Non-inertial reference frames question

In summary, the deceleration period causes Alpha Centauri to recede at faster than light speed to someone on the ship.
  • #1
Freixas
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Thread split from https://www.physicsforums.com/threads/can-an-observer-perceive-he-is-traveling-faster-than-light.955469/
Dale said:
GrantSB said:
What happens if he decelerates? Does it appear to recede?
There is no unique standard definition for the frame of reference for a non inertial spaceship. Please see the paper I posted earlier for an example of how to rigorously define such a concept.

Again, this is simply not B-level material

You cannot ignore it and answer the question. Depending on the arbitrary choices you make in the mathematical definition of the non-inertial frame you can get any answer you want.

Do you really need to absorb an advanced paper to answer the question? Let's view the problem this way:
  • The ship is traveling at a constant .999c relative to Earth just prior to reaching Earth
  • The ship puts on the brakes moments before reaching Earth, so that it is now at rest relative to Earth
We can calculate the distance to Alpha Centauri when traveling at a constant .999c and when at rest (both relative to Earth) without requiring advanced math, correct? We can make the deceleration period as small as we want (just as long as its not zero). Let's say it's 1 second of ship time. Then we could say that Alpha Centauri went from being 1 light month away (or whatever) to 4 light years away (or whatever) in 1 second of ship time. Odd things might happen during the 1 second deceleration, but it would appear (to someone on the ship) that Alpha Centauri receded at faster than light speed—or so it seems to me. I'm not saying that Alpha Centauri does anything of the sort—it is, of course, totally unaffected by whatever the ship does.

It is interesting that you say that arbitrary choices in the math can lead to arbitrary answers. For instance, if a machine on the ship during deceleration records Alpha Centauri's light, it doesn't seem like the question of whether the light is red-shifted or blue-shifted (and by how much) would be arbitrary nor should it depend on whatever choices you made in defining the non-inertial frame. If you have a real ship making a real breaking maneuver, it seems like the answer should be deterministic.

I'm sure I'm missing something, but it's not obvious (as usual). :confused:
 
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  • #2
Freixas said:
Odd things might happen during the 1 second deceleration, but it would appear (to someone on the ship) that Alpha Centauri receded at faster than light speed—or so it seems to me.

This is mixing distances in different frames (more precisely, different inertial frames), and mixing them with a time that is a proper time along a non-inertial worldline to boot. The result is not physically meaningful.

In fact, I would advocate an even more radical position than that: discard the concept of "distance" altogether. If an object is not co-located with you, there is no unique, well-defined answer to the question of how far away from you that object is. Calculations like the one you describe would simply be viewed as an illustration of this--you can get all kinds of weird, nonsensical answers to the question "how far away is it?" by switching frames in the middle of the calculation, so to speak.

Freixas said:
if a machine on the ship during deceleration records Alpha Centauri's light, it doesn't seem like the question of whether the light is red-shifted or blue-shifted (and by how much) would be arbitrary nor should it depend on whatever choices you made in defining the non-inertial frame

Correct. The observed redshift or blueshift of light is an invariant (since it's a direct observable). But the "distance" to Alpha Centauri is not.
 
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  • #3
m4r35n357 said:
In SR, all inertial frames can be considered on an equal footing. This doesn't not mean they are all identical or indistinguishable; in practice, we always select one frame to do the analysis, usually our own rest frame (this is a B-level thread so please don't shoot me mods!).

Labeling the frames makes explaining and understanding the problem simpler, and for example when there are just two frames in steady relative motion the problem is symmetrical, and we can look at it from a choice of two frames (train or station and so on). BUT, as soon as there are more than two frames in the problem, the symmetry is broken, so we tend to choose to pick one frame (an inertial one) and use that as a "master" (yes, like the Twin Paradox!). We do this to preserve our sanity.

Anyway, just wanted to check that you were aware of this; all inertial frames are on an equal footing, but that doesn't mean we can't single one out to make our lives easier.
I actually like Freixas’s suggestion on the grounds that it’s much easier to intuitively grasp the principle of relativity if you dispense with common examples like the Earth or a train as a reference frame and just use “observer A” or “observer B.” But then maybe that’s going too abstract and distracting from the fact that SR is a theory of physics, not merely a mathematical toy.
 
  • #4
PeterDonis said:
This is mixing distances in different frames (more precisely, different inertial frames), and mixing them with a time that is a proper time along a non-inertial worldline to boot. The result is not physically meaningful.

