The mass and energy of a vapor bubble in a turbulent ideal fluid

Valery P. Dmitriyev

REVIEW

The mass and energy of a vapor bubble in a turbulent ideal fluid

Valery P. Dmitriyev

General Physics · Submitted 21 Feb 2006 · 2 references

Pith's verdict: A short, internally consistent note that re-derives E=mc^2 by substituting definitions into the ideal gas law; the physical premise that turbulence energy acts as heat is unexamined and numerically implausible.

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arXiv IDphysics/0602142 (version 2)

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CitationValery P. Dmitriyev. "The mass and energy of a vapor bubble in a turbulent ideal fluid." arXiv:physics/0602142 (2006). https://pith.science/paper/FMOPNN2W

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Editor's note

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Nothing in this two-page note is obviously wrong algebraically, but the main result is a rearrangement of definitions gated by an unexamined analogy. The paper writes E = p0V, m = rho V*, then feeds the turbulence kinetic energy of the evaporated liquid volume into the ideal-gas equation of state and gets E = m c^2. It is transparent and honest: the author says the turbulence energy is 'in a way similar' to heat. That phrase carries the whole argument, and the paper never turns it into a physical coupling.

What is new is the specific packaging: combining Troshkin's 1990 turbulence wave speed with ideal-gas thermodynamics to produce an E=mc^2 lookalike for a bubble. As a formal exercise it is neat, and the paper is well-suited for a classroom or a philosophical note.

The soft spot is not the algebra; it is the identification itself. Using mass conservation, Eq. (5) reduces to c^2 = p0/rho_vapor, i.e. the turbulence wave speed must equal the vapor's thermal velocity scale. Nothing in the model forces that. For water at room temperature the required c is hundreds of meters per second, far above realistic turbulent fluctuation speeds. So the equality holds only for a special turbulence intensity; it is not a general property of turbulent ideal fluids. The paper does not claim it is general, but it also does not discuss the condition.

The citation pattern is clean: one relevant reference, no self-promotion. The paper is short, and every symbol is defined. It is not incoherent. It just does not deliver the advertised physics. I would not cite it in my own work, and I would not bring it to a reading group. But if it lands on an editor's desk, I would send it to a referee rather than desk-reject: it is clear enough and the central assumption is precisely identifiable, so a referee can quickly judge whether the analogy can be replaced by a real mechanism. My recommendation would be to give the author a chance to address the coupling problem and a numerical sanity check, then accept it only as a pedagogical or speculative note, not as a research result in fluid mechanics.

Referee report

Significance

If valid, the result would be a striking connection between fluid turbulence and a mass-energy relation of the Einsteinian form. The paper is transparent: it explicitly labels the turbulence-to-heat step as an analogy, and its algebra is correct. However, that analogy is the sole bridge between the liquid-side turbulence quantities and the vapor equation of state. Because the paper supplies no physical mechanism for the conversion of turbulent kinetic energy into molecular thermal energy, the central relation E = mc^2 is not actually derived. The result is conditional on an unverified equality between the turbulence wave speed and the thermal speed scale of the vapor, and with the given definitions the derivation reduces to a rearrangement of the assumed equality. The paper should therefore be assessed as a conjecture rather than a derivation.

Major objections (2)

[Note after Eq. (2)] The statement that turbulence energy in an ideal fluid is 'in a way similar' to heat energy is the only justification for substituting the turbulence kinetic energy K* of Eq. (3) into the ideal-gas relation pV = (2/3)K of Eq. (4). No mechanism is provided by which the kinetic energy of turbulent eddies in the liquid becomes the thermal kinetic energy of the vapor molecules. In an inviscid ideal fluid, turbulent kinetic energy does not dissipate into molecular thermal energy; the two forms of energy are physically distinct. Because Eq. (5) depends on this identification, the central result Eq. (8) is not derived from the stated assumptions unless this analogy is accepted without support.
[Eqs. (5)-(8)] Equation (8) is a restatement of the assumptions rather than a substantive result. Substituting the definitions of mass (6) and self-energy (7) into Eq. (5) yields E = mc^2 identically. The physical content lies entirely in the unproven equality p0V = rho_l V* c^2. Using mass conservation rho_l V* = rho_v V, this equality is equivalent to c^2 = p0/rho_v, which for an ideal vapor equals R_specific T. Thus the relation E = mc^2 holds only if the turbulence perturbation wave speed of Eq. (1) equals the thermal speed scale of the vapor. The paper gives no argument that the Reynolds-stress wave speed satisfies this condition; it is a constraint on the turbulence intensity, not a general relation. Without independent support, Eq. (8) does not follow for a general turbulent fluid.

