Q:

Something seems wrong with the Twins Paradox of Special Relativity. Consider the standard version of this thought experiment which lasts 20 years. But let’s assume the space traveling twin accelerates at a constant 1 g for the entire trip. This can be done as follows: his rocket ship has a main burner that accelerates him at a constant 1 g. He travels for 5 years with his main burner facing Earth; then he travels the next 10 years with his main burner facing away from Earth; and finally he travels the last 5 years with his main burner facing Earth. This procedure bring him to rest on Earth after being accelerated at 1 g for all 20 years of the trip. For 15 of the 20 years his velocity will be well above 50% the speed of light, so there should be plenty of time dilation between himself and his Earthbound twin. But how can this be? Both twins have the same relative velocity with repect to one another at all times during the trip. And both have the same acceleration with respect to each other at all times. Their two frames of reference a exactly equivalent. Consequently there should be no difference between the age of the twins when they meet again 20 years later.

- Pat Dolan (age 43)

Seattle, WA

- Pat Dolan (age 43)

Seattle, WA

A:

Pat- That's a nice question. As you say, the relative velocities and
accelerations of the two objects seem at first like they should just be
exact opposites of each other. Special relativity describes how space
and time are related for observers who do not accelerate and are not in
the presence of gravitational fields.

It is most convenient to analyze situations like these in just one nonaccelerating frame of reference. Only one twin, the stay-at-home one, is stationary in a single frame of reference. In this frame, one can add up all the time on his watch, and compare it with the time that accumulates on the traveling twin's watch, and compare it when they meet up. You can also do the calculation in any other nonaccelerating frame of reference, and you will always get the same answer, that the twin that accelerates will have less time on his watch when he meets his brother.

Special relativity also gives you the tools to analyze the situation from the traveling twin's perspective.

The traveling twin does not stay put in a single nonaccelerating frame of reference during his journey, however, so life is more complicated for him. Instead, at each moment of his journey, we can define a frame of reference that is moving along with the traveling twin. In this frame of reference, we can ask the question what time is read on his brother's watch. Then a moment later, we have to pick another frame of reference moving along with the traveling twin's new speed, and ask what time is read on his brother's watch. If we make a graph of the time read on the stay-at-home twin's watch in all of these different frames of reference as a function of the traveling twin's watch's reading, we will find it's not a straight line. On out outgoing part, the traveling twin thinks his stay-at-home twin's watch is running slow, just like his stay-at-home twin thinks the traveling twin's watch is runnning slow. On the turnaround part, the traveling twin thinks his stay-at-home brother's watch speeds up, and then on the return trip, he thinks the stay-at-home twin's watch runs slow again. It's a mess, but it gives the same answer as doing everything in just one frame.

In General Relativity. a broader class of reference frames is considered. These accelerating frames, like that of the traveler, have additional peculiar effects on time, but must agree with the Special Relativity answers when they are applicable (for example, in cases with no gravitational fields).

Your argument about the symmetry between the two twins assumes that only their relative motion, including acceleration, matters. It turns out that acceleration is not only a relative quantity -- you can devise experiments which give you different outcomes depending on the acceleration of the apparatus, without regard to the acceleration of anything else.

The idea that only relative motion can matter is known as Mach's principle. Einstein tried to follow it for a while, but ultimately dropped it in General Relativity.

Mike W. and Tom J.

It is most convenient to analyze situations like these in just one nonaccelerating frame of reference. Only one twin, the stay-at-home one, is stationary in a single frame of reference. In this frame, one can add up all the time on his watch, and compare it with the time that accumulates on the traveling twin's watch, and compare it when they meet up. You can also do the calculation in any other nonaccelerating frame of reference, and you will always get the same answer, that the twin that accelerates will have less time on his watch when he meets his brother.

Special relativity also gives you the tools to analyze the situation from the traveling twin's perspective.

The traveling twin does not stay put in a single nonaccelerating frame of reference during his journey, however, so life is more complicated for him. Instead, at each moment of his journey, we can define a frame of reference that is moving along with the traveling twin. In this frame of reference, we can ask the question what time is read on his brother's watch. Then a moment later, we have to pick another frame of reference moving along with the traveling twin's new speed, and ask what time is read on his brother's watch. If we make a graph of the time read on the stay-at-home twin's watch in all of these different frames of reference as a function of the traveling twin's watch's reading, we will find it's not a straight line. On out outgoing part, the traveling twin thinks his stay-at-home twin's watch is running slow, just like his stay-at-home twin thinks the traveling twin's watch is runnning slow. On the turnaround part, the traveling twin thinks his stay-at-home brother's watch speeds up, and then on the return trip, he thinks the stay-at-home twin's watch runs slow again. It's a mess, but it gives the same answer as doing everything in just one frame.

