Deriving Lorentz Transformations Using Calculus

[Ryder Rude]

TL;DR Summary

I'm trying to derive the Lorentz transformations using calculus. I was able to get to the spacetime interval using this method (which is the same as Lorentz transformations). Tell me if there's anything wrong with this and how to fix it:

We take an arbitrary spacetime point $(x,t)$ in any observer's reference frame $A$.

Let $(x(v),t(v))$ be the co-ordinates of this same event as seen from a frame $B$ moving at a velocity $v$ wrt $A$. As $v$ varies, the set of points $(x(v),t(v))$ constitute some curve $C$.

So if we transform the point $(x,t)$ in frame $A$ to a frame $B$ moving with a small velocity $dv$ wrt $A$, we get the transformed co-ordinates somewhere along the curve $C$ as:

$$x'=x+\frac{dx}{dv}dv$$

$$t'=t+\frac{dt}{dv}dv$$

where $\frac{dx}{dv}$ and $\frac{dt}{dv}$ are the derivatives of the $x$ and $t$ co-ordinates of the curve $C$ wrt the parameter $v$

Since space and time are homogenous, the above transformations should be linear in $x$ and $t$. So we let $\frac{dx}{dv}=px+qt$, $\frac{dt}{dv}=rx+st$, where $p, q, r, s$ are constants.

So,

$$x'=x+(px+qt)dv$$

$$t'=t+(rx+st)dv$$

The transformation of points of the form $(0,t)$ ($A's$ worldline) is:

$$x'=qtdv$$

$$t'=t+stdv$$

Its slope is $\frac{x'}{t'}=\frac{qdv}{1+sdv}$. This should be equal to $-dv$ (as $B's$ worldline has slope $dv$ in $A's$ frame, so $A's$ worldline has slope $-dv$ in $B's$ frame)

From $\frac{qdv}{1+sdv}=-dv$, we get $q=-1$ and $s=0$

Substituting, we get:

$$x'=x+(px-t)dv$$

$$t'=t+rxdv$$

The above transformation maps points of the form $(tdv,t)$ to the the time-axis (as $(tdv,t)$ is $B's$ worldline, it should get mapped to the time-axis after switching to $B's$ frame)

The x co-ordinate of the transformation of $(tdv,t)$ is:

$$x'=tdv+(ptdv-t)dv$$

Since this lies on the time axis, it should be zero.

$$tdv+(ptdv-t)dv=0$$

which gives:

$$pdv=0$$

implying:

$$p=0$$

Substituting for $p$, $q$, and $s$ in the derivatives, we get:

$$\frac{dx}{dv}=-t$$

$$\frac{dt}{dv}=rx$$

which give

$$\frac{dt}{dx}=-r\frac{x}{t}$$

This is the slope equation of the curve $C$. Depending on the sign of $r$, this kind of slope equation either describes a hyperbola or an ellipse.

The solution to that slope equation is:

$$tdt=-rxdx$$

Integrating:

$$\frac{t^2}{2}=-r\frac{x^2}{2}+k'$$

$$x^2-\frac{-1}{r}t^2=\frac{2k'}{r}$$$$x^2-ct^2=k$$

where $c=\frac{-1}{r}$, $k=\frac{2k'}{r}$

Our universe follows the hyperbolic physics as we have a positive $c$ (the invariant speed of light). We can call $x^2-ct^2$ the spacetime interval. This value remains fixed along the curve $C$.

Is the above correct?

 

[Ibix]

Any expression that looks like $x=Adv$ must be wrong. You need infinitesimals in all terms or in none. So you need to go back to where you wrote $x'=x+\frac{dx}{dv}dv$ and work out what that last term should actually be. I think you actually mean something like $x'=x+\int_0^v\frac{dx}{dv'}dv'$.

