Derive Lorentz Transformation by Visualizing Space-Time Coordinates

[Ryder Rude]

TL;DR Summary

I'm trying to have a simple english derivation of the Lorentz transformation. I'm trying to first get the transformations consistent with the principle of relativity (along with isotropy of space, homogeneity of spacetime, etc), and then rule out Gallilean transformation and rotations based on experimental evidence (existence of an invariant speed)

This approach is seeming intuitive to me as I can visualize what's going on at each step and there's not much complex math. But I'm not sure if I'm on the right track or if I'm making some mistakes. Here it is:

$A$ has set up a space-time co-ordinate system with some arbitrary event along his world-line as the origin. He assigns $(t,x)$ as the co-ordinates of the events around him. $A$ observes $B$ to be traveling at velocity $+v$. $B$ passes $A$ at the origin in $A's$ co-ordinates.

We need to find $(t',x')$ co-ordinates from $B's$ point of view, assuming he also sets up the same event as the origin as $A$ does (the event lies on both their worldlines)

Since all inertial frames are equivalent, $B$ must observe $A$ as moving with $-v$ speed.

If $A'$ worldline has co-ordinates $(t, 0)$ in $A's$ view, then the same worldline should be $(t,-vt)$ in $B's$ view, assuming absolute time (we keep the $t$ co-ordinate unchanged)

If we drop absolute time as a requirement, then $(t,0)$ from $A's$ frame can transform to $(\gamma t, -\gamma v t)$ in $B's$ frame. This is so the speed of $-v$ is still preserved. $\gamma$ is the stretching/squeezing factor and it should only depend on $v$ (because time is homogenous, so the stretching should be by a constant factor).

Now we know that $(t,0)$ from $A's$ view transforms to $(\gamma t, -\gamma vt)$ in $B's$ view. By symmetry, $(t,0)$ from $B's$ view transforms to $(\gamma t, \gamma vt)$ in $A's$ view.

So the transformation from $A's$ frame to $B's$ frame transforms points of the form $(\gamma t, \gamma vt)$ to $(t,0)$, and points of the form $(t,0)$ to $(\gamma t, -\gamma vt)$.

Now we look at a frame $C$ at rest relative to $A$. Its worldline is a verticle line in $A's$ frame parallel to the $t$ axis. Assuming the distance between $A$ and $C$ is $d$ in $A's$ frame, $B$ passes $C$ at the co-ordinate $(\frac{d}{v}, d)$ in $A's$ frame. Since this point lies on $B's$ worldline, it transforms to $(\frac{d}{\gamma v}, 0)$ in $B's$ frame.

This is where the intersection point of $B's$ and $C's$ worldlines gets transformed to. About the rest of the points on $C'$ worldline, if we shift the origin of $B$ to be the intersection point $(\frac{d}{v}, d)$, the situation of transforming $C's$ worldline to $B's$ frame is identical to the one where where we transformed $A's$ worldline to $B'$ frame.

So after transformation, the intersection point transforms to $(\frac{d}{\gamma v},0)$ and the rest of the points on $C's$ worldline transform to the line having the slope $-v$ containing the intersection point, and having the same stretching of $\gamma$ from the intersection point (same stretching as in the transformed version of $A's$ worldline, as $C$ is also moving at $-v$ relative to $B$, just like $A$)

So now we have the method to transform all the vertical lines (and hence every point) in $A's$ frame to $B's$ frame. Depending on the $\gamma (v)$ function, the transformation should be unique.

I don't know how to get the $\gamma (v)$ function, but is the rest of my above thinking correct or are there any holes in this? If there are issues, how can I fix this? I think $\gamma(v)<1$ should correspond to rotations, $=1$ to Gallilean transformation and $>1$ to Lorentz transformation.

 

[PeroK]

Summary:: I'm trying to have a simple english derivation of the Lorentz transformation. I'm trying to first get the transformations consistent with the principle of relativity (along with isotropy of space, homogeneity of spacetime, etc), and then rule out Gallilean transformation and rotations based on experimental evidence (existence of an invariant speed)

This approach is seeming intuitive to me as I can visualize what's going on at each step and there's not much complex math. But I'm not sure if I'm on the right track or if I'm making some mistakes. Here it is:

$A$ has set up a space-time co-ordinate system with some arbitrary event along his world-line as the origin. He assigns $(t,x)$ as the co-ordinates of the events around him. $A$ observes $B$ to be traveling at velocity $+v$. $B$ passes $A$ at the origin in $A's$ co-ordinates.

