Feeds:
Posts
Comments

Posts Tagged ‘Quantum Mechanics’

Antimatter and Dirac’s Equation

Posted in History, Nature, Physics, tagged , , , , on 01/04/2019| 1 Comment »

Yesterday I introduced Paul Dirac, number 10 in “The Guardian’s” list of the 10 best physicists. I mentioned that his main contributions to physics were (i) predicting antimatter, which he did in 1928, and (ii) producing an equation (now called the Dirac equation) which describes the behaviour of a sub-atomic particle such as an electron travelling at close to the speed of light (a so-called relativistic theory). This equation was also published in 1928.

The Dirac Equation

In 1928 Dirac wrote a paper in which he published what we now call the Dirac Equation.

The equation now known as the Dirac Equation describes the behaviour of an electron when travelling close to the speed of light. The equation now known as the Dirac Equation describes the behaviour of an electron when travelling close to the speed of light.

This is a relativistic form of Schrödinger’s wave equation for an electron. The wave equation was published by Erwin Schrödinger two years earlier in 1926, and describes how the quantum state of a physical system changes with time.

The Schrödinger eqation

The time dependent Schrödinger equation which describes the motion of an electron The time dependent Schrödinger equation which describes the motion of an electron

The various terms in this equation need some explaining. Starting with the terms to the left of the equality, and going from left to right, we have i is the imaginary number, remember i = \sqrt{-1}. The next term \hbar is just Planck’s constant divided by two times pi, i.e. \hbar = h/2\pi. The next term \partial/\partial t \text{ } \psi(\vec{r},t) is the partial derivative with respect to time of the wave function \psi(\vec{r},t).

Now, moving to the right hand side of the equality, we have
m which is the mass of the particle, V is its potential energy, \nabla^{2} is the Laplacian. The Laplacian, \nabla^{2} \psi(\vec{r},t) is simply the divergence of the gradient of the wave function, \nabla \cdot \nabla \psi(\vec{r},t).

In plain language, what the Schrödinger equation means “total energy equals kinetic energy plus potential energy”, but the terms take unfamiliar forms for reasons explained below.

Advertisements

Read Full Post »

The Pauli exclusion principle

Posted in Physics, Science, tagged , , , , , on 17/12/2013| Leave a Comment »

Electron configurations

In this blog, I discussed the “electron configuration” nomenclature which is so loved by chemists (strange people that they are….). Just to remind you, the noble gas neon, which is at number 10 in the periodic table, may be written as 1s^{2} \; 2s^{2} \; 2p^{6}. If you add together the superscripts you get 2+2+6=10, the number of electrons in neutral Helium. Titanium, which is at number 22 in the periodic table may be written as 1s^{2} \; 2s^{2} \; 2p^{6} \; 3s^{2} \; 3p^{6} \; 3d^{2} \; 4s^{2}. Again, if you add together the superscripts you get 2+2+6+2+6+2+2=22, the number of electrons in neutral Titanium. I explained in the blog that the letters s,p,d and f refer to “sharp, principal, diffuse” and “fine“, as this was how the spectral lines appeared in the 1870s when spectroscopists first started identifying them.

But, what I didn’t address in that blog on the electron configuration nomenclature is why do electrons occupy different shells in atoms? In hydrogen, the simplest atom, the 1 electron orbits the nucleus in the ground state, the n=1 energy level. If it is excited it will go into a higher energy level, n=2 or 3 etc. But, with a more complicated atom like neon, which has 10 electrons, the 10 do not all sit in the n=1 level. The n=1 level can only contain up to 2 electrons, and the n=2 level can only contain up to 8 electrons, the n=3 level can only contain up to 18 electrons, and so on. This leads to neon having a “filled” n=1 level (2 electrons), and a filled n=2 level (8 electrons), which means it does not seek additional electrons. This is why it is a noble gas.

Titanium on the other hand, with 22 electrons, has a filled n=1 level (2 electrons), a filled n=2 level (8 electrons), a partially filled n=3 level (8 electrons out of a possible 18), and a partially filled n=4 level (2 electrons out of a possible 32). Because it has partially filled n=3 and n=4 levels, and it wants them to be full, it will seek additional electrons by chemically combining with other elements.

What is the reason each energy level has a maximum number of allowed electrons?

It is all due to something called the Pauli exclusion principle.



Wolfgang Pauli, after whom the Pauli exclusion principle is named. In addition to this principle, he also came up with the idea of the neutrino.

Wolfgang Pauli, after whom the Pauli exclusion principle is named. He came up with the idea in 1925. In addition to this principle, he also came up with the idea of the neutrino.



The energy level n

Niels Bohr suggested in 1913 that electrons could only occupy certain orbits. I go into the details of his argument in this blog, but to summarise it briefly here, he suggested that something called the orbital angular momentum of the electron had to be divisible by \hbar \text{ where } \hbar = h/2\pi, \text{ } h being Planck’s constant. We now call these the energy levels of an atom, and we use the letter n to denote the energy level. So, an electron in the second energy level will have n=2, in the third energy level it will have n=3 etc.

As quantum mechanics developed over the next 15-20 years it was realised that an electron is fully described by a total of four (4) quantum numbers, not just its energy level. The energy level n came to be known as the princpical quantum number. The other three quantum numbers needed to fully describe the state of an electron are

  • its orbital angular momentum, l
  • its magnetic moment, m_{l} and
  • its spin, m_{s}

The orbital angular momentum l quantum number

As I mentioned above, spectroscopists noticed that atomic lines could be visually categorised into “sharp”, “principal”, “diffuse” and “fine“, or s,p,d \text{ and } f. It was found that the following correspondence existed between these visual classifications and the orbital angular momentum l. This is the second quantum number. l can only take on certain values from 0 \text{ to } (n-1). So, for example, if n=3, \; l \text{ can be } 0,1 \text{ or } 2.


spectroscopic name and orbital angular momentum
Spectroscopic Name letter orbital angular momentum l
sharp s l=0
principal p l=1
diffuse d l=2
fine f l=3



As this table shows, the reason a line appears as a “sharp” (s) line is because its orbital angular momentum l=0. If it appears as a “principal” (p) line then its orbital angular momentum must be l=1, etc.

