Can someone help?

[Jack_]

I have the following equation for a co-located SR light pulse.

If the light postulate is applied in the rest frame,

$$ \sqrt{x^2+y^2+z^2} = (ct) $$

When that light beam intersects this equation
$$ \frac{(x - vr/c)^2}{r^2\gamma^2} + \frac{y^2}{r^2} + \frac{z^2}{r^2} = 1 $$


We have.
$$ x'^2+y^2+z^2 = (ct')^2=r^2 $$

 

[Pete]

I have the following equation for a co-located SR light pulse.

If the light postulate is applied in the rest frame,

Please use terms that people understand, Jack. Guessing what you mean is really getting old.

What's a "co-located light pulse"?
What's "the light postulate", and what does it mean to apply it?

But, it look a lot like we've [post=2484036]been here before[/post], that you're about to embark on yet another round of repeating yourself.
Do you have any specific questions about AlphaNumeric's description of how an expanding spherical light shell maps under a lorentz transform to an expanding spherical light shell?

 

[James R]

Jack_:

You haven't asked a question. Your thread is titled "Can someone help?"

What do you need help with?

If you can't express yourself clearly, I'll close the thread.

 

[rpenner]

In the OP, you just demonstrated (in shorthand, so it is not clear you grasp the implications of your algebra) that if one inertial observer decides light can get between two events in four-space that an inertial observed in relative motion will also agree that light can travel between the same two events, even when his rulers and clocks don't square with those of the first observer.

Somewhat more generally, if we have

$$\begin{bmatrix}c\,\Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{bmatrix}
= \begin{bmatrix}
\gamma&-\frac{\gamma v_x}{c}&-\frac{\gamma v_y}{c}&-\frac{\gamma v_z}{c}\\
-\frac{\gamma v_x}{c} & 1+\frac{(\gamma-1) v_x^2}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2}\\
-\frac{\gamma v_y}{c}&\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2}&1+\frac{(\gamma-1) v_y^2}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2}\\
-\frac{\gamma v_z}{c}&\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2}&1+\frac{(\gamma-1) v_z^2}{v_x^2+v_y^2+v_z^2} \end{bmatrix}
\begin{bmatrix} c\,\Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{bmatrix}$$

Then $$\begin{eqnarray}
(c\,\Delta t')^2 - (\Delta x')^2 - (\Delta y')^2 - (\Delta z')^2 & = & ( \gamma c\,\Delta t & - & \frac{\gamma v_x}{c} \Delta x & - & \frac{\gamma v_y}{c} \Delta y & - & \frac{\gamma v_z}{c} \Delta z )^2 \\
& - & ( -\frac{\gamma v_x}{c} c\,\Delta t & + & \Delta x & + & \frac{(\gamma-1) v_x^2}{v_x^2+v_y^2+v_z^2} \Delta x & + &\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2} \Delta y & + &\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2} \Delta z )^2 \\
& - & ( -\frac{\gamma v_y}{c}c\,\Delta t & + &\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2} \Delta x & + & \Delta y & + & \frac{(\gamma-1) v_y^2}{v_x^2+v_y^2+v_z^2} \Delta y & + &\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2} \Delta z )^2 \\
& - & ( -\frac{\gamma v_z}{c}c\,\Delta t & + &\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2} \Delta x & + &\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2} \Delta y & + & \Delta z & + & \frac{(\gamma-1) v_z^2}{v_x^2+v_y^2+v_z^2} \Delta z )^2 \\
& = & ( \gamma^2 (c\,\Delta t)^2 & + & \frac{\gamma^2 v_x^2}{c^2} (\Delta x)^2 & + & \frac{\gamma^2 v_y^2}{c^2} (\Delta y)^2 & + & \frac{\gamma^2 v_z^2}{c^2} (\Delta z)^2 & - & 2 \gamma^2 v_x \Delta t \Delta x & - & 2 \gamma^2 v_y \Delta t \Delta y & - & 2 \gamma^2 v_z \Delta t \Delta z & + & 2 \frac{\gamma^2 v_x v_y}{c^2} \Delta x \Delta y & + & 2 \frac{\gamma^2 v_x v_z}{c^2} \Delta x \Delta z & + & 2 \frac{\gamma^2 v_x v_y}{c^2} \Delta y \Delta z) \\
