Derivation of Gamma: Is it Correct?

[epkid08]

\gamma = csc(cos^{-1}(v/c))

\gamma^{-1} = sin(cos^{-1}(v/c))

Is there something wrong with this derivation? If not, why isn't this the more formal way of writing it, it seems a lot more simple?

 

[Peeter]

I'm unclear what you mean by derivation, but I think you want the following hyperbolic trig relation:

<br /> \gamma = \cosh(\tanh^{-1}(v/c)) = \frac{1}{\sqrt{1-(v/c)^2}}<br />

 

[epkid08]

I'm unclear what you mean by derivation,

By derivation I just mean a derived form of.

but I think you want the following hyperbolic trig relation:

<br /> \gamma = \cosh(\tanh^{-1}(v/c)) = \frac{1}{\sqrt{1-(v/c)^2}}<br />

Yes, or the normal trig relation:
\gamma = csc(cos^{-1}(v/c))

In any case, why is this not used, as it's much more simple?

 

[Peeter]

I don't think that "simpler" version is correct. You need something that expresses the hyperbolic "distance" of spacetime. For example for a given event that occurs at time t and distance x, the length of a straight path to that event is:

<br /> s^2 = (ct)^2 - x^2<br />

It's this arc length that is invariant (constant) given other observer velocities, and therefore defines a hyperbola in coordinates ct, and x. With sines and cosines you have sine^2 + cosine^2 = 1 ... there's no minus sign there to express this hyperbolic relationship. You can do it with complex number angles but in that case you are really just saying "I need hyperbolic sine and cosine".

 

[epkid08]

I'm not so sure that it matters, I've tried it with a couple different examples.

Consider a spaceship traveling from Earth to the nearest star system outside of our solar system: a distance d = 4.45 light years away, at a speed v = 0.866c (i.e., 86.6 percent of the speed of light, relative to the Earth). The Earth-based mission control reasons about the journey this way (for convenience in this thought experiment the ship is assumed to immediately attain its full speed upon departure): the round trip will take t = 2d / v = 10.28 years in Earth time (i.e. everybody on Earth will be 10.28 years older when the ship returns). The flow of time on the ship and aging of the travelers during their trip will be slowed by the factor \epsilon = \sqrt{1 - v^2/c^2}, the reciprocal of the Lorentz factor. In this case \epsilon = 0.500 \, and the travelers will have aged only 0.500×10.28 = 5.14 years when they return.

t&#039; = t/(sin(cos^{-1}(v/c)))

Plugging in t=5.14, and v=.866c, yields the same as the normal time dilation equation.

 

[Antenna Guy]

t&#039; = t/(sin(cos^{-1}(v/c)))

Let u=v/c and see how many trig identities you can come up with.

Regards,

Bill

 

[Peeter]

I'm not so sure that it matters, I've tried it with a couple different examples.

I assumed without checking that the hyperbolic functions were required, but if I do the trig, you are right and the csc(cos^{-1}()) works too. I'd guess they aren't used since the hyperbolic functions are just more convienent than your identities (which makes logical sense to me since the metric is hyperbolic).

 

[robphy]

As suggested by others, the hyperbolic functions are more natural for special relativity.

Given two future-pointing unit-timelike 4-vectors (tangent to the observer worldlines) \hat t and \hat u,
\gamma = \hat t \cdot \hat u = \cosh(\theta)...
where \tanh\theta=v/c (as seen in Peeter's post).

Note that, using hyperbolic-trig functions, some of your Euclidean-intuition (via analogies) carries over into special-relativity. Observe that the "dot-product" is associated with the "cosine" function of the "angle-between" two vectors... and that the "slope" is a associated with a "tangent" function. The analogues don't stop there.

The Euclidean functions (with real-valued angles) don't enjoy such geometric interpretations in special-relativity... unless you are working solely in a hyperplane spanned by spacelike vectors.
(What would (say) the composition of velocities formula look like with with OP's formulation?)

 

[cryptic]

\gamma = csc(cos^{-1}(v/c))

\gamma^{-1} = sin(cos^{-1}(v/c))

Is there something wrong with this derivation? If not, why isn't this the more formal way of writing it, it seems a lot more simple?

