# Please show me my error

1. Jul 28, 2009

### eekf

I have been accused of crackpottery. I will therefore not make any statements in this post. Rather I would like to ask a simple mathematical question which lies at the root of the interpretation of SR.

If I have two equations:
1. x^2 + y^2 + z^2 - kt^2 = 0, and
2. x'^2 +y'^2 + z'^2 - kt'^2 = 0.

Let k=c^2, m=sqrt(1-v^2/k), x' = (x - vt)/m, y'=y, z'=z and t'=(t-vx/k)/m

Is the equation: x^2 + y^2 + z^2 - kt^2 = A(x'^2 +y'^2 + z'^2 - kt'^2), where A is a constant
a. valid for all points (x,y,z,t), or
b. only valid for points where x^2 + y^2 + z^2 - kt^2 = 0 and x'^2 +y'^2 + z'^2 - kt'^2 = 0?

If it is only valid for (b) above (as I was taught in university and in school), then (I believe) the equations derived for special relativity are only valid for points where x^2 + y^2 + z^2 - kt^2 = 0 and x'^2 +y'^2 + z'^2 - kt'^2 = 0, and not for all ponderable points (x,y,z,t) as currently being used.

Could somebody please prove me wrong?

2. Jul 28, 2009

### Staff: Mentor

x^2 + y^2 + z^2 - kt^2 = A(x'^2 +y'^2 + z'^2 - kt'^2)

substitute in 1. and 2. gives

0 = A 0

Which is trivially true for all A.

However, in the general case you have:
3) x^2 + y^2 + z^2 - kt^2 = s^2
4) x'^2 +y'^2 + z'^2 - kt'^2 = s^2

Substituting 3) and 4) in to the above gives

s^2 = A s^2

Which is true for A=1.

In both cases it is true for all (t,x,y,z) and (t',x',y',z').

3. Jul 28, 2009

### sylas

Equations are valid or not for different situations. Your first two equations seem to be given as relations for particular co-ordinates (t,x,y,z) and (t',x',y',z').

Then you follow by the Lorentz transformations relating (t,x,y,z) and (t',x',y',z').

The Lorentz transformations are a way of mapping co-ordinates between inertial frames.

The equations 1 and 2 are equations that hold for co-ordinates of points on a light path from the origin.

The A is equal to 1 for co-ordinates of any point... but to make sense of these equations we have to give the situations to which they apply.

Let (t,x,y,z) be co-ordinates in spacetime for some inertial frame. Suppose that there is a particle which moves from (0,0,0,0) with velocity v along the x axis. You can map co-ordinates from the original frame to the frame of the moving particle with the Lorentz transformations you have given.

The general result which then applies for ANY point is
$$x^2 + y^2 + z^2 - kt^2 = x'^2 +y'^2 + z'^2 - kt'^2$$​

To prove this just substitute the primed variables.
\begin{align*} x'^2 +y'^2 + z'^2 - kt'^2 & = ((x-vt)/m)^2 + y^2 + z^2 - k((t-vx/k)/m)^2 \\ & = \frac{x^2 - 2xvt + v^2t^2 - k(t^2 - 2vtx/k + v^2x^2/k^2)}{m^2} + y^2 + z^2 \\ & = \frac{x^2 + v^2t^2 - kt^2 - v^2x^2/k}{1-v^2/k} + y^2 + z^2 \\ & = \frac{(x^2 - kt^2)(1 - v^2/k)}{1-v^2/k} + y^2 + z^2 \\ & = x^2 + y^2 + z^2 - kt^2 \end{align*}​
... as required. Note that this works for any points, not for points limited to a light path.

It's very hard to sort out what you are really thinking, because you don't really describe what the equations represent. But if you think relativity is "currently being used" incorrectly in conventional courses or textbooks, then you are indeed making a mistake somewhere.

Cheers -- sylas

PS. DaleSpam wins the typing speed race this time....

