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Post Closed as "Duplicate" by John Rennie, Qmechanic
edited tags; edited title
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Qmechanic
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what What is the speed of one photon of a beam of light with respect to the other photon of the same beam?

Just made the formulas more readable. The text clearly needs an edit as well.
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Ali
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as according to special relativity the relative speed is u'=(u-v)/1-(uv)/c^2=0/0$u'=\frac{(u-v)}{1-(uv)/c^2}=\frac{0}{0}$ because v=c$v=c$ the speed of second photon with respect to any inertial observer and u=c spee$u=c$ speed of first photon with respect to same observer so u'$u'$ the speed of first with respect to second photon should also be c$c$ according to second postulate of special relativity but how

as according to special relativity the relative speed is u'=(u-v)/1-(uv)/c^2=0/0 because v=c the speed of second photon with respect to any inertial observer and u=c spee of first photon with respect to same observer so u' the speed of first with respect to second photon should also be c according to second postulate of special relativity but how

as according to special relativity the relative speed is $u'=\frac{(u-v)}{1-(uv)/c^2}=\frac{0}{0}$ because $v=c$ the speed of second photon with respect to any inertial observer and $u=c$ speed of first photon with respect to same observer so $u'$ the speed of first with respect to second photon should also be $c$ according to second postulate of special relativity but how

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what is the speed of one photon of a beam of light with respect to the other photon of the same beam?

as according to special relativity the relative speed is u'=(u-v)/1-(uv)/c^2=0/0 because v=c the speed of second photon with respect to any inertial observer and u=c spee of first photon with respect to same observer so u' the speed of first with respect to second photon should also be c according to second postulate of special relativity but how