- #1

RandyReukauf

- 4

- 1

- TL;DR Summary
- Einstein's equations lead to a completely degenerate result that causes confusion and raises questions.

I was reading Einstein's

In his equation (8), Einstein gives the Lorentz values for x′ and t′. In equation (8a), he states the condition that

$$

x'^2−c^2t'^2=x^2−c^2t^2.

$$ Substituting the Lorentz values for x′ and t′ into equation (8a) will eventually result in

$$

(0)\, x^2 + 2\gamma^2\,(0)\, vtx - (0)\,c^2\, t^2 = 0.

$$ (My math is below.) This is a completely degenerate result. To me, it is similar to claiming that ##1=2## because ##(0)1=(0)2##.

Assuming that my math is correct, the result creates a discrepancy for me. First, I don't believe that the laws of nature permit completely degenerate relationships. Second, it also seems unlikely that this is oversight which has lasted for over a century.

Can someone please clarify this discrepancy?

----------------------------------------------------------------------------------------

My source is

$$

x' = \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}}\,, \qquad

t' = \frac{t - \frac{v}{c^2} x}{\sqrt{1 - \frac{v^2}{c^2}}}\,.

$$ For brevity, let

$$

\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}

\implies

\gamma^2 = \frac{1}{1 - \frac{v^2}{c^2}}

\,.

$$ Then his equation (8) can be expressed in a more compact form as

$$

(A) \qquad

x' = \gamma (x - vt), \qquad

t' = \gamma (t - \frac{v}{c^2} x)

\,.

$$ Next, Einstein states that this equation satisfies the condition as given in his equation (8a),

$$

x'^2 - c^2 t'^2 = x^2 - c^2 t^2

.

$$

The degenerate result is obtained as follows.

First, place all terms in Einstein's equation (8a) on one side to form the new equation

$$

(1) \qquad x'^2 - x^2 - c^2 (t'^2 - t^2) = 0

\,.

$$ Second, substitute the values for ##x'## and ##t'## from (A) into the new equation,

$$ (2) \qquad

\big(\gamma (x - vt)\big)^2 - x^2

- c^2\big(\big(\gamma (t - \frac{v}{c^2} x) \big)^2 - t^2 \big)\big)

= 0

\,.

$$ Third, expand the squares,

$$ (3) \qquad

\gamma^2 (x^2 - 2vtx + v^2 t^2) - x^2

- c^2\big(\gamma^2 (t^2 - 2\frac{v}{c^2} tx + \frac{v^2}{c^4}x^2) - t^2 \big)

= 0

\,.

$$ Then rearrange the equation in terms of ##x^2##, ##tx##, and ##t^2##,

$$ (4) \qquad

(\gamma^2 - \,\gamma^2 \frac{v^2}{c^2} - 1 )\, x^2

+ 2\gamma^2\,(1-1)\, vtx

- (\gamma^2 - \gamma^2 \frac{v^2}{c^2} - 1 )\,c^2\, t^2

= 0

\,.

$$ This leads to

$$

(5) \qquad

\big(\gamma^2 (1 - \frac{v^2}{c^2}) - 1 \big)\, x^2

+ 2\gamma^2\,(1-1)\, vtx

- \big(\gamma^2 (1 - \frac{v^2}{c^2}) - 1 \big)\,c^2\, t^2

= 0

\,.

$$ Finally, substitute for ##\gamma^2##,

$$

(6) \qquad

\big(\,\frac{1}{1 - \frac{v^2}{c^2}} \,(1 - \frac{v^2}{c^2}) - 1 \big)\, x^2

+ 2\gamma^2\,(1-1)\, vtx

- \big(\,\frac{1}{1 - \frac{v^2}{c^2}}\, (1 - \frac{v^2}{c^2}) - 1 \big)\,c^2\, t^2

= 0

\,.

$$ The result is a completely degenerate equation where all variables are multiplied by zero,

\begin{aligned}

(7) \qquad &(1 - 1)\, x^2 + 2\gamma^2\,(1-1)\, vtx - (1 - 1)\,c^2\, t^2 = 0 \\

\implies &(0)\, x^2 + 2\gamma^2\,(0)\, vtx - (0)\,c^2\, t^2 = 0

\,.

