation $T$ of the fields provided
$$
F[\phi_T] = F[\phi]
$$
for all fields $\phi$. One the other hand, the equations of motion of the theory are said to be covariant with respect to the transformation $T$ provided if the fields $\phi$ satisfy the equations, then so do the fields $\phi_T$; the form of the equations is left the same by $T$.
For example, the action of a single real Klein-Gordon scalar $\phi$ is Lorentz-invariant meaning that it doesn't change under the transformation
$$
\phi(x)\to\phi_\Lambda(x) = \phi(\Lambda^{-1}x),
$$
and the equations of motion of the theory are Lorentz-covariant in the sense that if $\phi$ satisfies the Klein-Gordon equation, then so does $\phi_\Lambda$.
Also, I'd imagine that you'd find this helpful.
|
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What would happen to the Moon if Earth is turned into a black hole? Assume that all of sudden the Earth is turned into a black hole. And the moon revolves around the Earth (before turning into a black hole). What would happen to the Moon after earth changes to black hole will it be sucked to the black hole or continue to revolve the black hole?
|
$F= (GMm)/r^2$ where $M$ is mass of earth and $m$ is mass of moon and all factors are constant hence $F$ does not changes . As earth now has an infinite density hence only when moon hits the event horizon it would experience a greater $F$. So unless it happens moon would retain its orbit.
|
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Is a black hole a perfect black body? A black body absorbs all light/radiation in its reach. According to basic laws of physics, the more energy a body absorbs the more it can emit. Therefore, a black body absorbs all energy directed at it and also emits all energy that's been absorbed.
A black hole is known to absorb all sorts of every (light, radiation..). If it absorbs -all- energy around it, it should in addition emit all energy it has absorbed. But due to its vast gravity noting can "escape", and therefore nothing is actually emitted. Is a black hole still considered a perfect black body?
Disclaimer: I am no physicists, I am a senior year high school student with an interest for physics. I don't know the preferred terminology of things, especially not in English (2nd language), so if anything is unclear I will explain in comments below.
|
Yes, black holes are supposedly near-perfect black bodies. They emit thermal radiation called Hawking radiation, which, however, does not originate from beyond the event horizon, but is a consequence of the interaction of the strong gravitational field outside the horizon with the vacuum.
The process is sometimes described as the production of 'virtual' particle pairs due to quantum fluctuations, where one of the particles falls into the black hole, forcing the other one to become 'real'.
A graphic explanation why black holes are only near-perfect black bodies is that they cannot absorb photons with wavelength exceeding the black hole's size.
As a side note, due to the equivalence of gravitational and intertial effects due to acceleration, there's the related Unruh effect for accelerating observers.
|
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What is Anderson localization? Could someone give an example worked out in detail? What is Anderson localization, for someone with no previous knowledge on the subject?
I tried to read Anderson's original paper, but it was too terse for me. I have seen a couple of intuitive explanations, e.g. 50 years of Anderson localization on Physics Today. I also read "Localization of waves" by van Tiggelen, but it is more of a review with a lot of formulas and no deductions.
What I need is an introduction to the subject through one example of Anderson localization, worked out in detail.
You can skim over the tedious math. If you point out what needs to be done, I'll work it out (and if I get stuck I'll ask!).
(Related Phys.SE question: Introduction to Anderson localization)
|
A nice example due to Michael Berry is to look at a stack of transparency sheets. Collectively the stack makes an excellent reflector precisely because of the randomness of the gaps between the individual sheets. If the gaps were constant you would just have a photonic band gap. Finite 1D systems require finite randomness for all the modes to be localized. But in infinite random 1D systems all modes are localized no matter how small the randomness. Then localization length is the inverse of the lyapunov exponent which you can show is bounded away from zero in this case. Purely 1D systems are somewhat pathological, but can be realized in single-mode waveguides. The key point is that you can have localized modes (e.g., defect states in condensed matter) without Anderson localization. The latter is a coherent multiple scattering effect. So, go find a box of transparency film :-)
Sorry I missed that you wanted the gory details. Here's an experimental/computational/theoretical example for millimeter waves: http://mesoscopic.mines.edu/preprints/PhysRevB_76_085118.pdf
|
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Bose-Einstein condensate for general interacting systems There is Bose-Einstein condensate (BEC) for non-interacting boson systems. Can we prove the existence of BEC for interacting systems?
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Bogoliubov proved long, long ago that the condensate is stable against weak interactions. The interactions scatter some fraction of bosons out of the lowest-energy single-particle state ("depleting" the condensate), but off-diagonal long range order remains. For a nice introduction to Bogoliubov's theory see Ben Simon's lectures
http://www.tcm.phy.cam.ac.uk/~bds10/tp3.html
(lecture 9). "Weak" is relative though, and in fact real bose liquids with even decently strong interactions can retain a condensate. Bose condensation turns out to be a very robust phenomenon.
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Electric field near a conducting surface vs. sheet of charge I know perfectly well how to derive the magnitude of the electric field near a conductor,
$$E = \frac{\sigma}{\varepsilon_0}$$ and near a sheet of charge, $$ E = \frac{\sigma}{2\varepsilon_0} .$$
In fact, I can explain with clarity each step of the derivation and I understand why is one two times larger than the other. But here's what bothers me...
When I try to think about it purely intuitively (whatever the heck that actually means), I find it difficult to accept that a planar charge distribution with the same surface density can produce a different field.
Why should it care whether there's a conductor behind it or not... ?
I repeat, I understand Gauss' law and everything formally required, but I want to understand where my intuition went wrong.
EDIT: Thanks to you people, I developed my own intuition to deal with this problem, and I'm happy with it, you can see it posted as an answer!
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The answer is simple. It is all in the definition of sigma. In the case of a non-conducting sheet sigma means entire charge in a given area of the sheet meaning both surfaces and everything between them. In the conducting case it is just easier to think of sigma as being the charge on one surface not the sum of both as in the non-conducting case. Therefore, the conducting case looks twice as big simply because sigma is defined as half what it was before. Conceptually imagine for the non-conducting sheet defining sigma as the charge contained only in the upper half of the sheet. This redefinition of sigma will then give you the same answer as for the conductor.
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How long does a supernova last? Is a supernova over instantaneously? Or, does the (for want of a better word) explosion continue for a while? What is/are the order of timescales involved? What is the duration for which the supernova continues to release copious amounts of energy?
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Supernovae can take well over a week to reach maximum luminosity, and they stay rather bright for months after the peak. This just goes to show how much energy is involved in these event.
I was going to assemble a collage of light curves from my own research, but then I realized this has already been done at Wikimedia Commons:
These are rather idealized curves, but they do get the point across. In all cases, the ejecta is expanding at thousands of kilometers per second for most of the process. As the cloud of material thins out, its opacity drops and it is less able to heat up from any energy deposited on its interior surface. Add to that the fact that freely expanding gas will cool (think of air being let out of a pressurized tank).
In order to glow for so long, there must be an energy storage mechanism at work, slowly depositing energy into the gas so that it can give off light. For Type II supernovae, part of this energy is the latent heat of ionization of hydrogen - most of the hydrogen was initially ionized, and the electrons slowly recombine with the protons, giving off photons. Type I supernovae are defined as not showing signs of hydrogen in their spectra, so clearly this won't work. Instead, especially for Type Ia, energy is primarily obtained as radioactive byproducts of the original explosion decay. The most important decay chain is
$$ {}^{56}\mathrm{Ni} \stackrel{\text{6 days}}{\longrightarrow} {}^{56}\mathrm{Co} \stackrel{\text{77 days}}{\longrightarrow} {}^{56}\mathrm{Fe}, $$
and in fact some of the slopes of the piecewise linear Type Ia light curve can be attributed to these half-lives.
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What is the fastest a spacecraft can get using gravity-assist? Assuming normal spacecraft and space objects (no neutron stars, black holes, etc). To what speed can a spacecraft accelerate using gravity-assist?