& - & ( \gamma^2 v_x^2 (\Delta t)^2 & + & \frac{((\gamma-1) v_x^2 + V^2)^2}{V^4} (\Delta x)^2 & + &\frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} ( \Delta y)^2 & + &\frac{(\gamma-1)^2 v_x^2 v_z^2}{V^4} (\Delta z)^2 & - & 2 \gamma v_x \frac{(\gamma-1) v_x^2 + V^2}{V^2} \Delta t \Delta x & - & 2 \frac{\gamma (\gamma-1) v_x^2 v_y}{V^2} \Delta t \Delta y & - & 2 \frac{\gamma (\gamma-1) v_x^2 v_z}{V^2} \Delta t \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2) v_x v_y}{V^4} \Delta x \Delta y & + & 2 \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2) v_x v_z}{V^4} \Delta x \Delta z & + & 2 \frac{(\gamma-1)^2 v_x^2 v_y v_z}{V^4} \Delta y \Delta z) \\
& - & ( \gamma^2 v_y^2 (\Delta t)^2 & + &\frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} ( \Delta x)^2 & + & \frac{((\gamma-1) v_y^2 + V^2)^2}{V^4} (\Delta y)^2 & + &\frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} (\Delta z)^2 & - & 2 \frac{\gamma (\gamma-1) v_x v_y^2}{V^2} \Delta t \Delta x & - & 2 \gamma v_y \frac{(\gamma-1) v_y^2 + V^2}{V^2} \Delta t \Delta y & - & 2 \frac{\gamma (\gamma-1) v_y^2 v_z}{V^2} \Delta t \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2) v_x v_y}{V^4} \Delta x \Delta y & + & 2 \frac{(\gamma-1)^2 v_x v_y^2 v_z}{V^4} \Delta x \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2) v_y v_z}{V^4} \Delta y \Delta z) \\
& - & ( \gamma^2 v_z^2 (\Delta t)^2 & + & \frac{(\gamma-1)^2 v_x^2 v_z^2}{V^4} ( \Delta x)^2 & + &\frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} ( \Delta y)^2 & + & \frac{((\gamma-1) v_z^2 + V^2)^2}{V^4} (\Delta z)^2 & - & 2 \frac{\gamma (\gamma-1) v_x v_z^2}{V^2} \Delta t \Delta x & - & 2 \frac{\gamma (\gamma-1) v_y v_z^2}{V^2} \Delta t \Delta y & - & 2 \gamma v_z \frac{(\gamma-1) v_z^2 + V^2}{V^2} \Delta t \Delta z & + & 2 \frac{(\gamma-1)^2 v_x v_y v_z^2}{V^4} \Delta x \Delta y & + & 2 \frac{(\gamma - 1)((\gamma-1) v_z^2 + V^2) v_x v_z}{V^4} \Delta x \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_z^2 + V^2) v_y v_z}{V^4} \Delta y \Delta z) \\
& = & \gamma^2 & \times & ( 1 & - & \frac{v_x^2}{c^2} & - & \frac{v_x^2}{c^2} & - & \frac{v_x^2}{c^2} ) & \times & (c \, \Delta t)^2 \\
& - & & & ( - \frac{\gamma^2 v_x^2}{c^2} & + & \frac{((\gamma-1) v_x^2 +V^2)^2}{V^4} & + & \frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} & + & \frac{(\gamma-1)^2 v_x^2 v_z^2}{V^4} ) & \times & (\Delta x)^2 \\
& - & & & ( - \frac{\gamma^2 v_y^2}{c^2} & + & \frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} & + & \frac{((\gamma-1) v_y^2 + V^2)^2}{V^4} & + & \frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} ) & \times & (\Delta y)^2 \\
& - & & & ( - \frac{\gamma^2 v_z^2}{c^2} & + & \frac{(\gamma-1) v_x^2 v_z^2 }{V^4} & + & \frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} & + & \frac{((\gamma-1) v_z^2 +V^2)^2}{V^4} ) & \times & (\Delta z)^2 \\
& + & 2 \gamma v_x & \times & ( - \gamma & + & \frac{(\gamma-1) v_x^2 + V^2}{V^2} & + & \frac{ (\gamma-1) v_y^2}{V^2} & + & \frac{(\gamma-1) v_z^2}{V^2} ) & \times & \Delta t \Delta x \\
& + & 2 \gamma v_y & \times & ( - \gamma & + & \frac{(\gamma-1) v_x^2}{V^2} & + & \frac{ (\gamma-1) v_y^2 + V^2}{V^2} & + & \frac{(\gamma-1) v_z^2}{V^2} ) & \times & \Delta t \Delta y \\
& + & 2 \gamma v_z & \times & ( - \gamma & + & \frac{(\gamma-1) v_x^2}{V^2} & + & \frac{ (\gamma-1) v_y^2}{V^2} & + & \frac{(\gamma-1) v_z^2 + V^2}{V^2} ) & \times & \Delta t \Delta z \\
& + & 2 v_x v_y & \times & ( \frac{\gamma^2}{c^2} & - & \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2)}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2)}{V^4} & - & \frac{(\gamma-1)^2 v_z^2}{V^4} ) & \times & \Delta x \Delta y \\
& + & 2 v_x v_z & \times & ( \frac{\gamma^2}{c^2} & - & \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2)}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2)}{V^4} & - & \frac{(\gamma-1)^2 v_z^2}{V^4} ) & \times & \Delta x \Delta z \\
& + & 2 v_y v_z & \times & ( \frac{\gamma^2}{c^2} & - & \frac{(\gamma-1)^2 v_x^2}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2)}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_z^2 + V^2)}{V^4} ) & \times & \Delta y \Delta z \\
& = & (c\,\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2
\end{eqnarray}$$

