Algebraic Formalism of Quantum Mechanics

In the algebraic formalism, the physical system is described by its observables, viewed as self-adjoint elements in a certain *algebra and a state is a linear functional, if I understand right.

Can someone show a completely worked out example of constructing the Hilbert space using the GNS construction?

Starting from the knowledge of density matrices, pure and mixed states, can someone motivate the algebraic formalism with some completely worked-out example so that the idea gets clear? Then one will be able to appreciate the formal mathematics better.

I have not seen any reference doing this apart from doing things in an abstract way. It would be of great help if someone could work out examples and show what is happening and how to go to the algebraic formalism of QM just with the background of density matrices.

Answer

All right, let's work out the quantum toy model: a good old spin $1/2$...

The Hilbert space formulation of a spin $1/2$ is as follow:

1. 
   

   the Hilbert space is $\mathcal{H} := \mathbb{C}^2$

   
   



2. 
   

   the elementary observables are the Pauli matrices $\hat{\sigma}_1, \hat{\sigma}_2, \hat{\sigma}_3$, which satisfy $$\hat{\sigma}_i \hat{\sigma}_j = \delta_{ij} \mathbf{1} + i \varepsilon_{ijk} \hat{\sigma}_k$$

   
   

From there we can build the abstract *-algebra $\mathcal{A}$ generated by 3 elements $\sigma_1,\sigma_2,\sigma_3$ satisfying $\sigma_i \sigma_j = \delta_{ij} 1 + i \varepsilon_{ijk} \sigma_k$ and $\sigma _i^* = \sigma _i$. Turns out it is the 4-dimensional algbra $M_2(\mathbb{C})$ mentioned by @Phoenix87, a basis of which is $$1,\sigma_1,\sigma_2,\sigma_3$$

For any $a = \lambda 1 + \mu_i \sigma_i \in \mathcal{A}$, we define $$\hat{a} := \lambda \mathbf{1} + \mu_i \hat{\sigma}_i$$ Note that $\forall a,b \in \mathcal{A}, \widehat{ab} = \hat{a} \hat{b}$ by construction, and that $\left\{ \hat{a} \middle| a \in \mathcal{A} \right\}$ spans the full space of operators on $\mathcal{H}$ (namely, the space of $2\times 2$ complex matrices, which constitutes the defining representation of the abstract algebra $M_2(\mathbb{C})$).

Now, given a density matrix $\hat{\rho}$ on $\mathcal{H}$, we define a linear form $\rho: \mathcal{A} \rightarrow \mathbb{C}$ by $$\forall a \in \mathcal{A}, \rho(a) := \text{Tr} \hat{\rho} \hat{a}$$ We can check that $\rho$ is an algebraic state on $\mathcal{A}$ ie:

1. 
   

   $\rho(1) = 1$ ($\text{Tr} \hat{\rho} = 1$)

   
   
   


2. 
   

   $\forall a \in \mathcal{A}, \rho(a^*) = {\rho(a)}^*$ (for $\widehat{a^*} = \hat{a}^{\dagger}$)

   
   


3. 
   

   $\forall a \in \mathcal{A}, \rho(a^* a) \geq 0$ (we can write $\hat{\rho} = \sum_{k} p_{k} \left| \psi_{k} \middle\rangle\middle\langle \psi_{k} \right|$, so $\text{Tr} \hat{\rho} \hat{a}^{\dagger} \hat{a} = \sum_{k} p_{k} \left\langle \hat{a} \psi_{k} \middle| \hat{a} \psi_{k} \right\rangle \geq 0$)

   
   

So we can associate to each density matrix an algebraic state on $\mathcal{A}$. Moreover the map $\hat{\rho} \mapsto \rho$ has the following properties:

1. 
   

   It is injective: $\rho$ completely encodes the quantum state $\hat{\rho}$. Mathematically, this holds because for any $\left|e\right\rangle, \left|f\right\rangle \in \mathcal{H}$, there exists $a \in \mathcal{A}$ such that $\hat{a} = \left|f\middle\rangle\middle\langle e\right|$ (as stressed above, any operator on $\mathcal{H}$ can be written as $\hat{a}$ for some $a \in \mathcal{A}$), so $\left\langle e \middle| \hat{\rho} \middle| f \right\rangle$ can be read out from $\text{Tr} \hat{\rho} \hat{a} = \rho(a)$. Physically, all we use $\hat{\rho}$ for is to compute expectation values, and these are precisely what the linear form $\rho$ supplies: to each observable $a$, it maps its expectation value.

   
   


2. 
   

   It is also surjective: this can be seen by noting that any algebraic state $\tau$, being a linear form over $\mathcal{A}$, is entirely characterized by the 4 complex numbers $\tau(1), \tau(\sigma_1), \tau(\sigma_2), \tau(\sigma_3)$. Imposing normalization (condition 1. above), we get $\tau(1) = 1$. Condition 2. implies that the numbers $\tau_i := \tau(\sigma_i)$ are actually real. Finally, the positivity of $\tau$ (condition 3.) requires that, for any $a = \lambda\,1 + \mu_i \sigma_i \in \mathcal{A}$, $$\tau(a^* a) = \lambda_R^2 + \lambda_I^2 + \left\|\vec{\mu}_R\right\|^2 + \left\|\vec{\mu}_I\right\|^2 + 2 (\lambda_R \vec{\mu}_R + \lambda_I \vec{\mu}_I + \vec{\mu}_I \times \vec{\mu}_R) . \vec{\tau} \geq 0$$ (where the subscripts ${}_R$ resp. ${}_I$ denote real resp. imaginary part). Minimizing over $\lambda_R$ and $\lambda_I$, this is equivalent to the condition $$\forall \vec{u}, \vec{v} \in \mathbb{R}^3, (\vec{u}.\vec{\tau})^2 + (\vec{v}.\vec{\tau})^2 + 2 (\vec{u}\times\vec{v}).\vec{\tau} \leq \left\|\vec{u}\right\|^2 + \left\|\vec{v}\right\|^2$$ which is fulfilled iff $\vec{\tau}$ belongs to the unit ball of $\mathbb{R}^3$. We can then check that $$\hat{\tau} := \frac{1}{2} \left[ \mathbf{1} + \tau_i \hat{\sigma}_i \right]$$ is a well-defined density matrix on $\mathcal{H}$, which maps to the algebraic state $\tau$ (since $\text{Tr} \mathbf{1} = 2$ and $\text{Tr} \hat{\sigma}_i = 0$).

   
   

Thus, the space of density matrices over $\mathcal{H}$ can be identified one-to-one with the space of algebraic states over $\mathcal{A}$, so we can do physics without ever referring to $\mathcal{H}$: knowing $\mathcal{A}$, we can parametrize the space of states of our quantum theory by algebraic states, and, given one such algebraic state, we can compute the corresponding expectation values for measurements.

Reciprocally, if we start from a quantum theory formulated the algebraic way, the GNS construction allows to recover an Hilbert space formulation for it. This is important as this guarantees that, by using the algebraic formalism, we are not going to start constructing theories that have no counterpart in the Hilbert space formulation, ie. theories that wouldn't be admissible quantum theories in the usual sense.

For the algebra $\mathcal{A} = M_2(\mathbb{C})$, this works as follow. We need to first choose a state $\omega$ (the "vacuum"), characterized by $\vec{\omega}$ in the unit ball of $\mathbb{R}^3$ as described above. Then, we compute its "kernel" (I'm not completely sure it's called like that, but can't find a better term): $$\mathcal{K}_{\omega} := \left\{ a \in \mathcal{A} \middle| \omega(a^* a)=0 \right\}$$ Writing $a = \lambda\,1 + \mu_i \sigma_i \in \mathcal{A}$, we have $a \in \mathcal{K}_{\omega}$ iff $$\lambda_R^2 + \lambda_I^2 + \left\|\vec{\mu}_R\right\|^2 + \left\|\vec{\mu}_I\right\|^2 + 2 (\lambda_R \vec{\mu}_R + \lambda_I \vec{\mu}_I + \vec{\mu}_I \times \vec{\mu}_R) . \vec{\omega} = 0$$ Since we know that $\lambda_{R/I}^2 + 2 \lambda_{R/I} (\vec{\mu}_{R/I} . \vec{\omega}) + (\vec{\mu}_{R/I} . \vec{\omega})^2 \geq 0$ and $\left\|\vec{\mu}_R\right\|^2 + \left\|\vec{\mu}_I\right\|^2 - (\vec{\mu}_R .\vec{\omega})^2 - (\vec{\mu}_I .\vec{\omega})^2 - 2 (\vec{\mu}_R \times\vec{\mu}_I).\vec{\omega} \geq 0$ (see above), this becomes $$a \in \mathcal{K}_{\omega} \;\Leftrightarrow\; \lambda_{R/I} = - \vec{\mu}_{R/I} . \vec{\omega} \;\;\&\;\; (\vec{\mu}_R .\vec{\omega})^2 + (\vec{\mu}_I .\vec{\omega})^2 + 2 (\vec{\mu}_R \times\vec{\mu}_I).\vec{\omega} = \left\|\vec{\mu}_R\right\|^2 + \left\|\vec{\mu}_I\right\|^2$$

At this point, we need to distinguish cases. In the case where $\vec{\omega}$ is actually on the unit sphere (ie $\|\vec{\omega}\| = 1$), we find $$\mathcal{K}_{\omega} = \text{Span} \left\{ 1 - \omega_i \sigma_i, (u_j + i v_j) \sigma_j \right\}$$ where $\vec{u}, \vec{v}$ are some unit vectors such that $\vec{\omega}, \vec{u}, \vec{v}$ form an orthonormal basis. So, in this case, $\mathcal{K}_{\omega}$ is 2-dimensional, and therefore $\mathcal{H}_{\omega} := \mathcal{A} / \mathcal{K}_{\omega}$ is $4-2=2$-dimensional. $\mathcal{H}_{\omega}$ is spanned eg. by $\left [1\right ]$ and $\left [u_j \sigma_j\right ]$ (where $\left [\,\cdot\,\right ]$ denotes equivalence classes modulo $\mathcal{K}_{\omega}$).

On $\mathcal{H}_{\omega}$ we can define a scalar product $$\left\langle \left[a\right] \middle| \left[b\right] \right\rangle_{\omega} := \omega(a^*b)$$ and a representation $a \mapsto \hat{a}^{\omega}$ of the algebra $\mathcal{A}$ $$\hat{a}^{\omega} \left[b\right] := \left[ab\right]$$ (check that these are well-defined, ie that the result of each map does not depend on the representatives used to express its arguments, and that each map has the required properties to be a scalar product, resp. a representation of $\mathcal{A}$).

