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Boson in Superstring I'm confused about a point. Superstring sigma model is
$$
S=-\frac{T}{2}\int\mathrm{d}^2z \left[\eta^{ab}\partial_aX^\mu\partial_b X_\mu -i\bar\psi^\mu\rho^a\partial_a\psi_\mu \right],
$$
of course, the first term is in common with bosonic string one.
Then in addition to the bosonic string spectrum (the one coming from $X$s), that I have, as usual, I will have also the spectrum coming form $\psi$s. My questions are:
*
*What is the fate of the bosonic string spectrum in superstring? i.e. how should I interpret the dilaton $\Phi$, the graviton $g_{\mu\nu}$ and the 2-form $B_{\mu\nu}$ coming form bosonic string spectrum? Why all books refer to the dilaton, gravinton and 2-form as the ones coming from NS part of $\psi$s spectrum?
*After GSO projection the tachyon is cancelled form $\psi$s spectrum and the number of bosonic d.o.f. equals the number of fermionic ones. But this is referred again to the $\psi$s spectrum. If I consider also the $X$s spectrum I still have tachyon and extra boson that unbalance the d.o.f. counting.
Probabilly I make a mistake in my reasonment.
| Your confusion comes from thinking that going to superstrings simply means adding fermions in the spectrum. The spectrum is instead different. For bosonic string (let's focus on NN boundary conditions and open strings) you have something like:
$$\alpha' m^2=N-1$$
where N is the number operator of the transverse vibrational excitations of the bosonic string. In superstring you find:
$$\alpha' m^2=N_{bos}+N_{ferm}-a_{NS/R}$$
where $N_{bos}$ is the number operator of the string coordinates $X$, while $N_{ferm}$ is the one for $\psi$. The ordering constant and the integer/semi-interger nature of $N_{ferm}$ depends on whether you are in the Ramond of NS sector.
In summary, they are two different theories, for instance notice that one lives in 26 dimensions and the other one in 10.
A good suggested reading on string theory is "Basic Concepts of String Theory" by Blumenhagen, Lüst, Theisen.
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Energy diagram of a planet with fixed mechanical energy Consider the following energy diagram for the motion of a planet about a star.
The centrifugal potential curve can be represented once I fixed the angular momentum $\vec{L}$ of the planet. To fix the angular momentum vector of a planet in a orbit I have to fix not only its velocity vector $\vec{v}$ but I must fix this vector at a particular distance $\vec{r}$ from the point $O$. Is this correct?
Now suppose to fix the mechanical energy of the planet $E$ instead (i.e. the horizontal line cannot change). Suppose that I can change the angular momentum of the planet, for istance I can move approximately up or down the centrifugal potential curve. For fixed $E>0$ values, if $L$ grows, the minimum distance approached by the planet from the star increases too. But consider the ellipse situation, i.e. $E<0$. If I move up the centrifugal potential curve the perihelion gets further (and that seems right) but the aphelion gets closer to the star. I don’t see the reason of this last fact, if the planet rotates faster ($L$ increases) shouldn’t it get further both in the min and max distance in its orbit?
Is there something I am missing here?
| For your second case, you can change the angular momentum, but remember that you have fixed your total energy. You can't make the planet revolve arbitrarily fast or it will have more energy than allowed. By increasing the angular momentum without adding energy, you are circularizing the orbit.
To add, you might take a look at the Specific Orbital Energy equation
$$\epsilon = -\frac{1}{2}\frac{\mu ^2}{h^2}(1-e^2)$$
Considering the case where energy and masses are constant, we can rewrite this as
$$k = \frac{(1-e^2)}{h^2}$$
$$h^2 = k(1-e^2)$$
$$h^2 = k - ke^2$$
So for a given energy, angular momentum is at a maximum when eccentricity is 0 (or circular).
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Preference of Chirality I was interested to see that ,
$$
\gamma^5 \psi = \psi_R - \psi_L
$$
By the definition of chirality projection operator and that $\psi = \psi_R + \psi_L$.
since $\gamma^5 \psi$ pops up a lot in QED, I thought it was interesting that $\psi_R$ should necessarily by the positive quantity in this relation.
Is there theorem or book that may explain why we prefer $\psi_R - \psi_L$ rather than the reverse?
| We define positive chirality to be right-handed. Ultimately, this was an arbitrary sign choice (like the choice of which charges are negative versus positive), and (like the choice of charge sign) it was probably not the best choice. However, the choice of chirality, which is really just our choice to use right-handed coordinates, and which goes back originally to how Newton defined the polar angle in polar coordinates, is tied into all sorts of aspects of modern geometry and physics. Since the weak interactions are (unlike everything else in physics) not invariant under parity, it actually makes a difference there (and only there) which choice was made; and it's slightly inconvenient that the $W$ field only couples to the negative chirality fermions. However, fixing this would require undoing a lot of previous work, which would be unnecessarily confusing and not worth the small benefit of eliminating one minus sign. (We could put an extra negative sign in $\gamma_{5}$ and go no further, but this would introduce a minus sign in the relationship between chirality and helicity, which would probably not be an improvement. Changing the definition of helicity, in turn, would require us to go back and do everything in left-handed coordinates, changing the definition of the cross product, etc.--certainly not worth the trouble.)
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Is there any objective basis for the "south to north" directionality of magnetic fields? While explaining magnets to some students, I just realized a very basic thing I never thought about: We often say that magnetic fields have a "direction", that magnetic field lines "exit" the north pole of the magnet and go back around to "enter" (oversimplified, I know) the south pole. But is there anything that makes magnetic north and south actually different, other than being opposites? In electric current, the electrons are actually moving in one, specific direction, but I have no idea why we keep taking the north's side in this? Is there a difference between how magnetic north and south work, or is the special status of north just completely arbitrary?
| It is arbitrary. Just as the positive/negative convention for electrons is arbitrary.
The Earth's magnetic pole near the northern geographic pole is of course a south pole - because it attracts the north pole of a magnetic compass needle.
In the past the Earth's magnetic field has reversed, so if compasses had existed then, they would have pointed in the opposite direction.
Apart from this, electric currents are not always the result of electrons moving, the charge carriers can be positive ions.
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Atmospheric Pressure inside a closed room
Even though they’re too tiny to see, all the molecules of air in the atmosphere above your head weigh something. And the combined weight of these molecules causes a pressure pressing down on your body of 10,000 kg per square metre. This means that the mass of the air above the 0.1 square metre cross section of your body is 1,000 kg, or a tonne.
I would agree with the argument that the atmospheric pressure is a result of the weight of the air above me were I standing in an open area. I do not understand how, by this model of atmospheric pressure, the reason of atmospheric pressure can be explained in a closed room say.
Sourcehttp://www.physics.org/facts/air-really.asp
| This is a duplicate as far as atmospheric pressure goes.
As long as the container is not air tight there will come equalization of pressure. To understand why pressure equalizes one has to see the derivations of the ideal gas law, PV=RT using statistical mechanics, for example here. The attribute "law" is indicative of a thermodynamic law, which was observed to hold, not derived. Only after the understanding using statistical mechanics it could be derived.
Gas in an air tight container will keep the pressure it had when in equilibrium with the atmosphere unless temperatures change. The motions of the gas molecules exert an effective kinetic pressure on any surface they impinge on ( remember pressure is force over area) according to the ideal gas law.
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Why is Kinetic Energy = (-) Total Energy and Potential Energy = 2 $\times$ Total Energy? I came across this relation while reading on the Bohr atomic model. Are there any other forces for which these relations hold good?
| You've discovered the virial theorem.
The virial theorem tells us that for a bound system where the potential energy $V$ is given by an equation:
$$ V(r) \propto r^{-n} $$
The average kinetic energy $T$ and average potential energy $U$ are related by:
$$ 2T = -nU $$
For the electrostatic force $V(r) \propto r^{-1}$ so $n = 1$ and:
$$ 2T = -U \tag{1} $$
The total energy is $E = T + U$ so using equation (1) to substitute for $U$ gives us:
$$ E = -T$$
and substituting for $U$ gives:
$$ E = \tfrac{1}{2}U $$
The example you've found is for the electrostatic force, but exactly the same applies to the gravitational force. Indeed, the first evidence for dark matter was when Fritz Zwicky used the virial theorem to show that the velocities in a galaxy cluster were higher than they should be.
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Can $E=mc^2$ be derived from the Minkowski spacetime metric? $s^2=x^2+y^2+z^2-(ct)^2$? Can $E=mc^2$ be derived from the Minkowski spacetime metric? $s^2=x^2+y^2+z^2-(ct)^2$?
If so, how?
Can the equivalence of mass and energy be derived from the Minkowski spacetime metric?
Has someone done this somewhere? Please do share the link/derivation/proof!
Basically what I am asking is if the Minkowski spacetime metric is enough to imply the equivalence of mass and energy as stated in $E=mc^2$.
Thanks! :)
| Minkowski spacetime has the symmetries of the Poincaré group, which include the four spacetime translations. Noether's theorem then says that there are four conserved quantities, $p_0, p_1, p_2, p_3$, associated with these four symmetries. Typically $p_0$ is denoted by $E$. The structure of the Poincare group implies that these four quantities are related like the components of a four-vector. Thus $p_0^2 - p_1^2 - p_2^2 - p_3^2 = m^2$ is a relativistically invariant conserved quantity. This should be understood as the definition of $m$. If $m^2 > 0$, there exist observers such that $p_0 = E = m$, and $p_1 = p_2 = p_3 = 0$. But in physics we often consider systems such that $m = 0$, and then it is not possible to make $p_1=p_2=p_3=0$, so we need the more general formula to cover all interesting cases.
(I'm using units where $c = 1$.)
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Producing gravitational waves in labs Can gravitational waves be created on very small region of vacuum with quadruple movement of atom or subatomic particles?
| A explain this practically, you need to place a spherical mass of 0.5 kilometre diameter at the distance of 10 metre from LIGO sensor to detect its gravitational wave which may read the maximum of 1mm in the reading.
Simple answer would be Yes you can create a gravitational wave with quadruple movement of atom or subatomic particles Theoretically. But our technology is not upto the mark to prove it Practically
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How is a mass, suspended vertically by two springs in parallel, kept stable? Consider a mass suspended vertically from above by two springs in parallel with different spring constants. Wouldn't the tension be different in each spring? How is this system kept in equilibrium?
| The thing that the springs must have in common is their length $x$. This comes from the mass which is attached to the springs, having different lengths does not make sense in this setup.
From this you can compute the forces. Say the spring constants are $k_1$ and $k_2$. Then the net force exerted is $k_1 x + k_2 x$. In equilibrium, this matches the gravitational force $mg$. Then you have $mg = (k_1 + k_2) x$ which seems to have exactly one solution at
$$ x = \frac{mg}{k_1 + k_2} \,.$$
I do not think that anything is unstable here. Sure, the force that each spring exerts is different. They are $F_i = k_i x$. For different spring constants $k_i$, each spring contributes a different force. The stiffer spring (higher $k_i$) will also exert more force.
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Free Expansion Of and Ideal Gas We know that in free expansion of an ideal gas, no heat enters or leaves the system.
We also know that
$P_\text{initial}V_\text{initial}=P_\text{final}V_\text{final}$
is valid.
If heat exchange is zero, then we can call this process to be adiabatic.
Then why the following is not valid?
$P_\text{initial}{V_\text{initial}}^γ=P_\text{final}{V_\text{final}}^γ$
Also, if I am wrong above, are isothermal free expansion and adiabatic free expansion different?
| You can not classify free expansions into any of those categories as free expansion is not a reversible process and hence the intermediate states are not well defined. The equations are not working because they find the area under the p-v graph but here no such graph can be made.
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Kirchhoff's laws in phasor domain While analysing AC circuits, we write voltage, current etc all with complex numbers namely "phasors". While studying the same, I wondered if Kirchhoff's laws held good with current and voltage in their phasor form. And the internet said they did! They argued somewhat as follows:
$I=Re[\vec{I}]=Re[I_{max}e^{j(\omega t+\phi)}]$
Now, $\Sigma I=0$
[By normal Kirchhoff's law]
Or, $\Sigma Re[I_{max}e^{j(\omega t+\phi)}]=0$
Or, $Re[\Sigma I_{max}e^{j\omega t} e^{j\phi}]=0$
Now, $e^{j\omega t}\neq 0$
Therefore, $\Sigma I_{max}e^{j\phi}=0$
i.e. $\Sigma\vec{I}=0$
Here $I$ stands for scalar current and $\vec{I}$ for phasor current. Similar argument went on for the voltage law.
I didn't get what they did in the sixth step. The fact that real part of a complex number is zero doesn't always imply that the number itself is zero. Can anyone please explain (if this is correct at all!)? And I would be glad if anyone kindly provides any argument, appropriate and more lucid, for the same. Thanks.
P.S. Here is the link to what I found on the internet.
| Great observation! The solution lies in the fact that the complex number $\sum I e^{j\phi}$ is rotating with a speed $\omega$ counterclockwise in the complex plane.
Consider what the expression is telling you -
The real part of a complex number that is rotating counterclockwise in the complex plane is always zero or
if we take the projection of the rotating complex number in the real axis we see nothing.
We can therefore conclude that the complex number is zero, i.e.,
$\sum I e^{j\phi} = 0 + j \cdot 0$.
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When to consider friction as an impulsive force? Suppose a ball obliquely strikes a rough horizontal surface then it experiences a frictional impulse and conservation of linear momentum cannot be done on the horizontal direction.
Now consider another setup in which one block is resting on a rough horizontal surface and another block moving towards the 1st block collides with it.
Then is the momentum of the system conserved? I think that it should not be conserved because the value of friction acting on the system (combination of the 2 blocks) changes from zero to a non zero value. But in books I have seen that they apply conservation of linear momentum. Why do we consider frictional impulse in 1st setup and neglect frictional impulse in the second one?
| We consider friction to an impulsive force, in cases when normal force is impulsive. Here's how:We know that $f=\mu N$(only during slipping motion, for no slipping frictional force is equal to applied force RESISTING friction). Since friction is proportional to normal reaction, it will be impulsive only when normal force is impulsive.Thus, if in a situation there is a sudden change in normal force, friction will be impulsive( part of reason why a water ballon or mud ball burst or distort when thrown to ground, because of impulsive friction due to impulsivee normal force.) Therefore, in your question, since normal is impulsive in case 1 , momentum cant be conserved, but it can be done in case 2 because no extra force, tending to make normal force impulsive acts on system.
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Why do X-rays go through things? I always heard that the smaller the wavelength, the more interactions take place. The sky is blue because the blue light scatters. So why is this not true for X-rays, which go through objects so readily that we need often use lead to absorb it?
| You have to distinguish, which interactions take place, when electromagnetic radiation passes through a solid and interacts with it.
There is a nice plot on Wikipedia, showing the dielectric response of solids for different wavelengths/frequencies.
Basically, as the frequency gets higher, the wavelength becomes shorter, and the molecules or atoms are no longer able to follow the driving force that is transferred by the electromagnetic wave. Therefore in this picture the real part of the refractive index goes to $ 1 $, while the imaginary part, which leads to optical losses or absorption, goes to $ 0 $.
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Normal force and weight If two books are placed on top of each other on a table, why would it be incorrect to say that the weight on the upper book acts on the lower book? I thought that the weight of the upper book would act on the lower book, but my professor said that the normal force instead of the weight of the upper book would act on the lower book. I didn't understand his explanation.
| Both books are individually attracted by the whole mass of earth. That gives you the force $mg$. In principle also the books attract each other via gravity, but that force is so small that you can safely neglect it.
Let's start with the lower book. It sits on the table, gravity is pulling it downwards. The table then resists the compression by the book exerts a normal force onto the book. The gravitational pull goes downwards, the normal force goes upwards. Both forces cancel each other out exactly, the book is at rest.
Now add the upper book. The gravitational pull also be there for the second book. The lower book resists the compression by exerting a normal force onto the upper book. This keeps the upper book at rest.
The lower book is now pressed harder onto the table. To resist the compression, the table has to double its normal force now.
To sum up: The upper book has the following forces:
*
*Gravitational pull from the earth (down)
*Normal force from lower book (up)
The lower book has the following forces:
*
*Gravitational pull from the earth (down)
*Pressure from upper book (down)
*Normal force from table (up)
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May a point rotate about itself? Suppose we have two equivalent rigid cylinders. Cylinder 1 is moving (translating) with constant velocity of v. Cylinder 2 is rotating without slipping and its center’s velocity is constant and equal to v. So, the motion equations of both centers of cylinders are same (x=vt). If we consider centers of cylinders, their kinematics are same. My questions are: 1. What is the difference between these two points (centers of cylinders)? 2. Can we define rotation for a point about an axis that crosses that point?
| Rotations for infinitesimal points are not defined. A rotation is only defined when you have two or more points as a way to describe the fact that their relative distance remains constant.
