# Write down what metric this transformation preserves based on animation

https://www.desmos.com/calculator/u5qpd135uc

I made it because I wanted to compare and contrast it with the Lorentz boost.

The transformation should move a point to somewhere on the same green curve depending on the parameter $$b$$. And the red curves get mapped to each other.

How would I write out the transformation in concrete notation, and how would I determine the metric that this transformation preserves? (i.e. lorentz boost preserves the lorentz metric) Minkowski space.

Maybe it doesn't preserve anything $$-$$ I don't know at the moment. I'm working on an answer.

• Can you describe what the transformation does? I am guessing it should move a point to somewhere on the same green curve, depending on the parameter $b$? And such that the red curves get mapped to each other? – Milten Apr 15 at 19:59
• Yeah the transformation should move a point to somewhere on the same green curve depending on the parameter $b$. And the red curves get mapped to each other. I'll edit to include that – geocalc33 Apr 15 at 20:03
• The metric preserved is $x^2-y^2$ which is met in relativity under the form $x^2-c^2t^2$ or $x^2+y^2+z^2-c^2t^2$ (see this answer(math.stackexchange.com/q/3564572)) – Jean Marie Apr 15 at 20:37
• @Milten : you are right. I have had a too quick view on the Desmos graphic. – Jean Marie Apr 15 at 20:51
• @JeanMarie if you retract to linear coordinates, i.e. $u=\ln(x)$ and $v=\ln(t)$ it's isomorphic to the lorentz metric after re-scaling. So I think the metrics are the same in essence – geocalc33 Apr 15 at 21:10

I will forget about your parameter $$s$$, since it just scales $$b$$. So points are of the form: $$(x,t) = \left(\exp\left({\sqrt{n}e^{b}}\right),\exp\left({\sqrt{n}e^{-b}}\right)\right)$$ If we add a constant $$\Delta b$$ to $$b$$, we see that we can write the transformation as: $$(x, t)\mapsto (x',t') = \left(x^{\exp(\Delta b)}, t^{\exp(-\Delta b)}\right)$$ This transformation maps red curves to each other and points stay on the green curves.

"The metric it preserves" is just that points stay on the green curves. These curves are given by $$t=\exp\left(\frac{n}{\log x}\right)$$ So if $$(x,t)$$ satisfies the above for some $$n$$, then also $$t'=\exp\left(\frac{n}{\log x'}\right)$$. The interesting thing is that $$n$$ is constant. We can solve for $$n$$: $$n = \log x \cdot\log t$$ This is a preserved quantity of the transformation. I.e. you might say $$\log x \log t$$ is a preserved metric of $$(x,t)$$.

NOTE: I worked under the assumption $$x>1$$, $$t>1$$. In other regions, you may need to adjust the results.

EDIT, according to the discussion in the comments: Let $$(z,w) = \frac12(\log x + \log t, \log x - \log t)$$ I.e. we take log and rotate and scale. Then $$z^2 - w^2 = \log x\log t$$ so $$z^2-w^2$$ is invariant. With $$k=e^{\Delta b}$$, the transformation becomes: $$(z',w') = \frac12\left((k+\tfrac1k)z + (k-\tfrac1k)w, (k-\tfrac1k)z + (k+\tfrac1k)w\right)$$ Finally, if we substitute $$1/\sqrt{1-v^2} = (k+k^{-1})/2$$, we get: $$(z',w') = \frac{1}{\sqrt{1-v^2}}\left(z - vw, w- v z\right)$$ which is the Lorentz transformation with $$c=1$$. If we want to introduce $$c$$ into the original transformation, we could work backwards from here.

• I feel like if you retract to linear coordinates, i.e. $u=\ln(x)$ and $v=\ln(t)$ it's isomorphic to the lorentz metric after re-scaling? – geocalc33 Apr 15 at 21:03
• I also realise that we can just write $k:=e^{\Delta b}$. Then $(u',v')=(ku, v/k)$, and the preserved quantity is $uv$. This does indeed define a hyperbola, so I think you may be right about the connection to the lorentz metric. – Milten Apr 15 at 21:11
• We just need to rotate the $(u,v)$ system by $\pi/4$, i'd say. – Milten Apr 15 at 21:13
• To be concrete: $(z,w)=\frac12(u+v,u-v)$ preserves $z^2-w^2$. You might want to check if the tranformation on $(z,w)$ is the usual Lorents transformation. – Milten Apr 15 at 21:45
• Looks great (+1) – geocalc33 Apr 16 at 0:57