An alternative derivation of special relativity

I see 3D thinking in two dimensions.
A straight line is a special case of a dot.
A dot is a point in 3D space. (and in turn is a special case of a circle, thence a sphere)
To make it (the dot) into a straight line, you have to add the dimension of time (four dimensions) In taking the dot from from one set of coordinates to another takes t amount of time.

Even the straight line is a special case of the surface of a sphere, or toroid.

Archaea said:
in Re: An alternative derivation of special relativity
"« Reply #6 on: February 06, 2014, 08:33:50 AM
Which is just the classical Doppler effect... with an unfortunate minus sign that's RUINING MY THEORY
I'm probably a bit biased on this, however, since the Doppler effect is what I've based all my thinking around."

Imagine what an observer would hear/see if he had objects heading toward him from EITHER DIRECTION. Then imagine an observer on one of the objects, observing the other two objects.
The fun begins.
Mind you, the maths made my eyes glaze over..

MusicMan said:

I see 3D thinking in two dimensions.
A straight line is a special case of a dot.
A dot is a point in 3D space. (and in turn is a special case of a circle, thence a sphere)
To make it (the dot) into a straight line, you have to add the dimension of time (four dimensions) In taking the dot from from one set of coordinates to another takes t amount of time.

Even the straight line is a special case of the surface of a sphere, or toroid.

Archaea said:
in Re: An alternative derivation of special relativity
"« Reply #6 on: February 06, 2014, 08:33:50 AM
Which is just the classical Doppler effect... with an unfortunate minus sign that's RUINING MY THEORY
I'm probably a bit biased on this, however, since the Doppler effect is what I've based all my thinking around."

Imagine what an observer would hear/see if he had objects heading toward him from EITHER DIRECTION. Then imagine an observer on one of the objects, observing the other two objects.
The fun begins.
Mind you, the maths made my eyes glaze over..

Click to expand...

The mistake I think people make when visualizing this stuff is that they don't take their own perspective as part of the system. So when someone imagines three observers in a system, there really is four observers, the forth being the imaginer's perspective. I think it's this forth perspective which observes the speed of light to be something other than c in SR, and while this might seem to be not a problem, I think it is.

And maths is fun.

Archaea said:

The mistake I think people make when visualizing this stuff is that they don't take their own perspective as part of the system. So when someone imagines three observers in a system, there really is four observers, the forth being the imaginer's perspective. I think it's this forth perspective which observes the speed of light to be something other than c in SR, and while this might seem to be not a problem, I think it is.

And maths is fun.

Click to expand...

It's interesting to think about what massless particles experience in their reference frame.

_http://www.quantumtheorys.com/Distance-Time/Photon%20Kinematics.html

This shows that relative to the S frame, the S' frame with speed c possesses zero clock speed; therefore, relative to the S frame, the S' frame experiences in the distance-time of its speed c a single set of events located here-now. Along the X axis the events in this single set of events are given by x/c in Eq. (61). All other events located in this single set are on all axes parallel to the X axis and are also given by x/c of Eq. (61). However, relative to the S' frame, the ratio of distance to time is still D' = cT'. Therefore, theoretically S' would still experience a clock motion, and eventons would still travel at speed c relative to S'. Consequently, relative to itself, S' would experience not one, but many sets of events here-now.

Click to expand...

Personally I (via my favorite model) think massless particles do experience multiple worldlines of events relative to our here-now (aka the Cs), but the above view is not even the view of the above link (the link does like the single worldline of events).

Bluelamp said:

Archaea said:

The mistake I think people make when visualizing this stuff is that they don't take their own perspective as part of the system. So when someone imagines three observers in a system, there really is four observers, the forth being the imaginer's perspective. I think it's this forth perspective which observes the speed of light to be something other than c in SR, and while this might seem to be not a problem, I think it is.

And maths is fun.

Click to expand...

It's interesting to think about what massless particles experience in their reference frame.