4. Jul 28, 2009

### M Grandin

I also today happened to read about "Steven Bryan"s doubting established Einsteinian Special
Relativity (SR) and presenting own alternate solutions. The questions by "eefk" in this thread appear according to this.

Although not an "SR professional" I have since youth thought much about relativity and aquainted and also regarded the derivation and teaching of the subject may invite to misconceptions - like "eefk"s questions here indicate.

For instance I think they are talking about light as if it had magical quality generating that strange relations between coordinates. Instead of talking about a light ray /photon at speed C, I think they should ask more generally: If there is a speed C (not necessarily light speed) that measures equal in all inertial frames, what relations hold between coordinates?

5. Jul 28, 2009

### Staff: Mentor

But your derivation was definitely more complete!

That is a reasonable way to approach it. In that case pure Newtonian mechanics (Galilean relativity) is obtained if C is infinite and Special Relativity is obtained if C=c. Then it becomes simply a matter of experiment to determine the value of C. So far, experimentally it seems that C=c.

6. Jul 28, 2009

### matheinste

Wouldn't we get the same "strange relations holding between coordinates" but with the "magical" C replaced by this new speed which also measures equal in all inertial frames ? Or is that too simple an answer ?

Matheinste.

7. Jul 28, 2009

### M Grandin

Of course I also meant so, if I didn't misunderstand you. I just suggested an other less specialised way looking at it. I meant people might have got impression only light rays (electromagnetic waves) can be used to derive these relations between spatial and time coordinats. Light (EM-waves) are very fundamental and therefore people may be duped (without thinking closer on it) that only light rays, measuring equal observed speed C in different inertial frames, could result in the known SR coordinate relations.

I just meant it must be sufficient there exists a speed C having quality being the same in all inertial systems, to achieve the SR relations. Even if light not existed.

8. Jul 28, 2009

### matheinste

Hello m Granddin,

Are you thinking along the lines of there being another phenomenon having a constant velocity between inertial frames ?

As you know, Einstein's relativity postulates the constancy of light velocity between inertial frames. I believe I have seen it proved theoretically that there can be only one such constant velocity. This does not I suppose mean that in theory there may not be other, presumably massless, things that have the same constant velocity c in all inertial frames. However I stand to be corrected on both these points.

Matheinste

9. Jul 28, 2009

### diazona

Yes, I've seen this proof as well. I don't remember the details but I'm sure someone could look it up.
That's true. A while ago it was thought that neutrinos were massless, and that would mean that they, like photons, would travel at speed c in all inertial frames (and non-inertial frames too, I think). Of course, now we know that neutrinos have some tiny mass, so I believe (but don't quote me on this) that the only massless particle left is the graviton, if it actually exists. If it does, it would travel at speed c.

10. Jul 29, 2009

### eekf

Thank you for everybody replying, especially Sylas and DaleSpam.

Sylas, thanks for the algebraic breakdown, but I was not complaining about the algebra. This has to be correct as the original equations are used to derive the transform. The question is about where (which points) the transform is valid.

Let me try to explain by using Einstein's 1920 derivation as an example.

Einstein starts by describing the path light follows in the two reference frames:

1. x-ct = 0, and
2. x'-ct' = 0.
This limits the application of derived equations what follows to cases where x=ct and x'=ct'.

To use a generalisation as Sylas suggests can also not fly, as the next thing Einstein does is multiply the one side with a constant to scale, and doing the same for a light wave travelling the opposite direction.

3. $$\(x'-ct') = lambda.(x-ct)$$, and
4. $$\(x'+ct') = mu.(x+ct)$$

Then he adds and subtrats (3 and 4) and makes a substitution where
$$\a = lambda+mu/2$$, and
$$\b = lambda-mu/2$$
to get

5. x' = ax-bct, and ct'=act-bx

and then by stating x' = 0 at the origin of the moving frame

6. v = bc/a

If we use the generalisation, we would get

1. x-ct = s, and
2. x'-ct' = s.

s can then be interpreted as an constant offset in x or t. It could also be a complicated function of (t,x) - which I do not know how to interpret. As long as it is a function of either x or t, the light is not travelling at c.

