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I learned that things like the international space station or the moon are not "pulled" by the gravity of the earth but instead are following a geodesic (a straight line in curved spacetime). Gravity doesn't really exists it's only an illusion.

Fair enough, now why doesn't light also follow this 'straight line in curved space' just like the ISS? At first, I thought "oh, light is not affected by the space-time curvature", except it is! I've seen renders of black holes and light bends all over the place.

My question then is, shouldn't the astronauts in the ISS see the back of their own heads when looking into the direction of travel.

ps: I am a CS major, not a physicist (I'm guessing that's painfully obvious). I got my science from Veritasium's video Why gravity is NOT a force.

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    $\begingroup$ I like this youtube channel in general, but THIS video did nothing good for the popular science community. Of course, gravity DOES exist, it is a force, and it is NOT an illusion. (Jump in the air if you don't believe it). The mathematical description does NOT change these fundamental observations. $\endgroup$
    – Koschi
    Commented May 24, 2022 at 17:01
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    $\begingroup$ @Koschi While gravity definitely exists and is not an illusion, it is not quite a force. The force description is surely appropriate at weak gravitational fields, but once things get intense it works in a more subtle way $\endgroup$ Commented May 24, 2022 at 17:03
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    $\begingroup$ Why do you think light is not following a geodesic? $\endgroup$
    – BowlOfRed
    Commented May 24, 2022 at 17:11
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    $\begingroup$ The setup of the question includes the fragment: "Fair enough, now why doesn't light also..." Which seems to presume that light doesn't follow a geodesic. This presumption is wrong. OP doubles down on his confusion with the comment: "ISS astronauts don't see the back of their neck..." This illustrates OP's muddled thinking and the muddle that is this question. We can't really answer a muddle. The question should be closed or severely edited, in my opinion. In addition, an answer to a similar, less muddled, question already exists. $\endgroup$
    – hft
    Commented May 24, 2022 at 19:32
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    $\begingroup$ Pseudoforces aren't illusions, they're true real consequences of choice of reference frame. Forces differ from pseudoforces in that they are frame independent. Lots of things other than pseudoforces are frame dependent: time, length, velocity, work, energy, angle, volume... $\endgroup$
    – g s
    Commented May 24, 2022 at 19:54

5 Answers 5

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I think your question is an issue concerning what a straight line in a curved space should look like. To get some intuition, consider this funnel, respresenting a curved surface. https://www.youtube.com/watch?v=P5bf2BPgjS0 (You just have to imagine it being frictionless so that the coins don't approach the center)

From flat space, like tables and such, you have the idea that "straightness" is a property of a line. It has, especially, nothing to do with the velocity of an object following this line. Billiard balls follow the same straight lines no matter how fast they go.

If that were true for geodesic motion, a light ray and the ISS should trace out exactly the same path, and astronauts would get a good view on their bald spots, and I think that is the argument you are making, right?

But the truth is somewhat different. Think about the coin on the funnel surface in the video. If it started out with a higher or lower initial velocity, it would trace out a very different path. If you start it with velocity 0, it drops straight to the center, and at very high velocities, it will barely notice the curvature of the funnel (which is basically the way a light ray passes the earth).

This is just a short answer alluding to intuition. If you want it a bit more technical, consider the answers in Why would geodesics depend on velocity in general relativity?

Oh, and while we are at videos using curved surfaces, this one's a big hit and might help clarify some points: https://www.youtube.com/watch?v=MTY1Kje0yLg&t=373s

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    $\begingroup$ Oh I see, I forgot to factor in velocity... the geodesic is an orbit for the ISS because it has a specific velocity. For light the geodesic has a different shape. is that it? - I'm not a fan of videos that explain gravity using gravity (they make for an interesting demo but I think you should not be able to use the thing you are trying to explain to explain it) $\endgroup$
    – frankelot
    Commented May 24, 2022 at 17:26
  • $\begingroup$ Understandable. Maybe don't think too much about such devices "explaining" gravity but rather about giving you analogies to help you grow your intuition. To get an idea of what curvature does, you need a curved surface and something that makes sure that motion is "stuck" to that surface. $\endgroup$
    – TBissinger
    Commented May 24, 2022 at 17:58
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Here is a different way to look at it:

You have probably heard that if you shoot a bullet horizontally and if you drop a bullet, they will both hit the ground at the same instant. The same things hold for light: if you shoot a bullet, drop a bullet, and shine a laser horizontally, they will all hit the ground at the same time BUT the difference is that the dropped bullet will land at your feet in 1/4 second; the shot bullet will land 200 meters away in 1/4 second; and the laser light will land 75,000,000 meters away in 1/4 second. That is about 1/5 of the way to the moon. So the reason why astronauts can't see the back of their head is simply because the Earth is not big enough to give the light sufficient time to curve around, let alone fall to the ground. The only place where there is enough gravity to do this is at a black hole.

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Objects do not necessarily follow geodesics in space, but rather geodesics in spacetime. The difference is subtle, but important. In particular, objects with different velocities will have different geodesics to follow, because of the time factor. Light is moving so quickly that its geodesics do not orbit the Earth, merely bend a little bit as they pass by the Earth.

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In addition to the other answers, it is worth noting that all paths are sorted into spacelike, timelike, and lightlike paths, depending on whether they trace out positive, negative, or zero proper length relative to the (-, +, +, +) metric.

(Free-falling) particles follow timelike geodesics, and light follows null geodesics (spacelike geodesics can be thought of as "maximum proper distance" between two points).

Now, when solving the geodesic equation in the case of a spherically symmetric matter distribution, it turns out that there are no stable lightlike geodesic orbits, and the only (unstable) orbit is located at a radius $r=3GM/c^{2}$, which would put it inside of the Earth.

So, light does follow geodesics, but the types of geodesics that light follows are different than those that matter allows, and light's paths specifically don't allow for (stable) orbits.

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Because earth's radius is larger than the radius light would need to circle around a mass equivalent to the earth's.

The Schwarzschild radius for the earth's mass is 9mm, and the photon sphere is 1.5 times that, so you would need to compress the earth to a diameter smaller than 27mm for light to orbit around it at that radius.

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  • $\begingroup$ That doesn't answer the question though. Why does there have to be a specific radius? Why isn't light just following the geodesic (which orbits the earth if the ISS is anything to go by) $\endgroup$
    – frankelot
    Commented May 24, 2022 at 17:47
  • $\begingroup$ @frankelot Light is following its geodesic, which differs from massive bodies with different velocities $\endgroup$
    – Eletie
    Commented May 24, 2022 at 17:56
  • $\begingroup$ @frankelot - The geodesic does not only depend on the radius but also on the initial velocity, in order for the geodesic to be a circle you need the right radius for the right initial velocity and direction or vice versa. Since light always has c you can't tweak its velocity, so you have to chose the right radius, and for v=c that has to be the photon sphere of the dominant mass. $\endgroup$
    – Yukterez
    Commented May 24, 2022 at 18:03
  • $\begingroup$ Right, I forgot to factor in velocity. I'm trying to build an intuition that considers velocity now $\endgroup$
    – frankelot
    Commented May 24, 2022 at 21:27

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