How things appear to an accelerated observer

[Freixas]

I've been working on a Minkowsky spacetime diagram generator. The software is probably way overkill, but I'm retired and it keeps my brain active. I am no physicist, but I am a pretty good programmer.

Side note: if you have any interesting things to diagram on a 2D Minkowsky spacetime diagram, let me know. I need good test cases.

The software is complete enough that I can draw diagrams, but I'm still hunting for bugs. The software has the ability to define a problem with respect to one inertial frame and draw it with respect to another. It also allows for animations, and I can set up animations where every frame is drawn relative to a different inertial frame. This makes it possible to view things from the point of view of an accelerated observer in a variety of ways.

Here is a simple setup I tried. An observer accelerating at 1g travels 4 light years. At (0,0), the accelerating observer's velocity is 0. We'll call her Alice. There are two other observers, one at (0,0) (Bob)and one at (4,0) (Ted), both at rest with respect to each other and to Alice (at time 0). Here is a diagram that includes just Alice's worldline.

The solid colors represent where Alice sees Bob (black) and Ted (blue). The solid numbers represent the times she would see if Bob and Ted displayed huge clocks and Alice had a powerful telescope. The faded colors represent the calculated position (per Alice) of Bob and Ted, and the faded numbers are their calculated clock times. For example, Alice starts out seeing Ted 4 light years away and displaying a clock time of -4.

Here is an animation of the scenario, from Alice's point of view, from time 0 until she reaches Ted.

Since we are viewing this from Alice's point of view, her position doesn't change. I believe this animation is correct, but it surprised me. As Alice accelerates toward Ted, Alice sees Ted receding into the distance—at least, until his clock reaches 0.

Let me know if this animation looks wrong and in what way.

 

[Ibix]

I'm a bit confused about what's what. The Minkowski diagram looks fine, but I'm not sure what the gif is supposed to be showing me. Are you trying to represent what Alice actually sees? If so I'm not sure it's right for the eternally accelerating observer in your Minkowski diagram since she shouldn't see the blue guy at 4ly. Or are you trying to depict the distance to the blue guy in Alice's momentarily co-moving inertial frame? That looks more plausible, although I haven't checked the maths. What are the fainter dots in the animation?

 

[Freixas]

If so I'm not sure it's right for the eternally accelerating observer in your Minkowski diagram since she shouldn't see the blue guy at 4ly.

I don't understand your question, the part about seeing "the blue guy at 4ly." The blue guy is Ted and he is always visible to Alice. At time 0, she sees Ted as 4 LY away and with a clock reading of -4 years. At Alice's time 2.2 years, she reaches Ted, who now displays a clock time of 4.9 years.

Or are you trying to depict the distance to the blue guy in Alice's momentarily co-moving inertial frame?

Exactly. Every instant in the animation is from Alice's view.

What are the fainter dots in the animation?

The fainter dots represent where Alice calculates the positions of Bob and Ted. The fainter text represents the clock times she calculates fro them.

For example, at time 0, Alice sees Ted's clock reading -4 years, but calculates his clock (along her current line of simultaneity) to read 0. At that instant, of course, her inertial frame matches Bob's and Ted's.

 

[PeterDonis]

Since we are viewing this from Alice's point of view

How are you calculating "Alice's point of view". You can't just be reading it off a Minkowski diagram, because Alice is not an inertial observer, so any coordinate chart in which Alice is at rest cannot be a Minkowski coordinate chart. What chart are you using?

 

[Ibix]

I don't understand your question, the part about seeing "the blue guy at 4ly." The blue guy is Ted and he is always visible to Alice. At time 0, she sees Ted as 4 LY

Not if she's eternally accelerating and you are talking about what she actually sees through a telescope, since she sees the past and she wasn't 4ly from Ted in the past. I think you are saying "see" when you means "calculates based on observation and prediction of what Ted has been doing since the light now arriving left him". If so, I agree about 4ly.

Exactly. Every instant in the animation is from Alice's view.

There isn't a unique frame for Alice, so she doesn't really have a "point of view" without some further specification. I thought what you are doing is calculating the interval along the line Lorentz-orthogonal to Alice's worldline between Alice and Ted, but...

The fainter dots represent where Alice calculates the positions of Bob and Ted. The fainter text represents the clock times she calculates fro them.

