Crossing the brachistochrone- The spaceship in the brachistochrone

This is an example of how we may relate the journey of an object on the brachistochrone to the properties of a photon:

Suppose that a spaceship (black arrow) leaves point O, and travels accelerating on the brachistochrone OC΄O΄ (orange curve), towards point C΄. In the beginning of the journey, an observer on the spaceship emits a photon, which travels on the straight line ΟC΄ (black dotted line). What is the condition that the observer takes back the emitted photon at the end of the journey, at point C΄?

This is an analysis of the problem:

On one hand, we have the motion of the spaceship. Because the spaceship is accelerating, its path will be that of a brachistochrone, since this is the path of least action (time). Thus it will be the curve OC΄.

On the other hand, the emitted photon will travel on a straight line, since its speed (the speed of light) is constant, thus the photon does not have any acceleration. If we assume that the photon propagates as a spherical (or plane) wave, then the linear path Δr=OC΄ will be the radius of such a wave. Incidentally, the same distance can also be seen as a vector, whose origin is at point O, its end is at point C΄, and whose measure is Δr=cΔt.

The relationship between the curved and linear path OC΄ is that between an arc and a chord, respectively, joining the points OC΄. These two distances, as we have already seen, are given as follows. Calling ΔS the curved path (the arc) OC΄, Δr the linear path (the chord) OC΄, Δx the distance OC, and Δy the distance CC΄, we have

where R is the radius of the brachistochrone, so that

Thus the two distances are approximately equal to each other, so that we may write

An equivalent way to measure the same distances is to treat them as products of speed by time. In that sense, the distance ΔS will be equal to the speed v of the moving object multiplied by the time T it takes the object to travel the same distance, while the distance Δr will be equal to the speed of light c multiplied by the time Δt it takes light (the photon) to travel that distance. These times need not be the same, so that the speeds need not be the same. But because the two distances ΔS and Δr are approximately equal to each other, we can write,

This is the condition which relates the spatial and temporal properties (v, T) of the moving object to those (c,t) of the photon emitted by the same object.

Now the time T is in fact the time of the brachistochrone. This time is given by the formula

where

and

On the other hand, the speed of the emitted photon, which is the speed of light c, can also be given in relation to the wavelength λ and the period τ of the photon, so that

and

This formula relates the properties of the moving object (its speed v and the time T it takes for the object to complete the journey), to the properties of the photon emitted by the same object (the wavelength λ, or period τ, of the photon, in relation to the time Δt, which can be measured by the clock of an external observer, standing at point O).

Here, to simplify things, we will make the following assumptions. First of all the harmonics n of a photon, as we have already seen, are given as

where λ₀ is the photon’s wavelength before the spaceship begins to move, while λ_n is the photon’s wavelength when the spaceship is in motion. Here we have used the index ‘0’ (instead of ‘1’) for the first state, n=1.

The difference in the wavelengths is

If n is sufficiently big, then

Furthermore, if the distance Δr is measured in relation to the emitted photon, then if we call N₀ the number of photons (or wavelengths λ₀) which compose the distance Δr in the beginning, and N_n we call the number of photons (or wavelengths λ_n) which compose the distance Δr when the spaceship is moving (at some speed v_n, corresponding to the harmonic n), then we have

If we suppose that the distance Δr is of size comparable to the initial wavelength λ₀ of the photon, then it will be

Thus it will be

where the time of the brachistochrone was written as T (instead of ΔT) for simplicity.

Now if we also suppose that at the first harmonic of the photon, n=1, the spaceship travels the distance ΔS at the speed of light c, then calling v₀ the speed of the spaceship corresponding to the first harmonic, n=1, and v_n its speed corresponding to any harmonic n, and since we have identified the speed v₀ of the spaceship at the first harmonic, n=1, with the speed of light c, v₀=c, we have

This way we have identified the time T₀ of the brachistochrone corresponding to the first harmonic, n=1, with the period τ₀ of the emitted photon at the same harmonic, n=1.

Multiplying the previous equation by the harmonic n, we can also identify the time of the brachistochrone T_n, at any harmonic n, with the period of the emitted photon τ_n, at the same harmonic n,

Thus the speed v_n of the object, corresponding to some harmonic n, will be

and for the distance to be traveled, it will be

From the previous equations, solving for the wavelength λ_n and the period τ_n of the photon,

we see that what is ‘contracted’ or ‘delayed’ is not the distance to be travelled Δr, or the fixed time Δt, but the wavelength λ and the period τ of the photon, respectively.

But how is it is possible that the spaceship (moving faster than light) reaches the end of the journey, at point C΄, before the emitted photon reaches the same point?

If the photon is treated as a ‘point’ which travels all the distance in between, then this is impossible, because causality will be violated. In other words, if the photon is the carrier of the information (thus it also contains the information about the journey of the spaceship) then the spaceship cannot reach point C΄ before the information of its own arrival reaches the same point.

However, if the photon is treated as a disturbance of spacetime, produced by the oscillations of spacetime, then the emitted photon will be located wherever the spaceship is located at a given time. This is because, since the spaceship travels in spacetime, the observer on the spaceship can measure the oscillations of spacetime (in the form of photons) at the same point. The difference is that the period of the photon will change if the spaceship is in motion, because the period of the photon depends on the speed of the spaceship.

Thus, to answer the question in the original problem, the observer on the spaceship will always take back the reference, or ‘emitted,’ photon.

Here is a scheme which shows the difference in the first principles, if we consider spacetime as empty, so that photons are particles which travel in empty space on straight lines, or if we consider spacetime as the medium in which motion occurs, while photons are disturbances of the medium:

On one hand, we have that:

Spacetime is empty.
The observer emits a photon in spacetime.
The photon is a particle which travels on a straight line at the speed of light.
If the observer exceeds the speed of light, he/she will never take back the photon (and causality is violated).

On the other hand, we have that:

Spacetime is an oscillating medium.
The observer can measure the oscillations of spacetime in the form of photons.
The motion of the observer in spacetime disturbs spacetime.
The observer receives the disturbance as a photon with altered period.

As far as the previous condition of simultaneity is concerned,

it will be studied in more detail later on in this document.