If cos \phi = \frac {v} {c}

then sin \phi = \sqrt {1 - cos^2 \phi} = \sqrt {1 - \frac {v^2} {c^2} } = \frac {\sqrt {c^2 -v^2}} {c} = \frac {c&#039;} {c}

and \gamma = \frac {1} {sin \phi} = \frac {1} {\sqrt {1 - \frac {v^2} {c^2} } } = \frac {c} {\sqrt {c^2 -v^2}} = \frac {c} {c&#039;} .

 

[epkid08]

If cos \phi = \frac {v} {c}

then sin \phi = \sqrt {1 - cos^2 \phi} = \sqrt {1 - \frac {v^2} {c^2} } = \frac {\sqrt {c^2 -v^2}} {c} = \frac {c&#039;} {c}

and \gamma = \frac {1} {sin \phi} = \frac {1} {\sqrt {1 - \frac {v^2} {c^2} } } = \frac {c} {\sqrt {c^2 -v^2}} = \frac {c} {c&#039;} .

I don't understand. Are you agreeing or disagreeing?

 

[cryptic]

I don't understand. Are you agreeing or disagreeing?

I agree of course, but the consequence of this geometrical consideration is - the speed of light is in the moving frame $ c&#039; = \sqrt {c^2 - v^2} $ with respect to rest frame.

Einstein wrote in http://www.fourmilab.ch/etexts/einstein/specrel/www/"

An analogous consideration--applied to the axes of Y and Z--it being borne in mind that light is always propagated along these axes, when viewed from the stationary system, with the velocity $\sqrt{c^2-v^2}$ gives us ...

 

[JesseM]

I agree of course, but the consequence of this geometrical consideration is - the speed of light is in the moving frame $ c&#039; = \sqrt {c^2 - v^2} $ with respect to rest frame.

Einstein wrote in http://www.fourmilab.ch/etexts/einstein/specrel/www/"

An analogous consideration--applied to the axes of Y and Z--it being borne in mind that light is always propagated along these axes, when viewed from the stationary system, with the velocity \sqrt{c^2-v^2} gives us ...

You misunderstand Einstein here, he was talking about a coordinate system which was created by doing a Galilei transformation on an ordinary inertial frame--this is not a genuine inertial frame in SR. A while ago on this thread I provided my own summary of what Einstein was doing in that paper, I'll repost it here and put the parts related to the section you quote in bold:

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.

 

[cryptic]

You misunderstand Einstein here, he was talking about a coordinate system which was created by doing a Galilei transformation on an ordinary inertial frame--this is not a genuine inertial frame in SR. A while ago on this thread I provided my own summary of what Einstein was doing in that paper, I'll repost it here and put the parts related to the section you quote in bold:

1) This derivation does not work because x' and t are dependent variables x'=(V-v)t and

2) The assumption 1/2(tau0 + tau2) = tau1 can not be true

because x'/(V-v)+x'/(V+v) can not be equal to x'/(V-v)+x'/(V-v) unless v=0.

 

[Antenna Guy]

... 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)...

It seems to me that the word "sign" would have been more appropriate than "direction" here.

Also, I'm not quite sure I follow why the length of a vertical rod traveling horizontally might depend upon the magnitude of the velocity (although you imply later that it doesn't without actually mentioning so [Phi(v)=Phi(-v)=1]).

Regards,

Bill

 

[JesseM]

1) This derivation does not work because x' and t are dependent variables x'=(V-v)t and

That equation doesn't appear in what I wrote--are you substituting V for c or something? Please quote the section of what I wrote that contains the equation you're talking about so it'll be clear.

because x'/(V-v)+x'/(V+v) can not be equal to x'/(V-v)+x'/(V-v) unless v=0.

Again, where do the terms x'/(V-v)+x'(V+v) and x'/(V-v)+x'/(V-v) [which unless you've written it wrong is just equal to 2x'/(V-v)] appear in what I wrote?

 

[cryptic]

That equation doesn't appear in what I wrote--are you substituting V for c or something? Please quote the section of what I wrote that contains the equation you're talking about so it'll be clear.

I used original notation V instead of c.
You write for example following:"...its t' coordinate as a function of x' is t'(x') = x'/(c-v)".

And as You can see, it means: dx'=(c-v)dt'.

Again, where do the terms x'/(V-v)+x'(V+v) and x'/(V-v)+x'/(V-v) [which unless you've written it wrong is just equal to 2x'/(V-v)] appear in what I wrote?

It is just here, if You put Sx'=St'*(1/(c-v)).

"With a little algebra, this reduces to:

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

 

[JesseM]

I used original notation V instead of c.
You write for example following:"...its t' coordinate as a function of x' is t'(x') = x'/(c-v)".

And as You can see, it means: dx'=(c-v)dt'.

So what did you mean when you said "This derivation does not work because x' and t are dependent variables"? What specific step in the derivation do you think doesn't work because t' is a function of x'?

It is just here, if You put Sx'=St'*(1/(c-v)).

Why would you do that? Sx' and St' refer to the slopes of a 2D plane (slope along the x' axis and slope along the t' axis) in the 3D space defined by variables (x', t', tau), with the 2D plane being oriented in such a way as to include three points in this space corresponding to the x', t', and tau coordinates of three events: the event of the light being sent from the origin of the k frame, the event of the light being reflected by a mirror at rest in the k frame, and the event of the light returning to the origin of k. Why do you think these slopes would obey the relation Sx' = St'*(1/(c-v))?

 

[cryptic]

You do not need 2D-plane because x and t are dependent. That means, You need only one independent variable for this problem.

And further if You compare two formulas for xi, You can see that Einstein changed Beta^2 to Beta in the second set of formulas without any mathematical (or physical) reason.

He writes Beta instead of c^2/(c^2 - v^2) and this is Beta^2.

...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

 

[JesseM]

You do not need 2D-plane because x and t are dependent. That means, You need only one independent variable for this problem.

You're being very vague. Maybe you think I don't "need" it, but certainly there's nothing inherently wrong with graphing two dependent variables on different axes, so what specific step in my proof do you think is incorrect? Also, you're talking vaguely about x' and t' without specifying whether you're solely talking about pairs of x',t' coordinates that happen to lie on the path of the original light beam from the origin (it is only coordinates on this line that were meant to obey the relation t'(x') = x'/(c-v)), or whether you're talking about the x', t' coordinates of arbitrary events (including the event with tau-coordinate tau2 of the light that had been reflected by the mirror returning to the origin, which does not lie on the line of the original light beam and whose x' and t' coordinates would not obey the relation t'(x') = x'/(c-v), yet by construction this event must be part of the plane).

And further if You compare two formulas for xi, You can see that Einstein changed Beta^2 to Beta in the second set of formulas without any mathematical (or physical) reason.

He writes Beta instead of c^2/(c^2 - v^2) and this is Beta^2.

Please be specific. What original formula for xi contains c^2/(c^2 - v^2)? What is the new formula for xi in the "second set of formulas"?

 

[cryptic]

You're being very vague. ...

I'm talking about differential equation: 1/2(1/(c-v) + 1/(c+v))*(dtau/dt')
= dtau/dx' + (1/c-v)*(dtau/dt').
Here is x' a function of t' and in this case one must use chain rule for derivation i.e. d/dx'*dx'/dt'. So the equation changes to: 1/2(1/(c-v) + 1/(c+v))*(dtau/dt')
= (1/c-v)*(dtau/dt') + (1/c-v)*(dtau/dt')=(2/(c-v))*(dtau/dt').

Please be specific. What original formula for xi contains c^2/(c^2 - v^2)? What is the new formula for xi in the "second set of formulas"?

These are the original formulas in Einstein paper and in one of your previous posts (I quoted it).

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

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

 

[JesseM]

I'm talking about differential equation: 1/2(1/(c-v) + 1/(c+v))*(dtau/dt')
= dtau/dx' + (1/c-v)*(dtau/dt').

That's the equation he derives from the earlier equation 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)), under the assumption that x is "infinitesimally small". Are you claiming there is an error in the derivation, or do you agree that that equation follows from the earlier one?

Here is x' a function of t' and in this case one must use chain rule for derivation i.e. d/dx'*dx'/dt'.

The chain rule for derivation of what from what? By the way, note that although I avoided the use of calculus in my derivation by making use of the fact that if you zoom in on an infinitesimally small region of a smooth function it'll look like a flat plane, you can derive the equation above using calculus--see this thread as well as DrGreg's post #6 on this thread.

So the equation changes to: 1/2(1/(c-v) + 1/(c+v))*(dtau/dt')
= (1/c-v)*(dtau/dt') + (1/c-v)*(dtau/dt')=(2/(c-v))*(dtau/dt').

You just said something about the chain rule, but here you don't appear to have made use of the chain rule or any other calculus, you just did some algebra (and you forgot to carry over the 1/2 from the first equation, the final one should be (1/(c-v))*(dtau/dt'), not (2/(c-v))*(dtau/dt')).

