_{1}

^{*}

Although the formula of mass-energy equivalence was derived from the hypothesis that the speed of light in free space is constant, conversely, the purpose of this research is to show that a method of probabilistically determining whether the speed of light is constant is derived from this formula. By considering the formula of mass-energy equivalence to be a function of the energy of an object moving at speed V , the probability density function (PDF) of the energy can be obtained using the inverse function of this formula, if the speed of light obeys a probability distribution. The main result is that the PDF of the energy diverges to infinity at a certain energy value regardless of the PDF of the speed of light. Thus, when the speed calculated from this value enters a certain range of the speed of light as V increases stepwise from below 299 , 792 , 458 m/s, the PDF of the energy should increase abruptly. If not, then the speed of light is constant. This is the method of probabilistically determining whether the speed of light is constant. An experimental method is proposed to confirm this.

Albert Einstein published the theory of special relativity in 1905 [^{2}, where E is the energy of an object when it is moving, m is its mass while moving and c is the speed of light. This is derived from two hypotheses. One is that the speed of light in free space is constant for all observers, regardless of their relative motion or of the motion of the light source. This hypothesis is generally considered verified by the Michelson-Morley experiment, which shows the differences between the speed of light in the direction of motion of the earth and that in different directions are within experimental errors [^{2} to determine whether the speed of light is constant.

It assumes that the speed of light obeys a probability distribution.

The formula E = mc^{2} is also expressed as

E = m 0 1 − ( V c ) 2 c 2 , (1)

where m_{0} is the rest mass of the object, and V is its speed; V < c.

The assumptions are:

1) m_{0} and V are constant.

2) c obeys a probability distribution between c_{a} and c_{b}; c_{a} < c_{b}. Two probability distributions are adopted: a uniform distribution and a triangular distribution.

Then, the probability density function (PDF) of E is calculated by obtaining the inverse function of E as follows.

Step 1: Determining the inverse function of E

As 0 < V/c < 1, V/c is defined by sinθ (0 < θ < π/2).

From (1),

E = m 0 V 2 cos θ − cos 3 θ . (2)

Then, cosθ is represented as y:

c = V 1 − y 2 . (3)

Representing m_{0}V^{2} as P_{0}, the following third-order equation for y is obtained from (2):

f ( y ) = y 3 − y + P 0 E = 0. (4)

A positive value of E results from any value of θ on the interval 0 < θ < π/2. In other words, at least one of the three roots of f(y) is between 0 and 1. As the roots of f(y) are intersections of g ( y ) = y 3 − y = y ( 1 − y ) ( 1 + y ) and h(y) = −P_{0}/E, all three roots are real. Two of them are between 0 and 1 and the remaining root is negative. The two roots between 0 and 1 are denoted as y_{1} and y_{2}; y_{1} ≤ y_{2}. As 0 < y < 1, the negative root is neglected. Since y_{1} and y_{2} are determined by E, it is necessary to find out whether the inverse function of E, c º h(E), is a two-valued or single-valued function.

Step 2: Determining whether E is a two-valued or singled-valued function

From (1), E has the minimum 3 3 2 P 0 (P_{0} = m_{0}V^{2}) at c m ≡ 6 2 V . Three cases are examined on the basis of whether c_{m} is between c_{a} and c_{b}.

Case 1: c m ≤ c a

Then V ≤ 6 3 C a . As E increases monotonically between c_{a} and c_{b},

V 1 − y 1 2 < c a and c a ≤ V 1 − y 2 2 ≤ c b . (5)

Therefore, only y_{2} is accepted. In this case c = h(E) is a single-valued function (

Case 2: c a < c m < c b

Then 6 3 C a < V < 6 3 C b . As E is parabolic and downward convex between c_{a} and c_{b}, the following relationship occurs:

c a < V 1 − y 1 2 < V 1 − y 2 2 < c b . (6)

Both y_{1} and y_{2} are accepted and c = h(E) is a two-valued function (

V 1 − y 1 2 < c a or c b < V 1 − y 2 2 . (7)

Then c = h(E) is single-valued function (only y_{2} is accepted in

Case 3: c m ≥ c b

Then V ≥ 6 3 C b . As E decreases monotonically between c_{a} and c_{b},

c a ≤ V 1 − y 1 2 ≤ c b and c b < V 1 − y 2 2 . (8)

Therefore, only y_{1} is accepted. In this case c = h(E) is also a single-valued function (

Step 3: Calculation of PDF of E

From the PDF of c, f_{c}(c), and the inverse function of E, c = h(E), the PDF of the random variable E, f_{E}(E), can be obtained as

f E ( E ) = f c [ h ( E ) ] | d d E h ( E ) | (9)

where | d d E h ( E ) | is the Jacobian of the transformation. The absolute value should be taken since the PDF must be positive:

| d c d E | = | d d E h ( E ) | = | d c d θ | | d θ d E | (10)

