quantum mechanics - How to understand the entanglement in a lattice fermion system?

Topological insulator is a fermion system with only short-ranged entanglement, what does the entanglement mean here?

For example, the Hilbert space $V_s$ of a lattice $N$ spin-1/2 system is $V_s=V_1\otimes V_2\otimes...\otimes V_N$, where $V_i$ is the Hilbert space of the spin on site $i$. And the meaning of an entanglement state belongs to $V_s$ is clear — a state which can not be written as a direct product of the $N$ single spin states.

Now consider a spinless fermion system lives on the same lattice as spin-1/2, in the 2nd quantization framework, the fermion operators $c_i,c_j$ on different lattices $i,j$ do not commute with each other and the Hilbert space $V_f$ of the fermion system can not be written as a direct product of $N$ single fermion Hilbert spaces. Thus, how to understand the entanglement in this fermion system?

Mathematically, we can make a natural linear bijective map between $V_f$ and $V_s$, simply say, just let $\mid 0\rangle=\mid \downarrow\rangle,\mid 1\rangle=\mid \uparrow\rangle$. Thus, can we understand the entanglement of a fermion state in $V_f$ through its corresponding spin state in $V_s$?

Answer

Mark Mitchison is right. The concept of entanglement in systems of indistinguishable particles is more controversial than it is in the case of systems composed of distinguishable subsystems. You need to define first what do you mean by it when it comes, for example, to fermions. Do you mean entanglement between particles (connected with single Slater determinants), modes, pairing of states or whether a given state can be written as a convex combination of Gaussian states or sth completely else. You also should specify do you want to consider fermionic state with a fixed number of fermions (and then use the criteria from here) or just to fix the parity of the fermionic state and not the number of fermions, obtaining e.g. Gaussian states. This is also important, because even though physical states have a fixed number of fermions, Gaussian fermionic states are important approximations to physically non-trivial states, such as the superconducting BCS state. Of course then, the super-selection rule should also play a role somehow.

And about your question, you can find a nice definition of short-ranged entanglement in topological insulators in Sec. II of http://arxiv.org/pdf/1004.3835v2.pdf

quantum mechanics - Landau level degeneracy in symmetry gauge, finite system

As we know, Landau level degeneracy in a finite rectangular system is $\Phi/\Phi_0$, where $\Phi=BS$ is the total magnetic flux and $\Phi_0=h/q$ is the flux quanta. This can be easily derived using Landau gauge and assuming the periodic boundary condition.

However, if you choose the symmetric gauge, $l_z=x\hat p_y-y\hat p_x$ commutes with Hamiltonian, the corresponding quantum number $m$ is thus a good one. After some calculation, at last the energy level is written as $$E=\left[n+(m+|m|)/2+1/2\right]\hbar\omega.$$ I want to derive the degeneracy of Landau level for a finite system with radius $R$.

The thought I got is that because in a finite system the angular momentum is bounded, the maximum value of $l_z$ is $qBR^2$ when you consider the particle doing the circular motion classically. Hence the maximum $m=l_z/\hbar=2\Phi/\Phi_0$, and the degeneracy is thus at least doubled.

What we have learned is that the gauge choice won't change the observable effect, the finite degeneracy can obviously be observed. So what's wrong with my derivation? Or because they are different just because the system we considered is simply not equivalent?

quantum field theory - uncertainty deviations for vacuum astronomy

Since i've done this question, i've been trying to improve and make more precise the statements regarding cosmic squeezed states and how different uncertainties affect the vacuum energy, but as it turns out, it's more difficult than it seems, as i hope to make clear in this question

Intuitively, i can play with some algebra of the uncertainties and write a "minimum" of energy in function of them:

$$H = E^2 + B^2$$

$$H + \Delta H = (E+ \Delta E)^2 + (B + \Delta B)^2$$

$$\langle \Delta H \rangle = 2 ( \langle E \rangle \langle \Delta E \rangle + \langle B \rangle \langle \Delta B \rangle ) + \langle \Delta E^2 \rangle + \langle \Delta B^2 \rangle$$

In the vacuum, $\langle E \rangle = \langle B \rangle = \langle \Delta E \rangle = \langle \Delta B \rangle = 0$ so we can simplify:

$$\langle \Delta H \rangle = \langle \Delta E^2 \rangle + \langle \Delta B^2 \rangle$$

Now, if we accept the uncertainty principle applicability to both the total fields and its modes ( a tall proposition as we'll see) and assume a minimum uncertainty mode:

$$\Delta E = \frac{ \hbar }{ \Delta B}$$

then we are left with

$$\langle \Delta H \rangle = \langle \Delta E^2 \rangle + \frac{ \hbar^2}{\langle \Delta E^2 \rangle}$$

which has a minimum when $\Delta E = \Delta B = \sqrt{ \hbar}$. This is the equation that defines "true" electromagnetic vacuum in terms of the uncertainties of the vacuum.

Now, all of this is cute and nice, but all this applies to a single mode of radiation. As it becomes clear from this other question of mine stuff gets really hairy when we consider a real quantum field.

In any case, i would hope that one can ask in experimental, absolutely concrete terms is: given one region of the cosmos, what deviation from this "true" vacuum, as defined above, can we expect? note that since the electromagnetic spectrum that is relevant for astrophysical phenomena is huge (from gamma ray to microwaves) there might be different deviations at different wavelengths

Now, my question is: are there concrete, existing expressions for the effective field uncertainties in QFT in terms of the uncertainties of the modes that can be used for the purposes of this "vacuum astronomy"? Can we in practice, think in individual EM modes and their uncertainties as "isolated" from the rest of modes?

rotational dynamics - Rigid body rotation and translation about an arbitrary axis

If you have a disk which is rotating on a surface without slipping then we can view the motion in two different ways. Firstly as a rotation about the centre of mass of the disk along with a translational velocity of the centre of mass. Alternatively, the motion can be viewed as just a rotation about the instantaneous centre, in this example it is the point of contact between the surface and the disc.

Is it possible to select an arbitary axis (perpendicular to the plane of the disc) such that we can view the motion as a rotation about that axis plus a translational component? If so, is the angular velocity always the same regardless of the chosen axis?

Answer

Yes and Yes.

The rotational velocity ${\boldsymbol \omega}$ is shared among all parts of a rotating body. Linear velocity is obviously location dependent, so it usual we designate it with a letter or a symbol indicating the location where it is measured. For example, considering an arbitrary point A we state the transformation from the center C as

$$\mathbf{v}_A = \mathbf{v}_C + {\boldsymbol \omega} \times (\mathbf{r}_A - \mathbf{r}_C)$$

• 
  

  Lemma 1 - Knowing the rotational velocity and the linear velocity at any point, allows us the find the velocity at any other point with the equation above, and thus we know the entire state of motion.

  
  
  


• 
  

  Lemma 2 - Given the rotational motion and the location of the instant centre, allows us to find the velocity at any other point with the equation above, and thus we know the entire state of motion.

  
  

The is identical to Lemma 1 except the liner velocity is zero by definition at the instant centre.

• 
  

  Lemma 3 - Knowing the rotational velocity and the linear velocity at any point A, allows us to find the location B of the instant centre^** with the equation $$\mathbf{r}_{B} = \mathbf{r}_A + \frac{{\boldsymbol \omega} \times \mathbf{v}_A}{\|{\boldsymbol \omega}\|^2}$$

  
  



• 
  

  Proof - Consider the velocity of the arbitrary point A described as a rotation about the instant centre B with $\mathbf{v}_A = {\boldsymbol \omega} \times \mathbf{r}$ where $\mathbf{r}=\mathbf{r}_A-\mathbf{r}_B$ is the relative location of the point to the centre. Now consider the following $${\boldsymbol \omega} \times \mathbf{v}_A = {\boldsymbol \omega} \times ({\boldsymbol \omega} \times \mathbf{r}) = {\boldsymbol \omega} \left( {\boldsymbol \omega} \cdot \mathbf{r}\right) - \mathbf{r} \left( {\boldsymbol \omega} \cdot {\boldsymbol \omega}\right)$$

  
  

There is such an $\mathbf{r}$ to solve the above equation which has ${\boldsymbol \omega} \cdot \mathbf{r}=0$, or in layman's terms the location of the centre of rotation is perpendicular to the rotation axis. This value is $$\mathbf{r} = \mathbf{r}_A-\mathbf{r}_B=-\frac{{\boldsymbol \omega} \times \mathbf{v}_A}{{\boldsymbol \omega} \cdot {\boldsymbol \omega}}$$ _{Note that ${\boldsymbol \omega}\cdot {\boldsymbol \omega} = \| {\boldsymbol \omega} \|^2$. The triple vector product identity $\mathbf{a} \times ( \mathbf{b} \times \mathbf{c}) = \mathbf{b} ( \mathbf{a}\cdot \mathbf{c}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b})$ is used above also.}

_{** NOTE: In 3D, there isn't a unique instant centre point, but rather an axis of rotation. It is allowed to have linear velocity parallel to the axis of rotation and the location still is considered the instant centre.}

I encourage the reader to read more about the motion properties of a rigid body from this answer and more importantly from my full explanation of the terms twist and wrench which describe the geometry of rigid body mechanics.

everyday life - How does buoyancy work?

I realised, reading another Phys.SE question about balloons moving forwards in an accelerating car that I don't really understand how buoyancy works. Particularly concerning, for a SCUBA diver.

The top answers to that question seem to claim that balloons get their "sense of down" from a pressure differential. They continue: when a car accelerates, the air at the back of the car becomes more dense, and at the front less dense, changing the plane of the pressure differential and so also, the balloon's sense of up. I find that extremely hard to credit. However, I realised that I don't really know why less dense things float in more dense things.

I'm fairly sure it's something to do with displacement of heavier things by lighter things, and I think pressure acting on the lighter thing's surface has something to do with it, but that's about it.

