anuj, Author at Time Travel, Quantum Entanglement and Quantum Computing

Wave nature and speed of proton (particle)

anuj — Fri, 30 Jan 2026 20:24:21 +0000

High-Speed Protons and de Broglie Waves

If a proton is moving at high speed, does it affect its de Broglie wave nature?

Answer: Yes. A proton always has a wave description, but as its speed (and momentum) increase,
its de Broglie wavelength gets smaller. The wave nature doesn’t disappear—it becomes harder to
observe with everyday-sized apparatus.

1) de Broglie wavelength (core relation)

For any particle:

λ = h / p

where λ is the de Broglie wavelength, h is Planck’s constant, and p is momentum.

Non-relativistic proton

p = m v → λ = h / (m v)

Relativistic proton (high speed)

p = γ m v, γ = 1 / √(1 – v²/c²)

λ = h / (γ m v)

Key point: As speed increases, momentum increases, so λ decreases.

2) What “high speed” changes physically

The wave nature does NOT disappear. Quantum mechanics never “turns off.”
The wavelength becomes very small. At accelerator energies it can be far smaller than atoms or even nuclei.

3) Why fast protons often look “particle-like”

Wave behavior (diffraction/interference) is easiest to see when the wavelength is comparable to the size of
slits, gratings, or other structures:

If λ ≪ (size of apparatus), diffraction angles are tiny and interference fringes are extremely fine.

So the proton still has a wave description, but the wave effects become harder to detect
with typical instruments.

4) Relativity does not suppress quantum mechanics

Common misconception: “Relativistic particles become classical.”
Reality: Relativity increases momentum → wavelength shrinks → wave effects are hidden at accessible scales.

5) Phase vs group velocity (subtle but important)

For a relativistic de Broglie wave:

Phase velocity: v_phase = c² / v (> c)

Group velocity: v_group = v

No causality violation: information travels with the group velocity, not the phase velocity.

6) Why high-energy experiments still reveal “wave/quantum” structure

Even when λ is tiny, quantum behavior shows up strongly in scattering and diffraction-like measurements.
Higher energies probe smaller distances, revealing internal structure (e.g., quarks and gluons in the proton).

7) One-line takeaway

A fast proton is still a wave—just an extremely short-wavelength one.

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Invariant Subspaces — Definition & Examples

anuj — Tue, 27 Jan 2026 16:08:04 +0000

Invariant Subspaces — Definition & Examples

Invariant Subspaces

An invariant subspace is a subspace that a given linear transformation cannot “move you out of.”

Definition

Let V be a vector space and
T : V → V a linear transformation.
A subspace W ⊆ V is invariant under T if:

T(W) ⊆ W

Meaning: whenever w ∈ W, applying T keeps you inside W.

Intuition: Think of T as “dynamics.” If you start in an invariant subspace,
you never leave it.

Basic Examples

1) Trivial invariant subspaces

These exist for every linear operator:

{0}
The whole space V

2) Eigenspaces

If v is an eigenvector of T with eigenvalue
λ, then T v = λ v, and the span of v
is invariant. More generally, the eigenspace
E_λ = { v : T v = λ v } is invariant.

Example: For T = diag(2, 3) on ℝ²,
the x-axis is invariant (eigenvalue 2) and the y-axis is invariant (eigenvalue 3).

3) Upper-triangular matrix example

Let

T = [ [1, 1],
      [0, 1] ]

The subspace W = span{(1,0)} is invariant because
T(1,0) = (1,0).
But the y-axis is not invariant since
T(0,1) = (1,1) is not in span{(0,1)}.

Function Space Examples

4) Differential operator

Let T = d/dx.

The space of all polynomials is invariant.
The space of polynomials of degree ≤ n is invariant.
The space of even functions is not invariant (derivative of even is odd).
The space of odd functions is invariant under d²/dx².

5) Fourier / frequency subspaces

For periodic functions and T = d/dx, each Fourier mode
e^{ikx} spans an invariant subspace:

The span of e^{ikx} for fixed k is invariant.
This is why many PDEs “decouple” in Fourier space.

