Relativity of simultaneity doubt

[PAllen]

You are at the disadvantage of not seeing what was posted yesterday!

Oh well, but thread posts should not really take account of what is no longer present, even if I had seen it. Otherwise the thread becomes incomprehensible to a future reader.

 

[Dale]

Please focus only on what remains. I deleted the deleted stuff precisely to NOT discuss it. If you continue discussing the deleted stuff then I will have to delete the new discussion and keep the thread closed. Just leave Hans alone, he recognized the mistake and doesn’t need to be hassled about it further.

 

[italicus]

How you model simultaneity of the distant emission events is a separate question and is fundamentally one of convention, not physics. Frame dependent is not enough of a statement to capture the issue.

First of all, excuse me for my bad English.
I’d like to consider the following, which “is” physics, and there is no frame dependence, IMHO. Maybe I haven't well understood your observation, if this is the case please correct me.
Let us suppose that two light signals hit me ( enter my eyes) at the very same moment of my wristwatch, one coming from the explosion of a Supernova in Andromeda, and the other coming from an asteroid that hits the Moon.
Of course, I see them at the same wristwatch time. But the two events are not contemporary : the first one happened some 2.5 millions years ago ( apart from relative motion of Andromeda and our Galaxy...: let’s imagine that Andromeda and me are in the same inertial r.f. ), the second one happened a little more than a second ago.
The 4-spacetime interval of both events, each calculated from the happening event and the light entering my eyes, are both equal to zero.
Well, here I see physics at work, and no convention about simultaneity.
Is it possible that other inertial observers receive the light from the two events at the same moment (of their wristwatch ) ? Yes, it is. There are a lot.
I hope I explained my idea well.

 

[cianfa72]

More precisely, the only invariant statements that can be made about distinct events is whether one is in the causal future, past, or neither (acausal, "possibly now") from the other.

If the second event is not in the causal future or past then it is actually spacelike separated from the first, I believe.

 

[PAllen]

First of all, excuse me for my bad English.
I’d like to consider the following, which “is” physics, and there is no frame dependence, IMHO. Maybe I haven't well understood your observation, if this is the case please correct me.
Let us suppose that two light signals hit me ( enter my eyes) at the very same moment of my wristwatch, one coming from the explosion of a Supernova in Andromeda, and the other coming from an asteroid that hits the Moon.
Of course, I see them at the same wristwatch time. But the two events are not contemporary : the first one happened some 2.5 millions years ago ( apart from relative motion of Andromeda and our Galaxy...: let’s imagine that Andromeda and me are in the same inertial r.f. ), the second one happened a little more than a second ago.
The 4-spacetime interval of both events, each calculated from the happening event and the light entering my eyes, are both equal to zero.
Well, here I see physics at work, and no convention about simultaneity.
Is it possible that other inertial observers receive the light from the two events at the same moment (of their wristwatch ) ? Yes, it is. There are a lot.
I hope I explained my idea well.

Yes, of course, most of what you say is true. Let me further assume the asteroid strike is in the opposite direction from the Andromeda explosion. This ensures that the emission events are spacelike in separation. Then, for an observer moving near lightspeed relative to Earth in the direction away from Andromeda, passing you at the moment both signals arrive (thus also seeing them arrive at the same time), would conclude (using standard Minkowski coordinates) that the asteroid strike occurred before the Andromeda explosion. For any spacelike separated events there is no invariant time ordering of them.

 

[PAllen]

@cianfa72 asked "If the second event is not in the causal future or past then it is spacelike separated from the first, I believe."

Yes, in SR this is an unambiguous equivalence. In GR, there are some complications (e.g. there can be both timelike and spacelike geodesics connecting the two events; using the notion of causal future, past, or neither, resolves such ambiguities; the scenario where this can happen is CTCs). So I used the most general wording, which is certainly overkill in SR.

 

[italicus]

For any spacelike separated events there is no invariant time ordering of them.

