0
$\begingroup$

This question originates from my previous question here Results of two equivalent scenarios in SR

There is no specific need to visit my previous question to understand this one. I have made efforts to make this question clear and independent of previous one.

So what I understand is this: there are two in-line observers (H and H') that are connected by a tube and a third observer (T) travelling at velocity $v$ w.r.t to both H and H'. The motion is normal to tube i.e. the line joining H and H'.

Now, If H want to send a laser pulse to H', he does so by making sure he emits his laser pulse parallel to tube. But if T want to make sure his laser pulse reaches H', T must emit the light pulse at an certain angle $\theta$ when his position coincides with H. This is called transverse Relativistic Doppler effect.

What confuses me is this: T emitted the laser at angle $\theta=arctan(y/x)$ in his own frame. This angle will be seen as $\theta=arctan(y.\gamma/x)$ in the frame of H and H'. That is, the light will be emitted at some angle w.r.t the tube enclosing H and H'. If it is at some angle w.r.t to the tube it means it is not parallel to tube and hence must not reach H' (because it will strike the wall of tube at some time).

Is there anything I am missing?

$\endgroup$
1
$\begingroup$

Let me schematize the system. I think it is always a good approach when having those kinds of doubt, especially in special and general relativity where things might become rather counterintuitive. I will set $c=1$.

In the frame $S$, where the body $H$ is fixed at the center, motion laws are $$ X_H(t)=\left(\begin{array}{c}t\\0\\0\end{array}\right),\quad X_{H'}(t)=\left(\begin{array}{c}t\\L\\0\end{array}\right),\quad X_T(t)=\left(\begin{array}{c}t\\0\\vt\end{array}\right). $$ Here, $L$ is the length of the tube joining $H$ and $H'$ at the time $t=0$. As you can see, if $H$ wants to send a photon to $H'$, all it has to do is to send a photon along the $\hat x$ direction, that will move with law $x_\gamma(t)=(t,t,0)$. If $T$ wants to send a photon at $t=0$, it will have to send the same photon as $H$, aligned to the $\hat x$ direction.

Obviously, you want to know is $T$ has to send the photon with some angle in its frame. Let us then define the frame $S'$ where $T$ is at the origin: the Lorentz boost matrix will be \begin{align} \Lambda=\left(\begin{array}{ccc}\gamma&0&-v\gamma\\0&1&0\\-v\gamma&0&\gamma\end{array}\right), \end{align} where $\gamma=\gamma(v)$. You can easily calculate the laws of motion in $S'$, as you always have uniform motion: by multiplying the $X$'s with the $\Lambda$'s, you have

$$ X'_H(t)=\left(\begin{array}{c}\gamma t\\0\\-v\gamma t\end{array}\right),\quad X'_{H'}(t)=\left(\begin{array}{c}\gamma t\\L\\-v\gamma t\end{array}\right),\quad X'_T(t)=\left(\begin{array}{c}\frac t\gamma\\0\\0\end{array}\right). $$ As it is evident, the photon that you have to emit at $t=0$ cannot be aligned to the $\hat x'$ axis. We choose to parametrize time in the same way that is parametrized for $X'_T$: the photon must have motion law $$ X'_\gamma(t)=\left(\begin{array}{c}\frac{t}{\gamma}\\\cos\theta\frac{t}\gamma\\-\sin\theta \frac t\gamma\end{array}\right). $$ For $H'$ to receive $\gamma$, it is necessary that there exist two values of the time parameters such as $X'_{H'}(t_1)=X'_\gamma(t_2)$: this implies $$ t_2=t_1\gamma^2,\quad t_2=\frac{\gamma L}{\cos\theta},\quad t_2=\frac{\gamma^2 v}{\sin\theta}t_1. $$ You must select the angle $\theta$ in order for a solution to exist (that is nontrivial, as you have three equations of two unknowns): in this case, you must have $\sin\theta=v$. You can use the second equation to calculate $t_2$ and have the meeting point (space and time coordinates).

So you have a solution in this frame: you must emit with angle $\sin\theta=v$. Let us study this solution in the reference frame $S$: returning back through the matrix $\Lambda^{-1}$, you have $$ X_\gamma(t)=\left(\begin{array}{c}(1-v^2)t\\(1-v^2)t\\0\end{array}\right). $$ This is the 4-position of a photon traveling in the $\hat x$ direction, as predicted before.

I don't know how your guess can be interpreted. I don't understand what you would use for $y$ but you can try rethinking about your guess and confronting it with the schematization. It could look a little bit strange, but I think it's the best way to approach those problems.

EDIT: Typo fixing.

$\endgroup$
  • $\begingroup$ I will use $c$ to make my point clear. $\endgroup$ – Gaurav Goyal Feb 13 '17 at 19:08
  • $\begingroup$ I will use $c$ to make my point clear. The angle with which light pulse is emitted in T is $\theta=asin(v/c)$. This angle in the frame of H or H' will be $\theta=asin(\gamma.v/c)$. Am I correct till here? $\endgroup$ – Gaurav Goyal Feb 13 '17 at 19:25
  • $\begingroup$ No: in the frame of $H$ or $H'$ the photon is emitted parallel to the $\hat x$ axis, if the photon is emitted when $T$ is at $H$. I think that your sine transformation law is being applied incorrectly in this case. $\endgroup$ – Salvatore Baldino Feb 13 '17 at 19:50
  • $\begingroup$ Yes. That is where I think I am wrong. I am using geometric figures to convert sine between the two frames since I do not know the correct transformation. I am simply multiplying $\gamma$ with the side parallel to motion keeping side normal to motion as it is and ending up with statement just written in my previous comment. Can you please show the correct transformation of sine between inertial frames? $\endgroup$ – Gaurav Goyal Feb 13 '17 at 19:55
  • $\begingroup$ Let me clarify: if you want to understand how the angle changes with respect to the velocity, suppose to transform the vector $(ct,ct\cos \theta,ct\sin\theta)$ with a Lorentz boost with velocity $v'$ along the $\hat y$ axis, non necessarily equal to $v$: you get the vector $(ct(\gamma'-\gamma'\sin\theta),ct\cos\theta,ct(-\gamma'v'+\gamma'\sin\theta)$. This means that, in the moving system, the sine of the angle is given by $\sin\theta'=\frac{\sin\theta-v'}{1-v'\sin\theta}$. Now, in our case we start with $\theta=0$, so we get the results of our problem immediately. $\endgroup$ – Salvatore Baldino Feb 13 '17 at 19:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.