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Let's consider two cars A & B

Both cars are at rest in position O. Now both

cars started moving in opposite direction with

same velocity 2m/s for A & B in opposite direction

So the speed of car B relative A is = 2+2= 4m/s, right?

Now let's consider two photons X & Y moving in opposite

direction, from a starting point O. each photon have its own velocity, that is ’c’ so what is the speed of ’Photon B’ relative To

photon A?

i know the speed of photon from the starting point is ’c’

but just like in the case of cars the relative speed of Photon B from Photon A should be added, isn't it? therefore it becomes 2*c??

if not why! why is it said that the spotted of light is constant from every frame of reference?

(btw please don't include a 3rd observer here)

Hello Sreenath,

This is a perplexing fact about speeds close to the speed of light. Actually, the following formula applies to both of your cases:

w = (u + v) / (1 + u*v/c^2)

where w is the relative speed between the 2 bodies. The speed of A is u and the speed of B is v.

Let's use the formula with the 2 cars:

w = (2 m/s + 2 m/s) / (1 + (2 m/s)^2/(3x10^8 m/s)) = (4 m/s) / (1 + 4/9x10^16)

The value of c^2 is so large, that the term 4/9x10^16 is essentially zero. Therefore the result of the formula is that w = 4 m/s.

It's only when u and/or v is close in value to c that the result of the formula is surprising. Using the formula on your 2nd case:

w = (c + c) / (1 + c^2/c^2) = 2*c / 2 = c

Isaac Newton never had any data that could have given him any suspicion about such a result. Einstein didn't have any data to suggest this either, but he was brilliant.

I hope this helps,

Steve

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