The twin paradox

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INTRODUCTION

I came to know about the twin’s paradox with a Italian pocket sold with the Italian version or Scientific American “Le Scienze”

As the book did not explain the math behind it, I went to Youtube to search. I found again DrPhysicsA, who helped me create the page about EFE.

Dr PhysicsA has a BSc (physics) and PhD (nuclear physics) from King’s College, London.
Thanks to  “DrPhysicsA” I make good progress in applied and advanced math as well as understanding some of the more advanced Physics.

INDEX

 

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The lecture

DrPhysicsA explain the math behind the Twins paradox in this video

When Newton sstudied Cartesius had problems moving forward, it is said that starte to read the book from the beginning. I do the same.

As I need to go back now and then to compare I use the manuscript and add screenshots do do that. See below.

OBS. These are my notations. I recommend you to write your own nottations while listening to the lecture.

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Space-time

 
Today we’re going to look at space-time and a thing called the twins paradox and as usual we’re going to start at the very beginning so the question is what is space-time.
 
what is space-time?
As we understand it is three-dimensional that at least is what we observe. That means to say that in any area of space you can go left and right or up and down or back and forth three dimensions and we usually represent that by three dimensions of space or three coordinates of space
which we often
label x y and z and that is space.We now need to add to that time and that’s the fourth dimension but it’s very difficult to draw a picture with four dimensions. In fact it’s not impossible and even three dimensions can be rather confusing so let’s try and keep it simple and we’re just going to think of one dimension of time and one dimension of space so it will look like this time.
We’re going to draw upwards and space we’re going to draw as the x-axis we forget about y&z for the moment just to keep things simple. So time goes up and space goes along the x-axis and that is essentially a picture of space-time now.
Let’s compare the two if I stand still in space then my coordinates do not
change. I do not move along the x-axis or the y axis or the z-axis
I stand still and
therefore that point doesn’t change.
 
 
Some rules for space-time chart
 

If you standstill, time is ticking

 
But if we stand still in space-time then although my space dimension doesn’t
change, I stay hereon the x-axis (ed. points to the dot on the chart) of course time is passing by relentlessly.
 
 I’m moving that direction (ed. see arrow up) because although I’m not
moving in space I am moving in time if you standstill in the middle of a room,  time is nonetheless ticking by.
 
1. backwards in time is not allowed.
Now
there are some rules about space-time for a start. You can’t go backwards in time. So if we take our space-time chart any movement like that is forbidden because although it implies that you are moving
to the left in space it also suggests you’re going backwards in time and there is no known mechanism for doing that.
Although there are some theories that
suggest you might be able to find some
kind of hole that enables you to go back
to some point in the past there is no 
mechanism for doing that and so you
can’t have any mote movement through
space time which involves going
backwards in time.
 
Now space-time charts are usually drawn in such a way  that the speed of light is drawn at an angle of 45
degrees and the way you do that is you just make sure that the coordinates are appropriate.
So for example time might be

one year and the space might be one Lightyear which is the distance light travels in a year so light travels one light-year in one year and.
That’s how you organize the chart.
Or you could for example have time at one
second and distance three times ten to
the eighth meters  because light travels
at three times ten to the eighth meters
per second
The exact lightspeed is 299 792 458 m / s.
 
In space.com you read that the sun orbits the sun at about 149,597,870 km  from the sun. This equals i nlightminutes to 149,597,870,000 m / 299 792 458 m / s /60 sek/min =  8,316746358353462 min )
So the sunlight takes about 8 minutes to reach the earth.
 

So we usually arrange it (ed. the graph) just for presentational purposes so that the speed of light is at 45 degrees.

 

2. not allowed to exceed C in space-time

Now here’s another rule in about our space time chart here’s the chart time versus space.

You cannot have any kind of motion that goes faster than the speed of light. So for example something that goes like that is
going faster than the speed of light.
It suggests it’s getting somewhere faster than light could and you therefore cannot do it so that is not allowed.
Not only only are you not allowed to go
backwards in time. You’re not allowed to exceed the speed of light.

What this means
You can’t get to New York in five minutes (ed. as this means that you have to travel faster than light. You can get to any position in space time but at a later time than light.
 

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The twin paradox – constant speed

Now we’re going to use these charts to examine what is called
the twin paradox.

 
They won’t celebrate their tenth birthdays
 
Imagine that this is the earth on which there are two twins and let’s imagine they just both just been born and here comes a space rocket which is traveling at speed V and V is 3/5 of the speed of light and it just so happens that this rocket has a little hook underneath it and as it goes past at 3/5 the speed of light it picks up one of these twins and carries it along at 3/5 the speed of light for a period of five years If it is traveling for five years at 3/5 of the speed of light it will travel a total distance of three lightyears now where is fit traveled that distance three light-years in five years it meets another spacecraft coming the other way also at speed V equals 3/4 C and that spacecraft also has a hook on it and as it passes this spacecraft it grabs hold of the twin and brings the twin back and that will also take five years for the return journey and as it passes the earth it releases the hook and the twin falls back down next to its brother and they are reunited after ten years and you might think that they could together celebrate their tenth birthdays each but as we will see that is not the case.
 