For example, if a spacecraft is moving at relativistic speeds, it probably won't get seriously sped up by normal-density objects.
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There are other answers there, but I wanted to give an answer in terms of the velocity of the spacecraft, the velocity of the planet, and the surface gravity of the planet.
TL;DR The largest increase in speed possible in a gravity assist is
$$ \frac{2sgr}{gr + s^2} $$
where $s$ is the initial speed of the spacecraft relative to the planet, $r$ is the radius of the planet, and $g$ is the surface gravity of the planet.
A gravitational slingshot is when a spacecraft uses the gravity of a planet to gain speed relative to the solar system. In the reference frame of the planet, the spacecraft flies by it in a hyperbolic trajectory, with the same speed entering and leaving. Because the planet is moving, the speed of the craft relative to the solar system changes. This is demonstrated in the following gif from wikipedia.
Let's analyze the motion of the spacecraft relative to the planet. We place the center of the hyperbola on the origin, and the focus of the hyperbola is the planet's center, which we place at $(c, 0)$.
Since the speed is the same long before as long after the craft leaves the planet, the maximum velocity change comes if the change in angle is maximized, which happens when the craft just skims the planets surface. Thus, if the planet's radius is $r$, the vertex of the hyperbola is at $(c - r, 0)$.
The equation of the hyperbola is of the form
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$
with $a = c - r$, and $b^2 = c^2 - a^2$.
Now, the change in velocity of the craft with respect to the planet is twice the initial $x$ component of the velocity, or $2 s \frac{a}{c}$. To put this in terms of the physical properties of the planet and initial speed, we will analyze the gravitational acceleration at the vertex.
Let $s$ be the speed of the craft in the limit long before the craft reaches the planet. Initially, the craft travels at speed $s$ along the asymptote $y = \pm \frac{b}{a} x$. This asymptote is distance $b$ from the focus. Thus, the line segment between the craft and the planet sweeps out area at the rate $\frac{1}{2} b s$. By Kepler's Second law, the spacecraft sweeps out area at the same rate when it is at the vertex of the hyperbola, a distance of $c-a = r$ from the planet. So the speed $s_0$ of the craft at this time is given by $\frac{1}{2} b s = \frac{1}{2} r s_0$, which yields $s_0 = \frac{b}{r} s$.
Differentiating the equation for the hyperbola twice with respect to time, we get
$$ \left(x \frac{d^2x}{dt^2} + \left(\frac{dx}{dt}\right)^2\right) \frac{1}{a^2} = \left(y \frac{d^2y}{dt^2} + \left(\frac{dy}{dt}\right)^2\right) \frac{1}{b^2} $$
At the vertex, we have $x = a, \frac{dx}{dt} = 0, y = 0, \frac{dy}{dt} = s_0$, and the gravitational acceleration is entirely in the $x$ direction, so $\frac{d^2x}{dt^2} = g$, and this becomes
$$ g = s_0^2 \frac{a}{b^2} $$
Or, substituting in our formula for $s_0$
$$ g = s^2 \frac{a}{r^2} $$
Rearranging gives us $a$, and thus $c$ and $\frac{a}{c}$, in terms of $g$, $s$, and $r$.
$$\frac{a}{c} = \frac{a}{a + r} = \frac{1}{1 + \frac{r}{a}} = \frac{1}{1 + \frac{s^2}{gr}} $$
So the change of velocity is
$$ \frac{2sgr}{gr + s^2} $$
in the $x$-direction. If this direction is almost the same as the direction as the spacecraft's initial direction in the solar system reference frame, then this change in velocity equates to an absolute increase in speed.
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Parabolic motion (experiment) We performed a laboratory, performing six releases of a sphere with angles $15^\circ,30^\circ,45^\circ,60^\circ,75^\circ,40^\circ$ a parabolic movement, took five distances for each angle, the initial velocity was calculated $3.025~\text{m/s}$.
Then doing 5 tosses of the sphere with an angle of $90^\circ$ times were recorded and calculated initial velocity was $3.33~\text{m/s}$, the question is: Why are these speeds almost the same?
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I can think of two or three things.
*
*The whole experiment can be divided into two parts. In one part you calculate the initial speed by measuring distance. In the other part you calculate speed by measuring time. Assuming that your calculations are correct, that would suggest that there might be a difference in the accuracy of measuring distance and measuring time. Assuming that distance is more accurate than time, you can actually work out what the time should have been. You do this by plugging $3.025~\text{m/s}$ into the formula for the $90^\circ$ launch. This will give you the time you would have expected. Compare that to the actual time, by taking the difference, and see if that would be reasonable. (Google for "human reaction time", and see how it compares.)
*Since the time for $90^\circ$ is somewhat longer than expected, you must make sure that you didn't start your chronometer too soon. I haven't seen this experiment, and don't know if it makes a difference, but the chronometer should not be started at the moment of release of the spring (by the hand), but rather a bit later, at the moment of release of the ball (by the spring).
*I assume that the release point of the ball is on the same height as the landing plane. If your spring is sitting on top of the landing plane, then the release point will be slightly higher than the plane. In that case, all your formulas would need to be modified to include that difference.
PS: It is a bit awkward to respond since I can't see the experimental set-up in a diagram. If you have a link to a precise picture, that would be great.
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Can the effects of a person's mass upon the local gravitational field be detected and measured remotely? As the title suggests, Can the effects of a person's mass upon the local gravitational field be detected and measured remotely?
I am aware any mass produces and effects gravity but couldn't find anything in my searching if it is possible or theoretically possible to detect this effect remotely.
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The gravitational field of small objects can be measured. In fact as far back as 1797 Henry Cavendish measured the gravitational field from lead spheres. He used pairs of spheres of mass 158kg and 0.73kg, so for a person weighing say 70kg the same method is in principle possible, although in practice people are an inconvenient shape for doing the experiment and you'd probably have to kill them to get them still enough.
However you'd need to get pretty close for the field to be measurable. In the Cavendish experiment the distances used were around 0.2m. Once you get more than a few metres away I doubt any existing instrument would be sensitive enough to measure the gravitational field from a person.
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When does Thevenin's theorem not apply (modelling a power source with a ohmic internal resistance) Most physics text books say that a power source can be modelled as an EMF with a internal resistance. This is also know as Thevenin's theorem or Norton's theorem. However I have read in some sources that this is not always the case.
When does this model not apply/break down and for what reasons?
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When does this model not apply/break down and for what reasons?
Thevenin's theorem assumes linearity. So, if your circuit is non-linear, the theorem doesn't apply. However, one can linearize the circuit about an operating point and find a small-signal Thevenin equivalent circuit.
To find the Thevenin equivalent of a battery, you would (conceptually, at least) measure the open circuit voltage and then the short circuit current (not advisable in practice). The internal resistance is then just the ratio:
$R_{TH} = \dfrac{V_{OC}}{I_{SC}}$
Then, the voltage across the battery is:
$V_B = V_{OC} - I_B \, R_{TH}$
However, this model only approximates a real battery.
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Is light red shifted in optical tweezers? This is a question I put to my supervisor during my PhD many years ago, and never really got a satisfactory answer to.
In an optical tweezers, assume that a beam of light is used to move a glass bead. My question is whether the outgoing light is red-shifted. If it is not I cannot reconcile how energy is conserved in the system, as work has been done to move the bead?
Answers appreciated as always...
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Sure - the relativistic doppler effect means that light which is scattered off a moving object can be redshifted or blueshifted. And there can be more redshifted photons than blueshifted photons, or vice-versa, depending on where the object is, and how it's moving, relative to the center of the trap.
But since the object is moving much much much less than the speed of light, the frequency shift is completely negligible.
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Voltage of open circuit
A battery with emf $\varepsilon$ and internal resistance $r$ is connected
with a resistor $R$ in the following open circuit. What is the voltage $V_{ab}=V_a-V_b$?