For example, at time 0, Alice sees Ted's clock reading -4 years, but calculates his clock (along her current line of simultaneity) to read 0. At that instant, of course, her inertial frame matches Bob's and Ted's.

...then I am confused again. Can you sketch on the Minkowski diagram or state mathematically what calculations you are doing at some arbitrary time after t=0?

 

[Freixas]

How are you calculating "Alice's point of view". You can't just be reading it off a Minkowski diagram, because Alice is not an inertial observer, so any coordinate chart in which Alice is at rest cannot be a Minkowski coordinate chart. What chart are you using?

Yes, I am using Minkowsky spacetime diagrams. Any single frame of the animation is a Minkwosky spacetime diagram using Alice's instantaneous moving frame (IMF). The IMF changes for each frame.

Unfortunately, some of the animated diagrams that make it clearer are too big to post here. I may try to post a few stills when I get a chance.

But how I do it seems irrelevant. If you were Alice, would you see Ted recede as you accelerated towards him?

 

[PeterDonis]

Any single frame of the animation is a Minkwosky spacetime diagram using Alice's instantaneous moving frame (IMF). The IMF changes for each frame.

So how are you calculating where Bob and Ted are if you're changing frames all the time?

how I do it seems irrelevant.

It most certainly is not. Wrong calculations give wrong answers. Garbage in, garbage out.

 

[PeterDonis]

If you were Alice, would you see Ted recede as you accelerated towards him?

How would Alice measure the distance to Ted? You can't just "see" someone receding. You need to have some way of measuring their distance from you.

 

[Freixas]

Not if she's eternally accelerating and you are talking about what she actually sees through a telescope, since she sees the past and she wasn't 4ly from Ted in the past. I think you are saying "see" when you means "calculates based on observation and prediction of what Ted has been doing since the light now arriving left him". If so, I agree about 4ly.

I'll try to post some diagrams to make this clearer.

But I'd ask you the same question, I asked Peter: If you were Alice, would you see Ted recede as you accelerated towards him?

 

[PeterDonis]

If you were Alice, would you see Ted recede as you accelerated towards him?

How is Ted moving? Is he stationary in the spacetime diagram you posted in your OP?

 

[Freixas]

How would Alice measure the distance to Ted? You can't just "see" someone receding. You need to have some way of measuring their distance from you.

Let's assume a super-telescope with super-precise measurements. Let's say Ted is on a very large planet and Alice knows the exact diameter of the plant. Could Alice measure the diameter of the planet as seen through the telescope and determine Ted's distance from her using only the light received by her at that point in time?

 

[PeterDonis]

The solid numbers represent the times she would see if Bob and Ted displayed huge clocks and Alice had a powerful telescope. The faded colors represent the calculated position (per Alice) of Bob and Ted, and the faded numbers are their calculated clock times.

If Alice sees Ted's clock reading -4 years in a telescope when she is at (0, 0) in your diagram, then Ted's clock does not read -4 years when he is at the point where the blue dot is in your diagram, point (4, 0) (putting the $x$ coordinate first, then the $t$ coordinate). Light travels at a finite speed. So the light Alice is seeing from Ted at (0, 0), which shows Ted's clock reading -4, was emitted by Ted at (4, -4). At (4, 0), where the blue dot is, Ted's clock will read 0.

 

[PeterDonis]

Could Alice measure the diameter of the planet as seen through the telescope and determine Ted's distance from her using only the light received by her at that point in time?

No, because light travels at a finite speed, so she is not seeing the diameter of the planet at the time she receives the light, but the diameter of the planet at the time the light was emitted.

 

[Freixas]

How is Ted moving? Is he stationary in the spacetime diagram you posted in your OP?

Bob and Ted are moving inertially. As to whether he is stationary, relative to who? Bob and Ted view each other as having a relative velocity of 0 and view Alice as accelerating.

The initial diagram I posted is for an instant in time in which everyone has a relative velocity of 0. Alice is the only one accelerating.

 

[PeterDonis]

Bob and Ted are moving inertially.

That tells me nothing about their worldlines except that they are straight lines in a Minkowski diagram. I am asking which straight lines they are in the Minkowski diagram in your OP.

As to whether he is stationary, relative to who?

I said "stationary in the spacetime diagram in your OP". That would mean Ted's worldline would be a vertical line in that diagram. Is that the case?

