Undergrad How things appear to an accelerated observer

[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?

 

[PeterDonis]

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.

These values are correct in the original rest frame, the one diagrammed in your OP. But not in any other frame.

 

[PeterDonis]

This doesn't sound like a "a substantial fraction of the distance to Ted."

It's enough of one to make the condition @jartsa gave, that the effect of the velocity change should be much greater than the effect of the position change, not valid.

 

[PeterDonis]

I don't know what you guys are talking about

Wikipedia has a decent article on it:

https://en.wikipedia.org/wiki/Relativistic_beaming

It doesn't talk much about the effect on the apparent size of objects, though. The basic idea is this: suppose we have three observers, all co-located and all looking at the same distant object. Observer A is at rest relative to the distant object; observer B is moving towards the distant object at relativistic speed; and observer C is moving away from the distant object at relativistic speed. Then the following will be true:

Compared to observer A, observer B sees the object to be brighter, to be emitting light of higher frequency, and to have a smaller apparent size.

Compared to observer A, observer C sees the object to be dimmer, to be emitting light of lower frequency, and to have a larger apparent size.

The above is what @jartsa was calling the effect due to velocity.

However, note carefully that the above is what will be true at the instant when all three observers are co-located. As observer B moves towards the distant object, its apparent size will increase because it is getting closer; and as observer C moves away from the distant object, its apparent size will decrease because it is getting further away. This is what @jartsa was calling the effect due to position, and it is opposite from the velocity effect described above.

a one-word answer is not an explanation.

In your scenario, the effect due to velocity described above will be more than counteracted by the effect due to position. To put it another way, Alice is accelerating slowly enough that the gradual decrease in Ted's apparent size due to her increasing velocity is more than counteracted by the gradual increase in Ted's apparent size due to her decreasing distance from him.

For Ted's apparent size to decrease, Alice would have to accelerate extremely hard, at billions or trillions of g's, so that her velocity became relativistic very quickly, while her distance to Ted hardly changed at all. Then, during that initial period, where her velocity was going from zero to relativistic, Ted's appparent size to her would decrease because the velocity beaming effect would be changing very rapidly but the position effect would be negligible. But as soon as her velocity was relativistic, she would be moving quickly towards Ted, and Ted's apparent size would start to increase, because the velocity effect would be changing very slowly (once the velocity is relativistic the velocity beaming effect doesn't change much) while the position effect would be increasing very rapidly.

 

[PeterDonis]

When Alice is at Bob's position, she sees what Bob sees and both should calculate the same distance based on the same trigonometry

If Alice and Bob are both at rest relative to Ted (i.e., before Alice starts accelerating), then yes, they will both see Ted as having the same apparent size. Their calculation of his distance from that apparent size will only be valid because they have been at rest relative to Ted for the entire time that the light they are seeing was traveling from Ted.

As soon as Alice starts moving relative to Ted, however, she will no longer be at rest relative to Ted for the entire time the light she sees from Ted was traveling, so any calculation she makes of Ted's distance from his apparent size will no longer be valid.

 

[PeterDonis]

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?

You are missing the fundamental point that distance is frame dependent. (The only exception is a distance of zero, meaning the two objects are co-located; that is an invariant, true in all frames, but tells you nothing about distances to anything else.) The 4 LY that Alice and Bob calculate at the start (before Alice starts accelerating) are only valid in the frame in which Alice, Bob, and Ted are all at rest. But you could choose some other frame and make the distance anything you like (except zero since Alice and Ted are not co-located).

The same goes for the period while Alice is moving towards Ted; there is no such thing as "the" distance from Alice to Ted. You have to pick a frame. Then you can calculate the distance in that frame, but it won't be valid for any other frame.

 

[jartsa]

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?

Alice can calculate the distance that two light rays have traveled using the distance of the two points from where the rays left, and the angle between the rays.

Alice will calculate that the distance is large when she moves towards the light source. Then Alice can also easily calculate the time that the light has traveled. It's a long time if the distance is large.

We know that when Alice calculates the distance to the light source that is a extra short length-contracted distance.

According to Alice the light that Alice sees is old light that left a ruler on distant planet a long time ago. The planet is not a distant planet anymore according to Alice. Because it took a long time for the light to travel the long distance according to Alice, and during that time the planet moved at constant velocity towards Alice, according to Alice.