Quantum Mechanics Connection

In quantum mechanics, invariant subspaces often correspond to conserved sectors.
If a Hamiltonian respects a symmetry, the Hilbert space can decompose into invariant subspaces
that evolve independently.

Example idea: Symmetries related to angular momentum lead to invariant subspaces
labeled by quantum numbers (e.g., fixed ℓ).

Representation Theory View

A representation is irreducible if it has no nontrivial invariant subspaces.
Decomposing a space into invariant subspaces reveals “independent components” of an action.

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Mean Square Fluctuations of Energy (monoatomic gas)

anuj — Thu, 11 Dec 2025 03:19:23 +0000

Monoatomic Ideal Gas — Energy Fluctuations & Velocity Probabilities

Monoatomic Perfect Gas (N particles)

Below are (A) the mean-square fluctuation in energy in the canonical ensemble and (B) the probability that a chosen particle’s velocity component lies in the interval . Equations are provided in LaTeX and will be rendered by MathJax.

A. Mean-square fluctuation of the energy (canonical ensemble)

Start with the canonical partition function for the full system at temperature . Define . The canonical relations are

Using the thermodynamic identity (or equivalently using standard manipulations) one obtains the well-known relation

where is the heat capacity at constant volume.

For a classical monoatomic ideal gas

Relative fluctuation (useful scaling):

This shows energy fluctuations scale as and are negligible for macroscopic .

B. Probability that a particle’s lies in

In the canonical ensemble for a classical ideal gas the single-particle momentum/velocity components are independent and Gaussian. The one-component Maxwell–Boltzmann probability density for is

For a small interval (infinitesimal approximation), the probability that a chosen particle has in is

For the 3D speed (magnitude) distribution the Maxwell speed density is

so the probability a chosen particle’s speed lies in is .

C. Probability that any of the N particles has in the interval

Let

be the single-particle probability for that small interval. Assuming independent particles, the probability that none of the particles lies in the interval is . Thus the probability that at least one particle lies in the interval is

If is very small and then (expected number of particles in the interval).

D. Optional: brief canonical derivation of

From the partition function :

Hence

Noting and recognizing yields the relation .

E. Quick summary

for a monoatomic ideal gas.
Single-component velocity density: .
Probability (small interval): . For any of the N particles: .

If you’d like, I can also:

Provide the same page but with explicit numeric examples (choose ).
Show the derivation of the Maxwell distribution from the canonical single-particle Hamiltonian step-by-step.
Format this for printing (PDF-friendly) or convert to LaTeX source.

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The Universal Wave Function

anuj — Thu, 11 Dec 2025 03:17:11 +0000

World / Universal Wave Function — Explanatory Article

The World (Universal) Wave Function

The world wave function, also called the universal wave function, is the quantum-mechanical wave function that describes the state of the entire universe (all degrees of freedom: particles, fields, and even observers). Below is a concise, structured explanation with LaTeX-compatible equations for any renderer (MathJax, KaTeX, etc.).

1. Wave function in ordinary quantum mechanics

A wave function describes the quantum state of a system and evolves deterministically under the Schrödinger equation. For a nonrelativistic system:

Measurement in the standard (Copenhagen) picture is usually described as a non-unitary collapse of to one eigenstate; probabilities for outcomes are given by squared amplitudes, e.g. .

2. Extending the wave function to the whole universe

The universal wave function is a single wave function that contains every degree of freedom of the cosmos. Symbolically:

Here denotes the full set of coordinates (or field values, spins, etc.) for everything in the universe. Since nothing exists outside the universe to perform a collapse, evolves unitarily via the Schrödinger equation (or its quantum-field-theory / quantum-gravity generalization).

3. Many-Worlds / Everett perspective

In Everett’s interpretation, the universal wave function never collapses. Instead, apparent “collapse” corresponds to a branching structure of into decoherent components (branches) after interactions that entangle system and environment.

After decoherence, branches have negligible interference with each other and behave effectively like separate classical worlds. The Born-like rule for probabilities arises from the squared amplitudes (this is a subtle topic with varied derivations in the literature).

