Two-way tachyonic anti-telephone

Question

Consider the two-way tachyonic antitelephone where the speed at which message is transmitted is $a$. A person $A$ sends a message to $B$ which is moving away with a speed of $v$ with respect to $A$. And when $B$ receives a message $B$ transmits a message back to $A$. Since, both of them have the same device, in each of the persons frame the speed of transmission is $a$. Now, when $B$ is at a distance $L$ away from $A$ with respect to $A$, $A$ sends a message which takes a time $$t_1 = \frac{L}{a}$$ to reach $B$, at the same moment $B$ sends a message to $A$ with a speed $a$ relative to itself towards $A$, due to which the time taken by the message to reach $A$ in $A$'s frame will be $$t_2 = \frac{L(1-av)}{a-v} \equiv \frac{L}{a'}$$ where we have used the fact that now the speed of the transmission from $B$ towards $A$ relative to $A$ is $$a' = \frac{a-v}{1-av}$$ Hence, the total time measured by $A$ from sending the message to $B$ and getting a response is $$T=t_1+t_2=\left(\frac{1}{a}+\frac{1-a v}{a-v}\right) L$$

Now, the argument is that one can choose $a$ such that $T < 0$ and, hence, $A$ will receive a message from $B$ before he even sends one to $B$. However, if you choose $a$ such that $T < 0$ we see that it will $a' < 0$ which means that the reply from $B$ is actually going away from $A$, so $A$ will never receive a message from $B$. So I do not see where the paradox actually is?

Answer 1

Now, the argument is that one can choose $a$ such that $T < 0$ and, hence, $A$ will receive a message from $B$ before he even sends one to $B$. However, if you choose $a$ such that $T < 0$ we see that it will $a' < 0$ which means that the reply from $B$ is actually going away from $A$, so $A$ will never receive a message from $B$. So I do not see where the paradox actually is?

This doesn't work. The easiest way to see this is to simply use the Lorentz transform on some of the intermediate events on the worldline of the signal. You can parameterize these intermediate events with some frame independent affine parameter, $\lambda$.

If $\lambda$ increases along the signal's worldline from $B$ to $A$, that fact will hold in all frames. As $\lambda$ increases, the signal gets closer to $A$. In $B$'s frame $\lambda$ will increase with time but in $A$'s frame $\lambda$ will decrease with time, but in both frames the signal will get closer to $A$ with increasing $\lambda$.

For example, consider $B$ returning a tachyon signal to $A$. For concreteness let $A$ and $B$ pass each other at $t=t'=0$ with $A$ moving at $v=0.6$ in $B$'s frame (the unprimed frame) in units where $c=1$. In $B$'s frame the worldline for $A$ is $(t_A(t),x_A(t))=(t,0.6\ t)$. At time $t=5$ in $B$'s frame, $B$ will send a tachyon signal which travels at $u=6$. In terms of an affine parameter $\lambda$ the worldline for the signal is $(t_s(\lambda),x_s(\lambda))=(\lambda+5,6\lambda)$. Solving for $(t_A,x_A)=(t_s,x_s)$ we get $\lambda = 5/9$ which means that in all frames the affine parameter runs from $0$ to $5/9$. Now, Lorentz transforming to $A$'s frame (the primed frame), we get the worldline for the tachyon signal is $(t'_s(\lambda),x'_s(\lambda))=(-3.25 \lambda+6.25,6.75 \lambda-3.75)$. We can simply evaluate this worldline at say $\lambda = 2/9$ and $\lambda = 4/9$ to get $(t'_s(2/9),x'_s(2/9))=(5.52,-2.25)$ and $(t'_s(4/9),x'_s(4/9))=(4.81,-0.75)$. So in $A$'s frame the negative signal velocity means that the signal is going backwards in time as it approaches $A$.

Answer 2

Assume that the information transmitted is a bit, that Alice always repeats, and Bob always switches.

In Alice's point of view: she receives a call from Bob telling her a bit, and she calls him back repeating the same bit.

In Bob's point of view: he receives a call from Alice telling her a bit, and he calls her back and says the other bit.

You can then prove that both Alice and Bob pronounce both $0$ and $1$; this is a contradiction.

Answer 3

See my second paragraph for the tl;dr version of the paradox. @Dale has already given the correct answer, which is that the interpretation of negative speeds has to be done carefully. A better way to proceed would be to explicitly write down the spacetime coordinates of events: $S_A$ being the event when A sends the message, $R_B = S_B$ being the event where B receives the message and sends the reply, and $R_A$ being the event where A receives the reply. Start in A's frame of reference, figure out what the coordinates of $R_B$ are according to A, then transform to B's frame of reference. Then, working in B's frame of reference, figure out what the coordinates are of $R_A$ (where and when, according to B, A will receive the echo'd reply). Finally transform that back to A's frame, and you will find that, potentially, the time coordinate according to A of receiving the echo can be less than the original time coordinate at which A sent the first message.

To take an extreme case, suppose the FTL message is extremely fast, so that transmission time is negligible. If A sends the message at time $t=1$, then due to time dilation B will receive it at time $t' < 1$ (according to A, B's clock runs slow). But time dilation is symmetric, and indeed the whole scenario is symmetric. If B holds on to the message and sends it back at time $t' = 1$, A will receive it at time $t < 1$, i.e. before he sent it!

All speeds are relative, so if B's clock ticks slower according to A, then A's clock ticks slower according to B. This is the root cause of the tachyonic anti-telephone paradox.

Some students (even advanced students) have trouble accepting that time dilation is symmetric, but a very similar issue arises in ordinary Euclidean geometry, where relative lengths are "expanded" ($\frac{dL'}{dL} > 1$ and also $\frac{dL}{dL'} > 1$). This Euclidean equivalent and how it applies to the "twin paradox" is nicely illustrated by Ben Rudiak-Gould: