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From: hertz778@gmail.com (rhertz)
Newsgroups: sci.physics.relativity
Subject: Re: How Einstein missed his opportunity to derive Lorentz  in Point
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Date: Mon, 10 Feb 2025 21:50:15 +0000
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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.