Member Randy
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The Lorentz condition ( x’^2 – c^2 t’^2 = x^2 – c^2 t^2 ) is obtained from the spherical equations x^2 + y^2 + z^2 – (c*t)^2 = 0 and x’^2 + y’^2 + z’^2 – (c*t’)^2 = 0 by setting y’^2 + z’^2 = y^2 + z^2 and replacing them with their equalities (i.e., x^2 – c^2 t^2 = y^2 + z^2). The claim that the speed of light is the same in all reference frames is satisfied by the Lorentz condition because the final relationship [(0)(x^2 – c^2 t^2) = 0] is true regardless of the inputs. For example, let x = 50ct; substituting this into the Lorentz condition works because (0)(50^2(ct)^2 – ct) = 0. It is not the speed of light that is the same in all directions but that zero is zero in all directions. Essentially, the Lorentz condition is a tautology that works regardless of the inputs.
Are the laws of nature based upon tautologies?
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Sorry, the message didn’t post as it was written. Is there any way in this forum to post standard mathematical equations?
The message is posted below without HTML markups.
Perhaps we have different starting points because I cannot derive either the Lorentz or Galilean transformation from the equations without additional information. The equations, at least for the Lorentz transformation, can be either for a circle where r^2 = x^2 + y^2 or for a sphere where r^2 = x^2 + y^2 + z^2. Assuming a constant speed of light c where
x’ = α1 x + α2 t, y’ = y, z’ = z
t’ = β1 x + β2 t,
I arrive at the equation
(α1^2 – β1^2 c^2 – 1) x^2 – 2 (β1 β2 c^2 + α1^2 v) xt + (c^2 – β2^2 c^2 + α1^2 v^2) t^2 = 0
where v is the speed between reference frames. This equation is of the form A x^2 + B xt + C t^2 = 0 and cannot be solved without more information.
On the other hand, the values for the components from the Lorentz transformation are known as
α1 = β2 = γ and β1 = – γ v/c^2
where γ is the Lorentz factor. Substituting these values for the components produces
(0) x^2 – 2 (0) xt + (0) t^2 = 0
which is degenerate.
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Perhaps we have different starting points because I cannot derive either the Lorentz or Galilean transformation from the equations without additional information. The equations, at least for the Lorentz transformation, can be either for a circle where r^2 = x^2 + y^2 or for a sphere where r^2 = x^2 + y^2 + z^2. Assuming a constant speed of light c where
x’ = α<sub>1</sub>
x + α<sub>2</sub> t, y’ = y, z’ = zt’ = β<sub>1</sub>
x + β<sub>2</sub> t,I arrive at the equation
(α<sub>1</sub><sup>2</sup>
– β<sub>1</sub><sup>2</sup> c<sup>2</sup> – 1) x<sup>2</sup> – 2 (β<sub>1</sub> β<sub>2</sub> c<sup>2</sup> + α<sub>1</sub><sup>2</sup>
v) xt + (c<sup>2</sup> – β<sub>2</sub><sup>2</sup>
c<sup>2</sup> + α<sub>1</sub><sup>2</sup> v<sup>2</sup>) t<sup>2</sup> = 0where v is the speed between reference frames. This equation is of the form A x^2 + B xt + C t^2 = 0 and cannot be solved without more information.
On the other hand, the values for the components from the Lorentz transformation are known as
α<sub style=”background-color: transparent; font-family: inherit;”>1</sub>
= β<sub style=”background-color: transparent; font-family: inherit;”>2</sub> = γ and β<sub style=”background-color: transparent; font-family: inherit;”>1</sub>
= – γ v/c<sup style=”background-color: transparent; font-family: inherit;”>2</sup>where γ is the Lorentz factor. Substituting these values for the components produces
(0) x<sup>2</sup> – 2 (0) xt + (0) t<sup>2</sup> = 0
which is degenerate.
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To make sure I’m understanding your notation, is the r and ct in one reference frame such as the S frame and the R and cT in another reference frame such as the S’ frame?