# Some confusion regarding the equivalence principle

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.

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.