Recent content by dungonyu

pc= Eγ. So (mc^2-Eγ)^2=(m'c^2)^2+(Eγ)^2 then (mc^2)^2-2mc^2*(Eγ)+(Eγ)^2=(m'c^2)^2+(Eγ)^2 rearranging terms and cancelling (Eγ)^2 we get 2mc^2*(Eγ)= (mc^2)^2- (m'c^2)^2 if we assert Ei = mc^2 and Ef = m'c^2, 2mc^2*(Eγ)= (Ei)^2-(Ef)^2= (Ei-Ef)(Ei+Ef)=ΔE(2Ei-ΔE) so by dividing 2mc^2, which is...

I don't understand. Shouldn't I approximate the energy of the emitted photon with the mass of the nucleus? And after the emission, both the photon and the nucleus carry non zero momentum. I have enriched my steps in attempt. And I just wonder whether my first assumptions on Ei and Ef are correct

Homework Statement Consider a nucleus which is initially at rest and in an excited state with energy Ei. It then decays to a lower energy state with energy Ef by emitting a gamma-ray photon. Show that the energy of the photon is approximately given by Eγ≈ΔE−(ΔE)^2/(2mc^2) where ΔE = Ei - Ef...

urm. Ok, I see your point. And now I understand the arguement: with the disapperance of x-ct, there's also an disapperance of x'-ct'. so there should be some constants times the whole thing but separately such as x-ct=ax'-bct'. and that word "in general" satisfies my worry, since I've been...

Oh, in that little book, as you've said, using a light signal from both orgins of coordinate systems: x=ct; x'=ct'.(in the appendix of deriving Lorentz Transformation) so x-ct=0 and x'-ct'=0. Then Einstein guessed there's some relationship between the two equations, thus x'-ct'=λ(x-ct); and...

I like the derivation in Wikipedia. Starting with : x'= ax + bt and t'=dt + ex, for some constants a, b, d and e. And arguing it's linear, otherwise, there will be acceleration, not an inertial frame. I am satisfied with this derivation. And it uses the two postulates: special case for x=ct and...

Is your concern is two equations which is equal to 0 and then they are equated with a factor of some constants? yeah, I also admit that is strange. Just like a=0 and b=0, then arguing a=kb, for some constant k. but, indeed in a mathematical sense, k can be anything!