Special Relativity- Acceleration without Linear Algebra

[Battlemage!]

Homework Statement

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?

Homework Equations

Lorentz transformation in one spatial 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/c² dζ )​

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

§ 2.1

dx/dt = (dζ/dτ + v )/(1 + v/c² dζ/dτ)​

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

x = γ (ζ + vτ)

and

t = γ (τ + v/c² ζ )

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 d²x/dt², 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)]

d²x = dγ (dζ + vdτ) + γ (d²ζ + dvdτ + vd²τ) * by product rule

Right here is where I start to get problems. I have no idea what I am supposed to do with vd²τ, because I have never seen an acceleration with a " d²t" 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.

dt² = [γ (dτ + v/c² dζ] [γ (dτ + v/c² dζ] ==>

dt² = γ²dτ² + 2γv/c²dτdζ + γ²v²/c⁴dζ²


So, when I divide d²x by dt², I get the following, and have NO idea where to go from there (assuming I haven't messed up at the very beginning)





d²x/dt² = { dγ (dζ + vdτ) + γ (d²ζ + dvdτ + vd²τ)}/ { γ²dτ² + 2γv/c²dτdζ + γ²v²/c⁴dζ² }



d²x/dt² = { γ³vdv/c² (dζ + vdτ) + γ (d²ζ + dvdτ + vd²τ)}/ { γ²dτ² + 2γv/c²dτdζ + γ²v²/c⁴dζ² }



** I am assuming that dγ = d [ (1- v²/c²)^-1/2]

= (-1/2)(1- v²/c²)^-3/2 (-2vdv/c²)

= vdv/c² * (1- v²/c²)^-3/2

= γ³vdv/c²




Thanks to anyone who tries to help!

 

[Battlemage!]

Also, was this the correct place for this question? I can imagine where it might be at the beginning of upper classmen physics.

 

[Doc Al]

You're on the right track. I'm getting a headache from trying to follow your notation, so I'll use my own.

Start with the LT in differential form:

dx' = \gamma(dx - v dt)

dt' = \gamma(dt - v dx/c^2)

To get the velocity transformation (at least in one dimension), just calculate u' = dx'/dt'.

Once you get the velocity transformation, just repeat the process. Write the transformation in differential form (du' = etc.) and calculate a' = du'/dt'.

Here's a trick that might prove helpful:

dt' = \gamma(dt - v dx/c^2) = \gamma(dt - v u dt/c^2) = \gamma dt (1 - uv/c^2)

Good luck!

 

[Battlemage!]

Okay, I used your hint, but I did some other things too. Now, I made three assumptions, one of which is an assumption about mathematical technique (not sure if what I did was a "legal move"), and two were assumptions about the coordinate systems. I have reason to believe that what I did was correct, because I was able to derive Einstein's kinetic energy equation. However, I once derived the Lorentz factor incorrectly, even though I got the right answer, so I wouldn't be surprised if I just got lucky. I tried not to skip any steps, which makes it a lot longer, but I'm hoping that will help you see where I made mistakes. Anyway, please tell me if this is getting closer, or correct. Thanks!

Assumption 1

The first assumption is this: Because the Lorentz transformation is given by x' = γ (x - vt), and because the differential form of that transformation is dx' = γ (dx - vdt), I reason that you treat the v in gamma and the v in the transformation as constants. If you DIDN'T treat them as constants, then you'd have to use the product rule and the chain rule to take the differential of x' = γ (x - vt), and you'd end up with something more complex. I believe I have made the correct assumption here.
From there, to derive an acceleration relationship I started with the velocity transformation-u' = (u - v)/( 1 - uv/c² )

and followed the steps you outlinedAssumption 2(mathematical technique assumption)

Then I tried to find du'. This is where I made a mathematical assumption. I assumed that when taking the differential, I only had to "distribute" the "d" to the numerator of the fraction. However, because I am treating all v's as constant, dv = 0, and I get the following:

du' = du/ (1- uv/c² )
I then divided that by dt', which gave:du'/dt' = du*1/ (1- uv/c² ) * 1/γdt (1- uv/c² )
Assumption 3

There exists a situation in which u = v (for example, say that the particle is at rest in one of the two moving frames of reference). Therefore, (1- uv/c² ) = (1-v²/c²) = γ^-2. So, if I substitute that in, I get the following:du'/dt' = du/dt/γγ^-2γ^-2 = γ³du/dtI believe that the above is the acceleration relationship, at least for some special cases.
Now, the reason I think this is right: From this I derive E = mc² (γ - 1)
Work, W, is given by the integral of force over some displacement, right? So,

∫mdu'/dt' dx' = ∫mγ³du/dt dx

from 0 to v.
Dealing with the du/dt integral: For slow speeds, m is constant, so

W = m∫γ³du/dt dx

next was:

W = m∫γ³ d²x/dt² * dxW = m∫γ³ d²x/dt * dx/dt

W = m∫γ³ dv * v

W = m∫γ³ vdvPlugging in gamma gives:

W = m∫(1 - v²/c²)^-3/2 vdvusing substitution, u = 1 - v²/c² ==> -1/2c²du = vdv

which means that:

W = -1/2mc²∫u^-3/2 du
W = -1/2mc² (-2u^-1/2)

W = mc² [√(1 - v²/c²) ]

from 0 to v, which gives:
W = mc² [√(1 - v²/c²) - √(1 - 0²/c²)]

W = mc² [√(1 - v²/c²) - (1)]
Which, FINALLY, is Einstein's kinetic energy equation,

W = E = mc² (γ - 1)So, the questions are 1.) Did I derive the correct acceleration relationship?

du'/dt' =γ³du/dtand 2.) Did I derive the kinetic energy relationship correctly?

Or did I just get lucky?

Anyway, thanks a TON if you are patient enough to get through all this! I really do appreciate it.