4. Intuitive consequences & remarks

No external observer: There is no “outside” system to collapse the universal wave function.
Unitary evolution: evolves according to a universal Hamiltonian (or quantum-gravity law) without non-unitary collapse.
Branching and decoherence: When subsystems entangle with large environments, interference terms become effectively unobservable — giving the appearance of classical outcomes.
Probability interpretation: Probabilities are assigned to branches by their amplitude weights, but justifying why observers should use (Born rule) has been the subject of deep analysis and debate.
Huge, abstract object: The universal wave function is vastly high-dimensional and not directly computable in a literal sense — it’s a conceptual object that organizes quantum possibilities.

5. Simple branching diagram (visual)

6. Short FAQ

Q: Is the universal wave function proven?
A: The universal wave function is a theoretical construct. It follows from taking quantum mechanics (unitary evolution) literally for the whole universe — but interpretations differ on whether it is the best or only way to think about reality.

Q: Where does probability come from if everything happens?
A: In Many-Worlds, probability is associated with branch weights (amplitude squared). Explaining why agents should use these weights is non-trivial and has been addressed via decision-theoretic, symmetry, and envariance arguments in the literature.

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Partial-Sum Probabilities vs. Bell Correlations

anuj — Mon, 08 Dec 2025 22:45:19 +0000

Partial-Sum Probabilities vs. Bell Correlations — Illustrated

Partial-Sum Probabilities ⇄ Successive Measurements in Bell Experiments

Figure: Left — running-partial-sum Markov transitions; Right — Bell measurement settings and entangled state. (Generated diagram.)

Overview

Both the classical partial-sum process and the quantum Bell experiment produce probabilities for sequences of outcomes.
The useful analogy is that both systems evolve probabilities over a space of possible “paths” (integer sums vs. measurement outcomes). However, the
crucial difference is that Bell correlations require non-classical probability structure: no classical Markov chain or hidden-local-variable model can reproduce them.

1. The Partial-Sum (Markov) Model — short description

Suppose we draw integers uniformly from {1,…,N} with replacement and keep a running sum \(S_k\). The evolution is
classical and Markovian:

\(P(S_k = s) = \sum_{i=1}^{N} P(S_{k-1} = s-i)\cdot \frac{1}{N}.\)

To ask for the probability of reaching exactly \(n\) at some step is to condition on allowed paths through the integer-state lattice; paths that overshoot are excluded.
This process is classical: transition probabilities are nonnegative, normalized, and depend only on the previous state (Markov property).

2. Bell (CHSH-style) Experiment — short description

Two distant parties, Alice and Bob, share an entangled state \(\lvert\psi\rangle\). Each party chooses a measurement setting (Alice: \(a\) or \(a’\); Bob: \(b\) or \(b’\))
and obtains outcomes \(A,B\in\{+1,-1\}\). Quantum mechanics predicts joint probabilities

\(P(A,B \mid a,b) = \langle\psi \rvert \; \big(M_A^{(a)} \otimes M_B^{(b)}\big) \; \lvert\psi\rangle,\)

where \(M_A^{(a)},M_B^{(b)}\) are measurement operators (projectors).

The key empirical fact: for certain choices of \((a,a’,b,b’)\), the correlations violate the CHSH inequality and reach up to \(2\sqrt{2}\) (Tsirelson bound).

3. Classical Hidden-Variable / Markov-Factorization Assumption

A classical model with a hidden variable \(\lambda\) (and local Markov-like transitions) assumes the joint probability factorizes as:

\(P(A,B \mid a,b) \;=\; \int d\lambda \; \rho(\lambda)\; P(A\mid a,\lambda)\; P(B\mid b,\lambda).\)

This expresses locality and a classical probabilistic structure. If such a representation exists, all CHSH-type correlations obey the classical bound of 2.