 

[Freixas]

No, because light travels at a finite speed, so she is not seeing the diameter of the planet at the time she receives the light, but the diameter of the planet at the time the light was emitted.

Exactly. I am asking if she can measure the apparent distance, not the actual distance, the distance that Ted appears to be at the moment the light reaches Alice.

 

[Freixas]

That would mean Ted's worldline would be a vertical line in that diagram. Is that the case?

Both Bob's and Ted's worldline would be vertical at that instant in time. Since I am viewing things from Alice's viewpoint. their worldlines remain lines, but they don't remain vertical.

 

[Ibix]

But I'd ask you the same question, I asked Peter: If you were Alice, would you see Ted recede as you accelerated towards him?

I'm struggling to understand how you are defining "see" at the moment.

In terms of Alice's proper time $\tau$ I make it that the Alice's worldline (in Ted's rest frame) is$$\begin{eqnarray*}
x&=&\frac 1a\cosh(a\tau)-\frac 1a\\
t&=&\frac 1a\sinh(a\tau)
\end{eqnarray*}$$and that simultaneously (meaning along the radius of her hyperbola) Ted's coordinates are$$\begin{eqnarray*}
x&=&L\\
t&=&\frac La\tanh(a\tau)
\end{eqnarray*}$$and that the distance between them along that line is a mess. In the case that $a=1\mathrm{ly/y^2}\approx 1g$ and $L=4\mathrm{ly}$ it simplifies to $$\sqrt{1-\frac{10}{\cosh(\tau)}+\frac{25}{\cosh^2(\tau)}}$$which is monotonically decreasing, if that helps.

Edit: missed a few posts and have to do something else now - apologies if you've already explained your definition of "see".

 

[PeterDonis]

I am asking if she can measure the apparent distance, not the actual distance, the distance that Ted appears to be at the moment the light reaches Alice.

There is no such thing because light travels at a finite speed. So the light that is reaching Alice is telling her how Ted appeared in the past, not at the moment the light reaches her.

Both Bob's and Ted's worldline would be vertical at that instant in time.

You didn't read my question carefully. I asked how Bob's and Ted's worldlines appear in the diagram you gave in the OP. That diagram is in one particular Minkowski frame. If Bob's and Ted's worldlines are vertical at any instant in that diagram, they are vertical everywhere.

Since I am viewing things from Alice's viewpoint. their worldlines remain lines, but they don't remain vertical.

Your version of "Alice's viewpoint" is not a single frame but a mishmash of frames changing at every instant, so it is meaningless to talk about the shape of any worldline in your version of "Alice's viewpoint", since there aren't even any worldlines in it, only a series of snapshots, each of a different frame at a different instant.

If you think I am implying that your version of "Alice's viewpoint" is not a good way to analyze Alice's viewpoint, you are correct.

 

[Ibix]

OK, so I think I understand what you are doing now, @Freixas. Ted emits a pulse of light which arrives at Alice at some time when she has velocity $v$ in the frame shown in the Minkowski diagram in your OP. You transform to the inertial frame with velocity $v$ (so Alice is instantaneously at rest) and calculate the distance in this frame between Alice and Ted at the times in this frame of the emission event (dark blue dot) and reception event (pale blue dot). Right?

The problem is that using an instantaneously co-moving inertial frame like that gives you inconsistent results, essentially because Alice is rewriting her history: she is always stationary now and moving in the past, and what she said a moment ago about being stationary is forgotten. We have always been stationary with respect to Eastasia, you might say. So, for example, using your method the "distance now" when Ted's clock reads zero is 4ly, but the "distance when she sees his clock reads zero" is not 4ly - it's about as far to the right as you can go in your gif. So you don't have a consistent measurement of distance because of this rewriting of history.

What you need to do is have a consistent coordinate system - Rindler coordinates work for this case, or more generally radar coordinates (https://www.mrao.cam.ac.uk/~clifford/publications/abstracts/ced_twins.html) are very useful.

 

[Freixas]

Let me re-start by re-phrasing my basic question in a way that doesn't, for the moment, involve my diagrams.

• At time 0, Alice has velocity of 0 relative to Ted.
• At time 0, Ted is 4 LY from Alice.
• At time 0, she begins accelerating at 1g toward Ted.
• At time 0, she receives an image of Ted from light that left Ted four years prior. In this image, Ted is holding a BIG ruler.