We need to find $(t',x')$ co-ordinates from $B's$ point of view, assuming he also sets up the same event as the origin as $A$ does (the event lies on both their worldlines)

Since all inertial frames are equivalent, $B$ must observe $A$ as moving with $-v$ speed.

If $A'$ worldline has co-ordinates $(t, 0)$ in $A's$ view, then the same worldline should be $(t,-vt)$ in $B's$ view, assuming absolute time (we keep the $t$ co-ordinate unchanged)

If we drop absolute time as a requirement, then $(t,0)$ from $A's$ frame can transform to $(\gamma t, -\gamma v t)$ in $B's$ frame. This is so the speed of $-v$ is still preserved. $\gamma$ is the stretching/squeezing factor and it should only depend on $v$ (because time is homogenous, so the stretching should be by a constant factor).

Now we know that $(t,0)$ from $A's$ view transforms to $(\gamma t, -\gamma vt)$ in $B's$ view. By symmetry, $(t,0)$ from $B's$ view transforms to $(\gamma t, \gamma vt)$ in $A's$ view.

So the transformation from $A's$ frame to $B's$ frame transforms points of the form $(\gamma t, \gamma vt)$ to $(t,0)$, and points of the form $(t,0)$ to $(\gamma t, -\gamma vt)$.

Now we look at a frame $C$ at rest relative to $A$. Its worldline is a verticle line in $A's$ frame parallel to the $t$ axis. Assuming the distance between $A$ and $C$ is $d$ in $A's$ frame, $B$ passes $C$ at the co-ordinate $(\frac{d}{v}, d)$ in $A's$ frame. Since this point lies on $B's$ worldline, it transforms to $(\frac{d}{\gamma v}, 0)$ in $B's$ frame.

This is where the intersection point of $B's$ and $C's$ worldlines gets transformed to. About the rest of the points on $C'$ worldline, if we shift the origin of $B$ to be the intersection point $(\frac{d}{v}, d)$, the situation of transforming $C's$ worldline to $B's$ frame is identical to the one where where we transformed $A's$ worldline to $B'$ frame.

So after transformation, the intersection point transforms to $(\frac{d}{\gamma v},0)$ and the rest of the points on $C's$ worldline transform to the line having the slope $-v$ containing the intersection point, and having the same stretching of $\gamma$ from the intersection point.

So now we have the method to transform all the vertical lines (and hence every point) in $A's$ frame to $B's$ frame. Depending on the $\gamma (v)$ function, the transformation should be unique.

I don't know how to get the $\gamma (v)$ function, but is the rest of my above thinking correct or are there any holes in this? If there are issues, how can I fix this? I think $\gamma(v)<1$ should correspond to rotations, $=1$ to Gallilean transformation and $>1$ to Lorentz transformation.

Too many words, not enough maths!

You lost me when you introduced frame $C$. If it's at rest relative to $A$, then it is just frame $A$ really.

I think you next need to consider a composite transformation: $A$ to $B$ to $C$.

PS maths shouldn't be seen as a problem; but a lack of maths should. You cannot do physics properly if you have an aversion to mathematics!

PPS That said, you're on the right track.

 

[Ryder Rude]

Too many words, not enough maths!

You lost me when you introduced frame $C$. If it's at rest relative to $A$, then it is just frame $A$ really.

I think you next need to consider a composite transformation: $A$ to $B$ to $C$.

PS maths shouldn't be seen as a problem; but a lack of maths should. You cannot do physics properly if you have an aversion to mathematics!

I can follow the purely mathematical derivations, but in the end it just seems like magic and you're left unsatisfied. I can't visualize the complex steps. So I wanted a simpler approach.

By introducing frame $C$, I just meant to include another worldline of an object which would be at a constant distance from object $A$, and hence be at rest relative to it. I wanted to deduce how this worldline would transform to $B's$ frame. But yeah, it doesn't count as a different "frame", so my wording was wrong.

