Lorentz transformations for spacetime

[Cyosis]

We can write the set of equations:

$$\begin{align*}<br /> x=x'\cosh\phi+ct'\sinh \phi \\<br /> ct=x'\sinh\phi+ct'\cosh\phi<br /> \end{align*}$$

as

$$\begin{pmatrix}<br /> ct \\x<br /> \end{pmatrix}<br /> <br /> =<br /> <br /> \begin{pmatrix}<br /> \cosh \phi & \sinh\phi \\<br /> \sinh\phi & \cosh\phi<br /> \end{pmatrix}<br /> <br /> \begin{pmatrix}<br /> ct' \\ x'<br /> \end{pmatrix}$$

Or the inverted set of equations:

$$\begin{pmatrix}<br /> ct' \\ x'<br /> \end{pmatrix}<br /> <br /> =<br /> <br /> \begin{pmatrix}<br /> \cosh \phi & -\sinh\phi \\<br /> -\sinh\phi & \cosh\phi<br /> \end{pmatrix}<br /> <br /> \begin{pmatrix}<br /> ct \\ x<br /> \end{pmatrix}$$

 

[stevmg]

Cyosis -

Muchas gracias, Vielen Dank, Merci beaucoup, Thanks!

Steve G
Melbourne, FL

 

[stevmg]

By my "calculations"

$$\phi$$ = arcosh $$\gamma$$ or = arcsinh $$\gamma$$(v/c)

In order to make the matrices work out.

Confirmation:

cosh² - sinh² = gamma² - gamma²(v²/c²) = [ 1 - (v²/c²]/[ 1 - (v²/c²] = 1

Q.E.D.

This is a round about way to calculating the transformation as one could obtain arctanh v/c and then compute sinh (arctanh v/c) and cosh (arctanh v/c) and go from there by using the following identities - somehow.

sinh x = (ex - e-x)/2 equation 1
cosh x = (ex + e-x)/2 equation 2
sech x = 1/cosh x equation 3
csch x = 1/sinh x equation 4
tanh x = sinh x/cosh x equation 5
coth x = 1/tanh x equation 6
cosh² x - sinh² x = 1 equation 7
tanh² x + sech² x = 1 equation 8
coth² x - csch² x = 1 equation 9
sinh (x + y) = sinh x cosh y + cosh x sinh y equation 10
sinh (x - y) = sinh x cosh y - cosh x sinh y equation 11
cosh (x + y) = cosh x cosh y + sinh x sinh y equation 12
cosh (x - y) = cosh x cosh y - sinh x sinh y equation 13
sinh² x = [-1 + cosh (2 x)]/2 equation 14
sinh (2 x) = 2 sinh x cosh x equation 15
cosh2 x = [1 + cosh (2 x)]/2 equation 16
cosh (2 x) = cosh² x + sinh² x equation 17
arcsinh x = ln [x + (x² + 1)^1/2] equation 18
arccosh x = ln [x + (x² - 1)^1/2] equation 19
arctanh x = (1/2) ln [(1 + x)/(1 - x)] equation 20
arccoth x = (1/2) ln [(x + 1)/(x - 1)] equation 21
arcsech x = ln [[1 + (1 - x2)^1/2]/x] equation 22
arccsch x = ln [[1 + (1 + x2)^1/2]/|x|] equation 23
tanh (2 x) = 2 tanh x/(1 + tanh² x) equation 24
coth x - tanh x = 2 csch (2 x) equation 25
coth x + tanh x = 2 coth (2 x) equation 26

But this sure seems a very roundabout way.

 

[stevmg]

To anybody -

In the Minkowski 4-space, when one obtains the Minkowski norm of a vector, how is it that the t or ct vector, when squared is subtracted from the cartesian vecors (x, y, z) after they are squared and summed. In time-space, of course, the cartesian coordinates are squared and summed and subtracted from the square of the time component.

I do realize that the derivation of the Lorentz transformations is entirely algebraic. I've seen many derivations of this.

 

[stevmg]

Cyosis -

Would this be the correct representation with the hyperbolic functions?


$$\begin{pmatrix}<br /> ct' \\ x'<br /> \end{pmatrix}<br /> <br /> =<br /> <br /> \begin{pmatrix}<br /> \cosh \(arctanh(v/c) & -\sinh\(arctanh(v/c) \\<br /> -\sinh\(arctanh(v/c) & \cosh\(arctanh(v/c)<br /> \end{pmatrix}<br /> <br /> \begin{pmatrix}<br /> ct \\ x<br /> \end{pmatrix}$$

If this is the case, then all one would have to do is look up the arctanh of v/c or ($$\beta$$) and plug that into the cosh and sinh to get your matrix. That can be done on a scientific calculator directly.