4. CHSH inequality (derivation sketch)

Define correlators
\(\;E(a,b) = \sum_{A,B=\pm1} AB \, P(A,B\mid a,b).\)
Under the classical factorization with deterministic \(\pm1\) responses (or by convexity for probabilistic responses),
one can show the CHSH combination satisfies:

\(S \;=\; E(a,b) + E(a,b') + E(a',b) - E(a',b') \;\le\; 2.\)

The short intuitive proof: for a fixed \(\lambda\) and deterministic outcomes \(A(a,\lambda),B(b,\lambda)\in\{\pm1\}\),
the quantity
\[
Q(\lambda) = A(a,\lambda)\big[ B(b,\lambda) + B(b’,\lambda) \big] + A(a’,\lambda)\big[ B(b,\lambda) – B(b’,\lambda) \big]
\]
can only be \(\pm2\). Averaging over \(\lambda\) yields \(|S|\le 2\).

5. Quantum prediction violates the classical bound

For the two-qubit singlet state and spin (or polarization) measurements at appropriate angles one finds

\(S_{\text{quantum}} = 2\sqrt{2} > 2.\)

Therefore no model of the factorized classical form (and hence no classical Markov chain producing local factorized joint probabilities) can reproduce these correlations.

6. Why this forbids any classical Markov-chain representation

A Markov chain (or any classical sequential stochastic process) defines joint distributions built from nonnegative transition probabilities and local conditionalization on prior states.
If you attempt to represent the Bell scenario with a classical Markov chain / path model, you would need to assign joint probabilities
\(P(A,B\mid a,b)\) that simultaneously satisfy all measurement-setting marginals and the factorization/locality condition.
But because quantum correlations violate CHSH, no such global assignment of classical nonnegative transition probabilities exists.

Concretely:

Partial-sum Markov processes: probabilities are built from local, stepwise transition kernels (nonnegative, normalized).
Any classical hidden-variable or Markov description that respects locality must obey Bell (CHSH) bounds.\li>
Quantum correlations (experimentally verified) violate those bounds, so they cannot be written as expectations over local Markov transitions.

7. Side-by-side summary (compact)

Feature	Partial-Sum / Markov	Bell / Quantum
State space	Integer sums, classical lattice	Hilbert space (amplitudes)
Allowed transitions	Nonnegative transition kernel, Markov	Unitary + measurement (non-commuting)
Path weighting	Sum of nonnegative path probabilities	Amplitude interference (complex), not representable as simple path probabilities
Bell/CHSH	Always satisfies CHSH bound \(\|S\|\le2\)	Can achieve \(\|S\|=2\sqrt2>2\)

8. Intuition: interference & non-commutativity vs. classical conditioning

Classical path models add probabilities for disjoint paths. Quantum mechanics adds complex amplitudes that can interfere, producing correlations that cannot be decomposed into a convex mixture of local deterministic paths.
Mathematically this is tied to the non-commutativity of measurement operators and the fact that a global joint distribution for all possible measurement outcomes (for all settings) that is both local and reproduces quantum marginals does not exist.

9. Optional: short worked example (CHSH angles)

For the singlet state choose measurement directions such that
\(\theta_{a,b}=0^\circ,\; \theta_{a,b’}=90^\circ,\; \theta_{a’,b}=45^\circ,\; \theta_{a’,b’}=135^\circ\).
The quantum correlator for spin-1/2 is \(E(\alpha,\beta) = -\cos(\theta_{\alpha\beta})\).
Then

\(S = -\cos 0^\circ – \cos 90^\circ – \cos 45^\circ + \cos 135^\circ
= -1 – 0 – \tfrac{\sqrt2}{2} + \big(-\tfrac{\sqrt2}{2}\big)
= -2\sqrt2 \)

so \(|S|=2\sqrt2\), violating the classical limit of 2.

10. Final takeaway

The visual/structural analogy is useful: both systems manage probability flow across allowed paths (integer-lattice paths vs. sequences of measurement outcomes). But Bell correlations are fundamentally incompatible with any classical Markov-chain (or local hidden-variable) representation because they violate inequalities (CHSH) that any such classical model must obey. That violation is the signature of quantum non-classicality (entanglement + interference + non-commuting observables).