These are the original formulas in Einstein paper and in one of your previous posts (I quoted it).

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

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

Sorry, when I went to respond to your post it didn't include your own quote from my post in the text box, so I didn't notice you had quoted the context of the formulas already, my mistake. Anyway, notice that the first formula contains the unknown function a(v) and the second formula contains the unknown function Phi(v). I defined a(v) in this paragraph:

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').

And I defined Phi(v) in this one:

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)).

Naturally if Phi(v) = ac/squareroot(c^2 - v^2) and Beta is defined as c/squareroot(c^2 - v^2), this means Phi(v) = a*Beta. So, the second equation you quoted:

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

...becomes xi = (a*Beta)* Beta * (x - vt) = a * Beta^2 * (x - vt). And Beta^2 = c^2/(c^2 - v^2), so this gives:

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

And from the coordinate transformation between K and Kg we have x' = (x - vt), so this gives the first equation you quoted:

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

Is this OK, or do you still have an objection?

 

[jambaugh]

A note on the circle trig vs. hyperbolic trig relations.

The key relation is of course the invariant proper time:
d\tau^2 = dt^2 - dx^2
(in c = 1 units.)

You can thus parametrize the time and space components via:
dx = d\tau \sinh(\vartheta) and dt = d\tau \cosh(\vartheta).

The hyperbolic pseudo-angle \vartheta is a frame parameter which we can relate to the velocity relative to the initial frame (where \vartheta= 0) by:
\frac{v}{c}=\beta = \tanh(\vartheta)

Note this parametrization works because \cosh^2(\vartheta) - \sinh^2(\vartheta) = 1. We could however make use of circle trig functions via the identity:
1 = \sec^2(\theta) - \tan^2(\theta)
and thus parametrize frames via an angle:

dx = d\tau \tan(\theta) and dt = d\tau \sec(\theta)

Note then that the boost velocity is: \beta = sin(\theta)

You can view this another way by rewriting the proper time metric in the form:
dt^2 = d\tau^2 + dx^2
whence this circle-trig. angle denotes the mixing of proper time and space:
d\tau =dt \cos(\theta) and dx = dt \sin(\theta).

The problem with this regular trig approach is that the angle \theta does not linearly parametrize the one dimensional group of frame transformations. In short you don't get the correct velocity addition relationship by adding angles:
\beta_{1+2} \ne \sin(\theta_1 + \theta_2).
(Note also you must artificially restrict the angle to -\pi/4 &lt; \theta &lt; \pi/4)
This is because the (differential) frame time dt is not the invariant quantity under frame transformations. (Mathematically it is because the secant and tangent functions are not components of the exponential transformation matrix.)

You do get the correct velocity addition by adding pseudo-angles in the hyperbolic case:
\beta_{1+2} = \tanh(\vartheta_1 + \vartheta_2)
where \beta_1 = \tanh(\vartheta_1) and \beta_2 = \tanh(\vartheta_2)

I find that this helps clarify the distinction between Galilean and Special Relativites. Velocity is the linear parameter for Galilean boosts, this \vartheta psuedo-angle is the linear parameter for Lorentz boosts. And of course note that the small angle approximation also occurs in hyperbolic trig. giving us the small velocity non-relativistic limit:
\frac{V}{c} = \tanh(\vartheta) \approx \vartheta when |\vartheta|&lt;&lt; 1.

 

[cryptic]

The chain rule for derivation of what from what?

If one variable is a function of one another variable the chain rule must be used. Consequently, there is an error in Einstein derivation of SR.

...
Is this OK, or do you still have an objection?

No, it's all totally wrong! Einstein said a=Phi(v) and you try to define another "value" for this constant. Einstein himself found a=Phi(v)=1.

 

[JesseM]

If one variable is a function of one another variable the chain rule must be used.

Only if you are taking the derivative of the first variable with respect to the second variable. Where is Einstein doing that? Where am I doing that in my analysis of his proof?

No, it's all totally wrong! Einstein said a=Phi(v) and you try to define another "value" for this constant. Einstein himself found a=Phi(v)=1.

