Undergrad Relativity of simultaneity doubt

[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.

 

[PeroK]

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.

And, in general: $$\Delta t' = \frac{L/\gamma}{c-v} = \frac L c \sqrt{\frac{1 + \frac v c}{1 - \frac v c}}$$ So, it's always longer then $L/c$ if traveling towards the Moon; and shorter if you are traveling away from the Moon.

 

[italicus]

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.

Thank you Janus.
But note that I have told at the beginning of the post :

"When calculating the time that a signal emitted from the Moon takes to get to me..."

and I am on board a spaceship traveling at 0.8c towards the Moon (we have to imagine that my spaceship started with a speed of 0.8c suddenly, or crossed the Earth position already with that speed: no acceleration ! ). So I am interested in my propertime taken by the light signal emitted by the Moon to reach me, not the Earth. I simply divide the contracted distance, which is 0.75 ls , by 1 ls/s , and obtain 0.75s of my wristwatch time. This is the meaning of my last sentence :

"So the light takes less time” ...to reach me, of course! I don’t care of the Earth.

But you say that : “ the Earth is receding from the emission of the signal “ . Why? Earth and Moon are always at rest in the same reference frame. Please look at the following drawing, and tell me
what is wrong within it.

 

[PeroK]

Earth and Moon are always at rest in the same reference frame.

You'll need to explain what you mean by this.

 

[italicus]

You'll need to explain what you mean by this.

Well, the meaning should be clear : the line of universe of the Moon is parallel to the line of universe of the Earth, on the Minkowski diagram; they are always at the same distance of 1.25 ls. Time passes on the Moon at the same rate that on the Earth.

 

[PeroK]

Well, the meaning should be clear : the line of universe of the Moon is parallel to the line of universe of the Earth, on the Minkowski diagram; they are always at the same distance of 1.25 ls.

Not in every reference frame. Only in a reference frame where they are at rest. I.e. their mutual rest frame. Not in the rocket frame, for example.

 

[Dale]

Not in every reference frame. Only in a reference frame where they are at rest. I.e. their mutual rest frame. Not in the rocket frame, for example.

Well, the worldlines are parallel in all frames (ignoring the orbiting), but indeed the distance is not 1.25 ls in all frames.

 

[italicus]

Not in every reference frame. Only in a reference frame where they are at rest. I.e. their mutual rest frame. Not in the rocket frame, for example.

Do you want me to draw the Minkowski diagram referred to the spaceship? Things do not change, only the visual appearance of the drawing, because the plane of the drawing is “Euclidean”. But you are aware of this , no doubt. So, make an hyperbolic rotation of the t’ and x’ axes, draw the t and x axes as necessary, and you are done.

 

[Dale]

So I am interested in my propertime taken by the light signal emitted by the Moon to reach me, not the Earth.

This doesn’t make sense. Your proper time is only defined on your worldline. The signal emission is not on your worldline. So your proper time from the signal emission to anything is not defined.

All you can calculate is the coordinate time, and that coordinate time indeed does depend on the frame.

 

[italicus]

This doesn’t make sense. Your proper time is only defined on your worldline. The signal emission is not on your worldline. So your proper time from the signal emission to anything is not defined.

All you can calculate is the coordinate time, and that coordinate time indeed does depend on the frame.

It makes sense. My proper time is that signed by my wristwatch. But I just divide the contracted length 0.75 ls by c = 1 ls/s , and obtain 0.75s .

The coordinate time is calculated by the Earth observer, not by me. Between coordinate time and proper time there is the well known relationship : Deltat = gamma* Deltatau . (Sorry, I don’t know latex enough) . You can check that : 1.25 / 0.75 = 1.666... = gamma.

 

[vanhees71]

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

In GR everything holds only locally. In general it's not possible to synchronize all clocks globally.

 

[PeroK]

It makes sense. My proper time is that signed by my wristwatch. But I just divide the contracted length 0.75 ls by c = 1 ls/s , and obtain 0.75s .

If two events are simultaneous in your reference frame then the proper time that elapses on your watch between the two events is zero, as measured by you (*). In a different frame where the events are not simultaneous, there must be an elapsed proper time on your watch between the coordinate times of those events, as measured in that reference frame.