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Randomy generated numbers – Probability of Running Sums

anuj — Mon, 08 Dec 2025 05:58:19 +0000

Draws are independent and uniformly chosen from the set (with unlimited repetition).
Let denote the probability that at some time the running sum equals exactly .

1. Recurrence

With the convention for and , the probability satisfies the recurrence

(This comes from conditioning on the value of the first draw: if the first draw is then we must subsequently hit .)

2. Generating function

Let . Summing the recurrence gives the closed form

Equivalently,

The coefficient of in this rational function equals .

3. Combinatorial (explicit) formula

Another useful representation counts ordered compositions of into parts each in .
If denotes the number of such compositions with exactly parts, then

By inclusion–exclusion one can write

so a fully explicit formula is

4. Remarks and small cases

: for every (every draw is 1 so every positive integer is hit).
: with ; values can be computed quickly from the recurrence.
The generating function form is convenient for extracting asymptotics or for computing by partial fractions (roots of the denominator polynomial).

Recap (as before): draws are iid and uniform on . Let be the probability that some partial sum equals exactly .
With the convention for and ,

Numerical examples

Below are computed values of for and . (Remember .)

(each draw is 1 or 2 with probability )
n
0	1.000000000000
1	0.500000000000
2	0.750000000000
3	0.625000000000
4	0.687500000000
5	0.656250000000
6	0.671875000000
7	0.664062500000
8	0.667968750000
9	0.666015625000
10	0.666992187500
11	0.666503906250
12	0.666748046875

(each draw is 1, 2 or 3 with probability )
n
0	1.000000000000
1	0.333333333333
2	0.444444444444
3	0.592592592593
4	0.456790123457
5	0.497942386831
6	0.515775034294
7	0.490169181527
8	0.501295534217
9	0.502413250013
10	0.497959321919
11	0.500556035383
12	0.500309535772

(each draw is 1,2,3 or 4 with probability )
n
0	1.000000000000
1	0.250000000000
2	0.312500000000
3	0.390625000000
4	0.488281250000
5	0.360351562500
6	0.387939453125
7	0.406799316406
8	0.410842895508
9	0.391483306885
10	0.399266242981
11	0.402097940445
12	0.400922596455

Observations

For the probabilities converge quickly to (the table shows values oscillating about that limit).
For and the values for moderate tend to a limiting value (the limit exists and equals the reciprocal of the mean waiting-time to cross a large interval — this can be analyzed via renewal theory or by studying the generating function). Numerically you can see the values settle near about for and near for at the shown.

— End

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Perfect Gas with Dipole Moments – what can it model?

anuj — Wed, 03 Dec 2025 19:58:25 +0000

Can a Perfect Gas of Dipolar Molecules Model Stars or Planets?

Short answer: No — a “perfect gas with N molecules each having a dipole moment p” is not a good approximation for essentially any star or planet. But it is useful for a completely different class of systems (polar molecular gases, laboratory plasmas under special conditions, etc.).

Why Stars Cannot Be Modeled This Way

Stars are fully ionized plasmas, not molecular gases.

Temperatures: 10⁶–10⁷ K
Molecules cannot exist; even atoms are largely ionized.
Dipole moment p assumes neutral bound molecules — these are destroyed at stellar temperatures.
Stellar behavior is dominated by:
- Coulomb plasma interactions
- Radiation pressure
- Electron degeneracy pressure (white dwarfs)
- Nuclear reaction physics
- Global magnetic fields, not molecular dipoles

Conclusion: Stars contain no permanent molecular dipoles because they contain no molecules.

Why Planets Cannot Be Modeled This Way

Rocky planets

Matter exists as solids, molten rock, or ionic fluids.
Dipole orientation is irrelevant due to extremely high density.

Gas giants (Jupiter, Saturn)

Mainly H₂ and He, but at extreme pressures.
H₂ has a quadrupole moment, not a dipole.
Deep layers become metallic hydrogen — no molecular dipoles.