Einstein never says "a = Phi(v)", he just says "a is a function Phi(v) at present unknown", but it's clear from the context he was just speaking sloppily and actually meant that the two functions are related by a constant. If this isn't obvious to you, note that he writes:

Thus \eta = a \frac{c}{\sqrt{c^2 - v^2}} y

and then just a few lines later he writes:

\eta = \phi (v) y

So from the context, it's very clear he intends that \phi (v) = a \frac{c}{\sqrt{c^2 - v^2}}.

 

[cryptic]

Only if you are taking the derivative of the first variable with respect to the second variable. Where is Einstein doing that? Where am I doing that in my analysis of his proof?

Everywhere. Try to replace dx' with (c-v)dt' and you can see what the consequence is.

Einstein never says "a = Phi(v)", he just says "a is a function Phi(v) at present unknown", but it's clear from the context he was just speaking sloppily and actually meant that the two functions are related by a constant. If this isn't obvious to you, note that he writes:

and then just a few lines later he writes:

So from the context, it's very clear he intends that \phi (v) = a \frac{c}{\sqrt{c^2 - v^2}}.

Einstein says "a is a function Phi(v)" and not "a is a function of Phi(v)". That means clearly a=Phi(v) .

What he later writes is not consequence of mathematics but only an arbitrary construction to match his assumption. In addition he requires that Tau=0 if t=0 and in this case, a is an arbitrary constant without any relation to v and c.

 

[JesseM]

Everywhere. Try to replace dx' with (c-v)dt' and you can see what the consequence is.

I meant, where does he make a substitution that requires the chain rule, but he performs an incorrect substitution? It is always up to the person doing the proof to make a choice about when to perform a substitution, if you have dz/dx in an equation and you know z can be expressed as function of y and y can be expressed as a function of x, nothing requires you to use the chain rule and substitute (dz/dy)*(dy/dx) in for dz/dx, you only do it if it is helpful to the derivation. Similarly, in algebra if you have the equation z = 3y and you know y = 2x + 1, nothing requires you to substitute this in and get z as a function of x, you only do that if it is helpful for solving the problem.

As always, it's hard to have a discussion with you when your objections are always so vague. From now on I ask that if you think there is an error in the proof, you always point to a specific step where the conclusions don't follow from the initial assumptions. Presumably if there's an error, that means there's some specific place where he uses some equation or equations A, B and C to get another equation D, but you can show that D does not actually follow from A, B and C. If you can't do this, then you're essentially admitting that every step of the proof does follow logically from the steps before, so you should have no reasonable basis for objecting to the conclusions.

Einstein says "a is a function Phi(v)" and not "a is a function of Phi(v)". That means clearly a=Phi(v) .

And yet, I am confident you can't find a single equation where he substituted Phi(v) in for a; if you ignore his words and just look at the equations, he is totally consistent in always substituting according to the rule \phi(v) = a \frac{c}{\sqrt{c^2 - v^2}}. Unless you can find an actual inconsistency in the math, you're just quibbling over his choice of words (and the sloppiness could be the fault of the translator rather than Einstein), not over any substantive aspect of the proof itself.

What he later writes is not consequence of mathematics but only an arbitrary construction to match his assumption.

More vague language. What does he "later write", and what is the "arbitrary construction"? Remember, a and phi are unknown functions he's introducing so he can solve for them to find the factor in a coordinate transformation (the Lorentz transformation) which satisfies certain constraints (like the constraint that light must move at the same speed in both directions in the k frame, which is where he gets 1/2(tau0 + tau2) = tau1).

In addition he requires that Tau=0 if t=0 and in this case, a is an arbitrary constant without any relation to v and c.

How does your conclusion ('a is an arbitrary constant without any relation to v and c') follow from the fact that "that Tau=0 if t=0"? Again, a is not arbitrary, he's looking for the value of a that will satisfy constraints he expressed earlier, especially the idea that in both the K frame and the k frame, light moves at c in both directions.

 

[cryptic]

I meant, where does he make a substitution that requires the chain rule, but he performs an incorrect substitution?

It seams you simply don't want to accept obvious misconstruction in Einstein derivation of SR. I said Einstein used dx' in "(1/2)*St'*(1/(c-v) + 1/(c+v)) = Sx' + St'*(1/(c-v))" instead of (c-v)dt', possibly in order to confuse any critics (Sx' should be replaced with St'*1/(c-v).