The answer is $- \varepsilon$. "No current. There is no voltage change across R and r.". But I don't really understand why ... I was thinking intuitively it should be $0$? Then thinking of how to get 0, I was thinking ... $V_a = - \varepsilon$ since its on the negative terminal, the $V_b = + \varepsilon$ since its on the positive terminal. But $V_a - V_b = -2 \varepsilon$ ... how do I make sense of this?
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Think of the open as infinite resistance. Your current will be $I = E/(R+\infty) = 0$. Now, voltage dropping across every resistor is proportional to the current and resistance: $V(r) = I * r$. Since infinite resistance is infinitely larger than R, all $E$ will drop across $\infty$ and 0v is left to R. Particluarly, $V_R = 0\mathrm{v} * R = 0\mathrm{v}$. Meantime, $V_\infty = 0\mathrm{v} * \infty = E$. The only way to get a finite value with one of multipliers 0 is to multiply it by $\infty$. Actually, $0 * \infty$ is uncertainty. But here we know that it must be E since $V_\infty$ is the only place where all voltage, generated by supply, must be dropped.
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Forces as One-Forms and Magnetism Well, some time ago I've asked here if we should consider representing forces by one-forms. Indeed the idea as, we work with a manifold $M$ and we represent a force by some one-form $F \in \Omega^1(M)$. It sounded really natural, because if $v \in TM$ then $F(v)$ would be the work done on moving a particle located at $\pi(v)$ along the direction $v_p$ where $\pi : TM \to M$ is the projection $\pi(p,v_p) = p$.
That's pretty fine, however magnetic forces do no work. So in this framework every magnetic force would be zero automatically, since $F(v)=0$ for all $v \in TM$. This suggests that thinking of "force as the one-form that given a vector gives back the work done on changing the momentum of a particle on that direction" seems pretty limited.
Is this right? Magnetic forces really doesn't fit this representation of forces? How can we deal with this if we want yet to consider force as a one-form?
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I think the point is that the force of an EM Field acting on a particle is not just defined by the forms itself but more by the Lorzenz Force
$F= q(E +i_v B)$
Where $E$ is a one form, $B$ a two form where we insert the tangential vector of the particles curve with the inner derivative. Here you can see that the magnetic part is not contributing to the work integral since we have to insert the tangential vector again, which gives Zero.
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How can anything be hotter than the Sun? I've heard that if a space shuttle enters the atmosphere from a bad angle its surface will become so hot that it will be hotter than the surface of the Sun.
How can that be? It seems to an uneducated mind that Sun is really really hot, how could something seemingly minor such as a wrong angle of entrance to the earth's atmosphere could end up generating a heat hotter than the Sun?
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As for the question of whether anything can be hotter than the sun. The Sun is composed of plasma, an energetic phase of matter in which electrons get ripped off of atoms, and electrons and ions coexist in something that might best be described as an ionized gas.
According to this wiki page, the so-called Z machine has achieved temperatures on the order of $10^9\,\mathrm K$ (billions of Kelvin) which is even hotter than the Sun's core which is apparently at around $10^7\,\mathrm K$ according to the physical characteristics listed here.
Thanks to user Gugg who points out that apparently Brookhaven is recognized as having achieved the hottest man-made temperature on the order of $10^{12}\,\mathrm K$. See the link he provides down below.
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How can Ohm's law be correct if superconductors have 0 resistivity? Ohm's law states that the relationship between current ( I ) voltage ( V ) and resistance ( R ) is
$$I = \frac{V}{R}$$
However superconductors cause the resistance of a material to go to zero, and as I understand it, as $R \to 0$, $I \to \infty$. Does this present a problem for Ohm's law?
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In superconductivity we have current without voltage, and in Josephson effect we have a current without voltage. These are quantum systems where bizarre behavior is anticipated. The resistivity of the superconductor is given by $\rho=\mu_0\hbar/(2m_e)$, where the constants are understood.
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Polarization of sound Sound can't be polarized because the vibration of such type can't be polarized i.e, it can't be limited or controlled by any barriers and so polarization is not possible in them.
This is what my teacher answered me when i asked the question. But i didn't understand what did he mean by "the vibration can't be controlled or limited." Does the word cant be limited or controlled make sense here? Moreover can anybody explain in details and more clearly to me?
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I am not sure exactly what your teacher meant, but to me the simple explanation is simply that sound is a pressure wave. Pressure has no direction, only a single value (scalar) and sound waves are fast periodic modulations of the pressure.
A wave on a string can have polarization because the string can be distorted in more than one "direction". It can have an amplitude and a direction of distortion. Similarly, light has polarization because it is a distortion (or variation) of a vector-field (electric and magnetic fields are vector-fields).
Perhaps your teacher meant, that you can choose a direction for a wave on a string for example, thereby constraining it. So in this sense, if there is enough freedom for the wave (i.e. the wave is disturbing more than one direction) such that it can be constrained, it can carry polarization.
Hope this was helpful.
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Why does the quantum eraser seem to violate energy and momentum conservation? In the literature of the quantum eraser experiment it is argued that the change in statistics of the system from non-interference to interference is due to the erasing of "distinguishing information". The preparation of the experiment almost always introduces and then erases this information via a pair of unitary actions on the system. In the specific case that the interference effect is a second order effect of single photon interference, by destroying/creating the interference one is causing a physical disturbance to the system in that it changes the measurable statistics of the system, it destroys/creates the interference pattern that is collected at a single detector/screen. If it is distinguishing information that causes this physical change, if this physical change happens in the absence of a physical cause, is it not the case that this is a violation of the law of conservation of energy? There is no "driving force". It might only be the configuration of the statistics of the system that changes (from interference to non-interference or vica versa), but certainly this is a physical change that is measurable, so it should require energy and momentum to "push" the statistics in the right direction(away from or towards constructive interference). Distinguishing information does not provide this driving force.
And I know the generalized set of rules as to why quantum mechanics is suppose to be energy and momentum conserving, it has to do with the invariance of H under dislacements. This does not explain the specific case of quantum erasure which seemingly defies conservation.
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In the quantum eraser experiment, the interference is re-created logically, not physically. It is re-created by conditioning on some observable which is measured after the interference pattern has failed to appear on the detector. This doesn't require any energy.
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Is it possible to have topological degeneracy in 1D ? I mean to have q-fold degenerate ground states on a ring which could not be lifted by local perturbation.
If the answer is no, then what is the physical (or mathematical) reason against having such a state in 1D ?
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This may not be how people generally think about topological degeneracy, but if you define it to be any degeneracy or asymptotic degeneracy in the thermodynamic limit with the following two properties:
(1) it cannot be lifted by local perturbative Hamiltonians that respect the symmetry;
(2) it cannot be attributed to symmetry breaking, i.e. irreducible representation of dimension greater than 1;
then I suspect the answer to your question is affirmative. This definition makes it possible to speak of topological degeneracy between short-range entangled symmetric states.
My example is the Majumdar-Ghosh Hamiltonian for a spin-$\frac12$ chain (with periodic boundary condition and even number of sites),
\begin{equation}
H = \sum_j \left( \boldsymbol S_j \cdot \boldsymbol S_{j+1} + \frac12 \boldsymbol S_j \cdot \boldsymbol S_{j+2}\right),
\end{equation}
and my symmetry group is $SO(3)$. $H$ has two gapped, (exactly) degenerate ground states, the dimerized states with dimers on even and odd links, respectively. Both states are symmetric under $SO(3)$, so the degeneracy is not due to symmetry breaking. (By the way, both states are short-range entangled.) And I suspect that the degeneracy in the thermodynamic limit can survive any local perturbative Hamiltonian that respects $SO(3)$, on the following considerations.
First, any perturbations that respect not only $SO(3)$ but also the translation symmetry should leave the degeneracy intact, since the Lieb-Schultz-Mattis theorem is expected to apply. Secondly, arXiv:1208.0706 suggests that there is no quantum phase transition when one makes the nearest-neighbor coupling slightly different for even and odd links, which is one way to break the translation symmetry. So the degeneracy might in fact be robust against perturbations that respect just $SO(3)$ but not the translation symmetry.
