# Components of the 4-acceleration

1. Nov 25, 2007

### Lorna

1. The problem statement, all variables and given/known data

I read in my textbook that "in the instantanuous reference frame (IRF) $$[u^{'\gamma}]=(c,\overline{0})$$ imply that the components of the 4-acceleration in the IRF are $$[a^{'\gamma}]=(0,\overline{a'})$$"

2. Relevant equations

3. The attempt at a solution
I don't know why they got a' for the space-like components of the 4-acceleration. Isn't the accelaration = du/d$$\tau$$ if so then using $$[u^{'\gamma}=(c,\overline{0})$$ we get zero for both the time-like and space-like components of the 4-acceleration.

Last edited: Nov 25, 2007
2. Nov 26, 2007

### George Jones

Staff Emeritus
Think back to first-year calculus, and suppose $f\left(x\right) = x^3$. Then, $f\left(2\right) = 8$, and $8$ is a constant, so

$$\frac{d}{dx}8 = 0,$$

but

$$\frac{df}{dx}\left(2) = 12 \ne 0.$$

Just because the 4-velocity has a particular value (which could have zero spatial part) in a particular frame at a particular instant doesn't mean that the 4-velocity is constant. If the 4-velocity is not constant, then the 4-acceleration doesn't have to be zero.

Throw a ball in the air. When the ball reaches its greatest height, its (spatial) velocity is zero, but its (spatial) acceleration still has magnitude $g$

3. Nov 26, 2007

### robphy

In addition, note that the 4-acceleration of a particle is orthogonal to its 4-velocity. (Why?)

Last edited: Nov 26, 2007
4. Nov 27, 2007

### Lorna

That example was very helpful. thanks a lot.

Thanks robphy too.