Annus MirabilisEinstein in 1905
Reading preferences

Reading preferences

Keep every explanation and worked example on the page. Nothing moves or plays by itself, and the larger instruments load only when you ask.

Line length
Type size
Contrast
Paragraph spacing

special relativity

The equation explained

τ=a(t−vV2−v2x′),\tau = a\left( t - \frac{v}{V^2 - v^2} x' \right) ,
In wordsThe moving system's time is a factor still to be found times the stationary time less the speed of the moving system over the square of the velocity of light less the square of that speed, times the distance along the motion.OverviewThe moving system's time, at last: it runs with the stationary time, and it is set back by an amount that grows with the distance along the motion. One factor is still unknown and is settled later in the section.Full explanationThree facts are put together here. The rate of change across the motion is zero, so neither of the two transverse distances appears. The equation of the previous displays fixes the ratio of the two remaining rates of change. And the equations were assumed linear at the head of the section, so a function whose rates of change are constant is that constant times its argument, with nothing else. The constant in front, which the paper calls aa, is left standing: the paper says only that it is a function of vv still to be found, and φ(v)\varphi(v) is what it becomes on p. 900. The paper also fixes the zero of the moving system's time by putting τ=0\tau = 0 at t=0t = 0 at the origin of kk, which is why there is no constant added on the right.Every step∂τ∂y=0,∂τ∂z=0\frac{\partial \tau}{\partial y} = 0, \quad \frac{\partial \tau}{\partial z} = 0From the rays sent across the motion: neither transverse distance can appear in τ\tau.∂τ∂t=a\frac{\partial \tau}{\partial t} = aThe equations were assumed linear, so each rate of change is a constant. The paper names this one aa and leaves it to be found.∂τ∂x′=−vV2−v2a\frac{\partial \tau}{\partial x'} = - \frac{v}{V^2 - v^2} aThe equation of the previous display, with that constant put in: the two rates of change stand in a fixed ratio.τ=at−avV2−v2x′\tau = a t - a \frac{v}{V^2 - v^2} x'A linear function with these constant rates of change, and no constant term: the paper fixes τ=0\tau = 0 at t=0t = 0 at the origin of kk.τ=a(t−vV2−v2x′)\tau = a \left( t - \frac{v}{V^2 - v^2} x' \right)Take aa outside, which is the line as printed.Historian's marginThe linearity is asserted, not derived: the paper gives the homogeneity of space and time as its reason in one sentence (p. 898) and does not return to it. The factor aa is left unknown on purpose, and the paper spends the rest of the section, through the third system K′K' and a rod held across the motion, showing that it is one.Also

Read it in the paper