Absolutely correct !
But for now I’d leave GR aside. I’m aware of the complications that can be introduced here... !
This thread is about a doubt regarding the relativity of simultaneity , for which A.E. introduced the (not so happy...) example of two lightings that strike the embankment (and head and tail of the train...) : we are at a very basic level !
The drawing I reported was intended to show that , while the strikes are simultaneous for the Earth observer (OE=M in the drawing) , the aren’t such for the train observer (OT=M’ in the drawing) , because he is moving with speed v w.r.t. the embankment.

 

[cianfa72]

Sorry, CTCs what does it mean ?

 

[PAllen]

Sorry, CTCs what does it mean ?

closed timelike curve. These are present in many GR solutions. Many, but not all, physicists consider such solutions unphysical. For example, one of the interior regions of a spinning (Kerr solution) BH has them. However, it is generally believed that this section of the interior would never be physically realized because there would never be exact axial symmetry, and chaotically perturbing the collapse dynamics by a tiny amount would eliminate the CTC region. However, while I 'believe' this, to my knowledge, this has never been proven.

 

[PAllen]

Thank you for your answer. Lots to think about :)

1. First comment about simultaneity of flashes. Original idea,the starting thought-experiment describes situation in and out of the train. For the observer in the train flashes at the front (B) and at the back (A) are not simultaneous so in his reference frame he will not see them both as simultaneous (contrary to what Hans says).

Hans: "So both observers see two flashes at the same time." They don't. The point where the two flashes meet is outside OT's head towards the back of the train.

I read the explanation as that there is an observer in the train who visually sees the flashes arrive at the same time, this observation in the train is effectively colocated in space and time with the ground observer seeing the flashes arrive at the same time. However, the train observer models that the emission events are not simultaneous, while the ground observer models them as simultaneous. Neither of these statements about simultaneity represent anything that can be observed.

2. Relativistic beaming. As far as I know this effect amount of energy "perceived" (measured) by the observer so in this case my use of word "intensity" is wrong, thanks for pointing that out, however I don't quite get how the observer will measure changes of size of the object. If there is an orthogonal rod along y-axis at length of 1 meter we still see 1 meter, not less nor more in the train. So no "bigger" nor "smaller" flashes unless we define big and small in terms not size but intensity/amount of energy measured in time. I have just realized that Hans might use this colloquialism, however in our case this double meaning is very misleading.

Relativistic beaming is the effect that if you are moving towards an EM source at near the speed of light, then, compared to a colocated observer at rest with respect to the source, you measure an intensity (power per sq. centimeter, for example) on the order of $\gamma ^3$ times greater. The causes of the are a combination of aberration reducing the angular diameter of the object, and the Lorentz transform of amplitute of a wave (which implies energy). In the classical treatment, there is no coupling between wave amplitude and wave frequency. It happens that they both transform in the same way, so the Doppler factor applies to both. But classically, these are independently derived facts, which happen to be coupled in a quantum theory. Thus, looking at photons, you might think there is just Doppler and aberration. However, classically, Doppler has nothing to do with energy - that is determined by amplitude transformation.

3. Penrose-Teller rotation. In Minkowski space we have only three types of object that can be parametrized as simultaneous: a line in 2 dimensions, a plane in 3 d spacetime and finally a 3-d space when spacetime is 4 dimensional. All defined as orthogonal to the versor of time.

Now, however I see that Hans might use the term in 3-d space not in 4-d, and in this context "sphere of simultaneity" to an observer may mean a wavefront of e-m wave traveling from him in 3-d space, in vacuum this will indeed form a sphere in his rest frame and an ellipsoid in Minkowski space. I think there is a way of using this idea to explain events here, however in my opinion, this is a bit too much as it may complicate the picture in this quite difficult thought experiment. If I have time today I will re-read Hans explanations however I don't think 'spherical simultaneity" will add clarity to the picture, but of course I can be wrong.

Thank you again for your comments.

