Solution

Black body radiation, spectral density: ε (ν) dν = hνρ (ν) n (ν)

The photon energy, E = hν where h is Planck’s constant and ν is the photon frequency.

The density of states, ρ (ν) = Aνⁿ⁻¹ where A is a constant independent of the frequency and the frequency term is the scaling of surface area of an n-dimensional sphere.

The Bose–Einstein distribution,

n(v)$=\frac{1}{e\frac{hv}{kT}-1}$

where k is the Boltzmann constant and T is the temperature.

We let

$x=\frac{hv}{kT}$

and get

ε(x)$\propto =\frac{x^n}{e^x-1}$

We do not need the constant of proportionality (which is not simple to calculate in 4D) to find the maximum of ε (x). Working out the constant just tells us how tall the peak is, but we are interested in where the peak is, not the total radiation.

$\frac{d\epsilon }{dx}\propto \frac{n{x}^{n-1}}{e^x-1}-\frac{x^n{e}^x}{{\left(e^x-1\right)}^2}$

We set this equal to zero for the maximum of the distribution,

$\frac{x^{n-1}{e}^x}{{\left(e^x-1\right)}^2}\left(n\left(1-e^{-x}\right)-x\right)=0$

This yields x = n (1 − e^−x) where

$x=\frac{h{v}_{max}}{kT}$

and we can relate

${\lambda }_{max}=\frac{c}{v_{max}}$

and c being the speed of light.

This equation has the solution x = n +W (−ne⁻ⁿ) where W is the Lambert W function z = W (y) that solves ze^z = y (although there is a subtlety about which branch of the function). This is kind of useless to do anything with, though. One can numerically solve this equation using bisection/Newton–Raphson/iteration. Alternatively, one could notice that as the number of dimensions increases, e^−x is small, so to leading approximation x ≈ n. One can do a little better iterating this, x ≈ n − ne⁻ⁿ which is what we will use. Note the second iteration yields

$x\approx n-n{e}^{\left(n-n{e}^{-n}\right)}$

Number of dimensions, n	Numerical solution	Approximation
2	1.594	1.729
3	2.821	2.851
4 (the one we want)	3.921	3.927
5	4.965	4.966
6	5.985	5.985

Using the result above,

${\lambda }_{max}=\frac{hc}{kT{x}_{max}}=\frac{6.63 \times {10}^{-34}·3\times {10}^8}{1.38\times {10}^{-23}·6\times {10}^3·3.9}=616 nm$

616 nm is middle of the spectrum, so it will look white with a green-blue tint. Note, we have used T = 6000 K for the temperature here, as given in the question.

It would also be valid to look at ε (λ) dλ instead of ε (ν) dν.

Solution

First we will set up the equations of motion for our system. We will set one mass to be at position −x and the other to be at x, so the masses are at a distance of 2x from each other. Starting from Newton’s law of gravity:

$F=-\frac{G{m}^2}{{\left(2x\right)}^2}$

we can then use Newton’s second law to rewrite the LHS,

$m\ddot{x}=-\frac{G{m}^2}{4x^2}$

which we can simplify to

$\ddot{x}=-\frac{Gm}{4x^2}$

It is important that you get the right factor here depending on your choice for the particle coordinates at the start. Note there are other methods of getting this point, e.g. reduced mass.

We can now solve the second order ODE above. We will not show the whole process here but present the starting point and key results. We can write the acceleration in terms of the velocity. The initial velocity is zero and the initial position

$x_i=\frac{d}{2}$

So,

$v\frac{dv}{dx}=-\frac{Gm}{4x^2}\to \underset{0}{\overset{v}{\int }}vdv=-\frac{Gm}{4}\underset{x_i}{\overset{x}{\int }}\frac{dx}{x^2}$

and once the integrals are solved we can rearrange for the velocity,

$v=\frac{dx}{dt}=-\sqrt{\frac{Gm}{2}}\sqrt{\frac{1}{x}-\frac{1}{x_i}}$

Now we can form an expression for the total time taken for the masses to meet in the middle,

$T=\sqrt{\frac{2}{Gm}}\underset{0}{\overset{x_i}{\int }}\frac{dx}{\sqrt{\frac{1}{x}-\frac{1}{x_i}}}$

There are quite a few steps involved in solving this integral, for these solutions, we shall make use of the following (but do attempt to solve it for yourselves in full).

