# Minkowski diagram, what does observer see.

1. May 23, 2013

### 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.

Last edited: May 23, 2013
2. May 23, 2013

### 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?

3. May 23, 2013

### 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.

4. May 23, 2013

### 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.

5. May 23, 2013

### 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.

Last edited: May 23, 2013
6. May 23, 2013

### ghwellsjr

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).

7. May 23, 2013

### 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.

8. May 23, 2013

### ghwellsjr

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?

9. May 23, 2013

### ghwellsjr

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.

10. May 23, 2013

### 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?

Last edited: May 23, 2013
11. May 23, 2013

### 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.

12. May 23, 2013

### Myslius

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

13. May 23, 2013

### ghwellsjr

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.

14. May 23, 2013

### 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.

15. May 23, 2013

### 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?

Last edited: May 23, 2013
16. May 23, 2013

### ghwellsjr

Good.
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.
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.

17. May 23, 2013

### ghwellsjr

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.

Last edited: May 23, 2013
18. May 23, 2013

### 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?

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

19. May 23, 2013

### 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.

20. May 24, 2013

### ghwellsjr

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.