 

[PeterDonis]

Alice can calculate the distance that two light rays have traveled

Alice can also easily calculate the time that the light has traveled.

Both of these quantities are frame-dependent. Not only that, but by the method you describe, the "distance" and "time" calculated might not match those in any frame.

 

[PAllen]

Sorry I haven't read the whole thread, but I just wanted to discuss a bit about direct measurements by an observer on an arbitrary noninertial (potentially, wildly accelerating in different directions by different amounts, over their own time) world line, versus modeling by such an observer (I like the term modeling over computation, because computation must be done with respect to some model). In particular, how this relates to the use of momentarily comoving inertial frames (hereafter MCIF).

1) The arbitrarily moving observer and an MCIF will agree on any direct measurement that is effectively instant, e.g. camera picture, an observation of frequency of a light source (and thus Doppler), the angular size of something, etc.

2) They will not agree on any observation that takes any significant amount of time, e.g. radar ranging for distance determination, or even rate of change of any of the examples in (1).

3) Models of the direct measurements using the MCIF are typically absurd for the non-inertial observer. For example, consider observation of the change in angular size of some known body (size in its rest frame known), along with its color (also assuming it is known what color it is in its rest frame). Under the assumption that ones own motion is inertial and thus can be taken to be at rest, you can readily model the distance (at time of emission) and speed (at time of emission) of the body corresponding to your current observation, and thus also how long ago your current observation was emitted. However, none of these modeling assumptions make any sense for a non-inertial observer. For example, since your past motion is different from the past of the MCIF, a distance at emission valid for the past of of the MCIF cannot be sensible for a non-inertial observer, whose past positions are known to be very different from the MCIF. Thus, a relation of some observed angular size to e.g. 1 light year for the MCIF - of the body at time of emission - cannot possibly be sensible for the accelerated observer who would be in a completely different location a year ago. They would say, if the observed body was 1 ly away from the MCIF observer a year ago, and the MCIF observer was very far from me a year ago, that observed body could not possibly be 1 ly away from where I was a year ago.

 

[jartsa]

Both of these quantities are frame-dependent. Not only that, but by the method you describe, the "distance" and "time" calculated might not match those in any frame.

Well, I thought Alice's trigonometrical calculations would be correct in Alice's instantaneuosly inertial rest frame.

 

[PeterDonis]

I thought Alice's calculations would be correct in Alice's instantaneuosly inertial rest frame.

I suggest reading @PAllen's post #43.

 

[Freixas]

You are missing the fundamental point that distance is frame dependent.

I'm not going to agree that I am missing this fundamental point, but I'll agree that there are related problems.

The trigonometric calculation's problem is that, while the height of the ruler is not affected by velocity (as it is perpendicular to the direction of travel), the distance to the ruler is because it is in the direction of travel. The trigonometry assumes that opposite vs. adjacent are measured in the same frame, which is not going to be the case for most frames.

The simple proof is just to imagine that at time 0 Alice was co-located with Bob, saw exactly the same image of Ted as Bob, but was traveling at a constant 0.8c. Her distance to Ted would be 2.4 LY, not 4 LY, which is what Bob would say.

The calculation from my simulation has the problem that distances are in different frames when the light leaves Ted and when Alice receives it. I believe can use this method to determine what clock time Alice sees (as opposed to calculates) for Ted, but not for distances.

Thinking back to the formulas I have, if Alice can convert her time (tau) to rest time or distance (using the equations at https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html), then she it seems she could as easily calculate her distance to other points. But that's just the position I had as the calculated position (the dots with faded colors).

In the end, I wind up with just one position and two different times--the time seen and the time calculated along the IMF's line of simultaneity.

Luckily, none of this has any implications for my tool. It just does the math. If set things up incorrectly, it will apply the math incorrectly.

 

[PeterDonis]

The trigonometry assumes that opposite vs. adjacent are measured in the same frame

There is no such thing as "measured in a frame". Measurements are invariants; their results aren't frame-dependent. When physicists talk about something like a "distance measurement" being frame-dependent, they are really talking about different measurements (for example, measurements made using different rulers that are moving relative to one another).

