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Consider a 4-dimensional space $\mathrm{S}$ in which all objects move at the same speed $c_S$ but in different directions. For the objects A and B that move in directions $\overrightarrow{u_A}$ and $\overrightarrow{u_B}$, respectively, if $\Delta x$ is the distance that B moves in the (3-dimensional) space perpendicular to $\overrightarrow{u_A}$ during a time interval $\Delta T$, the distance it moves in the direction of $\overrightarrow{u_A}$ will be \begin{equation*} \sqrt{c_S^2 \Delta T^2 - \Delta x^2} = \sqrt{c_S^2 - v^2} \Delta T, \end{equation*} where $ v = \Delta x / \Delta T$. Let $\Delta \tau$ be the quantity such that \begin{equation*} c_S^2 \Delta T ^2 = v^2 \Delta T^2 + c_S^2 \Delta \tau ^2. \end{equation*}

Thus, if $\overrightarrow{u_A}$ is interpreted as constituting the coordinate temporal axis and the 3-dimensional space perpendicular to $\overrightarrow{u_A}$ as constituting the coordinate space, and if $c_S=c$ is the speed of light, then the last equation describes the motion of the object B in a special relativity spacetime traveling with speed $v$ during the coordinate time interval $\Delta T$, and that $\Delta \tau$ is the proper time interval experienced for B: \begin{equation*} \Delta \tau = \sqrt{1 - \frac{v^2}{c^2}} \Delta T. \end{equation*}

Is this a correct interpretation?

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    $\begingroup$ But the velocity in 3-dimensional space according to this interpretation is not $v$, it is the ratio between the 3-dimensional distance (first expression) and the temporal distance $\Delta x$ $\endgroup$
    – J. Delaney
    Commented May 24, 2023 at 11:32
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    $\begingroup$ There is indeed some sort of difficulty here, I see that you try to introduce the interpretation of $\overrightarrow{u_A}$ as time towards the end. But even in the first part you mention movement along that direction in a time interval $\Delta{T}$. im not sure if that's only confusing or a real inconsistency $\endgroup$
    – Amit
    Commented May 24, 2023 at 12:43
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    $\begingroup$ @J.Delaney It's not exclusive, $A$ is moving purely along the time direction meaning it is at rest, and $B$ moves wrt to $A$ a distance $\Delta x$ in coordinate time $\Delta T$ $\endgroup$
    – Amit
    Commented May 24, 2023 at 15:23
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    $\begingroup$ @robphy I think OP just means a decomposition of vector $\Delta \vec{x}$ with magnitude $c\Delta T$ that $B$ moves by into a component parallel to $u_A$ and one that's orthogonal/perpendicular to it. He calls $\Delta x$ the orthogonal component and calculates the parallel $\endgroup$
    – Amit
    Commented May 24, 2023 at 15:47
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    $\begingroup$ This site is for asking questions about mainstream physics, not for feedback about personal theories $\endgroup$
    – Dale
    Commented May 24, 2023 at 19:56

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