Time Dilation along Multiple Axes

[Reinhardt Walzer]

Been studying Special Relativity in Uni. and I've noticed that all examples of relativistic motion provided are motions only along a single axis, like the one below:

The particle's Reference Frame is moving only along the X axis in the example above.
In this case the Lorentz Transformation for Time is this: $t^{'}=\gamma(t-\frac{vx}{c^{2}})$.

But what about the case shown below:

How would I write the Lorentz Transformation for Time in this case?

• Like this: $t^{'}=\gamma(t-\frac{\vec v \cdot \vec r}{c^{2}})$ How would $\vec v \cdot \vec r$ even look like?
• Or would I need to have three different components of time for each axis: $t^{'}_{x}=\gamma(t-\frac{v_{x}x}{c^{2}})$ ; $t^{'}_{y}=\gamma(t-\frac{v_{y}y}{c^{2}})$ ; $t^{'}_{z}=\gamma(t-\frac{v_{z}z}{c^{2}})$

 

[weirdoguy]

How would even look like?

$\vec{v}\cdot\vec{r}=v_xx+v_yy+v_zz$

 

[A.T.]

But what about the case shown below:
View attachment 297437

How would I write the Lorentz Transformation for Time in this case?

https://en.wikipedia.org/wiki/Lorentz_transformation#Proper_transformations

 

[ergospherical]

Looks like you have it right in your first equation. Easy way to go about things is to let $\Lambda$ be the matrix representing the boost along the $x$ axis. What if you want a boost in an arbitrary direction $\mathbf{n}$? If $R$ is the rotation matrix taking $\mathbf{n}$ into $\hat{\mathbf{x}}$, then your boost is $R^{-1} \Lambda R$.

 

[Reinhardt Walzer]

Thank you for your input, @ergospherical @weirdoguy and @A.T. .
I would also want to know if knowing the Lorentz Transformation for Time in the Second Reference Frame would help me in transforming the trajectory of an object bouncing in the second frame to the first one.
Like this: I have an object P moving along the trajectory $\langle x^{'}({{t^{'}}})=1+{t'}^{3},y^{'}({t^{'}})={t^{'}}^{2},z^{'}({t^{'}})=1\rangle$ w.r.t. the Second Ref. Frame and I'd want to write this trajectory from the POV of an Observer standing in the Origin of the First Reference Frame (the proper/stationary frame).
Could I write this in a simple and elegant manner, that could be potentially simulated using desmos or geogebra etc. ?

 

[ergospherical]

Yep you can at least in principle find three equations defining the trajectory in S implicitly by making the coordinate transform $t' = \Lambda^0_0 t + \Lambda^0_1 x + \Lambda^0_2 y + \Lambda^0_3 z$, etc. for the given boost.

 

[Reinhardt Walzer]

Yep you can at least in principle find three equations defining the trajectory in S implicitly by making the coordinate transform $t' = at + bx + cy + dz$, $x' = \dots$ etc. for the given boost

And would I need something like this

$x = \gamma^{-1}_x x'$

(where $\gamma_x = \gamma(\vec v\cdot \hat x)$)
for every axis ?
Or could I use the same $\gamma$ for all axes?

 

[Nugatory]

Been studying Special Relativity in Uni. and I've noticed that all examples of relativistic motion provided are motions only along a single axis, like the one below:

We can always choose our axes so that the x-axis is parallel to the direction of motion. We don’t have to do this, but there’s no reason not to: it makes the math simpler. If we choose our axes some other way it complicates the formulas in ways that actually obscure the fundamentally simple and elegant physics, without providing any additional insight. (We do something similar teaching high school physics when we label the position of a dropped object by its height above the ground - we’ve quietly chosen our coordinates so that one axis is vertical, parallel to the direction of motion, so that the SUVAT equations take on their simplest and most informative form).

It is worth noting that the quantity $v^2$ and hence $\gamma$ is unaffected by our choice of axis. Taking a moment to prove this, and then considering the implications for the kinetic energy and momentum will go a long ways towards convincing yourself that the textbook examples really aren’t hiding anything here.

