- #1
larny
- 25
- 0
I have read the Einstein June 30, 1905 paper "On the Electrodynamics of moving bodies" but I can't see how he did the following (the
text I've quoted below is on page 6 in my .pdf version):
" 1/2 [ T(0,0,0,t) + T(0,0,0,t + x'/(c-v) + x'(c+v)] = T[x',0,0,t + x'/(c-v)]
Hence, if x' be chosen infinitesimally small,
1/2 [1/(c-v) + 1/(c+v)] dT/dt= dT/dx' + 1/(c-v) dT/dt "
I don't see how x'/(c-v) and x'/(c+v) becomes 1/(c-v) and 1/(c+v) when x' is chosen infinitesimally small.
I can do it by differentiation, eg. dT/x' = 1/(c-v) + 1/(c+v) etc. but I feel he did it by a more fundamental method such as
f'(x) = [f(x + h) - f(x)]/h but I can't see how to use this technique in this case.
Any assistance will be appreciated.
text I've quoted below is on page 6 in my .pdf version):
" 1/2 [ T(0,0,0,t) + T(0,0,0,t + x'/(c-v) + x'(c+v)] = T[x',0,0,t + x'/(c-v)]
Hence, if x' be chosen infinitesimally small,
1/2 [1/(c-v) + 1/(c+v)] dT/dt= dT/dx' + 1/(c-v) dT/dt "
I don't see how x'/(c-v) and x'/(c+v) becomes 1/(c-v) and 1/(c+v) when x' is chosen infinitesimally small.
I can do it by differentiation, eg. dT/x' = 1/(c-v) + 1/(c+v) etc. but I feel he did it by a more fundamental method such as
f'(x) = [f(x + h) - f(x)]/h but I can't see how to use this technique in this case.
Any assistance will be appreciated.