Classical Mechanics

Electron energy

What is an electronvolt (eV)?

The electronvolt (eV) is a unit of energy. It is the amount of kinetic energy gained by a single electron when accelerated through an electric potential difference of 1 volt.

Formally:

\[1\,\text{eV} = e \times 1\,\text{V}\]

where \(e = 1.602 \times 10^{-19}\) coulombs (the electron charge) and \(1\,\text{Volt} = 1\,\text{Joule per Coulomb (J/C)}\).

How do you convert eV to joules?

Using the definition above:

\[1\,\text{eV} = (1.602 \times 10^{-19}\,\text{C}) \times (1\,\text{J/C}) = 1.602 \times 10^{-19}\,\text{Joules}\]

This is a tiny amount of energy, which is why physicists use eV—it keeps the numbers manageable when working with atomic-scale phenomena.

What does “200 kV” mean in electron microscopy?

An electron is accelerated through a voltage difference of 200,000 volts. The kinetic energy gained is:

\[K = 200{,}000 \times 1.602 \times 10^{-19}\,\text{J} \approx 3.2 \times 10^{-14}\,\text{J}\]

That’s the actual kinetic energy per electron in a 200 kV transmission electron microscope.

Why do we need relativistic mechanics for electron microscopy?

At high voltages, electrons travel at significant fractions of the speed of light, where classical mechanics breaks down.

Classical prediction: Using \(\frac{1}{2}mv^2 = eV\), at 200 kV (200,000 volts):

\[v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2(1.602 \times 10^{-19})(2 \times 10^5)}{9.11 \times 10^{-31}}} \approx 2.65 \times 10^8\,\text{m/s}\]

That’s 0.88 times the speed of light (88% of \(c\)).

Problem: If you double the voltage to 400 kV, the classical formula predicts \(v = 1.25 \times c\)—faster than light, which is impossible.

What happens at relativistic speeds?

Einstein’s special relativity says that as particles approach the speed of light \(c\), their momentum and energy increase faster than Newtonian mechanics allows. The extra energy from the accelerating voltage doesn’t just make the electron move faster; it increases the electron’s momentum and effective inertia (relativistic mass).

Comparison of classical vs relativistic speeds:

Voltage	Classical speed	Relativistic speed	γ (Lorentz factor)
10 kV	0.19 c	0.19 c	1.02
100 kV	0.63 c	0.55 c	1.22
200 kV	0.88 c	0.70 c	1.39
400 kV	1.25 c (impossible)	0.87 c	1.78
1 MeV	— (nonsense)	0.94 c	2.96

Notice how speed saturates: at 1 MeV (million volts), the electron has nearly tripled its rest energy yet remains below light speed.

What do we mean by “momentum and energy grow faster”?

When you push a particle at relativistic speeds, its resulting velocity change gets smaller and smaller near the speed of light. Energy still increases as you add voltage, but velocity saturates.

The total energy includes both rest mass energy and kinetic energy:

\[E = \gamma m_0 c^2 = \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}}\]

where \(\gamma\) is the Lorentz factor:

\[\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}\]

At low speeds (\(v \ll c\)), \(\gamma \approx 1\) and classical mechanics works fine. At high speeds, \(\gamma\) grows rapidly:

At 0.7c (200 kV): \(\gamma = 1.39\)
At 0.87c (400 kV): \(\gamma = 1.78\)
At 0.99c: \(\gamma = 7.09\)

Example: How energy grows with speed

Speed \(v/c\)	γ	Total energy \(E/(m_0c^2)\)	Kinetic energy \(K/(m_0c^2)\)
0.5	1.15	1.15	0.15
0.7	1.40	1.40	0.40
0.9	2.29	2.29	1.29
0.99	7.09	7.09	6.09
0.999	22.4	22.4	21.4

The extra energy goes into increasing the electron’s effective mass rather than just its velocity.

Why does the electron have different “mass” at high energies?

The concept of “relativistic mass” \(m = \gamma m_0\) is sometimes used to describe how an electron becomes harder to accelerate at high speeds. However, modern physics prefers to say that the electron’s rest mass \(m_0 = 9.11 \times 10^{-31}\) kg is invariant, but its momentum and energy don’t follow the classical formulas.

