Graduate Solving Shapiro Delay Problem w/ Approximations: Q&A

[exmarine]

TL;DR

Without approximations, the integral blows up. Why?

I encountered a problem when calculating the Shapiro Delay for a radar signal from Earth to Venus on the other side of the sun. The integrand blows up at one of the integration limits, so I think that means the integral is infinite. So far, I have been unable to totally follow Zee’s (Einstein Gravity in a Nutshell) treatment, he glosses over it, and I think there might be some typos in there. But he does make an approximation at one point, which I did not do.

So I go back and review the successful calculation of the photon passing the sun (Eddington’s 1919 expedition, gravitational lensing, etc.) which turned out to be the same way Kevin Brown shows it (Reflections on Relativity). And sure enough, there is a similar approximation in that. I didn’t notice it at the time, but if I remove that approximation, then that integral also blows up.

Both approximations are “legitimate”, i.e., very small term(s) are ignored to facilitate the calculations. That does not seem right. Don’t we make approximations to help the math, but also to NOT significantly change the results? In both these cases, it looks like it is necessary to discard trivial factors in order to get correct results. I don’t think the Shapiro delay is actually infinite out here in the real world.

(1) Is the above correct? (2) If so, what is the explanation? Is our modelling, like the Schwarzschild metric, too simple or something? After all, the sun is not the only mass in the universe.

Thanks.

 

[Dale]

What is the specific integral and approximation, or a link?

 

[George Jones]

I am in a coffee shop on my way to work, and I don't have Zee at hand, but

The integrand blows up at one of the integration limits, so I think that means the integral is infinite.

is not necessarily true. For example,
$$\int_0^1 \ln x dx$$

(treated as an appropriate improper integral).

 

[exmarine]

I will see if I can type it here.

1/(1-eps) = 1+eps
sqrt(1-eps)=1-eps/2-eps^2/8...

where ϵ is the ratio of the Schwarzschild radius over the radius of closest approach to the sun. Zee does at least the first approximation, and Brown (I think) did at least the second for the gravitational lensing.

Try to adopt Zee’s notation for my integral for the Shapiro case – for just the earth-sun leg:

(Hmmmm I thought there was a semi-visual latex editor in here? I am just too dang old to try to remember all those commands, Unix type stuff. I will have to attach a pdf file for my integral for the Shapiro case. Sorry!)


where (rs) is the Schwarzschild radius of the sun, (r0) is the closest approach to the sun, and (rE) is the earth’s distance from the sun. The quantity in the radical in the denominator goes to 0 at (r0).

Maybe it would be simpler to note that if (dr/dt) goes to zero at closest approach to the sun, then (1/(dr/dt)) goes to infinity at closest approach?

 

[George Jones]

The integrand blows up at one of the integration limits

This happens in any situation where elapsed time is calculated between a "turning point" and some other point.

As an illustrative simple example, consider Newtonian gravity near the Earth's surface. Suppose a ball that is fired straight up from the Earth's surface reaches a maximum height of $h$. How long does it take the ball to reach maximum height?

Let $y$ be the height above the Earth's surface. Mechanical energy is conserved so

$k = \frac{1}{2} m \left( \frac{dy}{dt}\right)^2 + mgy,$

where $k$ is a constant and the zero for gravitational potential energy has been take to the surface of the Earth.

At maximum height, $y=h$, and speed equals zero, so $k = mgh$.

Using this in the above and rearranging gives the elapsed time

$\Delta t = \frac{1}{\sqrt{2g}} \int_0^h \frac{dy}{\sqrt{h - y}}.$

Note that the integrand is infinite at one of the limits of integration (the turning point).

In this case (unlike Zee's case), the integration can be expressed in closed form giving

$\Delta t = \frac{1}{\sqrt{2g}} \left[-2\sqrt{h-y} \right]_0^h = \sqrt{\frac{2h}{g}},$

which is the same result that constant acceleration kinematics gives.

In Zee's case, there isn't a closed form for the the integration, so Zee makes an approximation that does two things: 1) has a negligible effect on the answer; 2) removes the singularity at the turning point.

 

[exmarine]

Thanks!

 

[vanhees71]

I've given the Shapiro delay once as a problem in a general-relativity recitation. Maybe it helps:

https://itp.uni-frankfurt.de/~hees/art-ws16/sheet07.pdf
https://itp.uni-frankfurt.de/~hees/art-ws16/lsg07.pdf