Another spaceship and two points question....

[TomTelford]

Flattening... right.

 

[PeterDonis]

When the lecturer uses phrases like "the person on the train sees...", they are implying 'after removing the effects of doppler shift' or 'only considering the effects of SR' but they don't say it.

It is true that the word "see" is often used, not to mean what is actually directly seen with the eyes or a telescope or something similar, but what is calculated from direct observations by correcting for light travel time, Doppler shift, etc. This usage is indeed confusing, but unfortunately it's what we're stuck with; you just have to learn to deal with it.

what I'm trying to do now is to take it from what would REALLY be observed, all in, all effects and to be able to say "ok, this part is doppler so take that out and this part is SR or GR".

You can't separate "Doppler" from "SR". The correct Doppler shift formula includes SR effects.

Training tells me how to do the math but I still have to build a sense of those relationships in my head in order to make it work while I am flying. An intuitive sense of this much change in power produces this much change in groundspeed.

That's what I'm trying to do with these questions.

Unfortunately, this analogy with flying might do more harm than good, because in relativistic spaceflight there is no invariant quantity that is directly analogous to either indicated airspeed--the thing you actually see in the cockpit--or true groundspeed--your "real" speed in terms of getting from starting point to destination. Nor is there anything analogous to wind speed/direction, calibrated vs. indicated airspeed, etc., to apply as corrections to get from one to the other. Nor is "engine power" the best thing to think about as far as what you adjust to get there faster or slower. Relativistic spaceflight is too different, IMO, from flying airplanes in an atmosphere to make this analogy useful.

 

[PeterDonis]

The ship is moving at .5c as it passes clock A which reads noon and it "sees" clock B reading 11am. However the pilot thinks the distance to B is only .866 light hours not 1 lh away and that based on how quickly it observes time passing on clock B that it should read something like 12:07 if it were there right "now" in the ships frame as opposed to noon in clock A's frame?

You can't figure this out using time dilation/Doppler shift alone. You also have to take into account relativity of simultaneity--or, to put it another way, the fact that in the ship's rest frame, clocks A and B are not synchronized (they don't show the same readings at the same time in the ship's rest frame).

 

[Ibix]

Now I'm just trying to figure out which formula is used to solve which value being transformed.

The Lorentz transforms. It's always the Lorentz transforms. Sometimes they simplify to the time dilation and length contraction formulae, but until you get intuition for when, use the Lorentz transforms.

 

[Janus]

So moving on to the Lorentz part of this situation in case 2 is the following true:

The ship is moving at .5c as it passes clock A which reads noon and it "sees" clock B reading 11am. However the pilot thinks the distance to B is only .866 light hours not 1 lh away and that based on how quickly it observes time passing on clock B that it should read something like 12:07 if it were there right "now" in the ships frame as opposed to noon in clock A's frame?

While B, according to the ship, is 0.866 lh away "now" as the ship passes A, this doesn't mean that B was 0.866 lh away when the light the ship is seeing "now" left B.
Put another way, the light that arrives from B that arrives when the ship passes A had to have left B sometime before the ship passes A, and when the ship and B were much further apart than 0.866c. And, according to the ship, that light traveled at c relative to the ship, and thus took longer than 0.866 hrs to reach him, during which time clock B was running 0.866 as fast as his own.
In this example, the ship and B would be ~1.732 lh apart when the light leaves B, and when the ship is 0.866 lh short of passing A as measured by the ship. It will take the ship 1.732 hr to reach A and the same time for the Light from B to reach both A and the ship. During which time, clock B will advance by 1.732 x 0.866 = 1.5 hrs. ( from 11 am to 12:30 am) . So according to the ship, when it passes A and sees a time of 11:00 am on clock B, it would determine that it was actually 12:30 am at clock B.