It is always up to the person doing the proof to make a choice about when to perform a substitution, if you have dz/dx in an equation and you know z can be expressed as function of y and y can be expressed as a function of x, nothing requires you to use the chain rule and substitute (dz/dy)*(dy/dx) in for dz/dx, you only do it if it is helpful to the derivation. Similarly, in algebra if you have the equation z = 3y and you know y = 2x + 1, nothing requires you to substitute this in and get z as a function of x, you only do that if it is helpful for solving the problem.

No, it is a question of mathematics.

As always, it's hard to have a discussion with you when your objections are always so vague. From now on I ask that if you think there is an error in the proof, you always point to a specific step where the conclusions don't follow from the initial assumptions. Presumably if there's an error, that means there's some specific place where he uses some equation or equations A, B and C to get another equation D, but you can show that D does not actually follow from A, B and C. If you can't do this, then you're essentially admitting that every step of the proof does follow logically from the steps before, so you should have no reasonable basis for objecting to the conclusions.

It's not true. You try to runaround to avoid simple answers to my questions.

And yet, I am confident you can't find a single equation where he substituted Phi(v) in for a; if you ignore his words and just look at the equations, he is totally consistent in always substituting according to the rule \phi(v) = a \frac{c}{\sqrt{c^2 - v^2}}. Unless you can find an actual inconsistency in the math, you're just quibbling over his choice of words (and the sloppiness could be the fault of the translator rather than Einstein), not over any substantive aspect of the proof itself.

This is not "choice of words" because he said in original: "wobei a eine vorläufig unbekannte Funktion Phi(v) ist". That means: " where a is a function Phi(v) at present unknown". Mathematicaly I can use any value for a, as for example a=v⁵-c^-100.

More vague language. What does he "later write", and what is the "arbitrary construction"? Remember, a and phi are unknown functions he's introducing so he can solve for them to find the factor in a coordinate transformation (the Lorentz transformation) which satisfies certain constraints (like the constraint that light must move at the same speed in both directions in the k frame, which is where he gets 1/2(tau0 + tau2) = tau1).

a and Phi can not be unknown functions. They must come out as a mathematical consequence of solution of differential equation. I fear, you just try to start a hare. And if light is moving at the same speed in both directions in the k frame, it can not move with the same speed in both directions in another Frame because all frames have the same space.

How does your conclusion ('a is an arbitrary constant without any relation to v and c') follow from the fact that "that Tau=0 if t=0"? Again, a is not arbitrary, he's looking for the value of a that will satisfy constraints he expressed earlier, especially the idea that in both the K frame and the k frame, light moves at c in both directions.

If you are familiar with simple solutions of mathematical equations, you can see that any solution with a=f(v) satisfies the equation tau = a(t' - vx'/(c^2 - v^2)) if t'=0, Tau=0 and x'=0.

 

[JesseM]

It seams you simply don't want to accept obvious misconstruction in Einstein derivation of SR. I said Einstein used dx' in "(1/2)*St'*(1/(c-v) + 1/(c+v)) = Sx' + St'*(1/(c-v))" instead of (c-v)dt', possibly in order to confuse any critics (Sx' should be replaced with St'*1/(c-v).

You really are totally confused, St' and Sx' don't appear anywhere in Einstein's paper, they are terms I invented to refer to the slope of a plane in the 3D space with axes x', t', tau, with the specification that the plane pass through the three points corresponding the the x',t',tau coordinates of the three events of the light being sent out, the light being reflected by the mirror, and the light returning to the origin. Defining this plane was a trick I came up with to show how you could go from this equation:

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))

to this one:

1/2(1/(c-v) + 1/(c+v))*(dtau/dt') = dtau/dx' + (1/c-v)*(dtau/dt')

Einstein didn't explain how he got from the first equation to the second, so I was filling in the gaps in my own way. The basis for my trick is that he specified x' should be infinitesimally small, so the three events are infinitesimally close together; so even if tau(x',t') is some arbitrary curvy function, as long as it's differentiable, when you zoom in on an infinitesimally small neighborhood of this function it'll just look like a flat plane (similar to how the Earth looks flat on small scales even though it's really a curved surface) with slope Sx' = dtau/dx' along the x' direction and slope St' = dtau/dt' along the t' direction.

If you don't like the trick I used, you don't have to use it, there are other ways of going from Einstein's first equation above to his second one. I already gave you some links for some more calculus-oriented ways of doing it:

By the way, note that although I avoided the use of calculus in my derivation by making use of the fact that if you zoom in on an infinitesimally small region of a smooth function it'll look like a flat plane, you can derive the equation above using calculus--see this thread as well as DrGreg's post #6 on this thread.

