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So we know that the invariant interval in a two-dimensional spacetime in special relativity is given by

$$ s = -c^2t^2 + x^2 = -c^{'2}t^{'2} + x^{'2}$$

So this scalar should hold true in all frames. I'm trying to show that it is true under a spatial translation, but am not sure how to start.

I tried letting $x' \rightarrow x+\triangle x $. But am not sure how to continue from here. Would appreciate if someone could point me in the right direction.

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    $\begingroup$ The invariant interval is $\Delta s^2 = - c^2 (t_2 - t_1)^2 + (x_2 - x_1)^2$, not what you have written. Translational invariance is then the statement that if $(t_1,x_1) \to (t_1+a_1,x_1+a_2)$ and $(t_2,x_2) \to (t_2+a_1,x_2+a_2)$ then $\Delta s^2 \to \Delta s^2$. $\endgroup$
    – Prahar
    Commented Mar 27, 2016 at 5:41
  • $\begingroup$ Please note that Physics.StackExchange is not a homework help site. Please read this Meta post on asking homework-like questions and this Meta post for "check my work" problems. $\endgroup$
    – Kyle Kanos
    Commented Mar 27, 2016 at 11:11

1 Answer 1

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If anyone else is interested, I figured it out. Consider a 2D spacetime

$$\triangle s^2= -c^2(t_2 -t_1)^2 + (x_2 - x_1)^2 $$

A translation would mean $ x_1 \rightarrow x_1+x_0 $.

Now we go back to the unprimed frame and consider a 2-vector $\textbf{A}_i = (t_i,x_i) $. Let me also set c = 1.

To get the difference, $$\textbf{A}_2 - \textbf{A}_1 = (t_2-t_1,x_2-x_1)$$

Going to the primed frame would give $$\textbf{A}'_2 - \textbf{A}'_1 = (t_2-t_1,(x_2+x_0)-(x_1+x_0))$$ $$ = (t_2-t_1,x_2-x_1)$$

Therefore if we take the square of it, $$\triangle s^2 = \triangle s'^2$$

edit: didnt notice prahar's comment but yeah! thanks for pointing out my (mis)notation.

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