 

[vanhees71]

I've no clue, upon which ideas the derivation is based. If you start from the special principle of relativity (principle of inertia) any complete analysis, assuming Euclidicity of space for inertial observers and the corresponding symmetries (homogeneity and isotropy) you must come to the conclusion that there are (up to isomorphy) two possible spacetime structures: Galilei-Newton or Einstein-Minkowski spacetime. So either I don't understand the derivation or there's some explanation missing to the effect why you end up with Minkowski spacetime (btw. the finaly formula is dimensionally wrong).

 

[PeterDonis]

$$x^2-ct^2=k$$

I haven't analyzed your logic in detail, but I assume this is a typo; it should be $c^2$ in the second term on the LHS, not $c$.

 

[PeroK]

$$x^2-ct^2=k$$

where $c=\frac{-1}{r}$, $k=\frac{2k'}{r}$

Our universe follows the hyperbolic physics as we have a positive $c$ (the invariant speed of light). We can call $x^2-ct^2$ the spacetime interval. This value remains fixed along the curve $C$.

Is the above correct?

What happens to the mathematically valid option of the Galilean transformation?

What I think you've shown is:

Given spacetime coordinates of an event in one frame $(x', t')$, the spacetime coordinates in any other frame must satisfy:
$$x^2 - ct^2 = k$$
For some universal constant $c$ and some constant $k$, which is a function of $(x', t')$.

That seems to allow the Lorentz but not the Galilean transformation.

 

[Ryder Rude]

What happens to the mathematically valid option of the Galilean transformation?

What I think you've shown is:

Given spacetime coordinates of an event in one frame $(x', t')$, the spacetime coordinates in any other frame must satisfy:
$$x^2 - ct^2 = k$$
For some universal constant $c$ and some constant $k$, which is a function of $(x', t')$.

That seems to allow the Lorentz but not the Galilean transformation.

Actually, Gallilean transformation is the case when the constant $r$ in the post is 0, or when $c$ tends to infinity

 

[Ryder Rude]

I haven't analyzed your logic in detail, but I assume this is a typo; it should be $c^2$ in the second term on the LHS, not $c$.

It was only a variable re-naming there. We can substitute $c^2$ for $\frac{-1}{r}$ if we want.

 

[Ryder Rude]

I've no clue, upon which ideas the derivation is based. If you start from the special principle of relativity (principle of inertia) any complete analysis, assuming Euclidicity of space for inertial observers and the corresponding symmetries (homogeneity and isotropy) you must come to the conclusion that there are (up to isomorphy) two possible spacetime structures: Galilei-Newton or Einstein-Minkowski spacetime. So either I don't understand the derivation or there's some explanation missing to the effect why you end up with Minkowski spacetime (btw. the finaly formula is dimensionally wrong).

It uses principle of relativity and homogeneity of spacetime somewhere in the middle. Gallilean transformation is the case when the constant $r$ in the post is 0, or when $c$ tends to infinity

Also, the final formula was only a variable re-naming. I should have put $c^2$ in there.

 

[Ryder Rude]

Any expression that looks like $x=Adv$ must be wrong. You need infinitesimals in all terms or in none. So you need to go back to where you wrote $x'=x+\frac{dx}{dv}dv$ and work out what that last term should actually be. I think you actually mean something like $x'=x+\int_0^v\frac{dx}{dv'}dv'$.

That is the transformation of the point $(0,t)$. So $x$ is initially zero. And the change in $v$ is vanishingly small. Hence, the transformed $x$ is also very small.

 

[PeterDonis]

Any expression that looks like $x = A dv$ must be wrong.

That's not actually what his equations say. They say, basically, $x' = x + dx$, where $dx = (dx / dv) dv$. I don't see a problem with that as far as it goes.

 

[PeroK]

Actually, Gallilean transformation is the case when the constant $r$ in the post is 0, or when $c$ tends to infinity

The equation $x^2 = k$ makes no sense for the Galilean transformation. You have $x = x' + (dv)t'$. And squaring that doesn't give a constant.

In general $x$ is a coordinate. You can have an equation involving $x, t,x', t'$. But, an equation like $x^2 = k$ or $x^2 - ct^2 = k$ makes no sense where $x, t$ are your general coordinates.