$\underset{0}{\overset{1}{\int }}\sqrt{\frac{y}{1-y}}dy=si{n}^{-1}\left(1\right)=\frac{\pi }{2}$

Hence,

$T=\frac{\pi }{2}\sqrt{\frac{2x_i^3}{Gm}}=\frac{\pi }{2}\sqrt{\frac{d^3}{4Gm}}$

We can now rearrange for G and substitute in the values given in the question, don’t forget to convert the time into seconds.

$G=\frac{d^3}{4m}{\left(\frac{\pi }{2T}\right)}^2=6.67\times {10}^{-11} m^{3}k{g}^{-1}{s}^{-2}$

This is the generally accepted value for the gravitational constant of our universe as well.

Solution

We will write the period of oscillation of the pendulum at the surface of the Earth to be

$T_0=2\pi \sqrt{\frac{l}{g_0}}$.

At a height h above the surface of the Earth the period of oscillation will be

$T_h=2\pi \sqrt{\frac{l}{g_h}}$,

where g₀ and g_h are the acceleration due to gravity at the surface of the Earth and a height h above it respectively.

We can define τ₀ to be the total duration of the day which is 8.64 × 10⁴ seconds and equal to N complete oscillations of the pendulum at the surface. The lag is then τ_h which will equal N times the difference in one period of the two clocks, τ_h = NΔT, where ΔT = (T_h − T₀). We can now take a ratio of the lag over the day and the total duration of the day:

$\frac{{\tau }_h}{{\tau }_0}=\frac{N\left(T_h-T_0\right)}{N{T}_0}\Rightarrow {\tau }_h={\tau }_0\frac{T_h-T_0}{T_0}={\tau }_h={\tau }_0\left(\frac{T_h}{T_0}-1\right)$

Then by substituting in the expressions we have for the period of a pendulum at the surface and height h we can write this in terms of the gravitational constant,

${\tau }_h={\tau }_0\left(\sqrt{\frac{g_0}{g_h}}-1\right)$

[Award 1 mark for finding the ratio of the lag over the day and the total period of the day.]

The acceleration due to gravity at the Earth’s surface is

$g_0=\frac{GM}{R^2}$

where G is the universal gravitational constant, M is the mass of the Earth and R is the radius of the Earth. At an altitude h, it will be

$g_h=\frac{GM}{{\left(R+h\right)}^2}$

[Award 1 mark for finding the expression for the acceleration due to gravity at height h.]

Substituting into our expression for the lag, we get:

${\tau }_h={\tau }_0\left(\sqrt{\frac{{\left(R+h\right)}^2}{R^2}}-1\right)={\tau }_0\left(\sqrt{1+\frac{2h}{R}+\frac{h^2}{R^2}}-1\right)={\tau }_0\left(\frac{\sqrt{R^2+2hR+h^2}}{R}-1\right)={\tau }_0\left(\frac{R+h}{R}-1\right)$

This simplifies to an expression for the lag over a day. We can then substitute in the given values to find,

${\tau }_h=\frac{{\tau }_{0}h}{R}=\frac{8.64\times {10}^4 s·300 m}{8.3781\times {10}^6 m} =4.064 s\approx 4 s$

[Award 2 marks for completing the simplification of the ratio and finding the lag to be ≈ 4 s.]

Solution

We can imagine this as an inverted pendulum, with gravity acting from the centre of mass $\left(\frac{l}{2}\right)$ and at an angle θ from the unstable equilibrium point.

[Award 1 mark for a suitable diagram of the system.]

