Re: How Einstein missed his opportunity to derive Lorentz in Point §3.

I wanted to verify this approach using DeepSeek.
Even when there are small differences in the approach, like naming
terms, the results are coincident.
QUESTION: It's said that Einstein didn't need to use Taylor to derive
Lorentz transforms. Accepting that (x^2+y^2+z^2 - (ct)^2) =
(x'^2+y'^2+z'^2 - (ct')^2) =0 and that x'= k(x-vt) would do the trick.
Can you develop this approach?
DEEPSEEK AMSWER:
Absolutely! The approach you're describing is a clever and elegant way
to derive the Lorentz transformations without relying on Taylor
expansions or advanced mathematics. Let me walk you through this
step-by-step.
1. Key Assumptions
• Invariance of the speed of light: The speed of light c is the same in
all inertial frames.
• Linearity of transformations: The transformation between frames is
linear (this is reasonable for inertial frames).
• Relativity principle: The laws of physics are the same in all inertial
frames.
2. Setup
• Consider two inertial frames,  S and S′, where S′ is moving at a
constant velocity v relative to S along the x-axis.
• Assume the transformation for the x-coordinate is of the form:
x′ = k(x−vt)
where k is a constant to be determined.
3. Invariance of the Spacetime Interval for Light
• For light propagating along the x-axis, the spacetime interval in
frame S is:
x² − (ct)² = 0
• In frame S′, the interval must also be zero:
x′² − (ct′)² = 0
• Substitute x′=k(x−vt) into the equation for S′:
 [k(x − vt)]² − (ct′)² = 0
 k² (x − vt)² − (ct′)² = 0
4. Transformation for Time (t′)
• Assume the transformation for time is also linear and of the form:
t′ = at + bx
where a and b are constants to be determined.
5. Substitute t′ into the Interval Equation
• Substitute t′ = at + bx into the interval equation for S′:
• k² (x−vt)² − c² (at+bx)² = 0
• Expand both terms:
k² (x² − 2vxt + v²t²) − c² (a²t ² + 2abxt + b²x²) = 0
• Collect like terms:
(k²−c²b²) x ² + (−2k²v −2 c²ab) xt + (k²v²−c²a²) t² = 0
6. Equate Coefficients to Zero
• For the equation to hold for all x and t, the coefficients of x², xt,
and t² must each be zero:
1. k² − c²b² = 0 (coefficient of x ²)
2. −2k²v − 2c²ab = 0 (coefficient of xt)
3. k²v² − c²a² = 0 (coefficient of t²)
7. Solve for the Constants
• From the first equation:
k² = c²b²  --> b = k/c
• From the second equation:
−2k²v − 2c²ab = 0   -->  −2k²v − 2c²a(k/c) = 0
Simplify:
−2k²v − 2cak = 0  -->  −2k(kv + ca) = 0
Since k ≠ 0, we have:
kv + ca = 0    --> a = −kv/c
• From the third equation:
-2k²v² − c²a² = 0   --> k²v² = c²a²
Substitute a = −kv/c :
k²v² =c² (−kv/c)² --> k²v² = c² (k²v²/c²)
This simplifies to:
k²v² = k²v²
which is consistent.
8. Determine k Using the Relativity Principle
• The relativity principle requires that the transformation from S′ to S
should have the same form as the transformation from S to S′, but with v
replaced by −v.
• Applying this symmetry, we find:
k=1/ √(1− v²/c²)
This is the Lorentz factor γ.
9. Final Lorentz Transformations
• Substituting k=γ and the expressions for a and b, we obtain the
Lorentz transformations:
x′ = γ(x − vt),    t′ = γ (t−vx/c ²)
• The transformations for y and z remain unchanged:
y′ = y,   z′ = z
10. Summary
• By assuming linearity and using the invariance of the spacetime
interval for light, we derived the Lorentz transformations without
needing Taylor expansions or advanced mathematics.