I've been trying to get my head around k-space for a while. It's quite clear that I'm too stupid to understand Wikipedia's explanation, so in desperation, I asked ChatGPT all the dumb stuff that I'd found impossible to follow elsewhere. I know that this is just Physics 101 to most of you (though please feel free to flag any nonsense). What struck me was a) it often seemed able to guess what my next dumb question would be, even prompting me with it. b) it answered those dumb questions with far greater clarity than I expected and at the right level for me to comprehend. I know that the issue with authority remains a critical one. But I still found it a remarkable learning experience. Comments welcome.

I began with:

Explain k-space and its application to equipartition

ChatGPT said:

k-space (also called wave-vector space or momentum space) is a mathematical space where each point corresponds to a wavevector

k=(k_x,k_y,k_z),

which describes a sinusoidal mode of a field or particle in a system.

In physics—especially statistical mechanics, solid-state physics, and quantum mechanics—k-space provides a natural way to count and analyze the allowed energy modes of a system. This is crucial for understanding equipartition and why it fails for high-frequency modes (leading to the ultraviolet catastrophe).

---

✅ 1. What is k-space?

A system (like a box of gas, a vibrating solid, or electromagnetic radiation) can be decomposed into normal modes of oscillation:

ψ(r)=A e^ik⋅r.

Each allowed mode corresponds to one point in k-space.

For a box with periodic boundary conditions, the allowed values of k_x,k_y,k_z are discrete:

k_x=2πn_x/L, ky=2πn_y/L, kz=2πn_z/L.

Visually, k-space is like a 3D grid of allowed oscillation states.

Because the states are closely spaced when the box is large, one often treats them as continuous and counts states as densities:

Number of states in d³k=V/(2π)³ d³k.

---

✅ 2. Why k-space matters: counting the number of modes

Many physical quantities—energy, density of states, thermodynamic averages—require knowing how many allowed modes exist with wavenumbers between k and k+dk.

In 3D:

g(k) dk=V/(2π)² k² dk.

This function g(k) is the density of states in k-space.

---

✅ 3. Application to Equipartition

Equipartition theorem:
Each quadratic degree of freedom contributes 1/2 k_BT to the energy.

For classical waves (e.g., electromagnetic radiation or acoustic vibrations), each normal mode of frequency ω(k) has two quadratic terms (kinetic + potential), giving:

E_per mode=k_BT.

To find total energy:

1. Count how many modes exist at each k using k-space density of states.

2. Multiply by k_BT.

3. Integrate over all k.

---

⚠️ 4. What goes wrong (Ultraviolet Catastrophe)

Using equipartition with classical physics:

U=∫₀^∞g(k) (k_BT) dk.

But the density of states grows as:

g(k)∝k2,

so the integral diverges:

U→∞.

This is the ultraviolet catastrophe: classical equipartition predicts infinite energy because there are infinitely many high-frequency modes in k-space.

---

⭐ How quantum mechanics fixes it

Quantum mechanics says each mode has energy:

E=ℏω/(e^ℏω/kBT−1),

not simply k_BT.
At high frequencies (ℏω≫kBT), modes are frozen out (unoccupied).
Thus the k-space integral becomes finite: this is the origin of Planck’s law.

---

🔵 Summary

k-space is a representation where each point corresponds to an allowed wave mode.
Equipartition assigns energy to each mode—but because the number of modes grows as k², the classical prediction diverges.
Quantum mechanics modifies the energy per mode, preventing divergence.

---

If you'd like, I can also show:

• how k-space leads directly to the Rayleigh–Jeans law

• how density of states is derived from the geometry of k-space

• how equipartition applies differently in solids (Debye model)

Just tell me!

You said:

How does k-space lead to the Raleigh-Jeans Law

ChatGPT said:

Here is the clean, standard derivation of the Rayleigh–Jeans law starting directly from k-space.

---

🔷 1. Start with electromagnetic modes in a box

A cubic cavity of volume V=L³ supports standing electromagnetic waves.
Each mode is labeled by a wavevector:

k=(k_x,k_y,k_z), k_i=2πn_i/L.