And of course if you ever find yourself with a problem that is more naturally solved using axes that are not parallel to the direction of motion…. The posts above cover how the coordinate transformation machinery can be used.

 

[ergospherical]

Not quite no, there's only ever one $\gamma$ and it's a function of the magnitude of the relative velocity $|\mathbf{v}|$ only. What you ought to do is start with the boost along the $x$ axis\begin{align*}
\Lambda = \begin{pmatrix} \gamma & - \gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}
\end{align*}Then to get the boost you actually want, along the direction defined by a unit vector $\mathbf{n}$, you take the usual 3-matrix $\tilde{R}$ which rotates $\mathbf{n}$ into the $x$ direction, form the 4-matrix $R = \begin{pmatrix} 1 & \\ & \tilde{R} \end{pmatrix}$ and work out $\Lambda(\mathbf{n}) = R^{-1} \Lambda R$. Then the coordinate transformation in matrix notation is $\mathbf{x}' = \Lambda(\mathbf{n}) \mathbf{x}$.

 

[Reinhardt Walzer]

Not quite no, there's only ever one $\gamma$ and it's a function of the magnitude of the relative velocity $|\mathbf{v}|$ only. What you ought to do is start with the boost along the $x$ axis\begin{align*}
\Lambda = \begin{pmatrix} \gamma & - \gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}
\end{align*}Then to get the boost you actually want, along the direction defined by a unit vector $\mathbf{n}$, you take the usual 3-matrix $\tilde{R}$ which rotates $\mathbf{n}$ into the $x$ direction, form the 4-matrix $R = \begin{pmatrix} 1 & \\ & \tilde{R} \end{pmatrix}$ and work out $\Lambda(\mathbf{n}) = R^{-1} \Lambda R$. Then the coordinate transformation in matrix notation is $\mathbf{x}' = \Lambda(\mathbf{n}) \mathbf{x}$.

I see. How would this $R^{~}$ matrix look like? I've only been introduced to the classical rotation matrix:

Also does the empty space left in the matrix $R$ mean that those quadrants are occupated only by zeroes?

 

[Reinhardt Walzer]

We can always choose our axes so that the x-axis is parallel to the direction of motion. We don’t have to do this, but there’s no reason not to: it makes the math simpler. If we choose our axes some other way it complicates the formulas in ways that actually obscure the fundamentally simple and elegant physics, without providing any additional insight. (We do something similar teaching high school physics when we label the position of a dropped object by its height above the ground - we’ve quietly chosen our coordinates so that one axis is vertical, parallel to the direction of motion, so that the SUVAT equations take on their simplest and most informative form).

It is worth noting that the quantity $v^2$ and hence $\gamma$ is unaffected by our choice of axis. Taking a moment to prove this, and then considering the implications for the kinetic energy and momentum will go a long ways towards convincing yourself that the textbook examples really aren’t hiding anything here.

And of course if you ever find yourself with a problem that is more naturally solved using axes that are not parallel to the direction of motion…. The posts above cover how the coordinate transformation machinery can be used.

Thank you for providing context @Nugatory . Much appreciated!

 

[ergospherical]

That's a rotation matrix in two-dimensions, you need one in three. E.g. about the $z$ axis,
\begin{align*}
R = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & & & \\ 0 & & \tilde{R}^{(z)} & \\ 0 & & & \end{pmatrix} = \begin{pmatrix} 1 &0&0&0 \\ 0 & \cos{\phi} & -\sin{\phi} & 0 \\ 0 & \sin{\phi} & \cos{\phi} & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}
\end{align*}

 

[Reinhardt Walzer]

That's a rotation matrix in two-dimensions, you need one in three. E.g. about the $z$ axis,
\begin{align*}
R = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & & & \\ 0 & & \tilde{R}^{(z)} & \\ 0 & & & \end{pmatrix} = \begin{pmatrix} 1 &0&0&0 \\ 0 & \cos{\phi} & -\sin{\phi} & 0 \\ 0 & \sin{\phi} & \cos{\phi} & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}
\end{align*}

Alrighty then, and for your example in the previous post I would do the same but for the $x$ axis. And the vector $n$ would be like a position vector stemming from the origin point of the First Reference Frame, along which the "boost" is occurring.

