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Why is the density matrix of a system has this block form? In Ficek's paper (http://zon8.physd.amu.edu.pl/~tanas/spis_pub/pdf/04-joptb-S90.pdf), the density matrix of a two two-level atom system has a
block form like this. Why does it make sense to assume this ?
| The basis from the paper is
\begin{align}
\vert 1 \rangle = \vert g_1 \rangle \otimes \vert g_2 \rangle, \\
\vert 2 \rangle = \vert e_1 \rangle \otimes \vert e_2 \rangle, \\
\vert 3 \rangle = \vert g_1 \rangle \otimes \vert e_2 \rangle, \\
\vert 4 \rangle = \vert e_1 \rangle \otimes \vert g_2 \rangle.
\end{align}
The statement in the paper is that if the system is in the block form that you give to begin with, it will always remain in this block form. In particular, there is coherence only between the states for which both qubits are flipped. This implies that the unitary dynamics conserve the total qubit "parity"
\begin{equation}
P = \textrm{sign}\ S^z_1 S^z_2.
\end{equation}
The upper block has $P = 1$ while the lower block has $P = -1$. Indeed, the unitary part of the master equation (23) respects this symmetry. The non-unitary dynamics only move weight between the blocks, but do not induce coherence. You can easily check this by considering the action of the jump term $S^-_i \vert a \rangle \langle b \vert S^+_j$ term by term, i.e. for all $a,b = 1...4$ and $i,j = 1,2$.
| {
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What is positive rotation direction of a pulley in the Atwood machine? In the Atwood machine the mass $m_1$ hangs on the left and $m_2$ hangs on the right, with $m_2 > m_1$. When released from rest the system accelerates clockwise which we define to be the positive direction.
The pulley has non-negligible mass and also rotates clockwise. The tension $T_1$ on $m_1$ is in the positive direction, and the tension $T_2$ of $m_2$ is in the negative direction. Since $m_2 > m_1$, it follows $|T_2| > |T_1|$. The tension on the sides of the pulley generate torque. Why is there a positive net torque when simply combining $T_1$ and $T_2$ gives a negative result?
|
The tension on m2 (T2) is in the negative direction
Tension is a scalar. Assuming a "light" string, the tension is uniform and becomes a force (of equal magnitude) at both ends.
The force from the segment attached to m2 on the pulley is clockwise. The force from the segment attached to m1 on the pulley is counterclockwise.
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Why can’t quantum randomness be understood as epistemic? I often hear people say that quantum randomness is “true randomness”, but I don’t really understand it. Please bear with my question.
Before the development of quantum physics, randomness is understood as being “epistemic”. That is, things appear random because we couldn’t (or haven’t yet) take a measure.
This is also how probability theory was conceptualized by Kolmogorov.
My understanding is that quantum physics can also be described using standard measure-theoretic probability theory, or, in other words, an theory with merely “epistemic” randomness.
This leads to my question/confusion: in what sense is quantum randomness non-epistemic, given it can be described by standard probability theory? Is there any property of quantum randomness that shows it cannot be epistemic?
|
My understanding is that quantum physics can also be described using standard measure-theoretic probability theory, or, in other words, an theory with merely “epistemic” randomness.
Take the simple probability of a dice to come up with one of the six numbers. In principle if the distribution from 1 to 6 is not flat, it is a true indication that there is a bias in the dice, that one could find out by measuring, weighing accurately the dice.
The quantum mechanical probability cannot be predicted using classical probability distributions (even convoluted ones), because it is biased by the functional dependence on the boundary conditions in the solutions of the probability distributions which come from the quantum mechanical equations. It is the boundaries and the conservation laws that determine the probability distribution ( example).
| {
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Does the normal reaction on pull up bar change during the pull ups? Intuitively, I know the answer but I can't think of the right math.
I found this question but none of the answers were satisfying enough for me. Human body is not a rigid body so do how do we even apply $\Sigma F=ma_{net}$ over it?
|
Human body is not a rigid body so do how do we even apply $\Sigma
> F=ma_{net}$ over it?
It doesn't matter that the human body is not a rigid body. You only you need to apply Newton's second law to the acceleration of the center of mass (COM) of the human body. To simplify the problem, consider only vertical acceleration of the COM of body.
See the figures below showing free body diagrams of the body and the pull up bar, assuming symmetrical loading, while hanging and starting to accelerate upwards during the pull up.
While hanging, the total reaction force of the bar is simply the weight of the man, $Mg$ per Newton's 3rd law.
In order for the COM of the body to accelerate upward, there needs to be an upward force on the COM greater than $Mg$. To do this the man pulls down on the bar with a force $F>Mg$. Per Newton's third law the bar pulls up on the man with an equal and opposite reaction force of $F>Mg$ so that there is a net upward force on the COM of $F-Mg$ and an upward acceleration of $a=(F-Mg)/M$ on the COM.
Bottom line: The reaction force on the bar increases during the pull up.
Hope this helps.
| {
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Mental model of general relativity I am trying to visualize the curvature of space-time.
In almost all of the Yt videos on the topic, it's shown as depression in the space-time fabric. But what does the dimension into which space-time curves represent? If some mass creates more curvature in the fabric, what does it show?
| Unfortunately that "model" is complete and utter rubbish. You cannot learn anything from it, so please do not try!
Nobody who understands General Relativity uses it. It does not represent any equations or allow any calculations.
It needs to go away, now. Unfortunately (again) the internet will preserve it forever . . .
| {
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How to express $|m\rangle\langle n|$ in terms of ladder operators? Let us consider the Hamiltonian of a single harmonic oscillator, which is expressed in terms of creation/annihilation operators as
$H=\hbar \omega (a^{\dagger}a+1/2)$. The eigenstates of this Hamiltonian are the number states ($n\geq 0$)$$a^{\dagger}a|n\rangle=n|n\rangle.$$
My question is: how to express $|n\rangle\langle m|$ in terms of creation/annihilation operators, where $m$ and $n$ are two arbitrary integers?
| Combining the existing answers, since $|0\rangle\langle 0|=\prod_{k\ge1}(1-k^{-1}a^{\dagger}a)$ and $|n\rangle =\frac{1}{\sqrt{n!}}a^{\dagger n}|0\rangle$,$$\sum_{mn}c_{mn}|m\rangle\langle n|=\sum_{mn}\frac{c_{mn}}{\sqrt{m!n!}}a^{\dagger m}\prod_k(1-k^{-1}a^{\dagger}a)a^n.$$
| {
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Energy of a system executing forced oscillations In L&L's textbook of Mechanics (Vol. 1 of the Course in Theoretical Physics) $\S 22$ Forced oscillations, one finds the following statement:
\begin{equation}
\xi = \dot{x} + i \omega x, \tag{22.9}
\end{equation}
\begin{equation}
\xi = e^{i \omega t} \int\limits_{0}^{t} \frac{1}{m} F(t') e^{-i\omega t'} \mathrm{d} t' + \xi_{0}. \tag{22.10}
\end{equation}
Let us determine the total energy transmitted to the system during all time, assuming its initial energy to be zero. According to formula (22.10), with the lower limit of integration $-\infty$ and with $\xi(-\infty) = 0$, ...
The energy of the system is
\begin{equation}
E = \frac{1}{2} m \left(\dot{x}^{2} + \omega^{2} x^{2}\right) = \frac{1}{2}m \lvert \xi\rvert^{2}.\tag{22.11}
\end{equation}
However, at the beginning of this section, the Lagrangian is found to be
\begin{equation}
L = \frac{1}{2} m \left(\dot{x}^{2} - \omega^{2}x^{2}\right) + x F(t) \tag{22.1}
\end{equation}
where $F(t)$ is the external force driving the oscillations, so according to the definition of the energy one should have
\begin{equation}
E \equiv \frac{\partial L}{\partial \dot{x}} \dot{x} - L = \frac{1}{2}m \lvert \xi\rvert^{2} - x F(t).
\end{equation}
*
*Why is the term $x F(t)$ not being considered in the expression for the energy?
*How can one simply take $\xi(-\infty) = 0$? How can one be sure that the limit exists?
*How can one set the energy equal to zero at $t \to -\infty$? Could it not be the case that this limit is not well-defined?
| *
*The Lagrangian term $xF(t)$ is an interaction term between the system and the environment. The energy of the oscillator system itself is given by eq. (22.11).
2+3. L&L assume that the oscillator is at rest at $t=-\infty$.
| {
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Finding the Hermitian generator of a representation of a Symplectic transformation Consider a set of $n$ position operators and $n$ momentum operator such that
$$\left[q_{i},p_{j}\right]=i\delta_{ij}.$$
Lets now perform a linear symplectic transformation
$$q'_{i} =A_{ij}q_{j}+B_{ij}p_{j},$$
$$p'_{i} =C_{ij}q_{j}+D_{ij}p_{j}.$$
such that the canonical commutation relations are maintained
$$\left[q'_{i},p'_{j}\right]=i\delta_{ij}.$$
Any such symplectic transformation should be unitarily implemented due to the Stone-von Neumann theorem (right?)
$$U^{-1}q_{i}U =q'_{i},$$
$$U^{-1}p_{i}U =p'_{i}.$$
The question is: Assuming the coefficients $A$,$B$,$C$ and $D$ are given, is there a systematic way to calculate the generator $G$ of the unitary transformation $U=e^{-iG}$?
| *
*Classically, the symplectic group $Sp(2n, \mathbb{R})$ of dimension $n(2n+1)$ is the group of all linear time-independent canonical transformations (CTs)
$$z^{\prime I}~=~\sum_{J=1}^{2n}M^{I}{}_Jz^J.\tag{1}$$
The corresponding symplectic Lie algebra $sp(2n,\mathbb{R})$ is the set of all linear time-independent infinitesimal CTs, which have time-independent quadratic generating functions $$F(z)~=~\frac{1}{2}\sum_{I,J=1}^{2n}a_{IJ}z^Iz^J, \qquad a_{IJ}~=~a_{JI}.\tag{1}$$
A finite linear CT is of the form
$$z^{\prime I}~=~e^{\{F(z), ~\cdot~ \}}z^I,\tag{2}$$
where $\{\cdot, \cdot \}$ denotes the Poisson bracket.
*Quantum mechanically, the Poisson bracket $\{\cdot, \cdot \}$ is replaced with the commutator $\frac{1}{i\hbar}[\cdot, \cdot ]$, so a finite linear CT becomes
$$\hat{z}^{\prime I}~=~e^{\frac{1}{i\hbar}[F(\hat{z}), ~\cdot~ ]}\hat{z}^I~=~\hat{U}\hat{z}^I\hat{U}^{-1},\tag{3}$$
where
$$ \hat{U}~=~e^{\frac{1}{i\hbar}F(\hat{z})} \tag{4}$$
is a unitary operator with Hermitian generator
$$F(\hat{z})~=~\frac{1}{2}\sum_{I,J=1}^{2n}a_{IJ}\hat{z}^I\hat{z}^J, \qquad a_{IJ}~=~a_{JI}.\tag{5}$$
| {
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Find coefficient for pure and mixed states Consider a generic $2\times 2$ Hermitian matrix written as $$\rho =\alpha\sigma_0+\beta\hat{\vec n}\cdot\vec\sigma\quad ,$$ where $\hat{\vec n}$ is a unit vector and the coefficients are real numbers.
My question is this; how do I figure out what the coefficients have to be for a pure state and a mixed state. I know theoretically what I'm supposed to do, focusing on pure state I should find $$Tr[\rho^2]=1$$ and use the identity $(\vec a\cdot\vec\sigma)(\vec b\cdot\vec\sigma)=(\vec a\cdot\vec b)I+i(\vec a\times\vec b)\vec\sigma$. However somewhere along these lines I fail and can't get the right answer. I know an easy method would be to just calculate the matrix itself by expanding the pauli vector into its components but I am trying to avoid that if possible, any suggestions?
EDIT: Here is my full attempted solution
I start with the condition that $\mathrm{Tr}[\rho]=\mathrm{Tr}[\rho^2]=1$ for pure states. Starting with $\mathrm{Tr}[\rho]$ I can solve it according as $$\mathrm{Tr}[\alpha\sigma_0]+\mathrm{Tr}[\beta\hat{\vec n}\cdot\vec\sigma]=1\Longleftrightarrow \mathrm{Tr}[\alpha\sigma_0]=1\Longleftrightarrow \alpha=\frac{1}{2}\quad .$$ Where I used the fact that the Pauli vector is traceless. Now I try to find $\rho^2$.
$$\rho^2=(\alpha\sigma_0+\beta\hat{\vec n}\cdot\vec\sigma)(\alpha\sigma_0+\beta\hat{\vec n}\cdot\vec\sigma)=|\alpha|^2\sigma_o+2\alpha\beta\hat{\vec n}\cdot\vec\sigma+|\beta|^2(\hat{\vec n}\cdot\hat{\vec n})\sigma_0+i|\beta|^2(\hat{\vec n}\times\hat{\vec n})\vec\sigma\\
=|\alpha|^2+2\alpha\beta\hat{\vec n}\vec\sigma+|\beta|^2\sigma_0.\\
\Longrightarrow \mathrm{Tr}[\rho^2]=2|\alpha|^2+2|\beta|^2=1 \\
\Longleftrightarrow |\beta|^2=\frac{1}{4} \Longleftrightarrow\beta=\frac{1}{2}$$
Is this train of thought correct in any manner or have I made some baseless assumptions along the way?
| That seems to be right to me, but don't forget that $\beta = -\frac{1}{2}$ because you had absolute value in your equation and $\beta$ can be any real number. Also, another way to think about pure states is that they are points on the edge of the unit ball; that is to say, if when the density matrix is written in the form
$$\rho = \frac{1}{2}(I +r_x\sigma_x + r_y\sigma_y+r_z\sigma_z),$$
and if $r_1^2+r_2^2+r_3^2=1$, then the density matrix represents a pure state. If $r_1^2+r_2^2+r_3^2<1$, it is a mixed state. The form we're given is
$$\rho = \alpha\sigma_0 + \beta \hat{\vec{n}} \cdot \vec{\sigma}= \alpha I + n_x\beta\sigma_x + n_y\beta\sigma_y + n_z\beta\sigma_z = \frac{1}{2} ( 2\alpha I + 2n_x\beta\sigma_x + 2n_y\beta\sigma_y + 2n_z\beta\sigma_z)$$
Using this method, it is clear that $\alpha = \frac{1}{2}$, and we have that $r_1^2+r_2^2+r_3^2 = (2\beta n_x)^2 + (2\beta n_y)^2 + (2\beta n_z)^2 = 4\beta^2$. Since we want this to equal $1$, we have $1=4\beta^2 \implies \beta = \frac{1}{2},$ or $\beta= -\frac{1}{2}$.
Hope this helps.
| {
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Are water molecules at the surface closer or farther apart than the molecules inside? My lecturer says that since the energy of the molecules on the surface is higher (less negative), then at equilibrium there will be less molecules on the surface, hence the molecules on the surface are farther apart.
On the other hand, in Khanacademy, in this video for example , he says at 1:50 that the molecules will get a little more close to their neighbors.
And also in here "closer spacing at the surface".
And also there is this video:
On the third hand, my lecturer also said that water is not compressible. Doesn't it mean that the spacing between the molecules will stay the same?
| The water molecules on the surface have a vast empty space to move in on one side. This causes them to move further apart than inside the bulk molecules. This larger separation causes a force between surface molecules that makes the surface act like an elastic container. Hence the droplets on a window. Introducing pollution in the water might decrease the forces playing between the surface molecules and let the droplets decay into shapeless blobs.
| {
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Possible to stir a drink while maintaining a convex shape? The Wikipedia page for the Brouwer fixed point theorem has a cute example:
"no matter how much you stir a cocktail in a glass ... when the liquid has come to rest, some point in the liquid will end up in exactly the same place in the glass as before you took any action, assuming that the final position of each point is a continuous function of its original position, that the liquid after stirring is contained within the space originally taken up by it, and that the glass (and stirred surface shape) maintain[s] a convex volume."
(The kicker is that the same should not be said for a shaken cocktail.)
Question: is it possible to stir a cocktail while maintaining a convex shape? My impression from a classic calculus problem is that the sides of the surface would rise up above the center, breaking the convexity condition.
| I would suggest putting a lid on the cocktail that leaves not free surface.^^
If you have a free surface, you must make sure that it does not move in vertical direction at any point (that's what I want the lid for) because if it did, there would be a minimum of its height profile, where indeed the convexity of the fluid body is broken.
In principle you could try to create a stirring motion that causes purerly horizontal fluid motion... Not sure if that's possible.
| {
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Why do we consider special relativity for electrons consider their drift velocity is really small? I know that when looking at the electron and a current carrying wire in different inertial frames there will be different kind of forces acting between these two objects. And I know that it is due to length contraction which cause the charge density to change. But I was confused about something that is , we know that the drift velocity is really small compare to speed of light. Why do we still need to consider this special relativistic effect? (If I am not explaining myself clearly please forgive me, sorry)
| (a) If you ignore special relativity altogether, there is no length contraction and so no electrostatic force in the electron frame. Therefore the special relativistic effect, even if it is small, is larger than the (zero) non-relatvisitic effect. This is a general phenomenon: small effects can still be important if the effect that could have been there at zero-th order vanishes.
(b) Even though the force due to one non-relativist electron is small, a current carrying wire has a huge number of electrons. This huge number amplifies the effect so that it is macroscopically observable. Again, this is another general effect that can happen, where a small effect becomes large when multiplied by a large parameter, such as (in this case) the number of particles, or (in other cases) the length of time of the experiment.
| {
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$(α|0⟩ + β|1⟩)|0⟩$ in matrix/vector form I am currently working through superdense coding with bell states and have a question regarding this value:
$$(α|0⟩ \ + \ β|1⟩) \ |0⟩$$
I understand that $α|0⟩ \ + \ β|1⟩$ can be represented in vector format as $(α \ \ β)$ (vertically)
However, I'm not sure how to apply the outer $|0⟩$ ket? Would it be $(α \ \ β)×(1 \ \ 0)$ which would give $α$?
| This operation is what's called the tensor product. More formally it is denoted by
$$|\psi\rangle\otimes|\phi\rangle$$
but often the $\otimes$ is left out. When calculated numerically it is called the Kronecker product which can be calculated as follows for a 2 state system
$$\pmatrix{a\\b}\otimes\pmatrix{c\\d}=\pmatrix{ac\\ad\\bc\\bd}.$$
See also the wiki page on the Kronecker product for a more general computation. To get a little more intuition you can calculate all the combinations of $\{|0\rangle,|1\rangle\}\otimes\{|0\rangle,|1\rangle\}$ where $|0\rangle=\pmatrix{1\\0}$ and $|1\rangle=\pmatrix{0\\1}$. When you do this you will see that the result is a unit vector with the $1$ at all 4 different positions: the result gives all basis vectors of a 4 state system.