Classical momentum: \(p = m_0 v\) Relativistic momentum: \(p = \gamma m_0 v\)

At 200 kV, \(\gamma = 1.39\), so the electron’s momentum is 39% larger than the classical prediction. This affects wavelength calculations via the de Broglie relation \(\lambda = h/p\), which is critical for electron microscopy resolution.

Waves

The wave differential equation

What differential equation describes wave motion?

All waves (electromagnetic waves, matter waves, sound waves) satisfy the wave equation:

\[\frac{\partial^2 \psi}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2}\]

where:

\(\psi(x,t)\): wave function (displacement, field amplitude, or probability amplitude)
\(v\): wave propagation speed (speed of light \(c\) for electromagnetic waves)
\(\frac{\partial^2}{\partial x^2}\): second spatial derivative (how curved is the wave in space)
\(\frac{\partial^2}{\partial t^2}\): second time derivative (how fast is the oscillation accelerating)

Physical meaning: The spatial curvature determines the temporal acceleration. A wave that is more curved in space oscillates faster in time.

How do we verify that \(\psi = A\sin(kx - \omega t)\) is a solution?

The sinusoidal form \(\psi(x,t) = A\sin(kx - \omega t)\) satisfies the wave equation and gives the dispersion relation: \(\omega = vk\) (or equivalently, \(v = \omega/k = \lambda f\)). The same works for \(\cos(kx - \omega t)\).

How do we verify that \(\psi = Ae^{i(kx - \omega t)}\) is also a solution?

The complex exponential form \(\psi(x,t) = Ae^{i(kx - \omega t)}\) also satisfies the wave equation with the same dispersion relation \(\omega = vk\).

Why does the exponential form work mathematically?

Using Euler’s formula: \(e^{i(kx - \omega t)} = \cos(kx - \omega t) + i\sin(kx - \omega t)\). The wave equation is linear (no terms like \(\psi^2\) or \(\psi^3\)), so if \(\cos(kx - \omega t)\) and \(\sin(kx - \omega t)\) are both solutions, their linear combination is also a solution. The complex exponential is just the compact form of this combination.

Phase representation

What is phase?

Phase represents a wave’s position in its cycle, similar to how clock hands show time’s position within an hour. When studying waves in electron microscopy, there are two ways to describe wave behavior: traditional sine waves and complex exponentials.

How do sine waves represent waves?

The traditional sinusoidal form shows the wave’s actual oscillation:

\[\psi(x, t) = A \sin(kx - \omega t + \phi)\]

where:

\(A\): amplitude (wave height)
\(k\): wave number (spatial frequency)
\(\omega\): angular frequency (temporal frequency)
\(\phi\): phase offset (shifts wave left or right)

Check https://bobleesj.github.io/electron-microscopy-website/wave for interactive phase visualization.

How can we represent the same wave using complex exponentials?

The complex exponential form is mathematically equivalent but more convenient for calculations:

\[\psi(x, t) = Ae^{i(kx - \omega t + \phi)} = A(\cos(kx - \omega t + \phi) + i\sin(kx - \omega t + \phi))\]

Key insights:

The real part (\(\cos\)) and imaginary part (\(\sin\)) are 90° out of phase
The amplitude \(A\) becomes the radius in the complex plane
The phase \(\phi\) becomes the angle in the complex plane

Why is the complex form often more convenient?