Answer

Basic idea

Picture in your mind a deep ocean of water. Imagine a column of the water, going from the surface down to a depth $d$. That column of water has some weight $W$. Therefore, there is a downward force of magnitude $W$ on that column of water. However, you know the column of water is not accelerating, so there must be an upward force of magnitude $W$ pushing on that column. The only thing underneath the column is more water. Therefore, the water at depth $d$ must be pushing up with force $W$. This is the essence of buoyancy. Now let's do details.

Details

The weight $W$ of a column of water of cross-sectional area $A$ and height $d$ is

$$W(d) = A d \rho_{\text{water}}$$

where $\rho_{\text{water}}$ is the density of water. This means that the pressure of water at depth $d$ is

$$P(d) = W(d)/A = d \rho_{\text{water}}.$$

Now suppose you put an object with cross sectional area $A$ and height $h$ in the water. There are three forces on that object:

1. $W$: The object's own weight.


2. $F_{\text{above}}$: The force of the water above the object.


3. $F_{\text{below}}$: The force of the water below the object.

Suppose the bottom of the object is at depth $d$. Then the top of the object is at depth $d-h$. Using our results from before, we have

$$F_{\text{below}} = P(d)A=d \rho_{\text{water}} A$$

$$F_{\text{above}}=P(d-h)A=(d-h)A\rho_{\text{water}}$$

If the object is in equilibrium, it is not accelerating, so all of the forces must balance:

$\begin{eqnarray} W + F_{\text{above}} &=& F_{\text{below}} \\ W + (d-h) \rho_{\text{water}} A &=& d \rho_{\text{water}} A \\ W &=& h A \rho_{\text{water}} \\ W &=& V \rho_{\text{water}} \end{eqnarray}$

where in the last line we defined the object's volume as $V\equiv h A$. This says that the condition for equilibrium is that the weight of the object must be equal to its volume times the density of water. In other words, the object must displace an amount of water which has the same weight as the object. This is the usual law of buoyancy.

From this description I believe you can extend to the case of air instead of water, and horizontal instead of vertical pressure gradient.

special relativity - Spaceships connecting with spring (A deeply different version of Bell's spaceship paradox)

A very different version of Bell's spaceship paradox hit upon my mind for which I cannot find a convincing solution:

Two stationary spaceships are connected with a spring as shown in the figure below. Observer A is located somewhere in the midway of the spaceships at rest with respect to them. He sends two signals towards the spacecrafts so that the signals set them in motion simultaneously, having a constant velocity v with respect to observer A. Observer A claims that the spring has no tension during the motion of spaceships in the y-direction. However, observer B moves towards A in the x-direction with a velocity u. He claims that the initial length of the spring is $\alpha_u x_0$, moreover, he observes that the receipt of the signals are no longer simultaneous, and hence, the farther spacecraft moves first, and after a while $\Delta t=\frac {ux_0}{c^2\alpha_u}$, the nearer one moves. ($\alpha_u=\sqrt{1-u^2/c^2}$)

From the standpoint of B, the final length of the spring $L_u$ is thus calculated to be:

$$L_u=x_0\sqrt{\alpha_u^2+\frac{u^2v^2}{c^4}}$$

The difference in spring's length is:

$$\Delta L_u=x_0\sqrt{\alpha_u^2+\frac{u^2v^2}{c^4}}-\alpha_u x_0$$

If the spring constant is $K_u$ from the standpoint of B, the above difference exerts a spring force to the spaceships calculated as follows:

$$F_u=K_u\Delta L_u=K_u x_0 (\sqrt{\alpha_u^2+\frac{u^2v^2}{c^4}}-\alpha_u)$$

One observer detects tension, whereas the other one does not. What is wrong with my calculations?!

(Recall that both observers are inertial)

homework and exercises - Field theory: equivalence between Hamiltonian and Lagrangian formulation

Let $\mathscr{B}$ be a space of physics we have and $\mathscr{T}$ be the duration. Let $\mathscr{L}$ be a lagrangian density of the field such that the action is a functional of $\phi:\mathbb{R}^4\rightarrow\mathbb{R}$: $$S=\int_{\mathscr{B} \times\mathscr{T}}d^4x\mathscr{L}(\phi(t,\vec{x}),\partial_\mu\phi(t,\vec{x})).\tag{1}$$

We can then derive the equations of motion: $$\frac{\delta S}{\delta \phi}=0.\tag{2}$$

Otherwise we can define the hamiltonian density $$\mathscr{H}=\pi\dot\phi-\mathscr{L}=\mathscr{H}(\pi,\phi,\partial_i\phi)\tag{3}$$ whereas $$\pi=\frac{\partial\mathscr{L}}{\partial\dot\phi}\tag{4}$$ and $i=1.2.3$. Then the hamiltonian is a functional of $(\mathbb{R}^3\rightarrow\mathbb{R})$:

$$H(t)=\int_\mathscr{B} d^3x\mathscr{H}.\tag{5}$$ Let $\phi(t)$ and $\pi(t)$ be 2 functions $\mathbb{R}^3\rightarrow \mathbb{R}$. We define the Poisson bracket for 2 functionals $A[\phi,\pi]$ and $B[\phi,\pi]$: $$\{A,B\}=\frac{\delta A}{\delta\pi}\frac{\delta B}{\delta\phi}-\frac{\delta A}{\delta\phi}\frac{\delta B}{\delta\pi}\tag{6}$$ and we have the canonical relation (for $t,\vec{x},\vec{y}$ fixed): $$\{\pi(t,\vec{x}),\phi(t,\vec{y})\}=\delta^{(3)}(\vec{x}-\vec{y}).\tag{7}$$

How can we show that the equation of motions is now

$$\dot\pi=\{H,\pi\};\dot\phi=\{H,\phi\}~?\tag{8}$$

Answer

First, note that the EOM are the Euler-Lagrange equations:

$$\frac{\delta S}{\delta \phi}=\partial_\mu\left(\frac{\partial \mathscr L}{\partial\phi_{,\mu}}\right)-\frac{\partial \mathscr L}{\partial \phi}=\dot\pi+\partial_i\left(\frac{\partial \mathscr L}{\partial\phi_{,i}}\right)-\frac{\partial \mathscr L}{\partial \phi}$$ where I isolated the $\mu=0$ term, and used the definition of $\pi$.

Next, use $\mathscr L=\pi\dot\phi-\mathscr H$: $$\frac{\delta S}{\delta \phi}=\dot\pi+\partial_i\left(\frac{\partial \mathscr L}{\partial\phi_{,i}}\right)-\frac{\partial \mathscr L}{\partial \phi}=\dot\pi-\partial_i\left(\frac{\partial \mathscr H}{\partial\phi_{,i}}\right)+\frac{\partial \mathscr H}{\partial \phi}=\dot\pi+\frac{\delta H}{\delta \phi}$$ because $\pi$ and $\dot\phi$ are not functions of $\phi$.

Thus, we get $\dot\pi=-\frac{\delta H}{\delta\phi}=\{H,\pi\}$.

The other relation is easier: $$\{H,\phi\}=\frac{\delta \mathscr H}{\delta \pi}$$ Next, use $\mathscr H=\pi\dot\phi-\mathscr L$: $$\{H,\phi\}=\frac{\delta \mathscr H}{\delta \pi}=\dot\phi$$ because $\mathscr L$ is not a function of $\pi$.

ADDENDUM

Let $f(\phi,\pi)$ and $g(\phi,\pi)$ be two function on "Phase space". Then, by definition, $$\{f(\boldsymbol x),g(\boldsymbol y)\}\equiv\int\mathrm d\boldsymbol z\ \frac{\delta f(\boldsymbol x)}{\delta \phi(\boldsymbol z)}\frac{\delta g(\boldsymbol y)}{\delta \pi(\boldsymbol z)}-\frac{\delta g(\boldsymbol y)}{\delta \phi(\boldsymbol z)}\frac{\delta f(\boldsymbol x)}{\delta \pi(\boldsymbol z)}$$ where everything is evaluated at the same time $t$ (not writen for clarity). This means that $$\{H(t),\pi(t,\boldsymbol x)\}=\int\mathrm d\boldsymbol z\ \frac{\delta H(t)}{\delta \phi(\boldsymbol z)}\frac{\delta \pi(\boldsymbol x)}{\delta \pi(\boldsymbol z)}-\frac{\delta \pi(\boldsymbol x)}{\delta \phi(\boldsymbol z)}\frac{\delta H(t)}{\delta \pi(\boldsymbol z)}=\int\mathrm d\boldsymbol z\ \frac{\delta H(t)}{\delta \phi(\boldsymbol z)}\delta(\boldsymbol x-\boldsymbol z)$$ which equals $\frac{\delta H}{\delta \phi}$, as expected.

ADDENDUM II

Proof of $\frac{\delta H}{\delta \phi}=\frac{\partial \mathscr H}{\partial \phi}-\partial_i\left(\frac{\partial\mathscr H}{\partial\phi_{,i}}\right)$:

The dynamical variables of the Lagrangian are $\phi$ and $\partial_\mu\phi$. In the Hamiltonian formulation, we change $\partial_0\phi\leftrightarrow\pi$, so that the dynamical variables of the Hamiltonian are $\phi,\,\partial_i\phi$ and $\pi$.

With this in mind, the Hamiltonian is, by definition, $$H=\int\mathrm d\boldsymbol x\ \mathscr H(\phi(\boldsymbol x),\phi_{,i}(\boldsymbol x),\pi(\boldsymbol x))$$

Therefore, $$\delta H=\int\mathrm d\boldsymbol x\ \delta\mathscr H(\phi(\boldsymbol x),\phi_{,i}(\boldsymbol x),\pi(\boldsymbol x))$$ where $$\delta\mathscr H(\phi,\phi_{,i},\pi)=\frac{\partial \mathscr H}{\partial \phi}\delta\phi+\frac{\partial \mathscr H}{\partial \phi_{,i}}\delta\phi_{,i}+\frac{\partial \mathscr H}{\partial\pi}\delta\pi$$

Next, as we want $\frac{\delta H}{\delta \phi}$, we want to leave $\pi$ unchanged, so $\delta\pi=0$ (this is analogous to ordinary partial derivatives: when you calculate $\frac{\partial f(x,y)}{\partial x}$ you want to make a small displacement of $x$, while leaving $y$ unchanged)$\phantom{}^1$.

Anyway, in the integral over $\mathrm d\boldsymbol x$, we can integrate by parts the $\delta \phi_{,i}=\partial_i\delta \phi$ to make the derivative act on $\frac{\partial\mathscr H}{\partial\phi_{,i}}$:

$$\delta\mathscr H(\phi,\phi_{,i},\pi)=\frac{\partial \mathscr H}{\partial \phi}\delta\phi-\partial_i\left(\frac{\partial \mathscr H}{\partial \phi_{,i}}\right)\delta\phi+\text{surface terms}$$

Finally, back to $H$: $$\delta H=\int\mathrm d\boldsymbol x\ \left(\frac{\partial \mathscr H}{\partial \phi}-\partial_i\left(\frac{\partial \mathscr H}{\partial \phi_{,i}}\right)\right)\delta\phi$$ where I assumed that surface terms dont contribute. The expression in the parentheses is, by definition, the functional derivative of $H$ w.r.t. $\phi$.

$\phantom{}^1$: If we took $\delta\pi\neq 0$ and $\delta\phi=0$, we would get $\frac{\delta H}{\delta \pi}$ instead. All this is possible because the system is unconstrained, which is not true in some theories (such as the Dirac Lagrangian); in these cases, you can't use Poisson brackets, but Dirac brackets instead.

homework and exercises - Navier-Stokes system

I have to study this system which name is Navier-Stokes. Can you explain please what means that $p$, $u$ and $(u \cdot \nabla)u$. What represents in reality? Tell me please, how should I read the factor: $(u \cdot \nabla)u$? "$u$ multiplied with gradient applied to $u$ " ?

$(N-S)\begin{cases} -\mu \Delta u +(u \cdot \nabla)u+\nabla{p}=f &\mbox{in } \Omega, \\ \mbox{div }u=0 & \mbox{in } \Omega,\\ u_{\mid{\Gamma}}=0. \end{cases}$

one more question, what happens with with the system if $(u \cdot \nabla )u=0$ ? I found that the system describe the motion of a incompressible viscous fluid and it suppose the the motion is stationary but no slow, what means that stationary and that slow?

Answer

$(u \cdot \nabla)u$ is the so called advective acceleration term which arises when you consider the Navier-Stokes equations in an Eulerian frame of reference. It accounts for the effect that the we are following the particle as it moves around in the fluid, presumably to regions of the flow where the velocity is different. In contrast, if you consider the Navier-Stokes in Lagrangian coordinates, we are by definition tracking individual particles and therefore that term is not present. In large magnitudes, this term is highly-nonlinear and responsible for much of the more interesting behavior we see in fluid motion.

classical mechanics - What is the momentum mapping for a left translation group action?

This question is a follow-up to a previous question. In this survey of Symplectic Geometry by Arnol'd and Givental, the example of a constant-speed moving particle considered in the previous question was briefly mentioned on Page 62 as follows:

The action of the group by left translations on its cotangent bundle is Poisson. The corresponding momentum mapping $P:T^*G \to \mathfrak{g}^*$ coincides with the right translation of covectors to the identity element of the group.

We can take $G = \mathbb{R}$ to reduce to our example. Then $T^*G \cong \mathbb{R}^2$, $\mathfrak{g}^* \cong \mathbb{R}$, and the identity element of the group $\mathbb{R}$ is zero.

How does the momentum mapping $P: \mathbb{R}^2 \to \mathbb{R}$ coincide with the right translation of covectors to zero?

So I'm asking what exactly the map $P$ is and what Arnol'd and Givental mean by "the right translation of covectors to zero".

fluid statics - Hydrostatic pressure - doesn't density vary with depth?

Our class is learning about hydrostatic water pressure and we have been told that we can calculate the force of the liquid on an object at any depth using "the density x 9.8 x the depth". However, as the depth increases, wouldn't the density of the liquid increase because of the weight of the liquid above it compressing it? So should't there be something in the equation to account for the varying density? To me, "density x 9.8 x depth" seems like it is saying that the density will be constant...

Answer

The density does increase with depth, but only to a tiny extent. At the bottom of the deepest ocean the density is only increased by about 5% so the change can be ignored in most situations.

If you're dealing with these sorts of depths you also need to take temperature into account because the water temperature changes with depth and the density also changes with temperature.

lagrangian formalism - Is there any physics that cannot be expressed in terms of Lagrange equations?

A lot of physics, such as classical mechanics, General Relativity, Quantum Mechanics etc can be expressed in terms of Lagrangian Mechanics and Hamiltonian Principles. But sometimes I just can't help wonder whether is it ever possible (in the future, maybe) to discover a physical law that can't be expressed in terms of Lagrangian Equations?

Or to put it in other words, can we list down for all the physical laws that can be expressed in terms of Lagrangian equations, what are the mathematical characteristics of them( such as it must not contain derivatives higher than 2, all the solutions must be linear etc)?

Answer

Hamilton's dynamics occurs on a phase space with an equal number of configuration and momentum variables $\{q_i,~p_i\}$, for $i~=~1,\dots n.$ The dynamics according to the symplectic two form ${\underline{\Omega}}~=~\Omega_{ab}dq^a\wedge dp^b$ is a Hamiltonian vector field $$\frac{d\chi_a}{dt}~=~\Omega_{ab}\partial_b H,$$ with in the configuration and momentum variables $\chi_a~=~\{q_a,~p_a\}$ gives $${\dot q}_a~=~\frac{\partial H}{\partial p_a},~{\dot p}_a~=~-\frac{\partial H}{\partial q_a}$$ and the vector $\chi_a$ follows a unique trajectory in phase space, where that trajectory is often called a Hamiltonian flow.

For a system the bare action is $pdq$ ignoring sums. The Hamiltonian is found with imposition of Lagrangians as functions over configuration variables. This is defined then on half of the phase space, called configuration space. It is also a constraint, essentially a Lagrange multiplier. The cotangent bundle $T^*M$ on the configuration space $M$ defines the phase space. Once this is constructed a symplectic manifold is defined. Therefore Lagrangian dynamics on configuration space, or equivalently the cotangent bundle defines a symplectic manifold. This does not mean a symplectic manifold defines a cotangent bundle. The reason is that symplectic or canonical transformations mix the distinction between configuration and momentum variables.

As a result there are people who study bracket structures which have non-Lagrangian content. The RR sector on type IIB string is non-Lagrangian. The differential structure is tied to the Calabi-Yau three-fold, which defines a different dynamics.

classical mechanics - Hamilton-Jacobi formalism and on-shell actions

My question is essentially how to extract the canonical momentum out of an on-shell action.

The Hamilton-Jacobi formalism tells us that Hamilton's principal function is the on-shell action, which depends only on the coordinates $q$. Therefore, if $S$ refers to the off-shell action with Lagrangian $L$, a result for the canonical momentum is

$$\frac{\partial L}{\partial \dot q} = p = \frac{\partial S^\text{on-shell}}{\partial q}$$

In theory, when evaluating the on-shell action, one plugs in the EOM into $S$ and all that's left are boundary terms. If we illustrate this with the harmonic oscillator in 1D, one gets

$$\begin{align} S^\text{on-shell} &= \int_{t_i}^{t_f} L dt \\ &= \frac{m}{2} \int_{t_i}^{t_f} \left( \dot q^2 - \omega^2 q^2 \right) dt\\ &= \frac{m}{2} \int_{t_i}^{t_f} \left( \frac{d}{dt} (q \dot q) - q \ddot q - \omega^2 q^2 \right) \\ &= \frac{m}{2} \left[ q(t_f) \dot q(t_f) - q(t_i) \dot q(t_i) \right] \end{align}$$

Now, this depends on $\dot q$, whereas it should depend only on generalized coordinates. Also, the on-shell action depends on the boundary values of our coordinate $q$, so we can't differentiate w.r.t. $q(t)$ to get a conjugate momentum $p(t)$ for all times, like it is usually done in the Lagrangian formalism.

What am I not seeing?

string theory - Total Number of Dimensions in the Universe?

I have often heard that there are more than 4 (3 space and 1 time) dimensions of spacetime. What are the theories that say so, and how many does each predict? Has any experimental evidence been conclusive? And how are these extra-dimensional theories to agree with our daily observations?

general relativity - Can gravity be interpreted as the acceleration of spacetime towards an object?

Greetings StackExchange!

We were having a conversation with a peer about stupid ways of interpreting theories. We would go to and fro with an interpretations, but we would always find a way to disprove them. There was one, however, that we couldn't refute:

Gravity on a plane in a bidimensional space can be interpreted as the acceleration of spacetime towards an object.

We tried to invalidate in saying that the inertial mass and gravitational mass would be different, thus violating GR, but after some thought this couldn't be. The gravitational mass would accelerate spacetime towards it. If that object we were to be pushed, the pulling of spacetime would make it difficult for it to move.

We know for a fact this can't be true, and we would greatly appreciate any refutation of this thought. I can clarify if needed, or do a sketch. I apologise in advance as I'm not a native speaker.

Answer

Spacetime is not a thing so it can't accelerate. Spacetime is a manifold equipped with a metric.

However, in order to describe events in spacetime we construct coordinate systems, and coordinate systems can be accelerating. For example the Gullstrand-Painlevé coordinates describe the geometry around a black hole and they accelerate towards the black hole. So if you are stationary in the GP coordinates you are accelerating towards the black hole, or conversely if you are stationary with respect to the black hole you are accelerating outwards in the GP coordinates.