The magnetic moment quantum number m_{l}

The third quantum number is the magnetic moment m_{l}, which can only take on certain values. The magnetic moment only shows up if the electron is in a magnetic field, and is what causes the Zeeman effect, which is the splitting of an atom’s spectral lines when an atom is in a magnetic field. The rule is that the magnetic moment quantum number can take on any value from -l \text{ to } +l, so e.g. when l=2, \text{ } m_{l} can take the values -2, -1, 0, 1 \text{ and } 2 (5 possible values in all). If l=3 \text{ then } m_{l} \text{ can be } -3, -2, -1, 0, 1, 2, 3 (7 possible values).

The spin quantum number m_{s}

The final quantum number is something called the spin. Although it is only an analogy (and not to be taken literally), one can think of this as the electron spinning on its axis as it orbits the nucleus, in the same way that the Earth spins on its axis as it orbits the Sun. The spin can, for an electron, take on two possible values, either +1/2 \text{ or } -1/2.

Putting all of this together

Let us first of all consider the n=1 energy level. The only allowed orbital angular momentum allowed in this level is l=0, which means the only allowed values of m_{l} is also 0 and the allowed values of the spin are +1/2 \text{ and } -1/2. So, in the n=1 level, the only allowed state is 1s, and this can have two configurations, with the electron spin up or down (+1/2 or -1/2), meaning the n=1 level is full when there are 2 electrons in it. That is why we see 1s^{2} for Helium and any element beyond it in the Periodic Table. But, what about the n=2, n=3 etc. levels?

The number of electrons in each electron shell
State Principal quantum number n Orbital quantum number l Magnetic quantum number m_{l} Spin quantum number m_{s} Maximum number of electrons
1s 1 0 0 +1/2, -1/2 2
n=1 level Total = 2
2s 2 0 0 +1/2, -1/2 2
2p 2 1 -1,0,1 +1/2, -1/2 6
n=2 level Total = 8
3s 3 0 0 +1/2, -1/2 2
3p 3 1 -1,0,1 +1/2, -1/2 6
3d 3 2 -2,-1,0,1,2 +1/2, -1/2 10
n=3 level Total = 18
4s 4 0 0 +1/2, -1/2 2
4p 4 1 -1,0,1 +1/2, -1/2 6
4d 4 2 -2,-1,0,1,2 +1/2, -1/2 10
4f 4 3 -3,-2,-1,0,1,2,3 +1/2, -1/2 14
n=4 level Total = 32
5s 5 0 0 +1/2, -1/2 2
etc.



The astute readers amongst you may have noticed that the electron configuration for Titanium, which was 1s^{2} \; 2s^{2} \; 2p^{6} \; 3s^{2} \; 3p^{6} \; 3d^{2} \; 4s^{2}, suggests that the n=4 level starts being occupied before the n=3 level is full. After all, the n=3 level can have up to 18 electrons in it, with up to 10 electrons in the n=3, l=2 (d) state. In the n=3 level the (s) and (p) states are full, but not the (d) state. With only 2 electrons in the n=3, l=2 (d) state, the 4s state starts being populated, and has 2 electrons in it. Why is this?

I will explain the reason in a future blog, but it has to do with the “shape” of the orbits of the different states. They are different for different values of orbital angular momentum l.

Read Full Post »

The Bohr model of the atom

Posted in History, Physics, Science, tagged , , , , , , on 19/11/2013| 3 Comments »

In my blog on Niels Bohr, number 2 in The Guardian’s list of the 10 best physicists, I mentioned the Bohr model of the atom. In this blog I will go into more detail about this model, and how it agreed with the experimental results (the Rydberg formula) for the hydrogen atom.

Quantised orbits

In 1911, Rutherford had proposed that atoms have positively charged nuclei, with the negatively charged electrons orbiting the nuclei. One of the problems with this idea was that an orbiting electron would be accelerating, by virtue of moving in a circle. The acceleration is directed towards the centre of the circle. It was well known that when an electron is accelerated it radiates electromagnetic waves. Calculations showed that the orbiting electrons on Rutherford’s model should radiate away their energy in a few microseconds (millionths of a second), and spiral towards the nucleus. They clearly were not doing this, but why?

Bohr suggested in a paper in 1913 that electrons would somehow not radiate away their energy if they were orbiting in certain “allowed orbits”. If they were in these special orbits, the normal laws of EM radiation would not apply. He suggested that these allowed orbits were when the orbital angular momentum L could be written as


L = \frac{ n h }{ 2 \pi } = n \hbar \text{ (Equ. 1) }


(\hbar = \frac{ h }{ 2 \pi } \text{ where } h is Planck’s constant, and is given its own symbol in Physics at it crops up so often). What is orbital angular momentum? Well, it is the rotational equivalent of linear momentum. Linear momentum is defined as \vec{p} = m\vec{v} \text{ where } m \text{ is the mass and } \vec{v} \text{ is the velocity}. Notice, momentum is a vector quantity, this is important in doing calculations involving collisions, such as the ones I did in this blog.



20131119-070256.jpg

By analogy, orbital angular momenutm is defined as


\vec{L} = \vec{r} \times m \vec{v}


where \vec{r} is the radius vector of the orbit, which is defined as pointing from the centre of the orbit along the radius. For a circular orbit, where the radius vector is at right angles to the velocity vector, we can just write L=mvr where L is the magnitude (size) of the vector \vec{L}.