Naturally, $$ (c\,\Delta t')^2 - (\Delta x')^2 - (\Delta y')^2 - (\Delta z')^2 = (c\,\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2$$ is more powerful than your examples, because it holds for any two events as measured by any two inertial observers, while you take great pains to try to fix one point and time as the origin, use motion only parallel to the x axis, and work only with light.

 

[BenTheMan]

Pwnd

 

[Green Destiny]

Wow. That's some mean latexing.

 

[przyk]

$$\begin{bmatrix}c\,\Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{bmatrix}
= \begin{bmatrix}
\gamma&-\frac{\gamma v_x}{c}&-\frac{\gamma v_y}{c}&-\frac{\gamma v_z}{c}\\
-\frac{\gamma v_x}{c} & 1+\frac{(\gamma-1) v_x^2}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2}\\
-\frac{\gamma v_y}{c}&\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2}&1+\frac{(\gamma-1) v_y^2}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2}\\
-\frac{\gamma v_z}{c}&\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2}&1+\frac{(\gamma-1) v_z^2}{v_x^2+v_y^2+v_z^2} \end{bmatrix}
\begin{bmatrix} c\,\Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{bmatrix}$$

Then $$\begin{eqnarray}
(c\,\Delta t')^2 - (\Delta x')^2 - (\Delta y')^2 - (\Delta z')^2 & = & ( \gamma c\,\Delta t & - & \frac{\gamma v_x}{c} \Delta x & - & \frac{\gamma v_y}{c} \Delta y & - & \frac{\gamma v_z}{c} \Delta z )^2 \\
& - & ( -\frac{\gamma v_x}{c} c\,\Delta t & + & \Delta x & + & \frac{(\gamma-1) v_x^2}{v_x^2+v_y^2+v_z^2} \Delta x & + &\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2} \Delta y & + &\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2} \Delta z )^2 \\
& - & ( -\frac{\gamma v_y}{c}c\,\Delta t & + &\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2} \Delta x & + & \Delta y & + & \frac{(\gamma-1) v_y^2}{v_x^2+v_y^2+v_z^2} \Delta y & + &\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2} \Delta z )^2 \\
& - & ( -\frac{\gamma v_z}{c}c\,\Delta t & + &\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2} \Delta x & + &\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2} \Delta y & + & \Delta z & + & \frac{(\gamma-1) v_z^2}{v_x^2+v_y^2+v_z^2} \Delta z )^2 \\
& = & ( \gamma^2 (c\,\Delta t)^2 & + & \frac{\gamma^2 v_x^2}{c^2} (\Delta x)^2 & + & \frac{\gamma^2 v_y^2}{c^2} (\Delta y)^2 & + & \frac{\gamma^2 v_z^2}{c^2} (\Delta z)^2 & - & 2 \gamma^2 v_x \Delta t \Delta x & - & 2 \gamma^2 v_y \Delta t \Delta y & - & 2 \gamma^2 v_z \Delta t \Delta z & + & 2 \frac{\gamma^2 v_x v_y}{c^2} \Delta x \Delta y & + & 2 \frac{\gamma^2 v_x v_z}{c^2} \Delta x \Delta z & + & 2 \frac{\gamma^2 v_x v_y}{c^2} \Delta y \Delta z) \\
& - & ( \gamma^2 v_x^2 (\Delta t)^2 & + & \frac{((\gamma-1) v_x^2 + V^2)^2}{V^4} (\Delta x)^2 & + &\frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} ( \Delta y)^2 & + &\frac{(\gamma-1)^2 v_x^2 v_z^2}{V^4} (\Delta z)^2 & - & 2 \gamma v_x \frac{(\gamma-1) v_x^2 + V^2}{V^2} \Delta t \Delta x & - & 2 \frac{\gamma (\gamma-1) v_x^2 v_y}{V^2} \Delta t \Delta y & - & 2 \frac{\gamma (\gamma-1) v_x^2 v_z}{V^2} \Delta t \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2) v_x v_y}{V^4} \Delta x \Delta y & + & 2 \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2) v_x v_z}{V^4} \Delta x \Delta z & + & 2 \frac{(\gamma-1)^2 v_x^2 v_y v_z}{V^4} \Delta y \Delta z) \\
& - & ( \gamma^2 v_y^2 (\Delta t)^2 & + &\frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} ( \Delta x)^2 & + & \frac{((\gamma-1) v_y^2 + V^2)^2}{V^4} (\Delta y)^2 & + &\frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} (\Delta z)^2 & - & 2 \frac{\gamma (\gamma-1) v_x v_y^2}{V^2} \Delta t \Delta x & - & 2 \gamma v_y \frac{(\gamma-1) v_y^2 + V^2}{V^2} \Delta t \Delta y & - & 2 \frac{\gamma (\gamma-1) v_y^2 v_z}{V^2} \Delta t \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2) v_x v_y}{V^4} \Delta x \Delta y & + & 2 \frac{(\gamma-1)^2 v_x v_y^2 v_z}{V^4} \Delta x \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2) v_y v_z}{V^4} \Delta y \Delta z) \\
& - & ( \gamma^2 v_z^2 (\Delta t)^2 & + & \frac{(\gamma-1)^2 v_x^2 v_z^2}{V^4} ( \Delta x)^2 & + &\frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} ( \Delta y)^2 & + & \frac{((\gamma-1) v_z^2 + V^2)^2}{V^4} (\Delta z)^2 & - & 2 \frac{\gamma (\gamma-1) v_x v_z^2}{V^2} \Delta t \Delta x & - & 2 \frac{\gamma (\gamma-1) v_y v_z^2}{V^2} \Delta t \Delta y & - & 2 \gamma v_z \frac{(\gamma-1) v_z^2 + V^2}{V^2} \Delta t \Delta z & + & 2 \frac{(\gamma-1)^2 v_x v_y v_z^2}{V^4} \Delta x \Delta y & + & 2 \frac{(\gamma - 1)((\gamma-1) v_z^2 + V^2) v_x v_z}{V^4} \Delta x \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_z^2 + V^2) v_y v_z}{V^4} \Delta y \Delta z) \\
& = & \gamma^2 & \times & ( 1 & - & \frac{v_x^2}{c^2} & - & \frac{v_x^2}{c^2} & - & \frac{v_x^2}{c^2} ) & \times & (c \, \Delta t)^2 \\
& - & & & ( - \frac{\gamma^2 v_x^2}{c^2} & + & \frac{((\gamma-1) v_x^2 +V^2)^2}{V^4} & + & \frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} & + & \frac{(\gamma-1)^2 v_x^2 v_z^2}{V^4} ) & \times & (\Delta x)^2 \\
& - & & & ( - \frac{\gamma^2 v_y^2}{c^2} & + & \frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} & + & \frac{((\gamma-1) v_y^2 + V^2)^2}{V^4} & + & \frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} ) & \times & (\Delta y)^2 \\
& - & & & ( - \frac{\gamma^2 v_z^2}{c^2} & + & \frac{(\gamma-1) v_x^2 v_z^2 }{V^4} & + & \frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} & + & \frac{((\gamma-1) v_z^2 +V^2)^2}{V^4} ) & \times & (\Delta z)^2 \\
& + & 2 \gamma v_x & \times & ( - \gamma & + & \frac{(\gamma-1) v_x^2 + V^2}{V^2} & + & \frac{ (\gamma-1) v_y^2}{V^2} & + & \frac{(\gamma-1) v_z^2}{V^2} ) & \times & \Delta t \Delta x \\
& + & 2 \gamma v_y & \times & ( - \gamma & + & \frac{(\gamma-1) v_x^2}{V^2} & + & \frac{ (\gamma-1) v_y^2 + V^2}{V^2} & + & \frac{(\gamma-1) v_z^2}{V^2} ) & \times & \Delta t \Delta y \\
& + & 2 \gamma v_z & \times & ( - \gamma & + & \frac{(\gamma-1) v_x^2}{V^2} & + & \frac{ (\gamma-1) v_y^2}{V^2} & + & \frac{(\gamma-1) v_z^2 + V^2}{V^2} ) & \times & \Delta t \Delta z \\
& + & 2 v_x v_y & \times & ( \frac{\gamma^2}{c^2} & - & \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2)}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2)}{V^4} & - & \frac{(\gamma-1)^2 v_z^2}{V^4} ) & \times & \Delta x \Delta y \\
& + & 2 v_x v_z & \times & ( \frac{\gamma^2}{c^2} & - & \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2)}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2)}{V^4} & - & \frac{(\gamma-1)^2 v_z^2}{V^4} ) & \times & \Delta x \Delta z \\
& + & 2 v_y v_z & \times & ( \frac{\gamma^2}{c^2} & - & \frac{(\gamma-1)^2 v_x^2}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2)}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_z^2 + V^2)}{V^4} ) & \times & \Delta y \Delta z \\
& = & (c\,\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2
\end{eqnarray}$$

I hope you somehow got your computer to spit that out for you.

