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If a beam of light was shone horizontally, and simultaneously a stone was dropped from the same height, would they both hit the ground a the same time?

Of course on Earth they would not, but let's imagine a land mass large enough for the light not to fire off into space and away from the pull of the land's gravity.

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    $\begingroup$ According to einstein-online.info/spotlights/light_deflection, the deflection of light by gravity is twice as large as the "Newtonian" value; this would imply that the light would arrive first. However I am not sure that argument is valid, so I am posting this as a comment / starting point for the discussion. $\endgroup$ – Floris Feb 4 '15 at 15:13
  • $\begingroup$ On the Earth the light will move practically horizontally, and the stone will fall vertically, so what do you ask? Now, if you are not on the Earth, where is the height from which you drop the stone, and what means "ground"? $\endgroup$ – Sofia Feb 4 '15 at 15:14
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    $\begingroup$ @Sofia Obviously the height does not matter, neither does the term ground. It's hypothetical. $\endgroup$ – user59315 Feb 4 '15 at 15:22
  • $\begingroup$ @BrianBishop : Which question shall I read? Not the one asking when the two things will reach the ground? Instead of giving me orders, won't you consider formulating your question more clearly? $\endgroup$ – Sofia Feb 4 '15 at 15:27
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    $\begingroup$ @Sofia There is only one question. Floris seems to have got that. The question if fine as it stands I think. $\endgroup$ – user59315 Feb 4 '15 at 15:29
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Given that the stone is in free fall - that is, in a frame following a geodesic - it would have to observe the light emitted in its frame as behaving normally. This means that the stone would not observe the light bend away from a path radially outwards from it, which is a complex way of saying that it must hit the ground at the same time as the stone.

In the comments above, Floris mentioned how the deflection of light predicted by GR is twice what one expects from Newtonian calculations. This is true, however it is the measured deflection from the frame of a non-local observer that it refers to. So an observer standing on the ground would notice the discrepancy in the deflection of light, but the stone would not notice anything, since it is a local observer. If the stone were to see the photon hit the ground at a different time from it, that would violate the weak equivalence principle.

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    $\begingroup$ I don't know enough to mark this as correct or not.... $\endgroup$ – user59315 Feb 4 '15 at 17:46
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    $\begingroup$ @Brian Well, the weak equivalence principle (one of the principles on which GR is based) does require that the stone would see the photon hit the ground at the same time as it. However, there are other frames of reference one can approach the problem from. If all you care about is the stone's perspective, then my answer is correct for you. If it was another perspective you wanted considered, then that should be clarified. Although, if it's just a matter that I've said "weak EP, therefore it hits at the same time", then I can include the definition and expand my answer $\endgroup$ – Jim Feb 4 '15 at 18:29
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    $\begingroup$ If the rock has a hole in the middle and the light is actually bouncing between two perfect mirrors through the hole on the way down, then even though the rock is falling slowly it's supposed to think that the mirrors are bent enough such that the light should get deflected out of the hole? (Otherwise there's no way to reconcile the rock's "it lands at the same time as me" with the external observer's "the light falls out of the hole and stops when the rock has fallen one-half the hole diameter") $\endgroup$ – Rex Kerr Feb 5 '15 at 8:02
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    $\begingroup$ @RexKerr Can you pose this as a new related question? Two perfect mirrors and a falling stone would guarantee locality of the event (light hitting ground) and give us no problems with curvature. Then the Stone should see the light hitting the ground at the same time, while an external observer should see the light hitting the ground after half the time? $\endgroup$ – Falco Feb 5 '15 at 13:07
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    $\begingroup$ @Hypnosifl I have no issues with your interpretation. I assumed a land mass with no curvature and a constant gravity field, so the initial height of a stone and photon in the real world would have to be fairly low in order to make that approximation valid. But I used it because the root of the question is if light falls with other objects in the same gravitational field, not if in a realistic experiment the photon would actually hit the ground at the same time. I used the best set of assumptions to answer the spirit of the question as I understood it. $\endgroup$ – Jim Feb 6 '15 at 15:04
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Even if you just fire two balls horizontally from the same initial height with different horizontal velocities (lower than the velocity needed to escape or orbit), it's not true that they would both hit the ground at the same time if the tangential distance one of them travels is large enough that the curvature of the Earth must be taken into account. Just consider the Newton's cannonball thought-experiment which I discussed in this answer, in which we imagine a cannonball fired with a series of increasing velocities that increase the time it's able to remain above the ground because, although it is continuing to accelerate towards the center of the Earth at the same rate, the ground is continually curving away out from under it:

