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Schwarzschild supported thermal bath

Physical scope

Static test bath on the Schwarzschild exterior r>r_s, with asymptotic Killing-time normalization, exact lapse and acceleration, and a bounded weak-field expansion; no horizon or quantum-temperature claim.

Example

Let $m_{s} > 0$ be a source mass in kilograms, distinct from any kinetic particle mass, let $G_{N} > 0$ be Newton's constant in $m^{3} k g^{- 1} s^{- 2}$ , and let $c > 0$ be the speed of light in $m s^{- 1}$ . Set

$r_{s} : = \frac{2 G_{N} m_{s}}{c^{2}}, f (r) : = 1 - \frac{r_{s}}{r}, r > r_{s} .$

In physical length coordinates $x^{0} = c t$ , use the Schwarzschild exterior metric

$d s^{2} = - f (r) (d x^{0})^{2} + f (r)^{- 1} d r^{2} + r^{2} (d θ^{2} + \sin^{2} θ d ϕ^{2}),$

and select $ξ = \partial_{x^{0}}$ with the asymptotic normalization $N_{ξ} \to 1$ as $r \to \infty$ . The time orientation is chosen so $ξ$ is future-directed. Under the selected stationary-bath assumption, the local proper temperature satisfies

$T (r) = \frac{T_{\infty}}{\sqrt{1 - r_{s} / r}}, T_{\infty} : = T_{K} > 0.$

The static observer congruence has outward covariant acceleration component $a_{r} = r_{s} / (2 r^{2} f)$ , contravariant component $a^{r} = r_{s} / (2 r^{2})$ , and physical proper-acceleration magnitude

$c^{2} \sqrt{g_{a b} a^{a} a^{b}} = \frac{G_{N} m_{s}}{r^{2} \sqrt{f (r)}} .$

For a fixed $z_{0} \in (0, 1)$ and $0 < r_{s} / r \leq z_{0}$ , the weak-field expansion has the explicit bound

$∣ \frac{T (r)}{T_{\infty}} - 1 - \frac{G_{N} m_{s}}{c^{2} r} ∣ \leq \frac{3}{8} (1 - z_{0})^{- 5 / 2} {(\frac{r_{s}}{r})}^{2} .$

This is a conditional calculation in the specified exterior model. At $r = r_{s}$ , the chosen Killing field is null and the normalized stationary observers do not exist. The limiting divergence as $r ↓ r_{s}$ is not a constructed horizon state, a Hawking temperature, or a claim that every observer measures an infinite temperature.

Facts & Assumptions

Given: The Schwarzschild exterior metric and constants above, the chosen future Killing field $ξ$ , and the selected stationary thermal-bath assumption on $r > r_{s}$ .

[F1]

Conditional Tolman–Ehrenfest relation: for one selected connected stationary bath with smooth future timelike Killing field $ξ$ , positive proper temperature $T$ , and Killing $β = u / (k_{B} T)$ , the constant $T_{K} > 0$ satisfies $N_{ξ} T = T_{K}$ and $a = \nabla \log N_{ξ}$ . Its draft proof is conditional and its current Step 1 record is authoring readiness only.

[A1]

$m_{s} > 0$ , $G_{N} > 0$ , and $c > 0$ have the stated SI units; the metric is taken as the model geometry on $r > r_{s}$ , and the time orientation makes $\partial_{x^{0}}$ future-directed there.

[A2]

This example adopts the selected bath only on the connected exterior, with the chosen asymptotic normalization. No matter equation of state, support apparatus, experiment, or state construction is assumed.

Verification

technique · Compute the norm of the chosen Killing field, apply the conditional Tolman theorem, and differentiate the explicit lapse

1.1givenA1algebra

The metric coefficients are independent of $x^{0}$ , so $ξ = \partial_{x^{0}}$ is Killing. Its norm is $g (ξ, ξ) = - f (r)$ , and $f (r) > 0$ exactly on $r > r_{s}$ . Therefore it is future timelike on this domain and $N_{ξ} = \sqrt{f (r)}$ .

2.1givenA2F1step 1.1

The connected exterior, smooth positive lapse, selected stationary bath, and Killing inverse-temperature field meet [F1]. Hence $N_{ξ} T = T_{K}$ throughout the exterior, so $T = T_{K} / \sqrt{f}$ . Since $N_{ξ} \to 1$ as $r \to \infty$ , write this reference constant as $T_{\infty} = T_{K}$ .

2.2F1step 1.1algebra

From $a = \nabla \log N_{ξ}$ in [F1], $a_{r} = \partial_{r} (\frac{1}{2} \log f) = r_{s} / (2 r^{2} f)$ . Raising the index with $g^{r r} = f$ gives $a^{r} = r_{s} / (2 r^{2})$ , and then $g_{a b} a^{a} a^{b} = f^{- 1} (a^{r})^{2}$ , so $c^{2} ∣ a ∣ = c^{2} r_{s} / (2 r^{2} \sqrt{f}) = G_{N} m_{s} / (r^{2} \sqrt{f})$ .

3.1givenalgebra∎

Put $z = r_{s} / r$ . For $0 < z \leq z_{0} < 1$ , Taylor's theorem at $z = 0$ for $φ (z) = (1 - z)^{- 1 / 2}$ gives $φ (z) = 1 + z / 2 + φ^{''} (θ z) z^{2} / 2$ for some $θ \in (0, 1)$ , and $φ^{''} (y) = \frac{3}{4} (1 - y)^{- 5 / 2}$ . Thus the remainder is at most $\frac{3}{8} (1 - z_{0})^{- 5 / 2} z^{2}$ . Since $z / 2 = G_{N} m_{s} / (c^{2} r)$ , this proves the displayed bound.

Depends on

Conditional Tolman–Ehrenfest relation

Schwarzschild supported thermal bath

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources