How does an isolated observer know if they're accelerating?

[Dale]

More specifically how it follows from considering the entire worldline that the reference frame is non-inertial?

Certainly. For an inertial frame the metric is $ds^2=-c^2 dt^2+dx^2+dy^2+dz^2=-d\tau^2$. For a person’s reference frame their worldline can be written $r(t)=(c t, 0,0,0)$ so along the worldline the metric simplifies to $ds^2=-c^2 dt^2=-d\tau^2$.

The proper acceleration along a worldline $r$ is $a(t)=\frac{D^2}{d\tau^2}r(t)$. So for an inertial frame this reduces to $a(t)=0$ for all $t$.

Since for your observer $a(t)\ne 0$ for $t$ before colliding with the scale we can therefore unambiguously conclude from examining the observer’s entire worldline that the observer’s rest frame is not inertial.

Did you mean that if we consider the entirety of that person's path before and after hitting the scale, we'd conclude that that person's rest frame was non-inertial?

Yes

It's clear to me from what @PeterDonis said that local spacetime measurements will tell us that the frame is inertial

Only a partial frame covering the local region of spacetime around the observer’s worldline while $a(t)=0$

 

[Shirish]

I think you have some of the terminology backwards, but re-reading my last post I think I managed to write it slightly ambiguously.

Whatever you are doing (accelerating or not), you have a velocity. You can use that instantaneous velocity to define an inertial reference frame in which, just for a moment, you are at rest. This is the "Momentarily Comoving Inertial Frame" (MCIF). It's inertial by definition and is a perfectly normal inertial rest frame. It just gets a special name because you are instantaneously at rest in (or co-moving with) it. In another instant you may or may not be at rest in this frame.

If you look at your accelerometer and see it reads zero then you know that part of your personal rest frame is the same as an inertial frame - the part during the period when your accelerometer read zero. However, the whole thing is only inertial if your accelerometer always reads zero. Otherwise there have to be parts where it doesn't look like an inertial frame, so the whole thing isn't inertial.

You are correct that you can stitch together a string of parts of MCIFs to create a rest frame for you. However, you have to be very careful - the reason being the slippery "when" that I wrote in "when your accelerometer read zero" above. Different inertial reference frames have different notions of simultaneity, so you'll always find that your parts of spacetime that MCIF #1 calls "while you were at rest in MCIF #1" overlap with parts of spacetime that MCIF #2 calls "while you were at rest in MCIF #2", and there are parts of spacetime that no MCIF calls "while you were at rest in this MCIF". Care is needed when you stitch them together to make sure you don't end up with overlaps and missing areas, and you end up with something of a mess anyway. There are better ways to do it if you ever need to create non-inertial frames.

That makes it really clear! As for the free particle statement that I wrote, I guess I can restate it as:
a particle is free (in the context of an experiment) if its personal rest frame is inertial (not just if a part of it is inertial).
I could even more simply say that a particle is free if it has an accelerometer that reads zero throughout the experiment.

 

[Shirish]

You are correct that you can stitch together a string of parts of MCIFs to create a rest frame for you. However, you have to be very careful - the reason being the slippery "when" that I wrote in "when your accelerometer read zero" above. Different inertial reference frames have different notions of simultaneity, so you'll always find that your parts of spacetime that MCIF #1 calls "while you were at rest in MCIF #1" overlap with parts of spacetime that MCIF #2 calls "while you were at rest in MCIF #2", and there are parts of spacetime that no MCIF calls "while you were at rest in this MCIF". Care is needed when you stitch them together to make sure you don't end up with overlaps and missing areas, and you end up with something of a mess anyway. There are better ways to do it if you ever need to create non-inertial frames.

Hi again! I was attempting to understand the difficulties in stitching together MCIF's to specify a non-inertial frame, specifically on how we can encounter overlaps and missing areas. I'm not insisting that stitching MCIFs should be done - I'm sure the preprint you linked to gives a better description - just trying to get a better sense of the "overlap/missing area" problem.

Quick recap: In an experiment, a particle first decelerates for a while (non-zero accelerometer reading) and then enters an inertial state of motion (zero accelerometer reading). The particle's rest frame is definitely non-inertial. The previous discussion was on whether we can specify that particle's rest frame by stitching together its MCIFs. Both you and Dale pointed out that it's not at all recommended to do so.

