Add Velocities: Equation for Spaceships Going 99%c

[ilikescience94]

If a spaceship were going 99%c, and inside that spaceship was another spaceship going 99%c, and so on and so on. I know that the equation for adding 2 velocities is (v1+v2)/(1+v1*v2/c^2), so would the equation in this scenario be (v1+v2+v3+...vn)/(1+v1*v2*v3*...vn/c^n)? Or is there something else some other equation to do this?

 

[mathman]

I suggest you try adding in one at a time. I got for 3.

(v1 + v2 + v3 + (v1v2v3)/c^2)/(1 + (v1v2 + v1v3 + v2v3)/c^2)

 

[ghwellsjr]

If a spaceship were going 99%c, and inside that spaceship was another spaceship going 99%c, and so on and so on. I know that the equation for adding 2 velocities is (v1+v2)/(1+v1*v2/c^2), so would the equation in this scenario be (v1+v2+v3+...vn)/(1+v1*v2*v3*...vn/c^n)? Or is there something else some other equation to do this?

If you do the calculation on the original equation for v1=v2=c we get c as the correct answer:

(c+c)/(1+c²/c²) = 2c/(1+1) = 2c/2 = c

But on your equation for three terms it is:

(c+c+c)/(1-+c³/c³) = 3c/(1+1) = 3c/2 = 1.5c

Clearly, this simple example shows that your method does not work.

 

[ilikescience94]

I suggest you try adding in one at a time. I got for 3.

(v1 + v2 + v3 + (v1v2v3)/c^2)/(1 + (v1v2 + v1v3 + v2v3)/c^2)

Ya, that works, I'm a little stumped on 4

(v1+v2+v3+v4)/(1+(v1*v2*v3+v1*v2*v4+v2*v3*v4)/(c^3) works but I didn't involve (v1*v2*v3*v4)/c^3), so I don't think that's right.

 

[ghwellsjr]

If you are specifically wanting to answer your original question where all the relative speeds are the same, there is a much easier way to do the calculation. You can calculate the Relativistic Doppler Factor for the initial speed, then take it to any power you like and then do the reversed Relativistic Doppler calculation. For speeds expressed as a fraction of the speed of light, we use beta, β, and the formula for the RDF is (which I will call R):

R = √((1+β)/(1-β))

and the reverse formula is:

β = (R²-1)/(R²+1)

So let's say your initial speed is β = 0.267949. R will calculate to be 1.316074. If you raise that to the 4th power, you get 3 and if you then calculate the final speed you get 0.8c.

For speeds closer to c, you have to carry the arithmetic out to a lot of places past the decimal point.

For 99%c, β = 0.99 and R = 14.106736. If we raise that to the tenth power we get 312079600999 and if we calculate β we get 0.99999999999999999999997946478607c.

 

[yuiop]

If a spaceship were going 99%c, and inside that spaceship was another spaceship going 99%c, and so on and so on. I know that the equation for adding 2 velocities is (v1+v2)/(1+v1*v2/c^2), so would the equation in this scenario be (v1+v2+v3+...vn)/(1+v1*v2*v3*...vn/c^n)? Or is there something else some other equation to do this?

This is my stab at a generic formula, but to save typing I need to introduce some new notation. If n is the number of velocities we are adding together, then v(n,x) means the sum of all the possible unique combinations of x velocities multiplied together. For example, if there are n=4 velocities, then v(n:3) = (v4v3v2+v4v3v1+v4v2v1+v3v2v1). If n=x then there is only one unique combination as in for example v(4:4) = v4v3v2v1. If n<x then v(n:x)=0, so there is no need to include those combinations.

For 2 velocities:

$\frac{(v1+v2)}{1+v(2:2)/c^2}$

For 3 velocities:

$\frac{(v1+v2+v3)+v(3:3)/c^2}{1+v(3:2)/c^2}$

For 4 velocities:

$\frac{(v1+v2+...v4)+v(4:3)/c^2}{1+v(4:2)/c^2+v(4:4)/c^4}$

For 5 velocities:

$\frac{(v1+v2+...v5)+v(5:3)/c^2+v(5:5)/c^4}{1+v(5:2)/c^2+v(5:4)/c^4}$

I have not tested beyond 5 velocities, but there appears to be a pattern emerging, so if it holds, the equation for 9 velocities, for example, would be:

$\frac{(v1+v2+...v9)+v(9:3)/c^2+v(9:5)/c^4+v(9:7)/c^6+v(9:9)/c^8}{1+v(9:2)/c^2+v(9:4)/c^4+v(9:6)/c^6+v(9:8)/c^8}$

If the notation is difficult, comparing my equation for 3 velocities with the one given by mathman might help:

I suggest you try adding in one at a time. I got for 3.

(v1 + v2 + v3 + (v1v2v3)/c^2)/(1 + (v1v2 + v1v3 + v2v3)/c^2)

Just for reference, the number of possible unique combinations is given by the factorial formula n!/(x!(n-x)!) and some of these equations get very unwieldy very rapidly for large values of n.

Of course, if all the velocities are equal, then ghwellsjr's method is a lot simpler.

 

[mathman]

To simplify the calculation, I suggest that rather than one at a time, combine them using previous combinations. For example to get the sum for four, add the expressions for two pairs, etc.