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measure higher frequencies then half of Planck-frequency? The maximum frequency is defined by the Planck frequency $\omega_P$. Also there is the Shannon theorem which tells us that to lossless capture a signal, you have to sample it with minimum of the double frequency.
That states that you cannot measure frequencies higher than $\frac{\omega_P}{2}$.
Is this correct? I saw sites telling the highest frequencies measured are about $10^{30}$Hz which is under $\frac{\omega_P}{2}=9.27435\cdot10^{42}$Hz. So obviously there are some borders in test equipment and other physical boundaries, but could it be that in theory this is not possible to measure such high frequencies?
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The Nyquist-Shannon sampling theorem is about continuously sampling a waveform. This really matters when a waveform is a mix of many different frequencies. The theorem says you must sample at a rate double the highest frequency. With light you only need to take one "sample" (the energy of the photon or its momentum) to fully know its frequency:
$E = \frac{h c}{\lambda}$ and $p = \frac{h}{\lambda}$
In this sense, a light wave is much simpler than the sampling waves where the Nyquist limit matters.
I'm not sure you can learn about phase and polarization in this way and you certainly can't know the position (due to the uncertainty principle) if you very accurately measure the momentum. Besides this though, I don't think the Nyquist limit really applies to light.
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Probability of position in linear shm? The problem that got me thinking goes like this:-
Find $dp/dx$ where $p$ is the probability of finding a body at a random instant of time undergoing linear shm according to $x=a\sin(\omega t)$. Plot the probability versus displacement graph. $x$=Displacement from mean.
My work:
$$v=dx/dt=\omega \sqrt{a^2-x^2}$$
Probability of finding within $x$ and $x+dx$ is $dt/T$ where dt is the time it spends there and T$$ is the total period.
Therefore
$$dp=dt/T=\frac{dx}{\pi \sqrt{a^2-x^2}}$$
because $t=2\pi /\omega$ and the factor 2 is to account for the fact that it spends time twice in one oscillation. The answer matches the answer and also the condition that integration $-a$ to $a$ of $dp =1$.
But when i try to find p as a function of x to plot the graph I get
$$p=\frac{1}{\pi}\arcsin(x/a)+C.$$
But then I get stuck as there is no way to find $C$ (except the fact that for $C=0$ the probability at the mean position is $0$ and hence $C$ cannot equal 0) which I know of. So how can I get a restraint on $C$ to find its value and hence to properly graph it with the condition that the probability from $-a$ to $a$ be 1?
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The differential $dp(x)$ is the probability of finding the body in an interval of length $dx$ centered at $x$. The quantity $p$ you are looking for is the cumulative distribution function,
$$P(x)=\int_{-\infty}^x \frac{dp}{dx}(x) dx,$$
which is the probability that the particle will be to the left of the point $x$. Since the particle cannot be to the left of $-a$ you can fix $C$ by requiring that $P(-a)=0$. This will then give $P(0)=1/2$ as expected.
It's just a matter of being precise as to exactly what you are calculating.
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Is there a relativity-compatible thermodynamics? I am just wondering that laws in thermodynamics are not Lorentz invariant, it only involves the $T^{00}$ component. Tolman gave a formalism in his book. For example, the first law is replaced by the conservation of energy-momentum tensor. But what will be the physical meaning of entropy, heat and temperature in the setting of relativity? What should an invariant Stephan-Boltzmann's law for radiation take shape? And what should be the distribution function?
I am not seeking "mathematical" answers. Wick rotation, if just a trick, can not satisfy my question. I hope that there should be some deep reason of the relation between statistical mechanics and field theory. In curved spacetime, effects like particle production seems very strange to me, since they originate from the ambiguity of vacuum state which reflects the defects of the formalism. The understanding of relativistic thermodynamics should help us understand the high energy astrophysical phenomena like GRB and cosmic rays.
|
I don't know a definitive answer to your (really good) question, but here is a quote from an old textbook I have by Christian Moller ("The Theory of Relativity"):
Shortly after the advent of the relativity theory, Planck, Hassenoerl,
Einstein and others advanced separately a formulation of the
thermodynamical laws in accordance with the special principle of
relativity. This treatment was adopted unchanged including the first
edition of this monograph. However it was shown by Ott and indepently
by Arzelies, that the old formulation was not quite satisfactory, in
particular because generalized forces were used instead of the true
mechanical forces in the description of thermodynamical processes.
The papers of Ott and Arzelies gave rise to many controversial
discussions in the literature and at the present there is no
generally accepted description of relativistic thermodynamics.
So at least at the time that was written it was unresolved. I'd be interested if there are any more recent updates.
|
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How to determine the direction of medium's displacement vectors of a standing wave? Consider the following problem taken from a problem booklet. My questions are:
*
*What is displacement vector?
*And how to determine the direction of displacement vector at a certain point?
*Where is the position with zero displacement vector?
|
A standing wave is a wave that has nodes. The points of the wave go up and down in some places, and remain at zero at others (the nodes). The general form of a standing wave is a sine curve that remains at a fixed position, but its amplitude changes in time between $+A_0$ and $-A_0$. Specifially, there is a time where the wave form is completely flat.
(From Wikipedia)
The formula is something like
$$f(x) = A_0\cos(\omega t)\,sin(kx)$$
(not the most general form). Compare to a moving wave which has a fixed amplitude, but a changing offset, so it seems to move along the axis.
$$ f(x) = A_0\sin(\omega t + kx)$$
Now in your case you have a tube with air. Your waves don't go up and down (transversal), but back and forth (longitudinal). The nodes are points where the air doesn't move, anti-nodes are where the air moves maximally. Still, it can be described by the same equation. You can try to draw a sine-curve through your first figure. The $y$ value should be the air displacement at point $x$, at a fixed time ($t=0$ or $t=\pi/\omega$). The sine curve must cross the $x$ axis at the nodes, and have maxima and minima at the antinodes. There are two ways to draw the curve, which are mirrored along the $x$ axis. A positive displacement means that the air molecules are moved to the right (compared to where they should be at $t=(\pi/2)/\omega$), a negative displacement means they are moved to the left. You should be able to read off the correct displacement vectors from your drawing.
A little caveat: Don't confuse displacement and pressure, or speed. The nodes always have zero displacement, but the pressure there changes all the time. The points A, B, C, D (on the slopes of the curve) sometimes have zero displacement, when the waveform crosses the $x$-axis, but at that moment the air has the highest speed (change of displacement).
|
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Definition of electric charge and proper explanation Is there a definition of electric charge and proper explanation of it?
It is said "Electric charge is the physical property of matter that causes it to experience a force when close to other electrically charged matter." How is it though that matter can get charged?
Defining charge as the property of feeling a force with other charged matter seems circular. What is charge? Is there a non-circular definition / explanation?
|
I think every fundamental definition is kind of going in circle.
I would say an electric charge is something that obeys Maxwell's laws.
But to write those laws, you have to know $\vec{E}$ and $\vec{B}$ which need a definition of an electric charge.
At the end you just group things that look/react alike and named them.
The problem arise when you have to define what they have in common.
For example, you could say that a charge is something that can interact with a photon.
But what's a photon? It's a particle that carry interaction between charge.
The same happens when you try to define mass as something that resist to motion set by a force to be moved and than force like something applied to massive object to make them move
I don't see a way out of this circle. That's puzzling but you just have to live with it I guess
|
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Evolution principle of the physical laws I wanted to know if there is a physical theory that considers that the laws of physics undergo an evolutionary process. That see the law of physics or the absence of them, as something dynamic, and that with time they slowly converge to something we know today. A kind of simulated annealing of the physical laws.
|
The reason why the "evolution of physical laws" is not popular is because we can make sense of observations from a long time in the past with the laws we know (star formation and evolution comes to mind). One can of course consider an change in physics with time, but this would only add complications to a thoery that was perfectly able to explain most observations when it was static and therefore fall victim to Occam's razor.
|
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Finding surface tension of water at certain temperature and pressure The question is:
Using the Young-Laplace Equation (if applicable), find the surface tension (dynes/cm) for water at 20
degrees Celsius with 2.5 psi. Round to the nearest tenth.
