Another spaceship and two points question....

[TomTelford]

Sorry if this has been done elsewhere but I haven't seen it.

Case: At the start a craft with a pilot is collocated at a position with a clock A with a second clock B one light hour away synchronized with A. All are motionless relative to each other. Clock A reads noon and B reads 11am. Good so far? The craft accelerates but not fast say .01c and eventually reaches clock B and stops. All I want to know is if the following is true: The pilot agrees that clock A is now one hour behind B as opposed to the other way around at the beginning?

Thanks.

Tom.

 

[PeterDonis]

a clock A with a second clock B one light hour away synchronized with A

Clock A reads noon and B reads 11am. Good so far?

No. The two sentences quoted above contradict each other.

 

[TomTelford]

Sorry, to be more clear the pilot sees noon on Clock A and 11am on clock B. Better?

 

[PeterDonis]

the pilot sees noon on Clock A and 11am on clock B

Ok; since clock B is one light-hour away, the clock reading seen from clock B at the location of clock A will be one hour behind the reading seen at the same instant on clock A. Yes, this is fine.

And the question you are asking, then, is whether the clock reading seen from clock A at the location of clock B will similarly be one hour behind the reading seen at the same instant on clock B. Isn't the answer obvious?

 

[PeterDonis]

the pilot sees noon on Clock A and 11am on clock B

Note that this does not mean clock B is "one hour behind" clock A. Clocks A and B are synchronized; they have the same readings at the same times. So the reading "noon" on clock A happens at the same time (in the reference frame you are implicitly using, in which clocks A and B are at rest) as the reading "noon" on clock B. But someone co-located with clock A won't see the image of clock B reading "noon" until one o'clock by clock A.

In other words, you have to be careful to distinguish when events happen, in a given frame, with when those events are seen via light signals transmitted to distant locations.

 

[TomTelford]

Yes, but I'm laying bricks here and I want to be sure I'm not messing up anything.

The following then must be true that during the flight from Clock A to Clock B, Clock A would appear to the pilot to be running more slowly than Clock B to go from an apparent hour ahead to an hour behind. And that this effect would be true regardless of the effects of SR or GR.

 

[TomTelford]

I'll get to the meat of the question at this point.

In the case where the ship was not initially stationary but moving towards clock A in such a way as to be collocated with it when it read noon I believe Lorentz pops up at this point and says that clock B no longer says 11am as in the earlier case but something between 11am and noon. Am I heading in the right direction?

 

[PeterDonis]

The following then must be true that during the flight from Clock A to Clock B, Clock A would appear to the pilot to be running more slowly than Clock B to go from an apparent hour ahead to an hour behind

No. Once again, you are confusing when things happen with when they are seen.

During the pilot's flight, clocks A and B both run slow compared to the clock in the spaceship; and they both run slow by the same amount. The rates of clocks A and B are never different.

During the pilot's flight, the pilot sees clock B appearing to run faster than his own clock, and sees clock A appearing to run slower than his own clock. But these appearances are due to a combination of the changing light travel time from his ship to clocks A and B, and the relativistic Doppler effect caused by the ship moving towards clock B and away from clock A. When the pilot corrects for these effects, he finds, as stated above, that clocks A and B are both running slow relative to his own clock, and by the same amount.

 

[PeterDonis]

In the case where the ship was not initially stationary but moving towards clock A in such a way as to be collocated with it when it read noon I believe Lorentz pops up at this point and says that clock B no longer says 11am as in the earlier case but something between 11am and noon. Am I heading in the right direction?

No. Two observers who are co-located at a given instant will see the same light signals at that instant. So the ship will see the same image of clock B reading 11 am that an observer at rest relative to clock A sees.

What textbooks on SR have you read?

 

[Dale]

Yes, but I'm laying bricks here and I want to be sure I'm not messing up anything.

The problem is that you are asking other people if you are laying the bricks right, so if you communicate sloppily then we will give you incorrect feedback: correcting errors you didn’t intend and not noticing errors in your thoughts that were not expressed. Do make an effort, please.

I believe Lorentz pops up at this point and says that clock B no longer says 11am as in the earlier case

Because of your wording I cannot tell if you correctly understand that he receives the 11am signal and calculates that A and B are not synchronized in his frame or if you erroneously believe that the actual image that he receives is not the 11am image.

 

[PeterDonis]

I'm laying bricks here and I want to be sure I'm not messing up anything.