In fact, I would advocate an even more radical position than that: discard the concept of "distance" altogether. If an object is not co-located with you, there is no unique, well-defined answer to the question of how far away from you that object is. Calculations like the one you describe would simply be viewed as an illustration of this--you can get all kinds of weird, nonsensical answers to the question "how far away is it?" by switching frames in the middle of the calculation, so to speak.
I would rather say the result is not unique, but "physically meaningful" is somewhat subjective, and instead show by analogy, as I did, that there is really nothing weird or nonsensical about this FTL result. It is truly equivalent to apparent FTL motion of the moon if you turn your head quickly. Only a boost is a hyperbolic rotation instead of circular rotation.
 
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  • #5
PeterDonis said:
In fact, I would advocate an even more radical position than that: discard the concept of "distance" altogether. If an object is not co-located with you, there is no unique, well-defined answer to the question of how far away from you that object is.

You know, sometimes it sounds like you guys are pulling my leg. I mean, I don't think you are, but it sounds like it.

For example, if I asked the right expert about plotting me a course to Alpha Centauri (even if it said course involves a theoretical spaceship with a theoretical drive capable of making the trip in some reasonable amount of time), I would expect I would get some sort of answer. At least, I've never heard even a hint that I might get told "we can't plot a course because 'distance' is not well-defined." I mean, if you don't know how far away something is, how you can even begin to plot a course?

A flight to Alpha Centauri would certainly involve not just shifting inertial frames, but non-inertial frames. Plotting a course would be tricky, but I believe there are people who could do it with some reasonable degree of precision.

Another way to view this: Forget about a ship for a moment; I am on a planet. On this planet, it looks like Alpha Centauri (AC) is a light month away and approaching at .999c (or whatever). How do I know? I ask the astronomers on this planet. Astronomers always seem to like to tell you how far away things are and they seem to have ways of determining this. My planet is about to pass Earth, so I get on a spaceship and match speed with Earth in one second of my time. Assume I survive the acceleration change. I ask the astronomers on Earth how far away AC is and they tell me it's four light years away.

I won't argue that I might be comparing apples and oranges, but my personal experience is that AC appears to have receded from 1 light month to 4 light years in one second. Again, I have carefully used the words "appear" or "apparent". One second, the best minds on my planet tell me AC is 1 light month and literally the next second, the best minds on Earth tell me it is 4 light years.

So the cognitive dissonance here is the difference between what you're saying and what I'm used to hearing from astronomers regarding distances to things. In my example, each astronomer in question is in his own unchanging inertial frame. I change frames and witness and appear to witness a rapid apparent change in the distance to AC.

On the other hand, I agree with PAllen that the observation of apparent distance change to AC, while absolutely real, is not significant in the sense that AC did not experience any changes due to my deceleration and never exceeded the speed of light. It's just the effect of shifting frames.
 
  • #6
Freixas said:
On the other hand, I agree with PAllen that the observation of apparent distance change to AC, while absolutely real, is not significant in the sense that AC did not experience any changes due to my deceleration and never exceeded the speed of light. It's just the effect of shifting frames.
You misrepresent me here. I did not say or mean that this FTL coordinate velocity is absolutely real. It is neither more nor less real than the moon moving FTL when you turn your head quickly. However, neither FTL coordinate velocity is directly observed. In the ordinary rotation case, what is observed is a change in angle I see the moon in relation, say, to my nose. That is invariant. To arrive at FTL motion, I have to choose global coordinates in which my head is considered not to rotate and distances are defined in a particular way. Similarly, for the rocket, what is directly observed are changes in color and brightness of the star. To get to FTL motion, you have to adopt coordinates in which you are at rest, and distances are defined in a particular way (with other well motivated choices being available).

In sum, these coordinate dependent FTL motions, IMO, are in no way 'absolutely real', but neither are they useless, let alone nonsensical.
 
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  • #7
PAllen said:
"physically meaningful" is somewhat subjective

Agreed.

PAllen said:
there is really nothing weird or nonsensical about this FTL result. It is truly equivalent to apparent FTL motion of the moon if you turn your head quickly

Agreed. But many people feel a strong temptation to think of this apparent FTL motion as "real". Which is why I sometimes suggest a radical antidote against this particular temptation.

