Why isn't the relative y-velocity in this problem $.8c$?

Question

I am trying to solve the following problem:

A radioactive atom is moving in the $x$ direction in the laboratory at speed $0.3c$. It emits an electron having speed $0.8c$ in the rest frame of the atom. What will be the velocity of the electron in the laboratory when it is ejected in the $y$ direction in the atom's rest frame? (Find $u_y$, $u_x$, $|u|$, and $\theta$.)

I thought that since the atom is moving in the $x$ direction relative to the laboratory, then any motion in the $y$ direction relative to it must be the same for the laboratory. However, when I put $u_y = 0.8c$, the teacher marked it wrong; he refuses to explain why. Why is $u_y \neq 0.8c$?

Answer 1

To see why it is wrong it is easier to consider the case where the x direction of the initial particle is $w^{'}_{x} = 0.8c$ (velocity of radioactive particle) in the lab frame (indicated by prime). Now since the second particle was traveling at $u_{y} = 0.8c$ in the rest frame, if that was also how fast it was traveling in the lab frame then $u^{'}_{x} = u^{'}_{y} = 0.8c$ using pythagorus' theorem this means the velocity of the ejected particle is greater then the speed of light which indicates that something is wrong.

The source of the error comes about from the fact that there is time dilation between the reference frames. This is why you can never get above the speed of light. To work out the solution it is better to think about the world line that the ejected particle has, in the rest frame we get $\vec{x}(t) = (ct,0,u_{y} t)$. Here I'm giving time in units of length because that feels nicer to me and the transformation becomes nice and symmetric and I'll give that below, for a change in velocity $v$ the transformations are:

$y^{'} = y$

$x^{'} = \gamma \left(x-\frac{v}{c}(ct)\right)$

$(ct^{'}) = \gamma \left((ct)-\frac{v}{c}x\right)$

which means in the lab coordinates (where $v = -u_{x}$) this is $\vec{x}^{'}(t) = (\gamma ct,\gamma u_{x}t,u_{y} t)$.

At first glance this looks like the velocity in the y direction is still $u_{y}$ but t is the time coordinate for the original frame, not the lab frame. Instead we can see that it is $t^{'} = \gamma t$ by looking at the first term in $\vec{x}^{'}$ and we get:

$\vec{x}^{'}(t) = \left(ct^{'},u_{x}t^{'},\frac{u_{y}}{\gamma} t^{'}\right)$

So the correct speed is $\frac{u_{y}}{\gamma}$ which when we sub in the numbers comes to be $\frac{0.8c}{\sqrt{1-(0.3)^{2}}} = \frac{0.8}{\sqrt{0.91}}c$, so its slightly slower then what you would think, and this comes down to time dilation effects.

Main thing that might trip you up here is the sign for the coordinate change but that can be easily checked by working out what direction a stationary particle would be moving and check it would be moving in the correct direction)

(edit: Corrected subscript)

Answer 2

Link : Appearance of an angle of inclination on a horizontal rod moving upwards after a Lorentz Transformation. Equations (09), (11.1),(11.2) and the proof given in the footnote.

---

\begin{align} \textrm{velocity of Atom with respect to Lab : } \quad\boldsymbol{\upsilon} & =\left(\upsilon,0\right)=\left(0.30c,0\right) \tag{01}\\ \textrm{velocity of Electron with respect to Atom : } \quad\mathbf{u} &=\left(0,u\right)=\left(0,0.80c\right) \tag{02}\\ \textrm{velocity of Electron with respect to Lab : } \quad \mathbf{w} &=\left(w_x,w_y\right) \tag{03} \end{align}

\begin{equation} \mathbf{w}=\boldsymbol{\upsilon}+\dfrac{\mathbf{u}}{\gamma_{\upsilon}}=\left[\upsilon\,,\left(\!1\!-\!\frac{\upsilon^{2}}{c^{2}}\right)^{\!\!\frac12}\!\!u\right]\,,\quad \gamma_{\upsilon} = \left(\!1\!-\!\frac{\upsilon^{2}}{c^{2}}\right)^{\!\!\boldsymbol{-}\frac12} \tag{04} \end{equation}

---

EDIT

^{From Figure : Differential expression of Lorentz Transformation from the Atom system $\:\mathrm{S'}\equiv \{x'y't'\}\:$ to the Laboratory system $\:\mathrm{S}\equiv \{xyt\}\:$ \begin{align} \mathrm dx & =\gamma_{\upsilon}\left(\mathrm d x'+\upsilon\mathrm d t'\right)\vphantom{\left(\mathrm d t'+\dfrac{\upsilon}{c^{2}}\mathrm dx'\right)} \tag{01a}\\ \mathrm dy & =\mathrm dy'\vphantom{\left(\mathrm d t'+\dfrac{\upsilon}{c^{2}}\mathrm dx'\right)} \tag{01b}\\ \mathrm dt & = \gamma_{\upsilon}\left(\mathrm d t'+\dfrac{\upsilon}{c^{2}}\mathrm dx'\right) \tag{01c} \end{align} Now, the Electron is moving in $\:\mathrm S'\:$, the rest frame of the Atom, with velocity \begin{equation} \mathbf{u}=\left(\dfrac{\mathrm d x'}{\mathrm d t'},\dfrac{\mathrm d y'}{\mathrm d t'}\right)=\left(0,u\right) \tag{02} \end{equation} To find its velocity $\:\mathbf{w}\:$ in $\:\mathrm S\:$, the Laboratory, we divide (01a) and (01b) by (01c) \begin{align} w_{x} &=\dfrac{\mathrm dx}{\mathrm d t}=\dfrac{\left(\dfrac{\mathrm dx'}{\mathrm d t'}\right)+\upsilon}{1+\dfrac{\upsilon}{c^{2}}\left(\dfrac{\mathrm dx'}{\mathrm d t'}\right)}=\upsilon \tag{03a}\\ w_{y} &= \dfrac{\mathrm dy}{\mathrm d t}=\dfrac{\left(\dfrac{\mathrm dy'}{\mathrm d t'}\right)}{\gamma_{\upsilon}\left[1+\dfrac{\upsilon}{c^{2}}\left(\dfrac{\mathrm dx'}{\mathrm d t'}\right)\right]}=\dfrac{u}{\gamma_{\upsilon}} \tag{03b} \end{align} and finally \begin{align} w_{x} & =\upsilon \tag{04a}\\ w_{y} & = \dfrac{u}{\gamma_{\upsilon}}=\sqrt{1-\dfrac{\upsilon^{2}}{c^{2}}}\:u \tag{04b} \end{align} or \begin{equation} \mathbf{w}=\boldsymbol{\upsilon}+\dfrac{\mathbf{u}}{\gamma_{\upsilon}} \tag{05} \end{equation}}