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The four-velocity of a particle is $\vec{U}=\mathrm d\vec{x}/\mathrm d\tau$ and the four-acceleration is $d\vec{U}/d\tau$, where $(\mathrm d\tau)^2=-\,\mathrm d\vec{x} \cdot \mathrm d\vec{x}.$

I'll use the four-acceleration to see if I understand what this notation means. In some observer frame the four velocity can be described with the parametrization by time $$\vec{U}(t)=(U^0(t),U^1(t),U^2(t),U^3(t)),$$ where $U^0(t)=t$ is the time coordinate. But we can express time along the path of the particle in terms of the proper time $t=f(\tau)$. Then $d\vec{U}/d\tau$ is the value $d\vec{U}(f(\tau))/d\tau$.

Is this a correct understanding?

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Let us first fix the notation. Typically the arrow on top of a quantity is reserved for the spatial components. So that a vector which is not bold or with an arrow is assumed to be a 4-vector in the context of special relativity. Normally they come together with an index such as $U^\mu$.

Now I suggest you imagine a world-line for a particle as a time-like curve in Minkowski space-time, that is a curve on $\mathbb{R}^4$ equipped with a metric $$ds^2=-dt^2+dx^i dx^i$$ notice that $t$ is just a coordinate of space-time but with opposite sign and no reference to $\tau$ has been made yet. This curve may be parametrized by some variable $\lambda$, so that for now we have some curve $x(\lambda):\mathbb{R}\rightarrow \mathbb{R}^4$, as said I will refer to the components by using indices, so that its tangent vector can be written as $$U^\mu(\lambda) = \frac{dx^\mu}{d\lambda}:\mathbb{R}\rightarrow \mathbb{R}^4$$ The time-like condition just says that the norm of the tangent vector is negative at every point (in this metric signature), this corresponds to particles that travel slower than the speed of light as it should be.

The last step to get a "proper" description is to express the curve in terms of its arc-length, physically this is what we call proper-time, $\tau$. So in units where $c=1$ we are saying that $$U^\mu(\tau)U_\mu(\tau) = -1$$

Now the setting should be completely clear. If you want to compute 4-acceleration you need to take another proper-time derivative $$a^\mu = \frac{dU^\mu}{d\tau}$$ and that is all. This concepts might seem overly stated for the case of special relativity, but they become very relevant for general relativity, where the metric is more complicated.

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