Also rotational motion is shared for an entire body, meaning that all point on a body rotate the same. The idea of location for rotation only enters when linear velocity is considered as the location where such velocity is zero.
Your first case of a purely translating body, it can be said this is equivalent to a zero rotational speed located at infinity such that $v = \omega \cdot r = 0 \cdot \infty = \text{finite}$.
So the rotational motion (of the centers) is different in the two cases, and the translational motion is only equivalent at the central axis and nowhere else.
Appendix
If a cylinder is translating with velocity $v$ at the center and rotating by $\omega$ then it is said the motion is equivalent to an instantaneous rotation about a point $h = \frac{v}{\omega}$ above the center. For pure rolling to the right, the rotation is clockwise (negative) and the center of rotation is at $d=-r$, or at the contact point.
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Derivative with respect to a difference of independent variables I am dealing with an equation from nonlinear acoustics (Khokhlova-Zabolotskaya-Kuznetsov equation) where a strange term (for me as a mathematician) is used.
The equation looks like this
$$ \frac{\partial}{\partial\theta} \left( \frac{\partial V}{\partial\theta} - V \frac{\partial V}{\partial\theta} \right) = const \left( \frac{\partial^2 V}{\partial\xi^2} + \frac{1}{\xi} \frac{\partial^2 V}{\partial\xi^2} \right) $$
Here $V:(\mathbb{R}^+)^3 \to \mathbb{R}$, $V=V(r,x,t)$ and $\xi = r/const$, $\theta = const(t-x/const)$, $\sigma=x/const$
The question is: what does expression
$$ \frac{\partial V}{\partial\theta} = \frac{\partial V}{\partial (const (t-x/const) )}$$
mean? (of course one can just erase all $const$, they do not change anything)
| The problem is that your point view is too "mathematical". No offence, but every acoustician would jump to the ceiling hearing "one can just erase all const, they do not change anything". Oh, they do $-$ very much! Since one of them is the sound speed... But I get it, you solve that as a mathematical problem and we are undoubtedly grateful for such people.
The $\theta$ is called retarded time and its often used in problems involving spatio-temporal evolution such as radiation ("in a distance of $x$ you witness the radiated information of the source which is now past at the source").
Mathematically you can see that as a transformation to a set of variables with more appropriate features (usually less complicated terms).
Let's end it with a source recommendation. For such questions is that definitely Hamilton's and Blackstock's Nonlinear Acoustics, specifically chapter 3, where the KZK equation is derived as a special case of the second order wave equation.
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How do we know WHEN to get the result from quantum computers? So I always hear that you can't disturb quantum computers because this will ruin the particles superstate.
Well, how do we know WHEN to retrieve the result from the calculation?
How can we determine when the calculation is finished, without observing it?
| Even if you don't know the outcome of an operation, you know that you've applied that operation. So a quantum computer knows when to measure the same way it knows when to do anything else: it applies the operations it's told to, and measures when it reaches a 'measure' instruction.
Quantum algorithms are made up of operations. You apply them in order. Some of those operations are measurements. If the algorithm's 5th operation is a measurement, then you should measure after applying the fourth operation but before applying the sixth operation.
For example, here's a quantum circuit diagram with a measurement. It says to measure the second qubit after applying the controlled-not gate, but before applying the second Hadamard gate:
(You can re-order the measurement and second Hadamard gate without affecting the outcome, but the circuit as drawn specifies a particular order.)
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Why current in series circuit is the same? I have read in the internet that the charges do not have any other path to go and they must go through the same in a series circuit,hence the current is same.
It was quite convincing but what confused me was:
"A resistor is a passive two-terminal electrical component that implements electrical resistance as a circuit element. Resistors act to reduce current flow..."(according to the Wikipedia).
This means that the resistors slow down the rate of flow of charges. By definition, electric current is the rate of flow of charges. Then must not the current be reduced in a resistor even when the amount of charge is same?
|
Resistors act to reduce current flow...
I will give a counter-example to the claim that resistors "act to reduce current". Consider a $9V$ battery connected across a $100\Omega$ resistor; the battery current is $90mA$.
Now, connect another $100\Omega$ resistor in parallel with the first. The battery current increases to $180mA$.
Here's another counter-example. Adding a resistor in series with a current source will not reduce the current but will, instead, increase the voltage across the current source.
So, a blanket statement like "resistors act to reduce current" is as misleading as the ubiquitous "current takes the path of least resistance".
That fact is that a resistor has a voltage across that is proportional to the current through. You must apply this to your particular circuit arrangement to determine if adding a resistor will decrease, increase, or leave unchanged, a particular current.
I have edited the Wikipedia article to read "may be used to reduce current..." however, edits on electronic articles typically have a short half-life.
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Time taken for a layer of ice to form The book I have gives the following derivation:
Let the temperature of the atmosphere be $-\theta$ and the temperature of the water be $0$.
Consider unit cross sectional are of ice, if layer of thickness $dx$ forms in time $dt$ with $x$ thickness of ice above it,
heat released due to its formation is $dx\rho L$ where $L$ is latent heat. If this quantity of heat is conducted upwards in time $dt$,
$$dx\rho L=K\frac{\theta}{x}dt$$
Therefore, the time taken $$t=\frac{\rho L}{2K\theta}(x_{2}^2-x_{1}^2)$$
What I don't understand is why the same amount of time should be taken for the heat to be conducted and for a new layer of ice to be formed. In other words, why is it that the next layer of ice forms only after the heat is released into the atmosphere?
| A bit late maybe. A few google searches reveal that the speed at which heat propagates is infinite in the thermodynamics I am studying. So, the heat is transferred instantaneously and the time taken for a layer to form is equal to the time in which this heat is transferred.
I had thought that the heat is transferred at a finite speed as a layer is formed. So I had thought that the time taken for the heat to be transferred to the top as a layer is formed would be different from the time the layer took to form and if we take $dt$ as the time taken for the a small layer to form, there would be some delay for the heat to be conducted out of the ice and this time would obviously be greater than the the time it took for the ice to form itself.
I don't know if thats what the answers are trying to say but I decided to make an answer anyway.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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When only part of the surface an object is in contact with has friction, what is the normal force I should use? I have the following exercise:
A uniform rod of mass $M$ is given a horizontal velocity $v$ on a rough track as shown in the figure. The surface is rough on the right side of the origin $O$ and the surface is smooth and frictionless on the left side of the origin as shown in the figure. Express the velocity of the rod as a function of distance from the origin. Also find the distance before it comes to instantaneous rest.
I am not able to deduce what the force of friction on a small length $\mathrm{d}x$ of the rod should be. To get the friction on that part should I consider the normal reaction of that part only or of the whole rod ?
| You are correct : to calculate the friction force, you only need to consider the weight of that part of the rod resting on the rough surface, not the whole of it.
When the block overlaps the rough area by distance $x$, the normal reaction on that portion of the block is $\frac{Wx}L$ and the friction is $F={\frac{\mu Wx}L}$. The work done against friction in moving a short distance $dx$ is $Fdx$. The work done in moving distance $x \le L$ from the start position is ${\frac{\mu Wx^2}{2L}}$. When $x \gt L$ the friction force is $\mu W$ so the work done then is $\mu W(x-L)$.
Work done against friction gradually reduces kinetic energy to zero. The critical point is where $x = L$. If the block stops when $x \le L$ then
\begin{aligned}
\frac12Mv^2 &= \frac{\mu Wx^2}{2L} = \frac{\mu Mgx^2}{2L}\\
v^2 &= \frac{\mu gx^2}{L}\\
x &= v\sqrt{\frac L{\mu g}}
\end{aligned}
If the block stops at $x = L$ then $v_0^2 = \mu gL$. If the block starts with speed $v \gt v_0$ then it will stop where
\begin{aligned}
\frac 12M(v^2-v_0^2) &= \mu Mg(x-L)\\
v^2 &= v_0^2 + 2\mu g(x-L) = \mu gL + 2\mu g(x-L) = \mu g(2x-L)\\
x &= \frac L2 + \frac{v^2}{2\mu g}
\end{aligned}
Summary : If the block starts with speed $v_0 \lt \sqrt{\mu gL}$ then it will stop at after travelling a distance $x ={ v\sqrt{\frac L{\mu g}}= \frac{Lv}{v_0}}$. If it starts with speed $v_0 \gt \sqrt{\mu gL}$ then it will stop at after travelling a distance $x ={ \frac L2+\frac{v^2}{2\mu g} = (1+(\frac v{v_0})^2)\frac L2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Electric field inside a material I was thinking about the polarisation, and how the electric field behaves inside the material of permittivity greater than one.
I think to have understood what happens to D and P, but is not clear what happens to E.
Is the electric field inside the material bigger, remains constant, or is weaker?
| I assume you are talking about linear materials with dielectric constant greater than $1$.
Say you have a free charge distribution $\rho$ in vacuum, which produces an E field. Now you introduce linear material. Then the free charges will be "weakened" because they will by partially screened by charges of the dipoles sticking on them. However, the opposite amount of these bound charges must appear somewhere else, because the total bound charge is zero. Usually you will find the opposite amount of bound charges on the surface of the dielectric, which also produce an E field. So it's hard to say in general whether the E field will be smaller or larger.
Let's consider some special examples then.
(1) If it's known that there are no other bound charges except those sticking to the free charges, then the field will be weakened, like in the case of a dielectric-filled capacitor.
(2) If the surface bound charges are very far away, then they can be ignored and we can say the field will be weakened. Like in the case of extended dielectric media.
(3) If the bound charge distribution has symmetry, then we may conclude the answer easily. For example, in the case of a dielectric sphere, since the surface bound charges are spherically symmetric, their field inside is zero and their field outside cancels exactly of those of the volume bound charges. So one can conclude that when you introduce the dielectric sphere, the field inside is weakened, while the field outside remains the same.
| {
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"timestamp": "2023-03-29T00:00:00",
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How does electric charge flow in the typical experiment of scuffing your shoes across the carpet? I'm trying to understand the flow of electric charge. I have read about the triboelectric series, but I do not know where the rubber soles of shoes and carpet reside to know whether the shoes acquire a positive or negative charge. I am guessing that since rubber is harder, perhaps it looses electrons to the carpet like glass looses electrons to silk. I think this will effect the rest of my questions, so I have to start somewhere.
If this is true, then when I scuff my shoes, the area of carpet that I scuffed on will have an abundance of electrons, and acquire a negative charge. First, what will happen to this area of carpet? Will the electrons spread out more evenly across the entire carpet? Will they stay in that location? How will the charge dissipate? I imagine that, perhaps, if the air becomes more humid, that the electrons will flow through the moisture in the air and more evenly distribute, to the point of almost negligible charge imbalance.
Secondly, how does the charge move around on me? I thought the rubber soles are insulators, and resist the movement of electrons. Is it easier for electrons to move on the surface of the rubber? Will the electrons on the surface of my shoes and on me all shift around? How much movement are we talking about, on the scale of atoms and molecules?
Finally, if I walk down the hall, still carrying the charge and then touch a metal object, I get a shock. Why did that happen? Presumably, that metal object was neutral, so why would it want to gain a charge imbalance (perhaps a positive one to offset my positive charge)?
| It looks like I can answer at least one question. The charge imbalance in the carpet will dissipate into the air via ionized moisture, as described in this question: How does an object regains its neutrality after being charged by rubbing?
| {
"language": "en",
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Why an impact exerts so much force? If an object of velocity $v$ and mass $m$ moves towards a resting object of mass $M$, then if the object which is hit might break. Why?
What is the reason that a collision has more power than a statical force $F$ acting on this object?
I haven't found any literature where such things are modeled. Can someone give literature where such things are described mathematically?
| To understand this, use the definition of force $\frac{d{\bf p}}{dt} = {\bf F}$, namely the force is equal to the rate of change of momentum. Something like a collision can be very complicated to model, but the average force is approximately given by ${\bf F}_{average} = \frac{\text{change in momentum}}{\text{time taken}}$. Typically, in a collision, the change of momentum is just the initial momentum, since afterwards the bodies are at rest. So, for any collisions, if you assume that the time taken for the collision is about the same (and can be of the order milli to tenths of a second), you get bigger (average) forces when the momentum change is bigger, namely when the initial momentum is larger. As an aside, not really useful to think of this as a problem in quantum mechanics.
| {
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Rotation of a vector Is a vector necessarily changed when it is rotated through an angle?
I think a vector always gets changed because its projection will change, and also its inclination with axes will always change. However the direction may remain same. Kindly make things clear to me.
| Rotation of a 3-vector
We'll find an expression for the rotation of a vector $\mathbf{r}=(x_1,x_2,x_3)$ around an axis with unit vector $\mathbf{n}=(n_1,n_2,n_3)$ through an angle $\theta$, as shown in Figure .
The vector $\mathbf{r}$ is analysed in two components
\begin{equation}
\mathbf{r}=\mathbf{r}_\|+\mathbf{r}_\bot
\tag{01}
\end{equation}
one parallel and the other normal to axis $\mathbf{n}$ respectively
\begin{eqnarray}
&\mathbf{r}_\| &=(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n}
\tag{02a}\\
&\mathbf{r}_\bot &=(\mathbf{n}\times\mathbf{r})\times \mathbf{n}= \mathbf{r}-(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n}
\tag{02b}
\end{eqnarray}
If $\mathbf{r}$ is rotated to $\mathbf{r}^{\prime}$
\begin{equation}
\mathbf{r}^{\prime}=\mathbf{r}^{\prime}_\|+\mathbf{r}^{\prime}_\bot
\tag{03}
\end{equation}
then the parallel component remains unchanged
\begin{equation}
\mathbf{r}^{\prime}_\|=\mathbf{r}_\| =(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n}
\tag{04}
\end{equation}
while the normal component $\mathbf{r}_\bot =(\mathbf{n}\times\mathbf{r})\times \mathbf{n}$ is rotated by the angle $\theta$, so having in mind that this vector is perpendicular to $\mathbf{n}\times\mathbf{r}$ and of equal norm
\begin{equation}
\left\|(\mathbf{n}\times\mathbf{r})\times \mathbf{n}\right\|=\left\|\mathbf{n}\times\mathbf{r}\right\|
\tag{05}
\end{equation}
we find the expression, see Figure below
\begin{eqnarray}
\mathbf{r}^{\prime}_\bot &=& \cos\theta\left[(\mathbf{n}\times\mathbf{r})\times \mathbf{n}\right]+\sin\theta\left[\mathbf{n}\times\mathbf{r}\right]\nonumber\\
&=& \cos\theta\left[\mathbf{r}-(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n}\right]+\sin\theta\left[\mathbf{n}\times\mathbf{r}\right]\nonumber\\
&=& \cos\theta\;\mathbf{r}-\cos\theta(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\mathbf{n}+\sin\theta\left[\mathbf{n}\times\mathbf{r}\right]
\tag{06}
\end{eqnarray}
and so finally the vector expression
\begin{equation}
\bbox[#FFFF88,12px]{\mathbf{r}^{\prime}= \cos\theta \cdot\mathbf{r}+(1-\cos\theta)\cdot(\mathbf{n}\boldsymbol{\cdot}\mathbf{r})\cdot\mathbf{n}+\sin\theta\cdot(\mathbf{n}\times\mathbf{r})}
\tag{07}
\end{equation}
From this the $3\times3$ rotation matrix reads
\begin{equation}
\mathbb{A}\left(\mathbf{n}, \theta\right) = \text { 3D-rotation around axis} \:\:\mathbf{n}=\left(n_{1}, n_{2},n_{3}\right)\:\: \text{through angle} \:\:\theta
\end{equation}
\begin{equation}
=
\bbox[#FFFF88,12px]{
\begin{bmatrix}
\cos\theta+(1-\cos\theta)n_1^2&(1-\cos\theta)n_1n_2-\sin\theta n_3&(1-\cos\theta)n_1n_3+\sin\theta n_2\\
(1-\cos\theta)n_2n_1+\sin\theta n_3&\cos\theta+(1-\cos\theta)n_2^2&(1-\cos\theta)n_2n_3-\sin\theta n_1\\
(1-\cos\theta)n_3n_1-\sin\theta n_2&(1-\cos\theta)n_3n_2+\sin\theta n_1&\cos\theta+(1-\cos\theta)n_3^2
\end{bmatrix}}
\tag{08}
\end{equation}
| {
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Intensity fluctuations at the output of a single mode fiber coupled to a He Ne laser I have coupled a Thorlabs HNL050L-EC - HeNe, 632.8 nm, 5 mW, Polarized Laser to a 2 meter long single mode fiber patch chord using a Thorlabs F230-FC-B aspheric lens. While I am certainly able to obtain a pure single mode Gaussian at the output, the total output intensity seems to be fluctuating over time scales of about a second. In some sense, the mode appears to be "breathing". The aspheric lens has been mounted on a stable mount, and the fiber is at the correct wavelength. I have also verified that the fluctuations are over and above the intrinsic fluctuations from the laser itself. Has anyone had this issue before? If so, what is the cause and what could be the best way to work around it to get a stable single mode Gaussian output?
