Lorentz transformation for 3 frames

[arnau]

TL;DR

How to transform between the two moving frames, the properties of a third frame?

A particle is moving in the lab frame $S'$ at $\beta'_z$. I want to transform coordinates and momenta of the particle to a frame $S$ moving at $\beta_0$.

At time $t = t' = 0$:
$$z = \frac{z'} { \gamma_0 (1 - \beta'_z \beta_0) },\,
\gamma\beta_z = \gamma_0 ( \gamma'\beta'_z - \beta_0 \gamma' )$$

But what about at time $t \ne 0$ ?
I don't think it would be correct to do a normal lorentz transformation $z = \gamma_0 ( z' - \beta_0 c t' )$

Sources would be appreciated. Thanks in advance

 

[Nugatory]

[Before you post again, please please please take a moment to write your equations using LaTex: the link to the Latex Guide is right next to the box you used to enter your post. You will get better answers more quickly if people don't have to spend several minutes decoding what you wrote before they can start thinking about the question.]

But more important than that... You're asking about three frames, but in your problem there are only two frames: The lab frame which you are calling $S'$ in which the particle is moving with speed $\beta'_z$; and another frame $S$ whose origin is moving at speed $\beta_0$ relative to the origin of the lab frame.

The Lorentz transformations will convert the coordinates between the two frames and the velocity addition formula will give you the velocity of the particle in frame $S$.

 

[arnau]

[Before you post again, please please please take a moment to write your equations using LaTex: the link to the Latex Guide is right next to the box you used to enter your post. You will get better answers more quickly if people don't have to spend several minutes decoding what you wrote before they can start thinking about the question.]

But more important than that... You're asking about three frames, but in your problem there are only two frames: The lab frame which you are calling $S'$ in which the particle is moving with speed $\beta'_z$; and another frame $S$ whose origin is moving at speed $\beta_0$ relative to the origin of the lab frame.

The Lorentz transformations will convert the coordinates between the two frames and the velocity addition formula will give you the velocity of the particle in frame $S$.

Thanks, I couldn't find the guide earlier. Much nicer now.

By third reference frame I meant frame $S''$, where the particle is at rest.

At $t'=0$ the Lorentz transformation would be $z = \gamma_0z'$, but this would be incorrect I believe, because you would be be transforming what $S'$ sees.

For a particle at rest in $S''$ at position $z''$, at time $t=t'=0$:
$$z''=\gamma z = \gamma' z' \Rightarrow z = \frac{z'} { \gamma_0 (1 - \beta'_z \beta_0) } \ne \gamma_0z'$$

Maybe I'm missing something and doing an obvious mistake...

 

[arnau]

Thanks, I couldn't find the guide earlier. Much nicer now.

By third reference frame I meant frame $S''$, where the particle is at rest.

At $t'=0$ the Lorentz transformation would be $z = \gamma_0z'$, but this would be incorrect I believe, because you would be be transforming what $S'$ sees.

For a particle at rest in $S''$ at position $z''$, at time $t=t'=0$:
$$z''=\gamma z = \gamma' z' \Rightarrow z = \frac{z'} { \gamma_0 (1 - \beta'_z \beta_0) } \ne \gamma_0z'$$

Maybe I'm missing something and doing an obvious mistake...

If the particle were moving at $\beta_z' = \beta_0$, then we could do $z=\gamma_0z'$

 

[Nugatory]

By third reference frame I meant frame $S''$, where the particle is at rest.

But what do you need that frame for? You know the trajectory of the particle using the coordinates assigned by the lab frame $S'$. That's all you need to calculate coordinates of events and the trajectory of the particle in the moving frame $S$. If instead you want to know how to transform between $S''$ and $S$ you don''t need to involve $S'$, and if you want to know how to transform between $S''$ and $S'$ you don't need to involve $SS$. All you need is the velocity addition rule so that you have the correct relative velocity between the origins of frame $S$ and frame $S''$.

Maybe I'm not understanding what probem you're trying to solve here? What exactly are you looking for?

(As an aside, your convention here is a bit non-standard as unprimed coordinates are generally used for the frame in which you already know the coordinates, in this case the lab frame - your non-standard convention invites the mistake of getting the sign on the velocity backwards, but it's OK as long as you're careful and we all know what you're doing).

