If properly worded *this* would have been a good question. From the comments attached to the question we see that the questioner is really asking for *two* diagrams, one showing the point of view of *each* of the twins rather than a single diagram showing the coordinate systems of both. And by the ambiguous condition of “constant acceleration” he means constant acceleration as perceived by the stationary observer rather than constant proper acceleration as felt by the traveler.

Of course the case of constant proper acceleration would be more realistic in the sense that it just requires the traveler to experience a constant g-force, whereas constant observed acceleration requires an increasing applied force (which would actually become unbounded as the speed got closer and closer to c). But for a limited time it is possible to keep adjusting the applied force so as to create a constant acceleration relative to the Earth’s frame and in that case the relevant part of the world line (in any inertial frame) is a simple parabolic segment (rather than the hyperbolic segment that would correspond to constant *proper* acceleration).

With the assumption of constant accelerations in the stay-at-home inertial frame, the spacetime diagram in terms of stay-at-home coordinates is just this:

Here we have a parabolic segment taking the traveler from the start event to where he reaches a cruising speed of , followed by a straight line segment or the bulk of the trip, then a parabolic segment for deceleration, a vertical segment for time spent at the destination, another parabolic segment for acceleration back towards home, straight line for the cruise, and the final parabolic deceleration phase.

In this diagram the coordinates are [math]t_H[/math] for the time on the stay-at-home clock and [math]x_H[/math] for the position in the stay-at-home coordinate system, and we will use the name [math]x^{I}_{HT}[/math] for the function which gives the traveler’s position in stay-at-home coordinates in terms of the time [math]t_H[/math] that the stay-at-home observer perceives as concurrent with the traveler’s arrival at that position. (The superscript I on the function name is to indicate that this is what he infers rather than what he actually sees). So the graph of [math]x_H=x^{I}_{HT}(t_H)[/math] shows what the stay-at-home thinks is the position of the traveler when his (stay-at-home) clock shows time [math]t_H[/math]. This is one interpretation of the homie’s “point of view” but it is not what he actually *sees*.

What the homie actually *sees* is delayed by the light travel time from the traveler (just as what we see of a distant star many light years away is not what it is actually happening there now but what happened that many light years ago).

So to get the graph of what the homie actually sees we must look at the point on the previous graph that is the source of a light signal reaching home at time [math]t_H[/math].

We can get a graph of what the homie actually sees by tracing down each light-line from the [math]t_H[/math] axis to where it meets the [math]x_H=x^{I}_{HT}(t_H)[/math] graph and plotting the [math]x_H[/math] value of that event as [math]x^{O}_{HT}(t_{H})[/math] (with the superscript [math]O[/math] identifying the position actually observed at time [math]t_{H}[/math] rather than that which was inferred to be simultaneous).

Now let’s look at things from the point of view of the traveler.

The vertical axis now corresponds to the traveler’s clock time [math]t_T[/math] and the horizontal lines either to distances that we want to associate with that time. If we want to plot what is actually seen by the traveler then for each [math]t_T[/math] we plot the position coordinate corresponding to the distance from which the signal is coming (as determined, eg, by parralax). and if we want to plot where the traveller infers that the homie actually is at the time [math]t_T[/math] we attribute the distance of the source seen at time [math]t_T[/math] to the earlier time [math]t_T-\frac{|x_T|}{c}[/math]

What the traveler actually *sees* at any event on his worldline is exactly the same as what is seen by an inertial traveler whose world line passes through that event with zero relative velocity (ie for which the worldline is tangent to that of the traveler at that event). Such a tangential traveler sees the values of [math]x^O_{TTE}[/math] and [math]t^O_{TTE}[/math] corresponding to a time [math]x^O_{TTE}/c[/math] earlier in his own frame – so that [math]x^O_{TTE}=[/math] and [math]t^O_{TTE}[/math]

events that are seen by him at the time his clock shows time [math]t_T[/math] with position along that line corresponding to the distance he measures (eg by parallax) to that event; or, in the case of the inferred view those that are inferred to be happening simultaneously with that [math]t_T[/math] click of the clock or to those distance [math]x_T[/math] which he measures (eg by parallax) to whatever event we are talking about.

When the traveler’s clock reads time [math]t_T[/math] he is at the event for which [math]t_H[/math] is such that [math]\int_{0}^{t_H}\frac{1}{1-v(t)^2}dt=t_T[/math] where [math]v(t)=at,v_f,v_f-a(t-t_f),0,-a(t-t_r),-v_f,-v_f+a(t_h-t)…[/math]

To find [math]x^{I}_{TH}(t_T)[/math] we have to use the [math]t_T=[/math]constant simultaneity space for the traveler and find its intersection with the [math]x_H=0[/math] worldline of the homie

The light signal that the traveler is receiving from homie at this event can be seen from the above diagram to come from [math]t_H=[/math] and we have [math]x_H=[/math]

And