Mistakes in Einstein’s Derivation of Special Relativity Relations
Ref: Einstein’s 1916 book titled “Relativity: The Special and The General Theory” — Appendix I
The above derivation is based on corelating the distance and time of travel of a light signal in the two frames — stationary and moving.
The same is reproduced below with my intermittent observations in italics on the mistakes (placed between dotted lines).
A light-signal, which is proceeding along the positive axis of x, is transmitted according to the equation
𝑥 = 𝑐𝑡
Or,
𝑥 − 𝑐𝑡 = 0 . . .(1)
Since the same light-signal has to be transmitted relative to K’ with the velocity c, the propagation relative to the system K’ will be represented by the analogous formula
𝑥′ − 𝑐𝑡′ = 0 . . . (2)
Those space-time points (events) which satisfy (1) must also satisfy (2). Obviously this will be the case when the relation
(𝑥′ − 𝑐𝑡′) = 𝜆 (𝑥 − 𝑐𝑡). . . .(3)
is fulfilled in general, where 𝜆 indicates a constant; for, according to (3), the
disappearance of (𝑥 − 𝑐𝑡) involves the disappearance of (𝑥′ − 𝑐𝑡′).
If we apply quite similar considerations to light rays which are being transmitted along the negative x-axis, we obtain the condition
(𝑥′ + 𝑐𝑡′) = 𝜇 (𝑥 + 𝑐𝑡) . . . (4).
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Observations:
The above equation (4) is valid for a case when the event is located on (+)ve x side and the light rays are moving in (-)ve x-direction and vice versa because the two terms 𝑥 and 𝑐𝑡 are additive.
Now, the question is should the light signal be assumed moving against its own event; the light signal will never meet its supposed event. So, it is not clear as to what is this statement for. Or else, the author has something intuitive in his mind which he intends to incorporate in the derivation.
As shown in chapter 5 of my book, this inexplicable equation is not required for derivation of the relativity relations; however, its utility is limited only to working out relations for Lorentz Transformation Condition.
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By adding (or subtracting) equations (3) and (4), and introducing for convenience the constants 𝑎 and 𝑏 in place of the constants 𝜆 and μ, where
a = (𝜆 + 𝜇) / 2
and
b = (𝜆 — 𝜇) / 2
we obtain the equations
𝑥′ = 𝑎𝑥 − 𝑏𝑐𝑡
𝑐𝑡′ = 𝑎𝑐𝑡 − 𝑏𝑥 . . . (5)
We should thus have the solution of our problem, if the constants 𝑎 and 𝑏 were known. These result from the following discussion. For the origin of K’ we have permanently x’ = 0, and hence according to the first of the equations (5)
𝑥 = (bc/a)t
If we call 𝑣 the velocity with which the origin of K’ is moving relative to K, we then have
𝑣 = bc/a . . . (6)
The same value 𝑣 can be obtained from equations (5), if we calculate the velocity of another point of K’ relative to K, or the velocity (directed towards the negative x-axis) of a point of K with respect to K’. In short, we can designate 𝑣 as the relative velocity of the two systems.
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Observations:
The set of Equations (5) make statements about motion of a light signal, as observed from the two frames K and K’. Now, let us examine the statement
“For the origin of K’ we have permanently x’ = 0”.
It may be recollected that 𝑥′ is the distance travelled by the light signal (not the frame K’) in a time 𝑡′. Then how can this parameter be used for the frame K’?
Further, for the light signal, it is never 𝑥′ = 0 except at the beginning of its travel where we also concurrently have 𝑥 = 0, 𝑡 = 0 and 𝑡′ = 0. All these concurrent values, when substituted in the equations (5), result in zeros on both the sides, which is expected in accordance with the assumptions made.
So, working out of 𝑣 as above is incorrect.
It is added here that if the likes of equations (5) were not just for light but for any distance-time set, these objections wouldn’t have arisen.
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Furthermore, the principle of relativity teaches us that, as judged from K, the length of a unit measuring-rod which is at rest with reference to K’ must be exactly the same as the length, as judged from K’, of a unit measuring-rod which is at rest relative to K.
