Times at relativistic speeds

Consider an observer $A$ moving in a spaceship at a speed close to $c$, relative to an observer $B$. The $B$ knows that the time on $A$ runs slower, but in the inertial frame of $A$, $B$ is moving away at the same speed close to $c$, while $A$ is stationary. Would $A$ think the time in $B$ is slower?

• Do you expect any other answer than a simple "Yes"? – ACuriousMind Jan 26 '15 at 20:49

Absolutely YES!the term "relative to..." tells you everything!

• But if they then compare their clocks, one is going to have a different time than the other, so how would they know if their time is slower or faster? – George Jan 26 '15 at 20:57
• So,this is the problem of inertial and non-inertial REFERENCE FRAME,the observer with constant velocity can tell which one is true.A's clock runs slower,here B is sitting in a inertial reference frame,so whatever he says is true.also search for TWIN PARADOX. – user71065 Jan 26 '15 at 21:00
• note in case A has acceleration. – user71065 Jan 26 '15 at 21:14

Consider an observer $A$ moving in a spaceship at a speed close to $c$, relative to an observer $B$.

Alright.
Moreover, since the motion of $A$ with respect to $B$ is supposed to be characterized by some particular "speed" (even if it were not constant) therefore we may assume that $B$ is a member of an inertial system; namely having been and having remained at rest with respect to suitable additional participants (say observers $K$, $Q$ ...), where, let's say, $A$ moved from $B$ to $K$, then on to $Q$, etc.

In the following let's suppose that the "speed" which the members of $B$'s ($K$'s, $Q$'s) inertial system measure of $A$ is a constant: $v_B[~A~]$ which is necessarily smaller than $c$ (the "signal front speed", a.k.a. "speed of light in vacuum"), even though it may be close to $c$.

$B$ knows that the time on $A$ runs slower

That seems a nonsense assertion
("time on $A$ running slower" than what ??, by which method of comparison ??).

Correct is instead that the comparison of durations between $A$ and $B$ can (and, by the setup prescription above typically does) involve the "Lorentz factor" $$\sqrt{ 1 - \left( \frac{v_B[~A~]}{c}\right)^2 },$$

a.k.a. "$1/gamma$", whose value is necessarily less than 1. Specificly:

$$\tau A_{\circ B}^{\circ K} = \sqrt{ 1 - \left( \frac{v_B[~A~]}{c}\right)^2 } ~\times~ \tau B_{\circ A}^{\circledS K \circ A},$$

where

• "$A_{\circ B}$" denotes participant $A$'s indication of having been met and passed by participant $B$,
• "$A^{\circ K}$" denotes participant $A$'s indication of having been met and passed by participant $K$, and thus
• "$\tau A_{\circ B}^{\circ K}$" denotes $A$'s duration from having been met and passed by participant $B$ until having been met and passed by participant $K$,

• "$B_{\circ A}$" denotes participant $B$'s indication of having been met and passed by participant $A$,

• "$B^{\circledS K \circ A}$" denotes participant $B$'s indication simultaneous to participant $K$'s indication of having been met and passed by participant $A$, and thus
• "$\tau B_{\circ A}^{\circledS K \circ A}$" denotes the duration of $B$ between these two of its indications.

A detailed derivation of this equation can be found here (PSE/q/110316).

but in the inertial frame of $A$

Good!, that's consistent with my above assumption of speed $v_B[~A~]$ being constant. So, being a member of an inertial frame, $A$ is also at rest with respect to suitable additional participants (say observers $J$, $P$ ...), where, let's say, $B$ moved from $A$ to $J$, then on to $P$, etc.

These participants may therefore in turn measure the speed of $B$ (with respect to their, $A$'s, inertial frame): $v_A[~B~]$;
and necessarily it would be found that

$$v_A[~B~] = v_B[~A~],$$

thus both (mutual) speed values being constant.

$B$ is moving away at the same speed close to $c$, while $A$ is stationary.

Right.

Would $A$ think the time in $B$ is slower?

Surely, $A$ would not think such nonsense, either;
but, correspondingly to the measurements by $B$ along with $K$ etc., further comparisons between $A$ and $B$ of their durations can be obtained as

$$\tau B_{\circ A}^{\circ J} = \sqrt{ 1 - \left( \frac{v_A[~B~]}{c}\right)^2 } ~\times~ \tau A_{\circ B}^{\circledS J \circ B},$$

where

• "$B_{\circ A}$" denotes participant $B$'s indication of having been met and passed by participant $A$ (as had been mentioned above already),
• "$B^{\circ J}$" denotes participant $B$'s indication of having been met and passed by participant $J$, and thus
• "$\tau B_{\circ A}^{\circ J}$" denotes $B$'s duration from having been met and passed by participant $A$ until having been met and passed by participant $J$,

• "$A_{\circ B}$" denotes participant $A$'s indication of having been met and passed by participant $B$ (as had been mentioned above already),

• "$A^{\circledS J \circ B}$" denotes participant $A$'s indication simultaneous to participant $J$'s indication of having been met and passed by participant $B$, and thus
• "$\tau A_{\circ B}^{\circledS J \circ B}$" denotes the duration of $A$ between these two of its indications.