 #1
Cyneron
 3
 3
 Homework Statement:
 This isn't exactly a homework help problem, but it is something I am trying to figure out. I want to derive the Lorentz transformation for observed acceleration of some body between two frames in uniform motion in an arbitrary direction by differentiating the expression for the velocity transformation. I would like to show from the velocity and time transformations given below that the acceleration can be written as $$\vec a ' = \frac{\vec a}{\gamma^2 \left(1  \frac{\vec v \cdot \vec u}{c^2}\right)^2}  \frac{(\gamma  1) (\vec a \cdot \vec u) \vec u}{\gamma^3 u^2 \left(1  \frac{\vec v \cdot \vec u}{c^2} \right)^3} + \frac{(\vec a \cdot \vec u) \vec v}{\gamma^2 c^2 \left(1  \frac{\vec v \cdot \vec u}{c^2}\right)^3}.$$ Here ##\vec u## is the relative velocity and ##\gamma## is the Lorentz factor.
 Relevant Equations:

$$\vec r' = \vec r + \frac{(\gamma  1)}{u^2} (\vec r \cdot \vec u) \vec u + \gamma \vec u t.$$
$$\vec v' = \frac{1}{\gamma \left(1  \frac{\vec v \cdot \vec u}{c^2}\right)} \left[\vec v + \frac{(\gamma  1)}{u^2} (\vec v \cdot \vec u) \vec u  \gamma \vec u\right].$$
$$t' = \gamma \left(t  \frac{\vec r \cdot \vec u}{c^2}\right).$$
$$dt'  \gamma \left(1  \frac{\vec v \cdot \vec u}{c^2}\right) dt.$$
Edit: Ugh accidentally posted instead of previewing, this is a lot of latex to write to give my attempted solution, but I'll keep doing that. I am using the chain rule (or dividing the differential of ##\vec v'## by that of ##t'##). I get
$$d \vec v' = \frac{d \vec v \cdot \vec u}{\gamma c^2 \left(1  \frac{\vec v \cdot \vec u}{c^2}\right)^2} \left[\vec v + \frac{(\gamma  1)}{u^2} (\vec v \cdot \vec u) \vec u  \gamma \vec u\right] + \frac
{1}{\gamma \left(1  \frac{\vec v \cdot \vec u}{c^2}\right)} \left[d \vec v + \frac{(\gamma  1)}{u^2} (d \vec v \cdot \vec u) \vec u\right].$$
Then if I divide by ##dt'##, I get
$$\vec a' = \frac{\vec a \cdot u}{\gamma^2 c^2 \left(1  \frac{\vec v \cdot \vec u}{c^2}\right)^3} \left[\vec v + \frac{(\gamma  1)}{u^2} (\vec v \cdot \vec u) \vec u  \gamma \vec u\right] + \frac{1}{\gamma^2 \left(1  \frac{\vec v \cdot \vec u}{c^2}\right)^2} \left[\vec a + \frac{(\gamma  1)}{u^2} (\vec a \cdot \vec u) \vec u\right].$$
I am going to write
$$z = 1  \frac{\vec v \cdot \vec u}{c^2}$$
to make this less headache inducing. After a veritable mess of algebra, I arrive at this. I realize this may leave a lot out, but I can include them if someone can be certain that this line is not correct.
$$\vec a' = \frac{\vec a}{\gamma^2 z^2} + \frac{(\vec a \cdot \vec u) \vec v}{\gamma^2 c^2 z^3} + \frac{(\vec a \cdot \vec u)(\vec v \cdot \vec u) \vec u}{\gamma^2 u^2 c^2 z^3} + \frac{(\vec a \cdot \vec u) \vec u}{\gamma^3 u^2 z^2}.$$
I have two of the terms from the correct expression. I keep thinking that there is some way to simplify the other two into the one I am missing, but after hours I have not been able to piece it together.
$$d \vec v' = \frac{d \vec v \cdot \vec u}{\gamma c^2 \left(1  \frac{\vec v \cdot \vec u}{c^2}\right)^2} \left[\vec v + \frac{(\gamma  1)}{u^2} (\vec v \cdot \vec u) \vec u  \gamma \vec u\right] + \frac
{1}{\gamma \left(1  \frac{\vec v \cdot \vec u}{c^2}\right)} \left[d \vec v + \frac{(\gamma  1)}{u^2} (d \vec v \cdot \vec u) \vec u\right].$$
Then if I divide by ##dt'##, I get
$$\vec a' = \frac{\vec a \cdot u}{\gamma^2 c^2 \left(1  \frac{\vec v \cdot \vec u}{c^2}\right)^3} \left[\vec v + \frac{(\gamma  1)}{u^2} (\vec v \cdot \vec u) \vec u  \gamma \vec u\right] + \frac{1}{\gamma^2 \left(1  \frac{\vec v \cdot \vec u}{c^2}\right)^2} \left[\vec a + \frac{(\gamma  1)}{u^2} (\vec a \cdot \vec u) \vec u\right].$$
I am going to write
$$z = 1  \frac{\vec v \cdot \vec u}{c^2}$$
to make this less headache inducing. After a veritable mess of algebra, I arrive at this. I realize this may leave a lot out, but I can include them if someone can be certain that this line is not correct.
$$\vec a' = \frac{\vec a}{\gamma^2 z^2} + \frac{(\vec a \cdot \vec u) \vec v}{\gamma^2 c^2 z^3} + \frac{(\vec a \cdot \vec u)(\vec v \cdot \vec u) \vec u}{\gamma^2 u^2 c^2 z^3} + \frac{(\vec a \cdot \vec u) \vec u}{\gamma^3 u^2 z^2}.$$
I have two of the terms from the correct expression. I keep thinking that there is some way to simplify the other two into the one I am missing, but after hours I have not been able to piece it together.
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