If I'm right then principle of equivalence says that there is no difference between the experiments performed in a room present in a gravitational field and an identical room accelerating upwards with an acceleration equal to the gravitational fiend. Or in other words, there is no difference between gravity and the pseudo force experienced in accelerating frames.

However, this can't be correct. I thought of this experiment:

Consider a very long room in which a rectangular body of non-uniform mass-density is initially present at the top in a horizontal position. As time passes, it falls from there. Clearly, it will experience a torque due to its non-uniform mass if the room is in a uniform gravitational field. But, if the room is just accelerating upwards, then the body won't experience torque because no forces are actually 'acting' on it.

I think forces are absolute. If there is a force acting on the room, then you can say that the room is at rest and there is an imaginary force on you. But this imaginary force won't have the effects that a force actually 'acting' on you will have. So, gravity should be different from pseudo forces because unlike pseudo forces, gravity actually acts on a body.

EDIT: Here's another example. There is a maximum acceleration that the human body can withstand. Now, if at $t=0$, I'm at the top-most position of a very long room and am allowed to fall from there. If there is a gravitational field in the room greater than the maximum acceleration my body can withstand, then my body will be shattered to pieces due to the large acceleration. However, if there is no gravitational field in the room, instead the room is accelerating upwards with an acceleration greater than the one my body can withstand, then there should be no effect on my body whatever the acceleration of the room is, because my body is completely disconnected from the room. So, gravity has effects different from pseudo forces.


First, note that you feel the effect of gravity on your body only when there is some surface pushing it back in the direction opposite to gravitational acceleration (here, I am assuming that the space within which the experiment is carried out, is small; that is, local, which is where equivalence principle holds). When you are freely falling in a gravitational field, you do not feel any g-forces, and that is the reason why astronauts feel weightless. According to general relativity (in contrast to Newtonian mechanics), bodies falling under a gravitational force (only) are not accelerating. Such freely-falling bodies follow geodesics in spacetime. Only when the body is somehow prevented to keep going along that geodesic does it 'feel' something, and accelerates. And that 'feeling' is manifested in the form of some stress acting on your body. Now, to answer your two thought experiment questions:

1) Take a large sphere of mass M and a smaller sphere with mass m, where m$<$M. Perform the anecdotal Leaning Tower of Pisa experiment by dropping both the spheres from the same height. As you know, they will reach the ground at the same time, which means that both spheres were moving through space at the exact same rate (equal accelerations). This shows that two bodies, no matter what their masses are, fall at the same rate. Now, join the two spheres by a massive rod and place the apparatus in a horizontal position (like in your question), and drop it from the Tower's top again. Because of what we have just seen, the rod will also fall exactly the way the two spheres are falling. You can then extrapolate this argument to an experiment conducted with a rectangular body of non-uniform mass density, and the result will still be unchanged: there is no rotation. There will also be no rotation in the case of an upward-accelerating room. But as pointed out before, in order to compare a 'room with gravity' experiment with a 'room with acceleration' experiment, you need to allow the body to feel the reaction force on it, which is possible if, suppose, you place the body on the floor and then perform your experiment.

Another way to look at it is to note that the (proposed) axis of rotation through the body falls at the same rate as all other particles comprising the body. So there's no rotation with respect to that axis.

2) Assume for a minute that the room is in complete vacuum (except your body, of course). When freely falling in that situation, you will not feel any shattering forces at all. You will feel weightless. So now, it does not matter whether you are freely falling in the room, or are suspended in the room while it is accelerating upward. In both cases, you will 'feel' nothing. But yes, you will feel a crushing force on your body if both these experiments are conducted when you are standing on the floor of the room. In this case, the floor will act on your body with a reaction force. Gravity is not deadly here, it's the surfaces around you that act upon you with reaction forces when you are trying to go along your geodesic but instead being offered resistance by them.

The equivalence principle stands safe and well.

  • $\begingroup$ so, what if we had a 10000km hight man and put him in a free fall in a gravitational field he will feel some stretch; because he will experience more gravitational force on his legs than on his head (unless he falls upside down), this wont be the case if we put him in a box (it's better to be a big one) in the 2nd case he'll never feel such a stretch. does the equivalence principle hold true in this case? (I'm sorry if you have mentioned this but if you did then i still didn't get your answer) $\endgroup$ – Khaled Oqab Feb 3 '19 at 20:23
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    $\begingroup$ @KhaledOqab In the first paragraph, you can see a point I made about the importance of making measurements locally, that is, in small enough regions of spacetime. This prevents us from considering tidal forces. $\endgroup$ – Avantgarde Feb 3 '19 at 21:56

Clearly, it will experience a torque due to its non-uniform mass if the room is in a uniform gravitational field.

This should not be clear to you, because it's not true. I apologize if the next section beats you over the head with mathematics, I want to show you why Newton's laws don't say that, and then I want to give you some immediate physical insight afterwards.

Newton's laws and the center of mass

Consider a system of point masses $m_i$ at position vectors $\mathbf r_i$ experiencing external forces $\mathbf F_i = \mathbf F_i(\mathbf r_i)$ and internal forces $\mathbf G_{ij} = \mathbf G_{ij}(\mathbf r_i, \mathbf r_j)$ which should obey $\mathbf G_{ij} = -\mathbf G_{ji}$ consistent with Newton's third law. Newton's laws say that these must obey the equations, $$m_i ~\ddot {\mathbf r}_i = \mathbf F_i + \sum_j \mathbf G_{ij}.$$ Obviously one of the things that we like to do is to sum all of these equations up and define the center of mass by defining $M = \sum_i m_i$ and then defining $\mathbf R = \sum_i (m_i/M) ~\mathbf r_i,$ leading to the equation, $$M~\mathbf R = \sum_i \mathbf F_i,$$ with the $\mathbf G_{ij}$ dropping out due to their antisymmetry in their respective indices. If you've never seen the trick, use $q_{ij} = -q_{ji}$ to replace $\sum_{ij} q_{ij}$ with $\frac12 \sum_{ij} (q_{ij} - q_{ji}),$ then expand this into two sums and relabel the second one $i \leftrightarrow j$ (they're just names of indices, after all) to find after recombining, $\frac12 \sum_{ij} (q_{ij} - q_{ij}) = \sum_{ij} 0 = 0.$

Okay if you're solid on all of those, let's talk about torques about the arbitrary origin we've chosen.

Torques and angular momentums

We know these are defined for a force as taking the cross product between position and the force, so that suggests that above we need to try to work with those in Newton's laws, as $$m_i~\mathbf r_i\times \ddot{\mathbf r}_i = \mathbf r_i \times \mathbf F_i + \sum_j {\mathbf r_i \times \mathbf G_{ij}}.$$ We want to do something with both of these sides. The left hand side looks like the product of a thing with its second derivative, which looks like it might be related to a derivative of a product of a thing and its first derivative. Working it out we can actually see that for the cross product it's not just a relationship, it's an equality; the fact that any vector crossed with itself is 0 leads to$$\frac{d}{dt} (\mathbf v \times \dot{\mathbf v}) = \dot{\mathbf v}\times \dot{\mathbf v} + \mathbf{v} \times \ddot{\mathbf v} = \mathbf 0 + \mathbf v\times \ddot{\mathbf v}.$$ In turn defining the angular momentum about the origin $\mathbf L_i = m_i \mathbf r_i \times \dot{\mathbf r}_i$ and assuming $\dot m_i = 0$ as usual leads to the left hand side being just $\dot{\mathbf L}_i.$ For the right hand side, we can define $\tau_i = \mathbf r_i \times \mathbf F_i$ as the external torque on particle $i$, and $\mathbf L = \sum_i \mathbf L_i$ and $\mathbf T = \sum_i \mathbf \tau_i.$ We're ready to sum over $i$ to find,$$\dot {\mathbf L} = \mathbf T + \sum_{ij} \mathbf r_i \times \mathbf G_{ij}.$$

