Mass and energy: two accounts, one subtraction

Why does the shared unknown disappear?

Follow one checked elimination through five small steps. The equations below are the same semantic records used by the term inspector. The checker verifies the algebra exactly; it does not prove conservation, the light-energy transformation, or the unchanged-offset premise.

Modern teaching derivation · editorial and physics review pending. No rest energy is assigned to the body, and no mass-energy conclusion is used as an input.

5 of 5 steps supported by this selection. The unchanged-offset premise is in use.

1. 1. Subtract the complete accounts

   Algebra checked; supported under the selected premises.

   $$\begin{array}{rl}& H_0-H_1-\left(E_0-E_1\right)\\ & =\gamma L-L\end{array}$$

   $$\begin{aligned}& H_0 - H_1 - \left(E_0 - E_1\right) \\ &\qquad \mathrel{=} \gamma\,L - L\end{aligned}$$

   H before minus H after, less E before minus E after, equals gamma L minus L.

   Subtract the rest-frame equality from the moving-frame equality, including both right-hand sides.

   Why this step is allowed

   The outer minus sign applies to E before minus E after as a whole. Its expansion is minus E before plus E after. No absolute body energy is assigned.

   Open the mathematical tool behind this step →

   Inspect the exact algebra certificate

   For each equation, form left minus right. The result at this step is exactly the following linear combination of earlier residuals, with no floating-point tolerance:

   • 1 × The same emission in the moving description
   • -1 × The energy removed in the rest description

   Matching these residuals proves a conditional implication, not the truth of the starting equations.

2. 2. Put the two descriptions beside each other

   Algebra checked; supported under the selected premises.

   $$\begin{array}{rl}& H_0-E_0-\left(H_1-E_1\right)\\ & =L\left(\gamma -1\right)\end{array}$$

   $$\begin{aligned}& H_0 - E_0 - \left(H_1 - E_1\right) \\ &\qquad \mathrel{=} L\,\left(\gamma - 1\right)\end{aligned}$$

   H before minus E before, less H after minus E after, equals L times gamma minus one.

   Reorder the left-hand terms into before and after frame differences. Factor gamma L minus L on the right.

   Why this step is allowed

   This is still only a comparison between frame accounts. Nothing in these two algebraic steps identifies either difference as energy of motion.

   Open the mathematical tool behind this step →

   Inspect the exact algebra certificate

   For each equation, form left minus right. The result at this step is exactly the following linear combination of earlier residuals, with no floating-point tolerance:

   • 1 × Subtract the rest account from the moving account

   Matching these residuals proves a conditional implication, not the truth of the starting equations.

3. 3. Use the additional offset premise · the additional premise

   Algebra checked; supported under the selected premises.

   $$\begin{array}{rl}& K_0+C-\left(K_1+C\right)\\ & =L\left(\gamma -1\right)\end{array}$$

   $$\begin{aligned}& K_0 + C - \left(K_1 + C\right) \\ &\qquad \mathrel{=} L\,\left(\gamma - 1\right)\end{aligned}$$

   K before plus C, less K after plus C, equals L times gamma minus one.

   Replace each H minus E with its corresponding K plus C. Keep the same C in both places.

   Why this step is allowed

   This is the consequential premise. If the offset can change with emission, the two frame differences cannot be replaced by kinetic energies plus a shared constant. The earlier conservation comparison remains valid without it.

   Open the mathematical tool behind this step →

   Inspect the exact algebra certificate

   For each equation, form left minus right. The result at this step is exactly the following linear combination of earlier residuals, with no floating-point tolerance:

   • 1 × Subtract the two energy accounts
   • -1 × State the offset premise before emission
   • 1 × State the offset premise after emission

   Matching these residuals proves a conditional implication, not the truth of the starting equations.

4. 4. Now cancel the shared unknown

   Algebra checked; supported under the selected premises.

   $$K_0-K_1=L\left(\gamma -1\right)$$

   $$K_0 - K_1 = L\,\left(\gamma - 1\right)$$

   K before minus K after equals L times gamma minus one.

   Expand the brackets. The first C is added and the same C is subtracted, so they cancel exactly.

   Why this step is allowed

   The checker matches canonical quantity identities, not similar-looking letters. Different before and after offsets would remain in the equation rather than disappear.

   Open the mathematical tool behind this step →

   Inspect the exact algebra certificate

   For each equation, form left minus right. The result at this step is exactly the following linear combination of earlier residuals, with no floating-point tolerance:

   • 1 × Substitute both offset premises before cancelling anything

   Matching these residuals proves a conditional implication, not the truth of the starting equations.

5. 5. Give the result a short name

   Algebra checked; supported under the selected premises.

   $$\Delta K=L\left(\gamma -1\right)$$

   $$\Delta K = L\,\left(\gamma - 1\right)$$

   The kinetic-energy drop, K before minus K after, equals L times gamma minus one.

   Use the declared definition Delta K = K before minus K after.

   Why this step is allowed

   The exact kinetic-energy drop is established under the stated premises. Identifying the mass decrease still requires the separate low-speed coefficient argument; a finite-speed ratio is not that limit.

   Open the mathematical tool behind this step →

   Inspect the exact algebra certificate

   For each equation, form left minus right. The result at this step is exactly the following linear combination of earlier residuals, with no floating-point tolerance:

   • 1 × Cancel the shared offset
   • 1 × Name the before-minus-after difference

   Matching these residuals proves a conditional implication, not the truth of the starting equations.

The mass conclusion is a further step. The exact energy drop is not yet a mass decrease. The low-speed comparison supplies the next premise and limit.

Continue through the checked low-speed limit → Continue to the low-speed coefficient laboratory → Read the coefficient argument →

Why does the coefficient become a mass decrease?