Freixas said:
You know, sometimes it sounds like you guys are pulling my leg. I mean, I don't think you are, but it sounds like it.

I'm not pulling your leg, exactly, but I am (at least in this particular discussion) suggesting a viewpoint that, as I've already said, is radical. But, as I noted above, sometimes radical is what you need as an antidote to a strong temptation to make a mistake.

Freixas said:
if you don't know how far away something is, how you can even begin to plot a course?

You can certainly use a concept of "distance" in plotting a course. But it will be an abstraction, not something you can directly measure. If that abstraction makes it easier for you to plot a course, sure, go ahead and use it. But don't forget that it's an abstraction, not a "real thing".

The problem, as I noted above, is that many people just can't think of distance as an abstraction. That's not surprising: our brains are hard-wired to make certain assumptions about the world, and one of them is that "distance" is a "real thing". That's why SR is hard to grasp: everybody has to do at least some amount of reprogramming of their brains to override these hard-wired assumptions. My radical suggestion to discard the concept of "distance" altogether is really just an attempt to enable the reprogramming in people who are stuck on that particular point.
 
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  • #8
Let me make some additional points on this. It is a misleading, in my prior post, when I say FTL. What is really the case about these coordinate speeds is that they are faster than the numerical value of the speed of light in an inertial frame. However, within each of the coordinates I described, no body moves faster than the speed of light described in those coordinates. Thus a star may recede faster than the standard numeric value of c in such coordinates, but light emitted away by the the star will recede by a much greater value. In short, the standard value of c has no meaning in such coordinates. If we more consistently define FTL as faster than the speed of light in some direction in these coordinates, then no body travels FTL ever.
 
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  • #9
Freixas said:
I have carefully used the words "appear" or "apparent".

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).

Another "radical" suggestion (this one from the Usenet FAQ on the twin paradox, which often gets linked to here), then, is to not just discard the concept of "distance", but to replace it with the concept of "redshift/blueshift of light". That is a direct observable, and tells you immediately about a change in your motion relative to the object (assuming that the object is continuously emitting or reflecting light of known frequency/wavelength at emission, and that you can detect that light and measure its frequency or wavelength at detection). If, for some purposes, you end up wanting a "distance" number, you can always calculate it; but now it's clear that it's just a calculated number, not what you directly observe.
 
  • #10
PeterDonis said:
You can certainly use a concept of "distance" in plotting a course. But it will be an abstraction, not something you can directly measure.

Oddly enough, I think I understand what you're getting at, but it seems like more of a philosophical approach. All direct observations are a look into the past and any present measurements are extrapolations from that historical information. In addition, the whole concept of distance depends on a frame of reference; without a frame, there is no way to assign a value to a distance.

PeterDonis said:
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...

Actually, I was willing to ask the resident astronomer in each reference frame for their Alpha Centauri distance and accept their number as representing reality, so my standards of "reality" are low. I did mention on another thread that I approach this more from an engineer's viewpoint than a mathematician's.

At an absolute level of "reality", yes, the object might have exploded or radically altered course and who would know? On the other hand, in the future you will eventually find out, so maybe you could look back to an earlier moment and decide that the astronomer's distance calculation at the time was correct. So maybe today I couldn't say that Alpha Centauri appeared to change from 1 light month to 4 light years away, but eventually, I could confirm that this was "really" the case (for me in my particular travels through spacetime and for my level of comfort with "reality").
 
  • #11
Freixas said:
Actually, I was willing to ask the resident astronomer in each reference frame for their Alpha Centauri distance and accept their number as representing reality, so my standards of "reality" are low. I did mention on another thread that I approach this more from an engineer's viewpoint than a mathematician's.
Fine, but there is no reason to prefer this notion of distance to a radar distance you measure yourself. For noninertial motion, the results will be very different. For inertial motion they will be identical. I like to say that distance is well defined for inertial observers in SR because all reasonable definitions agree. In contrast, for non inertial observers in SR, and for any observer in GR, each reasonable definition of distance is different. A former science advisor here used to link to a cosmology calculator that gave six (? Several) different notions of distance, all different, to distant galaxies. All the different distances were used by cosmologists for different purposes, and none could be said to be real.
 
  • #12
Freixas said:
Do you really need to absorb an advanced paper to answer the question?
Yes.