Do you think this approach has any hope? I basically want to deduce the transformation of all the vertical lines in $A's$ frame to $B's$ frame, using principle of relativity, homogeneity/isotropy. Could you elaborate on your composing transform idea?

 

[PeroK]

I can follow the purely mathematical derivations, but in the end it just seems like magic and you're left unsatisfied. I can't visualize the complex steps. So I wanted a simpler approach.

By introducing frame $C$, I just meant to include another worldline of an object which would be at a constant distance from object $A$, and hence be at rest relative to it. I wanted to deduce how this worldline would transform to $B's$ frame. But yeah, it doesn't count as a different "frame", so my wording was wrong.

Do you think this approach has any hope? I basically want to deduce the transformation of all the vertical lines in $A's$ frame to $B's$ frame, using principle of relativity, homogeneity/isotropy. Could you elaborate on your composing transform idea?

There's not enough new information in frame C if it's just frame A.

Instead, try frame $C$ moving at some velocity $v_2$ relative to frame $B$, then compose the transformations $(t, x) \ \rightarrow \ (t',x') \ \rightarrow \ (t'', x'')$, where $C$ uses the double-primed coordinates.

 

[A.T.]

I can't visualize the complex steps.

If you want to visualize the difference between Galilean and Lorentz transformation, I would recommend using diagrams, or ideally animations, like in the video below, starting at time 0:50.

 

[Ryder Rude]

I just tried to calculate using my method in the post with $\gamma (v)=\frac{1}{\sqrt{1-v^2}}$. It does give the same values as Lorentz transformation

I tried transforming $(x,t)=(5,2)$, a random point, with $v=0.5$.

First, we calculate where the worldline $(0.5t,t)$ meets the line $(5,t)$. This intersection point is $(5,10)$ and it will map to $(0,\frac{10}{\gamma})$ after the transformation.

For now, I will shift this point to the origin (and shift the line $(5,t)$ to the $t$ axis) to perform stretching of $\gamma$ on the shifted version of the line.

The point $(5,2)$ lands at the point $(5-5,2-10)=(0,-8)$ after the shift. Now we change the velocity (slope) of this shifted line to $-0.5$ to change to moving frame's perspective. Now the point $(0,-8)$ lands at the point $(4,-8)$.

Now we stretch this line by $\gamma=\frac{1}{1-0.5^2}$. The point $(4,-8)$ now lands at $(4.62,-9.23)$

Now we finally shift the origin to $(0,\frac{10}{\gamma})=(0,8.66)$. The point $(4.62,-9.23)$ lands at $(4.62,-0.57)$

If we use the Lorentz transform formula on the point $(2,5)$, using $c=1$, $v=0.5$, we also get $(4.62,-0.57)$

I think deriving $\gamma (v)=\frac{1}{\sqrt{1-v^2}}$ should just be a matter of adding the requirement $T_{-v} (T_{v} (x,y))=(x,y)$, to the method in the post, right? This is equivalent to saying that we recover back the original point after we switch back to the original frame.

 

[PeroK]

I think deriving $\gamma (v)=\frac{1}{\sqrt{1-v^2}}$ should just be a matter of adding the requirement $T_{-v} (T_{v} (x,y))=(x,y)$, to the method in the post, right? This is equivalent to saying that we recover back the original point after we switch back to the original frame.

I think you must consider the composition of two transformations. Mathematically that makes sense since unless these transformations obey some composition rule, then you are missing vital information. If you don't specify a rule for compositions, you are not going to be able narrow down the possibilities to Gallilean and Lorentz transformations. Only looking at the inverse transformation will not be enough.

PS the rule I'm talking about is that if $L_1$ and $L_2$ are transformations of spacetime, then so is the composition $L_1L_2$.

 

[Ryder Rude]

I think you must consider the composition of two transformations. Mathematically that makes sense since unless these transformations obey some composition rule, then you are missing vital information. If you don't specify a rule for compositions, you are not going to be able narrow down the possibilities to Gallilean and Lorentz transformations. Only looking at the inverse transformation will not be enough.

PS the rule I'm talking about is that if $L_1$ and $L_2$ are transformations of spacetime, then so is the composition $L_1L_2$.

This works but the equations get complicated. Is there a more elegant alternative?