 

[Jack_]

Jack_:

You haven't asked a question. Your thread is titled "Can someone help?"

What do you need help with?

If you can't express yourself clearly, I'll close the thread.

Oh, I wanted to know if the OP is true.

In other words, since it is math, one would need to provide a counter example in math to refute it.

I have not seen that yet.

The OP was very specific mathermatically.

So, I was hoping someone could claim here are your equations and here is a specific example where there are false.

 

[Jack_]

In the OP, you just demonstrated (in shorthand, so it is not clear you grasp the implications of your algebra) that if one inertial observer decides light can get between two events in four-space that an inertial observed in relative motion will also agree that light can travel between the same two events, even when his rulers and clocks don't square with those of the first observer.

Somewhat more generally, if we have

$$\begin{bmatrix}c\,\Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{bmatrix}
= \begin{bmatrix}
\gamma&-\frac{\gamma v_x}{c}&-\frac{\gamma v_y}{c}&-\frac{\gamma v_z}{c}\\
-\frac{\gamma v_x}{c} & 1+\frac{(\gamma-1) v_x^2}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2}\\
-\frac{\gamma v_y}{c}&\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2}&1+\frac{(\gamma-1) v_y^2}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2}\\
-\frac{\gamma v_z}{c}&\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2}&1+\frac{(\gamma-1) v_z^2}{v_x^2+v_y^2+v_z^2} \end{bmatrix}
\begin{bmatrix} c\,\Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{bmatrix}$$

Then $$\begin{eqnarray}
(c\,\Delta t')^2 - (\Delta x')^2 - (\Delta y')^2 - (\Delta z')^2 & = & ( \gamma c\,\Delta t & - & \frac{\gamma v_x}{c} \Delta x & - & \frac{\gamma v_y}{c} \Delta y & - & \frac{\gamma v_z}{c} \Delta z )^2 \\
& - & ( -\frac{\gamma v_x}{c} c\,\Delta t & + & \Delta x & + & \frac{(\gamma-1) v_x^2}{v_x^2+v_y^2+v_z^2} \Delta x & + &\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2} \Delta y & + &\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2} \Delta z )^2 \\
& - & ( -\frac{\gamma v_y}{c}c\,\Delta t & + &\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2} \Delta x & + & \Delta y & + & \frac{(\gamma-1) v_y^2}{v_x^2+v_y^2+v_z^2} \Delta y & + &\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2} \Delta z )^2 \\
& - & ( -\frac{\gamma v_z}{c}c\,\Delta t & + &\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2} \Delta x & + &\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2} \Delta y & + & \Delta z & + & \frac{(\gamma-1) v_z^2}{v_x^2+v_y^2+v_z^2} \Delta z )^2 \\
& = & ( \gamma^2 (c\,\Delta t)^2 & + & \frac{\gamma^2 v_x^2}{c^2} (\Delta x)^2 & + & \frac{\gamma^2 v_y^2}{c^2} (\Delta y)^2 & + & \frac{\gamma^2 v_z^2}{c^2} (\Delta z)^2 & - & 2 \gamma^2 v_x \Delta t \Delta x & - & 2 \gamma^2 v_y \Delta t \Delta y & - & 2 \gamma^2 v_z \Delta t \Delta z & + & 2 \frac{\gamma^2 v_x v_y}{c^2} \Delta x \Delta y & + & 2 \frac{\gamma^2 v_x v_z}{c^2} \Delta x \Delta z & + & 2 \frac{\gamma^2 v_x v_y}{c^2} \Delta y \Delta z) \\
& - & ( \gamma^2 v_x^2 (\Delta t)^2 & + & \frac{((\gamma-1) v_x^2 + V^2)^2}{V^4} (\Delta x)^2 & + &\frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} ( \Delta y)^2 & + &\frac{(\gamma-1)^2 v_x^2 v_z^2}{V^4} (\Delta z)^2 & - & 2 \gamma v_x \frac{(\gamma-1) v_x^2 + V^2}{V^2} \Delta t \Delta x & - & 2 \frac{\gamma (\gamma-1) v_x^2 v_y}{V^2} \Delta t \Delta y & - & 2 \frac{\gamma (\gamma-1) v_x^2 v_z}{V^2} \Delta t \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2) v_x v_y}{V^4} \Delta x \Delta y & + & 2 \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2) v_x v_z}{V^4} \Delta x \Delta z & + & 2 \frac{(\gamma-1)^2 v_x^2 v_y v_z}{V^4} \Delta y \Delta z) \\
& - & ( \gamma^2 v_y^2 (\Delta t)^2 & + &\frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} ( \Delta x)^2 & + & \frac{((\gamma-1) v_y^2 + V^2)^2}{V^4} (\Delta y)^2 & + &\frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} (\Delta z)^2 & - & 2 \frac{\gamma (\gamma-1) v_x v_y^2}{V^2} \Delta t \Delta x & - & 2 \gamma v_y \frac{(\gamma-1) v_y^2 + V^2}{V^2} \Delta t \Delta y & - & 2 \frac{\gamma (\gamma-1) v_y^2 v_z}{V^2} \Delta t \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2) v_x v_y}{V^4} \Delta x \Delta y & + & 2 \frac{(\gamma-1)^2 v_x v_y^2 v_z}{V^4} \Delta x \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2) v_y v_z}{V^4} \Delta y \Delta z) \\
& - & ( \gamma^2 v_z^2 (\Delta t)^2 & + & \frac{(\gamma-1)^2 v_x^2 v_z^2}{V^4} ( \Delta x)^2 & + &\frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} ( \Delta y)^2 & + & \frac{((\gamma-1) v_z^2 + V^2)^2}{V^4} (\Delta z)^2 & - & 2 \frac{\gamma (\gamma-1) v_x v_z^2}{V^2} \Delta t \Delta x & - & 2 \frac{\gamma (\gamma-1) v_y v_z^2}{V^2} \Delta t \Delta y & - & 2 \gamma v_z \frac{(\gamma-1) v_z^2 + V^2}{V^2} \Delta t \Delta z & + & 2 \frac{(\gamma-1)^2 v_x v_y v_z^2}{V^4} \Delta x \Delta y & + & 2 \frac{(\gamma - 1)((\gamma-1) v_z^2 + V^2) v_x v_z}{V^4} \Delta x \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_z^2 + V^2) v_y v_z}{V^4} \Delta y \Delta z) \\
& = & \gamma^2 & \times & ( 1 & - & \frac{v_x^2}{c^2} & - & \frac{v_x^2}{c^2} & - & \frac{v_x^2}{c^2} ) & \times & (c \, \Delta t)^2 \\
& - & & & ( - \frac{\gamma^2 v_x^2}{c^2} & + & \frac{((\gamma-1) v_x^2 +V^2)^2}{V^4} & + & \frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} & + & \frac{(\gamma-1)^2 v_x^2 v_z^2}{V^4} ) & \times & (\Delta x)^2 \\
& - & & & ( - \frac{\gamma^2 v_y^2}{c^2} & + & \frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} & + & \frac{((\gamma-1) v_y^2 + V^2)^2}{V^4} & + & \frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} ) & \times & (\Delta y)^2 \\
& - & & & ( - \frac{\gamma^2 v_z^2}{c^2} & + & \frac{(\gamma-1) v_x^2 v_z^2 }{V^4} & + & \frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} & + & \frac{((\gamma-1) v_z^2 +V^2)^2}{V^4} ) & \times & (\Delta z)^2 \\
& + & 2 \gamma v_x & \times & ( - \gamma & + & \frac{(\gamma-1) v_x^2 + V^2}{V^2} & + & \frac{ (\gamma-1) v_y^2}{V^2} & + & \frac{(\gamma-1) v_z^2}{V^2} ) & \times & \Delta t \Delta x \\
& + & 2 \gamma v_y & \times & ( - \gamma & + & \frac{(\gamma-1) v_x^2}{V^2} & + & \frac{ (\gamma-1) v_y^2 + V^2}{V^2} & + & \frac{(\gamma-1) v_z^2}{V^2} ) & \times & \Delta t \Delta y \\
& + & 2 \gamma v_z & \times & ( - \gamma & + & \frac{(\gamma-1) v_x^2}{V^2} & + & \frac{ (\gamma-1) v_y^2}{V^2} & + & \frac{(\gamma-1) v_z^2 + V^2}{V^2} ) & \times & \Delta t \Delta z \\
& + & 2 v_x v_y & \times & ( \frac{\gamma^2}{c^2} & - & \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2)}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2)}{V^4} & - & \frac{(\gamma-1)^2 v_z^2}{V^4} ) & \times & \Delta x \Delta y \\
& + & 2 v_x v_z & \times & ( \frac{\gamma^2}{c^2} & - & \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2)}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2)}{V^4} & - & \frac{(\gamma-1)^2 v_z^2}{V^4} ) & \times & \Delta x \Delta z \\
& + & 2 v_y v_z & \times & ( \frac{\gamma^2}{c^2} & - & \frac{(\gamma-1)^2 v_x^2}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2)}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_z^2 + V^2)}{V^4} ) & \times & \Delta y \Delta z \\
& = & (c\,\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2
\end{eqnarray}$$