The force keeping the electron in orbit

From Classical Physics, Bohr argued that the force which was keeping the electron in orbit about the positively charged nucleus was the well known Coulomb force, given by


F = - \frac{ Z k_{e} e^{2} }{ r^{2} }


where k_{e} is the Coulomb constant (which determines the force between two 1 Coulomb charges separated by 1 metre), Z is the atomic number of the atom, e is the charge on an electron and r is the radius of the orbit. The minus sign is telling us that the force is directed towards the centre, whereas our definition of the radius vector is that it is away from the centre, so they are in opposite directions.

We can equate this to the formula for the centripetal force on any object moving in a circular orbit, so we can write


\frac{ m_{e} v^{2} }{ r } = \frac{ Z k_{e} e^{2} }{ r^{2} } \text{ (Equ. 2) }


where m_{e} is the mass of the electron and v is the speed of its orbit.

The Bohr radius

Re-arranging Equation 2 we can write


v = \sqrt{ \frac{ Z k_{e} e^{2} m_{e} r }{ m_{e}^2 r^{2} } }


which then allows us to write the angular momentum as


m_{e} v r = \sqrt{ Z k_{e} e^{2} m_{e} r } \text{ which (from Equ. 1) } = n \hbar


This allows us to write an expression for the “radius” of an electron’s orbit as


r_{n} = \frac{ n^{2} \hbar^{2} }{ Z k_{e} e^{2} m_{e} }


where n is the energy level of the electron. The so-called “Bohr radius” is the radius of an electron in the n=1 energy level for hydrogen (Z=1) and can be written


\boxed{ r_{1} = \frac{ \hbar^{2} }{ k_{e} e^{2} m_{e} } \approx 5.29 \times 10^{-11} \text{ metres } }


This is, indeed, about the size of a hydrogen atom.

The total energy of the electron

The total energy of the electron in its orbit is given by the sum of its kinetic energy and its potential energy. The kinetic energy is just given by 1/2 \; (mv^{2}). What about the potential energy? The potential energy can be found by using the relationship between work and force; back in this blog I said that work was defined as the force multiplied by the distance moved. Energy is the capacity to do work, and is measured in the same units, Joules. So we can derive the potential energy of an electron in orbit due to the Coulomb force as


P.E. = \int_{r}^{\infty} { F} dr = - \int_{r}^{\infty} \frac{ Z k_{e} e^{2} }{ r^{2} } dr


where dr is an incremental change in the radius. If we do this integration we get


P.E. = - \frac{ Z k_{e} e^{2} }{ r }


where the negative sign is telling us that we have to do work on the electron to increase its radius, or to put it another way that the force acts towards the centre but the radius vector acts away from the centre of the electron’s orbit. This means that the total energy E is given by


E = \frac{ 1 }{ 2 } m_{e} v^{2} - \frac{ Z k_{e} e^{2} }{ r }


But, from Equ. 2 we can write the kinetic energy as


\frac{ 1 }{ 2 } m_{e} v^{2} = \frac{ Z k_{e} e^{2} }{ 2r }


So then the total energy E can be written


E = \frac{ Z k_{e} e^{2} }{ 2r } - \frac{ Z k_{e} e^{2} }{ r } = - \frac{ Z k_{e} e^{2} }{ 2 r }
So, in the Bohr model, the energy of the n^{th} energy level is given by


\boxed{ E_{n} = - \frac{ Z k_{e} e^{2} }{ 2 r_{n} } \text{ or } -\frac{ Z^{2} (k_{e} e^{2})^{2} m_{e} }{2 \hbar^{2} n^{2} } }


In the case of hydrogen, where Z=1 we can write


\boxed{ E_{n} = -\frac{ (k_{e} e^{2})^{2} m_{e} }{ 2 \hbar^{2} n^{2} } \approx -\frac{ 13.6 }{ n^{2} } \text{ eV} }


This was in perfect agreement with the Rydberg formula for the energy levels of hydrogen, which had been experimentally derived by the Swedish physicist Johannes Rydberg in 1888. As I will show in a future blog, Bohr’s model was a “semi-empirical” model, in that it was a step along the way to the correct model. It was produced by using a mixture of classical physics and quantum mechanics, and Bohr did not understand why his condition that only orbits whose angular momentum were equal to n \hbar was true. The explanation was produced with the full theory of Quantum Mechanics in 1926 as a solution to Schrödinger’s wave equation for hydrogen.

Read Full Post »

Derivation of the Rayleigh-Jeans law – part 3 (final part)

Posted in History, Nature, Physics, Science, tagged , , , , , on 12/11/2013| 18 Comments »

In part 1 of this blog, I showed that the 3-dimensional wave equation for an electromagnetic (EM) wave can be written (ignoring the magnetic component \vec{B} as it is much smaller than the electric component \vec{E} ) as


\frac{ \partial^{2} E }{ \partial{x}^{2} } + \frac{ \partial^{2} E }{ \partial{y}^{2} } + \frac{ \partial^{2} E }{ \partial{z}^{2} } = \frac{ 1 }{ c^{2} } \frac{ \partial^{2} E }{ \partial{t}^{2} }


In part 2 I showed that, for EM waves in a cubic cavity with sides of length L, the only modes which can exist have to satisfy the equation


n_{x}^{2} + n_{y}^{2} + n_{z}^{2} = \frac{ 4 L^{2} }{ \lambda^{2} }


where n_{x} is the number of modes allowed in the cavity in the x-direction, etc. This is the so-called “standing wave solution to the wave equation for a cubical cavity with sides of length L”.

In this third part and final part of the derivation of the Rayleigh-Jeans law, I will calculate the total number N of allowed modes in the cavity given this standing wave solution, secondly I will calculate the number of modes per unit wavelength in the cavity, and finally I will calculate the energy density per unit wavelength and per unit frequency of the EM waves. This final part is the famous Rayleigh-Jeans law.

