# Spacetime in Special Relativity

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Then, how do you account for this comment:

(emphasis mine)

How does it "go away"?

I have understood that when the 2 clocks come back together, we all agree that their frequency will be the same again. How is it possible to obtain such a result accounting only on time dilation? There seems to be some inconsistency.

I don't understand what you don't understand. The frequency dilation and length contraction go away when the relative speed is zero. Both effects are speed dependent.

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When relative speed diminishes, you must agree that time dilation diminishes too. I hope.

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When relative speed diminishes, you must agree that time dilation diminishes too. I hope.

Yes, the clock frequencies will be closer in value.

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So time changed from a maximum dilated value to a less dilated value: time contracted. Or do we have a language issue here?

Edited by michel123456
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Could be a language issue. The actual contraction of time is in the frequency; I seem to recall that Einstein mentioned somewhere in his writings he regretted naming it time dilation. Time (phase) is the integral of the frequency. A clock measures the accumulated phase of the oscillator, just as an odometer is the accumulated distance of a measuring device (like a meter stick). The length of the meter stick and the frequency of the clock depend on your speed. They will return to your standard value when you are at rest. The integrated measurements (total distance, time reading on your clock) do not agree with the measurement made by someone who remained at rest.

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...and the dimension of the twin himself & his clock who were contracted during the travel.

Could be a language issue. The actual contraction of time is in the frequency; I seem to recall that Einstein mentioned somewhere in his writings he regretted naming it time dilation. Time (phase) is the integral of the frequency. A clock measures the accumulated phase of the oscillator, just as an odometer is the accumulated distance of a measuring device (like a meter stick). The length of the meter stick and the frequency of the clock depend on your speed. They will return to your standard value when you are at rest. (...)

Till this point we agree (what a relief!0

The integrated measurements (total distance, time reading on your clock) do not agree with the measurement made by someone who remained at rest.

There I am confused

I'll try to explain:

Planet A "at rest", spaceship B at Cap canaveral. B is at rest.

9,8,7,6,5,4,3,2,1,... psit, B is propulsed into space: acceleration.

B travels for a period of time T1 as observed from A (for time T2 as observed from B) till the middle of the travel. At this point, B operates a U-turn and comes back.

When she makes the U-turn, B is in the same FOR with A. I mean, at that moment, relative velocity between A & B is null. Otherwise, it is not a U-turn. (the same goes for an orbit)

Thus, at this moment, time dilation & length contraction have vanished, as observed by A. (they never existed as observed by B on herself)

The first part of the travel consisted of acceleration & deceleration ,as observed by A.

The come-back travel will consist of acceleration & deceleration as observed by B and smooth landing at Cap Canaveral as observed by both.

As seen from A, the travel consisted of 4 parts:

1. acceleration away

2. deceleration away

3. acceleration back

4. deceleration back.

As seen from B the travel consisted of:

5. acceleration away

6. deceleration away

7. acceleration back

8. deceleration back.

Measurements 1 from 8 will be different for both observers.

But I think that measurements 7 & 8 cancel out measurements 1 & 2 (they are exactly symmetric) and that 5 & 6 cancel out 3 & 4, so that both observers will agree at last on everything.

Edited by michel123456
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A U-turn involves an acceleration. You can let B be at rest for an arbitrary length of time at the turn-around point to eliminate that issue. But, since the details of the acceleration are unimportant to the concept, they are often taken to happen quickly and ignored. It only complicates the analysis.

The outward travel consists of an acceleration, motion at constant speed, and another acceleration. The return consists of the same. If we ignore the phase accrued in the acceleration phases by making the duration short, then all of the time dilation occurs during the motion at constant speed. This doesn't show up in your summary.

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A U-turn involves an acceleration. You can let B be at rest for an arbitrary length of time at the turn-around point to eliminate that issue. But, since the details of the acceleration are unimportant to the concept, they are often taken to happen quickly and ignored. It only complicates the analysis.

The outward travel consists of an acceleration, motion at constant speed, and another acceleration. The return consists of the same. If we ignore the phase accrued in the acceleration phases by making the duration short, then all of the time dilation occurs during the motion at constant speed. This doesn't show up in your summary.

