Crossing the brachistochrone- The rocket in space

Illustration of the ‘rocket in space:’ The rocket pushes ‘empty’ space backwards with a force F=-bv, while ‘empty’ space pushes the rocket forward with an equal but opposite force F=bv.

Once it was believed that the motion of a rocket in empty space would have been impossible, because the rocket would have had nothing to push (e.g. the air). But this was a misinterpretation of the true phenomenon, because, even in the simplest case of a car, it is not the exhaustion gases which push the air, so that the vehicle moves in the opposite direction, but it is the loss of mass of the vehicle, which changes the vehicle’s speed (thus also the vehicle’s momentum).

This aspect rises from the conservation of the momentum of the object. For example, if m₁ and v₁ is the initial mass and speed of the vehicle, and m₂ and v₂ is its final mass and speed, respectively, then the conservation of the momentum p gives

If the vehicle has lost mass in the form of burnt fuel, then it will be

so that the moving object increases its speed simply because it burns fuel (thus loses mass).

Still, if we use the notion of the force, the change of the moving object’s momentum will be equivalent to a force (a thrust) produced by the object. This is because the force is the derivative of the momentum with respect to time, according to Newton’s second law of motion. Also, according to Newton’s third law of action and reaction, if a rocket moving in empty space exerts a thrust on empty space, then empty space will push the rocket with an equal but opposite force. However, this raises the question of what nature is such a force, and, more importantly, how come empty space produces forces, if space is empty.

So let’s take the rocket- in- space example. If the rocket has a mass m, and moves at a speed v, using Newton’s second law, the equation of motion of the rocket will be

That the force the rocket produces is F=-bv, can be seen if we take the full expression (if the mass also changes) of Newton’s second law

where p=mv is the rocket’s momentum.

If the momentum is conserved, then

which gives us back the equation of motion earlier mentioned.

Here we should mention that the force F=-bv has the aspect of a resistance force, which opposes the motion of the rocket, if the medium is not empty. Such a force is usually considered to be proportional to the speed v of the moving object, where the constant of proportionality is b.

However, at the same time, this force can also be seen as the thrust which the rocket produces, and imposes on the medium. In such a sense this force will be a damping force, proportional again to the speed v of the object, where here the constant of proportionality b will be the damping constant.

Thus the constant b is related to a property of the medium (its resistance), as much as to a property of the object which moves in the medium (damping), since it represents the rate with which the object burns mass (in the form of fuel),

If we now rewrite the previous equation of motion in the following equivalent form,

we can identify the parameter γ with the damping factor, which we will often encounter hereafter.

The point here is that spacetime is not empty, but that it has a certain amount of energy, thus also a certain amount of mass associated to that energy. In the next section we will see that this mass is part of the moving object’s mass, in the form of its fuel content. Thus when the object burns its fuel in order to move, it consumes a part of the mass (or equivalently energy) which is stored in spacetime.

Returning now to the equation of motion,

we may apply the following solution,

Substituting the solution into the equation of motion, we take,

so that the proposed solution is valid as long as the damping factor γ is given by the previous expression.

Now with respect to the kinetic energy, we have by definition

so that in our case

Setting t=0 as initial condition, we have

so that the initial (total) energy will be

We may notice that the term

can be identified with the measure of the energy related to the damping, thus the damping energy.

More accurately, the damping energy E_d is given by the following integral,

But if in the last integral we treat the mass m of the rocket, as well as the damping factor γ, as constants, then we may write

so that we take an energy equation of the form

The equivalence between the two forms of energy, the kinetic energy E_k, and the damping energy E_d, can also be seen by the rate of change of the energies,

where

If the total energy of the system is conserved then one form of energy will be transformed into the other form, so that it will be

If we multiply and divide the term on the right hand side of the last equation by the kinetic energy, we take

The same result can be taken more directly if we simply multiply both terms in the equation of motion by v,

If we integrate the last differential equation, we will take back the expression for the kinetic energy in exponential form,

The same equation may be written without an index, by simply writing

In that form, because the energy decays with time due to the negative exponential, we may associate it with the (damping) energy which spacetime loses, so that the kinetic energy of the rocket moving in spacetime will more appropriately be given by the difference

We will meet again such a solution and energy equation, when the harmonic force ky will be included into the equation of motion, which is the case of the damped harmonic oscillator. There, when critical damping is applied, the equivalence between the forms of energy will become more obvious.

Notes:

We have already mentioned the aspect that spacetime contains an amount of energy. We shall see in the next section that such energy can be represented by a mass μ₀ of the wave of spacetime, different than the inertial mass m₀ of the object moving in spacetime.

Here we will make the comparison between the kinetic energy ΔE(t) which was previously mentioned, and the energy in relativity. The relativistic energy is given by the following expression,

where m is the relativistic mass (which presumably increases with the speed v), and m₀ is the ‘rest mass.’

Accordingly, the energy E_Ein=mc² (where the index ‘Ein’ stands for ‘Einstein’) is the total energy, while E₀ is the ‘rest energy.’ In that sense, the quantity

will be the kinetic energy.

This kinetic energy can be reduced to the classical kinetic energy E_k=1/2mv², at the limit of small speeds, in the following way. Taking the linear approximation of the Lorentz factor, if the speed v of the object is small compared to the speed of light c, v<<c, v≈0, we have

so that

The problem however with Einstein’s energy, E_Ein, is that it goes to infinity if the speed v of the moving object reaches the speed of light c,

Now we will see how this problem may be treated.

In order to compare the relativistic expression for the energy, to the energy in the exponential form

where the ¼ factor, which is not significant, was added in order to have a factor of ½ appear in the exponential.

Thus we take

We can rewrite the relativistic factor as

so that

We should not confuse the relativistic factor γ_L (Lorentz factor) with the damping factor γ=b/m, or β_L with b=dm/dt.

If β_L is sufficiently small, so that β_L=v/c≈0, v<<c, then the linear approximation of the exponential in the previous equation will be

The last approximation is based on the fact that if v is sufficiently small, then setting x=v/c, we have

Thus we take that

The last expression is the same as the relativistic expression

taking into account that in our expression E₀ refers to the total energy, while E₀ in the relativistic expression refers to the ‘rest energy,’ thus the minimum energy.

Therefore we take back the relativistic expression at the limit of small speeds.

On the other hand, as the speed v of the object approaches the speed of light c, we have that

Thus at the speed of light, v≈c, the energy is reduced just by a factor of 1/e (or 1/2e).

Additionally, at speeds greater than light, the exponential e^-x behaves like 1/x² (or 1/2x²),

so that for the energy we take

Thus the energy goes to zero only if the speed of the moving object goes to infinity, v→∞.

This is a graph comparing the difference between the two energies:

In the previous graph we see how the relativistic expression for the kinetic energy (red line) goes to infinity, as the speed v approaches the speed of light c (c=1 in the graph), while the kinetic energy in the exponential form approaches asymptotically the maximum value E₀ (E₀=1 in the graph), as the speed v approaches infinity.

This is an example of how the problem of the infinite energy in relativity can be treated, even if the speed v of the moving object exceeds the speed of light c.

In summary, by relating the damping factor γ to Lorentz factor γ_L, we came up with an expression for the energy of the moving object which can be defined even if the object moves faster than light.

This was the original purpose of this document. The relationship between the properties of a material object, and the properties of the medium (spacetime) in which the object moves, will be further explored, according to the wave-particle equivalence, in what follows.