Help with a wave equation derivation

In summary, the conversation discusses a derivation of an equation related to the formation of a superposition wave with a Doppler effect. The derivation involves applying a Lorentz transformation and using exponential algebra to simplify the equation. There is a typo in the original equation [1], but it can be reduced to the correct form [1c] by applying substitutions. The equivalence of [1b] and [2c] is confirmed, but the reasoning behind proceeding from [2] to [3] is unclear. There is also confusion about using trigonometric identities in the derivation.
  • #1
mysearch
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Hi,
Apologises if I have submitted this issue into the wrong Math forum. However, I was wondering if anybody could help me with 2 steps in a derivation of an equation. Simply by way of background, the derivation is linked to formation of a superposition wave subject to a Doppler effect

[1] [tex]\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)} [/tex]

So based on [1], the derivation states that [2] can be obtain by a process of multiplying exponents, rearranging and factoring.

[2] [tex]\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) [/tex]

However, so far, I have only managed to create a bit of a mess as I haven’t found a way of rationalising the different expressions in the exponentials. Would really appreciate any help that might confirm that step [2] is valid from [1] and/or any pointers as to how it is done. Thanks
 
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  • #2
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)
 
  • #3
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 ‘small 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] [itex]\beta=\frac{v}{c}=0[/itex]; [itex]\left(1-\beta \right)=1[/itex]; [itex]\gamma=\frac{1}{\sqrt{1-\beta^2}}=1[/itex]; [itex]\gamma \beta=0[/itex]

This allows [1] to be reduced to:

[1] [itex]\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)} [/itex]

[1a] [itex]\phi = Ae^{i \left( ct+r \right)k} - Ae^{i \left( ct-r \right)k} [/itex]

[1b] [itex]\phi = Ae^{i \left( \omega t+kr \right)} - Ae^{i \left( \omega t-kr \right)} [/itex]

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] [itex]e^{i \theta} = cos \theta + i sin \theta [/itex]

However, using only the ‘real’ part allows [1b] to be written as:

[1c] [itex]\phi = A cos \left( \omega t+kr \right) - A cos \left( \omega t-kr \right) [/itex]

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] [itex]sin(\theta)= \frac{e^{i\theta} – e^{-i\theta}}{2i} \Longrightarrow e^{i\theta} – e^{-i\theta}=(2i)sin(\theta)[/itex]

[TI-3] [itex]cos(\theta)= \frac{e^{i\theta} + e^{-i\theta}}{2} \Longrightarrow e^{i\theta} + e^{-i\theta}=(2)cos(\theta)[/itex]

Anyway, converting [2] to a stationary form using the substitutions in [0] above gives:

[2a] [itex]\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) [/itex]

[2b] [itex]\phi = Ae^{i \left( \omega t \right)}e^{i \left(kr \right)} -Ae^{i \left( \omega t \right)}
e^{-i \left(kr \right)} \Longrightarrow Ae^{\left( i \omega t +ikr\right)} - Ae^{\left( i \omega t - ikr\right)} [/itex]

[2c] [itex]\phi = Ae^{i \left( \omega t + kr\right)} - Ae^{i \left( \omega t - kr\right)} [/itex]

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] [itex]\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) [/itex]

[3] [itex]\phi = 2Ae^{ik \gamma \left( ct + \beta r \right)} sin \left[ k \gamma \left( \beta ct + r \right) \right] [/itex]

However, if [2] was changed to reflect a superposition of 2 additive waves it would presumably become:

[2’] [itex]\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) [/itex]

If so, the [TR-3] could then be used such that [3] would become:

[3’] [itex]\phi = 2Ae^{ik \gamma \left( ct + \beta r \right)} cos \left[ k \gamma \left( \beta ct + r \right) \right] [/itex]

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] [itex]\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][/itex]

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
 

1. What is a wave equation?

A wave equation is a mathematical equation that describes the behavior of a wave. It can be used to calculate various properties of the wave, such as its amplitude, frequency, and wavelength.

2. How is the wave equation derived?

The wave equation is derived from the principles of classical mechanics, specifically the equation of motion for a particle undergoing simple harmonic motion. It can also be derived from Maxwell's equations in electromagnetism.

3. What are the variables in a wave equation?

The variables in a wave equation depend on the specific type of wave being described. However, common variables include distance or position (x), time (t), amplitude (A), frequency (f), wavelength (λ), and velocity (v).

4. What are some practical applications of the wave equation?

The wave equation has many practical applications in various fields such as physics, engineering, and mathematics. It is used to study and understand the behavior of waves in different mediums, including sound waves, light waves, and water waves. It is also used in fields such as acoustics, optics, and signal processing.

5. Are there any limitations to the wave equation?

While the wave equation is a useful tool for understanding the behavior of waves, it does have some limitations. It assumes that the wave is traveling in a uniform medium and does not account for factors such as dispersion, damping, and nonlinear effects. In certain situations, more complex equations may be needed to accurately describe wave behavior.

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