# How Relative Is Relativity With Respect To Visual Effects?

This might seem naive, but here goes.

Imagine Albert and Rick, of equal mass, accelerate together to a significant fraction of light-speed (call it v) relative to Earth, enough to clearly see the effects of light aberration when looking at the stars. They expend 2 units of energy to do this. The two are now in an inertial frame of reference.

Can Albert now accelerate further so that his speed relative to Rick is v, expending 1 unit of energy?

How can you be in an inertial frame of reference, and still witness light-aberration effects?

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When Albert accelerates the second time, he experiences a large speed change relative to Rick, while speed change relative to stars is smaller. Also direction of light coming from Rick changes a lot, while direction of light coming from stars changes less.

When stars accelerate and Albert stays still, Albert can say that there is an aberration effect in the light coming from the stars. In my opinion this is not a relativistic effect, for the same reason that a car driving at 50 mph does not have a relativistic effect on a radar beam, it is too large an effect to be a relativistic effect.

Correction:

Let us consider a laser pointer, pointing into unchanging direction, accelerated two times. The second acceleration causes a smaller change of direction of the laser beam compared to the first acceleration.

Let us consider a laser pointer accelerated two times, after the first acceleration we turn the pointer so that the beam is NOT turned. Now the second acceleration causes the same amount of turning of the beam as the first acceleration.

Oh yes the first case is incorrect if the pointer points backwards.

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Ok, getting the gist of it now. I thought I had SR pegged down, but this scenario seemed a bit slippery :) – Per Jan 7 '12 at 17:54

The answer to the first question is yes. If you increase your energy by a factor of $n$, your velocity in the Earth reference frame will be $$v_{\text{new}} = c \left( 1 - \left( \frac{\sqrt{1-v^2/c^2}}{n - (n-1)\sqrt{1-v^2/c^2}} \right)^2 \right)^\frac{1}{2}.$$ So $$\lim_{n \rightarrow \infty} v_{\text{new}} = c.$$ I hope that answers your question.

As for the second question, you may wish to elaborate on the perceived paradox. As far as I know the light-aberration effect is something about inertial frames to begin with.

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It was the aberration that got me confused. Looking at the effect, the two could calculate their speed. So when Albert accelerates further, his increased speed relative the stars would not match the speed increase relative to Rick? – Per Jan 7 '12 at 9:28
Sorry for the delay. Yes that's right. Say, for example, that they are both travelling together at 0.9c, then surely Al can accelerate up to 0.9c relative to Rick. But he would not now be travelling at 1.8c relative to the stars (since 1.8c > c). Rather he would be travelling at some velocity $v_{new}$ such that 0.9c < $v_{new}$ < c relative to the stars. $v_{new}$ can be calculated from the above formula by entering v = 0.9c and n = 2, rendering $v_{new}$ approximately 0.96c. – Stephen Mc Ateer Mar 14 '12 at 23:17

As Stephen has answered, the answer to the first question is trivially yes. Albert and Rick perceive themselves to be at rest, so it would be nonsensical if one of them couldn't accelerate further.

Regarding the second part, you don't need to be in a noninertial reference frame to see relativistic aberration. You only need to be moving fast enough relative to the stars. Essentially, the reason is that the component of the light's path parallel to your motion is length-contracted, but the component perpendicular to your motion is not. That changes the angle at which you see the light.

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