In order to see how the points of the x’-axis appear as viewed from K, we only require to take a “snapshot” of K’ from K; this means that we have to insert a particular value of t (time of K), e.g. t = 0. For this value of t we then obtain from the first of the equations (5)
𝑥′ = 𝑎𝑥
Two points of the x’-axis which are separated by the distance Δx’ = 1 when measured in the K’ system are thus separated in our instantaneous photograph by the distance
Δ𝑥 = 1/a. . . .(7)
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Observations:
As already pointed out in the previous observations, equations (5) are for motion of a light signal with certain assumptions. This fact cannot be ignored while applying any boundary/special condition to the equations. For the light signal at 𝑡 = 0, the other concurrent statements are 𝑥 = 0, 𝑥′ = 0 and 𝑡′ = 0. By substituting these concurrent values, both sides of the equations become zero, quite in accordance with the assumptions.
Secondly, Eq.(7) has been obtained after application of Δ operator on both sides of the above result i.e. 𝑥′ = 𝑎𝑥 and thereby substituting Δ𝑥 = 1. Attention is invited to the fact that 𝑥′ is dependent on 𝑥 as well as 𝑡. We may also say that any relation between Δ𝑥′ and Δ𝑥 will necessarily involve terms of Δ𝑡. Therefore, Δ operator should be applied first, after which a value of Δ𝑡 can be substituted for any instant 𝑡.
If Δ operator is applied to the first of equation (5), the result would be as follows
Δ𝑥′ = 𝑎Δ𝑥 − 𝑏𝑐Δ𝑡
For Δ𝑡 = 0, this relation would result into Δ𝑥′ = 0, as for Δ𝑡 = 0, Δ𝑥 = 0.
Thus here is another unacceptable deviation from light to meter-rod.
It is reiterated again here that if the likes of equations (5) were not just for light but for any distance-time set, these objections wouldn’t have arisen.
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But if the snapshot be taken from K’(t’ = 0), and if we eliminate t from the equations (5), taking into account the expression (6), we obtain
𝑥′ = 𝑎 (1 − v²/c²)x
From this we conclude that two points on the x-axis separated by the distance 1 (relative to K) will be represented on our snapshot by the distance Δ𝑥′ = 𝑎 (1 − v²/c²). . . . (7a)
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Observations:
Similar to the previous observations.
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But from what has been said, the two snapshots must be identical; hence Δx in (7) must be equal to Δx’ in (7a), so that we obtain
a² = 1/ (1 — v²/c²). . . . (7b)
The equations (6) and (7b) determine the constants 𝑎 and 𝑏. By inserting the values of these constants in (5), we obtain the first and the fourth of the equations given in Section 11.
Thus we have obtained the Lorentz transformation for events on the x-axis. It satisfies the condition
𝑥′2 − 𝑐2𝑡′2 = 𝑥2 − 𝑐2𝑡2 . . . . (8a)
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Observations:
Coming to the start of derivation, multiplication of equation (1) with its conjugate and also the same operation for equation (2) will give 𝑥′2 − 𝑐2𝑡′2 = 𝑥2 − 𝑐2𝑡2 = 0. The zero result is obviously for light. For events other than those of light, the result is bound to be non-zero, as for these, 𝑥 − 𝑐𝑡 ≠ 0, 𝑥 + 𝑐𝑡 ≠ 0, 𝑥′ − 𝑐𝑡′ ≠ 0 and 𝑥′ + 𝑐𝑡′ ≠ 0. But these facts have been ignored when switching over to material bodies like the origin of moving frame and unit measuring-rod for finding out value of constants 𝑎 and 𝑏.
Further, substitution of 𝑥 with its value 𝑐𝑡, as per equation (1), in the above equations (8) lead to quite different results, as discussed in my book. The same would sometime be discussed here too.
Note: For entire details and much more, read my book “Refining Relativity Part 1 (The Special Theory)” available on Amazon. Also, do not forget to read various pages on my blog at “https://refiningrelativity.blogspot.com”.
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