So far you've messed up everything, so you might want to rethink this strategy.

What are you laying bricks for? What goal are you trying to reach?

 

[TomTelford]

That is exactly where I wanted to get to. There are two (at least) simultaneous effects going on while the pilot is flying; the Lorentz transform being a real effect and a doppler effect being virtual and self eliminating on the return trip.

 

[PeterDonis]

There are two (at least) simultaneous effects going on while the pilot is flying; the Lorentz transform being a real effect and a doppler effect being virtual and self eliminating on the return trip.

You might want to reconsider thinking of the Lorentz transform as a "real" effect. The Lorentz transform depends on your choice of reference frame; you have it coming out a certain way because you have implicitly chosen a reference frame in which clocks A and B are at rest. In a different frame the Lorentz transform would give you different answers.

But what observers see--what light signals reach them at what points in their travel through spacetime--is invariant; it doesn't depend on any choice of reference frame. So those things have a better claim to being "real".

This ties into what I said before, that you appear to have a basic confusion between when things happen--which requires you to choose a reference frame--and what observers see.

 

[TomTelford]

Apologies, I'm listening to a lecture on relativity and I would accuse the lecturer of using ambiguous language.

All I wanted to know was that in case 1 pilot sees Clock A showing one hour ahead of Clock B at the start and the reverse at the end.

In case 2 changing only that the craft is moving throughout and starts at Clock A at noon, what does he actually see is the time on Clock B when he passes Clock A? It may appear closer or running slower or whatever, don't care at this point. All I want to know is whether the time is still 11am or some other time.

 

[PeterDonis]

In case 2 changing only that the craft is moving throughout and starts at Clock A at noon, what does he actually see is the time on Clock B when he passes Clock A?

I've already answered that. Read post #9 again.

 

[Dale]

what does he actually see is the time on Clock B when he passes Clock A?

11 am

 

[TomTelford]

Thank you, I seem to have gotten my answer. Both cases show both clocks in the same configuration at the start. That's all I wanted to know.

 

[Ibix]

@TomTelford - when studying relativity, it's important to make a clear distinction between what you see and how you interpret that.

Whatever their state of motion, everyone passing clock A when it reads 12 will see clock B reading 11. It can't be any other way, because they are receiving the same light from it.

However, observers can differ in how they interpret this. They all need to correct for the travel time of light to calculate what clock B is showing now. And what Einstein showed was that if everyone uses the same procedure to do that correction, they come up with different answers. Someone sitting next to clock A, at rest, says clock B is correctly synchronised and one light hour away. Anyone moving with respect to clock A says that both clocks are ticking slowly, they weren't synchronised correctly, and the distance between them is less than a light hour.

It's true that a lot of sources use "see" rather sloppily, meaning either or both of "directly observe" and "interpret their observations to mean"n so be wary (or find a better source). Keep in mind that the Lorentz transforms relate different observers' interpretations, not their actual observations. If you want to know what they see directly with their eyes, you need to add the light travel time back in.

 

[PeroK]

Thank you, I seem to have gotten my answer. Both cases show both clocks in the same configuration at the start. That's all I wanted to know.

The travel time for a light signal from an event to an observer is irrelevant to the theory of SR. If you have clocks on Earth one kilometre apart, then you hear one chime three seconds before the other and see one about $3\mu s$ before the other. And, if you could neither see nor hear the distant clock, but got your friend to run from one to the other to tell you, say, when it was noon by that clock, then you might not receive that information for 5-6 minutes later.

It makes no difference, in other words, whether you look at a distant clock, hear it chime, or get someone to convey a message by foot, bicycle or carrier pigeon. This time delay is irrelevant to the time that the event took place in your reference frame.

Many people imagine that the light-signal-travel-time from one event to another is critical to time dilation and the theory of SR generally. Which it is not. It's important, therefore, to disentangle any observational time lag from the theory as soon as possible.

In my view, it's important immediately to start thinking about "reference frames", rather than "what one observers sees".

 

[TomTelford]

To all who have responded, thank you and I do understand the distinction.

It is in fact the ambiguity of language that I'm dealing with here. 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. Meanwhile I'm caught here visualizing a clock ahead and one behind and it seems contrary to what is being taught. Specifically the lecturer suggested that since both observers see time running slower for the other, and, if the both observers must agree on the time on the "stationary" clock once they are collocated at the end of the trip then somehow the "moving" observer was being magically thrust into seeing the future beforehand merely by being moving. That can't be true. I can do the math, that's not the problem. I am beginning to understand SR conceptually but 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".

