The physics of sneezing

July 7, 2020. I play around with very crude models of a sneeze. Including drag, the results are consistent with the WHO guidelines. But more importantly, they exhibit the fascinating physics lurking in this humble (if epidemiologically charged) phenomenon!

Ballistic droplets

When you sneeze, you eject a range of particles at a range of speeds. To start with, we will consider the simple case of ballistic droplets, where droplets are subject only to gravity. Suppose your mouth is at height $h$ above the ground. The time it takes for a droplet to fall to the ground $t_\text{G}$ under the influence of gravity is given by

\[h = \frac{1}{2}gt_\text{G}^2 \quad \Longrightarrow \quad t_\text{G} = \sqrt{\frac{2h}{g}}.\]

The range is the horizontal distance covered in this time. If you impart speed $v_0$ to the droplet, then the range $R$ is

\[R = v_0t_\text{G} = v_0 \sqrt{\frac{2h}{g}}.\]

Let’s plug in some numbers and see if the range is reasonable. Droplets can be emitted at speeds of up to $v_0 = 100 \text{ m/s}$. A reasonable height is $h \approx 2 \text{m}$, and $g = 9.8 \text{ m/s}^2$. So we get a range of

\[R = 100 \sqrt{\frac{2 \cdot 2}{9.8}} \text{ m} \approx 50 \text{ m}.\]

This is obviously much too large! It gets even worse if we change the firing angle.

Exercise 1. (a) Show that if fired at an angle $\theta$ to the horizontal, then the droplet reaches the top of the parabolic arc at time

\[t_\text{top} = \frac{v_0\sin\theta}{g}.\]

(b) Show that, for angle $\theta$, the time it takes for the droplet to fall is

\[t_\text{G} = t_\text{top} + \sqrt{\frac{2h}{g} + t_\text{top}^2}.\]

(c) Finally, give an expression for the range $R$.

(d) How far does a ballistic droplet travel for $\theta = 45^\circ$? Use $v = 100 \text{ m/s}$ and $h = 2\text{ m}$.

Stokes’ law

Obviously, we’ve missed something big. The WHO is not instructing us to stay $50$ m away from one another! In reality, droplets are subject to air resistance. This is a force directed against motion, and which increases with velocity. In the simplest case, the force is directly proportional to the velocity:

\[\mathbf{F}_\text{drag} = -\gamma \mathbf{v}.\]

Solving the problem in two dimensions is a little tricky, so I will use a very rough approximation, and see how far it moves in one dimension, but in the time $t_\text{G}$ we discussed in the last section. Newton’s second law is

\[m \dot{v} = F_\text{drag} = -\gamma v.\]

We can solve immediately using an exponential,

\[v(t) = v_0 e^{-\gamma t/ m}.\]

To find the position at time $t = t_\text{G}$, we have to integrate once:

\[x(t_\text{G}) = \int_0^{t_\text{G}} dt \, v(t) = v_0 \int_0^{t_\text{G}} dt \, e^{-\gamma t/ m} = \frac{mv_0}{\gamma} \left(1 - e^{-\gamma t_\text{G}/m}\right).\]

The particle slows down exponentially and gradually approaches the maximum distance of $mv_0/\gamma$. Let’s plug in some numbers and see what this maximum distance is. To find $m$, we need to know that droplets are mostly water, with density $ \rho_\text{water} \approx 10^3 \text{ kg/m}^3$ and have typical size $2r \sim 100 \,\mu\text{m}$, with mass

\[m = \rho_\text{water} \cdot \frac{4\pi}{3}r^3 \approx 5.2 \times 10^{-10} \text{ kg}.\]

Finally, we need to find $\gamma$. This is the trickiest part. For a slowly moving droplet, the drag is given by Stokes’ law:

\[\gamma = 6\pi \eta r \approx 1.7 \times 10^{-8} \text{ kg/s},\]

where $\eta = 1.8\times 10^{-5} \text{ kg/m s}$ is the viscosity of air and $r$ the droplet radius. This gives a maximum distance of

\[x_\text{max} = \frac{m v_0}{\gamma} \approx \frac{(5.2 \times 10^{-10}) 100}{(1.7 \times 10^{-8}) } \text{ m} \approx 3 \text{ m}.\]

This looks much better! If we plug in the time to ground from the previous section, $t_\text{G} = \sqrt{2h/g} \approx 0.6 \text{ s}$, we get more or less the same answer. This is still a bit larger than the WHO recommendations. You can check in the next exercise that the two-dimensional variant gives the same maximum distance.

Exercise 2. Let’s solve the problem in two dimensions, and include a nonzero firing angle for the heck of it. The velocity is denoted $\mathbf{v} = (v_x, v_y)$, with $\mathbf{v}_0 = v_0 (\cos\theta, \sin\theta)$. The equation of motion is

\[m \dot{\mathbf{v}} = -mh \hat{\mathbf{y}} - \gamma \mathbf{v}.\]

(a) Separate into $x$ and $y$ components. Solve for $v_x$, and show that

\[v_x(t) = v_0 \cos\theta e^{-\gamma t/mg}.\]

(b) Similarly, show that

\[v_y(t) = v_0\sin\theta e^{-\gamma t/mg} - \frac{mg}{\gamma}\left(1 - e^{-\gamma t/mg}\right).\]

(c) Integrate your result from part (a) to find

\[x(t) = \frac{m v_0 \cos\theta}{\gamma} (1 - e^{-\gamma t/mg}).\]

(d) Argue that the range $R \sim m v_0/\gamma$ by dimensional analysis.