We must now find the equations of motion of the system. For this we can use Newton’s second law $\overrightarrow{F}=m\overrightarrow{a}$ in its rotational form τ = Iα (torque = moment of inertia × angular acceleration). We have another equation for torque we can use as well

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\left|\overrightarrow{r}\right|\left|\overrightarrow{F}\right|\sin \theta \widehat{n}$

where r is the distance from the pivot to the centre of mass $\left(\frac{l}{2}\right)$ and F is the force, which in this case is gravity mg. We can then equate these giving

$\left|\overrightarrow{r}\right|\left|\overrightarrow{F}\right|\sin \theta =I\alpha$

Substituting in the given moment of inertia of the stick and that the angular acceleration

$\alpha =\frac{{\delta }^2\theta }{\delta t^2}=\ddot{\theta }$

We can cancel a few things and rearrange to get a differential equation of the form:

$\ddot{\theta }-\frac{3g}{2l}\sin \theta =0$

we then can take the small angle approximation sin θ ≈ θ, resulting in

$\ddot{\theta }-\frac{3g}{2l}\theta =0$

[Award 2 marks for finding the equation of motion for the system and using the small angle approximation.]

Solve with ansatz of θ = Ae^ωt + Be^−ωt, where we have chosen

${\omega }^2=\frac{3g}{2l}$

We can clearly see that this will satisfy the differential equation

$\dot{\theta }=\omega A{e}^{\omega t}-\omega B{e}^{-\omega t} and \ddot{\theta }={\omega }^{2}A{e}^{\omega t}+{\omega }^{2}B{e}^{-\omega t}$

Now we can apply initial conditions to find A and B, by looking at the two cases from the uncertainty principle

$\Delta x\Delta p=\Delta xm\Delta v\frac{\hslash }{2}$

Case 1: The stick is at an angle but not moving

At t = 0, θ = Δθ

θ = Δθ = A + B

At t = 0, $\dot{\theta }=0$

$\dot{\theta }=0=\omega A{e}^{{\omega }_0}-\omega B{e}^{-{\omega }_0}$, A=B

This implies Δθ = 2A and we can then find

$A=\frac{\Delta \theta }{2}=\frac{2\Delta x}{2l}=\frac{\Delta x}{l}$

So we can now write

$\theta =A\left(e^{\omega t}-e^{-\omega t}\right)=\frac{\Delta v}{\omega }\left(e^{\omega t}-e^{-\omega t}\right)$ or $\theta =\frac{2\Delta v}{\omega l}\sin h \omega t$

Case 2: The stick is at upright but moving

At t = 0, θ = 0

This condition gives us A = −B.

At t = 0, $\ddot{\theta }=\frac{2v}{l}$

This initial condition has come from the relationship between the tangential velocity, Δv which equals the distance to the centre of mass from the pivot point, $\frac{l}{2}$ and the angular velocity $\dot{\theta }$. Using the above initial condition gives us $\dot{\theta }=2\omega A$ where $A=\frac{\Delta v}{\omega l}$

We can now write

$\theta =A\left(e^{\omega t}-e^{-\omega t}\right)=\frac{\Delta v}{\omega }\left(e^{\omega t}-e^{-\omega t}\right) or \theta =\frac{2\Delta v}{\omega l}\sinh \omega t$

[Award 4 marks for finding the two expressions for θ by using the two cases of the uncertainty principle.]

Now there are a few ways we can finish off this problem, we shall look at three different ways. In each case when the stick has fallen on the ground $\theta \left(t_f\right)=\frac{\pi }{2}$.

Method 1

Take $\theta =\frac{2\Delta x}{l}\cosh \omega t$ and $\theta =\frac{2\Delta v}{\omega l}\sinh \omega t$, use $\theta \left(t_f\right)=\frac{\pi }{2}$ then rearrange for t_f in both cases. We have

$t_f=\frac{1}{\omega }\cos h{}^{-1}\frac{\pi l}{4\Delta x} and t_f=\frac{1}{\omega }\sinh {}^{-1}\frac{\pi \omega l}{4\Delta v}$

Look at the expression for cosh⁻¹ x and sinh⁻¹ x given in the question. They are almost identical, we can then approximate the two arguments to each other and we find,

$\Delta x=\frac{\Delta v}{\omega }$

we can then substitute in the uncertainty principle $\Delta x\Delta p=\frac{\hslash }{2}$ as $\Delta v=\frac{\hslash }{2m\delta x}$ and then write an expression of $\Delta x=\sqrt{\frac{\hslash }{2m\omega }}$, which we can put back into our arccosh expression (or do it for Δv and put into arcsinh).

