1
$\begingroup$

I stumbled across a paradox in special relativity for which I have found no answer. In the twin paradox, the loss of symmetry in time dilation is usually explained by the reversal of movement of the second twin. In the following thought experiment, this does not occur:

(And for the overzealous censors here: The following is an example that I made up myself. No one needs to calculate anything here. My question comes at the end of this post.)

My alien friend speeds past earth at 0,66c. From the moment he passes over my home, we send radar pulses every second. How many of my pulses reach him until he leaves the range of our radars, which is 15 light seconds? And how many of his radar pulses reach me? No matter how I calculate, the number of pulses that reach him seems to be fewer than the number of pulses that reach me, because he is "running away" from my signal. [My signals reach him every 3s, I get a signal every 1,66s.] This would, however, produce an asymmetry: In my world, I would receive more pulses, and in his world, he would receive more pulses, which should be impossible. Even if I include his time dilation, I cannot get a symmetrical number of pulses. [His time runs at 0,75 of mine, so I still get a signal every 2,21s.]

My question is the following: If there should not be two different worlds after this event (one in which I received more pulses and one in which he received more pulses) there must be something that I am missing, but what? Or is this really a paradox? If so, how does Relativity explain this paradox?

$\endgroup$
  • 1
    $\begingroup$ Didn't you post this homework question already? $\endgroup$ – CuriousOne Dec 28 '15 at 0:50
  • $\begingroup$ Imagine that you've put light-second markers along the path of the alien. At the instant the alien passes your marker 15, you claim the alien is 15 light-seconds distant. However, according the alien, the spacing between your light-second markers is less than 1 light-second. That is to say, you and the alien do not agree on the distance between your radars. Have you taken this into account? $\endgroup$ – Alfred Centauri Dec 28 '15 at 3:46
  • $\begingroup$ Draw the spacetime diagram and all will be clear. $\endgroup$ – WillO Dec 28 '15 at 7:05
1
$\begingroup$

You get different numbers of pulses because your setup is in fact asymmetrical. Here's why:

For easy algebra let's say the adimensional relative velocity between you and your friend is $\beta = \frac{v}{c}= \frac{2}{3}$.

You and your friend take your respective origins in space and time, $(x=0, ct=0)$ and $(x'=0, ct'=0)$, when your spaceships pass each other. At this moment you see the marker $R$ for your radar range set out at $x_R=15$ light-seconds away in your friend's direction of motion. You expect to observe your friend pass it at time $ct_R = \frac{x_R}{\beta}=\frac{3}{2}\times 15$s.

But let's see what does your friend see when he passes you at his time ct'=0. He definitely doesn't see the marker $R$ as you saw it at time $ct=0$ – due to relativity of simultaneity. Otherwise he would see 2 events that are simultaneous for you and $R$ in your rest frame also as simultaneous in his own rest frame, which cannot be. Instead he sees your marker $R$ as it is for you at some time $ct$ and at a position relative to him ($\gamma = \frac{1}{\sqrt{1-\beta^2}} = \frac{3}{\sqrt{5}}$ is the time dilation factor) $$ x'_R = \gamma (x_R - \beta ct) $$ We also know that, since his time is $ct'=0$, $$ ct'= \gamma(ct -\beta x_R) =0 \;\; \Rightarrow ct = \beta x_R $$ From this we find that he sees your marker R at position $$ x'_R = \gamma(x_R -\beta^2 x_R) = \frac{x_R}{\gamma} = 5\sqrt{5} ls < 15 ls $$ We could've gotten this result much more directly by invoking length contraction, but then it wouldn't have been very obvious what happens when and why (relativity of simultaneity).

Accordingly, your friend will see your marker pass by him at his time $ct'_R=\frac{x'_R}{\beta} = \frac{ct_R}{\gamma} < ct_R$. No wonder he sends fewer pulses than you do! He has less time to do so in the first place. So what gives?

Well, your friend is also going to complain of fewer pulses on your side if you guys decide to use his radar range marker $R'$, which he sets up at position $x'_{R'} = -15$ls. In this case, at $ct=0$ you see $R'$ closer than your marker $R$, at position $x_{R'} = -\frac{x_R}{\gamma}$, and expect to see it pass by you at a time $ct_{R'}=\frac{ct_R}{\gamma}$, before your friend passes your marker $R$. Now you have less time to send your pulses.

Fortunately this gives us a very simple idea on how to make the scenario symmetric between you and your friend: You send him pulses until you pass his marker $R'$ at your time $ct_{R'}=\frac{ct_R}{\gamma}=15\frac{\sqrt{5}}{2}$s, while he sends you pulses until he passes your marker $R$ at his time $ct'_R = \frac{ct_R}{\gamma}=15\frac{\sqrt{5}}{2}$s. This way both of you send and receive equal numbers of pulses.

Homework: How many pulses do you each send and receive in this latter case?

Moral: In relativity it doesn't pay to be too self-centered.

$\endgroup$
0
$\begingroup$

I guess the radars are identical, and you take the viewpoint of the earthling, but forget that the range of the alien's radar is length contracted in this particular frame.

$\endgroup$

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.