Minkowski diagram, what does observer see.

[Myslius]

Suppose we have a spacetime diagram like this:

Red lines indicating light travel from the moving object to the observer.

Object is moving at the speed of 0.8c. At this speed we have:
Lorentz factor 1/√(1 - v²/c²)=1/0.6=1.66(6)
Relativistic Doppler effect √((1 + β) / (1 - β)) = 3

My question is what does the observer see?
According to this diagram At time t=3 (let's say year) observer sees object 1 year old. At t=6, 2 years old. 6/2=3 and we get Doppler effect.

As far as i know moving at 0.8c speed in no matter what direction relative to the observer results in time ticking at the speed 0.6 (or 1.66(6) slower).

Suppose at t=6 (t'=2) object starts to move towards the observer (same speed).
At t=6, object should arrive to the observer, and be 3.6 years old.

But how's that possible?
How could it be that observer sees the object at time t=6 at both places: arrived, and just starting to making a turn?

I don't understand.

 

[ghwellsjr]

If the object makes a turn at its Proper Time of 2 years, why shouldn't the observer see that at his Proper Time of 6 years? What is the problem?

 

[Myslius]

Yes, object makes a turn at his proper time of two years. But the observer sees an object both: 1) making a turn and 2) arrived at destination at t=6 years.

 

[ghwellsjr]

The observer's Coordinate (and Proper) Time is 6 years and his Coordinate Location is 0 light-years. The object's Proper Time is 2 years but in the unprimed frame, its Coordinate Time is 3 1/3 years (not 6 years) and its Coordinate Location is 2 2/3 light-years. Read the coordinates off the graph.

 

[Myslius]

Yes, that's completely correct. But that doesn't answer my question. Or do you mean that observer sees an object at t=3 1/3, being an 2 years old and 2 2/3 light-years away? My question is what does the observer sees in time interval [0,6], the distance and object's time. So that's including light travel time.

 

[ghwellsjr]

Yes, that's completely correct. But that doesn't answer my question. Or do you mean that observer sees an object at t=3 1/3, being an 2 years old and 2 2/3 light-years away?

The observer cannot see the Coordinate Location unless someone has put up signposts or other landmarks to tell him the location and he cannot see the Coordinate Time unless someone has put a clock synchronized to the Coordinate Time at that location. We don't normally think of the Coordinates as being marked off with clocks and location markings but you could certainly do that if you want.

However, the observer can use radar measurements to determine how far away the object was at any particular time and those will match the Coordinates in his rest frame, but you didn't show him doing that.

Also, if the object was actually another observer, he could be making all the same observations and measurements of the first observer (except, of course, if he has turned around at his Proper Time of 2 years, things will look different, I'm talking about if he continued on in a straight line).

 

[Myslius]

No need any landmarks or rules in spacetime. You can calculate luminosity distance, and moving object might have a clock onboard and we might see his time.
Is is so hard to show a simple observed D(t) or observed t'(t) formula when v=0.8 That's all I'm asking.

 

[ghwellsjr]

No need any landmarks or rules in spacetime. You can calculate luminosity distance, and moving object might have a clock onboard and we might see his time.
Is is so hard to show a simple D(t) formula when v=0.8 or t(t') when v=0.8 That's all I'm asking.

It's not hard. D(t), or to be consistent with your drawing, d(t) = vt = 0.8t. But since you drew the line according to that function, you must have already been aware of it, correct?

 

[ghwellsjr]

No need any landmarks or rules in spacetime. You can calculate luminosity distance, and moving object might have a clock onboard and we might see his time.
Is is so hard to show a simple observed D(t) or observed t'(t) formula when v=0.8 That's all I'm asking.

I see you changed your question while I answered your previous version.

We can observe the Coordinates on a graph and the observers in the scenario can too if they know ahead of time what will happen (and everyone sticks to the plan) but they cannot see the Coordinate Locations and Times as they are happening. But as I said before, they can use radar to measure what happened (without knowing ahead of time what was going to happen) and then they can construct the graph later on.

 

[Myslius]

Thank you. So at time t=6 we see an object v*t=0.8*6 = 4.6 ly away. Is this consistent with statement that it takes some time for light to travel the distance? What about observed time?
If D observed is v*t, I guess t' observed is t'(t)=t/lorentz factor? Correct? Oh and should't D observed be also divided by lorentz factor?

 

[ghwellsjr]

Thank you. So at time t=6 we see an object v*t=0.8*6 = 4.6 ly away. Is this consistent with statement that it takes some time for light to travel the distance?

More like 4.8 ly away and it is consistent with the observer at rest sending a radar signal at his Proper Time of 1.2 years and receiving the echo at 10.8 years and the observer assuming that the signals traveled at c going and coming and calculating that at the midpoint between those two times the signal reflected off the target and was 1/2 the difference in those two times as a distance in light-years away. That's the radar measurement I was talking about.

 

[Myslius]

I changed my answer yet again. We are discussing quite fast. My apologies. And thank you for fast replies.

 

[ghwellsjr]

Thank you. So at time t=6 we see an object v*t=0.8*6 = 4.6 ly away. Is this consistent with statement that it takes some time for light to travel the distance? What about observed time?
If D observed is v*t, I guess t' observed is t'(t)=t/lorentz factor? Correct? Oh and should't D observed be also divided by lorentz factor?

More edits.

The observer doesn't necessarily know that the speed is v, he can use successive radar measurements to calculate the speed based on the assumption that the signals travel a c with respect to himself. He doesn't have to know about the Lorentz factor, he just looks at the time on the object's clock compared to the time he calculated based on his radar measurements. It will turn out that all his measurements correspond to what we would calculate with the Lorentz Transformation process. Also, there is no length contraction going on in this scenario. If you had a size for the object, then it would be length contracted and the observer could measure that also using radar.

 

[Myslius]

More like 4.8 ly away and it is consistent with the observer at rest sending a radar signal at his Proper Time of 1.2 years and receiving the echo at 10.8 years and the observer assuming that the signals traveled at c going and coming and calculating that at the midpoint between those two times the signal reflected off the target and was 1/2 the difference in those two times as a distance in light-years away. That's the radar measurement I was talking about.

Yes, 4.8. I'm wondering how you calculated 10.8 years. d(1.2) is 0.96 ly. Object is hit at time t 1.2 + 0.96/ (1 - 0.8) = 6 years.
But we don't see it yet, D(6) = 4.8, it takes 4.8 years for light to reach us. So we see a reflection at 4.8+6=10.8.
Ah yes, makes perfect sence.

In a set up experiment v might be determined by looking at Doppler effect.
To summarize the answer:
observed D(t)=vt
observed t'(t)=t/lorentz factor.
That's not including radar.

 

[Myslius]

For radar measuring technique we include t_e (time we shoot a photon)
Observed D using radar:
D = (t_e * v) / (1 - v/c)
Time when photon comes back:
t = te + (2/c) * (t_e * v) / (1 - v/c)
Radar measures time ((t_e * v) / (v - v²/c))/LorentzFactor

Is that correct?

 

[ghwellsjr]

More like 4.8 ly away and it is consistent with the observer at rest sending a radar signal at his Proper Time of 1.2 years and receiving the echo at 10.8 years and the observer assuming that the signals traveled at c going and coming and calculating that at the midpoint between those two times the signal reflected off the target and was 1/2 the difference in those two times as a distance in light-years away. That's the radar measurement I was talking about.

Yes, 4.8. I'm wondering how you calculated 10.8 years. d(1.2) is 0.96 ly. Object is hit at time t 1.2 + 0.96/ (1 - 0.8) = 6 years.
But we don't see it yet, D(6) = 4.8, it takes 4.8 years for light to reach us. So we see a reflection at 4.8+6=10.8.
Ah yes, makes perfect sence.

Good.