Freixas said:
We can calculate the distance to Alpha Centauri when traveling at a constant .999c and when at rest (both relative to Earth) without requiring advanced math, correct?
Not exactly. You can calculate the distance in the two inertial frames without requiring advanced math. But neither inertial frame is the ships non inertial frame. The question is specific to the (ill defined) non inertial frame, the ship’s point of view. So to answer it a definition of the frame is needed. It is not impossible to do, there just is no standard method.

Freixas said:
We can make the deceleration period as small as we want
The period of the deceleration is not the issue. I generally assume instantaneous deceleration unless told otherwise.

Freixas said:
Then we could say that Alpha Centauri went from being 1 light month away (or whatever) to 4 light years away (or whatever) in 1 second of ship time.
No, you cannot. You cannot say anything about distances in the ship frame at all without properly defining the ship’s frame. That is the whole point. The statements you made cannot be checked to determine if they are true or false without a proper definition of the ship’s frame

Freixas said:
For instance, if a machine on the ship during deceleration records Alpha Centauri's light, it doesn't seem like the question of whether the light is red-shifted or blue-shifted (and by how much) would be arbitrary nor should it depend on whatever choices you made in defining the non-inertial frame.
This is correct. The outcome of the frequency measurement is an invariant. The distance is not.

Freixas said:
Plotting a course would be tricky, but I believe there are people who could do it with some reasonable degree of precision.
Sure, I am one of them. Those people (including me) would know to use a definite reference frame. We might choose a particular inertial frame, or we might go ahead and properly define the ship’s frame (e.g. using the method I posted earlier)

Freixas said:
my personal experience is that AC appears to have receded from 1 light month to 4 light years in one second.
No, that is not your personal experience at all. That is the experience of two different groups of inertial observers, neither of which represents your non inertial frame.

Freixas said:
In addition, the whole concept of distance depends on a frame of reference; without a frame, there is no way to assign a value to a distance.
Precisely. And since there is no standard definition of the frame of reference for a non inertial observer, there is therefore no standard concept of distance for them.
 
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  • #13
Freixas said:
All direct observations are a look into the past and any present measurements are extrapolations from that historical information.

No, any present measurements are things you directly observe. Calculations of things like "how far away is Alpha Centauri" are extrapolations from measurements.

Freixas said:
the whole concept of distance depends on a frame of reference

The usual concept does, but there are ways to define "distance" that do not. For example, define "distance" as arc length along a particular spacelike curve. Or define "distance" as half the round-trip light travel time.
 
  • #14
Freixas said:
We can calculate the distance to Alpha Centauri when traveling at a constant .999c and when at rest (both relative to Earth) without requiring advanced math, correct? We can make the deceleration period as small as we want (just as long as its not zero). Let's say it's 1 second of ship time. Then we could say that Alpha Centauri went from being 1 light month away (or whatever) to 4 light years away (or whatever) in 1 second of ship time.
The problem is the relativity of simultaneity. What you mean by "now" in the phrase "where Alpha Centauri is now" is a radically different thing just before and just after the deceleration if you try to naively stitch together the relevant inertial frames. In fact, if you are heading towards Alpha Centauri when you slam on the brakes, "now" on Alpha Centauri after you stop is before "now" was before you stop - if you use that naive method. That kind of thing is problematic.

One solution, presented in the paper Dale linked, is to define distance to be half the round trip time of a radar pulse divided by c.
 
  • #15
Dale said:
No, that is not your personal experience at all. That is the experience of two different groups of inertial observers, neither of which represents your non inertial frame.

Ok, this seems like a strange statement. Where do the two observers come from? Is there something preventing me from being the observer who shifts from one reference frame to another?

Let me point out the confusion as it appears to non-physicist trying to understand your responses. If I ask what happens to distances as I shift from one reference to another and you say that it depends on understanding a complex mathematical formula that requires making certain assumptions where said assumptions affect the answer, then it sounds to me (the non-physicist) that you're saying that if I jumped on a ship and actually shifted from one reference frame to another, what I "observed" would depend on some mathematicians assumptions rather than being a fixed and predictable result.

Nor does it seem like it should be that complicated. The Lorentz transformations are used to convert distances (lengths) from one inertial frame to another. I start in one inertial frame and end up in another. I stipulate that my personal time spent changing frames is 1 second as measured by my clock. Just from the Lorentz transforms, I should know that a point 1 light month in the starting frame is now 4 light years in the ending frame. I would know this in advance, in fact, just by using the transform. The change occurs in 1 person second of time (instantaneously is even better). Case closed, it seems. But apparently not...
 