Naturally, $$ (c\,\Delta t')^2 - (\Delta x')^2 - (\Delta y')^2 - (\Delta z')^2 = (c\,\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2$$ is more powerful than your examples, because it holds for any two events as measured by any two inertial observers, while you take great pains to try to fix one point and time as the origin, use motion only parallel to the x axis, and work only with light.

We do not need this matrix since the example is in standard configuration.

To handle the equations, you solve both for y² + z².

Then use the quadratic formula to solve for t.

Once you see the solution for t, you will note it is the LT equation for t' and t in the context of the stationary frame which of course all the equations operate in.

So, if the OP is false, then LT is false.

 

[Jack_]

Pwnd

What does this mean?

 

[Jack_]

Please use terms that people understand, Jack. Guessing what you mean is really getting old.

What's a "co-located light pulse"?
What's "the light postulate", and what does it mean to apply it?

But, it look a lot like we've [post=2484036]been here before[/post], that you're about to embark on yet another round of repeating yourself.
Do you have any specific questions about AlphaNumeric's description of how an expanding spherical light shell maps under a lorentz transform to an expanding spherical light shell?

I stated the light postulate in the frame mathematically. It is clear.
A co-located light pulse means the origins of the frames are co-located and a light pulse is emitted thence.


Where is AN's math? I did not see it.

Does it mean what is simultaneous to O will also be simultaneous to O'? It sounds like it. How does that work with R of S? Do you know?

 

[Green Destiny]

What does this mean?

It's basically a rudimentary replacement of saying, you've been ''learned'', using bad english.

 

[Jack_]

In the OP, you just demonstrated (in shorthand, so it is not clear you grasp the implications of your algebra) that if one inertial observer decides light can get between two events in four-space that an inertial observed in relative motion will also agree that light can travel between the same two events, even when his rulers and clocks don't square with those of the first observer.