In a set up experiment v might be determined by looking at Doppler effect.

Yes, with the appropriate formula you can calculate its speed easily if it is traveling in line with you. It becomes more complicated if it is traveling with some parallel motion with respect to you. But we assume this is not the case when we draw a spacetime diagram.

To summarize the answer:
observed D(t)=vt
observed t'(t)=t/lorentz factor.
That's not including radar.

Why do you insist on including "observed"? It's a combination of an assumption (about the speed of light), a measurement (or observation), and a calculation. As long as you understand that "observed" means that and not what we normally mean we say that someone observed something, then it's OK. But keep in mind that we could transform the scenario into another frame in which the Coordinates and the Lorentz factor are different and yet the radar measurements and Doppler observations remain the same, so what would you say about what the observers observe in this case with respect to D(t) and t'(t)? Remember, there's no preferred (or special) reference frame in Special Relativity, not even one in which an observer is at rest.

 

[ghwellsjr]

For radar measuring technique we include t_e (time we shoot a photon)
Observed D using radar:
D = (t_e * v) / (1 - v/c)
Time when photon comes back:
t = te + (2/c) * (t_e * v) / (1 - v/c)
Radar measures time ((t_e * v) / (v - v²/c))/LorentzFactor

Is that correct?

The Proper Time (the observer's time on his clock) is used to measure the time at which he shoots a photon and the time at which he receives the reflected photon. If we call the first one t_e and the second one t_r, then (using units where c=1):

D = (t_r - t_e)/2

and

T = (t_r + t_e)/2

where T is the time according to the observer's Proper Time that the distance, D, applies and these two coordinates are what he would plot on a diagram.

Assuming that the observer can see the Proper Time, t', on the target's clock at his own Proper Time at t_r, then T/t' = gamma. EDIT: This is assuming that the clocks were synchronized when they were colocated and that they remain inertial.

 

[Myslius]

I agree with #16 and #17 posts and I have no doubts or questions about it. You helped me to clarify few things. Thank you. Do you have any comments on the following quote?

I believe the universe is 13.7 billion years old as per a clock on Earth. A clock at rest with respect to the CMB and away from any major gravitational sources would be SLIGHTLY faster. About 10 years faster by my estimate.

Would a time tick faster if we were at rest with respect to the CMB?

 

[ghwellsjr]

No, I don't have any comments.

The last time I commented on the CMB, I got it wrong so I leave that topic to others. See post #10 and following in this thread.

 

[ghwellsjr]

Suppose we have a spacetime diagram like this:

Red lines indicating light travel from the moving object to the observer.

Object is moving at the speed of 0.8c. At this speed we have:
Lorentz factor 1/√(1 - v²/c²)=1/0.6=1.66(6)
Relativistic Doppler effect √((1 + β) / (1 - β)) = 3

My question is what does the observer see?
According to this diagram At time t=3 (let's say year) observer sees object 1 year old. At t=6, 2 years old. 6/2=3 and we get Doppler effect.

As far as i know moving at 0.8c speed in no matter what direction relative to the observer results in time ticking at the speed 0.6 (or 1.66(6) slower).

Suppose at t=6 (t'=2) object starts to move towards the observer (same speed).
At t=6, object should arrive to the observer, and be 3.6 years old.

But how's that possible?
How could it be that observer sees the object at time t=6 at both places: arrived, and just starting to making a turn?

I don't understand.

I was looking back over your original post and I see that you edited it after I responded to it. You added the comment that I have put in bold and now I can see what you were asking. Your drawing doesn't show the object turning around but if it did, you would have to show it coming back at 0.8c instead of 1.0c as it appears you were assuming. As such, it would not arrive back at its age of 3.6 years old but rather at 4.0 years old and it would not arrive back at t=6 but rather at t=6 2/3 years. So the observer does not see the object at time t=6 at both places. At t=6, he sees it turn around and during the course of the next 8 months, he sees it coming back to him and he sees it aging by 24 months during that time which is in agreement with the Relativistic Doppler factor of 3.

