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Two persons A, B are uniformly accelerating in the positive z direction by amount $g$.

$A$ flashes two pulses of light with an interval $T_a$.

The time interval between two flashes of light in $A$ frame is $T_a$ and correspondingly this event has a time interval $T_b$ in $B$ frame.

We can prove that $T_b$>$T_a$ using only kinematics and the universality of speed of light( Kleppner and Kolenkow have done in their mechanics book)

By Equivalence principle the situation will remain same if the acceleration is is replaced by a constant gravitational field pointing downwards.

We know therefore $T_b$>$T_a$ if we've two stationary observers $A$ and $B$ in a gravitational field pointing downwards, and $A$ flashes light in interval $T_a$.

Question: I understand that the time intervals will be different for the two frames but the frames should both be in the gravitational field,that's how our argument is constructed.

The authors say ( I shorten it) that as a consequence of the time delay we can observe red shift (of light coming from stars having large gravitational field) on the earth.

But from the way the logic was constructed we can only say that any observer inside the nearly uniform gravitational field will observe the red shift.

How come the authors talk about observing red shift on earth which is almost outside of the gravitational field from which the light is coming? Can anyone please help me.


1 Answer 1


You can imagine that there's another person C a short distance above B, and D above C, and so on, and this chain of people extends all the way to Earth.

B doesn't have a flashlight: instead, they just let the light from A pass them by, while passively measuring its spectrum. The effect from C's point of view is just as though B is shining a flashlight that emits a redder spectrum than A's. The same local constant-gravitational-field argument shows that C's spectrum is more redshifted than B's, D's more than C's, etc.

Once you get close to Earth, the direction of the acceleration flips and the spectra start to blueshift, but since Earth's gravity is less than the emitting star's, it doesn't completely reverse the effect and you get a net redshift.

  • $\begingroup$ Thank you I think I get it. I'll take some time to read it again :) $\endgroup$
    – Kashmiri
    Feb 24, 2021 at 12:49
  • $\begingroup$ Observer C is not in the field, because only A and B are in the lab which is accelerating upwards, so how will he observe more red shift? $\endgroup$
    – Kashmiri
    Mar 1, 2021 at 6:33
  • 1
    $\begingroup$ @Kashmiri C is also accelerating upwards. It's a very tall lab with many floors. Consecutive floors are close enough together that you can approximate spacetime as flat, but the lab as a whole covers a region of spacetime that is significantly curved. $\endgroup$
    – benrg
    Mar 1, 2021 at 6:52
  • $\begingroup$ Yes I agree. Now what will a person just outside of the lab who is not accelerating observe? $\endgroup$
    – Kashmiri
    Mar 1, 2021 at 7:01
  • $\begingroup$ @Kashmiri I'm not sure what you mean by outside the lab. If you mean someone freefalling past one of the floors of the lab, then they'll see a Doppler shift that depends on their instantaneous velocity. If they're instantaneously at rest relative to the floor then they'll see the same thing as a person standing on the floor. If you mean someone very far above A and B, just extend the lab upward until they're inside it. $\endgroup$
    – benrg
    Mar 1, 2021 at 7:09

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