Flow(#2) – Plateau- Rayleigh Instability

August 9, 2019December 2, 2020 / 153armstrong / Leave a comment

When you wake up in the morning and open/close your faucet when you brush your teeth, you might have noticed that it undergoes a transition between a smooth jet to a dripping flow like so :

When the velocity of the fluid exiting the faucet is high, it appears smooth for a longer time before it breaks into droplets:

But when you make the velocity of the fluid exiting the faucet low, it seems to form droplets much earlier than before.

Here’s the water breaking into smaller droplets shot in slow motion:

Notice that just by changing the exit velocity of the water you can control when the droplets form.

You can also control the nature of the droplets that form by changing fluids. Here’s how it looks like if you use water as the fluid (left), pure glycerol(center) and a polymeric fluid(right).

What is causing a jet of fluid to form droplets?

A simple answer to this is perturbations on the surface of the fluid. What does that mean?

Initially the fluid is just falling under the influence of gravity. And velocity of any freely falling object increases as it falls:

But the surface tension of the fluid holds the molecules of the fluid together as they fall down.

Therefore depending on the initial velocity of fluid, the surface tension of the fluid and the acceleration you get a characteristic shape of the jet as it falls down:

This is what you observe as the fluid exits the faucet.

Perturbations

Just after exiting the faucet, there are tiny perturbations on the surface of this fluid as it falls down. This is apparent when you record the flow at 3000fps:

Source: engineerguy

Those tiny perturbations on the surface of the fluid grow as the fluid falls down i.e the jet becomes unstable.

And as a result the fluid jet breaks down to smaller droplets to reach a more thermodynamically favorable state. This is known as the Plateau-Rayleigh instability.

Source

It takes different fluids different time scales to reach this instability. This depends on the velocity of the fluid, the surface tension and the acceleration it experiences.

And some viscous fluids like honey are also able to dampen out these perturbations that occur on their surface enabling them to remain as fluid thread for an extended time.

A note on inkjet printers

By externally perturbing the fluid instead of making the fluid do its own thing, you can make droplets of specific sizes and shapes.

This engineerguy video explains how this is used in inkjet printers in grand detail. Do check it out.

We started today by trying to understand why water exiting a faucet behaves the way it does. Hopefully this blog post has gotten you a step closer to realizing that. Have a great day!

Sources and more:

* This is a topic that is home to a lot of research work and interesting fluid dynamics. If you like to explore more take a look into the mathematical treatment of this instability here.

Cooking up a Lorentz invariant Lagrangian

November 23, 2018December 30, 2020 / 153armstrong / 1 Comment

Let’s consider a scalar field, say temperature of a rod varying with time i.e $T(x,t)$ . (something like the following)

w1hldkr

We will take this setup and put it on a really fast train moving at a constant velocity $v$ (also known as performing a ‘Lorentz boost’).

train-animated-gif-3

Now the temperature of the bar in this new frame of reference is given by $T^{'}(x^{'}, t^{'})$ where,

$x^{'}(x,t) = \gamma \left( x - vt \right)$

$t^{'}(x,t) = \gamma \left( t - \frac{v x}{c^{2}} \right)$

Visualizing the temperature distribution of the rod under a Length contraction.

Temperature is a scalar field and therefore irrespective of which frame of reference you are on, the temperature at each point on the rod will remain the same on both the frames i.e

$T^{'}(x^{'}, t^{'})= T(x, t)$

Therefore we can say that Temperature (a scalar field) is Lorentz invariant. Now what other quantities can we make from T that would also be Lorentz invariant ?

Is $\nabla . T^{'}(x^{'}, t^{'})= \nabla . T(x, t) ?$

Well, let’s give it a try:

$\frac{\partial T}{\partial x} = \frac{\partial T^{'}}{\partial x^{'}} \frac{\partial x^{'}}{\partial x} + \frac{\partial T}{\partial t^{'}} \frac{\partial t^{'}}{\partial x}$

$\frac{\partial T}{\partial x} = \frac{\partial T^{'}}{\partial x^{'}} \gamma - \frac{\partial T}{\partial t^{'}} \gamma v$

$\frac{\partial T}{\partial x} = \gamma \left( \frac{\partial T^{'}}{\partial x^{'}} - \frac{\partial T}{\partial t^{'}} v \right)$

