Second Opinion Needed: Time of Light Pulse Arrival

[Orodruin]

What is the time difference in the reference frame of A between the sending of the signal in A's rest frame and the event on the world line of A that is simultaneous with the event of B receiving the signal?

Answer = 6 years ... (i hope)

3.75 years (γγ\gamma times your last answer).

The easier way of seeing this is: The distance between B and the signal decreases at speed 1.6c and so the result is 6/1.6 = 3.75 years. Everything in the problem refers to frame A and so there is no reason to involve anything from B (except its velocity relative to A).

 

[robphy]

Thank you Orodruin for your phasing of the two questions and you subsequent answers. They clarified how I should be posing the question. Would below be a better way to phase it?

Two observers, A and B, are traveling towards each other at a relative speed of 0.6c.

In A's frame of reference, when B is 6 light-years away, A sends a photon to B.

What is the time difference in the reference frame of B between the event on the world line of B that is simultaneous with the sending of the signal in A's rest frame and the event of B receiving the signal?

Answer = 3 years ... (i hope)

What is the time difference in the reference frame of A between the sending of the signal in A's rest frame and the event on the world line of A that is simultaneous with the event of B receiving the signal?

Answer = 6 years ... (i hope)

In A's frame of reference, when B is 6 light-years away, A sends a photon to B. What is the time difference between this event and B receiving the signal in the reference frame of B?

Answer = 7.5 years ... (i hope)

Here's a diagram I made with my GeoGebra tool: https://www.geogebra.org/m/HYD7hB9v# .
It's an ordinary Minkowski diagram with light-clock diamonds supplementing worldlines (and other segments on a spacetime diagram)
(See my Insight for details: https://www.physicsforums.com/insights/relativity-rotated-graph-paper/ )

It's probably similar to what @Jeronimus drew in his earlier post #44
assuming the situation there is the same as is quoted above---I didn't follow this thread from the beginning.

Since you chose a relative velocity with a rational Doppler k-factor,
you could easily sketch this on graph paper and obtain precise values--by counting boxes--, assuming decent drafting skills, fairly easily.
(With merely rational relative velocities, a little more drafting skill is needed..
or else a little more abstraction by thinking in terms of square-intervals rather than counting ticks on a worldline.)

Note: no need to draw another spacetime diagram for the other frame.
But if you want to, you could easily perform the Lorentz transformation visually
by restacking the corresponding transformed diamonds on the graph paper.

With this spacetime diagram, you can then use it (in particular, the triangles) to "explain" various approaches to get the results...
use "spacetime trigonometry" [shameless pitch] or 4-vector methods or "relativistic formulas" and "Lorentz transformations" or k-calculus methods or radar methods.

 

[Herman Trivilino]

That isn't what I said - I didn't say the light travel time wasn't real or is a trick, I said time dilation itself is real and not a trick - there is a logical difference. I understand the article wasn't exactly denying that either but I just said it could be misleading - in the sense that it could lead people to that conclusion if they didn't already know better. I don't know how that wasn't clear.

Right. I didn't mean to imply that you had said that. I was in effect offering a response to the "someone" who may have been mislead.

As soon as I saw the Bondi quote under the title of the Sci Am article and began to read it I immediately thought about and compared that analysis of the twin trip to one employing an exchange of (Doppler-shifted) light signals between the twins. When first encountered that analysis seems to gloss over time dilation and the point I was making is that time dilation is part of the explanation of the Doppler shift.

More to the point, the time differences explained using light signal travel time are a direct consequence of the fact that the speed of the signals is independent of the speed of either twin.

That is perhaps at least part of what @Ibix referred to when he responded:

Actually, you can adjust time dilation freely by changing your interpretation of what causes the Doppler effect. You can't make the Doppler effect go away, nor can you make differential aging (as in the twin paradox) go away. In some senses they are more fundamental than time dilation and length contraction.

 

[Jeronimus]

Thank you Orodruin for your phasing of the two questions and you subsequent answers. They clarified how I should be posing the question. Would below be a better way to phase it?

