Relativistic velocity addition formula from Bayes Theorem

[Daniel Audet]

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?

[xyzt]

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?

 

 

Nope, especially since it isn't true that 0 < (1+v/c)/2 < 1 since, in reality , 0.5 < (1+v/c)/2 < 1

[Daniel Audet]

Nope, especially since it isn't true that 0 < (1+v/c)/2 < 1 since, in reality , 0.5 < (1+v/c)/2 < 1

 

Here is the proof:

 

-c < v < c

=> -1 < v/c < 1

=> 0 < 1+ v/c < 2

=> 0 < (1+v/c)/2 <1

[swansont]

Here is the proof:

 

-c < v < c

=> -1 < v/c < 1

=> 0 < 1+ v/c < 2

=> 0 < (1+v/c)/2 <1

 

v > 0 so while v > -c is true, and so is (1+v/c)/2 > 0, so is (1+v/c)/2 > 0.5

[xyzt]

Here is the proof:

 

-c < v < c

=> -1 < v/c < 1

=> 0 < 1+ v/c < 2

=> 0 < (1+v/c)/2 <1

Nope, the formula assumes v>0. swansont pointed out the same thing.

[Daniel Audet]

Nope, the formula assumes v>0. swansont pointed out the same thing.

 

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

[xyzt]

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

Yes, so? It has nothing to do with the subject being discussed, nor with the fact that two people have explained to you that you are wrong and why you are wrong, yet you persist.

[Daniel Audet]

Yes, so? It has nothing to do with the subject being discussed, nor with the fact that two people have explained to you that you are wrong and why you are wrong, yet you persist.

 

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."

[xyzt]

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."

 

 

Not in the formula you cited. In the formula you cited, 0<v<c. Time to give it a rest.

Besides, what you are doing isn't physics. Google "numerology".

Edited May 4, 2013 by xyzt

[swansont]

Again, for any speed v, (1+v/c)/2 > 0 is true, but so is (1+v/c)/2 > 0.5

 

The rapidity article is sloppy as they are referring to vectors without proper vector notation or analysis — it mixes vectors and scalars.

 

Just because you can constrain a variable to be between 0 and 1 does not make it a probability.

[Daniel Audet]

Not in the formula you cited. In the formula you cited, 0<v<c. Time to give it a rest.

Besides, what you are doing isn't physics. Google "numerology".

 

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.

 

Again, for any speed v, (1+v/c)/2 > 0 is true, but so is (1+v/c)/2 > 0.5

 

The rapidity article is sloppy as they are referring to vectors without proper vector notation or analysis — it mixes vectors and scalars.

 

Just because you can constrain a variable to be between 0 and 1 does not make it a probability.

 

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.

[xyzt]

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.

Numerology never leads to physics. Besides, the formula is valid for the simplified case of same direction velocities, you guessed right.

[Daniel Audet]

Numerology never leads to physics. Besides, the formula is valid for the simplified case of same direction velocities, you guessed right.

 

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

[xyzt]

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

 

 

[tex]w=\frac{u-v}{1-uv/c^2}[/tex]

 

And no, you cannot do any numerology with it.

[Daniel Audet]

[tex]w=\frac{u-v}{1-uv/c^2}[/tex]

 

Thanks for prooving my point.

[xyzt]

Thanks for prooving my point.

I learned very early on that you can never explain physics to a crackpot, so , I guess that you "won", whatever your "point" is.

Edited May 5, 2013 by xyzt

[Phi for All]

I learned very early on that you can never explain physics to a crackpot, so , I guess that you "won", whatever your "point" is.

!

Moderator Note

Let's focus on the ideas and not the people who have them, to keep things civil.

[uncool]

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?

For one thing, it's false.

 

Let's say we have that P(A and B) = 0.3, P(A and not B) = 0.2, P(not A and B) = 0.3, P(not A and not B) = 0.2. Then P(A) = 0.5, so v1 = 0c; P(B|A) = P(B and A)/P(A) = 0.3/0.5 = 0.6, so v2 = 0.1c. v3 is then 0.1c, so according to your formula, P(A|B) = 0.6. But P(A|B) = P(A and B)/P(B) = 0.3/(P(A and B) + P(not A and B)) = 0.3/(0.3 + 0.3) = 0.5.