 

[Fredrik]

Stevemg, the Minkowski bilinear form g (I won't call it a scalar product, or an inner product, since it isn't positive definite) is defined by

$$g(x,y)=x^T\eta y$$

I usually take

$$\eta=\begin{pmatrix}-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{pmatrix}$$

to be the definition of $\eta$, but you could choose the opposite sign if you want to. This wouldn't change any of the physics. It would just change a few signs here and there. The two different "norms" (they're not really norms, since they're not positive definite) are just different choices of what sign to use in the definition of $\eta$.

My definition above is more common in GR books, and the opposite sign is more common in QFT books. Some people prefer to write $$\langle x,y\rangle$$ instead of g(x,y).

Since you have taken an interest in the concept of rapidity, you may find the notes I made for myself useful. Note that I'm using units such that c=1. Keeping that c around is like putting something sharp in one of your shoes before you go out running. It's a pain with no benefits. This is what I wrote in my notes:

In special relativity, we define the rapidity $$\phi$$ of a particle moving with velocity v by $$\tanh\phi=v$$. Note that the rapidity and the velocity are approximately the same when the velocity is small. To be more precise, in the limit $$\phi\rightarrow 0$$, we have $$v=\phi+O(\phi^2)$$.

The concept of rapidity is useful because it's easier to "add" rapidities than velocities. If the velocity of frame B in frame A is v, and the velocity of a particle in frame B is v', then the velocity of the particle in frame A is

$$v\oplus v'=\frac{v+v'}{1+vv'}$$

If the rapidity of frame B in frame A is $$\phi$$, and the rapidity of a particle in frame B is $$\phi'$$, then the rapidity of the particle in frame A is just $$\phi+\phi'$$. This follows from the velocity addition formula above, the definition of rapidity, and the identity

$$\tanh(\phi+\phi')=\frac{\tanh\phi+\tanh\phi'}{1+\tanh\phi\ \tanh\phi'}$$

The definition of rapidity also implies that

$$\cosh\phi=\gamma=\frac{dt}{d\tau}$$

$$\sinh\phi=\gamma v=\frac{dx}{d\tau}$$Proof:

$$v^2=\tanh^2\phi=\frac{\sinh^2\phi}{\cosh^2\phi}=\frac{\cosh^2\phi-1}{\cosh^2\phi}=1-\frac{1}{\cosh^2\phi}$$

$$\frac{1}{\cosh^2\phi}=1-v^2=\frac{1}{\gamma^2}$$

$$\cosh\phi=\gamma$$

$$\sinh^2\phi=\cosh^2-1=\gamma^2-1=\frac{1}{1-v^2}-\frac{1-v^2}{1-v^2}=\frac{v^2}{1-v^2}=\gamma^2v^2$$

$$\sinh\phi=\gamma v$$

$$\tau=\int\sqrt{dt^2-dx^2}=\int dt\sqrt{1-\dot x^2}=\int\frac{dt}{\gamma}$$

$$\frac{d\tau}{dt}=\frac 1 \gamma$$

$$\frac{dt}{d\tau}=\gamma=\cosh\phi$$

$$\frac{dx}{d\tau}=\frac{dt}{d\tau}\frac{dx}{dt}=\gamma v=\sinh\phi$$

These results enable us to express an arbitrary proper and orthochronous Lorentz transformation as a hyperbolic rotation by an "angle" equal to the rapidity.

$$\Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}=\begin{pmatrix}\cosh\phi & -\sinh\phi\\ -\sinh\phi & \cosh\phi\end{pmatrix}$$

Cyosis -

Would this be the correct representation with the hyperbolic functions?$$\begin{pmatrix}<br /> ct' \\ x'<br /> \end{pmatrix}<br /> <br /> =<br /> <br /> \begin{pmatrix}<br /> \cosh \(arctanh(v/c) & -\sinh\(arctanh(v/c) \\<br /> -\sinh\(arctanh(v/c) & \cosh\(arctanh(v/c)<br /> \end{pmatrix}<br /> <br /> \begin{pmatrix}<br /> ct \\ x<br /> \end{pmatrix}$$

If this is the case, then all one would have to do is look up the arctanh of v/c or ($$\beta$$) and plug that into the cosh and sinh to get your matrix. That can be done on a scientific calculator directly.

Yes, but if you look at my last equation before the quote, I'm sure you'll see an easier way.