Does this argument work for rotations (we're only talking about two space axes here)?:

Suppose in frame $A$, with line $y$ as the vertical axis and $x$ as the horizontal axis, we observe a line $l$ in space of slope $\frac{x}{y}=v$.

Since the axes are orthogonal, the same line $l$ should have slope equal to $\frac{1}{v}$ in the frame $A'$ with $x$ as the vertical axis and $y$ as the horizontal axis (we just switched the axes).

Now using "principle of relativity" for rotations, we could argue that all rotated frames are equivalent. Hence, in the rotated frame with line $l$ as the vertical axis, the transformed versions of $y$ and $x$ axes of frame $A$ should have slopes $v$ and $\frac{1}{v}$ respectively.

Now that we know that the transformed version of the horizontal $x$ axis should have slope $\frac{1}{v}$ in the rotated frame, we could combine this with the method in the post to get $\gamma (v)=\frac{1}{\sqrt{1+v^2}}$.

Is this argument ok?

 

[PeroK]

This works but the equations get complicated. Is there a more elegant alternative?

It shouldn't get complicated. If you get the mathematcs right, then it is fairly simple and quite elegant.

Can you post your working?

Note that there are many parts of physics (classical Electromagnetism, for example) where the maths does get complicated. This derivation should be very simple by comparison.

Does this argument work for rotations (we're only talking about two space axes here)?:

Suppose in frame $A$, with line $y$ as the vertical axis and $x$ as the horizontal axis, we observe a line $l$ in space of slope $\frac{x}{y}=v$.

Since the axes are orthogonal, the same line $l$ should have slope equal to $\frac{1}{v}$ in the frame $A'$ with $x$ as the vertical axis and $y$ as the horizontal axis (we just switched the axes).

Now using "principle of relativity" for rotations, we could argue that all rotated frames are equivalent. Hence, in the rotated frame with line $l$ as the vertical axis, the transformed versions of $y$ and $x$ axes of frame $A$ should have slopes $v$ and $\frac{1}{v}$ respectively.

Now that we know that the transformed version of the horizontal $x$ axis should have slope $\frac{1}{v}$ in the rotated frame, we could combine this with the method in the post to get $\gamma (v)=\frac{1}{\sqrt{1+v^2}}$.

Is this argument ok?

I don't follow this argument at all. You appear to have simply pulled $\gamma (v)=\frac{1}{\sqrt{1+v^2}}$ out of a hat.

Once you have the 1D form of the Lorentz Transformation, you can apply the principle of relativity for rotations to obtain the general 3D Lorentz Transformation for a boost in any direction. That is going to get more complicated.

You must stick to 1D initially if you want to keep it simple.

 

[PeroK]

Just to summarise where you are. From the principle of relativity, you have shown that:
$$x' = \gamma(v)(x - vt) \ \ \text{and} \ \ t' = \gamma(v)(t - \alpha(v)x)$$
As the general relationship between inertial reference frames. There are two unknown functions of $v$, namely $\gamma(t)$ and $\alpha(t)$.

My suggestion is to look at the composition of two transformations (boosts in the same direction) and assume that must also be a valid transformation.

First, we do this by explicitly composing two boosts as above. To do this, we can let:
$$x' = \gamma(v_1)(x - v_1t) \ \ \text{and} \ \ t' = \gamma(v_1)(t - \alpha(v_1)x)$$
$$\ \ x'' = \gamma(v_2)(x' - v_2t') \ \ \text{and} \ \ t'' = \gamma(v_2)(t' - \alpha(v_2)x')$$
Where we have boosts $v_1$ and $v_2$. Now, with a bit of algebra you should obtain:
$$x'' = \gamma(v_1)\gamma(v_2)[1 - v_2\alpha(v_1)][x - (\frac{v_1 + v_2}{1 - v_2 \alpha(v_1)})t]$$
And a similar result for $t''$. When we put the two equations for $x'', t''$ together we find that:
$$v_1\alpha(v_2) = v_2 \alpha(v_1)$$
Now, here is a neat trick. This must hold for all velocities $v_1, v_2$. If we imagine fixing $v_1$ and varying $v_2$, this equation implies that $\alpha(v_2)$ is a constant times $v_2$.