Two observers, A and B, are traveling towards each other at a relative speed of 0.6c.

In A's frame of reference, when B is 6 light-years away, A sends a photon to B.

What is the time difference in the reference frame of B between the event on the world line of B that is simultaneous with the sending of the signal in A's rest frame and the event of B receiving the signal?

Answer = 3 years ... (i hope)

What is the time difference in the reference frame of A between the sending of the signal in A's rest frame and the event on the world line of A that is simultaneous with the event of B receiving the signal?

Answer = 6 years ... (i hope)

In A's frame of reference, when B is 6 light-years away, A sends a photon to B. What is the time difference between this event and B receiving the signal in the reference frame of B?

Answer = 7.5 years ... (i hope)

And a final question:

Two observers, A and B, are traveling relative to each other at a relative speed of 0.6c.

In A's frame of reference, there is a ruler that is 4.8 light years long.

In A's frame of reference how long would it take a photon of light to travel the length of the ruler?

Answer = 4.8 years ... ( I hope)

In B's frame of reference there is a ruler. Said ruler measures 4.8 light long in A frame's of reference. How long would it take a photon to travel the length of the ruler in B's frame of reference?

Answer = 6 years ... (i hope... still feeling shaky)

I wasn't sure how you arrived at 3 and 6 years, going by the description you gave.

For example:

"In A's frame of reference, when B is 6 light-years away, A sends a photon to B.

What is the time difference in the reference frame of B between the event on the world line of B that is simultaneous with the sending of the signal in A's rest frame and the event of B receiving the signal?

Answer = 3 years ... (i hope)"

and...

"In A's frame of reference, when B is 6 light-years away, A sends a photon to B.

What is the time difference between this event and B receiving the signal in the reference frame of B?

Answer = 7.5 years ... (i hope)"

...are equivalent descriptions to me. Both would result in 7.5years, not 3 years. After drawing it, i realized that in the former case, you are asking for the time difference between the event of a light signal reaching B that was sent by A when he observed B to be at 6 lightyears away of him (light blue/magenta filled circle), and the event when A sent a signal towards B, when B observed A to be at 4.8 lightyears away. Or alternatively, when A observed B to be at 3.75 lightyears away (the blue filled circle).

Here is a more complete drawing of the situation, which will hopefully shed some light on this

 

[david316]

In the interests of clarifying things, and saving my sanity, I thought I'd offer some background on how this problem came about and hopefully clarify where the ambiguities may have been introduced.

It all started when this Scientific American article (https://www.scientificamerican.com/article/time-and-the-twin-paradox-2006-02/) came up in discussion - it claims to "explain" the twin paradox in terms of the time it takes for light to reach each observer. Now I claimed that this is misleading, because while the time delays might be interesting to consider in terms of what you'd actually "see", they are not relevant to the effects derived in Special Relativity. That is, time dilation and other effects are "real" and not just tricks due to how long it takes light to reach observers from distance events.

I made the claim... not sure if I should admit it...

"During the trip to the star both the traveller and homebody would view each other clocks as running slow. But when the traveller gets to the star the travellers clock would have elapsed 6 years and the travellers clock would have elapsed 10 years. This doesn’t make sense although it must be correct. It only makes sense if you say that it is not possible to know two events at the same time which is the basis of space-time. Hence you need to take into account the time traveled in order to observe the clocks or else it doesn’t work. "

lets pretend I said this:

"During the trip to the star both the traveller and homebody would view each other clocks as running slow. But when the traveller gets to the star the travellers clock would have elapsed 6 years and the travellers clock would have elapsed 10 years. This doesn’t make sense although it must be correct. It only makes sense if you take into account the time traveled in order to observe the clocks or else it doesn’t work. "

Then, I constructed an example with all sorts of problems to try and demonstrate this... now I am lying awake at night thinking about SR...

 

[Herman Trivilino]

Consider the event "traveler arrives at star" and the event "home clock reads 10 years". Saying that the first one happened when the second one happened is not a meaningful assertion. The two events occur in different places, so if they're simultaneous to one of the twins they won't be simultaneous to the other.

