# Why the stay-at-home twin is not considered to be accelerating?

The key point to take away from the twin paradox is that the time difference is due to the moving twin's change of frame.

The acceleration is essentially irrelevant except for the fact that the moving twin must decelerate and accelerate so that they can change their frame.

You can easily set up a twin paradox without any accelerations. I use the values from above.

Set up a stationary observer on earth, another at the turning poiint and another at twice the distance of turning point. All three synchronise their clocks to read 0.

The moving twin_1 accelerates before the experiment starts in such a way that she is travelling at V when she passes earth, and she passes when the earth clock reads 0. She looks at the earth clock and sets her clock to 0.

A similarly moving twin_2 passes the point twice the turning point distance away, travelling at V towards earth.

The two moving travellers meet at the turning point. The stationary person's clock reads 5 but twin_1's clock reads only 4.5.

twin_2 sets her clock to be the same at twin_1, or 4.5. twin_2 continues to earth and arrives at earth when earth says it is 10. However, twin_2's clock reads 9, made up of the 4.5 which she set, plus the 4.5 for her journey time, totalling 9.

So, whereas the earth bound twin says 10 years have elapsed since twin_1 left, the total travel time measured by the two moving twins is only 9 years.

So, we have the same time difference but there have been no accelerations.

Acceleration has nothing to do with the twin paradox other than it is necessary for the moving twin to accelerate and decelerate to change her frame. The twin paradox is about changing frames.

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haushofer and dayalanand roy
Frodo said:
If I see a light 186,000 miles away flash at exactly noon, I observe that it flashed a second before noon.
I like that convention. This issue is always a source of confusion.
It is one which I thought was in general use so I am very surprised by those who appear never to have heard of it or recommend it not be used. All here seems to cite wiki as their authority so I shall too. See Special relativity and go to Measurement versus visual appearance to find (my underlines):
Scientists make a fundamental distinction between measurement or observation on the one hand, versus visual appearance, or what one sees.
Energy has one meaning in normal life and a different, formal meaning in science. So does observe.

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dayalanand roy
A.T.
Science Advisor
Acceleration has nothing to do with the twin paradox other than it is necessary for the moving twin to accelerate and decelerate to change her frame. The twin paradox is about changing frames.
Acceleration doesn't affect the clock rate directly, but it makes no sense to present it as irrelevant but also necessary. This just confuses people.

The analogy I would use is a car at constant speed:

proper acceleration <-> changing the cars direction
proper time <-> traveled distance

Changing direction doesn't directly affect the rate at which distance is traveled, but it can affect the total traveled distance between two points.

dayalanand roy, etotheipi and Dale
Dale
Mentor
The acceleration is essentially irrelevant except for the fact that the moving twin must decelerate and accelerate so that they can change their frame.
I wouldn’t go that far. It is irrelevant in the calculation of the final difference in age, but it is relevant in breaking the symmetry. Any scenario which avoids acceleration must break the symmetry in some other way.

dayalanand roy
The Wikipedia page is a good start. Two or three of the editors in the history have usernames that are suspiciously familiar from here.
Thank you.

I have discovered The Equation of Motion in Rindler Space

Consider a frame travelling at velocity v relative to a fixed frame, and at (local) acceleration a as depicted by the top pair of frames below.

We can instead consider this moving frame to be at rest relative to the stationary frame, where v and a are zero; as long as we consider that the frame experiences a gravitational field determined by the rate of acceleration. We now use Rindler coordinates and the Lorentz transformation equations does not have a term v in them. This is depicted by the lower pair of frames below.

In the diagram , the top pair of frames are in conventional mode. The lower two frames are in Rindler coordinates.

Acceleration doesn't affect the clock rate directly, but it makes no sense to present it as irrelevant but also necessary. This just confuses people.
If, as in the twin paradox, the acceleration causes a change in speed then this does does affect the clock rate.

I think therefore your statement is both incorrect and very misleading.

99% of the explanations of the twin paradox say "It is caused by the fact that the travelling twin accelerates" but they never give a calculation showing how it comes about, nor provide an equation relating the age difference to the rate and duration of the acceleration experienced. I suggest the statement is therefore as meaningless as saying "It is caused by the fact that the travelling twin is wearing a bikini".

The twin paradox is explained by the fact that the moving twin changes their frame of reference. That is the essential kernel of the solution - everything else is second order. It is a simple application of the Lorentz transformation equations to get the resultant time difference.

Secondly, no-one citing acceleration as the cause ever shows how acceleration can account for the fact that both twins see each each other age more slowly than themselves during the entire time. This can only be resolved by invoking a frame change. It has nothing to do with acceleration.

Perhaps we could ask the original poster dayalanand roy , who has obviously worked on the subject, whether he found the change of frame explained things to him.

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A.T.
Science Advisor
Acceleration doesn't affect the clock rate directly, ...
If, as in the twin paradox, the acceleration causes a change in speed then this does does affect the clock rate.

I think therefore your statement is both incorrect and very misleading.
What I'm refering to is the clock hypothesis:

https://en.wikipedia.org/wiki/Time_dilation#Clock_hypothesis
The clock hypothesis is the assumption that the rate at which a clock is affected by time dilation does not depend on its acceleration but only on its instantaneous velocity.

Note that I explicitly wrote "directly".

dayalanand roy and Dale
In the thread Twin paradox not including accelerations, it is wrong where? Andrew Kirk, a Science Advisor, Homework Helper, Insights Author and Gold Member states something identical to my assertion, namely (my underlines):
In the Twin Thought Experiment, acceleration is not the key explanation of why the travelling twin ages less than the non-travelling twin. Rather, it is the fact that the inertial frame of the travelling twin has changed. The inertial frame while going out is very different from the one while coming back. That reason applies whether we are considering the original thought experiment (in which there is acceleration) or the one you describe (in which there is no acceleration).

In both cases the time measurement that is delivered back to the home twin (in your case the measurement is by the amount of fading of the photo) is based on two very different inertial frames, and hence is less than the measurement based on the home twin, which relates to only one inertial frame.

In summary, it is the number of different inertial frames that contribute to each time measurement, rather than the acceleration, that allows to see why the travelled and the non-travelled time measurements differ.

PeterDonis
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2019 Award
In the thread Twin paradox not including accelerations, it is wrong where? Andrew Kirk, a Science Advisor, Homework Helper, Insights Author and Gold Member states something identical to my assertion
And if you go read my post #60 in that same thread, you will find this, in response to a similar statement by another poster:

it's important to understand that this is a convention, not required by the laws of physics. And it's not what causes the two clocks to have different readings at the end--the cause of that is the different lengths of the two paths through spacetime.
Here the "this" that is a convention is what Andrew Kirk described as "the inertial frame changing". And, as my quote just above says, that is not what causes the twins' clocks to have different readings at the end; the cause of that is the different lengths of the paths they take through spacetime.

dayalanand roy
Janus
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Science Advisor
Gold Member
99% of the explanations of the twin paradox say "It is caused by the fact that the travelling twin accelerates" but they never give a calculation showing how it comes about, nor provide an equation relating the age difference to the rate and duration of the acceleration experienced. I suggest the statement is therefore as meaningless as saying "It is caused by the fact that the travelling twin is wearing a bikini".
If you were only concerned with how much much the "traveling twin" ages during the trip according to the "stay at home twin" or according to the traveling twin, then you don't have to consider the acceleration of the traveling twin. The stay at home twin gets his answer by measuring time dilation for the traveling twin, and th traveling twin gets his answer by measuring length contraction between his twin and turn around point.

Acceleration of the traveling twin does become important when you are concerned with how the traveling twin arrives at the conclusion that his stay at home twin ends up being older than he is upon his return.
During the outbound leg, the traveler would say that his twin aged more slowly than he did, and would say the same during the return leg. It during his acceleration at turn around that he will say that his stay at home twin aged more rapidly.
There is an equation for how an accelerating observer measures non-local clocks:
$$T = \frac{t}{\sqrt{1-\frac{2ah}{c^2}}}$$

t is the time rate as measured by the observer's own clock
T is the time rate on a clock a distance of 'h' from the observer in the direction of the acceleration.
a is the magnitude of the acceleration.

In essence, clocks in the direction of the acceleration run fast according to the accelerating observer at a rate that depends on how far away they are and the magnitude of the acceleration.

It does not matter if the distant clock doesn't share the observer's acceleration (though this will mean that the distance between the two doesn't remain constant over the period of acceleration, and this, in turn, has to be factored in when calculating the total time difference over the acceleration period).

dayalanand roy and Sagittarius A-Star
A.T.
Science Advisor
Here the "this" that is a convention is what Andrew Kirk described as "the inertial frame changing". And, as my quote just above says, that is not what causes the twins' clocks to have different readings at the end; the cause of that is the different lengths of the paths they take through spacetime.
I'm not sure if there is an ultimate answer to those "what is the best explanation/reason" questions.

To me both, "changing frames" and "paths through spacetime" are abstractions, which rely on conventions and geometrical interpretations. But the difference/asymmetry in elapsed proper-times between two meetings, doesn't rely on such. So there must be something else that differentiates the twins, that also doesn't rely on such. And that is the difference in proper acceleration.