From c = V/sinθ and y = cosθ,

| d c d θ | = P 0 m 0 | − cos θ 1 − cos 2 θ | = P 0 m 0 y 1 − y 2 . (11)

From (2) and y = cosθ,

| d θ d E | = P 0 E 2 1 sin θ | 1 − 3 cos 2 θ | = P 0 E 2 1 1 − y 2 | 1 − 3 y 2 | . (12)

From (10), (11) and (12),

| d c d E | = | d d E h ( E ) | = P 0 P 0 m 0 y ( 1 − y 2 ) 1 − y 2 | 1 − 3 y 2 | 1 E 2 . (13)

From (9) and (13), the PDF of E can be obtained:

f E ( E ) = f c [ h ( E ) ] | d d E h ( E ) | = f c [ h ( E ) ] P 0 P 0 m 0 y ( 1 − y 2 ) 1 − y 2 | 1 − 3 y 2 | 1 E 2 (14)

As 0 < y < 1, f_{E}(E) diverges to infinity at y = 1 3 . From (3), c = 6 2 V . This value of c is equal to c_{m}. Then E = 3 m 0 c m 2 .

Step 4: Setting of parameters

As the value of m_{o} has no qualitative effect on the relationship between E and c, m_{o} is set equal to 1. The values of V, c_{a} and c_{b} are set arbitrarily, depending on whether c_{m} is within the range of c: c a ≤ c ≤ c b .

Case 1: c m ≤ c a

The object speed and limits of c are set to V = 0.5, c_{a} = 2.2 and c_{b} = 4.2. Then c_{m} ≈ 0.6124.

Case 2: c a < c m < c b

The object speed and limits of c are set to V = 2, c_{a} = 2.2 and c_{b} = 4.2. Then c_{m} ≈ 2.4495. _{m}, (2.3 < c_{m}), and higher than c_{m}, (3.7 > c_{m}).

Case 3: c m ≥ c b

The object speed and limits of c are set to V = 2, c_{a} = 2.2 and c_{b} = 2.4. Then c_{m} ≈ 2.4495. In

When the speed of light is assumed to be variable, its probability distribution is unknown. Although this study only examines two probability distributions for it, the probability distribution of E has certain characteristics. If the distribution of c is uniform, the probability density of E is always maximum at minimum E and decreases monotonically regardless of whether c_{m} is within the range of c or not. If the distribution of c is triangular and c m ≤ c a , the distribution of E is also triangle-like with the vertex moving rightward with c. If the distribution of c is triangular and c m ≥ c b , the probability distribution of E changes from trapezoid-like shape to triangle-like shape. In both cases the vertex of the probability distribution of E moves together with the probability distribution of c. In contrast with these cases, if c a < c m < c b , the PDF of E is always maximum at minimum E in either distribution of c. This is because the PDF of E diverges to infinity at minimum E in either distribution of c from Equation (14). In practice, the PDF of E at minimum E increases even faster as the calculation step becomes smaller. This equation shows that if c a < c m < c b , the PDF of E is always maximum at minimum E: E = 3 m 0 c m 2 regardless of the distribution of c (that is, diverges to infinity). This suggests that even if the distribution of c is unknown, E will rapidly increase as soon as c_{m} enters a certain range of c as the speed V of an object increases.

On this basis, the following method is proposed to detect any range of c. As the speed of light is defined as 299,792,458 m/s º c_{L}, c a ≤ c L ≤ c b .

If

c m = 6 2 V < c a ≤ c L ≤ c b (15)

then

V < 6 3 c L . (16)

For example, the speed of one thousand electrons or protons is increased stepwise by an accelerator from below 6 3 c L to c_{L}. For each step, a frequency distribution of E will be obtained. As V is increased from below 6 3 c L , c_{m} will

enter the range of c at the critical value of V. Then the probability density of minimum E will increase sharply. Since the speed of light has been measured with very fine precision [_{m} exceeds c_{b}, the probability density of minimum E will decrease abruptly. If these phenomena are observed, then c is variable. If not, then c is constant.

If it is possible that the speed of light in free space is variable, then a probabilistic method to detect the variability is applicable. This assumes that c obeys a probability distribution. From mass-energy equivalence, the PDF of E can be obtained

using the inverse function of E. The energy is minimum at c m = 6 2 V , and the PDF of E diverges to infinity at E = 3 m 0 c m 2 . Thus, when c_{m} enters the range of c as V is increased stepwise from below 6 3 c L , the PDF of E increases abruptly

regardless of the PDF of c. If this is observed by accelerating a beam of electrons or photons, it will show that c is variable; otherwise, c is constant.

Mark Kurban, M.Sc., from Edanz Group (http://www.edanzediting.com/ac) edited a draft of this manuscript.

The author declares that there is no conflict of interests regarding the publication of this paper.

Osaka, M. (2019) A Probabilistic Method to Determine Whether the Speed of Light Is Constant. Applied Mathematics, 10, 51-59. https://doi.org/10.4236/am.2019.102005