# Two expression for relativistic acceleration

1. Jul 11, 2011

### Pi-Bond

Suppose a particle in frame S moves with acceleration $a_{x}$ and velocity $u_{x}$ at a given instance in the x-direction. I wanted to find the acceleration in a frame S' moving with velocity v in the positive x-direction with respect to frame S. To do this I used the following approach:

$a_{x}=\frac{du_{x}}{dt}$ and $a'_{x}=\frac{du'_{x}}{dt'}$

Using the chain rule,

$a'_{x} = \frac{du'_{x}}{dt'} = \frac{du'_{x}}{du_{x}} \frac{du_{x}}{dt'} = \frac{du'_{x}}{du_{x}} \frac{du_{x}}{dt} \frac{dt}{dt'} = a_{x} \frac{du'_{x}}{du_{x}} \frac{dt}{dt'}$

Using the velocity transformation,

$\large \frac{du'_{x}}{du_{x}} = \frac{1- \frac{v^{2}}{c^{2}} }{ 1 - \frac{u_{x} v} {c^{2}} }$

Similarly from the Lorentz transformations:

$\large \frac{dt}{dt'} = \frac{\sqrt{1-\frac{v^{2}}{c^{2}}} } {1 - \frac{u_{x} v} {c^{2}}}$

Thus,

$\large a'_{x} = a_{x} \frac{(1- \frac{v^{2}}{c^{2}})^{3/2}}{(1-\frac{u_{x}v}{c^{2}})^{3}}$

Now I know this formula is correct, as it listed in Resnick's and French's introductory books on Special relativity. However in W. Rindler's book on the subject, the author shows the relativistic acceleration as:

$\large a'_{x} = \gamma^{3} a_{x}$

How come there are two formulas for this quantity, one of which does not even refer to the speed of the particle? I have posted my working so that maybe someone can understand and help discriminate between these formulas.

2. Jul 11, 2011

### bcrowell

Staff Emeritus
3. Jul 11, 2011

### Pi-Bond

I am aware of four-vectors but not of the acceleration four vector (I have only done a first year basic course on relativity, and the books have not mentioned it till yet either). Both results are right though, aren't they?

4. Jul 11, 2011

### George Jones

Staff Emeritus
Is the particle (instantaneously) at rest in in the primed frame in Rindler's treatment?

5. Jul 11, 2011

### Pi-Bond

Rindler says: "Let S' be the the instantaneous rest frame of P at some time t..."

6. Jul 11, 2011

### George Jones

Staff Emeritus
What does this say about $u_x$ and $v$?

7. Jul 11, 2011

### Pi-Bond

"..u=v and u'=0 at $t_{0}$ , but u and u' vary while v is constant"

He is using different notation, with $u=u_{x}$ and so on. He formulates his proof by finding $\frac{dt}{dt'}$ and then diffrenciating the velocity transformation with respect to dt'. I don't understand his proof very well, but I have seen the gamma cubed formula at other places.

8. Jul 11, 2011

### George Jones

Staff Emeritus
Rindler's result is a special case of your more general result.

If S' moves with speed v with respect to S, and if the particle is at rest in S', then the particle moves with speed v in S, i.e., Rindler looks at the special case u_x = v. Use this in

9. Jul 11, 2011

### Pi-Bond

I see. Thanks for the help again!