Crossing the brachistochrone- The wet sponge example

Here it would be helpful to clarify some points with respect to the meaning of inertia (and the inertial mass m₀).

As we previously saw, if we take Newton’s second law of motion, and consider that the (inertial) mass m₀ is a constant, we have

But if the mass changes, then we have that,

The quantities m₀ (presumably the inertial mass) and v₀ (for example the final speed) will have to be determined by some boundary conditions. Therefore it has to be justified whether the inertial mass m₀ refers to the initial (total) mass of the object in motion, or to its final mass, as well as what is the difference between the inertial mass m₀, which is constant, and the mass m(t), which changes.

For this purpose we will use the example of the wet sponge.

So let’s imagine that we have a tank filled with water. We will call the mass of the water μ₀. We also have a sponge, which initially is dry, so that its initial mass is m₀. Supposedly this will be the sponge’s inertial mass. Then we sink the sponge into the tank of water, so that the sponge absorbs a quantity of water. If we call m_f the mass of the absorbed water, then the total mass m of the sponge will be

The problem here is that we want to measure the inertial mass m₀ of the sponge. The way to do this is in two steps. First we measure the mass m of the sponge filled with water. Then we drain the water off the sponge, and measure the mass of the drained water, thus the mass m_f of the absorbed water. Therefore the inertial mass of the sponge will be the difference,

One might wonder why we didn’t measure the inertial mass m₀ of the sponge (e.g. by weighing it), in the first place. The answer is that we cannot directly measure the mass of the sponge <i>if we use the mass of the water as the reference unit of measurement</i>.

Another way to put this, is to say that we cannot measure the inertial mass m₀ of an object in free fall, because, from the equation of motion m₀a=m₀g, the inertial mass disappears. This is to say that the falling object is embodied in the gravitational field (the ‘water’) which makes the object fall.

Therefore we may say that the inertial mass m₀ of the sponge (or of any object) is what is left after all the other masses (the ‘water’) have been excluded.

Now if we call M₀ the total mass of the water together with the sponge, we can write the following equation for the masses,

where μ is the mass of water remaining in the tank, after a quantity m_f of water has been absorbed by the sponge.

Here, in order to return to the case of a rocket in space, we can identify the mass μ₀ with the total mass in a region of spacetime, m_f will be the amount of fuel of the spaceship (the amount of energy the spaceship has absorbed in the form of fuel), m₀ will be the spaceship’s inertial mass (its mass without the fuel), while μ will be the mass left to spacetime.

The rate of change of the mass m of the spaceship will be,

Thus what changes is not the inertial mass m₀ of the spaceship, but the mass m_f of the fuel it contains. In such a sense, the total mass m of the spaceship is reduced, as the spaceship burns fuel.

If the spaceship burns all its available fuel m_f, then it will be left with its inertial mass m₀,

On the other hand, the rate of change of the mass μ of spacetime will be,

Thus spacetime loses mass μ to the degree the spaceship burns fuel m_f. If the spaceship makes use of all the available energy stored in a region of spacetime, then the mass μ₀ of spacetime will have been transformed into the fuel m_f of the spaceship,

Thus the total mass m_f of fuel the spaceship consumes cannot be greater than the total mass μ₀ available in a region of spacetime.

Notes:

The previous conclusion may be shown graphically in the following way. Firstly, we need to determine if the damping factor γ,

is constant, or if it changes with time.

The simplest case is to assume that all related quantities are constant, so that the damping factor will also be constant,

In that case, the damping factor will be defined with respect to the inertial mass m₀.

Still, we are left with a mass m(t) which changes with time,

Another way to keep the damping factor constant, is to keep the rate b(t)/m(t) constant,

so that for the mass m(t) we take that,

In this expression, since the mass m(t) increases exponentially with time, the quantity m₀ will be the initial minimum mass (presumably the inertial mass).

If we assume that the mass m(t) decreases with time, then we may use a negative sign in the exponential, so that the quantity m₀ will be the maximum mass.

However, in the general case, if we consider a damping factor γ(t) which changes with time,

then with respect to the mass m(t) we take that,

In order to deal with the integral on the right hand side of the previous equation, we may simply set

so that the last expression takes the form

This is the related graph:

In this graph we have set t₀=1, so that the function

takes its maximum value m₀ at some boundary time t=t₀.

Thism₀ will refer to the maximum mass available to the object, thus it can also be identified with the total mass μ₀ of spacetime. Therefore if we replace m₀ by μ₀, and also set

we take

If additionally we set v₀=c, so that the function takes its maximum value μ₀ at v=c, we take

Such transformations are indicative of the fact that the mass m(t) of the object, although it takes its maximum value μ₀ at the speed of light, v=c, can still be defined even if the speed of the object is greater than the speed of light, v>c. This can happen if the mass μ₀ of the wave of spacetime is a mass ‘per oscillation,’ or per wavelength of the wave, so that there is more mass, thus also energy, available as fuel to the object moving in spacetime.