Amongst other things I am a pilot. I deal with relativity every time I fly. I need to know how fast I'm going. Okay, relative to what? Well, the ground ideally but I can't do that directly. So I look at the airspeed indicator but it tells me only the indicated airspeed relative to the air. First I have to convert to calibrated airspeed to remove effects of the design of the system and that has to do with the speed of sound and hydrostatic pressure. Then I have to convert that to true airspeed based on atmospheric conditions. Then I have to convert that to ground speed based on what the air mass is doing relative to the ground. 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. I do want to know what the relationship is between doppler and SR (or any other effects) during high speed flight as they are all going to happen AND to be able to calculate both. But I needed to know a couple of "absolutes" first and the questions above dealt specifically with the values perceived under specific circumstances not the rates of change of time during the action. SR deals with effects on rates of change but I needed to set the stage BEFORE asking the question of what happens once movement begins in order to understand it better.

 

[PeroK]

I am beginning to understand SR conceptually but 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".

There is only SR, which is "flat" spacetime. If you have a source of gravity, then you have curved spacetime and that is a whole different ballgame.

In particular, you do not have "Doppler" and "SR". You have SR, a theory of spacetime, which describes and explains (among other things):

Time dilation, length contraction and the relativity of simultaneity (between inertial reference frames)
The Lorentz transformation (of events between frames)
Relativistic velocity addition
The (relativistic) Doppler effect
Energy-momentum and its transformation
Particle collisions, and the energy-mass relationship.
etc.

 

[TomTelford]

Apologies again. Then I meant distinguishing between the elements of SR as you have listed them.

 

[TomTelford]

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?

 

[Ibix]

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.

I recommend looking up Minkowski diagrams, also called spacetime diagrams. They're essentially displacement-vs-time graphs, but since we interpret relativity as being about spacetime we take these seriously as a "slice" through spacetime. Or, at least, the best Euclidean representation of such a slice.

Once you've drawn a few from different frames you can build an intuition for how they transform. You can always add light rays to them and work out when they arrive at some observer if you want to know what they actually see.

I wrote a simple Javascript tool for drawing Minkowski diagrams, with animated frame changes. It works better on a laptop/desktop than a touchscreen, I'm afraid, but you can at least look at the "canned" scenarios if you want. www.ibises.org.uk/Minkowski.html

 

[Ibix]

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?

I make it that clock B reads 12.30 "now", according to the ship as it passes clock A. How did you get your 12.07?

 

[PeroK]

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?

I suggest you rephrase this:

In the ship frame:

Clocks A and B are moving towards the ship at $0.5c$. They are 0.866 light hours apart in this frame. When Clock A reaches the ship, Clock A reads noon and clock B reads 11am (as "seen" from the ship).

(This is because clocks A and B are 1 light hour apart and synchronised in their rest frame.)

If we have another observer, X, at rest in the ship's frame, a distance of 0.866 light hours in front of the ship, then Clock A reaching the ship is simultaneous with clock B reaching X. X sees clock B read ... (calculation required).

 

[TomTelford]

Ibix,

Now I'm just trying to figure out which formula is used to solve which value being transformed. I was using t' of the distant event as being 1.14 times the 1 hour difference from the rest frame. At this point I was just trying to get a sense that the "now" of the distant clock would show a later time than the "now" of the rest frame.

I promise I will learn how to say these things using your phraseology.

 

[PeroK]

Ibix,

Now I'm just trying to figure out which formula is used to solve which value being transformed. I was using t' of the distant event as being 1.14 times the 1 hour difference from the rest frame. At this point I was just trying to get a sense that the "now" of the distant clock would show a later time than the "now" of the rest frame.

Hint: first you have to work out where Clock B was (in the ship frame) when it read 11am. Or, more importantly, when!

That's if you insist on using your "delayed-signal-kinematics" approach!

 

[TomTelford]

Oh, last thing...

Time dilation and distance contraction are independent of direction, right? Distances behind you are equally contracted as those in front (or to the side) ?

 

[PeroK]

Oh, last thing...

Time dilation and distance contraction are independent of direction, right? Distances behind you are equally contracted as those in front (or to the side) ?

Length contraction is the same "front" and "back" in the direction of motion. But there is no length contraction in the directions orthogonal to the direction of relative velocity.

 

[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.