The ledger subtraction gives an energy difference. To identify inertia, we must take a low-speed limit and add the Newtonian meaning of mass. A single finite-speed calculation cannot replace either step.

Modern teaching derivation · editorial and physics review pending. Exact algebra and local Taylor coefficients are checked, not the truth of the physical premises.

Fix L > 0 and c > 0, hold both fixed, and take v/c toward zero through |v/c| < 1. Division by v² uses v ≠ 0.

The mass-decrease conclusion follows under all selected premises. The mathematical limit remains distinct from its physical interpretation. Changing the retained order does not change that limit.

1. Expand the factor, not an assumed body energy

   Supported under the stated conditions.

   $$\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$$

   $$\gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^{2}}}$$

   The Lorentz factor equals one divided by the square root of one minus the squared speed ratio.

   The expansion variable is the dimensionless ratio β = v/c. The positive radical is analytic near zero. Exact rational arithmetic gives the following coefficients of γ − 1; these are not fitted to plotted samples.

   Exact coefficients of each power of β, through order eight

   Power	Coefficient	Current truncation
   β0	0	Retained
   β1	0	Retained
   β2	1/2	Retained
   β3	0	Beyond selected order
   β4	3/8	First nonzero term omitted
   β5	0	Beyond selected order
   β6	5/16	Beyond selected order
   β7	0	Beyond selected order
   β8	35/128	Beyond selected order

   $$\gamma -1\approx \frac{1}{2}{\beta }^2$$

   $$\gamma-1\approx \frac{1}{2}\beta^{2}$$

   Keep powers through 2 in the dimensionless speed ratio beta, which means v divided by c. This is an approximation, not a finite-speed equality.

   The first nonzero omitted term has coefficient 3/8 and power 4. A local order statement is not a numerical error bound at an arbitrary finite speed.

   Open the Taylor-expansion tool →

2. Read the energy-of-motion coefficient

   Supported under the stated conditions.

   $$\Delta K=L\left(\gamma -1\right)$$

   $$\Delta K = L\,\left(\gamma - 1\right)$$

   The kinetic-energy drop, K before minus K after, equals L times gamma minus one.

   Using the already checked ledger argument, divide by the fixed positive emitted energy L. The first nonzero coefficient is exactly 1/2. Multiplying back by L makes the leading term one half (L/c²)v². This remains conditional on the unchanged-offset premise.

   $$\Delta K\approx 0.5L{\left(\frac{v}{c}\right)}^2$$

   $$\Delta K \approx 0.5\,L\,\left(\frac{v}{c}\right)^{2}$$

   The exact kinetic-energy drop is approximately one half L times the squared ratio v over c, at small speed.

   Inspect how the unknown body energies and shared offset were removed →

3. Divide only away from zero speed

   Supported under the stated conditions.

   $$m_{\mathrm{proxy}}:=\frac{2\Delta K}{v^2}$$

   $$m_{\mathrm{proxy}} := \frac{2\,\Delta K}{v^{2}}$$

   The finite-speed proxy is defined as twice the exact kinetic-energy drop divided by v squared.

   $$\begin{array}{rl}& \frac{2\Delta K}{v^2}=\frac{L}{c^2}\frac{2(\gamma -1)}{{\beta }^2},\\ & \beta =\frac{v}{c},v\ne 0\end{array}$$

   $$\begin{aligned}&\frac{2\Delta K}{v^2}=\frac{L}{c^2}\,\frac{2(\gamma-1)}{\beta^2},\\ &\beta=\frac{v}{c},\quad v\ne 0\end{aligned}$$

   For nonzero v, twice the kinetic-energy drop over v squared equals L over c squared times the dimensionless quotient twice the difference gamma minus one, divided by beta squared. The unchanged-offset premise is required for the kinetic interpretation.

   This exact rescaling uses β = v/c and v ≠ 0. It does not cancel a zero denominator. The finite-speed proxy retains its speed dependence; it is not yet the mass decrease.

4. Take the limit, not the value at zero

   Supported under the stated conditions.

   $$\underset{\beta \to 0}{\lim }\frac{2(\gamma -1)}{{\beta }^2}=1$$

   $$\lim_{\beta\to0}\frac{2(\gamma-1)}{\beta^2}=1$$

   The two-sided limit of the dimensionless quotient is exactly 1. The original quotient has no value at beta equal to zero.

   The constant and first-order coefficients of γ − 1 vanish exactly. Dividing by β² shifts the series by two powers; multiplying by two leaves constant term 1. All remaining terms tend to zero locally. This establishes the two-sided limit.

   At β = 0 the original quotient is still undefined. The analytic continuation has a value there, but it is not a measurement of the finite-speed proxy at zero.

   Inspect the exact regularized coefficients

   Coefficients from power zero through power 6: 1, 0, 3/4, 0, 5/8, 0, 35/64. The nonzero β² coefficient is why the finite-speed ratio is not identically its limit.

5. The move: read the coefficient as a lost mass

   Supported under the stated conditions.

   $$m_{\mathrm{loss}}=\frac{L}{c^2}$$

   $$m_{\mathrm{loss}} = \frac{L}{c^{2}}$$

   The positive inertial mass decrease equals L divided by c squared.

   The Newtonian low-speed meaning of inertial mass says that the coefficient of v² in the kinetic-energy decrease is one half the mass decrease. Comparing it with one half L/c² identifies the positive mass loss. It does not assign either absolute body energy or assume a rest-energy formula.

   Without that Newtonian premise, the mathematical limit survives but the identification as a mass decrease does not. Removing the unchanged-offset premise instead breaks the connection between the ledger difference and kinetic energy.

Compare the exact drop, finite-speed proxy and limit at 0.6c in the laboratory →

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