What is Gravity?

Here you will find an answer — and a challenge for you!
Let’s derive gravity from quantum principles in four steps.

Step 1: Quantum Particle

The quantum uncertainty principle states that the position \( r \) and momentum \( p \) of a particle cannot be simultaneously measured with arbitrary precision:

\[ \Delta r \Delta p \geq \frac{\hbar}{2} \]

Where \(\Delta r\) is the uncertainty of the position \( r \), \(\Delta p\) is the uncertainty of the momentum \( p \) and \( \hbar \) is the Planck constant divided by \( 2 \pi \). If we reduce the uncertainty in \( r \), the uncertainty in \( p \) increases. This reflects the particle’s finite information content.

Additionally, if we make several measurements \(\{r, p\}\), they would be uncorrelated and exhibit random scatter in both position and momentum. A quantum particle does not follow a classical trajectory.

Step 2: Macroscopic Object

Let us now consider a macroscopic object—a large collection of bound quantum particles, such as the Moon. Its position and momentum, \(\{r, p\}\), can be regarded as some sort of average of the individual particles’ \(\{r_i, p_i\}\). Since the position and momentum of the macroscopic object are derived from the finite information content of its constituent particles, the macroscopic object also has a finite amount of information—and thus uncertainties \(\Delta r\) and \(\Delta p\).

Unlike a quantum particle, a macroscopic object exhibits deterministic evolution of \( r \) and \( p \); they are not random.

While \( r \) and \( p \) are deterministic, their uncertainties \(\Delta r\) and \(\Delta p\) impose limits on their precision. When the object is at position \( r \), the minimum change in its position cannot be smaller than \(\Delta r\). The same logic applies to momentum \( p \) and its uncertainty \(\Delta p\).

Thus, the uncertainty becomes the magnitude of the change in each variable. Let’s assume the following equation for the quantum uncertainty of a macroscopic object:

\[ \Delta r \Delta p = \frac{\hbar}{2} \]

This is similar to the quantum uncertainty principle, but since \( r \) and \( p \) are deterministic, the inequality does not apply.

Step 3: A Property of Space

Let us shift the origin of the coordinate system to the center of the macroscopic object. For clarity, let’s refer to the macroscopic object as "mass \( M \)" when distinguishing it from other objects.

Consider a test object moving toward mass \( M \). As the test object approaches \( M \), its position \( r \) decreases by \(\Delta r\), and the uncertainty \(\Delta r\) diminishes correspondingly. This reduction in \(\Delta r\) corresponds to an increase in \(\Delta p\). Consequently, the magnitude of the test object’s momentum \( p \) increases toward mass \( M \).

Assuming the product \(\Delta r \Delta p\) is positive and constant, both \(\Delta r\) and \(\Delta p\) must be negative as the test object approaches \( M \).

This observation reveals that the presence of a macroscopic object induces properties in space analogous to the only known natural phenomenon governing the dynamics of objects in empty space based solely on their mass: gravity.

Step 4: Newtonian Gravity

The acceleration experienced by a test object of mass \( m \) as it approaches an object of mass \( M \) is described by Newtonian gravity:

\[ ma = -\frac{GMm}{r^2} \]

where \( a \) is the acceleration of the test object and \( G \) is the gravitational constant. However, if we solve the uncertainty equation for a macroscopic object, we find a different solution than that of Newtonian gravity. What are we missing?

To obtain the same solutions, we need a modification:

\[ \Delta r \Delta p = F(t) \]

With an appropriate choice of \( F(t) \), this equation and the Newtonian equation yield the same solution. Hence, \( F(t) \) is the precise formulation for some sort of average referred to in Step 2.

I adopt \( F(t) \) to ensure that the uncertainty of the macroscopic object is consistent with Newtonian gravitational behavior. This expression is not derived from first principles; rather, its significance lies in showing how quantum mechanics and gravitation are connected.

Some Insights

Since \( F(t) \) is different from zero, the last equation implies that the macroscopic object can never be at rest with respect to any inertial coordinate system, since the changes in both \( r \) and \( p \) are non-zero. This implies that the object is always experiencing acceleration with respect to any inertial coordinate system.

The Answer

Gravity is the effect of the quantum uncertainty of macroscopic objects. The intrinsic coupling among \( r \), \( p \), \(\Delta r\), and \(\Delta p\) with respect to a macroscopic object is exactly gravity.

A Challenge for You!

Now that you know what gravity is, would you derive \( F(t) \) from fundamental principles?

Providing a sound derivation for \( F(t) \) is the last piece needed to understand the connection between gravity and quantum mechanics. Would you provide the final piece of the quantum-gravity puzzle?

Further Study

I worked these steps in detail, and also a connection to General Relativity, in my article: "Conjectural Connection Between Quantum Mechanics and Gravitation" [https://doi.org/10.5281/zenodo.17682882]

Sharing

Let others know about this connection between Quantum Mechanics and Gravity! Share the link with your physicist friends.

Please let me know at nicolas@corpusk.info if you publish a derivation for \( F(t) \)!