If, however, the two events occur at the same place, then such a statement is meaningful. Because it will be true for all observers.

 

[david316]

Consider the event "traveler arrives at star" and the event "home clock reads 10 years". Saying that the first one happened when the second one happened is not a meaningful assertion. The two events occur in different places, so if they're simultaneous to one of the twins they won't be simultaneous to the other.

If, however, the two events occur at the same place, then such a statement is meaningful. Because it will be true for all observers.

Accepted. What I meant is:

During the trip to the star both the traveller and homebody would view each other clocks as running slow. But when the traveller gets to the star, in the travellers frame of reference his clock would have elapsed 6 years. In the event on the world line of the home body that occurs simultaneously with the event of the traveller arriving at the star, the homebody's clock would have elapsed 10 years. So both view each others clocks running slow, and at the event described above, both clocks read 6 and 10 years. It only makes sense if you accept that in order for each observer to observe the others clock, the observation between the observers is limited by the speed of light and the distance required to travel in the respective rest frame.

 

[david316]

Is the statement below correct, in the context of SR not GR ... I'm not liking my odds...

"Two people will only disagree on the time of a single clock if you limit information travel by the speed of light and one person is traveling relative to the other. "

 

[russel_]

Is the statement below correct, in the context of SR not GR ... I'm not liking my odds...

"Two people will only disagree on the time of a single clock if you limit information travel by the speed of light and one person is traveling relative to the other. "

I suspect someone will take exception to the notion of a "single clock"! I say this because the observers have different clocks (different frames), and while a single event could be used to set both to "0", I don't think that really gets at the matter. Maybe the statement below would be as good? Here the "clock" is replaced with an event, and you need two of them so the observers can make a comparison (i.e., the spacing and coincidence of events is what defines our underlying notion of time).

"Two people will only disagree on the simultaneity of two events at different locations if you limit information travel by the speed of light and one person is traveling relative to the other."

And I hazard to say that, at least as I understand it, this is not true: they can disagree even after taking into account the time it took the information to travel. They literally observe (not just "see") parts of space arriving at various times differently than the other does. We'll see what others say...

 

[david316]

I suspect someone will take exception to the notion of a "single clock"! I say this because the observers have different clocks (different frames), and while a single event could be used to set both to "0", I don't think that really gets at the matter. Maybe the statement below would be as good? Here the "clock" is replaced with an event, and you need two of them so the observers can make a comparison (i.e., the spacing and coincidence of events is what defines our underlying notion of time).

"Two people will only disagree on the simultaneity of two events at different locations if you limit information travel by the speed of light and one person is traveling relative to the other."

And I hazard to say that, at least as I understand it, this is not true: they can disagree even after taking into account the time it took the information to travel. They literally observe (not just "see") parts of space arriving at various times differently than the other does. We'll see what others say...

Yip. In my sleep deprived state I think I agree with what you think I think. i.e.

"In the framework of special relativity, two people will only disagree on the simultaneity of two events at different locations if you limit information travel by the speed of light and one person is traveling relative to the other."

I still think this true but I accept my knowledge in this area has been proved to be vastly lacking and hence won't be surprised if its wrong!

 

[robphy]

Here is a more complete drawing of the situation, which will hopefully shed some light on this

The relative velocity on the left looks like 6/10.
The relative velocity on the right looks like 5/8... is that what you want?

 

[Jeronimus]

Yip. In my sleep deprived state I think I agree with what you think I think. i.e.

"In the framework of special relativity, two people will only disagree on the simultaneity of two events at different locations if you limit information travel by the speed of light and one person is traveling relative to the other."

I still think this true but I accept my knowledge in this area has been proved to be vastly lacking and hence won't be surprised if its wrong!

Two observers traveling at ANY given speed relative to each other, will NEVER agree on the simultaneity of two space separated events. If observer A observers two events to be simultaneous, hence their spacetime location withing his reference frame is x1, t1 and x2, t2 with x1 ≠ x2 (space separated) and t1=t2 (simultaneous), THEN observer B will _always_ get t1' ≠ t2' (NOT-simultaneous).