 

[The_Duck]

A simpler way of doing this is to define the "rapidity" $\eta = \tanh^{-1} (v/c)$. Velocities don't add straightforwardly, but rapidities do. Then to add a bunch of velocities, you can first compute the total rapidity

$\eta = \tanh^{-1}(v_1/c) + \tanh^{-1}(v_1/c) + \tanh^{-1}(v_1/c) + ...$

and then compute the total velocity from the total rapidity

$v = c \tanh \eta = c \tanh [\tanh^{-1}(v_1/c) + \tanh^{-1}(v_1/c) + \tanh^{-1}(v_1/c) + ... ]$

which is a general formula for adding any number of velocities.

 

[yuiop]

If you like playing around with formulas, another generic formula for multiple velocity addition, (based on the definition of the hyperbolic tangent) is:

$v = \frac{(1+v1)(1+v2)...(1+vn) - (1-v1)(1-v2)...(1-vn)}{(1+v1)(1+v2)...(1+vn) + (1-v1)(1-v2)...(1-vn)}$

using units where c=1.

 

[ilikescience94]

This is my stab at a generic formula, but to save typing I need to introduce some new notation. If n is the number of velocities we are adding together, then v(n,x) means the sum of all the possible unique combinations of x velocities multiplied together. For example, if there are n=4 velocities, then v(n:3) = (v4v3v2+v4v3v1+v4v2v1+v3v2v1). If n=x then there is only one unique combination as in for example v(4:4) = v4v3v2v1. If n<x then v(n:x)=0, so there is no need to include those combinations.

For 2 velocities:

$\frac{(v1+v2)}{1+v(2:2)/c^2}$

For 3 velocities:

$\frac{(v1+v2+v3)+v(3:3)/c^2}{1+v(3:2)/c^2}$

For 4 velocities:

$\frac{(v1+v2+...v4)+v(4:3)/c^2}{1+v(4:2)/c^2+v(4:4)/c^4}$

For 5 velocities:

$\frac{(v1+v2+...v5)+v(5:3)/c^2+v(5:5)/c^4}{1+v(5:2)/c^2+v(5:4)/c^4}$

I have not tested beyond 5 velocities, but there appears to be a pattern emerging, so if it holds, the equation for 9 velocities, for example, would be:

$\frac{(v1+v2+...v9)+v(9:3)/c^2+v(9:5)/c^4+v(9:7)/c^6+v(9:9)/c^8}{1+v(9:2)/c^2+v(9:4)/c^4+v(9:6)/c^6+v(9:8)/c^8}$

If the notation is difficult, comparing my equation for 3 velocities with the one given by mathman might help:



Just for reference, the number of possible unique combinations is given by the factorial formula n!/(x!(n-x)!) and some of these equations get very unwieldy very rapidly for large values of n.

Of course, if all the velocities are equal, then ghwellsjr's method is a lot simpler.

Could you or someone elaborate slightly on the (3:3) notation you used?

 

[yuiop]

Could you or someone elaborate slightly on the (3:3) notation you used?

It is probably not worth the effort trying to understand the notation as the method I outlined in #9 is more straightforward and the method mentioned by TheDuck in #8 is vastly superior again. For me it was just a curiosity that velocity addition could involve a method that involved finding all the possible permutations* of the velocities.

Your choice of (3:3) is not particularly instructive as there is only one unique way of choosing 3 items from a total of 3 items so that v(3:3) = (v1*v2*v3).

The first step of evaluating for example v(4:2) is to find all 6 unique ways of choosing a subset with 2 velocities from the total choice of 4 velocities with no repetitions, duplications and where order is not important. If we have a list of 4 velocities, (a,b,c,d), then this step can be automated by entering "subsets (a,b,c,d) with length 2" into the Wolfram alpha website.

The result is: {a, b} | {a, c} | {a, d} | {b, c} | {b, d} | {c, d}

The final step is to multiply all the elements in each subset and then add all the subsets together so that:

v(4:2) = (a*b)+(a*c)+(a*d)+(b*c)+(b*d)+(c*d)

For v(3:2) there are 3 velocities (a,b,c) and the possible unique combinations are {a, b} | {a, c} | {b, c} so v(3:2) = (a*b)+(a*c)+(b*c).

I have already given an example for v(4:3) so hopefully that is enough to be going on with.

I am interested if your interest is mathematical or perhaps you are trying to create an algorithm for a software project? If that is the case, the simplest approach is probably to just iteratively apply the standard relativistic velocity addition one velocity at a time in a loop. A similar approach would work just fine in a spreadsheet.

*More precisely "combinations" rather than permutations.

 

[ilikescience94]

If you like playing around with formulas, another generic formula for multiple velocity addition, (based on the definition of the hyperbolic tangent) is:

$v = \frac{(1+v1)(1+v2)...(1+vn) - (1-v1)(1-v2)...(1-vn)}{(1+v1)(1+v2)...(1+vn) + (1-v1)(1-v2)...(1-vn)}$

using units where c=1.

9O was just trying as much as you are to try to find answer for why we are here, but I came up short , but i sure like your answer better than john john lennons all you need is love and jesus's all ypu need is to nelieve that I am the one of the true one.

I was just using (3:3) as an example not as the basis btw.

Thank you for answereing as well. I Reallly can't show you a fair amount of gratitude in this universe as I am sure you know,but I hope I hope I show you in a less enlightend universe for you what it all means.

I came here in hopes of this answer and you gave me this answer, thank you.