Continuing, let's say I draw a spacetime diagram of the inertial frame that the particle ultimately ends up being at rest in (call that IRF $R_0$):

The curved line is the worldline of the particle in $R_0$. My guess is that the MCIF at any point $p$ of the particle's worldline can be specified as follows: the time axis of the MCIF should be parallel to the tangent to the worldline at $p$. Also, we can fix the origin of the MCIF at $p$.

So now the sequence of MCIFs constructed in this way at every point of the particle's worldline uniquely specify that worldline.

In order to construct the worldline given the sequence of MCIFs, can I not reverse the above procedure - i.e. the time axes of the MCIFs will give me a sequence of tangents and I can construct the worldline using those?

In this scenario, how will I run into the overlap / empty area problem?

 

[PAllen]

Hi again! I was attempting to understand the difficulties in stitching together MCIF's to specify a non-inertial frame, specifically on how we can encounter overlaps and missing areas. I'm not insisting that stitching MCIFs should be done - I'm sure the preprint you linked to gives a better description - just trying to get a better sense of the "overlap/missing area" problem.

Quick recap: In an experiment, a particle first decelerates for a while (non-zero accelerometer reading) and then enters an inertial state of motion (zero accelerometer reading). The particle's rest frame is definitely non-inertial. The previous discussion was on whether we can specify that particle's rest frame by stitching together its MCIFs. Both you and Dale pointed out that it's not at all recommended to do so.

Continuing, let's say I draw a spacetime diagram of the inertial frame that the particle ultimately ends up being at rest in (call that IRF $R_0$):

View attachment 264230​

The curved line is the worldline of the particle in $R_0$. My guess is that the MCIF at any point $p$ of the particle's worldline can be specified as follows: the time axis of the MCIF should be parallel to the tangent to the worldline at $p$. Also, we can fix the origin of the MCIF at $p$.

So now the sequence of MCIFs constructed in this way at every point of the particle's worldline uniquely specify that worldline.

In order to construct the worldline given the sequence of MCIFs, can I not reverse the above procedure - i.e. the time axes of the MCIFs will give me a sequence of tangents and I can construct the worldline using those?

In this scenario, how will I run into the overlap / empty area problem?

consider a zigzag world line, and you will see the problem in spades. That is, moving in the +x direction, then the -x, then +x, then -x.

 

[Shirish]

consider a zigzag world line, and you will see the problem in spades. That is, moving in the +x direction, then the -x, then +x, then -x.

Why can I not use the same procedure in that case? I guess I can still trace out the tangents and reconstruct the worldline?

 

[Ibix]

In this scenario, how will I run into the overlap / empty area problem?

Here are two diagrams of a slightly simpler scenario - a standard Twin Paradox. In each case the green line is the ship and the red lines represent a pair of planets. The ship travels most of the way from one planet to the other, then turns round and goes home. Both planets hold parties at the time of turnaround (according to their frame), marked by red crosses on their worldlines.

On the left hand diagram, I've marked in grey the slice of spacetime that is "during" the outbound leg according to the frame where the ship is at rest outbound. It's a slanted slice on this diagram because the Lorentz transforms tell you that a line of constant $t'$, say $t'=T'$, satisfies $T'=\gamma(t-vx/c^2)$, which can be rearranged as $t=T'/\gamma+vx/c^2$, clearly the equation of a line with gradient $v/c^2$. The right hand diagram is the same, except I've marked the slice of spacetime that is "during" the inbound leg.

Note that the party on the left hand planet is not during the outbound leg or the inbound leg, and the party on the right hand planet is both during the inbound leg and during the outbound leg. Note also there are regions of spacetime that are before the first leg but after the second leg, and vice versa. This is all clearly unhelpful. If you naively take those two slices of the spacetime diagram and glue them together edge to edge, you are missing bits and double counting bits. It's analogous to printing two overlapping sections of Google maps and then gluing them together edge to edge and wondering why this town appears to have two High Streets.

Moving towards your problem, imagine that instead of having two legs the journey has three - outbound, rest, and return. Then four legs, then five, six and so on - basically constructing a polygonal approximation to your smooth curve. The problem doesn't go away as you add more line segments, and it never does. Even if you take the limit of infinitesimally short legs, each leg must be slightly non-parallel compared to the next one and therefore the infinitesimally thin slices of spacetime "during" that leg don't fit together.

That's the problem.

 

[Shirish]

Moving towards your problem, imagine that instead of having two legs the journey has three - outbound, rest, and return. Then four legs, then five, six and so on - basically constructing a polygonal approximation to your smooth curve. The problem doesn't go away as you add more line segments, and it never does. Even if you take the limit of infinitesimally short legs, each leg must be slightly non-parallel compared to the next one and therefore the infinitesimally thin slices of spacetime "during" that leg don't fit together.