This might take some time to see. One possibility to see this is to fix $v_1$ and differentiate with respect to $v_2$. This gives:
$$v_1 \alpha'(v_2) = \alpha(v_1)$$
Hence we can see that $\alpha'(v_2)$ is constant, and $\alpha(v_2) = kv_2$ for some constant $k$.

I don't see any way you can avoid the above mathematics. It must be done, one way or another. In any case, we now have:
$$x' = \gamma(v)(x - vt) \ \ \text{and} \ \ t' = \gamma(v)(t - kvx)$$
Finally, you can now use the special case where $v_2 = -v_1$ to show that:
$$\gamma(v) = \frac{1}{\sqrt{1 - kv^2}}$$

 

[Ryder Rude]

It shouldn't get complicated. If you get the mathematcs right, then it is fairly simple and quite elegant.

Can you post your working?

Note that there are many parts of physics (classical Electromagnetism, for example) where the maths does get complicated. This derivation should be very simple by comparison.
I don't follow this argument at all. You appear to have simply pulled $\gamma (v)=\frac{1}{\sqrt{1+v^2}}$ out of a hat.

Once you have the 1D form of the Lorentz Transformation, you can apply the principle of relativity for rotations to obtain the general 3D Lorentz Transformation for a boost in any direction. That is going to get more complicated.

You must stick to 1D initially if you want to keep it simple.

Let me explain the argument clealry. Here I'm only talking about what co-ordinates different observers would assign to 2D space around them (no time involved). Suppose two observers have set up co-ordinate systems with the same origin, to observe 2D space around them. The vertical axis of observer $B$ is at an angle $\theta$ to the vertical axis of observer $A$ such that $\tan {\theta} =v$

Suppose $A$ has the lines $y$ and $x$ as his vertical and horizontal axes, and $B$ has the lines $y'$ and $x'$.

The line $y'$ has slope $v$ in $A's$ frame.

If there's a third observer $C$ whose vertical axis is $x$ (same as the horizontal axis of $A$), then $y'$ (the vertical axis of $B$) has the slope $\frac{1}{v}$ in $C's$ co-ordinates. This implies that, in $B's$ co-ordinates, the line $x$ (the horizontal line of $A's$ frame) should have slope $\frac{1}{v}$ (since all rotated frames are equivalent).

Using the method in the post, we find that points of the form $(x,0)$ transform to:

$$x'=\gamma x$$

$$y'=\frac{x}{v}(\frac{1}{\gamma}-\gamma)$$

This has the slope $\frac{x'}{y'}=\frac{\gamma ^2 v}{1-\gamma ^2}$. We equate this slope to $\frac{1}{v}$ to get $\gamma (v)=\frac{1}{\sqrt{1+v^2}}$

 

[PeroK]

What is the state of relative motion of A, B and C?

 

[Ryder Rude]

What is the state of relative motion of A, B and C?

There's no motion involved. It's just a similar problem where $A$ and $B$ have set up co-ordinate systems to assign to 2D space. You can visualize it as observers just standing on a 2D plane and measuring co-ordinates of points around them using a ruler. $C$ is basically just $A$ except he uses $A's$ horizontal axis as his vertical axis (he assigns flipped versions of the ordered assigned by $A$). The co-ordinate transformation involved here is simple rotation, and that is what we derive in the end by $\gamma (v) =\frac{1}{\sqrt{1+v^2}}$

 

[PeroK]

There's no motion involved.

Then what is $v$?

 

[Ryder Rude]

Then what is $v$?

Sorry for the confusing naming. Actually, $B's$ vertical axis is titled relative to $A's$ vertical axis at an angle $\theta$, and $\tan{\theta}=v$

 

[PeroK]

Sorry for the confusing naming. Actually, $B's$ vertical axis is titled relative to $A's$ vertical axis at an angle $\theta$, and $\tan{/theta}=v$

So what on Earth has that to do with the Lorentz Transformation?

 

[Ryder Rude]

So what on Earth has that to do with the Lorentz Transformation?

It's a very similar problem if you think about it. Even the $\gamma (v)$ just differs by a sign. Do you think my argument works for this problem?

 

[PeroK]

It's a very similar problem if you think about it. Even the $\gamma (v)$ just differs by a sign. Do you think my argument works for this problem?