The total number of modes in our cubic cavity

In order to calculate the total number of allowed modes N in our cubic cavity we need to sum over all possible values of n_{x}, n_{y} \text{ and } n_{z}. To do this we use a mathematical trick of working in “n-space”, that is to say we determine the volume of a sphere where the x-axis is given by n_{x}, the y-axis by n_{y} and the z-axis by n_{z}. The value of n = \sqrt{ n_{x}^{2} + n_{y}^{2} + n_{z}^{2} }. We can determine the value of N by considering the volume of a sphere with radius n, which is of course just


N = \frac{ 4 \pi }{ 3 } n^{3}


But, as we can see in the diagram below, if we sum over n for an entire sphere we will be including negative values of n_{x}, n_{y} \text{ and } n_{z}, whereas we only have positive values of each.



20131108-140657.jpg

To correct for this, to only consider the positive values of n_{x}, n_{y} \text{ and } n_{z}, we just need to divide the volume above by 8, as the part of a sphere in the positive part of the diagram is one eighth of the total volume. But, we also need to make another correction. Light can exist independently in two different polarisations at right angles to each other, so we need to double the number of solutions to our standing wave equation to account for this .We therefore can write


N = ( \frac{ 4 \pi }{ 3 } n^{3} ) ( \frac{ 1 }{ 8 } ) 2 = \frac{ \pi }{ 3 } n^{3} = \frac{ \pi }{ 3 } ( n_{x}^{2} + n_{y}^{2} + n_{z}^{2} )^{3/2} = \frac{ \pi }{ 3 } ( \frac{ 4 L^{2} }{ \lambda^{2} } )^{3/2}


which gives the number of modes in the cavity as


\boxed{ N = \frac{ 8 \pi L^{3} }{ 3 \lambda^{3} } }

The number of modes per unit wavelength

The expression above is the total number of modes in the cavity summed over all wavelengths. The number of modes per unit wavelength can be found by differentiating this expression with respect to \lambda, i.e. we find \frac{ dN }{ d\lambda }.


\frac{ dN }{ d\lambda } = \frac{ d }{ d\lambda } ( \frac{ 8 \pi L^{3} }{ 3 \lambda^{3} } ) = - 3 \frac{ 8 \pi L^{3} }{ 3 \lambda^{4} } = - \frac{ 8 \pi L^{3} }{ \lambda^{4} }


The minus sign is telling is that the number of modes decreases with increasing wavelength.

We can also derive the number of modes per unit wavelength in the cavity volume by dividing by the volume of the cavity


\frac{ \text{Number of modes per unit wavelength} }{ \text{cavity volume} } = - \frac{ 1 }{ L^{3} } \frac{ dN }{ d\lambda } = - \frac{ 1 }{ L^{3} } \frac{ 8 \pi L^{3} }{ \lambda^{4} } = -\frac{ 8 \pi }{ \lambda^{4} }

The energy per unit volume per unit wavelength and per unit frequency

Because the matter and radiation are in thermal equilibrium with each other, we can say that the energy of each mode of the EM radiation is E = kT where k is Boltzmann’s constant and T is the temperature in Kelvin of the radiation. This comes from the principle of the Equipartition of Energy.We write the energy per unit volume (also called the energy density) with the symbol u so we have that the energy per unit volume per unit wavelength is given by


\frac{ du }{ d\lambda } = \frac{ 1 }{ L^{3} } \frac{ dE }{ d\lambda} = \frac{ 1 }{ L^{3} } ( \frac{ dN }{d \lambda } ) kT = \frac{ 8 \pi }{ \lambda^{4} } kT

To write this in terms of frequency we remember that


\lambda = \frac{ c }{ \nu } \text{ so } \frac{ d \lambda }{ d \nu } = - \frac{ c }{ \nu^{2} } \rightarrow \frac{ 1 }{ \lambda^{4} } = \frac{ \nu^{4} }{ c^{4} }


and, from the chain rule we can write that


\frac{ du }{ d \nu } = \frac{ du }{ d\lambda } \frac{ d\lambda }{ d\nu } \text{ and } \frac{d \lambda }{ d \nu } = -\frac{ c }{ \nu^{2} }


(the minus sign is just telling is that as \lambda \text{ increases } \nu \text{ decreases )}.



This gives us that


\frac{ du }{ d \nu} = (8 \pi kT) \frac{ \nu^{4} }{ c^{4} } ( \frac{ c }{ \nu^{2} } )


So, finally we have the Rayleigh-Jeans law, that the energy density of the radiation is given by


\boxed{ \frac{ du }{ d \nu } = \left( \frac{ 8 \pi kT }{ c^{3} } \right) \nu^{2} }


So, using Classical Physics, we find that the energy density is proportional to the frequency squared (\frac{ du }{ d\nu } \propto \nu^{2}), which means the energy density plotted as a function of frequency should look like the purple curve below.


The so-called "ultraviolet" catastrophe".

The purple curve shows the so-called “ultraviolet” catastrophe”, because the energy density \frac{ du }{ d \nu } \propto \nu^{2}.




Of course, physicists already knew that the energy density of blackbodies followed the blackcurve, not the purple curve. If it were to follow the purple curve (the Rayleigh-Jeans law) the blackbody would get brighter and brighter at shorter and shorter wavelengths, the so-called ultraviolet catastrophe. In a future blog I will outline how Max Planck resolved this problem, and in so doing heralded in the dawn of Quantum Mechanics, an entirely new way of thinking about the sub-atomic world.



Part 1 of this blog is here.


Part 2 of this blog is here.

