Momentum Bonus

Breaking large insurmountable tasks into smaller manageable subtasks enables hyper-productivity. I found that completing a high volume of small tasks results in a larger area under the productivity curve due to a ‘momentum bonus’ when compared to completing fewer monolith tasks.

Let’s prove it mathematically.

Definitions. Momentum is a mass in motion $p=mv$. Let’s define velocity to be productivity and mass to be the magnitude of the task being completed.

Momentum is conserved so if two objects $A$ and $B$ collide, elastically, $m_av_a + m_bv_b = m_av_a' + m_bv_b'$ where $v'$ represents their posterior velocities, respectively. In the inelastic case $m_av_a + m_bv_b = (m_a+m_b)v$ kinetic energy is not conserved and lost to the environment. The analogy in the context of productivity is switching from task $A$ to task $B$ after either $A$ has been completed or not. The reason I model context switching as a collision is because it can be classified as elastic or inelastic. Kind of neat.

For a quick refresher:

• Elastic collisions have both momentum and kinetic energy conserved e.g. throwing billiard balls against the edge of a pool table and it comes back just as fast
• Inelastic collisions only have momentum conserved e.g. you trying to push your friend off a cliff (their kinetic energy is converted mostly to friction on the ground)

The coefficient of restitution is designed to quantify collision elasticity

The value of $e$ means that

• $e= 0$ is perfectly inelastic
• $0< e < 1$ is real-world inelastic
• $e = 1$ is perfectly elastic
• $e > 1$ occurs in special cases where the collision results in an explosion
• $e = \infin$ when there is a perfect explosion of a rigid system

Back to productivity.

Case 1: You fully completely task $A$ and move onto $B$ (elastic)

Since $m_av_a + m_bv_b = m_av_a' + m_bv_b'$ you’re posterior velocities remain unchanged i.e. you haven’t hit the wall yet. At this point you’ll have the maximum momentum prior to starting task $B$. God speed.

Case 2: You don’t fully complete task $A$ and move onto $B$ (inelastic)

This case falls under the condition where $0\le e < 1$ which means your posterior velocity will decrease from loses to the environment hence $m_av_a + m_bv_b = (m_a+m_b)v$. By definition we know that $v \le v_a, v_b$. We can re-arrange to get

If we decrease either $m_a$ or $m_b$ whilst increasing $v_a$ or $v_b$ then the final velocity $v$ increases. The real-world analogy is have maximum productivity on the smallest task possible results in a larger final velocity for your next collision i.e. task $C$. Unfortunately we don’t live in a world where $v\rightarrow \infin$. Productivity always feels like it tapers at a physical maximum. There’s many different methods to increasing $v_a$ and $v_b$, namely:

• Great sleep
• Physical exercise
• Standing desk
• Frequent breaks
• Deep focus/flow state triggered from deep curiosity

Once you’ve applied everything from this list (and your own personal hacks not included in there) the only thing left to do is to start shaving down $m_a$ and $m_b$ to non-zero values.

We always strive to have our days structured as a sequence of case 1 victories, but reality is way more brutal. I find myself frequently in case 2 territory, but now with a theoretical framework—a new attack angle.