So if observer A checks his clock, and his clock displays whatever count. Let's assume 0 as you seem to like that number. At the same time, A observes B's clock to show 0 THEN observer B will never ever for eternity observe his clock to display a count of 0 while A's clock displays 0 as well. (unless the clocks are damaged or drugs were involved)

When we say observe, we mean measure the incoming information we get locally and then calculate the spacetime positions of events based on this information. If for example a light beam reaches us, then we can calculate the spacetime location of when and where the lightbeam was shot towards us, given we get the maths and physics right and we have sufficient additional information.
Or, if you follow Einstein. We would place an infinite amount of observers at every location, collect the event data they registered after some time, and then create some nice spacetime diagrams we would map all spacetime events they registered within it.

Observation is not about what we see, but about what we measure and calculate.

 

[Jeronimus]

The relative velocity on the left looks like 6/10.
The relative velocity on the right looks like 5/8... is that what you want?

Not. That would be a disaster if it was the case.

But i don't think it is.

The white observer in the left x-t diagram at the spacetime location x=6ls t=0s will be crossing the t axis in 10 seconds according to the diagram. So we have 6ls /10s = 0.6c

The Blue clock in the right diagram is located at x=-4.8ls, t = 0 and will be crossing the t axis within 8 seconds as you can see from its worldline. -4.8ls/8s = -0.6c

Looks good to me.

 

[david316]

Two observers traveling at ANY given speed relative to each other, will NEVER agree on the simultaneity of two space separated events. If observer A observers two events to be simultaneous, hence their spacetime location withing his reference frame is x1, t1 and x2, t2 with x1 ≠ x2 (space separated) and t1=t2 (simultaneous), THEN observer B will _always_ get t1' ≠ t2' (NOT-simultaneous).

So if observer A checks his clock, and his clock displays whatever count. Let's assume 0 as you seem to like that number. At the same time, A observes B's clock to show 0 THEN observer B will never ever for eternity observe his clock to display a count of 0 while A's clock displays 0 as well. (unless the clocks are damaged or drugs were involved)

When we say observe, we mean measure the incoming information we get locally and then calculate the spacetime positions of events based on this information. If for example a light beam reaches us, then we can calculate the spacetime location of when and where the lightbeam was shot towards us, given we get the maths and physics right and we have sufficient additional information.
Or, if you follow Einstein. We would place an infinite amount of observers at every location, collect the event data they registered after some time, and then create some nice spacetime diagrams we would map all spacetime events they registered within it.

Observation is not about what we see, but about what we measure and calculate.

Can I re-phase the statement.

"Two people traveling relative to each other will disagree on the simultaneity of two events at different locations. The fundamental principe underlying this is that the speed of light is constant in all frames of reference."

And as I believe you implied above, if two observers somehow knew the spacetime location of an event in each others frame of reference and they know their relative speed, they could both agree at what space time location the event occurred in each others location.

 

[Herman Trivilino]

And as I believe you implied above, if two observers somehow knew the spacetime location of an event in each others frame of reference and they know their relative speed, they could both agree at what space time location the event occurred in each others location.

By spacetime location do you mean spacetime coordinates? For example, an event occurred at t=5 and x=2 in my rest frame, and the same event occurred at t'=3 and x'=0 in your rest frame? Regardless of our relative speed we will both agree on all of that.

 

[Jeronimus]

Can I re-phase the statement.

"Two people traveling relative to each other will disagree on the simultaneity of two events at different locations. The fundamental principe underlying this is that the speed of light is constant in all frames of reference."

And as I believe you implied above, if two observers somehow knew the spacetime location of an event in each others frame of reference and they know their relative speed, they could both agree at what space time location the event occurred in each others location.

Yes, and that is what the Lorentz transformations are about. Their basic forms are very simple too.