$t_f=\frac{1}{\omega }\cosh {}^{-1}\frac{\pi l}{4\Delta x}$

where $\Delta x=\sqrt{\frac{\hslash }{2m\omega }}$ and $\omega =\sqrt{\frac{3g}{2l}}$.

Method 2

In this next method, when you get to the inverse hyperbolic functions, you can take an expansion of their natural log forms in the tending to infinity limit. To first order both functions give ln 2x, we can then equate the arguments and find Δx or Δv in terms of the other and use the uncertainty principle. This would give the time taken as,

$t_f=\frac{1}{\omega }\ln \frac{\pi l}{2\Delta x}$

where $\Delta x=\sqrt{\frac{\hslash }{2m\omega }}$ and $\omega =\sqrt{\frac{3g}{2l}}$.

Method 3

Rather than using hyperbolic functions, you could do something like above and do an expansion of the exponentials in the two expressions for t_f or we could make life even easier and do the following.

Disregard the e^−ωt terms as they will be much smaller than the e^ωt terms. Equate the two expressions for $\theta \left(t_f\right)=\frac{\pi }{2}$ and then take the natural logs, once again arriving at an expression of

$t_f=\frac{1}{\omega }\ln \frac{\pi l}{2\Delta x}$

where $\Delta x=\sqrt{\frac{\hslash }{2m\omega }}$ and $\omega =\sqrt{\frac{3g}{2l}}$.

This method efficiently sets B = 0 when applying the initial conditions.

[Award 2 marks for reaching an expression for t using one of the methods above or a suitable alternative that gives the correct units for time.]

Then, by using one of the expressions above for time, substitute in the values and find that t = 10.58 seconds.

[Award 1 mark for finding the correct time value of t = 10.58 seconds.]

• If you’re a student who wants to sign up for the 2025 edition of PLANCKS UK and Ireland, entries are now open at plancks.uk

The post PLANCKS physics quiz – the solutions appeared first on Physics World.

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They chat about the upcoming high-luminosity upgrade to the LHC (HL-LHC), which will be completed in 2030. The interview explores long-term strategies for particle physics research and the challenges of managing large international scientific organizations. Thomson also looks back on his career in particle physics and his involvement with some of the field’s biggest experiments.

 

The post Mark Thomson looks to the future of CERN and particle physics appeared first on Physics World.

Air traffic controllers asked Flight 5342 to switch runways when coming in to land. Black box recordings from the jet and the military helicopter it collided with will help reveal whether this contributed to the accident.

Oil spills can pollute large volumes of surrounding water – thousands of times greater than the spill itself – causing long-term economic, environmental, social and ecological damage. Effective methods for in situ capture of spilled oil are thus essential to minimize contamination from such disasters.

Many oil spill cleanup technologies, however, exhibit poor hydrodynamic stability under complex flow conditions, which leads to poor oil-capture efficiency. To address this shortfall, researchers from Harbin Institute of Technology in China have come up with a new approach to oil cleanup using a vortex-anchored filter (VAF).

“Since the 1979 Atlantic Empress disaster, interception and adsorption have been the primary methods for oil spill recovery, but these are sensitive to water-flow fluctuation,” explains lead author Shijie You. Oil-in-water emulsions from leaking pipelines and offshore industrial discharge are particularly challenging, says You, adding that “these problems inspire us to consider how we can address hydrodynamic stability of oil-capture devices under turbulent conditions”.