Edit: It is also distrubitive so $$\big(|\alpha\rangle+|\beta\rangle\big)\otimes|\gamma\rangle=|\alpha\rangle\otimes|\gamma\rangle+|\beta\rangle\otimes|\gamma\rangle\\
|\alpha\rangle\otimes\big(|\beta\rangle)+|\gamma\rangle\big)=|\alpha\rangle\otimes|\beta\rangle+|\alpha\rangle\otimes|\gamma\rangle$$
and this should allow you to calculate the tensor product product in question.
| {
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How does a gravity defying stand comes in equilibrium? The gravity defying wine bottle stand are now becoming very common these days.
Can someone explain in detail the physics involved here?
Here is a video that gives details about its making.
|
The gravity defying wine bottle stand are now becoming very common these days.
The bottles do not defy gravity. The normal force exerted on the support (contact area between the wood and the table) equals weight of the two bottles plus weight of the wood. The bottles would tip over if there is some torque about one of the edges of the area of support.
The trick is to place center of gravity of the system within the area bounded by the support (check the figure below). In that case there will be no torque and the system will remain at rest, i.e. the bottles will not tip over.
Source: H. D. Young, R. A. Freedman, "University Physics with Modern Physics in SI Units", 15th ed., 2019.
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How do you estimate the core temperature of a star? Given a star's mass, radius and average composition (e.g. 90% H, 10% He), is there a formula to estimate the core temperature of that star?
I only found one for a lower bound but that wasn't very accurate.
| There are approximate formulas for this, but to do this right you really need an equation of state for matter in the stellar core. Writing that down is really, really hard, which is why predicting the temperature and density profiles for stellar cores is really, really hard.
| {
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Intuition for the negative sign of gravitational potential energy The gravitational potential energy is introduced to us as $U = mgy$. We usually set $U = 0$ on ground level and hence, for an object at height $y$, we have a potential energy equal to $U = mgy$.
I have adopted this convention when getting the potential for a system. For instance, a pendulum of mass $m$ attached to a massless string. I simply set $U = 0$ on the ground level. Hence, the height of the mass pendulum can be written as $y = l(1-\cos\theta)$.
When reading books by Marion and Thornton or Morin, they sometimes set the $U=0$ at the top of the system. For instance, consider a pulley with mass $m_1$ and $m_2$ attached at the left and right side of the string around it. They set $U=0$ at the center of the pulley and simply give the potential of the system as something like $U = -mgy_1 - mgy_2$ without any explanation.
They could have either simply set the coordinate system such that $+y$ is set to be downwards... but the potential should be invariant regardless of the coordinate system used, right?
The only reason I can think of for the negative sign is that near the top of the pulley, the potential should be greater than when it is far away from the pulley ($0 > -mgy_1$). Is this right? I have also read online that when you define $U = mgy$, then $+y$ must point away from the center of the earth. Could someone provide a good intuition about this?
TL;DR Having a hard time in getting the right sign for the potential when writing the lagrangian of the system.
| This is because, when near earth, at the surface, when we define U=0, as we move some height(h) up, potential energy increases by amount $mgh$.
Similarly, when we go down, potential increases by amount $-mgh$.
So when dealing with Gravitational force, which is attractive in nature.
We define U=0 at infinity (You can define it 0 at any location as only change in potential energy matters), when we go some distance up, the potential energy increases. Say, from Earth we were at distance $R_1$, and we moved to distance $R_2$.
The change in potential energy is $-GM_1M_2(\frac{1}{R2} - \frac{1}{R1})$. Since, $R_2 > R_1$, $\Delta U$ is positive, as expected.
| {
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How can satellites change direction without any medium in space? How can satellites change direction without any medium in space? How do spaceships move in space if there is no medium?
How does Newton's third law of motion work in space?
| You do not need a medium to push against. Several principles are used to change direction of satellites.
*
*Newton's third law is used by thrusters and cold gas engines. Every action (particles leave thruster of the satellite) has a reaction. E.g. The satellite moves the opposite direction. $\vec p_r=-\vec p_a = m_a \cdot \vec v_a$ follows conservation of momentum. We want to save fuel. Hence we want to reduce mass $m_a$. As a result the fuel has to be leave the satellite a high velocity by e.g. combustion or acceleration by electric force.
*Magnetic torquers leverage Lorentz force of a magnetic field of e.g. the Earth. An electric coil creates a magnetic moment $\vec m$. The interaction with Earth's magnetic field of $T_e\approx 50\,\mu T$. A torque $\tau = \vec m \times\vec B $ is created.
*Reaction wheels are as well not based on Newton's law and stabilise the orientation of the satellite. The physical principle is a gyroscope (spinner).
| {
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Difference between the wave forms in the water and in the Young double slit experiment We can observe when we cause a slight disturbance at two points on the water surface which is intially totally undisturbed , it will form water waves which would look like as shown in below image:
we can observe that there are constructive and destructive interferences at some places and also which lies in between these type of interferences (that is between fully destructive and fully constructive) . We notice that it doesnt need screen to show the interference effects at all , so why in YDSE we need screen to show the interference patterns, is it because we cant be able to observe the interference being happen in air or any other medium from our naked eyes?
| The screen is used simply because it makes the interference effect easier to see and to record and analyse. As is the case with the picture of the interference of water waves which you included in your question, the light interferes with itself everywhere. The use of a screen allows you to see the interference along a straight line, measure the fringes at a fixed distance from the slits, photograph them, and so on.
| {
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Why do we use different differential notation for heat and work? Just recently started studying Thermodynamics, and I am confused by something we were told, I understand we use the inexact differential notation because work and heat are not state functions, but we are told that the '$df$' notation is only for functions and that the infinitesimal heat and work are 'not changes is anything' surely they can be expressed as functions of something? and they are still changes as they do change? What is the thermodynamic reason for describing them as not being changes in anything?
| I believe the good way to present thermodynamics is through the formalism of differential geometry.
When the thermodynamic process is reversible it can be described as a curve on a manifold of equilibrium states (because each intermediate step is equilibrated). Then $\delta W = -p dV$ and $\delta Q = T dS$ are differential forms - covectors tangential to the manifold of equilibrium states. However, (in general) they are not exterior derivatives $d f$ of any state function $f$. There is no function of state $W$ such that $dW = \delta W$.
| {
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Principle of Equivalence for non-uniformly accelerated system Consider a frame being accelerated according to the law of gravity, $a=-K/r^2$ where $K$ is some constant and $r$ is measured from some point. Since here the acceleration is exactly the form of gravity, can the equivalence principle be extended to include such non-uniform acceleration and thus extending the applicability of the principle to non-uniform gravitational fields?
PS: I have no familiarity with general relativity, the motivation for this question arises from reading; Kleppner D, Kolenkow R. An Introduction to Mechanics
| The equivalence principle holds locally, that is, over domains for which changes in position do not significantly change the reading of an accelerometer. While one can come up with certain arrangements of matter such that certain arrangements of acceleration are indeed indistinguishable, one should be careful with them and not assume sort of extended equivalence principle that holds for all $\vec r$. Consider for example a body at a L1 Lagrange point, whose accelerometer reads zero but which nonetheless is time dilated relative to a point distant from the masses.
With a suitably powerful and well programmed rocket, we could imposed on ourself $a = -K/(r+B)^2$ where r is the displacement from the starting point, $a$ is the proper acceleration (including gravity). But light, causality, and the rest of the universe would not care what our acceleration would be if we went somewhere else. Jeff in his rocket still measures the apple falling on Isaac's head, whatever Jeff's proper acceleration would be if he were flying past the tree.
| {
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How to find a bend curve in 2D? I am looking for a solution on how to find the (approxmiate) shape when bending a rigid-flex circuit board. Please see the abstract sketch below. I have two solid objects ($A$, $B$) which are connected by a thin and flexible but non-stretchable strip of material.
Given that I know
(i) the position snf rotation of the fixed parts $A$ and $B$ in 2D space and
(ii) the length $d$ and thickness $w$ of the flexible connecting strip
how can I calculate the shape of the flexible strip when it is bent?
Any help would be highly appreciated, approximate solutions (which may assume $w=0$) are fine.
| The curvature is proportional to the local bending moment.
According to the picture, it seems that the strip is firmly held by the parts A and B, in a way that it can not slide. In that case, the situation is pure bending and the curve is an arc of circle. See http://emweb.unl.edu/NEGAHBAN/Em325/11-Bending/Bending.htm
The situation is different if a strip is firmly held in A (by a vise for example), and we bend it by pushing the other end with a hand. In this case, the bending moment at the hand is zero (as the strip is free to slide there) while it is maximum at the vise. The curvature is not constant and the form is not simple, except for small displacement (angle $\alpha \approx 180^\circ)$
| {
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Non-Comulative nature of Mass-Radius curves of Neutron Stars For finding the mass-radius curve of neutron stars, we can solve TOV Equations which are constraint equations got by solving Einsteins equations. The boundary conditions are $m(r=0)=0$ and $\rho(r=0)=\rho_c$. Then we put a physical condition that for $r\rightarrow r_*$(radius of star), $\rho\rightarrow 0$, and by this we can get the limiting radius $r_*$ and mass $m_*$ and plots like
and for density
I understood and took it from MIT OCW
One can also do the same process for Modified Gravities and more involved $P(\rho)$ matter equations
*
*I have seen many places where the mass-radius curve looks like
This curve is from the paper "Neutron stars in Einstein-Λ gravity: the cosmological constant effects
G. H. Bordbar1,
, S. H. Hendi1, and B. Eslam Panah"
*
*Confusion: Why isn't the mass of neutron star cumulative as radius increases, i.e. it must be $0$ at $r=0$ and reach its maximum at $r=r_*$? Maybe I am not getting the right interpretation of these curves. So, I would be grateful for some explanation or references.
| Your non-cumulative curves relate the total mass of the star to its surface radius. Different points represent different stars.
Your cumulative curves relate enclosed mass of a star versus radius of the shell enclosing the mass. Different points represent different radii within a given star.
| {
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Do the components of a force written for a purpose actually exist? On an inclined plane if you put a box, the force of gravity $mg$ is written as sum of two forces $mg\sin\theta$ and $mg\cos\theta$ where $\theta$ is the angle the incline is making with earths surface. Do these forces $mg\sinθ$ and $mg\cosθ$ actually work on the object?
|
Do these forces $mg\sinθ$ and $mg\cosθ$ actually work on the object?
Only the component of the force of gravity, $mg\sin\theta$, does work on the object. That work is then
$$W=Fd=mgd\sin\theta$$
Where $d$ is the length of the incline plane traveled by $m$. Since $\sin\theta=h/d$ where $h$ is the height of the plane,
$$W=mgh$$
Which tells us the work done by gravity depends only on the initial and final vertical position of $m$, regardless of the path between the two positions. This is a property of conservative forces, of which gravity is one.
Hope this helps.
| {
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Does the Schrödinger equation apply to spinors? I was reading about Larmor precession of the electron in a magnetic field in Griffiths QM when I came across the equation
$$
i\hbar \frac{\partial \mathbf \chi}{\partial t} = \mathbf H \mathbf \chi,
$$
where $\mathbf\chi(t)$ is a 2D vector that represents only the spin state and does not include information of the wave function. The Hamiltonian is
$$
\mathbf H = - \gamma \mathbf B \cdot \mathbf S = - \frac{\gamma B_0 \hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}
$$
for a uniform magnetic field $\mathbf B = B_0 \hat k$. Why should these spinors also obey the Schrödinger equation? The book does not provide any further information as to why this should hold.
| The correct wave equation for an electron is the famous Dirac equation (in fact it is correct for all spin $\frac{1}{2}$ particles).
The non-relativistic limit of the Dirac equation is called Pauli equation. The derivation can be found here. In this equation $|\psi \rangle$ is a two-component spinor wavefunction.
Pauli equation (general)
$${\displaystyle \left[{\frac {1}{2m}}({\boldsymbol {\sigma }}\cdot (\mathbf {\hat {p}} -q\mathbf {A} ))^{2}+q\phi \right]|\psi \rangle =i\hbar {\frac {\partial }{\partial t}}|\psi \rangle }$$
If you substitute $\mathbf {A}=0$ above you get back the Schrödinger equation. In fact, a relativistic wave equation for any spin should under some limit give back the Schrödinger equation be it Klein–Gordon equation or Proca action or Rarita–Schwinger equation.
| {
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Hall sensor for electric(!) field? Is it (in principle) possible to measure the strength of an electic(!) field with a hall sensor?
I think so, for the following reasons:
*
*The hall sensor is a conductor. If we place an conductor in an electic field, charges will rearrange so that there will be no electric field in the interior of the conductor (~Faraday's cage).
*The field that is created by the new charge distribution in the hall sensor is opposite to the surrounding field, but has the same strength.
*Following the usual argumentation, we get a voltage across the hall sensor that is proportional to the field inside the hall sensor, which has the same strength as the surrounding field that we want to measure.
Is this line of thought correct? If so, what are technical difficulties that make the hall effect not suitable to measure electric fields?
| It seems you do not actually consider a Hallsensor wich needs current flowing trough, but just a conducting sheet put in an electric field measuring the difference between two sides. But first the sheet will disturb the field, and second how will you measure it without a closed circuit of some kind?
| {
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Why is the force on a rigid pendulum directed radially? Consider a typical presentation of a simple pendulum: a point mass $m$ attached to a rigid rod of length $L$, which is free to rotate around a pivot. Newton's equations are
$$\begin{gather}
mg \cos\theta - F = -m L \dot{\theta}^2 \\
-mg \sin\theta = m L \ddot{\theta},
\end{gather}$$
where $F$ is the force from the rod. It seems that it's universally assumed that this force points in the radial direction, like it would if the rod was replaced by a string - though of course the force from the rod can point outwards as well as inwards, while a string can only pull and not push.
Why is this? A rod is rigid, so it can transmit shear stress, so in principle the force from the rod could have a tangential component too. This is easily illustrated by considering a situation where the rod is attached to a motor which makes the system turn with a constant angular velocity $\omega$. We are forced to introduce a tangential component $F_\theta$ into the equations
$$\begin{gather}
mg \cos\theta - F_r = -mL \omega^2 \\
-mg \sin\theta + F_\theta = 0
\end{gather}$$
because something has to balance the tangential component of the weight. But this only works because we have the constraint that $\dot{\theta}$ - if we do this for the pendulum, the system becomes undetermined since we're introducing one more unknown.
What is the difference between this case and the pendulum? What condition allows us to assume that the force from the rod is radial for the pendulum? And as a bonus question, is there a simple way to explain this to an introductory physics course?
|
What is the difference between this case and the pendulum? What condition allows us to assume that the force from the rod is radial for the pendulum?
The condition that explains both results is that the rod is assumed to be massless. Since the rod is assumed to be massless then both the net force and the net torque acting on the rod must be zero at all times, regardless of the acceleration.
Specifically, consider the torques acting on the rod at the axis of the hinge or motor. Since the rod is massless the torque from gravity is 0. That means that the torque from the mass must be equal and opposite from the torque from the hinge or motor.
In the case of the pendulum the torque from the hinge is zero, so the torque from the mass is also zero and therefore the tangential force from the mass is zero.
In the case of the motor the torque from the motor is non-zero, so the torque from the mass is equal and opposite and therefore the tangential force from the mass is non-zero.
| {
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Mathematical Definition of Point Source Wikipedia describes a mathematical definition of a point source as "a singularity from which flux or flow is emanating".
The usual definition in Physics describes it just as a source whose dimensions are negligible in comparison to another variable you're relating it to, which leads to well-behaved phenomena such as isotropic emission.
Well, if there's something I wouldn't expect from a singularity, that's well-behaving. Given that, how does that math definition of point source makes sense? How to phenomenologically relate it, lucidly, it with physics?
| "Point" sources are typically represented by $\delta$-functions.
For example, an electron with charge $e$ located at $\vec{r}_e$ must have charge density
$$\rho(\vec{r}) = \left\{{0,\, \vec{r}\not=\vec{r}_e\atop\infty,\, \vec{r}=\vec{r}_e}\right.,$$
and we therefore write that
$$\rho(\vec{r})=e\,\delta^{3}(\vec{r}-\vec{r}_e),$$
so
$$Q=\int_V\rho(\vec{r})\,d^3r=e,$$
as required
And note that the $\delta$-function also possesses various additional useful mathematical properties, e.g.,
$$ \nabla^2\left(\frac1{|\vec{r}-\vec{r}\,'|}\right)=-4\pi\,\delta^3(\vec{r}-\vec{r}\,'),$$
useful when evaluating typical E & M expressions involving $\nabla^2$.
| {
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Is there a quantum of entropy - and how large is it? Google Scholar lists quite a number of papers when searching for "Quantum of entropy" or "entropy quantization". Most are old.
Some of these papers mention k as quantum of entropy, others k/2, still other log 2 k (see H.S. Leff, https://link.springer.com/article/10.1007/s10701-007-9163-3).
Also the Google book search gives many results: see https://www.google.de/search?tbm=bks&hl=de&q=%22quantum+of+entropy%22
What is the present status?
Update: As the answers show, entropy is usually (= almost always) not a multiple of a smallest value. But is there a smallest possible total entropy in a physical system?
| In physics, we normally define entropy as $S = k \ln \Omega$, where $k$ is Boltzmann's constant and $\Omega$ is the number of microstates consistent with the observations you've made of the system. For the rest of this answer, I'll work in units with $k=1$ (you can multiply all the expressions below by $k$ if you want).
The smallest possible value of entropy is $0$, since if you know the microstate exactly then $\Omega=1$ so $S = \ln \Omega=\ln 1 = 0$.
If you had knew the system was in one of two microstates, and you assume there is an equal chance for the system to be in either one -- which is an assumption normally made in thermal physics -- then the smallest non-zero value of entropy would be $\ln2$. However, this is not really a "quantum" of entropy, since the next allowed value of entropy would be $\ln 3$, which is not an integer multiple of $\ln 2$.
There is a more general definition of entropy, where you don't assume each microstate is equally likely. The more general definition of entropy is
\begin{equation}
S = - \sum_k p_k \ln p_k
\end{equation}
where $p_k$ is the probability of microstate $k$. If we have two states, with probabilities $p$ and $1-p$, then the entropy is
\begin{equation}
S = - p \ln p - (1-p) \ln (1-p)
\end{equation}
Here's a plot of that function from wolfram alpha; you can see the entropy can take on any value from $0$ to $\ln 2$ and is distinctly not quantized
| {
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Would our orbit really remain the same if the sun were a black hole of equal mass? There seems to be an idea floating around that the sun could be replaced by anything of equal mass with no consequence to our orbit.