This should be part on any explanation. Then you use the tools like spacetime intervals, to compute how much proper time will elapse for either of them, etc.

dayalanand roy
PeterDonis
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2019 Award
I'm not sure if there is an ultimate answer to those "what is the best explanation/reason" questions.
Probably not, but there is a key distinction to be made between actual measurable quantities and abstractions. See below.

To me both, "changing frames" and "paths through spacetime" are abstractions
"Changing frames" is, since "frames" are (at least in any sense in which a single observer can be said to "change" frames). But "paths through spacetime" are not: the length of each twin's path through spacetime is directly measured by the clock carried by that twin.

the difference/asymmetry in elapsed proper-times between two meetings, doesn't rely on such
The difference in elapsed proper times is the difference in path lengths through spacetime.

there must be something else that differentiates the twins, that also doesn't rely on such. And that is the difference in proper acceleration
I agree with @Dale's position on this in post #54.

dayalanand roy and Dale
A.T.
Science Advisor
The difference in elapsed proper times is the difference in path lengths through spacetime.
- The difference in elapsed proper times is what their clocks measure.
- The difference in their proper accelerations is what their accelerometers measure.

These are direct measurements, that do not rely on the notion of spacetime.

I agree with @Dale's position on this in post #54.
Me to.

dayalanand roy and Dale
Dale
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I am sure feeling the love here!

dayalanand roy
Dale
Mentor
I'm not sure if there is an ultimate answer to those "what is the best explanation/reason" questions.
I think that is probably true. There are a lot of different ways that students learn and sometimes one way or another really “clicks” for a particular student. For me it was the geometric explanation, so I have a particular affinity to that one and tend to push it more than others.

dayalanand roy
Dale
Mentor
The twin paradox is explained by the fact that the moving twin changes their frame of reference. That is the essential kernel of the solution - everything else is second order.
I wouldn’t say this either. You can use a single non-inertial frame to represent the traveling twin without any change of reference frame. In fact, that is a more direct (though less familiar) approach to the problem. Also, it is possible to do everything in a geometrical approach without even introducing reference frames at all.

dayalanand roy and vanhees71
There is an equation for how an accelerating observer measures non-local clocks:
$$T = \frac{t}{\sqrt{1-\frac{2ah}{c^2}}}$$
Yes. Via the general 1st order approximation formula ...
$$(1-x)^n \to 1-nx$$
... this equation can also be approximately written as:
$$T/t \approx 1 + \frac {ah} {c^2} = 1 + \frac {\Delta\phi} {c^2}$$
This formula can be easily derived in SR. The rocket turns by 180 degrees, so that the front end is directed towards the "stay at home" twin. Then the rocket engine is switched on, and the rocket is uniformly accelerating. At the front of the rocket is a lamp, which sends a short light pulse. At the rear end of the rocket with lenght Δh is a sensor, which receives this light pulse. I will show, that it is received blue-shifted.

First, I define a “co-moving” inertial reference frame.

The accelerated rocket shall have in this frame the velocity Zero at the point in time, when the light-pulse is sent. The light needs approximately
Δt ≈ Δh / c until it reaches the sensor. After that time, the sensor has approximately the velocity

v = a * Δt ≈ a * Δh / c.

The sensor moves with that velocity into the light, that was sent out, when the lamp had the velocity Zero in the defined inertial frame. For small “v”, the formula for the classical Doppler effect can be used:
$$f(received) / f(sent) \approx 1 + v/c = 1 + \frac {a * \Delta h} {c^2} = 1 + \frac {\Delta\phi} {c^2}$$
In the accelerated rest frame of the sensor, it is not a Doppler effect, but time-dilation between different pseudo-gravitational potentials Φ of lamp and sensor.

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Dale
PeterDonis
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The difference in their proper accelerations is what their accelerometers measure.
Yes, agreed, the proper accelerations are directly measurable.

The reason I prefer focusing on the difference in path lengths (the geometric explanation) is that it is the only explanation that generalizes to all cases. In flat spacetime you can have scenarios where both twins have nonzero proper acceleration but their elapsed times are not the same. In curved spacetime you can even have scenarios where both twins have zero proper acceleration but their elapsed times are not the same. There is no general rule involving proper accelerations that will always work. Looking at the spacetime geometry is the only technique that will always work.

dayalanand roy
A.T.
Science Advisor
Yes, agreed, the proper accelerations are directly measurable.

