World Science Scholars
5.1 The Mathematics of Speed

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Sajith Perumbilavil
If we also consider a case, where V and W are both C, then (V+W)/(1+VW/C^2) gives a value C. But if the two beams are moving in the same direction, then: (V-W)/(1-VW/C^2) gives the undefined value (0/0). How do we interpret this?
Cristino Pérez Ramos
He was not looking!! Haha
Samyak Sanghvi
V=W THEN IT BECOMES (V-V)/(1-(V/C)^2)=0 (as v/c is not equal to 1 for them to cancel out the denominator)
Istvan Joba
@Sajith: I would interpret the result 0/0 as the question of "what's the speed of another object while I'm traveling at the speed of light" doesn't make sense in special realitivity. Why? When you measure the speed of another object (its "relative speed") you need to reference it against your "base speed" which you set to zero. If you travel at the speed of light you would need to make the speed of light your reference speed and set it to zero. By setting the speed of light to zero you are breaking the basic law of special relativity: "the speed of light is always a constant ~300m m/s for every observer". If you ignore this rule you're not working with the theory of special relativity, but a new theory. Good luck proving it with real world experiments! :) I found this article very helpful

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