Ice giants (Uranus, Neptune)

Contain polar molecules (H₂O, NH₃, CH₄), but in supercritical or ionic phases.
Dipoles do not behave as free ideal-gas dipoles.

Conclusion: Planetary interiors are too dense and too hot for ideal dipole-gas assumptions.

When the “Perfect Gas with Dipoles” Model Is Useful

This model applies to molecular physics, not astrophysics.

Dilute polar gases (HCl, HF, H₂O vapor)
Dielectric susceptibility calculations
Statistical mechanics of orientable dipoles
Weak-field polarization in low-density gases

This leads to results like the Langevin–Debye law for orientational polarization.

Summary Table

System	Molecules?	Permanent Dipoles?	Gas-Like?	Suitable for “Ideal Gas of Dipoles”?
Stars	None	None	Plasma	No
Gas giants (deep layers)	Metallic hydrogen	None	Fluid/Metal	No
Gas giants (upper atmosphere)	H₂ gas	Quadrupole only	Yes	No
Ice giants	Ionic/supercritical fluids	Dipoles present	No	No

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Spin versus Angular Momentum

anuj — Tue, 02 Dec 2025 16:10:09 +0000

Angular Momentum, Spin, and the Particle in a Box

Clarifying Angular Momentum in a Particle in a Box

The statement “a particle in a box has no angular momentum” refers specifically to the absence of orbital angular momentum as a good quantum number. This does not apply to spin, which is an intrinsic form of angular momentum independent of geometry.

1. Two Kinds of Angular Momentum

(A) Orbital Angular Momentum

The operator is:

It depends on spatial coordinates and exists only when the system has rotational symmetry.

(B) Spin Angular Momentum

Spin is an internal degree of freedom and does not depend on the potential’s shape or boundary conditions.

2. What “No Angular Momentum” Really Means

A rectangular (1D or 3D) infinite potential well has no rotational symmetry. Therefore:

This means neither nor are conserved or define good quantum numbers. The particle’s orbital motion is described only by the quantum numbers .

Thus: a particle in a box has no conserved orbital angular momentum.

3. But the Particle Can Still Have Spin

The spin degree of freedom is completely unaffected by placing the particle in a box.

The full Hilbert space becomes:

The spatial eigenfunctions are the usual box states:

The spin state is independent:

The complete state is therefore:

If no magnetic fields or spin–orbit coupling are present:

Spin angular momentum is fully conserved inside the box.

4. What If the Particle Is “Already Rotating”?

This phrase has two possible interpretations:

Case 1 — The particle has orbital angular momentum before entering the box

The initial state might be something like:

When placed in a rectangular box:

rotational symmetry is lost,
orbital angular momentum is no longer conserved,
the state becomes a superposition of box eigenstates.

The angular momentum “scrambles” because the box does not allow rotations.

Case 2 — The particle has spin

Spin remains a well-defined, conserved quantum number. The full state can be:

Spin exists independently of the geometry and survives unchanged inside the box.

5. Summary Table

Quantity	Exists in a Box?	Depends on Geometry?	Conserved?
Orbital Angular Momentum	Not a good quantum number	Yes	No
Spin Angular Momentum	Always	No	Yes (unless external fields)
Total	Not conserved (because is not)	Yes	Only for spherical potentials

In Summary

A particle in a rectangular/1D box does not have conserved orbital angular momentum because the geometry breaks rotational symmetry.
But the particle’s spin remains fully intact and unaffected by the boundary conditions.

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Is a Space Elevator Feasible?

anuj — Tue, 02 Dec 2025 01:24:44 +0000

Is a Space Elevator Feasible?

Short answer: Not with today’s materials—but theoretically yes, if a breakthrough material emerges.

A space elevator is a proposed structure that extends from Earth’s surface to geostationary orbit (GEO, approximately 35,786 km up), with a counterweight beyond GEO to keep the cable taut using centrifugal force. In theory it would allow payloads to climb into orbit without rockets, greatly reducing cost. Feasibility depends almost entirely on a single factor: the tether material.