Read Full Post »

Derivation of the Rayleigh-Jeans law – part 2

Posted in History, Nature, Physics, Science, tagged , , , , , , , on 04/11/2013| 34 Comments »

In part 1 of this blog I showed that the 1-dimensional (let’s say the x-direction) wave equation for the electric field \vec{E}(x,t) = E \sin(kx - \omega t) is


\frac{ d^{2}E }{dx^{2} } = \frac{ 1 }{ c^{2} } \frac{ d^{2} E }{dt^{2} }


and that this can be generalised to the 3-dimensional wave equation


\frac{ \partial^{2} E }{ \partial x^2 } + \frac{ \partial^{2} E }{ \partial y^{2} } + \frac{ \partial^{2} E }{ \partial z^{2} } = \frac{1}{c^2} \frac{ \partial^{2} E }{ \partial t^{2} }

By analogy to the 1-D electric field, the 3-D electric field (that is, travelling in any direction) can be written as


\vec{E}(r,t) = E \sin( k\vec{r} - \omega t )


The direction \vec{r} of the electric field can be split into its x,y and z components so that \vec{r} = \vec{x} + \vec{y} + \vec{z}, and the wavenumber k can be split into its x,y and z-components such that k = \sqrt{k_{x}^2 + k_{y}^2 + k_{z}^2} (k_{x},k_{y} \text{ and } k_{z} are not generally the same value, unless the wave is travelling at 45 degrees to each of the axes).

We can then write


\vec{E}(r,t) = \vec{E}(x,y,z,t) = E \sin (k_{x}x + k_{y}y + k_{z}z - \omega t)

Radiation in a cubical cavity

Remember, for blackbody radiation we need to enclose the radiation in a cavity with perfectly reflective walls, so that the radiation and the matter (thin gas in our example) can come into thermal equilibrium with each other with no loss of energy. For simplicity, the cavity will be a cube where each side will be of length L.



20131029-164434.jpg


We are going to define the vertical direction as the z-direction, the direction to the right as the y-direction, and the direction towards us as the x-direction, as shown in the figure above. We need the electric field to be zero at the walls, because if it were non-zero the electric field would impart energy to the cavity walls and lose energy itself. If the electric field needs to be zero at the walls, this means its x,y and z-components E_{x}, E_{y} \text{ and } E_{z} also need to be zero at the walls.

Thus, only electric fields with suitable wavelengths can exist in the cavity. This is analogous to standing waves on a string, if we have two fixed ends (such as a guitar string), only certain wavelengths can exist. In the diagram below I show the first four possible “modes” for the electric field between the two walls a distance L apart [first four in the sense that the one with the lowest frequency (longest wavelength) is the yellow curve, then the next lowest frequency is the blue curve, then the green, and finally the black].



20131028-150419.jpg

In the graph above, the x-axis is in units of L, so 1 corresponds to L=1. The yellow curve is the lowest frequency (longest wavelength) wave that can exist between the two walls. Again, using the analogy of standing waves on a string, this would be the fundamental wave, the lowest note the string could produce. The equation which describes this yellow curve is y=\sin( \frac{ \pi x }{ L }), but remember we can also write, for any wave, y=\sin(k x) where k is the wavenumber which we introduced in part 1 of this blog. k is the number of waves per unit wavelength, and is related to the wavelength via the equation k = \frac{ 2\pi }{ \lambda } \text{ where } \lambda is the wavelength.

So, if k for the green curve is equal to \frac{ \pi }{ L }, we can write \lambda = \frac{ 2\pi }{ k } = \frac{ 2\pi L }{ \pi } = 2L, so half of the wave fits between the two walls.

For the blue curve, we have y = \sin ( \frac{ 2 \pi x }{ L } ), and so k = \frac{ 2 \pi }{ L } which leads to \lambda = \frac{ 2 \pi }{ k } = \frac{ 2 \pi L}{ 2 \pi } = L, and as we can see the blue curve has one complete wavelength between the two walls.

For the green curve, we have y = \sin ( \frac{ 3 \pi x }{ L } ), and so k = \frac{ 3 \pi }{ L } which leads to \lambda = \frac{ 2 \pi }{ k } = \frac{ 2\pi L }{ 3 \pi } = \frac{ 2L }{ 3 }; the wavelength of the green curve is 2/3rds of the distance L, or to put it another way there are one and a half waves of the green curve between the walls.

Finally, for the black curve, we have y = \sin( \frac{ 4\pi x }{ L } ) and so k = \frac{ 4 \pi }{ L } which leads to \lambda = \frac{ 2\pi L }{ 4\pi } = \frac{ 2 L }{ 4 } = \frac{ L }{ 2 }, so the wavelength is half the distance between the two walls, meaning there will be two complete waves between the walls.

I have only shown the first four possible waves, but of course in reality we can have y=\sin( \frac{ 5\pi x }{ L } ), y=\sin( \frac{ 6\pi x }{ L } ) etc., so in general we can write y=\sin( \frac{ n \pi x }{ L } ) where n=1,2,3,4,5,6 etc. In theory, n \text{ can take any value from } 1 \text{ to } \infty.

So, in the x-direction we can write that


E_{x}=E \sin ( \frac{ n \pi x }{ L }).

Although the wavelength of the x,y and z-components of the EM are all the same, there may not be the same number of wavelengths in the x,y and z-directions between the walls. For example, if the wave is travelling wholly horizontally then there may be no electric field component in the vertical (z) direction, and if it is travelling horizontally at e.g. 30^{\circ} to the left hand wall then the distance the wave is travelling in the y-direction will be \sqrt{ 3 } times the distance it travels in the x-direction. Because of this, although we can write similar equations for E_{y} \text{ and } E_{z}, the values of n will be different.