  • #16
Freixas said:
If I ask what happens to distances as I shift from one reference to another and you say that it depends on understanding a complex mathematical formula that requires making certain assumptions where said assumptions affect the answer, then it sounds to me (the non-physicist) that you're saying that if I jumped on a ship and actually shifted from one reference frame to another, what I "observed" would depend on some mathematicians assumptions rather than being a fixed and predictable result.

No, what we're saying is that you don't observe distances. What you actually observe are things like redshifts/blueshifts of light signals and round-trip light travel times. These are invariants, independent of any choice of reference frame. But they aren't distances. Distances are things you calculate from direct observations like these; and how you calculate them depends on your reference frame. And what reference frame you use is an arbitrary choice; nothing in the physics tells you what frame to use.

Freixas said:
The Lorentz transformations are used to convert distances (lengths) from one inertial frame to another. I start in one inertial frame and end up in another. I stipulate that my personal time spent changing frames is 1 second as measured by my clock. Just from the Lorentz transforms, I should know that a point 1 light month in the starting frame is now 4 light years in the ending frame.

Yes, but none of these things are things you directly observe. They are all things you calculate, first in one frame, then in the other. Or you could construct some non-inertial frame in which you were at rest the whole time (but, as already noted, there is not one unique way to do this), and calculate distances in that frame. Or you could calculate distances in some other frame, such as the frame in which the CMBR is isotropic, in which you are never at rest during any portion of the process.

Which of these calculations you care about is up to you; but all of them are calculations. None of them are things you directly observe, so none of them are things about which you can say that it should be a "fixed and predictable result".
 
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Freixas said:
... what I "observed" would depend on some mathematicians assumptions rather than being a fixed and predictable result...
This is again the issue of using "observer" as an synonym for "reference frame".
 
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  • #18
Freixas said:
Where do the two observers come from?
The astronomers you mentioned.

Freixas said:
Is there something preventing me from being the observer who shifts from one reference frame to another?
No, but if you do so then neither inertial frame represents your viewpoint. I.e. a non inertial frame is not the same as a momentarily comoving inertial frame.

Freixas said:
Let me point out the confusion as it appears to non-physicist trying to understand your responses. If I ask what happens to distances as I shift from one reference to another and you say that it depends on understanding a complex mathematical formula that requires making certain assumptions where said assumptions affect the answer, then it sounds to me (the non-physicist) that you're saying that if I jumped on a ship and actually shifted from one reference frame to another, what I "observed" would depend on some mathematicians assumptions rather than being a fixed and predictable result.
What you directly measure is invariant. The problem is that you as a non physicist have a mistaken notion that you directly measure distance. You do not, you calculate distance based on your direct measurements. Since you calculate it the result clearly depends on the calculation. In an inertial frame the calculation is easy and there is a well defined approach, in a non inertial frame it is neither easy nor standard.

Again, the problem is that you are incorrectly believing that distance is directly measured. It is calculated from measurements.

Freixas said:
Nor does it seem like it should be that complicated.
Did you read and understand the paper? You should also read and understand the first two chapters of Sean Carroll’s lecture notes. The lecture notes explain the mathematical framework, but the primary problem is clearly identified in the paper.
 
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  • #19
I think there's some fine hair-splitting going on and I understand it's part of your desire to be precise. But you are talking to a non-physicist for whom a Neil deGrasse Tyson simplification might be more appropriate.

If I walk up a hundred astrophysicists and ask them how far to Alpha Centauri, they are all going to say something around 4 light years, particularly if I am the stupid administrator who pays their salaries. They aren't likely to say that distance is an abstract concept or that it depends on how I decide to calculate it. I mean, if a highly respected physicist asked them the same question, they might go into technicalities, but not to a lay person and probably not to a high school student (B level).

If I asked someone like Neil deGrasse Tyson on a PBS show how long it would take me to get to Alpha Centauri (from my perspective) if I could speed up to .999c in one second, he'd probably bring up length contraction—even though Alpha Centauri would appear to be moving toward me at .999c, it would seem to me to take far less time than 4 years to get there and so I could reasonably assume (at my non-physicist level) that during my very short acceleration time, Alpha Centauri appeared to move a lot closer to me—light years closer. Tyson might clarify that Alpha Centauri didn't move at all, but that space contracted rapidly along my direction of motion due to the rapid acceleration. He might also talk about how long someone on Earth would think my trip would take and that it would be far longer than what I would experience.

