# Electromagnetic radiation bending on Earth

Most articles say that a radiowave is able to propagate itself beyond the horizon because it is reflected off by the ionosphere (and the Earth itself).

But do radio waves also get bent according to the Earth's curvature due to gravity thus transmitting beyond the horizon without need for the bounce effect?

• Yes, they do, but the effect is extremely small and not measurable with regular radio. Some of the best evidence for the validity of general relativity comes from time delay experiments in the solar system. These experiments can not measure the curvature of spacetime directly, but they can measure delays of radio signals that are being sent across the solar system (typically it's a Venus radar experiment or a time delay from a planetary probe). Around the Earth the GPS system can detect these derivations from a flat metric, too. – CuriousOne Dec 21 '14 at 1:22
• So the radio waves are bent by a small amount. But is this enough to follow the curvature of the Earth? Disregarding attenuation, could the same signal continue in a constant 'orbit' around the planet? :-) – ChaimKut Dec 21 '14 at 1:53
• Exactly, there is bending but it is very small. Now, if you want to observe large bending, then you need to look at the propagation of radio waves or light (!) around a black hole. Hollywood just released a beautiful special effect in "Interstellar" which shows what that bending looks like. That is what strong gravity does to radio waves, too, and astronomers are looking for these kinds of signals all over the universe, with radio telescopes and with optical observations. Look for "weak lensing" and "strong lensing", which are the terms describing the bending of signals around heavy object – CuriousOne Dec 21 '14 at 2:06
• There is a significant bending from diffraction. See this tutorial mike-willis.com/Tutorial/PF7.htm, or look up the knife edge effect. – mmesser314 Dec 21 '14 at 4:13

Calculating the path that a light ray takes in a gravitational field is a complicated business, but for the special case of a light ray coming from infinity and escaping to infinity there is a convenient approximate formula for the angle, $\theta$, the light ray is deflected:
$$\theta \approx \frac{4GM}{r_0 c^2}$$
Where $M$ is the mass of the deflecting object and $r_0$ is the distance of closest approach. If we feed in the mass of the Earth and set $r_0$ to the smallest value it can have (the radius of the Earth) we get a deflection of $\theta \approx 2.8 \times 10^{-9}$ radians, or about $0.0000002$ degrees.