# B Twin Paradox with different speeds

1. Aug 11, 2016

### sqljunkey

Hi,
I'm trying to understand the twin paradox and time dilation.

Someone told me that if observer A and observer B are traveling apart from each other with each having a uniform speed of .4c they will have the same age when they return back at the point of origin. Even though I'm unsure of this outcome, because observer A would think he's standing still and B is moving away and so when B returns at the origin A should be older, but I'm going to assume what this person told me is right and they both will be the same age when they return at point of origin because they changed directions.

I'm assuming they are moving with uniform speed and that no acceleration is involved. They are traveling in complete vacuum.

To extend this I'm adding an extra observer C who is standing at the point of origin. I'm assuming that observer C will see both A and B with the same age when they return from this trip.

If observer A was standing still and observer B was the one moving, when B comes back he would be 8 years and observer A and C would be 10 years old.

Now I will change the original example so observer A is traveling at a much slower speed. To be annoying lets say he moves a total of 1 meter back and forth in the 10 year duration. So the first 5 years he moves away from the origin with a uniform of .1 meters per year and then the next 5 years he will move back with the same uniform velocity.

Traveler B on the other hand moves with very fast speeds so that together they move apart with a combined uniform velocity of .8c. Just like the above example, they are both traveling apart at .8c and they both change their directions. So just as the above example, due to the principle of relativity they should be the same age when they return at the origin. I'm assuming they both get there at the same time.

Will observer C see them both being the same age as in the above example? For example both will be 8 years instead of just B being younger and A being 10 years old?

I assuming the second example is identical to the first example.
Thanks!

2. Aug 11, 2016

### Staff: Mentor

Then they cannot return to the origin.

3. Aug 12, 2016

### Ibix

Presuming that you actually mean that the travellers turn around instantaneously rather than that they do not accelerate (to address Dale's comment) then no, your situations are not identical. You seem to be trying to use a Newtonian intuition to reason about relativistic experiments, so you have neglected the relativity of simultaneity and the relativistic velocity addition formula.

I recommend looking up the Lorentz transforms. Try to work out the coordinates if the key points in your experiment (departures, turnarounds, returns) in C's rest frame and then use the Lorentz transforms to determine the times in the various moving frames.

4. Aug 12, 2016

### vanhees71

You don't need Lorentz transforms, just a trajectory of each observer, say $\vec{x}_a(t)$ and $\vec{x}_b(t)$, where I express the trajectory in terms of the space-like components of an arbitrary inertial reference frame (supposed we talk about special relativity only and not general relativity). Then the aging is given by the proper time of each observer, which you can simply compare:
$$\tau_a=\int_{t_1}^{t_2} \mathrm{d} t \sqrt{1-\vec{v}_a^2(t)/c^2}, \quad \tau_b=\int_{t_1}^{t_2} \mathrm{d} t \sqrt{1-\vec{v}_b^2/c^2}.$$
Then it becomes a no-brainer to compare the different values of these proper times :-).

5. Aug 12, 2016

### Ibix

Indeed. But it looks to me like, in both threads the OP has started, he or she is trying to use Newtonian intuition on relativistic problems. Although I agree that one can completely avoid the Lorentz transforms in predicting the outcome of the twin paradox, I think in this case that the LTs will help more - introducing concepts like the lack of absolute simultaneity and leading to the velocity transforms.

6. Aug 14, 2016

### m4r35n357

A bit of advice; the twin "paradox" is already as simple as it can be (as below). I seriously doubt that you will gain any insights by making it more complicated.

Assuming traveler B goes 4 ly away at 0.8c (which takes in 5 years in A's frame) then returns . . . while A stays put.
$$\tau_A ^2 = 5^2 - 0^2 = 5^2 \implies \tau_A = 5$$
$$\tau_B ^2 = 5^2 - 4^2 = 3^2 \implies \tau_B = 3$$ for each half of the journey. Now double those figures for the return journey and see that A ages 10 years whilst B ages 6 years.

Once you understand this you can knock yourself out with lengthy application of the Lorentz Transform, agonize about acceleration, gravity, the equivalence principle or whatever, but you had better make sure you get the answers above ;)