# If gravity were a "real" force, then how would I be able to tell if I'm falling or accelerating in space or on Earth?

Background:

I think it would be helpful for laypersons like myself to understand how, in practice, a "real" force differs from a pseudo-force. Virtually all explanations (eg, on this stack, Wikipedia, Quora, great Veritasium video) of the difference between a pseudo-force and a real force that use the freefalling/accelerating reference frame thought experiment, focus on how the falling/accelerating object cannot tell whether it's falling/accelerating in space or on Earth, but fail to explain how this differs from a real force.

As I see it, an argument that gravity is not a real force should proceed as follows:

1. Here is how a falling/accelerating object would behave were gravity a real force.
2. A falling/accelerating object does not behave this way;
3. Hence, gravity is not a real force.

Based on this great answer, I believe that the difference between a real force such as say, the electromagnetic force, and gravity, is that the object's mass does not affect its acceleration under gravity, while the object's mass (or rather, the ratio of mass to the strength of the field) does affect its acceleration under the electromagnetic force. I interpret this to mean that if I drop iron objects of various masses from equal height, then they will accelerate toward the Earth at the same rate (reaching the ground at the same time), whereas if I place them an equal distance from a strong magnet, then they will accelerate toward the magnet at rates proportional to their masses (reaching the magnet at different times).

Additionally, I believe that the implication behind the thought experiment that one cannot tell the difference between weightlessness and freefall, or acceleration and gravity, is that with respect to a real force, one would be able to tell the difference. This begs the question ... how?

Question:

With focus on how gravity is not a real force, rather than on how gravity is a pseudo-force, I believe that the below questions are essentially equivalent, so answering any one of them should be sufficient, but I may be wrong about that:

• If gravity were a "real" force, then how would I be able to tell if I'm weightless in space or free-falling to Earth?
• If gravity were a "real" force, then how would I be able to tell if my ship is accelerating "up" in space or I'm on Earth being accelerated "down" by gravity?
• If I'm wearing a ferromagnetic (chainmail) suit while my non-magnetic (plastic) ship flies nearby a magnetar, then how would I be able to tell whether my ship is accelerating away from the magnetar, or I'm being pulled toward it?
• If I wake up in a hospital bed in an unknown spaceship, and I feel that the bed is pushing up against me / I'm pressing down on the bed, then how would I be able to tell whether I'm now Wolverine and have a metal skeleton while the ship is made of plastic (magnet below the ship), or if the ship is made of metal (magnet above the ship)?

Note: Force gradients (tidal forces) are ignored in the original thought experiment, and should be ignored here too.

• If gravity was a "real" force... By this you seem to mean if gravity was different than it is, what would it be like? We can't answer questions like that. On the other hand, for a look a what a pseudoforce or fictitious force is all about, see Coriolis Force: Direction Perpendicular to Rotation Axis Visualization. It isn't about gravity, but it should help. Gravity is a fictitious in the same sense that the Coriolis force is. Oct 11 '20 at 4:43
• $$F=\frac{G M_1 M_2}{R^2}$$ gives the force between 2 masses when one mass is the earth you get gravity. By knowing F=ma and dividing out the mass we get the acceleration of GM2/R^2 where M2 is earth much much larger in mass than a typical object. The acceleration is typically 9.8m/s^2 for small R's. Gravity imo is a real force unlike the centrifugal force. If i had a vibrating gyroscope in space it would register this acceleration Oct 11 '20 at 14:35
• @ChemEng No, it wouldn't. Acceleration due to gravitation cannot be sensed by any local experiment. This is true even in Newtonian mechanics, and it is a big problem with regard to navigating in space using accelerometers. Oct 14 '20 at 5:22
• i believe an accelerometer measures the net force on the object. If there is a weight force and a normal force then yes it will not measure an acceleration but if there is only the force of weight then it will measure the acceleration Oct 14 '20 at 14:27

I think it would be helpful for laypersons like myself to understand how, in practice, a "real" force differs from a pseudo-force.