It seems to me that if the mass of the sun were confined to a single point that the local geometry of space would be different, therefore our orbit and those of the other planets would also be different.
*
*Am I mistaken to think of this as analogous to distribution of weight on an elastic surface? - (A more confined distribution of mass would create a deeper gravity well, so to speak).
*I'm thinking that angular momentum would need to be conserved and that might result in frame dragging which may need to be accounted for, but I'm not sure and I don't know how to calculate it (yet).
*Anything I'm overlooking?
P.S. I know black holes don't come in our size.
Thanks!
| Yes, the number of gravitational degrees of freedom remains the same if the Sun were magically turned into a black hole. Now apply that to disk-like interaction, and you have focused, super strong gravitation. Count.
| {
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Is angular momentum just a convenience? I'm wondering whether angular momentum is just a convenience that I could hypothetically solve any mechanics problems without ever using the concept of angular momentum.
I came up with this question when I saw a problem in my physics textbook today. In the problem, a puck with known velocity hits a lying stick. The puck continues without being deflected, and the stick starts both linear and angular motion. There are three unknowns: velocity of puck and stick after collision, and the angular speed of the stick. So, we need three equations: conservation of linear momentum, kinetic energy, and angular momentum.
So, for instance, is it possible to solve this problem without using angular momentum?
Also, how would a physics simulator approach this problem?
|
I'm wondering whether angular momentum is just a convenience that I could hypothetically solve any mechanics problems without ever using the concept of angular momentum.
If your criterion for something being a convenience is that you could solve problems without it then everything in physics is just a convenience.
There are an infinite number of possible mathematical formulations. So, in principle, it should be possible to convert any mathematical problem into a different formulation that avoids the use of any specific concept that you would like to avoid (or at least hides it so that it is not apparent that you are using the concept).
That said, angular momentum is conserved and it is related (by Noether's theorem) to the fact that the laws of physics are symmetric under spatial rotation. Both conserved quantities and symmetries are very important in modern physics. So even if you classify it as a convenience, it is one of the most important and pervasive conveniences in physics.
| {
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Barometric equation with different species The barometric equation gives the pressure dependence of a perfect gas in a gravitational potential. In particular
$$
P(h) = P_0e^{-\frac{mgh}{k_BT}}
$$
where $m$ is the mass of molecules and $T$ the temperature of the gas.
What would happen if we had an atmosphere composed of different species? I would expect it to be layered with heavier gases at the bottom and lighter ones on top with transitions between gases looking something like a $tanh(h)$. Can this be derived from statistical mechanics relatively easily?
| If we suppose ideal gases, remember that the pressure of an ideal gas depends on the number of molecules, however, it has an influence on the distribution of the gas with a gravitational potential. However, you can substitute in your equation the molar mass with the average molar mass. This molar mass $M$ can be found in the following equation, which you can find in Wikipedia:
$$M = \sum _i x_i M_i$$
where $x_i$ is the mole fraction of the $i_{th}$ compound and $M_i$ the molecular mass of that $i_{th}$ compound. So, for example, air gives us a molar mass of $28.7 g/mol$. Substituting the formula you gave will give the result you wanted.
| {
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How to calculate velocity vector from scalar angular velocity and position vector in 2D? I would like to know, if I have an angular velocity as scalar, how can I calculate the velocity vector.
I know that the product of angular velocity and the length of the distance gives the speed, but I would like to know vector of that.
I tried solving this problem many times, but honestly, I dont even know what I did, Im complete beginner, and couldnt really find a solution, or answer.
| I assume you want to find the linear velocity if you know the angular speed. You need to realize that all the points on a rotating rigid body have the same angular speed and angular velocity but different linear speeds and velocities. A point on a rigid body that describes a circle of twice the radius as another point will have twice the linear speed because it covers twice as much distance in the same amount of time. The linear speed $v$ is related to the angular speed $\omega$ by $v=\omega r$.
The corresponding vector equation is the cross product $\vec v=\vec \omega \times \vec r$. The direction of the velocity vector is perpendicular to the plane defined by $\vec \omega$ and $\vec r$ using the right hand rule. Put the thumb of your right hand perpendicular to the plane of rotation and match the circulation of your four fingers to the circulation of the rotating body. The velocity vector at a point is tangent to the circle at that point.
Example
The plane of rotation is $xy$ and have position vector $\vec r$ in that plane. The angular velocity is perpendicular to the plane.
$\vec \omega=\{0,0,\omega\}$ and $\vec r=\{r \cos\theta,r\sin\theta,0\}$. Then $$\vec v=\vec \omega\times \vec r=\{0,0,\omega\}\times \{r \cos\theta,r\sin\theta,0\}=\omega r\{-\sin\theta,\cos\theta,0\}.$$Both the velocity and position vectors are 2D.
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How can light produce electric and magnetic field when there are no accelerating charged particles? If we see light as a wave, especially in vaccum, there is nothing there, no particles, yet light has an electric and magnetic field. How can this be possible?
| As pointed out in a comment by @PhysicsDave, there is the supposition of the existence of a field acting as a mediator of electric interaction and magnetic interaction.
The fields extends over spatial distance. That is how the interaction is envisioned. The interaction being mediated does not require particles in the inbetween space. The field extends over spatial distance.
Maxwell envisioned the electromagnetic interaction as medidated by an entity that in the absence of a source of electrostatic force or magnetic force is in a uniform state.
A source of electrostatic force is then envisioned as inducing a stressed state of that entity. That stressed state is then thought of as the mediator of the interaction.
When the source changes the stressed state changes. The electromagnetic field has the property that the rate of change of the stressed state tends to persist.
Propagation of transversal waves:
-You need some form of elasticity. If the entity is pushed into a stressed state the entity tends to return to the uniform state
-Rate of change should persist. With persistence of rate of change: when the entity is changing in the diraction towards uniform state, at some rate, then it will overshoot the zero point, and will go on towards a stressed state in the opposite direction.
Those two properties, elasticity and persistence of rate-of-change, give the electromagnetic field the ability of wave propagation. There is no necessity for particles to be present in the space; the electromagnetic field itself supports waves propagation.
| {
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Has it been experimentally proven that energy causes gravity? I know that under general relativity energy and mass are equivalent under $E=mc^2$. But has it been experimentally proven that energy alone causes gravity, for example, does a nuclear reaction generate gravity independent of the mass of the reactor alone? Is a kugelblitz possible?
| The parametrized post-Newtonian (PPN) formalism is a generalized way of exploring gravity theories, including general relativity. In the older "beta-delta" parametrization, three of the parameters ($\beta_1$, $\beta_2$, and $\beta_3$) describe how much gravity is produced by kinetic energy, gravitational energy, and internal energy respectively. In addition, there's another parameter $\beta_4$ that describes how much gravity a given amount of pressure creates; this is important for photons, since photons have a pressure equal to their energy density (up to a factor of $c$.) The case $\beta_1 = \beta_2 = \beta_3 = \beta_4=1$ corresponds to all three types of energy creating the same amount of gravity as conventional mass does, given the conversion factor $E = mc^2$; this is what is predicted by general relativity.
In terms of the other PPN parameters mentioned in that article, we have:
$$
\begin{align*}
(\beta_1 - 1) &= \frac{\gamma - 1}{2} + \frac{\alpha_3}{4} + \frac{\zeta_1}{4} \\
(\beta_2 - 1) &= - \frac{\beta - 1}{2} + \frac{3 (\gamma - 1)}{2} + \frac{\zeta_2}{2} \\
(\beta_3 - 1) &= \zeta_3 \\
(\beta_4 - 1) &= (\gamma - 1) + \zeta_4
\end{align*}
$$
From current observational bounds on gravity (such as the tracking of space probes in the solar system, the perihelion shift of Mercury, the behaviors of pulsars, etc.) these parameters are bounded to around the following orders of magnitude:
\begin{align*}
|\gamma - 1| &\lesssim 10^{-5} & |\beta - 1| &\lesssim 10^{-4} & \alpha_3 &\lesssim 10^{-20} \\
\zeta_1 &\lesssim 10^{-2} & \zeta_2 &\lesssim 10^{-5} & \zeta_3 &\lesssim 10^{-8} & \zeta_4 \lesssim 10^{-2}
\end{align*}
So to within an order of magnitude, these parameters suggest that $\beta_1$ and $\beta_4$ are constrained to be equal to 1 to within a few percent. In other words, we're pretty sure that kinetic energy and pressure create the same amount of gravity that mass do to within a few percent. The gravitational effects of gravity itself and of internal energy are even more tightly bounded.
Caveat: I'm playing a bit fast and loose with these bounds. In reality, they were all established via a series of interdependent experiments, and it's possible that the published bounds are interdependent on one another in a way that allows for larger values. Still, this hopefully gives you a feel for how this question has been experimentally addressed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does amplitude really go to infinity in resonance? I was recapping the forced oscillations, and something troubled me. The equation concerning forced oscillation is:
$$
x=\frac{F_0}{m(\omega_0^2-\omega^2)}\cos(\omega t)
$$
I don't understand why this equation predicts that the amplitude will approach infinity as $\omega$ approaches $\omega_0$. One can come up with the argument that in the actual world, there are damping forces, friction etc. The trouble is, however, even in the ideal world, the amplitude wouldn't approach infinity as the spring's restoring force will catch the driving force at some point, and the system will stay in equilibrium.
What I'm wondering is
*
*Is my suggestion in the last paragraph correct?
*If it is correct, what assumption led us to the erroneous model of $x$?
*If it is not correct, what am I missing?
| transfer this differential equation
$${\frac {d^{2}}{d{t}^{2}}}x \left( t \right) +{\omega_{{0}}}^{2}x
\left( t \right) ={\frac {F\cos \left( \omega\,t \right) }{m}}$$
to Laplace domain (with the initial conditions $~x(0)=0~,\dot{x}(0)=0~)$ you obtain
$$x(s)={\frac {Fs}{m \left( {s}^{2}+{\omega}^{2} \right) \left( {s}^{2}+{
\omega_{{0}}}^{2} \right) }}
$$
and back to time domain (the solution)
$$x(t)=-{\frac {F\cos \left( \omega\,t \right) }{m \left( {\omega}^{2}-{
\omega_{{0}}}^{2} \right) }}+{\frac {F\cos \left( \omega_{{0}}t
\right) }{m \left( {\omega}^{2}-{\omega_{{0}}}^{2} \right) }}
$$
the limit of x(t) for $~\omega=\omega_0~$ is: (l'hospital approach)
$$\lim x(t)\bigg|_{\omega\mapsto\omega_0}=\frac{\partial_\omega \left[-F \left( \cos \left( \omega\,t \right) -\cos \left( \omega_{{0}}t
\right) \right)
\right]}{\partial_\omega\left[m \left( {\omega}^{2}-{\omega_{{0}}}^{2} \right)\right]}\bigg|_{\omega\mapsto\omega_0}=\frac 12\,{\frac {F\sin \left( \omega_{{0}}t \right) t}{m\omega_{{0}}}}
$$
thus x(t) for $~\omega=\omega_0~$ is not infinity
$$\frac{F}{m}\cos(\omega\,t)\overset{\text{Laplace}}{=}{\frac {Fs}{m \left( {s}^{2}+{\omega}^{2} \right) }}$$
| {
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Maxwell's eq-meaning of del's cross and dot product? In maxwell's eq there is del whose cross and dot products exist.
So what is del in cross vs dot product.
What's the difference when it's just a partial differential operator.
| They aren't real cross/dot products, it is a notational trick.
$\nabla × $ is called the curl.
$\nabla \cdot$ is called the divergence.
$\nabla \cdot \vec{F} = \frac{\partial F_{x}}{\partial x} \hat i + \frac{\partial F_{y}}{\partial y} \hat j + \frac{\partial F_{z}}{\partial z} \hat k$
$\nabla × \vec{F} = (\frac{\partial F_{z}}{\partial y} - \frac{\partial F_{y}}{\partial z})\hat i +(\frac{\partial F_{x}}{\partial z} - \frac{\partial F_{z}}{\partial x})\hat j +(\frac{\partial F_{y}}{\partial x} - \frac{\partial F_{x}}{\partial y})\hat k $
Curl and divergences computation is the same as a cross product or dot product, but instead multiplying the vectors, you differentiate the component.
| {
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Lagrangian first integral I want to extremize $$\int dt \frac{\sqrt{\dot x ^2 + \dot y ^2}}{y}.$$
I have thought that, since the Lagrangian $L(y, \dot y, \dot x)$ is $t$ dependent only implicitly, that i could use the fact that $$L(z,z') \implies L - z' \partial L / \partial z' = c.$$
So $$L - y' \partial L / \partial y' = c_1,$$
$$L - x' \partial L / \partial x' = c_2$$
But these two equations, when we substitute the values and arrange it, give us
$$dy/dx = c_3 \implies y = c_3 x +b.$$
This is certainly wrong, the answer is supposed to be a circle equation. Even so we can solve it another way, i am still confused: Why did we got the wrong answer using the above two equation? If, for example, the Lagrangian was $\int dt \sqrt{\dot x ^2 + \dot y ^2}$, we could use the above approach to get the answer (in this case, a line is the right answer).
| with
$$L=\frac{\sqrt{\dot x^2+\dot y^2}}{y}$$
and because L is not a function of x you obtain that
$$\frac{\partial L}{\partial \dot x}=\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}\,y}=\text{constant}$$
from here
$$\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}\,y}\mapsto
\frac{1}{\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,y(x)}=\text{constant}$$
or
$$\sqrt{1+\frac{dy}{dx}^2}\,y(x)=k^2$$
| {
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Why is torque defined as $\vec{r} \times F$? Here I cannot convince myself myself that it is units because the torque is defined to be in units of Newton meter is a reiteration of the law stated above. Why was it not $r^2 \times F$ or $r^3 \times F$ or $r^2 \times F^2$ etc. The argument "in our experience how much something rotates depends on the lever length and the force applied" is really insufficient. Can someone outline a more rigorous proof or motivation?
| Torque is change of angular momentum: $$ \vec{\tau} = \frac{d\vec{L}}{dt}$$
Angular momentum is defined as $$ \vec{L} = \vec{r} \times\vec{p} $$
Using the chain rule:
$$ \vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times\vec{p})$$
$$\frac{d}{dt}(\vec{r} \times\vec{p}) = \frac{d \vec{r}}{dt}\times \vec{p}+ \vec{r} \times\frac{d \vec{p}}{dt} = \vec{0} + \vec{r} \times\frac{d \vec{p}}{dt} \tag{1}$$
Remembering that $|\vec{a} \times \vec{b} |= |\vec{a}||\vec{b}| \sin \theta$ and that the velocity vector is parallel to the linear momentum vector we get:$$ \frac{d \vec{r}}{dt}\times \vec{p} = \vec{v} \times \vec{p} = \vec{v} \times m\vec{v} = m|\vec{v}| |\vec{v}| \sin \theta \ \hat r=m|\vec{v}| |\vec{v}| \sin 0 \ \hat r = \vec{0}$$
$\theta$ is the angle between two vectors, which is $0$ for any vector with itself.
So we get for torque, using the result in $\text{(1)}$:
$$ \vec{\tau} = \frac{d\vec{L}}{dt} = \vec{r} \times\frac{d \vec{p}}{dt} = \boxed{\vec{r} \times\vec{F}}, \text{ since } \vec{F}=\frac{d \vec{p}}{dt} $$
But the question then reduces to why angular momentum is defined as $ \vec{L} = \vec{r} \times\vec{p} $. I think this has to do with Noether's theorem, that this quantity is conserved when a system stays the same under a change of angle.
| {
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Space-time continuum expansion I still don't understand how the expansion of the universe works.
If the universe is made up of an infinite number of points that make up space-time, then how can space expand or stretch. Common sense dictates that in order for space to be stretched, the length between points must increase, the minimum length. But when we have an infinite number of points, changing the length between them simply does not make any sense. It wouldn't affect anything.
So maybe space-time has a finite volume with some fundamental length?
I understand that the minimum length can be so small (below the Planck length) that we cannot notice the graininess of the spacetime and think that it is a continuum-like. But this breeds contradictions.
| An infinite number of points is not a definite number. Don't think of it like a normal number.
No matter how small a distance you make between two any points there is always an infinite number of points between them. There's no cutoff for that. So when you imagine two points spreading apart there are an infinite number of points between those and always wil be. There will never be a gap because there are an infinite number of points between any finite gap.
Put another way, there is no way to select two points which are not themselves separated by an infinite numebr of points.
Common sense dictates that in order for space to be stretched, the length between points must increase, the minimum length.
Common sense is fine for everyday things in normal human experience. When you start think about infinite anything you need to move beyond common sense because human experience (our rules of thumb we call common sense) don't ever have to deal with infinite. We do have a tool for managing infinities and it's called math.
| {
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What is the angular momentum of a particle rotating around an axis in 3D?
What would be the angular momentum of the particle at position $r_i$ in the diagram above?
The vector from the axis of rotation is $R_i$ and the tangential velocity is $v_i$ so the magnitude of angular momentum should be $|R_i|m_i|v_i|$ which gives the vector $R_i \times (m_iv_i)$ since $R_i$ and $v_i$ are at right angles in the diagram.
However, if you look at most resources (for example or another example or another example) they all take each particle's angular momentum to be $r_i \times (m_iv_i)$ instead of $R_i \times (m_iv_i$). Which is correct and why?
|
they all take each particle's angular momentum to be ri×(mivi) instead of Ri×(mivi). Which is correct and why?
I can understand your confusion since both terms do not seem to be the same. I think the main point can be extracted, for example, from the second ressource you cite
The total angular momentum of the body (about the origin) is written ...
Therefore, both can be correct, depending on the point you're interested in.
The second ressource states out quite clearly that they consider the angular momentum wrt. the origin. Hence, their result is correct, although it is not what you might have "expected", namely something that matches with $I\omega = mr^2 \omega$. Here,, the angular momentum is described using the angular velocity $\omega$, which indeed requires some rotational axis. In more detail, when speaking about $\omega$ you implicitly define a rotational axes. Probably this caused the misconception that angular momentum is defined wrt. an axis(?)
Summing up:
*
*Angular momentum is defined with respect to a point, not an axis, which can be seen in the formula $\vec{L} = r \times m\vec{v}$ since both quantities are vectors (defined wrt. some origin). In this definition there does not occur any axis.
*However, you implicitly define an axis once you introduce the angular velocity $\omega$. So, once you describe the angular momentum using the angular velocity, there is some axis present, although the angular momentum is still defined wrt. a point.
| {
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What frequency of cord shaking maintains the same vertical motion for a point on the cord after increasing the wave speed on the cord? I'm studying for my upcoming AP Physics 1 exam but can't figure out this problem
A student shakes a horizontally-stretched cord, creating waves. The graph above shows the vertical position $y$ as a function of time $t$ for a point on the cord. The student then tightens the cord so that waves on it will travel faster than before. How should the student now shake the cord to make the graph of $y$ versus $t$ for the point look the same as above?