I don't understand what your are getting at here. One traditional analysis of the case of a moving sphere is to consider that light always travels at c in some chosen frame, but that no relativistic effects are present (relativity of simultaneity, length contraction, or time dilation). Then, you find that due to which light rays (simplifying to a geometric optics approximation) arrive at the same time, the sphere is elongated to an ellipse. However, then assuming that what was a sphere in its rest frame must be considered as a squashed in a frame in which it is moving (due to length contraction), then, redoing the arrival time analysis, you find an exact sphere is what will be observed. Member @A.T., I believe, has links to nice visualizations of all this.

Penrose-Terrell rotation is a completely general phenomenon in special relativity which can be most elegantly approached in terms of aberration, because then you need not consider arrival times or simultaneity, and can immediately see that it applies to objects of any shape. The analysis, in brief, is as follows.

First, what is aberration? In SR, it is simply the way the angle of a light ray transforms in SR under a Lorentz transform. Conceptually, this has nothing in common with the Bradley derivation, which requires a corpuscular model of light as well as the false assumption that light speed depends on the motion of the source (even though the two different formulas are currently indistinguishable due to observational precision limitations). The general rule is simple. Consider two colocated observers, A and B. Suppose B is moving in the +x direction relative to A. Suppose A observes a light ray (using geometric optics approximation, rather than plane waves) arriving at some $\theta$ relative to the x axis. Then B will observe this light ray arriving at an angle closer to the +x axis. The amount depends on speed, and in the limit of near c, even a ray A perceives as arriving from near the direction of the -x axis, B will observe as coming from near the +x axis.

Consider an arbitrary object moving relative relative to observer B as described above, but at rest relative to A. In the rest frame of A, you can analyze the image without considering simultaneity or arrival times. The situation is completely static. But then, in B rest frame, the set of rays arriving cannot be different. Their angles can be different, as can the frequency and energy. Thus, if you want to know the image B sees when the object looks orthogonal to x, per B, you compute the image per A when the object is back towards -x by the appropriate amount. For an object not too large in angular size, aberration then simply rotates this image towards +x, so it appears to come orthogonal to x per B. This means B sees an image of the object as if it is rotated towards B. In particular, even for a ruler with markings, the image looks just like the ruler was simply turned towards B, and this 'explains' the decrease in angular size (length contraction). Especially, if you imagine B able to image the object continuously despite high speed, it would look like the object, of whatever shape, is rotating as it moves, without any first order change in size or shape.

 

[PeterDonis]

here I see physics at work, and no convention about simultaneity.

Yes, there is a simultaneity convention implicit in what you said, specifically, here:

the two events are not contemporary : the first one happened some 2.5 millions years ago ( apart from relative motion of Andromeda and our Galaxy...: let’s imagine that Andromeda and me are in the same inertial r.f. ), the second one happened a little more than a second ago.

You cannot derive these statements just from the light signals that you received; light signals contain no inherent "how long it took the signal to get to you". That "how long" is frame dependent--i.e., it depends on your choice of simultaneity convention. When you say the Andromeda signal took 2.5 million years to get to you, but the one from the Moon took only a little more than a second, you have adopted a particular simultaneity convention. Other such conventions would give different propagation times for the two light signals.

 

[italicus]

When you say the Andromeda signal took 2.5 million years to get to you, but the one from the Moon took only a little more than a second, you have adopted a particular simultaneity convention. Other such conventions would give different propagation times for the two light signals.

I am a bit perplexed here. Isn’t the speed of light equal to c , for all IO ? This is the second postulate of SR . The signal from Andromeda takes 2.5 million years to get to me, because d = ct, hence t = d/c . Which particular simultaneity convention have I adopted? The easiest and most natural one is that by Einstein, who writes : “ When you say that a train has arrived at 7.00 o’clock , you are just stating that the arrival of the train is simultaneous to such and such position of your clock hands” (something like that...I don’t remember exactly!).
How is it possible that another simultaneity convention gives a different propagation time ? But, first of all : which other simultaneity convention could be adopted here?

