Ignatowsky relativity

Vladimir Ignatowski (1875-1942) was a Russian physicist. “In 1910 he was to first who tried to derive the Lorentz transformation by group theory only using the relativity principle (postulate), and without the postulate of the constancy of the speed of light.” K M Browne gave a simplified derivation in the European Journal of Physics, 39 (2018) 025601, from which the key steps are presented below, followed by the corresponding steps for a dual transformation, switching space and time.

This is a derivation of the Ignatowsky transformation in which the axes x, y, and z are taken to represent space axes r_x, r_y, and r_z with time t. The relativity postulate is taken to be: a valid relativistic transformation must be identical in all inertial frames.

Step 1. To find a valid transformation, we take the usual inertial reference frames S and S′ (the latter moving at velocity v in the +x direction relative to the former) for which intra-frame space is Euclidean but inter-frame space (measured from one frame to the other) may be non-Euclidean. Linear equations are necessary so that an event in one frame appears as a single event, without echoes, in the other. Initial conditions are x′ = x = 0 when time t′ = t = 0. We expect the generalised x equation to be the Euclidean equation x′ = x − vt with an added multiplier, and if time is the fourth dimension, then the time equation will be similar but with two additional multipliers. The second of these, n, having the dimension of inverse velocity squared, is required to make the equation dimensionally correct. The y and z coordinates are not expected to be affected by x and t. The generalised transformation and its inverse are then

$x' = \lambda_{x} (x - vt); y' = \lambda_{y}y; z' = \lambda_{z} z; t' = \lambda_{t} (t - nvx), (1a)$

$x=\frac{x'}{\lambda _{x}}+vt;\; y=\frac{y'}{\lambda _{y}};\; z=\frac{z'}{\lambda _{z}};\; t=\frac{t'}{\lambda _{t}}+nvx;\; (1a.1)$

$x=\frac{x'}{\lambda _{x}}+v(\frac{t'}{\lambda _{t}}+nvx);\; y=\frac{y'}{\lambda _{y}};\; z=\frac{z'}{\lambda _{z}};\; t=\frac{t'}{\lambda _{t}}+nv(\frac{x'}{\lambda _{x}}+vt);\; (1a.2)$

$x(1-nv^2)=\frac{x'}{\lambda _{x}}+\frac{vt'}{\lambda _{t}};\; y=\frac{y'}{\lambda _{y}};\; z=\frac{z'}{\lambda _{z}};\; t(1-nv^2)=\frac{t'}{\lambda _{t}}+\frac{nvx'}{\lambda _{x}};\; (1a.3)$

$x = \frac{\frac{x'}{\lambda_{x}}+ \frac{vt'}{\lambda_{t}}}{1-v^{2}n}; y = \frac{y'}{\lambda_{y}}; z = \frac{z'}{\lambda_{z}}; t = \frac{\frac{t'}{\lambda_{t}}+ \frac{nvx'}{\lambda_{x}}}{1-v^{2}n}, (1b)$

where λ_x, λ_y, λ_z and λ_t are independent dimensionless coefficients to be determined. The constant n may be positive or negative, but is assumed initially to be positive. Omitting the v term in the time equation leads to ambiguity in the sign of λ_t. Mathematically, transformation (1b) is exactly the same as (1a) but will have the same form as (1a) if, and only if, the transformation is relativistic.

Step 2. Relativity is forced by equating coefficients of the right-hand side terms of the matching equations in (1a) and (1b) remembering that v is reversed (i.e. negative) in (1b). Then,

$\lambda_{x} = \lambda_{t} = \frac{1}{\sqrt{1-v^2n}}\; \textup{and} \; \lambda_{y} = \lambda_{z} = 1.$

Now, for the transformation to have real solutions, λ_x and λ_t must be real, which means that v² < n⁻¹. There will be, therefore, a limit for the speed v of the S′ frame with respect to the S frame in order for x′ and t′ to remain real. Calling this unspecified limiting speed V, where V² = n⁻¹, transformation (1a) now becomes the Ignatowsky transformation:

$x'=\frac{x-vt}{\sqrt{1-v^2/V^2}}; y'=y; z'=z; t'=\frac{t-vx/V^2}{\sqrt{1-v^2/V^2}}.\; (2)$

Note that v ≤ V. This is not the Lorentz transformation, but a valid generalised inertial frame transformation, according to our first postulate.