_http://www.quantumtheorys.com/Distance-Time/Photon%20Kinematics.html

This shows that relative to the S frame, the S' frame with speed c possesses zero clock speed; therefore, relative to the S frame, the S' frame experiences in the distance-time of its speed c a single set of events located here-now. Along the X axis the events in this single set of events are given by x/c in Eq. (61). All other events located in this single set are on all axes parallel to the X axis and are also given by x/c of Eq. (61). However, relative to the S' frame, the ratio of distance to time is still D' = cT'. Therefore, theoretically S' would still experience a clock motion, and eventons would still travel at speed c relative to S'. Consequently, relative to itself, S' would experience not one, but many sets of events here-now.

Click to expand...

Personally I (via my favorite model) think massless particles do experience multiple worldlines of events relative to our here-now (aka the Cs), but the above view is not even the view of the above link (the link does like the single worldline of events).

Click to expand...

My understanding is that the above quote you posted is based on the mainstream theory of SR, so the massless particle is traveling at speed c, and is following a light like path. My thinking, however, is that every observer observes themselves to be at the origin of their own reference frame, which naturally leads to the question of what light observes.

The C's said somewhere that there's no space or time for light but there is gravity, which seems to fit in well with mainstream SR. But the question then becomes what does light observe the speed of light to be? does it observe it to be c or does it observe it to be 0. If the answer is 0 then it would make sense that there is no space and time for light since light is the means by which space and time are observed.

Archaea said:

Bluelamp said:

Archaea said:

The mistake I think people make when visualizing this stuff is that they don't take their own perspective as part of the system. So when someone imagines three observers in a system, there really is four observers, the forth being the imaginer's perspective. I think it's this forth perspective which observes the speed of light to be something other than c in SR, and while this might seem to be not a problem, I think it is.

And maths is fun.

Click to expand...

It's interesting to think about what massless particles experience in their reference frame.

_http://www.quantumtheorys.com/Distance-Time/Photon%20Kinematics.html

This shows that relative to the S frame, the S' frame with speed c possesses zero clock speed; therefore, relative to the S frame, the S' frame experiences in the distance-time of its speed c a single set of events located here-now. Along the X axis the events in this single set of events are given by x/c in Eq. (61). All other events located in this single set are on all axes parallel to the X axis and are also given by x/c of Eq. (61). However, relative to the S' frame, the ratio of distance to time is still D' = cT'. Therefore, theoretically S' would still experience a clock motion, and eventons would still travel at speed c relative to S'. Consequently, relative to itself, S' would experience not one, but many sets of events here-now.

Click to expand...

Personally I (via my favorite model) think massless particles do experience multiple worldlines of events relative to our here-now (aka the Cs), but the above view is not even the view of the above link (the link does like the single worldline of events).

Click to expand...

My understanding is that the above quote you posted is based on the mainstream theory of SR, so the massless particle is traveling at speed c, and is following a light like path. My thinking, however, is that every observer observes themselves to be at the origin of their own reference frame, which naturally leads to the question of what light observes.

The C's said somewhere that there's no space or time for light but there is gravity, which seems to fit in well with mainstream SR. But the question then becomes what does light observe the speed of light to be? does it observe it to be c or does it observe it to be 0. If the answer is 0 then it would make sense that there is no space and time for light since light is the means by which space and time are observed.

Click to expand...

It's based on mainstream SR but most mainstream physicists don't tend to think it's appropriate to even think about a reference frame for a massless particle (via limiting SR to only inertial frames). Yeah the multiple worldlines via just mainstream SR might not be the best way to think about it. You could think one worldline experienced all at once via SR and then more worldlines via quantum transactions branching off from any point on the previous worldlines. You couldn't represent this by a single universe state so it would be more like changing universe information rather than spacetime travel.

Archaea said:

x' = Ax + Bt
t' = Cx + Dt

Since the speed of light is constant in all reference frames we can say:

Δx'/Δt' = Δx/Δt = c

where Δx is the distance light travels for some amount of time Δt, and c is the speed of light. Because light travels as waves we can make Δx equal to the wavelength (λ) and Δt equal to the inverse of the frequency (1/f).

So we can write the transformations as:

Δx' = AΔx + BΔt
Δt' = CΔx + DΔt

And divide the top equation by the bottom one:

Δx'/Δt' = c = (AΔx + BΔt)/(CΔx + DΔt)
c(CΔx + DΔt) = AΔx + BΔt (Multiplying both sides by CΔx + DΔt)
(cC - A)Δx + (cD - B)Δt = 0 (Collecting terms)
c2C - cA + cD - B = 0 (dividing by Δt, and substituting c for Δx/Δt)

So this gives the equation:

c2C - B + c(D - A) = 0

Click to expand...

