- #1

Perspicacious

- 76

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Perspicacious
- Start date

- #1

Perspicacious

- 76

- 0

- #2

- 6,548

- 1,917

§ 2. On the Relativity of Lengths and Times

[tex]t_B - t_A = \displaystyle \frac{r_{AB}}{c-v}[/tex] and [tex]t'_A - t_B = \displaystyle \frac{r_{AB}}{c+v}[/tex]

(see http://www.fourmilab.ch/etexts/einstein/specrel/www/ ),

light velocity is c.

2. Any ray of light moves in the ``stationary'' system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body.

[tex]t_B - t_A = \displaystyle \frac{r_{AB}}{c-v}[/tex] comes from [tex]c(t_B - t_A) = r_{AB} + v(t_B - t_A)[/tex],

where [tex]c(t_B - t_A)[/tex] is the distance that light travelled, and [tex]v(t_B - t_A)[/tex] is the distance that the moving-observer (the rod) travelled.

Note that all of these variables are measured in the "stationary system".

- #3

Perspicacious

- 76

- 0

- #4

JesseM

Science Advisor

- 8,518

- 16

First Einstein assumes you have to coordinate systems defined by

measurements on a system of rigid measuring rods and clocks, K and k,

with all the spatial axes parallel and with k moving at velocity v along

K's x-axis. K's coordinates are (x,y,z,t) while k's coordinates are (xi,

eta, zeta, tau). Then he defines a new coordinate x'=x-vt, and says "it

is clear that a point at rest in the system k must have a system of

values x', y, z, independent of time". Although he doesn't state it this

way, what he has effectively done is to introduce a *third* coordinate

system Kg, with y,z,t coordinates identical to K, but with x' coordinate

given by x-vt. To make this a little clearer, I'm going to modify his

notation and say that coordinate system Kg uses coordinates x',y',z',t',

with these coordinates related to K's coordinates x,y,z,t by a Galilei

transform:

x'=x-vt

y'=y

z'=z

t'=t

An important thing to notice is that, unlike k and K, Kg doesn't

necessarily correspond to the measurements of any observer using a

system of measuring-rods and clocks; it isn't really an inertial

reference frame at all. So, the postulate that all observers must

measure the speed of light to be c in their own rest frame doesn't apply

to Kg. In fact, since we know light must travel at c in both directions

in K, and Kg is related to K by a Galilei transform, it must be true

that light travels at (c-v) in the +x' direction of Kg, and (c-v) in the

-x' direction of Kg.

So now a light beam is sent from the origin of k at tau0, reflected by a

mirror at rest in k at tau1, and returned to the origin at tau2. As

Einstein said, any point which is at rest in k must also be at rest in

this new coordinate system which I am calling Kg, so neither the point

of origin nor the mirror are moving in Kg. So if the origin of Kg

coincides with the origin of k, and if we say the mirror is at position

x'=xm' in Kg, then since light travels at (c-v) in the +x' direction of

Kg, the light will take time xm'/(c-v) to reach the mirror in Kg, and

since light travels at (c+v) in the -x' direction of Kg, it will take an

additional time of xm'/(c+v) to return to the origin. Thus, if the light

is emitted at t'=t0' in Kg's frame, it is reflected at t'=t0' +

xm'/(c-v), and returns to the origin at t'=t0' + xm'/(c-v) + xm'/(c+v)

So, if k's coordinate tau is expressed as a function of Kg's coordinates

like tau(x',y',z',t') then we have:

tau0 = tau(0, 0, 0, t0')

tau1 = tau(xm', 0, 0, t0' + xm'/(c-v))

tau2 = tau(0, 0, 0, t0' + xm'/(c-v) + xm'/(c+v))

Now, since k represents the actual measurements of a non-accelerating

observer using his measuring-rods and clocks, we know that light must

travel at c in both directions in this coordinate system, and since the

origin and the mirror are at rest in k, the light must take the same

amount of time to reach the mirror as it takes to be reflected back to

the origin in k. So, this gives 1/2(tau0 + tau2) = tau1, which

substituting in the expressions above gives

1/2[tau(0, 0, 0, t0') + tau(0, 0, 0, t0' + xm'/(c-v) + xm'/(c+v))] =

tau(xm', 0, 0, t0' + xm'/(c-v))