Step 3. To derive the Galilean transformation from (2), set t′ = t, which from (2) requires that V = ∞, i.e., n = 0. In that case, 0 ≤ v ≤ ∞, and there is no maximum speed.

Step 4. Add Einstein’s second postulate: light is always propagated in empty space with a definite velocity c. In that case, c = V, the maximum speed. The constant c is the conversion factor from time to space.

The term V was originally introduced as a limiting value of v, that is a limiting relative speed of all inertial frames and inertial masses. In each model, V is also the speed of light in a vacuum associated with that particular model.

This is a derivation of the Ignatowsky transformation for 3D time in which x, y, and z are taken to represent time axes t_x, t_y, and t_z with with stance r.

We have already established that for the Galilean transformation has no maximum speed, and so no minimum pace (the pace may be zero). But for the Lorentz transformation the maximum speed is c, so the minimum pace is 1/c. In both cases, the pace may be infinite.

Step 1. To find a valid transformation, we take the usual inertial reference frames S and S′ (the latter moving at lenticity u in the +x direction relative to the former) for which intra-frame time is Euclidean, but inter-frame time (measured from one frame to the other) may be non-Euclidean. Linear equations are necessary so that an event in one frame appears as a single event, without echoes, in the other. Initial conditions are x′ = x = 0 as r′ = r = 0. We expect the generalised x equation to be the Euclidean equation x′ = x − ur with an added multiplier, and if stance is the fourth dimension, then the stance equation will be similar but with two additional multipliers. The second of these, k, having the dimension of inverse lenticity squared, is required to make the equation dimensionally correct. The y and z coordinates are not expected to be affected by x and r. The generalised transformation and its inverse are then

$x'=\lambda _{x} (x-r/(ku));\; y'=\lambda _{y} y;\; z'=\lambda _{z} z;\; r'=\lambda _{r}(r-x/u),\; \; (1a)$

$x=\frac{x'}{\lambda _{x}}+r/(ku);\; y=\frac{y'}{\lambda _{y}};\; z=\frac{z'}{\lambda _{z}};\; r=\frac{r'}{\lambda _{r}}-x/u,\; \; (1a.1)$

$x=\frac{x'}{\lambda _{x}}+\frac{r'}{\lambda _{r}}+x/(ku^2);\; y=\frac{y'}{\lambda _{y}};\; z=\frac{z'}{\lambda _{z}};\; r=\frac{r'}{\lambda _{r}}+\frac{x'/u}{\lambda _{x}}+r/(ku^2),\; \; (1a.2)$

$x(1-\frac{1}{ku^2})=\frac{x'}{\lambda _{x}}+\frac{r'}{\lambda _{r}};\; y=\frac{y'}{\lambda _{y}};\; z=\frac{z'}{\lambda _{z}};\; r(1-\frac{1}{ku^2})=\frac{r'}{\lambda _{r}}+\frac{x'/u}{\lambda _{x}},\; \; (1a.3)$

$x= \frac{\frac{x'}{\lambda _{x}}+\frac{r'}{\lambda _{r}}}{1-\frac{1}{ku^2}} ;\; y=\frac{y'}{\lambda _{y}};\; z=\frac{z'}{\lambda _{z}};\; r= \frac{\frac{r'}{\lambda _{r}}+\frac{x'/u}{\lambda _{x}}}{1-\frac{1}{ku^2}} ,\; \; (1b)$

where λ_x, λ_y, λ_z and λ_r are independent dimensionless coefficients to be determined. The constant k may be positive or negative, but is assumed initially to be positive. Omitting the u term in the stance equation leads to ambiguity in the sign of λ_r. Mathematically, transformation (1b) is exactly the same as (1a) but will have the same form as (1a) if, and only if, the transformation is relativistic.

