Spacetime invariance algebraic proof

[john t]

TL;DR

Phillip Harris of University of Sussex offers an algebraic proof of invariance of spacetime interval. Need help with steps he omits.

In Phillip Harris' (U. Sussex) post on special relativity he includes on p. 45 an algebraic proof of invariance of spacetime intervals. He starts with the definition S^2 =c^t^2 - x^2 -y^2 -z^2, he inserts the Lorentz transform expressions fot t and x, and he does some algebra to show that one gets the above expression back again. He skips a bunch of steps and I am going nuts trying to get his result. I attach a screen shot of the equations. I am ok with the first of them. Note, the y and z terms were cut off in his post, but they are pretty simple. Would someone kindly show me the algebraic steps.

John Thompson [email address redacted by the Mentors}

 

[kuruman]

Summary:: Phillip Harris of University of Sussex offers an algebraic proof of invariance of spacetime interval. Need help with steps he omits.

In Phillip Harris' (U. Sussex) post on special relativity he includes on p. 45 an algebraic proof of invariance of spacetime intervals. He starts with the definition S^2 =c^t^2 - x^2 -y^2 -z^2, he inserts the Lorentz transform expressions fot t and x, and he does some algebra to show that one gets the above expression back again. He skips a bunch of steps and I am going nuts trying to get his result. I attach a screen shot of the equations. I am ok with the first of them. Note, the y and z terms were cut off in his post, but they are pretty simple. Would someone kindly show me the algebraic steps.

John Thompson

If you cannot figure it out when you have the entire derivation, why do you think we can explain it to you with only pieces of it? Based on the fragment you posted, the only thing I can think of is that you might have missed that $$\gamma^2 \left(1-\frac{v^2}{c^2}\right)=1.$$

 

[john t]

If you cannot figure it out when you have the entire derivation, why do you think we can explain it to you with only pieces of it? Based on the fragment you posted, the only thing I can think of is that you might have missed that $$\gamma^2 \left(1-\frac{v^2}{c^2}\right)=1.$$

Thanks. The identity you wrote is an obvious one, but it does not help me understand the path to the last equation. If the full article would help you do so here is the url - https://web.stanford.edu/~oas/SI/SRGR/notes/srHarris.pdf

I am just a 76 year old legally blind guy trying to fill in some blanks in my understanding, and I was requesting assistance only with the algebra for the steps the author left out.

 

[Ibix]

The first line to the second is just expanding the brackets from the first line and then collecting terms with $x'^2$ and $t'^2$. The second to the third line just cancels as @kuruman says, and also cancels the two $x't'$ terms. That's it.

 

[kuruman]

Thanks. The identity you wrote is an obvious one, but it does not help me understand the path to the last equation. If the full article would help you do so here is the url - https://web.stanford.edu/~oas/SI/SRGR/notes/srHarris.pdf

I am just a 76 year old legally blind guy trying to fill in some blanks in my understanding, and I was requesting assistance only with the algebra for the steps the author left out.

I understand. For future reference: If you see terms or symbols completely disappear in an equation, that means one one of two things: (a) in additions something was added to its negative giving zero; (b) in multiplications something was multiplied by its inverse giving a factor of 1 which is omitted in the product. Here you have both. Watch the step by step development
$$\gamma^2\left\{c^2{t'}^2 \left(1-\frac{v^2}{c^2}\right) +2vt'x'-2vt'x'-{x'}^2\left(1-\frac{v^2}{c^2}\right) \right\}$$First eliminate the two terms adding to give zero.
$$\gamma^2\left\{c^2{t'}^2 \left(1-\frac{v^2}{c^2}\right) +0-{x'}^2\left(1-\frac{v^2}{c^2}\right) \right\}$$Next remove the "0" that adds nothing and distribute $\gamma^2$$$c^2{t'}^2 \left(1-\frac{v^2}{c^2}\right)\gamma^2 -{x'}^2\left(1-\frac{v^2}{c^2}\right)\gamma^2$$Next replace the product of $\gamma^2$ times its inverse with "1"$$c^2{t'}^2 \times 1- {x'}^2\times 1$$Finally remove the factor of 1 that doesn't change anything as a factor.$$c^2{t'}^2- {x'}^2.$$ As @Ibix said, "That's it."

I included all the gory details to show you the rationale behind these two algebraic transformations. Please be mindful that quite often you might encounter them going backwards, i.e. add and subtract a number or multiply and divide by the same number. These are useful when grouping terms into recognizable entities.

 

[john t]

Thanks, Kurman, Ibix et. al. Once I gathered the t^2 and x^2 terms I got to the result. Instead of 3 steps it took me about 9 - guess that's why I am a chemist and not a physicist.

 

[Ibix]

Speed, and few steps, comes with practice.

 

[PeroK]

Thanks, Kurman, Ibix et. al. Once I gathered the t^2 and x^2 terms I got to the result. Instead of 3 steps it took me about 9 - guess that's why I am a chemist and not a physicist.

The author may not necessarily have done it in three steps, but you don't want the text to become bogged down with every line of algebra written out explicitly. Especially as you get to more advanced texts, there may be quite a bit of intermediate algebra to do if you want to fully check the result for yourself.

 

[vanhees71]

Maybe it helps to rewrite the Lorentz transformation in terms of the rapidity, $\eta$, defined by
$$\beta=\tanh \eta \; \Rightarrow\; \gamma=\cosh \eta, \quad \gamma \beta=\sinh \eta.$$
Then the Lorentz boost reads, in matrix vector notation
$$\underline{x}'=\hat{\Lambda} \underline{x}=\begin{pmatrix} c t' \\ x' \end{pmatrix}=\begin{pmatrix} \cosh \eta & -\sinh \eta \\ -\sinh \eta & \cosh \eta \end{pmatrix} \begin{pmatrix} c t \\ x \end{pmatrix}.$$
Then you have
$$ x^{\prime \mu} x^{\prime \nu} \eta_{\mu \nu} = \underline{x}'^{\text{T}} \hat{\eta} \underline{x}' = \underline{x}^{T} \hat{\Lambda}^T \hat{\eta} \hat{\Lambda}\underline{x},$$
and now
$$\hat{\Lambda}^{T} \hat{\eta} \hat{\Lambda}=\begin{pmatrix} \cosh \eta & -\sinh \eta \\ -\sinh \eta & \cosh \eta \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \cosh \eta & -\sinh \eta \\ -\sinh \eta & \cosh \eta \end{pmatrix} = \begin{pmatrix} \cosh^2 \eta -\sinh^2 \eta & -\cosh \eta \sinh \eta + \sinh \eta \cosh \eta \\ -\cosh \eta \sinh \eta + \sinh \eta \cosh \eta & \cosh^2 \eta-\sinh^2 \eta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 &1 \end{pmatrix}=\hat{1},$$
and thus
$$\underline{x}' \cdot \underline{x}'=\underline{x} \cdot \underline{x}.$$
The same calculation also holds if you take two arbitrary space-time vectors,
$$\underline{x}' \cdot \underline{y}'=\underline{x} \cdot \underline{y},$$
i.e., the Lorentz transformations leave the Minkowski product invariant.