# Finding Lorentz acceleration transformation for arbitrary direction

• Cyneron
In summary, the conversation discusses using the chain rule to find the expression for ##\vec a'## and the process of simplifying it. After a lot of algebraic manipulation and substitution, the correct expression is found to be $$\vec a' = \frac{\vec a}{\gamma^2 \left(1 - \frac{\vec v \cdot \vec u}{c^2}\right)^2} - \frac{(\gamma - 1) (\vec a \cdot \vec u) \vec u}{\gamma^3 u^2 \left(1 - \frac{\vec v \cdot \vec u}{c^2} \right)^3} + \frac{(\vec a \ Cyneron Homework Statement This isn't exactly a homework help problem, but it is something I am trying to figure out. I want to derive the Lorentz transformation for observed acceleration of some body between two frames in uniform motion in an arbitrary direction by differentiating the expression for the velocity transformation. I would like to show from the velocity and time transformations given below that the acceleration can be written as$$\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}.$$I have two of the terms from the correct expression. I keep thinking that there is some way to simplify the other two into the one I am missing, but after hours I have not been able to piece it together. Last edited: Cyneron said: I get$$\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##.

Cyneron
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.

Last edited:
rudransh verma and TSny

## 1. What is the Lorentz acceleration transformation?

The Lorentz acceleration transformation is a mathematical formula that describes how the acceleration of an object changes when observed from different reference frames in special relativity. It takes into account the effects of time dilation and length contraction.

## 2. Why is it important to find the Lorentz acceleration transformation for arbitrary direction?

It is important to find the Lorentz acceleration transformation for arbitrary direction because it allows us to accurately calculate the acceleration of an object when observed from any reference frame, regardless of its direction of motion. This is necessary for making precise measurements and predictions in special relativity.

## 3. How is the Lorentz acceleration transformation derived?

The Lorentz acceleration transformation is derived using the Lorentz transformation equations, which describe how time and space measurements change between two reference frames moving relative to each other at a constant velocity. By applying these equations to the acceleration of an object, we can derive the Lorentz acceleration transformation.

## 4. Can the Lorentz acceleration transformation be applied to any type of motion?

Yes, the Lorentz acceleration transformation can be applied to any type of motion, including linear, circular, and rotational motion. It is a general formula that takes into account the effects of relative motion and is applicable to all types of acceleration.

## 5. How does the Lorentz acceleration transformation affect our understanding of motion and gravity?

The Lorentz acceleration transformation is a fundamental concept in special relativity, which has greatly expanded our understanding of motion and gravity. It allows us to accurately describe and predict the behavior of objects moving at high speeds, and has led to the development of theories such as the theory of general relativity, which explains the effects of gravity on a large scale.

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