Each allowed mode corresponds to one point in k-space.

---

🔷 2. Count the number of modes using k-space geometry

The allowed k-points form a 3D grid.
In the large-box limit, this becomes continuous, and the density of states in k-space is

V/(2π)³

The number of states with wavenumber between k and k+dk is the volume of a shell in k-space:

g(k) dk=V/(2π)³ ⋅ 4πk² dk=V/2π² k²dk.

For electromagnetic waves we must multiply by 2 for two polarizations:

g_EM(k) dk=V/π² k²dk.

---

🔷 3. Convert from wavenumber kkk to frequency ν

Light obeys

ν=ck/2π.

So

k=2πν/c, dk=2πcdν

Insert these into the density of states:

g_EM(ν)dν=V/π² (2πν/c)²(2π/c)dν.

Simplify:

g_EM(ν)dν=8πVν²/c³ dν.

This is the classic k-space density of photon states per frequency.

---

🔷 4. Apply classical equipartition to each mode

Equipartition theorem says every quadratic degree of freedom contributes

1/2 k_BT.

Each electromagnetic mode has two quadratic terms (electric + magnetic), so the total energy per mode is:

ε=kBT.

So energy in the frequency interval dνd\nudν is:

dU=ε g_EM(ν)dν=k_BT⋅8πVν²/c³ dν.

Divide by volume to get the spectral energy density:

u(ν)dν=dU/V=8πν²/c³ k_BT dν.

---

🔷 ⭐ This is the Rayleigh–Jeans Law

u(ν)=8πν²/c³ k_BT

This predicts that energy density grows as ν².

---

🔥 5. Ultraviolet divergence from k-space

Because the number of modes grows as:

g(ν)∝ν²,,

integrating u(ν) to high frequency gives:

∫₀^∞u(ν)dν=∞.

This is the ultraviolet catastrophe—purely a result of k-space mode counting + classical equipartition.

Planck solved this by replacing classical equipartition with quantized energy levels.

---

If you'd like, I can also derive:

• Rayleigh–Jeans law using ω instead of ν,

• the Planck law using the same k-space method,

• or a graphical comparison showing how quantization fixes the catastrophe.

You said:

Can k-space be represented in spherical coordinates?

ChatGPT said:

Yes — k-space is very naturally represented in spherical coordinates, and in fact this is often the most convenient way to count states because many physical systems depend only on the magnitude of the wavevector:

k=∣k∣=(k_x²+k_y²+k_z²)^0.5..

---

✅ k-space in spherical coordinates

We define:

• k = radial distance in k-space

• θ = polar angle

• ϕ = azimuthal angle

So a point in k-space is:

k=(k,θ,ϕ)..

The volume element becomes:

d³k=k²sin⁡θ dk dθ dϕ.

This is exactly analogous to spherical coordinates in ordinary real space — except now the “space” is the space of wavevectors.

---

🔷 Why spherical coordinates are useful

Most systems (e.g., photons, phonons, free electrons) have energy that depends only on the magnitude of the wavevector:

E(k),ω(k),ε(k),etc.

So instead of summing over a 3D grid of (k_x,k_y,k_z), we sum over shells in spherical k-space.

The number of states in a thin spherical shell of radius k and thickness dk is:

g(k) dk=V/(2π)³∫d³k=V/(2π)³ (4πk²) dk.

This simplifies to:

g(k) dk=V/2π² k²dk.

With polarization for photons:

g_EM(k) dk=V/π² k²dk.

This expression (arising directly from spherical coordinates in k-space) is the starting point for:

• the Rayleigh–Jeans law,

• the Planck spectrum,

• the Debye model for lattice vibrations,

• electron density of states in solids.

---

🔷 Geometric picture

In k-space:

• All allowed wavevectors fill a 3D grid.

• Modes with the same magnitude k lie on a sphere of radius k.

• Counting states becomes counting the volume of a spherical shell.

This is much simpler than counting states in Cartesian coordinates.

---

⭐ Summary

Yes — k-space is commonly expressed in spherical coordinates.