Somewhat more generally, if we have

$$\begin{bmatrix}c\,\Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{bmatrix}
= \begin{bmatrix}
\gamma&-\frac{\gamma v_x}{c}&-\frac{\gamma v_y}{c}&-\frac{\gamma v_z}{c}\\
-\frac{\gamma v_x}{c} & 1+\frac{(\gamma-1) v_x^2}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2}\\
-\frac{\gamma v_y}{c}&\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2}&1+\frac{(\gamma-1) v_y^2}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2}\\
-\frac{\gamma v_z}{c}&\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2}&\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2}&1+\frac{(\gamma-1) v_z^2}{v_x^2+v_y^2+v_z^2} \end{bmatrix}
\begin{bmatrix} c\,\Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{bmatrix}$$

Then $$\begin{eqnarray}
(c\,\Delta t')^2 - (\Delta x')^2 - (\Delta y')^2 - (\Delta z')^2 & = & ( \gamma c\,\Delta t & - & \frac{\gamma v_x}{c} \Delta x & - & \frac{\gamma v_y}{c} \Delta y & - & \frac{\gamma v_z}{c} \Delta z )^2 \\
& - & ( -\frac{\gamma v_x}{c} c\,\Delta t & + & \Delta x & + & \frac{(\gamma-1) v_x^2}{v_x^2+v_y^2+v_z^2} \Delta x & + &\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2} \Delta y & + &\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2} \Delta z )^2 \\
& - & ( -\frac{\gamma v_y}{c}c\,\Delta t & + &\frac{(\gamma-1) v_x v_y}{v_x^2+v_y^2+v_z^2} \Delta x & + & \Delta y & + & \frac{(\gamma-1) v_y^2}{v_x^2+v_y^2+v_z^2} \Delta y & + &\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2} \Delta z )^2 \\
& - & ( -\frac{\gamma v_z}{c}c\,\Delta t & + &\frac{(\gamma-1) v_x v_z}{v_x^2+v_y^2+v_z^2} \Delta x & + &\frac{(\gamma-1) v_y v_z}{v_x^2+v_y^2+v_z^2} \Delta y & + & \Delta z & + & \frac{(\gamma-1) v_z^2}{v_x^2+v_y^2+v_z^2} \Delta z )^2 \\
& = & ( \gamma^2 (c\,\Delta t)^2 & + & \frac{\gamma^2 v_x^2}{c^2} (\Delta x)^2 & + & \frac{\gamma^2 v_y^2}{c^2} (\Delta y)^2 & + & \frac{\gamma^2 v_z^2}{c^2} (\Delta z)^2 & - & 2 \gamma^2 v_x \Delta t \Delta x & - & 2 \gamma^2 v_y \Delta t \Delta y & - & 2 \gamma^2 v_z \Delta t \Delta z & + & 2 \frac{\gamma^2 v_x v_y}{c^2} \Delta x \Delta y & + & 2 \frac{\gamma^2 v_x v_z}{c^2} \Delta x \Delta z & + & 2 \frac{\gamma^2 v_x v_y}{c^2} \Delta y \Delta z) \\
& - & ( \gamma^2 v_x^2 (\Delta t)^2 & + & \frac{((\gamma-1) v_x^2 + V^2)^2}{V^4} (\Delta x)^2 & + &\frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} ( \Delta y)^2 & + &\frac{(\gamma-1)^2 v_x^2 v_z^2}{V^4} (\Delta z)^2 & - & 2 \gamma v_x \frac{(\gamma-1) v_x^2 + V^2}{V^2} \Delta t \Delta x & - & 2 \frac{\gamma (\gamma-1) v_x^2 v_y}{V^2} \Delta t \Delta y & - & 2 \frac{\gamma (\gamma-1) v_x^2 v_z}{V^2} \Delta t \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2) v_x v_y}{V^4} \Delta x \Delta y & + & 2 \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2) v_x v_z}{V^4} \Delta x \Delta z & + & 2 \frac{(\gamma-1)^2 v_x^2 v_y v_z}{V^4} \Delta y \Delta z) \\
& - & ( \gamma^2 v_y^2 (\Delta t)^2 & + &\frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} ( \Delta x)^2 & + & \frac{((\gamma-1) v_y^2 + V^2)^2}{V^4} (\Delta y)^2 & + &\frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} (\Delta z)^2 & - & 2 \frac{\gamma (\gamma-1) v_x v_y^2}{V^2} \Delta t \Delta x & - & 2 \gamma v_y \frac{(\gamma-1) v_y^2 + V^2}{V^2} \Delta t \Delta y & - & 2 \frac{\gamma (\gamma-1) v_y^2 v_z}{V^2} \Delta t \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2) v_x v_y}{V^4} \Delta x \Delta y & + & 2 \frac{(\gamma-1)^2 v_x v_y^2 v_z}{V^4} \Delta x \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2) v_y v_z}{V^4} \Delta y \Delta z) \\
& - & ( \gamma^2 v_z^2 (\Delta t)^2 & + & \frac{(\gamma-1)^2 v_x^2 v_z^2}{V^4} ( \Delta x)^2 & + &\frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} ( \Delta y)^2 & + & \frac{((\gamma-1) v_z^2 + V^2)^2}{V^4} (\Delta z)^2 & - & 2 \frac{\gamma (\gamma-1) v_x v_z^2}{V^2} \Delta t \Delta x & - & 2 \frac{\gamma (\gamma-1) v_y v_z^2}{V^2} \Delta t \Delta y & - & 2 \gamma v_z \frac{(\gamma-1) v_z^2 + V^2}{V^2} \Delta t \Delta z & + & 2 \frac{(\gamma-1)^2 v_x v_y v_z^2}{V^4} \Delta x \Delta y & + & 2 \frac{(\gamma - 1)((\gamma-1) v_z^2 + V^2) v_x v_z}{V^4} \Delta x \Delta z & + & 2 \frac{(\gamma - 1)((\gamma-1) v_z^2 + V^2) v_y v_z}{V^4} \Delta y \Delta z) \\
& = & \gamma^2 & \times & ( 1 & - & \frac{v_x^2}{c^2} & - & \frac{v_x^2}{c^2} & - & \frac{v_x^2}{c^2} ) & \times & (c \, \Delta t)^2 \\
& - & & & ( - \frac{\gamma^2 v_x^2}{c^2} & + & \frac{((\gamma-1) v_x^2 +V^2)^2}{V^4} & + & \frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} & + & \frac{(\gamma-1)^2 v_x^2 v_z^2}{V^4} ) & \times & (\Delta x)^2 \\
& - & & & ( - \frac{\gamma^2 v_y^2}{c^2} & + & \frac{(\gamma-1)^2 v_x^2 v_y^2}{V^4} & + & \frac{((\gamma-1) v_y^2 + V^2)^2}{V^4} & + & \frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} ) & \times & (\Delta y)^2 \\
& - & & & ( - \frac{\gamma^2 v_z^2}{c^2} & + & \frac{(\gamma-1) v_x^2 v_z^2 }{V^4} & + & \frac{(\gamma-1)^2 v_y^2 v_z^2}{V^4} & + & \frac{((\gamma-1) v_z^2 +V^2)^2}{V^4} ) & \times & (\Delta z)^2 \\
& + & 2 \gamma v_x & \times & ( - \gamma & + & \frac{(\gamma-1) v_x^2 + V^2}{V^2} & + & \frac{ (\gamma-1) v_y^2}{V^2} & + & \frac{(\gamma-1) v_z^2}{V^2} ) & \times & \Delta t \Delta x \\
& + & 2 \gamma v_y & \times & ( - \gamma & + & \frac{(\gamma-1) v_x^2}{V^2} & + & \frac{ (\gamma-1) v_y^2 + V^2}{V^2} & + & \frac{(\gamma-1) v_z^2}{V^2} ) & \times & \Delta t \Delta y \\
& + & 2 \gamma v_z & \times & ( - \gamma & + & \frac{(\gamma-1) v_x^2}{V^2} & + & \frac{ (\gamma-1) v_y^2}{V^2} & + & \frac{(\gamma-1) v_z^2 + V^2}{V^2} ) & \times & \Delta t \Delta z \\
& + & 2 v_x v_y & \times & ( \frac{\gamma^2}{c^2} & - & \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2)}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2)}{V^4} & - & \frac{(\gamma-1)^2 v_z^2}{V^4} ) & \times & \Delta x \Delta y \\
& + & 2 v_x v_z & \times & ( \frac{\gamma^2}{c^2} & - & \frac{(\gamma - 1)((\gamma-1) v_x^2 + V^2)}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2)}{V^4} & - & \frac{(\gamma-1)^2 v_z^2}{V^4} ) & \times & \Delta x \Delta z \\
& + & 2 v_y v_z & \times & ( \frac{\gamma^2}{c^2} & - & \frac{(\gamma-1)^2 v_x^2}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_y^2 + V^2)}{V^4} & - & \frac{(\gamma - 1)((\gamma-1) v_z^2 + V^2)}{V^4} ) & \times & \Delta y \Delta z \\
& = & (c\,\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2
\end{eqnarray}$$

