Time between two observer in special relativity

Question

Let $A$ and $B$ be two observers and suppose they meet when both their clock are zero. Suppose that $B$ moves with constant velocity $v$ in relation to $A$ then we have $$t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} t'$$

where $t$ is the time for observer $A$ and $t'$ the time for observer $B$.For simplicity suppose that $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=0.5$ then when the clock of $B$ is $t'=1$ the clock of $A$ is $t=2$.

From the point of view the observer is moving with velocity $-v$ and so by symmetry

$$ t'=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} t$$

So when the clock of observer $A$ is $t=2$ the clock of $B$ is $t'=4$

That is if the clock of $A$ run faster than $B$, the clock of $B$ run faster than the clock of $B$ by symmetry.

Isn't this a contradiction ?

Answer 1

In one equation, $\Delta t$ is the time, as measured by Observer A, between two events at which Observer A is present.

In the other equation, $\Delta t$ is the time, as measured by Observer A, between two events at which Observer B is present.

If you choose to use the same symbol to denote two different things, then it's very easy to derive a contradiction. You don't need relativity for this. Just set $x=3$ and $x=7$. Then from $x=x$, you can derive $3=7$. This is not a cause for concern.

Update: Now that you've edited the question to replace $\Delta t$ with $t$, the answer remains exactly the same. You've used the same letter $t$ to mean one thing in one equation and a different thing in the other. So this answer still applies if you just drop the $\Delta$s.

(Relatedly: You say that "$t$ is the time for observer $A$". The time of what for Observer $A$? If your equations are correct, then it must be the time of one event in one equation and the time of a different event in the other.

Answer 2

At first it seems to have only 2 variables: $t_A$ and $t_B$ and a contradition: ($t_A > t_B$ and $t_B > t_A$). But there are 4 variables: $t_A$, $t_B$, $t'_A$ and $t'_B$.

$t_A > t_B$, measurements of both clocks at A frame.
$t'_B > t'_A$, measurements of both clocks at B frame.

So, there is no contradiction.

Answer 3

This is why the Loedel Diagram was invented. It is the Minkowski Diagram drawn by the the observer who sees A and B moving at $\pm v'$ with $v = 2v'/(1+v'^2)$:

A sees B's clock running slower and B sees A's clock running slower because when they are separated, their planes of simultaneity are different.

That they both see the other person's "t" coordinate as less than their's is the same phenomenon that occurs when two cars are driving opposite the directions on the highway: they both see the other with a larger "left" coordinate than their own.

Answer 4

Since you edited your question, I shall edit my answer to suit.

Instead of using $t$ and $t'$ I'll use $t_A$ and $t_B$ for clarity.

So in some birds eye frame of reference, you have $A$ who is stationary, and $B$ who is moving with velocity $v$ away from $A$.

Lets say they agree to measure the time difference after $\tau$ seconds has elapsed on their individual clocks.

In their individual reference frames:

$A$:

A sees B moving with velocity $v$, away from B and $A$ measures the time between $t_1$ and $t_2$ as $\Delta t_A$.

$B$:

$B$ in their reference frame sees $A$ moving away at velocity $-v$ and like $A$, measures the time after $t$ second has elapsed on his clock $\Delta t_B$

You'll notice that each observer measures the same time elapsed between $t_1$ and $t_2$, in their own frame of reference which is $\tau$.

The amount of time that has elapsed on $A$'s clock in comparison to $B$'s clock, according to $B$, is $$ \Delta t_A = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\Delta t_B \tag{1} $$ Because $B$ was moving faster, $A$'s clock sped up according to $B$.

The amount of time elapsed for $B$'s clock, according to $A$ is $$ \Delta t_B = {\sqrt{1-\frac{v^2}{c^2}}} \Delta t_A \tag{2} $$

The dilation formula just compares the two according to the other in a specific reference frame. You cannot switch them around like that as you've done in your contradiction as they're unique to the reference frame.

So basically, $A$ being stationary, $A$'s clock runs faster than $B$ according to $B$ and $B$'s clock moving, runs slower according to $A$.

Answer 5

It is not a contradiction because the observers have different notions of synchroneity, as seen in this diagram