Finally, we want to confirm that the thus obtained Hilbert space/representation reproduces the $\mathcal{H}$ we started from at the very beginning. Indeed, denoting by $\left|\phi_{\omega}\right\rangle$ the eigenvector of $\omega_i \hat{\sigma}_i$ with eigenvalue $+1$ (ie spin $+1$ in the direction of $\vec{\omega}$), $\mathcal{H}_{\omega}$ can be identified with $\mathcal{H}$, via the map $$\begin{array}{ccc}\mathcal{H}_{\omega} & \rightarrow & \mathcal{H}\\ \left[a\right] & \mapsto & \hat{a} \left|\phi_{\omega}\right\rangle \end{array}$$ which is well-defined, bijective and matches the scalar products and representations on both sides.

This can be checked directly, or by observing that $\omega(a) = \left\langle \phi_{\omega} \middle| \hat{a} \middle| \phi_{\omega} \right\rangle$ (as the algebraic state $\omega$ corresponds to the density matrix $\left|\phi_{\omega}\middle\rangle\middle\langle \phi_{\omega}\right|$ via the identification discussed in the previous section). For example, this implies that $\mathcal{K}_\omega$ consists of the $a \in \mathcal{A}$ such that $\left\langle \hat{a} \phi_{\omega} \middle| \hat{a} \phi_{\omega} \right\rangle = 0$, ie $\hat{a} \left|\phi_{\omega}\right\rangle = 0$, and therefore $\left[a\right] = \left[b\right] \Leftrightarrow a-b \in \mathcal{K}_{\omega} \Leftrightarrow \hat{a} \left|\phi_{\omega}\right\rangle = \hat{b} \left|\phi_{\omega}\right\rangle$.

In other words, we have been able to entirely reconstruct $\mathcal{H}$ just by knowing the algebra $\mathcal{A}$ (choosing some algebraic state $\omega$ on $\mathcal{A}$ to "seed" $\mathcal{H}_{\omega}$). Crucially, this reconstruction didn't assume any a priori knowledge of $\mathcal{H}$ (the definitions of $\mathcal{K}_{\omega}$, $\mathcal{H}_{\omega}$, $\left\langle\cdot\middle|\cdot\right\rangle_{\omega}$ and $a \mapsto \hat{a}^{\omega}$ were purely in terms of 1/ the operations in $\mathcal{A}$, ie addition, multiplication and adjoint, and 2/ the state $\omega$).

But what would have happened if we had chosen a state characterized by a $\vec{\omega}$ in the interior of the unit ball (ie $\|\vec{\omega}\| < 1$)? In this case, $\mathcal{K}_{\omega}$ would be just $\{0\}$ so $\mathcal{H}_{\omega}$ would be 4-dimensional. One can check that the representation created by the GNS construction can then be identified with $\mathcal{H} \oplus \mathcal{H}$ via: $$\begin{array}{ccc} \mathcal{H}_{\omega } \approx \mathcal{A} & \rightarrow & \mathcal{H}^{(1)} \oplus \mathcal{H}^{(2)}\\ \left[a\right] = a & \mapsto & \hat{a} \left( \cos \eta \left|\phi ^+_{\omega }\right\rangle ^{(1)} + \sin \eta \left|\phi ^-_{\omega }\right\rangle ^{(2)} \right) \end{array}$$ where $\eta \in \left[0,\frac{\pi }{2}\right[$ is defined by $\cos^2 \eta := \frac{1 + \|\omega \|}{2}$ and $\left|\phi ^{\pm }_{\omega }\right\rangle$ denote the eigenvectors of $\frac{\omega _i}{\|\omega \|} \hat{\sigma }_i$ with eigenvalue $\pm 1$ respectively.

It is a general property of the GNS construction that the representation arising from a mixed state will typically be a direct sum of independent representations (aka. superselection sectors), while pure states lead to irreducible representations (ie. representations that cannot be split as a direct sum of simpler representations).

Since density matrices are now $4\times 4$ matrices, does it means that we suddenly have more quantum states that we had on just $\mathcal{H}$? Well, actually no. Density matrices on $\mathcal{H}^{(1)} \oplus \mathcal{H}^{(2)}$ can be written as block matrices: $$\hat{\rho }^{\omega } = \left( \begin{array}{cc} \hat{\rho }^{(1)} & \hat{\gamma } \\ \hat{\gamma }^{\dagger} & \hat{\rho }^{(2)} \end{array} \right)$$ with $\hat{\rho }^{(1)}, \hat{\rho }^{(2)}$ positive semi-definite and $\text{Tr} \hat{\rho }^{(1)} + \text{Tr} \hat{\rho }^{(2)} = 1$ (as well as suitable bounds on $\hat{\gamma }$ for $\hat{\rho }^{\omega }$ to be positive semi-definite). But since observables have the form: $$\hat{a}^{\omega } = \left( \begin{array}{cc} \hat{a} & 0 \\ 0 & \hat{a} \end{array} \right)$$ expectation values (which are the only physically measurable quantities) are simply given by $$\text{Tr} \hat{\rho }^{\omega } \hat{a}^{\omega } = \text{Tr} \left[ \left(\hat{\rho }^{(1)} + \hat{\rho }^{(2)}\right) \hat{a} \right]$$ So we don't have any new quantum state, we just have many different density matrices corresponding to the same quantum state, namely the state that on $\mathcal{H}$ was represented by the density matrix $\hat{\rho } = \hat{\rho }^{(1)} + \hat{\rho }^{(2)}$. In this sense, $\mathcal{H}_{\omega }$ describes just the same quantum theory as $\mathcal{H}$, but with some extra redondancy in the representation of quantum states.

To be honest, this toy model works a bit too well:

1. 
   

   all algebraic states can be represented as density matrices on $\mathcal{H}$

   
   


2. 
   

   all algebraic states give rise, via the GNS construction, to one or more copies of $\mathcal{H}$

   
   

Propositions 1. and 2. are in fact equivalent: they express the fact that there is only one "folium" of states on $M_2(\mathbb{C})$. The folium of an algebraic state $\omega$ is defined as the space of algebraic states that can be represented as density matrices of the GNS representation $\mathcal{H}_{\omega }$ arising from $\omega$. If $\omega$ is pure, all states in its folium will have one or more copies of $\mathcal{H}_{\omega }$ as their GNS representation. But in general an algebra of observables can admit many disjoint foliums, or, equivalently, many inequivalent representations.

For example, if we were to redo the analysis for the *-algebra generated again by 3 self-adjoint elements $\sigma_1,\sigma_2,\sigma_3$ but now only assuming the weaker relations $\left[\sigma_i, \sigma_j\right] = 2i \varepsilon_{ijk} \sigma_k$, aka the universal enveloping algebra of the Lie algebra $\mathfrak{so}(3) = \mathfrak{su}(2)$, we would find infinitely many inequivalent representations, corresponding to all possible spins $j$.

In the case of ordinary quantum mechanics with an algebra of observables generated by exponentiated position and momentum operators (to avoid issues with unbounded operators), the Stone–von Neumann theorem implies that there is only one folium of "sufficiently regular" states, although if we drop the regularity condition more exotic states and associated GNS representations can be constructed. In QFT, on the other hand, there is a plethora of distinct foliums. Moreover, on curved spacetimes, there may be no way to single out a preferred vacuum state/folium and associated representation: that's where the algebraic formalism truly shines, because it allows to discuss all foliums at once, without having to specialize to an arbitrarily chosen one as we would have to do to get an Hilbert space formulation.

cipher - Some messed up words - Clue Six

_{<---Previous clue}

---

The door swings open, to reveal [redacted] waiting there for you. Your jaw drops. You had no idea that this was so important, that she would get involved.

[redacted] steps towards you and hands you a piece of paper. It reads:

ho an head ded ho if an ac

But you don't immediately focus on it. Instead, you focus on the [redacted] family leader, A [redacted] C [redacted].

"What are you doing here?" you ask.
But instead of replying, she just smiled, and whispered, "You should go work on that now."
And stepping back, she disappears through a trapdoor.

You go to inspect the floor where she disappeared, but there's nothing. Those [redacted] scientists are good. You settle down on the floor in defeat, staring gloomily at the paper in your hand.

---

_{Next clue-->}

Answer

And the answer is

wormwood

All you have to do is to

add up the alphabetical value of all letters in the word to produce a new letter.

general relativity - Can a deformable object "swim" in curved space-time?

Possible Duplicate:
Swimming in Spacetime - apparent conserved quantity violation

It is well known that a deformable object can perform a finite rotation in space by performing deformations - without violating the law of conservation of angular momentum since the moment of inertia can be changed by the deformations of the object, see e.g. this Phys.SE question.

It is also well known that in flat space-time, it is not possible for a deformable object to displace it's center of gravity by performing deformations, see e.g. this Phys.SE question.

However in curved space-time can a deformable object swim through space by performing deformations?

quantum field theory - Complex integration by shifting the contour

In section 12.11 of Jackson's Classical Electrodynamics, he evaluates an integral involved in the Green function solution to the 4-potential wave equation. Here it is:

$$\int_{-\infty}^\infty dk_0 \frac{e^{-ik_0z_0}}{k_0^2-\kappa^2}$$

where $k$ and $z_0$ are real constants.

Jackson considers two open contours: one above and one below the real axis. I understand that in order to use Jordan's lemma, when $z_0 < 0$ we have to close the contour in the upper half of the complex plane whereas if $z_0 > 0$ we have to close the contour in the lower half plane.

What I don't understand is why it's OK to consider contours above and below the real axis when the original integral is along the real axis. As I understand it, the necessity to deal with poles like this also arises a lot in QFT, so perhaps it is well understood from that point of view.