 

[PAllen]

I am a bit perplexed here. Isn’t the speed of light equal to c , for all IO ? This is the second postulate of SR . The signal from Andromeda takes 2.5 million years to get to me, because d = ct, hence t = d/c . Which particular simultaneity convention have I adopted? The easiest and most natural one is that by Einstein, who writes : “ When you say that a train has arrived at 7.00 o’clock , you are just stating that the arrival of the train is simultaneous to such and such position of your clock hands” (something like that...I don’t remember exactly!).
How is it possible that another simultaneity convention give a different propagation time ? But, first of all : which other simultaneity convention could be adopted here?

How do you know d?

There is no fundamental reason to use a reference frame based on you being at rest. Suppose you consider that some object of cultural significance should be the basis of coordinates, and that object has high speed relative to you. Then you might conclude the asteroid colliding with the moon happened at the same time as the Andromeda explosion (or even before it).

 

[italicus]

There is no fundamental reason to use a reference frame based on you being at rest

Well, the reason for me is that it is the simplest reference frame! Nobody moves with respect to himself, no ? If I am sitting in a fast (not accelerating) train, I prefer the reference frame of the train for my observation of physical phenomena, and say that the landscape is moving towards me.
Galileo implicitly suggests me this : “ Shut yourself under deck of a big ship...” and let butterflies fly, drops of water fall vertically from a vase...

{ Dear friends, my life in this moment is simultaneous with 01:20 of Sept 24 ...Excuse me, I am tired and will go to sleep. See you later ! }

 

[PAllen]

Well, the reason for me is that it is the simplest reference frame! Nobody moves with respect to himself, no ? If I am sitting in a fast (not accelerating) train, I prefer the reference frame of the train for my observation of physical phenomena, and say that the landscape is moving towards me.
Galileo implicitly suggests me this : “ Shut yourself under deck of a big ship...” and let butterflies fly, drops of water fall vertically from a vase...

{ Dear friends, my life in this moment is simultaneous with 01:20 of Sept 24 ...Excuse me, I am tired and will go to sleep. See you later ! }

Yes and no. Do astronomers each use their own personal rest frames, with all different relative velocities depending on latitude and longitude? In fact, they rarely use Earth an Earth centric frame (GPS does, but not astronomers). Instead they may use sun centric frame (actually, solar system barycenter is most common for solar system phenomena), Milky Way centric frame, or CMB frame, for different purposes. All would have different definitions of simultaneity.

 

[PeterDonis]

the reason for me is that it is the simplest reference frame!

So when you want groceries, you don't go to the grocery store, you make the grocery store come to you? And move the entire Earth in the process?

Most people's "simplest reference frame" for everyday life is not one in which they are always at rest.

 

[italicus]

So when you want groceries, you don't go to the grocery store, you make the grocery store come to you? And move the entire Earth in the process?

Ahahah…this is irony for a smile, of course!
But You know very well, from classical mechanics, that the system “Earth + me” can be considered an isolated system, at least for the scope of buying groceries….
The friction forces that my shoes apply to the Earth push it back, it is an internal force of the system,no? And the reaction force exerted by Earth on me pushes me forward. So, why doesn’t the Earth rotate backwards when I push it? Because its mass is enormous wrt mine! (about 6*10^24 kg, against 60 kg!). Every motion is relative, isn’t it?
But let’s go back to relativity of simultaneity in SR. I am still waiting to be told what different kind of conventional simultaneity could be adopted, so that the time for a signal emitted by the Moon to reach me is different from that calculated using c, universal constant.

 

[PeterDonis]

You know very well, from classical mechanics, that the system “Earth + me” can be considered an isolated system

So what? You still move relative to the Earth when you get groceries. If you are going to pick a reference frame to describe that relative motion, you have two choices: a frame in which you are at rest, or a frame in which the Earth is at rest. The vast majority of people choose the latter.