The key identifier is that an inertial-force is proportional to an object’s mass. Forces which are proportional to mass can be made to appear or disappear by judicious choice of reference frame.

Experimentally, this is easy to detect using an accelerometer. If the accelerometer detects an acceleration then the object is subject to a net real force. If the accelerometer reads zero then the net real force is zero and any remaining acceleration is due to an inertial force.

Here is how a falling/accelerating object would behave were gravity a real force.

If gravity were a real force then an accelerometer attached to a free falling object near the surface of the earth would read $$9.8\text{ m/s}^2$$ downwards.

A falling/accelerating object does not behave this way;

An accelerometer attached to a free falling object near earth’s surface reads 0. Anyone can verify this with a typical smart phone.

Hence, gravity is not a real force.

Hence, the $$9.8\text{ m/s}^2$$ acceleration is not due to a real force.

If gravity was a "real" force, then how would I be able to tell if I'm weightless in space or free-falling to Earth?

As described above, if gravity were a real force then it would be detected by an accelerometer. So someone far from any gravitational sources would have an accelerometer reading of 0, and someone free falling to earth would have an accelerometer reading of $$9.8\text{ m/s}^2$$ downwards.

• You are the only one so far to answer the actual question, thank you +1. Just to clarify then, this means that flying by a magnetar while wearing a metal suit, my accelerometer would temporarily read >0m/s^2 toward the magnetar until I hit the bottom of the ship, after which it would read 0m/s^2, while if the ship accelerated away, then it would read 0m/s^2 on the way down, and >0m/s^2 away from the magnetar after hitting bottom? Oct 11 '20 at 23:44
• If I understand the magnetar scenario correctly, then yes.
– Dale
Oct 11 '20 at 23:53

Your first two questions basically have the answer "you can't". The fact that "weightlessness" is often confused with free fall (eg in orbit) highlights this fact.

For the 3rd and 4th question we can tell the difference because there are objects which do not respond to magnetism. So if you have a plastic key in your pocket you could take it out and observe its behaviour. Otherwise you could look at how your blood (which is nonmagnetic) behaves. Options of this type are not available for gravity.

Gravity is a real force. If you are asking if it is an emergent property of the curvature of space-time, you are correct. However in a local frame, it behaves as what you would expect from a real force (that is, $$F = \frac{dp}{dt}$$).

If gravity was a "real" force, then how would I be able to tell if I'm weightless in space or free-falling to Earth?

Once again, gravity is a real force and you can't tell. That is called the principal of equivalence.

If gravity was a "real" force, then how would I be able to tell if my ship is accelerating "up" in space or I'm on Earth being accelerated "down" by gravity?

You can't. Again, the principal of equivalence.

If I'm wearing a ferromagnetic (chainmail) suit while my non-magnetic (plastic) ship flies nearby a magnetar, then how would I be able to tell whether my ship is accelerating away from the magnetar, or I'm being pulled toward it?

If this turn away/acceleration from the magnetar is very sharp it's possible you may not be able to tell (from inside your ship). But I guess that because magnetic force is several orders of magnitude greater than the "usual inertial force", you may feel a difference in favour of the magnetic force (also, if you have magnetic/metal objects in the ship though, you will notice them deflect much more than objects non magnetic).

If I wake up in a hospital bed in an unknown spaceship, and I feel that the bed is pushing up against me / I'm pressing down on the bed, then how would I be able to tell whether I'm now Wolverine and have a metal skeleton while the ship is made of plastic (magnet below the ship), or if the ship is made of metal (magnet above the ship)?

The force you would feel is the bed holding you up. Granted it may feel like you're being pinned down with other objects not so much (if you dropped a metal object from the side of the bed, and then tried with another plastic object you'd notice).

The point of the equivalence prinicipal, and this is the most important point, there are no experiments/observations that will distinguish a uniformly accelerated frame of reference from a uniform gravitational field from inside that frame of reference, whereas clearly in the last two example you provide above, you can.

I'd say that gravity is a force. But in special relativity there are two kinds of force. Electromagnetism, normal forces, and the like are four-vector forces. Centrifugal and Coriolis forces can't be described by four-vectors; they're "tensor forces", the tensor in question being the metric tensor. When you add GR, gravity also turns out to be a tensor force. This distinction doesn't exist in Newtonian physics, where all of these forces are described by vectors.

The easiest way to tell the difference is probably to look for time dilation.

If you're standing on a planet (which is not rotating, to keep things simple), and attracted to it by (tensor) gravity, countered by the (vector) normal force of the ground, then ideal clocks at different heights will tick at different rates. If gravity was a vector force, the clocks would tick at the same rate.

Similarly, if you're pressed into your bed by the upward acceleration of a rocket ship, clocks at different heights will tick at different rates, but if there's a magnet in the floor attracting your adamantium skeleton, the clocks will tick at the same rate.

The answer to the first 2 of your questions is - you can't tell the difference...the only reason you could is by looking around or feeling the wind as you free fall. Floating in space is an inertial frame, and so is free falling

I think you have doubts about gravity not being a force... read this- it gives proof (in a very analogous way) that gravity is not a force and is a curvature in space time... it might conquer your doubts:-

Look at it this way,

All of us and the universe is moving forward in time and so everything is moving forward in the space time continuum. Think of gravity not as a force, but a curvature in the 4d rubber sheet of space time that slows down things travelling in the space time continuum. Why? Well heres an ANALOGY (notice I have bolded 'analogy' as it is important). In the 2d space time demo ( the one where people drop balls on a rubber sheet) when you drop balls on a rubber sheet, the rubber sheet curves towards the earth. So smaller balls move down to the lowest possible location of the rubber sheet (after that they don't climb over the 'hill curvature' created by rubber). And since the whole rubber sheet represents space and time, it has also stopped moving in time...That is very analogous to the way gravity slows down time. An object in an inertial frame of reference (I'm just going to call that 'O.I.F.R' from now on) is an object that is not affected by gravity...in other words its an object that climbs over the hill curvature I mentioned before. Derek in veritasium said that,(yes I watched the video too) because of the Normal force, an O.I.F.R sees something in gravity accelerating upwards. And that's what you see in the 2d space time analogy... When a smaller ball get to the minimum point it stops right? But an O.I.F.R sees it as accelerating away from it because an O.I.F.R climbs over the 'hill curvature' after reaching the bottom and keeps moving while the object in gravity stays still....

I hope your doubts were conquered... The reason I bolded 'analogy' was because the whole thing was an anology...not the real thing...the real thing happens in 4d space and is governed by Einstein's field equations...

By the way Dereks video on veritasium was what helped me understand this thing fully...I had questions like yours but the video helped me a lot...hope my explanation helped you

A pseudo-force or "fictitious force" is a force that arises from using a particular reference frame. In certain reference frames, fictitious forces will disappear - these reference frames are called inertial reference frames. The term "fictitious force" is somewhat confusing because a fictitious force will still have real effects if we are measuring position, velocity and acceleration relative to a non-inertial reference frame.

Gravity differs from the other three fundamental forces (electromagnetic force, strong force and weak force) because it exerts a force on every object that is proportional to the object's inertial mass, and so gives all objects the same acceleration (as long as we stick to close by or "local" observations). Other fundamental forces act on different fundamental particles in different ways - and not at all on some particles. In this respect, gravity is like frame-dependent forces such as centrifugal force or coriolis force - which is why it is described as a pseudo-force or fictitious force.

The equivalence principle says that you cannot tell the difference between accelerated motion and an equivalent gravitational field with local measurements. However, if you can see out of your spaceship then you could use non-local effects to distinguish between being on the surface of the earth and accelerating through space. For example, you could look for anisotropies in the cosmic microwave background radiation.