Then he goes from this to the equation 1/2(1/(c-v) + 1/(c+v))*(dtau/dt')

= dtau/dx' + (1/c-v)*(dtau/dt'), which also confused me for a little

while because I didn't know what calculus rule he was using to go from

the last equation to this one. But then I realized that if you just

ignore the y' and z' coordinates and look at tau(x',t'), then since he

says "if x' is chosen infinitesimally small", you can just assume tau is

a slanted plane in the 3D space with x',t' as the horizontal axes and

tau as the vertical axes. The general equation for a slanted plane in

these coordinates which goes through some point xp', tp', and taup would be:

tau(x',t') = Sx'*(x' - xp') + St'*(t' - tp') + taup

Where Sx' is the slope of the plane along the x' axis and St' is the

slope of the plane along the t' axis. If we say this plane must go

through the three points tau0, tau1 and tau2 earlier, then we can use

tau0's coordinates for xp', tp' and taup, giving:

tau(x',t') = Sx'*x' + St'*t' + tau0

So, plugging in tau1 = tau(xm', t0' + xm'/(c-v)) gives

tau1 = Sx'*xm' + St'*(t0' + xm'/(c-v)) + tau0

And plugging in tau2 = tau(0, t0' + xm'/(c-v) + xm'/(c+v)) gives:

tau2 = St'*(t0' + xm'/(c-v) + xm'/(c+v)) + tau0

So plugging these into 1/2(tau0 + tau2) = tau1 gives:

1/2(tau0 + St'*(t0' + xm'/(c-v) + xm'/(c+v)) + tau0 ) = Sx'*xm' +

St'*(t0' + xm'/(c-v)) + tau0

With a little algebra, this reduces to:

(1/2)*St'*(1/(c-v) + 1/(c+v)) = Sx' + St'*(1/(c-v))

And since tau(x',t') is just a plane, of course St' = dtau/dt' and Sx' =

dtau/dx', so this gives the equation 1/2(1/(c-v) + 1/(c+v))*(dtau/dt') =

dtau/dx' + (1/c-v)*(dtau/dt') which Einstein got.

He then reduces this to dtau/dx' + dtau/dt'*(v/(c^2 - v^2)) = 0, which

is just algebra. He also says that light moves along the y'-axis and

z'-axis of Kg at velocity squareroot(c^2 - v^2), which isn't too hard to

see--a light beam moving vertically along the zeta-axis of k will also

be moving vertically along the z'-axis of Kg since these coordinate

systems aren't moving wrt one another, but in K the light beam must be

moving diagonally since k is moving at v in k, so if you look at a

triangle with ct as the hypotenuse and vt as the horizontal side, the

vertical side must have length t*squareroot(c^2 - v^2), and since z'=z

and t'=t the light beam must also travel that distance in time t' in

Kg's coordinate system. The same kind of argument shows the velocity is

also squareroot(c^2 - v^2) in the y'-direction.

Since tau(x',y',z',t') is a linear function (ie tau(x',y',z',t') = Ax' +

By' + Cz' + Dt'), then from this you can conclude that if dtau/dx' +

dtau/dt'*(v/(c^2 - v^2)) = 0 for a light ray moving along the x'-axis,

tau(x',t') must have the form tau = a(t' - vx'/(c^2 - v^2)) where a is

some function of v (so that dtau/dt' = a and dtau/dx' = -av/(c^2 - v^2),

which means dtau/dx' = (-v/(c^2 - v^2))*dtau/dt').

Next he says that in the k coordinate system the light ray's position as

a function of time would just be xi(tau)=c*tau, so plugging that

expression for tau in gives xi=ac(t' - vx'/(c^2 - v^2)). But in system

Kg, this light ray is moving at velocity (c-v), so its t' coordinate as

a function of x' is t'(x') = x'/(c-v). Plugging this in gives xi =

ac(x'/(c-v) - vx'/(c^2 - v^2)) = a*(c^2/(c^2 - v^2))*x'.