Answer

Suppose we want to analyse the problem of a forced harmonic oscillator. Denote as $\phi(t)$ the time dependent position of the oscillator. The oscillator experiences two forces, the spring force $-k\phi(t)$ and an external force $F_{\text{ext}}(t)$. Newton's law says

$$ \begin{align} F(t) &= m a(t) \\ -k \phi(t) + F_{\text{ext}}(t) &= m \ddot{\phi}(t) \\ F_{\text{ext}}(t)/m &= \ddot{\phi}(t) + (k/m) \phi(t) \\ j(t) &= \ddot{\phi}(t) + \omega_0^2 \phi(t) \tag 1 \end{align} $$

where $\omega_0$ is the free oscillation frequency and $j(t)\equiv F_{\text{ext}}/m$. We use the following Fourier transform convention: $$ \begin{align} f(t) &= \int_\omega \tilde{f}(\omega) e^{i\omega t} \frac{\mathrm d\omega}{2\pi} \\ \tilde{f}(\omega) &= \int_t f(t) e^{-i\omega t}~\mathrm dt . \end{align} $$

With this convention on Eq. $(1)$, and defining $$\omega_{\pm} \equiv \pm \omega_0,$$ we find $$ \tilde{\phi}(\omega) = \frac{\tilde{j}(\omega)}{\omega_0^2-\omega^2} = \frac{-\tilde{j}(\omega)}{(\omega-\omega_+)(\omega-\omega_-)}. \tag 2 $$ From Eq. $(2)$ we see that the Green's function is $$\tilde{G}(\omega) = \frac{-1}{(\omega-\omega_+)(\omega-\omega_-)}$$ which has poles on the real axis. If we want to compute $\phi(t)$ we do a Fourier transform

$$ \phi(t) = \int_\omega \frac{-\tilde{j}(\omega)e^{i\omega t}}{(\omega-\omega_+)(\omega-\omega_-)} \frac{\mathrm d\omega}{2\pi} = \int_\omega e^{i\omega t}\tilde{j}(\omega) \tilde{G}(\omega)\frac{\mathrm d\omega}{2 \pi}. \tag{*} $$

This integral is tricky because of the poles on the axis. The solution everyone knows is to push the poles off the axis by adding an imaginary part to $\omega_{\pm}$, or by moving the contour above or below the real axis, but what does this actually mean physically? How do we choose which direction to push the poles or move the contour?

In a real system, we always have some damping. In our oscillator model, this could come in the form of a velocity dependent friction $F_{\text{friction}} = -\mu \dot{\phi}(t)$. Defining $2\beta = \mu/m$, the equation of motion becomes

$$\ddot{\phi}(t) + 2\beta \dot{\phi}(t) + \omega_0^2\phi(t) = j(t) . \tag 3$$

Fourier transforming everything again leads to Eq. $(2)$ but now with \begin{equation} \omega_{\pm} = \pm \omega_0' + i\beta \end{equation} where \begin{equation} \omega_0' = \omega_0\sqrt{1-(\beta/\omega_0)^2}. \end{equation}

Therefore, we see that adding damping moves the poles a bit toward the origin along the real axis, but also gives them a positive imaginary component. In the limit of small damping (i.e. $\beta \ll \omega_0$), we find $\omega_0' \approx \omega_0$. In other words, the frequency shift of the poles due to the damping is small. So let's ignore that and focus on the added imaginary part.

Ok, suppose we want to do the integral $(*)$ in the case that $j(t)$ is a delta function at $t=0$. In that case, $\tilde{j}=1$ (I'm ignoring units) and we have $$ \phi(t) = \int_\omega \frac{e^{i\omega t}}{(\omega-\omega_+)(\omega-\omega_-)} \frac{\mathrm d\omega}{2\pi} $$ As you noted, for $t<0$ you have to close the contour in the lower plane in order to use Jordan's lemma. There aren't any poles in the lower half plane, so we get $\phi(t<0)=0$. This makes complete sense: the driving force is a delta function at $t=0$ and there shouldn't be any response of the system before the driving happens. This means that our introduction of friction imposed a causal boundary condition to the system! For $t>0$, you close in the upper half plane where there are poles, and so you get some response out of the integral.

In many cases, you don't naturally have damping in the system. For example, the Green's function from the question, $$\int^{\infty}_{-\infty}~\mathrm dk_0 \frac{e^{−ik_0 z_0}}{k_0^2 − \kappa^2}$$ doesn't have any damping and thus the poles sit on the real axis. So what you do is just bump the contour a bit up or down, or equivalently add $\pm i \beta$ to the poles (most people write $i \epsilon$ instead of $i \beta$), then do the integral, and finally take $\beta \rightarrow 0$. In doing this, you're solving the problem in the presence of damping (or anti-damping), and then taking the damping to zero in the end to recover the no-damping case.

Choosing to push the contour up or down, or equivalently choosing the sign of $\pm i \beta$, corresponds to imposing either friction or anti-friction, causal or anti-causal boundary conditions. If you pick the "causal" boundary condition, you find that the response of the system to a delta function in time and space is an outgoing spherical wave which starts at the delta function source. This gives you the so-called "retarded Green's function". If you pick the other condition, you find that the solution for a point source is actually an incoming spherical wave which converges right onto the point of the source. This gives you the so-called "Advanced Green's function".

The thing is, you can solve a problem using either Green's function. You're "allowed" to push the contour up or down (or add $+i\beta$ or $-i\beta$ to the poles) because you invented that as a trick to do the integral; it's not representing a real factor in your physical system. Of course, in problems where there is damping, the choice is made for you. When you have damping, you can't have fields at infinity; they'd be damped away by the time they interact with your sources.

I hope this was helpful, and I really hope if someone finds mistakes they'll jump in and fix 'em.

lateral thinking - 3 ants walking in the desert

3 ants are walking on a straight path in the desert, one behind the other.

The sun is setting, so the temperature is bearable.

• The 1^st one says : There're 2 ants behind me.


• The 2^nd one says : There's 1 ant in front of me, and 1 behind me.


• The 3^rd one says : There're 2 ants in front of me, and 1 behind me.

How is it possible ?

Answer

It could be that

The third ant is lying

homework and exercises - What affects the damping of a spring?

What variables affect the damping of a spring executing simple harmonic motion?

What are the independent variables, and what variables would need to be controlled in an experiment?

I'm attempting to complete an investigation where I measure the decrease in amplitude of a damped spring, and to prove the relationship between variables in the motion.

Thanks!

Answer

You should already be familiar with damping. It simply refers to the fact that if you set a spring going, it eventually stops.

The wikipedia article should cover most of what you want to know.

Any particular spring may be damped for all sorts of reasons. Any way the spring can lose energy contributes to damping, so it could be lost internally to heat due to stressing the material, or externally to heat via friction, or externally to an electromagnetic field, to some sort of mechanical dashpot, etc.

If you were designing an experiment to study damping, you could be interested in a number of different particular things. You will have to make your own choice about what the most interesting thing to study is. For example, would you like measure the damping ratio? The overall magnitude of the damping effect? The time it takes your spring to reduce its amplitude to 1/2 its previous value?

If you have a particular effect picked out, do you want to know how it depends on the load on the spring, or the material of the spring, or whether the spring is in a vacuum, etc. There are many possible things to study, so your question has no definite answer. I'm sure you can find something particular you find worthwhile.

general relativity - How to prove that a spacetime is maximally symmetric?

In Carroll's book on general relativity ("Spacetime and Geometry"), I found the following remark:

In two dimensions, finding that $R$ is a constant suffices to prove that the space is maximally symmetric [...] In higher dimensions you have to work harder

Here, $R$ is the Ricci scalar. This begs the following questions:

• How does ones prove that a $d$-dimensional spacetime is maximally symmetric?


• If the general case is very complicated, then is there a simpler way to obtain the result in four dimensions?

Answer

Two general methods come to mind:

1. Prove that the Riemann tensor takes the form of equation 3.191, i.e. $$R_{abcd} = \frac{R}{d(d-1)}(g_{ac}g_{bd}-g_{ad}g_{bc}) $$ If you are handed a metric, this should in principle be a straightforward calculation. If the metric is actually maximally symmetric, the calculation of the Riemann tensor usually turns out to be easier than usual, especially if you use a high-tech method like the Cartan formalism with vielbeins and spin connections (see Appendix J of Carroll).


2. Find the maximal number of Killing vectors. For a manifold with dimension $d$, it admits a maximum of $\frac12 d(d+1)$ Killing vectors (these are explained in section 3.8 of Carroll). This technique is usually easier if you have a good sense of what the isometries of the metric are and can basically guess all the Killing vectors. For example, in flat Minkowski space it is fairly obvious that boosts, rotations and translations are all symmetries of the metric. So you write down the vectors corresponding to flowing in the direction of these transformations, and can easily check that they satisfy Killing's equation, $\nabla_{(a}\xi_{b)}=0$, or $£_\xi g_{ab}=0$, where $\xi^a$ is the Killing vector.

In practice, constructing maximally symmetric spaces is easier than it sounds. Generally, you start with the manifold $\mathbb{R}^{n,m}$ with a flat metric of signature $(n,m)$ (i.e. there are n spacelike coordinates with a $+dx_i^2$ contribution to the line element and $m$ coordinates with a $-dy_j^2$ contribution). It is fairly straightforward to prove that this is maximally symmetric. Then, you define a submanifold $\mathcal{S}$ as the locus of points that are some fixed distance from the origin. For example, if we started with $\mathbb{R}^{3,0}$, which is just Euclidean $3$-space, we say $\mathcal{S}$ is the set of all points such that $$x^2+y^2+z^2=r_0^2$$ for some fixed radius $r_0$. This of course defines a 2-sphere, which is a maximally symmetric space of one less dimension from $\mathbb{R}^{3,0}$. You can infer from this construction that the submanifold $\mathcal{S}$ is going to be maximally symmetric, because we broke the full group of isometries of $\mathbb{R}^{n,m}$ down into the ones that leave the origin fixed. So we started with $\frac12 d(d+1)$ isometries ($d=n+m$), and lost all the translations, of which there are $d$, so we are left with $\frac12 d(d+1)-d = \frac12(d-1)d$ isometries, which is the maximal number for $d-1$ dimensions.

Since you asked about 4D maximally symmetric spacetimes, there are basically three things you can do. The trivial one is just Minkowski space, $\mathbb{R}^{3,1}$. The next thing we can do is start with five dimensional Minkowski space, $\mathbb{R}^{4,1}$, and pick out all the points that are a fixed spacelike distance from the origin, $$x^2+y^2+z^2+w^2-t^2 = r_0^2 $$ (here $(x,y,z,w)$ are the spatial coordinates). The induced metric on the submanifold is de Sitter space, $dS_4$, the 4D spacetime with constant positive curvature.

Finally, you can begin with $\mathbb{R}^{3,2}$ the Euclidean space with $3$ spacelike directions $(x,y,z)$ and $2$ timelike directions $(t_1,t_2)$. This time we consider all points that are a fixed timelike distance from the origin, $$x^2+y^2+z^2-t_1^2-t_2^2=-r_0^2 $$ The induced metric for this submanifold is anti-de Sitter space, $AdS_4$, which is the 4D spacetime of negative constant curvature.