An analogy is often made with an observer flowing in a river - if you are stationary wrt the water you are moving wrt the river banks as the river carries you along. For this reason the GP coordinates are often described as the river model.

But, but, but, it is essential to be clear what is going on here. In the river model we are using an accelerating coordinate system i.e. it's the coordinate system that is accelerating not spacetime. As I mentioned at the outset, spacetime is not a thing and to attempt to ascribe properties like acceleration to it is meaningless.

newtonian mechanics - Why is the total energy of an orbiting system negative?

Assume it's an circular orbit. Object A orbits around object B. Take object B as frame of reference.

.$E=KE_a + GPE$

.$E=\frac 12m_av_a^2 +(-\frac {GM_bm_A}r)$

.$E=\frac 12m_a(GM_br)+(-\frac {GM_bm_a}r)$

.$E=-\frac {GMm}{2r} < 0$

What does negative total energy at any instant of time mean?

Answer

Negative energies are totally fine, because you had to pick a zero-point for energy. In your calculation you picked it to be at infinity. You could have chosen the zero-point for potential energy in such a way that your system had zero energy, or whatever. Only changes in energy are meaningful, in general.

Consider this: what happens if you add energy to this system? It gets closer to zero, and zero for us is the point where the particle is at rest, but is infinitely far away from the other particle. So negative energy represents the fact that to "free" the particle from the central potential requires you to add energy. This comes up a lot in quantum mechanics--the ground state energy of the hydrogen atom is -13.6 eV.

electromagnetism - Is magnetism an electrostatic or electrodynamic effect in the rest frame of the affected charge?

In textbooks, such as[1,2], magnetism is taught to be a consequence of relativistic length contraction. The magnetic force is derived in the special case of an infinite straight wire, by finding the total line charge density $\lambda_{tot}=\lambda_++\lambda_-$, and evaluation it as if it was a static charge. The Wikipedia page on the subject[3] states that "The chosen reference frame determines if an electromagnetic phenomenon is viewed as an effect of electrostatics or magnetism". I have found that this is not correct. The textbook analysis will only work in the special case of parallel motion next to an infinite wire.

This can easily be seen by considering perpendicular motion of the test charged with respect to the infinite wire. If we let the test charge move along the x-axis, and the charged wire run along the y-axis, the Lorentz transformations will be independent of y. All charges in the wire will have the same x' and t' coordinate, and there will be no length contraction along the y-axis. The line charge density will not change. $\gamma_-=\gamma'_-$ and $\gamma_+=\gamma'_+$. Hence the wire is electrostatically neutral in the rest frame of the test charge, and no force occurs.

The problem can be resolved by using the dynamic (retarded) electric field in both frames. (eq. 1) See my in-depth analysis in this link: https://drive.google.com/file/d/1HITikNdOX-IbxHmQVZVKQLATOrNXheXp/view?usp=sharing

$$E_D=\frac{q(1-v^2/c^2 )}{4 \pi_0r^2 (1-v^2/c^2 sin^2 ( \theta ))^{(3/2)}} \mathbf{\hat{r}} \tag{1}$$

I come to the conclusion that magnetism must be understood as an electrodynamic phenomenon in the rest frame of the test particle, and as a combined effect of field retardation and relativistic length contraction.

Question: From my analysis, it seems obvious that the magnetism is an electrodynamic effect, and yet I have found no mention of it in either textbooks or on the internet. There seems to be a common agreement that it is an electrostatic effect.

Am I missing something here, or is there really a wide spread misconception about this?

References
1 R. Feynman, The Feynman lectures on physics volume II, chapter 13.6

https://www.feynmanlectures.caltech.edu/II_13.html
2 David J. Griffiths, Introduction to electrodynamics, third edition, chap. 12.3.1
[3] Wikipedia: Classical electromagnetism and special relativity
https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Relationship_between_electricity_and_magnetism

Does dark matter originate from matter falling into a black hole?

Is it possible that due to the fact that dark matter is clumped around black holes, that the dark matter used to be matter, fell into a black hole, and simply became dark?

Furthermore, if the total "dark mass" of a galaxy is changing over time, maybe this is related to the amount of mass being consumed by the black hole(s) within it.

These are just some thoughts from the coffee table, harsh criticism welcomed.

Answer

This wouldn't work because the dark matter wouldn't be able to escape the black hole. Also, the measured mass of black holes contradicts this.

general relativity - Gravitational time dilation - Kip Thorne example

Please find in the images an example from Kip Thorne’s book ‘black holes & time warps’ on Gravitational Time Dilation. I do not understand the last sentences where the doppler effect results in the gravitational time dilation conclusion. I perceive it in this sense more as velocity time dilation?

quantum mechanics - How can particles travel in a straight line?

A particle can be set off in a certain direction by giving them momentum. Momentum is a vector, so the particle heads off in a specific direction. But the wave function of the particle allows it to obtain other momentum values, which would steer the particle on a different path. How then can we "shoot" electrons and other particles in straight lines? How can they maintain their momentum in the face of quantum uncertainty?

electrostatics - Why are the two outer charge densities on a system of parallel charged plates identical?

One of the ways examiners torture students is by asking them to calculate charge distributions and potentials for systems of charged parallel plates like this:

the ellipsis is meant to indicate any number of additional plates could be inserted where I've placed the ellipsis. The plates are assumed to be large enough that edge effects can be neglected i.e. they are effectively infinite.

In practice these problems are solved by the (frequently tedious) application of some simple rules, one of which is:

The charge densities on the two outermost surfaces (labelled $Q_{ext}$ in my diagram) are identical.

My question is how we can prove the statement that the two charge densities, $Q_{ext}$, are equal?

The problem is that our system of plates can have an arbitrary number of plates with arbitrary charges and spacings. Presumably one could with enough effort grind through a massive calculation where the plate charges and positions are all variables, but this seems an awfully complicated way of approaching an exceedingly simple observation. Surely there has to be a better way?

I am guessing that there must be some general argument one can make that means we get the same external charges regardless of the details of plates charges and spacings. Can anyone describe such an approach?

For bonus marks another of the starting points students are taught is that if any plate in the setup is earthed then the exterior charges are zero. If this doesn't emerge naturally from whatever approach is used to answer my first question I would also be curious to see how this could be proved.

quantum field theory - Is there a simple explanation for Schwinger's relation $g=2+frac{alpha}{pi}+{cal O}(alpha^2)$ for the $g$-factor of the electron?

Schwinger has on his grave (it seems) the relation between the g-factor of the electron and the fine structure constant:

$$g~=~2+\frac{\alpha}{\pi}+{\cal O}(\alpha^2)$$

Did Schwinger or somebody else ever give a simple explanation for the second term of the right hand side? The 2 appears from Dirac's equation. The second term is due to the emission and absorption of a photon. Is there a simple way to see that this process leads to the expression $\frac{\alpha}{\pi}$?

quantum mechanics - Can spin be related to a shift in angle?

If $\hat{T}(\Delta x) = e^{-\frac{i}{\hbar}\hat{p}\Delta x}$ is the spatial translation operator, then there exists a function $f$ from $\mathbb{R}$ to the ket space $V$ such that $\hat{T}(\Delta x) f(x) = f(x+\Delta x)$. Namely, the function that sends $x$ to the position eigenstate $|x\rangle$.

Similarly if $\hat{U}(\Delta t) = e^{-\frac{i}{\hbar}\hat{H}\Delta t}$ is the time evolution operator (for time-independent Hamiltonians), then there exists a function $f$ from $\mathbb{R}$ to $V$ such that $\hat{U}(\Delta t) f(t) = f(t+\Delta t)$. Namely, the function that sends $t$ to $|ψ(t)\rangle$, the state of a particle at time $t$.

And similarly if $\hat{R}_z(\Delta\theta) = e^{-\frac{i}{\hbar}{\hat{L}_z\Delta \theta}}$ is the orbital rotation operator, then there exists a function $f$ from $\mathbb{R}$ to $V$ such that $\hat{R}_z(\Delta \theta)f(\theta) = f(\theta+\Delta\theta)$. Namely the function which sends $\theta$ to $|r,\theta,\phi\rangle$, an eigenstate of the position operator $\hat{\theta}$ in spherical coordinates.

But my question is, if $\hat{R}_z(\Delta\theta) = e^{-\frac{i}{\hbar}\hat{J}_z\Delta \theta}$ is the intrinsic rotation operator, then does there exist a function $f$ from $\mathbb{R}$ to $V$ such that $\hat{R}_z(\Delta \theta)f(\theta) = f(\theta+\Delta\theta)$? By intrinsic rotation operator I mean the rotation operation related to spin angular momentum.

I suspect the answer is no, because there is no operator corresponding to $\theta$, the parameter of the intrinsic rotation operator, since in quantum mechanics it doesn't really make sense to think of spin as a particle rotating about its own axis. But then again, there is no time operator in non-relativistic quantum mechanics, and yet the time evolution operator satisfies the property.

In any case, assuming that the answer to my question is no, I'd like a formal proof that there cannot exist such an $f$.

EDIT: I posted a follow-up question here.

astrophysics - Why can't friction in an accretion disk radiate away enough of the angular momentum do create inward spiral?

As almost every source that I'd found explains, it was a challenge for scientists to understand the mechanism by which the total angular momentum in the accretion disc was conserved, while taking the latter (total conservation of angular momentum) in an accretion disc as a matter of fact. However, clearly, if such a large amount of friction as that it can generate enough light to outshine a thousand of Milky Ways (and a lot more X-ray radiation and heat) is present, it will radiate away enough of the accretion disk's energy for it to slow down and drop orbits, which is essentially losing angular momentum to radiation? Or is there any other way we know the total angular momentum of the accretion disc has to be conserved?