But if you are willing to accept the trick that a differentiable function of two variables will look like a flat plane if you zoom in on an infinitesimal region, then please explain why you think the slope of the plane on each axis should obey the relation Sx' = St'*1/(c - v). Where are you getting that from? Can you quote the prior equation or equations that lead you to this conclusion?

It is always up to the person doing the proof to make a choice about when to perform a substitution, if you have dz/dx in an equation and you know z can be expressed as function of y and y can be expressed as a function of x, nothing requires you to use the chain rule and substitute (dz/dy)*(dy/dx) in for dz/dx, you only do it if it is helpful to the derivation. Similarly, in algebra if you have the equation z = 3y and you know y = 2x + 1, nothing requires you to substitute this in and get z as a function of x, you only do that if it is helpful for solving the problem.

No, it is a question of mathematics.

So are you arguing I am obligated to make certain substitutions even if they aren't useful to what I'm trying to prove? Suppose I have a simple problem like finding the second derivative of z, d/dx(dz/dx), given x. Suppose I already have z as a function of x, z(x) = 9x^5. Obviously it's pretty simple to find the second derivative of z with respect to x here just by differentiating 3x^5 twice with respect to x, giving 60x^3. But suppose I also happen to know z as a function of y, z(y) = y^3, and y as a function of x, y(x) = 3x^2. Are you saying that instead of just differentiating z with respect to x to find dz/dx, and then differentiating dz/dx with respect to x again, I'm somehow obligated to substitute (dz/dy)*(dy/dx) in for dz/dx, and then differentiate that with respect to x? This would be a totally unnecessary step of course, since I will end up getting exactly the same answer for the second derivative of z with respect to x, 60x^3, but the calculations will be slightly more complicated.

When two expressions are equivalent, of course that always means you have a choice of which one to use in your equations. Since d/dx(dz/dx) = d/dx[(dz/dy)*(dy/dx)], I have the choice of whether I want to use z(x) to find the value of dz/dx and then differentiate the result with respect to x, or if I want to use z(y) and y(x) to find the values of dz/dy and dy/dx, then multiply them together, then differentiate the result with respect to x. Both paths must necessarily lead to the same answer, so you're free to use either.

As always, it's hard to have a discussion with you when your objections are always so vague. From now on I ask that if you think there is an error in the proof, you always point to a specific step where the conclusions don't follow from the initial assumptions. Presumably if there's an error, that means there's some specific place where he uses some equation or equations A, B and C to get another equation D, but you can show that D does not actually follow from A, B and C. If you can't do this, then you're essentially admitting that every step of the proof does follow logically from the steps before, so you should have no reasonable basis for objecting to the conclusions.

It's not true. You try to runaround to avoid simple answers to my questions.

I genuinely don't understand what your broad statements mean most of the time; for example, I had no idea that your statement about Einstein failing to use the chain rule was supposed to be somehow related to the equation I gave for the slopes Sx' and St' until you said:

Einstein used dx' in "(1/2)*St'*(1/(c-v) + 1/(c+v)) = Sx' + St'*(1/(c-v))" instead of (c-v)dt', possibly in order to confuse any critics (Sx' should be replaced with St'*1/(c-v)

So, even if you think your meaning is obvious, please do me the courtesy of being as explicit as possible about which steps of the proof you think are wrong because when you just speak in generalities it isn't obvious to me which steps your comments are supposed to be related to. OK?

Would you agree with my general comment that any mathematical proof is valid as long as each step follows from the ones before it? So in order to show an error in the proof, you need to show an equation that does not follow mathematically from the previous ones?

Unless you can find an actual inconsistency in the math, you're just quibbling over his choice of words (and the sloppiness could be the fault of the translator rather than Einstein), not over any substantive aspect of the proof itself.

This is not "choice of words" because he said in original: "wobei a eine vorläufig unbekannte Funktion Phi(v) ist". That means: " where a is a function Phi(v) at present unknown". Mathematicaly I can use any value for a, as for example a=v⁵-c^-100.

