- #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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