As you can probably see, I am not very advanced in this area of physics, so a simple as possible answer (quickly deducable or intuitive equations - or anything up to A-level (maybe basic university) physics - count as being simple) would be largely appreciated.

Answer

This is a curious question, since stars, black holes, white dwarfs, quasars etc do accrete matter and do find a way to transfer orbital angular momentum outwards whilst transferring mass inwards and that process does occur because of viscosity, but the angular momentum is not transported by radiation as you suggest.

The energy radiated comes about because of the inward mass flow, not vice-versa. The amount of angular momentum lost in the form of light is totally negligible. To carry angular momentum the light would need to be circularly polarised. Circular polarisation of light from sources such as quasars is quite limited (e.g. see Hutsmekers 2010) and even if the light were 100% polarised it could only carry $\hbar$ of angular momentum per photon. Thus, the angular momentum per energy unit would be $\omega^{-1}$, where $\omega$ is the frequency of the light, compared to the angular momentum per energy unit of a mass in a circular orbit of $2\Omega^{-1}$ (e.g. see here), where $\Omega$ is the orbital frequency. Since $\omega \gg \Omega$, then the angular momentum of the light is irrelevant - a body in orbit that loses energy just by emitting light loses a negligible amount of angular momentum and therefore cannot move inwards by this mechanism alone. To put it another way, a hot body in orbit that radiates away its thermal energy just turns into a cold body in the same orbit.

To fall into a lower orbit then a torque needs to be exerted because the specific angular momentum of an object in orbit is $\propto r^{1/2}$. Angular momentum is transferred outwards by torques associated with viscosity - at the microscopic level this is still poorly understood. Material in closer orbits moves faster than material further out. Turbulence or magneto-rotational instabilities may be capable of supplying the required viscosity. Angular momentum can also be lost from the system as a whole via winds from the disc surface or rotating jets propelled along the rotation axis; but not by electromagnetic radiation.

lightning - Why should you stay in the car during thunderstorms?

So there appears to be quite a bit of misinformation on the web as to why people should stay in their cars during a thunderstorm. So I'd like to clear some things up. One such non-nonsensical answer is that cars have rubber tires which insulate you from the ground. I believe this contributes little to nothing to the actual reason.

The "correct" answer appears to be because the car acts like a Faraday cage. The metal in the car will shield you from any external electric fields and thus prevent the lightning from traveling within the car.

However, what happens if you have an imperfect Faraday cage around you? Say for example, you had a window open. I think the car would still protect you a) because it still acts as a Faraday cage, albeit a bit not perfect and b) because electrons will travel the path of least resistance which would be through the body of the car and not through you.

Now going with my b) reasoning, wouldn't you be just as safe standing next to a giant conductive pole (i.e. a lightning rod)? Wouldn't the lightning just go through the lightning rod and you'd be 100% safe?

Also a side question: lightning is essentially just an huge electric arc from the clouds to the ground, correct?

Answer

Yes, if it is not a plastic covered car it is an effective Faraday cage.

If the tires are such that the car is insulated electrically, if it is hit it will take some time to discharge to the ground, but still the passengers would be safer than standing next to it outside. I have learned that modern tires are particularly constructed so that the static charge generated by the friction on the road is discharged so that would also help. ( in olden times they used to have chains trailing from the trucks in order to discharge the static. Recently I saw a car with a discharger too, trailing on the road!).

lightning is essentially just an huge electric arc from the clouds to the ground, correct?

Wrong, the current actually may start from the ground. That is the rational of the lightning rods, to create a path for a current to be generated by the potential difference to the cloud and to meet the current from the clouds in a prefered location instead of a random one. It is not wise to stand next to a rod, read in the link the amount of power dissipated by a bolt.

The average peak power output of a single lightning stroke is about one trillion watts — one "terawatt" ($10^{12}$ W ), and the stroke lasts for about 30 millionths of a second — 30 "microseconds".[18]

And it is not wise to stand, because you may also give rise to leaders that will meet the lightning path. If in the open it is best you fall on the ground as much sheltered as possible.

A colleague once was about 20 meters from a lightning bolt, and he was so shocked by the sound and fury, it took him a week to come down to normal.

quantum field theory - Identically vanishing trace of $T^{munu}$ and trace anomaly

Let us consider a theory defined by an action on a flat space $S[\phi]$ where $\phi$ denotes collectively the fields of the theory. We will study the theory on a general background $g_{\mu\nu}$ and then we will set the metric to be flat.

The Euclidean partition function of the theory in the presence of an external source is

$$Z[J] = \int [d\phi] e^{-S -\int d^d x\, J \, \mathcal{O}}\tag 1$$

where $\mathcal{O}$ can be either an elementary or a composite field (in what follows we will take it to be the trace of the energy-momentum tensor).

Now, a very well-known result is that a traceless energy momentum tensor implies conformal invariance; indeed, under a conformal transformation $g_{\mu\nu} \rightarrow f(x)g_{\mu\nu}$ such that

$$\partial_{(\mu}\epsilon_{\nu)} = f(x) g_{\mu\nu}$$

the action transforms like

$$\delta S = \frac{1}{d}\int d^d x T^{\mu}_\mu \partial_\rho \epsilon^\rho$$

Now, another well-known result states that in a generic background metric the expectation value of $T^\mu_\mu$ is not zero but depends on the Weyl-invariant tensors and the Euler density, that is

$$\langle T^\mu_\mu \rangle = \sum a_i E_d - c_i W_{\mu\nu\rho....}^2$$

where $\langle T^\mu_\mu \rangle$ is usually defined by the variation of the connected vacuum functional $W = \log Z[J]$ under variations of the metric.

First question. Is $\langle T^\mu_\mu \rangle$ calculable in the usual way using the partition function? That is setting $\mathcal{O} = T^\mu_\mu$ in Eq.(1) we compute

$$\langle T^\mu_\mu \rangle = \frac{\delta}{\delta\, J} Z[J]\Big|_{J=0}\tag 2$$

Second question. If the answer of the first question is YES, then I would expect the $\langle T^\mu_\mu \rangle$ computed as the variation of the connected vacuum functional $W[J]$ is the same as the one computed in Eq.(2). Is this true?

Third question.

There are two ways the classical traceless condition can be realized:

1. on-shell; then, $T^{\mu}_\mu$ is not identically zero but it is so once you apply the equation of motion, e.g. $\lambda \phi^4$ theory in d=4.


2. $T^\mu_\mu$ is identically zero; that is, you don't need to use the equation of motion (e.g. massless scalar field in d=2 on a curved background)

In the first case, since $T^\mu_\mu$ vanishes on the equation of motion, I agree that it may receive quantum corrections through the coupling of the theory to a curved space; everything is ok.

In the second case instead, namely $T^\mu_\mu$ identically zero, I am unable to compute its expectation value from Eq.(1) and Eq.(2) since $\mathcal{O}=0$ identically; that is, the RHS of Eq.(2) is zero, since Z[J] is actually J-independent. This would imply $\langle T^\mu_\mu \rangle=0$.

Is it still true that the theory enjoys an anomaly on a curved space background? I would say YES, since the anomaly depends only on the central charges. How to resolve this apparent contradiction?

quantum mechanics - Why not use the Lagrangian, instead of the Hamiltonian, in nonrelativistic QM?

Undergraduate classical mechanics introduces both Lagrangians and Hamiltonians, while undergrad quantum mechanics seems to only use the Hamiltonian. But particle physics, and more generally quantum field theory seem to only use the Lagrangian, e.g. you hear about the Klein-Gordan Lagrangian, Dirac Lagrangian, Standard Model Lagrangian and so on.

Why is there a mismatch here? Why does it seem like only Hamiltonians are used in undergraduate quantum mechanics, but only Lagrangains are used in quantum field theory?

Answer

In order to use Lagrangians in QM, one has to use the path integral formalism. This is usually not covered in a undergrad QM course and therefore only Hamiltonians are used. In current research, Lagrangians are used a lot in non-relativistic QM.

In relativistic QM, one uses both Hamiltonians and Lagrangians. The reason Lagrangians are more popular is that it sets time and spacial coordinates on the same footing, which makes it possible to write down relativistic theories in a covariant way. Using Hamiltonians, relativistic invariance is not explicit and it can complicate many things.

So both formalism are used in both relativistic and non-relativistic quantum physics. This is the very short answer.

differential geometry - Why is the Taub-NUT instanton singular at $theta=pi$?

Consider the following metric

$$ds^2=V(dx+4m(1-\cos\theta)d\phi)^2+\frac{1}{V}(dr+r^2d\theta^2+r^2\sin^2\theta{}d\phi^2),$$

where $$V=1+\frac{4m}{r}.$$

That is the Taub-NUT instanton. I have been told that it is singular at $\theta=\pi$ but I don't really see anything wrong with it. So, why is it singular at $\theta=\pi$?

EDIT:: I have just found in this paper that the metric is singular "since the $(1-\cos\theta)$ term in the metric means that a small loop about this axis does not shrink to zero at lenght at $\theta=\pi$" but this is still too obscure for me, any clarification would be much appreciated.

quantum mechanics - Demonstration that vibrating basic particles constitute non-vibrating individuals

I am a dilettante in physics; I ask for pardon for my confusion-causing (if any) terminology usage, and also for my imprecise choice of question tags.

I know that basic particles of any individual stopping vibrating contradicts Heisenberg's uncertainty principle, so the particles must whenever vibrate. Then how could basic particles constitute a seemingly non-vibrating individual? Is it a natural limitation of human eyesight or ...? Is there any scientific explanation for it?

A demonstration not using analogy or common sense is seeking after.

Answer

It is certainly true that as I type this the atoms in my fingers are vibrating, and so are the atoms in my keyboard. Yet I can still type. There are several reasons we don't perceive any vibration.

Firstly, the vibrational frequencies are around $10^{12}$ to $10^{14}$Hz, that's from a trillion times a second to 100 trillion times a second. Since we can't even see the 50Hz refresh on our TV screens it shouldn't be any surprise that at such high frequencies we can't detect any vibration of the atoms in us.

Secondly the vibrations are all in random directions, so on average they balance each other out. On average we aren't vibrating.

Lastly, and this is really the important point, atoms are quantum objects. We tend to visualise a vibrating atom as a little ball flying to and fro on the end of a spring but this isn't a good description of what a vibrating atom looks like. The vibration means the wavefunction of the atom spreads out, so in effect it just becomes a little blurry. Since this blurriness is on a scale of around 0.1 nm it is completely undetectable in everday life.

newtonian mechanics - Which is the right explanation for rocket motion?

What actually causes a rocket to move? Is it the pressure in the rocket engine or the amount and velocity of mass that is being ejected out.

The reason I am asking is, I found these two explanations for a rocket motion. Are they same or different. If same, is there any correlation between them?

mass - Law of conservation of matter

If scientist have made small particles of matter then why do we still haw the law of conservation of matter? Is it because the few particles don't make a noticeable difference in our life?

Answer

For one thing, you're probably thinking of the law of conservation of mass. Calling it "conservation of matter" is technically inaccurate because matter isn't a quantitative property of a physical system, and only such properties have conservation laws.

Now with that detail out of the way: there isn't really a law of conservation of mass, either. But you can have an effective law of mass conservation, if you can safely ignore any processes that convert particles with one mass to particles of a different mass. In daily life, we can always do that, because the kinds of processes that involve particles of different masses just don't occur under normal (for us) conditions. But if you're dealing with high-energy particles, there are lots of processes where the masses of the inputs and outputs differ. That's how we can create bits of matter, by taking advantage of those processes.

fluid dynamics - Derivation of Kelvin's circulation theorem

In the derivation of Kelvin's circulation theorem, I take the material derivative of circulation, or

$$\begin{align} \dfrac{D\Gamma}{Dt} = \dfrac{D}{Dt}\oint_C \vec{u} \cdot d\vec{\ell}. \end{align}$$

Moving the material derivative inside the integral gives

$$\begin{align} \dfrac{D\Gamma}{Dt} = \oint_C \dfrac{D\vec{u}}{Dt} \cdot d\vec{\ell} + \oint_C \vec{u} \cdot \dfrac{D(d\vec{\ell})}{Dt}. \end{align}$$

Given that $d\vec{\ell} = ds \hspace{1mm} \vec{t}$, the second integral of the equation above becomes

$$\begin{align} \oint_C \vec{u} \cdot d\vec{u}. \end{align}$$

How does one arrive at this equation? My attempt would be to use the chain rule so that $$\begin{align} \dfrac{D(ds \vec{t})}{Dt} = ds\dfrac{D\vec{t}}{Dt} + \vec{t}\dfrac{D(ds)}{Dt}. \end{align}$$

From here, I expand each material derivative as

$$\begin{align} \dfrac{D\vec{t}}{Dt} = \dfrac{\partial \vec{t}}{\partial t} + u_t\dfrac{\partial \vec{t}}{\partial s}+ u_n\dfrac{\partial \vec{t}}{\partial n}\\ \dfrac{D(ds)}{Dt} = \dfrac{\partial ds}{\partial t} + u_t\dfrac{\partial ds}{\partial s}+ u_n\dfrac{\partial ds}{\partial n}. \end{align}$$

I am unsure how these two relations will simplify so that I get $$\dfrac{D(d\vec{\ell})}{Dt} = d\vec{u}.$$

quantum mechanics - Is it theoretically possible to reach $0$ Kelvin?

I'm having a discussion with someone. I said that it is -even theoretically- impossible to reach $0$ K, because that would imply that all molecules in the substance would stand perfectly still.

He said that this isn't true, because my theory violates energy-time uncertainty principle. He also told me to look up the Schrödinger equation and solve it for an oscillator approximating a molecule. See that it's lowest energy state is still non-zero.

Is he right in saying this and if so, can you explain me a bit better what he is talking about.

Answer

By the third law of thermodynamics, a quantum system has temperature absolute zero if and only if its entropy is zero, i.e., if it is in a pure state.

Because of the unavoidable interaction with the environment this is impossible to achieve.

But it has nothing to do with all molecules standing still, which is impossible for a quantum system as the mean square velocity in any normalized state is positive.

quantum mechanics - Don't all waves transport mass?

How do matter waves not transport mass? I know that matter waves are associated with moving sub-atomic particles(which is insignificant for macroscopic particles). If a wave is associated with a moving particle, then how can this kind of wave not involve transport of mass? As detailed in this paragraph written by Glenn Elert in his Physics Hypertextbook:

"The nature of the waves associated with elementary entities are probability waves — unitless numbers, numerical ratios. They tell you the probability of finding a particular particle at a particular place and time and nothing else. They do not measure the value of any physical quantity. The waves themselves carry no mass, no charge, no energy, no momentum, no angular momentum, no information of any sort other than the likelihood of existence. In essence, they carry information only. Nothing else."

I am really confused, any help is greatly appreciated.

Answer

One must distinguish the underlying quantum mechanical framework from the emergent classical mechanics and electrodynamics framework when discussing waves.

In classical mechanics wave equations are solutions of differential equations which depend on the $(x,y,z,t)$ variables of massive ideal particles which are derivable from differential equations . These equations are called wave equations from the everyday terminology which first arose from water waves, and water waves can be described by a differential equation, called a wave equation. The solutions are sinusoidal functions which fit the waves in $(x,y,z,t)$ . Energy is transported by these waves, but there is not much mass motion. Classical electromagnetic waves also transport energy.

The confusion comes from the de Broglie interpretation which is applicable to the quantum mechanical level of elementary particles, atoms , molecules ... The de Broglie wavelength:

$$\lambda = \frac{h}{p}$$

This labeling as "matter waves" is what causes the confusion and should not really be used. The diffraction experiments with electrons show an associated wave interference pattern with the deposition of electrons on screens ( and measuring instruments) , and the de Broglie wavelength which depends on the momentum is verified by experiment. BUT the wave differential equation which describes accurately the behavior of the electrons is a second order differential equations which only predicts probabilities, according to the postulates of quantum mechanics.

So it is not that the electron is spread out all over the available phase-space space. It is the probability of the electron to be found at $(x,y,z,t)$ that controls the diffraction patterns of the experiment.

Electron buildup over time

One always measures a whole electron, in any experiment, not fractions of it. Each dot is one whole electron in this single electron double slit experiment. It is the probability of finding it that is "waving" and generating the interference pattern. The end result looks like a matter wave, but it is a bad terminology, because no matter is waving.

Edit incorporating part of comments:

From matter transport the question in comments became of whether a quantum mechanical particle has a specific reality, trajectory.

Obviously classical trajectories exist otherwise bullets would not be sure to find the target. Mathematical functions of classical mechanics describe classical trajectories with great accuracy, macroscopically , i.e at dimension where the Heisenberg uncertainty principle is automatically fulfilled because of the large numbers involved. The HUP is in a way a mnemonic of quantum mechanical behavior, because it is the direct result of the existence of non zero commutator relations which are at the basis of quantum mechanical theory. The HUP gives an envelope in the two non commuting variables, for example momentum and position, which are relevant for trajectories, where the probabilistic nature of quantum mechanics dominates and any particular boundary problem solutions will have to be bounded by.

To understand the difference between basic uncertainty and experimental error, take this example: the line drawn by a compass and a lead pencil, which with mathematics we idealize as a perfect circular trajectory. This is not true "experimentally" as a microscope would show all the little bits of lead dispersed by the pencil . This is due to the experimental construct of the pencil and has an experimental error sigma(x). Since we are talking of a mathematical function we are free to imagine the in reality the circle is perfect on as small errors as one can make. This holds true for the mathematical functions of classical mechanics.

What happens at the quantum mechanical level, when $\hbar$ becomes commensurable with the variables measured, the uncertainty is inherent to the way that nature behaves,

The bubble chamber picture of an electron is a good example. It is curving in a helix in a magnetic field, and the little dots that make up the track are small interactions with the hydrogen atoms, tiny dots of kicked off electrons ionizing even more atoms and making up the dots. The continuous energy loss reduces the radius of the theoretical spiral to the effect seen. This is the macroscopic picture and we call it the trajectory of the electron. We have found out that one cannot go to a microscopic detection that will give a specific trajectory within the HUP volume, the electron is within a fuzzy circle, which is the probability of finding the electron within the circle, the non-zero-probability-of-interaction circle. Thus there are no trajectories in the quantum mechanical frame, just loci of probability for an interaction of the particle under examination.

homework and exercises - Time period of simple pendulum with varying mass

How do you find time period as a function of time for a simple pendulum that is in the form of a hollow sphere that is filled with mercury and there is a hole in the bottom through which the mercury is constantly falling at a fixed rate? I tried creating a function for the time period to check how it varies with the mass of mercury, but I had too many variables.

EDIT: This question was a concept based multiple choice question in my homework book. The option were time period becomes erratic, time period increases, time period decreases, time period first increases then decreases.

higgs - A question about spontaneous symmetry breaking at macroscopic limit

The question relates to this post.

