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.

Several years ago I was sitting at the back of a classroom supporting a newly qualified science teacher. The lesson was going well, a pretty standard class on Hooke’s law, when a student leaned over to me and asked “Why are we doing this? What’s the point?”.

Having taught myself, this was a question I had been asked many times before. I suspect that when I was a teacher, I went for the knee-jerk “it’s useful if you want to be an engineer” response, or something similar. This isn’t a very satisfying answer, but I never really had the time to formulate a real justification for studying Hooke’s law, or physics in general for that matter.

Who is the physics curriculum designed for? Should it be designed for the small number of students who will pursue the subject, or subjects allied to it, at the post-16 and post-18 level? Or should we be reflecting on the needs of the overwhelming majority who will never use most of the curriculum content again? Only about 10% of students pursue physics or physics-rich subjects post-16 in England, and at degree level, only around 4000 students graduate with physics degrees in the UK each year.

One argument often levelled at me is that learning this is “useful”, to which I retort – in a similar vein to the student from the first paragraph – “In what way?” In the 40 years or so since first learning Hooke’s law, I can’t remember ever explicitly using it in my everyday life, despite being a physicist. Whenever I give a talk on this subject, someone often pipes up with a tenuous example, but I suspect they are in the minority. An audience member once said they consider the elastic behaviour of wire when hanging pictures, but I suspect that many thousands of pictures have been successfully hung with no recourse to F = –kx.

Hooke’s law is incredibly important in engineering but, again, most students will not become engineers or rely on a knowledge of the properties of springs, unless they get themselves a job in a mattress factory.

From a personal perspective, Hooke’s law fascinates me. I find it remarkable that we can see the macroscopic properties of materials being governed by microscopic interactions and that this can be expressed in a simple linear form. There is no utilitarianism in this, simply awe, wonder and aesthetics. I would always share this “joy of physics” with my students, and it was incredibly rewarding when this was reciprocated. But for many, if not most, my personal perspective was largely irrelevant, and they knew that the curriculum content would not directly support them in their future careers.

At this point, I should declare my position – I don’t think we should take Hooke’s law, or physics, off the curriculum, but my reason is not the one often given to students.

A series of lessons on Hooke’s law is likely to include: experimental design; setting up and using equipment; collecting numerical data using a range of devices; recording and presenting data, including graphs; interpreting data; modelling data and testing theories; devising evidence-based explanations; communicating ideas; evaluating procedures; critically appraising data; collaborating with others; and working safely.

Science education must be about preparing young people to be active and critical members of a democracy, equipped with the skills and confidence to engage with complex arguments that will shape their lives. For most students, this is the most valuable lesson they will take away from Hooke’s law. We should encourage students to find our subject fascinating and relevant, and in doing so make them receptive to the acquisition of scientific knowledge throughout their lives.

At a time when pressures on the education system are greater than ever, we must be able to articulate and justify our position within a crowded curriculum. I don’t believe that students should simply accept that they should learn something because it is on a specification. But they do deserve a coherent reason that relates to their lives and their careers. As science educators, we owe it to our students to have an authentic justification for what we are asking them to do. As physicists, even those who don’t have to field tricky questions from bored teenagers, I think it’s worthwhile for all of us to ask ourselves how we would answer the question “What is the point of this?”.

The post ‘Why do we have to learn this?’ A physics educator’s response to every teacher’s least favourite question appeared first on Physics World.

As proclaimed by the United Nations, 2025 is the International Year of Quantum Science and Technology, or IYQ for short. This year was chosen because it marks the 100th anniversary of Werner Heisenberg’s development of matrix mechanics – the first consistent mathematical description of quantum physics.

Our guest in this episode of the Physics World Weekly podcast is the Turkish quantum physicist Mete Atatüre, who heads up the Cavendish Laboratory at the UK’s University of Cambridge.

In a conversation with Physics World’s Katherine Skipper, Atatüre talks about hosting Quantour, the quantum light source that is IYQ’s version of the Olympic torch. He also talks about his group’s research on quantum sensors and quantum networks.

• There is much more about Heisenberg’s mathematical breakthrough in quantum physics here: “Return to Helgoland: celebrating 100 years of quantum mechanics”.