Well, I didn't use the Young-Laplace equation, not sure if needed though. What I did was use the Eötvös rule and its special case for water to solve the question. The equation is:
$$\gamma = 0.07275\;\frac{N}{m}\;\times\;(1-0.002\times(T-291K))$$
What I did was convert 20 Celsius to Kelvin (293K) and then put it in the equation to get:
$$\gamma = 0.07275\;\frac{N}{m}\;\times\;(1-0.002\times(293K-291K))= 0.072459\frac{N}{m}$$
However, I think I may be wrong as this does not account for pressure at all. Which ends up becoming about $72.46\frac{dynes}{cm}$ Am I right or wrong? And is there a better/correct way of doing this?
|
Using the Young-Laplace Equation (if applicable)
Basically a trick question trying to get you to equate the pressure in the question with $\Delta P$ in the Young-Laplace equation.
The actually pressure dependence of water's surface tension is given in On the Evaluation of the Surface Tension-Pressure Coefficient for Pure Liquids
The rate of change of surface tension with respect to pressure is $7 \times 10^{-8} cm$ near atmospheric pressure. So since the question says "Round to the nearest tenth", the pressure effect is insignificant.
|
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Recommendation for a physics history/non-fiction book I know that there are a lot theses being published on lives of physicists. Is there a history/non-fiction book that tracks the development of a problem chronologically? Like pieces of a puzzle.
I would like it to be mathematical and trying to get into the heads of people trying to solve that problem.
Something like a case study.
|
There is hardly a book covering all physics, but for particular subjects there is some. For example:
Jammer: The Conceptual Development of Quantum Mechanics.
Whittaker: A History of The Theories of Aether and Electricity.
|
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North and south of magnetic field
The current I is flowing upward in the wire in this figure. The direction of the magnetic filed due to the current can be determined by the right hand rule.
Can we determine the north and the south of the magnetic field produced by the current I by using a hand rule?
|
You're right with the right hand rule. It's accepted because it agreed with the observations. Placing a magnetic needle (compass) in the influence of the (theorized) magnetic field lines, the compass deflects in the direction of the field indicating the curl. The direction how we twist our fingers show the direction of field. Since we've theorized that the lines of force start at the north pole and end at south pole, of course it can be (it already is) determined by the rule...
So, for a curling magnetic field, there's no specific NS poles. It's curled along the direction of lines.
|
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Does relativistic mass have weight? If an object was sliding on an infinitely long friction-less floor on Earth with relativistic speeds (ignoring air resistance), would it exert more vertical weight force on the floor than when it's at rest?
|
My previous answer proved to be wrong. Energy density ("relativistic mass") does contribute to gravity - and the fact that the object is moving at relativistic speeds does affect the space-time around it.
There is an interesting document that explains the problem in further context.
Besides, when we think about it, if energy density didn't contribute to the force of gravity, photons, mass of which consists of their energy density, would not be affected by gravity. But we have provided many times, that the lightwaves - thus photons - are affected by gravity.
|
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How do you determine the heat transfer from a $p$-$V$ diagram? I doubt this question has been addressed properly before, but if there are similar answers, do direct them to me.
I am currently studying the First Law of Thermodynamics, which includes the p-V diagram and of course, $\Delta U = Q_{to} + W_{on}$.
My question is: how do I determine the heat supplied to the system from a given $p$-$V$ diagram? I know $W_{on} = -p(V_2-V_1)$, but is there a way to find $Q_{to}$ for a given path?
|
If you have a path on $p-V$ diagram that is $p=F(V)$, then using
$$
dU=\delta Q-pdV \implies \delta Q=dU+pdV
$$
NOTE MINUS SIGN as $pdV$ is work done BY the system. $Q$ is the total heat received by the system (it is negative if system releases heat). Assume we are dealing with an ideal gas with $f$ degrees of freedom per particle ($f=3$ for monatomic gas). Then $U$ is given by:
$$
U=\frac{f}{2}nRT=\frac{f}{2}pV
$$
So,
$$
dU=\frac{f}{2}(pdV+Vdp)
$$
$$
\delta Q=\frac{f+2}{2}pdV+\frac{f}{2}Vdp
$$
then using $p=F(V)$ and $dp=F'(V)dV$
$$
\delta Q=(\frac{f+2}{2}F(V)+\frac{fV}{2}V F'(V))dV
$$
Which you can just integrate along the path, care has to be taken though for the sign of $\delta Q$ along the path - if you are looking for total heat in instead of net heat be careful if $\delta Q$ becomes negative along the path as that will just give net heat...
|
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Angular momentum conservation while internal frictional torque is present So this appears in a problem which looks simple enough in its context; It's something like this:
Two discs, A and B, are mounted coaxially on a vertical axle. The discs have moments of inertia $I$ and $2I$ respectively about the common axis. Disc A is imparted an initial angular velocity $2\omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc B is imparted an angular velocity $\omega$ by the same spring compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.
When disc B was brought in contact with A, they acquire a common angular velocity in time $t$. The average frictional torque by the other during this period is: ?
The answer ($2I\omega/3t$) was obtained by applying angular momentum conservation and is exactly what confused me. How can we apply angular momentum conservation when friction is present?
|
How can we apply angular momentum conservation when friction is present?
Why not? If we have a closed system, momentum and angular momentum are conserved. In this case, the full system is disk A and disk B, and there are no external forces, so the system is closed. There are internal forces, namely in this case, friction, but that doesn't matter.
You might be mixing this up with conservation of mechanical energy, which is not conserved if there is friction. (but total energy still is - if you include heat loss etc.)
(Note that this implies that the exact form of the frictional force is irrelevant.)
|
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Donors/Acceptors in Metal Oxides Can anyone explain to me why most articles describe chromium as an acceptor in titanium dioxide? In TiO2, titanium has the charge state Ti$^{4+}$ and oxygen has the charge state O$^{2-}$. When Cr substitutes for Ti, it does so as Cr$^{3+}$. Now, at first glance, Cr has atomic number 24 and Ti 22. Cr therefore has two more valence electrons and is a donor. In TiO2, Cr$^{3+}$ actually has three more valence electrons than the Ti$^{4+}$ ([Ar]$3d^34s^0$ vs [Ar]$3d^04s^0$). It should therefore be a donor, right? The thing is, it forms a deep impurity level near the valence band. TiO2 has an energy gap of around 3.2 eV, and the impurity state is about 1.0 eV from the valence band maximum. To me, that makes it a deep donor. For some reason, journals almost always describe it as an acceptor. Can someone help me make sense of this?
My understanding has always been simply this: more electrons than host $\Rightarrow$ donor, fewer electrons than host (more holes) $\Rightarrow$ acceptor. The position of the impurity level, to my (perhaps incorrect) knowledge, does not determine whether or not the impurity is actually a donor or acceptor, but rather whether it is a recombination center or trap. We can have localized states near the middle of the bandgap that are technically donors/acceptors but function as recombination centers, so I'm not sure what I'm missing here.
|
A 3+ ion substituting for a 4+ ion is called p-type doping. Since it only contributes 3 electrons there is one missing, called a hole. This hole will readily accept an electron, so it is an acceptor. As you say this will also give an impurity band just above the valence band. That Cr gives fewer electrons to the lattice than Ti has to do with splitting of the bands in the lattice.
|
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Physical interpretation of Poisson bracket properties In classical Hamiltonian mechanics evolution of any observable (scalar function on a manifold in hand) is given as
$$\frac{dA}{dt} = \{A,H\}+\frac{\partial A}{\partial t}$$
So Poisson bracket is a binary, skew-symmetric operation
$$\{f,g\} = - \{f,g\}$$
which is bilinear
$$\{\alpha f+ \beta g,h\} = \alpha \{f, g\}+ \beta \{g,h\}$$
satisfies Leibniz rule:
$$\{fg,h\} = f\{g,h\} + g\{f,h\}$$
and Jacobi identity:
$$\{f,\{g,h\}\} + \{g,\{h,f\}\} + \{h,\{f,g\}\} = 0$$
How to physically interpret these properties in classical mechanics? What physical characteristic each of them is connected to? For example I suppose the anticommutativity has to do something with energy conservation since because of it $\{H,H\} = 0$.
|
The physical interpretation is integrability conditions being satisfied on the manifold. From the first equation, if you would take A not depending on 't' explicitly then dA/dt = [A,H]. The Poisson bracket contains in it the dynamics involved in canonically conjugate variables and in classical mechanics, we can measure them simultaneously. Apart from this, laws of conservation can be explicitly seen in this representation.
