# Why can't we half-close our eyes?

Can we half-close our eyes such that they don't vibrate?

If not, then why we are unable to half-close our eyes properly?

Alternatively, why isn't it possible to keep the upper eyelid close to the lower eyelid without it vibrating?

The motion of the eyelid is driven by the levator palpebrae superioris, i.e. elevating muscle of upper eyelid, and it can be positioned to the intermediate, half-closed positions. At least I can do it.

Yes, this muscle tends to shake or oscillate simply because it's a very weak muscle. It's similar with the muscles controlling the motion of the little finger (or ring finger). Vibration and "noise" is present in all muscles but the stronger muscle you have, the more accurately the random motion is averaged out.

If you connected $N$ eyelid muscles so that they move in unison, the noisy vibrations would be reduced by the factor of $\frac{1}{\sqrt{N}}$. This is the well-known scaling of the statistical errors in physics etc.

I would think that this question was totally appropriate at the Physics Stack Exchange.

• What is the number N in this context? A muscle, even a small one as that of the eyelid, contains countless number of muscle fibers, and even more myofibrils ... Jan 16, 2017 at 11:22
• Roland, it isn't supposed to matter how much $N$ is. You may take $N$ to be the number of fibers. The point is that what matters is the ratio $N_1/N_2$ between two situations. The eyelid muscle is a lot thinner ($N_1/N_2$ is large) than the biceps so the relative oscillations are square root of lot times more pronounced in eyelids. The oscillations only appear in the middle because the eyelids are "supported" in both extreme positions. When a person with Parkinson's pushes his hand on the table, it cannot vibrate up and down... Jan 16, 2017 at 14:12
• That doesn't sound quite right to me. For very large N1, the statistical error (and therefore muscle tremor) should vanish, and then it's irrelevant if some other N2 is even larger than N1. So the 1/root(N) law should be important only when N is quite small. I would guess that the key factor is how many neurons innervate the muscle; very small muscles have few motor neurons, which makes for a small N. Can you elaborate and provide references? Jan 16, 2017 at 20:20
• Dear Roland, the number of neurons innervating an elementary muscle fiber was assumed to be the same while comparing the quivers of $N_1$ and $N_2$ nerves. It's plausible that the source of the oscillations is the small number of neurons around the muscles, not the thin mechanical part of the muscle. In that case, $N$ in my answer should be interpreted as the thickness of the nerves innervating the muscle. From a broader viewpoint, it's a detail. No, I don't have references, this was being answered as a physics problem which can be done even if no one else was solving it before. Jan 18, 2017 at 7:56
• I think this is not the cause, the muscle is already very much larger in terms of number of myosin bridges compared to the scale at which you'd expect oscillations of noticeable amplitude. And also, the motion is controlled by 2 muscles for the upper eyelid, and 1 for the lower but at a different angle than the levator.
– Joce
Jan 21, 2017 at 21:38

The upper eyelid position is governed by two muscles (tarsus sup and levator palpebrae sup) working against one another, but they are at a 90 degree angle: When you are closing your eyelid, it is the tarsus that contracts and reduces its length. If you try to maintain the upper lid in a half-closed position, you need to counteract with the levator muscle. But because of the nearly 90 degree angle, a little force of the levator needs to be counterbalanced with a strong levator force, simply because of trigonometry: You see above (upside down compared to eyelid muscles) that for a constant force of the levator (F, down), the force of the tarsus to the left and right need to be doubled when you go from 30 to 20 degree angle of the "eyelid". These variations grow even larger when the angle goes flatter, in fact, it is a $1/\tan \alpha$ singularity [EDIT]as $\alpha$ goes to zero[/EDIT]. Tough maths problem, and also tough control problem for your nerves!

And definitely, this question is very much physics!

• Very interesting elaboration, thanks. But could you please be a little bit more specific which singularity appears near 30 or 20 degrees in tangent etc.? You surely don't dispute that tangent of 30 degrees is regular, finite, and of order one, do you? Jan 22, 2017 at 9:23
• @LubošMotl If $|F|=1$, then $|F_L|=|F_R|=1/(2 \tan \alpha)$. It is $1/\tan$ which is singular, as you well know.
– Joce
Jan 23, 2017 at 12:58
• Sorry, $\tan\alpha$ is singular for $\alpha=\pi/2$ or 90 degrees. None of your angles is 90 agrees, is it? Similarly, $1/\tan\alpha$ diverges for $\alpha=0$ but your $\alpha$ isn't zero. So there's nothing singular. Are you playing a shell game trying to sell an argument that obviously doesn't work or am I overlooking something? Jan 24, 2017 at 14:25
• OK, maybe this was unclear: we are looking for nearly closed eye, so nearly straight polyline joining $F_R$ and $F_L$, hence $\alpha$ small. I realise you have taken the angles in the diagrams as the only values considered, they are examples.
– Joce
Jan 24, 2017 at 17:12