- #1
kent davidge
- 933
- 56
I read that ##\delta^4(x-y)## is invariant under Lorentz transformations. I was trying to show myself this, so I procceded as follows.
The following integrals are both equal to 1 ##\int \delta^4(x-y) d^4 x## and ##\int \delta^4 (x'-y') d^4x'## so I assume they are equal to one another, as long as we integrate them over all ##\mathbb R^4##. Now ##d^4 x = d^4 x'## because the Lorentz metric has ##\sqrt {| \eta |} = 1##. So I conclude that ## \delta^4(x-y) = \delta^4(x'-y') ##. Is this correct?
The following integrals are both equal to 1 ##\int \delta^4(x-y) d^4 x## and ##\int \delta^4 (x'-y') d^4x'## so I assume they are equal to one another, as long as we integrate them over all ##\mathbb R^4##. Now ##d^4 x = d^4 x'## because the Lorentz metric has ##\sqrt {| \eta |} = 1##. So I conclude that ## \delta^4(x-y) = \delta^4(x'-y') ##. Is this correct?