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.

## Best Answer

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 theexternal fieldwas 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

applythe 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,

stronglyencourage 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.