As Alice accelerates toward Ted, does Ted's ruler appear to grow bigger or smaller?

@Ibix @PeterDonis <--I'm hoping this tags my response so it notifies the two of you.

 

[Freixas]

OK, so I think I understand what you are doing now, @Freixas. Ted emits a pulse of light which arrives at Alice at some time when she has velocity in the fame in your OP. You transform to the inertial frame with velocity (so Alice is instantaneously at rest) and calculate the distance in this frame between Alice and Ted at the times of the emission event (dark blue dot) and reception event (pale blue dot). Right?

You are right about the emission event.

The pale blue dot is just the distance to Ted calculated along Alice's instantaneous line of simultaneity, the points in space which she would assign (at that moment) the same time coordinate as appears on her clock.

The problem is that using an instantaneously co-moving inertial frame like that is that you get inconsistent results,

I'll defer to you on the general principle. But bear with me. Let's say that the pale blue dot falls under your "inconsistent results" umbrella. Let's focus on the solid blue dot—it is solid because it represents something directly observed, not calculated. I cannot picture how Alice's observations can be inconsistent.

Alice receives an image from Ted where Ted is holding a BIG ruler and a BIG clock. She reads Ted's clock. She calculates Ted's distance using basic trigonometry (she knows the actual size of the ruler). None of this can be ambiguous and, at this point, I believe both can be calculated.

 

[PeterDonis]

• At time 0, Alice has velocity of 0 relative to Ted.
• At time 0, Ted is 4 LY from Alice.

Both of these claims are frame-dependent. If they are true in one frame (I assume you intend them to be true in the frame that is shown in the diagram in your OP), they will not be true in any other frame.

• At time 0, she begins accelerating at 1g toward Ted.

This could be phrased in a frame-independent way, by specifying the time on Alice's clock when she starts accelerating. In general, things that can be phrased in a frame-independent way should be, since all of the actual physics must be frame-independent.

• At time 0, she receives an image of Ted from light that left Ted four years prior.

The "four years prior" is also frame-dependent; if it is true in the frame shown in the diagram in your OP (which I assume is what you intended), it will not be true in any other frame. (Also, "at time 0" here could mean the time by Alice's clock, which would be a frame-independent way of specifying the reception event itself.)

• In this image, Ted is holding a BIG ruler.

Oriented how? Along the direction of relative motion, or perpendicular to it?

 

[PeterDonis]

Let's focus on the solid blue dot—it is solid because it represents something directly observed, not calculated.

What direct observation does the solid blue dot represent? Remember that we have already said that the distance from Ted to Alice at a given instant of time cannot be directly observed, because light has a finite speed.

 

[PeterDonis]

Alice receives an image from Ted where Ted is holding a BIG ruler and a BIG clock. She reads Ted's clock. She calculates Ted's distance using basic trigonometry (she knows the actual size of the ruler).

She can calculate a "distance" from this data, but it will not have any physical meaning, because, as I have already pointed out several times now, light travels at a finite speed.

 

[Ibix]

Let's focus on the solid blue dot—it is solid because it represents something directly observed, not calculated.

The clock measurement seen is directly observed, yes. Taking into account your question about a ruler held by Ted, the angular subtense of the ruler (I assume it's being held perpendicular to the line between them) is also directly observed. Converting any of that into a distance requires making assumptions about frames, though, and that's where you need to be careful.

 

[Freixas]

Converting any of that into a distance requires making assumptions about frames, though, and that's where you need to be careful.

Let's say that every second Alice measures the size of Ted's ruler and makes the trigonometric calculation to get a number, which may or may not represent a distance. As she accelerates toward Ted, does the value she calculate get bigger or smaller?

Or, even simpler, will Ted appear to shrink in the telescope?

 

[PeterDonis]

will Ted appear to shrink in the telescope?

This one is easy: no.

 

[Freixas]

I said I'd post some other diagrams. It's pretty easy with my tool to switch the viewpoints and change various parameters.

I've re-drawn the diagram from the point of view of Bob. I've captured three frames, at Alice's time .5, 1, and 1.5 years. I've added some features that make it harder to read, so I removed Bob's data and just focused on Alice and Ted.

The magenta axes are Alice's instantaneous moving frame's (IMF) axes. The grid is Alice's IMF's grid. The red dot is her position. The red hyperbola is her worldline.