One important factor to note is that, Poisson brackets are valid only for exact differentials and they follow the canonical transformations. In fact, canonical transformations are nothing but invariance of Poisson brackets.
|
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Assuming space is infinite can our observable universe be an island amongst an archipelego? According to recent measurements our observable universe is roughly 93 billion light years in diameter; also it appears (according to WMAP measurements) that spacetime is flat.
Supposing space is infinite.
It seems to me that it isn't outside logical possibility that there is another observable universe completely outside of our observational range and so far away it has no appreciable effect on the curvature of our universe.
Note, I'm not using the word universe here as everything in space.
|
As noted above in comments, I'm not competely sure I understand the question. But anyway, I'll give it a shot.
The answer is model-dependent. The standard cosmological model at the moment is the Lambda-CDM model. This model has various parameters. Depending on these parameters, the spatial curvature can be positive, negative, or zero. Observation puts (model-dependent) bounds on the spatial curvature: What is the curvature of the universe? Given these bounds, we can put (model-dependent) bounds on the size of the universe: Size of the universe . We then find that the universe is much larger than our own observable region. Therefore the (model-dependent) answer to the question (as I construe it) is yes: there are other regions of the universe in which observers would have observable regions that don't intersect our observable region (and probably never will, given the acceleration of expansion).
|
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What do people actually mean by "rolling without slipping"? I have never understood what's the meaning of the sentence "rolling without slipping". Let me explain.
I'll give an example. Yesterday my mechanics professor introduced some concepts of rotational dynamics. When he came to talk about spinning wheels he said something like:
"If the wheel is rolling without slipping, what's the velocity of the point at the base of the wheel?? It is... zero! Convince yourself that the velocity must be zero. Since if it wasn't zero, the wheel wouldn't be rolling without slipping. So the wheel is rolling without slipping if and only if the point at the base has velocity zero, i.e. if and only if the tangential speed equals the speed of the center of mass."
Well, what I really don't understand is this: is the "rolling without slipping" condition defined as "Point at the base has zero speed"? If not, what's the proper definition for that kind of motion?
Looking across the internet, I have found more or less the same ideas expressed in the quotation. Furthermore, if it was a definition, then it would be totally unnecessary to say "convince yourself" and improper to talk about necessary and sufficient conditions.
I'd like to point out that I'm not really confused about the mathematics behind this or with the meaning of the condition above. What puzzles me is why are those explanations always phrased as if the condition $v'=0$ (where $v'$ is the relative velocity beetween the point at base and the surface) is some necessary and sufficient condition to be "rolling without slipping". Seems to me that this is exactly the definition of "rolling without slipping" and not an "iff".
Any help is appreciated, thanks.
|
watch the following video for a great explanation:
http://www.youtube.com/watch?v=xbXsSEtbkzU
and read this article for the interesting causes of rolling resistance/friction: //www.school-for-champions.com/science/friction_rolling.htm
|
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Why does increasing the temperature of a thermistor decrease it's resistance? Surely, upon an increase in temperature, the atoms within the thermistor would vibrate with more energy and therefore more vigorously, hence making the electrons flowing through the electric circuit more likely to collide with one of the atoms, so increasing resistance.
However, the effect of temperature on a thermistor is contrary to this. I can't understand how it can be.
It's analogous to running across a playground: if everyone is still you're less likely to collide with someone, however if everyone is constantly moving from left to right then a collision is more likely.
So why does an increase in temperature decrease the resistance of a thermistor?
|
Using your playground example....
Imagine if you had to pass a message (electricity) across the playground, when cold you would have to stretch between each fixed person to pass this message. When hot, more people fill the gaps, the message is easier to pass.
Hope this helps :)
|
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Does inertia increase with speed? I have heard that when the speed of the object increase, the mass of the object also increase. (Why does an object with higher speed gain more (relativistic) mass?)
So inertia which is related to mass, increase with speed?
So, if I accelerate on a bus, my mass will increase and my inertia will increase for a while on the bus, until the bus stops?
|
It depends on how you define inertia. http://en.wikipedia.org/wiki/Inertia#Interpretations. The answer to your question is 'yes' if we interpret inertia as p=mv, and no if we interpret it as F=ma.
|
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How does an earthen pot keep water cool? I understand that evaporative cooling takes place thanks to small pores contained in the pot and that allow some water to go through and evaporate. However I couldn't understand clearly whether water inside the pot stays at its original temperature or would it cool further?
If it will become cooler then how?
|
Imho, this process is driven not by energy considerations but by kinetic considerations. That should be why it naively seems weird that water absorbs heat from a cooler object and evaporates. Note: This is an explanation I came up with on-the-fly and have no references to back up with.
Since the earthen pot has small pores, water "flows" through those (probably by capillary action) to the outside surface of the pot. There, it encounters a relatively dry environment and it is kinetically encouraged to leave the surface of the pot and "evaporate". Presumably, while it is on the surface of the pot, it doesn't/can't interact much with the atmosphere and so it needs to take the necessary latent heat from the pot1,2. If that happens, it would help keep the earthen pot cool. Since the water evaporates, by Le Chatelier's principle, the reaction equilibrium gets a strong forward push and the water keeps evaporating.
[1] This seems like the weakest part of the argument. It would be interesting if someone could comment on this aspect.
[2] Since this is a surface effect, the cooling rate should now be propotional to the surface area of the pot. This would mean that smaller pots cool better than larger pots due to a better surface/volume ratio. This claims seems quite easily testable.
|
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Comparing Static Frictions In this figure, which of the static frictional forces will be more?
My aim isn't to solve this particular problem but to learn how is static friction distributed . Since each of the rough-surfaces are perfectly capable of providing the $-1N$ horizontal frictional force but why don't they ? This is kind of ambiguity that who will provide a bigger share in total static friction. And as the surface have different $\mu$, so we can't even invoke symmetry.
|
The contact forces with two blocks are $N_1 = m_1 g + m_2 g$ for the bottom block (to the floor) and $N_2 = m_2 g$ for the top block (to the 1st block).
The available traction is $F^\star_1 = \mu_1 (m_1+m_2)\,g$ and $F^\star_2 = \mu_2 m_2\, g$ or
$$ \begin{pmatrix}F_1^\star\\F_2^\star\end{pmatrix} = \begin{bmatrix}1&-1\\0&1\end{bmatrix} ^{-1} \begin{pmatrix}\mu_1 m_1\,g\\\mu_2 m_2\,g\end{pmatrix} = \begin{pmatrix}\mu_1 (m_1+m_2)\,g\\\mu_2 m_2\,g\end{pmatrix} $$
The balance of horizontal forces is
$$ \boxed{ \begin{pmatrix}P_1\\P_2\end{pmatrix} - \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{pmatrix}F_1\\F_2\end{pmatrix} = \begin{pmatrix}m_1 \ddot{x}_1\\m_2 \ddot{x}_2\end{pmatrix}} $$
where $P_1$, $P_2$ are any applied forces on the blocks (in your case $P_1=1N,\; P_2=0N$) and $F_1$, $F_2$ are the friction forces. Here comes the fun part:
Assume blocks are sticking and solve for the required friction $F_1$, $F_2$ when $\ddot{x}_1=\ddot{x}_2=0$
$$ \begin{pmatrix}F_1\\F_2\end{pmatrix}_{stick} = \begin{bmatrix}1&-1\\0&1\end{bmatrix} ^{-1} \begin{pmatrix}P_1\\P_2\end{pmatrix} = \begin{pmatrix}P_1+P_2\\P_2\end{pmatrix} $$
Find the cases where required friction exceeds traction
$$ \begin{pmatrix}F_1\\F_2\end{pmatrix}_{stick} > \begin{pmatrix}F_1^\star\\F_2^\star\end{pmatrix} = \begin{pmatrix}\mu_1 (m_1+m_2)\,g\\\mu_2 m_2\,g\end{pmatrix} $$
For those cases set $F_i = F_i^\star$ otherwise set $\ddot{x}_i = \ddot{x}_{i-1}$ and solve the balance of horizontal forces.