Therefore, we are going to write


E_{x} = E \sin ( \frac{ n_{x} \pi x }{ L }), E_{y} = E \sin ( \frac{ n_{y} \pi y }{ L }) \text{ and } E_{z} = E \sin ( \frac{ n_{z} \pi z }{ L })


where n_{x}, n_{y} \text{ and } n_{z} are related to the number of waves between the walls in each of the x,y and z-directions via k_{x}=\frac{ n_{x} \pi }{ L }, k_{y}=\frac{ n_{y} \pi }{ L } \text{ and } k_{z}\frac{ n_{z} \pi }{ L } in each of the spatial directions.

Remember the electric field in 3-D can be written as


\vec{E} = E\sin(k_{x}x + k_{y}y +k_{z}z - \omega t) \text{. But } k_{x}x = \frac{ n_{x} \pi x }{ L } \text{ etc. and } \omega = 2\pi \nu = \frac{ 2 \pi c }{ \lambda }


so we can write the electric field in 3-dimensions as


\vec{E} = E \sin( \frac{ n_{x} \pi x }{ L } + \frac{ n_{y} \pi y }{ L } + \frac{ n_{z} \pi z }{ L } - \frac{ 2 \pi c t }{ \lambda } )


You may remember from trigonometry that \sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B) (the so-called compound angle formula. Although it is pretty tedious to prove it (see for example this link), we can write


\sin(X+Y+Z) = \sin(X)\cos(Y)\cos(Z) + \cos(X)\sin(Y)\cos(Z) + \cos(X)\cos(Y)\sin(Z) - \sin(X)\sin(Y)\sin(Z)


(where we have written X,Y \text{ and } Z to represent \frac{ n_{x} \pi x }{ L } etc.). To find \sin(X+Y+Z-T) (where T= \frac{ 2 \pi c t }{ \lambda }) is even more tedious, and also involves working out \cos(X+Y+Z), but if you go through it you will find (eventually!) that


\sin(X+Y+Z-T) = \sin((X+Y+Z)-T) = \sin(X+Y+Z)\cos(T) -\cos(X+Y+Z)\sin(T)


where \cos(X+Y+Z) can be written


\cos(X+Y+Z) = \cos(X)\cos(Y)\cos(Z) - \sin(X)\sin(Y)\cos(Z) - \sin(X)\cos(Y)\sin(Z) - \cos(X)\sin(Y)\sin(Z)


and so


\sin(X+Y+Z-T) = \sin(X)\cos(Y)\cos(Z)\cos(T) + \cos(X)\sin(Y)\cos(Z)\cos(T) + \cos(X)\cos(Y)\sin(Z)\cos(T) - \sin(X)\sin(Y)\sin(Z)\cos(T) - \cos(X)\cos(Y)\cos(Z)\sin(T) + \sin(X)\sin(Y)\cos(Z)\sin(T) + \sin(X)\cos(Y)\sin(Z)\sin(T) + cos(X)\sin(Y)\sin(Z)\sin(T)


Remember, however, that the electric field has to be zero at the walls, and in particular at the origin where X=Y=Z=0. This means that all the terms with \cos(X) \text{ or } \cos(Y) \text{ or } \cos(Z) will disappear, and so the above expression will simplify to


\sin(X+Y+Z-T) = - \sin(X)\sin(Y)\sin(Z)\cos(T)


But, unlike the values of X=0,Y=0 \text{ and } Z=0 which are determined by the origin of the cavity, the value of T=0 is completely arbitrary, so we can just shift it by \omega = \pi/2 (one quarter of a period) to give us \cos(T+\pi/2) = -\sin(T) and then we have


\sin(X+Y+Z-T) = \sin(X)\sin(Y)\sin(Z)\sin(T)

Replacing our X,Y,Z \text{ and } T we have


\vec{E} = E\sin(\frac{ n_{x} \pi x }{ L }) \sin(\frac{ n_{y} \pi y }{ L }) \sin(\frac{ n_{z} \pi z }{ L }) \sin(\frac{ 2 \pi ct }{ \lambda })
Then,using the 3-dimensional wave equation we had earlier


\frac{ \partial^{2} E }{ \partial x^2 } + \frac{ \partial^{2} E }{ \partial y^{2} } + \frac{ \partial^{2} E }{ \partial z^{2} } = \frac{1}{c^2} \frac{ \partial^{2} E }{ \partial t^{2} }
we will need to work out \frac{ \partial^{2} E }{ \partial x^{2} } etc.


\frac{ \partial E }{ \partial x } = \frac{ \partial }{ \partial x } E\sin(\frac{ n_{x} \pi x }{ L }) = E \frac{ n_{x} \pi }{ L } \cos(\frac{ n_{x} \pi x }{ L })
and
\frac{ \partial^{2} E }{ \partial x^{2} } = -E (\frac{ n_{x} \pi }{ L })^{2} \sin( \frac{ n_{x} \pi x }{ L }) = -E_{x} (\frac{ n_{x} \pi }{ L })^{2}


If we do the same thing for \frac{ \partial^{2} E }{ \partial y^{2} }, \frac{ \partial^{2} E }{ \partial z^{2} } \text{ and } \frac{ 1 }{ c^{2} } \frac{ \partial^{2} E }{ \partial t^{2} } we get


(\frac{ n_{x} \pi }{ L })^{2} + (\frac{ n_{y} \pi }{ L })^{2} + (\frac{ n_{z} \pi }{ L })^{2} = \frac{ 1 }{ c^{2} } (\frac{ 2 \pi c }{ \lambda })^{2} = (\frac{2 \pi }{ \lambda })^{2}


which simplifies to


n_{x}^{2} + n_{y}^{2} + n_{z}^{2} = \frac{ 4 L^{2} }{ \lambda^{2} }.


This is the so-called “standing wave solution to the wave equation for a cubical cavity with sides of length L”. In part 3 of this blog I will explain what this solution means in terms of the number of allowed EM-wave modes that can exist in the cavity.



Part 1 of this blog can be found here, and Part 3 is here.