In General Relativity. a broader class of reference frames is considered. These accelerating frames, like that of the traveler, have additional peculiar effects on time, but must agree with the Special Relativity answers when they are applicable (for example, in cases with no gravitational fields).

Your argument about the symmetry between the two twins assumes that only their relative motion, including acceleration, matters. It turns out that acceleration is not only a relative quantity -- you can devise experiments which give you different outcomes depending on the acceleration of the apparatus, without regard to the acceleration of anything else.

The idea that only relative motion can matter is known as Mach's principle. Einstein tried to follow it for a while, but ultimately dropped it in General Relativity.

Mike W. and Tom J.

*(published on 10/22/2007)*

Q:

My question is about time dliation. I often hear theories that if an astronaut goes off into space at 99.9% of the speed of light for 10 years, as measured by his watch, then when he returns he'll find more time has passed locally hear on Earth.
I assume that 99.9% of the speed of light in this scenario is measured "relative" to the Earth (maybe this is the bit I'm getting wrong). If that's the case then isn't it equally valid to say that earth is travelling at 99.9% of the speed of light relative to the astronaut and that when the Earth returns to the astronaut it will be the Earth which has observed less time passing.
I'd like to add that I found your answers here to be very interesting, informative and complete and I appreciate you taking the time to share your expertise.

- Neil (age 44)

UK

- Neil (age 44)

UK

A:

We've moved your question to be a follow-up to one that comes closer to it.

Your assumptions are right, but you're missing one key step. When the traveler accelerates back toward earth, he isn't using an inertial reference frame. The speed-up of the earth's clocks that he sees due to accelerating toward them is just enough to make him agree with the stay-at-home about what their age ratio is when he returns.

Mike W.

*(published on 11/21/2013)*

Q:

Since the twins will not be able to read the other's clock, let us replace the clocks with a light emitting strobe that emits a signal precisely every 1 second. From the frame of reference of the twin on Earth (Say A), he sees that the duration between each light signal keeps expanding as the other twin (Say B) accelerates away from A, until he sees no light signal when B reaches speed of light. Till this, I am able to understand. But once B starts to decelerate, stop, and accelerates back towards A, the light signals will now start coming in at 1 sec intervals (when B is stationary), and then start coming in at increasing frequency until B reaches A. Now, my contention is that when A & B meet again, the number of light signals emitted by both A & B will be exactly same. Relativity seems to be just an illusion or perception of time dilation, and not a real phenomenon. Is this correct?

- Subra (age 42)

Chennai, India

- Subra (age 42)

Chennai, India

A:

No, it's not correct. You've assumed that the correct description of how things behave is what we call Galilean Relativity, but that turns out not to be true.

In the real world, the twin B who is leaving will *not* reach the speed of light, as seen by the stay-at-home, A. No matter how long B's rocket engine fires, she will only get close to the speed of light. This behavior is directly confirmed by all sorts of particle accelerators. In this flat-spacetime picture, where we ignore cosmological effects, the signals from B will never cease reaching A.

The rate at which the signals from B reach A will be not far from what you estimate when B isn't moving too fast relative to A. When B is moving at a significant fraction of c, the type of calculation you made just doesn't work, it doesn't describe the real world.

The simple effects of the sort you're describing were known since Romer used them to find the speed of light from the observations of Jupiter's moons, whose rotations make a sort of clock. That was before 1700 AD. The world just turns out to be different from that.

Mike W.

*(published on 11/19/2014)*

Q:

While reading on the twin paradox I found many (internet) sources say it is not a paradox because one of the twins can be viewed as the accelerating one and the other twin as stationary. I wondered what determined whom of the twins is the accelerating one, i.e. what is acceleration?One answer I saw was that the effects of acceleration can be observered in one but not the other twin, using e.g. an accelerometer. Which leads to the next question: how come the accelerometer would show acceleration in the twin moving away from earth, but not on the earthbound twin?This I have not found an answer to. What I can only speculate is that, since acceleration is a change in velocity relative to some reference, there must be some universal reference frame for location. In this frame the Earth's velocity does not change (much), while the velocity of the twin in the rocket does. Is this true? If not, than what is acceleration and how does one define which twin is accelerating?