Tyson is not going to tell the audience that they need to read an advanced paper with a lot of math. I mean, this is a B-level thread, after all. Perhaps we need an "L" level (for lay person).
 
  • #20
Freixas said:
If I asked someone like Neil deGrasse Tyson on a PBS show how long it would take me to get to Alpha Centauri (from my perspective) if I could speed up to .999c in one second, he'd probably bring up length contraction—even though Alpha Centauri would appear to be moving toward me at .999c, it would seem to me to take far less time than 4 years to get there and so I could reasonably assume (at my non-physicist level) that during my very short acceleration time, Alpha Centauri appeared to move a lot closer to me—light years closer. Tyson might clarify that Alpha Centauri didn't move at all, but that space contracted rapidly along my direction of motion due to the rapid acceleration. He might also talk about how long someone on Earth would think my trip would take and that it would be far longer than what I would experience.

An important aspect of learning SR (and, in fact, it is stressed in the original 1905 paper) is to look clearly at what we mean by measurements of time and distance. Given that a spaceship's acceleration doesn't physically affect the rest of the universe, it must be the subtleties of a distance measurement than brings Alpha Centauri closer to the Earth for the accelerating spacecraft . And, again, its the simultaneity issue that's at the heart of the matter.
 
  • #21
Freixas said:
If I walk up a hundred astrophysicists and ask them how far to Alpha Centauri, they are all going to say something around 4 light years, particularly if I am the stupid administrator who pays their salaries.
They've adopted the inertial reference frame in which Earth and Alpha Centauri are approximately at rest. That's fine. It's like when a cop asks you how fast you were going you adopt the reference frame of the ground without commenting on it - it's conventional to do so and doing otherwise is likely to cause more problems than it solves.

But what answer would a pilot give? Do you mean the airspeed or the ground speed? One is important for not falling out of the air; the other is important for how long it takes to get somewhere; and both are important for how much fuel the plane'll need. There isn't a single answer. It gets worse if we switch to rocket pilots.

So, yes, we can give simple answers. But you have to stick to questions that have simple answers. In relativity, that means using inertial frames only. "How far away is Alpha Centauri, as viewed from an accelerating rocket?" needs a clear definition of distance, because there isn't a unique one and different definitions (and methodologies for measuring it) can give different answers. So if we don't point out that distance is a derived quantity and just give an answer based on parallax (as the OP requested), then someone else gives an answer based on radar measurements and they turn out to be different, what is your 'stupid administrator' going to think? Are they going to realize independently that distance is a conventional quantity and any answer is possible, or are they going to think we physicists don't know what we're talking about?
Freixas said:
Tyson is not going to tell the audience that they need to read an advanced paper with a lot of math.
If they ask a question that can't be answered in straightforward terms, I hope he'd say "there isn't a unique answer to this question, it depends how you measure it". Maybe he'd give an answer (radar coordinates aren't too bad - I can do the maths for an instantaneous stop in my head if the Doppler factor is nice). But I hope he'd point out that the world is not as simple a place as our naive intuitions might like.
 
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  • #22
Freixas said:
... on a PBS show...
One common purpose of this forum happens to be clearing up misconceptions stemming form overly simplified pop-sci descriptions.
 
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  • #23
Freixas said:
I think there's some fine hair-splitting going on and I understand it's part of your desire to be precise. But you are talking to a non-physicist for whom a Neil deGrasse Tyson simplification might be more appropriate.

If I walk up a hundred astrophysicists and ask them how far to Alpha Centauri, they are all going to say something around 4 light years, particularly if I am the stupid administrator who pays their salaries. They aren't likely to say that distance is an abstract concept or that it depends on how I decide to calculate it. I mean, if a highly respected physicist asked them the same question, they might go into technicalities, but not to a lay person and probably not to a high school student (B level).

If I asked someone like Neil deGrasse Tyson on a PBS show how long it would take me to get to Alpha Centauri (from my perspective) if I could speed up to .999c in one second, he'd probably bring up length contraction—even though Alpha Centauri would appear to be moving toward me at .999c, it would seem to me to take far less time than 4 years to get there and so I could reasonably assume (at my non-physicist level) that during my very short acceleration time, Alpha Centauri appeared to move a lot closer to me—light years closer. Tyson might clarify that Alpha Centauri didn't move at all, but that space contracted rapidly along my direction of motion due to the rapid acceleration. He might also talk about how long someone on Earth would think my trip would take and that it would be far longer than what I would experience.

