Relative Simultaneity: Comparing 2 Incompatible Calculation Methods

[Marilyn67]

TL;DR Summary

I used two calculation methods and found two completely different results ...
I don't understand !

Hello,

Hope you had a merry Christmas,

In order to deepen some knowledge on the notion of relative simultaneity, I studied a graph taken from a video, and the ages of the twins being very very rounded, I decided to make precise calculations using two methods different.

I would like to point out that this is not a recurring question like those that often come up on this great "classic" of forums.

These two methods both seem consistent to me, yet they don't give me the same result at all.

My question:

Is either of the two methods wrong, and if so, why ?

We know that in relative movement, the situation is symmetrical, that is not the problem, I understood it for a long time.

Here are the starting assumptions, with the attached graph :

A twin leaves Earth in the direction of Proxima Centauri.

We set the distance D = 4 light years to simplify Its speed represents 2/3 of C. (speed is considered constant over the vast majority of the route, to simplify).

The Lorentz factor is about 1.3416.

Passing next to Proxima Centauri, the spacecraft twin does not stop, it continues its course indefinitely at uniform speed.

For the Earth, it takes 6 years to travel this distance. OK.

For the spacecraft , 4.47 years have passed, and the astronaut is 4.47 years more when passing Proxima Centauri, OK. (6/1.3416)
(Small yellow and green cones represent champagne bottles (birthdays) )

For the astronaut, his simultaneity line (dotted line on the graph) indicates that at this event, his brother on Earth is a little over 3 years more (other event), OK.

How many exactly ?

This is where I used two different calculation methods :

First method :

It consists in saying that for the astronaut aged 4.47 years more, his brother on Earth moves away from him at 2/3 of C, and is therefore affected by the Lorentz factor ɣ = 1.3416.

I find 3.33 years more by simply doing 4.47 / 1.3416.
This result seems fair to me.

Second method :

We have been taught that two simultaneous events in R '( spacecraft frame of reference) moving at a speed v relative to R (earth frame of reference) are separated by a time interval :

Δt '= - ɣ (v / c²). (X'a-X'b)

So I find Δt '= - 1.3416 (2/9). (4 - 0) = -1.1925 years.
In the end, on this line of simultaneity, the terrestrial twin would therefore have 4.47 - 1.19 = 3.28 years more.

You will tell me, the difference is not important ...

So I redid the same calculation with a much larger Lorentz factor (tending to infinity).

What seems incredible to me is that with the first method, the terrestrial twin sees his age approaching 0 years more with the first method, (he has hardly aged), OK, whereas with the second method, Δt 'tends towards -∞, in other words, an arbitrarily distant time in the past, even before the departure of the spaceship, and even before the birth of the terrestrial twin !

It seems crazy to me.

Which result is true, which result is false, and why ?

However, I am sure that the second formula is correct, so I don't understand.

Thank you in advance for your answers !

Cordially,

Marilyn

 

[PeterDonis]

First method :

It consists in saying that for the astronaut aged 4.47 years more, his brother on Earth moves away from him at 2/3 of C, and is therefore affected by the Lorentz factor ɣ = 1.3416.

I find 3.33 years more by simply doing 4.47 / 1.3416.
This result seems fair to me.

Good, because it's correct.

Second method :

We have been taught that two simultaneous events in R '( spacecraft frame of reference) moving at a speed v relative to R (earth frame of reference) are separated by a time interval :

Δt '= - ɣ (v / c²). (X'a-X'b)

Where have you been taught this?

 

[Marilyn67]

Hello PeterDonis,

We find the formula here for example (sorry, it's in French) :

https://fr.wikipedia.org/wiki/Simultanéité#Simultanéité_et_diagramme_d'espace-temps

 

[Orodruin]

Summary:: I used two calculation methods and found two completely different results ...
I don't understand !

So I find Δt '= - 1.3416 (2/9). (4 - 0) = -1.1925 years.

The Earth twin is not 4 ly away in R’ at this simultaneity in R’. How would the Earth twin get that far in 4.47 years when traveling at 2c/3?

You also need to double check your other inputs.

 

[PeterDonis]

We find the formula here for example

Ok. Now: which two simultaneous events in R' are you plugging into this equation? And what values of X' are you using for those events?

 

[robphy]

We set the distance D = 4 light years to simplify Its speed represents 2/3 of C. (speed is considered constant over the vast majority of the route, to simplify).

The Lorentz factor is about 1.3416.

Obligatory suggestion to use
3/5 or 4/5 or even 5/13,
which have rational Doppler factors
and lead to Pythagorean triples, nicer time-dilation factors, and simpler arithmetic.
(This is not so for 1/3, 1/2, 2/3, 9/10.)
If you want close to 99%, try 99/101.

 

[anuttarasammyak]

Summary:: I used two calculation methods and found two completely different results ...
I don't understand !

Δt '= - ɣ (v / c²). (X'a-X'b)

So I find Δt '= - 1.3416 (2/9). (4 - 0) = -1.1925 years.

I calculate it
$$-\gamma (v/c) \frac{X'_a-X'_b}{c}=-\gamma (v/c) \frac{X_a-X_b}{\gamma c}=-\frac{v}{c}\frac{X_a-X_b}{c}=-8/3$$years.

 

[anuttarasammyak]

Summary:: I used two calculation methods and found two completely different results ...
I don't understand !

