4
$\begingroup$

As I understand it, light that is emitted from a source is not imparted with the motion of the source and so always follows a "straight line". If this is correct, I am having a difficult time conceiving how the Lunar Laser Ranger experiments can detect photons.

In this experiment, a pulse of light is aimed towards a retro-reflector and it is reflected back. It takes about 2.5 seconds round trip and due to diffraction, the returning pulse covers a circle of approximately 20kms in diameter.

The Earth's orbital velocity is approximately 30km/s and in the time the pulse takes to make the round trip, the detector would be 75kms further along the Earth's orbital path. If my first paragraph is accurate, and the light pulse is not imparted with the Earth's orbital velocity, then how is detection of the returning light pulse achieved?

I hope this makes sense and thank you.

$\endgroup$
2
$\begingroup$

Interesting question. Taking a stab at it - not absolutely sure this is correct, but let the comments begin.

In the frame of reference of Earth, the light travels straight out to the reflector, and straight back. You are asking about the case where an observer is in a reference frame that is moving with respect to Earth/moon, and the picture would have to look like this:

enter image description here

In such a frame of reference, it will seem that Earth is shooting the light "slightly ahead", and that the retro-reflection is no longer quite "retro". In other words - the reflector appears to act like a mirror rather than a corner cube.

Why will it look to the observer that the Earth is deliberately shooting to "miss" the Earth? It is of course because the people on Earth see the moon in the wrong place. The light of the moon that arrives on Earth left a while ago, and will appear to be incident at an angle on the Earth's surface (as observed by the moving observer). This is a weak version of the "relativistic beaming effect".

Then there's the issue that the reflected beam does not appear to be retro-reflected; I think this can be explained if you consider that the angle of the reflector will no longer be 90 degrees as seen in that frame of reference. You might want to try to do that calculation yourself - and interesting way to calculate Lorentz contraction.

$\endgroup$
  • $\begingroup$ Thanks Floris! I wanted to clarify to be sure I can properly interpret your answer. You are referring to an observer being in motion relative to the Earth-Moon system, I was referring to the motion of the Earth-Moon system relative to the sun. My understanding is that a photon emitted from a laser station with respect to the Earth-Moon system would not carry to motion of the Earth-Moon system relative to sun. So, an observer stationary with respect to the Earth-Moon system would see the photon with a tangential velocity of 30km/s. Is that accurate? $\endgroup$ – Rama Set May 8 '16 at 16:58
  • $\begingroup$ *laser stationary with respect to $\endgroup$ – Rama Set May 8 '16 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.