I think I get it now.

Many, many thanks @ergospherical !

 

[DrGreg]

For what it's worth, the Lorentz boost for a velocity $\textbf{v}$ in an arbitrary direction is given by$$
\left(
\begin{array}{c}
ct' \\ \hline \\ \textbf{x}' \\ \\
\end{array}
\right)
=
\left(
\begin{array}{c|c|c|c}
\gamma & -\gamma \textbf{v}^T \\
\hline
\\
-\gamma \textbf{v} & \textbf{I}
+ (\gamma - 1)\frac{\textbf{v} \textbf{v}^T}{\textbf{v}^T \textbf{v}}
\\
\\
\end{array}
\right)

\left(
\begin{array}{c}
ct \\ \hline \\ \textbf{x} \\ \\
\end{array}
\right)
$$ where $$
\gamma = \frac{1}{\sqrt{1 - \frac{\textbf{v}^T \textbf{v}}{c^2} } }
$$ But in practice it's often more convenient to choose axes aligned to the velocity.

 

[Reinhardt Walzer]

For what it's worth, the Lorentz boost for a velocity $\textbf{v}$ in an arbitrary direction is given by$$
\left(
\begin{array}{c}
ct' \\ \hline \\ \textbf{x}' \\ \\
\end{array}
\right)
=
\left(
\begin{array}{c|c|c|c}
\gamma & -\gamma \textbf{v}^T \\
\hline
\\
-\gamma \textbf{v} & \textbf{I}
+ (\gamma - 1)\frac{\textbf{v} \textbf{v}^T}{\textbf{v}^T \textbf{v}}
\\
\\
\end{array}
\right)

\left(
\begin{array}{c}
ct \\ \hline \\ \textbf{x} \\ \\
\end{array}
\right)
$$ where $$
\gamma = \frac{1}{\sqrt{1 - \frac{\textbf{v}^T \textbf{v}}{c^2} } }
$$ But in practice it's often more convenient to choose axes aligned to the velocity.

Very interesting!
Thank you!

 

[PeroK]

Alrighty then, and for your example in the previous post I would do the same but for the $x$ axis. And the vector $n$ would be like a position vector stemming from the origin point of the First Reference Frame, along which the "boost" is occurring.

I think I get it now.

Many, many thanks @ergospherical !

Note that there is a subtlety here. If the boost is along a coordinate axis (e.g. the x-axis), then the two frames can agree upon the x-y-z axes. This is because at time $t = 0, t' = 0$, the points $(0, 1, 0)$ and $(0, 0, 1)$ transform to the points $(0, 1, 0)'$ and $(0, 0, 1)'$ in the primed frame. In other words, if we imagine a rigid system of coordinates axes in each frame then these will coincide at the spatial origin at $t = t' = 0$.

If, however, the boost is along a different direction (e.g. between the x and y axes), then the coordinate axes of the of the primed frame will not coincide with the coordinate axes of the unprimed frame (good exercise).

Therefore, we need a convention for how we set up the coordinate system in the primed frame. The convention that is implied by using the rotation matrix (and its inverse) to establish the coordinate tranformation is to take the direction of relative motion to be the same in each frame. In other words, if the origin of the primed frame is moving at angle $\theta$ to the x-axis (as measured in the unprimed frame), then the origin of the primed frame is taken to be moving at an angle $\pi + \theta$ as measured in the primed frame. I.e. with the same angle but in the opposite direction.

Or, put more simply, the convention is that if the three-velocity of the origin of the primed frame is $\vec v$ as measured in the unprimed frame, then the three-velocity of the origin of the unprimed frame is taken to be $-\vec v$ as measured in the primed frame.

 

[vanhees71]

In more simple words: The Lorentz-transformation matrix in #14 defines a rotation-free Lorentz boost in direction of the boost velocity $\vec{v}$.

It's important to keep in mind that the composition of two rotation-free Lorentz transformations in non-collinear directions is not a rotation-free Lorentz transformation but the composition of a rotation-free Lorentz transformation and a rotation (the socalled "Wigner rotation").