The post International Year of Quantum Science and Technology: our celebrations begin with a look at quantum networks and sensors appeared first on Physics World.

Each year, the International Association of Physics Students organizes a physics competition for bachelor’s and master’s students from across the world. Known as the Physics League Across Numerous Countries for Kick-ass Students (PLANCKS), it’s a three-day event where teams of three to four students compete to answer challenging physics questions.

In the UK and Ireland, teams compete in a preliminary competition to be sent to the final. Here are some fiendish questions from past PLANCKS UK and Ireland preliminaries and the 2024 final in Dublin, written by Anthony Quinlan and Sam Carr, for you to try this holiday season.

Question 1: 4D Sun

Imagine you have been transported to another universe with four spatial dimensions. What would the colour of the Sun be in this four-dimensional universe? You may assume that the surface temperature of the Sun is the same as in our universe and is approximately T = 6 × 10³ K. [10 marks]

Boltzmann constant, k_B = 1.38 × 10⁻²³ J K⁻¹

Speed of light, c = 3 × 10⁸ m s⁻¹

Question 2: Heavy stuff

In a parallel universe, two point masses, each of 1 kg, start at rest a distance of 1 m apart. The only force on them is their mutual gravitational attraction, F = –Gm₁m₂/r². If it takes 26 hours and 42 minutes for the two masses to meet in the middle, calculate the value of the gravitational constant G in this universe. [10 marks]

Question 3: Just like clockwork

Consider a pendulum clock that is accurate on the Earth’s surface. Figure 1 shows a simplified view of this mechanism.

A pendulum clock runs on the gravitational potential energy from a hanging mass (1). The other components of the clock mechanism regulate the speed at which the mass falls so that it releases its gravitational potential energy over the course of a day. This is achieved using a swinging pendulum of length l (2), whose period is given by

$T=2\pi \sqrt{\frac{l}{g}}$

where g is the acceleration due to gravity.

Each time the pendulum swings, it rocks a mechanism called an “escapement” (3). When the escapement moves, the gear attached to the mass (4) is released. The mass falls freely until the pendulum swings back and the escapement catches the gear again. The motion of the falling mass transfers energy to the escapement, which gives a “kick” to the pendulum that keeps it moving throughout the day.

Radius of the Earth, R = 6.3781 × 10⁶ m

Period of one Earth day, τ₀ = 8.64 × 10⁴ s

How slow will the clock be over the course of a day if it is lifted to the hundredth floor of a skyscraper? Assume the height of each storey is 3 m. [4 marks]

Question 4: Quantum stick

Imagine an infinitely thin stick of length 1 m and mass 1 kg that is balanced on its end. Classically this is an unstable equilibrium, although the stick will stay there forever if it is perfectly balanced. However, in quantum mechanics there is no such thing as perfectly balanced due to the uncertainty principle – you cannot have the stick perfectly upright and not moving at the same time. One could argue that the quantum mechanical effects of the uncertainty principle on the system are overpowered by others, such as air molecules and photons hitting it or the thermal excitation of the stick. Therefore, to investigate we would need ideal conditions such as a dark vacuum, and cooling to a few milli­kelvins, so the stick is in its ground state.

Moment of inertia for a rod,

$I=\frac{1}{3}m{l}^2$

where m is the mass and l is the length.

Uncertainty principle,

$\Delta x\Delta p\ge \frac{\hslash }{2}$

There are several possible approximations and simplifications you could make in solving this problem, including:

sinθ ≈ θ for small θ

${\cosh }^{-1}x=ln \left(x+\sqrt{x^2-1}\right)$

and

${\sinh }^{-1}x=ln \left(x+\sqrt{x^2+1}\right)$

Calculate the maximum time it would take such a stick to fall over and hit the ground if it is placed in a state compatible with the uncertainty principle. Assume that you are on the Earth’s surface. [10 marks]

Hint: Consider the two possible initial conditions that arise from the uncertainty principle.

• Answers will be posted here on the Physics World website next month. There are no prizes.
• 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 – how do you measure up against the brightest physics students in the UK and Ireland? appeared first on Physics World.