Step 2. Relativity is forced by equating coefficients of the right-hand side terms of the matching equations in (1a) and (1b) remembering that u is reversed (i.e. negative) in (1b). Then,

$\lambda _{x}=\lambda _{r}= \frac{1}{\sqrt{1-1/(ku^2)}}\; and \; \lambda _{y}=\lambda _{z}=1.$

Now, for the transformation to have real solutions, λ_x and λ_r must be real, which means that u² > k⁻¹. There will be, therefore, a limit for the pace u of the S′ frame with respect to the S frame in order for x′ and r′ to remain real. Calling this unspecified limiting pace U, where U² = k⁻¹, transformation (1a) now becomes the Ignatowsky transformation:

$x'=\frac{x-rU^2/u}{\sqrt{1-U^2/u^2}}; y'=y; z'=z; r'=\frac{r-x/u}{\sqrt{1-U^2/u^2}}.\; (2)$

Step 3. To derive the Galilean transformation from (2), set x′ = x, which from (2) requires that U = 0, i.e., k = ∞.

Step 4. Add Einstein’s second postulate: light is always propagated in empty time with a definite lenticity k. In that case, k = U, the minimum pace.

The term U was originally introduced as a limiting value of u, that is a limiting relative speed of all inertial frames and inertial masses. In each model, U is also the pace of light in a vacuum associated with that particular model.

What follows is a derivation of what could be called the dual-Ignatowsky transformation in which x, y, and z are taken to represent time axes t_x, t_y, and t_z with with stance r.

$x'=\lambda _{x}(x-ur);\; y'=\lambda _{y}y;\; z'=\lambda _{z}z;\; r'=\lambda _{r}(r-mux),\; \; (1a)$

$x=\frac{\frac{x'}{\lambda _{x}}+\frac{mux}{\lambda _{x}}} {1-mu^2};\; y=\frac{y'}{\lambda _{y}};\; z=\frac{z'}{\lambda _{z}};\; r=\frac{\frac{r'}{\lambda _{r}}+\frac{x'/u}{\lambda _{x}}}{1-mu^2},\; \; (1b)$

where λ_x, λ_y, λ_z and λ_r are independent dimensionless coefficients to be determined. The constant m may be positive or negative, but is assumed initially to be positive. Omitting the u term in the stance equation leads to ambiguity in the sign of λ_r. Mathematically, transformation (1b) is exactly the same as (1a) but will have the same form as (1a) if, and only if, the transformation is relativistic.

Step 2. Relativity is forced by equating coefficients of the right-hand side terms of the matching equations in (1a) and (1b) remembering that u is reversed (i.e. negative) in (1b). Then,

$\lambda _{x}=\lambda _{r}= \frac{1}{\sqrt{1-mu^2}}\; and \; \lambda _{y}=\lambda _{z}=1.$

Now, for the transformation to have real solutions, λ_x and λ_r must be real, which means that u² < m⁻¹. There will be, therefore, a limit for the pace u of the S′ frame with respect to the S frame in order for x′ and r′ to remain real. Calling this unspecified limiting pace U, where U² = m⁻¹, transformation (1a) now becomes the dual Ignatowsky transformation:

$x'=\frac{x-ur/U^2}{\sqrt{1-u^2/U^2}}; y'=y; z'=z; r'=\frac{r-ux}{\sqrt{1-u^2/U^2}}.\; (2)$

Step 3. To derive the dual-Galilean transformation from (2), set x′ = x, which from (2) requires that U = ∞, i.e., m = o.

Step 4. Add Einstein’s second postulate in dual form: light is always propagated in empty time with a definite lenticity c. In that case, c = U, the maximum pace. The constant c is the conversion factor from space to time.

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