Thus, although the time dependence is different, the maximum distance is the same as the one-dimensional case up to the cosine. This agrees with the dimensional analysis. Incidentally, this maximum distance is largest for a horizontally fired droplet, unlike pure ballistic motion where $\theta = 45^\circ$ is optimal.

Reynolds number and quadratic drag

Our answer is a wee bit too big, so perhaps something was wrong with our drag force. In fact, to use Stokes’ law, the droplet needs to be moving slowly. When it moves fast, there is different physics at play, and the drag is quadratic in $v$. To tell the difference, we have to quantify whether the fluid feels sticky (linear drag) or bumpy (quadratic drag) to the droplet. In general, both types of drag are present, but we will make the approximation of considering only whichever one is dominant.

The Reynolds number is the ratio of bumpiness to stickiness, defined as

\[\text{Re} = \frac{\rho v D}{\eta}.\]

where $D = 2r$ is the diameter of the droplet and $\rho$ is the density of the fluid. When $\text{Re} < 1$, Stokes’ law applies, but otherwise we should use quadratic drag. Air has density around $\rho = 1.2 \text{ kg/m}^3$. The other parameters appeared in the preceding section, and plugging them all in gives

\[\text{Re} = \frac{\rho v D}{\eta} = \frac{1.2 \cdot 100 \cdot 10^{-4}}{1.8 \times 10^{-5}} \approx 700.\]

Stokes’ law is definitely not valid! We should be using the quadratic form for the drag force,

\[\mathbf{F}_\text{drag} = - \alpha v^2 \hat{\mathbf{v}},\]

In two dimensions, this cannot be solved analytically, at least if we want everything as a function of time. (See the Appendix for a different approach.) To simplify, let’s solve the toy one-dimensional version of the problem. The equation of motion is

\[F_\text{drag} = m\dot{v} = - \alpha v^2.\]

This is easily solved by separation of variables:

\[-\frac{dv}{v^2 } = \frac{\alpha}{m} dt \quad \Longrightarrow \quad v(t) = \frac{m}{\alpha t + mv_0^{-1}}.\]

Integrating gives a logarithm:

\[x(t) = \int_0^{t} dt' \, v(t') = \frac{m}{\alpha}\log\left[1 + \frac{\alpha v_0 t}{m}\right].\]

It seems there is no longer any maximum distance! This is because $v^2$ shrinks faster than $v$ at small $v$, so that at slow speeds the droplet experiences much less drag. Finally, $\alpha$ is to a good approximation given by

\[\alpha \approx \frac{1}{2}\rho A = \frac{1}{2}\pi \rho r^2,\]

where $\rho$ is the density of air. (In fact, there is dimensionless drag coefficient sitting out front of this which depends on the Reynolds number, but we’ll ignore it.)

Linear and quadratic drag

Of course, there is a maximum distance — once the particle slows down enough, the linear drag begins to dominate! This will happen around $\text{Re} = 1$, or

\[\text{Re} = \frac{2\rho v_\text{L} r}{\eta} = 1 \quad \Longrightarrow \quad v_\text{L} = \frac{\eta}{2\rho r} \approx 0.15 \text{ m/s}.\]

We can find the time $t_\text{L}$ at which this happens using our equation for velocity in the quadratic drag regime:

\[v(t_{\text{L}}) = \frac{m}{\alpha t_{\text{L}} + mv_0^{-1}} = v_{\text{L}} \quad \Longrightarrow \quad t_{\text{L}} = \frac{m}{\alpha}\left(v_{\text{L}}^{-1} - v_0^{-1}\right).\]

Plugging this into our formula for distance, along with the recipe for $\alpha$, gives

\[x(t_\text{L}) = \frac{2m}{\pi \rho r^2}\log\left(\frac{v_0}{v_\text{L}}\right).\]

We can plug all our numbers in and see if we get a more sensible result. Note that this won’t be the full answer, just how far the droplet moves before Stokes’ drag kicks in:

\[x(t_\text{L}) \approx \frac{2(5.2 \times 10^{-10})\text{m}}{1.2 \pi(5 \times 10^{-5})^2}\log\left(\frac{100}{0.15}\right) \approx 0.7 \text{ m}.\]

This is looking very good! To see how much further the droplet travels, we take our results for Stokes’ drag and replace $v_0$ with $v_\text{L}$. This leads to a maximum distance

\[x_\text{max} = \frac{mv_\text{L}}{\gamma} \approx 4.5 \text{ cm}.\]

In total, the droplet travels around $75 \text{ cm}$, which is pretty close to the WHO guidelines!

Exercise 3. (a) Show that the range as a function of droplet size $r$ is

\[R(r) = \frac{2m}{\pi \rho r^2} \left[\log\left(\frac{2 \rho r v_0}{\eta}\right) + \frac{1}{24}\right].\]

(b) Show that for small particles, this model gives nonsensical results, and explain why the physical reasoning breaks down.


Our simple model gives reasonable results in accord with WHO recommendations, at least for large particles. For smaller particles the model is obviously broken. But as the world ponders the epidemiological significance of the ballistic range of a sneeze, it seems timely to explore some of the basic physics as well!

Appendix: Bernoulli’s approach to polynomial drag

Here, I review Bernoulli’s beautiful solution to the problem of polynomial drag, one of the most thorny mathematical problems of ballistics. Or at least, I will add an appendix soon!