I How things appear to an accelerated observer

[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).

 

[PeterDonis]

the same setup but from Alice's viewpoint:

Alice is not at rest in this diagram (her worldline is not a vertical line), so I don't see how this represents "Alice's viewpoint", for the reason I've already given: changing frames from event to event on Alice's worldline does not give "Alice's viewpoint", since it does not result in a single valid coordinate chart covering the region of spacetime of interest, in which Alice is at rest.

 

[PeterDonis]

red is Alice's worldline.

Isn't Alice supposed to have been at rest, co-located with Bob, until she starts accelerating at time $t = 0$ in the "Bob's viewpoint" diagram?

If so, that diagram only shows her worldline correctly after $t = 0$, not before it. Before it her worldline should be a vertical line co-located with Bob's.

 

[jartsa]

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.

Nah. Light ray goes from a point on the object to the eye of Alice. No change of angle of the light ray in this case. I mean light ray that reaches Alice's eye must have a certain direction.

 

[PAllen]

Nah. Light ray goes from a point on the object to the eye of Alice. No change of angle of the light ray in this case. I mean light ray that reaches Alice's eye must have a certain direction.

No, look up relativistic aberration. If a distant object suddenly starts moving rapidly toward you, then when light emitted right after the change reaches you, it will have a more horizontal angle, which looks like it got smaller, thus further away.

 

[PeterDonis]

Nah.

The relativistic beaming effect is frame independent; it doesn't matter whether you consider the source or the observer to be moving.

 

[jartsa]

No, look up relativistic aberration. If a distant object suddenly starts moving rapidly toward you, then when light emitted right after the change reaches you, it will have a more horizontal angle, which looks like it got smaller, thus further away.

Relativistic beaming effect is frame independent. But the change of the size of the image works as I explained earlier. Let me explain second time:

Let's say a distant object emits rays equally to all directions. If said distant object suddenly starts moving rapidly toward you, then when light emitted right after the change reaches you, more of those emitted rays reach you ("headlight effect"), which would have the effect of rays with more different angles reaching you, if there was not the effect of rays becoming more parallel.

The maximum angle between those rays emitted by the object that reach you depends on the size of the object and the distance. Not velocity of the object.

 

[PAllen]

The maximum angle between those rays emitted by the object that reach you depends on the size of the object and the distance. Not velocity of the object.

Distance is frame dependent. Consider light reaching you from right after the far away object has started moving rapidly towards you. Consider another observer, colocated with you, moving with the same direction and speed as the far away object, after it changed speed. That observer sees the 'standard' angular diameter of the object. By aberration (derivable directly from Lorentz transform), you must see a smaller angular diameter for the object than that colocated, comoving observer.

 

[PeterDonis]

Relativistic beaming effect is frame independent. But the change of the size of the image

Must also be frame independent, since it's part of relativistic beaming.

 

[lemelinm]

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.

View attachment 294211​

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.

View attachment 294212​

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.

Could you tell me what software yoiu are talking about?

 

[jartsa]

Distance is frame dependent. Consider light reaching you from right after the far away object has started moving rapidly towards you. Consider another observer, colocated with you, moving with the same direction and speed as the far away object, after it changed speed. That observer sees the 'standard' angular diameter of the object. By aberration (derivable directly from Lorentz transform), you must see a smaller angular diameter for the object than that colocated, comoving observer.

If the other observer is standing still next to me and starts co-moving with the faraway object when she sees that the far away object has started moving, then the observer-object system is put into motion in a non-Born-rigid way, so the observer sees a change in that system.

I guess your idea was that there will be no change, because effects of object starting to move and observer starting to move cancel out. But there is a change.

I had to change the scenario, because the original is just too confusing for me. I hope the basic idea did not change too much.

 

[PeterDonis]

I had to change the scenario, because the original is just too confusing for me.

It's not your scenario and it's not your thread, so if you cannot respond to the scenario as the OP of the thread formulated it, you should not try to change the scenario; you should just not post in the thread since you have nothing of interest to contribute.