Tyson is not going to tell the audience that they need to read an advanced paper with a lot of math. I mean, this is a B-level thread, after all. Perhaps we need an "L" level (for lay person).

No, he won't, but that's only because he isn't discussing a situation that is far removed from the intuitive understanding of distance (he won't be discussing distance as measured in non-inertial frames to lay people, actually).

You measure distance by looking at the ends of a rod and comparing it to something. That means light signals from the ends of the rod come to your eye, and you calculate the distance based on those light signals. In the event that you are at rest with respect to what you are measuring and are co-local with it, the process is easy and matches the "lay-person" intuition. Otherwise things get more complicated, which is why an operational definition of distance (and simultaneity) is required. In the case of measuring the distance of something in a non-inertial reference frame, you've basically broken the machine and have to choose exactly what you mean by "distance" in order to get something meaningful.

This is what these guys are trying to tell you, and furthermore, they're trying to tell you that when you've brought in non-inertial reference frames, you've blasted the complexity into another level.

Think of it like this:

When you're right next to an object and it's at rest with respect to you, measuring the distance is as easy as looking at the ends of the object and comparing it to a rod.

When the object is moving very fast, but uniformly, with respect to you, now you've opened up a huge can of worms: you are timing light signals, essentially, and because of the finite and constant speed of light, you're getting a shortened distance compared to what the object would be if it were at rest with respect to you. So, what distance means has now changed. Before it was absolute in a sense, and now it is speed dependent.

Now if you add in non-uniform motion, you are complicating it a further order of magnitude, except now you have many different ways to define what you mean by "distance."

Some info:

https://books.google.com/books?id=L...nal+definition+of+distance+special+relativityhttp://www.francescopoli.net/Classe_V_file/CLIL - The special theory of relativity - part 1.pdf

https://www.taylormade.com.au/pdf/relativity.pdf

Now, with uniform speed, you can calculate distance with an easy equation: the Lorentz transformation equations. With non-inertial situations, you cannot. It is more complicated. You don't seem to have a problem with length contraction, which is in fact a calculation that involves some algebra that many people are intimidated by. Why wouldn't a non-inertial situation complicate things further?
 
  • #24
Freixas said:
If I walk up a hundred astrophysicists and ask them how far to Alpha Centauri, they are all going to say something around 4 light years

We're not talking about what a hundred astrophysicists would say. We're talking about what the OP asked about. The OP's question is not just a question about "how far away is Alpha Centauri". He specifically asked about the observer seeing himself appear to move faster than light. Pointing out how Neil deGrasse Tyson would explain distances in different inertial frames does not answer that question. The question is not just about distances.
 
  • #25
Freixas said:
I mean, this is a B-level thread, after all
I have split this I-level hijack into its own thread that is appropriately labeled.
 
  • #26
PeterDonis said:
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
Freixas said:
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
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
PAllen said:
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
PeterDonis said:
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
PAllen said:
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
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
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
PAllen said:
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
Ibix said:
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.
 
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1. What is a non-inertial reference frame?

A non-inertial reference frame is a coordinate system in which Newton's laws of motion do not hold true. This can occur when the frame is accelerating or rotating.

2. How is a non-inertial reference frame different from an inertial reference frame?

In an inertial reference frame, an object will remain at rest or in uniform motion unless acted upon by an external force. In a non-inertial reference frame, an object may appear to accelerate even if no external forces are acting on it due to the frame's own acceleration or rotation.

3. What are some examples of non-inertial reference frames?

Examples of non-inertial reference frames include a car turning around a corner, a person standing on a spinning merry-go-round, and a spaceship accelerating in outer space.

4. How do non-inertial reference frames affect the measurement of physical quantities?

In non-inertial reference frames, the measurement of physical quantities such as velocity and acceleration may appear to be different than in an inertial reference frame. This is because the frame's own acceleration or rotation can affect the measurement.

5. Why is it important to consider non-inertial reference frames in scientific experiments?

It is important to consider non-inertial reference frames in scientific experiments because they can affect the accuracy of measurements and the validity of conclusions. Understanding and accounting for these frames is crucial for accurate and reliable results in physics and other scientific fields.

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