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Special Relativity- Acceleration without Linear Algebra

  1. Jul 11, 2008 #1
    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/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)


    t = γ (τ + v/c2 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 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 = γ22 + 2γv/c2dτdζ + γ2v2/c42

    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)}/ { γ22 + 2γv/c2dτdζ + γ2v2/c42 }

    d2x/dt2 = { γ3vdv/c2 (dζ + vdτ) + γ (d2ζ + dvdτ + vd2τ)}/ { γ22 + 2γv/c2dτdζ + γ2v2/c42 }

    ** 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!
  2. jcsd
  3. Jul 12, 2008 #2
    Also, was this the correct place for this question? I can imagine where it might be at the beginning of upper classmen physics.
  4. Jul 13, 2008 #3

    Doc Al

    User Avatar

    Staff: Mentor

    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:

    [tex]dx' = \gamma(dx - v dt)[/tex]

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

    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:

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

    Good luck!
  5. Jul 14, 2008 #4
    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 outlined

    Assumption 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/dt

    I 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' = 3du/dt dx

    from 0 to v.

    Dealing with the du/dt integral: For slow speeds, m is constant, so

    W = mγ3du/dt dx

    next was:

    W = mγ3 d2x/dt2 * dx

    W = mγ3 d2x/dt * dx/dt

    W = mγ3 dv * v

    W = mγ3 vdv

    Plugging in gamma gives:

    W = m(1 - v2/c2)-3/2 vdv

    using substitution, u = 1 - v2/c2 ==> -1/2c2du = vdv

    which means that:

    W = -1/2mc2u-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/dt

    and 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.
    Last edited: Jul 14, 2008
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