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I have trouble understanding why the second postulate of Special Relativity implies the invariance of spacetime interval $\Delta s^2=-c^2\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2$.

Suppose we have two frames $O$ and $O'$. $O'$ moves at constant velocity $v$ relative to $O$. At initial time $t=t'=0$, observers at both frames emit a light beam in the x-direction. According to the second postulate, both observers agree that the speed of light is constant:

$$c=\frac{\Delta x}{\Delta t}=\frac{\Delta x'}{\Delta t'}$$

However, I don't understand how it follows that

$$-c^2\Delta t^2+\Delta x^2=-c^2\Delta t'^2+\Delta x'^2$$

Qmechanic
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John Davies
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  • Your formula $c=\frac{\Delta x}{\Delta t}$ applies only for light, or anything traveling at the speed of light. Using these relations yields zero for both sides of the equation. A zero result that is always true for light. For slower objects, try the Lorentz transformations to see what you get. Remember to transform both x and t. Also remember that $\Delta x$ refers to two positions and $\Delta t$ refers to two times. It is easiest to make x=x'=0 at t=t'=0 as your first location and time. If not, you must transform both positions and both times separately. – Ken Mellendorf May 12 '23 at 15:40

1 Answers1

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The basic idea is the following: If something moves with lightspeed in O, that means ${\Delta s}^2 = c^2 {\Delta t}^2 - {\Delta x}^2 - {\Delta y}^2 - {\Delta z}^2 = 0$. If something moves with lightspeed in O', that means ${\Delta s'}^2 = c^2 {\Delta t'}^2 - {\Delta x'}^2 - {\Delta y'}^2 - {\Delta z'}^2 = 0$. By Einstein's second postulate, those two conditions are equivalent. In Invariance of the relativistic interval it is proven that that implies ${\Delta s}^2 = \lambda {\Delta s'}^2$ for some $\lambda$. Now, following Landau-Lifshitz, one can argue $\lambda = 1$ the following way: Because of the homogeneity and isotropy of space $\lambda$ can only depend on the absolute value of the relative velocity of O' to O. Now assume you have 3 systems 1, 2 and 3. Then you have ${\Delta s_{1}}^2 = \lambda(v_{12}) {\Delta s_2}^2 = \lambda(v_{12}) \lambda(v_{23}) {\Delta s_3}^2$ but also ${\Delta s_1}^2 = \lambda(v_{13}) {\Delta s_3}^2$, so $ \lambda(v_{12}) \lambda(v_{23}) = \lambda(v_{13})$. The right hand side of the last equation obviously depends on the angle beetween $\vec v_{12}$ and $\vec v_{23}$ (since $\vec v_{13} = \vec v_{12} + \vec v_{23}$), but the left hand side doesn't. Therefore $\lambda(v_{13})$ must be independent of $v_{13}$ --> $\lambda$ is constant. Using that in the equation above, one gets $\lambda^2 = \lambda$ or $\lambda = 1$ ($\lambda$ must not be 0 since ${\Delta s'}^2$ is not always 0).

Tarik
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