waldofj
2009-08-26 14:13:02 UTC
“Here is the calculation of the 'speed composition formula' from the
Given two frames of reference K and K', where K' is moving with the
speed v along the positive x-axis of K. Origins aligned at t = t' = 0,
axes parallel.
Let an object move along the x-axis of K with the speed u, so that it
is at the events E_0 : x_0 = 0, t_0 = 0 and E_1 : x_1 = ut_1, t_1
E_0: x'_0 = 0, t'_0 = 0
E_1: x'_1 = gamma(x_1 - vt_1) = gamma.t_1(u-v)
t'_1 = gamma(t_1 - vx_1/c2) = gamma.t_1(1 - uv/c2)
w = (x'_1 - x'_0)/(t'_1 - t'_0) = (u-v)/(1 - uv/c2)”
Paul, I cannot see how the foregoing shows that the speed composition
formula was derived from the Lorentz transforms or vice versa. You
‘calculate’ a speed but the speed of what? The facts are that AE
derived the speed formula from the work of Fizeau and Lorentz derived
the transforms from the work of Voigt who derived it from the work of
Doppler (Voigt ‘On Doppler’s Principle’ (‘Ueber das Doppler’sche
Prinzip’, Nachr. Ges. Wiss. Goettingen 2, 1887). If you want to
establish a relationship between the two formulas you should do it in
terms of v1, v2, v3, t and c. I doubt it can be done because the two
formulas have different grandfathers.
Peter Riedt
The facts are that AE
derived the speed formula from the work of Fizeau
no, he derived it from the Lorentz transformationGiven two frames of reference K and K', where K' is moving with the
speed v along the positive x-axis of K. Origins aligned at t = t' = 0,
axes parallel.
Let an object move along the x-axis of K with the speed u, so that it
is at the events E_0 : x_0 = 0, t_0 = 0 and E_1 : x_1 = ut_1, t_1
E_0: x'_0 = 0, t'_0 = 0
E_1: x'_1 = gamma(x_1 - vt_1) = gamma.t_1(u-v)
t'_1 = gamma(t_1 - vx_1/c2) = gamma.t_1(1 - uv/c2)
w = (x'_1 - x'_0)/(t'_1 - t'_0) = (u-v)/(1 - uv/c2)”
Paul, I cannot see how the foregoing shows that the speed composition
formula was derived from the Lorentz transforms or vice versa. You
‘calculate’ a speed but the speed of what? The facts are that AE
derived the speed formula from the work of Fizeau and Lorentz derived
the transforms from the work of Voigt who derived it from the work of
Doppler (Voigt ‘On Doppler’s Principle’ (‘Ueber das Doppler’sche
Prinzip’, Nachr. Ges. Wiss. Goettingen 2, 1887). If you want to
establish a relationship between the two formulas you should do it in
terms of v1, v2, v3, t and c. I doubt it can be done because the two
formulas have different grandfathers.
Peter Riedt
The facts are that AE
derived the speed formula from the work of Fizeau
it's in section 5 of his 1905 paper
personally I find Paul's use of symbols confusing
I’m on my bicycle going down the road at 20 mph, I lob a baseball in
front of me going 20 mph relative to me. Question: how fast is the
baseball going relative to the ground?
I will derive two formulas, one based on the Galilean transformation
(it's easier to follow and demonstrates my methodology)
and one based on the Lorentz transformation.
Actually I’ll be working with the reverse transformations.
The coordinate system associated with the ground will be the un-primed
coordinate system using (t,x,y,z). the coordinate system associated
with me on my bicycle will be the moving system with the primed
coordinates (t’, x’, y’, z’) . The baseballs’ velocity is described
in my system as dx’/dt’ which I will call v2. My velocity with respect
to the ground (the velocity between the two coordinate systems) is v1.
What I want to find is dx/dt, the velocity of the
baseball with respect to the ground, which I will call, v.
gamma = 1 / ((1 – (v^2 / c^2))^ ½).
I want to get dx/dt (which is v) from dx’/dt’ (which is v2)
I do this by multiplying as follows
dx/dt = dx’/dt’ * dx/dx’ * dt’/dt
(I will make a simplification right now. It works out that dealing
with dt’/dt is incredibly difficult. I will instead divide by it’s
dx/dt = (dx’/dt’ * dx/dx’) / dt/dt’
(keeping in mind I'm using v1 as the relative
speed between the twp frames)
z = z’
y = y’
t = t’
x = x’ + (v1 * t’)
ok, some differential calculus now
dx/dx’ = 1 + (v1 * dt’/dx’)
dx/dx’ = 1 + (v1 / v2)
multiply dx’/dt’ (which is v2) by dx/dx’, and divide by dt/dt’
you get
dx/dt = (v1 + v2) / dt/dt’
dx/dt is v (the ball's velocity w.r.t. the ground)
v = v1 + v2 is the end result.
z = z’
y = y’
x = (x’ + (v1 * t’)) * gamma
t = (t’ + (v1 * x’ / c^2)) * gamma
dx/dx’ is very similar to the above equation
dx/dx’ = (1 + (v1 / v2)) * gamma (the gamma term being the only
difference)
multiplied by dx’/dt’ (which is v2) and divided by dt/dt’
v = ((v1 + v2) * gamma) / dt/dt’
dt/dt’ = (1 + ((v1 / c^2) * dx’/dt’)) * gamma
v = ((v1 + v2) * gamma) / ((1 + (v1 x v2 / c^2)) * gamma)
the gamma’s divide out and we have
v = (v1 + v2) / (1 + (v1 * v2 / c^2))
the ball's velocity w.r.t. the ground is
v = v1 + v2
v = 20 + 20
v = 40
v = (v1 + v2) / (1 + (v1 * v2 / c^2))
v = (20 + 20) / (1 + (20 * 20 / c^2))
v = 39.999999999999964314721319384703
==========================================
Babbling idiot.
Why did the cretin Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
You don't know shit from shinola.
no one answers it so I’ll answer it now. The two points A and B are in
the moving coordinate system. Einstein never said the speed of light
from A to B is c-v, he just uses c – v without explaining where he got
it. Same for c + v. In this scenario he has a beam of light released
from point A to point B and reflected back to point A. C – v and c + v
are the respective closing speeds as seen from the stationary system.
These times are of course different. When he states the times are the
same he is referring to times as seen from the moving system (i.e. ½
(tau1 + tau3) = tau2). He then goes on to derive the LTE to show how
to reconcile these two different points of view.
Btw, the LTE reconciles these two different points of view by showing
that the clocks that are synchronized in one system are unsynchronized
from the point of view of the other system, in other words, the
relativity of simultaneity.
It’s obvious you don’t agree with any of this and I’m not trying to
convince you one way or the other, I’m just explaining what Einstein
actually said.