# I Acceleration of a relativistic particle

1. Feb 13, 2019 at 12:15 PM

### kent davidge

In an inertial frame, consider that a particle's position and the time measured by a clock in this frame are respectively, $(t,x)$.

Suppose there's another frame, moving with constant speed $v_R$ with respect to the frame described above.

The particle acceleration is given in the first frame by $d^2 x / dt^2$. In the second frame I would expect it would be given by $d^2 x' / dt'^2$. I know the relation between the primed and unprimed coordinates: $$t' = \gamma (t - v_R x) \\ x' = \gamma (x - v_R t)$$ But in expressing $d^2 x' / dt'^2$ by chain rule, etc... in terms of $(t,x)$ I got an ugly expression with several terms. But when I evaluate the vector quantity $(d^2 t' / dt^2, d^2 x' / dt^2)$ I get a beautiful expression for $a'$, namely $a' = \gamma a$, aside from the fact that that vector is Lorentz invariant.

What bothers me, however, is that in the last expression I used $t$, instead of $t'$, and it seems senseless to use the time $t$ measured from the first frame to get a quantity measured in the second frame.

If the particle was not relativistically moving, then it would be easy: $a' = d^2 x' / dt'^2 = d^2 x' / dt^2 = a$. What can I do?

2. Feb 13, 2019 at 12:17 PM

### Orodruin

Staff Emeritus
If you do not write down the expression, we have no way of determining whether it is correct or not.

It is not. The 4-acceleration is the second derivative of the spacetime vector with respect to proper time, not any coordinate time.

3. Feb 13, 2019 at 12:23 PM

### kent davidge

It is not what? The Lorentz invariance I mentioned can be seen here $(d^2 t' / dt^2, d^2 x' / dt^2) = (- \gamma v_R a, \gamma a)$. Its inner product with itself gives $a^2$. On the other frame, it's $(d^2 t / dt^2, d^2 x / dt^2) = (0, a)$ and its inner product with itself is $a^2$.
But the most natural time available on the frame is the time measured by a clock moving with the frame. What justifies me to take the derivatives with respect to a time measured by a clock on the particle's frame?

4. Feb 13, 2019 at 12:34 PM

### Orodruin

Staff Emeritus
It is not Lorentz invariant because $dt$ is not Lorentz invariant. That is why the 4-acceleration (and 4-velocity) are defined using the proper time, not the time of any particular frame. Your $a$ is frame dependent.

The most natural time available is the proper time of the particle, not an arbitrary coordinate time.

5. Feb 13, 2019 at 12:47 PM

### pervect

Staff Emeritus
An old possibly useful related thread

"Magnitude of proper acceleration in terms of three vectors"

The main thing I'd take away from this thread is that it is rather messy to get proper acceleration from the coordinate accelerations in an inertial frame. In contrast, it's much simpler to use a 4-vector approach, from the 4-velocity u (the derivative of 4-position with respect to proper time), and the 4-acceleration a, the derivative of the 4-velocity with respect to proper time.

Note that much of the thread has an analysis by me that is probably wrong :(, though I never did track down exactly where I went wrong.

For the 1 space + 1 time case, rapidity methods can be quite useful, fundamentally due to the fact that rapidities add linearly (unlike velocities). It's much less useful in the 3 space + 1 time case, though.

6. Feb 13, 2019 at 2:10 PM

### PeroK

The correct expressions for the three-acceleration are given here:

https://en.wikipedia.org/wiki/Acceleration_(special_relativity)

7. Feb 13, 2019 at 2:59 PM

### robphy

In analogy to the curvature of a plane curve,
the proper acceleration is the derivative of the rapidity (angle of the tangent) with respect to proper time (arc length).
$a=\frac{d\theta}{ds}=\frac{\ddot x}{\sqrt{1-\dot{x}^2}^3}=\gamma^3 \ddot{x}$

8. Feb 13, 2019 at 3:35 PM