Minor objections (3)

[Abstract] The abstract says the relationship 'can be obtained,' which overstates the conditional nature of the argument; a phrase such as 'is conjectured under the stated analogy' would more accurately reflect the content.
[Eq. (6)] The definition of the bubble mass as rho_l V* should explicitly state that the mass of liquid evaporated equals the mass of vapor in the bubble (rho_l V* = rho_v V); without that equality, the relation between V* and V is left implicit.
[References] The paper cites a single reference for the turbulence wave speed relation (1); a brief derivation or further context would improve reproducibility for readers unfamiliar with Troshkin's result.

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The paper

The paper claims

A vapor bubble's self-energy equals its vapor mass times the square of the turbulence wave speed.

The reading

The paper argues that for a vapor bubble in an ideal turbulent fluid, the work needed to create the bubble—its self-energy—equals the vapor mass times the square of the turbulence perturbation wave speed, $E=mc^2$. The derivation follows from identifying the kinetic energy of turbulent fluctuations with the internal kinetic energy of an ideal gas, then balancing gas pressure against fluid pressure. If this holds, a fluid-mechanical system reproduces the relativistic mass-energy relation, with turbulence playing the role of heat.

What carries the argument

The load-bearing identity is the equation-of-state analogy $pV=\frac{2}{3}K$, where $K$ is the turbulent kinetic energy of the fluid volume that becomes vapor. This connects the turbulence wave speed $c$ from Eq. (1) to gas pressure, allowing the bubble's self-energy $E=p_0V$ to be expressed directly as $mc^2$.

Extended reading notes

Core claim

The paper's central claim is that the self-energy of a vapor bubble, defined as the work performed against the fluid pressure to create it, satisfies $E=mc^2$, where $m$ is the mass of vapor in the bubble and $c$ is the speed of the turbulence perturbation wave given by Eq. (1). This follows from three identifications: the kinetic energy of the fluid volume that evaporates into the bubble is $K^*=\frac{3}{2}\rho V^*c^2$; the vapor obeys the ideal-gas relation $pV=\frac{2}{3}K$; and mechanical equilibrium sets gas pressure equal to fluid pressure. Combining these with $m=\rho V^*$ and $E=p_0V$ yields $E=mc^2$.

Load-bearing premise

The derivation assumes that the kinetic energy of turbulent fluctuations in the liquid becomes the internal kinetic energy of the vapor molecules, so turbulence in an ideal fluid behaves exactly like heat; no mechanism is given for this conversion.

Editorial extensions

If this is right

If correct, the energy required to create a vapor bubble is precisely the mass of vapor times the square of the turbulence wave speed.
The bubble's volume is fixed by the background pressure through $p_0V=\rho V^*c^2$, giving a measurable relation between bubble size, vapor density, and turbulence intensity.
The turbulence perturbation wave speed plays the role of $c$, making the bubble's self-energy formally identical to the rest energy of a massive object.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

One could test this by measuring the work needed to create cavitation bubbles in a turbulent flow and comparing it with independently measured turbulence fluctuation velocities to compute $c$ from Eq. (1).
The derivation suggests a broader family of systems—any continuum whose fluctuation kinetic energy satisfies $pV=\frac{2}{3}K$—will display mass-energy relations of the same form.
If the turbulence-as-heat analogy is taken seriously, it implies an effective temperature for the turbulent fluid, which could be compared with the kinetic theory of gases.