Spontaneous symmetry breaking is somehow a volume effect, that in-principle only happens at infinity large system. Weinberg in the second volume of his QFT used a chair demostrated the lost of rotational symmetry due to large scale of the chair.

p. 163

We do not have to look far for examples of spontaneous symmetry breaking. Consider a chair. The equations governing the atoms of the chair are rotationally symmetric, but a solution of these equations, the actual chair, has a definite orientation in space. Here we will be concerned not so much with the breaking of symmetries by objects like chairs, but rather with the symmetry breaking in the ground state of any realistic quantum field theory, the vacuum.

p.164-165

Spontaneous symmetry breaking actually occurs only for idealized systems that are infinitely large. The appearance of broken symmetry for a chair arises because it has a macroscopic moment of inertia $I$, so that its ground state is part of a tower of rotationally excited states whose energies are separated by only tiny amounts, of order $\frac{{\hbar}^2}{I}$. This gives the state vector of the chair an exquisite sensitivity to external perturbations; even very weak external fields will shift the energy by much more than the energy difference of these rotational levels. In consequence, any rotationally asymmetric external field will cause the ground state or any other state of the chair with definite angular momentum numbers rapidly to develop components with other angular momentum quantum numbers. The states of the chair that are relatively stable with respect to small external perturbations are not those with definite angular momentum quantum numbers, but rather those with a definite orientation, in which the rotational symmetry of the underlying theory is broken. For the vacuum also,

To a very good approximation, the chair is described by quantum electrodynamics $$L= -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} + i \bar{\psi}(\gamma^{\mu}D_{\mu}-m) \psi$$ The rotation of chair corresponds to $SO(3)$ group as a subgroup of the Lorentz group $SO(1,3)$. If vacuum is similar with the chair, the rotational symmetry could also break together with global rigid $O(N)$ symmetry. The standard argument that Higgs boson being a scalar, is to keep Lorentz invariance. If the $SO(3)$ rotational symmetry is lost, then this argument is not necessarily valid. Is there any other rationalization for the Higgs boson being scalar?

If vacuum is not similar with the chair, why only global rigid $O(N)$ symmetry is broken but not $SO(3)$? Is that a pure guess then confirmed experimentally?

quantum field theory - Measuring and calculating free quark masses

Particle data book contains masses of the free quarks. I wonder, how do experimentalists determine the masses of the free quarks even though they are trapped inside hadrons (except perhaps in quark-gluon plasma)? Can quark masses be theoretically calculated and how much do they agree with experiments?

Answer

Nobody has seriously calculated theoretically a quark mass from first principles. So there is no issue of agreement with experiment. They are parameters in experimental fits, but sometimes remarkably consistent across a broad range of experiments-- and the QCD/EW calculations using them as inputs. If someone pretends to know their origin, he/she is bluffing.

Constituent quark masses have been fitted to hadron masses using simple potential models in the style of additive nuclear models. They range from 0.34Gev to 177GeV and are not too informative, as they are mostly glue for the light quarks, i.e. they arose out of chiral symmetry breaking---see below. They are not of any use in QCD/EW perturbative calculations, and, culturally, they have a life of their own, of possible assistance to diffractive scattering studies.

The more fundamental current quark masses that are used in the fundamental lagrangian of the SM are extracted from DIS fits (deep inelastic scattering), for the heavy ones; but, mostly for the light ones, they are extracted, sometimes with the aid of lattice theories, out of chiral perturbation theory on the masses of the pseudoscalar mesons, via Dashen's formula of chiral symmetry breaking.

They range from 2MeV to 175GeV. For example, stringing several such formulas together one finds fits such as the popular one, $$\frac{m_u+m_d}{2m_s}=\frac{m_\pi^2}{2m_K^2-m_\pi^2}\approx \frac{1}{25}~.$$

solid state physics - Macroscopic quantum phenomena and tunneling

Wikipedia's article on macroscopic quantum tunneling says

Quantum phenomena are generally classified as macroscopic when the quantum states are occupied by a large number of particles (typically Avogadro's number) or the quantum states involved are macroscopic in size (up to km size in superconducting wires).

To comply with copyright laws, the following is an edited paraphrase of this reference, pp6-7.

http://assets.cambridge.org/97805218/00020/sample/9780521800020ws.pdf

The term “dynamical degrees of freedom” should be used carefully. Imagine a baseball moving through a wall without being compressed. Certainly, this phenomenon can be called a macroscopic tunneling; since the ball is a collection of atoms, the number of degrees of freedom is comparable to the number of atoms.

Macroscopic tunneling depends on the number of microscopic degrees of freedom like the positions of constituent atoms. Collective degrees of freedom are superior: they are singled out by rearranging the microscopic ones.

Are there any circumstances under which the ball could pass through the wall via macroscopic quantum tunneling, or is this wishful thinking?

On what basis do we trust Conservation of Energy?

I'm happy to accept and use conservation of energy when I'm solving problems at Uni, but I'm curious about it to. For all of my adult life, and most of my childhood I've been told this law must hold true, but not what it is based on.

On what basis do we trust Conservation of Energy?

Answer

Let me expand a bit on Manishearth's answer. There's an idea going back a long time called the principle of stationary action. See http://en.wikipedia.org/wiki/Principle_of_stationary_action for a description that isn't too mathematical. In the 18th and 19th centuries century the mathematicians Lagrange and Hamilton found ways of using this to describe mechanics. Then in the early 20th century the mathematician Emmy Noether discovered that in Lagrangian mechanics if a symmetry of the equations existed this meant there was a corresponding conservation law. As Manishearth says, one example of this is that time symmetry means that energy must be conserved.

Strictly speaking, the symmetry involved is "shift symmetry of time". This means that if I do an experiment, the time I do it doesn't matter so I'd get the same result tomorrow as I do today. If this is true Noether's theorem means that energy must be conserved.

Experimentally we find that repeating experiments does indeed give the same results, and we also find that everything observed so far obeys Lagrangian mechanics. This suggests that energy is indeed conserved. Strictly speaking this is an experimental observation not a proof, but few doubt that the principle applies as the universe would be a strange place if it didn't.

Wikipedia has lots of articles on Langrangian Mechanics and Noether's theorem, but they're a bit intimidating for the non-mathematician. If you're interested in knowing more Googling should find you plenty of more accessible articles.

simulations - Is there a mass spectrometer software that can measure the mass of a proton

I spent a lot of hours looking for mass spectrometer software (or Cathode-ray simulator) that can measure the mass of a proton or any sub atomic particle. Until now I only found software that work on isotopes, organic mixtures and other compounds.

For my school project, I need a computer simulation to measure the mass of a proton (I chose the idea not my school) so any other way of measuring a proton will not be useful to me.

Thanks in advance.

general relativity - Black Hole: mass density or energy density?

Recently, I read a quora post in which the OP asked the following question: Can a black hole be formed from accelerating a body and increasing its relativistic mass to the level of a Schwarzschild black hole? This question had me confused because of an apparent conflict of principles:

1. 
   

   On the one hand, a well received answer to the quora post suggests that formation of black holes are only dependent upon the invariant rest mass, instead of the relativistic mass. Therefore, acceleration of any body to arbitrary speed would not cause a black hole to form. (See: https://www.quora.com/Is-it-possible-to-accelerate-matter-so-much-that-it-becomes-a-black-hole-in-one-frame-of-reference-and-not-in-another)

   
   
   


2. 
   

   On the other hand, a quick search on wikipedia indicates that kugelblitz (black hole formed from concentration of energy, possibly with zero rest mass) is theoretically possible.

   
   

So my questions are:

1. 
   

   Are there theoretical models for how a kugelblitz could possibly form? (The wikipedia page mentioned Wheeler's geons. But I do not have access to his paper and cannot tell if geon solutions are actually black hole solutions in any sense)

   
   



2. 
   

   If the answer to question 1 is yes, energy can indeed create black holes. So, from some reference frame A moving rapidly away from the earth (or really anything for that matter), will the earth have enough energy density to become a black hole in A?

   
   

Personally I believe that both answers should be no. But I'd love to hear from experts who know a lot more to judge.

Answer

A fast moving mass will not become a black hole. See the existing question If a 1kg mass was accelerated close to the speed of light would it turn into a black hole? for a discussion of this.

However suppose you take two fast moving 1kg masses moving in opposite directions and collide them. In the centre of mass frame you have the original rest masses, i.e. 2kg, but you have also concentrated all the kinetic energy into a small volume at the point of collision. So at the moment of collision we have everything at rest and a very high energy density. If you make the speeds of the 1kg masses high enough this will form a black hole.

I have to confess I don't know if this argument applies to light beams because unlike the masses a light beam has no rest frame. You cannot argue that the point in between two colliding light beams is a rest frame for the collision. However it seems plausible that by concentrating the light from multiple beams into a small volume you could create a black hole.

If it turns out that neutrinos do travel at faster than lightspeed, how will the success of special relativity be explained?

As per in the title. If it turns out that neutrinos do travel faster than the speed of light, how will the success of special relativity be explained? My apologies if this has been asked before; I've just browsed through few questions, and I couldn't find one that matches (although it was slightly touched on in one answer).

Answer

The various theoretical options are very different in nature, and the answer to this question almost defines the option.

1) Relativity is wrong, there is objective absolute time, Lorentz symmetry is emergent (as in electrodynamics before Einstein), and going faster than light doesn't create any time-loop paradoxes.

2) Relativity is valid, but neutrinos, not photons, travel at the actual relativistic speed-limit; photons have a small mass, or photons are confined to a braneworld while neutrinos can take shortcuts through hyperspace.

3) Relativity is valid, neutrinos are tachyons (or the OPERA result manifests some other sort of spacelike correlation), and something (Deutsch's multiverse, Hawking's CTC instability, weakness and/or uncontrollability of the effect, QM itself actually comes from CTCs...) prevents it from creating a time-loop paradox.

It should be extremely difficult to turn any of these 3 options into a fully working theory capable of reproducing the standard model. The odds are still heavily against OPERA having actually observed FTL neutrinos.

particle physics - Why is the spectrum of the $beta$-decay continuous?

the spectrum of the Gamma and Alpha decays are both discrete, i.e. the $\alpha$-particles and the $\gamma$-rays take on only discrete values when emitted from a decaying nucleus.

Why is it then, that the $\beta^{\pm}$ can take on continuous values?

The main thing that distinguishes the beta decay from the other two is, that it is a three body problem, i.e. the nucleus does not only decay into an electron/positron, but also into a electron-neutrino/antineutrino. I don't see yet how this immediately implies that the spectrum of the electrons is continuous though.