\end{aligned}

*Simple Derivation of the Lorentz Transformation*which is Appendix I in his book*Relativity: the Special & the General Theory*. (Online copies can be found at Bartleby's and the Gutenberg Project websites.) I came across an interesting but confusing result by using the values from one equation in another equation.In his equation (8), Einstein gives the Lorentz values for x′ and t′. In equation (8a), he states the condition that

$$

x'^2−c^2t'^2=x^2−c^2t^2.

$$ Substituting the Lorentz values for x′ and t′ into equation (8a) will eventually result in

$$

(0)\, x^2 + 2\gamma^2\,(0)\, vtx - (0)\,c^2\, t^2 = 0.

$$ (My math is below.) This is a completely degenerate result. To me, it is similar to claiming that ##1=2## because ##(0)1=(0)2##.

Assuming that my math is correct, the result creates a discrepancy for me. First, I don't believe that the laws of nature permit completely degenerate relationships. Second, it also seems unlikely that this is oversight which has lasted for over a century.

Can someone please clarify this discrepancy?

----------------------------------------------------------------------------------------

My source is

*Appendix I: Simple Derivation of the Lorentz Transformation*in Einstein's book*Relativity: the Special and the General Theory*. A completely degenerate result is obtained by substituting Einstein's equation (8) into his equation (8a). Einstein's equation (8) specifies the values of ##x'## and ##t'## as$$

x' = \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}}\,, \qquad

t' = \frac{t - \frac{v}{c^2} x}{\sqrt{1 - \frac{v^2}{c^2}}}\,.

$$ For brevity, let

$$

\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}

\implies

\gamma^2 = \frac{1}{1 - \frac{v^2}{c^2}}

\,.

$$ Then his equation (8) can be expressed in a more compact form as

$$

(A) \qquad

x' = \gamma (x - vt), \qquad

t' = \gamma (t - \frac{v}{c^2} x)

\,.

$$ Next, Einstein states that this equation satisfies the condition as given in his equation (8a),

$$

x'^2 - c^2 t'^2 = x^2 - c^2 t^2

.

$$

The degenerate result is obtained as follows.

First, place all terms in Einstein's equation (8a) on one side to form the new equation

$$

(1) \qquad x'^2 - x^2 - c^2 (t'^2 - t^2) = 0

\,.

$$ Second, substitute the values for ##x'## and ##t'## from (A) into the new equation,

$$ (2) \qquad

\big(\gamma (x - vt)\big)^2 - x^2

- c^2\big(\big(\gamma (t - \frac{v}{c^2} x) \big)^2 - t^2 \big)\big)

= 0

\,.

$$ Third, expand the squares,

$$ (3) \qquad

\gamma^2 (x^2 - 2vtx + v^2 t^2) - x^2

- c^2\big(\gamma^2 (t^2 - 2\frac{v}{c^2} tx + \frac{v^2}{c^4}x^2) - t^2 \big)

= 0

\,.

$$ Then rearrange the equation in terms of ##x^2##, ##tx##, and ##t^2##,

$$ (4) \qquad

(\gamma^2 - \,\gamma^2 \frac{v^2}{c^2} - 1 )\, x^2

+ 2\gamma^2\,(1-1)\, vtx

- (\gamma^2 - \gamma^2 \frac{v^2}{c^2} - 1 )\,c^2\, t^2

= 0

\,.

$$ This leads to

$$

(5) \qquad

\big(\gamma^2 (1 - \frac{v^2}{c^2}) - 1 \big)\, x^2

+ 2\gamma^2\,(1-1)\, vtx

- \big(\gamma^2 (1 - \frac{v^2}{c^2}) - 1 \big)\,c^2\, t^2

= 0

\,.

$$ Finally, substitute for ##\gamma^2##,

$$

(6) \qquad

\big(\,\frac{1}{1 - \frac{v^2}{c^2}} \,(1 - \frac{v^2}{c^2}) - 1 \big)\, x^2

+ 2\gamma^2\,(1-1)\, vtx

- \big(\,\frac{1}{1 - \frac{v^2}{c^2}}\, (1 - \frac{v^2}{c^2}) - 1 \big)\,c^2\, t^2

= 0

\,.

$$ The result is a completely degenerate equation where all variables are multiplied by zero,

\begin{aligned}

(7) \qquad &(1 - 1)\, x^2 + 2\gamma^2\,(1-1)\, vtx - (1 - 1)\,c^2\, t^2 = 0 \\

\implies &(0)\, x^2 + 2\gamma^2\,(0)\, vtx - (0)\,c^2\, t^2 = 0

\,.

\end{aligned}