It doesn't work for this problem. But, another way to characterise Lorentz boosts is to use a concept called rapidity. You can think of a boost as a hyperbolic rotation in 4D spacetime. The key quantities in that case are not the standard trig functions, but their hyperbolic versions: $\sinh, \cosh, \tanh$.

So, there is an anology between rotations in 3D space and Lorentz boosts in 4D spacetime. You should should give yourself some credit for stumbling on this. You ended up with the wrong sign in the gamma factor, because you have consider spatial rotations (governed by the trig functions), whereas velocity-based boosts are governed by the hyperbolic trig functions.

If you assume spacetime has a hyperbolic geometry, then you can generate a lot of results much more easily than by purely algebraic means. In fact, some people think SR should be taught this way.

But, if you are starting from first principles and trying to show, from homogeneity and isotropy alone, that spacetime must be either: a) Euclidean space with absolute time or b) Minkowski 4D Spacetime, then you can't start with the assumptions about hyperbolic Minkowski geometry.

You might also be interested to look up the Wigner rotation.

All this said, I don't think it helps with the derivation of the Lorentz transformation from first principles.

 

[Ryder Rude]

It doesn't work for this problem.

My derivation is in post #11. What part of it doesn't work though? In the end of the derivation, if I equate the expression to $\frac{-1}{v}$ instead of $\frac{1}{v}$, I do arrive at Lorentz transformation instead of rotations.

In $C's$ frame, the line $y'$ (the vertical line of $B's$ frame) has a slope $\frac{1}{v}$. Principle of relativity implies that the slope of $x$ ($C's$ vertical line) in $B's$ frame should also have an absolute value of $\frac{1}{v}$. The sign can change as sign is only direction. So I can equate it to $\frac{1}{v}$ and $\frac{-1}{v}$ one by one, and then choose the $\frac{-1}{v}$ case based on experimental evidence.

One big problem is that the frame $C$, in context of spacetime, is a faster than light frame (because the time axis of frame $C$ is the same as the space axis of frame $A$). But I think special relativity does not rule out faster than light objects. I have read than faster than light objects are still speculated in physics. Special relativity only rules out 'slower than light object' reaching light speed, right?

 

[PeroK]

My derivation is in post #11. What part of it doesn't work though?

It's simply not relevant, because you are deriving a relationship between angles in rotated spatial frames with no inertial motion. You are trying to derive a relationship between frames which have relative inertial motion.

These two exercises are very different.

 

[Ryder Rude]

It's simply not relevant, because you are deriving a relationship between angles in rotated spatial frames with no inertial motion. You are trying to derive a relationship between frames which have relative inertial motion.

These two exercises are very different.

Here I'm making the same argument in the context of motion:

Using the method in the post, we first derive the transformation of points of the form $(x,0)$ (points on the horizontal axis of a frame) in terms of $\gamma (v)$.

It is : $x'=\gamma x$, $t'=\frac{x}{v}(\frac{1}{\gamma}-\gamma)$

Its slope is equal to $\frac{x'}{t'}=\frac{\gamma ^2 v}{1-\gamma ^2}$

The worldline of $B$ (the frame we're transforming to) has a slope $v$ in $A's$ frame. If we consider a frame $C$ whose time axis is the same as $A's$ space axis and whose space axis is the same as $A's$ time axis, then $B'$ worldline has slope $\frac{1}{v}$ in $C's$ frame ($C$ is the faster than light inertial frame here).

Now principle of relativity implies that $C's$ worldline (which is the space axis of $A$) seen in $B's$ frame also has slope of absolute value $\frac{1}{v}$. The sign of the slope can change as sign only refers to direction.

So we have $\frac{\gamma ^2 v}{1-\gamma ^2}= \frac{1}{v}$ or $=\frac{-1}{v}$

This gives $\gamma=\frac{1}{\sqrt{1+v^2}}$ or $\gamma=\frac{1}{\sqrt{1-v^2}}$. We accept the latter formula based on experimental evidence.

This requires the existence of a faster than light inertial frame $C$. But I've read that the existence of those frames is still speculated and special relativity does not prohibit those frames.

 

[PeroK]

Here I'm making the same argument in the context of motion:

Using the method in the post, we first derive the transformation of points of the form $(x,0)$ (points on the horizontal axis of a frame) in terms of $\gamma (v)$.