(A) With fewer shakes per second than before
(B) With the same number of shakes per second as before
(C) With more shakes per second than before
(D) The answer cannot be determined without knowing the wavelength of the waves.
My intuition would tell me that increasing the speed of the waves would cause the point to oscillate at a faster rate vertically, thus fewer shakes per second than before are needed to maintain the same frequency for the particle's oscillation. However, the correct answer is B, so I really need a thorough explanation as to why the answer is B.
| You already have a nice, brief and concise answer but I'd like to add more to it.
From the "fundamental" equation for waves
$$c = \lambda f \tag{1}\label{1}$$
where $c$ is the speed of wave propagation, $\lambda$ the wavelength and $f$ the temporal frequency, we can see that if you change the speed of propagation then the right-hand side must also change for the equality to hold. There are two quantities that could change in an arbitrary way to provide the result we need, so this is an under-determined problem.
What comes to rescue is the fact that the frequency of oscillation depends solely on the external excitation force (the student that shakes the chord in this case) (please keep in mind that this is the case for linear waves). Thus, even if the speed of propagation changes, this won't affect the frequency. So, what has to change is the wavelength, which will change in order to accommodate the necessary changes for the equality to hold.
To provide some more intuition, consider the fact that we refer to a specific point in space. Thus, wavelength, which encodes information of the spatial variations of the wave, is completely irrelevant to our problem.
| {
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Clarification on Searle's method to determine thermal conductivity
Here In Searle's method to determine the thermal conductivity of a good conductor, the heat transfer rate of the conductor along a displacement(thickness) d is given by:
Q1/t=KD(T1-T2)/d.
=>Q1=KD(T1-T2)t/d
And the heat transferred to the cold water is given by:
Q2=mS(T3-T4)
I can follow up to this point, but my problem arises when it’s said that,
Q1=Q2
My argument is that, Q1 is the heat transferred along the d displacement, not the entire rod. If T1 and T2 was placed at the beginning and end of the rod respectively, only then we could say Q1=Q2. But this is not specified. All the sources say that d can be any known distance on the rod. But clearly anything other than d=the entire length of the rod wouldn't give the correct value, right? Should this clarification be made?
| Consider a proof by contradiction: $Q_1\neq Q_2$, such that $Q_1=Q_2+Q_3$, where $Q_3$ must be stored on the right side of the rod. (We can show this with an energy balance of the right side of the rod.) We consider the sensible heating of that region: $Q_3=mc\frac{dT}{dt}$, where $m$ is the mass, $c$ is the specific heat capacity, $T$, is the temperature, and $t$ is time. At steady state, the temperature must be constant ($\frac{dT}{dt}=0$), so $Q_3=0$ and we have in fact $Q_1=Q_2$.
| {
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In general relativity, assuming a spherical uniform mass distribution, what is the total energy value of the gravitational field inside the sphere? In Newtonian mechanics, assuming a spherical uniform mass distribution, the total gravitational potential energy (gravitational self-energy) inside the sphere is
$$U_{gs}=-\frac35\frac{GM^2}R.$$
*
*In general relativity, assuming a spherical uniform mass distribution, what is the total energy value of the gravitational field inside the sphere?
*In general relativity, assuming a spherical uniform mass distribution, does "total energy value of the gravitational field inside the sphere" equal the "gravitational self-energy"?
*This question was added after 1 answer.
In general relativity, I have seen many articles about the difficulty of defining the energy of the gravitational field.
The point of my question is,
*In a weak gravitational field or in near-flat spacetime, can the energy density of the gravitational field be taken from the total gravitational potential energy obtained from Newtonian mechanics, or from the gravitational self-energy, as a (good?) approximation?
| In general relativity, there is no gauge invariant local definition of the energy of a gravitational field. In simpler terms, there is not a well-defined way to define the total energy contained in a finite region of space. So, while you can take a non-relativistic limit and recover Newtonian gravity, it is not possible to unambiguously generalize the idea of the non-relativistic gravitational energy contained in a region, to GR.
For an asymptotically flat spacetime (such as a lump of matter sitting in isolation), you can define a mass or energy associated with the space. But, you cannot localize this energy to a region.
| {
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Most likely velocity vector of ideal gas?
$$\left( \text{probability of a molecule having velocity }\vec{v}\right) \propto e^{-mv^{2}/2kT}$$
So the most likely velocity vector for a molecule in an ideal
gas is zero. Given what we know about Boltzmann factors, this result should hardly be surprising.
Schroeder's Thermal Physics, page-232.
First of all I don't understand how is that true. I can see that average velocity should be zero. But why should most likely velocity vector should be zero?
And if that is true then, after that we derive the speed distribution. And it goes to zero for $v=0$. So that means that we won't find any molecule with zero speed (at rest in container's frame).
But that also means that there is no molecule with zero velocity!
SO then how could zero velocity be the most likely velocity if there is no molecule with zero velocity??
| The distribution is
$$
w(v_x,v_y, v_z)=Ae^{-\frac{m(v_x^2+v_y^2+v_z^2)}{2k_BT}},
$$
where A is the normalization constant that can be easily found by integration (see Gaussian integral):
$$
\int dv_xdv_ydv_z w(v_x,v_y, v_z) = 1.
$$
The average velocity along any direction is zero, e.g.,
$$
\langle v_x\rangle = \int dv_xdv_ydv_z v_x w(v_x,v_y, v_z)=0
$$
This can be shown either by direct integration by parts or from the symmetry reasons: the odd function is integrated in symmetric limits (from $-\infty$ to $\infty$).
On the other hand, the speed, i.e., the magnitude of this vector, is not zero:
$$
v=\sqrt{v_x^2+v_y^2+v_z^2},
\langle v\rangle =\int dv_xdv_ydv_z v w(v_x,v_y, v_z)>0,
$$
since the integrand is positive everywhere. However, usually one prefers to calculate $\langle v^2\rangle$, since in this case integration is easier.
But why should most likely velocity vector should be zero?
The most likely velocity corresponds (by definition) to the peak of the distribution, which here is at $v_x=v_y=v_z=0$, so it is zero as well.
| {
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Expressing Maxwell's equations in tensor form using Electromagnetic field strength tensor I have yet another derivation question from Carroll's General Relativity textbook. Given the electromagnetic field strength tensor is of the form: $$ F_{\mu\upsilon} =
\left(
\begin{matrix}
0 & -E_1 & -E_2 & -E_3\\
E_1 & 0 & B_3 & -B_2\\
E_2 & -B_3 & 0 & B_1\\
E_3 & B_2 & -B_1 & 0\\
\end{matrix}
\right)
= -F_{\upsilon\mu}$$
The Maxwell's equations are expressed in component notation:
$$
\bar{\epsilon}^{ijk}\partial_jB_k - \partial_0E^i = J^i\\
\partial_iE^i = J^0\\
\bar{\epsilon}^{ijk}\partial_jE_k + \partial_0B^i = 0\\
\partial_iB^i = 0.$$
Given that the field strength tensor can be written in the two tensor equations $F^{0i} = E^i$ and $F^{ij} = \bar{\epsilon}^{ijk}B_k$, how do I reduce the last two equations to the form, $$ \partial_{\lambda} F_{\mu \nu} + \partial_{\mu} F_{\nu \lambda}+ \partial_{\nu} F_{\lambda \mu} = 0,\qquad \mu,\nu,\lambda=0,1,2,3 $$
| Since $F_{ij} = \epsilon_{ijk}B^k$ one has $\epsilon^{lij} F_{ij} = \epsilon^{lij}\epsilon_{ijk}B^k = 2\delta_{lk} B^k$, hence $B^l = \frac{1}{2} \epsilon^{lij} F_{ij}$.
The third Maxwell's equation in OP's question can be expressed with the field strength tensor $F_{\mu\nu}$ according to
\begin{align}
0 &= \epsilon^{ijk} \partial_j E_k + \partial_0 B^i \\
&= -\epsilon^{ijk} \partial_j F_{k0} + \frac{1}{2} \epsilon^{ikl} \partial_0 F_{kl}\\
&= -2\epsilon^{ijk} \partial_j F_{k0} + \epsilon^{ijk} \partial_0 F_{jk}\\
&=-\epsilon^{ijk} \partial_0 F_{jk} + \epsilon^{ijk} \partial_j F_{k0} - \epsilon^{ijk} \partial_j F_{0k}\ .
\end{align}
The Bianchi identity, i.e., the relation $\partial_\mu F_{\alpha\beta}+ \partial_\alpha F_{\beta\mu}+ \partial_\beta F_{\mu\alpha} =0$, can be compactly arrange with a four-dimensional totally antisymmetric Levi-Civita symbol $$\epsilon^{\alpha\beta\gamma\delta} \partial_\beta F_{\gamma\delta}=0$$
with the property $\epsilon^{0\beta\gamma\delta}=\epsilon^{ijk}$.
Replacing the three-dimensional Levi-Civita symbol we obtain
\begin{align}
0&=-\epsilon^{ijk} \partial_0 F_{jk} + \epsilon^{ijk} \partial_j F_{k0} - \epsilon^{ijk} \partial_j F_{0k}\\
&=\epsilon^{i0jk} \partial_0 F_{jk} + \epsilon^{ij0k} \partial_j F_{k0} + \epsilon^{ijk0} \partial_j F_{0k}\\
&=\epsilon^{i\beta\gamma\delta} \partial_\beta F_{\gamma\delta}\ .
\end{align}
The still missing case $\alpha=0$ follows immediately from the last Maxwell's equation:
\begin{align}
0&=\partial_i B^i\\
&=\epsilon^{ijk} \partial_i F_{jk}\\
&=\epsilon^{0\beta\gamma\delta}\partial_\beta F_{\gamma\delta}\ ,
\end{align}
which eventually reduces to the Bianchi identity.
PS: note the sign for covariant and contravariant indices, e.g. $F_{0i}=\eta_{00}\eta_{ij}F^{0j}=-F^{0i}$.
| {
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Is it possible to visualise red shift? If a picture of a star or galaxy hurtling away from Earth is taken, does it appear red despite it being a different colour? Would a blue coloured star moving away from us appear red to us or vice versa? If so how do scientists understand if say, the red colour of a star is due to it having a cooler surface temperature (red supergiants like betelgeuse) or if it is due to the red shift?
| To detect the redshift of distant objects, we can use the fact that (to the best of our knowledge) the laws of physics are the same everywhere. This means the spectral lines of elements (both absorption and emission) will be the same at the location of the star as they are on earth.
We can measure absorption and emission spectra of elements here on earth and then compare them to the spectra we receive from distant objects. The amount the lines are shifted must then be due to the redshift (or blueshift if the object is relatively close and moving towards us).
You can see this visualised in this picture from Wikipedia
| {
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Conservation of energy and work done by a torque Suppose you let a solid roll down an incline without slipping, from height $h$. My textbook gives the following conservation of energy relation
$$mgh = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2.$$
Why do we not have to include the work done by the static friction (nonconservative force) on the left side? I know it is supposed to do zero work, as there is no motion where it acts, but it is the only force providing a torque and $\omega$ is obviously increasing, so in my opinion it should be doing (rotational) work.
| You're correct it does zero work because before the motion and after the motion the incline is in the same location. There can be no energy extracted from it. The only energy available is the gravitational position energy from the height of the ball.
The rotational energy added to the solid comes at the expense of the kinetic energy it would otherwise have. On a frictionless plane, the solid would have no rotational energy at the bottom, but more kinetic energy. The frictional forces in this problem can affect the energy balance but not the energy total.
| {
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What happens when the universe runs out of fuel? After some X billion years, one would think the stars in the entire universe will run out of hydrogen. What would happen next? Is there any way to get hydrogen out of heavy metals (extreme fission)? Just curious.
| Well, fate of the universe has a lot of possible scenarios, from which periodical expansion/contraction is very unlikely, because universe expands in an accelerated fashion, and it doesn't looks that it will change, unless dark energy will run-out too, which is highly unlikely due to quantum mechanical laws. So most probable scenarios of fate would be Big Rip and Big Freeze. As for Big Freeze,- it's about thermodynamic equilibrium, when all supermassive black holes will finally evaporate due to Hawking radiation in about $10^{100}\text{years}$ and universe will become a cold soup of sparse gas of photons and leptons. The good thing is that there's a probability that some place in this "particle fog" can quantum tunnel into a new inflating universe, after about $\Large {10^{10^{10^{56}}}}$$\text{years}$. Btw, keep in mind that this is not the same as repeating inflation model, it's more like a multiverse theory,- spawning child branches of new universes in some "root tree" of process.
| {
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Partition Function Question I've been looking at calculating the Internal energy of a non-isothermic van de Waals gas, and in doing so have been researching the free energy and hence the partition function necessary to calculate that.
The problem is that the partition function cannot be computed in any formulae that I have come across due to the very large value of $N$ being ~ 1e^20 particles. This means that the free energy and hence the internal energy cannot be calculated, and I cannot solve my problem.
For example using the formula:$$Z = \frac{V^N}{N!{h^{3N}}}{\left({\frac{2{\pi}m}{\beta}}\right)}^{\frac{3N}{2}}$$
There are several incomputable terms which must be approximated. The term $N!$ can be approximated using Stirling's approximation but even in that form it still contains a term that is to the power of N just like many other terms, which cannot be computed and must be further approximated. For example, approximating the term $h^{3N}$, it can be seen to approach zero when N approaches $\infty$. This term alone for a large number of particles N ~ 1e^20 seems to completely break the equation and the same for the terms ${\left({\frac{2{\pi}m}{\beta}}\right)}^{\frac{3N}{2}}$ and $V^N$ when approximated.
My question is how do scientists, engineers, mathematicians, etc, actually compute this partition function for a large number of particles, I have seen that it is possible in many papers but I have never seen a successful substitution of values, and have clearly never been able to successfully complete it myself.
| In my experience, any useful thermodynamic quantities involve taking the natural log of $Z$ rather than the absolute value of $Z$ itself. For example the Helmholtz free energy is
$$F=-kT\ln Z$$
and internal energy is
$$U=kT^2\frac{\partial\ln Z}{\partial T}$$
It's fine that $Z$ is usually incommputably large. It is a measure of the number of microstates available to a system, and for most systems, this should be an incredibly large number (practically infinite). It must be this way if things like the second law of thermodynamics are to be obeyed!
| {
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Complex Coupling Strength in Light-Matter Interaction Hamiltonian The quantised electric field operator is given by :
$$ \hat{\mathbf{E}}(\mathbf{r},t) = i\sum_{\xi}E_{\xi}\left(\mathbf{u}_{\xi}(\mathbf{r})a_{\xi}-\mathbf{u}^*(\mathbf{r})a_{\xi}^{†}\right) $$
where the complex pre-factor is sometimes absorbed into the mode function $\mathbf{u}(\mathbf{r})$ such that $\mathbf{f}(\mathbf{r}) = i\mathbf{u}(\mathbf{r})$ and therefore:
$$ \hat{\mathbf{E}}(\mathbf{r},t) = \sum_{\xi}E_{\xi}\left(\mathbf{f}_{\xi}(\mathbf{r})a_{\xi} + \mathbf{f}^*(\mathbf{r})a_{\xi}^{†}\right) $$
The interaction of a two level quantum system with a single mode of the electric field is then given by $\mathbf{D}\cdot\mathbf{E}(\mathbf{r}_0,t)$ where $\mathbf{D} = \mathbf{D}_{12}\left(\sigma_+ + \sigma_-\right)$ is the dipole operator. The overall interaction can then be written as:
$$ \mathbf{D}\cdot\mathbf{E}(\mathbf{r}_0,t) = \left(\sigma_++\sigma_-\right) \left(g_{\xi}a_{\xi} + g_{\xi}^*a^{†}_{\xi}\right)$$
where $g_{\xi}=E_{\xi}\mathbf{D}_{12}\cdot\mathbf{f}(\mathbf{r}_0) = iE_{\xi}\mathbf{D}_{12}\cdot\mathbf{u}(\mathbf{r}_0)$. If the mode is assumed to be a plane wave, $e^{i\mathbf{k}\cdot\mathbf{r}}$, in the same direction as the dipole moment and $\mathbf{r}_0=0$ then $g_{\xi}=iE_{\xi}|\mathbf{D}_{12}|$.
However, $g_{\xi}$ is usually assumed to be real, so that the interaction hamiltonian can be written in the form $g_{\xi}\left(\sigma_+ + \sigma_-\right)\left(a_{\xi} + a_{\xi}^†\right)$. This doesn't seem to be true here, despite being used all over the place.
I guess my questions are:
(1) Is there a mistake in my logic?
(2) If not, is there a reason the factor of $i$ can be ignored in $g_{\xi}$ and are there any consequence to this?
(3) Also, any other information related to this question would really be appreciated.
| You can redefine the ladder operators to absorb the phase. The mapping
$$a \rightarrow e^{i\phi} a$$
is canonical, since it preserves the commutation relations.
So there is no mistake in your logic, but the dynamics only depend trivially on the phase. Note, however, that your Hamiltonian may need to be adapted in certain cases, such as for complex dipole moments.
| {
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Is it possible the Black Holes to be pure deformations in the fabric of spacetime and not an effect of super-dense matter? Is there any theory in the literature that supports this hypothesis that BHs in their center do not have a super-dense matter singularity but are pure deformations in the fabric of spacetime itself or vacuum space, possible caused after supernova or other violent event or maybe preexisted as features or defects of spacetime or vacuum space long before any matter creation in the Universe?
|
Is there any theory in the literature that supports this hypothesis that BHs in their center do not have a super dense matter singularity but are pure deformations in the fabric of spacetime itself or vacuum space
It's the General Theory of Relativity. The Schwarzschild solution for non-rotation, non-charged black holes is a vacuum solution, that is, the space is empty (vacuum): There's no matter and no radiation and no electric fields etc. anywhere.
The singularity itself is not a part of such spacetimes / solutions of Einstein's field equations. This means it is not correct and it makes no sense so say "there's a singularity of inifitit dense matter at the center of a BH", because the center is not a part of the space-time.
| {
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Non-parallel light diffraction Does light (and in general any kind of wave) diffract only when the wave fronts are parallel? Like if you did the double slit experiment when the waves were coming from a point source close to the diffraction grating, would the light still diffract?
| Yes it does. A common version on the double slit experiment is, in fact, the one you describe. You can, for instance, install a narrow hole with the primary light source behind it. This hole will act as the source for the experiment, and you can put it as close as you like from the double slit, so the wavefront will be more or less spherical.
