## Main Question or Discussion Point

hi............

In the twin paradox, we consider that one of the twins went out of the surface of earth ACCELERATING , then suddenly TURNED BACK and then DECELERATED back towards the earth with some relativistic velocity………. Why is it so?
What if the twin goes on a round trip around the earth and returns to the same place….. will the paradox be still effective?
If not, please do explain the reason behind this……….

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CompuChip
Homework Helper
That's an interesting variation. Usually we just consider that they go out from earth at constant velocity (of, say, 0.99c) and they come back at (the same) constant velocity. So almost everywhere along the journey, the velocity is constant and SR holds.

The point is, that to turn around from outbound to inbound, there has to be an acceleration. It cannot be instantaneous, so it has to be some (short but) finite time, and during that time special relativity is not valid - which allows you to resolve the paradox.

If you go on a round trip around earth, you are actually accelerating all the time... you're undergoing centripetal acceleration to stay in a circular orbit (remember that acceleration is not just change of magnitude of velocity, but also change of direction of velocity) and in fact SR will be nowhere valid because (considering the earth at rest) clearly one of the twins is accelerating (or, considering the rotation of the earth) they are in different gravitational fields and you should be using general relativity in the first place.

tom.stoer
Here it becomes slightly more difficult b/c you may want to consider the gravitational field.

First consider a formula for proper time along a curve C valid both in SR and GR

$$\tau_C = \int_C d\tau$$

The difference of proper times for the two twins measured along two curves C and C' connecting two spacetime points A and B is

$$\Delta\tau_{C,C^\prime} = \Delta\tau_{C_{A\to B}, C^\prime_{A\to B}} = \int_{C_{A\to B}} d\tau - \int_{C^\prime_{A\to B}} d\tau$$

Of course one can evaluate this in flat space, but it is possible to take a gravitational field into account.

As a first step suppose that the twin surrounding the earth is at constant altitude along his journey and that he is travelling closed to speed of light. Then one can safely neglect the effect of the gravitational field and use flat Minkowski spacetime. One can parameterize the curves as follows:

First twin at rest

$$C: x^\mu = (t,R,0,0)$$

Second twin

$$C^\prime: {x^\prime}^\mu = (t,R\cos\,\omega t,R\sin\,\omega t,0)$$

with

$$\omega = \frac{2\pi}{T} = \frac{v}{R}$$

Now one can change integration variables from τ to t, i.e. integrate dτ along the curve from from t=0 to t=T and solve for the two line integrals.

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ghwellsjr
Gold Member
hi............

In the twin paradox, we consider that one of the twins went out of the surface of earth ACCELERATING , then suddenly TURNED BACK and then DECELERATED back towards the earth with some relativistic velocity………. Why is it so?
What if the twin goes on a round trip around the earth and returns to the same place….. will the paradox be still effective?
If not, please do explain the reason behind this……….
Actually, Einstein used exactly the scenario you described when he introduced the Twin Paradox, although he didn't call it that. You can read about it at the end of section 4 of his famous 1905 paper:

http://www.fourmilab.ch/etexts/einstein/specrel/www/

In fact, he concluded that a clock at the equator would tick slower than a clock at one of the poles, ignoring gravity.

tom.stoer
btw.: the calculation as rather simple:

$$\tau = \int d\tau = \int_0^T dt \sqrt{\dot{x}_\mu \dot{x}^\mu}$$

Now we calculate this for the two curves

$$C: \dot{x}^\mu = (1,0,0,0) \Rightarrow \dot{x}_\mu \dot{x}^\mu = 1$$

$$C^\prime: {\dot{x}^\prime}^\mu = (1,-R\omega\sin\,\omega t,R\omega\cos\,\omega t,0) \Rightarrow \dot{x}_\mu \dot{x}^\mu = 1-v^2$$

That means we find for the two proper times

$$\tau_C = T \Rightarrow \frac{2\pi R}{v}$$

$$\tau_{C^\prime} = \sqrt{1-v^2} \, T \Rightarrow \sqrt{1-v^2} \, \frac{2\pi R}{v}$$

where the last expression is valid for a closed curve.

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D H
Staff Emeritus
What if the twin goes on a round trip around the earth and returns to the same place….. will the paradox be still effective?
If not, please do explain the reason behind this……….
Minor aside: Stop with the dots, please. They detract from, not add to, what you are saying.