Locally I believe any maximally symmetric space will look like one of the spaces constructed using this embedding technique. In some cases however, there can be nontrivial topological features that cause the maximally symmetric space to differ from the embedded manifolds we just constructed. One example is for $AdS_4$: as it stands, the manifold we constructed has closed timelike curves (arising from moving around in the $t_1$-$t_2$ plane). These can be removed by "unwinding the time direction," which mathematically means we go to the simply connected universal cover, which is topologically different from the $AdS_4$ we just constructed, but locally looks exactly the same.

electromagnetism - Why is the return current through a printed circuit board's ground plane concentrated below the circuit trace?

The Wikipedia article on ground plane says

In addition, a ground plane under printed circuit traces [the paths that the circuit currents take] can reduce crosstalk between adjacent traces. When two traces run parallel, an electrical signal in one can be coupled into the other through electromagnetic induction by magnetic field lines from one linking the other; this is called crosstalk. When a ground plane layer is present underneath, it forms a transmission line with the trace. The oppositely-directed return currents flow through the ground plane directly beneath the trace. This confines most of the electromagnetic fields to the area near the trace and consequently reduces crosstalk.

but unfortunately there are no citations. Page 5 of this document says the same thing ("Return Current Flow is directly below the signal trace") and gives a formula that the magnitude of the return current at point $x$ on the ground plane is proportional to $\cos(\theta)/r$, where $r$ is the distance between point $x$ and the circuit trace and $\theta$ is the angle that point $x$ makes relative to the vertical. But the document gives no explanation justifying this formula.

My naive intuition is that the return current would want to avoid the circuit trace, since opposite currents repel, and so it would spread out and increase crosstalk. Why does the return current instead concentrate below the circuit trace?

Answer

I just asked a professor this question, and he thinks that the concentration of return current below the circuit trace is actually just an AC effect. The rapidly oscillating electric and magnetic fields induce a fictitious image current on the other side of the conductor, and the real and image wires together act as a transmission line. But this only works if the fields oscillate fast enough - in the DC limit, the magnetostatic repulsion between the trace and the return current does indeed cause the return current to spread out far away from the circuit trace.

electromagnetism - What limits the maximum sustainable surface charge density of a sphere in space?

Suppose I charge a sphere and leave it in vacuum for 10 years. After that time, I want its surface charge density to be in the order of 10^5C/m^2. Would that be possible? Would it depend on the material used and how? Would adding or removing electrons make a difference (positively vs negatively charged)?

Answer

If you take, for example, a perfect metal sphere then it has a work function that is the energy required to remove an electron from the metal to infinity. If you start charging the sphere by adding electrons to it then the work function decreases, and above some limiting charge the work function falls to zero. This means any more electrons you add to the sphere immediately escape again. This is an example of a phenomenon is called field emission.

I've chosen the example of a metal sphere since it's nice and simple, but this will apply to any object, and it means that there is a maximum charge that can be sustained on any object regardless of how close to perfectly it has been made.

If you keep the charge below the level where field emission occurs, and keep the object in a vacuum, and mask it from any light with energy of greater than the work function, and keep it at a temperature below which thermionic emission occurs, then the sphere will stay charged forever.

dimensional analysis - Functions and Length Scales

Regretfully I have to start with an apology as I fear I might be unable to express the question rigorously.

Often reading physics papers the concept of "length scale" is used, in statements such as "over this length scale, the phenomenon can be characterized by an exponential decay", or "the increase in X is virtually linear over such and such time scale". The Nobel Laureate De Gennes seems to me a virtuoso in this particular art.

I am able to follow some reasoning, but I am not so sure I understand fully their methods. For example, let us imagine I have got a model characterizing the change of a certain quantity, $Y$, versus time $t$: $$Y = e^{-A/t}$$ where $A$ is a constant. The function tends to zero for small times, it is first concave and then becomes convex. One expects to be able to characterize the "length scales" at which the transition occurs, in relation to the constant $A$. How can that be done? I tried to calculate the second derivative, which equals $$Y'' = e^{-A/t} \frac{A^2}{t^4} - 2 e^{-A/t} \frac{A}{t^3}$$ so imposing the conditions it equals zero I get the equation $$ \frac{A}{t^4} = \frac {2}{t^3}$$ suggesting the conclusion the concave-convex transition occurs at "time scales in the order of A", is this correct?

But what truly puzzles me is how to characterize, for example, the time scale over which the functions is "almost flat", in the initial concave region. For example, if one fixes $A = 100$ and plot the function for a maximum $t = 5$, intuitively one suspects there must be a way to say "over such and such length scale, the function is flat". I wonder if to give sense to this statement one should specify which variations in the functions can be considered negligible.

Thanks a lot for any help on this I appreciate rather misty question.

quantum mechanics - Why do lines in atomic spectra have thickness? (Bohr's Model)

Consider the atomic spectrum (absorption) of hydrogen.

The Bohr's model postulates that there are only certain fixed orbits allowed in the atom. An atom will only be excited to a higher orbit, if it is supplied with light that precisely matches the difference in energies between the two orbits.

But how precise does 'precisely' mean. Of course, if we need energy $E$ to excite the electron to a higher energy level, and I supply a photon with just $E/2$ I would expect nothing to happen (since the electron cannot occupy an orbit between the allowed ones). But what if I supplied a photon with energy $0.99E$, or $1.0001E$ or some such number. What will happen then?

I think that the electron should still undergo excitation precisely because the lines we observe in the line spectrum have some thickness. Which means that for a given transition, the atom absorbs frequencies in a certain range.

Is my reasoning correct? If not, why? How does Bohr's model explain this? How about modern theory?

If I'm right, what is the range of values that an atom can 'accept' for a given transition?

Answer

According to Bohr model, the absorption and emission lines should be infinitely narrow, because there is only one discrete value for the energy.

There are few mechanism on broadening the line width - natural line width, Lorentz pressure broadening, Doppler broadening, Stark and Zeeman broadening etc.

Only the first one isn't described in Bohr theory - it's clearly a quantum effect, this is a direct consequence of the time-energy uncertainty principle:

$$\Delta E\Delta t \ge \frac{\hbar}{2}$$

where the $\Delta E$ is the energy difference, and $\Delta t$ is the decay time of this state.

Most excited states have lifetimes of $10^{-8}-10^{-10}\mathrm{s}$, so the uncertainty in the energy sligthly broadens the spectral line for an order about $10^{-4}Å$.

gravity - Gravitationally bound systems in an expanding universe

This isn't yet a complete question; rather, I'm looking for a qual-level question and answer describing a gravitationally bound system in an expanding universe. Since it's qual level, this needs a vastly simplified model, preferably one in a Newtonian framework (if this is even possible) that at least shows the spirit of what happens. If it isn't possible, and there is some important principle that any Newtonian model will miss, that is probably the first thing to point out. Otherwise, ...

I'm thinking of a setup that starts something like this:

A body of mass $m$ orbits another of mass $M \gg m$ (taken to be the origin) in a circle of radius $R$ (and consequently with a period $T = 2 \pi \sqrt{ R^3 / GM }$). At time $t = 0$, space begins to slowly expand at a constant Hubble rate $H$ (where "slowly" means $ HT \ll 1 $, so that the expansion in one period is negligible compared to the orbital radius).

Now here is where I'm not sure how to phrase the problem. I'm thinking instead of something like the "bead on a pole" problems. Then the bead has generalized Lagrangian coordinates relative to a fixed position on the pole, but the pole is expanding according to some external function $a(t)$, a la FLRW metric (For a constant Hubble rate, $H = \dot a / a$ and so $ a(t) = e^{Ht} $). So, we can call the bead's distance from the origin in "co-moving units" the Lagrangian coordinate $q$. Then the distance of the bead from the origin is $d(t) = q(t) a(t)$, giving us a time dependent Lagrangian.

Basically, I need a 3D version of this that will work for the gravitationally bound system described. I also hope that the problem can be solved as an adiabatic process, which might mean changing coordinates to be around the original position relative to the more massive body at the origin, rather than co-moving coordinates.

EDIT 1: I won't claim that I fully understood the details of Schirmer's work, but I think one of the big takeaway points is that the cosmological expansion damages the stability of circular orbits and causes them to decay. I finished my very hand-waivey "Newtonian" model of a solar system in an expanding universe, and I don't think it captures the essence of this part of the GR solution. The model does have several sensical properties:

• There is a distance at which bound solutions are impossible and the two bodies are guaranteed to expand away from each other


• There is still a circular orbit (I haven't checked that it is stable) whose radius is modified by a term involving the Hubble constant.


• Using solar parameters, this shift is entirely negligible; it is larger (though still small) on a galactic scale.

Here is the model:

Without this scaling factor, the Lagrangian would be $$ \mathcal{L} = \frac{1}{2} m \left( \dot r^2 + r^2 \dot \theta^2 + r^2 \sin^2 \theta \, \dot \phi^2 \right) + \frac{GMm}{r} $$ Adding in the scale factor, all the kinetic terms will pick up a factor $e^{2Ht}$ and the potential a factor $e^{-Ht}$: $$ \mathcal{L} = \frac{1}{2} m \left(\dot r^2 + r^2 \dot \theta^2 + r^2 \sin^2 \theta \, \dot \phi^2 \right) e^{2Ht}+ \frac{GMm}{r} e^{-Ht} $$ (I have not added in any derivatives of the scaling factor; this would mean I was just changing coordinates and not adding in an externally controlled expansion of space). Since the problem is spherically symmetric, the total angular momentum is conserved and the motion lies in a plane (the expansion of space does not change this) at $\theta = \pi/2$. This reduces the effective Lagrangian to $$ \mathcal{L} = \frac{1}{2} m \left( \dot r^2 + r^2 \dot \phi^2 \right) e^{2Ht} + \frac{GMm}{r} e^{-Ht} $$ Now let's change coordinates to $r = \eta e^{-Ht}$. This represents a point that is stationary with respect to the origin (i.e. the massive body that the smaller body orbits). Then $\dot r e^{Ht}= \dot \eta - H \eta $, so the Lagrangian becomes $$ \mathcal{L} = \frac{1}{2} m \left((\dot \eta - H \eta)^2 + \eta^2 \dot \phi^2 \right) + \frac{GMm}{\eta} $$ This looks like a Lagrangian that has been modified slightly by the parameter $H$. After these manipulations, $\phi$ is still a cyclic coordinate, and its conjugate momentum $p_\phi = m\eta^2 \dot \phi = L$ is still conserved. The momentum conjugate to $\eta$ is $$ p_{\eta} = m (\dot \eta - H \eta)$$ So, the equation of motion for $\eta$ is $$ m \frac{d}{dt} (\dot \eta - H \eta) = m (\ddot \eta - H \dot \eta) = - mH(\dot \eta - H \eta) + m \eta \dot \phi^2 - \frac{GMm}{\eta^2} $$ Canceling terms and substituting in the equation of motion for $\phi$, this becomes $$ \ddot \eta = H^2 \eta + \frac{L^2}{m^2 \eta^3} - \frac{GM}{\eta^2} $$ So, the expansion of space shows up as a force term proportional to $\eta$. For large $\eta_0$ at time $t = 0$, there is a solution $\eta(t) = \eta_0 e^{Ht}$, where $\dot \eta = H \eta$ agrees with Hubble's law (the corresponding comoving coordinate is $r(t) = \eta_0$). There is also an a highly unstable orbital solution for large $\eta$, where gravitational force just balances cosmological expansion. This new term also shifts the location of the "stable orbit" slightly (I haven't checked that it actually is stable in this situation). The location of the stable circular orbit when $H = 0$ is $$ \eta_0 = \frac{L^2}{GMm^2} $$ Then let $$ \eta = \eta_0 + \delta \eta = \eta_0 \left( 1 + \frac{\delta \eta}{\eta_0} \right)$$ To lowest order, the shift in location of the stable orbit is $$ \frac{\delta \eta}{\eta_0} = \frac{H^2}{GM/\eta_0^3 - H^2} $$ For the orbit of the earth around the sun, this shift would be entirely negligible- about 15 pm. Taking the mass to be the mass of the Milky way and the distance to the distance of the Sun from the center, the shift would be a bit larger - a fractional amount of about 2e-7, which is several hundred AU - but still largely small.

Answer

I wrote an unpublished paper on this once, and presented the poster at an AAS meeting. You can actually do some relatively simple model building with this if you look at Schwarzschild-de Sitter solutions of Einstein's equation.${}^{1}$

Once you have the exact solution, you can then look at circular orbits, and do the usual stability analysis that you use to derive the $r > 6M$ limit that you get in the Schwarzschild solution. What you will find is that, in the Schwarzschild-de Sitter solution, you will get a cubic equation. One solution will give you a $ r < 0$ value, which is unphysical, but you will also get an innermost stable circular orbit and an outermost stable circular orbit. The latter represents matter getting pulled off of the central body by the cosmology.

There are also exact solutions that involve universes with nonzero matter densities other than the cosmological constant with a central gravitating body. These solutions generically don't have a timelike killing vector nor globally stable orbits. I ran numerical simulations of these orbits and published them in my dissertation's appendices.

EDIT: stack exchange seems to be killing hotlinks, but search for my name on the arxiv if you're interested.

${}^{1}$EDIT 2: you can find the Schwarzschild-de Sitter solution by replacing $1 - \frac{2M}{r}$ everywhere with $1-\frac{2M}{r} - \frac{1}{3}\Lambda r^{2}$. It should be easy enough to prove that that solution satisfies Einstein's equation for vacuum plus cosmological constant.

homework and exercises - Mass hanging from spring: potential energy

I have encountered a problem with a mass $m$ hanging from a spring. I want to find the maximum distance it will be stretched from the spring's equilibrium. At the maximum distance the spring is stretched, the mass will be at rest so the weight will cancel out the spring force, so we should have,

$$ky = mg$$

or

$$y = \frac{mg}{k}$$

Where $k$ is the spring constant, $y$ is the maximum distance, $m$ is the mass and $g$ is gravitational acceleration. But, if you look at this problem through the conservation of energy, you would conclude energy is conserved, since we have only a spring force and gravity acting. If we define our origin to be the spring's equilibrium, then the initial energy of the system is zero. The energy at all other times must be zero, so at maximum distance, we have,

$$\frac{ky^2}{2} + mg(-y) = 0$$

Which yields a different answer of

$$y = \frac{2mg}{k}$$

What's going on? Where am I going wrong in approaching this problem?

Answer

What's going on? Where am I going wrong in approaching this problem?

Carefully consider your 3rd statement:

At the maximum distance the spring is stretched, the mass will be at rest so the weight will cancel out the spring force

At maximum distance, the mass will certainly be at rest but is it necessarily true that the weight and spring force cancel?

Put another way, does that fact that an object is (instantaneously) at rest imply that the acceleration of the object is zero?

newtonian gravity - Can a balloon float into space? (+orbital velocity)

After watching the recent "space jump" a question arose. Why can a balloon not float into space? Can one be made/designed to do this?

Next, everything in orbit is falling back to earth. It only maintains its altitude with speed, about 26,000mph give or take. This number means what, 20,000+ mph relative to what?

If the shuttle or station was to stop or approach zero mph it would fall to Earth, right? The balloon was at 128,000 feet and was only traveling at the Earth's rotational velocity of 1600mph. As viewed from the ground zero mph since it came down almost in the same stop. So what happens between say 130,000 feet and space that you need to increase your speed by 24,000mph?

Last if we were at a high enough orbit and "stopped", where we would not fall back to Earth, would we be able to watch the earth leave us. Could we wait there for a year for the Earth to return? How far out would this be?

electromagnetism - Are the field lines the same as the trajectories of a particle with initial velocity zero

Is it true that the field lines of an electric field are identical to the trajectories of a charged particle with initial velocity zero? If so, how can one prove it?

The claim is from a german physics book from Nolting "Grundkurs theoretische Physik 3 - Elektrodynamik" page 51, let me quote:

Man führt Feldlinien ein und versteht darunter die Bahnen, auf denen sich ein kleiner, positiv geladener, anfangs ruhender Körper aufgrund der Coulomb-Kraft (2.11) bzw. (2.20) fortbewegen würde.

In english:

One introduces field lines and means trajectories along which a small, positively charged, initially resting body moves due to the Coulomb-foce (2.11) resp. (2.20).

2.11 is just the coulomb law, 2.20 is $F = q E$.

(If someone has a better translation, feel free to edit it).

I don't see why this should be true. So it would be great to see a proof or a counterexample with solved equations of motion.

For a magnetic field this claim is obviously wrong since the Lorentz Force depends linearly on the velocity.

Are there other physical fields where the claim is analogously true?

Edit: The answers show that the claim is not true in general but holds in the special case of a highly viscous medium. Is this also the case for moving charged cotton along the field lines in air, as shown in this animation: http://www.leifiphysik.de/web_ph09_g8/grundwissen/01e_feldlinien/01e_feldlinien.htm ?

Do you have any references or more details for this viscous media limit?

Do you have any computational counter example why it doesn't hold in general or a simulation which shows that?

differential geometry - Dimensions of strings in string theory

In the above image taken from wikipedia, at the string level the strings have been shown as some loops, the article in wikipedia says that in string theory the particles at lower level are broken down into one dimensional strings, but I understand that only a straight line can be one dimensional, how are these loop like strings still said to be one dimensional?

Answer

the article in wikipedia says that in string theory the particles at lower level are broken down into one dimensional strings, but I understand that only a straight line can be one dimensional, how are these loop like strings still said to be one dimensional ?

Maybe this will help:

In mathematics, the dimension of an object is an intrinsic property independent of the space in which the object is embedded. For example, a point on the unit circle in the plane can be specified by two Cartesian coordinates, but one can make do with a single coordinate (the polar coordinate angle), so the circle is 1-dimensional even though it exists in the 2-dimensional plane.

Take a string and distort it, the position of the points on the string are into one to one correspondence with the stretched string, thus there is one string coordinate that describes the position of the points on the string, whatever its shape. Given the shape, one can find a mathematical function that will reduce the two or three dimensional description possible for a distorted string's points into one variable, as with the simple example of a circle above.

general relativity - What was the Law of Gravity better explained by?

In mechanics, our professor made the declaration that "all laws of physics" have been disproven. He mentioned several examples including the Law of Gravity, mentioning briefly that it is better explained by Einstein's theory of General Relativity. He also mentioned Bohr's model of the electron, and how it was better explained by other models, which I've since learned.

Now, haven taken Modern Physics as part of a Physics 3 class, I'm still scratching my head wondering how the theory of general relativity explains attraction between large objects.

Has has Einstein's theory of General Relativity truly explained it or did I misunderstand? If so, how?

Answer

You probably learned about Einstein's Theory of Special Relativity in your Modern Physics class. It is Einstein's Theory of General Relativity that provides a more accurate description of what we normally call gravity.

Basically, general relativity explains gravity not as an interaction between two bodies but as a warp in space-time in the presence of matter. Rather than thinking of gravity as a force, general relativity treats gravity as a change in space and time itself. It is more of a theory of geometry than of forces and attractions. But that is in general relativity, not special relativity.

Why do we get saturation current in photoelectric effect

So apparently if you increase the number of photons emitted per second to infinity, the photocurrent will approach a limiting value called the 'saturation current'

I feel like we shouldn't get a saturation current because as one electron is emitted, another will immediately (or close to immediately) fill its place just like in a normal current - thus electrons can pretty much be continuously emitted.

Can anyone explain why we get saturation current? Thanks :D

pattern - What is a Dotted Word™?

_{This is based on the What is a Word/Phrase™ series of Phrase™ and Word™ puzzles, started by JLee.}

---

If a word follows a certain rule, then I call it a Dotted Word™.

Use the example word lists below to find the rule. Each word can be tested for whether it is a Dotted Word™ without depending on other words in these lists.

Dotted Word™, Not Dotted Word™
ONLINE, DIGITAL
HOWDY, HEY
URGES, INSTINCTS
NASAL, BARITONE
FROG, TOAD
POLAND, UKRAINE
WATER, LUBRICATE

FOLIO, DOCUMENT
COBRAS, PYTHONS
REBIND, UNSEAL
PRATTLER, BABBLER
LUAU, FIESTA
RECEPT, RECEIVING
KNIFE, SPOON
FRAMES, BOXES
LANDLADY, DUCHESS

Hint:

The number of possible Dotted Words™ is relatively small.

Hint 2:

This is a very elementary puzzle. You don't need more than one hint.

Answer

The last hint and also Gareth's answer make it clear that ...

... all Dotted Words can be split into symbols for chemical elements. But this is not a sufficient condition, because some of the non-dotted words, but not all, can also be "elementarily split".

The distinguishing trait is ...

