Finding Lorentz acceleration transformation for arbitrary direction

  • Thread starter Cyneron
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  • #1
Cyneron
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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.
 
Last edited:

Answers and Replies

  • #2
TSny
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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 was able to reduce the above equation to the expression that you want to get, namely

$$\vec a ' = \frac{\vec a}{\gamma^2 \left(1 - \frac{\vec v \cdot \vec u}{c^2}\right)^2} - \frac{(\gamma - 1) (\vec a \cdot \vec u) \vec u}{\gamma^3 u^2 \left(1 - \frac{\vec v \cdot \vec u}{c^2} \right)^3} + \frac{(\vec a \cdot \vec u) \vec v}{\gamma^2 c^2 \left(1 - \frac{\vec v \cdot \vec u}{c^2}\right)^3}.$$

So, check over your work starting from your expression for ##a'## that I quoted above.

You can verify that your final answer is incorrect by considering the case where ##\vec v## and ##\vec u## are perpendicular, so that ##\vec v \cdot \vec u = 0##.
 
  • #3
Cyneron
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Thanks for your input. I'm still picking away at this, but thus far none of my approaches have gotten me anything that obviously takes me to my goal. I know I'm overlooking something. I'll keep trying.

Edit: Okay, I think I'm on the right track. I'm making some substitutions of expressions you can make from the Lorentz factor and it looks like I'm going to be able to cancel the ##\vec v \cdot \vec u## terms.

Final edit: Yes I got it, took some expanding, rearranging and then using ##\gamma c^2 - \gamma u^2 = \frac{c^2}{\gamma}## at the end. Thanks for letting me know I was right up to that point previously.
 
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