Hopefully, this all makes sense to you now.

 

[Myslius]

Diagram with turnaround.

 

[Myslius]

Yes, it makes perfect sense. One more thing that doesn't make sense for me is the expansion of the universe. As far as i know space itself do expand. But I heard nothing about the time. I got a feeling like time is excluded from the expansion of the universe. Do you know any framework/work where expansion of time is included? Maybe there is any Minkowski-like diagrams for that?

 

[Myslius]

Friedmann equation.

The thing that i don't get is that there is H²=constant * ρ relationship.
Suppose we expand a 2D space at the constant speed. Density decreases by R^-2 law, H by R^-1 law.
But shouldn't in 3D space the relation between H and ρ be as H³=constant * ρ, where density decreases by R^-3 law and H by R^-1 law?

As i mentioned T is missing so obscure formula would be T*H³=constant * ρ + curvature.

I know this is a question of cosmology. So it might be better to ask elsewhere. But I'm trying to look from relativity perspective and here people know it the best.

 

[PeterDonis]

Suppose we expand a 2D space at the constant speed. Density decreases by R^-2 law, H by R^-1 law.
But shouldn't in 3D space the relation between H and ρ be as H³=constant * ρ, where density decreases by R^-3 law and H by R^-1 law?

I think you're missing the point of the Freidmann equation. It's not an equation based on geometric considerations like the one you raise. It's derived from the Einstein Field Equation--it's a dynamic equation, telling how the gravity created by matter and energy is related to the expansion rate.

Put another way: the argument you give would apply if I took the universe at some instant of time and scaled everything in it by some factor. But the universe at some later instant of time is *not* identical to the universe at the earlier instant, but scaled by some factor. More precisely, it isn't if there is any matter and energy in the universe, because the matter and energy creates dynamics: it *changes* the spatial geometry of the universe as time passes. That change invalidates your logic.

 

[WannabeNewton]

As far as i know space itself do expand. But I heard nothing about the time. I got a feeling like time is excluded from the expansion of the universe. Do you know any framework/work where expansion of time is included? Maybe there is any Minkowski-like diagrams for that?

What does it even mean for time to expand? As far as the FRW universe is concerned, the flow of time is future-directed, that is all.

 

[Myslius]

The discussion of the following subject is purely hypothetical.
Equation: T*H³=constant * ρ + curvature.
You didn't get my point. At specific time there is specific density and specific H value. I never said that the universe scaled down by applying some scale factor IS identical. I'm saying that the expansion of the universe is the result of time going and vice versa (time going is the result of the expansion). There is no distinction between two of them. Basically time can't tick if there's no expansion. I will not go into any details because: I don't understand it, I have no mathematical framework, theoretical speculations aren't tolerated usually, some more reasons. But i will gladly read any work made on such subject.

 

[PeterDonis]

I'm saying that the expansion of the universe is the result of time going and vice versa (time going is the result of the expansion). There is no distinction between two of them. Basically time can't tick if there's no expansion.

I'm not sure what you mean by this. There is a way of interpreting it that makes it true, as far as it goes: the FRW solutions to the Einstein Field Equation cannot be static with a zero cosmological constant, so if the cosmological constant is zero the universe must indeed be expanding or contracting. So one could, for this limited class of solutions, interpret "time can't tick if there's no expansion" as meaning that the solutions can't be static.

However, I'm not sure the above is what you meant, because you seem to me to be making a far more general and sweeping statement, which is false. There are plenty of solutions to the EFE which are not expanding or contracting; there are even FRW solutions with nonzero cosmological constant which are not expanding or contracting (although it's true that they are unstable against small perturbations). So as a general statement, it is not true that there has to be expansion (or contraction) for time to tick.

Also, I'm not clear how your claim about the expansion being required for time to tick relates to the claim you made about how H^2 and rho should be related. I don't see that these claims have anything to do with each other.