... that all elements used in the dotted words must be of the same period, i.e. they must occur on the same row in the periodic table:

O·N·Li·Ne: Oxygen (8), Nitrogen (7), Lithium (3), Neon (10): period 2
Ho·W·Dy: Holmium (67), Tungsten (74), Dysprosium (66): period 6
U·Rg·Es: Uranium (92), Roentgenium (111), Einsteinium (99): period 7
Na·S·Al: Sodium (11), Sulfur (16), Aluminium (13): period 3
Fr·Og: Francium (87), Oganesson (118): period 7

Po·La·Nd: Polonium (84), Lanthanum (57), Neodymium (60): period 6
W·At·Er: Tungsten (74), Astatine (85), Erbium (68): period 6
F·O·Li·O: Fluorine (9), Oxygen (8), Lithium (3), Oxygen (8): period 2
Co·Br·As: Cobalt (27), Bromine (35), Arsenic (33): period 4
Re·Bi·Nd: Rhenium (75), Bismuth (83), Neodymium (60): period 6
Pr·At·Tl·Er: Praseodymium (59), Astatine (85), Thallium (81), Erbium (68): period 6
Lu·Au: Lutetium (71), Gold (79): period 6
Re·Ce·Pt: Rhenium (75), Cerium (58), Platinum (78): period 6
K·Ni·Fe: Potassium (19), Nickel (28), Iron (26): period 4
Fr·Am·Es: Francium (87), Americium (95), Einsteinium (99): period 7

La·Nd·La·Dy: Lanthanum (57), Neodymium (60), Lanthanum (57), Dysprosium (66): period 6

(There are several possible divisions for some words, but only one that makes it count as a Dotted Word.)

Why are they called Dotted Words?

No, it has nothing to do with Lewis dots, which would describe the electron configuration in an atom and therefore would refer to the columns in the periodic table.

They are called Dotted words, because some people call a dot a period.

classical mechanics - Why does my door shut faster when the window is open?

I've noticed that if I shut my door when the window is open in a room, the door will tend to shut faster. If I shut the door when the window is closed with a normal force it will not fully close as if the air is cushioning it. I think when the window is open the door doesn't stop or slow down at all, it just slams.

I was wondering if this was to do with the pressure of the room or something. My hypothesis was that there is more room for the air to escape or there is some sort of convection? Just to note it is definitely not wind because I checked.

Here is a graphical simulation of the phenomenon:

Can somebody explain what is going on?

Answer

As the door nears the door frame there reaches a point where the door, for a moment, effectively seals off the air in the room from the air outside the room. This only happens for a moment, since most doors aren't 100% air proof. When this happens, as the door continues to close it decompresses the air inside the room, because the volume of the room increases as the door continues to close but the amount of air inside the room doesn't change because the room is briefly sealed off from outside the room. Thus there is an air pressure difference across the door, with the greater pressure coming from outside the room. This greater pressure slows the door down right before it closes.

On the other hand, with a window open air is let into the room and so even though the volume of the room increases as the door closes the air pressure from outside the window pushes air into the room to keep the air pressure inside the room about the same as outside. No pressure air pressure difference is found across the door and thus it does not slow down.

It is also possible that a fan or something inside the building could be creating a lower air pressure inside the building and thus there is a small air flow from the window into the building, which would push the door to close faster one the above-described brief sealing of the room happens, thus increasing the volume of the "slam!".

riddle - Word of the Day #2

Test run, take two. The first run went well, so let's keep it up!

For your second riddle, things might just get trickier -
With some common bond, things might just get stickier!
When I'm with a group I'll be holding you tight;
When alone, I'll protect you in the absence of light.

What am I?

Answer

I'm not too sure, but I'll guess:

Rubber

With some common bond, things might just get stickier!

Rubber cement (a type of adhesive)

When I'm with a group I'll be holding you tight;

Rubber band (group = band)

When alone, I'll protect you in the absence of light.

Rubber (i.e. condom, "protection" at night time, alone = no other word)

newtonian mechanics - Does the opening angle of the cone matter?

When discussing orbital mechanics, you learn that all orbits roughly follow an ellipse which is obtained as the intersection of a cone with an inclined plane, creating conic sections.

Below is a plot of different mathematical variables of a cone. I am trying to figure out if the ellipses that orbits follow are a special rule where the opening angle (at the top of the diagram) of the cone is 90 degrees, or if it is irrelevant what that angle is.

Answer

The angle is not important when mentioning elliptical orbits. Let's consider two cones with $\theta_{1}$ and $\theta_{2}$ respective; with $\theta_{1} < \theta_{2}$ (not shown, but take my word).

Both cones have an ellipse, who's center is $y$ from the top of the cone. The solution to the orbit equation in it's most general form is:

$r(\phi) = \frac{\ell^2}{m^2\gamma}\frac{1}{1+e\cos\phi}$

and is $y$ independent.

The cone with the the larger angle ($\theta_{2}$) has an ellipse that is larger by a certain factor $ A = \frac{tan\theta_{2}}{tan\theta_{1}}$. That is if $r_{1}(\phi)$ is associated with $\theta_{1} $ and $r_{2}(\phi)$ is associated with $\theta_{2}$ then,

$$ \theta_{1}\rightarrow \theta_{2} \Rightarrow r_{1}(\phi) \rightarrow Ar_{1}(\phi)=r_{2}(\phi). $$

It also follows that if cone 2 has a different angle than cone 1, there exists a y for cone - 1 such that $r_{1}(\phi)=r_{2}(\phi)$. But since the solution is independent of y, then it does not matter what $\theta$ you choose.

classical mechanics - Finding generalized coordinates when the implicit function theorem fails

Given some coordinates $(x_1,\dots, x_N)$ and $h$ holonomic constraints, it should always be possible to reduce the coordinates to $n=N-h$ generalized coordinates $(q_1,\dots, q_n)$. This is guaranteed by the implicit function theorem (its application of reducing coordinates is briefly mentioned here).

For a constraint $F(x, y) = 0$ concerning only two coordinates, the implicit function theorem states that a function $g$ wherefore $y = g(x)$ exists in a neighbourhood of a point $(x_0, y_0)$ satisfying the constraint only if

1. $F$ is continuously differentiable, and



2. $\frac{\partial F}{\partial y}(x_0,y_0) \neq 0$.

However, this is not always the case.

Example 1: unit circle

As pointed out by Emilio Pisanty, an easy example is a particle that must move on the unit circle in the $(x, y)$ plane: $F(x, y) = x^2 + y^2 - 1 = 0$ with derivative $\frac{\partial F}{\partial y}(x, y) = 2y$. When trying to eliminate $y$ as a coordinate, there are two local solutions: either $y = \sqrt{1-x^2}$ or $y = -\sqrt{1-x^2}$, thus one can easily describe the motion of the particle locally using the Lagrangian. Unfortunately, there is a problem with two points, $(1, 0)$ and $(-1, 0)$, because the derivative $\frac{\partial F}{\partial y}$ becomes zero, thus the points joining the solutions of $y$ violate the second condition and we're unable to describe the global motion of the particle (the particle might "hop" from one solution to the other)!

Of course the problem in the example above is easily solved by choosing other initial coordinates, namely polar coordinates $(r, \theta)$. The constraint becomes $F(r, \theta) = r - 1 = 0$, which satisfies both conditions.

Example 2: quadrilateral

Imagine two pendulums hanging from a fixed ceiling tied together by a rigid string:

Choosing $\alpha_1$ and $\alpha_2$ as initial coordinates, there is only one constraint: the $d$-string, the $l$-strings and the ceiling should always form a quadrilateral with given lengths for all sides.

With some uninteresting geometry, one can find a relation between $\alpha_1$ and $\alpha_2$, as done here in a revision. The point is that, when given some $\alpha_1$, there are again two possible solutions for $\alpha_2$: $\alpha_2 = \phi_1(\alpha_1) + \phi_2(\alpha_1)$ or $\alpha_2 = \phi_1(\alpha_1) - \phi_2(\alpha_1)$, where $\phi_1$ and $\phi_2$ are defined as in the revision, but are unimportant. Here is an illustration of the two possibilities:

The constraint can be written as $F(\alpha_1, \alpha_2) = |\alpha_2 - \phi_1(\alpha_1)| - \phi_2(\alpha_1) = 0$, which violates the first condition, or it can be written as $F(\alpha_1, \alpha_2) = (\alpha_2 - \phi_1(\alpha_1))^2 - \phi_2(\alpha_1)^2 = 0$, which violates the second condition. Again, I'm not able to describe the global motion of the system.

In the previous example this could be solved by choosing different initial coordinates, but I don't know whether this can be done here. This brings me to my questions.

Questions

• Is it always possible to choose initial coordinates in such a way that all constraints satisfy the conditions of the implicit function theorem?


• If so, is there a systematic way of finding them?


• If not, is it still possible to describe the global motion of the system using the Lagrangian?

Answer

Generally speaking, given a set of coordinates $x_1,\ldots,x_N$ under a set of $h=N-n$ holonomic constraints of the form $F_j(x_1,\ldots,x_N)=0$, you won't be able to find a subset $x_1,\ldots,x_n$ of your original coordinates that will function globally as generalized coordinates: the implicit-function theorem tells you that locally you are guaranteed such a subset, but in general that subset won't work everywhere.

This is kind of the main reason why we work with generalized coordinates $q_1,\ldots,q_n$, because they allow us to re-parametrize the available configuration space in a way that requires fewer coordinate charts. However, even those are not a full solution, because generally speaking the constraints will define the available coordinate space as a manifold which might require two or more charts in its minimal atlas, i.e. you are never guaranteed the existence of a set of coordinates which will work globally. Some examples:

• A particle in a plane confined to the unit circle is best parametrized via the polar angle, but even this isn't perfect (it fails to reproduce the periodicity when taken literally).


• For something where it's more clear that coordinate patches just won't work, consider e.g. a particle in 3D confined to the surface of a double torus: a single coordinate patch will simply be unable to cope with the nonzero genus of the surface.

This is why analytical mechanics, when done right (such as e.g. in the style of V.I. Arnol'd) works abstractly with the configuration space as a manifold and acknowledges that coordinate patches can only ever be local - and that that's OK. This means that wherever your configuration space is a regular manifold, you can always find a chart that works, and you can solve the dynamics in an explicit coordinate-dependent fashion there; if the solution wanders off the edge of the chart, then you can always use a separate chart where things are just fine.