Read Full Post »

Blackbody radiation and the “ultraviolet catastrophe”

Posted in History, Physics, Science, tagged , , , , on 21/10/2013| 12 Comments »

In Physics, we refer to an object which radiates perfectly as a blackbody. The name is a little misleading, as there is nothing about a blackbody which is black in colour. An example of a blackbody is a heated cannon ball, or the surface of stars like the Sun. It has been known since time immemorial that if you heat a metal object, such as a cannon ball, it will start to glow; it actually starts giving off its own visible light rather than just reflecting light. We now know that, even at room temperature, all objects are giving off their own light, but it is infrared light, a type of light to which are eyes are not sensitive and is thus invisible unless we use a thermal imaging camera.

Wien’s Displacement Law

When an object glows, we can analyse which parts of the spectrum are giving off how much energy. To put it another way, how bright the object is at different wavelengths. When we do this for any object which is radiating like a blackbody we find that the spectra all have the same shape, independent of the temperature of the object or its composition.

The figure below shows the spectra for three different blackbodies. The only difference between the three is their temperature. One is at a temperature of 3,000K, the second one is at 4,000K and the third one is at 5,000K. These three curves explain why an object does not give off visible light when it is at low temperature. 3,000K is pretty hot (nearly 2,700 C), but an object at this temperature is barely giving off any visible light, the peak of the curve is in the infrared. If we came down to room temperature, about 300K, there would be no radiation short-wards of 700nm, so no visible light is to be seen from a room temperature object. These three curves also explain why an object changes colour as we heat it, it goes from red-hot to white-hot to blue-hot. It is why stars have different colours, the cool ones appear red, the hotter ones white and the hottest of all appear blue.

 

Blackbody spectra for three different temperatures, 3000K (red), 4000K (green) and 5000K (blue).

Blackbody spectra for three different temperatures, 3000K (red), 4000K (green) and 5000K (blue). Notice the wavelength of the peak of each curves moves as the temperature changes. This is Wien’s displacement law. The x-axis is in nanometres, on this scale the visible part of the spectrum is from about 400 to 700 nm.

 

As you can see from the figure, the size of the curve changes as we change the temperature, and also the wavelength of the peak changes, moving to shorter wavelengths (to the left) as we increase the temperature. The scale on the x-axis is nanometres, so on this scale the kind of wavelengths we can see with our eyes (the visible part of the spectrum) is from about 400 to 700 nm.

In 1893 the German physicist Wilhelm Wien came up with a mathematical formula which allowed one to calculate the wavelength of this peak if one knew the temperature. This equation is known as Wien’s displacement law. It is an empirical law, in the sense that Wien had no physical explanation for why the equation worked, but it did. The physics behind why Wien’s law works did not come until seven years later in 1900 with the work of Max Planck.

We can state Wien’s displacement law as

 

\lambda \text{ (in metres) } = \frac{2.898 \times 10^{-3} }{T}

 

where T is the temperature in Kelvin. Another law, called the Steffan-Boltzmann law, gives the total amount of power per unit area (Watts per metre squared) that is being emitted at all wavelengths by the blackbody (mathematically this is the same as the area under these curves).

 

P = \sigma T^{4}

 

where \sigma is known as the Steffan-Boltzmann constant, and once again T is the temperature of the object in Kelvin.

The “ultraviolet catastrophe”

To understand the physics of blackbody spectra, physicists imagined the electromagnetic waves bouncing around inside a closed box, what is often referred to as a cavity. The cavity has a small hole in it, which allows the radiation to escape, this is the radiation we observe. We can see from the curves above that the spectrum (either expressed against wavelength \lambda or against frequency \nu (one can go from wavelength to frequency and back by using the wave equation. For electromagnetic radiation, this can be written

 

c = \lambda \ \nu
where c is the speed of light (3 \times 10^{8} \text{ m/s}).

The electromagnetic (EM) radiation is produced by an EM field in the cavity. But, there are EM waves with different frequencies. The ensuing EM field is due to the superposition (addition) of all the possible EM waves in the cavity. We can think of the EM waves like pieces of string between two fixed points (the sides of the cavity). So we can draw an analogy to the musical notes that e.g. a guitar string can produce. With a piece of string, only certain standing waves can exist between two fixed points, you cannot get vibrations of any wavelength. Let us suppose the fixed points are a distance L apart. As the vibrations need to be zero at the fixed ends, the only wavelengths which can be produced by such a string have to have wavelengths \lambda so that L = \frac{n\lambda}{2} where n=1,2,3,4....... \text{ etc }.

The same was believed to be true for EM waves inside a cavity. The EM waves had to have a complete number of half-cycles within the box. Each of these allowed vibrations of the EM field is called a mode.

The number of modes of a particular frequency \omega in the frequency interval \omega to \omega + d\omega (where d\omega is a small increment) is found to be

N(\omega)d\omega \propto \omega^{2}d\omega

(this comes about because we can think of the vibrations of the EM field like a harmonic oscillator).
In classical physics, in the 19th century, a theory called the equipartition theorem had been developed, and this stated that each mode would have an energy of E = \frac{1}{2}kT where k is Boltzmann’s constant. This equipartition theorem had allowed physicists to work out the kinetic energy of molecules of a gas at any given temperature. This was an attempt to apply this equipartition law to EM radiation, which were thought of as waves.

If the energy density (which is the same as the power per unit area) of each mode is \frac{1}{2}kT, the energy \epsilon as a function of frequency in a small frequency interval from \omega \text{ to } \omega+d\omega would be given by
\epsilon(\omega) \propto \omega^{2} kT d\omega

This is the so-called Rayleigh-Jeans law, and notice that is says that the energy density in a frequency interval is proportional to the square of the frequency. If we plot the energy density at a particular frequency \epsilon (\omega) as a function of frequency \omega we get a graph which looks like this

 

The so-called "ultraviolet" catastrophe".

