The Porthole View, Looking Forward

Suppose an astronaut blasts off from Earth and travels to Alpha Centaurus. The question most often considered is how time appears to pass on the ship, as viewed from Earth (or Alpha Centaurus), and how time appears to pass on Earth and Alpha Centaurus as viewed from the ship.

On this page, we’re going to put that aside, and consider how things actually look to the traveler. In particular, how far away does the end point appear to be, looking through a telescope, at each point on the trip?  In comparison with our common Earthly experience, the answer to this question is strange!

But first, we need to discuss a few preliminaries. (As usual we will be assuming c = 1 throughout this page.)

Expectations

Living on Earth, traveling only at (relatively) slow speeds and dealing only with nearby objects, we have formed a very simple expectation: If we move toward something, we expect it to appear to get closer to us. In this section I’ll go over the equations which show how an image grows during acceleration as we view it through an Earthly camera. Feel free to skip this section – the relativistic parts start in the next section.

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 distance to the object is given by:



The kind of camera doesn’t affect this. A TV camera, a telescope used to make a prime-focus photograph, a pinhole camera, and a human eye all behave the same way. The simplest of these is a pinhole camera: it uses no lens, but just passes the light through a tiny hole and projects it on a flat (or curved) focal plane. The smaller the pinhole, the sharper the image, until diffraction effects take over. A pinhole camera is shown in figure 1.

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 real star at real interstellar distances using a real pinhole camera would be a dot the size of the pinhole you used!  But we’ll pretend that the star we’re looking at is amazingly huge, so we can obtain a reasonable image even when it’s several light-years away.

At non-relativistic velocities and terrestrial distances, motion doesn’t affect the image. If we’re moving toward the image at fixed velocity v, and our distance from the object at time t₀ = 0 is l₀, then the distance and image size will be given by



and at the point where f = l the image size will be equal to the object size, as is well-known to anyone who’s ever done any macro photography. If we start out stationary, and then accelerate toward something while filming it, then we’ll have:



Eq (9) is a little messy and could probably be simplified. None the less, it’s obvious that dI/dt is always positive – the image size just grows as we accelerate toward the object. This is, of course, exactly what we expect: If we accelerate toward something, it gets closer, and if it gets closer, it looks closer, too.

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

Images at Relativistic Velocities

Let’s assume our pinhole camera is moving at (relativistic) velocity v to the right. We define



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 (wrongly!), from simple trigonometry, that the image size in the moving camera must be contracted by a factor of 1/γ.
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, the film is traveling to the right. The result is that the distance traveled by the rays of light between the pinhole and the film plane is less than the “nominal” length of the camera. (Note that this is a “classical” effect – the image would be distorted by this even if the camera’s length were not contracted.) With this in mind, we can draw the correct picture, in Figure 3.

Figure 3 -- Correct View of Abberation:

We can now see that the sum of the distance traveled by the light as it moves from the pinhole to the film plane, and the distance traveled by the film plane in that same time, must be the length of the camera body as measured in the “stationary” frame. So, we have:



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_m, to the stationary frame, l_s, is inversely proportional to the ratio of the image size in the moving camera to the image size in a stationary camera, and so we have:



Note, in particular, that if the camera is moving to the right, the star will appear to be farther away than it is when the camera is stationary. If the camera is moving at 0.866c, then the apparent distance to the star will be increased by a factor of about 3.73. An astronaut starting at Earth and traveling toward Alpha Centaurus at 0.866c who uses the size of the image in a telescope to determine the distance to the star will find that it’s about 15 lightyears away, not 4 lightyears! But we also know that the astronaut will make it to Alpha Centaurus in just 2.3 years of “ship time”. So, during that time, the image in the telescope must appear to be approaching at about 6.5c. Can that possibly be correct? In the next two sections we’ll redo this calculation by transforming the event coordinates into the ship’s frame of reference, and we’ll see if the results agree.

Finding the Image Distance by the Lorentz Transform

The light which forms the image was emitted at a particular time at a particular location in space, and we can find the coordinates of that event easily enough in the stationary frame. If we transform the coordinates to the frame of the ship, we’ll know how far away the star was in the frame of the ship when the light was emitted. For a distant star or planet, we can safely pretend that the object is actually a flat disk; clearly, neither the shape nor size of its image would be affected by its motion if it were moving directly toward or away from us. So, the image size will be the same as if it were an image of a star which was stationary in the ship’s frame of reference at the location where the light was emitted. So, all we need are the coordinates; let’s find them.
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 A be the event when the light arrives, and E be the event when it is emitted. Then in the stationary frame, we have (recalling that c=1),



(Of course, if the light came from a location l_s away, it must have been emitted l_s time units in the past.)
The camera’s frame is moving to the right at v. Then we have



And, since E_m = (-l_m, l_m), we find that



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

Image Velocity Again

As time goes by the camera will approach the star, and the image in the camera will necessarily grow. From the rate at which the image grows, an observer can tell how fast the object appears to be approaching. We know how l_s changes as the camera moves toward its destination, and from that we can determine the apparent image velocity. Let l₀ be the initial value of l_s. Then, once again using eq (17), we have:



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

If v=0.866c, then γ=2 and the apparent velocity will be 6.46c which agrees with the calculation we made using our pinhole camera, above.