The Lorentz transformation formulas to get x' and t' are

x' = γ(x-vt)
t' = γ(t-vx/c²)

to get back, you would use the inverse Lorentz transformations formulas

x = γ(x'+vt')
t = γ(t'+vx'/c²)

γ=1/(√1-v²/c²) = 1.25 for v=0.6c

x: space position of event (where is it located)
t: time position of event (when does it happen)
v: The relative velocity
c: The speed of light

see more here https://en.wikipedia.org/wiki/Lorentz_transformation

Those formulas can be derived mathematically just by the two postulates of SR

1) The laws of physics are the same in every inertial frame of reference (no IFR is special compared to another)
2) The speed of light is always c in a vacuum absent of gravity (independent on which inertial frame of reference frame you observe a beam of light, you will always observe it traveling at c, independent of the velocity of the emitting source)

(note that the formulas also depend on where you choose to locate the positive and negative x-axis within the two reference frames/coordinate systems)

And just by knowing how to apply those formulas, you can create the simulation of the twin paradox i programmed, i already linked at an earlier post. Scroll back if you missed it or click here https://www.physicsforums.com/threads/second-opinion-needed.906688/page-3#post-5710416
Try to apply the formulas by taking any event x, t on the left x-t diagram in my simulation at any time you wish and then see if you get the right x', t' values for the right diagram for the event.

Take an event in the right x-t diagram of my simulation and apply the inverse Lorentz transformation formulas and see if it maps properly in the left diagram.

For example, let's take the white filled circle in the screenshot i made.

It is located at x=6ls, t=0s in the left diagram

x' = 1.25*(6ls - (-0.6c)*0) = 7.5ls
t' = 1.25*(0 - (-0.6c)*6ls/c²) = 4.5s (you won't have to deal with c² as -0.6c*6cs (lightsecond/ls) /c² shortens to -0.6*6s)

Now that makes me look like a liar, because in the right diagram, the white filled circle is always located at x=0, t=0 (or x'=0 , t'=0 if you wanted to be in accord with how this is usually defined - in my simulation it is x and t for both sides)

But that is intentionally. The observer in the right diagram can always draw the x-t diagram such that he is in the middle, and map all events relative to that.

Check where the teal filled circle is in the right diagram. It is at x' = -7.5ls, t' = -4.5ls

So i could have drawn the right x-t diagram such that the teal circle is at x'=0, t'=0 and instead the white circle would be at x' =7.5ls and t'=4.5s as calculated with the Lorentz transformations but since i wanted to draw it from the perspective of the traveling twin, hence him mapping all events relative to him as he "travels", this seemed the more appropriate way to do the simulation.

Think of the right diagram as the traveling twin (white filled circle) drawing x-t diagrams periodically along his travel. In those x-t diagrams he maps the spacetime location of every event relative to himself while placing himself at x=0, t=0 within the diagram.
Afterwards he creates an video animation of those diagrams. The right diagram is what you would get.

And that is all you need basically to create the simulation i created. That, and some petty coding skills like mine, willing to stare at code for hours without writing a line.

 

[david316]

Yes, and that is what the Lorentz transformations are about. Their basic forms are very simple too.

The Lorentz transformation formulas to get x' and t' are

x' = γ(x-vt)
t' = γ(t-vx/c²)

to get back, you would use the inverse Lorentz transformations formulas

x = γ(x'+vt')
t = γ(t'+vx'/c²)

γ=1/(√1-v²/c²) = 1.25 for v=0.6c

x: space position of event (where is it located)
t: time position of event (when does it happen)
v: The relative velocity
c: The speed of light

see more here https://en.wikipedia.org/wiki/Lorentz_transformation

Those formulas can be derived mathematically just by the two postulates of SR

1) The laws of physics are the same in every inertial frame of reference (no IFR is special compared to another)
2) The speed of light is always c in a vacuum absent of gravity (independent on which inertial frame of reference frame you observe a beam of light, you will always observe it traveling at c, independent of the velocity of the emitting source)

(note that the formulas also depend on where you choose to locate the positive and negative x-axis within the two reference frames/coordinate systems)

And just by knowing how to apply those formulas, you can create the simulation of the twin paradox i programmed, i already linked at an earlier post. Scroll back if you missed it or click here https://www.physicsforums.com/threads/second-opinion-needed.906688/page-3#post-5710416
Try to apply the formulas by taking any event x, t on the left x-t diagram in my simulation at any time you wish and then see if you get the right x', t' values for the right diagram for the event.