If you're talking about more complex diffraction gratings, typically with thousands of slit instead of just two, things are a bit more complicated. Diffraction happens no matter the shape of the wavefront, but interferences behind the grating are a mess with non-parallel light. So, for practical reasons, a source of parallel light is usually used.
| {
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Introduction to magnetohydrodynamics does anybody have any reference books for introduction to magnetohydrodynamics? I want to dive into this topic and I don´t know about any good reference.
| If you want a very quick introduction to Magnetohydrodynamics (MHD) I would recommend Reitz, Milford and Christie's 'Foundations of Electromagnetic Theory'.
If you are willing to take a leisurely path that will build your foundations you can choose Arnab Rai Choudhuri's 'Physics of Fluids and Plasmas'. The author does a phenomenal job in starting from the microscopic equations of motion and building the equations of neutral fluids and plasmas. The book has two sections, divided evenly between neutral fluids and plasmas with plenty of examples from astronomy to illustrate the principles.
More classic works would be the books by a) T G Cowling (Magnetohydrodynamics), b) Lymon Spitzer (Physics of Fully Ionized Gases) and c) S. Chandrasekhar (Plasma Physics).
Of these, only the first book uses SI units.
| {
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Angular Momentum and Coefficient of Restitution If there was a situation where you had two rods pinned in the center and the left rod having an initial angular velocity $\omega_1$, and the right rod was at rest. I am wondering what the final angular velocities would be if there was a coefficient of restitution during the impact between the rods, $e = 0.8$, i.e. can you apply the traditional coefficient of restitution relation for angular velocities?
Does $$e = \frac{\omega_2' - \omega_1'}{\omega_2 - \omega_1}$$ still apply?
Then you could use the conservation of the total angular momentum equation to find the angular velocity of each rod where
$$ H_1 + H_2 = H_1' + H_2' $$
$$ I_1\omega_1 = I_1\omega_1' + I_2\omega_2' $$
| The problem is that angular momentum is not conserved in this scenario. In order to define angular momentum in a system, there needs to be an origin. If we put the origin at the center of the first rod, $M_1$, then it is clear that any force that the pin at $M_1$ exerts does not result in any torque in the system. However, we then have a problem because $M_2$ will exert a torque on the system; the pin of the second rod holds it in place and exerts forces so that it doesn't fly upwards; instead, it only rotates.
| {
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Does the standing wave equation proof require $\ell=Nλ$? Consider two identical sources $S_1$ and $S_2$ of waves, separated by a distance $\ell$ (as shown in the figure).
The sources produce waves in opposite directions(and towards each other). Now, suppose we wish to derive the equation for the standing wave produced.
Let us take the origin at $S_1$. The equation of the wave due to $S_1$ is:-
$$ y_1=A\sin(wt-kx_1)$$
where $x_1$ is the distance from $S_1$.
Now the equation of the wave due to $S_2$ is:-
$$ y_2=A\sin(wt-kx_2)$$
where $x_2$ is the distance from $S_2$. Note that here we take $x_2>0$.
Now, applying the principle of superposition, we get:-
$$ y=y_1+y_2=A\sin(wt-kx_1)+A\sin(wt-kx_2)$$
Now, observe $x_2=\ell-x_1$, so we get:-
$$ y=y_1+y_2=A\sin(wt-kx_1)+A\sin(wt-k(\ell-x_1))=A\sin(wt-kx_1)+A\sin(wt+kx_1-k\ell)$$
Using $\sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{D-C}{2}\right)$, we get:-
$$y=2A \cos\left(kx-\frac {k\ell}{2}\right)\sin\left(wt-\frac{k\ell}{2}\right)$$
Note, that here we replace $x_1$ with $x$ since $x=x_1$ as the origin is at $S_1$.
However, the standard equation of stationary waves in such a case is given as $y=2A \cos(kx)\sin(wt)$. Using the equation we just derived, $\ell$ must =Nλ (where N is a non-negative integer) so that $k\ell=\frac{2π}{λ}.Nλ=2Nπ$ and now,
$$y=2A \cos\left(kx-\frac {2Nπ}{2}\right)\sin\left(wt-\frac{2Nπ}{2}\right)=2A (-\cos(kx))(-\sin(wt))=2A\cos(kx)\sin(wt))$$ as required.
Therefore, our proof of the standard equation of stationary waves requires $\ell=Nλ$. However, looking at the proof in my textbook, there is no discussion of this and the author uses $y=y_1+y_2=A\sin(wt-kx)+A\sin(wt+kx)$. A superficial explanation is given that a wave traveling in the opposite direction is taken with a '+kx' with no deeper reasoning. Although, my proof helps explain the origin of the '+kx' term (in the wave traveling in opposite direction), it adds a rather restrictive condition i.e. $\ell=Nλ$. Thus, I am skeptical about it.
Please help me out with two things:-
*
*Is my proof correct? If not, then what is the reasoning behind the '+kx'?
*If my proof is correct, then why isn't the $\ell=Nλ$ condition more commonly seen?
| I think that you're missing boundary conditions. Anytime you want to solve partial derivative equations, you have to provide boundary conditions, or the solution is undefined.
To get a standing wave, the most commun situation is to use "strict boundary conditions", where the wave's amplitude has to be zero on both extremities. If you take $y(x,t)=2A\cos(\omega t)\cos(kx)$ with $y(0,t)=y(l,t)=0$, you naturally get the quantization condition similar to the one you mentioned.
Another possibility is "periodic boundary conditions": the wave's amplitude must be equals on both ends, but not necessarily zero.
So far it's mostly math, so choosing the proper boundary conditions is where physics starts.
| {
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Does gravitation really exist at the particle level? As I understand, we usually talk about gravity at a macro scale, with "objects" and their "centre(s) of mass". However, since gravity is a property of mass generally (at least under the classical interpretation), it should therefore apply to individual mass-carrying particles as well.
Has this ever been shown experimentally? For example, isolating two particles in some manner and then observing an attraction between them not explained by other forces.
To pose the question another way, let's say I have a hypothesis that gravitation is only an emergent property of larger systems, and our known equations only apply to systems above some lower bound in size. Is there any experiment that reasonably disproves this hypothesis?
| Here is an easy way to grasp how difficult it would be to make the direct measurement you propose.
Take two protons and place them one centimeter apart. They will exert a certain tiny amount of gravitational attraction, which we measure by some magic means, and a certain amount of electrostatic repulsion, which we will also measure.
Now, how far apart would we need to separate those two protons in order for the strength of the electrostatic force they experience to diminish to the point where it is as weak as the gravitational force they experience when they are one centimeter apart? Answer: 1.8 light years.
This means that when performing experiments where we have to account for electrostatic forces between individual subatomic particles, those forces will be stronger than the gravitational forces between them by a factor of (1.8 light years/1 centimeter).
And that means that we have no hope of ever, ever directly measuring the force of gravitational attraction between two protons in an experiment: the electrostatic force will utterly overwhelm that experiment.
| {
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What is the instant velocity? The velocity is the variation rate of the position correct? So does it make sense to talk about velocity without time?
| Average speed is defined as passed-distance-over-passed-time:
$$v_\text{average}=\frac{\Delta s}{\Delta t}.$$
In other words, choose a point on your path. Then choose one more point. Plug in the difference in distance and time between them.
If your speed varies in-between those two chosen points, then the average speed will not show it clearly. Any variation - any information - is "lost", so to say, in-between the chosen points. For a more accurate speed that more accurately follows the actual speed variation during the trip, we can thus simply move the two points closer (make measurements more quickly after one another).
Soon we have moved the points so close together that we basically can't distinguish them - so close that they to all practical purposes overlap so that the two points appear as just one point. The average speed between them can then be thought of as the speed in just that point.
In that specific case we might then rename the average speed to instantaneous speed. We might even invent new notation for this specific situation:
$$v_\text{instantaneous}=\frac{\mathrm ds}{\mathrm dt}.$$
When we say instantaneous, we are idealising this scenario. We are imagining that the "average" indeed is found in just a point. Mathematically we might say that we let the two points go towards each other (in fact we let the moments in time go towards each other which means that their difference goes towards zero) so that the value of average speed goes towards the value of instantaneous speed at its limit:
$$v_\text{instantaneous}=\lim_{\Delta t\to 0}\left(\frac{\Delta s}{\Delta t}\right).$$
When considering what your speedometer shows in your car, you are right that we in reality never can measure instantaneous speed with perfect precision. But as long as the measurements happen over a very short period of time, then it counts as instantaneous to all practical purposes.
| {
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Can't understand a statement about motion From the book where I am studying motion, It says
Motion is a combined property of the object under study and the observer. There is no meaning of rest or motion without the viewer.
I know that, for an object, it can be said that 'it is moving' in one frame of reference, and it can be said that 'it is at rest' in another frame of reference, but the sentence I mentioned above seems somewhat confusing. How can a phenomena be a property of two things? Also, how is it that, when there is no one to see, the topic of motion and rest is irrelevant? I don't know exactly what the second sentence is trying to say, provided that my understanding of the second sentence is wrong. I need assistance.
| I think the second sentence is nonsense if taken literally. The temperature and pressure of the air in my room is the result of the kinetic energy- ie the motion- of the air molecules. You do not need to posit the existence of an observer to understand that.
| {
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How does a sliding object stop moving by the effect of kinetic friction, if kinetic friction is constant and Fk can't be greater than Fapp on its own The kinetic friction is constant. So if say, I apply a force of 20N on an object and the kinetic friction is 13N, then how does the object stop moving after some seconds.
I believe that kinetic friction is constant, it cannot increase more than the applied force by changing its value on its own like static friction?
So does the object stop because the effect of applied force decreases over time? How does this happen due to kinetic friction (keeping in mind kinetic friction is constant - has a fixed value)? How does this work?
How does an object remain balanced on a frictionless surface and keep moving with a constant velocity without any friction to balance the forces (the resultant net force is equal to 0.)
| There are very tiny bumps and ridges along the surfaces of contact.
What we call "Force of friction" is a sort of overall estimate on a large-scale of how all these bumps colliding with other surface bumps are causing deceleration.
The coefficient of friction of surfaces has been proven experimentally.
But what's going on are really tiny bumps on the surface are colliding with each other which slowly decelerate the object and this deceleration decreases.
So you're losing momentum overtime.
| {
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Absence of velocity in energy conservation
Here we see a rod at rest hinged about a point. We want to know the angular speed of the rod when it becomes vertical as shown in the figure.
The solution which is given in the books goes like this.
Taking the black line as reference level,they say that initially,the rod had only potential energy which is $Mg\frac{L}{2}$. And they say that the final energy which will be present is $\frac{1}{2}I\omega^2$. And then they establish the equation $Mg\frac{L}{2}=\frac{1}{2}I\omega^2$. But I don't understand one thing. Since the center of mass of the rod here is moving,it surely has a linear velocity $v$. As we see in transrotational motion,where we use $\frac{1}{2}mv_{\mathrm{CM}}^2+\frac{1}{2}I_{\mathrm{CM}}\omega^2$. Why hasn't it been used in this case? Shouldn't the equation be $Mg\frac{L}{2}=\frac{1}{2}I\omega^2+\frac{1}{2}Mv^2$ where $v$ is the linear velocity of center of mass?
| When you're using the moment of inertia, you have to specify a reference point or axis. Then $\frac{1}{2}I\omega^2$ is the kinetic energy of the system in a rotation around that point.
So you have two equivalent ways to procede:
*
*Use $I_G$, the moment of inertia with the center of mass as a reference point. Since this point moves, then yes, it happens as you describe, you have to add $\frac{1}{2}mv(G)^2$ to the kinetic energy.
*Use $I$, the moment of inertia with the extremity of the bar as a reference point. Since this point doesn't move, you don't add anything to the kinetic energy.
The relation between $I$ and $I_G$ is exactly given by expressing the same kinetic energy with both methods. It's the parallel axis theorem.
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What is the plane of a concave or convex mirror? What is the plane of a concave or convex mirror? Is it along its principal axis or its area? I am convinced that the plane should be along the area of mirror but i am not sure about it. I tried to find it in books but didn't got any answer. Some books have questions like a point object is moving in a circle in a perpendicular plane to the mirror so find radius of image. Such questions are confusing to solve if we don't know where would be the plane of mirror.
| There is no such particular definition for plane of mirror, in fact this term is very ambiguous since spherical mirrors aren't planar. "Plane of mirror" - these words are just used to express questions (like plane of paper or plane of the plane mirror). Physics doesn't defines that this particular plane is called the "plane of mirror".
In your question-
Some books have questions like a point object is moving in a circle in a perpendicular plane to the mirror so find radius of image.
The plane of motion of the point object is perpendicular to the Principle Axis of the mirror.
Generally, we take principle axis of the mirror as our reference for all calculations and observations.
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Showing $[\hat{A},\hat{B}] = i\mathbb{1} \Leftrightarrow [\hat{A},\hat{B}^{\textstyle n}] = i\,n\,B^{\textstyle n-1}$ Actually it is self redundant to show having in mind $[\hat{A},\hat{B}^{\textstyle n}] = \,n\,B^{\textstyle n-1}\,[\hat{A},\hat{B}]$ but I suppose it is not supposed to solve it this way. Instead I guess the primal relation $[\hat{A},\hat{B}] = i\mathbb{1}$ has to be exploited. Expanding is not getting me further:
$(1) \quad[\hat{A},\hat{B}] = \hat{A}\,\hat{B}-\hat{B}\,\hat{A} = i\,\mathbb{1}$
$(2)\quad[\hat{A},\hat{B}^{\textstyle n}] = \hat{A}\,\hat{B}^{\textstyle n}- \hat{B}^{\textstyle n}\,\hat{A} = i\,n\,B^{\textstyle n-1}$
Briefly how the one follows from the other?
| Note that $-\hat{B}^n \hat{A} = -\hat{B}^{n-1}\hat{B}\hat{A} = \hat{B}^{n-1}(i\mathbb{1}-\hat{A}\hat{B})$ and then you can continue with
$-\hat{B}^{n-1}\hat{A}\hat{B}=-\hat{B}^{n-2}\hat{B}\hat{A}\hat{B}=\hat{B}^{n-2}(i\mathbb{1}-\hat{A}\hat{B})\hat{B}$.
Can you see the pattern?
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Eyes shut, can a passenger tell if they’re facing the front or rear of the train? Suppose you’re a passenger sitting in one of the carriages of a train which is travelling at a high, fairly steady speed. Your eyes are shut and you have no recollection of getting on the train or the direction of the train’s acceleration from stationary. Can you tell whether you’re facing the front or the back of the train?
This isn’t a theoretically perfect environment - there are undulations, bends and bumps in the track. Not a trick question - you cannot ask a fellow passenger!
Edit: This is intentionally lacks rigorous constraints. Do make additional assumptions if it enables a novel answer.
| The simple answer is no, you would not be able to know. In fact you would not even be able to know that you are traveling at a constant speed. To you, the train could be completely still (assuming that the environment is perfect. Otherwise you could tell by for example sound). Even when the train starts to accelerate or decelerate you would still not be able to know which direction you are traveling. You could only know if someone tell you "the train is now decelerating" and by then observing how your body would react that you could tell which direction you are facing. For example, someone tell you "The train is now accelerating" and you notice that your body pushes into the seat (resisting the movement of the train accelerating 'forward') then you would know that you are facing the direction that the train are traveling in.
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The effective action in the linear sigma model I am reading the section 11.4 of Peskin and Schroeder's book (page 373), and there is a step I could not follow.
To calculate the effective action of linear sigma model, the determinant of $\frac{\delta^{2}\mathcal{L}}{\delta\phi^{i}\delta\phi^{j}}$, and in peskin's book, it shows that
$$\frac{\delta^2\mathcal{L}}{\delta\phi^i\delta\phi^{j}}=-\partial^2\delta^{ij}+\mu^2\delta^{ij} - \lambda[(\phi_{\mathrm{cl}}^k)^2\delta^{ij} + 2\phi_{\mathrm{cl}}^i\phi_{\mathrm{cl}}^j].\tag{11.67}$$
Then, orient the coordinates to make the $\phi^{i}_{cl}$ points in the Nth direction,
$$\phi^{i}_{cl}=(0,0,...,0,\phi_{cl}),\tag{11.68}$$
and the operator in the RHS of the first equation is just a KG operator $(-\partial^{2}-m^{2}_i),$ where
$$m^2_i=\begin{cases} \lambda\phi^2_{cl}-\mu^2\quad acting\; on\;\eta^1,...,\eta^{N-1};\\3\lambda\phi^2_{cl}-\mu^2\quad acting\; on\;\eta^N.\end{cases}\tag{11.69}$$
I think for all $\eta^i$, the value of $m^2_i$ should equal to $3\lambda\phi^2_{cl}-\mu^2$, how the $\lambda\phi^2_{cl}-\mu^2$ is obtained?
| Simply read the diagonal N×N matrix $$ -\partial^2\delta^{ij}+\mu^2\delta^{ij} - \lambda[ v^2 \delta^{ij} + 2\phi_{\mathrm{cl}}^i\phi_{\mathrm{cl}}^j],
\tag{11.67}$$
where, to allay your confusion, I call $\phi_{cl}=v$, so $(\phi_{\mathrm{cl}}^k)^2=v^2$.
The non-gradient part of this diagonal matrix is, for $i,j= 1...,N-1$,
$$(\mu^2-\lambda v^2, \mu^2-\lambda v^2,...,\mu^2-\lambda v^2),$$
whereas the last entry (only!) in the diagonal is
$$
\mu^2-\lambda v^2-2\lambda v^2= \mu^2-3\lambda v^2.
$$
| {
"language": "en",
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Does it take light longer to reach me if I'm moving away? I'm having trouble grasping the intuitions behind the second postulate of special relativity, particularly what it implies.
*
*For example, imagine that a laser pointer is aimed at me at 1 lightsecond away. Then, I move away from it at a speed of 0.5c. Does the laser still take 1 second to reach me?
I presume the answer is yes. In that case, consider the following case:
*A moves at 0.5c and B is stationary. A shoots a laser from its frame of reference where it is at rest, and measures the time it took the light to travel 1 lightsecond in front of A. This time is also measured by B, but in his frame of reference the light travels a longer distance because A is also moving (thus to be ''1 lightsecond in front of A'' requires light to travel more than 1 lightsecond). Would they measure the same time?
If the answer to question 1 is true, that would mean the answer to question 2 is also true (correct me if this is a non-sequitur), which violates the second postulate of special relativity.
I think I have some gaps in my understanding.
|
For example, imagine that a laser pointer is aimed at me at 1 lightsecond away. Then, I move away from it at a speed of 0.5c. Does the laser still take 1 second to reach me?
It takes 2 seconds according your buddy that you leave behind standing still. Because the buddy sees the distance shrinking at rate 0.5 c.
According to said buddy during those 2 seconds your clock has proceeded 2 seconds multiplied by time dilation factor, so it took 1.732 seconds of time according your clock. You will agree that it took 1.732 second.