Naturally, $$ (c\,\Delta t')^2 - (\Delta x')^2 - (\Delta y')^2 - (\Delta z')^2 = (c\,\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2$$ is more powerful than your examples, because it holds for any two events as measured by any two inertial observers, while you take great pains to try to fix one point and time as the origin, use motion only parallel to the x axis, and work only with light.

I have a question for you in your equations.

I want to understand how well you know math and logic.

When you select points from a frame and apply LT and then create equations off the LT results, you are supposed to indicate to the audience with those "LT results" they are bound variables. So, for example, if x' is calculated by LT based on a free variable x, then it is a bound variable based on the free variable x.

So if you list an equation of t' and x' based on x and t, then x' and t' are not free variables and this can lead to incorrect results and proofs.

Differently, when you write an equation based x' and t' that makes them appear as free variabes and they are in fact bound based on another domain set, you are committing a logical fallacy in proof theory.

This fallacy pretends the domain of the bound equation is only the variables of the and parameters of the equation when in fact the variables are bound by some other domain set and equation.

You do know this right?

 

[Jack_]

So rpenner, for example,
when we write
$$ x'^2+y^2+z^2 = (ct')^2=r^2 $$

and we have selected the domain set from the stationary frame, mathematical logic commands we provide an audit trail of our work. If we do not, we are a fraud.

Hence, assuming the light sphere in the moving frame acquired a radius r,

$$ x'^2+y^2+z^2 = r^2 $$

Finally, by LT,

$$ x' = ( x - vt )\gamma $$

So, if we use x' based on coordinates in the stationary frame, mathematical logic commands we correctly write out work.

Thus,

$$ (( x - vt )\gamma)^2+y^2+z^2 = r^2 $$

So, your equation of $$ x'^2+y^2+z^2 = r^2 $$ based on points selected in the stationary frame is a logical fallacy.

The correct equation is
$$ (( x - vt )\gamma)^2+y^2+z^2 = r^2 $$

 

[Pete]

I stated the light postulate in the frame mathematically. It is clear.

No, Jack. You are anything but clear.

A co-located light pulse means the origins of the frames are co-located and a light pulse is emitted thence.

Great. Do you realise that your shorthand is very obscure? That no one except yourself uses "co-located light pulse" to mean that?

Where is AN's math? I did not see it.

You've got to be joking, Jack. Do you just not read people's responses to you? Do you just not read your own threads?
Remember the [thread=99729]Twins thread[/thread] you seem to think is a relativity killer? Go read it. In the first 50 posts, your misunderstandings were clearly explained mathematically maybe half a dozen times.

Does it mean what is simultaneous to O will also be simultaneous to O'? It sounds like it.

No, Jack. If you ever understand why, perhaps we can all move on.

 

[rpenner]

I know that those that copy whole posts twice without responding to the content at a level of detail which requires whole copies are boorish.

I know that I selected no points. I know that the Lorentz Transform is a homogeneous transformation and is not dependent on the choice of origin. I named the difference in coordinates between two events in four-dimensional Minkowski space, but left the selection of the events up to the reader.
I am not concerned with x or x' but only with $$\Delta x$$ and $$\Delta x'$$.
As you have not identified the subject matter, of course you have nothing to say on free versus bound variables. In the expression $$x^2 - y^2$$, neither variable is logically bound, the value of the expression depends on x and on y, but neither is bound. Consequently the expression is still valid if x = y. Even the pair of events which is the actual topic of discussion, is fully free.

I could bind variables by writing $$\forall A \forall B (c \Delta t'_{AB})^2 - (c \Delta x'_{AB})^2 - (c \Delta y'_{AB})^2 - (c \Delta z'_{AB})^2 = (c \Delta t_{AB})^2 - (c \Delta x_{AB})^2 - (c \Delta y_{AB})^2 - (c \Delta z_{AB})^2$$ and then everything would be bound, but that is already implied by my free statement and the axiom or rule of generalization.