The blue line is Ted's worldline.

The yellow line is the worldline of a photon that travels from Ted to Alice's current position. The dark blue dot is the originating event on Ted's worldline that launched the photon. The matching label gives Ted's clock time (which no one seems to get upset about) and Ted's distance from Alice (which seems to bug people because it's measured using an IMF).

The cyan line is Alice's IMF line of simultaneity, all the the coordinates in space which have the same time value relative to the IMF. The faded blue dot is where the line of simultaneity crosses Ted's worldline. The associated label gives Ted's clock time (calculated) and Ted's distance from Alice (calculated and relative to Alice's IMF).

I should also note that the grid lines on the final shot have been auto-scaled by 10 or else they would have been too dense.

I also have these same views from Alice's point of view. I had to re-scale and I dropped Bob's axes as they weren't useful and were obscuring things.

 

[jartsa]

As Alice accelerates toward Ted, does Ted's ruler appear to grow bigger or smaller?

On Alice's retina the image of the ruler gets smaller as Alice moves faster towards said ruler. So Alice could very well have an experience somewhere in her visual brain area about of a receding ruler.

I know the image gets smaller because ... it's just a basic thing about light rays.

Oh yes, we need to postulate that the ruler is so far away that Alice only travels a very small fraction of said distance during a time during which she changes her velocity quite a lot. So that only the change of velocity has an effect on the image formed on her retina, not the change of her position.

 

[PeterDonis]

I know the image gets smaller because ... it's just a basic thing about light rays.

What "basic thing about light rays" do you have in mind here?

 

[jartsa]

What "basic thing about light rays" do you have in mind here?

Relativistic beaming. Which means light rays become more parallel. Which means that a lens focuses those rays on a smaller spot.

 

[PeterDonis]

Relativistic beaming.

Ah, ok. But I'm still not sure why you added this in your earlier post:

Oh yes, we need to postulate that the ruler is so far away that Alice only travels a very small fraction of said distance during a time during which she changes her velocity quite a lot. So that only the change of velocity has an effect on the image formed on her retina, not the change of her position.

First, if the velocity is relativistic, so that beaming is significant, the change of position must also be significant; you can't make the latter small without making the former small.

Second, even if you focus on the period of initial acceleration, so that possibly the velocity change could be large before the velocity itself becomes relativistic (even though this will only be true for a limited time), for that the criterion is not that the distance to the ruler is large (at the start). It is that the proper acceleration times the distance is large, so that the distance over which the proper acceleration changes the velocity by a large amount is much smaller than the total distance to the ruler.

Unfortunately this will not be true in the scenario as given, since Alice's acceleration is 1 g, which is equivalent to about 1 inverse light-year, and Ted's initial distance from Alice (in the original rest frame, the one shown in the diagram in the OP) is 4 light years, so the product of proper acceleration and distance is only 4. Or, to put this another way, it will take about a year by Alice's clock for her speed relative to Ted to become relativistic so that beaming is significant, and by that time she will already have covered a substantial fraction of the distance to Ted.

 

[Freixas]

Or, to put this another way, it will take about a year by Alice's clock for her speed relative to Ted to become relativistic so that beaming is significant, and by that time she will already have covered a substantial fraction of the distance to Ted.

Hmm...this is interesting. I don't know what you guys are talking about, but when Alice's clock reads 1 year, she has traveled 0.6 LY in 1.2 Y rest time and has a velocity of 0.8c.

I'm using the equations at https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html. He has a handy little table that says that at Alice's time 1 Y, the rest time is 1.9 Y, the rest distance traveled is only .56 LY and her velocity would be 0.77c, close enough because I'm rounding. This doesn't sound like a "a substantial fraction of the distance to Ted."

 

[Freixas]

This one is easy: no.

Easy-peasy, but a one-word answer is not an explanation.

I'm willing to accept that my method of determining the distance Alice would calculate for Ted is wrong. I still believe she could calculate a distance (based on trig) and that this calculated value could be predicted knowing only the basics that I've described.

When Alice is at Bob's position, she sees what Bob sees and both should calculate the same distance based on the same trigonometry and come up with a distance of 4 LY. When Alice reaches Ted, she should easily calculate a distance of 0 LY. What is the proper method to determine the results of Alice's calculation at an arbitrary point in between?