For the spacecraft , 4.47 years have passed, and the astronaut is 4.47 years more when passing Proxima Centauri, OK. (6/1.3416)
(Small yellow and green cones represent champagne bottles (birthdays) )

Just to share not decimal approximate but exact values for OP setting
$$\gamma = 1/\sqrt{1-2^2/3^2}=\frac{3}{\sqrt{5}}\approxeq 1.3415$$
The pilot's clock when he arrives shows year
$$6/\gamma = 2\sqrt{5} \approxeq 4.472$$
At the same time for the pilot the Earth clock shows year
$$2\sqrt{5}/\gamma = 6/\gamma^2=\frac{10}{3} \approxeq 3.333$$

Imagine that the Earth clock network covers Proxima Centauri, the pilot see the clock there shows 6 years. He/She judges that the Earth clock network has systematic mal synchronization per distance for he/she. The Earth is $$2\sqrt{5}*\frac{2}{3}=\frac{4\sqrt{5}}{3}\approxeq 2.9814$$ light years away for he/she. The accumulated errors of synchronization to the Earth should be
$$6-\frac{10}{3}=\frac{8}{3}$$

 

[PeroK]

I calculate it
$$-\gamma (v/c) \frac{X'_a-X'_b}{c}=-\gamma (v/c) \frac{X_a-X_b}{\gamma c}=-\frac{v}{c}\frac{X_a-X_b}{c}=-8/3$$years.

This looks right. It's relative to the time of $6$ years in the Earth-Proxima frame when the rocket reaches Proxima. In the rocket frame a clock on Proxima reads $6$ years and a clock on Earth reads $6 - \frac 8 3 = \frac {10} 3$ years. Which is consistent with the time-dilation calculation.

 

[Orodruin]

If you want close to 99%, try 99/101.

98% ...

 

[PeroK]

Second method :

We have been taught that two simultaneous events in R '( spacecraft frame of reference) moving at a speed v relative to R (earth frame of reference) are separated by a time interval :

Δt '= - ɣ (v / c²). (X'a-X'b)

So I find Δt '= - 1.3416 (2/9). (4 - 0) = -1.1925 years.
In the end, on this line of simultaneity, the terrestrial twin would therefore have 4.47 - 1.19 = 3.28 years more.

I think there are several misunderstandings in this calculation, I'm sorry to say.

It's probably better to look at a correct calculation:

Suppose we have two synchronised clocks at rest relative to each other, a proper distance $D$ apart. Now consider a frame in which the clocks are moving at speed $v$ in the direction between them. In this frame, the clocks are not syncronised. The leading clock shows an earlier time than the other clock. The difference is $$\Delta t = \frac{Dv}{c^2}$$
Alternatively, if we take $D' = \frac D \gamma$ as the length-contracted distance between the clocks in the frame in which they are moving, we have: $$\Delta t = \frac{\gamma D'v}{c^2}$$
We can now apply this to your scenario as follows:

Suppose we have a clock on Earth and a clock on Alpha-Proxima, synchronised in the Earth-AP frame. In the rocket frame, the Earth clock is the leading clock and records an earlier time than the AP clock:
$$\Delta t = \frac{Dv}{c^2} = 4(\frac 2 3) = \frac 8 3 \ \text{years}$$
When the rocket reaches AP, the AP clock reads $6$ years and the Earth clock reads
$$t = 6 - \frac 8 3 = \frac{10}{3} \ \text{years}$$ (As measured in the rocket frame).

This is consistent with the time dilation calculation in the rocket frame, where the Earth clock has advanced by $$t = \frac {t'}{\gamma} = \frac{D}{v\gamma^2} = \frac 6 {9/5} = \frac {10} 3 \ \text{years}$$

 

[PeroK]

PS Note that when the rocket started its journey, as measured in its frame of reference, the Earth clock read $t = 0$ and the AP clock read $t = +\frac 8 3$ years.

During the journey, both the Earth and AP clocks advanced by $\frac {10} 3$ years. And, therefore, at the end of the journey, the Earth clock read $t = \frac {10} 3$ years and the AP clock read $6$ years. Again, as measured in the rocket frame. And, again, consistent with the previous calculations.

 

[Marilyn67]

Hello,

Excuse me if I am answering you so late, but it was very late at my house (France) last night

Thank you all for your constructive interventions !

You all made me understand that I was making 3 serious mistakes by applying my second method :

1/ I did not apply the correct spatial coordinates, (those from the point of view of the spacecraft ) which amounts to using the coordinates of the Earth-Proxima Centauri reference frame without removing the Lorentz factor from the equation (I had already seen this formula WITHOUT the Lorentz factor and I understand why now, because I inverted [Xa, Xb] with [X'a, X'b] !

2/ Applying the formula correctly (and without making a numerical error, like me), Δt' represents the difference between the clock of Proxima Centauri and that of the Earth from the point of view of the spacecraft when it is at the Proxima Centauri point, and NOT (as I thought) the difference between the spacecraft 's clock and that of the Earth under the same conditions.

3/ I made a college error with dimensionless values of v and c by attributing 2/9 instead of 2/3 to the quotient v/c² ...!

Without you, I wouldn't risk finding the right result !

So the second method is good also, as long as you use it correctly !
I am reassured but I am ashamed !

Cordially,

Marilyn