Example 1, All slipping:
$$ \begin{pmatrix}P_1\\P_2\end{pmatrix} - \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{pmatrix}\mu_1 (m_1+m_2)\,g\\\mu_2 m_2\,g\end{pmatrix} = \begin{pmatrix}m_1 \ddot{x}_1\\m_2 \ddot{x}_2\end{pmatrix} $$
to be solved for $\ddot{x}_1$ and $\ddot{x}_2$
Example 2, All sticking:
$$\begin{pmatrix}P_1\\P_2\end{pmatrix} - \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{pmatrix}F_1\\F_2\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$$
to be solved for $F_1$ and $F_2$
Example 3, Bottom slipping, top sticking:
$$ \begin{pmatrix}P_1\\P_2\end{pmatrix} - \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{pmatrix}\mu_1 (m_1+m_2)\,g\\ F_2\end{pmatrix} = \begin{pmatrix}m_1 \ddot{x}_1\\m_2 \ddot{x}_1\end{pmatrix} $$
to be solved for $F_2$ and $\ddot{x}_1$.
The matrix $A=\begin{bmatrix}1&-1\\0&1\end{bmatrix}$ is the connectivity matrix, and it can be expanded if you have more blocks. See my full solution here of similar problem in more detail: https://physics.stackexchange.com/a/79182/392
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|
Optics alignment of a confocal scanning microscope I am facing a challenge in my project regarding optical alignment.
See the figure:
The challenge is with the vertical optical system alignment. I considered placing a mirror and check back if the image and source coincide. But since the light is too low on power (less bright).
How can I align it so well that everything is absolutely vertical and good?
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A low tech version of a point-like source at the stage plane is to use a pinhole (very small, depending on magnification of your objective, say 1um - 10um). Then illuminate the back side of the pinhole using a laser. You will get a quasi-point source, of course not good for determining PSFs but maybe enough for alignement...
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How can a car's engine move the car? Newton's First Law of Motion states that an object at rest or uniform motion tends to stay in that state of motion unless an unbalanced, external force acts on it.
Say if I were in a car and I push it from the inside. It won't move. So how is the engine of a car capable of moving the car?
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The car's engine tries to make the wheels turn. However, the wheels encounter friction against the road so they cannot just spin. As the road has much higher inertia than the car, it will not move when the wheels want to turn. Instead, it is the car that moves.
The end effect is that the engine pushes against the road, just as you do when you push the car: your feet are on the road, allowing you to push the car forward. In the case of the car, the wheels are on the road and they can push the car.
Pushing the car assumes you have enough force, and the road is not too slippery. This also applies to the engine: if it does not have enough power, or the road is too slippery (icy), it cannot push the car.
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{
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|
Quantum Mechanical Operators in the argument of an exponential In Quantum Optics and Quantum Mechanics, the time evolution operator
$$U(t,t_i) = \exp\left[\frac{-i}{\hbar}H(t-t_i)\right]$$
is used quite a lot.
Suppose $t_i =0$ for simplicity, and say the eigenvalue and eigenvectors of the hamiltionian are $\lambda_i, \left|\lambda_i\right>$.
Now, nearly every book i have read and in my lecture courses the following result is given with very little or no explanation:
$$U(t,0) = \sum\limits_i \exp\left[-\frac{i}{\hbar}\lambda_it\right]\left|\lambda_i\right>\left<\lambda_i\right|$$
This is quite a logical jump and I can't see where it comes from, could anyone enlighten me?
|
Starting with:
$$U(t,t_i) = e^{\frac{-i}{\hbar }H(t-t_i)}$$
If $t_i=0$:
$$U(t,0) = e^{\frac{-i}{\hbar }Ht}$$
Using the identity: $\sum\limits_i \left|\lambda_i\right>\left<\lambda_i\right|=\mathbb{I}$
$$U(t,0) = \sum\limits_i e^{\frac{-i}{\hbar }Ht}\left|\lambda_i\right>\left<\lambda_i\right|$$
Since the exponential of an operator is (by Taylor expanding): $e^H=\mathbb{I}+H+\frac{1}{2}H^2+\dots$
And: $H\left| \lambda_i \right> =\lambda_i \left| \lambda_i \right>$
You should be able to see that:
$$U(t,0) = \sum\limits_i e^{\frac{-i}{\hbar }\lambda_it}\left|\lambda_i\right>\left<\lambda_i\right|$$
|
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Why the magnetic flux is not zero? If $\vec{\mathbf B}=B\vec{\mathbf a}_z$, compute the magnetic flux passing through a hemisphere of radius $R$ centered at the origin and bounded by the plane $z=0$.
Solution
The hemisphere and the circular disc of radius $R$ form a closed surface, as illustrated in the figure; therefore, the flux passing through the hemisphere must be exactly equal to the flux passing through the disc.
The flux passing through the disc is
$$\Phi=\int_S\vec{\mathbf B}\cdot\mathrm d\vec{\mathbf s}=
\int\limits_0^R\int\limits_0^{2\pi}B\rho\,\mathrm d\rho\,\mathrm d\phi
=\pi R^2B$$
The reader is encouraged to verify this result by integrating over the surface of the hemisphere.
According to Maxwell's equations the magnetic flux over a closed surface must be zero, why in this case does not happen?
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According to Maxwell's equations the magnetic flux over a closed surface must be zero.
In this case the hemispherical surface in question is not a closed surface, it is an open surface.
If we consider the closed surface (the hemispherical section And the circular base)
the total flux passing through will be zero.
Using this information it is clear that the flux leaving thought the hemisphere will be equal in magnitude and opposite in sign to the flux entering through the circular base.
|
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|
Induced current using a reference system bound with a moving charge Suppose we have a charge moving at velocity $\mathbf{v}$ in the same plane of a square wire.
If I sit in a reference frame where the square wire is still, since the charge is moving with velocity $\textbf{v}$ in this coordinate system, I will see an induced current in the wire.
$$\textbf{B} = \frac{\textbf{v}}{c^2} \times \textbf{E} $$
$$ \frac{d\phi_B}{dt} \neq 0 $$
Now, what If I choose a reference system where the charged particle is at its origin?
According to this frame, since the the electric charge(and its electric field) is static, $\text{rot}\,\textbf{E}$ will be zero.
$$\nabla \times\mathbf{E} = 0$$
But this means that there is no induced current.
Are my assumptions right? If not, how should I estimate the induced current in a reference system bound with a moving charge at its origin?
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In these type of questions, the following effects usually exist, but are usually missed:
*
*The electric potential(and ergo the electric field), that corresponding points of the wire feel, varies. This effect will polarize the wire which is always negligible, and we are going to ignore it as well.
*in a moving wire, the electric charges will feel the Magnetic Force(which doesn't exist in this case).
$$ q \left( \mathbf v \times \mathbf B \right) $$
*Special Relativity will affect these questions as well. The length contraction is one effect; the other is the relativistic Electromotive Force (relativistic Ohm's law as well). In the latter we will have a term including $ \mathbf v \left(\mathbf v . \mathbf E \right)$(more on this below).