Acceleration at Non-Relativistic Speeds

Now we come to the surprising part!
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 linear in the velocity: it increases with the first order of the velocity. We can, therefore, expect that as we accelerate from a standing start, things in front of us will appear to recede: Objects toward which we accelerate will appear to get farther away. This is completely at odds with our everyday experience!
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 stationary frame, is constant. This means the proper acceleration of the ship will increase as time goes by, but until the ship’s velocity becomes large we can ignore that effect. So, assuming the ship starts with velocity 0 and accelerates at rate a directly toward a star whose initial distance is l₀, and using eq (23), we have:



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 positive – the apparent distance to the target is increasing.  But long, long before the ship arrives at the goal, the image will start to approach; at that point, we’ll still have x << l₀ , so at the point where the derivative crosses zero and the image starts to approach, we must have



where t_c is the time of “crossover”, when the image stops receding and starts to approach.

An Example: From Earth to the Moon at 1 Gravity

Suppose we’re in a rocket ship situated on or near Earth, looking at the Moon through a telescope. We start the engine, and accelerate toward the Moon at 1g. For simplicity, ignore Earth’s gravity and ignore the Moon’s gravity. Furthermore, let’s assume we keep right on accelerating until we zip past the Moon, just grazing its edge as we go by. (Or you could assume we charge headlong into it – perhaps we’re using a robot probe in that case. Either way, we don’t slow down.) What do we see? What’s the apparent distance of the image of the Moon as seen in the telescope? Let’s cast it all in light-seconds. Then we have:



In other words, the trip lasts 8,779 seconds, and for the first 1.26 seconds, the moon appears to recede from us. Only after that does it start to get closer, as seen in the telescope.
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 apparent distance to the Moon and subtracting it from the total distance which was to be traveled. Note that the two curves are apparently identical.

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 backwards about 25 feet, as determined by watching the image of the Moon in the telescope! But to put this in perspective, the moon is about 235,000 miles away at that point, so the motion represents just 2 millionths of a percent of the distance to the moon. It won’t actually be possible to see it in any real telescope, sad to say.

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:

A More Extreme Example:  Blastoff for Alpha Centaurus

Suppose we accelerate at 1 gravity toward Alpha Centaurus, 4 lightyears away.  Suppose further that we have a telescope powerful enough to resolve an actual image of the star.  What will we see?

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 0.1c, which translates to γ=1.005.  So, at least for the first 30 days we can expect our previous results to be applicable.

Note first that for low-speed travel, t_c = al₀ = the initial distance to the goal.  In this case, acceleration in ly/year² is about 1, and l₀ is 4 years.  So, we would expect the star to appear to recede for the entire duration of the trip!  Of course, as already noted our approximation is incorrect when v approaches c; with a more accurate formula we would expect a more reasonable result.  But let's see what the formulas we've got show for the first month of the voyage.  We'll do everything in terms of light-years, since that's easiest in this case; then we have



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 four times the speed of light!  In fact, this is obvious from a glance at eq (31): at time 0, velocity = distance = 0, γ=1, and the formula is exact: the apparent recession rate of the star is al₀, which, if we plug in everything in terms of light-years, is slightly larger than 4c.
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 images which appear to move faster than c.  In fact, since we've ignored γ throughout and worked with the low-speed limiting cases, the results we've obtained have actually been "classical".  As always, on the other hand, to measure the velocity of an actual object in a particular frame of reference, it's necessary to find the coordinates of two events on the object's timeline and divide the coordinate time difference into the spatial difference.  And that ratio is never larger than c, as long as the frame of reference is inertial.

The Time Seen on a Distant Clock

As we've already seen elsewhere on this site, as you accelerate toward a distant clock, time at that clock seems to run faster (see, in particular, the accelerating twins and the revolving astronaut).  However, that's just what you compute if you try to figure out what time it is according to that clock "right now".  If you look at the clock with a telescope, it doesn't seem to zoom ahead, at all; it just continues ticking normally (as adjusted for Doppler shift).  That seems somewhat paradoxical.

Now we have the solution to this paradox.  The clock doesn't appear to zoom ahead, but it does appear to zoom away.  And, as it continues to read about the same time (just ticking ahead "normally"), as it moves farther away, we realize that it's also actually moving ahead of the reading on our own clock!  For, if we look at distant clock, and we want to know the current time on that clock, we must take the time we see on its face and add to it the time it took the light to get from that clock to us.  So, when the clock appears to be zooming away from us (faster than light) as its reading continues to advance, we realize that the "current time" we see on that clock, adjusted for the speed-of-light delay of the image, must also be zooming ahead.

Remaining Questions

There are a number of things we'd like to look at, if we ever can find the time.

• 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.

But for now, ça suffit.

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Plot files are linked to from the captions on the graphs.
Page created, 10/6/05.  Updated with addition of "Time Seen on a Distant Clock", 10/7/05