What will happen in a twin paradox like situation if acceleration is not involved?

Question

I have known about the twin paradox for some time and as far as I know that it's not a paradox because one of the twins accelerate in his return journey.

But if this is not the case, then what will they see when they meet each other? If by some means the twins synchronize their clocks when the space twin reaches his ending point and then on his return journey both of the twins' clocks start at $0$ from the beginning of the return journey, then the effect of acceleration will be removed from the time taken for the return journey.

So they when they meet each other and can compare each others' clocks. What will they see then? A will see B's clock to run slow and B will see A's clock to run slow. Will they see that both their clocks has measured the same time for the return journey?

Here A is the twin who goes from point X to point Y at uniform velocity and B is the twin who stays at point X.

Edit: Suppose A sends light signal from Y to X and the light signal reaches X and is immediately reflected by a mirror. The it will go back to Y where it will once again be reflected back to X. In the above case, the light will take 2 years to go from X to Y and back to X. So A will figure out that light takes 1 year to cover the distance XY. When light reaches B for the second time (3 years after the first signal), he will start his clock from $0$.

And also after 3 years A will start his return journey with starting his clock from $0$. So both of the clocks will start from $0$ exactly 3 years after the first signal was sent.

Answer 1

$B$ is the twin who remains on earth. $A$ is the twin who starts at $Y$ and travels to earth.

There are two frames here: The $B$-frame (which is $B$'s frame throughout and also $A$'s frame before he starts his journey) and the $A$-frame (which is $A$'s frame once the journey starts).

Your procedure synchronizes the clocks in the $B$ frame. They are not synchronized in the $A$ frame.

Here is the story in the $B$-frame:

1. When both clocks read $0$, $A$ starts his journey.

2. The journey takes $1$ year. During that year, $A$'s clock runs at 1/2 speed, but $B$'s runs normally.

3. Therefore when $A$ arrives, $A$'s clock says $1/2$, and $B$'s clock says $1$.

Here is the story in the $A$ frame (where it's $B$ who does all the traveling):

1. When $A$'s clock says $0$, $B$'s clock says $3/4$, and he's already partway along his journey.

2. The remainder of the journey takes half a year. During that time, $B$'s clock runs at half speed (so that it advances by 1/4 of a year) while $A$'s run normally.

3. Therefore when $B$ arrives, $A$'s clock says $1/2$ and $B$'s clock says $1$.

I arrived at all this not by thinking about time dilation or clocks running slow, but by meditating on the geometry of the spacetime diagram. Only after I understood it did I translate it into the language of "clocks running slow". In other words, m4r35n357 basically gave you good advice.

Answer 2

The problem is they can't synchronize their clocks unless they are in the same place. In relativity, you need to keep in mind that there is no "universal now".

Answer 3

Acceleration is absolutely not necessary to understand the "Twin Paradox". Furthermore, it is a significant impediment to understanding the "Twin Paradox". The "Twin Paradox" is a great example of a really basic concept badly taught over many decades.

If you learn about the spacetime interval, $\delta \tau^2 = \delta t^2 - \delta x^2$, you will know what all the participants' clocks say, at every stage of the process. Specifically, here is the time shown on the traveler's clock:

$$\delta \tau^2 = (1 / v^2 - 1) \delta x^2$$

Anyone with basic algebra skills will be able to derive this in two lines starting from the Spacetime Interval together with $v = \delta x / \delta t$ (taking $c = 1$). So why make them worry about acceleration on top of that?

If you get hung up on distractions like "acceleration", "clocks running slow", "time dilation" and puzzle over things happening "at the same time" you will never gain an understanding.

[EDIT] following @Not_Einstein's comment, perhaps I should add that for the non-traveler, the time elapsed is simply $\delta \tau = \delta t$, the coordinate time, since $\delta x = 0$. I had not considered this useful (visual clutter), but I am adding it now anyway.