P.S Please drop a comment if you require any further details to diagnose this issue.
| Some other effects that might be at play:
1. Reflections from the end-faces of the fiber causing interference
2. Brillion Scattering
3. Check to see if in fact the fiber you're using has a cut-off wavelength shorter than the wavelength you're actually using.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is it true that $\frac{d}{dt}\int_S \mathbf{B} \cdot d \mathbf{a}$ goes to zero if the amperian loop delimiting $S$ contracts indefinitely? I suppose to have an ordinary magnetic field: in the answer I'm not interested to involve Dirac delta: the integral goes to zero. I want to focus on another point: an infinitesimal physical quantity can have a finite time derivative? Of course derivative of zero is zero, but this flux is never strictly zero, and this trouble me because the step
$$
\frac{d}{dt}\int_S \mathbf{B} \cdot d \mathbf{a} \to 0
$$
(when the surface connected to the amperian loop can be taken indefinitely small) is used when we exploit Maxwell equations to fix boundary conditions on the discontinuity between two media. Maybe I'm getting flustered in the slightest thing, but this confuse me and I can't get to the bottom of this problem. How could I see clearly this passage?
| You're right that a function can be "small" at a point but have a "large" derivative at that point. But maybe the confusion is that you're imagining the surface $S$ shrinking in time, so that it's only "small" at one instant. But the surface doesn't shrink in time - you're taking the limit where it's "small" at all times. And if a function is always small, then its derivative is also always small, because $d/dt\ (\epsilon \Phi(t)) = \epsilon\, d\Phi/dt$, where $\Phi = \int_s {\bf B} \cdot d{\bf S}$. So you don't even have to exchange the time derivative with the line integral if you don't want to - the smallness of the loop integral will make $\Phi(t)$ infinitesimal for all times, so it's time derivative will also be infinitesimal.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is the energy operator special? Only the energy operator controls the time dependence of a quantum system, but not the others, why is that?
| The way I like to understand this is in terms of generators of translation. A well known result from classical mechanics (see Goldstein) is that momentum is a generator of translation in the canonical coordinate conjugate to that momentum. For example, linear momentum generates space translations, and angular momentum generates rotations. In Hamiltonian mechanics it is possible to treat time as a true coordinate instead of a parameter, and when this is done, it turns out that the Hamiltonian is the momentum conjugate to time! Thus, energy, or more generally the Hamiltonian, is the generator of time translations which gives the time dependence.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If we threw a baseball from the ISS, could we deorbit the ball? Clearly this is a hypothetical question.
Say we bring a star baseball player into NASA, prep them appropriately for a mission in space, and fly them up to the International Space Station. They go on a spacewalk with a baseball, and at the apoapsis (highest, slowest point in the orbit) throw it retrograde as hard as they can. Could they decelerate the baseball enough that its periapsis (lowest, fastest point in the orbit) dips into Earth's atmosphere enough to de-orbit the ball over time?
(Let's assume that the ball must de-orbit within about 10 years or less. 10,000 years is too long. Also, let's neglect any loss in mobility that a space suit might cause.)
| On a practical note, there was a space walk a few years back when they replaced a failed ammonia pump that was the size of a refrigerator. The astronaut simply gave it a swift push away from the station knowing that it would soon deorbit fast enough that it wouldn't be a collision hazard.
PS- If you have Kerbal Space Program, this would be a fun thing to test.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Gimbal Lock: why is it a problem? I was watching the video Gimble Lock - Explained, by videodumper, about the gimbal lock problem. I understood that during rotations it could happen that one DOF disappear. Looking at the middle part of the video (min 3.22) can be seen that two planes are stuck in only one plane. For this reason performing rotations along X axis is the same of performing rotations along Z axis.
Why are these two planes stuck together? For my imagination it does not seem so complicated to separate them and to continue having 3 DOFs. But I think that something eludes me...
| The planes are not deliberately stuck together - they just happen to coincide when one rotation (by 90 degrees) has dragged one plane of rotation to coincide with another. After that, you can no longer distinguish between rotation about the two axes whose planes coincide - so you have gone from three degrees of freedom to just two.
When this happens, you can no longer describe an arbitrary motion for the next moment in time - there are certain directions of motion that cannot be described (if you think of rotation as a vector pointing "somewhere" in space, you cannot reach every direction in space with just two basis vectors).
Even when you "unlock", and there are once again three distinct directions, two of the basis vectors point in almost the same direction - which means that certain rotations can only be described by the superposition of very large (and nearly opposite) rotations about the two axes that are nearly parallel.
This just makes the problem of describing motion with these axes ill-conditioned. And that is what is called gimbal lock.
| {
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"url": "https://physics.stackexchange.com/questions/254082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Vertical filaments during Atomic bomb explosion What are those vertical filaments that form in the air whenever an atomic bomb is detonated? They are clearly visible in this movie at 3:08.
|
They are smoke rocket trails. Before each test blast, technicians fired these rockets up in the air, leaving large smoke trails that rose well above the bomb's mushroom.
When the atomic blast's shockwave arrived, they moved the trails. Scientists at observation stations could instantly see the effect of the shockwave, hitting, moving and deforming the smoke columns. They were able to measure the speed of the shockwave, as well as the shape and the pressure in relation to the space.
Source
| {
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Spinning top fixed point I have seen many explanations about the movement of a spinning top. The explanations were in a varied level, from basic newtonian mechanics to Lagrangian formalism. But I do not understand why some people consider different fixed points. In same cases it is the point of contact with the surface and others consider some point in the "middle" of the spinning top. My question is whether this ambiguity is a misinterpretation (of those authors), a free choice to describe the movement or a difference caused from different spinning tops?
| Spinning of a top about a fixed point is different from spinning in space or about center of mass. When a top lean certain angle it slips like a slanting ladder because of lack of sufficient frictional force. nutation is observed in the case of a top when there is slip at the contact point.
| {
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Is a falling leaf an example of a chaotic system? Let´s assume is a wind still day in autumn. When a little change is made in the initial motion of a leaf at the time it falls off a tree, the resulting path of motion of the leaf is very different from the path that would develop if these changes wouldn´t have been made.
All the leafs though reach the ground within a maximum radius (wich is a function of the height of the tree) caused by chance effects. Can we, because we know that the leaves land within a certain area, still say that a falling leaf is a chaotic system? Or do we have to consider an infinite high tree, and consider the combined system of the air and the falling leave?
| Chaos is typically phrased as a sensitivity to perturbation in initial conditions (amongst other important things things). You can have a statistical distribution describing the final destination of leaves in general, when the path taken by any individual leaf is deemed chaotic.
As an example, consider common strange attractors. Its easy to see that there is a statistical distribution in locations where a particle ends up, but predicting where any one particle ends up with any degree of accuracy is impossible after enough distance in orbit around a strange attractor.
| {
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Find unitary for given rotations on Bloch sphere I want to characterize a unitary by given rotations on the Bloch sphere.
I know, that when I send in the State $|\Psi\rangle =\begin{pmatrix}1\\0 \end{pmatrix}$, I get the state $U|\Psi\rangle=\begin{pmatrix}\cos \theta\\ e^{i\varphi}\sin \theta \end{pmatrix}.$
When I send in the state $|\Psi'\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1 \end{pmatrix}$, I get the state $U|\Psi'\rangle=\begin{pmatrix}\cos \theta'\\ e^{i\varphi'}\sin \theta' \end{pmatrix}.$
Rotations, the way I know them on the bloch sphere are defined by an axis $\vec{n}$ and an angle $\alpha$: $$U(\vec{n},\alpha)= \cos(\theta/2)\mathrm 1-i\sin(\theta/2) \vec{n}\vec{\sigma}$$
Shouldn't that be enough information to find the unitary $U$?
So basically I have 4 free parameters here.. I don't really know how to deal with this problem. I guess I could just try angles and axis and solve this as an optimization problem numerically. But shouldn't there be a more analytical way?
It would be very nice if someone could help me with that problem.
| Your 2x2 unitary is mostly determined by its action on the state vector $\left(\begin{array}{c}1 \\0\end{array}\right)$. This is because 1) in a 2-d Hilbert space, for any given vector there is a single other orthogonal vector (up to a phase factor), and 2) a unitary map preserves orthogonality. Once $|u\rangle = U\left(\begin{array}{c}1 \\0\end{array}\right)$ is known, $|v\rangle = U\left(\begin{array}{c}0 \\1\end{array}\right)$ is also known as the orthogonal of $|u\rangle$ up to a phase factor, which means $U$ is completely determined by its action on a complete basis set.
$$
\\
$$
Indeed, a unitary can always be parametrized as
$$
U = \left(\begin{array}{cc} a & -b^* \\ b & a^* \end{array}\right)
$$
with $a, b \in \mathbb C$, $|a|^2 + |b|^2 = 1$. In your case, its action on $\left(\begin{array}{c}1 \\0\end{array}\right)$,
$$
\left(\begin{array}{cc} a & -b^* \\ b & a^* \end{array}\right)\left(\begin{array}{c}1 \\0\end{array}\right) = \left(\begin{array}{c} a \\ b \end{array}\right)
$$
allows the identification
$$
a = \cos\theta
$$
$$
b = e^{i\phi}\sin\theta
$$
which implies
$$
U = \left(\begin{array}{cc} \cos\theta & -e^{- i\phi}\sin\theta \\ e^{i\phi}\sin\theta & \cos\theta \end{array}\right)
$$
(Note that requiring $a=\cos\theta$ imposes $a = a^*$ for this case.)
This is then easily rearranged in the $\left(\cos\frac{\alpha}{2}\right) {\hat I}- i \left(\sin\frac{\alpha}{2}\right) \left({\vec n}\cdot{\hat{\vec \sigma}}\right)$ form:
$$
U = \left(\begin{array}{cc} \cos\theta & -e^{-i\phi}\sin\theta \\ e^{i\phi}\sin\theta & \cos\theta \end{array}\right) =
$$
$$
= \left(\cos\theta\right) {\hat I} + \left(\sin\theta\right) \left(\begin{array}{cc} 0 & -e^{-i\phi} \\ e^{i\phi} & 0 \end{array}\right) =
$$
$$
= \left(\cos\theta\right) {\hat I} + i \left(\sin\theta\right) \left(\sin\phi\right) {\hat \sigma}_x - i \left(\sin\theta\right) \left(\cos\phi\right) {\hat \sigma}_y
$$
which identifies
$$
\alpha = 2\theta,\;\; n_x = - \sin\phi,\;\;n_y = \cos\phi
$$
In this case $n_z=0$ and the unitary represents a rotation of angle $2\theta$ around a vector ${\vec n}$ in the x-y plane.
| {
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AdS boundary global vs Poincare' Is the global boundary of AdS the same of the boundary written in Poincare' coordinates?
| Here's what is written on wikipedia (it may be helpful):
Global coordinate:
$AdS _{n}$ is parametrized in global coordinates by the parameters $ (\tau ,\rho ,\theta ,\varphi _{1},\cdots ,\varphi _{n-3}$) as:
\begin{cases}X_{1}=\alpha \cosh \rho \cos \tau \\X_{2}=\alpha \cosh \rho \sin \tau \\X_{i}=\alpha \sinh \rho \,{\hat {x}}_{i}\qquad \sum _{i}{\hat {x}}_{i}^{2}=1\end{cases}
where $\hat {x}_{i}$ parametrize a $S^{n-2}$ sphere. i.e. we have $\hat {x}_{1}=\sin \theta \sin \varphi _{1}\dots \sin \varphi _{n-3}$, $\hat {x}_{2}=\sin \theta \sin \varphi _{1}\dots \cos \varphi _{n-3}$ etc. The ${AdS} _{n}$ metric in these coordinates is:
$ds^{2}=\alpha ^{2}(-\cosh ^{2}\rho \,d\tau ^{2}+\,d\rho ^{2}+\sinh ^{2}\rho \,d\Omega _{n-2}^{2})$
where $\tau \in [0,2\pi ]$ and $\rho \in \mathbb {R} ^{+}$. Considering the periodicity of time $\tau$ and in order to avoid closed timelike curves (CTC), one should take the universal cover $\tau \in \mathbb {R}$. In the limit $\rho \to \infty$ one can approach to the boundary of this spacetime usually called ${AdS} _{n}$ conformal boundary.
With the transformations $r\equiv \alpha \sinh \rho$ and $t\equiv \alpha \tau$ we can have the usual ${AdS} _{n}$ metric in global coordinates:
$ds^{2}=-f(r)\,dt^{2}+{\frac {1}{f(r)}}\,dr^{2}+r^{2}\,d\Omega _{n-2}^{2}$
where $f(r)=1+{\frac {r^{2}}{\alpha ^{2}}}$
Poincaré coordinates:
By the following parametrization:
\begin{cases}X_{1}={\frac {\alpha ^{2}}{2r}}(1+{\frac {r^{2}}{\alpha ^{4}}}(\alpha ^{2}+{\vec {x}}^{2}-t^{2}))\\X_{2}={\frac {r}{\alpha }}t\\X_{i}={\frac {r}{\alpha }}x_{i}\qquad i\in \{3,\cdots ,n\}\\X_{n+1}={\frac {\alpha ^{2}}{2r}}(1-{\frac {r^{2}}{\alpha ^{4}}}(\alpha ^{2}-{\vec {x}}^{2}+t^{2}))\end{cases}
the ${AdS} _{n}$ metric in the Poincaré coordinates is:
$ds^{2}=-{\frac {r^{2}}{\alpha ^{2}}}\,dt^{2}+{\frac {\alpha ^{2}}{r^{2}}}\,dr^{2}+{\frac {r^{2}}{\alpha ^{2}}}\,d{\vec {x}}^{2}$
in which $0\leq r$. The codimension 2 surface $r=0$ is Poincaré Killing horizon and$r\to \infty$ approaches to the boundary of ${AdS} _{n}$ spacetime, so unlike the global coordinates, the Poincaré coordinates do not cover all ${AdS} _{n}$ manifold. Using $u\equiv {\frac {r}{\alpha ^{2}}}$ this metric can be written in the following way:
$ds^{2}=\alpha ^{2}\left({\frac {\,du^{2}}{u^{2}}}+u^{2}(\,dx_{\mu }\,dx^{\mu })\right)$
where $x^{\mu }=(t,{\vec {x}})$. By the transformation $z\equiv {\frac {1}{u}}$} also it can be written as:
$ds^{2}={\frac {\alpha ^{2}}{z^{2}}}(\,dz^{2}+\,dx_{\mu }\,dx^{\mu })$
| {
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'Hovering' light rays on the edge of a black hole According to Prof. Hawking, light rays will 'hover' on the edge of a black hole. If this is true, and the light 'stops' on the edge, how can the electric/magnetic fields which, constitute the light, continue their self-perpetuating state?
What does Hawking mean? His quote is,
the boundary of the black hole, the event horizon, is formed by rays of light that just fail to getaway from the black hole. Instead, they stay forever, hovering on the edge of the black hole.
The Theory Of Everything
| Hawking, I believe, is referring to a more metaphorical 'hovering'. As light, or anything, approaches the event horizon, it becomes more and more redshifted---it's motion appearing to go slower and slower and slower, approaching zero apparent velocity to an outside observer (approximately) infinitely far away. Anything falling into a BH, thus appears to end up 'hovering' just outside of it.
From the perspective of the object falling into the BH, or an observer traveling nearby/similarly to it, nothing special happens. To the infalling observer, time still seems to pass normally... everything is the same. So there is no problem with the electromagnetic wave itself behaving (basically) normally as it approaches the blackhole.
There are lots of questions and material about this and related subjects on physics.stackexchange which might be helpful.
Aside: Apologies for the extremely pedantic and unhelpful comments you received on your question.
| {
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Does a box containing photons have more inertia than an empty box? A box containing photons gravitates more strongly than an empty box, and thus the equivalence principle dictates that a box containing photons has more inertia than an empty box. The inescapable conclusion seems to be that we can ascribe the property of inertia to light. Is this a correct deduction?
| Yes, mass and energy are equivalent. A more competent relativist might be able to give you the complete description, but to first order you can say that the mass of an object is simply the total energy in its volume divided by c^2. That mass is equivalent to the inertial mass by the weak equivalence principle, which is a cornerstone of GR.
That is to say, the answer is yes by the weak equivalence principle.
| {
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Pressure on the sides of a container? Sorry if this is an incredibly basic question for these categories. Basically, I don't understand these types of problems. I'm sure it's something really simple I'm missing.
Let's say there's an open swimming pool with width $w$, length $l$, depth $d$, and density $\rho$ (equal to water's density).
So basically, I'm quite sure the formula I should be using for this is $P = \rho gd$. This turns into $\frac{F}{A}=\rho gd$
How would I find the force of the water exerted onto the sides ($w$ and $l$)? In a problem like this, what would the area $A$ represent? If I wanted to find the pressure on a $w$ side, would I use $w\times h$? I try this, but it doesn't work. I get $F = \rho gwh^2$. But this answer is double the actual answer. It seems like I'd need to integrate (that's what I tried first), but it didn't work either.
There has to be some really basic concept I don't understand.
| It can be shown by integration that total force affecting whole pool side wall of width $w$ is :
$$F_{wall} = \frac {\rho gw~d^2}{2}$$
Divide both sides by wall area $A = w\cdot d$, and you'lll get water pressure affecting whole pool side :
$$P_{wall} = \frac {\rho gd}{2}
$$
So average pressure at whole wall will be water pressure at middle depth $d/2$,
or we can simply state that in a uniform pressure gradient
$$P_{wall} = \frac {P_{min}+P_{max}}{2}$$,
in your case $P_{min} = 0 ~\text{Pa}$ when $h=0$
| {
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Propagating a Gaussian wavepacket backwards in time So, I'm following the MIT OCW lectures on 8.04 quantum mechanics by Prof. Allan Adams. I have the expression for the probability distribution of a gaussian wavepacket for a free particle situation. No initial momentum is imparted. This is a non-relativistic treatment.
$$\mathbb{P}(x,t) = \frac{1}{\sqrt{\Pi}} \frac{1}{\sqrt{a^{2}+(\frac{\hbar}{2ma})^2t^{2}}} e^{\frac{-x^{2}}{2(a^{2}+(\frac{\hbar}{2ma})^2t^{2})}}$$
Say at time $t=5$, I calculate the gaussian form. If I then ask what would the gaussian have looked like at time $t=-5$, the answer would be the same, because of the quadratic factors of $t$. Basically, as you decrease $t$ from $t=5$, the gaussian gets tighter till $t=0$, and then disperses again for negative times.
If someone at time $t=-5$ had indeed actually prepared a gaussian wavepacket of the form we found above at time $t=5$, and propagated the system forward in time, that person would have to update $a$ in the expression for $\mathbb{P}(x,t)$ and would find that the gaussian disperses instead of getting tighter.
There seems to be a contradiction here. How do I reconcile the two scenarios?
| Yes, the gaussian wavepacket can get narrower as the time passes indeed. It's a matter of phases. You know that a gaussian wavepacket is the superposition of plane waves, each one having a precise wavevector. So it really depends on how you "prepare" this superposition, i.e. on how you set the phase of each chromatic component. If at $t=0$ all the plane waves have the same phase, the wavepacket will be maximally narrow and will get wider and wider both in the past and in the future, simply because the various chromatic components will go more and more out-of-phase (recall that the angular velocity depends on $k^2$, so it's different for every chromatic component). On the other hand, if, at $t=0$, the various chromatic components have their relative phases arranged in a particular way, then, as the time passes, the phase difference will get smaller and smaller. When all of them will have the same phase you will have maximum localization, and, after that, broadening i.e. increasing delocalization.
| {
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Why do liquids exert pressure on the sides of a container? What makes a liquid push against the walls of a container if the liquid is completely static?
I was thinking a comparable situation would be a bin full of baseballs. Unless the balls were perfectly stacked they would be rolling off one another and the walls of the bin would stop them. Is it correct to assume the same is happening in liquid on a larger scale, or is something else going on?
It seems like if that was the case the pressure on the walls would be much less than on the bottom.
| In a liquid (or any fluid), the molecules are in random motion (best to say is vibration). So each molecule is vibrating and hence collides with each other. Likewise the molecules in contact with the container also collides with the container walls. Assuming perfect elastic collision, the collided molecules are pushed backwards as insisted by Newton's third law. Under static condition (i.e., no external force acting on the liquid) the liquid exerts force on the wall perpendicular to the surface area. The amount of how much force molecules exert on a unit area is what we call pressure. The force exerted by molecules under no external force will be equal to the reaction force exerted by the container walls. So pressure is actually a consequence reaction force.
Additional : The pressure exerted by solids is a little different from that of fluids as in the case of solids, due to tight interactions of atoms/ molecules, the solid body feels a tension that causes it to deform. But in the case of fluids, the pressure on them sets them in motion, as they have no definite shape to be deformed.
| {
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Why does the interference pattern change with given relation when the source slit changes? Why should the dependence relation be like $${\frac{s}{S}}<{\frac{\lambda}{d}}$$ for the interference condition to be seen?
Where $s$ is the width of the source slit and $S$ is the distance between the source slit and the double slit. $\lambda$ is the wavelength of light and $d$ is the distance between double slits.
Why is it that the interference condition is not seen when the relation is equal or just greater than it?
| Let $L$ be the distance from the double slit to a screen, $S$ be the distance from a source slit to a double slit, and $s$ be the width of the source slit.
The interference pattern produced by a source point is like the cross section of a Fresnel diffraction pattern: the fringes get closer together as the distance from the center increases. The whole pattern scales up linearly as the ratio $L/S$ of the distance from double slit to screen and source slit to double slit increases. Also, if the source point is moved laterally by a distance $x$, the interference pattern moves laterally (in the opposite direction) by a distance $X$ where $$X = x L/S.$$
If two such interference patterns are superimposed but offset by a distance $X/2$, the result will be a pattern in which the fringes separated by a distance $X$ are completely blurred. The fringes whose separation is larger will be blurred slightly, but will still be visible.
If the source is a slit of width $s$, then it is effectively a continuum of point sources across the slit. It includes points with separations ranging from $0$ to $s$, so all features in the resulting diffraction pattern corresponding to fringe separations smaller than $X$ will be blurred severely.
If $X/2$ is equal to or greater than the spacing between the central fringes, then even the central fringes are blurred. So, if the source slit is too wide, there is no fringe visibility at all.
This is the basis of stellar interferometry: the angular width of a star can be measured by the width of the non-blurred portion of a diffraction pattern formed with light from the full width of the star. If you want to know the math in detail, this is a good source.
| {
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Vertical circular motion/How can tension be negative?
This is a rock tied to a string spinning vertically.
Here,
$T+mgsin\theta = mv_1^2/r => T = mv_1^2/r-mgsin\theta$
Suppose I give it a velocity $v$ at the bottom.
1) At what angle $\theta$ will the tension become zero?
2) If the velocity ends up $=0$ at $\theta = 0$, then the tension $T = m0^2/2-mg$ which would end up giving tension a negative value. How is this possible?
3) If the velocity at any point ends up zero, does the tension necessarily have to end up equalling zero as well?
|
[...] which would end up giving tension a negative value. How is this possible?
It isn't. If you set zero speed $v=0$, then you will no longer have circular motion, and the object will accelerate downwards. A non-zero speed $v$ is a requirement for circular motion to happen, because a radial acceleration towards the center can be present only as such. Otherwise it would be like assuming that the object would continue moving around the center even if you stop pulling in the string which obviously isn't the case.
| {
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Wave speed derivation
The wave speed derivation approximates the wave as a circle. It uses that to know that $$a=\frac{v^2}{R}$$However, numerous functions can approximate the wave. A straight line, $x^2$, $x^3$, etc. If I used those I would get a different equation for a. So why is a circle the correct approximation choice?
| The wave can be any shape $f(x)$. But when you focus in on a sufficiently small element of the curve, you can do a Taylor expansion about the point $x_0$:
$$f(x) = f(x_0) + (x-x_0)f' + \frac{(x-x_0)^2}{2!}f'' + ...$$
As the distance $(x-x_0)$ gets smaller, higher order terms can be neglected. If you consider a point with horizontal slope, then $f'=0$ and the first significant term is the quadratic term.
The Taylor expansion of a circle of the right radius happens to match that exactly; and this gives certain "nice" mathematical properties that makes the rest of the calculation easier. But note that if the real function was of the form $x^3$, meaning that the curvature will change with position, it will still have a certain value of curvature at a particular point - and therefore there will still be a circle that "matches" the curve at that point.
| {
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Covariant derivative of a covariant derivative I'm trying to find the covariant derivative of a covariant derivative, i.e. $\nabla_a (\nabla_b V_c)$.
This is something I've taken for granted a lot in calculations, namely I though that by the Leibniz rule we just have:
$$\nabla_a (\nabla_b V_c) = \partial_a(\nabla_b V_c) - \Gamma_{ab}^{d}\nabla_c V_d - \Gamma_{ac}^{d} \nabla_d V_c$$
However when we prove that the covariant derivative of a $(0,2)$ tensor is the above, we use the fact that the covariant derivative satisfies a Leibniz rule on $(0,1)$ tensors: $\nabla_a(w_b v_c) = v_c\nabla_a(w_b) + w_b\nabla_a(v_c)$. However $\nabla_a$ on it's own is not a tensor so how do we have the above formula for it's covariant derivative?
| Easy way
Let me first state the straight-forward way to do this computation.
$$
\langle \nabla_a \nabla_b V, \partial_c\rangle =
\partial_a \langle \nabla_b V, \partial_c \rangle - \langle \nabla_aV, \nabla_a \partial_c\rangle = \partial_a (\nabla_bV)_c - (\nabla_bV)_d \Gamma_{ac}^d
$$
First equality follows from compatibility, second equality uses definition of Levi-Civita symbols.
Hard way
You are suggesting a roundabout way to do this, which formalizes to the following:
$$
\nabla_a\nabla_bV = \nabla_a\left[~(\nabla_cV\otimes dx^c)[\partial_b]~\right]
= \nabla_a\left[~C(\nabla_cV\otimes dx^c \otimes \partial_b)~\right]
= C [\nabla_a (\nabla_cV\otimes dx^c \otimes \partial_b)]
$$
where $$
C: T_pM \otimes T_pM \otimes T^*_pM \to T_pM, ~~ w\otimes z\otimes V \to
z[V]w
$$
is the contraction map of the last two arguments. Covariant derivative on mixed-type tensors commute with contractions (used in the last equality). Observe the expression within $C[ \cdots ]$ is a covariant derivative of a mixed tensor, which you can compute with the Leibneiz rule, and use your favorite component-wise formulas.
| {
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Can light be a spinor? A recent discovery suggests that photons can have half-integer spins. This seems to contradict the well understood notion that photons are vector (1-form) fields
What does this mean for the fundamental picture of electromagnetic propagation?
| It probably does not mean anything. That paper concerns the quantization of electromagnetic waves in less than three spatial dimensions. In fact, there are a number of decades-old results showing that it is often possible to evade the spin-statistics relationship in lower-dimensional systems. While these kinds of results (including this new one) may be very interesting theoretically and may have applications to the quantization of excitations in two-dimensional condensed matter systems (not pure photons, but coupled excitations involving the charge density of the material and the electromagentic field), it is not going to change anything we know about how physical photons propagate in three-dimensional space.
| {
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Flat space Solution of Einstein Field Equation Does a trace-free energy-momentum tensor $T_{\mu}^{\mu} = 0$ ensure that the Einstein's field equations have a flat space solution?
| The Einstein field equations
$$
R_{\mu\nu}~-~\frac{1}{2}Rg_{\mu\nu}~=~8\pi GT_{\mu\nu}
$$
for zero stress energy means that the Ricci Curvature $R_{\mu\nu}$ is proportional to the metric with $R_{\mu\nu}~=~\frac{1}{2}Rg_{\mu\nu}$. This is called an Einstein spacetime, and for a constant Ricci scalar $R~=~R_{\mu\nu}g^{\mu\nu}$ this is a spacetime of constant curvature, such as a 4-sphere.
By taking the trace of this Stress energy it is not hard to show that
$$
8\pi G\left(T_{\mu\nu}~-~\frac{1}{2}Tg_{\mu\nu}\right)~=~R_{\mu\nu}.
$$
The traceless condition $T~=~0$ just means the Ricci tensor is propotional to the stress energy, but the Ricci scalar is zero.
| {
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What does it mean that the laws of physics are time reversible?
The Universe, as far as we can tell, only operates according to laws
of physics. And just about all of the laws of physics that we know are
completely time-reversible, meaning that the things they cause look
exactly the same whether time runs forward or backward.
*
*http://www.sciencealert.com/physicists-just-found-a-link-between-dark-energy-and-the-arrow-of-time
I am not sure I understand this, how can we reverse a law in time? So for example how does a time-reversible law and a time-irreversible law look like? What is the difference between them?
| This is not a useful answer!
1. The statement: "There is no force/law that makes the cup rise" is completely misleading because it deals in likelihood (probability) not law.
2. The statement, "Even though, the mathematical formulas may allow the backward flow of events, there are no such forces that would cause it" is another irrelevant statement about probability.
3. The mathematical formulas are equivalent to the law, because the law NEVER violates the math. For purposes of explanation, the formula and law are the same.
The true issue in reversibility is, IF a series of events is run backwards (not will they run backwards), are laws violated or not? The answer is no there is no violation. Therefore physical law is time reversible.
| {
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What does a voltmeter actually measure? For time varying fields (even quasistatic ones) the electric field is given by
$${\bf E} = - \nabla \Phi - \frac{\partial {\bf A}}{\partial t}$$
So what does a voltmeter measure? Does it measure a difference in $\Phi$ between two points $a$ and $b$, or does it measure $\int_a^b {\bf E} \cdot d{\bf l}$ or does it measure something else?
| Although both answers above have valid correct facts, they are both missing the original point in the question. The op tried to ask what happens in the electrodynamic case when the line integral of the electric field is path dependant.
In this case the line integral would give a different value for each trajectory of the electric field.
Above was mentioned (Walter lewins classes were used as reference) that in the end the voltmeter would measure the line integral of the electric field (the nonconservative electric field in this case).
This is correctBut it is important to mention that Walter Lewins explanation was for the voltage in a line element without volume. In the general case of a volume element, the voltmeter (at least the classic one based in a galvanometer) measures an average voltage around a cross sectional area perpendicular to all current lines.
Imagine having two cross sectional surfaces in a resistance. Then you would calculate each line integral from each origin point in the initial cross sectional surface following the current lines to the landing cross sectional surface.
There is a line integral for each point in the initial cross sectional surface, you would have to average all the line integrals over the cross sectional surface, and that is what you area measuring with a classic voltmeter. You are measuring the average work per unit charge done in the volume element you are considering, limited by the two cross sectional surfaces normal to the current lines.
| {
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Analysis of a system consisting of a leaking tank of water The departure point is this problem:
A water tank on wheels is moving over an horizontal trail with negligible friction. There is a small opening in one of the walls, at a depth of $h$ below the tank's water level. The cross-section area of the opening is $A$. The initial masses of the tank and the water are $M$ and $m_0$. What is the initial acceleration of the cart?
Can we consider the water at the top of the tank to be stationary? If so, then it is pretty straightforward to find the velocity at which the fluid exits the opening. Then I would guess the acceleration could be estimated by looking at momentum variations. However, this is a varying mass system, so the mass also varies. This ends up being similar to the rocket equation, which involves solving a system of differential equations.
Is there a simpler way to solve this kind of problem (in other words, can you obtain the value of the acceleration without having to solve differential equations)?
| Using Torricelli's law (https://en.wikipedia.org/wiki/Torricelli%27s_law) you get $v=\sqrt{2gh}$ irregardless of how big the opening is. Now you can calculate how much mass is leaving the tank at any time by multiplying the volume that is leaving the tank by its density: $$\Delta m=A\Delta s\cdot\rho$$ with $\Delta m$ the mass leaving the tank at any second, $\Delta s$ the distance travelled by the water in a second and $\rho$ the density of the water. Dividing by the timestep and letting the timestep be really small you get:
$$\frac{\Delta m}{\Delta t}=A\frac{\Delta s}{\Delta t}\rho$$
$$\frac{dm}{dt}=Av\rho=A\rho \sqrt{2gh}$$
Using the momentum of the stream to calculate the force applied to the cart (assuming the velocity doesn't change in the beginning)
$$F=\frac{dp}{dt}=\frac{d(mv)}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}=A\rho\sqrt{2gh}^2+0$$
$$F=A\rho g h$$
And finally applying the force to the cart:
$$a=\frac{F}{m}=\frac{2A\rho gh}{M+m_0}$$
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Where does $\hat{P}\psi(x) = -i\hbar \partial_x \psi(x)$ come from? It's a very basic question, where does the relation $$\hat{P}\psi(x) = -i\hbar \partial_x \psi(x)$$ for any square integrable $\psi(x)$ come into existence? Some texts I found states that the above relation comes as a consequence of momentum being defined as generator of translation. But what is the basis of this definition? If momentum were defined to be generator of other form of symmetry, then it wouldn't have had the form as it does now.
In some other text, it's the other way around. Namely the action of momentum on a wavefunction is defined to be $$\hat{P}\psi(x) = -i\hbar \partial_x \psi(x)$$ and thence it leads to momentum being the generator of translation.