For your configuration, though, you don't need to get particularly fancy, because the core topology of the problem is generally the same as that of the unit circle mentioned above. This can be seen by rephrasing the constraint in the form $F(\alpha_1,\alpha_2) = d$ where $$ F(\alpha_1,\alpha_2) = \bigg\|(D-l\cos(\alpha_2),-\sin(\alpha_2))-(l\cos(\alpha_1),-l\sin(\alpha_1))\bigg\|. $$ How do you analyze this? Well, have a look at a contour plot of $F(\alpha_1,\alpha_2)$ on the $\alpha_1,\alpha_2$ plane (taking $l=D/2$ for definiteness):

Depending on what $d$ is, your system is confined to one of the contour lines in the plot, and for the most part, these are topological circles of the boringest, most vanilla kind. This means that you can find a single angle-type generalized coordinate that parametrizes the entirety of the configuration-space manifold modulo the same periodicity problems that you get for a particle confined to the unit circle.

This isn't to say that finding that coordinate will be easy, and indeed (i) its analytical form is likely to be quite messy and (ii) this messiness will be reflected and amplified in the Euler-Lagrange equations of motion. In this specific instance I don't think there's anything to be gained by trying to find such a coordinate (as opposed to just using $\alpha_1$ and $\alpha_2$ as independent variables on the charts where they work, obtaining and solving the Euler-Lagrange EOMs on that parametrization, and just being aware of when one might need to switch charts) but ultimately it's a personal choice.

---

Now, that said, there are some (very specific) situations where the coordinate space might fail to be a manifold, essentially because $F(\alpha_1,\alpha_2)$ fails to have a nonzero gradient. For your problem, this occurs e.g. at the $d=0.9D$ contour line when $l=0.95D$, which looks like this,

i.e. it has a crossing at $\alpha_1=\alpha_2=0$, where the two arms are 'crossed', and you can get out of the configuration by pulling either arm away and making the other one follow, i.e. via two distinct routes. This type of configuration is the worst you'll encounter on the systems you've mentioned ─ but, really, you almost certainly never encounter it, which is why it's generally OK to dismiss the possibility as a pathological corner case for the mathematicians to worry about.

If you do insist on handling it, then what will generally happen from a physicist's perspective is that the system will tend to have some inertia, and that will tend to carry it on the same 'branch' it's going. (Or, in other words, you parametrize the figure-of-eight curve in a regular way, with a self-crossing which you then ignore.) This will work for all cases except those where the system is at the crossing with zero velocity, in which case it will stay there or only arrive asymptotically (but, if you really push things, yes: the lagrangian mechanics are undefined after you reach that kind of a pathological point).

Or, put another way, this kind of problem isn't really a problem.

classical mechanics - How far does a trampoline vertically deform based on the mass of the object?

If a baseball is dropped on a trampoline, the point under the object will move a certain distance downward before starting to travel upward again. If a bowling ball is dropped, it will deform further downwards. What is the nature of the relationship between the magnitude of this deformation and the object's mass? Linear? Square? etc.

Edit: I would like to add that the heart of what I'm asking is along the lines of this: "If a small child is jumping on a trampoline and the trampoline depresses 25% towards the ground, would an adult who weighs slighly less than four times as much be safe from depressing it all the way to the ground?"

Answer

It depends also on the shape of the object. If you assume the trampoline is circular, and the object is much smaller (like a point mass) then you can start developing the equations. You have to know the initial tension of the trampoline, and also assume the material non-elastic but supsended by perfect strings in a radial direction (with known stiffness).

After some math the static deflection (with pre-tension) obeys the following:

$$ \frac{W}{k\, R}=\tan\theta+\left(\frac{F_{0}}{k\, R}-1\right)\sin\theta $$

If $R$ is the radius of the trampoline then the dip is $H=R\,\tan\theta$ and so $\theta$ is the angle from horizontal that the cone makes. For any given angle $\theta$ the trampoline supports weight $W$ (given above) given total stiffness of $k$ and pre-tension of $F_{0}$.

So the above will give you the weight $W$ it will support given a dip $H$. It is the reciprocal of what you want, but it is solvable.

If the trampoline has $N$ linear springs each with stiffness rate of $k_i$ then the total stiffness (springs in parallel) is $k=k_i\,N$. To define the pre-tension $F_{0}$ assume that the free radius of the trampoline surface is $R$, but the springs are located at $R_0$ then the pre-tension is $F_{0}=k\,(R_0-R)$.

Example: A trampoline of 12 feet in diameter needs $F_{0}=100$ lbs total of pulling to string into a 12.5 foot ring. The stiffness is $k=\frac{100}{6}$ in pounds per inch. To dip the trampoline by 5 feet $\tan(\theta)=\frac{H}{R}=\frac{5\times12}{12\times12}$. Plug these into the above and you should get $W=115.4$ lbs.

I know I am going to confuse some people because I am treating radial quantities such as stiffness and loads as linear, but it works out (just use cylindrical coordinates).

Approximation

Small angle approximation (weight < 10% pre-tension, cone angle < 6°)

$$W = F_{0}\,\frac{H}{R}$$

Example: Using the same numbers as above a $W=10$ lbs weight will dip $H = (12\times12)\frac{W}{100} = 14.4$ inches. The full solution above gives $13.1$ inches

Theory

The deformed shape of the trampoline is a perfect cone. The distance from the center to where the springs start in the deformed state is always equal to $R$. The extension of the springs tension is then $F=F_{0}+k\,\left(\frac{R}{\cos\theta}-R\right)$ which needs to be balanced by the weight as $W = F\,\sin\theta$. The pieces come together to make the equation shown above.

Update

Solution is corrected for the fact the radial distance is constant, not the surface area of the trampoline. As the trampoline deforms its perimeter crumples and folds on itself like a napkin when lifted from the center.

quantum mechanics - Pressure in Harmonic Oscillation

Classical Harmonic oscillator's energy depends on temperature as it equals $k_B$$T/2$. However, quantum harmonic oscillator energy is $(n+1/2)hf$. So, when T=0, quantum predicts motion.

I have been wondering what would happen if there is change in pressure? If the pressure is changed, does the quantum and classical prediction differ?

special relativity - Does light itself experience time dilation?

This seems weird for me. Every time I see light, I will think, "Hey, that light travels from the past far away from us!"

Correct me if there is misconception.

Answer

In special relativity, time dilation is determined by the Lorentz factor $$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$$ Say the photon travels some time $T$ in our frame of reference. It is obviously travelling at the speed of light. We see that the Lorentz factor diverges as we go to the speed of light. Thus, in order for the photon's time in our frame to remain finite, its proper time must tend to zero. This is the colloquial "time stops at the speed of light."

gravity - Is the gravitational constant $G$ a fundamental universal constant?

Is the gravitational constant $G$ a fundamental universal constant like Planck constant $h$ and the speed of light $c$?

mathematics - Ernie and the Bubbles of Lucretia

"So, what do you know about milking spiders?", asked Ernie as he passed a fresh cup of coffee to me.

Good grief!, I thought and gave an involuntary shudder as I glanced at the milk jug sitting between our two cups I didn't even know spiders had udders!!!. "Think I'll have black coffee this morning", I replied.

Ernie gave an amused snort. "Not that kind of milking." he reassured me. "I spent half of last week getting samples of fresh silk directly from Lucretia's spinarettes. Lucretia is a Vanuatuan Orb Web Spider - and it is claimed that, weight for weight, her web is 10 times stronger than steel!"

I looked over at the terrarium - Lucretia was as big as the palm of my hand appeared to be sulking. "Non-venomous of course" continued Ernie, "but they can be a bit bitey when they are stressed". I noticed that Ernie's fingers and thumbs were dotted with numerous red wealts. "But I can't complain - all great research requires a little sacrifice - and thanks to my analysis of Lucretia's silk I have managed to synthesize a material even better than nature can create - an artificial silk that is transparent as glass, 1000 times as strong as steel, chemically inert, and what's more - if you expose it to a special catalyst it sets as hard as diamond!"

When we had finished our coffee he asked me to help test his new silk applicator gun. "When you press the red button it will squirt out a thread of silk - when you stop it will spray a little cloud of catalyst to rigidize it".

I squeezed the button, but instead of a thread, the device produced a small bubble. "Oops" said Ernie, "I didn't expect that - must be air in the reservoir". We examined the bubble. It was about as big as an orange with a negligibly thin wall thickness, but to my surprise, when I picked it up it was as rigid as a crystal ball. Ernie thought for a moment then smiled. "Actually, I think bubbles will be even more useful than rigid threads. I can make special flasks for an experiment I have planned."

Ernie asked me to blow a second bubble, this time much bigger. "It has to have a certain exact volume for the experiment" he said, "I'll tell you when to stop." I pressed the button again and this time held it down - a large bubble inflated from the applicator. When Ernie nodded his head I let go of the trigger, but the bubble continued to expand until it was bigger than a beach-ball. Ernie examined the gun. "Oops - didn't expect that - needs a bit of lubricant on the trigger mechanism". He fiddled with the applicator then claimed it was fixed.

But now the volume of the sphere was much too large for Ernie's experiment. "By my measurement it is exactly one meter in diameter" he said, "Much too big, but if we drill a small hole into it then we can blow another bubble inside - that will reduce the remaining volume in the sphere until it is exactly right. I'll tell you when to stop." I followed his instructions, but this time, although I kept holding the button down, the interior bubble stopped growing too soon. "Oops - didn't expect that, small electrical fault in the secondary trigger mechanism." He fiddled with the applicator then claimed it was fixed.

This meant the exterior bubble's remaining volume was still too large. "Keep adding more bubbles into the large sphere until I tell you to stop - I think we can still get it right" he said. I blew a second interior bubble inside the big sphere - when I had finished it formed the biggest sphere that could fit inside the exterior bubble alongside the first interior bubble.

"The remaining volume is still too big", said Ernie. So I blew a third interior bubble. When I had finished it formed the largest sphere that could fit inside the exterior bubble alongside the first two interior bubbles.

"Volume is still too big", said Ernie. So I blew a fourth one. When I had finished it formed the biggest sphere that could fit inside the large bubble alongside the first three interior bubbles.

"Aha", said Ernie, "now the remaining volume in the exterior sphere is exactly correct for my experiment. It's an interesting coincidence that the ratio of surface areas of the first interior bubble to the third interior bubble is exactly the same as the ratio of surface areas of the second interior bubble to the fourth interior bubble".

