- #1

Ryder Rude

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- 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?

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?