Which one is the correct one? How was such action of momentum on wavefunction historically developed?
| Ab initio the momentum operators can be constructed using de Broglie Plane waves
In one dimension, using the plane wave solution of the Schrodinger equation,the wave function
Psi = exp. i (kx -wt) ,
if one takes the partial derivative w.r. to x of the wave function
delta/delta x (Psi) = ik. Psi
and using de-Broglie relation p = hbar . k we get
delta/delta x (Psi) = i p/hbar . Psi
The above relation suggests the operator equivalence of momentum:
p-operator = -ihbar. Delta/deltax
so the momentum value p is a scalar factor, the momentum of the particle and the value that is measured, is the eigenvalue of the momentum operator.
As the partial derivative is a linear operator the momentum operator is also linear,
(one can think of momentum as generator of translational symmetry)
and because any wave function can be expressed as a superposition of other possible states
when this momentum operator acts on the entire superimposed wave, it furnishes the momentum eigenvalues for each plane wave component.
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Determination of the electric field of charged object using Gauss's law when we determine the electric field of a charged rod of infinite length,we consider a circular cylinder to be the gaussian surface for convenience.In the case of a charged sheet we choose the gaussian surface to be a cylinder going through the sheet,again for convenience.
yes it is easy to calculate the flux if we choose the gaussian surface to be a cylinder in the above cases?but how can we calculate the flux if we choose an arbitrary surface to be the gaussian surface?
| When you choose an arbitrary gaussian the flux is still given by the charge inside the surface. However it would be useless in calculating the electric field unless you are able to write
$$\int\vec E\cdot d\vec A=\int EdA=E\int dA.$$
That is, Gauss law is useful when the surfaces elements are always parallel, antiparallel or perpendicular to the electric field and the magnitude of the field is constant along finite surfaces so that it can be put out of the integral.
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A Formula for the Phase Difference Between the Electric and Magnetic Wave Oscillations? A) Is there a formula for the phase difference between the electric and magnetic field oscillations, in vacuum, in an electromagnetic wave emitted from an antenna, as a function of the frequency the distance from the antenna?
B) Does the formula depend on the antenna type and on the direction of the radiation?
| The electric and magnetic fields are always in-phase if the wave can be treated as a plane wave (which simply means it cannot be too close to the source), and in vacuum or any medium with linear response, such as air.
Boundary conditions of wave guides change this relationship, and must be solved for each specific case. If the wave guide is large enough, you will only see effects near the surfaces.
Transmission through a conductor results in phase-lag of the magnetic field, and a rapid extinction of the propagating wave; see skin depth.
The methods used are nicely described here: Chapter 9: Electromagnetic Waves - MIT OpenCourseWare
Electromagnetic plane wave: electric and magnetic fields are always in phase.
You can study the mathematical proof, based on Maxwell's equations for an arbitrary plane wave, here.
So the direct answers are:
(A) No, except that they are always in phase in the "far field";
(B) No, see (A).
In the near field, for a dipole antenna, see Chapter 10: Antennas and Radiation - MIT OpenCourseWare, especially section 10.2.
This is why you ordinarily only need to consider the electric field with radio or light transmission, except in a wave guide, or non-linear media.
Occasionally somebody claims that the electric and magnetic fields are out of phase for circular polarization. This is not quite correct: the quarter wave plate accepts linearly polarized light at some angle wrt the fast axis of the QWP; the orthogonal components of the electric field vector under go different amounts of optical delay, resulting in two distinct electromagnetic waves, with the same phase delay in their electric and magnetic fields. See this animation of circular polarization; the input light is linearly polarized, with the red and blue representing the two components wrt the fast axis of the quarter wave plate, QWP.
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The relation between anomalous dimensions and renormalization constants I am trying to understand the general strategy and technical details of calculating $\beta$-function at higher orders. $\beta$-function is the anomalous dimension of the coupling constant and there is a complete set of anomalous dimensions corresponding to different fields, propagators (gauge field), and vertices.
The anomalous dimension corresponding to a renormalization constant could be defined as
$$\gamma=-\mu^2\frac{d \log Z}{d\mu^2}$$
And in the minimal subtraction scheme, one could expand the renormalization constants as
$$Z=1+\sum_{i=1}^\infty \frac{z_i (a_s,\xi)}{\epsilon^i}$$
where $\xi$ is the gauge parameter, i.e. we do not need to fix the gauge before computing the anomalous dimension, and the number of space-time dimensions $D=4-2\epsilon$. Now, let's consider the anomalous dimension of the coupling constant, and assume that scale dependence of the corresponding renormalization constant $Z_{a_s}$ happens through $a_s$ and $\xi$. How does the following relation hold?
$$-\beta(a_s)=\left(-\epsilon + \beta(a_s) \right) a_s \frac{\partial \log Z_{a_s}}{\partial a_s} $$
What about the anomalous dimension of $\xi$. What would the relation be?
note: Please refer to http://arxiv.org/abs/hep-ph/0405193v3 for the conventions; This article http://link.springer.com/article/10.1007%2FBF01079292 also contains valuable points such as (2.4) and (2.6) which I believe are related to the problem at hand.
| In $\overline{MS}$ the $\beta$ function does not depend on the gauge parameter. This means that the dependence on $\mu$ in $Z$ only comes from the coupling constant $a \propto \mu^{-2 \epsilon}$.
For general $Z(a_s,\xi)$, the relation is as follows (eq. 21 in the Chetyrkin paper):
$$-\gamma =\left(-\epsilon + \beta(a_s) \right) a_s \frac{\partial \log Z}{\partial a_s}+\gamma_3(a_s,\xi)\xi \frac{\partial \log Z}{\partial \xi}$$
| {
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Tractrix - velocity pointing to pulling point It is said the tractrix is the curve described by a mass being pulled by a string, where the end of the string being pulled moves with constant speed, and the mass suffers a friction force. What is the physics explanation for why in the tractrix the velocity is always aligned with the string pulling the mass. Why if $h(t)=(h_x(t),h_y(t),h_z(t))$ is the position of the mass, and $j(t)$ is the position of the start of the string pulling the mass, then the velocity is always aligned with the string, that is $h'(t) = k (j(t) - h(t) )$ holds, where $k>0$ is some constant. Can this be derived by for instance stating the forces applied on the mass and then using $F = m a$ or some other physical argument ?
| You can imagine it as you pulling your dog (object) with a rope (string), while walking on a straight line, but the dog don't want to follow where you are pulling him and just want to stay in place. As you move, the tension of the rope felt by the dog is always directed toward you, making the dog move slightly (but unwillingly) toward the direction of that tension.
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Why does Bernoulli's equation only apply to flow along a streamline that is in viscid, incompressible, steady, irrotational? I am learning about hydrofoil on this website.
In a later video I watched, I learned that in the process of deriving Bernoulli's equation, $$constant=P/d+gh+1/2v^2$$ has to multiplied through by density. To keep the left side constant, fluid density has to be constant and thus is incompressible.
But what about other qualities like in-viscid, and irrotational? What do they mean? And why are they necessary?
| It can also be derived from Euler's equation of motion of a fluid element $dm$ moving (translating but NOT rotating) along a flow line through a conduit:
That equation (a balance of forces acting on the fluid element) is:
$$\frac{dp}{\rho g}+\frac{vdv}{g}+dh+\frac{d\sigma_w}{\rho g}=0$$
The fourth term is the shear stress term for a viscous fluid. For an inviscid fluid that term becomes zero, so:
$$\frac{dp}{\rho g}+\frac{vdv}{g}+dh=0$$
Integrated between two points along a flow line and assuming incompressibility ($\rho = \text{constant}$), we get:
$$\int_{p_1}^{p_2}\frac{dp}{\rho g}+\int_{v_1}^{v_2}\frac{vdv}{g}+\int_{h_1}^{h_2}dh=0$$
$$\implies \frac{p_2-p_1}{\rho g}+\frac{v_2^2-v_1^2}{2g}+(h_2-h_1)=0$$
Slightly reworked:
$$\frac{p_2}{\rho}+\frac12 v_2^2+gh_2=\frac{p_1}{\rho}+\frac12 v_1^2+gh_1$$
| {
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Muon lifetime determination My colleagues and I performed several experiments to determine the lifetime of the muon (from secondary cosmic rays) using scintillator detectors coupled to multi-channel analysers. The results invariably showed a muon lifetime lower than the standard 2.2 microseconds. Apart from poor statistics,and assuming no faults in the equipment used, what other factors could be responsible for the discrepancy?
| I know that you are explicitly asked about not equipment related answers.
But when I learned something from experimental physics then that you should always consider equipment flaws.
I could imagine a scenario where the events on which you trigger to start/stop the clock have different rise times depending on where they take place in the scintillator, creating an observational error.
One other thing you could check is that your measured lifetimes are normally distributed. If they are not, you can think of things like applying a power transformation to your data.
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Coherence length of a single photon If I pass individual photons through a M-Z interferometer with equal arms I will observe interference (eg only one detector will respond). As I increase the path length of one arm I will observe the two detectors responding alternately as I pass through each phase cycle. Eventually I suspect that at a certain point, the interference will disappear and the two detectors will respond with equal probability. What determines this point and what does this tell us about the 'length' of an individual photon. What does QM predict when the path difference is greater than this?
| By length maybe you mean wavelength. A single photon traveling at the speed of light and oscillating at a certain frequency will oscillate through one cycle every wavelength or say 500 nm. As you increase the length of one arm of the experiment the interference will go in and out of phase every one half cycle or every 250 nm.
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Relativity paradox with mirrors and light pulses
Consider two very short light pulses emitted from the centre (C) of two mirrors A and B (as shown in the diagram).
From the point of view of the lab frame, the apparatus is all moving to the left at velocity v.
Imagine there is also an electron near the centre of the apparatus, which is stationary in the apparatus frame and therefore also moving with velocity v to the left according to the lab frame.
The short light pulses (much shorter than the apparatus length) bounce off mirrors A and B and return and strike the electron.
This situation has similarities with the Michelson-Morley experiment.
According to the frame moving with the apparatus, the pulses take an equal time to bounce off the mirrors and arrive back at C. Therefore the EM waves cancel and there is no net radiation pressure exerted on the electron.
According to the lab frame, the light pulse emitted to the left has less distance to travel overall and so arrives at C before the pulse that was emitted to the right. Therefore the first pulse accelerates the electron by exerting a radiation pressure on it.
Does the electron accelerate or not? :)
(I'm looking for derivations/proofs showing both frames' interpretations)
| In the lab frame both pulses arrive at C at the same time. The reason is that the distances traveled are the same (they do not reach A and B simultaneously). The distances of paths CA and BC are equal, the same happens with the paths AC and CB. The distance CAAC is equal to CBBC.
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Meaning of the phase space in statistical physics I have a silly question about the phase space. I am confused with the meaning of points in phase space. Does the each point in phase space represent concrete particle of the system, or does it represent the whole state of the system? Our teacher told us, that we use the phase space to describe the development of each particle. It is not right, isn't it?
| The quickest way to understand phase space is to read a phase diagram. It may be a PV diagram, or mixture fraction-temperature diagram. So read carefully what x axis represents and what y axis represents. Take PV diagram for example. each particle on the diagram is a pair of pressure (P) and volume (V). It describes the gas system state at that moment. If you, somehow, change (compress or expand) the system state, i.e. its pressure and volume change, the particle moves to another location in the phase space. The confusing part may be the term "particle". It is not one particle in the gas. It is a point in the phase space.
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Why are usually 4x4 gamma matrices used? As far as I understand gamma matrices are a representation of the Dirac algebra and there is a representation of the Lorentz group that can be expressed as
$$S^{\mu \nu} = \frac{1}{4} \left[ \gamma^\mu, \gamma^\nu \right]$$
Usually the representations used for them are the Dirac representation, the Chiral representation or the Majorana representation.
All of these are 4x4 matrices. I would like to know what the physical reason is that we always use 4x4, since surely higher dimensional representations exist.
My guess is that these are the smallest possible representation and give spin half fermions as the physical particles, which are common in nature. Would higher dimensional representations give higher spin particles?
| Thats a nice question. To answer this lets start with clifford algebra generated by $\gamma$ matrices.
\begin{equation} \gamma_{\mu}\gamma_{\nu}+ \gamma_{\mu}\gamma_{\nu}=2\eta_{\mu\nu} \end{equation}
with $\mu,\nu=0,1,2,\cdots N$
with the metric signature $\eta_{\mu\nu =}\text{diag}(+,-,-,-,\cdots,-)$. Using $I$ and $\gamma_{\mu}$ we can construct a set of matrices as follow
\begin{equation} I, \gamma_{\mu},\gamma_{\mu}\gamma_{\nu}\quad(\mu<\nu), \gamma_{\mu}\gamma_{\nu}\gamma_{\lambda}\quad(\mu<\nu<\lambda),\cdots,\gamma_{1}\gamma_{2}\cdots\gamma_{N} \end{equation}
There are \begin{equation} \sum_{p=0}^{N}\binom{N}{p} = 2^{N} \end{equation}
such matrices. Lets call them $\Gamma_{A}$, where $A$ runs from $0$ to $2^{N}-1$.
Now let $\gamma_{\mu}$ are $d\times d$ dimensional irreducible matrices. Our goal is to find a relation between $d$ and $N$. To this end lets define a matrix
\begin{equation} S = \sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1}Y\Gamma_{A} \end{equation}. Where $Y$ is some arbitary $d\times d$ matrix. It is follows that
\begin{equation} (\Gamma_{B})^{-1}S\Gamma_{B} = \sum_{A=0}^{2^N-1}(\Gamma_{A}\Gamma_{B})^{-1}Y\Gamma_{A}\Gamma_{B} =\sum_{C=0}^{2^N-1}(\Gamma_{C})^{-1}Y\Gamma_{C}=S \end{equation}
Where we have used $\Gamma_{A}\Gamma_{B}=\epsilon_{AB}\Gamma_{C}$, with $\epsilon_{AB}^{2}=1$
Hence \begin{equation}S\Gamma_{A}=\Gamma_{A}S\end{equation}
Since $S$ commutes with all the matrices in the set, by Schur's lemma we conclude that $S$ must be proportional to the identity matrix so that we can write
\begin{equation} S = \sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1}Y\Gamma_{A} = \lambda I \end{equation}
Taking trace we get
\begin{eqnarray} \text{Tr} S & = & \sum_{A=0}^{2^N-1} \text{Tr} Y = \lambda d\\
\Rightarrow \lambda & = & \frac{2^{N}}{d}\text{Tr} Y
\end{eqnarray}
or \begin{equation} \sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1}Y\Gamma_{A} = \frac{2^{N}}{d}\text{Tr} Y \end{equation}
Taking the $(j; m)$ matrix element of both sides of last equation yield
\begin{equation} \sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1})_{jk}(\Gamma_{A})_{km} = \frac{2^{N}}{d}\delta_{jm} \delta_{kl} \end{equation}
where $j; k; l; m = 1; 2;\cdots; d$ and we have used the fact that Y is an
arbitrary $d \times d$ matrix. If we set $j = k; l = m$ and sum over these
two indices, that gives
\begin{equation} \sum_{A=0}^{2^N-1} \text{Tr}[(\Gamma_{A})^{-1}] \text{Tr}[\Gamma_{A}] = 2^{N}\end{equation}
There are two cases to consider, namely, $N$ even and $N$ odd. For $N = 2M$ (even), $\text{Tr} \Gamma_{A} = 0$ except for $\Gamma_{0} = 1$ for which $\text{Tr} \Gamma_{0} = d$. Which gives
\begin{equation} d^2 = 2^N\qquad \text{or} \quad \boxed{d = 2^{N/2}} \end{equation}
This is the main result. For four dimensional Minkowski space time $N=4$ cosequntly the dimension of irreducible representation is $d = 2^{4/2} =4$.
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Is a spacetime of constant positive curvature just a 4-hypersphere? In discussions of basic cosmological models, I don't see "spacetime of constant positive curvature" described more simply as a "4-hypersphere". What am I missing?
| There are different descriptions of Spacetime according to General Relativity.
Look at the De-Sitter-Space. It is a mathematical concept of Spacetime with a positive curvature. It is a submanifold of Minkowski-Space.
there is also an Anti-De-Sitter-Space, which has a negative curvature. It plays a role in some cosmological theories (like Inflation).
The curvature of Space is also addicted to the Hubble Constant and the measured density parameters.
It can not be said much about the global topology of the Universe but it seems as if it is flat.
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Can we write the wave function of the living things? If yes then how? In quantum mechanics we studied that everything has a wave function associated with it.My question is can we write down the wave functions of things. Then how we can write down the wave functions of the things like animals, human eye, motion of snake etc.
| There are 37.2 trillion cells in a typical human body, (probably a good few more in mine ;), then in each cell there are 20 trillion atoms, then you have to obtain the wave function for each of the electrons.......
