Summary of what's gone on before. In the last two days, unbeknownst to me at the time, we first showed that the Lorentz/Fitzgerald contraction only happens in the direction of the velocity of a frame. Coupling this with rotational symmetry, yesterday, it was shown how the vector version of the Lorentz transform could be derived. Yesterday's treatment needs to be repeated from end to start to truly count as a derivation though, I suppose.
Picking up where I left off yesterday, the second half of the vector Lorentz transform, the expression for time in the primed frame can be written as
$t^{\prime} = \gamma\left(t - \vec{v}\cdot\vec{r}\right)$
Once again, this shows that only the component of the position vector parallel to the velocity gets to play. However, here you should notice that the velocity vector is not normalized, not divided by it's magnitude. In other words, while only the component of the position vector parallel to the velocity gets to play, the entire magnitude of the velocity gets to play. This is also noted in the notes where the expression for $\gamma$ in this case is
$\gamma = \left(1 - v^2\right)^{1/2}$
Finally, in the section 1.2 ending on page 9, it's observed that what will become known as the spacetime line element is invariant between frames
$x^2 + y^2 + z^2 - t^2 = x^{\prime 2} + y^{\prime 2} + z^{\prime 2} - t^{\prime 2}$.
Consequently, we also just introduced hyperbolic geometry.
Index Notation
The first really interesting thing here is the totally-antisymmetric tensor. I think this might be called the Levi-Civita symbol elsewhere. I need to look it up.
Here's the definition:
$\epsilon_{ijk} = -\epsilon_{ikj} = -\epsilon_{jik} = -\epsilon_{kji}$.
Then, it only remains to define the value of $\epsilon_{123} = 1$ and we get that
$\epsilon_{123} = \epsilon_{231} = \epsilon_{312} = 1$, $\epsilon_{132} = \epsilon_{321} = \epsilon_{213} = -1$
With our new symbol, we can write down the vector cross product more concisely,
$\left(\vec{A} \times \vec{B}\right) = \epsilon_{ijk}A_jB_k$
Hunh, and now I have a question.
QUESTION: Why is swizzling the indices between the following two versions of the same equation 'convenient'?
$\left[\vec{A}\times\left(\vec{B} \times \vec{C}\right)\right]_i = \epsilon_{ijk}\epsilon_{klm}A_jB_lC_m$
and the 'more convenient version'
$\left[\vec{A}\times\left(\vec{B} \times \vec{C}\right)\right]_i = \epsilon_{ijk}\epsilon_{lmk}A_jB_lC_m$
Memorize the following:
$\epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$
QUESTION: I'm guessing the second 'more convenient' version above was handy in deriving the Levi-Civita to delta function identity above?
Picking up where I left off yesterday, the second half of the vector Lorentz transform, the expression for time in the primed frame can be written as
$t^{\prime} = \gamma\left(t - \vec{v}\cdot\vec{r}\right)$
Once again, this shows that only the component of the position vector parallel to the velocity gets to play. However, here you should notice that the velocity vector is not normalized, not divided by it's magnitude. In other words, while only the component of the position vector parallel to the velocity gets to play, the entire magnitude of the velocity gets to play. This is also noted in the notes where the expression for $\gamma$ in this case is
$\gamma = \left(1 - v^2\right)^{1/2}$
Finally, in the section 1.2 ending on page 9, it's observed that what will become known as the spacetime line element is invariant between frames
$x^2 + y^2 + z^2 - t^2 = x^{\prime 2} + y^{\prime 2} + z^{\prime 2} - t^{\prime 2}$.
Consequently, we also just introduced hyperbolic geometry.
Index Notation
The first really interesting thing here is the totally-antisymmetric tensor. I think this might be called the Levi-Civita symbol elsewhere. I need to look it up.
Here's the definition:
$\epsilon_{ijk} = -\epsilon_{ikj} = -\epsilon_{jik} = -\epsilon_{kji}$.
Then, it only remains to define the value of $\epsilon_{123} = 1$ and we get that
$\epsilon_{123} = \epsilon_{231} = \epsilon_{312} = 1$, $\epsilon_{132} = \epsilon_{321} = \epsilon_{213} = -1$
With our new symbol, we can write down the vector cross product more concisely,
$\left(\vec{A} \times \vec{B}\right) = \epsilon_{ijk}A_jB_k$
Hunh, and now I have a question.
QUESTION: Why is swizzling the indices between the following two versions of the same equation 'convenient'?
$\left[\vec{A}\times\left(\vec{B} \times \vec{C}\right)\right]_i = \epsilon_{ijk}\epsilon_{klm}A_jB_lC_m$
and the 'more convenient version'
$\left[\vec{A}\times\left(\vec{B} \times \vec{C}\right)\right]_i = \epsilon_{ijk}\epsilon_{lmk}A_jB_lC_m$
Memorize the following:
$\epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$
QUESTION: I'm guessing the second 'more convenient' version above was handy in deriving the Levi-Civita to delta function identity above?
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