As some extra information let me tell that the distance between the point were the beam was emitted and the point were the beam was absorbed is 1.732 lightseconds according to you.
| {
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Could a glass cup containing a vacuum rise into the air? https://what-if.xkcd.com/6/ has been mentioned here before, but I'm questioning whether or not the glass cup with the bottom half as a vacuum would rise at all.
To start with, a vacuum exerts no force. Any perceived "sucking" is actually external pressure pushing into the vacuum. So the only force that could be lifting the glass would be buoyancy of the air around the cup.
Let's ballpark it with a drinking cup that can hold about 500ml. If we consider the cup as an open-top cylinder, an internal radius of 3.8 cm and height of 11 cm gets us a 499ml volume, and the internal surface area is 308 cm^2. Based on a quick Google search, glass is about 2000x more dense than air, so in order for the glass to rise it would need to displace 2000x as much air as there is glass. That suggests that if the glass was sealed (by a weightless forcefield at the top) and was completely empty instead of having some water in it, the total volume of glass would have to be less than 0.5ml, resulting in an average width of 0.016mm. That's thinner than a human hair.
Given that glass cups are significantly thicker than human hair1, is there any truth to the conclusions of that "What If?" Is there some effect that I've misunderstood or underestimated that significantly changes the situation? Or should we conclude, like https://physics.stackexchange.com/a/33642/79374 did with the other vacuum cup, that Randall Munroe either miscalculated or was greatly exaggerating?
1 Citation needed
| It's not just buoyancy because the slug of liquid can move. Your "force field" at the top of the glass modifies the premise of the XKCD scenario. If you look closely at the illustrations, his slug of liquid moves downward as the glass moves upward. The atmosphere pushes down on the slug of liquid and up on the glass, moving them both. But the center of mass of the system would not move.
| {
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Is sand in a vacuum a good thermal insulator? My reason for thinking that sand in a vacuum would be a good insulator is that heat cannot be conducted in a vacuum, and the area of contact between adjacent grains of sand is very small, which means heat would transfer between grains relatively slowly. Is this correct, or is there something I'm missing?
Also, the sand is there instead of pure vacuum for structural support.
| Powder filled vacuum (I don't know if sand per se is used for this purpose) is in fact used in cryogenic insulation, and can be better than vacuum alone, because vacuum alone is susceptible to radiant heat transfer and also a type of heat conduction that happens because the mean free path of the remaining molecules (in the less than perfect vacuum) is large, so the individual molecules can bounce between the inner and outer walls like ping pong balls, transferring heat. The powder interrupts these transfer modes, while having a very low conductivity itself.
| {
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Matrix form of fermionic creation and annihilation operators in two-level system I'm trying to find the matrix form of fermionic creation and annihilation operators in two-level systems from this text. I understand that for one site, the operators take the form:
$$
f_{0}=\left(\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right), \quad f_{0}^{\dagger}=\left(\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right),
$$
where
$$
\begin{aligned}
f_{0}|1\rangle &=|0\rangle = \left(\begin{array}{l}
1 \\
0
\end{array}\right), & f_{0}|0\rangle=0 \\
f_{0}^{\dagger}|1\rangle &=0, & f_{0}^{\dagger}|0\rangle=|1\rangle=\left(\begin{array}{ll}
0 \\
1
\end{array}\right)
\end{aligned}
$$
For two sites, I was able to deduce
$$
f_{0}^{\dagger}=\left(\begin{array}{l11l}
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\end{array}\right), \quad f_{0}=\left(\begin{array}{l11l}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{array}\right),
f_{1}^{\dagger}=\left(\begin{array}{l11l}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
\end{array}\right), \quad f_{1}=\left(\begin{array}{l11l}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}\right),
$$
which allow for these rules as indicated by the text
$$
\begin{aligned}
f_{0}^{\dagger}|0,0\rangle &=|1,0\rangle ; \quad f_{0}^{\dagger}|1,0\rangle=0 \\
f_{0}|1,0\rangle &=|0,0\rangle ; \quad f_{0}|0,0\rangle=0 \\
f_{0}|0,1\rangle &=0 ; \quad f_{0}^{\dagger}|1,1\rangle=0 \\
f_{1}^{\dagger}|0,0\rangle &=|0,1\rangle ; \quad f_{1}|0,0\rangle=f_{1}|1,0\rangle=0 \\
f_{1}^{\dagger}|1,0\rangle &=-|1,1\rangle ; \quad f_{1}|0,1\rangle=|0,0\rangle \\
f_{1}^{\dagger}|0,1\rangle &=f_{1}^{\dagger}|1,1\rangle=0 ; \quad f_{1}|1,1\rangle=-|1,0\rangle\\
f_{0}^{\dagger}|0,1\rangle &=|1,1\rangle ; \quad f_{0}|1,1\rangle=|0,1\rangle
\end{aligned}
$$
where $|0,0\rangle =\left(\begin{array}{l}
1 \\
0 \\
0 \\
0 \\
\end{array}\right), |1,0\rangle = \left(\begin{array}{l}
0 \\
1 \\
0 \\
0 \\
\end{array}\right), |0,1\rangle = \left(\begin{array}{l}
0 \\
0 \\
1 \\
0 \\
\end{array}\right), |1,1\rangle = \left(\begin{array}{l}
0 \\
0 \\
0 \\
1 \\
\end{array}\right)$.
My question is: am I thinking about this the right way? And what is the general formula of the operators for when there are $n$ sites instead? Is there some material that discusses this? Thank you!
| Briefly: you have to order your sites and add a string $\eta_{\alpha}$ of operators in front of the creation and annihilation operators
$$
\overline{f}_{\alpha}=\eta_{\alpha}f_{\alpha}, \qquad \overline{f}_{\alpha}^{\dagger}=\eta_{\alpha}f_{\alpha}^{\dagger}, \qquad \eta_{\alpha}=\prod_{\beta=1}^{\alpha-1}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}_{\beta}
$$
The point is that your single site operators $f_{\alpha}=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}_{\alpha}$ and $f_{\alpha}^{\dagger}$, obey the right anticommutation rules on the site $\alpha$, but they commute of the on different sites.
You can see that $\overline{f}_{\alpha}$ and $\overline{f}_{\alpha}^{\dagger}$ , thanks to the string $\eta_{\alpha}$ we have attached to them, obey the right anticommutation relations
$$
\{\overline{f}_{\alpha}^{\dagger}, \overline{f}_{\beta}\} = \delta_{\alpha\beta} \qquad \{\overline{f}_{\alpha}, \overline{f}_{\beta}\}=0
$$
This implies that when we costruct a state from the vacuum
$$
|\alpha,\beta,\gamma\rangle = \overline{f}_{\alpha}^{\dagger}\overline{f}_{\beta}^{\dagger}\overline{f}_{\gamma}^{\dagger}|0\rangle
$$
this is antisymmetric under exchange of two indices: that is what we want from a fermionic state.
| {
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Points where electric field is zero when charges are present at vertices of a regular polygon There is a $n$-sided regular polygon with a charge $q$ at each vertex. I know that there are $n$ points, other the center of the polygon, where the electric field is zero. But why is this so? Is there a general way to prove it?
PS: I know some questions related to my question have been asked, but none of them gives me a satisfactory reason why there should be a total of $n+1$ neutral points in space for such a charge distribution.
| *
*For what it's worth, 2D Morse theory (with the assumption that all critical points for the electric potential are non-degenerate) yields that
$$c_1-c_0~=~n-1\qquad\text{and}\qquad c_2~=~0,$$
where
$$\begin{align} c_0~:=~& \#{\rm minimum~pts}, \cr
c_1~:=~& \#{\rm saddle~pts}, \cr
c_2~:=~& \#{\rm maximum~pts},\end{align}$$
cf. e.g. my Phys.SE answer here.
*So under these assumptions$^1$ OP's claim for the regular $n$-polygon$^2$
$$\begin{align}V(z)~=~&\sum_{j=1}^n \left|z-\exp\frac{2\pi i j}{n}\right|^{-1}\cr
~=~&\sum_{j=1}^n \left(\left(x-\cos\frac{2\pi j}{n}\right)^2+\left(y-\sin\frac{2\pi j}{n}\right)^2\right)^{-1/2}\cr
~=~&\sum_{j=1}^n\left(1+r^2-2r\cos\left(\theta-\frac{2\pi j}{n}\right)\right)^{-1/2}
\end{align} $$
that the number of critical points are $c_2+c_1+c_0=n+1$ would follow if we can show that $c_0=1$, i.e. that the center is the only local minimum.
$^1$ It is not difficult to see that the center $r=0$ is a non-degenerate critical point:
$$ V(r,\theta)~=~\left\{
\begin{array}{rl}
n\left(1+\frac{r^2}{4}\right) +{\cal O}(r^3) &\text{minimum pt for } n\geq 3,\cr 2+\frac{r^2}{2}\left(1+3\cos 2\theta\right)+{\cal O}(r^3)&\text{saddle pt for } n=2. \end{array}\right. $$
$^2$ We assume that $n\geq 3$. If $n=1$, then $c_0=0=c_1$. If $n=2$, then $c_0=0$ and $c_1=1$.
| {
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How are these Covariant Derivative Identities found? In David Tong's Gauge Theory notes on page 137 near eq. (3.30) he makes use of the following expressions for the covariant derivative $D_{\mu}$
$$\frac{1}{2}[\gamma^{\mu},\gamma^{\nu}]D_{\mu}D_{\nu}=\frac{1}{4}[\gamma^{\mu},\gamma^{\nu}][D_{\mu},D_{\nu}]\tag{1}$$
and
$$e^{-ikx}e^{D^2}e^{ikx}=e^{(D_{\mu}+ik_{\mu})^2}\tag{2}$$
I'm guessing the first is just a change of dummy indices in the second term of the commutator, but I don't see how the indices are dummy.
The second expression I'm more confused about. It looks like $x^{\mu}$ is acting like a generator of translation in momentum space, but I'm not sure.
| By Leibnitz' rule
$$
e^{-ikx} \partial_x \{e^{ikx} f(x)\} = e^{-ikx}\{f(x)(\partial_x e^{ikx})+ e^{ikx}(\partial_x f)\}\\
= e^{-ikx}\{ f(x) (ik e^{ikx})+ e^{ikx}(\partial_x f)\}\\
= ik f(x) + \partial_x f(x)
=(\partial_x +ik)f(x).
$$
As $f(x)$ can be anything, we have
$$
e^{-ikx}\partial_x e^{ikx}= \partial_x+ik.
$$
| {
"language": "en",
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Why do not all the semimetals turn into the excitonic insulators? I want to know the condition for forming the excitonic insulator.
When the binding energy of the exciton, $E_b$, overcomes the band gap, the system becomes the excitonic insulator. If so, all semimetals should meet this condition and be supposed to turn into excitonic insulators, like 1T-TiSe2.
Why isn't it thought so? Is there what I must take into account?
|
When the binding energy of the exciton, Eb, overcomes the band gap, the system becomes the excitonic insulator.
What is meant here is the direct gap, i.e., the gap between the conduction and valence band states located at the same point of the Brillouin zone (having the same quasi-momentum). On the other hand, in most semimetals the band overlap happens for the states having very different quasi-momenta (se image below, image source is here) - that is, the gap for the electron and the hole bound into an exciton might be still bigger than the binding energy.
| {
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On the equivalence of the Schrödinger and Heisenberg (and all other) pictures The Schrödinger and Heisenberg (and, indeed, infinitely many other pictures) are often referred to as equivalent descriptions of quantum dynamics in a given system. I'm wondering two things in particular:
*
*What exactly do we mean by equivalent?
*What is the proof of their equivalence?
Elaborating briefly now, I think that the answer to (1) is simply that they reproduce the same predictions (i.e. probabilities) for all observable experiments. That then begets question number (2); how does one prove that this is indeed the case? In particular, it's usually shown (quite simply) in textbooks that expectation values are preserved under different shifts to different pictures. However, that is not sufficient to say that they are equivalent. Is there a most general statement/proof of this commonly made (and admittedly intuitively expected) claim?
| (1) Indeed, you say it correctly: both representations ("pictures") are equivalent because there is a one-to-one correspondence between their predictions. More precisely, there is an isomorphism whereby any solution in the Schrödinger picture admits one solution and only one solution in the Heisenberg picture.
(2) Regarding the proof, it is based on the uniqueness of the solution of the linear differential equations as in the case of the Schrödinger equation. If $|\psi(t,\boldsymbol{x})\rangle$ is a solution of the Schrödinger equation with given initial conditions, then it follows that $\hat{U}(t)|\psi(0,\boldsymbol{x})\rangle$ is also a solution of the Schrödinger equation with the same initial conditions (where $\hat{U}(t)$ is the unitary group associated with the Hamiltonian):
$$\frac{\text{d}\hat{U}}{\text{d}t} = -\frac{i}{\hbar}\hat{H}\hat{U}, \qquad \qquad \hat{U}(0) = \mathbf{1}$$
| {
"language": "en",
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In Hamilton-Jacobi theory, how is the new coordinate $Q$ time-independent when Hamilton's principal function separates? Following the notation in Goldstein, the solution to the Hamilton-Jacobi equation is the generating function $S$ for a canonical transformation from old variables $(q,p)$ to new variables $(Q,P)$ where the new Hamiltonian is $K=0$ and the new momentum is the integration constant $P=\alpha$. Therefore, $\dot{Q}=\dot{P}=0$. If the old Hamiltonian is not a function of time, the solution may be expressed as
$$
S(q,\alpha,t)=W(q,\alpha) - \alpha t.\tag{1}
$$
What's confusing me is that, by definition, $S$ is a type 2 generating function where the new variables are constant in time. However, the new coordinate $Q$ from the type 2 generating function above is expressed as
$$Q=\frac{\partial S}{\partial P} = \frac{\partial S}{\partial \alpha} = \frac{\partial W}{\partial \alpha} - t.\tag{2}$$
Therefore giving
$$\dot{Q} = -1 \neq 0.\tag{3}$$
What am I missing here?
| The culprit of OP's question seems to be that Goldstein considers 2 different generators for a type-2 canonical transformation (CT) in Hamilton-Jacobi theory:
*
*Hamilton's principal function $S(q,P,t)$. Here the Kamiltonian $K\equiv 0$ is zero. Therefore all the new phase space variables $(Q_S^i,P_j)$ are constants of motion (COM).
*Hamilton's characteristic function $W(q,P)$. The latter assumes that the Hamiltonian $H(q,p)$ has no explicit time dependence, and that the Kamiltonian $K(P)$ does not depend on the new coordinates and time $t$. In fact Goldstein in section 10.3 effectively assumes that the Kamiltonian $K(P)=P_1$ is the 1st new momentum $P_1$. As a consequence the 1st new position $Q^1_W$ is identified as time $t$ (up to a possible shift) rather than a COM.
When both methods apply, under the identification
$$S(q,P,t)~=~W(q,P)-P_1t,\tag{A}$$
we conclude that the new positions in the 2 pictures are related as
$$ Q_S^i ~=~ \frac{\partial S}{\partial P_i}
~\stackrel{(A)}{=}~\frac{\partial W}{\partial P_i} -\delta^i_1 t~=~Q_W^i-\delta^i_1 t .\tag{B}$$
TL;DR: The main point is that the 2 CTs produce slightly different sets of new positions $Q_S^i$ and $Q_W^i$, where we for clarity have decorated them with a subscript $S$ and $W$, respectively.
| {
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Spin Connection Vanishes? I'm trying to reproduce a result for the components of the spin connection in FRW spacetime. The formula for the spin connection $\Gamma_{\mu}$ is
$$\Gamma_{\mu} = \frac{1}{2} \Sigma^{a b} L_{a}^{\nu} L_{b \nu, \; \mu}$$
Where $\Sigma^{ab} = \frac{1}{4}[\gamma^{a},\gamma^{b}]$ are Lorentz group generators and $L_{a}^{\mu}$ is the vierbein. If I try to use this formula for the Conformally Flat FRW spacetime with metric $$d s^{2}=a^{2}(t)\left[-d \tau^{2}+\left(d x^{1}\right)^{2}+\left(d x^{2}\right)^{2}+\left(d x^{3}\right)^{2}\right],$$
I know I should get the following answer:
$$\Gamma_{t}=0, \quad \Gamma_{x^{i}}=-\frac{\dot{a}}{2 a} \gamma^{i} \gamma^{0}$$
based on a paper I've been reading https://doi.org/10.1103/PhysRevD.30.2573 (I have access through my University), but I keep getting zero for all components of $\Gamma_{\mu}$ . It seems to me that the problem lies in the fact that the Lorentz group generators are antisymmetric, whilst the vierbeins are symmetric. I don't know how to remedy this, clearly I'm going wrong somewhere. Any advice/direction is welcome!
| I think this is good exercise for me. Please tell me if I made any mistakes.
The vierbein are
$$
L_0^\mu=\frac{1}{a(t)}\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\quad
L_1^\mu=\frac{1}{a(t)}\begin{pmatrix}0\\1\\0\\0\end{pmatrix},\quad...
$$
The nonzero Chrstoffel symbols are
$$
\Gamma^t_{tt}=3\frac{\dot a}{a},\quad
\Gamma^t_{ii}=\frac{\dot a}{a}, \quad
\Gamma^i_{it}=\Gamma^i_{ti}=\frac{\dot a}{a}.
$$
To compute for example $\Gamma_1$, we need $L_{a,1}^\mu$. They are
$$
L_{a=0,1}^\mu=\begin{pmatrix}0\\\dot a/a^2\\0\\0\end{pmatrix},\quad
L_{a=1,1}^\mu=\begin{pmatrix}\dot a/a^2\\0\\0\\0\end{pmatrix},\quad
L_{a=2\text{ or }3,1}^\mu=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}.
$$
So $\Gamma^\nu_a\Gamma_{b\nu,i=1}$ is (you may forgot the minus sign here)
$$
\begin{pmatrix}
0&\dot a/a&0&0\\
-\dot a/a&0&0&0\\
0&0&0&0\\
0&0&0&0
\end{pmatrix}.
$$
After contraction with $\Sigma^{ab}$, this gives the correct answer.
| {
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Relativistic ship and base on Earth talking Let say humans make it possible to travel so close to the speed of light that time dilation for the ship can achieve 100 times the time of a standing still object. So one of that ships starts its journey and after a while something goes wrong and they have to call the base on Earth. The chat consists of two sentences that the pilot on board says in 1 minute. My question is why his sentences the people on the base should listen for almost two hours? To not have problems with signal shift due to ship relative motion regarding Earth let assume the ship is orbiting Earth at a large but constant distance.
| Let's say there are two events:
A. The pilot starts their message
B. The pilot finishes their message
In the rocket frame, which we'll call $S^{\prime}$, the time between those events is $\Delta t_{AB}^{\prime}$ as you've said in the question. To work out the time between the message being first received and ending on Earth, the $S$ frame, we'll need to Lorentz transform.
\begin{equation}
\Delta t_{AB} = \gamma \Delta t_{AB}^{\prime}
\end{equation}
where
\begin{equation}
\begin{split}
\gamma &= \left(1 - \frac{v^{2}}{c^{2}}\right)^{-\frac{1}{2}}
&= 100
\end{split}
\end{equation}
from the information you've given about the relative velocity of the ship and someone standing still on Earth. Therefore $\Delta t_{AB} = 100 $min.