What the calculation of distance from angle that you describe does assume is that Alice and Ted are at rest relative to each other during the entire time the light is traveling. I have already pointed this out. To put it more precisely, if Alice and Ted are at rest relative to each other (and moving inertially--I did not include this part previously, but it's also necessary), then their common state of rest picks out a particular inertial frame, the one in which they are both at rest, and either one can calculate the distance to the other using trigonometry in the way you describe. To be even more precise, the distance so calculated is the invariant length along a spacelike curve that goes from one to the other and is orthogonal to both of their worldlines; that is the frame-independent way of describing it.

But as soon as either one moves relative to the other, or as soon as either one accelerates and becomes non-inertial, the output of the trigonometry calculation no longer has any physical meaning as a distance. The reason is that there is no longer any spacelike curve that satisfies the invariant requirements I stated just now.

The simple proof is just to imagine that at time 0 Alice was co-located with Bob, saw exactly the same image of Ted as Bob, but was traveling at a constant 0.8c.

If she is traveling at 0.8c relative to Bob when they are co-located, she will not see exactly the same image of Ted as Bob, because of the relativistic beaming effect and the relativistic Doppler shift. I have described in post #38 how her observations would differ from Bob's (Bob would be "Observer A" in that post, and Alice would be either "Observer B" or "Observer C" depending on which direction she is moving).

Her distance to Ted would be 2.4 LY, not 4 LY, which is what Bob would say.

This distance is not something she could calculate using trigonometry as you describe. (Note also that it is not a "distance" that satisfies the invariant requirements I stated above: the spacelike curve that has that length is orthogonal to Alice's worldline at the point where Alice is co-located with Bob, but it is not orthogonal to Ted's worldline at the point where it crosses it--which is not the same as the point where the spacelike curve between Bob and Ted that has length 4 LY crosses Ted's worldline. Note that Bob's curve is orthogonal to both Bob's and Ted's worldlines, and so does meet the invariant requirements I gave above.)

I believe can use this method to determine what clock time Alice sees (as opposed to calculates) for Ted

I don't understand what "method" you would use for this. You certainly can't use trigonometry. The way a relativity physicist would normally do this would be to use a single frame, such as the one shown in the diagram in your OP, to assign coordinates to events, and then calculate the appropriate null worldlines for light rays from Ted to Alice. (In simple enough scenarios you can even get pretty good results by just drawing the null worldlines as 45 degree lines on the diagram.) The time on Ted's clock when a given light ray leaves Ted is the time Alice will see on Ted's clock when the same light ray reaches Alice.

Thinking back to the formulas I have, if Alice can convert her time (tau) to rest time or distance

These will be coordinate times and distances in the original rest frame, which in your case is the frame shown in the diagram in your OP. However, to use those formulas to obtain coordinate times from Alice's clock times, you would have to know Alice's clock times. As you have set up the scenario in your OP, you don't; you know coordinate times and distances in the original rest frame (since you know the equations of the worldlines of Alice, Bob, and Ted in that frame), but you have to calculate Alice's clock times. (Bob's and Ted's clock times are simple because they are always at rest in the given frame, so their clock times are the same as coordinate time in that frame.)

 

[PeterDonis]

none of this has any implications for my tool. It just does the math.

There's still the question of what math it is doing, though. So far I am not convinced it's doing the correct math for anything meaningful.

 

[PeterDonis]

it seems she could as easily calculate her distance to other points.

You say you are not missing the fundamental point I made, but this statement indicates that you still are. There is no such thing as "her distance to other points" without specifying a frame. Since you did not specify a frame in the statement quoted above, that statement is meaningless.

And specifying a different frame for each point on Alice's worldline does not count as specifying a frame. You have to pick one frame to cover the entire extent of Alice's worldline that you are interested in (i.e., where she is sending or receiving any light signals of interest or experiencing any other events of interest). If Alice is accelerating and you want her to always be at rest in your chosen frame, you have to pick a single non-inertial frame (the obvious one for constant proper acceleration is Rindler coordinates, as I think @Ibix pointed out in an earlier post--but since Alice only starts accelerating at a particular point, you would have to use Rindler coordinates only after that point).

 

[PeterDonis]

You have to pick one frame to cover the entire extent of Alice's worldline that you are interested in

And that same frame also has to cover the entire extend of any other worldlines that you are interested in, in this case Bob's and Ted's.

 

[Freixas]

I don't understand what "method" you would use for this.