The way I understand the two body decays is, that the initial nucleus spontaneously decays into the two bodies, i.e. a smaller nucleus and a gamma or alpha particle. As the energy levels in both the nuclei are quantized, only certain values for the energy of the photon and the Helium core are allowed, sine Energy and momentum need to stay conserved.

Does the third particle change things in the way that there basically are two things (i.e. the electron and the neutrino in the beta decay) that are not restricted by an inner energy level hierarchy in the way the nuclei are, thus allowing the energy given by the nuclei to be split arbitrarily (and continuously) between the electron and the neutrino?

If this is the wrong explanation, please correct me.

Answer

At first, consider two particles decay:

$A\rightarrow B + e^-$ Where A is initially rest. So $\vec{p_B}+\vec{p_{e^-}}=0$

now $$\begin{align} \frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released} \\ \frac{p_{e^-}^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released} \tag{1} \end{align}$$

see here you have uncoupled equation (equ.1) for $p_{e^-}$ .. So, solving above (equ.1) you will get a fixed $p_{e^-}$. Hence the energy($\sqrt{p_{e^-}^2+m_{e^-}^2}$) of the $\beta$ particle is always fixed in the two body $\beta$ decay. (you can find $p_{B}$ too using the momentum conservation formula)

At first, consider three particles decay:

$A\rightarrow B + e^-+\nu_e$ Where A is initially rest. So $\vec{p_B}+\vec{p_{e^-}}+\vec{\nu_e}=0$

so $$\begin{align} \frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}+p_{\nu_e}&=E_{released} \\ \frac{(\vec{p_{e^-}}+\vec {p_{\nu_e}})^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}+p_{\nu_e}&=E_{released} \tag{2} \end{align}$$

Now see in contrast to the two particles decay equation, here we have five unknowns $\big{(}p_{e^-},p_{\nu},p_{B},\theta\ (\rm the\ angle\ between\ \vec{p_{e^-}},\vec{p_{\nu}}),\phi\ (\rm the\ angle\ between\ \vec{p_{e^-}},\vec{p_{B}})\big)$ but four coupled equations *. so we can't solve them uniquely. That's also what happens physically. You will get different values of $p_{e^-},p_{\nu},p_{B},\theta,\phi$ satisfying the four coupled equations. Hence different $\beta$ particles will have different energy($\sqrt{p_{e^-}^2+m_{e^-}^2}$) maintaining the statistics of decay process. Hence the continuous spectra.

*The four coupled equations are equ.2 and three equations which we can get by taking Dot products of $\vec{p_{e^-}},\vec{p_{\nu}}\rm\ and\ \vec{p_{B}}$ with the momentum conservation($\vec{p_B}+\vec{p_{e^-}}+\vec{\nu_e}=0$ ) and remember they lie in a plane so two angles ($\theta\rm\ and\ \phi$) are sufficient.

Can't quantum teleportation be superluminal some percentage of times?

I apologize if this is a really silly question.

In the (textbook) quantum teleportation algorithm, in the step right after Alice has measured her system but before she has sent her classical information to Bob, she is about to send one of the following values: 00,01,10,11.

What if Bob doesn't want to wait and simply takes a guess? Wouldn't there then be superluminal communication 25% of the time?

Answer

This is really a subtle point. You are right that in 25% of the cases, Bob will randomly chose the "correct" measurement basis and thus get the correct value.

However, there is no way for Bob to know when he has actually chosen the right basis and when he has chosen the wrong basis, so his measurement outcome does not contain more information that a random coin-toss.

It is only when the information from Alice (regarding which basis to measure in) has reached him, that he can make use of his earlier (75% erroneous) measurements.

It is in this sence that information cannot propagate faster than the speed of light.

newtonian mechanics - Positive and Negative works

When we displace a body from ground to a height we do positive work against gravity and gravity too does work but negative. Do they cancel each other? Or does the external force do extra work?

Answer

Since the body is at rest at the beginning and at the end of the motion, the total change in kinetic energy is zero. This means that the total work done on the body is zero, which means that the work done by the lifting force and gravity canceled each other out. In other words, $$\Delta K = W$$ or the change in kinetic energy of a body is equal to the work done on it.

Now, sometimes it is useful to separate work done by conservative forces (like gravity) and non-conservative forces, so $$\begin{align} \Delta K &= W_{cons} + W_{non-cons} \\ &= F_{cons}\Delta x + W_{non-cons} \\ &= -\Delta U + W_{non-cons} \end{align}$$ where the subscript $cons$ means conservative and $non-cons$ means non-conservative, with $F_{cons}$ referring to the sum of all conservative forces. These forces act through a distance $\Delta x$. The symbol $\Delta U$ refers to a change in potential energy, which is equal to the negative of the work done by a conservative force. If we define the total mechanical energy of a body as the sum of kinetic and potential energy $$E = K + U$$ then we can write $$\Delta E = \Delta K + \Delta U = W_{non-cons}$$ So, the total change in mechanical energy of a body is equal to the work done by non-conservative forces. If all of the forces on a body are conservative, then total mechanical energy is conserved.

standard model - Why doesn't a changed particle ever lose energy by interacting with others by radiation of virtual photons? Are all virtual photons exchanged?

I've had it explained to me in a separate post that charged particles are constantly exchanging virtual particles with other charged particles and their energy is a steady state. How it is a surety that all of the virtual photons sent out by the charged particle will be exchanged with other charged particles? Are the particles sent in a directed manner toward other charged particles? How much do we know about this interaction/mechanism? Are some virtual particles just radiated away, leaving the charged particle with less energy?

astronomy - Why can I never see any stars in the night sky?

I have always lived near a large city. There is a stark contrast between the picture linked below for example, and what I see with the naked eye.

Sometimes I can see a few stars here and there, but usually they can be counted on my fingers. I'm looking up at the night sky right now, and there isn't a single visible star!

Question: What creates this large gap in what I can see?

Here's the photograph I was talking about: http://apod.nasa.gov/apod/ap120123.html

Answer

This is a simple and clear issue, with a unique answer. I see other replies mentioning weather conditions, dark adaptation and so on. That's just so much hand waving, given that the first thing you said was "I've always lived in somewhat large cities".

The core problem here, by a very wide margin, is light pollution if you live in a large city. This is the one factor, above everything else, that affects your ability to see the stars.

Here's a light pollution map:

http://www.jshine.net/astronomy/dark_sky/

The white zones are the worst, and they are in the middle of the cities. Black zones are the best.

Here's a somewhat better (but not perfect) comparison of a dark sky versus light polluted sky (your picture was taken with a very long exposure that doesn't look very realistic):

The dome of light above the city is very visible if imaged from afar:

Long exposure pictures in cities will reveal the orange skyglow, which is the main reason why you can't see the stars - it's like noise masking off the faint light from the distant objects:

Light pollution affects primarily the observations of faint objects, such as nebulae or distant galaxies. Bright objects such as the Moon, the big planets, or some of the bright stars, are not affected by light pollution.

Using a telescope with a large aperture alleviates the effects of light pollution to some extent, but it cannot work miracles. A dark sky is always better.

Usually a 1 hour drive away from the city will bring you in a place with dark sky, free of light pollution - but it depends on several factors. In such a place you should be able to see the Milky Way with your naked eye. The Andromeda galaxy also is visible with the naked eye if the sky is dark enough.

newtonian mechanics - Do centripetal and reactive centrifugal forces cancel each other out?

In order for a body to move with uniform velocity in a circular path, there must exist some force towards the centre of curvature of the circular path. This is centripetal force. By Newton's Third Law, there must exist a reactive force that is equal in magnitude and opposite in direction. This is the reactive centrifugal force.

My question is simple, and it is probably the result of lack of common sense but here it goes: In uniform circular motion, why don't these forces simply cancel each other out? If they did, how would we know they exist in that situation?

When I swing a rock tied to a rope, I feel the centrifugal force, but not the centripetal force. In this situation how can the reactive force be greater than the force itself?

Answer

Some of the confusion about centripital, centrifugal, and reactive forces is just vocabulary. It can be easier to understand if we consider a similar example without rotation.

Suppose you are floating in space near a rocket. A rock is tied to the rocket with a thread. When the engine starts, the rocket pulls on the thread and exerts a force on the rock. The rock accelerates.
$T = m_{rock} * a_{rock}$

The reaction force is the equal and opposite force the rock exerts on the rocket. The rock pulls on the thread and reduces the acceleration of the rocket.
$F = F_{rocket} - T$
The reaction force doesn't cancel anything. It is just a force that added to all the other forces on the rocket.

Since you are floating near the rocket, you see the rocket move. The pilot seated in the rocket finds it more convenient to adopt point of view where the rocket stays at rest. At time $t_0$, the seat is right under him. At time $t_1$, the seat is still right under him.

For the pilot to use laws like $F = ma$, he must redefine acceleration so that $a = 0$. This means he must redefine force so that $F = 0$.

The changes in definition are not big. Everything is consistent if he adds a fictitious acceleration $a_{fict}$ to all accelerations, where $a_{fict} = -a_{rock}$. He adds a fictitious force $f_{fict} = ma_{fict}$ to all forces.

He is saying "$f_{fict}$ acts on everything in the universe, causing everything to accelerate backword with acceleration $a_{fict}$. The total acceleration of rocket+rock is $0$ because of the additional force $F$ from the engine.

Fictitious forces do not cancel anything. They just change your point of view, or frame of reference.

---

Returning to circular motion, suppose you are floating near a rocket that is stationary but spinning. A rock is tied to the rocket, and rotates around the rocket.

The rocket engine provides the centripetal force that keeps the rock moving in a circle.

The reaction force holds the rocket stationary.

Centrifugal force is useful to an ant on the rock. The ant finds it useful to adopt a frame of reference where the rock is stationary. This is a more complex case, because $a_{centrifugal}$ depends on position.