Inspired by the natural world

You and colleagues believe that the answers to oil spill challenges could come from nature – arguably the world’s greatest scientist. They found that the deep-sea glass sponge E. aspergillum, which lives at depths of up to 1000 m in the Pacific Ocean, has an excellent ability to filter feed with a high effectiveness, selectivity and robustness, and that its food particles share similarities with oil droplets.

The anatomical structure of E. aspergillum – also known as Venus’ flower basket – provided inspiration for the researchers to design their VAF. By mimicking the skeletal architecture and filter feeding patterns of the sponge, they created a filter that exhibited a high mass transfer and hydrodynamic stability in cleaning up oil spills under turbulent flow.

“The E. aspergillum has a multilayered skeleton–flagellum architecture, which creates 3D streamlines with frequent collision, deflection, convergence and separation,” explains You. “This can dissipate macro-scale turbulent flows into small-scale swirling flow patterns called low-speed vortical flows within the body cavity, which reduces hydrodynamic load and enhances interfacial mass transfer.”

For the sponges, this allows them to maintain a high mechanical stability while absorbing nutrients from the water. The same principles can be applied to synthetic materials for cleaning up oil spills.

The VAF is a synthetic form of the sponge’s architecture and, according to You, “is capable of transferring kinematic energy from an external water flow into multiple small-scale low-speed vortical flows within the body cavity to enhance hydrodynamic stability and oil capture efficiency”.

The tubular outer skeleton of the VAF comprises a helical ridge and chequerboard lattice. It is this skeleton that creates a slow vortex field inside the cavity and enables mass transfer of oil during the filtering process. Once the oil has been forced into the filter, the internal area – composed of flagellum-shaped adsorbent materials – provides a large interfacial area for oil adsorption.

Using the VAF to clean up oil spills

The researchers used their nature-inspired VAF to clean up oil spills under complex hydrodynamic conditions. You states that “the VAF can retain the external turbulent-flow kinetic energy in the low-speed vortical flows – with a small Kolmogorov microscale (85 µm) [the size of the smallest eddy in a turbulent flow] – inside the cavity of the skeleton, leading to enhanced interfacial mass transfer and residence time”.

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The research is published in Nature Communications.

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Anomalous Hall crystal

“Several theory groups have speculated that analogues of these structures could emerge in quantized anomalous Hall systems, giving rise to a type of TEC termed an ‘anomalous Hall crystal’,” Folk explains. “This structure would be insulating, due to a frozen-in electronic ordering in its interior, with dissipation-free currents along the boundary.”

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In a vast majority of cases, the twisted structures studied by the team had moiré patterns that were very disordered. Moiré patterns occur when two lattices are overlaid and rotated relative to each other. Yet out of tens of thousands of permutations of twisted graphene stacks, one structure appeared to be different.

Exceptionally low levels of disorder

“One of the stacks seemed to have exceptionally low levels of disorder,” Folk describes. “Waters shared that one with our group to explore in our dilution refrigerator, where we have lots of experience measuring subtle magnetic effects that appear at a small fraction of a degree above absolute zero.”

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“We observed the first clear example of a TEC, in a device made up of bilayer graphene stacked atop trilayer graphene with a small, 1.5° twist,” Folk explains. “The underlying topology of the electronic system, combined with strong electron-electron interactions, provide the essential ingredients for the crystal formation.”

After decades of theoretical speculation, Folk, Waters and colleagues have identified an anomalous Hall crystal, where the quantum Hall effect emerges from an in-built electronic structure, rather than an applied magnetic field.

Beyond confirming the theoretical possibility of TECs, the researchers are hopeful that their results could lay the groundwork for a variety of novel lines of research.

“One of the most exciting long-term directions this work may lead is that the TEC by itself – or perhaps a TEC coupled to a nearby superconductor – may host new kinds of particles,” Folk says. “These would be built out of the ‘normal’ electrons in the TEC, but totally unlike them in many ways: such as their fractional charge, and properties that would make them promising as topological qubits.”

The research is described in Nature.

The post Anomalous Hall crystal is made from twisted graphene appeared first on Physics World.

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