In this case the first two will not have any effect. On the other hand, if you calculate the last one precisely, it should give the same result as the one acquired in the other reference frame.
To calculate the Emf in a moving wire one has to calculate an integral proportional to $$\oint \left(\gamma \left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right ) - \left ({\gamma-1} \right ) ( \mathbf{E} \cdot \mathbf{\hat{v}} ) \mathbf{\hat{v}} \right).\mathbf{dl}$$, where the integral goes around the wire in an instant of time. This comes from the relativistic field transformations.
Now looking at this formula and getting back to our problem, since $\nabla \times \mathbf{E}=0$ and $\mathbf{B}=0$ the first two terms of the integral are zero; unlike the last term which will give non-zero answer.
|
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|
Convergence of Light on the Retina So, I've learned about lens ray diagrams-but the problem I'm having is that when ray diagrams are drawn for a point of an image, they converge to another point, but there are two problems that I see with regards to our eyes:
1) In the simplest drawn case it's at least three rays converging to the same point, but our eye would only have once receptive cell at a point, how does it take in all of them?
2) At any given time, isn't there more than one point being sent to the same point on the back on the retina? We draw 2D cases in high school physics, but I feel like there should be more than one point that gets sent to the same point. In any case, part one is the more pressing matter.
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Geometric optics can be confusing. On your first question: you have to keep in mind that the three rays one usually draws are just example rays highlighting the divergent cone of light emitted from a point and hitting the lens and subsequently being focused via a converging cone onto a unique point in the image plane. The receptive cell at that point receives the total energy of the infinitely many light rays contained in the light cone.
Your second question boild down to "is there a 1:1 mapping between points in the subject plane and points in the image plane. The answer is "yes". This can be seen from the fact that for each point imaged the central ray (out of the three rays commonly drawn) follows a straight line path through the center of the lens, thereby creating a 1:1 correspondence between points in the subject plane and points in the image plane (your retina).
|
{
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How does relativity explain gravity, without assuming gravity I have seen the "objects pull down on space-time" explanations, but they assume a "pull down" force themselves. Could anyone explain the space-time explanation without assuming gravity in the first place?
|
but they assume a "pull down" force themselves.
The images of flat sheets "pulled down" where the planets are do not reflect the fact that the curvature of spacetime is an intrinsic curvature that is measured by geodesic deviation.
What has been done, in order to help visualize the spatial curvature, is to take a two dimensional spatial slice and then embed that into a fictional, flat 3D space where the intrinsic curvature of the slice is represented as an extrinsically curved 2D surface.
A good example of how this is done for a spherically symmetric static star can be found in the book "Gravitation" on page 613:
Therefore, depict 3-space only as it is at one time, t = constant.
Moreover, at any one time the space itself has spherical symmetry.
Consequently, one slice through the center, $r=0$, that divides the
space symmetrically into two halves (for example, the equatorial
slice, $\theta = \pi/2$ ) has the same 2-geometry as any other such
slice (any selected angle of tilt, at any azimuth) through the center.
Therefore limit attention to the 2-geometry of the equatorial slice.
The geometry on this slice is described by the line element
$$ds^2 = [1-2m(r)/r]^{-1}dr^2 + r^2d\phi^2.$$
Now one may embed this two-dimensional curved-space geometry in the
flat geometry of a Euclidean three-dimensional manifold.
Read more using Google Books.
|
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|
Stable Nuclei - Deviation from equal protons and neutrons While studying the semi-empirical mass formula for nuclei, I came across an "asymmetry term" whose function, as far as I understand, is to build in the fact that nuclei "prefer" to have equal numbers of protons and neutrons. This is explained by the Pauli exclusion principle; the neutrons and protons are distinguishable, and hence have fill separate energy levels. Hence, if the number of neutrons and protons is close, the nuclei will have a smaller energy.
Yet, it is observed that large stable nuclei have more neutrons than protons. Why is this so? If nuclei prefer to have equal protons and neutrons, shouldn't the stable nuclei lie along the N=Z line? (in the image)
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Well, look at the other terms...in particular consider the Coulomb term as that one depends only on the proton number.
Walecka's book writes it as
$$ E_3 = \frac{3}{5} \frac{Z (Z-1)}{4\pi R_C} e^2 \approx a_3 \frac{Z^2}{A^{1/3}} \,.$$
This term is strictly positive and grows rapidly as the atomic number increases, while being slightly decreased by growing neutron number.
At some point this contribution begins to outweigh the energy demands of the neutron degeneracy.
|
{
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|
Why does sound move faster in solids? I know that the molecules are closer together in solids, and I know thicker springs also respond carry waves faster than thinner springs, but for some reasons I can't understand why.
The molecules will have a larger distance to move before colliding with another molecule, but in a thicker medium wouldn't that time just be spent relaying the message between multiple atoms? Why is the relaying between a lot of tight knit atoms faster than one molecule moving a farther distance and colliding with another?
|
The difference between solids and gases appears in the momentum conservation equation:
$\rho\frac{d\vec v}{dt}=\vec S$ where $\vec S$ is a source term that expresses the rate at which momentum is exchanged between neighboring volumes, a "restoring force".
In gases, $\vec S=-\vec \nabla p$, where the pressure $p$ relates to density and temperature through equations of state, which describe the interaction between gas molecules. In statistical physics, this term is written as a collision operator. It means that the transmission of the disturbance, or the restoring force, is driven by collisions between molecules. These collisions are rare in average, so each molecule travels a long way before transferring its momentum.
In solids, $\vec S$ is the electrostatic force on an atom that is displaced from its equilibrium position. A very small displacement is enough to imply a large force because of the proximity of other atoms repelling each other. Also, the distance at which atoms "feel" each other is generally higher in solids than in gases (charge interaction instead of dipole interaction). Overall, atoms in solids don't need to travel long before they transmit their momentum.
That's why the disturbance propagation is faster in solids than gases.
|
{
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|
Tensor equations in General Relativity In the context of general relativity it is often stated that one of the main purposes of tensors is that of making equations frame-independent.
Question: why is this true?
I'm looking for a mathematical argument/proof about this fact.
|
In the context of general relativity it is often stated that one of the main purposes of tensors is that of making equations frame-independent. Question: why is this true?
Actually this isn't quite true. General relativity doesn't have frames of reference (except locally, which is trivially true because GR is the same as SR locally). A better way of saying this would be:
The purpose of tensors is to make equations coordinate-independent.
The idea is that when we assign coordinates to something, that's just a name. The laws of nature should be expressible in a manner such that the names don't matter.
I'm looking for a mathematical argument/proof about this fact.
A tensor is defined as something that transforms in a certain way under a change of coordinates. Since the transformation of tensors is well-defined, it follows that a tensorial equation retains the same form under a change of coordinates.
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{
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|
Why are the magnetic moment and the angular moment related? Why are the magnetic moment and the angular moment related? I've always read everywhere that they are related but found nowhere a satisfactory explanation of the cause
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Magnetic moment, angular momentum, and charge are related, because the magnetic field is how the electromagnetic interaction carries angular momentum.
If there were an intrinsic relationship between magnetic moment and angular momentum, you would expect the neutrino to have a magnetic moment. The current PDG reports an upper limit $\mu_\nu < 29\times10^{-12}\,\mu_B$ from experiments with reactor neutrinos, quite different from the electron's magnetic moment $\mu_e \approx 2\mu_B$. Note that an electron-type neutrino will spend part of its time as an $e^-$-$W^+$ loop (and similar for neutrinos with contributions from the other flavors), which will give it some miniscule magnetic moment whose predicted value I don't know. However at tree level the neutrino's coupling to the photon is exactly zero, which means it carries angular momentum without magnetic moment.
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{
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"url": "https://physics.stackexchange.com/questions/65822",
"timestamp": "2023-03-29T00:00:00",
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