I am still waiting to be told what different kind of conventional simultaneity could be adopted, so that the time for a signal emitted by the Moon takes a time different from that calculated using c, universal constant.

There is no such convention, because changing your simultaneity convention, i.e., changing which inertial frame you use, does not change the speed of light. What it changes is the distance between you and the Moon (and Andromeda), so that the time you calculate for the light to travel at speed $c$ is different. This is basic Special Relativity, and if you don't understand it, I strongly suggest taking the time to work through an SR textbook, such as Taylor & Wheeler's Spacetime Physics.

 

[PeterDonis]

why doesn’t the Earth rotate backwards when I push it?

In a frame in which you are at rest, it does.

 

[italicus]

So what? You still move relative to the Earth when you get groceries. If you are going to pick a reference frame to describe that relative motion, you have two choices: a frame in which you are at rest, or a frame in which the Earth is at rest. The vast majority of people choose the latter.There is no such convention, because changing your simultaneity convention, i.e., changing which inertial frame you use, does not change the speed of light. What it changes is the distance between you and the Moon (and Andromeda), so that the time you calculate for the light to travel at speed $c$ is different. This is basic Special Relativity, and if you don't understand it, I strongly suggest taking the time to work through an SR textbook, such as Taylor & Wheeler's Spacetime Physics.

When calculating the time that a signal emitted from the Moon takes to get to me, I have assumed that the Moon and me are at rest in the same reference frame. I have specified this clearly for Andromeda, but it applies to the Moon too. So the distance is fixed, and t is a little bit greater than 1 second. It is obvious (better: relativistic ) that if I am in a spaceship flying towards the Moon at speed v = kc (referred to the rest frame said before), k<1, the distance contracts , becoming L_c = L/gamma . So the light takes less time.

As concerns my knowledge of SR…I have known and studied that book, and many others, since its publication.
But now it’s time for me to leave this discussion, the air is becoming heavy.
I don’t want to be deleted , like Hans. My intervention was just to propose the Minkowski diagram. Thank you for the attention.
Greetings to everyone.

 

[italicus]

In a frame in which you are at rest, it does.

And this is what I told you : in my reference frame , the train I am sitting in , the landscape comes towards me.
Bye bye Peter , it has been a pleasure to discuss with you.

 

[Dale]

Every motion is relative, isn’t it?

No, not every motion is relative. Only linear velocity is relative. All other motions are not relative

So the distance is fixed, and t is a little bit greater than 1 second.

In other frames length contraction makes the distance different and t can be arbitrarily large or arbitrarily small.

 

[cianfa72]

But simultaneity of distinct events is never an observable, per se.

Definitely, since an observable is a frame-indipendent thing while simultaneity of distinct events is not.

 

[vanhees71]

No, not every motion is relative. Only linear velocity is relative. All other motions are not relative

I'd say to define "motion" you need some (local) reference frame with respect to which you describe the motion of a body. It doesn't make sense to say, a body is in "absolute motion".

 

[Dale]

I am a bit perplexed here. Isn’t the speed of light equal to c , for all IO ? This is the second postulate of SR . The signal from Andromeda takes 2.5 million years to get to me, because d = ct, hence t = d/c . Which particular simultaneity convention have I adopted? The easiest and most natural one is that by Einstein, who writes : “ When you say that a train has arrived at 7.00 o’clock , you are just stating that the arrival of the train is simultaneous to such and such position of your clock hands” (something like that...I don’t remember exactly!).
How is it possible that another simultaneity convention gives a different propagation time ? But, first of all : which other simultaneity convention could be adopted here?

