Question

相对论速度叠加公式我自己推导的错哪了?

Answer 1

速度加速度质量动量能量力的洛仑兹变换公式推导顺便都给你

1

首先给出坐标的洛仑兹变换公式

x'=γ(x-vt) x=γ(x'+vt)

y'=y y=y'

z'=z z=z'

t'=γ(t-vx/c^2) t=γ(t'+vx’/c^2)

2

推导速度的洛仑兹变换公式

由t'=γ(t-vx/c^2) t=γ(t'+vx’/c^2)可知

dt/dt'=1/γ(1-vUx/c^2)=γ(1+vUx'/c^2)此式备用

Ux'=dx'/dt'=(dx'/dt)(dt/dt')

dx'/dt=d[γ(x-vt)]/dt=γ(Ux-v)再带入(dt/dt')=1/γ(1-vUx/c^2)

Ux'=γ(Ux-v)/γ(1-vUx/c^2)=(Ux-v)/(1-vUx/c^2)

同理Uy'=Uy/γ(1-vUx/c^2)=Uy/γ(1-vUx/c^2)

Uz'=Uz/γ(1-vUx/c^2)=Uz/γ(1-vUx/c^

把v换成-v,带'与不带'的量互换就可以得到逆变换

结论

Ux'=(Ux-v)/(1-vUx/c^2) Ux=(Ux'+v)/(1+vUx'/c^2)

Uy'=Uy/γ(1-vUx/c^2) Uy=Uy'/γ(1+vUx'/c^2)

Uz'=Uz/γ(1-vUx/c^2) Uz=Uz'/γ(1+vUx'/c^2)

3

推导加速度的洛仑兹变换公式

a_x'=dUx'/dt'=(dUx'/dt)(dt/dt')=

{d[(Ux-v)/(1-vUx/c^2)]/dt}(dt/dt')=

{[(dUx/dt)(1-vUx/c^2)-(Ux-v)(-dUx/dt)v/c^2)]/(1-vUx/c^2)^2}/γ(1-vUx/c^2)=a_x/γ^3(1-vUx/c^2)^3

a_y'=dUy'/dt'=(dUy'/dt)(dt/dt')=

{d[Uy/γ(1-vUx/c^2)]/dt}(dt/dt')=

{[(dUy/dt)γ(1-vUx/c^2)-γUy(-dUx/dt)v/c^2)]/γ^2(1-vUx/c^2)^2}/γ(1-vUx/c^2)=[a_y+a_xUyv/(c^2-vUx)]/γ^2(1-vUx/c^2)^2

同理a_z'=[a_z+a_xUzv/(c^2-vUx)]/γ^2(1-vUx/c^2)^2

结论

a_x'=a_x/γ^3(1-vUx/c^2)^3

a_y'=[a_y+a_xUyv/(c^2-vUx)]/γ^2(1-vUx/c^2)^2

a_z'=[a_z+a_xUzv/(c^2-vUx)]/γ^2(1-vUx/c^2)^2

a_x= a_x'/γ^3(1+vUx'/c^2)^3

a_y=[a_y'-a_x'Uy'v/(c^2+vUx')]/γ^2(1+vUx'/c^2)^2

a_z=[a_z'-a_x'Uz'v/(c^2+vUx')]/γ^2(1+vUx'/c^2)^2

4

质量能量变换公式

m=m0/(1-UU/cc)^(1/2)

m'=m0/(1-U'U'/cc)^(1/2)

m'=m(1-UU/cc)^(1/2)/(1-U'U'/cc)^(1/2)

其中(1-U'U'/cc)=1-(Ux'Ux'+Uy'Uy'+Uz'Uz')

带入速度变换公式可以得出

(1-U'U'/cc)^(1/2)=(1-UU/cc)^(1/2)/γ(1-vUx/c^2)

结论

m'=m γ(1-vUx/c^2)

m =m'γ(1-vUx/c^2)

再由E=mc^2 E'=m'c^2可以得到

E'=E γ(1-vUx/c^2)

E =E'γ(1-vUx/c^2)

5

动量能量变换公式

Px'=m'Ux'=mγ(1-vUx/c^2)*(Ux-v)/(1-vUx/c^2)=

γ(mUx-mv)=γ(Px-Ev/c^2)

Py'=m'Uy'=mγ(1-vUx/c^2)*Uy/γ(1-vUx/c^2)=Py

Pz'=Pz

E'=Eγ(1-vUx/c^2)=γ(E-EvUx/c^2)=γ(E-EvUx/c^2)=γ(E-vPx)

结论

Px'=γ(Px-Ev/c^2) Px=γ(Px'+E'v/c^2)

Py'=Py Py=Py'

Pz'=Pz Pz=Pz'

E' =γ(E-vPx) E =γ(E'+vPx')

6

力的洛仑兹变换公式

fx'=dPx'/dt=(dPx'/dt)(dt/dt')=[dγ(Px-Ev/c^2)/dt](dt/dt')=

γ[dPx/dt-(v/c^2)dE/dt](dt/dt')

dPx/dt是fx,dE/dt是fx的做功功率dE/dt=fxUx+fyUy+fzUz

带入可得

fx'=γ[fx-(v/c^2)(fxUx+fyUy+fzUz)]/γ(1-vUx/c^2)=

fx-(fyUy+fzUz)v/(c^2-vUx)

fy'=(dPy'/dt)(dt/dt')=(dPy/dt)(dt/dt')=fy/γ(1-vUx/c^2)

fz'=(dPz'/dt)(dt/dt')=(dPz/dt)(dt/dt')=fz/γ(1-vUx/c^2)

结论

fx'=fx-(fyUy+fzUz)v/(c^2-vUx)

fy'=fy/γ(1-vUx/c^2)

fz'=fz/γ(1-vUx/c^2)

fx= fx'+(fy'Uy'+fz'Uz')v/(c^2+vUx')

fy= fy'/γ(1+vUx'/c^2)

fz= fz'/γ(1+vUx'/c^2)