What you have neglected here is dimensionality. Your equations do not show special relativity is wrong, your equations are the same, the constants simply are not dimensionless like the normal derivation. By removing the c from ct you introduce this issue. If we take your first equation.

x' = Ax + Bt

We see that Bt must have the dimensions of length (because it is equated to x), so B must have the dimensions of length over time. Therefore it can be re-expressed as a dimensionless constant times a speed c. B=Pc. With this and Cc=Q (for similar reasons) we have:

x' = Ax + Pct
ct' = Qx + Dct

Follow this though and you get the same as the normal derivation but with the changes I described.

c2C - B + c(D - A) = 0
cQ - cP + c(D - A) = 0
Q-P=A-D

This is the same form as the paper. Your derivation is just as correct as the normal one but your constants are different. Your final line doesn't match theirs because your A isn't their A and so on but mathematically they are the same.

The normal derivation is not in any way flawed because of this.

What you have neglected here is dimensionality. Your equations do not show special relativity is wrong, your equations are the same, the constants simply are not dimensionless like the normal derivation. By removing the c from ct you introduce this issue. If we take your first equation.

x' = Ax + Bt

We see that Bt must have the dimensions of length (because it is equated to x), so B must have the dimensions of length over time. Therefore it can be re-expressed as a dimensionless constant times a speed c. B=Pc. With this and Cc=Q (for similar reasons) we have:

x' = Ax + Pct
ct' = Qx + Dct

Follow this though and you get the same as the normal derivation but with the changes I described.

c2C - B + c(D - A) = 0
cQ - cP + c(D - A) = 0
Q-P=A-D

This is the same form as the paper. Your derivation is just as correct as the normal one but your constants are different. Your final line doesn't match theirs because your A isn't their A and so on but mathematically they are the same.

The normal derivation is not in any way flawed because of this.

Click to expand...

I think your right. A lot of what I did in the earlier posts of this thread is messy IMO. However, I still think the changes perceived in space-time correspond to the Doppler effect.

Here are two playlists of quantum mechanics lectures on YouTube from MIT:

https://www.youtube.com/playlist?list=PLUl4u3cNGP61-9PEhRognw5vryrSEVLPr
https://www.youtube.com/playlist?list=PLUl4u3cNGP60QlYNsy52fctVBOlk-4lYx

I think these lectures are pretty good for learning about some of this stuff. :)

Here's an update on what I've been working on lately.

In order for there not to be any funny business when rotating in a plane of one time coordinate and one space coordinate we need to break up the Schrödinger equation into two separate operators:

Pt = -iћ(∂/∂t)

And,

H = P2/2m + V(x)

So the Schrödinger becomes:

-Ptψ = Hψ

The justification for breaking the Schrödinger equation into two separate operators is that the left hand side always commutes with the operators x, y, z as well as Px, Py, and Pz. Whereas the right hand side only commutes with Px, Py, and Pz for free particles, and never commutes with x, y, z.

When H has no time dependence, [Pt, H] = 0 which means Pt and H have common eigenfunctions. In order to recover the results of the Schrödinger equation, the eigenvalues of both operators would have to be the same for each common eigenfunction. Interestingly, if H is time dependent, then [Pt, H] ≠ 0, so there are no common eigenfunctions and the Schrödinger equation shouldn’t work.


We can now define our space-time rotation without having to worry about funny things happening with commutators. The time evolution operator is:

U = exp[-(iPtt)/ћ)]

And the translation operator in the x direction is:

Tx = exp[-(iPxx)/ћ)]

If we follow Wikipedia: Using the time evolution operator instead of a translation operator, we get:

t' = r cos(θ+dθ) = t - xdθ + ⋯
x' = r sin(θ+dθ) = x + tdθ + ⋯

Following on we get:

Rxt(dθ) = U(-xdθ) Tx(tdθ)

So:

Rxt(dθ) = exp[-(iPt(-xdθ))/ћ)] exp[-(iPx(tdθ))/ћ]
= exp[-i(Ptx-Pxt)dθ)/ћ)

And then solving some differential equation to get:

Rxt(θ) = exp[-i(Ptx-Pxt)θ)/ћ]

From this It seems to make sense to create three new operators (Λx, Λy and Λz) which represent rotations on the (x, t), (y, t) and (z, t) planes respectively. In component form these operators are:

Λx = Ptx - Pxt
Λy = Pty - Pyt
Λz = Ptz - Pzt

These operators represent the conserved quantity when transforming between inertial reference frames. Also note that these operators don’t commute with either Pt or H.