This is quite straightforward. Suppose we have events $b_1 = (t_1,x_1)$ and $b_2=(t_2,x_2)$ in units where $c=1$ in some inertial frame, and suppose further that $b_1$ and $b_2$ are both on the past light cone of the origin such that $b_1=(-d_1,-d_1)$ and $b_2=(-d_2,d_2)$ where $d_i>0$. Then, by the Lorentz transform, in an inertial frame moving at $v$ with respect to the original we have: $$b'_1=(-d'_1,-d'_1)= \left( -\frac{ d_1 + d_1 v}{\sqrt{1-v^2}},-\frac{d_1 + d_1 v}{\sqrt{1-v^2}} \right) $$ $$b'_2= (-d'_2,d'_2)= \left( -\frac{d_2 - d_2 v}{\sqrt{1-v^2}},\frac{d_2 - d_2 v}{\sqrt{1-v^2}} \right) $$

So clearly we can adopt the simultaneity convention of any inertial frame we like and $t_1 \ne t'_1$ in general. Note, in particular that $d_1 = 2500000 \text{ years}$ and $d_2 = 0.0000003 \text{ years}$ does indeed specify one frame with its specific simultaneity convention. For example, there is another frame where $d'_1 = d'_2$, which is the frame moving at $v=(d_1-d_2)/(d_1+d_2)$ with respect to the unprimed frame. In all of these other frames the light from both events arrives at the origin at the same time.

 

[Dale]

I'd say to define "motion" you need some (local) reference frame with respect to which you describe the motion of a body. It doesn't make sense to say, a body is in "absolute motion".

I guess that is a matter of semantics. But clearly proper acceleration is not relative, it is invariant. If you want to exclude proper acceleration from "motion" then you could make that claim. The issue would be that many people will consider proper acceleration to be motion, so you would have to repeatedly explain your meaning of the word "motion" that excludes proper acceleration as motion.

I would prefer to not argue the point that proper acceleration is a type of motion, and rather argue that only velocity is relative and all higher derivatives of velocity are invariant, being directly measurable by an accelerometer.

I would not use the term "absolute" at all. Only "relative" and "invariant".

 

[vanhees71]

I'm not sure I understand this statement. For time-like and light-like separated events the time-ordering is preserved under orthochronous Lorentz transformations, and that's why two caually connected must be either time-like or light-like separated.

Proof: Let $x$ and $y$ be two four vectors with $z=x-y$ time-like or light-like, i.e.,
$$z_{\mu} z^{\mu}=t^2-\vec{z}^2 \geq 0.$$
with units used, where $c=1$. We assume $t>0$.

In another inertial frame of reference the same four-vector has components $z'=\Lambda z$ with a orthochronous Lorentz transformation $\Lambda$, i.e., ${\Lambda^0}_0>1$. From this you have
$$t'={\Lambda^0}_{0} t - {\Lambda^0}_j x^j,$$
where latin indices run from 1 to 3 ("spatial components"). For easier discussion, let's write
$$t'={\Lambda^0}_{0} t - \vec{\lambda} \cdot \vec{x},$$
where $\vec{\lambda}$ are the components ${\Lambda^0}_j$.

Further the Lorentz transformation fulfills
$$\eta^{\rho \sigma} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}_{\sigma}}=\eta^{\mu \nu}.$$
For $\mu=\nu=0$ you get
$${\Lambda^0}_{0} = \sqrt{1+\vec{\lambda}^2}$$
and thus
$$t' \geq {\Lambda^0}_{0} t - |\vec{\lambda}||\vec{x}|.$$
Further we have $t \geq |\vec{x}|$ by assumption and thus finally
$$t' \geq ({\Lambda^0}_0 - |\vec{\lambda}|) t=(\sqrt{1+\vec{\lambda}^2}-|\vec{\lambda}|) t>0.$$

For space-like separated events the time-ordering is not preserved under orthochronous Lorentz transformations, and you can always find some reference frame, where these two events are simultaneous. So let's assume this time $z \cdot z<0$. Now consider a pure (rotation free) boost. Then
$$t'=\gamma (t-\vec{\beta} \cdot \vec{x}).$$
Obviously we can find a vector $\vec{\beta}$ such that $t'=0$. E.g., we can choose $\vec{\beta}=\beta \vec{x}/|\vec{x}|$. Then
$$\beta=\frac{t}{|\vec{x}|},$$
and since by assumption $|\vec{x}|>t$ we have $0<\beta<1$, i.e., we have constructed a proper boost-Lorentz matrix transforming to an inertial reference frame, where $t'=0$, i.e., where the events $x$ and $y$ are simultaneous.