Original abstract

The mass of a bubble in a fluid can be taken as the mass of the vapor in it. The self-energy of the bubble is defined as the work performed against the pressure of the fluid in order to create the bubble. Taking the vapor to be an ideal gas the relationship between the self-energy, the mass of the bubble and the speed of the perturbation wave in a turbulent ideal fluid can be obtained.

The compiled record

The paper's own sentences, arranged, kept beside the review and separate from it. Each label is drawn from a closed vocabulary; the number beside it is the probability the classifier gave that label. Numbers in the tables are the authors', checked by code. No sentence in this record was written by a model.

Compiled by classifier-1.13.0 via TypeSafe API, question bank abstract-p1-20260923, 2026-09-23: 1 request, 2,905 input tokens.

What the abstract itself states, as probabilities: claims first 0.03 · states limitation 0.10 · human participants 0.02 · controlled comparison 0.01 · randomized assignment 0.01 · states quantitative result 0.05 · compares with prior results 0.04 · mentions code or data availability 0.02.

The abstract, sentence by sentence

arXiv listing abstract v2. Results and the main claim are underlined; show every role.

The mass of a bubble in a fluid can be taken as the mass of the vapor in it. The self-energy of the bubble is defined as the work performed against the pressure of the fluid in order to create the bubble. Taking the vapor to be an ideal gas the relationship between the self-energy, the mass of the bubble and the speed of the perturbation wave in a turbulent ideal fluid can be obtained.

background contribution method result interpretation main claim

What the classifier reports for each sentence. The number is the probability it gave the label.
#	sentence	role	the authors' own work	states the main conclusion	strength
s0	The mass of a bubble in a fluid can be taken as the mass of the vapor in it.	background 0.54	0.18	0.05	hedged possibility 0.57
s1	The self-energy of the bubble is defined as the work performed against the pressure of the fluid in order to create the bubble.	method 0.49	0.25	0.05	asserted 1.00
s2	Taking the vapor to be an ideal gas the relationship between the self-energy, the mass of the bubble and the speed of the perturbation wave in a turbulent ideal fluid can be obtained.	result 0.52	0.80	0.59	asserted 0.91

What kind of paper the classifier reports this is: theoretical result 1.00 · evidence from proof or derivation 1.00.

What was examined, and what was not

sources read

arXiv listing abstract v2

not examined

full text
figures and tables
references and citing sentences
relations to other papers

levels run

P abstract

not run

S sections
T body sentences
C citations
R cross-paper relations

Every quoted sentence on this record is the authors'. Every label is from Pith's closed vocabulary, with the probability shown beside it. Every number in the measurements table is the authors', checked by code. Record as JSON

References

Works cited (2)

[1]

arXiv:physics/0602142v2 [physics.gen-ph] 22 Feb 2006 The mass and energy of a vapor bubble in a turbulent ideal ﬂuid Valery P. Dmitriyev Lomonosov University P.O.Box 160, Moscow 117574, Russia ∗ (Dated: 1 February

View on Pith Pith review arXiv 2006
[2006]

The mass of a bubble in a ﬂuid can be taken as the mass of the vapo r in it. The self-energy of the bubble is deﬁned as the work performed against the pressu re of the ﬂuid in order to create the bubble. Taking the vapor to be an ideal gas the relationship b etween the self-energy, the mass of the bubble and the speed of the perturbation wave in a turbule...

View on Pith 1990

Discussion

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[1] [1]

arXiv:physics/0602142v2 [physics.gen-ph] 22 Feb 2006 The mass and energy of a vapor bubble in a turbulent ideal ﬂuid Valery P. Dmitriyev Lomonosov University P.O.Box 160, Moscow 117574, Russia ∗ (Dated: 1 February

View on Pith Pith review arXiv 2006

[2] [2006]

The mass of a bubble in a ﬂuid can be taken as the mass of the vapo r in it. The self-energy of the bubble is deﬁned as the work performed against the pressu re of the ﬂuid in order to create the bubble. Taking the vapor to be an ideal gas the relationship b etween the self-energy, the mass of the bubble and the speed of the perturbation wave in a turbule...

View on Pith 1990