The Porthole View, Looking Forward |

On this page, we’re going to put that aside, and consider how things actually

But first, we need to discuss a few preliminaries. (As usual we will be assuming

If we look at something in a telescope, we can tell from the size of the image how far away it is. If we move toward it, we expect it to appear to get closer, and we expect the image in the telescope to get larger. Taking a picture of something is equivalent, but a bit simpler (there’s just one lens, whereas in a telescope there are two: objective and eyepiece). So, suppose we take a picture of something, and suppose we have:

We find that the size of the image on the film will be given by:

And, conversely, given a particular image and assuming we know the actual size of the object, the

The

Figure 1 -- A
Simple (Stationary) Pinhole Camera: |

The picture is not very realistic, of course – the only image of a star you could obtain with a

At non-relativistic velocities and terrestrial distances, motion doesn’t affect the image. If we’re moving toward the image at fixed velocity

and at the point where

Eq (9) is a little messy and could probably be simplified. None the less, it’s obvious that dI/dt is always

When the accelerations and velocities are relativistic, however, the situation is completely different, as we shall see!

Viewed from the “stationary” frame, the camera’s length, along its line of motion, is contracted by a factor of 1/γ. Its width is unaffected, and in particular, when the film is oriented perpendicular to the line of motion, the size of the image on the film as measured in the stationary frame is identical to the size of the image as measured in the frame comoving with the camera. We’re tempted to immediately draw a picture as in Figure 2:

Figure 2 --
Obvious but Incorrect Picture of Abberation: |

From this we would conclude

To see that this is wrong, consider a “ring” of light approaching the camera from the edges of the star (which is conveniently shown as a circle in our illustrations, which makes visualizing the “ring” of light a bit easier). The “ring” shrinks down to zero size as it passes through the pinhole, then expands again as it travels onward to the film plane. The amount it expands while traveling from the pinhole to the film depends on how far it travels. But while it’s traveling to the left, from the pinhole to the film,

Figure 3 --
Correct View of Abberation: |

We can now see that the

Equation (16) is what we might call the “effective focal length” of the moving camera. Conversely, the ratio of the apparent distance of the object in the moving frame, l

Note, in particular, that if the camera is moving to the right, the star will appear to be

Let the coordinates in the stationary frame be (t,x) and the coordinates in the moving frame be (τ,ξ).

Let’s set the origin at the point where the light arrives at the moving camera (and at a stationary observer immediately adjacent to the camera). Let

(Of course, if the light came from a location

The camera’s frame is moving to the right at

And, since

which is identical to eq (17), which leads us to conclude that our geometric derivation of the abberation formula for this case was correct.

The minus sign in eq (25) is, of course, because the image is approaching, and the distance to it is decreasing.

If v=0.866

Before we begin, let’s glance back at eq(17). For small velocities, γ is about 1 (to first order), and if we start at zero velocity, our distance to an object toward which we accelerate is initially unchanging (to first order). But eq(17) has a term which is

Let’s run the equations through to get a more precise answer, and then we’ll see how this apparently absurd result can actually be correct. To keep it simple, we’ll assume initially that an astronaut views the image from a spaceship whose nominal acceleration is fixed. That is, the acceleration of the ship, as viewed from a

To start with let’s assume v << 1 so we can replace eq (29) with an approximation. If x(t) is the distance the ship has traveled at time t, and v(t) is its velocity at time t, then we’ll have

Differentiating, we see

This clearly starts out

where t

In other words, the trip lasts 8,779 seconds, and for the first 1.26 seconds, the moon appears to

This implies that the effect is too small to be noticed, unless you’re using very sensitive instruments and you’re specifically looking for it; even then it would not be easy to see. Let’s plot the apparent distance and see if we can learn anything more about how this behaves. In Plot 1, we show the distance traveled during the trip. The red line shows how the ship appears to move as viewed from Earth, and the green dotted line shows the distance traveled as observed by the astronaut, by taking the

Plot 1: |

But now, let’s blow up the first few seconds of the trip, in plot 2. This is very different! During the first 1.25 seconds, we see that the ship appears to have moved

Plot 2: |

Finally, let’s look at how far away from its “correct” position the image of the Moon appears to be throughout the trip. In plot 3 we show the “offset” of the moon versus distance the spaceship has traveled. This reaches a maximum magnitude of about 26 miles, when the ship has traveled about 80,000 miles along its course. After that, the offset gradually declines until, at the final approach to the moon, apparent and real positions agree.

Plot 3: |

Our eq (29), above, which gives the apparent distance to the goal during acceleration, assumed constant acceleration viewed from the "stationary" frame. Proper acceleration wasn't assumed constant, so that equation won't apply for large velocities. However, after a month of accelerating at 1g we'll still be traveling at less than 0.1

Note first that for low-speed travel,

We plug eq (34) into eq (29) to obtain the apparent motion of the ship during the first 30 days, as determined by looking at Alpha Centaurus through a telescope. This is shown in plot 4.

Plot 4: |

Note that the green line (apparent distance moved) goes down a lot faster than the red line (actual distance moved) goes up. The star is obviously running away from us very rapidly indeed! Let's plot the apparent velocity, from eq (31), again for the first 30 days of the trip; see plot 5.

Plot 5: |

This is remarkable! The star appears to be receding from us at

There is, of course, no violation of special relativity here. First, we're not dealing with an inertial frame. Second, it's well known that visual abberation can frequently lead to

Now we have the solution to this paradox. The clock doesn't appear to zoom ahead, but it

- How big is the effect if I just get up and walk across the
room? Small, of course, but
*how*small?

- We'd like an exact solution with fixed
*proper*acceleration, and then we'd like to apply the solution to the accelerating twins page to obtain the porthole view in that case. (That would include a full solution to the entire trip to Alpha Centaurus, not just the first 30 days.)

- What happens if the moving astronaut uses a radar gun to measure
the distance to the goal? A back-of-the-envelope scribble
indicates the radar results agree with the visual distance we've been
looking at here (as they must, of course!) but it would be interesting
to work through the details and see how the two measurements fit
together.

- How does eq. (32b) relate to the distance measured by a radar gun? The fact that the critical time is equal to the total distance to the target when v=0 seems tantalizingly significant.

Page created, 10/6/05. Updated with addition of "Time Seen on a Distant Clock", 10/7/05