Relativity of simultaneity doubt

[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.

 

[Dale]

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?

Yes, you can measure proper time from A to B, but that has nothing whatsoever to do with the light pulse when you said “I am interested in my propertime taken by the light signal”. That is what I am objecting to.

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 .

That single event, event D, is the only event for which a proper time related to the light pulse may be defined. Since it is a single event, no proper time duration may be calculated at all related to the light.

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.

Sure $\vec D = (t_D, x_D) = (0.69,0.56)$ and $\vec D’ = (t’_D, x’_D) = (0.42,0)$.

 

[italicus]

Yes, you can measure proper time from A to B, but that has nothing whatsoever to do with the light pulse. I thought that is what you were discussing.

That single event, event D, is the only event for which a proper time related to the light pulse may be defined. Since it is a single event, no proper time duration may be calculated at all related to the light.

Sure $\vec D = (t_D, x_D) = (0.69,0.56)$ and $\vec D’ = (t’_D, x’_D) = (0.42,0)$.

Oh, at last, we have agreed on something! I see that you have also determined t’_d and x’_d=0, thanks. But now we are in OT a lot!
Buona notte.

 

[Dale]

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.

See:

So I am interested in my propertime taken by the light signal emitted by the Moon to reach me

Which sure sounds like you are talking about your proper time from C to D. This is what I objected to, and why I stated that your proper time is undefined. If that was a mistake then we can move on.

 

[cianfa72]

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).

Sorry, so that basically means that in GR the notion of spacelike separated events is not well defined ? Does it make sense in GR only locally (i.e. in a limited region of spacetime) ?
Thank you.

 

[PeterDonis]

there can be both timelike and spacelike geodesics connecting the two events

Geodesics? Or just curves? In Godel spacetime, for example, there are CTCs through every event, but AFAIK none of them are geodesics.

 

[cianfa72]

Geodesics? Or just curves? In Godel spacetime, for example, there are CTCs through every event, but AFAIK none of them are geodesics.

Some thead ago we said it is "better" to employ geodesics only (not just generic curves) in order to define the separation type for a couple of events.

 

[PAllen]

Geodesics? Or just curves? In Godel spacetime, for example, there are CTCs through every event, but AFAIK none of them are geodesics.

I’m not sure about the possibility of CTC being geodesic, but that doesn’t change my point. What is important is that events being connected by a spacelike geodesic no longer guarantees that one is not in the causal future of the other. Only light cones can be used to define causal structure in a general GR manifold. (I know you know all this, I just needed to clarify my point in context of the CTC not being geodesic).

 

[cianfa72]

What is important is that events being connected by a spacelike geodesic no longer guarantees that one is not in the causal future of the other.

Yet we cannot claim those events are actually spacelike separated since there is a timelike geodesic connecting them too.

Only light comes can be used to define causal structure in a general GR manifold.

it should read light cones, I think.

 

[PAllen]

Yet we cannot claim those events are actually spacelike separated since there is a timelike geodesic connecting them too.

Actually, the whole point of the post is that the timelike curve may not be a geodesic, but it does not matter. Existence of any timelike path between events means they are causally connected. The distinction from SR is that existence of a spacelike geodesic is no longer sufficient to decide the issue.

it should read light cones, I think.

Corrected, thanks.

 

[Mister T]

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.

There is one more thing. You would need to know what time it is here on Earth when that light signal left the moon. For that you need a simultaneity convention.

You may want to step back a bit and ask yourself how it's possible that time dilation is symmetrical. That is, how is it possible each of two moving observers will conclude that the other's clock is running slow?

 

[PAllen]

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

Please do study post #85 and the associated diagram. It shows exactly why a rapidly approaching rocket would measure the moon farther away than andromeda using a radar bounce to each emission event.

 

[PeterDonis]

What is important is that events being connected by a spacelike geodesic no longer guarantees that one is not in the causal future of the other.

Yes, agreed.

 

[PeterDonis]

Some thead ago we said it is "better" to employ geodesics only (not just generic curves) in order to define the separation type for a couple of events.

That was for cases where issues like the presence of closed timelike curves do not apply. In spacetimes where CTCs are present, as @PAllen has pointed out, the fact that two events are connected by a spacelike geodesic does not imply that there is no causal curve (timelike or lightlike) connecting them. So one's intuitive picture of what "spacelike separation" means doesn't even work in such a spacetime.

 

[LBoy]

Are you possibly confused by 'see' versus 'model'? If two detectors are colocated at some moment (present at the same event), irrespective of their relative state of motion, it is physically impossible and absurd to claim that one receives two signals at that event and the other does not.

The Minkowski diagram (Italicus post 30) shows that the two signal reception events by OT and OE are not "colocated at the same event", we are dealing with two different events: OE receives a lightning strike signal, and OT receives two signals. These events are separated both temporally and spatially.

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. 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. The only further statement that can be made is that if two observers use the same convention meeting certain properties (e.g. the Einstein convention) for assigning simultaneity to spacelike separated events, and one is in motion relative to the other, then they will disagree on simultaneity assignment. But simultaneity of distinct events is never an observable, per se.

The second issue I have is the existence of simultaneity of events (what you call, I believe, modeling simultaneity). Here I use a mathematical definition: events are simultaneous to an observer if they are orthogonal to its time vector in Minkowski space. And maybe it's just a matter of name (you used the term "convention") but to me it's not a convention, in a simple SR model the straight line, plane or R3 containing simultaneity events are precisely defined for the observer.I s here something more to that?

 

[vanhees71]

I'd define the simultaneity of events wrt. an observer in a frame-independent way by: two events with spacetime fourvectors $x$ and $y$ are simultaneous wrt. an observer with four-velcity $u$ if $u \cdot (x-y)=0$.