 

You forgot about P(B).

=Uncool-

Edited May 5, 2013 by uncool

[ydoaPs]

Aside from the whole not meeting the requirement of probability theory that zero is a possible value for probability, it still doesn't follow from the probability calculus at all.

 

P(B|A) = P(B'|A') is wrong.

 

P(B|A) = P(B'|A') = 1 - P(B'|A') would mean that all of those formulas must take the value of 1/2. So, since, according to the OP, P(B|A) = P(B'|A') = 1 - P(B'|A') = 1 - P(B|A') = (1+v₂/c)/2, 1/2=(1+v₂/c)/2. This means 1=1+v₂/c. Obviously, this means v₂/c=0. c is a constant, so v₂=0.

 

v3 = (v1 + v2)/(1 + v1 v2/c^2), so v₃=v₁.

 

Using the information in the OP (along with the derived result that v₃=v₁) and Bayes's Theorem, we get P(B)=1/2.

 

So, P(A)=P(A|B)x(P(B)/P(B|A))=P(A|B) showing A and B are independent which kind of blows tying them into relativity.

 

 

Going back to v₃=v₁, v₁+(v₁²)/c²=v₁+v₂ which gives v₁=c

 

So, our velocities that are completely independent have fixed values of v₁=c, v₂=0, and v₃=c. That doesn't sound much like relativity.

[Daniel Audet]

For one thing, it's false.

 

Let's say we have that P(A and B) = 0.3, P(A and not B) = 0.2, P(not A and B) = 0.3, P(not A and not B) = 0.2. Then P(A) = 0.5, so v1 = 0c; P(B|A) = P(B and A)/P(A) = 0.3/0.5 = 0.6, so v2 = 0.1c. v3 is then 0.1c, so according to your formula, P(A|B) = 0.6. But P(A|B) = P(A and B)/P(B) = 0.3/(P(A and B) + P(not A and B)) = 0.3/(0.3 + 0.3) = 0.5.

 

You forgot about P(B).

=Uncool-

 

 

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.

 

Aside from the whole not meeting the requirement of probability theory that zero is a possible value for probability, it still doesn't follow from the probability calculus at all.

 

P(B|A) = P(B'|A') is wrong.

 

P(B|A) = P(B'|A') = 1 - P(B'|A') would mean that all of those formulas must take the value of 1/2. So, since, according to the OP, P(B|A) = P(B'|A') = 1 - P(B'|A') = 1 - P(B|A') = (1+v₂/c)/2, 1/2=(1+v₂/c)/2. This means 1=1+v₂/c. Obviously, this means v₂/c=0. c is a constant, so v₂=0.

 

v3 = (v1 + v2)/(1 + v1 v2/c^2), so v₃=v₁.

 

Using the information in the OP (along with the derived result that v₃=v₁) and Bayes's Theorem, we get P(B)=1/2.

 

So, P(A)=P(A|B)x(P(B)/P(B|A))=P(A|B) showing A and B are independent which kind of blows tying them into relativity.

 

 

Going back to v₃=v₁, v₁+(v₁²)/c²=v₁+v₂ which gives v₁=c

 

So, our velocities that are completely independent have fixed values of v₁=c, v₂=0, and v₃=c. That doesn't sound much like relativity.

 

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.

[uncool]

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 haven't changed anything in what I used. I didn't use anything that was primed. Further, P(B|A') is not necessarily equal to P(B'|A). So no, my example still shows that your reasoning is incorrect.

=Uncool-

Edited May 6, 2013 by uncool

[Daniel Audet]

You haven't changed anything in what I used. I didn't use anything that was primed. Further, P(B|A') is not necessarily equal to P(B'|A). So no, my example still shows that your reasoning is incorrect.

=Uncool-

 

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.

[uncool]

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.

The fact that you've added this requirement to the usual probability stuff is unclear. Is there any reason for us to assume it, other than for your formula to work out?

=Uncool-

[Daniel Audet]

The fact that you've added this requirement to the usual probability stuff is unclear. Is there any reason for us to assume it, other than for your formula to work out?

=Uncool-

 

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.