- Ben Steemers (age 29)

Nethetlands

- Ben Steemers (age 29)

Nethetlands

A:

If the acceleration is caused by anything other than gravity (e.g. rocket engines) it is indeed directly measurable with accelerometers of all types, without requiring any chosen reference frame. So for that type of problem the usual story is ok. If, on the other hand, the important acceleration (where the spaceship quits leaving and starts to return) is due to slinging around a star or planet, via gravity, then that story about feeling accelerations breaks down. The effects of uniform gravity cannot be detected in any local experiment, and the gravity here can be very nearly uniform. So the local quantity that has direct physical meaning is not acceleration but only acceleration with respect to a locally free-falling object. That has some sort of cause, like the rocket engines, so it's not surprising that it has measurable effects.

Yet the results still work. The twin who goes out and slings around the heavy planet does come back less aged than the stay-at home. The full description requires general relativity. The traveling twin spends time "down" near the heavy planet, and that slows her clocks, as seen by the "up" stay-at-home.

Mike W.

p.s. Please follow up if this doesn't quite answer your question.

*(published on 09/06/2016)*

Q:

Hi Mike,Thank you very much for taking the time to answer my question.A very interesting line of thought on the uniform gravity in the slingshot scenario. The conclusion you draw from this is "So the local quantity that has direct physical meaning is not acceleration but only acceleration with respect to a locally free-falling object. That has some sort of cause, like the rocket engines, so it's not surprising that it has measurable effects. "Yet this does not completely answer my question. Do you mean by 'locally free falling object' an object locally to the travelling ship? If this is so and you also mean that whatever causes the change in velocity of the ship does not effect this free-falling object, than I think we can boil your answer down to: change in velocity has a cause and can be measured relative to a locally free-falling object.If this free-falling object is what we use as our reference point for velocity, than what does one choose to be the locally free-falling object in the sling-shot scenario? Could the twin back on earth not see the twin in the space-ship as his locally free-falling object? If he does than the twin on earth would be accelerating away from, and eventually back to, the twin in the space-ship. Unless the reason why this cannot be is that the twin in the space-ship has a cause for it's change in velocity (the propulsion of the rocket), while the twin on earth does not (or it does have a cause, but this cause has a lesser effect). In which case, can we make a definition for any cause that changes velocity, not specific to rocket engines? (i.e. if X is true, velocity must change)What it all boils down to is, can we point to something specific that will always tell us which of the two twins is the accelerating one and which one the static, even if one of the two would be e.g. swimming in the ocean whilst the other is standing on the shore?Kind regards

- Ben Steemers (age 29)

Netherlands

- Ben Steemers (age 29)

Netherlands

A:

"Could the twin back on earth not see the twin in the space-ship as his locally free-falling object? "

No, because the other twin is not local. She's far away.

We cannot in general say who is accelerating. We have a choice of coordinates that allows any object to be described as accelerating or not. What we can say is whether any object is accelerating with respect to what it would do if it were in free-fall.

So yes, change in velocity with respect to free-fall has some specific cause. Change in velocity in other frames does not have to have a cause, since you can pick different frames accelerating with respect to each other. For example on Earth, we usually pick the Earth-frame rather than the free-fall frame. We say that somebody falling is accelerating, even though they feel nothing different thatn they would in a much bigger field or no field. We say that we aren't accelerating, even though we feel forces (on my feet, I'm using a standing desk) and can measure effects of those forces like different clock rates in the basement and attic. Such effects are alway present for paths other than free-fall and never for free-fall.

In the standard version of the twin story, gravity is left out, i.e. the gravitational field is assumed to be uniform, which is treated most simply as uniformly zero. That means we call call anybody's local free-fall "stationary" and it doesn't matter who we pick. In my slingshot version, however, the gravity isn't uniform so we need a more complicated analysis.

Mike W.

*(published on 09/12/2016)*

Q:

I have heard two things said in Physics which appear to contradict each other. First I have heard it said that if I stay on earth and you go on a trip going near the speed of light for many years (my years), when you come back, I will have aged much more than you. But i have also heard it said that velocity is only relative to something. Well if that's the case, when you go on your trip, from your point of view, you can be standing still and i am the one moving near the speed of light. So why am I not the one aging more slowly?

- Larry Elterman (age 64)

02138

- Larry Elterman (age 64)

02138

A:

You can try our old answers and follow up if you need more.

Mike W.

*(published on 08/06/2017)*