——–

$\frac{\partial T}{\partial t} = \frac{\partial T^{'}}{\partial x^{'}} \frac{\partial x^{'}}{\partial t} + \frac{\partial T^{'}}{\partial t^{'}} \frac{\partial t^{'}}{\partial t}$

$\frac{\partial T}{\partial t} = - \frac{\partial T^{'}}{\partial x^{'}} \gamma v + \frac{\partial T^{'}}{\partial t^{'}} \gamma$

$\frac{\partial T}{\partial t} = \gamma \left( - \frac{\partial T^{'}}{\partial x^{'}} v + \frac{\partial T^{'}}{\partial t^{'}} \right)$

Clearly, **

$\frac{\partial T}{\partial x} + \frac{\partial T}{\partial t} \neq \frac{\partial T^{'}}{\partial x^{'}} + \frac{\partial T^{'}}{\partial t^{'}}$

But just for fun let’s just square the terms and see if we can churn something out of that:

$\left( \frac{\partial T}{\partial x} \right)^{2} = \gamma^{2} \left( \frac{\partial T^{'}}{\partial x^{'}} - \frac{\partial T^{'}}{\partial t^{'}} v \right)^{2}$

$\left( \frac{\partial T}{\partial t} \right)^{2} = \gamma^{2} \left( - \frac{\partial T^{'}}{\partial x^{'}} v + \frac{\partial T^{'}}{\partial t^{'}} \right) ^{2}$

We immediately notice that:

$\left( \frac{\partial T}{\partial t} \right)^{2} - \left( \frac{\partial T}{\partial x} \right)^{2} = \gamma^{2} \left[ \left( - \frac{\partial T^{'}}{\partial x^{'}} v + \frac{\partial T^{'}}{\partial t^{'}} \right) ^{2} - \left( \frac{\partial T^{'}}{\partial x^{'}} - \frac{\partial T^{'}}{\partial t^{'}} v \right)^{2} \right]$

$\left( \frac{\partial T}{\partial t} \right)^{2} - \left( \frac{\partial T}{\partial x} \right)^{2} = \left( \frac{\partial T^{'}}{\partial t^{'}} \right)^{2} - \left( \frac{\partial T^{'}}{\partial x^{'}} \right)^{2}$

Therefore in addition to realizing that $T$ is Lorentz invariant, we have also found another quantity that is also Lorentz invariant. This quantity is also written as $\partial_{\mu} T \partial^{\mu} T$ .

** There is a very important reason why this quantity did not work out. This post was inspired in part by Micheal Brown’s answer on stackexchange . I request the interested reader to check that post for a detailed explanation.

Euler-Lagrange equations – Field Theory

November 19, 2018 / 153armstrong / Leave a comment

Consider a function $\phi(x)$ where $x$ is a point in spacetime. If we assume that the Lagrangian is dependent on $\phi(x)$ and its derivative $\partial_{\mu} \phi(x)$ . The action is then given by,

$S = \int d^{4}x \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x))$

According to the principle of least action, we have:

$\delta S = 0 = \int d^{4}x \delta \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x))$

$\delta \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x)) = \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \delta(\partial_{\mu} \phi)$

According to product rule,

$\frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \partial_{\mu} (\delta\phi) + \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \right) \delta\phi = \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \delta\phi \right)$

Therefore,

$\delta \mathcal{L}(\phi(x) , \partial_{\mu} \phi(x)) = \left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \right) \right)\delta \phi + \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \delta\phi \right)$

$\delta S = \int d^{4}x \left[ \left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \right) \right)\delta \phi + \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \delta\phi \right) \right]$

The second term in that equation is zero because the end points of the path are fixed i.e

20181118_210223

This then leads to the Euler-Lagrange equation for a field as :

$\frac{\partial \mathcal{L}}{\partial \phi} = \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial( \partial_{\mu} \phi)} \right)$

Parallax method, 61-Cygni and the Hipporcas mission

November 14, 2018November 24, 2018 / 153armstrong / Leave a comment

It is trivial for most astronomy textbooks to illustrate the parallax method as follows:

This is absolutely fascinating, but it was really hard to find actual images of stars in books that illustrate this.