This effect is known as time dilation. Special relativity predicts that, in order for the speed of light to remain constant in all inertial frames, time has to run at different rates for observers moving with different velocities.
EDIT:
As WillO has pointed out in the comments of your original question and both answers given, the ambiguity as to whether the ship is orbiting or travelling towards/away from Earth is actually quite important. The answer given here is based on the assumption that the ship is orbiting Earth, however in the other two cases you must consider the relativistic Doppler effect as detailed in the answer below.
| {
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Why I cannot write the time evolution operator $e^{-i(T+V)t}$ as the product of operators $e^{-iTt}e^{-iVt}$ To calculate the wave equation of a time-independent Hamiltonian we use:
$$
\Psi_{i}(r,t)=e^{-iH^{0}t}\psi_{i}(r,0).
$$
We also know that the time-independent Hamiltonian $H^{0}=T+V$ is given to the sum of kinetic and potential energies, in an isolated atom.
However, we cannot write $e^{-iH^{0}t}$ as the product of the operator, $e^{-iTt}e^{-iVt}$. Why is that?
| When we put the Hamiltonian $H$ in an exponential like this, we are not assuming the $H$ represents a scalar value. The exponential notation used in this case is just shorthand for the Taylor series equivalent for the exponential function:
$e^{-iHt} \equiv 1 + (-iHt) + \frac{(-iHt)^2}{2!} + \frac{(-iHt)^3}{3!} + \cdots$
In many contexts we treat $H$ as a matrix, which means we are then dealing with powers of the matrix $H$.
If you think of $H$ as the sum of two matrices, one each for $T$ and $V$, and then substitute them into the power series shown above, you will have to compute powers of $T+V$. When you do this, you have to remember that matrices don't commute in general, so the order of the terms matters. Because they don't commute in general, you cannot simply assume that every term in the Taylor series can be condensed down to some complex number multiplied by powers of $T$ and $V$ like $T^nV^m$. Terms might look more like $TVT^2V$, etc. Until you are familiar with it, it's tempting to try and use this shorthand the way we did in calculus for real/complex areguments. However, since the exponent may involve quantities whose products don't commute, we cannot say that the exponential involving $H$ is equal to the product of an exponential involving $T$ and another exponential involving $V$.
| {
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Can we have motion in systems where inertia is neglected? According to Newton's law:
$$ \sum F=ma$$
So, if we have some acceleration, it's because we have a certain amount of motion in our systems.
This makes me confused if inertia was neglected. Are there cases where we can have motion when inertia is neglected? Maybe when there is a friction so we have velocity, but this friction will be already balanced with other forces, so it shouldn't move. Or viscous systems?
Adding to the question after reading the comments:
Can example about that: cells are viscoelastic media, there, the inertia can be neglected with respect to viscous forces, but still the cell can move. According to biological processes happens in the cells, we must worry about the stresses at the surfaces coming from for example the actin-myosin contractility. So we can say that motion comes from deformations?
Also, if we have a viscous material or rod so that we can neglect inertia, and we pull this material, we can say that $F_{ext}=\text{rate of deformation} \times viscosity$, and the rate of deformation depend on the velocity and the shape of the rod will involve with time?
| We sometimes assume a body has negligible mass, such as a rope in a pulley system where the pulley and weights on the ends of the rope have much greater mass than the light rope. This simplifies the evaluation of the motion. See a basics physics textbook for examples.
| {
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What is the origin of the inertia? Is there any explanation why it is harder to move an object with more mass than an object with lesser mass? What kind of force is opposing our force? Is it finalized currently and well known what the origin of inertia is, or not yet 100%? I read some explanations that are linked to general relativity and Mach's principle, but can someone please tell me in the case of current physics, it is well understood?
| Currently, there is no scientific consensus on why inertia exists, or why the geodesic through spacetime is the least path of energy (ultimately leading to: why do we need to expend energy to move an object away).
Although, we know how much force is needed, we don't know why, and so this question may be well suited for either opinions, philosophy, or just to take it as a given property of mass (the resistance to change in motion).
As for "why it is harder to move an object with more mass than an object with lesser mass", I would be cheating you if I say $F=ma$ because you could question why that specific equation should apply to nature. I can only say observation until we discover more fundamental truth.
By fundamental truth, I am indicating a possible theory or law that dictates the formation of present physics laws and tells why nature works. If your curiosity is satisfied with that theory, congratulations! But, if you are an extremely curious person, you might ask, "Why this set of principles", and ultimately, one of two scenarios will happen.
*
*You will keep asking "why" questions and never reach a final answer (unless you believe the statement by Hawking that asking why or who caused the Big Bang to happen makes no sense because there was no time existing before it)
or, 2) You will have to allow something to be true and accept that humans can never get a final answer.
| {
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Is there a fundamental lower bound to resistor precision? The resistor noise index provides a lower bound to how precise a resistor can be measured, even using ideal instrumentation. Effectively, it even makes no sense to define the resistance any more precise than given by its noise index. Intuitively, this can be understood as more probe charge (either due to higher current or longer integration time) also influences the resistor more.
However, by increasing the volume of the resistor, one can reduce the noise index: Below in Fig. 1, both resistances are equal to $R$, but the right assembly uses four identical resistors each contributing uncorrelated noise. It can be shown, that the 4 resistor assembly has a twice lower total excess noise and thus lower noise index. Therefore, the resistance of the right side assembly can be defined twice more precise than the single resistor on the left side. By quadrupling the volume of the resistor and leaving everything else unchanged, the excess noise drops to half demonstrating the relation $n\propto V^{-1/2}$.
My question is:
How far can this be taken ? One could always add more resistors achieving extremely precise assemblies. But fundamentally, to probe the resistance electrically one relies on electron-electron interaction which are quantum mechanical processes. So I figure that the uncertainty principle might sneak into the backyard somehow.
Figure 1: Using 4 identical resistors, one can make an assembly with the same resistance, but twice lower excess noise
| Resistor excess noise is poorly understood. There is, as far as I know, no verified model that can yield the answer you seek.
This is a fine example of a physical effect that is observable with simple lab equipment but remains a mystery.
| {
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Can we say that Black hole is a Black body? I want to know if a black hole is a black body.I got this doubt while studying photoelectric effects.
I am an 11th grader.I would be glad to have your help.
| The Hawking radiation theoretically emitted by a black hole has a black body spectrum. So in a sense we could say that a black hole is a black body, although at a really really low temperature (a tiny fraction of a degree above absolute zero). But there is nothing deep about this - it is really just a coincidence of terminology.
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Why are fields described as force divided by mass or charge? I have read that application of force on a body from a distance, like gravitational or electrostatic force is a two-step process, first, the field is created by the body, then, the application of force on the second body by the field. I want to know why the expression for gravitational field is given as F/m or why the expression for electric field is given as F/q?
|
Why are fields described as force divided by mass or charge?
Because they follow from the classical universal law of gravitation and Coulomb's law.
The force that each of two masses or charges experience is due to the gravitational or electric field generated by the other mass or charge.
For gravity the magnitude of the force, where the centers of the masses are separated by distance $r$, and $G$ is the universal gravitational constant, is given by the universal law of gravitation
$$F=G\frac{m_{1}m_{2}}{r^2}$$
So if one is interested in the magnitude of the gravitational field $g$ due to $m_2$ that causes a gravitational force $F$ on $m_1$, it is given by $F/m_1$ or
$$g=G\frac{m_2}{r^2}$$
Similarly, the electrical force between two point charges is given by Coulombs law where $k$ is the Coulomb constant
$$F=k\frac{q_{1}q_2}{r^2}$$
And the electric field E due to $q_2$ that causes a force $F$ on $q_1$ is
$$E=k\frac{q_2}{r^2}$$
Hope this helps.
| {
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Are two states with the same measurement probabilities necessarily equal up to unitary equivalence? Let $\rho$ and $\rho'$ be $n\times n$ density matrices, and suppose that for every observable $A$ and every $\lambda$ in the spectrum of $A$ we have
$$
\text{tr}(\rho P_{\lambda})=\text{tr}(\rho' P_{\lambda}),
$$
where $P_{\lambda}$ is the orthogonal projection onto the $\lambda$-eigenspace of $A$. Does it then necessarily follow that $\rho'=U\rho U^{\dagger}$ for some unitary $U$?
| Actually $\rho=\rho'$. Indeed, specializing $A= P= |\psi\rangle \langle \psi|$ for $||\psi||=1$, the hypothesis implies $\langle\psi| (\rho-\rho')\psi\rangle =0$. Linearity permits to relax the requirment $||\psi||=1$. By polarization, in turn, it implies $\langle\psi| (\rho-\rho')\phi\rangle =0$ for every pair of vectors $\psi,\phi$. Fixing $\phi$ and choosing $\psi:= (\rho-\rho')\phi$, we conclude that, for every vector $\phi$
$$|| (\rho-\rho')\phi||^2 =0 $$
which means $\rho-\rho'=0$.
The result is valid also for infinite dimensional Hilbert spaces.
| {
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The value of $g$ in free fall motion on earth When we release a heavy body from a height to earth. We get the value of $g=9.8 \ ms^{-2}$. Now, I'm confused about what it means. For example, does it mean that the body's speed increases to $9.8$ every second? Or, does it mean that the speed of the body is $9.8 \ m/s$?
| It means the speed of the falling body increases with 9.8 m/s each second.
| {
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Solving 3D Kepler Problem substitution goes wrong I'm trying to arrive at the effective potential equation in Kepler Problem using Routh reduction method. We can procede in two ways, either using polar coordinates in the plane where the orbit happens or using spherical coordinates. I'm having trouble with this last one. I'm gonna follow steps taken in this Wikipedia page. Recalling,
\begin{gather*}
\mathcal{L}(r, \dot{r}, \theta, \dot{\theta}, \dot{\phi}) = \frac{1}{2}m \left(\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2 (\theta) \dot{\phi}^2\right) - V(r).\tag{1}
\end{gather*}
Because $\phi$ is cyclic, its momentum conjugate is conserved
\begin{gather*}
p_{\phi} = \frac{\partial \mathcal{L}}{\partial \dot{\phi}} = mr^2 \sin^2 (\theta) \dot{\phi} = L_z = cte.\tag{2}
\end{gather*}
Now, consider the Routhian
\begin{gather*}
\mathcal{R}(r, \dot{r}, \theta, \dot{\theta}) = \frac{1}{2} \frac{p_{\phi}^2}{mr^2\sin^2 (\theta)} - \frac{1}{2}m \left(\dot{r}^2 + r^2 \dot{\theta}^2 \right) + V(r).\tag{3}
\end{gather*}
Now consider $\theta$ Lagrange equation, which is equivalent to the conservation of the modulus of the momentum
\begin{align*}
m(2r\dot{r}\dot{\theta} + r^2\ddot{\theta}) &= \frac{p_{\phi}^2\cos(\theta)}{mr^2\sin^3(\theta)} \tag{4} \\
m^2r^4\dot{\theta}^2 + \frac{p_{\phi}^2}{\sin^2(\theta)} &= L^2 = cte.\tag{5}
\end{align*}
However, if I substitute in the Routhian, it does not work properly
\begin{gather*}
\mathcal{R}(r, \dot{r}, \dot{\theta}) = \frac{1}{2} \frac{L^2}{mr^2} - \frac{1}{2}m \left(\dot{r}^2 + 2r^2 \dot{\theta}^2 \right) + V(r).\tag{6}
\end{gather*}
| Substituting some conserved quantity is not how the Routhian mechanics works. If your Lagrangian is $\mathcal{L}(r,\dot{r},\theta,\dot{\theta},\phi,\dot{\phi})$ (we do not exclude $\phi$ here as generalization for other possible potential), we always have a Routhian $R(r,\dot{r},\theta,\dot{\theta})$ given by
$$R(r,\dot{r},\theta,\dot{\theta})=p_{\phi}\dot{\phi}-\mathcal{L}$$
which satisfies the equations of motion
\begin{align}
{d \over dt}\bigg({\partial R \over \partial \dot{r}}\bigg) & = {\partial R \over \partial r} \\
{d \over dt}\bigg({\partial R \over \partial \dot{\theta}}\bigg) & = {\partial R \over \partial \theta}
\end{align}
You can also choose some other Routhians like $R(r,\dot{r},\phi,\dot{\phi})=p_{\theta}\dot{\theta}-\mathcal{L}$, and it will satisfy
\begin{align}
{d \over dt}\bigg({\partial R \over \partial \dot{r}}\bigg) & = {\partial R \over \partial r} \\
{d \over dt}\bigg({\partial R \over \partial \dot{\phi}}\bigg) & = {\partial R \over \partial \phi}
\end{align}
So as pointed out in the wiki page, the choice of generalized coordinates to obtaining the corresponding Ruthian is arbitrary. However, some can probably make solving the problem easier. In the example given in your question, since $p_\phi$ is conserved, there will be $2$ equations of motion $R(r,\dot{r},\theta,\dot{\theta})$ have to satisfy with only $2$ degrees of freedom left. This is not the case if we choose $R(r,\dot{r},\phi,\dot{\phi})$. However, you cannot have the Routhian with the substitution for the total angular momentum $L$ since we do not have $p_\phi\dot{\phi}={L^2 \over mr^2}$.
| {
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Is there a version of the Einstein field equations that uses the Riemann curvature tensor instead of the Ricci curvature tensor? I understand that the Einstein field equation uses the Ricci curvature tensor, Ricci curvature scalar, and stress-energy momentum tensor. But is there a way to form an equation that uses the Riemann curvature tensor and the stress-energy momentum tensor?
| If you mean an equation such that the stress energy tensor at a point of the spacetime determines the Riemann tensor at that point, then the answer is negative for physical reasons. Such an equation would imply that the spacetime is flat outside the gravitational sources. This makes no sense because the gravitational field propagates outside its sources.
| {
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Double slit experiment: Are electrons interacting with other electrons to create a wave? Assume a double slit experiment with electrons and no observer (light source). Can the wave-like behavior and resulting interference pattern be explained by the single electron that is being shot, doesn't really travel to the detector, but interacts with other electrons in the medium (e.g. air) between source (electron gun) and target (detector), and this creates the wave? I imagine it as if the shot electron is repelled from other electrons and they again repel other electrons and so forth.
Furthermore, in case of a light source acting as an observer. Could they interact electromagnetically with all these electrons, between the source and target, in a way that the electrons are not moving freely and repel each other. But, acts a contiguous block and the shot electron hits the first electron, that transfers the energy to the second electron, then to the next until the last electron hits the detector. Similarly to Newton's cradle?
| No, the experimental evidence does not support the idea that the interference patterns are created in the way you suggest. Aside from the fact that experiments have been performed in a vacuum, which rules out the idea that the incoming electrons are interacting with other particles in their path, diffraction experiments have also been performed with other types of particles, including neutrons, for example, which don't interact in the way that charged particles do and yet the interference effects are still produced. The effects also seem to be largely independent of the material surrounding the slits.
| {
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Why are positrons traveling backward in time in Feynman diagrams? We draw the positron as if it's traveling backward in time. Why? The momentum of the positron is drawn opposite to the time-direction of the diagram. I don't see this having any effect on the calculation. We still arrive at the result $k=p_1+p_2$, and not $k=p_1-p_2$ like the diagram would indicate. So, if anything, the diagram is misleading? $p_1$ and $p_2$ are the momenta of the incoming electron and positron, $k$ is the momentum of the virtual photon.
| The arrow direction in a Feynman diagram does not correspond to the time direction. There are conventions in which times flows from left to right, i.e. the time on the left is always earlier than the time on the right, although these are not the most common ones. Regardless of the convention, the arrow direction is just used to distinguish different kind of particles, for examples electrons from positrons or electrons from holes.
The momentum of the particle is also usually independent from the arrow direction, although also this may depend on the convention used. For example in non relativistic many body perturbation theory, the momentum usually refers to the one of the state on the band structure. Then the electron has the same momentum, while the hole has opposite momentum compared to the one written in the diagram.
On the time direction. Real particles never travel backward in time. For virtual particles instead this can happen. For example you can have an initial particles which goes from say $t_0$ to $t_1<t_0$, then interacting with a photon for example, and propagate from $t_1$ to $t_2>t_0$. Indeed $t_1$ would be an integrated variable, and all the possible values of $t_1$ need to be considered. In the convention where time flows from left to right, and for the case $t_2<t_1$, there would be a line from center to the left, and a second line going back from the left to the right. In both lines the direction of the arrow would just depend on the particle kind, and not on the time flow.
However, also in case of virtual particles, one should be careful when saying that something goes back in time. It is a fashinating concept, but also one which generates a lot of confusion.
| {
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Breaking down of 2nd law of thermodynamics Do you know a scenario where the second law of thermodynamics breaks down?
| The second law breaks down when some of the assumptions underlying this law (or the thermodynamics itself) are broke.
Among what is typically cited as "violations" of the second law are:
*
*Violation of the laws of thermodynamics in small systems. Thermodynamics and statistical physics are applicable in thermodynamic limit that is for systems with a huge number of particles. While in some cases this number may be much smaller than the Avogadro number ($N_A\sim 10^{24}$), violation of the thermodynamics in systems with a finite number of degrees-of-freedom is not surprizing. This is equally true for ratchets and billiards.
*Entropy decrease in some systems, notably in living systems. This is again not surprizing, since these are open systems, which exchange energy and matter with the environment. Thus, while the entropy of the system might be decreasing, this is accompanied by the increase of entropy in the environment, and the net entropy is growing. See this answer and this answer for more background.
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The Absorption of light I just want to know, how can we explain the phenomena of : 'absorption of light' based on the wave aspect of light ( light is an electromagnetic wave). In fact, light has an electromagnetic field, so it has an electric field which interact with atoms's electrons, I think at this point, something might be happened and it lead to the absorption phenomena, but I don't know what is it .
Thanks .
| Every light absorption has a preceding light emission. Electrons emit photons after an excitation. And that is where we should start.
These electrons are elementary electric charges and are, last but not least, elementary magnets. To the photons they bequeath both the electric and the magnetic field. And they do so in perfect harmony; photons have very different energy contents, but the ratio of the electric to the magnetic field is always a constant.
