I think I understand this. https://users.sussex.ac.uk/~waa22/relativity/Complete_Derivation_files/derivation.pdf Is there more to it than assuming that time is not an absolute thing and the speed of light is the absolute thing?
we arent helping u figure it out urself
isnt time absolute as well?
according to special theory of relativity, no \(\color{#0cbb34}{\text{Originally Posted by}}\) @Florisalreadytaken isnt time absolute as well? \(\color{#0cbb34}{\text{End of Quote}}\)
@Florisalreadytaken what do you mean by "more to it" I mean any other assumptions?
what about this \[ (c t)^{2}=(v t)^{2}+\left(c t_{0}\right)^{2}\] \[c^{2} t^{2}=v^{2} t^{2}+c^{2} t_{0}^{2} \] \[\Rightarrow c^{2} t_{0}^{2}=\left(c^{2}-v^{2}\right) t^{2} \] \[\Rightarrow t_{0}^{2}=\left(1-\frac{v^{2}}{c^{2}}\right) t^{2} \] \[t \sqrt{1-\frac{v^{2}}{c^{2}}}=t_{0} \] \[\Rightarrow t=\frac{t_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\]
Where do you get this from ? \[(c t)^{2}=(v t)^{2}+\left(c t_{0}\right)^{2}\]
If a spherical light wave is produced at the origin of the S and S' inertial frames at t = 0, the statement about the constant velocity of light in different inertial frames can be expressed using that formula
I'm confused. What do you mean by spherical? Is it as in spherical wave, like light originated from a point source?
yes, thats right
what does have to do with the derivation. I was thinking more of a photon bouncing from one mirror to another back and forth.
because it emerges from one point and then when it hits the mirror, it reflects and so on
still not clear how it matters. We are thinking this is in photon level aren't we? Is there any importance in thinking about spherical wave fronts?
anyway let's just forget about what I said -- what you got on that pdf file is very well explained, and, based on what you're asking for, I don't believe there's anything more.
My suspicion was if it is a toned down version of a more advanced version. Ok thanks for help.
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