Similarly, if light is going in the eta-direction then eta(tau)=c*tau,

so plugging in tau = a(t' - vx'/(c^2 - v^2)) gives eta=ac(t' - vx'/(c^2

- v^2)). In Kg this ray is moving at squareroot(c^2 - v^2) in the

y'-direction, so t'(y')=y'/squareroot(c^2 - v^2), and plugging this in

gives eta= a(y'/squareroot(c^2 - v^2) - vx'/(c^2 - v^2))...since x'=0

for this ray, this reduces to eta=(ac/squareroot(c^2 - v^2))* y'. The

relation between zeta and z' is exactly the same, so we have:

tau = a(t' - vx'/(c^2 - v^2))

xi = a*(c^2/(c^2 - v^2))*x'

eta = (ac/squareroot(c^2 - v^2))* y'

zeta=(ac/squareroot(c^2 - v^2))* z'

Since the relation between Kg coordinates (x',y',z',t') and K

coordinates (x,y,z,t) is just x'=x-vt, y'=y, z'=z, t'=t, we can plug in

and simplify to get:

tau = Phi(v) * Beta * (t - vx/c^2)

xi = Phi(v) * Beta * (x - vt)

eta = Phi(v) * y

zeta = Phi(v) * z

Where where Beta = c/squareroot(c^2 - v^2) = 1/squareroot(1 - v^2/c^2),

and Phi(v) = ac/squareroot(c^2 - v^2) (since a was an undetermined

function of v in the first place he just writes Phi(v)).

To find Phi(v), he then imagines a coordinate system K' which is moving

at -v relative to k (and unlike Kg, this coordinate system is supposed

to correspond to the measurements made on a system of measuring-rods and

clocks, so it's a valid inertial reference frame). *He uses

(x',y',z',t') for the K' coordinate system, but since I've already used

that for Kg, let's call the K'-coordinates (x",y",z",t"). Then the

transform from k-coordinates to K'-coordinates would have to be:

t" = Phi(-v) * Beta(-v) * (tau + v*xi/c^2) = Phi(v)*Phi(-v)*t

x" = Phi(-v) * Beta(-v) * (xi + v*tau) = Phi(v)*Phi(-v)*x

y" = Phi(-v) * eta = Phi(v)*Phi(-v)*y

z" = Phi(-v) * zeta = Phi(v)*Phi(-v)*z

If K' is moving at -v in the k-system, and k is moving at +v in the

K-system, then K and K' should really be the same system, so

Phi(v)*Phi(-v) should be 1. Then he argues that if you have a rod of

lenght l lying along the eta-axis of k and at rest in that system, then

if you transform the coordinates of its ends into the K-system, you find

that its length in the K-system is l/Phi(v), and by symmetry he argues

that the length of a vertical rod moving horizontally can only depend on

the magnitude of the velocity and not the direction, so l/Phi(v) =

l/Phi(-v), which means Phi(v) = Phi(-v)...combining with Phi(v)*Phi(-v)

= 1 which he obtained earlier, he concludes that Phi(v)=1, which gives

him the Lorentz transform.

- #5

rbj

- 2,227

- 9

A while ago on this usenet thread I posted my own attempt to follow Einstein's derivation of the Lorentz transform in his original 1905 paper. Here it is:

are you Daryl McCullough? or no, what is a different post in that thread? i thought this post from Daryl was interesting (answering essentially why it's [itex]8 \pi[/itex] instead of [itex]4 \pi[/itex], where that extra factor of 2 comes from). i just thought it would be cool to point that out to Daryl if you're the same.

- #6

Share:

- Replies
- 3

- Views
- 594

- Replies
- 66

- Views
- 5K

- Replies
- 1

- Views
- 384

- Last Post

- Replies
- 1

- Views
- 421

- Replies
- 5

- Views
- 347

- Last Post

- Replies
- 34

- Views
- 1K

- Replies
- 42

- Views
- 1K

- Replies
- 76

- Views
- 3K

- Replies
- 28

- Views
- 494