That's the problem.

Awesome explanation! So what I understood from this: suppose $M_1$ is the comoving inertial frame with the rocket in the first leg (considering only "comoving" for now since this is a simple example in which the MCIF is the same throughout the first leg) and $M_2$ is the comoving inertial frame in the second leg.

In $M_1$, the part of spacetime that corresponds to when the rocket was at rest in it, is a plane that is bounded by two horizontal lines (one for $t=0$ and the other for $t=T$). Due to the Lorentz transformation, that plane transforms into the "slanting up" plane in the planet's spacetime diagram.

Similarly for $M_2$ the spacetime portion during which the rocket was at rest in it transforms into the "slanting down" plane.

So if I didn't know the rocket's worldline and only had these slanting up and slanting down planes to work with (the information given by the comoving inertial frames), and had to reconstruct the rocket's worldline with this info, I'd have to account for the overlaps and missing areas as you've shown clearly.

For this very simple example, it seems easy enough to do, but for more complicated motions in which a particle frequently and erratically changes its motion, this method will be a complete pain in the neck. Hope I got most of it right!

 

[PAllen]

Why can I not use the same procedure in that case? I guess I can still trace out the tangents and reconstruct the worldline?

You will get multiple labeling. That is, there will be large regions assigned to be simultaneous to 2 or more points on the origin world line. That means they will have multiple coordinates assigned to them , which is not a valid coordinate system.

 

[Grasshopper]

My intuitive understanding of the original problem (please correct me if I'm wrong):

If I choose a coordinate system in which the scale and I are both accelerating upwards at 9.8 m/s$^2$, then there would be a zero on the scale. That's why if you're not standing on the scale there is a zero on the scale. When however, I step on the scale, first I lift one foot and put on it, then I raise the other foot, and place it on the scale. I see this as being similar to someone in an accelerating rocket (accelerating w.r.t. the "fixed stars") holding a ball suddenly releasing the ball. The ball will appear to accelerate "down" from the observer on the rocket, because the force holding it at rest with the rocket is temporarily removed when the ball is let go. When the ball hits the back of the rocket, it once again will be accelerating along with the rocket.

Likewise, when you step onto a scale, first you place one foot on it, and then another, and in so doing, temporarily you remove part of your body from the Earth (or all, if you jump on the scale), which is the the thing causing you to accelerate at 9.8m/s$^2$ upwards in the coordinate system chosen here. Then, when the accelerating Earth and scale meet your foot again when both are firmly placed on the scale, the spring inside the scale is depressed, and you and the scale are once again accelerating at the same rate.

However, if you remain still, the spring remains depressed, because it is the thing "pushing" you "up", and that is why a steady reading is made.

How close is that to accurate? What specifically is misunderstood?

*My biggest concern with that explanation is that it doesn't account for (I guess you would call it) the temporary dynamic force when you first jump on the scale. If I jump on a scale, for a moment the reading is heavier than my weight, then it stabilizes. But maybe that's just the system adjusting for the additional energy from inside my body which was used for me to jump.

 

[Ibix]

If I choose a coordinate system in which the scale and I are both accelerating upwards at 9.8 m/s$^2$, then there would be a zero on the scale.

The coordinate system has no effect on the reading on the scale. It's either zero or it isn't. Given the rest of your post, you seem to be imagining a circumstance where both you and the scale are undergoing proper acceleration. Thus the scale will not be showing zero - it will be showing whatever the weight of the plate you stand on is. Most household weighing scales are zeroed at this position, but this is because you are interested in the excess weight (i.e., the weight of whatever is on the plate) and not in using it as an accelerometer. If you want to use a bathroom scale as an accelerometer you need to hold it vertically and zero it like that (or drop it and zero it while it's in freefall - conceptually better, but not recommended for practical reasons), then place it on the floor and it will give you a positive reading.

When however, I step on the scale, first I lift one foot and put on it, then I raise the other foot, and place it on the scale. I see this as being similar to someone in an accelerating rocket (accelerating w.r.t. the "fixed stars") holding a ball suddenly releasing the ball.

No. When you lift one foot off the floor you are still supported by the other foot. This isn't like a ball being dropped which, neglecting air resistance, is in freefall.

Then, when the accelerating Earth and scale meet your foot again when both are firmly placed on the scale, the spring inside the scale is depressed, and you and the scale are once again accelerating at the same rate.