So let’s start by drawing a very rough space-time diagram just to show what’s going on and then we’ll do it slightly more accurately so that we can understand.
 
The first twin that remains on earth
what is happening is that the first twin the one that remains on earth throughout will start at the year naught and stay still basically up to year 10 and so the first twin doesn’t move along the space axis at all just stays still on the earth we’ll imagine for the moment that the earth is not moving and so the space-time direction for this twin is straight up they do not move through space they just move steadily
through time for a period of 10 years.
 
The second twin that leaves the earth
Now the other twin is picked up by the spacecaft and is whizzed out to this point here (ed on tthe right side of the image above)
and then the the return spacecraft picks them up and brings them back here and the total distance that they travel is three
lightyears.
 
 
twin number one
So just to recap twin number one stays
still on the earth and just travels
through time for ten years.
Twin number two
travels at 3/5 the speed of light to this point outwards three light-years and then travels back at 3/5 the speed of light and rejoins their twin on earth.
 
Now what has actually happened.
I explained that a moving clock appears to run slow. and I’m just going to explain that again here so that we get the full picture
 
The mirror clock
I said that you could invent a very simple clock it would consist of two mirrors and what you do is you simply bounce
light up and down between the two
mirrors and every time the light beam
hits a mirror it essentially causes a counter to move on and since the
distance is known and since the speed of light is invariant we know that the time to travel from the bottom mirror to the
top mirror is going to be the distance D
divided by C the speed of light.
And that becomes in a sense our time measure one movement from the bottom mirror to the top mirror constitutes a time interval which equals D divided by C.
 
The mirror clock on a rocket, as seen from earth
 
Now suppose that apparatus is in a rocket that is going past the earth so here’s the earth with an observer on the earth and here’s the rocket
and here’s the mirror in the rocket.
 
Now the rocket is moving at velocity vand for these purposes v will have to be pretty fast approaching the speed of light maybe 3/5 the speed of light.
 
So  the light leaves the bottom mirror on its way up to the top mirror but before it gets there the rocket has moved so that by the time it gets to the top mirror that top mirror is now here
(ed. in the middle of the image above)
inside the rocket and then it’s reflected but by the time it gets to the bottom mirror the rocket has moved to here
(ed. in the right side of the image above)
and there’s the bottom mirror.
So to the observer on the earth what the light beam has appeared to do is not one up and down but gone like that so how do the two observers one of them inside the rocket and one of them on the earth compare these two events.
 
The person inside the rocket says the light simply went up like that.
 
The person on the earth says no it didn’t it went like that
(ed. like the image above).
 
some geometry
Now we can do some geometry. 
What we’re going to do is we’re going to create a right-angled triangle
d
this is distance d
that was the distance we had here. (ed Between the mirrors on the rocket)
d’
this is going to be distance we’re going to call it D Prime and of course this is also distance d.
 
vt’
What is this distance here well it’s the velocity of the rocket which is V times the time it took from to get from here to here which we’re going to call T Prime.
 
t’
it’s the time which the observer on the earth will reckon that the rocket took to get from here to here.
 well now we can do some basic Pythagoras.
d’2 equals …
 
as we know that
d=ct
then we can write:

The following operations are:

  1. Adding vt’ on both side and flipping side,
    OBS! (ct2) should be (ct)2 in the first line above.
  2. Dividing both side with c2
  3. Square root of both side, gives finally this equation with the
  4. Lorentz transform (sqr root (1-v2/c2)

Lorentz transform
is defined as a “
lineartransformations from a coordinate frame in spacetime to another frame, that moves at a constant velocity relative to the former. (Wiki )

The equation says that the bigger the velocity v, the bigger is the time difference between t and t’.

 
 
d’ = …. if v= 3/5C … the relationship between the time interval for the person in the spacecraft and the time interval for the person observing
that clock but from the earth … will be…4/5 of the
time that the person on the earth thinks
it ought to be.
 
If five years have gone past on earth only 4 years
will appear to have gone past in the
spacecraft.
 
communication between the
two twins
Let us suppose that there is some communication between the
two twins.
The communication travels at
the speed of light.
 
It cannot travel faster.
There can be no instantaneous
communication.
( ed. from left to right)
 
 Sso let’s suppose that there is some radio waves carrying picture signals
but actually …
 
 
 
This case is actually not realistic as noone will survive  going from still stand to v=4/5 C. But a clock will.
 

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The twin paradox –  speed with accelerations.

 
A the twin in the rocket will accelerate at start and end of the travel on both directions, the perspective changes.
 
Acceleration slows clock down.
 
 General relativity laws says that if a clock spends 40 years of its life in a (ed. Accelerating)
spacecraft to arrive back on earth, the clock will be
 back on earth 59.000 years into the future.
 

Twins won’t see each other again.

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Section 5

 

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Section 6

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conclusion

 

The two examples are difficult to compare as in the first constant velocity case the twin in the rocket traveled 10 years.
I the second case they talk about 40 years. How many years later would the twin be back on earth with acceleration to v=3/5C and back to v=0?

If calculation 1/4 of 59.000 years is ok, then that would equal to 14700 years on earth. Then the twins won’t met again in both cases

 

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