When I said "you're just quibbling over his choice of words" I wasn't talking about the translation (I just mentioned the possibility of a translation problem in a parenthetical aside), I was saying you were quibbling over the fact that he might have made a mistake in his wording (which I acknowledge is a possibility), which is irrelevant to the mathematical validity of the proof if he never actually used the substitution phi(v)=a to get any equation. Do you agree that if he had just written "where a is a function of v at present unknown", and later said something like "and we'll define phi(v) as a*c/sqrt(c^2 - v^2)", then all the mathematics in this last section of the paper would be perfectly valid? (leaving aside your objection to a step earlier in the paper)

a and Phi can not be unknown functions. They must come out as a mathematical consequence of solution of differential equation.

It is perfectly valid to write down an unknown function which must satisfy certain constraints, and then use those constraints to find what this function actually is.

And if light is moving at the same speed in both directions in the k frame, it can not move with the same speed in both directions in another Frame because all frames have the same space.

Are you familiar with how to use the Lorentz transform to switch between the coordinates of different inertial frames in SR? If you know the Lorentz transform, it's easy to see that if you pick two events in one frame that lie along the worldline of a light beam--say, x=0,t=0 and x=cT,t=T--then if you find the coordinates of these same events in another inertial frame using the Lorentz transform, and find the distance between the events in the new frame and divide by the time between the events in the new frame, you'll still find that (change in position)/(change in time) = c. I can provide a simple example of this if you like.

How does your conclusion ('a is an arbitrary constant without any relation to v and c') follow from the fact that "that Tau=0 if t=0"? Again, a is not arbitrary, he's looking for the value of a that will satisfy constraints he expressed earlier, especially the idea that in both the K frame and the k frame, light moves at c in both directions.

If you are familiar with simple solutions of mathematical equations, you can see that any solution with a=f(v) satisfies the equation tau = a(t' - vx'/(c^2 - v^2)) if t'=0, Tau=0 and x'=0.

The equation tau=a(t' - vx'/(c^2 - v^2)) must be satisfied at all coordinates, not just tau=0 and t'=0 and x=0. The fact that tau=0 when t'=0 and x'=0 is just a consequence of this equation, a special case. It isn't supposed to tell you how to determine a, you need some additional conditions (like the speed of light being constant in both directions in both frames) in order to find a.

 

[cryptic]

It is not important what the notation is, if you write Sx or if you write ∂/∂x it is the same. What I try to say is, Einstein's differential equation is not correct because he considers space coordinate and time coordinate as two independent coordinates.

Einstein equation is:

\frac{1}{2}\left(\frac{1}{c-v}+\frac{1}{c+v}\right) \frac{\partial\tau}{\partial t}= \frac{\partial \tau}{\partial x&#039;}+\frac{1}{c-v}\frac{\partial\tau}{\partial t}.

Because x'=(c-v)t we can write ∂x'=(c-v)∂t . If we insert this in Einstein's equation we obtain:

\frac{1}{2}\left(\frac{1}{c-v}+\frac{1}{c+v}\right) \frac{\partial\tau}{\partial t}= \frac{1}{c-v}\frac{\partial\tau}{\partial t} +\frac{1}{c-v}\frac{\partial\tau}{\partial t},

with the solution v=-c/2.

As we can see Einstein equation can not be used to derive Gamma.

 

[JesseM]

It is not important what the notation is, if you write Sx or if you write ∂/∂x it is the same. What I try to say is, Einstein's differential equation is not correct because he considers space coordinate and time coordinate as two independent coordinates.

Einstein equation is:

\frac{1}{2}\left(\frac{1}{c-v}+\frac{1}{c+v}\right) \frac{\partial\tau}{\partial t}= \frac{\partial \tau}{\partial x&#039;}+\frac{1}{c-v}\frac{\partial\tau}{\partial t}.

Because x'=(c-v)t we can write ∂x'=(c-v)∂t .

Einstein's differential equation above is a general relation between arbitrary sets of tau, x', and t' coordinates; if you find the correct expression for tau as a function of x' and t' (it ends up being given by the function \tau = \frac{1}{\sqrt{1 - v^2/c^2}} [t&#039; - \frac{v*(x&#039; + vt&#039;)}{c^2}]), so you can find dtau/dx' and dtau/dt' at any event (x',t'), then the differential equation will always be satisfied. In contraxt, x'=(c-v)*t' is not a general relation between arbitrary x' and t' coordinates, it's an equation of motion for a single light pulse: only combinations of x' and t' coordinates that lie along this light pulse's worldline will satisfy the equation x'=(c-v)t'. There are plenty of other perfectly valid combinations of x' and t' where x is not equal to (c-v)t'.