This is the proper motion of 61-Cygni, a binary star system over a span of couple of years.

61 Cygni showing proper motion at one year intervals

But Bessel discovered that in addition to this proper motion, 61-Cygni also wobbled a little bit from side to side because of the parallax during observation.

The following is a plot of the motion of 61 Cygni – A which beautifully elucidates the proper motion and the effect of parallax (i.e the wiggle of the blue line with respect to the mean free path)

In addition, if you would like to actually play around with data for yourself, the The Hipparcos Space Astrometry Mission might interest you a lot. The mission was Launched in August 1989 and successfully observed the celestial sphere for 3.5 years before operations ceased in March 1993 employ

The documentation and the catalogue are fairly clear , instructive and easy to use. Have fun!

Video – 61-Cygni and it’s importance
More proper motions of stars animations can be found here.
1838: Friedrich Bessel Measures Distance to a Star

Using Complex numbers in Classical Mechanics

April 18, 2018April 18, 2018 / 153armstrong / Leave a comment

When one is solving problems on the two dimensional plane and you are using polar coordinates, it is always a challenge to remember what the velocity/acceleration in the radial and angular directions ( $v_r , v_{\theta}, a_r, a_{\theta}$ ) are. Here’s one failsafe way using complex numbers that made things really easy :

$z = re^{i \theta}$

$\dot{z} = \dot{r}e^{i \theta} + ir\dot{\theta}e^{i \theta} = (\dot{r} + ir\dot{\theta} ) e^{i \theta}$

From the above expression, we can obtain $v_r = \dot{r}$ and $v_{\theta} = r\dot{\theta}$

$\ddot{z} = (\ddot{r} + ir\ddot{\theta} + i\dot{r}\dot{\theta} ) e^{i \theta} + (\dot{r} + ir\dot{\theta} )i \dot{\theta} e^{i \theta}$

$\ddot{z} = (\ddot{r} + ir\ddot{\theta} + i\dot{r}\dot{\theta} + i \dot{r} \dot{\theta} - r\dot{\theta}\dot{\theta} )e^{i \theta}$

$\ddot{z} = (\ddot{r} - r(\dot{\theta})^2+ i(r\ddot{\theta} + 2\dot{r}\dot{\theta} ) )e^{i \theta}$

From this we can obtain $a_r = \ddot{r} - r(\dot{\theta})^2$ and $a_{\theta} = (r\ddot{\theta} + 2\dot{r}\dot{\theta})$ with absolute ease.

Something that I realized only after a mechanics course in college was done and dusted but nevertheless a really cool and interesting place where complex numbers come in handy!

Prof.Ghrist at his best!

July 28, 2017July 28, 2017 / 153armstrong / Leave a comment

Screenshot from 2017-07-28 10:55:27

To understand why this is true, we must start with the Fundamental Theorem of Vector calculus. If $F$ is a conservative field ( i.e $F = \nabla \phi$ ), then

$\int\limits_{A}^{B} F.dr = \int\limits_{A}^{B} \nabla\phi .dr = \phi_{A} - \phi_{B}$

What this means is that the value is dependent only on the initial and final positions. The path that you take to get from A to B is not important.

Screenshot from 2017-07-28 11:29:46

Now if the path of integration is a closed loop, then points A and B are the same, and therefore:

$\int\limits_{A}^{A} F.dr = \int\limits_{A}^{A} \nabla\phi .dr = \phi_{1} - \phi_{1} = 0$

Now that we are clear about this, according to Stokes theorem the same integral for a closed region can be represented in another form:

$\int_{C} F.dr = \int\int_{A} (\nabla X F) .\vec{n} dA = 0$

From this we get that Curl = $\nabla X F = 0$ for a conservative field (i.e $F = \nabla \phi$ ). Therefore when a conservative field is operated on by a curl operator ( $\nabla X$ ), it yields 0.

Bravo Prof.Ghrist! Beautifully said 😀

On the direction of the cross product of vectors

May 8, 2017May 8, 2017 / 153armstrong / Leave a comment

One of my math professors always told me:

Understand the concept and not the definition

A lot of times I have fallen into this pitfall where I seem to completely understand how to methodically do something without actually comprehending what it means. And only after several years after I first encountered the notion of cross products did I actually understand what they really meant. When I did, it was purely ecstatic!