The reason I prefer focusing on the difference in path lengths (the geometric explanation) is that it is the only explanation that generalizes to all cases.
Well, as soon you point out the directly measurable different proper accelerations (break in symetry), people tend to get the wrong idea, that proper acceleration directly affects the clock rate (in contradiction to the clock hypothesis). And that's where you need geometry, to explain how accelerations can affect the total proper time, without directly affecting the clock rate. Like in the analogy I used in post #53.

dayalanand roy
Mister T
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For example, suppose both A and B are standing side by side. Their relative speed is zero. A starts moving and his speed increases from zero to 100 m/s in 1 second. His acceleration is 100 m /s 2. Relativity says that both are moving in relation to each other. So B's speed also increases from zero to 100 m /s. Thus, B should also be accelerating at 100 m /s2. So, why can't B too said to be accelerating.
Another thought I had on this issue. Suppose you have a third spaceship C moving away from B with an acceleration of 200 m/s2. Do you think that now, all of a sudden, the acceleration of B is 200 m/s2 instead of 100 m/s2.

Note that this difficulty does not arise with speed. If A and B are moving relative to each other with a speed of 100 m/s, and B and C are moving relative to each with a speed of 200 m/s, there are no contradictions.

dayalanand roy
Nugatory
Mentor
If, as in the twin paradox, the acceleration causes a change in speed then this does does affect the clock rate.
A change in speed does not affect the clock rate. If it did we could identify a true rest frame of the universe - it would be the only frame in which a clock at rest in that frame is unaffected.
Or consider that right now you are moving at almost ##c## using a frame in which a charged particle in a linear accelerator is at rest; you are moving at a few kilometers per second using a frame in which someone on the opposite side of the earth is at rest; and you aren’t moving much at all in a frame in which your computer screen is at rest while you’re reading this. Which is the speed that is affecting your clock rate?

Velocity-based time dilation, the stuff they talk about in introductory presentations where they say that a moving clock ticks slow, is a consequence of the relativity of simultaneity - and has next to nothing to do with why that the traveller ages less than the stay-at-home twin. One way of seeing this is that the stay-at-home twin’s clock is dilated relative to the traveller’s clock at every moment of the journey - yet the traveller ages less.

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dayalanand roy, vanhees71 and PeterDonis
PAllen
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Velocity-based time dilation, the stuff they talk about in introductory presentations where they say that a moving clock ticks slow, is a consequence of the relativity of simultaneity - and has next to nothing to do with why that the traveller ages less than the stay-at-home twin. One way of seeing this is that the stay-at-home twin’s clock is dilated relative to the traveller’s clock at every moment of the journey - yet the traveller ages less.
I have to disagree with this, as a general statement. Consider the traveling twin to start out with some outgoing velocity, and uniformly accelerate toward the stay at home twin until they reunite. Then by any reasonable coordinate choice for the traveler, the home twins clock runs uniformly faster between meetups. There is no period of time dilation at all.

dayalanand roy
Nugatory
Mentor
I have to disagree with this, as a general statement.
As a general statement, yes, you're right. I was speaking in terms of the most common intro presentation of the twin paradox, the one in which the traveller heads out at a constant velocity relative to stay-at-home, turns around quickly, and then returns at a constant velocity. In this case, the time dilation formula says what I said... and indeed that naive application of the time dilation formula is what makes this a "paradox".

Another subtlety that I glossed over: I said "every moment", but if we assume an instantaneous turnaround we can't apply the time dilation formula at every moment - it's not defined at the moment of the instantaneous turnaround. We dig our way out of that rathole by assuming a very large but not infinite acceleration applied for a very small but not instantaneous turnaround so that we always have an usable MCIF in which the traveller is momentarily at rest.

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dayalanand roy
Another thought I had on this issue. Suppose you have a third spaceship C moving away from B with an acceleration of 200 m/s2. Do you think that now, all of a sudden, the acceleration of B is 200 m/s2 instead of 100 m/s2.

Note that this difficulty does not arise with speed. If A and B are moving relative to each other with a speed of 100 m/s, and B and C are moving relative to each with a speed of 200 m/s, there are no contradictions.
Thanks. Nice thought.

Because B's accelerometers read zero at all times.

Imagine being in a train travelling next to another train at the same speed. You can see the other train's passengers through your window. One train puts on its brakes. According to the passengers on both trains, the previously stationary passengers in the other train start to move. This is coordinate acceleration, where the velocity of something relative to you changes. However, only one set of passengers will feel a jolt and be pushed back in their chairs - that's proper acceleration.

In the twin paradox only one of the twins feels proper acceleration. It's the proper acceleration that's important because coordinate acceleration is an effect of your choice of what "at rest" means whereas proper acceleration is actually felt.
Thanks. Nice explanation of coordinate and proper acceleration. Now I am getting the point. The problem is, in most of the books dealing with SR or GR, that I have read, and as far as I remember, only the term acceleration is used, not proper acceleration. More so, doesn't it mean that, "motion is relative" this statement is not fully applicable in case of motion experiencing proper acceleration?