Therefore, the proper time that elapses on your watch between events that are not both on your worldline is not an invariant. If, however, the two events are both on your worldline, then the proper time measured by your watch is the proper time between the events and is an invariant.

In this thread you seem to be using non-standard terminology. Perhaps that's a problem with translation, but generally it makes it hard to know what you are trying to say.

(*) In which case, your watch is measuring the coordinate time (in your rest frame) for the two events.

 

[Dale]

My proper time is that signed by my wristwatch.

Your wristwatch was present at the event that the signal reached you, but not when the event that the signal was emitted.

The coordinate time is calculated by the Earth observer, not by me.

You have coordinates too. And the time in those coordinates is a coordinate time, not a proper time. You might be thinking that your coordinate time is proper time but it is not.

You seem to have a misunderstanding of what proper time is. Proper time is the integral of the spacetime interval along a specified worldline. It is directly measured by a clock traveling on that worldline. As such, it is not defined anywhere off the specific worldline on which it is defined. So your proper time is only defined on your worldline.

The emission on the moon is an event that is not on your worldline so your proper time between that even and any other event is undefined.

 

[italicus]

To everybody.
Most probably there is a problem of translation, I am italian and am making great efforts to write a decent English, on a subject which is difficult by its own. I think you should appreciate this, anyway I apologise for language mistakes.

This said, I’ll repeat what I’m trying to explain, in a simpler way if possible.

I am on board of a spaceship, which crosses the Earth at a certain instant, that we consider instant t=t’=0 for Earth and me (look at the Minkowski diagram, please. Otherwise, my spaceship leave the Earth at that instant, with that immediate speed) The speed of the spaceship is 0.8c wrt Earth. The ship is directed toward the Moon, which is at rest wrt the Earth, at a distance of 1.25 ls . All right , until to now? So the Moon is on my worldline, Dale : axis t’ on the drawing.
Due to my relativistic speed, the distance Earth-Moon , which in their common rest reference frame is L = 1.25 ls , appears contracted to me : L_c = L/γ = 0.75 ls.
How long does it take a light signal, emitted by the Moon, to cover this contracted distance ? Divide L_c by c=1ls/s , and obtain 0.75s of my time.
That's all.

As far as the conceptual difference between proper time and coordinate time is concerned, my wristwatch time is my propertime, and of course is also a coordinate time if I consider events wrt to my reference frame, to which I have given spacetime coordinates (Italicus, t’, x’). I speak of “coordinate time” for (Earth, t, x) , when I consider events wrt Earth. But t is also the “proper time for Earth observer”.
I hope I have clarified my point of view.

Best regards to everybody.

 

[PeroK]

Due to my relativistic speed, the distance Earth-Moon , which in their common rest reference frame is L = 1.25 ls , appears contracted to me : L_c = L/γ = 0.75 ls.
How long does it take a light signal, emitted by the Moon, to cover this contracted distance ? Divide L_c by c=1ls/s , and obtain 0.75s .
That's all.

First, you are assuming that the signal is emitted by the Moon when you pass the Earth in your rocket frame. The situation would be different in the Earth-Moon frame, where the signal would be emitted after you have passed the Earth.

The signal travels the contracted initial distance from the Moon to your ship in $0.75s$, as measured by you. In this scenario, however, the light signal from the Moon to your rocket would take only $0.25s$, as measured in the Earth-Moon frame.

 

[PAllen]

Attached is a diagram. It is all drawn in the andromeda, earth, moon rest frame. However, calculations for the rocket are still readily done in the frame using invariant interval computations. For more convenient scale, Andromeda is considered to be at distance 10 in Earth frame, and moon at distance 2, but no matter what the ratio is, the same thing is is possible - you just need higher rocket speed.

I show the rocket world line along with the andromeda, Earth and moon world lines. These are all dashed lines. Light paths are solid lines. I also show 2 rocket simultaneity dashed lines, which, of course, are consistent with mid point between send and receive times for rocket. I show the light paths the rocket would need to send to measure distance to Andomeda and the moon.

As you can see clearly, the rocket considers the moon to be at a distance of 4.9 when the moon signal is sent, the signal needing to be sent 4.9 seconds before meeting of earth, rocket and two signals - per the rocket. Similarly, per the rocket, the andromeda signal must be sent 3.96 seconds before the meeting to cover the distance of 3.96 and arrive at the meeting. Thus, for the rocket the time ordering of signal transmission events and distances to them are reversed relative to Earth frame.

