(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I have been trying for a month to derive a relationship between the acceleration of a particle in a coordinate system at rest and one moving at some velocity. I have not taken Linear Algebra yet, so I can't do anything with matrices (as I often see relativity presented). I am trying to derive an acceleration without using four-vectors or matrices or any Linear Algebra. The problem is that I keep getting equations that seem unreasonably complicated. a.) Is it possible to derive an acceleration in special relativity using only underclassman math, and b.) How do I go about doing it?

2. Relevant equations

Lorentz transformation in one spacial and one time dimension, where τ is the time coordinate and ζ is the position coordinate for the moving system :

§ 1.1

dx = γ (dζ + vdτ)

§ 1.2

dt = γ (dτ + v/c^{2}dζ )

Lorentz velocity transformation then equals dx/dt, ==>

§ 2.1

dx/dt = (dζ/dτ + v )/(1 + v/c^{2}dζ/dτ)

*I actually started with these, and then took the differentials:

x = γ (ζ + vτ)

and

t = γ (τ + v/c^{2}ζ )

3. The attempt at a solution

In getting the velocity transformation, I just took the differentials of the x to ζ and t to τ transformation, and then divided them. This doesn't seem to work for an acceleration though (and I think it CAN'T work for it), but that is what I've tried. An acceleration can be written as d^{2}x/dt^{2}, so I am assuming that can be read as "the differential of the differential of x DIVIDED BY the differential of time squared." If that is the case, I have the right idea but the wrong mathematical technique. I am getting something that appears to be unreasonably complex. Here is what I've been doing:

Differentials of § 1.1:

d(dx) = d [γ (dζ + vdτ)]

d^{2}x = dγ (dζ + vdτ) + γ (d^{2}ζ + dvdτ + vd^{2}τ) *by product rule

Right here is where I start to get problems. I have no idea what I am supposed to do with vd^{2}τ, because I have never seen an acceleration with a " d^{2}t" in it.

The next problem I have is that when I divide the differential of the displacement equation with the square of the time equation I cannot simplify it in any meaningful way (at least to my eyes). I will carry out that operation so maybe someone can help.

dt^{2}= [γ (dτ + v/c^{2}dζ] [γ (dτ + v/c^{2}dζ] ==>

dt^{2}= γ^{2}dτ^{2}+ 2γv/c^{2}dτdζ + γ^{2}v^{2}/c^{4}dζ^{2}

So, when I divide d^{2}x by dt^{2}, I get the following, and have NO idea where to go from there (assuming I haven't messed up at the very beginning)

d^{2}x/dt^{2}= { dγ (dζ + vdτ) + γ (d^{2}ζ + dvdτ + vd^{2}τ)}/ { γ^{2}dτ^{2}+ 2γv/c^{2}dτdζ + γ^{2}v^{2}/c^{4}dζ^{2}}

d^{2}x/dt^{2}= { γ^{3}vdv/c^{2}(dζ + vdτ) + γ (d^{2}ζ + dvdτ + vd^{2}τ)}/ { γ^{2}dτ^{2}+ 2γv/c^{2}dτdζ + γ^{2}v^{2}/c^{4}dζ^{2}}

** I am assuming that dγ = d [ (1- v^{2}/c^{2})^{-1/2}]

= (-1/2)(1- v^{2}/c^{2})^{-3/2}(-2vdv/c^{2})

= vdv/c^{2}* (1- v^{2}/c^{2})^{-3/2}

= γ^{3}vdv/c^{2}

Thanks to anyone who tries to help!

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# Homework Help: Special Relativity- Acceleration without Linear Algebra

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