How can Lorentz transforms be inverted?

[john t]

TL;DR

algebra for inverting lorentz transforms?

John

How are the Lorentz transforms inverted from x' = gamma(x -vt) and t' =gamma(t - vx/c^2)
to the equations x =gamma(x'+vt') and t = gamma(t'+ vx'/c^2) ? The closest explanation I have seen on line is to change the non-primes to primes and reverse the direction of v). But what is the algebra? I have tried and failed to do the inversion by solving for t in the 2nd eqn above and plugging into the 1st eqn.

John

 

[Ibix]

All you need to do is write the Lorentz gamma factors explicitly as $\gamma=1/\sqrt{1-v^2/c^2}$ and then it's just a pair of simultaneous equations.

Post your working so far and we can see where you went wrong. You can write maths with LaTeX - see the guide linked below the reply box.

 

[PeroK]

Summary:: algebra for inverting lorentz transforms?

John

How are the Lorentz transforms inverted from x' = gamma(x -vt) and t' =gamma(t - vx/c^2)
to the equations x =gamma(x'+vt') and t = gamma(t'+ vx'/c^2) ? The closest explanation I have seen on line is to change the non-primes to primes and reverse the direction of v). But what is the algebra? I have tried and failed to do the inversion by solving for t in the 2nd eqn above and plugging into the 1st eqn.

John

It should work out however you do it, of course, but perhaps a good way is to start with:
$$\gamma(x' + vt')$$
And show that this is equal to $x$, using the LT for $x'$ and $t'$.

 

[john t]

All you need to do is write the Lorentz gamma factors explicitly as $\gamma=1/\sqrt{1-v^2/c^2}$ and then it's just a pair of simultaneous equations.

Post your working so far and we can see where you went wrong. You can write maths with LaTeX - see the guide linked below the reply box.

Thanks, but LaTex is not working as its instructions say, and I downloading of mactex was disallowed on my mac. Could I put my work in a word document using MathType for eqns and paste the doc in reply. I have severe glaucoma and have trouble sorting these things out on my mac.

 

[Ibix]

I can't help you if you type in Word, since I've nothing on my phone that can display it. Others may be able to help.

You don't need to download anything to use LaTeX here. Do you see the mathematical expressions PeroK and I have typed? If so, just type $1/\sqrt{1-v^2/c^2}#[/color]# and hit the Preview button. You should get a version of your post that includes the expression$1/\sqrt{1-v^2/c^2}##.

 

[john t]

I can't help you if you type in Word, since I've nothing on my phone that can display it. Others may be able to help.

You don't need to download anything to use LaTeX here. Do you see the mathematical expressions PeroK and I have typed? If so, just type $1/\sqrt{1-v^2/c^2}$ and hit the Preview button. You should get a version of your post that includes the expression $1/\sqrt{1-v^2/c^2}$.

I attached my efforts as a pdf of my work in the mac program called Notes. At the top I show the transforms giving x' and t' on the left (the derivations of whch I understand) and the desired inversions on the right. Under the line on the left I show my work, which then shifts over to the right to end in a confused mess.

 

[Ibix]

Do you know how to solve simultaneous equations? If not, I'd suggest looking that up. What you are doing is moving in the right direction but you've failed to eliminate x' between the two equations.

 

[PeterDonis]

LaTex is not working as its instructions say

How so? Feel free to PM me to avoid cluttering up this thread. If there are issues with the LaTeX handling or with the help instructions we want to know about them.

 

[john t]

Do you know how to solve simultaneous equations? If not, I'd suggest looking that up. What you are doing is moving in the right direction but you've failed to eliminate x' between the two equations.

Thanks Ibix. Yes I know how to solve simultaneous equations, but I am still having difficulty. What 2 equations do you refer to? The original 2 equations do not both have x' in them, so I cannot see how one eliminates it. I guess the common-sense answer to my question is to consider that neither reference frame is preferred, so the inverted equations must have the same form, except for the direction of the velocity. But it would be satisfying to get the algebraic proof.

 

[vanhees71]

Well, you simply invert the matrix. You can do this by brute force. We have (setting $c=1$ for convenience)
$$\begin{pmatrix}t' \\ x' \end{pmatrix}=\begin{pmatrix} \gamma & -\gamma v \\ -\gamma v &\gamma \end{pmatrix} \begin{pmatrix} t \\ x \end{pmatrix}.$$
In short-hand notation: $\hat{x}'=\hat{\Lambda} \hat{x}$. You can invert the matrix $\hat{\Lambda}$ in any of the known brute-force ways. Here I'd say for a $2 \times 2$ matrix, the most simple way is in terms of determinants,
$$\hat{\Lambda}^{-1} = \frac{1}{\mathrm{det} \hat{\Lambda}} \begin{pmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{pmatrix}=\begin{pmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{pmatrix}.$$
A more clever way is to remember that Lorentz transformations leave the Minkowski product invariant and that thus
$$\hat{\Lambda}^{-1} = \hat{\eta} \hat{\Lambda}^{\text{T}} \hat{\eta},$$
where $\hat{\eta}=\mathrm{diag}(1,-1)$ are the components of the Minkowski pseudometric and the T at the matrix indicates the transpose (which is irrelevant here since pure boosts are represented by symmetric matrices).

The result is also very easy to understand: If the inertial frame $\Sigma'$ moves with velocity $v$ in $x$ direction wrt. $\Sigma$ and thus $\Sigma$ moves with $-v$ with respect to $\Sigma'$. Thus the inverse transformation must also be a Lorentz transformation but just the one where $v$ is substituted by $-v$.

