Daniel Audet

This is an indication that the relativistic velocity addition formula is a special case of Bayes formula. I have no physical interpretation of this mathematical truth. All I know is that the (well known) change from velocities to rapidities (v -> 1/2 ln((1+v/c)/(1-v/c))) that makes the relativistic velocity addition formula additive is mathematically the same change (suggested by Alan Turing) from probability to weight of evidence (p -> ln(p/(1-p))) that makes the Bayes formula additive.

Your example was P(A and B) = 0.3, P(A and not B) = 0.2 and P(not A and B) = 0.3. But this is not a counter-example since it doesn't meet the hypothesis P(B|A) = P(B'|A'). To make it fit the hypothesis I changed your example to P(A and B) = 0.3, P(A and not B) = 0.2 and P(not A and B) = 0.2.

You are right what I wrote is false. What I should have written is P(A) = 1 - P(A') = (1+v1/c)/2 P(B|A) = P(B'|A') = 1 - P(B|A') = 1 - P(B'|A) = (1+v2/c)/2 and your example becomes P(A and B) = 0.3, P(A and not B) = 0.2, P(not A and B) = 0.2, P(not A and not B) = 0.3. Then v1 = 0c, v2 = 0.1c and v3 = 0.1c and everything works out well. Sorry about that. You spotted the same problem as uncool did. What I should have written is P(A) = 1 - P(A') = (1+v1/c)/2 P(B|A) = P(B'|A') = 1 - P(B|A') = 1 - P(B'|A) = (1+v2/c)/2 that will satisfy every formula checker. Thanks for pointing it out.

Thanks for prooving my point.

Then would you please tell me what is the relativistic velocity addition formula for paralell (and opposite) velocities?

The relativistic velocity addition formula is in the wikipedia Rapidity page in the form since v3 = (v1 + v2)/(1+v1 v2/c^2) is algebraically equivalent to ln((1+v3/c)/(1-v3/c)) = ln((1+v1/c)/(1-v1/c)) + ln((1+v2/c)/(1-v2/c)). Beside, if the formula was only valid for v > 0 that would mean only valid for paralell and same direction velocities. Although I agree with you that my proposition if not physics since it is not even wrong. What I point out is a resemblance in the math and sometime following what the math suggest can lead to physics. I don't "just constrain a variable to be between 0 and 1" read my post again. I not only show that (1+v/c)/2 looks like a probability but also that it behave like a probability by obeying the Bayes theorem. The trick is to realise that the change from velocities to rapidities that makes the relativistic velocity addition formula additive is the same as Alan Turing's change from probability to weight of evidence that makes the Bayes formula additive.

From wikipedia Rapidity : "The speed of light c being finite, any velocity v is constrained to the interval -c<v<c and the ratio v/c satisfies -1<v/c<1."

You are wrong, and I may be not even wrong, but I am not wrong. Look at wikipedia "Rapidity" http://en.wikipedia.org/wiki/Rapidity

Here is the proof: -c < v < c => -1 < v/c < 1 => 0 < 1+ v/c < 2 => 0 < (1+v/c)/2 <1

I recently noticed a mathematical ressemblance between special relativity and bayesian probability: If A and B are events such that P(A) = 1 - P(A') = (1+v1/c)/2 P(B|A) = P(B'|A') = 1 - P(B'|A') = 1 - P(B|A') = (1+v2/c)/2 where v1 and v2 are relativistic one dimentional velocities then according to Bayes formula P(A|B) = (1+v3/c)/2 where v3 = (v1 + v2)/(1 + v1 v2/c^2) witch is the relativistic velocity addition formula. So the math suggest that the quantity 0 < (1+v/c)/2 < 1 is a probability since it behave like probability. This might indicate a fundamental connection between probability theory, special relativity and hyperbolic geometry. Anyone care?