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Velocity addition expressed as gamma 
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#1
Dec1310, 08:46 PM

P: 2,258

if object A is moving with gamma=x in one direction and
object B is moving with gamma=y in the opposite direction then what is the gamma factor for object A as it would be calculated by object B? I know the velocity addition rule for relativity but I cant find anything on the net about velocity expressed as gamma. 


#2
Dec1310, 09:39 PM

P: 429

x = sqrt(1  (vA / c)^2), so rearranging gives x^2 = 1  (vA / c)^2 (vA / c)^2 = 1  x^2 vA / c = sqrt(1  x^2) Likewise for y, we would have vB / c = sqrt(1  y^2) and similarly for how B measures A, we would have vBA / c = sqrt(1  z^2) So applying the relativistic velocity addition formula, we get vBA / c = (vA / c + vB / c) / (1 + (vA / c) (vB / c)) sqrt(1  z^2) = [sqrt(1  x^2) + sqrt(1  y^2)] / [1 + sqrt(1  x^2) sqrt(1  y^2)] 1  z^2 = [(1  x^2) + 2 sqrt(1  x^2) sqrt(1  y^2) + (1  y^2)] / [1 + 2 sqrt(1  x^2) sqrt(1  y^2) + (1  x^2) (1  y^2)] 1  z^2 = [2  x^2  y^2 + 2 sqrt(1  x^2) sqrt(1  y^2)] / [2  x^2  y^2 + x^2 y^2 + 2 sqrt(1  x^2) sqrt(1  y^2)] 1  z^2 = 1  x^2 y^2 / [2  x^2  y^2 + x^2 y^2 + 2 sqrt(1  x^2) sqrt(1  y^2)] z^2 = x^2 y^2 / [2  x^2  y^2 + 2 sqrt(1  x^2) sqrt(1  y^2)] z^2 = x^2 y^2 / [1 + sqrt(1  x^2) sqrt(1  y^2)]^2 z = x y / [1 + sqrt(1  x^2) sqrt(1  y^2)] 


#3
Dec1310, 10:08 PM

P: 2,258

good lord.
well thank you but I think there might be a few errors in there. http://en.wikipedia.org/wiki/Lorentz_factor 


#4
Dec1310, 10:13 PM

P: 429

Velocity addition expressed as gamma



#5
Dec1310, 10:21 PM

P: 2,258



#6
Dec1310, 10:30 PM

P: 429

Oops, you're right, I used the inverse of gamma, sorry. I so used to thinking of time dilation and length contraction as sqrt(1  (v/c)^2) rather than the inverse of gamma. Okay, so we would need to use the inverse for x, y, and z from what I had before, giving
1 / z = (1 / x) (1 / y) / [1 + sqrt(1  1 / x^2) sqrt(1  1 / y^2)] z = x y [1 + sqrt(x^2  1) sqrt(y^2  1) / (x y)] z = x y + sqrt(x^2  1) sqrt(y^2  1) 


#7
Dec1310, 10:47 PM

P: 2,258



#8
Dec1310, 10:49 PM

P: 2,258

so for very big x and y its close to xy?
I had that vague impression from something that i read somewhere but couldnt find it anywhere 


#9
Dec1310, 10:52 PM

P: 429




#10
Dec1310, 11:46 PM

P: 429

x = 1 / sqrt(1  (v / c)^2) 1 / x = sqrt(1  (v / c)^2) (1 / x)^2 = 1  (v / c)^2 (v / c)^2 = 1  (1 / x)^2 v / c = sqrt(1  1 / x^2) So for Wolfram, you'd want to enter sqrt(1  1/z^2) = (sqrt(1  1/x^2) + sqrt(1  1/y^2)) / (1 + sqrt(1  1/x^2) sqrt(1  1/y^2)) http://www.wolframalpha.com/input/?i=sqrt(1++1%2fz%5e2)+ but Wolfram gives the solution for z as the square of what I gave within a square root form. 


#11
Dec1410, 02:01 AM

P: 3,967

> z = x y + sqrt[(x^2  1)(y^2  1)] > z = x y + sqrt[x^2 y^2  x^2  y^2 + 1] > z = x y + sqrt[x^2 y^2 (1  1/x^2  1/y^2 + 1/(x^2 y^2))] which for large x and y approximates to 2xy (as you said). 


#12
Dec1410, 01:45 PM

P: 2,258

I entered:
solve z = sqrt(1/(((sqrt(1  1/x^2) + sqrt(1  1/y^2)) / (1 + sqrt(1  1/x^2) sqrt(1  1/y^2)))^2  1))and got: z = sqrt(2 x^2 y^2+2 sqrt(((x1) (x+1))/x^2) x^2 y^2 sqrt(((y1) (y+1))/y^2)x^2y^2+1) 


#13
Dec1410, 02:45 PM

P: 429

z = sqrt(2 x^2 y^2+2 sqrt(((x1) (x+1))/x^2) x^2 y^2 sqrt(((y1) (y+1))/y^2)x^2y^2+1) z = sqrt[2 x^2 y^2 + 2 x y sqrt(x^2  1) sqrt(y^2  1)  x^2  y^2 + 1] If we square what we got earlier, then z = x y + sqrt(x^2  1) sqrt(y^2  1) z^2 = x^2 y^2 + 2 x y sqrt(x^2  1) sqrt(y^2  1) + (x^2  1) (y^2  1) z^2 = x^2 y^2 + 2 x y sqrt(x^2  1) sqrt(y^2  1) + x^2 y^2  x^2  y^2 + 1 z^2 = 2 x^2 y^2 + 2 x y sqrt(x^2  1) sqrt(y^2  1)  x^2  y^2 + 1 and then taking the square root of z, then, we get z = sqrt[2 x^2 y^2 + 2 x y sqrt(x^1) sqrt(y^2  1)  x^2  y^2 + 1] the same as before. Wolfram just won't take it out of square rooted form. 


#14
Dec1710, 11:50 PM

Sci Advisor
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P: 4,139

rapidity (theta, the Minkowskiangle) can be your friend...
tweak your intuition for euclideantrigonometry... let x=cosh(theta_A)... where v_A=tanh(theta_A) let y=cosh(theta_B) you want... z=cosh( theta_A  (theta_B)) .... the gammafactor for a relative rapidity =cosh( theta_A + theta_B ) =cosh( theta_A ) cosh( theta_B) + sinh( theta_A ) sinh( theta_B) = x y + sqrt(x^21) sqrt(y^21) .... as gravuniverse got in (#2 and #6) where I used cosh^2(theta)sinh^2(theta)=1 to write sinh^2(theta)=cosh^2(theta)1 and that cosh(theta)=cosh(theta). 


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