The so-called “ultraviolet catastrophe”. The black curve shows the observed blackbody curve, the red curve shows the prediction of the Rayleigh-Jeans law.

The black curve in this plot shows the observed spectrum of a blackbody (note, this plot is against frequency not wavelength, so it is backwards from the figure above with the three different spectra. In this plot against frequency, frequency increases to the right, which means we are moving to shorter wavelength).

Now look at the red curve, it is going off to infinity!!! That is to say, the classical theory (the Rayleigh-Jeans law) predicts that there will be an infinite number of modes allowed in the cavity as we go to higher and higher frequency. This means the curve will go off up to infinity as shown, because the power per unit area at each frequency \epsilon(\omega) is proportional to \omega^{2}, so more and more energy exists in each frequency interval as we go to higher and higher frequencies (shorter and shorter wavelengths). The total area under the curve, the power per unit area, of the blackbody would be infinite and it would be infinitely bright at ultraviolet wavelengths (high frequencies)!!!!! This clearly doesn’t happen, the theory had completely broken down.

This disagreement of the observed behaviour with the prediction of the Rayleigh-Jeans law came to be known as the ultraviolet catastrophe. It showed that there was something deeply flawed in our understanding of physics. Max Planck would solve the problem in 1900, and usher in a complete overthrowing of our understanding of physics, bringing in the age of quantum physics and leading to the demise of classical physics.

 

Addendum

Here is the first part of a series of blogs where I will derive the Rayleigh-Jeans law from first principles, after which I will derive the Planck law.

Read Full Post »

The 10 best physicists – no. 2 – Niels Bohr

Posted in History, Physics, Science, tagged , , , on 13/09/2013| 5 Comments »

At number 2 in The Guardian’s list of the ten best physicists is Niels Bohr.

20130619-151948.jpg

Bohr is best known for his work on the allowed orbits an electron in an atom can have. This is the so-called “Bohr model”, and in 1922 he won the Nobel Prize in Physics for this suggestion that electrons can only exist in certain allowed orbits, which naturally explained the line spectra of the elements.

Bohr’s brief biography

Niels Bohr was born in Copenhagen in 1885, the second of three children. His father, Christian Bohr, was a professor of physiology at the University of Copenhagen. His mother, Ellen Adler, came from a wealthy Danish Jewish family. At 18, Bohr enrolled at Copenhagen University where he studied physics. In 1905, whilst still an undergraduate, Bohr won the first prize in a competition sponsored by the Royal Danish Academy of Arts and Sciences. In May 1911 he obtained his PhD for a thesis on the electron theory of metals.

In the same year, Bohr took up a post-doctoral research position at The University of Manchester, working in Ernest Rutherford’s research team. He arrived in Manchester just as Rutherford was proposing the theory that atoms contained small, positively charged nuclei; where nearly all the atom’s mass was concentrated. However, by 1912 Bohr had returned to his native Denmark where he got a job teaching medical students. In 1913 he published his first paper suggesting what is now known as the “Bohr model” of the atom.

He returned to Manchester in 1914, and spent two more years working with Rutherford, as a Reader in the Physics Department. Then, in 1916 a Professorship in Theoretical Physics was created for him at the University of Copenhagen. In 1918 Bohr started trying to establish an institute of theoretical physics in Copenhagen, the institute opened in 1921 and became known as the Niels Bohr Institute. Apart from fleeing from Nazi occupied Denmark in 1943, Bohr spent the rest of his career as the Director of this institute he had established, and died in 1962 at the age of 77.

Bohr’s contributions to Physics

When Rutherford proposed his model of the atom with a positively charged nucleus and the electrons in orbit about it, a problem arose. Classical physics predicted that an electron in orbit, because it is constantly accelerating through changing its direction, should be constantly radiating. As a consequence, it should lose its energy and spiral in towards the nucleus. Calculations showed that this should happen in millionths of a second, meaning all atoms would be unstable.

In 1913, Bohr suggested that electrons could only exist in certain allowed orbits. He suggested that these orbits were “quantised”, and that the angular momentum of electron orbits had to be equal to L=n\hbar where n is an integer (1,2,3 etc.) and \hbar \text{ is } h/2\pi where h is Planck’s constant.

Bohr then suggested that electrons could be excited from one orbit to another, either by gaining some (or all) of the energy of an incoming electron, or by absorbing all of the energy of an incoming photon. When electrons were excited to a higher energy level, they would quickly jump back down to the lowest available orbit, and in so doing would emit light (photons) of particular wavelengths.

This “Bohr model” was able to naturally explain the observed spectra of hydrogen (which only has one electron and one proton), and of singly-ionised helium, which again only has one electron. Its success led to Bohr being awarded the Nobel Prize in 1922, the citation reading “for his services in the investigation of the structure of atoms and of the radiation emanating from them”.

Bohr played a central role in the probabilistic interpretation of Quantum Mechanics. The so-called “Copenhagen interpretation”, of which he was the main champion, was that physics was not able, under the laws of quantum mechanics, to give us any more than the probabilities of the outcomes of experiments. This was in stark contrast to e.g. Einstein, who believed that that nature was inherently deterministic not probabilistic.

Bohr was instrumental in the establishment of CERN, the European Centre for Particle Physics Research. He was very much one of the elder-statesmen of the Physics community, and a period at his Institute in Copenhagen became almost essential in the career of any theoretical physicist.

Do you think Niels Bohr deserves to be in this list of the ten best physicists?

Read more

You can read more about Niels Bohr and the other physicists in this “10 best” list in our book 10 Physicists Who Transformed Our Understanding of the Universe. Click here for more details and to read some reviews.

IMG_2529

Ten Physicists Who Transformed Our Understanding of Reality is available now. Follow this link to order

Read Full Post »

Older Posts »