# Why is the value of thrust for a perfectly spherical event equal to ${\frac{1}{2}}$?

In a particle collision the thrust is defined by:

$$T = max_{\hat{n}} \frac {\sum_i \left| \hat{p}_i . \hat{n}_i \right|}{\sum_i \left| \hat{p}_i \right|}$$

where $\hat{n}$ is the unit vector that maximizes the ratio of the sums ($\hat{n}$ is called the thrust axis) and the sum is over all the particles emerging from the collision.

In some notes I have, it is stated simply that the value of thrust for a perfectly spherical event is ${\frac{1}{2}}$. Why is this?

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Say what? Please provide more context so we can make sense of the question. What sort of event are you talking about? Thrust of what? Note that the standard meaning of "thrust" refers to a force, which has units. So a thrust can't be a pure number like $1/2$. – Michael Brown Sep 9 '13 at 14:19
@LoganM I see. That makes sense of it. – Michael Brown Sep 9 '13 at 16:13

It seems to me that the convention you're using for thrust is the one used in this lecture, namely that for an event with momenta $\mathbf p_i$, the thrust is $T= \max_{|\mathbf n| = 1} \frac{\sum_i |\mathbf p_i \cdot \mathbf n|}{\sum_i |\mathbf p_i|}$. In the limit of a spherical event, the momenta are uniformly distributed spherically, so the sums become integrals: $T= \max_{|\mathbf n| = 1} \frac{\int_{S^2} |\mathbf p \cdot \mathbf n| d \mathbf p}{\int_{S^2} |\mathbf p|}$. Here I've only considered an event where all the momenta have the same magnitude, but if we have some distribution of momenta with spherical symmetry then the magnitude of the momentum factors out of the numerator and denominator and thus is irrelevant.
The integral in the numerator is independent of $\mathbf n$ by rotational invariance. Thus we can pick $\mathbf n$ to point along the $z$-axis, giving (for a perfectly spherical event) $$T = \frac{\int_0^\pi \sin \theta |\cos \theta| d \theta \int_0^{2 \pi} d\phi}{\int_0^\pi \sin \theta d \theta \int_0^{2 \pi} d\phi} = \frac{\int_0^{\pi/2} \sin \theta \cos \theta d\theta - \int_{\pi/2}^\pi \sin \theta \cos \theta d\theta}{\int_0^\pi \sin \theta d \theta} = \frac{1/2 + 1/2}{2}=\frac12$$