Why on earth is the direction of cross product orthogonal ? Like seriously…

I mean this is one of the burning questions regarding the cross product and yet for some reason, textbooks don’t get to the bottom of this. One way to think about this is :

It is modeling a real life scenario!!

The scenario being :

When you try to twist a screw (clockwise screws being the convention) inside a block in the clockwise direction like so, the nail moves down and vice versa.

tumblr_inline_o5ew5ogSZ11srob4n_400

i.e When you move from the screw from u to v, then the direction of the cross product denotes the direction the screw will move.

tumblr_inline_o5ewlziDu81srob4n_250

That’s why the direction of the cross product is orthogonal. It’s really that simple!

Another perspective

Now that you get a physical feel for the direction of the cross product, there is another way of looking at the direction too:

Displacement is a vector. Velocity is a vector. Acceleration is a vector. As you might expect, angular displacement, angular velocity, and angular acceleration are all vectors, too.

But which way do they point ?

crossproduct

Let’s take a rolling tire. The velocity vector of every point in the tire is pointed in every other direction. BUT every point on a rolling tire has to have the same angular velocity – Magnitude and Direction.

How can we possibly assign a direction to the angular velocity ?

cross-product-simple

Well, the only way to ensure that the direction of the angular velocity is the same for every point is to make the direction of the angular velocity perpendicular to the plane of the tire.
Problem solved!

Types of Damping

May 30, 2015July 12, 2019 / 153armstrong / Leave a comment

Damping is an influence within or upon an oscillatory system that has the effect of reducing, restricting or preventing its oscillations.

There are 4 types of damping:(in the order of the animations shown)

1. Under Damped System.

The system oscillates (at reduced frequency compared to the undamped case) with the amplitude gradually decreasing to zero.

2. Critically Damped System.

The system returns to equilibrium as quickly as possible without oscillating.

3. Over Damped System.

The system returns to equilibrium without oscillating.

4. Un-Damped System.

The system oscillates at its natural resonant frequency

( Sources: xmdemo, timewarp,wikipedia)

ISS is the third brightest object in the night sky!

May 28, 2015July 12, 2019 / 153armstrong / Leave a comment

It often comes as a surprise to people when i tell them that the space station can be seen with the naked eye! Flying at 400 km above your head, the ISS looks like a really fast moving airplane in the sky.

The ISS isn’t brighter than the day sky and hence cannot be seen during the day. But in the night, it’s the third brightest object in the sky! It reflects the sunlight off the solar panels on its surface.

Spot the station!

If you would like to see the ISS for yourself, NASA ‘s Spot the station! is at your disposal. Register with your email address/mobile number and every time the ISS passes by your town/city, you will get a notification with the time, duration and inclination.

Have fun!

(Extras: There are mobile phone apps which you could use too, like the ISS detector satellite detector for android. At the end of the day all that matters is what is convenient to you.

ISS tracker– Real time tracking of the ISS)

Source: The Atlantic

Why don’t rain drops kill you?

May 21, 2015July 12, 2019 / 153armstrong / Leave a comment

Pardon me for being melodramatic. Think about it, rain drops fall from thousands of feet in the air and yet we hardly twitch when one falls on us.This defies all logic, if you drop a penny from the top of a building and let it fall on you, we know that it hurts!
Have the angels cast a magical spell on the rain drops to spare us from the pain?

Terminal Velocity.

When you drop something in air, it does not accelerate forever. Molecules in air constantly bombard with the object, exerting an upward force. This is known as air resistance or drag.

As the object gains velocity there comes a time when the force of the air resistance is enough to balance the force of gravity, so the acceleration stops and the raindrop attains terminal velocity.

Terminal Velocity is the maximum velocity an object can travel in air!

The Angel’s mystical spell.

Rain drops are only 0.5 mm-4 mm in diameter. Their terminal velocity is only about 10 m/s ( 20 mph ) That’s the maximum speed that they can travel in air irrespective of their initial height. Also the mass of the rain drop is about a few milligrams.

Hence, the force that it exerts on the body is really small, small enough that we find the experience pleasurable and soothing.

Ecstasy Shots

A poetry of logical ideas

classical mechanics