Actually, it may well be that you cannot describe a wavefunction for a macroscopic object, like a human body. In the study of quantum mechanics, we are usually presented with the exercise of writing a wave equation for a single microscopic particle, an electron, proton and so on.
But a macroscopic object is "joined" to it's surroundings by entanglement, rather than the single electron wavefunctions we are used to deal with, which does not need to take account of this.
If two (or more) systems are entangled, such as the parts of our body and their surroundings, as in this case, then we cannot describe the wave function directly as a product of separate wavefunctions, as I implied incorrectly in my first line.
However, by the use of Reduced Density Matrices, as pointed out by Mitchell Porter below, we can describe entangled states. With the number of wave functions involved, this would theorically possible, but in practice, not a feasible option.
Incidentally, this may be one reason why the STAR TREK, "beam me aboard" transporter system may be rather difficult to achieve, but that is probably covered elsewhere on this site.
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Birefringent filter, optical path length difference? In 'The Light Fantastic' by Kenyon, I.R. (p424), it is said that for a birefringent material inclined at Brewster's angle and who's optical axis lies in the plane of the plate, we have an optical path length difference between the ordinary and extraordinary waves of:
$$\Delta s= \frac{\Delta n t}{\sin(\phi_B)}$$
Where $\Delta n=n_0-n_e$, $t$ is the thickness of the plate and $\phi_B$ is Brewster's angle. My question is where does this equation come from and have any assumptions been made deriving it? It seems to be assuming that no refraction of either the ordinary or extraordinary wave occurs at the surface of the birefringent material, when infact I think they should refract by different amount.
| Note I am the OP.
The first and foremost thing to note about this equation is that it is not exact and is based on approximations.
For the situation described above, once the light has entered the filter it splits into two, one that experiences the ordinary refractive index $n_0$ and the other that experiences the refractive index $n_e$ (which may be different from actual extraordinary refractive index of the crystal). We assume that the two beams follow the same path through the beam, at an angle dictated by:
$$\sin(\phi_B)=n\sin(\theta)$$
Where $n$ is the mean refractive index experienced by the two beams. The optical path length therefore between the two beams after passing through the filter is:
$$\Delta s=(n_0-n_e) \frac{t}{\cos(\theta)}$$
We now note that at Brewster's angle we have:
$$\theta+\phi=\pi/2$$
which therefore gives us:
$$\Delta s= \frac{\Delta n t}{\sin(\phi_B)}$$
As required.
Reference
Svelto, O. 2010. Principles of Lasers. 5th ed. Translated from Italian by D.C.Hanna. New York: Springer (p286-287)
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Merging black holes makes them less dense, so According to What is exactly the density of a black hole and how can it be calculated? (more specifically, John's answer here made me think: if you merge a whole load of chunks of an element heavier than iron (to prevent them from fusing), the resulting object would either be more dense than a black hole of the same mass, or would become less dense by becoming a black hole.
So which one of these would happen, in this hypothetical situation? Or would neither happen, but some completely different situation? Both seem impossible to me, since such heavy objects would have no way to prevent gravity from crushing them down (which implies it must become a black hole), but if a black hole would form, it would require gravitational energy input in order to become less dense. So that would exclude both possibilities, right?
Of course this situation would never occur in real life, but this hypothetical situation would have no angular momentum in the system, so no mass would be ejected.
| If I understand you correctly you are concerned that a black hole somehow manages to become less dense than the matter that made it, as if it somehow expands against its own gravity to increase its volume.
However a black hole event horizon is not an object - it is just a place in spacetime. Although we can calculate a density by calculating the volume inside the event horizon this density is of no physical significance. The matter inside the event horizon is not uniformly distributed, as it is in a ball of iron, so all we are calculating is an average density.
Anything falling into a black hole rapidly reaches the singularity at the centre where the density is infinite (actually it's undefined, but let's save that complication for another day). So inside the event horizon you have empty space with a singularity at the centre. While there's nothing to stop you calculating an average density for this object your result doesn't have any special meaning.
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Check dimensions of the integral of a function I and a colleague are arguing about the dimensions of:
$$\int_0^x f(x) dx $$
in this particular case $[f(x)]=m^2/s^3$ and $[x]=m$.
Does it follow that $[\int_0^x f(x) dx]=m^2/s^3$ or $[\int_0^x f(x) dx]=m^2/s^3m$?
| The dimensions of the integral are simply those of $f(x)dx$, so in this case they would be $m^2/s^3 \times m = m^3/s^3$.
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What's the minimum time scale for a candle to be lit in order for it to establish a steady state? My Understanding
When a candle is lit initially, it goes through a few stages (see, e.g. this explanation from the National Candle Association):
*
*Wick burns.
*Crusted wax on wick melts, evaporates and burns.
*Base of wick warms, nearby wax melts.
*Capillary action "pulls" melted wax near base of wick up through the wick to where it evaporates and burns.
If I extinguish the candle, then the wax solidifies.
My Question
Suppose I lit a candle briefly, then extinguished it right away. The wick would burn some of the wax that was in it, but wouldn't have time to pull max up it to replenish the wax that got burned.
Suppose I did this repeatedly. Eventually the candle wouldn't light as well, right?
For standard household candles, what's a minimum amount of time to let the candle burn before extinguishing it so that it replenishes its wax supply and doesn't burn too much of the wick off? A few seconds? A few dozen seconds? A minute or two? How do you know -- that is, how did you estimate this time scale?
If that's too broad, then what factors affect this time scale?
| Factors affect the time (i.e. from start to the time of equilibrium state where the following processes are sustainable: flame heat up wax, liquefied wax being pulled up, evaporated, wax vapor mixes with air, burn and produce heat) are followings,
- size of wick (the larger, the more heat it can produce)
- size of wick (for capillary flow)
- wax type (melting temperature, surface tension etc.)
- environment temperature (hot or cold)
- environment oxygen concentration
You didn't quantify how much is to too much for burning out wick. So the constraint is not well defined. If we know these, a detailed calculation can be laid out for estimating this time.
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How to pour water from a bottle as fast as possible? When one pours water out of a bottle, it first flows smoothly but then a pressure 'blockage' develops and the pouring becomes interrupted and turbulent, so that the water comes out in splashes. This seems to slow down the flow of water from the bottle.
What is the optimal way to pour the water so that it completely empties fastest?
Possible strategies:
*
*Holding the bottle at a certain angle
*Wildly shaking the bottle
*Squeezing the bottle
*Other...
It probably depends on the shape of the opening and/or the bottle itself, but we shall assume this beautiful example of a standard water bottle:
CLARIFICATION
The question is asking how to pour the water the fastest, so no straws, hole insertion and evaporating lasers allowed...
| I assume you are not worried about the few drops that are always left in the bottle after pouring out the water. The reason I make this assumption is that without "evaporating lasers" being allowed every method suggested would need to wait hours for the drops to naturally evaporate. Even then, there would still technically be a tiny bit of water vapor in the air in the "empty" bottle.
Given this assumption I will propose the fastest way to empty 99% of the water in the bottle. With modern technology the fastest way to empty the bottle would be mounting it inside a metal projectile that will be launched from a rail gun. Point the opening of the bottle to the back of the rail gun and fire the projectile. According to https://en.wikipedia.org/wiki/Railgun the "General Atomics Blitzer system" (a railgun) can exert over 60,000 g force. With this kind of force the vacuum created behind the water as it comes out of the bottle will be negligible. Given say a 20 cm tall bottle and 60,000 g force. We know for constant acceleration that d=a*t^2/2. Therefore the water at the bottom of the bottle will take roughly 0.000368 seconds to travel the 20cm to the tip of the bottle. Of course this is to be taken as more of an "order of magnitude" calculation with the exact value determined with experimentation :)
| {
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Friction-free rolling/sliding on an inclined plane Suppose a sphere is rolling down an inclined plane. There is no friction. The body will not roll and undergo just a translation. But why is this so?
If we consider the axis to be along the point of contact, then there would be a torque which will cause it to rotate but in reality the body won't rotate. Why is this so?
| In such a hypothetical situation in which there is no friction between the sphere and plane, there can be no tangential force acting on the sphere, and hence no torque. The only force acting on the sphere would therefore be its weight, and the component of that force acting perpendicularly to the plane would be responsible for its translation down the plane, without rolling.
| {
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If the speed of light is constant, why can't it escape a black hole? When speed is the path traveled in a given time and the path is constant, as it is for $c$, why can't light escape a black hole?
It may take a long time to happen but shouldn't there be some light escaping every so often?
I'm guessing that because time is infinite inside a black hole, that this would be one possible reason but wouldn't that mean that we would require infinite mass?
What is contradicting with measuring black holes in solar masses, what means they don't contain infinite mass.
So how can this be?
| The picture I always liked is for an observer free-falling into the black hole, when they're just outside of the event horizon, it looks like the event horizon is propagating outward at nearly the speed of light. After the observer falls just inside, the event horizon now looks like it's propagating outward at greater than the speed of light.
| {
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Problem with magnetic field due to relative motion We know that, moving charge produces magnetic field in the surrounding space.
Consider this scenario :
A charge 'q' is moving with a constant speed 'v' in the direction of positive x axis of a coordinate frame 'A'.
As a result, there exists magnetic field everywhere in the space.
Now, consider another frame 'B' moving with speed 'v' in the direction of positive x axis (same velocity as that of the charge). Thus the velocity of the charge as seen in 'B' frame is zero. Thus there should be no magnetic field produced. Could someone explain this to me?
| Electric and magnetic fields are not relativistically invariant. What you measure will depend on the frame of reference you are in.
In your example, the moving charge in frame A will be responsible for both an electric field and a magnetic field.
In frame B where the charge is stationary, then an observer would only see a static electric field.
Exactly the situation you propose is used as an example on the relevant wikipedia page https://en.m.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity
| {
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Definition of anomalous symmetry in Hamiltonian formalism In the Lagrangian path-integral formulation of QFT, an anomalous symmetry is defined to be a symmetry of the action which is not a symmetry of the measure of the path integral, and therefore not a symmetry of the partition function. How do we define an anomalous symmetry in the Hamiltonian formulation of QFT, where this is no path integral or partition function?
| According to this paper, the Hamiltonian interpretation of anomalies is that one cannot formulate any Gauss-like law to constrain the physical states in the anomalous theories.
*
*Luis Alvarez-Gaumé and Philip Nelson, Hamiltonian Interpretation of Anomalies, Comm. Math. Phys. Volume 99, Number 1 (1985), 103-114.
| {
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Vlasov equation, Maxwell distribution I have the Maxwellian distribution:
$$f(v)=n\left(\frac{m}{2\pi kT}\right)^{\frac{3}{2}}\exp\left(-\frac{mv^2}{2kT}\right)$$
I have to show that it is a solution to the Vlasov equation:
$$\frac{\partial f}{\partial t}+\vec{v}
\cdot \text{grad}(f)+\frac{q\vec{E}}{m}\cdot \text{grad}_v(f)=0$$
Since $f(v)$ depends on the velocity $v$ only, I assume that the first two terms are $0$. However, when I differentiate over $v$, I get something which is not $0$. So, am I on the right path? If not, any idea what can be done?
| when you put the Maxwell equation in the vlasov equation, you calculate the averages and that is how the terms
$\left\langle \frac{\partial f}{\partial t}\right\rangle =0 $ since the distribution is not dependent on time and
$\left\langle v.\nabla f\right\rangle =0$ because distribution is uniform on an average.
similarly if you differentiate the third term you will get the term
$\left\langle E.v\right\rangle$ which will equate to zero since on the average velocity in the distribution do not change
I think this will help
EDIT:
Regarding your comment that exponent also contain the electrostatic potential $\phi$. I would like to add that the exponential term containing the potential will look like
$n=n(0)\exp\left(\frac{e\phi}{kT}\right)$.
This term is independent of velocity hence the velocity derivative will vanish. Also if the system is in equilibrium the total number of charge particles will be constant which leads to
$$\left\langle\frac{\partial n}{\partial t}+v.\frac{\partial n}{\partial x}\right\rangle=\left\langle\frac{\partial n}{\partial t}+\frac{\partial n.v}{\partial x}\right\rangle=0$$
which is just the conservation of charge i.e. number of particles changing within volume $dv$ will be equal to the current flowing through the enclosed surfaces.
| {
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What is the angular velocity of the electron? An electron has angular momentum. Shouldn't it also have angular velocity?
Ignoring the g-factor (just for the order of magnitude approximation) and the fact that an electron is not a sphere the electron's angular velocity should be around:
$$ \omega \approx \frac{\mu}{er^2} $$
or about 0.01 to 10^17 rad/s depending on whether the radius is the classical radius, the compton wavelength, or the planck length.
Is there some "average" angular velocity that can be assigned to the electron?
| the spin is assumed to be an intrinsic property unrelated to rotation, as it is assumed usually that the electron is truly elementary and does not have any size. The same happens with the expansion of space into... the nothingness, not necessarily into another spatial dimension. If you can accept that you are a long way into understanding physics.
| {
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Relationship Between Magnetic Dipole Moment and Spin Angular Momentum I am reading Introduction to Quantum Mechanics 1st edition by David J. Griffiths and I have a couple questions about this section on page 160.
A spinning charged particle constitutes a magnetic dipole. Its magnetic dipole moment $\mu$ is proportional to its spin angular momentum S: $$ \mathbf{\mu} = \gamma\mathbf{S}$$ the proportionality constant $\gamma$ is called the gyromagnetic ratio.
Taking the magnetic dipole moment to be a vector in $\mathbb{R}^3$, what is S is referring to? I have not yet seen any vector in $\mathbb{R}^3$ defined as the spin angular momentum in the text, only spinors that give the general state of, for example, a spin-1/2 particle as $$\chi = \begin{pmatrix}a\\b\end{pmatrix} = a\chi_+ + b\chi_-$$ using the spin up and spin down eigenstates as basis vectors.
The section continues:
When a magnetic dipole is placed in a magnetic field $\mathbf{B}$, it experiences a torque, $\mathbf{\mu \times B}$, which tends to line it up parallel to the field (just like a compass needle). The energy associated with the torque is $$H = -\mu\cdot\mathbf{B}$$ so the Hamiltonian of a spinning charged particle, at rest in a magnetic field $\mathbf{B}$, becomes $$H = -\gamma\mathbf{B\cdot S}$$ where $\mathbf{S}$ is the appropriate spin matrix.
What is the mathematical meaning of this dot product $\mathbf{B\cdot S}$ of a vector in $\mathbb{R}^3$ with a 2x2 matrix (in the case of spin 1/2)?
| $\mathbf{S}$ is the spin operator. It is a vector operator that acts on spinors. It will have three components $(S_x, S_y, S_z)$ and for example if you take the $z$ axis as your spin measurement axis, you define spin up and down as the two eigenstates of $S_z$.
It can be shown that in matrix form $S_i$ is proportional to the Pauli matrix $\sigma_i$.
Finally, $\mathbf{S}\cdot\mathbf{B} = S_xB_x + S_yB_y + S_zB_z$. Note that in matrix form each component of $\mathbf{S}$ is a $2\times2$ matrix, so $\mathbf{S}\cdot\mathbf{B}$ is a $2\times2$ matrix too.
| {
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Can I recirculate water from an open reservoir to the bottom of a bigger, closed one, without a pump? A fountain head pumps water out of the main tank into a 'pond' reservoir. Can the water recirculate back into the main tank without the help of another pump?
I'm sorry if this a dumb question. I'm guessing it would not function as the diagram shows, as the pressure of the water in the main tank would not let any water in at the bottom,right? Any solutions? (not requiring additional pumps)
| The water level in the pond must be the same as in the tank, so:
:-)
| {
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System rigidity What is the meaning of system rigidity in mechanics? I can't understand how to classify the system as rigid or not, and what is the effect of rigidity on the whole system.
If you know anything about system rigidity... Please let me know
| In the materials science context, rigid means hard, resisting deformation. It is the opposite of elastic.
If two material bodies are forced sideways against each other and one or both are elastic like rubber, it/they will deform, allowing them to squeeze past each other without much damage being done.
However, if both bodies are hard or rigid like stone, this will prevent them moving past each other. They do not deform in response to the forces between them. If there is enough force pushing them sideways they will grind against each other, causing the contact surfaces to wear down or break away until they can pass.
If the forces push the bodies together, the deformation of elastic bodies increases the area of contact, which spreads out the force. Hard surfaces deform very little, so the force remains concentrated at the contact point, causing rock or stone to fracture and break off.
| {
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How to find entropy production of opening the windows in the winter? Every time you open the windows in the winter (to bring in healthy and fresh air), room's warm air goes outside into the atmosphere.
How to find entropy production of opening the windows in the winter?
The room's volume is denoted by $V_r$ and its temperature by $T_r$, and atmospheric volume is denoted by $V_a$ and its temperature by $T_a$.
| Step 1) Find a model for the gas in the room. We might as well take an ideal gas for this demonstration; I think this is generally an alright model for the air since it's mostly inert nitrogen.