 

[arnau]

Ok so this is for a simulation code where at the beginning of the code the user gives the parameters in lab frame $S$. They are then immediately boosted to $S'$ frame. All the calculations are done in the co-moving $S'$ frame. For some outputs at certain required times, we transform back to lab frame, hence the non-standard convention with the primes.

This is not my code, and this is what it does:
The user gives $z'$ and $\beta'_z$ of the particle in lab frame.
Then the code does
$$z = \frac{z'}{\gamma_0(1-\beta_0\beta'_z)}$$
to boost the parameters before starting the simulation.

At first I did not understand this transformation, as I was expecting $z=\gamma_0(z'-\beta_0ct')$.
I think I now understand it: it is in fact a transformation from this "third frame" $S''$
$$ \gamma'(z'-\beta_z ct') = z'' = \gamma(z-\beta_z ct),$$ and with $t = t' = 0$, we get
$$z = \frac{z'}{\gamma_0(1-\beta_0\beta'_z)}.$$

Do you agree with this explanation?So now I am trying to add a feature to this code where it can print the location of the particle in the lab frame $S'$; we only have the coordinates and momenta in $S$ since that is the frame in which all the computations are done.
I was going to do $$z'=\gamma_0(z+\beta_0ct),$$ but I am unsure if this is correct. I think it is only valid if the particle is moving at $\beta'_z=\beta_0$ ( $\beta_z=0$), which is only the case at some particular points in the simulation, but not the case in general. What do you think? Am I overthinking it and it actually is $z'=\gamma_0(z+\beta_0ct)$?

Thanks

 

[vanhees71]

It's easier to work with covariant objects. So let's start with the four-velocity
$$u'=\gamma' \begin{pmatrix} 1 \\ \beta' \end{pmatrix}$$
with $\beta'=\beta_z'$ and $\gamma'=1/\sqrt{1-\beta^{\prime 2}}$.
Then the Lorentz boost
$$\Lambda_1=\begin{pmatrix} \cosh y_1 & -\sinh y_1 \\ -\sinh y_1 & \cosh y_1 \end{pmatrix}$$
with $y_1=\mathrm{artanh} \beta'$ brings you to
$$u''=\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\Lambda_1 u'.$$
Proof: You have $\cosh y_1=\gamma'$, $\sinh y_1=\beta' \gamma'$. Then
$$u''=\gamma' \begin{pmatrix} \gamma'(1-\beta{\prime 2}) \\ 0 \end{pmatrix}=\begin{pmatrix}1 \\ 0 \end{pmatrix}.$$
Then you boost with
$$\Lambda_2=\begin{pmatrix} \cosh y_2 & \sinh y_2 \\ \sinh y_2 & \cosh y_2 \end{pmatrix}$$
with $y_2=\mathrm{artanh} \beta_0$ to the final frame. Indeed you get
$$\Lambda_2 u''=\gamma \begin{pmatrix} 1 \\ \beta_0 \end{pmatrix}.$$
The direct Lorentz from the primed to the unprimed coordinates is
$$\Lambda_2 \Lambda_1 =\begin{pmatrix} \cosh(y_1-y_2) & -\sinh(y_1-y_2) \\ -\sinh(y_1-y_2) & \cosh(y_1-y_2) \end{pmatrix}.$$
The boost velocity is
$$\beta_{\text{boost}}=\tanh(y_1-y_2)=\frac{\sinh y_1 \cosh y_2-\sinh y_2 \cosh y_2}{\cosh y_1 \cosh y_2-\sinh y_1 \sinh y_2} = \frac{\tanh y_1-\tanh y_2}{1-\tanh y_1 \tanh y_2}=\frac{\beta'-\beta_0}{1-\beta' \beta_0}.$$

 

[arnau]

... the code the user gives the parameters in lab frame $S$. They are then immediately boosted to $S'$ frame. All the calculations are done in the co-moving $S'$ frame. For some outputs ...

Sorry this should be:
... the code the user gives the parameters in lab frame $S'$. They are then immediately boosted to $S$ frame. All the calculations are done in the co-moving $S$ frame. For some outputs ...