Take an event in the right x-t diagram of my simulation and apply the inverse Lorentz transformation formulas and see if it maps properly in the left diagram.

For example, let's take the white filled circle in the screenshot i made.

It is located at x=6ls, t=0s in the left diagram

x' = 1.25*(6ls - (-0.6c)*0) = 7.5ls
t' = 1.25*(0 - (-0.6c)*6ls/c²) = 4.5s (you won't have to deal with c² as -0.6c*6cs (lightsecond/ls) /c² shortens to -0.6*6s)

Now that makes me look like a liar, because in the right diagram, the white filled circle is always located at x=0, t=0 (or x'=0 , t'=0 if you wanted to be in accord with how this is usually defined - in my simulation it is x and t for both sides)

But that is intentionally. The observer in the right diagram can always draw the x-t diagram such that he is in the middle, and map all events relative to that.

Check where the teal filled circle is in the right diagram. It is at x' = -7.5ls, t' = -4.5ls

So i could have drawn the right x-t diagram such that the teal circle is at x'=0, t'=0 and instead the white circle would be at x' =7.5ls and t'=4.5s as calculated with the Lorentz transformations but since i wanted to draw it from the perspective of the traveling twin, hence him mapping all events relative to him as he "travels", this seemed the more appropriate way to do the simulation.

Think of the right diagram as the traveling twin (white filled circle) drawing x-t diagrams periodically along his travel. In those x-t diagrams he maps the spacetime location of every event relative to himself while placing himself at x=0, t=0 within the diagram.
Afterwards he creates an video animation of those diagrams. The right diagram is what you would get.

And that is all you need basically to create the simulation i created. That, and some petty coding skills like mine, willing to stare at code for hours without writing a line.

So for example. If there are two observers in the same rest frame. Both with clocks that are synced. One observer instantly accelerates to 0.6c (I know this isn't possible but humour me). The clock signals are sent between the observers at the speed of light. When one observer receives a clock signal he can derive what the other observer will be seeing on their clock at the event on the other observers world line that is simultaneous with when said observer receives clock signal. e.g. When observer A receives a clock signal he can derive what observer B will be seeing on their clock at the event on observers B world line that is simultaneous with when observer A receives the clock signal. This calculation will give observer A the correct time that observer B will be seeing on their clock at the event on observers B world line that is simultaneous with observer A receiving said clock signal.

 

[Jeronimus]

So for example. If there are two observers in the same rest frame. Both with clocks that are synced. One observer instantly accelerates to 0.6c (I know this isn't possible but humour me). The clock signals are sent between the observers at the speed of light. When one observer receives a clock signal he can derive what the other observer will be seeing on their clock at the event on the other observers world line that is simultaneous with when said observer receives clock signal. e.g. When observer A receives a clock signal he can derive what observer B will be seeing on their clock at the event on observers B world line that is simultaneous with when observer A receives the clock signal. This calculation will give observer A the correct time that observer B will be seeing on their clock at the event on observers B world line that is simultaneous with observer A receiving said clock signal.

I won't try to fully decipher this because it might blow my mind. But i believe i know what you are trying to understand, so i will give you an example more or less based on the above.

Two observers, A and B are at rest relative to each other. B is 5 lightseconds away of B. Their clocks are synced, let's say that both display 0 seconds.

To keep it simple, We place A in the middle of our x-t diagram, hence A's spacetime location is x1=0ls, t1=0s while B is located at x2= 5ls, t2=0s. Both their clock counts display 0 seconds at those spacetime locations.
B located at x=5ls, t=0s with a clock count of 0s is an event. Let's call this event E2.

Now A accelerates near instantaneously to 0.6c. What will be the clock count on B's clock he will observe post-acceleration?