During absorption, however, the photons have almost no chance of hitting an electron that is in exactly the same excited state as the emitting electron. In addition to the electron being raised to a higher energy level in the atom, there is often re-emission of infrared photons and excitations within the atomic compound.
In fact, light has an electromagnetic field, so it has an electric field which interacts with atoms's electrons.
More than that. A photon interacts with both its electric and magnetic field with the material it hits. The best example is the polarised photons that emerge from an antenna rod. If these photons hit a straight antenna rod, we speak of an electrical antenna, but if they hit a circular antenna, we speak of a magnetic antenna.
| {
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Will quantum events ever occur on a macro-scale rather than a vacuum? Michio kaku says there's a chance we'll wake up on Mars tomorrow https://www.theatlantic.com/science/archive/2018/10/beyond-weird-decoherence-quantum-weirdness-schrodingers-cat/573448/
In this post, it is shown that quantum decoherence in the macro world occurs almost instantly, except in vacuum and at low temperatures. Therefore, in the macro world, we cannot pass through walls. Nor can you suddenly wake up on Mars the next day. But Michio Kaku says that if we wait longer than our cosmic lifetimes, we could get through a wall one day or wake up on Mars the next day. Was Michiokaku right? Wrong? Is it possible that such events are possible because there is a possibility that our body does not interact with all other oxygen or photons and thus exhibits quantum coherence, just as there is a possibility that oxygen particles in a room only gather in the opposite direction? Will quantum events ever occur on a macro-scale rather than a vacuum?
| I will add that there are everyday examples of quantum effects on the macro scale here on earth. My favorite is the laser, where simulated emission is purely quantum but those emission events do not average out to zero; instead they add together and are blindingly obvious on the macro scale.
| {
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Interpretation of Feynman Slash Notation I'm self-learning Relativistic Quantum Mechanics and was playing around with the Dirac equation when I noticed something.
I was trying to interpret the meaning of ${\not} \partial$. So since I can represent a 3-vector by a traceless hermitian matrix: $$\vec{x}=x^i \vec{e}_i \to X = x^i \sigma _i$$
Can I represent a 4-vector using gamma matrices?
$$\vec{x}=x^\mu \vec{e}_\mu \to X = x^\mu \gamma _\mu$$
So in the case of a covector:
$$x_\mu \vec{e}^\mu \to X = v_\mu \gamma ^\mu \equiv {\not} v$$
Therefore, ${\not} \partial$ is just the matrix representation of the covector $\partial _\mu$
| Yes, of course; people do this routinely. Read your texts on.
The essence of the map is its invertibility through
$$\operatorname{Tr} \left(\gamma^\mu\gamma^\nu\right) = 4\eta^{\mu\nu},$$
so that
$$v_\mu ~~ \mapsto ~~ {\not} v \equiv v_\mu \gamma ^\mu ~~
\mapsto~~ \tfrac{1}{4}\operatorname{Tr} \left( {\not} v\gamma^\nu\right)=v^\nu.$$
You know how to lower the free index.
This parallels the elementary ubiquitous spinor map in 3D,
$$\vec{x}~ \mapsto ~X \equiv \vec x \cdot \vec \sigma ~\mapsto~ \tfrac{1}{2}\operatorname{Tr} X\vec \sigma = \vec x. $$
| {
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Why is angular momenum related to the spin? What I know about spin ½ particles is that they are represented by spinors, and thus, you need to apply a 720° rotation in order for the spinor to return to its original value. Spin 1 particles are vectors, and their transformation is trivial.
Also, spin ½ have $\frac{\hbar}{2}$ or $-\frac{\hbar}{2}$ angular momentum, while spin 1 have $\hbar$ or $0$ or $-\hbar$ angular momentum.
But why are these related? Why should a particle that transforms like a vector have integer angular momentum, while spinors have half-integer angular momentum? Is it just an observation?
| A very simplified, hand-wavey explanation is to say that in the same way that linear momentum is related to one over the wavelength of the wavefunction as you move each linear direction, so the angular momentum (spin) is related to one over the 'wavelength' (really we need a new word, like 'waveangle' for this idea) of waves as you rotate in each angular direction. (Have a look at some pictures of the spherical harmonics to see what this looks like. We fit a number of sine-waves 'around' a sphere.) If you spin around in a full circle, you have to be able to fit a whole number of waves into that circle to avoid any discontinuity, so spin is quantised.
With a vector, spinning 360 degrees has to get you back where you started, so the number of angular 'wavelengths' has to be a whole number. With a spinor, spinning 360 reverses sign, and you have to turn 720 degrees to get back where you started. Fitting an odd whole number of angular 'wavelengths' into 720 degrees means fitting a half-integer number into 360 degrees.
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it? I have read that water (or any other liquid) cannot be compressed like gases and it is nearly as elastic as solid. So why isn’t the impact of diving into water equivalent to that of diving on hard concrete?
| Well, depends on the way someone dives in and the altitude.
You might have wondered at some point, why does it hurt when you fall into the water with your belly, but when you fall in with your feet almost vertically to the water you feel no pain. On the internet you can definitely find some answer on this question, for example here: https://www.scienceabc.com/sports/why-does-water-feel-like-concrete-when-you-belly-flop-into-it.html
So as someone dives in, by making contact, pressure is indeed exerted on the surface of the water (or any other fluid of similar physical properties). However, unlike ground, water has the property to change its shape (deformation) based on its density viscuosity, which is a process that uses time. As a result, when contact is made by some large and surface (like the belly), water might not have time to deform and columns of water can get trapped beneath the surface temporarily exerting back force to the body following the action-reaction law.
Consequently, the less surface you use and the lower altitude (thus slower diving) you jump from, the less probability you have to get hurt by diving in, as water gets enough time to deform and make its way out. The viscuosity of water is the property that counters its incompressibility (reciprocal to its elasticity) so that you can dive in without pain.
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"answer_count": 13,
"answer_id": 4
} |
Finding Locally flat coordinates on a unit sphere I know this is more of a math question, but no one in the Mathematics community was able to give me an answer, and since physicists are familiar with General Relativity, I thought I might get an answer.
Imagine a unit sphere and the metric is:
$$ds^2 = d\theta ^2 + \cos^2(\theta) d\phi^2$$
I want to find Locally Flat Coordinates (I think they're called Riemann Normal Coordinates) on the point $(\frac{\pi}{4}, 0)$, so what I need are coordinates such that the metric would reduce to the Kronecker Delta and the Christoffel Symbols should vanish. I start by the following translation:
$$\theta' = \theta - \frac{\pi}{4}$$
then do the following substitution by guessing:
$$\frac{f(\theta')}{\cos(\theta)} d\phi' = d\phi$$
And the condition is $f(0)$ should be 1, so the metric becomes:
$$ds^2 = d\theta' + f^2(\theta')d\phi'$$
And it is a matter of finding $f(\theta')$. I calculate the Christoffel Symbols:
$$\Gamma^{\lambda}_{\mu\nu} = \frac{1}{2} g^{\lambda \alpha}(\partial_{\mu}g_{\alpha \nu} + \partial_{\nu}g_{\mu \alpha} - \partial_{\alpha}g_{\mu \nu})$$
And make them vanish.
So what I get is:
$$\frac{f'(0)f(0)}{f^2(0)} = 0$$
Obviously, $f(\theta')=\cos(\theta')$ is a solution which is the thing I know is correct. However, there are infinite functions that satisfy the above conditions. Are all of these functions eligible to make the new coordinates Riemann normal coordinates?
| starting with components of the unit sphere :
\begin{align*}
&\begin{bmatrix}
x \\
y \\
z \\
\end{bmatrix}=\left[ \begin {array}{c} \cos \left( \phi \right) \sin \left( \theta
\right) \\ \sin \left( \phi \right) \sin \left(
\theta \right) \\ \cos \left( \theta \right)
\end {array} \right]
\end{align*}
from here
\begin{align*}
&\begin{bmatrix}
dx \\
dy \\
dz \\
\end{bmatrix}=\underbrace{\left[ \begin {array}{cc} \cos \left( \phi \right) \cos \left( \theta
\right) &-\sin \left( \phi \right) \sin \left( \theta \right)
\\ \sin \left( \phi \right) \cos \left( \theta
\right) &\cos \left( \phi \right) \sin \left( \theta \right)
\\ -\sin \left( \theta \right) &0\end {array}
\right]}_{\mathbf J}\, \left[ \begin {array}{c} d\theta \\ d\phi
\end {array} \right]
\end{align*}
and the metric
\begin{align*}
&\mathbf{G}=\mathbf J^T\,\mathbf J=\left[ \begin {array}{cc} 1&0\\ 0& \left( \sin
\left( \theta \right) \right) ^{2}\end {array} \right]
\end{align*}
now we are looking for the transformation matrix $~\mathbf{T}~$ that transformed the metric to unit matrix
\begin{align*}
&\mathbf{T}^T\,\mathbf{G}\,\mathbf T=\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}\quad\Rightarrow\quad
\mathbf{T}=\begin{bmatrix}
1 & 0 \\
0 & \frac{1}{\sin(\theta)} \\
\end{bmatrix}
\end{align*}
hence
\begin{align*}
&\begin{bmatrix}
dx \\
dy \\
dz \\
\end{bmatrix}\mapsto \underbrace{\mathbf{J}\,\mathbf T}_{\mathbf{T}_n}\,
\left[ \begin {array}{c} d\theta \\ d\varphi
\end {array} \right]
\end{align*}
and the neue metric is:
\begin{align*}
&dx^2+dy^2+dz^2\mapsto d\theta^2+d\phi^2
\end{align*}
where $~\mathbf{T}_n~$ is a function of $~\theta~,\phi~$
\begin{align*}
&\mathbf{T}_n(\theta=\pi/4~,\phi=0)=
\left[ \begin {array}{cc} \frac 12\,\sqrt {2}&0\\ 0&1
\\ -\frac{1}{2}\,\sqrt {2}&0\end {array} \right]
\end{align*}
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/718912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How does a decelerating universe agree with this version of the Friedmann-Robertson-Walker formula? I was watching a video discussing dark matter, and he presented the following simplified version of the Friedmann-Robertson-Walker equation:
$$ \left(\frac{\dot{a}(t)}{a(t)}\right)^2 = \frac{8\pi G}{3} \rho(t) -\frac{k}{a(t)^2}$$
where $a$ is the scale factor, $\rho$ is the density, $k$ is a constant, and $G$ is the gravitational constant.
He said that if the $k$ factor is more negative than the $G$ factor, the universe expansion would have negative deceleration. However, the L.H.S term is squared so it can never be negative. So, what did he mean by negative?
| The video is "Dark Energy" by DrPhysicsA, and the claim in question is at 19:45.
It's a mistake. It would have been correct to say that if $ρ(t)>0$ at all times and $k\le 0$, then the RHS is positive at all times, so there is no time at which $a'(t)=0$, so $a'(t)$ can't switch sign, so if the universe is expanding at any time then it's expanding at all times. But he definitely says that the RHS being negative means the universe is contracting, which is wrong.
It isn't the only mistake in the video. At 21:30 he says that in the critical case, the scale factor asymptotically approaches a maximum value. That's quite wrong and I don't see how it can be explained away as mere blackboard nervousness. If it were true, Einstein wouldn't have needed the cosmological constant at all: he could have just said that we are in a critical-density universe at a very late time.
At 38:30, and again at 1:04:10, he draws a sudden transition from radiation to matter dominance with a kink in the graph of $a(t)$ as though $a''(t)\gg 0$ briefly at that time. In a correct graph, there are no sudden changes in $a''(t)$ and it's negative at all times prior to $Λ$ dominance (much later).
At the end of the video he repeats the common misconception that light from galaxies with a recession velocity greater than $c$ won't reach us. He only brings up this idea when he starts talking about dark energy; he doesn't seem to realize that even in the $Λ=0, k=0$ models that he discussed earlier, galaxies more distant than $c/H$ have a recession velocity larger than $c$, and yet the light from them reaches us (there is no cosmological horizon in those models).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/719178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why do we give so much importance to energy, i.e., the conserved quantity under time symmetry? In almost all equations—from GR to QFT—energy conservation is a tool for solving those equations, but we know that energy on large scales is not conserved.
Why do we still use this (not) conserved quantity in our fundamental laws of physics?
| As a complement to Nickolas's great answer, note that throughout physics, conservation laws are so useful that even approximate conservation laws can be of enormous interest. Conservation laws let you deduce aspects of the behavior of a physical system, without needing to actually solve equations. This is tremendously useful for (a) building intuition, (b) quickly solving problems, (c) providing ways to check if there are mistakes in the solutions to the equations.
Some examples of approximate conservation laws include:
*
*Mass (which is conserved to a very good approximation in non-relativistic physics, but is not conserved in special relativity)
*Kinetic energy is approximately conserved in collisions which are approximately-but-not-exactly elastic. "Elastic" collisions in a freshman college physics lab will often be analyzed as if kinetic energy were exactly conserved.
*Baryon number (this is used in particle physics)
Energy is not an exact conservation law in general in GR (although in some cases it is). However, even in GR, energy is approximately conserved locally -- meaning, on time scales short compared to the scale on which the gravitational field is changing in time. In fact, in ordinary circumstances in a lab, this conservation law holds to extremely high precision. That's why energy conservation is still useful, even though it isn't exact.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/719642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How can Entropy be maximal when it is undefined everywhere else? This question is about classical thermodynamics.
I learned that when an isolated system is not in equilibrium, its thermodynamic variables such as Entropy are undefined.
I also learned that when an isolated system is in equilibrium, its Entropy is maximized.
However, both statements together don't make sense. How can a value be maximized, when everywhere else it is undefined? It's not smaller everywhere else, it's undefined! It can't be maximal or minimal because there isn't anything else nearby to compare it with.
So how am I supposed to interpret these statements?
| As you wrote, entropy is undefined in non-equilibrium conditions. However, one can compare entropy in different equilibrium states of the same system. That is a meaningful comparison, and in particular, one can compare the entropy value in conditions characterized by the same values of the thermodynamic variables but different constraints. That is where we look for the maximum of the entropy to establish which of the various systems corresponds to the equilibrium state spontaneously reached when constraints are relaxed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/719989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 6,
"answer_id": 4
} |
Angular momentum of a planet about apogee
Angular momentum of a planet about its apogee is maximum at __________
Now, I do know that
*
*Angular momentum of a planet around the focus of the elliptical orbit (the sun) is conserved due to gravity being a central force. (Side note: this gives Kepler’s second law.)
*The formula of speed of a planet at any point on its orbit which is a distance $r$ away from the star ($a$ is the length of the semimajor axis): $$v(r)=\displaystyle\sqrt{\frac{GM(2a-r)}{ar}}$$
*Angular momentum about a point is defined as $\vec L=m\vec v\times \vec r$.
So I can’t see any easy way out to determine the answer. I was thinking that it may be at perigee because of the large distance but got confused because of the decreasing speed as the planet moves from apogee to perigee. So is the only way out for me now to do some coordinate-bashing of the ellipse? Or is there any easier way? Intuitive (but convincing) reasoning is more than welcome.
| I just found out an explanation: For local extrema of $\vec L$, we must have $\dfrac{d\vec L}{dt}=0$ but we know $\dfrac{d\vec L}{dt}=\vec \tau$ so that $$\tau_{\text{about apogee}}=0$$ which happens at only the apogee and perigee. It is easy to notice that apogee gives the minimum (because $\vec r=\vec 0$) so that $\vec L_{\text{apogee}}$ is maximised for the perigee.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/720152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How conservation of energy looks like in the moving frame? It is obvious that a car accelerates by converting its chemical energy in the fuel to produce kinetic energy to accelerate itself.
If the energy is lost into friction and heat, the car will slow down.
But from the point of view of the car, it always appears to be stationary, which means its kinetic energy is zero. So what is the thing that the fuel actually does in the car frame?
| Let' say there are two earths, on one of those earths a car is accelerating.
Now the car says that the kinetic energy of the earth that the car is not pushing increases by a huge amount, and the kinetic energy of the earth that the the car is pushing increases by the same huge amount plus the energy of the burned fuel. Or it increases by the same huge amount + force * distance.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/720630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What does an area represent in a spacetime diagram? If I have a spacetime diagram with $t$ on the vertical axis and $x$ on the horizontal axis, does calculating an area in this diagram have any physical significance?
The reason I ask is because I'm trying to figure out if the $u$ in all the $\cosh(u)$ and others hyperbolic trigonometry functions has a physical meaning.
| The name itself of "hyperbolic functions" is due to the fact that the functions $cosh(t)$ and $sinh(t)$ play, in the parametric representation of the equilateral hyperbola $x^{2}-y^{2}= a^{2}$
the same role as the functions $cos(t)$ and $sin(t)$ for the circle
$x^{2}+y^{2}=a^{2}$
The parametric representation of circle is: $x=a\cos(t) , y= a \sin(t)$
and, for the hyperbola $x=a\cosh(t), y=a\sinh(t)$
as it is easy to see using the relation: $cosh(t)^{2}-sinh(t)^{2}=1$
The geometrical significance of the parameter $t$ in both cases, for the circle and for the hyperbola, is identical.
If we designate $S$ the area of a portion of the circle and by $S_{0}$ the area of the entire circle ($S_{0}=\pi a^{2}$), we have:
$t= 2\pi \frac{S}{S_{0}}$
Let us now assume that S denotes the area of an analogous sector of the equilateral hyperbola. We have $S= area\, OMN - area\, AMN$, M a point on the heperbolus and N its projection on the axis (Ox) and A the point of intersection of the hyperbola with the axis (Ox).
$S=\frac{1}{2}xy-\int_{a}^{x} ydx=\frac{1}{2}x\sqrt{x^{2}-a^{2}}-\int_{a}^{x}\sqrt{x^{2}-a^{2}}\,dx$
$S=\frac{1}{2}a^{2}\ln(\frac{x}{a}+\sqrt{\frac{x^{2}}{a^{2}}-1}\,)$
If now we put, denoting again by $S_{0}$ the area of the circle
$t= 2\pi \frac{S}{S_{0}}=\ln(\frac{x}{a}+\sqrt{\frac{x^{2}}{a^{2}}-1}\,) $
we easily find that:
$e^{t}=\frac{x}{a}+\sqrt{\frac{x^{2}}{a^{2}}-1}\,\,\,\,\;, e^{-t}=\frac{x}{a}-\sqrt{\frac{x^{2}}{a^{2}}-1}$
hence, adding term to term and multiplying by $\frac{a}{2}$:
$x=\frac{a}{2}(e^{t}+e^{-t})=a\cosh(t)$
$y=\sqrt{x^{2}-a^{2}}=\sqrt{a^{2}\cosh^{2}(t)-a^{2}}=a\sinh(t)$
Reference: Higher Mathematics Course,Volume I, V.Smirnov
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/720934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
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