maajdl said:
Equation [1] was obtained by applying a Lorentz transformation from a 'stationary frame' {x',t') to a 'moving frame' {x,t}. If you go back to the 'stationary frame', you will easily see the equivalence of [1] and [2] . The equivalence of [1] and [2] in the moving frame is then obvious.
note1: there is a small typo in [1]
note2: remember the Lorentz transformation: x' = γ (x + β c t) and c t' = γ (c t + β x)
Many thanks for the helpful reply in #2. Your suggestion about collapsing [1] back to the stationary frame allowed me to spot a mistake I was making with my exponential algebra, it has been awhile! The following equations are primarily for confirmation and for future cross reference. However, I would like to know exactly where you think the ‘s
mall typo’ in [1] lies as it is correct to the source. As suggested, switching back to the stationary frame, i.e. v=0, allows the following substitutions:
[0] \beta=\frac{v}{c}=0; \left(1-\beta \right)=1; \gamma=\frac{1}{\sqrt{1-\beta^2}}=1; \gamma \beta=0
This allows [1] to be reduced to:
[1] \phi = Ae^{i \left( ct+r \right)k/\gamma \left(1+\beta \right)} - Ae^{i \left( ct-r \right)k \gamma \left(1-\beta \right)}
[1a] \phi = Ae^{i \left( ct+r \right)k} - Ae^{i \left( ct-r \right)k}
[1b] \phi = Ae^{i \left( \omega t+kr \right)} - Ae^{i \left( \omega t-kr \right)}
The second version is just to highlight the normal format of a standard wave equation within the exponential, which I believe can be transposed back into a trigonometric form using the following identity:
[TI-1] e^{i \theta} = cos \theta + i sin \theta
However, using only the
‘real’ part allows [1b] to be written as:
[1c] \phi = A cos \left( \omega t+kr \right) - A cos \left( \omega t-kr \right)
I don’t know where you think the typo was in [1], as the solution in [1c] looks OK to me, except I don’t understand why the two wave amplitudes are subtracted as this is a description of superposition wave. While I am still investigating this issue, I suspect it is because the following identity [TI-2] is used later in the derivation - see [2]/[3] below, although the substitution looks incorrect to me:
[TI-2] sin(\theta)= \frac{e^{i\theta} – e^{-i\theta}}{2i} \Longrightarrow e^{i\theta} – e^{-i\theta}=(2i)sin(\theta)
[TI-3] cos(\theta)= \frac{e^{i\theta} + e^{-i\theta}}{2} \Longrightarrow e^{i\theta} + e^{-i\theta}=(2)cos(\theta)
Anyway, converting [2] to a stationary form using the substitutions in [0] above gives:
[2a] \phi = Ae^{ik*1 \left( ct + 0*r \right)} \left( e^{ik*1 \left( 0+r \right)} - e^{-ik*1 \left( 0+r \right)} \right) \Longrightarrow Ae^{i \left( \omega t \right)} \left( e^{i \left(kr \right)} - e^{-i \left(kr \right)} \right)
[2b] \phi = Ae^{i \left( \omega t \right)}e^{i \left(kr \right)} -Ae^{i \left( \omega t \right)}<br />
e^{-i \left(kr \right)} \Longrightarrow Ae^{\left( i \omega t +ikr\right)} - Ae^{\left( i \omega t - ikr\right)}
[2c] \phi = Ae^{i \left( \omega t + kr\right)} - Ae^{i \left( \omega t - kr\right)}
So, as you indicated, the equivalence of [1b] and [2c] appears to be confirmed. However, I still don’t really understand why the derivation proceeds from [2] as follows, which appears to be dependent on [TR-2], but then seems to lose the complex value
in [3]:
[2] \phi = Ae^{ik \gamma \left( ct + \beta r \right)} \left( e^{ik \gamma \left( \beta ct+r \right)} - e^{-ik \gamma \left( \beta ct+r \right)} \right)
[3] \phi = 2Ae^{ik \gamma \left( ct + \beta r \right)} sin \left[ k \gamma \left( \beta ct + r \right) \right]
However, if [2] was changed to reflect a superposition of 2 additive waves it would presumably become:
[2’] \phi = Ae^{ik \gamma \left( ct + \beta r \right)} \left( e^{ik \gamma \left( \beta ct+r \right)} + e^{-ik \gamma \left( \beta ct+r \right)} \right)
If so, the [TR-3] could then be used such that [3] would become:
[3’] \phi = 2Ae^{ik \gamma \left( ct + \beta r \right)} cos \left[ k \gamma \left( \beta ct + r \right) \right]
Finally, using [TR-1], I would have thought that the remaining complex exponential could also be replaced by taking only the ‘rea partl’, such that [3’] would become:
[4] \phi = 2A cos \left[ k \gamma \left( ct + \beta r \right) \right] cos \left[ k \gamma \left( \beta ct +r \right) \right] \Longrightarrow 2A cos \left[ \gamma \left( \omega t + \beta k r \right) \right] cos \left[ \gamma \left( \beta \omega t + kr \right) \right]
Anyway, as stated, much of this is purely for future reference, but would appreciate the correction of the 'maths' on my part. I still need to take a closer look at the Lorentz transformation in this wave process and while I already have some issues to resolve within this derivation, it is presumably best if I raise these in another PF forum, e.g. general physics? Again, I really appreciate the help. Thanks