3. $$\(x'-ct'+s(lambda-1)) = lambda.(x-ct)$$, and
4. $$\(x'+ct'+s(mu-1)) = mu.(x+ct)$$

5. $$\x' = ax - bct - as - s$$, and
$$\ct' = act - bx + bs$$.

If we now take x' as 0 for the origin of the moving frame as originally, we get
ax = bct+s(a+1), which would give v = bc/a if and only if s is not a function of x or t.

If you can generalise using s, s should not depend on x or t. If it does, the v used in the equations is incorrect.

Please check my math. How should we select s to make it independent of x and t?

What am I doing wrong?

11. Jul 29, 2009

### eekf

Sorry for my last garbled submission. I was trying to use the Latex editor. Obviously I have a lot to learn:}

To summarise my objections:
The starting statements x-ct= 0 and x'-ct' limits the derivation to point (t,x) where x=ct and (t',x') where x'=ct'.
By using a generalisation x-ct=s (not proposed by Einstein), we can make the equations valid for more points, but this places a constraint on s that it should be independent of (t,x) and (t',x'): it should therefore be constant throughout all (t,x). If it is not, v is not constant, and becomes a function of x and/or t.
By taking a few sample points to calculate s as proposed by Sylas, gives me an s that changes on (t,x,y,z). It is therefore my proposition that if the transforms are used for any point where x^2+y^2+z^2-(ct)^2 != 0, we should be careful to use v as a constant velocity for the transforms.

12. Jul 29, 2009

### cristo

Staff Emeritus
Where are the statements "x=ct and x'=ct'" the starting point for a derivation? The derivation of the Lorentz transformation starts out by demanding the the transformations are linear, and so look something like x'=At+Bx and t'=Ct+Dx.

Having derived the Lorentz transformation, taking the limit v->c then says that if x=ct then x'=ct'; i.e. c is a constant speed.

13. Jul 29, 2009

### eekf

Sorry Cristo.

I used the derivation as explained by Einstein in his 1920 publication: Relativity: The Special and General Theory, Appendix I. (I also worked through his 1912 and 1905 derivations, but as stated in my posting, in this case I worked from his 1920 publication.

If you can point me to Lorentz's derivation, I would really appreciate it and love to look at it.

14. Jul 30, 2009

### eekf

Thanks Cristo.

I found an excellent document at http://www.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf
which derives the equations which at first glance seems to only make two assumptions:
1. If frame O' moves at v in the frame O, frame O moves in O' at -v.
2. Performing the transformation from O at v gives you the points in O and from O' at -v gives you the same point as your original in O.

15. Jul 31, 2009

### eekf

I have worked through the indicated document in detail now, and have found one fundamental assumption (or belief) which is required (for this derivation). I have not found any mathematical errors.

1. The fundamental assumption is that space and time can be combined into a single continuum. Although not the complete implication, it a priori assumes that in the case investigated:
s^2 = x^2+y^2+z^2+kt^2 = A(x'^2+y'^2+z'^2+kt'^2), where A can be set, and is normally set to 1. The reason I introduce an A here, is because the assumption made in the derivation is that the relative distance between two events in one reference frame must depend only on the relative distance between the events in the other frame linearly. It does not require them to be the equal, although it is assumed they are equal in the derivation. I believe A=1 to be implied, though not stated.

2. The problematic dx/dt = c is then be resolved as a required scaling of the t-axis with respect to the length axes, without limiting the case to (t,x,y,z) where x^2+y^2+z^2-(ct)^2=0. It is also not done explicitly, but is introduced as an observed requirement that for Lorentz frames the (x,x') and (t,t') elements of the transformation matrix must be equal. This leads to a required constant accross reference frames, which is first named a and later identified through observation/experiment to be a = -c^2.

It therefore assumes special relativity as used today, and derives the transformation required. I have no problem with these mathematics.