There is a very simple way to resolve the twin paradox using general relativity:

Let's put the non-traveling twin at the South pole to take the rotation of the Earth out of the picture. Then the non-traveling twin is in free-fall, here orbiting the sun. So the world line of this twin is a spacetime geodesic. The traveling twin must accelerate and decelerate, he is not in freefall, ergo the world line of the traveling twin is not a geodesic.

In general relativity, because of the signature of the Minkowski/Lorentz metric, geodesics maximize arc length between their end points.

The twins start at a single point and re-unite at a single point in spacetime. The arc length of the world line non-traveling twin, the geodesic, is longer than that of the traveling twin. The arc lengths are the time, proper time, experienced by each of the twins. Ergo the traveling twin is younger.

...and the dimension of the twin himself & his clock who were contracted during the travel.

Till this point we agree (what a relief!0

There I am confused

I'll try to explain:

Planet A "at rest", spaceship B at Cap canaveral. B is at rest.

9,8,7,6,5,4,3,2,1,... psit, B is propulsed into space: acceleration.

B travels for a period of time T1 as observed from A (for time T2 as observed from B) till the middle of the travel. At this point, B operates a U-turn and comes back.

When she makes the U-turn, B is in the same FOR with A. I mean, at that moment, relative velocity between A & B is null. Otherwise, it is not a U-turn. (the same goes for an orbit)

Thus, at this moment, time dilation & length contraction have vanished, as observed by A. (they never existed as observed by B on herself)

The first part of the travel consisted of acceleration & deceleration ,as observed by A.

The come-back travel will consist of acceleration & deceleration as observed by B and smooth landing at Cap Canaveral as observed by both.

As seen from A, the travel consisted of 4 parts:

1. acceleration away

2. deceleration away

3. acceleration back

4. deceleration back.

As seen from B the travel consisted of:

5. acceleration away

6. deceleration away

7. acceleration back

8. deceleration back.

Measurements 1 from 8 will be different for both observers.

But I think that measurements 7 & 8 cancel out measurements 1 & 2 (they are exactly symmetric) and that 5 & 6 cancel out 3 & 4, so that both observers will agree at last on everything.

Kinematically this is true. But special relativity requires more. Special relativity very specifically requires an inertial reference frame. The Lorentz transformations relate to measurements in two inertial reference frames in relative motion. But because B can clearly detect his acceleration (a force tries to push him to the back of his rocket ship) his reference frame is not inertial. This breaks the perceived symmetry and if one wants to use special rather than general relativity, one is forced to use the reference frame of A.

Inertial reference frames are special. Given any inertial reference frame, any frame in uniform motion with respect to it is inertial. More importantly the converse is true -- given one inertial frame, the set of all inertial frames is just the set of frames in uniform motion with respect to the given inertial frame.

This begs the obvious question as to whether any truly inertial frame exists. The answer is "probably not". General relativity avoids the by replacing a global inertial frame with a local frame in free fall. There are, of course, as with Newtonian mechanics, frames that are sufficiently close to inertial to permit accurate modeling.

A lot of special relativity confusion can be avoided by recognizing just how special inertial frames really are.

Edited by DrRocket
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The Twins Paradox can also be explained using special relativity. It involves time dilation and the Dopper effect. And the turn-around of the traveling twin is the key. It isn't brief, but I find it the easiest explanation for my small brain to understand. I wrote up a version on my website: Link: http://www.marksmodernphysics.com/ Click on It's Relative, Archives, Twins Paradox

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In this diagram, the twin travel is represented with the blue trajectory O-B-A.

As far I can see, all mathematical analysis of the twin travel represents the trajectory O-B-C. There is no mathematical account of the U-turn in point B.

Everything is calculated as if the twin traveled twice the distance d, that is 2d, which is the path length*, no matter of what happened at point B. In fact the path is correct, but displacement is null.

IMHO there is an obvious physical difference between A and C: they have different spacetime coordinates. The conclusion we usually extract from calculations is those that would happen if A and C were in contact. But they are not.

If you prefer, there should be a difference between the equations describing an object getting away and the equations describing an object approaching.

