How is the relativistic velocity superposition formula proved? What s wrong with my own proof? 20

The scale effect of time and displacement you have considered repeatedly.

It's f=v+u'

u'= u√(1-u^2/c^2)

So f=v+u (1-u2 c 2).

Just give it a try, the theory of relativity is difficult, hehe, no.

Don't be surprised.

Let's assume k and k'System axes.

Parallel, k'The relative K macro moves in the X direction, and the relative K velocity is u

k'An object in the system is along x'Directional motion, relative k'The speed of the system is v',t'The displacement in time is x' =v't'

Transformed by Lorentz, the displacement of the motion of the body in the k system is.

x = x'+ut')/1-u^2/c^2) =v't'+ut')/1-u^2/c^2)

The time that the motion of the object in the k-system is elapsed.

t = t'+ux'/c^2)/√1-u^2/c^2) =t'+uv't'/c^2)/√1-u^2/c^2)

Then the velocity of the object in the k system is.

v = x / t = v't'+ut')/1-u^2/c^2)]/t'+uv't'/c^2)/√1-u^2/c^2)] v' +u)/(1 + uv'/c^2)

According to the coordinate transformation relation in the Lorentz transform, x'=γ（x-ut），t'= (t-ux c 2), by velocity is the first derivative of the coordinates versus time.

Get v'=dx'/dt'=d (x-ut) d (t-ux c 2), where time t is the independent variable.

u is a constant, so the equation can be reduced to v'=(dx-udt) (dt-udx c 2), fraction.

Divide the top and bottom by dt at the same time to get v'=(v-u) (1-uv c 2), i.e. the formula for the superposition of velocity in the x-direction. The same can be said for the direction of y and z.

Let's assume k and k'The axis of the system is parallel, k'The relative k system moves in the x direction, and the relative k system velocity is u

k'An object in the system is along x'Directional motion, relative k'The speed of the system is v'，t'The displacement in time is x' =v't'

By the Lorentz transformation, the displacement of the motion of the object in the k system is.

x = x'+ut')/1-u^2/c^2) =v't'+ut')/1-u^2/c^2)

The time that the motion of the object is experienced in the k-system.

t = t'+ux'/c^2)/√1-u^2/c^2) =t'+uv't'/c^2)/√1-u^2/c^2)

Then the velocity of the object in the k system is.

v = x / t = v't'+ut')/1-u^2/c^2)]/t'+uv't'/c^2)/√1-u^2/c^2)] v' +u)/(1 + uv'/c^2)

v(x)=dx dt= (dx-ut) ( (dt-udx c 2)) =(dx dt-u) (1-(dx dt)u c 2) =(v(x)-u) (1-v(x)u c 2) The same can be said for v(y),v(z).

Let the object be relative to the k system, k'Department and k'The velocities relative to the k system are u, u, respectively'and v, according to the Lorentz transform.

x'= (x - vt), t'= (t-vx c 2), is the coefficient of expansion).

Differentiation on both sides of the equation:

dx’=γ(dx - vdt)

dt’=γ(dt-vdx/c2)

Divide by two formulas: u'=dx'/dt'=(u-v)/(1-uv/c^2)

Let the object be relative to the k system, k'Department and k'The velocities relative to the k system are u, u, respectively'and v, according to the Lorentz transform.

x’=γxvt)

t’=γt-vx/c^2)

is the swelling number of matter).

Separately cover the yard branch on both sides of the formula differentiation:

dx’=γdx

vdt)dt’=γdt-vdx/c2)

Divide by two formulas: u'Mold demolition = dx'/dt'=(u-v)/(1-uv/c^2)

v(x)=dx dt= (dx-ut) ( (dt-udx c 2)) =(dx dt-u) (1-(dx dt)u c 2) =(v(x)-u) (1-v(x)u c 2) The same can be said for v(y),v(z).