by granpa
 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.
P: 429
 Quote by granpa 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.
Well, we would have

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)]
 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
P: 429

 Quote by granpa good lord. well thank you.
No problem. I do try to be detailed.
 P: 2,258 http://www.wolframalpha.com/ use the 'expand' command
 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)
 P: 2,258
 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
P: 429
 Quote by granpa 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
That should have been z = x y + sqrt(x^2 - 1) sqrt(y^2 - 1), sorry again, I editted it. For very big x and y, it's close to 2 x y.
P: 429
 Quote by granpa
Looks like that inverse gamma thing might be getting both of us now, unless that's my fault. For gamma, you'd have

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.
P: 3,967
 Quote by grav-universe 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)
I get the same result as you got in #6 independently, and I think I have the gammas the right way up, so ....

--> 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).
 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(((x-1) (x+1))/x^2) x^2 y^2 sqrt(((y-1) (y+1))/y^2)-x^2-y^2+1)
P: 429
 Quote by granpa 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(((x-1) (x+1))/x^2) x^2 y^2 sqrt(((y-1) (y+1))/y^2)-x^2-y^2+1)
Right, that's it. That works out to

z = sqrt(2 x^2 y^2+2 sqrt(((x-1) (x+1))/x^2) x^2 y^2 sqrt(((y-1) (y+1))/y^2)-x^2-y^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.
 Sci Advisor HW Helper PF Gold P: 4,139 rapidity (theta, the Minkowski-angle) can be your friend... tweak your intuition for euclidean-trigonometry... 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 gamma-factor 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^2-1) sqrt(y^2-1) .... as grav-universe 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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