Does Time Dilation Occur for Both Twins in the Twin Paradox?

[stevmg]

Two twins A and (evil) B. Twin A remains "stationary" while evil B goes at near light speed to-and-fro. Ignoring General Relativity (acceleration/deceleration) evil B experiences time dilation and as result, when they meet up again, B's clock is behind A's clock.

Look at it from B's point of view, B is stationary and A goes to-and-fro. Why won't A's clock be behind B? Why doesn't A experience time dilation?

 

[PeterDonis]

Before posing any questions about the twin paradox, I highly recommend reading the Usenet Physics FAQ article on it:

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

The short answer to your question is that A and B are not symmetric; A is at rest in the same inertial frame the whole time but B is not. However, there are *lots* of subtleties, which are addressed in the article.

 

[phinds]

... evil B experiences time dilation

No such thing. No one ever "experiences" time dilation, it is something observed by a different reference frame. For example, you, right now as you read this, are traveling at almost the speed of light from some reference frame. Do you fell dilated?

 

[ghwellsjr]

Two twins A and (evil) B. Twin A remains "stationary" while evil B goes at near light speed to-and-fro. Ignoring General Relativity (acceleration/deceleration) evil B experiences time dilation and as result, when they meet up again, B's clock is behind A's clock.

Look at it from B's point of view, B is stationary and A goes to-and-fro. Why won't A's clock be behind B? Why doesn't A experience time dilation?

A's clock will be behind B.

 

[PeterDonis]

A's clock will be behind B.

If I'm understanding the scenario correctly, A remains at rest in the same inertial frame the whole time while B does not. That means A's clock will show more elapsed time than B's when they meet up again.

 

[Dale]

Ignoring General Relativity (acceleration/deceleration)

You can certainly ignore General relativity, but you cannot ignore acceleration.

 

[ghwellsjr]

A's clock will be behind B.

If I'm understanding the scenario correctly, A remains at rest in the same inertial frame the whole time while B does not. That means A's clock will show more elapsed time than B's when they meet up again.

I understood him to be describing two different scenarios. In the first, he said A is stationary and B travels to and fro. In the second he said B is stationary and A travels to and fro. I took him at his word. But, until he clarifies, we won't know. We get ill-defined scenarios like this all the time, don't we?

 

[DrGreg]

In the first, he said A is stationary and B travels to and fro. In the second he said B is stationary and A travels to and fro.

It seems pretty clear to me that he was asserting that these are two different ways of describing the same scenario. Of course, they are not.

 

[ghwellsjr]

It seems pretty clear to me that he was asserting that these are two different ways of describing the same scenario. Of course, they are not.

And, of course, that's the way I answered his question.

 

[Nugatory]

Ignoring General Relativity (acceleration/deceleration)...

You don't need GR to deal with acceleration and deceleration. Special Relativity handles them just fine, as long as there's no gravity.

 

[tom.stoer]

Look at it from B's point of view, B is stationary and A goes to-and-fro.

You can't look at it from B's point of view (in SRT) b/c he does not define an inertial frame ;-)

It's better to introduce an inertial frame with coordinate time t and define the proper times τ_A and τ_B. A and B follow two worldlines C_A and C_B. For their proper times you find

\tau[C_{A,B}] = \int_{C_{A,B}} d\tau = \int_{t_1}^{t_2} dt \sqrt{1-\vec{v}_{A,B}^2(t)}

The two curves are defines such that the they intersect at t₁ and t₂. At t₂ A and B can compare their proper times τ_A and τ_B.

Iff A remains at rest in the inertial frame (with coordinate time t), then his proper time and the coordinate time t coincide, i.e.

\tau[C_{A}] = {t_2}-{t_1}

And for B you may define a circular path C_B with constant speed |v_B| along C_B. Then you find

\tau[C_{B}] = \sqrt{1-v_{B}^2}\cdot({t_2}-{t_1}) = \sqrt{1-v_{B}^2}\cdot\tau[C_{A}]

Now the situation is not symmetric for A and B b/c their curves aren't (in any inertial frame).

But if the curves (other curves, of course) are symmetric (in some inertial frame) then the proper times are identical.