The method of drawing a 45° line from Alice's current position through Ted's worldline. The intersection gives me the clock time that Alice sees. I was also using this to calculate how far Ted was from Alice. Using this method for the clock time still seems valid. Using it for distance does not.

These will be coordinate times and distances in the original rest frame, which in your case is the frame shown in the diagram in your OP. However, to use those formulas to obtain coordinate times from Alice's clock times, you would have to know Alice's clock times. As you have set up the scenario in your OP, you don't; you know coordinate times and distances in the original rest frame (since you know the equations of the worldlines of Alice, Bob, and Ted in that frame), but you have to calculate Alice's clock times. (Bob's and Ted's clock times are simple because they are always at rest in the given frame, so their clock times are the same as coordinate time in that frame.)

No, I use Alice's clock time to derive the others. For example, my animation is driven by tau, Alice's time, not t, the rest time.

The equations come from https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html. The author provides equations to calculate t, d, v, and gamma from tau.

You say you are not missing the fundamental point I made, but this statement indicates that you still are. There is no such thing as "her distance to other points" without specifying a frame. Since you did not specify a frame in the statement quoted above, that statement is meaningless.

I think you keep missing my point. I've stated repeatedly that, for Alice, everything is relative to her IMF, so I have, in fact, specified a frame.

Consider a problem in which Alice has a constant relative speed of 0.8c relative to Bob, who is at rest. I can diagram this situation and then flip it so Alice is at rest. The diagrams are equivalent; there is no "right" or "wrong" one. They are two views of the same thing.

Now let's give Alice a constant acceleration. We diagram this from Bob's point of view and I can show everything in a single diagram. But I can also create an equivalent view from Alice's point of view by using her various IMFs combined with animation. Any single frame (used in the sense of a movie frame, not an inertial frame) of the animation has a mapping back to Bob's rest view. Any single frame of the animation maps to any other frame of the animation. Rather than just being a diagram from Alice's point of view, these would be diagrams from Alice's point of view at a specific instance in time, at which she has a well-defined inertial frame.

 

[Ibix]

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

This one is easy: no.

Are you sure? I get an initial decrease in angle subtended by the ruler and then an increase. That does seem counter intuitive, so analysis is below in case you can see an error.

Working in Ted's frame, let Ted and the ruler lie at $x=L$ and extend from $y=-h$ to $y=+h$. At time $t$ Alice lies at position $x_a=\frac 1a\sqrt{1+a^2t^2}-\frac 1a$ with velocity $v=at/\gamma$, where the Lorentz gamma factor is $\gamma=\sqrt{1+a^2t^2}$. Sketch:

A light ray arriving at Alice from the top of the ruler travels parallel to the three vector $(x_a-L,-h,0)$. We can write a four momentum (up to a multiplicative constant we don't care about) for the light ray, then, as $p=(\sqrt{(x_a-L)^2+h^2},x_a-L,-h,0)^T$which is manifestly null. Then we can transform that into Alice's instantaneous rest frame:$$p'=\left(
\begin{array}{cccc}
\gamma&-v\gamma&0&0\\
-v\gamma&\gamma&0&0\\
0&0&1&0\\
0&0&0&1\end{array}\right).p$$Finally, we can take the ratio of the $y'$ and $x'$ components of $p'$ to get the tangent of $\theta'$, the half-angle subtended by the ruler for Alice expressed as a function of coordinate time in Ted's rest frame. That's rather messy:$$
\tan(\theta')=-{{ha}\over{{{\sqrt{a^2\,t^2+1}\,\left(-a\,L+\sqrt{a^2\,t^2+1}-1
\right)}}-at\,\sqrt{a^2\,L^2-\sqrt{a^2\,t^2+1}\,\left(2\,a\,L
+2\right)+2\,a\,L+a^2\,t^2+a^2\,h^2+2}}}$$
However, one thing is not messy. You can calculate the initial rate of change of the tangent of Alice's measured subtense, which is $$\left.\frac d{dt}\tan\left(\theta'(t)\right)\right|_{t=0}=-ha\frac{\sqrt{L^2+h^2}}{L^2}$$which is manifestly negative for any acceleration and size of object.

As always, I could have made a mistake, but I don't see where. A Maxima batch file to do all of this is spoilered below.