enter image description here

If you fired a cannonball with a velocity just shy of being able to complete a full orbit without ever hitting the ground, then the time before it hit the ground would be just shy of the time for a full orbit, and that's going to be a lot more than the time for a cannonball to hit the ground if it's just dropped from the same height with zero tangential velocity. So by analogy, if the light makes a path that curves around some significant portion of the circumference of the event horizon of a black hole before "hitting" it, I'm pretty sure the time could be larger. A further complication here is that in general relativity it's not really clear what time coordinate we should use--in the Newtonian case it's assumed we're using some inertial frame's time coordinate, but in general relativity you can only have "local inertial frames" defined in very small neighborhoods of spacetime (see the equivalence principle), all coordinate systems covering large regions of curved spacetime are non-inertial and the definition of simultaneity in non-inertial frames is basically a matter of arbitrary choice (though there is still an objective frame-independent fact about whether one event occurs in the future light cone of another, so in some cases it might actually be true that because of the way the light curves around the event horizon before crossing it, the event of the light crossing the horizon might be in the future light cone of the event of the other object crossing it--not sure about this though, maybe someone else can comment).

But there is a special case which is more straightforward, the one where the spacetime separation between the event of the rock and the light beam being released and the event of them crossing the horizon is infinitesimal, so the whole thing can be observed from the perspective of the local inertial frame of a freefalling observer (Jim may have implicitly assumed this sort of case in his answer, I'm not sure). From this perspective, both the rock and the light ray should move in straight line in the horizontal direction while the event horizon rushes upwards to meet them at the speed of light (the event horizon of a black hole is a null surface, so it must move at the speed of light in local inertial frames), and since neither the rock nor the light ray has any vertical component to their velocity in this frame, the event horizon should hit both of them at the same time in the coordinates of this frame.

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    $\begingroup$ True, but the OP specifically requested we consider a land mass large enough for the curvature not to be apparent. $\endgroup$ – John Rennie Feb 4 '15 at 17:09
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    $\begingroup$ @JohnRennie: Technically he said "not fire into space", so this still applies. However, it could be considered that he mispoke and meant an infinitely large flat plane with non-infinite/wierd "gravity" straight toward it from all points above. $\endgroup$ – Mooing Duck Feb 4 '15 at 17:59
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    $\begingroup$ @John Rennie - As Mooing Duck said, I interpreted "not fire into space" just to mean the gravity was high enough that the light would bend down and hit the source of gravity, which wouldn't happen on Earth if the light was aimed parallel to the ground...I didn't think it was specifically there being no noticeably curvature between the point the light was emitted and the point it hit the surface. $\endgroup$ – Hypnosifl Feb 4 '15 at 19:19
  • $\begingroup$ The horizontal component of an objects velocity does not affect the vertical (gravitational) velocity. All cannonballs would fall to earth with the speed of ~9.8m/s^2, regardless of the initial horizontal velocity imparted on the object. Any cannon ball that achieves orbit does not fit the parameters of the original question. $\endgroup$ – SeanC Feb 5 '15 at 21:29
  • $\begingroup$ @Sean Cheshire - Although the cannonball is always experiencing an acceleration due to gravity of 9.8 m/s^2 in the radial direction, the fact that the surface is curving away from it means it doesn't have to approach the surface a rate of 9.8 m/s^2, as should be obvious from the orbital case. Also, all paths that hit the ground can be understood as subsections of possible elliptical orbits that the ball would take if all the Earth's mass were concentrated at a point at the center, see the second diagram I added to my answer here. $\endgroup$ – Hypnosifl Feb 5 '15 at 21:55
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Not to be pedantic, but which one lands first according to whom??? Simultaneity is relative for events that are not at the same position, so you can check that you dropped/shone simultaneously or the order of landing (if you set up so both land at the same place), but not both unless you manage to get the light to curve around and land "below" where it left from. From the way you asked the question I'm imagining a semi-infinite sheet of mass with a uniform gravitational field, so such curving won't work. Since at least one pair of events you want to check time ordering/simultaneity for is spatially separated, the time ordering will change depending on which observer you ask.

For an analysis of one observer that is interesting to consider, see Jim's answer.