*as observed from A.

Edited by michel123456
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In this diagram, the twin travel is represented with the blue trajectory O-B-A.

As far I can see, all mathematical analysis of the twin travel represents the trajectory O-B-C. There is no mathematical account of the U-turn in point B.

Everything is calculated as if the twin traveled twice the distance d, that is 2d, which is the path length, no matter of what happened at point B. In fact the path is correct, but displacement is null.

IMHO there is an obvious physical difference between A and C: they have different spacetime coordinates. The conclusion we usually extract from calculations is those that would happen if A and C were in contact. But they are not.

If you prefer, there should be a difference between the equations describing an object getting away and the equations describing an object approaching.

The spacetime diagram shown is one drawn from the rest frame of the stay at home twin. There are two other frames that should have their S-T diagrams shown also: The rest frame of the traveling twin on his outbound leg, and his rest frame durng the return leg. It is this transistion between these two rest frames at point B by the traveling twin which is responsible for the breaking of symmetry between Stay at Home twin and Traveling twin.

For example, here is set of S-T diagrams for this scenario in which I have added a couple of things:

For one, I added events to each world line marking equal time intervals (they could be seconds, days, years, but they are the same from world line to world line.)

For another, I added a light signal sent from the traveling twin to the stay at home twin at the moment he turns around.

The first diagram is from the rest frame of the stay at home twin.

Note that the traveling twin's time (tB), always lags behind the stay at home twins time (tA). When they meet up again, tA=8 and tB=7.36 also note that the traveling Twin turns around at tA=4 according to this frame and that the light signal from the Traveling twin is sent a little before Bt = 4 and arrives about halfway between tA=5 and tA=6.

Same setup, but drawn from the rest frame of the traveling twin on his Outbound leg:

This time we note that is tA that lags behind tB during the outbound leg of the Traveling twin. In fact, when the traveling twin reaches the turn around point, it is sometime between tA=3 and tA=4, not tA=4 like it was in the previous diagram. The traveling twin still reaches the turn around point a little before tB=4 and the light signal still would reach the stay at home twin halfway between tA=5 and tA=6.

The traveling twin does his u-turn and starts his return leg, as shown in this diagram:

Again, note that is still a little before tB=4 when he completes the turn and sends the light signal back to the stay at home twin and that signal still arrives between tA=5 and tA=6. However, whereas is was between tA=3 and tA=4 before he turned around, the instant after he turned around it is now between tA=4 and tA=5 (assuming negligible turn around time) .

Time does run slower for the stay at home twin than for the traveling twin in this frame, but it never makes up for that "forward jump" and when the traveling twin meets up again with the stay at home twin, he will see his clock read 7.36 to his twin's clock reading of 8, exactly the same as his twin does in the first diagram.

Edited by Janus
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The Twins Paradox can also be explained using special relativity. It involves time dilation and the Dopper effect. And the turn-around of the traveling twin is the key. It isn't brief, but I find it the easiest explanation for my small brain to understand. I wrote up a version on my website: Link: http://www.marksmodernphysics.com/ Click on It's Relative, Archives, Twins Paradox

The explanation in terms of special relativity is quite simple"

-- Special relativity is explicitly a theory regarding physics as described in inertial reference frames

-- In the twin paradox only the reference frame of the non-traveling twin is inertial.

Ergo analysis in the reference frame of the non-traveling twin is the correct analysis.

Note: Thw traveling twin experiences measurable acceleration at take-off, and at return, and similarly at the distant star, and therefore his reference frame is not inertial. The problem is not symmetric.

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The spacetime diagram (...)

Can you do that on a Loedel diagram?

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Can you do that on a Loedel diagram?

The advantage of Loedel diagram is that the scales in both frames are the same. However, this only works if there are only two frames. If you add a third frame, it has to converted in a way so that its scales are not the same as the other two, thus removing the advantage. IOW, there is no good reason to do a three frame Loedel diagram, because a three frame Minkowski diagram is simpler and easier to work with.

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I was asking because i struggled with it, entering weird situations. I thought maybe someone already did that.

Now I have trouble with a simple Minkowski one. I'll start another thread in speculations.

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