Spoiler: Maxima code

/* Alice's position, velocity, and gamma factor at Ted's frame's coordinate time t */
xalice:sqrt(1+a^2*t^2)/a-1/a;
v:diff(xalice,t);
gamma:ratsimp(1/sqrt(1-v^2));

/* Lorentz transform to Alice's instantaneous rest frame from Ted's */
LT:matrix([gamma,-v*gamma,0,0],[-v*gamma,gamma,0,0],[0,0,1,0],[0,0,0,1]);

/* Vector from tip of ruler to Alice... */
dx:xalice-L;
dy:-h;

/* ...hence four-vector of incoming light from the ruler tip */
assume(a>0);
u:ratsimp([sqrt(dx^2+dy^2),dx,dy,0]);

/* Four vector expressed in Alice's instantaneous rest frame coordinates */
uprime:LT.u;

/* Derive the implied angle */
tanthetaprime:uprime[3][1]/uprime[2][1];

/* Calculate the derivative at t=0 */
substitute(0,t,diff(tanthetaprime,t));

 

[Freixas]

As always, I could have made a mistake, but I don't see where.

Peter will, I'm sure, give you the definitive answer, but I can try to give you my layman's answer.

Imagine that as Alice travels towards Ted, she passes a string of observers, all of whom are motionless relative to Ted. As she passes each of them, she sees the same height h as they do. As the rest observers' distance from Ted decreases, h increases, so Alice should see an increasing h.

Peter says relativistic beaming means that she doesn't see the same thing as the rest observer. You can determine if that applies to h. The article he pointed me to only talks about it causing luminosity and frequency changes.

Since you are getting the same results I did, perhaps you are making the same mistake. I measured Alice's distance to a point in the past where Ted emitted a photon. I measured this distance using Alice's instantaneous frame at the time she received the photon, so I was creating a distance by mixing events occurring at different times and in different instantaneous frames.

 

[PeterDonis]

I get an initial decrease in angle subtended by the ruler and then an increase. That does seem counter intuitive

It's only counterintuitive to the extent that relativistic beaming is counterintuitive. Your analysis is one way of describing and calculating the effect of relativistic beaming.

I see now that my intuitive, heuristic analysis of relativistic beaming in an earlier post does need one correction: since at the very start of Alice's motion, the only effect present is the "velocity effect", there will always be some period of time at the very start where Ted's angular size will in fact decrease. What we need to estimate is how long that period will be for the parameters given in the OP. We are given that $a$ is 1 g, or about $10$ meters per second squared, and $L$ is 4 light years, but we are not told what $h$ (half the length of Ted's ruler) is. If we assume it's about 1 meter, i.e., $h = 1$, then the initial rate of change from your formula is approximately (using the fact that $h << L$ to simplify the formula):

$$
\left.\frac d{dt}\tan\left(\theta'(t)\right)\right|_{t=0} \approx - \frac{h a}{L} = - 2.5 \times 10^{-16}
$$

which is pretty small. I expect that the second derivative will be positive so I would expect that the time until $\theta'$ starts increasing is pretty short for these parameters.

 

[PeterDonis]

Peter says relativistic beaming means that she doesn't see the same thing as the rest observer. You can determine if that applies to h.

It doesn't change the value of $h$ at the source, but it does change the apparent size of the object. That should be clear from my earlier post where I initially described the effects of relativistic beaming. However, as I just posted in #54, my initial heuristic analysis of relativistic beaming left out one aspect.

 

[Freixas]

It doesn't change the value of at the source, but it does change the apparent size of the object.

Got it.

Apparent size wasn't what I was going for, so it's an interesting side issue, but irrelevant to my simulation. Some time in the past, I do remember using some physics simulator or viewing some video where I was told that rapid acceleration would make an object appear to recede, so when I saw the effect in my simulation, I thought it looked odd, but within the realm of crazy physics things. The simulator or video or whatever used a low value for c to make the effect more apparent.

 

[jartsa]

Got it.

Apparent size wasn't what I was going for, so it's an interesting side issue, but irrelevant to my simulation. Some time in the past, I do remember using some physics simulator or viewing some video where I was told that rapid acceleration would make an object appear to recede, so when I saw the effect in my simulation, I thought it looked odd, but within the realm of crazy physics things. The simulator or video or whatever used a low value for c to make the effect more apparent.

Well, there is no apparent size change of an object according to Alice when said object abruptly accelerates towards Alice.