 

[italicus]

The signal travels the contracted initial distance from the Moon to your ship in 0.75s, as measured by you

Yes, and this is what I care of.
Frankly speaking, I don’t understand this :
The situation would be different in the Earth-Moon frame, where the signal would be emitted after you have passed the Earth.
.....
In this scenario, however, the light signal from the Moon to your rocket would take only 0.25s, as measured in the Earth-Moon frame.

@PAllen : I have to study your diagram, thanks.

 

[PeterDonis]

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.

In other words, you have adopted the simultaneity convention of that frame. That counts as choosing a simultaneity convention. So your statement that you did not choose any such convention was incorrect.

 

[PeroK]

Frankly speaking, I don’t understand this :
The situation would be different in the Earth-Moon frame, where the signal would be emitted after you have passed the Earth.
.....
In this scenario, however, the light signal from the Moon to your rocket would take only 0.25s, as measured in the Earth-Moon frame.

That's a result of the relativity of simultaneity, which is what this thread was supposed to be all about!

 

[Dale]

Most probably there is a problem of translation, I am italian and am making great efforts to write a decent English, on a subject which is difficult by its own. I think you should appreciate this, anyway I apologise for language mistakes.

Then maybe you should consider paying attention when people tell you that you are using words wrong.

I have labeled some events on your diagram, and explicitly added the light pulse that we are discussing.

Your wristwatch is on the worldline that goes from event A to event B. Your proper time is not defined for any events that are on any other line besides the line AB. In particular, your proper time is not defined for event C nor is it defined for event E. The only event on the light path for which your proper time is defined is event D. It makes no sense to speak of your proper time between C and E, it does not exist.

Due to my relativistic speed, the distance Earth-Moon , which in their common rest reference frame is L = 1.25 ls , appears contracted to me : L_c = L/γ = 0.75 ls.
How long does it take a light signal, emitted by the Moon, to cover this contracted distance ? Divide L_c by c=1ls/s , and obtain 0.75s of my time.
That's all.

Now, let the Earth frame be unprimed and the spaceship frame be primed, and use units of time in seconds so that c=1. Then $\vec A = (t_A,x_A) = (0,0)$ and $\vec C = (t_C,x_C) = (0,1.25)$ and $\vec E = (t_E,x_E)= (1.25,0)$.

Now, if we do a Lorentz transform into the primed frame moving at 0.8 c we find $\vec C' = (-1.67,2.08)$ and $\vec E' = (2.08,-1.67)$. So the time according to the rocket is $t'_E - t'_C = 3.75 \ne 0.75$.

Note that in the primed frame the distance between the Earth and the moon is indeed 0.75 ls, but it is incorrect to assume that the time for a light pulse to go from one to the other is simply that distance divided by c. In fact, it takes considerably longer because the Earth and moon are moving in the primed frame.

 

[italicus]

@Dale

you said this :

"The only event on the light path for which your proper time is defined is event D. It makes no sense to speak of your proper time between C and E, it does not exist."

I have never spoken of my proper time between C and E, which is a light path. I have spoken of my proper time between A and B , which is my worldline. Can I watch my wristwatch , during my trip to the Moon?
Do we want to determine at what time, measured by an Earth observer, the signal coming from the Moon meets the rocket ? Your additions to my diagram help. The meeting happens at the event D . A simple reasoning goes like this ( see for example the first chapter of the book by Morin on Relativity, already cited, but I don’t remember where! ) :

The distance Earth- Moon is L= 1.25 ls .

This distance is covered in part by the rocket, that has a speed of 0.8c wrt Earth ; and in part by the signal, that has speed c , obviously, and direction opposed to that of the rocket. So, the instant of meeting , as calculated by the Earth observer, that I'll call T_d, is given by the following simple equation :
L - cT_d = vT_d

which means : T_d = L/(c+v) = (1.25 / (1 + 0.8) ) s = 0.6944 s

this is the time of meeting, event D, measured by an Earth observer.

What time is it where you are? Here in Italy it is 01:23 of Saturday : time to go to bed for me. Good night.