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I understand how the Lorentz contraction works but was thinking about how that would manifest itself as perspective of a distant object by two different observers. One is traveling at high velocity towards say, a distant planet, and another is stationary with respect to the planet. At the moment of observation both observers are the same distance from the planet.

Would the moving observer see a larger planet compared to the stationary observer? The consistency between time slowing down for the moving observer, as perceived by the stationary observer, and the length contraction of the moving observer would have them both agreeing on how long the journey to the distant planet took (in their relative reference frames) but would the observers see different sized planets as the angle subtended by the planet would be different at different distances?

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  • $\begingroup$ If the observers are at the same place at the same time, what they will actually see depends only on the light rays arriving at that place and time. This must of course be the same for both observers. What they will infer about the distance those light rays must have traveled, the angles between them, and how long the journey took is another matter. Are you asking about the immediate perception or about the inference? $\endgroup$ – WillO Jan 10 at 21:05
  • $\begingroup$ Assuming you are describing the following situation: two reference frames related by a boost along an axis (the x-axis for simplicity), with a spherical object at rest in one of the two frames (say in frame $K_1$). Then the observer of frame $K_2$ will see length contracted along the x-direction: he will see an ellipsoid, not a sphere, as he will see the diameter of the sphere parallel to the x-axis contracted. $\endgroup$ – Luthien Jan 10 at 21:17
  • $\begingroup$ @WillO Yes, what they actually perceive at the same instant when both are at the same location. Does one perceive a larger planet than the other? $\endgroup$ – Phil H Jan 10 at 21:36
  • $\begingroup$ @Luthien Yes an ellipsoid which he sees as a circle when traveling directly towards it but because the distance from it is contracted, it would infer a larger subtended angle and a larger circle. Is there an explanation for why this would or wouldn't be the case? $\endgroup$ – Phil H Jan 10 at 21:41
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    $\begingroup$ en.wikipedia.org/wiki/Relativistic_aberration. Also spacetimetravel.org/bewegung/bewegung5.html (and similar) for some generated images. $\endgroup$ – BowlOfRed Jan 10 at 22:47
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enter image description here

Apparently the “moving” towards a distant planet observer will “see” that apparent size of this planet is smaller, than the same planet as seen by “stationary” wrt to this planet observer.

For the sake of clarity let’s consider how will look pictures as taken by “stationary” photocamera (Fig. 1) and “moving towards the planet one” (Fig.2) at the moment, when their apertures coincide.

We can analyze this problem from the frame of the planet or from the frame of the observer. The picture should not depend on the chosen frame.

In the frame of the photo camera the planet is moving (Fig. 3) ; but we must bear in mind light - time correction. The rays of light always travel faster than the planet. That means that at the picture the planet will not appear where it actually is, but when it was in the past, when it’s apparent size was smaller. While rays of light are moving towards the aperture, the planet is moving closer.

In the frame of the planet the photocamera moves; but distance between its aperture and the film Lorentz – contracts plus the film moves further while rays travel from the aperture towards film. At the moment when “moving” and “stationary” apertures coincide, the same rays, or the same information goes through the aperture (Fig.2). Rays from the edges of the planet to aperture and from aperture to the film form similar triangles. However, the “moving” camera is shorter; hence image on it will be smaller.

It will be exactly of the same size as in the first scenario, but smaller than on the picture as taken by the “stationary” camera.

By the way. There is the relativistic aberration formula:

$$\tan(\phi) = \frac{u_y'}{u_x'} = \frac{u_y}{\gamma(u_x+v)} = \frac{\sin(\theta)}{\gamma(v/c + \cos(\theta))}$$

While light travels distance $ct$, "moving" observer travels distance $vt$, so in the "classical" case aberration angle $\tan \theta = ct/vt$.

What $\gamma$ does in the denominator of the relativistic equation? Distance $vt$ in relativistic case turns into $vt \cdot \gamma$, it is not $\gamma$ times shorter, but $\gamma$ times longer, because effect of aberration is "tied" with moving observer.

Since measuring rod of "moving" observer becomes $\gamma$ times shorter, he measures horizontal distances with "squashed" ruler and distant object appear to him even more further away.

The Feynman Lectures - Relativistic Effects in Radiation

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  • $\begingroup$ Very good, I posted a specific example using the relativistic aberration and Lorentz contraction formulas. I guess the classical consideration in reducing image size overshadows the smaller increase due to relativistic effects for my specific example. I assume that will always be the case. $\endgroup$ – Phil H Jan 11 at 16:08
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Given your clarification in the comments, the first part of my comment is the relevant answer: If the observers are at the same place at the same time, what they will actually see depends only on the light rays arriving at that place and time. This must of course be the same for both observers.

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  • $\begingroup$ Intuitively that makes a lot of sense. With that thought, how do we make a light diagram work for the moving observer? Could the explanation be that the observer's eyeball has contracted in the direction of travel, hence the angle subtended by the light from the planet on his retina would be identical to that of the stationary observer. Does that sound reasonable? $\endgroup$ – Phil H Jan 10 at 21:56
  • $\begingroup$ @philh : In his own frame, the "traveler" is stationary and his retina is not contracted. $\endgroup$ – WillO Jan 10 at 22:17
  • $\begingroup$ @WillO, the image formed depends not only on the light arriving at the same point and the same time, but on the observer interpreting the direction of the light being identical. I don't believe the direction will be the same for a moving and a non-moving observer. $\endgroup$ – BowlOfRed Jan 10 at 23:58
  • $\begingroup$ @BowlOfRed : I am taking each observer to be spatially zero-dimensional. To interpret the direction of the light, it seems to me you'd need a retina with some spatial extent. But if we allow that, then it becomes impossible to interpret the question, since the observers cannot agree that all parts of A's retina were in the same place at the same time as all the corresponding parts of B's retina. So I think that any fair interpretation of the question (with A and B unambiguously co-present at a single event) requires zero dimensionality and therefore prohibits directional interpretation. $\endgroup$ – WillO Jan 11 at 0:30
  • $\begingroup$ @BowlofRed: Of course I might be overlooking a possible alternative interpretation, and would be happy to have one. $\endgroup$ – WillO Jan 11 at 0:31
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Here is my understanding using a specific example.

enter image description here

There is consistency in the math between classical and relativistic corrections in the image received by A with the difference due to the Lorentz contraction. In both cases, as stated in Albert's previous answer, the moving observer sees a smaller image of the planet compared to the stationary observer.

In summary, there is an increase in image size of an approaching planet at relativistic velocities due to Lorentz contraction but this is overshadowed by the classical consideration of a moving observer receiving the image of the planet when it was at a much further distance. The overall effect being that a moving observer sees a smaller approaching planet than a stationary observer.

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