Central force motion

Now we want to try the same "antisymmetry trick" on the second half; under a $\frac12 \sum_{ij}$ symbol we have $\mathbf r_i \times \mathbf G_{ij} - \mathbf r_i \times \mathbf G_{ji}$ and under $i\leftrightarrow j$ relabeling this becomes $$\dot {\mathbf L} = \mathbf T + \frac12 \sum_{ij} (\mathbf r_i - \mathbf r_j)\times \mathbf G_{ij}.$$ Now it's not 100% of all possible systems, but in the largest class of systems that we care about, the interaction force $\mathbf G_{ij}$ points along the line connecting $j$ and $i$. This is true for the gravitational force, for the Coulomb force, or even if we make this thing out of very rigid massless struts connecting the little masses. So in the fast majority of cases (but not all!) we have $\mathbf G_{ij}\propto \mathbf r_i - \mathbf r_j$ and the latter term is 0. We have just $\dot {\mathbf L} = \mathbf T.$ These are known as "central forces", and I'm going to assume that your entire mass can be regarded as a bunch of point masses held together by massless struts and gravitational self-interaction and electromagnetic forces, all the forces are "central" in the sense that they act between two masses pointing along the line connecting them. As the above argument shows, they also therefore cannot generate net torque. (That's not what you were interested in anyway, you thought that the external field was going to torque these things, but I guess I'm just saying that external gravity also can't easily influence the constituent parts to torque each other.)

A uniform gravitational field

Whew! Okay, math rant is almost done! Now we just need to apply the above equations. Consider if $\mathbf F_i = m_i \mathbf g$ for some uniform gravitational acceleration $\mathbf g$. Then these two crucial equations are: $$ M~\ddot{\mathbf R} = \sum_i m_i~\mathbf g = M~\mathbf g,\\ \dot {\mathbf L} = \sum_i m_i ~\mathbf r_i \times \mathbf g = M~\mathbf R \times \mathbf g.$$ Do you see where I'm going here? Use that first to substitute into the second to find $\mathbf R \times \ddot{\mathbf R}$ which we know is just $\frac d{dt} (\mathbf R \times \dot{\mathbf R})$ and so we can integrate once, $\mathbf L = M~\mathbf R \times \dot{\mathbf R} + \mathbf C_0.$ However we also have $\mathbf R = \frac 12~\mathbf g~t^2 + \mathbf C_1~t + \mathbf C_2.$

We can use our choice of reference frame to set $\mathbf C_1 = \mathbf C_2 = \mathbf 0$ and in this special reference frame where the center of mass starts off at rest at the origin, we find that $\mathbf R \propto \dot{\mathbf R}$ and therefore $\mathbf L = \mathbf C_0.$ The angular momentum about the starting point for the center of mass is, in fact, a constant, no matter how the mass is non-uniformly spread.

Physical insight

In retrospect this should not really surprise you all that much. You know that everything falls at the same rate: fill up two water bottles, one full-up, one half-full, drop them side by side, and you'll notice that within experimental error they will hit the ground at the same time when dropped side-by-side. One has nearly twice the mass of the other, but their falling profiles are identical.

Now you are proposing that if you put a thin, massless rod between them to "connect" them, then as if by magic, one of them will want to fall faster than the other and they will not land side-by-side. But what is this rod going to do? It's going to communicate forces horizontally. And what are you claiming it does? Well if they start falling differently vertically from how they were otherwise going to fall, then it must communicate a vertical force. So that's the tension between your pre-experimental intuitions and how experiments show the world actually works.

I strongly, strongly encourage you to try the experiment with the plastic water bottles, or something similar where you've got two very different masses dropped side-by-side but you don't care about them breaking when they hit the floor. (You might try a coin alongside a water bottle for example.) Build up this intuition, it can serve you very well.

