Error in Concepts: Intro to Mechanics Kolenkow & Kleppner

[Ramanathan k s]

TL;DR

While reading "The Equivalence Principle and the Gravitational Red Shift" in the book (Introduction to Mechanic by Kolenkow & Kleppner) I'm having doubts about how they arrived at the result gL/(c^2).
They are treating the spacetime curve as a parabola, but is it a parabola?
And I think there is some serious conceptual error in the analysis.
please help me

 

[Dale]

They are treating the spacetime curve as a parabola, but is it a parabola?

Every differentiable curve is a parabola to 2nd order about any given point.

 

[Ramanathan k s]

Every differentiable curve is a parabola to 2nd order about any given point.

But what will be the actual curve in this case?

 

[Dale]

But what will be the actual curve in this case?

It is actually a hyperbola. However, that does not make the derivation wrong. It just makes the derivation a derivation to 2nd order. This is not a "conceptual error", it is a valid approximation.

 

[vanhees71]

It's still a somewhat hand-waving argument, but it's correct given the approximations made.

A simple argument is to use the exact Schwarzschild solution, of which we only need
$$\mathrm{d} s^2=(1-2m/r) c^2 \mathrm{d} t^2 + \ldots$$
Now a wave is emitted from a source at rest at $r=r_0$. The time it takes from one maximum of the wave to the next is
$$\frac{2 \pi}{\omega_{\text{em}}} = \mathrm{d} \tau (r_0)=\sqrt{1-2m/r_0} \mathrm{d} t_{\text{em}}.$$
Then it's received by an observer at rest at $r_0+h$, and there
$$\frac{2 \pi}{\omega_{\text{obs}}}=\mathrm{d} \tau (r_0+h) = \sqrt{1-2m/(r_0+h)} \mathrm{d} t_{\text{obs}}.$$
Now $\mathrm{d} t_{\text{obs}}=\mathrm{d} t_{\text{em}}$, from which
$$\frac{\omega_{\text{obs}}}{\omega_{\text{em}}}=\sqrt{\frac{1-2m/r_0}{1-2m/(r_0+h)}} \simeq 1-\frac{m h}{(r_0-2m)r_0}.$$
Now $m=G M/c^2=R_{\text{S}}/2$ is half the Schwarzschild radius. For $r_0 \gg R_{\text{S}}$, where the Newtonian limit is valid, we have
$$\frac{\omega_{\text{obs}}}{\omega_{\text{em}}}=1-\frac{m h}{r_0^2}=1-\frac{G M/(c^2 r_0^2)} h =1-\frac{g h}{c^2}.$$

 

[Ramanathan k s]

It's still a somewhat hand-waving argument, but it's correct given the approximations made.

A simple argument is to use the exact Schwarzschild solution, of which we only need
$$\mathrm{d} s^2=(1-2m/r) c^2 \mathrm{d} t^2 + \ldots$$
Now a wave is emitted from a source at rest at $r=r_0$. The time it takes from one maximum of the wave to the next is
$$\frac{2 \pi}{\omega_{\text{em}}} = \mathrm{d} \tau (r_0)=\sqrt{1-2m/r_0} \mathrm{d} t_{\text{em}}.$$
Then it's received by an observer at rest at $r_0+h$, and there
$$\frac{2 \pi}{\omega_{\text{obs}}}=\mathrm{d} \tau (r_0+h) = \sqrt{1-2m/(r_0+h)} \mathrm{d} t_{\text{obs}}.$$
Now $\mathrm{d} t_{\text{obs}}=\mathrm{d} t_{\text{em}}$, from which
$$\frac{\omega_{\text{obs}}}{\omega_{\text{em}}}=\sqrt{\frac{1-2m/r_0}{1-2m/(r_0+h)}} \simeq 1-\frac{m h}{(r_0-2m)r_0}.$$
Now $m=G M/c^2=R_{\text{S}}/2$ is half the Schwarzschild radius. For $r_0 \gg R_{\text{S}}$, where the Newtonian limit is valid, we have
$$\frac{\omega_{\text{obs}}}{\omega_{\text{em}}}=1-\frac{m h}{r_0^2}=1-\frac{G M/(c^2 r_0^2)} h =1-\frac{g h}{c^2}.$$

actually the result is $$\ 1+\frac{g h}{c^2}.$$

 

[Nugatory]

actually the result is $1+\frac{g h}{c^2}$

Can’t be, the ratio of frequencies has to be less than one for upwards-moving light to be redshifted.

 

[vanhees71]

I think my sign is correct, because for $h>0$ there must be red shift, because the "naive photon" looses energy moving "upwards".

 

[Sagittarius A-Star]

Even if that specific example is formally correct as an approximation, I would be careful with the book "Mechanics by Kolenkow & Kleppner" regarding SR, see:
https://www.physicsforums.com/threa...t-treatment-of-sr-oblique-coordinates.919465/