If an LED is dropped from a great height >> R_Earth and the LED falls inertially towards the surface of the Earth:- I predict the relativistic doppler shift (ie actual observed shift from the Earth's surface) remains unchanged as the LED falls.

True/false and/or comments welcome.

Thank you.

-C2

I just can't resist...

Is the effect the same if a FEATHER is dropped from the same height?

I've just got some everlasting LED torches (solar cell included) - so one of them - they're easier to see than a feather - but a feather would do - if you insist.

As I said, just teasing.

In an real world situation, the velocities between the two would be less than needed to provide a measurable relativistic Doppler shift, so 'true' is my answer.

It does depend on high high the LED is dropped from, and how long you can take to measure it, but I said "real world"...

I propose 'real' to the (unreal) extent that an on-going Pound-Rebka experiment would confirm/refute the prediction.

This is a counter-intuitive prediction (always the best) from my interpretation of Trout's approach to GR. Could well be wrong. If your physics isn't predictive then it isn't physics.

This is unlikely to be a good experiment.

LED's have wide spectral peaks -- about 5% -- so you would need a huge effect to see the peak move conclusively.

But how do you expect to separate the time-dilation effects (tends to slow down, an effect which starts at zero) from the altitude (GR) effects (tend to speed up, but decreases to zero at ground), from the geometric doppler shift (tends to speed up, an effect which starts at zero)?

For a short fall, v = gt, h = H - gt^2/2, and v << c, H << Radius of Earth << c^2/g
we have classical (geometric) shift Δf/f = c/(c-v) - 1 ≈ v/c = +gt/c
we have a SR shift (geometric + time dilation) of Δf/f = sqrt((c+v)/(c-v)) - 1 ≈ v/c = +gt/c

And that effect is smaller than 5%. But to see just the time-dilation effect you need
sqrt((c+v)/(c-v)) - c/(c-v) ≈ -v^2/c^2 = -g^2t^2/c^2

And for GR, measured from the ground, we have Δf/f = +gh/c^2 = +gH/c^2 -g^2t^2/2c^2

Putting it altogether, the prediction is Δf/f ≈ +gH/c^2 + gt/c - 3g^2t^2/2c^2

And that rises to a maximum at t = c/(3g) that goes to (about) zero at t = 2c/(3g) which is long after the experiment is valid.


Notes:

c/(c-v) ≈ 1+v/c+v^2/c^2+v^3/c^3+v^4/c^4+v^5/c^5+O(v^6)
sqrt((c+v)/(c-v)) ≈ 1+v/c+v^2/(2 c^2)+v^3/(2 c^3)+(3 v^4)/(8 c^4)+(3 v^5)/(8 c^5)+O(v^6)
gamma = 1/sqrt(1 - v^2/c^2) ≈ 1+v^2/(2 c^2)+(3 v^4)/(8 c^4)+(5 v^6)/(16 c^6)+(35 v^8)/(128 c^8)+(63 v^10)/(256 c^10)+O(v^12)
1/gamma = sqrt(1 - v^2/c^2) ≈ 1-v^2/(2 c^2)-v^4/(8 c^4)-v^6/(16 c^6)-(5 v^8)/(128 c^8)-(7 v^10)/(256 c^10)+O(v^12)

I was rather hoping that the effect of observing the LED through the field would cancel out the effects of the increasing relative velocity as the LED (inertially) approaches the observer. It looked 'self-evident' for a while - many thanks for your post.
-C2.