Comparing both forms with practical examples:

Adding waves with different phases:
- Sine form (messy):
  
  \[A\sin(kx + \phi_1) + A\sin(kx + \phi_2) = 2A\cos(\frac{\phi_1 - \phi_2}{2})\sin(kx + \frac{\phi_1 + \phi_2}{2})\]
- Complex form (simple):
  
  \[Ae^{i(kx + \phi_1)} + Ae^{i(kx + \phi_2)} = Ae^{ikx}(e^{i\phi_1} + e^{i\phi_2})\]
Multiplying waves (when light passes through multiple materials):
- Sine form (complicated):
  
  \[A_1\sin(kx + \phi_1) \cdot A_2\sin(kx + \phi_2) = \frac{A_1A_2}{2}[\cos(\phi_1 - \phi_2) - \cos(2kx + \phi_1 + \phi_2)]\]
- Complex form (just add exponents):
  
  \[A_1e^{i(kx + \phi_1)} \cdot A_2e^{i(kx + \phi_2)} = A_1A_2e^{i(2kx + \phi_1 + \phi_2)}\]
Separating amplitude and phase changes:
- In electron microscopy, a sample changes both amplitude and phase
- Complex form clearly shows both: \(\psi = Ae^{i\phi}\)
- Phase changes just modify \(\phi\)
- Amplitude changes just modify \(A\)

How do we capture a moment in time?

The electron wave travels through space and time, described by \(\psi(x,y,t) = Ae^{i(k_xx + k_yy - \omega t + \phi_0)}\). When taking a measurement at \(t=0\), we freeze this motion into a snapshot: \(\psi(x,y) = Ae^{i(k_xx + k_yy + \phi_0)}\).

As the electron wave interacts with the sample, it accumulates additional phase:

The sample contributes \(\phi_{sample}(x,y)\), while the wave’s propagation through space contributes \(k_xx + k_yy\). These combine to give the total phase \(\phi_{total}(x,y) = k_xx + k_yy + \phi_0 + \phi_{sample}(x,y)\). The wave’s amplitude transforms to reflect the local intensity: \(A \rightarrow \sqrt{I(x,y)}\).

Final wave function:

\[\psi(x, y) = \sqrt{I(x, y)} e^{i \phi_{total}(x,y)}\]

where:

\(I(x, y)\): local intensity (measurable)
\(\phi_{total}(x,y)\): total phase including:
- Wave propagation (\(k_xx + k_yy\))
- Initial phase offset (\(\phi_0\))
- Sample-induced phase shift (\(\phi_{sample}(x,y)\))

How do you represent the complex wave function?

The complex wave function can be expressed in terms of its real and imaginary components using Euler’s formula:

\[\psi(x, y) = \psi_{real}(x, y) + i \psi_{imag}(x, y)\]

where:

\(\psi_{real}(x, y) = \sqrt{I(x, y)} \cos(\phi(x, y))\)
\(\psi_{imag}(x, y) = \sqrt{I(x, y)} \sin(\phi(x, y))\)

This representation is useful for numerical simulations and computations involving wave propagation and interactions.

What is the magnitude of the complex wave function?

The magnitude of the complex wave function is given by:

\[|\psi(x, y)| = \sqrt{\psi_{real}^2(x, y) + \psi_{imag}^2(x, y)} = \sqrt{I(x, y)}\]

This shows that the magnitude corresponds to the square root of the intensity, which is a directly measurable quantity in experiments.

Does the phase contribute to the magnitude?

No, the phase \(\phi(x, y)\) does not affect the magnitude of the wave function. The magnitude depends solely on the intensity \(I(x, y)\). The relationship can be proven mathematically:

\[\begin{split}\begin{align} |\psi(x, y)|^2 &= \psi(x, y) \cdot \psi^*(x, y) \\ &= \sqrt{I(x, y)} e^{i\phi(x, y)} \cdot \sqrt{I(x, y)} e^{-i\phi(x, y)} \\ &= I(x, y) e^{i\phi(x, y) - i\phi(x, y)} \\ &= I(x, y) e^{0} \\ &= I(x, y) \end{align}\end{split}\]

Alternatively, using the real and imaginary components:

\[\begin{split}\begin{align} |\psi(x, y)|^2 &= \psi_{real}^2(x, y) + \psi_{imag}^2(x, y) \\ &= (\sqrt{I(x, y)} \cos(\phi(x, y)))^2 + (\sqrt{I(x, y)} \sin(\phi(x, y)))^2 \\ &= I(x, y) (\cos^2(\phi(x, y)) + \sin^2(\phi(x, y))) \\ &= I(x, y) \cdot 1 \\ &= I(x, y) \end{align}\end{split}\]