You are always accelerating at the same rate as the scale. If it were otherwise, you'd start moving relative to one another. A weighing scale will work perfectly well if you mount it flush with the floor and are pushed arbitrarily slowly onto it on roller skates.

If I jump on a scale, for a moment the reading is heavier than my weight, then it stabilizes.

The weighing scale simply measures the compression of a spring, which we interpret as a measure of weight via Hooke's Law. If you jump on to a scale the spring has to bring you to rest, which requires more compression than merely your weight.

 

[Ibix]

So if I didn't know the rocket's worldline and only had these slanting up and slanting down planes to work with (the information given by the comoving inertial frames), and had to reconstruct the rocket's worldline with this info, I'd have to account for the overlaps and missing areas as you've shown clearly.

You seem to have understood the point about the planes, yes. However, the but I've quoted isn't quite right. The problem with a "coordinate system" that assigns multiple labels to an event is that you can't do maths with it.

The example I gave of two streetmaps showing overlapping parts of a town is relevant - we've assigned two locations on the map to every point in the overlap region. Say you want to know how far apart two places are. In some cases you can just measure. In other cases you need to measure then subtract out the extra distance from the overlap. Try phrasing that mathematically! It can be done, but the book keeping is awkward.

Then imagine that you have infinitely many maps, all overlapping each other to different extents in different places and with some gapsn which is what the "stitched MCIF" approach gives you. You just can't work with that.

 

[Shirish]

You seem to have understood the point about the planes, yes. However, the but I've quoted isn't quite right. The problem with a "coordinate system" that assigns multiple labels to an event is that you can't do maths with it.

The example I gave of two streetmaps showing overlapping parts of a town is relevant - we've assigned two locations on the map to every point in the overlap region. Say you want to know how far apart two places are. In some cases you can just measure. In other cases you need to measure then subtract out the extra distance from the overlap. Try phrasing that mathematically! It can be done, but the book keeping is awkward.

Then imagine that you have infinitely many maps, all overlapping each other to different extents in different places and with some gapsn which is what the "stitched MCIF" approach gives you. You just can't work with that.

So what I meant by the part you'd bolded was that, if I'm a planet observer and draw the parts of spacetime in $M_1$ and $M_2$ (the outbound and inbound comoving frames) in my spacetime diagram, they'd look like this:

​

Green is the part of spacetime in outbound comoving frame corresponding to the ship being at rest in it, red is for the inbound comoving frame. To trace the rocket's potential worldline in the green zone, I need to draw a line segment that starts from the lower edge and ends at the upper edge and has a specific slope (since I know the speed of the outbound rocket / $M_1$ frame relative to me).

Similarly for the red zone, I have a whole bunch of line segments that trace the rocket's potential worldline in the red zone. Out of all those potential worldline candidates in the green and red zones, the only ones that stay within their zones, don't cross over and don't leave gaps/overlaps are like this:

This is doable since this is a simple scenario, and as you said, would quickly become untenable for even mildly complicated situations.

 

[Grasshopper]

The coordinate system has no effect on the reading on the scale. It's either zero or it isn't. Given the rest of your post, you seem to be imagining a circumstance where both you and the scale are undergoing proper acceleration. Thus the scale will not be showing zero - it will be showing whatever the weight of the plate you stand on is. Most household weighing scales are zeroed at this position, but this is because you are interested in the excess weight (i.e., the weight of whatever is on the plate) and not in using it as an accelerometer. If you want to use a bathroom scale as an accelerometer you need to hold it vertically and zero it like that (or drop it and zero it while it's in freefall - conceptually better, but not recommended for practical reasons), then place it on the floor and it will give you a positive reading.
No. When you lift one foot off the floor you are still supported by the other foot. This isn't like a ball being dropped which, neglecting air resistance, is in freefall.

You are always accelerating at the same rate as the scale. If it were otherwise, you'd start moving relative to one another. A weighing scale will work perfectly well if you mount it flush with the floor and are pushed arbitrarily slowly onto it on roller skates.

The weighing scale simply measures the compression of a spring, which we interpret as a measure of weight via Hooke's Law. If you jump on to a scale the spring has to bring you to rest, which requires more compression than merely your weight.

So is the scale simply measuring the force that keeps me from falling through the floor?

 

[Ibix]

So is the scale simply measuring the force that keeps me from falling through the floor?

Yes.

 

[Ibix]

This is doable since this is a simple scenario, and as you said, would quickly become untenable for even mildly complicated situations.

The reasoning in this post seems valid, although I'd say it's backwards since the observable thing is the rocket's worldline, rather than its inertial frames.