And that is why Alice may say that the extra small object that she sees moving towards her was far away when the light she sees was emitted.

Alice says it's a distance effect, not the beaming effect that makes the image of the object small.

 

[Freixas]

Thanks for all the assistance from everyone: @Ibix, @PeterDonis, @jartsa, and @PAllen.

Here is a corrected simulation. It is not as interesting. It represents Alice's viewpoint as she accelerates toward Ted. The faded positions are gone, the solid times still represent the Bob's and Ted's times seen by Alice, the faded colors represent Bob's and Ted's times as calculated by Alice.

With a few changes to the diagram, I can show a more typical view of the same data. Because of size constraints, I won't post an animation, but I will just capture the scene where Alice's tau = 0.2. The black axes are Bob's axes, green is Bob's worldline, blue is Ted's worldline, and red is Alice's worldline.

The cyan line is Alice's line of simultaneity. The yellow lines represent light and where they intersect Bob's and Ted's worldlines is where the light that Alice sees at tau 0.2 originates (and the source of the "Time seen" value).

Here it is from Bob's viewpoint:

And the same setup but from Alice's viewpoint:

Every intersection in one diagram has an equivalent match in the other. Although the animation might seem completely different from the two stills, it is about 98% the same thing.

Probably no one is curious, but I'll show the basics of the script used to create the diagram—the key elements, anyway.

There are three observers and they are defined this way:

let bob = [observer ]; let ted = [observer origin (4, 0)]; let alice = [observer acceleration 1];

Here's how the animation is set up:

let travelTau = dToTau(4, alice); animation reps: 1; let tau = 0 to travelTau step travelTau/500; let curAliceFrame = [frame observer alice at tau tau];

dToTau() is a function used to convert a distance traveled (4 LYs in this case) to a tau (local time) value for a given observer (Alice). This drives the simulation, which has 500 steps. I then grab Alice's instantaneous moving frame (IMF) at the point her local time equals the variable "tau", the one driving the animation.

let simulLine = [line axis x curAliceFrame offset tau]; let bobPos = intersect(simulLine, bob); let tedPos = intersect(simulLine, ted); let alicePos = intersect(simulLine, alice);

This line identifies the location of everyone along Alice's line of simultaneity. To get Bob's and Ted's clock readings as they arrive by light, I use

let light1 = [line angle 45 through alicePos]; let light2 = [line angle -45 through alicePos]; let bobLightLoc = intersect(light1, bob); let tedLightLoc = intersect(light2, ted);

Then I draw the positions as labeled events, where the label is the time seen. I add additional labels for the time calculated and I include an event dot for Alice:

event location: bobPos, text: "Time seen " + bobLightLoc.t, bobStyle; label location: bobPos, text: "Time calc " + bobPos.t, bobStyleFaded; event location: tedPos, text: "Time seen " + tedLightLoc.t, tedStyle; label location: tedPos, text: "Time calc " + tedPos.t, tedStyleFaded; event location: alicePos, text: tau, aliceStyle;

To show the animation so that it appears as a one-dimensional world (with the time dimension coming from
the actual time of the animation), I use this:

frame frame: [frame velocity curAliceFrame.v origin (0, tau) < curAliceFrame];

Basically, I modify Alice's IMF (only for drawing purposes) so that Alice stays in one spot. I'm also viewing everything from her point of view. The only things displayed are things occurring on her line of simultaneity.

 

[PeterDonis]

there is no apparent size change of an object according to Alice when said object abruptly accelerates towards Alice.

Not immediately, no, because of the finite speed of light. But once the light from the time the distant object abruptly accelerated reaches Alice, she will see an apparent size change.

 

[PeterDonis]

It represents Alice's viewpoint as she accelerates toward Ted.

What do you mean by "Alice's viewpoint"? What frame does it represent?

Also, what do "time seen" and "time calc" mean? I would guess, from those terms, that "time seen" means "the time Alice sees on Bob's/Ted's clock in the light signal she receives from Bob/Ted at the given event on Alice's worldline", and "time calc" means "the time Alice calculates that Bob's/Ted's clock reads at the event on Bob's/Ted's worldline that is simultaneous, in Alice's frame, with the given event on Alice's worldline". But I would like you to confirm that (or explain what the terms mean if what I've just said is not what they mean).