Step 2) Apply the first law of thermodynamics, ignore the work done in opening the window. The first law now states
$$ \Delta S = \int \limits_{T_0}^{T_f} \frac{dE}{T}$$
because the room has rigid walls and thus $dV = 0$.
Step 3) The average energy in the room associated with an ideal gas as a function of temperature is $E(T)=c_v n R T$.
Step 4) Plug in $dE(T) = c_v n R\: dT$ and integrate:
$$ \Delta S = c_v n R \ln\left(\frac{T_f}{T_0}\right) $$
| {
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Is anisotropic radioactivity really impossible? What if the nucleus has a magnetic moment, and also the electron shell has one? I suspect, in this case, the orientation of the nucleus could be "fixed" by the electron shell.
Maybe a mono-crystal of such a material would have an anisotropic radioactivity. Thus, its radioactivity wouldn't be the same in all directions.
I think, relative many requirements should be fulfilled:
*
*The nucleus must be radioactive.
*It must have a magnetic moment.
*The electron shell must have also a magnetic moment.
*The orientations of the atoms in the crystal structure should be unidirectional.
Is it possible? Does any such crystal already exist? Maybe the crystal of a radioactive isotope of some ferromagnetic element could fulfill all of these?
If not, why not?
| The idea that you propose is possible. In fact, there is a very famous example of your setting: the Wu experiment.
Chien-Shiung Wu at the US National Bureau of Standards prepared a thin surface of ${}^{60}Co$. This isotope decays by beta decay, producing one electron and one antineutrino. Due to the small magnetic moment of the nuclei, they had to cool the surface to 0.003 K, and then the sample was magnetized in an uniform magnetic field.
The result was quite amazing: the emitted electrons showed a preference to be in the direction of the nuclear magnetic moment, what proves that parity symmetry is violated. But that's another story...
More info: Wu experiment (wikipedia)
| {
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Ball spontaneously rolling down hill I'm trying to remember a problem in classical mechanics involving a special surface that allows a ball to roll to the top and lose all it's momentum in finite time.
This leads to some interesting problems with time reversibility, as it implies the ball will spontaneously roll down the surface.
I'm not looking for an explanation, so much as a name and link to study it some more.
| I believe the analysis of the Norton Dome is flawed (as many philosophers thought experiments). The ball does not stay at rest and start to move spontaneously in the absence of any force. If there were no forces it will stay there forever. The reason it starts to move is some small perturbations. They could be either external (random variations in pressure around, or just nonisotropic temperature fluctuations; there are plenty of choices). So if you had full information of the system and its surroundings you should be able to predict (in theory, not in practice) which way the ball would move and when. The system is deterministic.
| {
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Why do we believe baryon asymmetry? The question is that how can we make sure that our universe is baryon asymmetric? I mean, is it possible that there are antimatter domains beyond some very large scale. Yes, if that kind of domains exist, the scale must be very large. But does such constraint bring any problems such that we prefer to believe the baryon asymmetry?
| So, there are several possible ways the universe could be baryon symmetric:
*
*A region of the universe where antimatter dominates. There is a problem with this theory, though - 30 years' worth of scientific research has calculated just how far away this type of region would have to be, and from these calculations it is considered very unlikely that any part of the observable universe, at least, would have this sort of region, and there isn't really any solid evidence for this theory, so these regions might not exist at all.
*The second possibility is that antimatter repels matter instead of attracting it gravitationally; however, this is in conflict with the theory of relativity.
So then there are two explanations for the universe being baryon asymmetric:
*
*An electric dipole movement, or EDM. If this was present in any fundamental particle, it would violate parity and time symmetries, therefore allowing matter and antimatter to decay at different rates. However, no EDM has been detected to date.
*There's something we're missing about the laws of the early universe.
Hope this helps!
| {
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The analytical result for free massless fermion propagator For massless fermion, the free propagator in quantum field theory
is
\begin{eqnarray*}
& & \langle0|T\psi(x)\bar{\psi}(y)|0\rangle=\int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}+i\epsilon}e^{-ik\cdot(x-y)}.
\end{eqnarray*}
In Peskin & Schroeder's book, An introduction to quantum field theory
(edition 1995, page 660, formula 19.40), they obtained the analytical
result for this propagator,
\begin{eqnarray*}
& & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}+i\epsilon}e^{-ik\cdot(x-y)}=-\frac{i}{2\pi^{2}}\frac{\gamma\cdot(x-y)}{(x-y)^{4}} .\tag{19.40}
\end{eqnarray*}
Question: Is this analytical result right? Actually I don't know
how to obtain it.
| In the following, I will carefully deal with Wick rotation. In the
end, I have found that I was confused.
The integration is
\begin{eqnarray*}
& & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{1}{k^{2}+i\epsilon}e^{-ik\cdot x}\\
& = & \frac{1}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\int_{-\infty}^{\infty}dk_{0}\frac{1}{k_{0}^{2}-(E_{k}-i\epsilon)^{2}}e^{-ik_{0}t}\\
& \equiv & \frac{1}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\times\mathrm{I}
\end{eqnarray*}
with
\begin{eqnarray*}
\mathrm{I} & = & \int_{-\infty}^{\infty}dk_{0}\frac{1}{k_{0}^{2}-a^{2}}e^{-ik_{0}t}\\
a & = & E_{k}-i\epsilon=\sqrt{m^{2}+\mathbf{k}^{2}}-i\epsilon
\end{eqnarray*}
Now we will use Wick rotation to calculate $\mathrm{I}$. Note that
$\pm a$ are two singularities of the integrand. Consider following
contour. The radii of coutours $l_{5},l_{6}$ are both $R$ and $R\rightarrow\infty$.
According to contour integral theorem, we can see
\begin{eqnarray*}
\mathrm{I} & = & \int_{-\infty}^{\infty}dk_{0}\frac{1}{k_{0}^{2}-a^{2}}e^{-ik_{0}t}\\
& = & \int_{l_{1}}dz\frac{1}{z^{2}-a^{2}}e^{-izt}+\int_{l_{2}}dz\frac{1}{z^{2}-a^{2}}e^{-izt}\\
& = & \bigg(\int_{l_{5}}dz\frac{1}{z^{2}-a^{2}}e^{-izt}+\int_{l_{3}}dz\frac{1}{z^{2}-a^{2}}e^{-izt}\bigg)+\bigg(\int_{l_{4}}dz\frac{1}{z^{2}-a^{2}}e^{-izt}+\int_{l_{6}}dz\frac{1}{z^{2}-a^{2}}e^{-izt}\bigg)\\
& & \bigg[\text{note: set }z=ik_{E}^{0}\text{ in }l_{3},l_{4}\text{ and combine }l_{5},l_{6}\bigg]\\
& = & (-i)\int_{-\infty}^{\infty}dk_{E}^{0}\frac{1}{(k_{E}^{0})^{2}+a^{2}}e^{tk_{E}^{0}}+\int_{l_{6}}dz\frac{1}{z^{2}-a^{2}}(e^{-izt}+e^{izt}),\ \bigg[\text{set }z=Re^{i\phi}\text{ in }l_{6}\bigg]\\
& = & (-i)\int_{-\infty}^{\infty}dk_{E}^{0}\frac{1}{(k_{E}^{0})^{2}+a^{2}}e^{tk_{E}^{0}}\\
& & -iR\int_{0}^{\frac{\pi}{2}}d\phi e^{i\phi}\frac{1}{R^{2}e^{2i\phi}-a^{2}}(e^{-itR\cos\phi+tR\sin\phi}+e^{itR\cos\phi-tR\sin\phi})\\
& \equiv & (-i)\int_{-\infty}^{\infty}dk_{E}^{0}\frac{1}{(k_{E}^{0})^{2}+a^{2}}e^{tk_{E}^{0}}+\mathrm{II}
\end{eqnarray*}
with
\begin{eqnarray*}
\mathrm{II} & = & -iR\int_{0}^{\frac{\pi}{2}}d\phi e^{i\phi}\frac{1}{R^{2}e^{2i\phi}-a^{2}}(e^{-itR\cos\phi+tR\sin\phi}+e^{itR\cos\phi-tR\sin\phi})
\end{eqnarray*}
Actually, I do not know how to prove $\mathrm{II}=0$ as $R\to\infty$.
But if $\mathrm{II}\neq0$ as $R\to\infty$, then we can not simply
obtain
\begin{eqnarray*}
\int_{-\infty}^{\infty}dk_{0}\frac{1}{k_{0}^{2}-a^{2}}e^{-ik_{0}t} & = & (-i)\int_{-\infty}^{\infty}dk_{E}^{0}\frac{1}{(k_{E}^{0})^{2}+a^{2}}e^{tk_{E}^{0}}
\end{eqnarray*}
I am just confused at this point.
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Accessibility by reversible processes and the Second Law of Thermodynamics One common way of motivating the existence of Entropy as a state function is the following. Let us take the Clausius/Kelvin-Planck statement of the second law, from which we can deduce Clausius' theorem
$$\oint \frac{\delta Q}{T} \le 0,$$
where equality holds if and only if the cyclic process is reversible.
This of course means that the quantity
$$\int_C \frac{\delta Q}{T} $$
is path independent for reversible paths $C$, and so it defines a function of state which we call Entropy.
But this only seems to hold on the presumption that all states within our state space is mutually accessible through reversible processes, i.e. given any two states $A$ and $B$ in our state space, there exists some reversible process $A\rightarrow B$ and some reversible process $B\rightarrow A$. I don't see why this is necessarily true. Is this taken to be an additional (and apparently implicit) assumption? Or is this assumption provable? Or is it not actually needed to define entropy this way?
|
Is this taken to be an additional (and apparently implicit) assumption?
You are correct.
Take two arbitrary points $A,B$ on the $PV$ (or any other) plane, and draw an arbitrary curve connecting them: you have just defined a reversible transformation connecting $A$ and $B$.
This is because every point in the $PV$ (or any other) plane represents an equilibrium state, so every continuous set of points (such as the curve you drew) represents a reversible transformation.
The existence of reversible processes is one of the postulates of thermodynamics. Of course, a reversible process is an idealization, because it would require that the system is in equilibrium at every instant during the process, which is clearly absurd, because if the state variables are changing then clearly there is no equilibrium. This is why we talk of "quasi-static" transformations, in which an infinite number of infinitesimal steps is performed in such a way that the system is always in equilibrium.
Regarding your last question, it is actually possible to define entropy in another way in statistical mechanics. Between 1872 and 1875 Boltzmann formulated the equation
$$S=k \log(\Omega)$$
where $k$ is a constant with dimensions of $J/K$ and $\Omega$ is the number of microstates corresponding to a the macrostate of the system.
This definition is in some way more fundamental than the thermodynamic one, as it gives the connection between the microscopic and the macroscopic description of Nature.
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How to choose proper measurement operator? Let's assume I have two states inside the Bloch sphere, at radial vectors $r_1$ and $r_2$ respectively $(r_1<r_2<1)$. Their angular location is same. These are like:
\begin{equation}
\rho =
\begin{pmatrix}
\frac{1+r_1 \cos\theta}{2} &\frac{r_1 \exp(-i\phi)\sin\theta}{2} \\
\frac{r_1 \exp(i\phi)\sin\theta}{2} &\frac{1-r_1 \cos\theta}{2} \nonumber
\end{pmatrix}
\end{equation}
and another state as
\begin{equation}
\rho' =
\begin{pmatrix}
\frac{1+r_2 \cos\theta}{2} &\frac{r_2 \exp(-i\phi)\sin\theta}{2} \\
\frac{r_2 \exp(i\phi)\sin\theta}{2} &\frac{1-r_2 \cos\theta}{2} \nonumber
\end{pmatrix}
\end{equation}
I know that measurement can increase radial vectors but I don't know the procedure to find the suitable measurement operators which can relate the above states $\rho$ and $\rho'$.
| When don't you condition on the result, measurement of a qubit can only decrease its purity (you end up with less information than you started with).
When you do condition on the result, measurement of a qubit will make it 100% pure but there are two possible results. One possible result is along the measurement axis you measured. The other is against the measurement axis (well... along it, but negative-ward instead of positive-ward). This will invert the angles instead of only changing the radius.
So I'm not sure how you could do what you want (increase radius without affecting angles) without post-selecting. Measure the qubit along the axis component of its Bloch sphere representation, retry the whole experiment up to that point until you get the "along" answer instead of the "against" answer, then continue.
Here's an example circuit, where I setup a qubit to have a mixed state pointing towards the -X state then post-select it into the -X state. But keep in mind that it's the post-selection that's entirely responsible for the final state; states along the Z axis would also end up at -X in this case.
If you want to decrease the radius, that's easy. use the qubit to control small/misaligned rotations on other qubits (that's what I do in the above circuit). Partial measurements also work, as long as you don't condition on the result.
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Out-of-plane electronic band structure for 2D materials? There's been much recent interest in 2D materials since they can form monolayer-thick films. Since their crystal structure is periodic along the in-plane directions, the electronic band structure along these directions is quite well understood and can be formulated using approaches used for conventional bulk semiconductors, such as Si.
How can we think of the out-of-plane electronic band structure (which is not periodic) for 2D materials?
I need help bridging the techniques used in solid-state physics for bulk crystalline materials to the calculation of band structure along a non-periodic direction. My initial thought is that the notion of band structure along the out-of-plane direction is ill-conceived.
The reason why I'm interested in the concept of out-of-plane band structure, is that devices have been proposed and fabricated that have electron transport (tunneling) from one 2D layer to another, and the concept of band structure is quite useful in applying conventional techniques to calculate current, rates, etc. [For example, this 2014 paper discusses tunneling between different layers of 2D materials, but uses the in-plane band structure for the out-of-plane direction.]
| You're right, the usual language of band theory doesn't apply in the out-of-plane direction. The system really becomes a quantum well in that direction, so you will have a discrete spectrum of energy levels and there won't be any dispersion with $k_z$ ($dE/dk_z = 0$) if $z$ is out of the plane.
The relevant tunneling processes in that paper are basically occurring between a double quantum well formed by the two 2D layers with the barrier in between. The depth of each well depends on the band structure; where the electrons sit in the 2D Brillouin zone (what momentum or equivalently wavevector $\vec{k}$ they have in the plane) and the 2D band structure will dictate their energies. In the case of semiconductors, they're assuming that only the lowest valleys of the conduction bands (highest peaks of valence band) will have appreciable electron/hole density, and by tuning the those to the same level by gating, then tunneling occurs. There's no dependence on $z$ momentum because (except for tunneling processes) the electrons are localized in the 2D planes and there just isn't any $z$ momentum.
| {
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Is there any effect on mechanical waves by electromagnetic waves (and vise versa)? Do electromagnetic waves like light and gravitational waves (due to moon for instance) affect on mechanical waves like sound?
Can sound change the path of light?
| Yes, sound waves in a gas, liquid or solid can affect the light passing through it, as the motion of the atoms due to sound waves changes the atomic spacing, and this changes the index of refraction slightly. So the light would be diffracted and some amount of the light would experience a frequency shift up and a frequency shift down by the sound wave frequency.
The other possibility is that of electromagnetic waves creating sound waves directly. Here, the effect is much smaller: when radiation like light shines on a surface it exerts a pressure known as "radiation pressure" though again this is a very small effect.
In principle, though, if a wide-beam high powered laser were to operate in a repeating pulsed mode (with a power level low enough that it does not ionize or otherwise cause phase transitions like melting, etc) then the laser would cause a pulsed mechanical vibration in the surface due to the changing radiation pressure, thereby generating sound waves.
| {
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Force of water hitting a wall If you had a 8" pipe with 500psi stream of water exiting it and hitting a wall at 90 degrees 8 feet away, what would the force of the water on the wall be?
Thank you all.
Non-mathematician.
| Actually. the velocity can be determined as follows.
V = sq. root of 2gh where: V = velocity in ft./sec. g = acceleration const. 32.2 ft per sec per sec. And h = head in feet of liquid.
h = P * 2.31/SG where: P = pressure in psi. SG = liquid specific gravity.
With the above information you can now calculate F.
| {
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Torque on electric dipole placed in non uniform electric field When electric dipole placed in non uniform electric field, what is the approach to calculate torque acting on it? Can it be zero?
| If the dipole is small enough, then the force on dipole would be:
$$\vec{F}=\nabla(\vec{p}.\vec{E})$$
and consequently the torque would be:
$$\vec{F} \times \vec{r}=\nabla(\vec{p}.\vec{E}) \times \vec{r}$$
where r is the length of the dipole
| {
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Relation between entropy and internal energy I am confused as to what is the relation between entropy and internal energy. Entropy is always presented as a measure of the randomness in a system. So when we supply heat to a well insulated system say ideal gas in a container with fixed boundary, the internal energy and temperature increase, which implies that the motion of gas particles increases and hence the system becomes more chaotic and thus entropy increases. But if we take the same system, and supply heat isothermally and reversibly, the defnition of entropy change ΔS=Q/T , says that the entropy of system would increase(at the cost of equal entropy drop in surroundings). But for an ideal gas, internal energy is only a function of temperature and so internal energy remains constant here,no change in average kinetic energy of gas particles takes place, so where does the chaos come from to increase entropy of the system.