The exact experiment you suggest has been performed multiple times with ever-increasing accuracy. There is an added complication that because gravity is involved in that one must look to general relativity as well as special relativity for the answer.

Suppose you place an atomic clock in an airplane and fly the airplane on a trip around the world. General relativity tells us that the clock in the plane will run a bit faster than an Earthbound clock because the plane will be flying at some altitude above the surface of the Earth. Special relativity tells us that an eastbound flight will lose time because the plane is going faster than a clock on the Earth's surface that is moving due to the Earth's rotation; a westbound flight will gain time because the plane is going slower than a clock on the Earth's surface. Add the general and special relativistic effects together and the westbound flight will gain time while the eastbound flight can gain or lose time depending on the altitude at which the plane flies.

This is exactly what the Hafele–Keating experiment (first conducted in 1971) observed.

The point is, that to turn around from outbound to inbound, there has to be an acceleration. It cannot be instantaneous, so it has to be some (short but) finite time, and during that time special relativity is not valid - which allows you to resolve the paradox.
Why do you think that SR cannot handle acceleration?

If you go on a round trip around earth, you are actually accelerating all the time... you're undergoing centripetal acceleration to stay in a circular orbit (remember that acceleration is not just change of magnitude of velocity, but also change of direction of velocity) and in fact SR will be nowhere valid .....
A traveler in orbit does not accelerate at all.

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D H
Staff Emeritus
The point is, that to turn around from outbound to inbound, there has to be an acceleration. It cannot be instantaneous, so it has to be some (short but) finite time, and during that time special relativity is not valid - which allows you to resolve the paradox.
Nonsense. The clock hypothesis says that only velocity, not acceleration, affects time dilation. The only thing special relativity can't handle is gravitation.

tom.stoer
The clock hypothesis says that only velocity, not acceleration, affects time dilation. The only thing special relativity can't handle is gravitation.
Exactly.

The equations I used above are perfectly valid both in GR and SR; of course for a round trip accelaration is present, nevertheless the calculation is valid and one finds the well known factor for time dilation.

Here it becomes slightly more difficult b/c you may want to consider the gravitational field.

First consider a formula for proper time along a curve C valid both in SR and GR

$$\tau_C = \int_C d\tau$$

The difference of proper times for the two twins measured along two curves C and C' connecting two spacetime points A and B is

$$\Delta\tau_{C,C^\prime} = \Delta\tau_{C_{A\to B}, C^\prime_{A\to B}} = \int_{C_{A\to B}} d\tau - \int_{C^\prime_{A\to B}} d\tau$$

Of course one can evaluate this in flat space, but it is possible to take a gravitational field into account.

As a first step suppose that the twin surrounding the earth is at constant altitude along his journey and that he is travelling closed to speed of light. Then one can safely neglect the effect of the gravitational field and use flat Minkowski spacetime. One can parameterize the curves as follows:

First twin at rest

$$C: x^\mu = (t,R,0,0)$$

Second twin

$$C^\prime: {x^\prime}^\mu = (t,R\cos\,\omega t,R\sin\,\omega t,0)$$

with

$$\omega = \frac{2\pi}{T} = \frac{v}{R}$$

Now one can change integration variables from τ to t, i.e. integrate dτ along the curve from from t=0 to t=T and solve for the two line integrals.
Could you explain why you think that we can safely neglect the effects of gravity when one twin travels in orbit at relativistic speeds?

tom.stoer
Could you explain why you think that we can safely neglect the effects of gravity when one twin travels in orbit at relativistic speeds?
Both twins have to stay approximately at the same altitude; then for speed closed to c gravitational effects are small; for both twins (nearly) at rest but at different altitudes it's just the other way round

If we consider the case of polar orbits (without any proper acceleration) and include the gravitational effects then by approximation we get:
$$\Large \frac{\tau_S}{\tau_E} = 1+\frac{GM}{Rc^2} - \frac{3GM}{2rc^2}$$
or in geometric units:
$$\Large \frac{\tau_S}{\tau_E} = 1+\frac{M}{R} - \frac{3M}{2r}$$

G = 6.6726E-11
M = 5.9742E+24 kg (0.004435407 m)
R = 6378000.1 m

So in this case the traveling clock is faster if r − R > 3189km otherwise it is slower.

Obviously if the traveler is at a low circular orbit traveling relativistically he must be accelerating.

tom.stoer