Re: Relativistic vector addition/subtraction of velocities
- From: Greg Egan <gregegan@xxxxxxxxxxxxxxx>
- Date: Sun, 24 Dec 2006 02:08:36 +0000 (UTC)
In article <1166013405.172967.103330@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Chalky" <chalkyspam@xxxxxxxxxxxxxxxx> wrote:
Consider an expanding sphere of shrapnel which propagates at speed v
from the source of an explosion.
If the trajectories of shrapnel particles a and b are separated by an
angle theta, what is the velocity of particle b relative to particle a?
At relativistically insignificant velocities the answer seems simple:
The tangential velocity Vt is v sin theta
The radial velocity Vr is v ( 1 - cos theta)
And the resultant vector magnitude is sqrt { Vr ^ 2 + Vt ^ 2}
= sqrt { [ v (1 - cos theta)] ^ 2 + [v sin theta] ^ 2}
= v sqrt { 1 - 2 cos theta + cos ^ 2 theta + sin ^ 2 theta }
= v sqrt { 2 - 2 cos theta}
= sqrt (2 Vr)
However, I am not so sure of the appropriate formulae when v is
relativistically significant.
Using the relativistic addition of velocities formula w = (u + v)/(1 +
uv/c^2), I still get:
Vt = v sin theta, but
This is not correct! Aberration means that you can't take two directions
that are orthogonal in one frame, and then assume that they will remain
orthogonal when you transform to a different frame.
Vr = v (1 - cos theta) / (1 - {[v/c] ^ 2} cos theta)
Thus, if v/c = s , I get
Vt = 0 when theta = 0, ~ sc/1.414 when theta = 45, and sc when theta =
90 degrees. Similarly,
Vr = 0 when theta = 0, ~ 0.414sc/(1.414 - s ^ 2) when theta = 45, and
sc when theta = 90 degrees.
These functions seem well behaved as s tends to 1 giving
Vt = 0 when theta = 0, ~ 0.707c when theta = 45, and c when theta = 90
degrees. Similarly,
Vr = 0 when theta = 0, ~ c when theta = 45, and still c when theta = 90
degrees.
Nevertheless, I am still stuck for what the total magnitude of the
vector is. Presumably it can't still be sqrt {2 Vr}, since that would
be greater than c for most of the angular distribution, when s tends to
1. Can anyone help out?
Chalky
If you want only the magnitude of the relative velocities between pieces
of shrapnel, by far the simplest method (IMHO) is to calculate the
relevant 4-velocities then take the Lorentzian dot product between them.
In everything that follows, I'm going to give v in units where c=1; if
you want to translate back to your usage, this is your s. (Most people
who do a lot of relativistic calculations get tired of putting in the
c's, and just adopt units where c=1 for all the algebra.)
If we pick some inertial frame with coordinates {t,x,y}, an object that's
stationary in that frame has a 4-velocity:
u_1 = (1,0,0)
This says it's just moving along the t axis at 1 second per second of its
proper time.
An object with an ordinary velocity of v in the x direction has a
4-velocity:
u_2 = (1,v,0) / sqrt(1-v^2)
which gives the change of t coordinate (i.e. 1/sqrt(1-v^2)) and x
coordinate (i.e. v/sqrt(1-v^2)) for every second of its proper time.
If we want to know the magnitude of the relative velocity of two objects,
we take the Lorentzian dot product:
(t1,x1,y1) . (t2,x2,y2) = -t1*t2 + x1*x2 + y1*y2
of their 4-velocities, which in this case is:
dp_{12} = u_1.u_2 = -1/sqrt(1-v^2)
and we see that we can recover v as:
v = sqrt(1 - 1/dp_{12}^2)
Now, for your shrapnel, a piece moving at theta=0 (along the x-axis) has
4-velocity:
u_3 = (1,v,0) / sqrt(1-v^2)
whereas a piece moving at arbitrary theta has 4-velocity:
u_4 = (1, v cos theta, v sin theta) / sqrt(1-v^2)
Remember, the 4-velocity gives you the rates of change of all coordinates
wrt proper time, and the ordinary velocity is just the rate of change of
the *spatial* coordinates wrt the *coordinate time*, so you can confirm
the correct ordinary velocities in the {t,x,y} frame just by dividing out
the t component of the 4-velocity.
Now, this gives us:
dp_{34} = u3.u4 = (-1 + v^2 cos theta) / (1-v^2)
and then the magnitude of the relative velocity between the pieces of
shrapnel is:
sqrt(1 - 1/dp_{34}^2)
Now, if you want individual velocity components of the shrapnel at angle
theta transformed into a frame in which the shrapnel at angle zero is
stationary, then the simplest way to do *that* is to apply a Lorentz
transformation to the 4-velocity u_4.
The Lorentz transformation:
| 1/sqrt(1-v^2) -v/sqrt(1-v^2) 0 |
L = |-v/sqrt(1-v^2) 1/sqrt(1-v^2) 0 |
| 0 0 1 |
is a boost in the -ve x direction, that puts everything into the rest
frame {t',x',y'} of the shrapnel moving along the x axis.
u_3' = L u_3 = (1, 0, 0)
u_4' = L u_4 = ((1-v^2 cos theta)/(1-v^2),
-v (1-cos theta)/(1-v^2),
v sin theta / sqrt(1-v^2))
To get the ordinary velocities associated with u_4' in the {t',x',y'}
frame, we divide out the t' component, giving:
v_{x'} = -v (1-cos theta) / (1-v^2 cos theta)
v_{y'} = v sqrt(1-v^2) sin theta / (1-v^2 cos theta)
In other words, your Vr is correct (though the signs we've used are
opposite) but your Vt is not.
If you check, you'll find that:
sqrt(1 - 1/dp_{34}^2) = sqrt(v_{x'}^2 + v_{y'}^2)
i.e. both of the approaches I've described give the same answer for the
magnitude of the relative velocity.
.
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