How are the two related?
| The chaos comes from by changing of volume or pressure of the system. The average kinetic energy doesn't change, but number of collisions increases (if pressure increase) or length of paths increases (if volume increases).
| {
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What is the significance of the Inverse-square law? Considering its occurrences in various fields like Electrostatics, Gravitation, Acoustics etc. how does the law bind these topics together?
| For point sources of a field or energy source, such as a charged particle, a gravitational body (which acts like a point source), or a loudspeaker on top of a tall column, the geometry of the problem controls how energy and fields distribute themselves in space. At all points that are an equal distance from a point source, the energy or field strength is constant, assuming that the point source is radiating equally in all directions. The locus of all points that are an equal distance from a point source describes a spherical surface, whose area is 4*pi*R^2, where R is the distance from the point source. If you double your distance from a point source, the sphere that has the new radius of 2R has 4 times the area of the preceding sphere, so the field or energy from the point source covers 4 times its original area, meaning that it is 1/4 the previous strength. Thus, the energy or field falls off in a 1/R^2 fashion, due to the geometry of the situation.
| {
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Prove that an electron in a hydrogen atom doesn't emit radiation According to electrodynamics, accelerating charged particles emit electromagnetic radiation.
I'm asking myself if the electron in an hydrogen atom emits such radiation. In How can one describe electron motion around hydrogen atom?, Murod Abdukhakimov says the total momentum of the electron is zero, hence it does not emit radiation. Could someone prove this statement ?
It may be an obvious question, but I can't figure out why the total momentum of the electron should always be zero, in any energy state.
| You have your "prove" in the wrong place. The way to prove that ground-state electrons in hydrogen atoms don't emit radiation is the following:
*
*Construct a sample of ground-state neutral hydrogen atoms.
*Place this sample near a detector which is sensitive to the sort of EM radiation you expect.
*Die of old age waiting for a signal, because ground-state hydrogen doesn't emit radiation.
This experimental evidence demonstrates that classical electromagnetism, in which accelerating charges emit radiation, does not describe the hydrogen atom.
| {
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Physical meaning of enthalpy I've been reading about thermodynamics and reached the topic about enthalpy . I've understood its derivation but I don't understand its physical meaning ... Also I don't understand why they have divided by the mass of gas to get to the specific enthalpy equation . what's the use of it? I know the meaning of all state variables the enthalpy contains but I can't see the benefit of combining them together to have the enthalpy ..
| Enthalpy, H, is H = U + PV where U is internal energy, P is pressure, and V is volume. Specific enthalpy, h (enthalpy per unit mass), is h = u + pv where u is internal energy per unit mass, P is pressure, and v is specific volume (inverse of density). Physically, enthalpy represents energy associated with mass flowing into and out of an "open" thermodynamic system. An "open" system is one with mass transfer in and out, in contrast to a "closed" system where there is no mass transfer. From a Lagrangian viewpoint (following a fixed mass) pv is work; from an Eulerian viewpoint (considering a fixed region) pv is energy. Most developments of fluid dynamics use the Eulerian viewpoint. Specific heat at constant pressure is based on enthalpy; specific heat at constant temperature is based on internal energy. For an ideal gas, enthalpy- like internal energy- is a function of temperature only. Engineering texts on thermodynamics provide more details. A texbook I like is an old one, Elements of Thermodynamics and Heat Transfer, by Obert and Young, because of the clarity of the definitions and explanation of concepts.
| {
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Why don't the magnetic dipole moments in a neutron cancel out? This may be a silly question. I thought magnetic dipoles were dependent on electric charge, so why in a neutron do the dipole moments not just cancel each other out?
| The magnetic moment of the neutron is not due to circulating charge. Instead it is due to the combined magnetic moments of the partons inside it.
The inside of a hadron is a ferociously complicated place, but let's take the simple model of a hadron as made up of three quarks. The quarks have a magnetic moment due to their spin. This magnetic moment is an intrinsic property of the quarks and not due to rotation in any classical sense.
So the problem is to find the lowest energy state configuration of the three dipoles in a neutron, and this turns out to be non-zero. If fact it's approximately:
$$ \mu_n = \tfrac{4}{3} \mu_d − \tfrac{1}{3} \mu_u $$
where $\mu_d$ and $\mu_u$ are the intrinsic magnetic moments of the down and up quarks respectively.
| {
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Quantum master equation and off-diagonal terms I have a couple of related questions
*
*What is exactly the difference between the quantum master equation and the regular master equation? My understanding is that the normal master equation is used to find a "vector" of state probabilities (like in a regular Markov chain), whereas in the quantum master equation one finds the density matrix. Is this correct? If this is the case, how does the "transition matrix" look, in the quantum case?
*I'm also a bit confused about the off-diagonal elements in a density matrix. As the density matrix is self-adjoint it can be diagonalizable in some orthonormal basis. So why do we speak about off-diagonal elements? Is it because the density matrix $\rho$ can be time dependent (and orthonormal basis stop being so as time evolves $\Rightarrow$ off-diagonal elements appear)?
| For the first question on master equation, it turns out that there are loads of equations from different fields that are respectively being called master equations, but they are not related in any certain ways. The Markov Chain master equation and the quantum master equation are one example of this. It seems like you understand the Markov Chain master equation, so I'll just explain the quantum master equation. Simply put, the quantum master equation is a generalization of the Schrodinger's equation to general open quantum systems. For example, if we have a quantum communication channel, where we want to distribute a pair of entangled photons to two different people over a long distance, this channel would obviously be an open system since the photons are coupled to the environment. To address this, we would need to use the quantum master equation.
You can understand the quantum master equation this way. Suppose you consider both the environment and the system you are interested in, then the total system is essentially closed, and you can write down time evolution of the density matrix of the entire system
$$\dot{\rho} = - \frac{i}{\hbar}\left[H_{tot}, \rho\right] = - \frac{i}{\hbar}\left[H_{env} + H_{sys} + H_{coupling}, \rho\right]$$
The quantum master equation is is result of tracing out the environment degrees of freedom and leaving only the system.
| {
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Running Constant Values At Very Low Temperatures From Wikipedia Coupling Constants, using QED as an example. I realise that the one-loop beta function in quantum chromodynamics is negative.
If a beta function is positive, the corresponding coupling increases with increasing energy. An example is quantum electrodynamics (QED), where one finds by using perturbation theory that the beta function is positive. In particular, at low energies, α ≈ 1/137, whereas at the scale of the Z boson, about 90 GeV, one measures α ≈ 1/127. Moreover, the perturbative beta function tells us that the coupling continues to increase, and QED becomes strongly coupled at high energy. In fact the coupling apparently becomes infinite at some finite energy.
My questions are based on pure curiosity (and a total lack of experimental experience, so my apologies if this combination displays naivety on my part).
Have we tested coupling constants at the lowest temperature/energy to confirm a reduction at the far end of the energy scale from the LHC?
It may be that low temperature experiments have to take any changes in values as a matter of routine, to correspond to theoretical predictions, so "yes, of course!!!" is an perfectly acceptable answer.
If a reduction has been observed in the value of any arbitrary constant, at these extremely low temperatures, can we compare this to conditions if the "heat death of the universe" scenario is true and predict what effects will occur as the temperature drops?
| The fine structure constant $\alpha\approx\frac1{137}$ appears in the Coulomb force between fundamental charges:
$$
\alpha\hbar c = e^2/4\pi\epsilon_0,
\quad\text{so}\quad |E_\text{Coulomb}| = \frac{e^2}{4\pi\epsilon_0} \frac1r = \frac{\alpha\hbar c}{r}
$$
Quantum electrodynamics is pretty well tested down into the radio frequencies, with techniques like magnetic resonance, and radio frequencies correspond to micro-eV photons. This is zero temperature as compared to the LHC.
| {
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Electric field dependence on distance How can it be proved that for a point charge, $E$ is proportional to $$1/r^2$$ using the concept of Electric field lines (or lines of force)? I tried to show that if field lines are close, then magnitude of Electric field is higher. But, I couldn't show the given dependence.
| This is a much more deeper question then it looks in first glance. The simple logic given by @Anthony B is not enough for proving the inverse square law. There are numerous experiments that have been done to verify this law. There is a collection of the experimental works in this review.
In earlier days Cavendish and Coulomb have performed experiments with conducting hemispheres and torsion spring, which proved the inverse square law.
Procs and deBrolgie have postulated that if the photons have rest mass then there will be deviations from inverse square law. However the estimates of the rest mass of photons are really low.
If there is a deviation from inverse square law then there will be critial situation for the physics.
| {
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Is measure relative velocity the same for both observer n particle A particle is moving at velocity v. A stationary observer tries to measure its velocity. From the observer reference frame, he will measure a shorter distance travel as compared to what the particle will measure due to length contraction. The observer will also measure a greater time compared to what the particle will measure. Won't this mean that the observer measured velocity of the particle is different from what the particle measure?
Edit: I think the problem here is that the particle can't measure its own velocity as it must be wrt something. But I thought about what if it is moving from point x to point y. Surely it will be able to measure its velocity through that.
|
In Figure 1 the system $\bbox[#99FFFF,2px]{S'\equiv O'x'y'z'}$ is moving with velocity $\mathbf{b}=(b,0,0)$ relatively to $\bbox[#E0E0E0,2px]{S\equiv Oxyz}$.
In Figure 2 we build two systems $\bbox[#E0E0E0,2px]{\Sigma\equiv K\rm{uvw}}$ and $\bbox[#99FFFF,2px]{\Sigma'\equiv K'\rm{u'v'w'}}$ as shown therein.
The system $\bbox[#E0E0E0,2px]{\Sigma\equiv K\rm{uvw}}$ is built from $\bbox[#99FFFF,2px]{S'\equiv O'x'y'z'}$ by reversing the axes $\:O'x',O'y'\:$. The system $\bbox[#99FFFF,2px]{\Sigma'\equiv K'\rm{u'v'w'}}$ is built from $\bbox[#E0E0E0,2px]{S\equiv Oxyz}$ by reversing the axes $\:Ox,Oy\:$.
In Figure 3 we see these two built systems alone.
The situation in Figure 3 is physically identical to that of Figure 1, so there is no good reason not to accept that $\bbox[#99FFFF,2px]{\Sigma'\equiv K'\rm{u'v'w'}}$ is moving with velocity $\:\mathbf{b}=(b,0,0)\:$ relatively to $\bbox[#E0E0E0,2px]{\Sigma\equiv K\rm{uvw}}$ and, returning back to Figures 2 and 1 with this series, to accept that $\bbox[#E0E0E0,2px]{S\equiv Oxyz}$ is moving with velocity $\:-\mathbf{b}=(-b,0,0)\:$ relatively to $\bbox[#99FFFF,2px]{S'\equiv O'x'y'z'}$.
| {
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If a electrical current can induce a magnetic field, can the reverse be true? From my GCSE studies, near to the end of a module, I was told that when an electrical current flows through a conductor, a magnetic field will be induced. Can a magnetic field create an electrical current?
I ask this because I hear about "free energy" that uses a magnetic core to move a copper coil to create a current (which in turn, I guess, will aid the magnetic field in creating movement). For this to be possible, wouldn't the magnet need to be moved along the coil? In the example, the magnet is stationary.
I guess my actual question is (based on the answer to the original question) can a stationary magnet induce a current in a conducting wire?
| A stationary magnet will not induce a current, but a magnet that is moving will induce a current
Currents are induced by the time derivative of the magnetic field (see faradays law) EM induction
Now there is the fun idea of relativity in that everything is in relative motion so viewed from a moving frame you will see a current but viewed from the frame where the magnet is at rest you will not see a current. Induction and Relativity
| {
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Measuring different components of spin simultaneously I'm reading Griffiths Introduction to QM and I'm having trouble understanding why you can't simultaneously measure the x,y and z components of spin. I know that the uncertainty principle prevents this but I still don't see why.
Griffiths' example is that if we have a particle in its up state, $\chi_+$ then we know the z-component of its spin is $\frac{\hbar}{2}$. If we then measure the x-component, then we're suddenly left with a 50-50 probability of the x-component being $\frac{\hbar}{2}$ or $-\frac{\hbar}{2}$. First off, why is it a 50-50 chance? If the state of the z-component is $\chi^z$ then $$ \chi^z=a\chi_+ ^z + b\chi_- ^z$$ and the x-component is $$ \chi^x =\dfrac{a+b}{\sqrt{2}}\chi_+ ^x +\dfrac{a-b}{\sqrt{2}}\chi_- ^x$$
If the z-component is in its upstate, does $\chi^z$ collapse to $$\chi^z = \chi_+ ^z$$
and so $a=1$ and then $b=1$. Therefore, there is a 50-50 chance of the x-component to be in its up or down state. Is this why it is 50-50 or am I understanding it wrong?
Next, if the particle is in its up state, then shouldn't the x-component also be in the x-component up state i.e $\frac{\hbar}{2}$ or does its up and down state 'reset' every time we measure? If it does reset, does it mean that once I measure for the x-component I lose knowledge about the z-component? So I'm left with a definite x-component but now I only have a 50-50 probability of knowing if it's spin up or down in the z-component? If it even does reset, what causes it? Is it just because of the uncertainty principle?
| Mutually non-commuting operators cannot have simultaneous eigenstates, namely the eigenstates of the former must by all means be expressed as a linear combinations of (all) the eigenstates of the latter. In the case at hand, given ${|+\rangle}_z$ as eigenstate of the operator $S_z$, the following must hold:
$$
{|+\rangle}_z = c_1 {|+\rangle}_x + c_2 {|-\rangle}_x
$$
and likewise for the other component ${|-\rangle}_z$, only with different coefficients. Exploiting the commutation relations and the $\mathfrak{su}(2)$ Lie-algebra one finds out that $c_1 = \pm c_2 = 1/\sqrt{2}$ (signs may be inverted though).
| {
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Can an inhabitant of a spacetime region measure its curvature tensor? So, lets say that I am an ant living on a 2-D spherical surface that is stretching to the equator...like half a sphere. I can not describe this surface in terms of the outside coordinates only someone living in the outside world can do this. So, can I really determine that I am in fact living on a spherical and not flat surface? If I try to measure the phi coordinate change as I move over the R coordinate (phi is the angle that runs from 0 to 2pi and R runs from 0 to rpi where r is radius of a sphere) at two separate places as I move to the north I can measure that this distance is getting smaller. I hope that I am clear enough about what I mean. But, would not the measuring stick also get smaller? Making it imposible for the inhabitant of this world to determine its geometry. So how can someone living in this world determine its geometry while his measuring equpment distorts acording to this?
| So as I was going through smart stuff of general relativity, still confused with this problem of detecting the curvature I realised how not very smart I was. In my question I am confused because I think that the stick shoul shrink as unit distance of a coordinate shrinks but of course it wont....sphere does not care if you and how you draw your coordinates. According to my incorrect thinkink on this matter behaviour of a stick would depend on a way in which I drew my coordinates...so if I chose that north pole is where the equator is stick would sudenly shrink because of that choice and that is just nonsense! So stick would be curved but that is it...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why can we see the moon when it is between the Earth and the Sun? A rather stupid question, why can we see the moon when it is between the Earth and the Sun?
| The moon circles around the earth, so half the time it is between the earth and the sun and half the time the earth is between the sun and the moon.
Therefore also the moon rises and sets, the same way the sun rises and sets.
If it's midnight (your are on the opposite site of the earth than the sun) and the moon is betweem the sun and the earth you can't see it.
You can see a half-moon in the late evening or early morning, tough.
| {
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Extension of Schrödinger's cat thought experiment My question is quite simple. In the thought experiment of Schroedinger's cat: When the scientist measures the state of the cat, its wavefunction collapses into either the alive or dead state. But wouldn't then the scientist in turn be in a superposition of measuring dead respectively alive until someone opens the door to the laboratory and asks the scientist about the outcome of the experiment (and therefore measures the state of the system)?
| This is an excellent question and stresses one of the weird features of quantum mechanics.
Indeed, the scientist would in turn be in a superposition. And we could even measure this if we'd be able to maintain coherence of such large systems.
Ultimately, your question is asking for the solution of the Measurement problem: Why don't we see any superpositions? There are no cats running around half dead and half alive. Neither do we see scientists being in a superposition of having measured this or the other.
However, the measurement problem has remained an unsolved problem. All we have are some interpretations.
| {
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"url": "https://physics.stackexchange.com/questions/266606",
"timestamp": "2023-03-29T00:00:00",
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