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  • $\begingroup$ Maybe if the point of contact of the light beam was noted and called point B. If the light arriving at point B triggered a second light to beam back in the direction of the stone, and the stone was dropped from twice the height of the light, they would both touch down at the exact same time?? Maybe:-) $\endgroup$ – user59315 Feb 4 '15 at 17:35
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    $\begingroup$ @BrianBishop But you can't send the return signal from where the outgoing signal arrived, you're already on the ground. The return signal needs to leave from somewhere else, so you're screwed ;) (you might try and contrive some situation where you can send the return signal from where the outgoing arrived - it won't work, but you're welcome to try!) $\endgroup$ – Kyle Oman Feb 4 '15 at 22:09
  • $\begingroup$ I mean send the return signal from the original height just above the light impact point. $\endgroup$ – user59315 Feb 4 '15 at 22:10
  • $\begingroup$ So you need some time to get the message saying "go" from the arrival point to the return-departure point, no? $\endgroup$ – Kyle Oman Feb 4 '15 at 22:11
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    $\begingroup$ @DavidConrad well met sir! $\endgroup$ – Kyle Oman Feb 5 '15 at 22:30
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A land mass large enough for light to not escape is a tricky proposition though. Effectively, you are asking for a black hole.

And I'm speculating here, someone else will have to check the math, but I think the light can "orbit" the black hole forever, exactly at the event horizon, while a massive object will always fall inwards. So if your starting level is above the event horizon, the light always escapes the gravity. If the starting level is exactly at the horizon and your "ground" was below it, I'd say the stone hits first, the light still never does, but it also never escapes.

If the starting position is below the event horizon, there is no such geometric direction that you can call "horizontal" - all possible directions for the light ray point inwards, and there actually is a reference frame in which the ray is actually directly vertical, so it seems to me that it will beat the stone... At least if they are racing to the singularity itself. If it is to some other "ground"... well... in the same way as you can not shine the light "horizontally", you can not define a target "ground", or at least it would not be a space-like surface, but a time-like one, or a "moment", not a "place". So I am getting totally confused.

Very nice question though, and something I should try to think about more.

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  • $\begingroup$ Light will orbit a black hole at the photon sphere, which is 1.5 times the radius of the event horizon. While normally, escape velocity can be interpreted as escape speed, for a black hole, the speed of light escape velocity is actually a velocity. It is the speed of light directed radially outwards that will escape. Any radius less than the 1.5 times the radius of the event horizon, a horizontally directed photon will not escape and will fall into the black hole $\endgroup$ – Jim Feb 6 '15 at 15:30
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At the same time with writing my answer about black holes and other mumbo jumbo, I am also intuitively feeling that something must be wrong with it. There must be a way to visualize the equivalence principle also applying for light and it should not involve us falling in a black hole...

If we assume uniform gravity and a perfectly flat surface, we can decrease the height we run our experiment from so the time it takes the stone to fall is smaller and smaller, thus also limiting how far the light can go, so obviously we do not need a surface as huge as to create a black hole or even make its curvature obvious...

Well, it seems we have to get really precise, but it is a thought experiment anyway, so why not?

A micro-meter height gives us a 135km distance, a nanometer gets it down to 4.3 km (see formula).

(You can play around with the calculation in google :p or maybe wolfram alpha if you are less lazy than me.)

Those distances may seem small enough, but a planet's curvature over them is still huge... the vertical displacement of our "ground" if it were not flat but had a radius like our earth at those 4.3km away for the nanometer height test is almost 3 meters. (see formula)

In the end, if we had some super flat surface and a super-uniform gravity field so it looked like it was perpendicular to it over a sufficient length, I guess we could see a light ray falling to the surface same as a stone. But then we could also say that we have an infinitely massive gravity source an infinite distance away from us, or who knows what... is that just like if we are at a black hole event horizon again?

I never imagined a thought experiment could fail before, but here we are, folks.

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  • $\begingroup$ "But then we could also say that we have an infinitely massive gravity source an infinite distance away from us, or who knows what... is that just like if we are at a black hole event horizon again?" More like, we could say that we are in a uniformly accelerating, non-inertial reference frame. Maybe I should have thought about it from that direction instead... $\endgroup$ – user72443 Feb 5 '15 at 11:13
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    $\begingroup$ You could have a large (compared to the distance traveled by the light) planar slab of matter. It would have a uniform gravitational field without needing infinite mass. $\endgroup$ – Floris Feb 5 '15 at 22:18