It is : $x'=\gamma x$, $t'=\frac{x}{v}(\frac{1}{\gamma}-\gamma)$

It's slope is equal to $\frac{x'}{t'}=\frac{\gamma ^2 v}{1-\gamma ^2}$

The worldline of $B$ (the frame we're transforming to) has a slope $v$ in $A's$ frame. If we consider a frame $C$ whose time axis is the same as $A's$ space axis and whose space axis is the same as $A's$ time axis, then $B'$ worldline has slope $\frac{1}{v}$ in $C's$ frame ($C$ is the faster than light inertial frame here).

Now principle of relativity implies that $C's$ worldline (which is the space axis of $A$) seen in $B's$ frame also has slope of absolute value $\frac{1}{v}$. The sign of the slope can change as sign only refers to direction.

So we have $\frac{\gamma ^2 v}{1-\gamma ^2}= \frac{1}{v}$ or $=\frac{-1}{v}$

This gives $\gamma=\frac{1}{\sqrt{1+v^2}}$ or $\gamma=\frac{1}{\sqrt{1-v^2}}$. We accept the later formula based on experimental evidence.

This requires to existence of a faster than light inertial frame $C$. But I've read that the existence of those frames is still speculated and special relativity does not prohibit those frames.

What I think you've done is as follows:

1) Using Euclidean geometry, you have found a relationship between slopes of lines that has a factor $k^2 = 1$. And, it turns out that in fact $k = 1$ for Euclidean geometry.

Note: Euclidean geometry is characterised in this case by the distance between points is $\sqrt{t^2 + x^2}$.

2) You then note that the second solution $k = -1$ gives the gamma factor in the Lorentz Transformation.

3) What you've actually done (I think) is "discovered" the concept of an alternative hyperbolic geometry, where $k = -1$ and the distance between two points is $\sqrt{t^2 - x^2}$.

What you have not shown, as far as I can see, is that this hyperbolic geometry applies to the transformation of spacetime between inertial frames of reference.

We know that $\gamma = \frac{1}{\sqrt{1-v^2}}$ is the answer we want. Does that mean that any sequence of logic and algebra that produces this answer is correct? What you did was quite clever, but ultimately I think you took a Euclidean relationship, swapped the signs and noticed that you had the Lorentz gamma factor.

 

[PeroK]

PS another way to product the Lorentz Transformation is as follows:

1) Assume the distance between a point and the origin is $d^2 = t^2 - x^2 - y^2 - z^2$. And, extend this in the obvious way to the distance between any two points.

2) Assume that distance between two points is inertial frame-invariant.

From this you can deduce the Lorentz Transformation - and all of SR.

This is closer to what you have done than a direct derivation from first principles.

 

[PeroK]

PPS I think I see exactly what you've done. You didn't actually need to swap the space and time axes - that was just a limiting case of the following alternative derivation of the Lorentz transformation.

1) Assume an inertial boost is a rotation of (in this case) 2D hyperbolic spacetime.

That's essentially what you've done. And, lo and behold, out pops the Lorentz Transformation!

In summary, if you described more precisely what you are doing, then it's all perfectly valid. But, it's not quite a derivation from first principles, as there is nothing to justify assumption 1). Either as a rotation in spacetime or - in the extreme case - swapping time and space axes.

 

[Ryder Rude]

What I think you've done is as follows:

1) Using Euclidean geometry, you have found a relationship between slopes of lines that has a factor $k^2 = 1$. And, it turns out that in fact $k = 1$ for Euclidean geometry.

Note: Euclidean geometry is characterised in this case by the distance between points is $\sqrt{t^2 + x^2}$.

2) You then note that the second solution $k = -1$ gives the gamma factor in the Lorentz Transformation.

3) What you've actually done (I think) is "discovered" the concept of an alternative hyperbolic geometry, where $k = -1$ and the distance between two points is $\sqrt{t^2 - x^2}$.

What you have not shown, as far as I can see, is that this hyperbolic geometry applies to the transformation of spacetime between inertial frames of reference.

We know that $\gamma = \frac{1}{\sqrt{1-v^2}}$ is the answer we want. Does that mean that any sequence of logic and algebra that produces this answer is correct? What you did was quite clever, but ultimately I think you took a Euclidean relationship, swapped the signs and noticed that you had the Lorentz gamma factor.