This gives:

d³k=4πk² dk,

and leads directly to standard mode-counting formulas used in blackbody radiation, phonons, electrons, and any system whose energy depends only on k=∣k∣.

---

If you want, I can show:

• a diagram of spherical shells in k-space,

• how the spherical form leads to the Debye density of states,

• or a comparison between Cartesian and spherical counting.

You said:

Why doesn't the number of states V/(2π)^3 have dimensions?

ChatGPT said:

Great question — and the key is that

V/(2π)³

is not the number of states.
It is the density of states in k-space, and it does have dimensions.

Let’s unpack this carefully.

---

✅ 1. What is V/(2π)³?

It is the factor that tells us how many allowed k-states occupy a small k-space volume element:

number of states=V/(2π)³ d³k.

So it is a density:

(states)  /  (k-space volume).olume).

---

✅ 2. Dimensions of each quantity

◾ Volume VVV

[V]=(length)³=L3.

◾ Wavevector k

A wavevector has dimensions of inverse length:

[k]=L⁻¹.

Thus the k-space volume element has dimensions:

[d³k]=(L⁻¹)³=L⁻³.

---

🔷 3. Dimensions of the factor V(2π)³

The denominator (2π)³ is dimensionless.

The numerator V has dimension L³.

So:

[V/(2π)³]=L³..

It is not dimensionless.

---

🔷 4. Why does the final number of states become dimensionless?

Because you always multiply this factor by d³k:

V/(2π)³ d³k.

Now check the dimensions:

• V∼L³

• d³k∼L⁻³

So:

[V]⋅[d³k]=L³⋅L⁻³=1

→ dimensionless, as it must be, because you’re counting states.

---

⚠️ Common misconception

People sometimes say “V/(2π)³ is the number of states per unit k-space volume,” but they forget that k has inverse-length units, so the density must carry units.

It is:

V/(2π)³ with units L³.

This exactly cancels the L⁻³ from d³k.

---

⭐ Summary

• V/(2π)³ does have units: it scales like L³.

• It is a density of states in k-space (states per k-volume).

• When you multiply by d³k, the length units cancel, giving a dimensionless count of states, as required.

---

If you'd like, I can also explain:

• why the 2π appears at all in the first place,

• how the density changes for periodic vs. fixed boundary conditions,

• or how this leads directly to the Debye and Rayleigh–Jeans densities of states.

You said:

Why does 2π appear in the first place?

ChatGPT said:

The appearance of 2π in the density of states

V/(2π)³

comes directly from the wavelength–wavevector relationship and the boundary conditions used to define allowed wave modes in a box.
It’s not arbitrary — it follows from Fourier analysis and the structure of standing waves.

Let’s go step by step.

---

🔷 1. Allowed wavevectors come from boundary conditions

For a particle or wave in a cubic box of side L, the standard choice is periodic boundary conditions:

ψ(x+L)=ψ(x).

A plane wave inside the box looks like:

ψ(x)=e^ikxx.

Periodic boundary conditions require:

e^ikx(x+L)=e^ikxx⇒e^ikxL=1.

This is only true if:

k_xL=2πn_x, n_x∈Z.

So the allowed values are:

k_x=2πn_x/L.

This is where 2π first appears.

---

🔷 2. The spacing between allowed k-values is

Δk_x=2π/L.

In 3D:

Δk_xΔk_yΔk_z=(2π/L)³.

Thus each allowed k-state occupies one “cell” in k-space of volume:

(Δk)³=(2π/L)³=(2π)³/V.

Flip this over:

density of states in k-space=1/(Δk)³=V/(2π)³.

That is exactly the famous factor.

---

🔷 3. But why is it 2π and not something else?

Because Fourier modes are periodic with period 2π.
A plane wave has phase:

e^i(kx)=e^i(2πx/λ).

The connection between wavelength λ and wavenumber k is:

k=2π/λ.

This is built into:

• Fourier transforms

• complex exponentials

• orthogonality relations

• momentum quantization in quantum mechanics

If you used a convention like e^inx/L instead, the 2π would go somewhere else (into the Fourier transform normalization). The physics is unchanged — the 2π is a convention that makes formulas come out clean.