One should, however, note that this is not really "obvious" (we are only used to it from Newtonian physics!): It's an implication of the full symmetry group, including also isotropy of space in addition to the equivalence of all inertial frames (the special principle of relativity).

 

[Ibix]

The original 2 equations do not both have x' in them, so I cannot see how one eliminates it.

Apologies - that was a typo. Start with the forward transforms and either eliminate x to solve for t or eliminate t to solve for x.

 

[john t]

Well, you simply invert the matrix. You can do this by brute force. We have (setting $c=1$ for convenience)
$$\begin{pmatrix}t' \\ x' \end{pmatrix}=\begin{pmatrix} \gamma & -\gamma v \\ -\gamma v &\gamma \end{pmatrix} \begin{pmatrix} t \\ x \end{pmatrix}.$$
In short-hand notation: $\hat{x}'=\hat{\Lambda} \hat{x}$. You can invert the matrix $\hat{\Lambda}$ in any of the known brute-force ways. Here I'd say for a $2 \times 2$ matrix, the most simple way is in terms of determinants,
$$\hat{\Lambda}^{-1} = \frac{1}{\mathrm{det} \hat{\Lambda}} \begin{pmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{pmatrix}=\begin{pmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{pmatrix}.$$
A more clever way is to remember that Lorentz transformations leave the Minkowski product invariant and that thus
$$\hat{\Lambda}^{-1} = \hat{\eta} \hat{\Lambda}^{\text{T}} \hat{\eta},$$
where $\hat{\eta}=\mathrm{diag}(1,-1)$ are the components of the Minkowski pseudometric and the T at the matrix indicates the transpose (which is irrelevant here since pure boosts are represented by symmetric matrices).

The result is also very easy to understand: If the inertial frame $\Sigma'$ moves with velocity $v$ in $x$ direction wrt. $\Sigma$ and thus $\Sigma$ moves with $-v$ with respect to $\Sigma'$. Thus the inverse transformation must also be a Lorentz transformation but just the one where $v$ is substituted by $-v$.

One should, however, note that this is not really "obvious" (we are only used to it from Newtonian physics!): It's an implication of the full symmetry group, including also isotropy of space in addition to the equivalence of all inertial frames (the special principle of relativity).

Thank you very much vanhees71. The matrix inversion will do it.

 

[PeroK]

Thank you very much vanhees71. The matrix inversion will do it.

For the record, the algebra goes like this
$$\gamma[x' + vt'] = \gamma[\gamma(x - vt) + v\gamma(t - vx/c^2)] = \gamma^2[(1- v^2/c^2)x -vt + vt] = x$$
And similary for $t$.

 

[robphy]

Following up on @PeroK's post,
for the record, here's the rapidity version [where $v=\tanh\theta$ and $\gamma=\cosh\theta$]
(to show the analogy to Euclidean trigonometry).
[I'll absorb the $c$'s.]

$$\begin{pmatrix}t' \\ x' \end{pmatrix}=\begin{pmatrix} \gamma & -\gamma v \\ -\gamma v &\gamma \end{pmatrix} \begin{pmatrix} t \\ x \end{pmatrix}.$$
...
$$
\hat{\Lambda}^{-1} = \frac{1}{\mathrm{det} \hat{\Lambda}} \begin{pmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{pmatrix}=\begin{pmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{pmatrix}.
$$

is
$$
\begin{align*}
\begin{pmatrix}t' \\ x' \end{pmatrix}
\
&=\begin{pmatrix} \cosh\theta & -\cosh\theta \tanh\theta \\ -\cosh\theta \tanh\theta &\cosh\theta \end{pmatrix} \begin{pmatrix} t \\ x \end{pmatrix}\\
&=\begin{pmatrix} \cosh\theta & -\sinh\theta \\ -\sinh\theta &\cosh\theta \end{pmatrix} \begin{pmatrix} t \\ x \end{pmatrix}.
\end{align*}
$$
and
$$
\hat{\Lambda}^{-1} =
\begin{pmatrix} \cosh\theta & \sinh\theta \\ \sinh\theta &\cosh\theta \end{pmatrix},
$$
which could be interpreted as saying the inverse boost (in one spatial dimension) is using $-\theta$ for $\theta$, analogous to a Euclidean rotation in the plane.

$$\gamma[x' + vt'] = \gamma[\gamma(x - vt) + v\gamma(t - vx/c^2)] = \gamma^2[(1- v^2/c^2)x -vt + vt] = x$$

is
$$\cosh\theta[x' + \tanh\theta\ t'] = \cosh\theta[\cosh\theta(x - \tanh\theta\ t) + \tanh\theta \cosh\theta(t - \tanh\theta\ x)] = \cosh^2\theta[(1- \tanh^2\theta)x -\tanh\theta\ t + \tanh\theta\ t] = x$$
since $\gamma^2(1-(v/c)^2) \ =\ \cosh^2\theta(1-\tanh^2\theta) \ =\ \cosh^2\theta-\sinh^2\theta \ =\ 1$.

Actually, it might be better to write
$$\begin{align*}
\cosh\theta[x' + \tanh\theta\ t']
&=
\cosh\theta\ x' + \sinh\theta\ t'\\
&=
\cosh\theta (\cosh\theta\ x - \sinh\theta\ t) + \sinh\theta(\cosh\theta\ t - \sinh\theta\ x)\\
&=
(\cosh\theta^2-\sinh^2\theta)x +( -\cosh\theta \sinh\theta + \sinh\theta\cosh\theta) t\\
&=
(1 )x +( 0 ) t\\
&=x
\end{align*}
$$