First let's check where the instance of B with a clock count of 0 seconds will be located post-acceleration as measured/calculated by A.

x' = γ(x-vt)
t' = γ(t-vx/c²)

γ=1/(√1-v²/c²) = 1.25 for v=0.6c

so we get

x' = 1.25*(5ls-0.6c*0s) = 6.25ls
t' = 1.25(0-0.6*5s) = -3.75s

So for A who accelerated to 0.6c near instantaneously, the instance of B with a clock count of 0s now has a spacetime position of x'=6.25ls and t'= -3.75s . This instance of B is not simultaneous to A anymore. Let's call this event E2'

We might want to know which instance of B is simultaneous to A now.

Let's see what we have. We have t' =0 (the time location of the instance of B in A's rest frame), γ = 1.25, v=0.6c, x=5ls (the location of B in B's rest frame - All instances of B are located at x=5ls in B's rest frame... he stays at 5ls at all times obviously)

using t' = γ(t-vx/c²) and solving for t, we get:

t= t'/γ + vx/c² = 3s - That gives us the time location of the instance of B in B's rest frame, which A will observe to be at t'=0 (simultaneous). This also tells us that the clock count of this instance of B will be 3 seconds at this location.
In B's rest frame, this instance is located at x=5ls, t=3s with a clock count of 3s. Let's call this event E1.

Now that we have the time location of the instance we are looking for, we can solve for x'. We already have t'=0.

x' = γ(x-vt) = 1.25(5ls-0.6c*3s) = 4ls

So now we know, that A will observe the instance of B which is simultaneous to him, to be at x' = 4ls, t' =0s with a clock count of 3 seconds. This is event E1'

But let's see what the simulation tells us

check if the calculations match the screenshot of the simulation, download and run the simulation yourself (remember to enter 0.6c as Vmax before you start it)

I

 

[Orodruin]

Two observers traveling at ANY given speed relative to each other, will NEVER agree on the simultaneity of two space separated events

This is only true in 1+1 space-time dimensions. In 1+3, it is true if you make the separation along the relative direction of motion non-zero. (In 1+1 there is only one spatial direction to move in.)

 

[Herman Trivilino]

So for example. If there are two observers in the same rest frame. Both with clocks that are synced. One observer instantly accelerates to 0.6c (I know this isn't possible but humour me).

Just have the two observers pass each at speed $0.6c$ and agree to synchronize their clocks when they do so.

The clock signals are sent between the observers at the speed of light. When one observer receives a clock signal he can derive what the other observer will be seeing on their clock at the event on the other observers world line that is simultaneous with when said observer receives clock signal. e.g. When observer A receives a clock signal he can derive what observer B will be seeing on their clock at the event on observers B world line that is simultaneous with when observer A receives the clock signal. This calculation will give observer A the correct time that observer B will be seeing on their clock at the event on observers B world line that is simultaneous with observer A receiving said clock signal.

You mean A sends a light signal and B receives the signal. B can then figure out what B's clock must have read when A sent the signal.

And of course A will not agree with B. A will conclude that B's clock had a different reading when A sent the signal. It is not the clock-readings themselves that they disagree on. All they disagree on is the notion of "when".

 

[david316]

Just have the two observers pass each at speed $0.6c$ and agree to synchronize their clocks when they do so.

Sure.

You mean A sends a light signal and B receives the signal. B can then figure out what B's clock must have read when A sent the signal.

And of course A will not agree with B. A will conclude that B's clock had a different reading when A sent the signal. It is not the clock-readings themselves that they disagree on. All they disagree on is the notion of "when".

No. After some arbitrary distance A sends a light signal to B. When B receives the signal he can use the maths of Lorentz transformation to figure out what A's clock would have read. A will agree that his clock read what B calculated.

 

[Herman Trivilino]

After some arbitrary distance A sends a light signal to B.

Yes, that's what I was thinking, too.

When B receives the signal he can use the maths of Lorentz transformation to figure out what A's clock would have read.

Yes. But let's say that the value B calculates for A's clock-reading is $t_A$. Moreover B calculates that the reading on his own clock was $t_B$ when that signal was sent.

A will agree that his clock read what B calculated.

Yes. But A won't agree that B's clock read $t_B$ when A's clock read $t_A$.