But with the standard Fourier basis, the natural spacing in k-space is:

2π/L.

---

🔷 4. Summary

The factor 2π appears because:

1. Allowed wavevectors in a finite box satisfy

   k_i=2πn_i/L

   from the requirement e^ik/L=1.

2. This gives a spacing in k-space of

   Δk=2π/L.

3. Each quantum state occupies a volume

   (Δk)³=(2π)³/V.

4. Therefore the density of states is

   V/(2π)³.

Ultimately, 2π is baked into Fourier analysis and the definition of the wavevector k=2π/λ.

---

If you'd like, I can also show you:

• how the result changes if you choose non-periodic boundary conditions,

• the full derivation from Fourier transforms,

• or how this connects to quantum mechanical momentum spacing p=ℏk.

You said:

Expand e^(ikx) into trig functions

ChatGPT said:

To expand

e^ikx

into trigonometric functions, we use Euler’s formula:

e^iθ=cos⁡θ+isin⁡θ.

Set θ=kx, giving:

e^ikx=cos⁡(kx)+isin⁡(kx).

That’s the full expansion.

---

🔍 Optional: Real and imaginary parts

• Real part: ℜ[e^ikx]=cos⁡(kx)

• Imaginary part: ℑ[e^ikx]=sin⁡(kx)

This is why a plane wave can always be represented as a combination of cosine and sine waves.

---

If you want, I can also expand e^−ikx, express plane waves as sums/differences of sinusoids, or show how this relates to Fourier transforms.

You said:

Show how it relates to Fourier transforms

ChatGPT said:

The identity

e^ikx=cos⁡(kx)+isin⁡(kx)

is not just a mathematical curiosity — it is the foundation of Fourier transforms. Here’s how they are connected in a clean, intuitive way.

---

🔷 1. Fourier transforms use complex exponentials as basis functions

A function f(x) can be decomposed as a superposition of waves:

f(x)=∫_−∞^∞f^~(k) e^ikx dk.

This works because the complex exponential behaves like a pure oscillation — it is the combination of cosine (even part) and sine (odd part):

e^ikx=cos⁡(kx)+isin⁡(kx).

Thus a Fourier transform is really the decomposition of a function into cosines and sines of varying wavelengths and phases.

---

🔷 2. Real Fourier series arise by splitting exponentials into trig functions

A Fourier series for a periodic function can be written in two forms:

Complex exponential form

f(x)=∑_n=−∞^∞c_ne^i2πnx/L.

Trig (sine–cosine) form

f(x)=a₀+∑_n=1^∞[a_ncos⁡(2πnx/L)+b_nsin⁡(2πnx/L)].

The equivalence comes directly from:

e^i2πnx/L=cos⁡ (2πnx/L)+isin⁡ (2πnx/L).

Thus:

• the real part of the exponential gives the cosine term;

• the imaginary part gives the sine term.

The Fourier coefficients relate as:

c_n=1/2 (a_n−ib_n), c_−n=1/2 (a_n+ib_n).

Splitting the exponential form gives the trig form.

---

🔷 3. Continuous Fourier transform uses the same logic

The continuous transform is:

f^~(k)=∫_−∞^∞f(x)e^−ikxdx, f(x)=∫_−∞^∞f^~(k)e^ikxdk.

Using the expansion:

e^ikx=cos⁡(kx)+isin⁡(kx),

this becomes:

f(x)=∫_−∞^∞f^~(k)[cos⁡(kx)+isin⁡(kx)]dk.

If f(x) is real, the sine and cosine parts combine to give the familiar cosine transform (even part) and sine transform (odd part).

---

🔷 4. Why complex exponentials are preferred

Complex exponentials make Fourier transforms:

• much easier to manipulate (differentiation, convolution, phase shifts).

• more symmetric (positive and negative k handled naturally).

• compact (no separate sine and cosine coefficients).

Using sines and cosines directly is possible, but it explodes the algebra — complex exponentials tidy everything.