 

[Dale]

I'm not sure I understand this statement.

Which statement? Is this post directed to me or to someone else?

If it is directed to me then I agree that "For time-like and light-like separated events the time-ordering is preserved under orthochronous Lorentz transformations" but I am not sure what would make you believe that I disagreed and needed a proof of that fact.

 

[vanhees71]

I guess then it's sure that I didn't understand the statement. What have time-like vectors to do with simultaneity/synchronization conventions?

 

[Dale]

I guess then it's sure that I didn't understand the statement.

Which statement?

 

[vanhees71]

I guess that is a matter of semantics. But clearly proper acceleration is not relative, it is invariant. If you want to exclude proper acceleration from "motion" then you could make that claim. The issue would be that many people will consider proper acceleration to be motion, so you would have to repeatedly explain your meaning of the word "motion" that excludes proper acceleration as motion.

I would prefer to not argue the point that proper acceleration is a type of motion, and rather argue that only velocity is relative and all higher derivatives of velocity are invariant, being directly measurable by an accelerometer.

I would not use the term "absolute" at all. Only "relative" and "invariant".

Of course, also accelerated motion is motion, and proper acceleration is a frame-independent four-vector, but this doesn't imply that you can describe motion without a reference frame in an absolute way. There's no absolute motion since there's no absolute reference frame. You are right, you can abandon the term "absolute" from the discussion of motion at all.

 

[vanhees71]

Which statement?

The statement in #60 following your two formulae. I don't understand, what two light-like vectors have to do with simultaneity conventions. If two events are time- or light-like separated they are not simultaneous in any frame. The time-order is only frame-dependent for space-like separated events, and there's always a frame, where both are simultaneous. All this is of course assuming we talk about SR, not GR.

 

[cianfa72]

The time-order is only frame-dependent for space-like separated events, and there's always a frame, where both are simultaneous. All this is of course assuming we talk about SR, not GR.

Why what you said about time-order does not extend from SR to GR ?

 

[Dale]

The statement in #60 following your two formulae.

So clearly we can adopt the simultaneity convention of any inertial frame we like and $t_1 \ne t'_1$ in general.

I am not sure what is unclear about it. Clearly $$-d_1 \ne -\frac{ d_1 + d_1 v}{\sqrt{1-v^2}}$$ except for $v=0$. This means that there is only one frame where the light from Andromeda takes 2500000 years to reach earth. So specifying that the light from Andromeda takes that long to reach Earth implicitly identifies the frame you are using, and hence defines the simultaneity.

 

[Janus]

When calculating the time that a signal emitted from the Moon takes to get to me, I have assumed that the Moon and me are at rest in the same reference frame. I have specified this clearly for Andromeda, but it applies to the Moon too. So the distance is fixed, and t is a little bit greater than 1 second. It is obvious (better: relativistic ) that if I am in a spaceship flying towards the Moon at speed v = kc (referred to the rest frame said before), k<1, the distance contracts , becoming L_c = L/gamma . So the light takes less time.

If you are in a spaceship flying towards the Moon, measuring how long it takes for a light signal to travel from Moon to Earth, you measure the signal as traveling at c relative to yourself, and the Earth as moving at kc away from the point of emission.
So for example, if we assume the Moon is exactly 1.25 light sec from Earth, In the rest frame of the Earth-moon system, the light takes 1.25 sec.
If you were traveling at 0.8c towards the Moon, then the Earth-Moon distance is 0.75 light sec by your measure. However, since the Earth is receding from the emission of the signal, it will take 0.75/(1-0.8) = 3.75 sec between signal leaving Moon and arriving at Earth by your measure; a longer time than that measured in the Earth-Moon rest frame.