1. The Core Problem: No Known Material Is Strong Enough

Key requirements: extremely high tensile strength, very low density, and the ability to exist without weak points for tens of thousands of kilometers.

Required tensile strength

Roughly 60–120 GPa strength-to-weight ratio is often cited for an Earth-to-GEO tether.

Best materials today

Carbon nanotubes (CNTs): lab-scale flakes/fibers show 50–100 GPa in ideal samples, but we cannot manufacture continuous macroscopic cable at required lengths. Real-world CNT fibers are far weaker (≈ 1–6 GPa).
Graphene: extremely strong in small flakes, but impossible today to scale into kilometer-length ribbons or cables.
Kevlar, Zylon, Dyneema, steel: all orders of magnitude too weak for an Earth-based elevator.

Bottom line: the necessary material for an Earth-based elevator does not exist yet.

2. Earth-Based Elevator: Probably Not This Century

Even if a perfect material were discovered, there are many practical complications beyond raw strength:

Radiation damage and degradation in the space environment.
Micrometeorite and orbital debris impacts that could sever or damage a tether.
Weather, lightning, and atmospheric effects along the lower portion of the cable.
Dynamic instabilities (vibrations, oscillations — the “guitar string” problem).
Enormous manufacturing, deployment, and maintenance challenges.
Political and legal risk — a 36,000 km cable is an international liability.

Conclusion: Earth elevator is theoretically possible but not practically feasible with current technology or near-term foreseeable technology.

3. Lunar / Martian Space Elevators: Much More Feasible

Lower gravity and (in the Moon’s case) near-absence of atmosphere make extraterrestrial elevators dramatically easier:

Lunar elevator

Tensile strength requirements are much lower — often < 5 GPa, which is achievable with today’s high-performance polymers and fibers.
No atmosphere → no weather, no lightning, and far fewer material degradation mechanisms.
No orbital debris problem compared with low Earth orbit.
Deployment and operations are far simpler and much more feasible in the near term.

Short verdict: A lunar space elevator could be built today using materials like Zylon or ultra-high-performance polymers; it is one of the most feasible near-term megastructure concepts.

4. What If We DO Invent a Suitable Material?

If we could manufacture carbon nanotube fiber or graphene ribbons at macroscopic scales and with the necessary reliability:

Launch cost per kilogram could plummet — some estimates suggest reductions of up to ~95% relative to current rocket costs for certain cargos.
Continuous cargo transport becomes possible, enabling routine movement of power, materials, infrastructure, and satellites.
Stations or platforms along the tether (at different altitudes) could enable new industries and services.
Access to space could become as routine as accessing a high-rise building — enabling large-scale space industrialization.

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Angular momentum for particle in a box

anuj — Tue, 25 Nov 2025 20:23:36 +0000

angular momentum energy levels

https://stationarystates.com/basic-quantum-theory/angular-momentum…article-in-a-box/

Energy Levels of a Particle in a Box with Angular Momentum

1. Particle in a 1D Box

In 1D, angular momentum doesn’t exist in the usual sense because rotation requires at least two dimensions.
Energy levels remain:

2. Particle in a 2D or 3D Box

For a 2D rectangle or 3D cube, the Schrödinger equation separates in Cartesian coordinates:

Angular momentum is not conserved in a cubical box. Energy depends on quantum numbers along each axis:

3. Particle in a Spherical Box

If the box is spherically symmetric, angular momentum L is a good quantum number.
The Schrödinger equation in spherical coordinates:

Separate variables:

where Y_\ell^m are spherical harmonics and \ell is the angular momentum quantum number.
Energy levels include a centrifugal term:

Here, are the zeros of spherical Bessel functions.
Larger angular momentum () increases energy because the wavefunction is “pushed outward”.

4. Key Takeaways

1D box: Angular momentum is irrelevant; energy levels are unchanged.
Rectangular/cubical box: Energy depends on quantum numbers along each axis, not angular momentum.
Spherical box: Higher angular momentum quantum number \ell raises the energy.

Intuition: Higher angular momentum → particle “rotates” more → less probability near the center → higher energy.

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