  • $\begingroup$ +1 Ok, so no torque acts on non-uniform bodies in a gravitational field. Would you explain the other experiment I've mentioned in the edit? $\endgroup$ – Dove Apr 22 '17 at 5:13
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    $\begingroup$ @Dove: Briefly, acceleration isn't actually harmful (or even perceptible) to humans, as long as every part of our body is accelerated at the same rate. What's harmful to us is uneven acceleration, causing different parts of our body to be squeezed together or pulled apart. When we say something like "humans can stand up to X gees of acceleration", what we really mean is that we can tolerate having an external surface (e.g. an airplane seat) push against our body with a force equal to X times our mass times 9.81 m/s/s. $\endgroup$ – Ilmari Karonen Apr 22 '17 at 11:41
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    $\begingroup$ What @IlmariKaronen said. I gave you the experiment for the bottles side-by-side, the same holds for bottles above-and-below or in any configuration: if two masses start with the same velocity in a uniform acceleration, $$\mathbf r_{0,1}(t)=[x_{0,1}+v_x~t,~~y_{0,1}+v_y~t,~~z_{0,1}+v_z~t-\frac12at^2]$$ then the displacement between them $\mathbf r_1(t)-\mathbf r_0(t)=[x_1-x_0,~y_1-y_0~,z_1-z_0]$ is constant. Engineering-wise this means the strains are constant which means there are no added stresses, see this answer for the basic idea behind those. $\endgroup$ – CR Drost Apr 22 '17 at 16:39

I remember asking the same question at uni. My example was about a pole orbiting in free fall. does it spin?

The answer is that you are (sort of) right. The key phrase you are missing from the law is that its only true "locally"

In your first example of the long box, or lets say unequal bar bell. there is no torque, the heavy side doesnt fall faster than the lighter

In your second, the human body is only 'ripped apart' if unequal forces apply to its component parts, either because something resists its movement, which in the classic example nothing does untill it hits the box, or if the gravitational gradient is high enough that your legs are pulled harder than your head. in which case the experiment is no longer "local" as the field is significantly non uniform.

from wikipedia:

Locality eliminates measurable tidal forces originating from a radial divergent gravitational (e.g., the Earth) upon finite sized physical bodies. The "falling" equivalence principle embraces Galileo's, Newton's, and Einstein's conceptualization. The equivalence principle does not deny the existence of measurable effects caused by a rotating gravitating mass (frame dragging), or bear on the measurements of light deflection and gravitational time delay made by non-local observers.

So basically yes, the law is incorrect for real life situations because gravitational forces are not uniform in real life. Where as its easy to imagine uniformly accelerating boxes.

In its defence the genius of the law is much deeper. I certainly don't fully understand the implications, but you can see they are profound when it comes to stop watches on trains, people on top of mountains, transmitting time signals from gravity wells and the like.


The non-uniform body will experience torque both in gravitational field and inside an elevator, but only from the point of view of inertial frame. In the reference frame where the body is always at rest, the net torque on the body is zero; the gravity force will be cancelled by inertial force -ma.

There is no limit to acceleration that human can survive if the acceleration is due to gravity. Some limit applies only to situations where the acceleration is due to mechanical contact forces of the seat or other body that makes the human accelerate through mechanical contact.

  • $\begingroup$ But what about black holes? Humans can't withstand the large gravitational forces of black holes. $\endgroup$ – Dove Apr 22 '17 at 8:23
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    $\begingroup$ @Dove you're confusing tidal forces with uniform gravitational field. Principle of equivalence only considers the latter. $\endgroup$ – Ruslan Apr 22 '17 at 8:28
  • $\begingroup$ But all the cells of the human body will have to move in the direction of force. The internal forces are what that keep the cells together while movement. They could be very weak compared to the large external gravitational force even if it is not tidal. $\endgroup$ – Dove Apr 22 '17 at 8:32
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    $\begingroup$ Ján, I would add a word to your statement for clarity: "There is no limit to acceleration that human can survive if the acceleration is due to uniform gravity." Without saying uniform gravity a confusion like the one asked in the comment arises. @Dove, the thing is that near black holes, gravity is not uniform. You are being pulled more by your feet than by your head, and as such your feet will accelerate faster and you body ripped apart. The keywords here are uniform or non-uniform gravity $\endgroup$ – Steeven Apr 22 '17 at 8:56
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    $\begingroup$ @Dove That does not matter, because they will all be accelerated the same. Only if the gravitational field is different over the whole body do you get problems. $\endgroup$ – Graipher Apr 22 '17 at 11:00

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