Somewhere along the line you've mixed up $x, t$ as functions of $x', t'$ and $dv$, and $x, t$ as your coordinates.

 

[Ryder Rude]

The equation $x^2 = k$ makes no sense for the Galilean transformation. You have $x = x' + (dv)t'$. And squaring that doesn't give a constant.

In general $x$ is a coordinate. You can have an equation involving $x, t,x', t'$. But, an equation like $x^2 = k$ or $x^2 - ct^2 = k$ makes no sense where $x, t$ are your general coordinates.

Somewhere along the line you've mixed up $x, t$ as functions of $x', t'$ and $dv$, and l$x, t$ as your coordinates.

If you use $r=0$ in the equation I derived:

$rx^2+t^2=k$

which implies:

$t^2=k$

This perfectly matches with the concept of absolute time. It means that the curve traced by the $(x,t)$ co-ordinates of an event, as $v$ varies, satisfies $t^2=k$, meaning time is same for all observers.

The equation $x^2-c^2t^2=k$ (there is a typo in the post. It should be $c^2$) represents the curve traced by the $x$ and $t$ co-ordinates of a fixed event as $v$ varies.

 

[PeroK]

If you use $r=0$ in the equation I derived:

$rx^2+t^2=k$

which implies:

$t^2=k$

This perfectly matches with the concept of absolute time. It means that the curve traced by the $(x,t)$ co-ordinates of an event, as $v$ varies, satisfies $t^2=k$, meaning time is same for all observers.

The equation $x^2-c^2t^2=k$ (there is a typo in the post. It should be $c^2$) represents the curve traced by the $x$ and $t$ co-ordinates of a fixed event as $v$ varies.

That all works out then. Very interesting!

 

[Ryder Rude]

This is also assuming differentiability of the transform wrt v. How could you justify that? We can at least expect it to be continuous wrt v. If we assume it to be continuous at every point, could the existence of Left Hand and Right Hand derivatives be justified at every point?

Aren't there only three cases for the LHDs/RHDs?:

1. The slopes narrow down (converge) as we approach points (in which case the limit exists)

2. The slopes blow up to infinity as we approach points (this is not physically possible anyway)

3. The slopes keep oscillating up and down, not narrowing down to any value. Could this case be disregarded?

 

[PeroK]

This is also assuming differentiability of the transform wrt v. How could you justify that? We can at least expect it to be continuous wrt v. If we assume it to be continuous at every point, could the existence of Left Hand and Right Hand derivatives be justified at every point?

Aren't there only three cases for the LHDs/RHDs?:

1. The slopes narrow down (converge) as we approach points (in which case the limit exists)

2. The slopes blow up to infinity as we approach points (this is not physically possible anyway)

3. The slopes keep oscillating up and down, not narrowing down to any value. Could this case be disregarded?

It's not unreasonable to assume differentiablity here.

 

[Ryder Rude]

It's not unreasonable to assume differentiablity here.

Is it only because physical processes are supposed to behave nicely?

 

[PeroK]

Is it only because physical processes are supposed to behave nicely?

It's not easy to construct a function that is continuous but not differentiable, except at a finite (or countable) set of points. Like $|x|$ at $x = 0$. A continuous function that isn't piecewise differentiable is fairly pathological.

If you are interested then the Weierstrass function, for example, is continuous everywhere and differentiable nowhere. That sort of function would be assumed to be too unphysical to represent something in this case.

That said, you could study things like fractal spacetimes - but that's beyond my knowledge. That would be related to ideas for quantising spacetime.

 

[vanhees71]

I think the answer to this is simply that we use the math we use, because it works well in the sense of describing nature, i.e., the assumption that the symmetries of spacetime that are continuously connected to the identity transformation form a Lie group is simply used, because it is so successful for centuries!

 

[Nugatory]

Is it only because physical processes are supposed to behave nicely?

I'd say it's because the math that describes a physical process has to behave as nicely as the process itself... but yes, "safe to assume" is pragmatism, not mathematical rigor.