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Positive-energy spin–size inequality

Statement

Fix c>0 and a future unit observer n in an affine Minkowski spacetime as in Oriented time-oriented affine Minkowski space (the spacetime primitives) and Norms of four-velocity and four-acceleration, and the observer rest-space decomposition. Identify the positive-definite rest space n⊥ isometrically with R3, with its Euclidean norm ∣⋅∣ and Lebesgue measure d3x. Define the closed Euclidean ball by B‾ℓ(0):={x∈R3:∣x∣≤ℓ}. Let e:R3→[0,∞) be measurable with e∈L1(R3), and let j:R3→R3 be measurable with each component ji∈L1(R3) for i=1,2,3. Thus j is vector-valued integrable and every vector integral below is the componentwise Lebesgue integral. Assume e and j are both zero almost everywhere outside B‾ℓ(0) for some ℓ>0, and that ∣j(x)∣≤e(x)/c almost everywhere. Define E0:=∫R3e(x) d3x>0,M:=E0/c2,s∗:=∫R3x×j(x) d3x. Then s∗ is absolutely integrable and ∣s∗∣≤ℓMc,equivalentlyℓ≥∣s∗∣Mc. The hypotheses are the full mathematical content of this estimate; it is not a bound for a quantum point particle or for a source whose energy density or momentum density fails them.

Facts & Assumptions

Given: The affine Minkowski spacetime, a future unit observer and its Euclidean rest space, c>0, and integrable e,j satisfying the stated positivity, support and pointwise bound.

[A1]

The rest densities are Lebesgue-integrable, vanish almost everywhere outside B‾ℓ(0), satisfy e≥0 and ∣j∣≤e/c almost everywhere, and have E0=∫e>0.

[F1]

The affine Minkowski model has signature (−+++); the orthogonal complement of a future unit timelike vector is a positive-definite Euclidean rest space, with the Euclidean inner product and norm of R3. Oriented time-oriented affine Minkowski space (the spacetime primitives); Norms of four-velocity and four-acceleration, and the observer rest-space decomposition; The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn

[F2]

The cross product has components (x×y)1=x2y3−x3y2, (x×y)2=x3y1−x1y3, (x×y)3=x1y2−x2y1. The cross product in R3

[F3]

L1 means measurable with finite integral of the modulus; real function integrals are defined by positive and negative parts. A vector-valued map into R3 is integrable when each coordinate function is in L1, and its integral is the componentwise integral. The class L1(μ) of integrable functions; Integrable real and complex functions, and their integrals

[F4]

The nonnegative integral is monotone and positively homogeneous, the Lebesgue integral is linear on L1, and ∣∫f∣≤∫∣f∣ for f∈L1. Monotonicity and nonnegative homogeneity of the nonnegative integral; The Lebesgue integral is linear on L1(μ); The modulus of an integral is bounded by the integral of the modulus

Proof

technique · direct
1.1givenA1F1F2F3F4algebra

By the Statement, e and each component of j belong to L1; the norm and cross-product conventions are those of [F1] and [F2]. The component formula of [F2] expands to ∣x×y∣2=∣x∣2∣y∣2−(x⋅y)2≤∣x∣2∣y∣2. For almost every x in the support ball, ∣x∣≤ℓ and ∣j(x)∣≤e(x)/c, so ∣x×j(x)∣≤ℓe(x)/c. From e,j∈L1 in [F3], monotonicity and homogeneity in [F4] give ∫∣x×j∣ d3x≤ℓE0/c<∞, establishing absolute integrability of s∗.

2.1givenA1F3F4F5step 1.1algebra

If s∗=0 the bound is immediate. Otherwise set u=s∗/∣s∗∣, so ∣u∣=1. Linearity, the integral triangle inequality [F4] and Cauchy–Schwarz [F5] give ∣s∗∣=u⋅s∗=∫u⋅(x×j(x)) d3x≤∫∣u⋅(x×j(x))∣ d3x≤∫∣x×j(x)∣ d3x≤ℓE0c=ℓMc. This proves the first inequality.

3.1givenstep 2.1algebra∎

Since M=E0/c2>0 and c>0, divide the inequality of step 2.1 by Mc to obtain ℓ≥∣s∗∣/(Mc).

Remarks

This is a conditional material inequality. Positivity of e and the pointwise energy-current bound ∣j∣≤e/c are load-bearing hypotheses, not consequences of the definition of a quantum or point particle.

Depends on

Dependency tree · 2 levels

Sources