I think my derivation is from the first principles with an extra assumption: There exist faster than light objects whose worldline is the same as the space axis of some other (slower than light) frame. And also that the principle of relativity applies to the inertial frames of these faster than light objects as well (which is implicit in the statement : '*All* inertial frames are equivalent')

Once we add the existence of these inertial frames as an assumption, then the two cases of $\gamma (v)=\frac{1}{\sqrt{1+v^2}}, \frac{1}{\sqrt{1-v^2}}$ arise naturally only using principle of relativity (and homogeneity of spacetime)

Also, if $B$ is moving at velocity $v$ in $A's$ frame, then we find that $A$ moves at $-v$ in $B's$ frame (a sign change). Interestingly, if we apply this same sign change rule to the transformation from a 'faster than light' frame to a 'slower than light' one, we would be able to directly consider the case where we equate the slope to $\frac{-1}{v}$.

Suppose $B$ is moving at $v$ in $A's$ frame, and $C$ is the 'faster than light' version of frame $A$ (with space and time axes of $A$ interchanged)

Now since $B$ is moving at $\frac{1}{v}$ in $C's$ frame, the sign change rule directly implies that $C$ is moving at $\frac{-1}{v}$ in $B's$ frame. So this sign change rule is giving us the correct transformation right away without needing to consider the other case. I'm not sure if this is just a co-incidence though.

 

[Ryder Rude]

But, it's not quite a derivation from first principles, as there is nothing to justify assumption 1). Either as a rotation in spacetime or - in the extreme case - swapping time and space axes.

I would say that assuming 'faster than light' inertial frames and applying the same old principle of equivalence of all inertial frames seems better than directly assuming hyperbolic geometry.

 

[PeroK]

I think my derivation is from the first principles with an extra assumption: There exist faster than light objects whose worldline is the same as the space axis of some other (slower than light) frame.

Yes, but you didn't need to do that. Swapping the axes was an extreme case, but it amounts to the same thing as taking a boost to be a rotation in spacetime. You effectively took the limit as $v \rightarrow 1$.

 

[PeroK]

So this sign change rule is giving us the correct transformation right away without needing to consider the other case. I'm not sure if this is just a co-incidence though.

It's not a coincidence. Your sign change is, in a nutshell, taking the geometry to be hyperbolic. And that is what these boost transformations are: rotations in hyperbolic spacetime.

It's all good stuff, by the way.

 

[Ryder Rude]

It's not a coincidence. Your sign change is, in a nutshell, taking the geometry to be hyperbolic. And that is what these boost transformations are: rotations in hyperbolic spacetime.

It's all good stuff, by the way.

But the sign change I was talking about in that quote is present even in Galilean relativity. There too the relative velocity between two frames changes from $v$ to $-v$ when we switch frames. I just tried to apply the same rule to transformations between 'faster than light' and 'slower than light' frames.

I guess you're saying that applying the same rule there is the same as assuming hyperbolic geometry. In that case, I'm kinda assuming hyperbolic geometry.

Though just assuming frames with swapped axes is not the same as assuming hyperbolic geometry, as that assumption still let's us consider the two cases of Euclidean and Minkowski rotations.

 

[PeroK]

I would say that assuming 'faster than light' inertial frames and applying the same old principle of equivalence of all inertial frames seems better than directly assuming hyperbolic geometry.

It's simply unnecessary to assume FTL. Especially since these FTL frames are disallowed by the Lorentz transformations that have been deduced. Try here, for example:

https://en.wikipedia.org/wiki/Rapidity

 

[Ryder Rude]

It's simply unnecessary to assume FTL. Especially since these FTL frames are disallowed by the Lorentz transformations that have been deduced. Try here, for example:

https://en.wikipedia.org/wiki/Rapidity

I read somewhere that Lorentz transformations actually disallow matter being accelerated to FTL. Particles which are already moving FTL can still be hypothesized. Here are some related links:

https://physics.stackexchange.com/q...-it-has-always-been-travelling-faster-t/55871

https://en.m.wikipedia.org/wiki/Faster-than-light

It is stated that these particles would cause causality problems, but that wouldn't be the first time the universe defied our everyday logic.