You have confused predicate logic with real algebra. What you call a fallacy is a symptom of your incorrect learning and not a statement of objective truth.

The statement $$x \in \left\{ \frac{-a-\sqrt{a^2+4b}}{2} \, , \, \frac{-a+\sqrt{a^2+4b}}{2} \right\} \rightarrow x^2 + a x = b$$ is a statement of propositional logic with no bound variables.

 

[Pete]

I want to understand how well you know math and logic.

Jack, you have a great opportunity at sciforums to learn from some very well educated people.

Why are you not open to learning? Why do you approach any disagreement as a fight? Why so antagonistic?

Why not approach disagreements as an opportunity for learning?
Maybe it will be the other poster who learns something, maybe it will be you, maybe both.

If you don't consider the possibility that you're sometimes wrong, you'll never grow.



It's pretty sad, really. It's very clear that you have a lot to learn, and that rpenner and AlphaNumeric (maybe even myself) could help you... but you just don't want to. You don't want to become right... you just want be right.

 

[Jack_]

I know that those that copy whole posts twice without responding to the content at a level of detail which requires whole copies are boorish.

I know that I selected no points. I know that the Lorentz Transform is a homogeneous transformation and is not dependent on the choice of origin. I named the difference in coordinates between two events in four-dimensional Minkowski space, but left the selection of the events up to the reader.
I am not concerned with x or x' but only with $$\Delta x$$ and $$\Delta x'$$.
As you have not identified the subject matter, of course you have nothing to say on free versus bound variables. In the expression $$x^2 - y^2$$, neither variable is logically bound, the value of the expression depends on x and on y, but neither is bound. Consequently the expression is still valid if x = y. Even the pair of events which is the actual topic of discussion, is fully free.

I could bind variables by writing $$\forall A \forall B (c \Delta t'_{AB})^2 - (c \Delta x'_{AB})^2 - (c \Delta y'_{AB})^2 - (c \Delta z'_{AB})^2 = (c \Delta t_{AB})^2 - (c \Delta x_{AB})^2 - (c \Delta y_{AB})^2 - (c \Delta z_{AB})^2$$ and then everything would be bound, but that is already implied by my free statement and the axiom or rule of generalization.

You have confused predicate logic with real algebra. What you call a fallacy is a symptom of your incorrect learning and not a statement of objective truth.

The statement $$x \in \left\{ \frac{-a-\sqrt{a^2+4b}}{2} \, , \, \frac{-a+\sqrt{a^2+4b}}{2} \right\} \rightarrow x^2 + a x = b$$ is a first-order logical statement with no bound variables.

Ah, what is your domain for x'? Since you are mathematical, you will reveal it. If you fail to, you are committing mathematical fraud.

Further, let me try to explain.

Suppose you have the general equation x' + 2 = y.

Once would suppose at some y would be even.

But, if x' = 2x+1, where x is a natural number, y can never be even. x' is a bound variable.

That is why you must disclose your domain set or you commit a logical fallacy.

So, the correct equation is

(2x+1) + 2 = y where x is a natural number.

Now, we know y cannot be an even number.

What is your domain set rpenner?

 

[Jack_]

Jack, you have a great opportunity at sciforums to learn from some very well educated people.

Why are you not open to learning? Why do you approach any disagreement as a fight? Why so antagonistic?

Why not approach disagreements as an opportunity for learning?
Maybe it will be the other poster who learns something, maybe it will be you, maybe both.

If you don't consider the possibility that you're sometimes wrong, you'll never grow.



It's pretty sad, really. It's very clear that you have a lot to learn, and that rpenner and AlphaNumeric (maybe even myself) could help you... but you just don't want to. You don't want to become right... you just want be right.

I am trying to learn.

These are very very smart people.

If points are selected from the rest frame and x' after LT calculation is used in an equation, is x' a free variable over the reals or bound by some quantifier?

I think this is a reasonable question for these very smart people.

 

[Jack_]

I know that those that copy whole posts twice without responding to the content at a level of detail which requires whole copies are boorish.

I know that I selected no points. I know that the Lorentz Transform is a homogeneous transformation and is not dependent on the choice of origin. I named the difference in coordinates between two events in four-dimensional Minkowski space, but left the selection of the events up to the reader.
I am not concerned with x or x' but only with $$\Delta x$$ and $$\Delta x'$$.
As you have not identified the subject matter, of course you have nothing to say on free versus bound variables. In the expression $$x^2 - y^2$$, neither variable is logically bound, the value of the expression depends on x and on y, but neither is bound. Consequently the expression is still valid if x = y. Even the pair of events which is the actual topic of discussion, is fully free.

I could bind variables by writing $$\forall A \forall B (c \Delta t'_{AB})^2 - (c \Delta x'_{AB})^2 - (c \Delta y'_{AB})^2 - (c \Delta z'_{AB})^2 = (c \Delta t_{AB})^2 - (c \Delta x_{AB})^2 - (c \Delta y_{AB})^2 - (c \Delta z_{AB})^2$$ and then everything would be bound, but that is already implied by my free statement and the axiom or rule of generalization.

You have confused predicate logic with real algebra. What you call a fallacy is a symptom of your incorrect learning and not a statement of objective truth.

The statement $$x \in \left\{ \frac{-a-\sqrt{a^2+4b}}{2} \, , \, \frac{-a+\sqrt{a^2+4b}}{2} \right\} \rightarrow x^2 + a x = b$$ is a statement of propositional logic with no bound variables.

I could bind variables by writing $$\forall A \forall B (c \Delta t'_{AB})^2 - (c \Delta x'_{AB})^2 - (c \Delta y'_{AB})^2 - (c \Delta z'_{AB})^2 = (c \Delta t_{AB})^2 - (c \Delta x_{AB})^2 - (c \Delta y_{AB})^2 - (c \Delta z_{AB})^2$$ and then everything would be bound, but that is already implied by my free statement and the axiom or rule of generalization.

You could do this but LT works by selecting points from a frame and mapping them to another.

So, your binding is false since x' is existentially quantified by the selection of x or x is existentially quantified by the selection of x'.

You can easily prove this to yourself since the LT matrix is invertible and is not the identity matrix.