Non-inertial reference frames question

  • I
  • Thread starter Freixas
  • Start date
  • #26
71
16
It is worth noting that, if we take those words at face value, they are the wrong words. What actually appears to your eyes, or to your telescopes or other instruments, is not a drastic change in "distance" in a short time. All that actually appears to your eyes and instruments is a drastic change in the redshift of light coming from Alpha Centauri--from "strongly redshifted" (because you were moving away from AC at close to the speed of light) to "no shift" (in the idealized case where Earth and AC are at rest relative to each other).
Actually, in the presented case of rapid deceleration from very fast to 0 (relative to Earth) cause the image of AC to shrink so as to make it appear further away (in addition to blue shifting)? In a spacetime diagram, I notice the distance the light has traveled as the ship arrives at Earth in the frame where it is still moving is much less than the distance it appeared to have traveled in the frame where it has stopped at Earth. That suggests to me the image of AC would "zoom away" to make it look as if it receded from you at that moment. It points to the fact that things you are moving away from appear to be closer to you and receding from you at a slower speed than they actually are (when considering the appearance from observing the light), while things you're moving toward appear further away and appear to approach you faster than they are (potentially faster than c).
 
  • #27
Mister T
Science Advisor
Gold Member
2,587
840
We can make the deceleration period as small as we want (just as long as its not zero).
The obvious way to have zero acceleration is to simply note the readings on the clocks when the ship passes by Earth.
 
  • #28
PAllen
Science Advisor
2019 Award
8,162
1,423
I have not read the different scenarios discussed, but the general rule for aberration is:

Suppose B and C are colocated, with B at rest relative to A, and C approaching A at high speed. Then C will observe A to subtend a smaller angle than B.

In the same scenario with C receding from A, C will see A subtending a larger angle than B.
 
  • #29
PeterDonis
Mentor
Insights Author
2019 Award
30,718
9,700
In the same scenario with C receding from A, C will see A subtending a larger angle than B.
Hm--did I get this backwards in post #27?
 
  • #30
PAllen
Science Advisor
2019 Award
8,162
1,423
Hm--did I get this backwards in post #27?
I think so. The image compression on approach is a large part of the relavistic beaming effect, increasing brightness. This is a larger impact than blueshift.
 
  • #31
PeterDonis
Mentor
Insights Author
2019 Award
30,718
9,700
The image compression on approach is a large part of the relavistic beaming effect, increasing brightness. This is a lrger impact than blueshift.
Yep, you're right, I took a look at the formulas. I've deleted my previous post that was in error.
 
  • #32
Ibix
Science Advisor
Insights Author
6,833
5,682
When the rocket stops instantaneously, the last measurement it may make that is unambiguously made when it was moving is the position of Alpha Centauri on the surface of the past light cone of the deceleration event. If it stops at distance ##d## as measured in Centauri's rest frame, this should be $$d\sqrt{\frac{c+v}{c-v}}$$At some point after that, all measurements should agree that the distance is ##d##. But how it happens depends on the measurement process.

If you use the angular size of the star, that changes instantaneously at the deceleration event. Since there's a finite speed of light, this means that you interpret the change from one inertial frame to the other as something that happened discontinuously on the surface of the past light cone of your deceleration event. So, the distance changed before you decelerated. The time at which this occurred is either ##-d/c## or ##-fd/c##, where ##f## is the Doppler factor (the square root above), depending on which frame you include the surface of the light cone in.

If you use the radar method, the first measurement unambiguously made after stopping is the first radar pulse sent out after the stop. So the last pulse to return before deceleration establishes that Alpha Centauri was at distance ##fd## at ship's time ##-fd/c##. The first pulse sent out after the stop establishes that Alpha Centauri was at distance ##d## at ship's time ##+d/c##. Dolby and Gull show that pulses sent before the deceleration that return afterwards show a linear change in distance in the intervening period. So this method says that Alpha Centauri slows before the deceleration event and stops after it.

The radar method is actually building one possible non-inertial frame. The angular size method is giving you a physical justification for when to switch inertial frames. The radar method is better, to my way of thinking, because it's assignment of times is never problematic. Stitching together inertial frames leads to the ship asserting that some negative times happened twice in some places, so there's nasty book keeping hidden under a simpler exterior.
 
  • #33
PAllen
Science Advisor
2019 Award
8,162
1,423
Note that radar coordinates are globally well behaved only if, for the defining observer, motion is inertial for all time before some event, and also inertial for all time after some event. Given this constraint, no matter what the world line does in between gives you well behaved global coordinates. However, if this condition is not met, radar coordinates may have no more consistent coverage than Fermi-normal coordinates (which are what you get from stitching MCIF together). For example, for eternal uniform acceleration, the coverage of radar coordinates and Fermi-normal coordinates are identical. Note that Rindler coordinates are just Fermi-normal coordinates with a translation to make x=0 the horizon rather than a given accelerating observer.
 
  • #34
Ibix
Science Advisor
Insights Author
6,833
5,682
Note that radar coordinates are globally well behaved only if, for the defining observer, motion is inertial for all time before some event, and also inertial for all time after some event.
Or just inertial on average, I think. For example I can swing my radar set around my head on a string for all eternity and I should get good coordinates everywhere - if I look on a large enough scale the deviation from the Minkowski frame of my and my radar set's joint centre of mass should be negligible. Or is there something I'm missing?
 
  • #35
PAllen
Science Advisor
2019 Award
8,162
1,423
Or just inertial on average, I think. For example I can swing my radar set around my head on a string for all eternity and I should get good coordinates everywhere - if I look on a large enough scale the deviation from the Minkowski frame of my and my radar set's joint centre of mass should be negligible. Or is there something I'm missing?
That sounds plausible. I did not consider oscillating situations when deriving the rule I stated.
 
  • Like
Likes Ibix

Related Threads on Non-inertial reference frames question

Replies
3
Views
917
Replies
14
Views
5K
Replies
3
Views
22K
Replies
4
Views
4K
Replies
3
Views
11K
Replies
9
Views
4K
  • Last Post
Replies
13
Views
2K
Replies
12
Views
1K
  • Last Post
Replies
15
Views
3K
  • Last Post
Replies
2
Views
3K
Top