A few days later I had a phone message from Ernie (who had left town for a few days). His experiment had gone well but now he needed a second 'reaction chamber'. Could I pop around to the lab and blow one more sphere so it would be ready for his return? It needed to have exactly the same interior volume as the the remaining space in the one we made earlier. "I've mixed some more spider silk, but there isn't much to spare so I think you need to blow it the right size on the first attempt. I've also added a size dial on the applicator so you can stop at exactly the diameter you need (to the nearest micron)."

The problem is that Ernie failed to say exactly what diameter that should be! Can you help me work it out?

Answer

If the first bubble has diameter $n$, then the second bubble must have diameter $1-n$.

In the space remaining, there is always space for two bubbles of the same size that are both the largest bubbles that can fit into that space. (A proof to come once I'm not on my phone.)

We're also given that $\pi d_1^2 / \pi d_3^2 = \pi d_2^2 / \pi d_4^2$. Simplifying this gives that $(d_1 / d_2)^2 = (d_3 / d_4)^2$, which implies that $d_1 / d_3 = d_2 / d_4 = x$ as diameters can't be negative.

Therefore, $d_1 = xd_3 = xd_4 = d_2$, and the first two bubbles both have a diameter of 0.5 m.

Then, we have a diagram as above, and solve the equation $(0.5-r)^2 + 0.25^2 = (0.25+r)^2$. The solution is 1/6, so the diameter of the remaining two bubbles is 1/3 m.

Then, we calculate the cube root of $1^3 - 2(1/2)^3 - 2(1/3)^3$ to get a final diameter of 0.877606 m.

riddle - If you only have half of me, then I am of no use

People think a part of me is sticky, but I'm not.
I do have a sticky friend though.

I can create almost anything,
but I'm not very creative.

After I have done my thing, you can't tell that I did it.
To be honest, it wasn't my creation to begin with.

What am I?

Hint:

People think a part of me is sticky, but I'm not.
I do have a sticky friend though.
 
I can create almost anything,
but I'm not very creative.
 
After I have done my thing, you can't tell that I did it.

To be honest, it wasn't my creation to begin with.

Answer

You are:

Ctrl+C and Ctrl+V (Copy and Paste)

If you only have half of me, then I am no use

Why copy something if you can't paste it? And you can't paste something without a copy

People think a part of me is sticky, but I'm not

Copy doesn't do anything on itself.

I do have a sticky friend though

Copy is best friends with Paste.

I can create almost everything

You can copy pretty much all text

But I'm not very creative

But copy doesn't generate anything by itself

After I have done my thing, you can't tell that I did it

You can't distinguish between copied text and the original

To be honest, it wasn't my creation to begin with

Of course it wasn't. It's a copy!

Hint:

It's a copy of the original

special relativity - Photons moving relative to photons

I know we can't move at the speed of light, but if we were to travel on a photon how fast would we see other photons going? The speed of light is constant... So do photons see other photons moving at the speed of light?

Answer

According to Special Relativity, an object moving with speed c in one frame of reference moves with speed c in all frames of reference.

The immediate consequence is this: a photon has no frame of reference.

To appreciate this fully, consider that, no matter what speed you travel with respect to something else, you are at rest with respect to yourself.

But, and again, according to SR, if you were to have speed c in any frame of reference, there would be no frame of reference in which you were at rest with respect to yourself.

The bottom line is that asking a question along the lines of your question presumes that there exists a frame of reference that moves with speed c in which observations can be made.

But, there isn't!

quantum mechanics - What is $hat{p}|xrangle$?

Trying to solve a QM harmonic oscillator problem I found out I need to calculate $\hat{p}|x\rangle$, where $\hat{p}$ is the momentum operator and $|x\rangle$ is an eigenstate of the position operator, with $\hat{x}|x\rangle = x|x\rangle$. It turns out that $\hat{p}|x\rangle$ should be equal to $-i\hbar\frac{ \partial |x\rangle }{\partial x}$. But what does this even mean? I'm not sure what to do with an expression like that. If I try to find $\langle x | \hat{p} | x_0 \rangle$, I end up with $-i\hbar \frac{\partial}{\partial x} \delta(x-x_0)$, which I'm pretty sure is not allowed even in physics, where we would scoff at phrases like "$\delta$ is really a distribution and not a function".

Edit: Motivation. I have solved Heisenberg's equation and found, for a Hamiltonian $H = p^2/2m + m\omega^2x^2/2$, that $\hat{x}(t) = \hat{x}_0 \cos(\omega t) + \frac{\hat{p}_0}{m\omega} \sin(\omega t)$. I am given that the initial state is $|\psi(0)\rangle = |x_0\rangle$, i.e., an eigenstate of position, and I have a hunch that by finding $\hat{x}(t)|x_0\rangle$ I might find a way to get $|\psi(t)\rangle$. But this is not what I need help with; I am now curious as to what $\hat{p}|x\rangle$ might be, never mind my original problem.

Answer

The formula you quote is sort of correct, but I would encourage you to flip it around. More specifically, the old identification of momentum as a derivative means that $$\langle x |\hat p|\psi\rangle=-i\hbar\frac{\partial}{\partial x}\langle x |\psi\rangle,$$ and from this you can 'cancel out' the $\psi$ to get $$\langle x |\hat p=-i\hbar\frac{\partial}{\partial x}\langle x |.$$ (More specifically, these are both linear functionals from the relevant Hilbert space into $\mathbb C$ and they agree on all states in the Hilbert space, so they must agree as linear functionals.) This is the form that you actually use in everyday formulations, so it makes the most sense to just keep it this way, though you can of course take its hermitian conjugate to find an expression for $\hat p|x\rangle$.

Regarding your discomfort at the derivative of a delta function, I should assure you it is a perfectly legitimate object to deal with. It is best handled via integration by parts: for any well-behaved function $f$ $$ \int_{-\infty}^\infty \delta'(x)f(x)\text dx = \left.\delta(x)f(x)\vphantom\int\right|_{-\infty}^\infty-\int_{-\infty}^\infty \delta(x)f'(x)\text dx = -f'(0). $$ In essence, this defines a (different) distribution, which can legitimately be called $\delta'$. It is this distribution which shows up if you substitute $|\psi\rangle$ for $|x_0\rangle$ in my first identity: $$\langle x |\hat p|x_0\rangle=-i\hbar\delta'(x-x_0).$$

geometry - Plant 9 trees in 10 rows of 3

"Tree-planting" puzzles are also known as "points and lines" puzzles. The English puzzle author and mathematician Henry Ernest Dudeney was very fond of them. In 1917, Dudeney published a collection of puzzles called "Amusements in Mathematics", which also contains the following classic puzzle:

Sir Isaac Newton's tree-planting puzzle: A farmer wants to plant 9 trees in such a way that there are 10 rows of 3 trees. How does he do this?

Note: the different rows are not allowed to be collinear.

P.S. I know of two different solutions. If you know any more that would be great and if you are able to prove that there aren't any more solutions that would be even greater. I consider two solutions different when there is no graph isomorphism possible between them.

Answer

A third solution. I've been searching systematically for them (I've eliminated all the combinatorial solutions other than ABC, A14, A25, A36, B15, B26, B34, C16, C24, C35 - it is the only one with geometric solutions).

To see that it is distinct from the others, note that it has a tree with 7 immediate neighbors.

I believe that is all of them, though. Once I double check to make sure I haven't missed something, I try to put something together to show it.

statistical mechanics - Deriving the Boltzmann distribution using the information entropy

I was going through my lecture notes and I found something I could not quite understand. First, it starts by deriving an expression for the information entropy (as used in physics?):

Let $p_i$ be the probability of finding the system in microstate $i$. With the total number of accessible Microstates $N(A)$, this can be written $p_i = \frac{1}{N}$ for all microstates compatible with the macrostate $A$. We can write $\ln N(A) = -1\cdot\ln p_i = -\left(\sum_{i = 1}^{N(A)}p_i\right)\ln p_i$ due to the normalization of the $p_i$. [...]

Therefore, we can write for the information entropy of a macrostate $A$: $$S(A) = -k_\mathrm{B} \sum_{i = 1}^{N(A)}p_i\ln p_i$$

Later, it tries to derive the Boltzmann distribution for the ideal gas:

We will do so by finding the extremum of $$\phi = S(A) - \lambda_1\left(\sum_i p_i - 1\right) - \lambda_2 \left(\sum_i p_i E_i - E_\mathrm{avg}\right) $$ using the method of Lagrange multipliers. With $S(A) = -k_\mathrm{B} \sum_{i = 1}^{N(A)}p_i\ln p_i$.

It goes on to find the correct formula.

My question is, why this expression for the entropy $S(A)$ can be used, even though for the second example the $p_i$ are obviously not constant and equal to $\frac{1}{N}$?

Answer

You can actually derive the Gibbs entropy from purely mathematical concerns and the properties of probability. The properties we require entropy to have are:

1. Extensivity - the entropy of two independent systems, considered as a whole, should be the sum of the entropies of the individual systems $$S(A\cap B) = S(A) + S(B).$$


2. Continuity - the entropy should be a smoothly differentiable function of the probabilities assigned to each state.


3. Minimum - the entropy should be zero if and only if the system is in a single state with probability $1$.



4. Maximum - the entropy should be maximized when every state is equally probable.

It follows from probability theory that, when $A$ and $B$ are independent, then $$P(A\cap B) = P(A)\, P(B).$$ The unique continuous function that changes multiplication to addition is the logarithm, and which base is chosen is a matter of convention. Requirements 1 and 4 also imply that the entropy increases when states become less probable, requiring the constant of proportionality to be negative. Since $A$ and $B$ are microscopic random variables, the macroscopic entropy has to be an expectation value that averages over the microstates. Therefore \begin{align} S & = \langle -\ln(p_i) \rangle \\ & = -k\sum_{i=1}^N p_i \ln(p_i). \end{align}

In physics we choose the constant of proportionality to be $k_B$, Boltzmann's constant, and assign it units $\operatorname{J} \operatorname{K}^{-1}$, Joules per Kelvin, in order to match Clausius's formula for classical entropy. When all of the $p_i$ are equally probable, the formula reduces to the Boltzmann entropy.

You get the classical canonical ensembles and their corresponding distributions when you maximize the entropy of a system that is interacting with a 'bath' in a way that constrains the average value of a parameter (e.g. energy, volume, particle number) without specifying the value that parameter takes. The MB distribution comes, as the questioner saw, when the average energy is constrained but the total energy is allowed to vary; total energy would be fixed by adding a Lagrange multiplier of the form $\lambda_E (E_i - E_{\mathrm{tot}})$, producing the microcanonical ensemble.