Suppose we are moving two electrons in opposite direction both having a speed of 1.6e8 meter per sec.When they will cross each other then their relative velocity will be 3.2e8 ms_1.Isn't it faster than speed of light?
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1$\begingroup$ Possible duplicate of If I run along the aisle of a bus traveling at (almost) the speed of light, can I travel faster than the speed of light? $\endgroup$– SmarthBansalCommented Nov 21, 2018 at 5:30
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$\begingroup$ Possible duplicate of physics.stackexchange.com/q/7446 $\endgroup$– PM 2RingCommented Nov 21, 2018 at 5:30
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$\begingroup$ Also see physics.stackexchange.com/q/192891 and the links on that page. $\endgroup$– PM 2RingCommented Nov 21, 2018 at 5:34
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$\begingroup$ no research effort $\endgroup$– m4r35n357Commented Nov 21, 2018 at 9:57
2 Answers
The problem is that although your frame of reference(the electron) is moving at speeds close to light, you are using GALILEAN TRANSFORMATION. You should use LORENTZ TRANSFORMATION to get the right relative speed.
Below is the Relativistic Velocity Transformation : $$ u=\frac{\acute{u}+v}{1+\acute{u}v/c^2 }\ $$ Where,
u=velocity of particle as measured from frame S
$\acute{u} $=velocity of particle as measured from frame $\acute{S}$
v=velocity of frame $\acute{S}$
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2$\begingroup$ With common questions like this, it's better to search for an existing duplicate than to write a fresh answer. $\endgroup$– PM 2RingCommented Nov 21, 2018 at 5:38
Just to add to the previous answer:
Lorentz transformations were constructed with the idea that no matter which frame you’re in, the speed of light is always mapped to the same number. This is relatively non trivial to imagine. So in all frames, the speed of light is preserved, whereas other speeds (less than the speed of light of course) get mapped to some other value. It turns out that this is a correct way to switch between frames as per STR, and using the velocity addition formula, derived from the Lorentz transformation, tells us that relative speed between two photons will always be the ‘speed of light’.