# Does a constant 4-acceleration implies constant acceleration?

Before asking the question I explain here my line of thoughts:

I'm almost sure that a constant acceleration $\bf{a}$ implies constant 4-acceleration $w^\mu$. So I assume constant acceleration, then, in the rest frame of the particle where its velocity is zero we have $w^\mu=(0,\bf{a})$, thus, if $\bf a$ is constant, the quantity $w^\mu w_\mu$ will be constant aswell and Lorentz-invariant. If there's a mistake you're free to point it out.

Anyway, assuming what I've said true I ask the opposite question: does constant four acceleration implies constant acceleration?

I didn't come up with a satisfying yes, I tried to think in analogy of what I've written above, but I only obtained that just in the comoving reference frame i have constant acceleration, but then in another reference frame where $w^\mu$ is function both of the velocity and the acceleration of the particle, how can I know they don't modify themselves in order to keep the initial assumption of constant 4-acceleration valid?

The norm of the four-acceleration (i.e. the proper acceleration) defined by:

$$A^2 = g_{\alpha\beta} a^\alpha a^\beta$$

is a Lorentz scalar and therefore the same for all observers. However there is nothing to stop me from choosing some bizarre coordinates in which the four acceleration is time dependent but this is cancelled out by the time dependence in the metric to give a constant proper acceleration. So in this sense a constant proper acceleration does not imply a constant four acceleration or vice versa.

• Thank you for your clear explanation. So we can say "out loud" that what we mean when we talk about constant 4-acceleration is totally unrelate with what usually image when we say constant acceleration? – Run like hell Jan 16 '17 at 22:35
• @Runlikehell: constant four-accleration is a somewhat vaguely defined term. One interpretation is that if we write $\mathbf a = (a^0, a^1, a^2, a^3)$ then the values of all the components $a^\mu$ are constant. But the values of $a^\mu$ depend on the coordinates we choose and we can always find coordinates in which the components won't be constant. If we are using this sense of the word constant then yes constant four-acceleration and constant proper acceleration are unrelated. – John Rennie Jan 17 '17 at 6:17
• With constant I meant that it doesn't change in time, that the whole vector stays the same not its components, the components changes in different coordinate system, that's why I thought that if you can write $w^\mu=(0, \bf{a})$ with $a$ constant, in a sistem of coordinates, since both the whole 4-vector and the 3-vector are constant in a sistem of coordinates I thought that the whole 4-vectore would stay like that in every coordinate system. I probably didn't have a clear mind and I wrote that part about the sum $w^\mu w_\mu$ which may be useless and misleading. – Run like hell Jan 17 '17 at 10:06

No. Suppose a relativistic mass accelerates at constant proper acceleration $a$. Its rapidity after a proper time $\tau$ is $\phi=\phi_0+\frac{a\tau}{c}$. The speed is $v=c\tanh\phi$, so $\frac{dv}{d\tau}=a\text{sech}^2\phi$ falls over time, as does $\frac{dv}{dt}=\gamma^-1\frac{dv}{d\tau}=a\text{sech}^3\phi$.

If \begin{align} \mathbf{v} & =\textrm{velocity 3-vector} \tag{01a}\\ \mathbf{a} & =\textrm{acceleration 3-vector} \tag{01b}\\ \mathbf{U} &=\textrm{velocity 4-vector} \tag{01c}\\ \mathbf{A} &=\textrm{acceleration 4-vector} \tag{01d} \end{align} then \begin{align} \mathbf{A} & = \gamma_{v}\left(\gamma_{v}\,\mathbf{a}+\dfrac{\mathrm{d}\gamma_{v}}{\mathrm{d}t}\mathbf{v},\dfrac{\mathrm{d}\gamma_{v}}{\mathrm{d}t}c \right) \tag{02a}\\ \Vert\mathbf{A}\Vert^{2} & = -\gamma_{v}^{4}\left[\Vert\mathbf{a}\Vert^{2}+\left(\gamma_{v}^{2}-1\right) \left(\dfrac{\mathrm{d}v}{\mathrm{d}t}\right) ^{2} \right] \tag{02b} \end{align} You have all answers from equations (02).

Note that in the rest frame of the particle $\:(\mathbf{v}_{0}\equiv \boldsymbol{0}, \gamma_{v}=1,\mathbf{a}=\mathbf{a}_{0})\:$

\begin{align} \mathbf{A}_{0} & = \left(\mathbf{a}_{0},0 \right) \tag{03a}\\ \Vert\mathbf{A}_{0}\Vert^{2} & = -\Vert\mathbf{a}_{0}\Vert^{2} \tag{03b} \end{align}

• Thank you for your clear explanation. So we can say "out loud" that what we mean when we talk about constant 4-acceleration is totally unrelate with what usually image when we say constant acceleration? – Run like hell Jan 16 '17 at 22:35
• @Run like hell : I think so. By the way, your question would be more clear if it was clarified that the constancy concerns the 4- and 3-vectors and not their norms. – Frobenius Jan 16 '17 at 22:43
• Yes I think that the part where i wrote the norm $w^\mu w\_nu$ may be useless and misleading. I probably don't have a clear mind at the moment, it looks like that my first line of thought is right. the whole 4-vector w^\mu constant in a reference frame( due to the constancy of $a$) implies that it is constant in every reference frame, but the opposite is not true. That's how I read your answer, but I may be biased by my interpretation and misunderstood your answer. – Run like hell Jan 17 '17 at 10:12