---

🔷 Summary

Euler’s formula

e^ikx=cos⁡(kx)+isin⁡(kx)

is the bridge between:

➜ complex exponential form of Fourier transforms

➜ sine–cosine form (classical Fourier decomposition)

Thus the complex exponential basis used in Fourier transforms is simply a compact way to encode all sine and cosine components of a function.

---

If you’d like, I can show:

• how differentiation in real space becomes multiplication in k-space,

• how Fourier transforms relate to the uncertainty principle,

• or how the Fourier transform is used in quantum mechanics (momentum representation).

You said:

So the 2π factor arises because there are 2π radians in a full wave cycle?

ChatGPT said:

Exactly — that’s the deepest and simplest reason.

The factor of 2π appears everywhere in wave physics and Fourier analysis because:

One full oscillation corresponds to 2π radians of phase.

Here’s what that means and why it affects the allowed k-values and the density of states.

---

✅ 1. A plane wave oscillates by 2π radians per wavelength

A plane wave is written as:

e^ikx.

Its phase is:

ϕ=kx.

After one wavelength λ, the wave must return to the same value:

e^ik(x+λ)=e^ikx.

This means the phase change over one wavelength is exactly one full cycle, i.e., 2π:

kλ=2π.

Therefore:

k=2π/λ.

This alone explains why 2π appears in all wavevector formulas.

---

✅ 2. Boundary conditions → allowed k-values contain 2π

For a box of length LLL with periodic boundaries:

ψ(x+L)=ψ(x)

requires:

e^ikL=1=e^i2πn.

So:

kL=2πn⇒k=2πn/L.

Again the 2π is directly tied to the number of radians in a cycle.

---

✅ 3. Every “cell” in k-space has volume (2π/L)³

The allowed k-values are spaced by:

Δk=2π/L.

So each quantum state occupies a k-space volume:

(Δk)³=(2π/L)³.

Then the density of states becomes:

1/(Δk)³=V/(2π)³.

This is why the Rayleigh–Jeans, Planck, Debye, and electron density-of-states formulas have (2π)³.

---

🔷 4. If a different convention were used, the 2π would move elsewhere

Some older textbooks define the Fourier transform without the 2π:

k=1/λ,

or distribute 2π symmetrically in the transform pair.

Those conventions shift where the 2π appears, but the physics is the same.
Modern physics uses the convention:

k=2π/λ

because it makes plane waves take the clean form e^ikx with kx measured in radians.

---

⭐ Bottom line

Yes — the entire reason the 2π factor appears is because a full wave oscillation corresponds to 2π radians.
This single fact propagates through:

• the definition of wavenumber

• the spacing of allowed states

• Fourier transforms

• mode counting in k-space

• the Rayleigh–Jeans and Planck laws

• the Debye model of solids

• momentum quantization p=ℏk

It all starts from one cycle = 2π radians.

---

If you want, I can also show how the picture changes if we define the Fourier transform using cycles instead of radians.

The only thing I ever knew about k-space was its use in MRI, where there's a Fourier transform of image measurements. And why the 2π factor. (glad the AI gave such a straightforward answer to that) The rest of it may as well be spaniels howling in Urdu. (though I dimly grasped the use of Fourier analysis when I volunteered my PC years ago for a SETI crowd computing project that analyzed signals from...somewhere)

22 minutes ago, TheVat said:

The only thing I ever knew about k-space was its use in MRI, where there's a Fourier transform of image measurements. And why the 2π factor. (glad the AI gave such a straightforward answer to that) The rest of it may as well be spaniels howling in Urdu.

For me (and my point of interest here is low temperature thermodynamics) the crossing the Rubicon moment was the confirmation of my instinctive feel that spherical coordinates were a much better framework for these essentially circular functions than Cartesian coordinates. Trying to bang square pegs into round holes and all that. I'm a lot more comfortable mit Dackeln, die auf Deutsch heulen.

13 hours ago, sethoflagos said:

Dackeln

Makes sense, given they're basically cylindrical. And cylindrical coordinates being a 3D extension of polar coordinates.