These new operators have some fun commutators, which I'll get to in my next post. :)

Alright, now it's time for some fun with commutators.

The commutators between the three new Λ operators are:'

[Λx, Λy] = iћLz
[Λy, Λz] = iћLx
[Λz, Λx] = iћLy

Where the operators (Lx, Ly, Lz) are the angular momentum operators. The commutators between the angular momentum operators and the Λ operators are:


We can define two sets of creation and annihilation operators as:

Λ1+ = Λx + iLy
Λ1- = Λx - iLy

And

Λ2+ = Lx + iΛy
Λ2- = Lx - iΛy

These sets of creation and annihilation operators have similar commutation relations as the creation and annihilation operators for angular momentum:

[Λ1+, Λ1-] = [Λ2+, Λ2-] = 2ћΛz
[Λz, Λ1+] = ћΛ1+
[Λz, Λ2+] = ћΛ2+

And:

[Λz, Λ1-] = -ћΛ1-
[Λz, Λ2-] = -ћΛ2-

All these commutators need to be checked.

Since Lz commutes with Λz they both share a common eigenfunction. Hopefully we'll be able to use these creation and annihilation operators to find the eigenvalues of Λz.

Hi Archaea,

I agree with MysterMan10, your issue is with neglecting the c as a factor of B and D.

Since these are no longer constants you need to consider how the Δ interacts with them, you have treated them as if they were dimensionless constants.

I would suggest possibly reading this:
http://en.wikipedia.org/wiki/Differential_(infinitesimal)

edwardthefirst said:

Hi Archaea,

I agree with MysterMan10, your issue is with neglecting the c as a factor of B and D.

Since these are no longer constants you need to consider how the Δ interacts with them, you have treated them as if they were dimensionless constants.

I would suggest possibly reading this:
http://en.wikipedia.org/wiki/Differential_(infinitesimal)

Click to expand...

That's true, I'm not used to keeping track of dimensions when doing this stuff, which is a problem. The same thing happens with the Λ operators I'm playing with now. They don't seem to have well defined units, unless we either throw in a velocity somewhere or say something like "time and space have the same dimensions."

At any rate what I'm interested in doing now is seeing if it's possible to see how particles change when transforming between inertial reference frames (which is just an ordinary rotation). I think dimensions are going to be a problem throughout all of this. :)

Also, I tired looking at the Wikipedia page you linked to, but it said the page didn't exist.

The link got broken. This is the right one:

http://en.wikipedia.org/wiki/Differential_(infinitesimal)

Thanks Monotonic :)

Ok, I've read through Wikipedia's infinitesimal derivatives page and I've re-read some of this thread and I agree that I was wrong about finding the transformations for SR, because I neglected dimensions. I still think, however, that we can't treat the following two equations as simultaneous equations:

A - C = B - D
A + C = B + D

I can't really think of why not at the moment however...

Another problem with the paper is that it say's that x = ct and x' = ct' (pages 4 and 5) and then it goes on to say (on page 5) that:

x/t = -Bc/A = v
x'/t' = Bc/A = -v

Where v is the relative velocity of the two observers, but v should be equal to c.

This is a reason why I think that we should treat x and x' as the observed wavelength of a light ray, and we should treat t and t' as the observed frequency of the same light ray.

Hi Archaea,

But to answer your question, my understanding is that v is the relative velocity of the two reference frames. I don't think there is any reason it needs to be c. For that to be the case, it would imply that the two reference frames are always moving away from each other at the speed of light.

As an aside, my background is in computer science, not physics so please bear that in mind when reading my answers. I'm no expert, just someone who finds this fascinating. :)