 Solution

ID: past-exam-of-the-symmetry-and-relativity-course-of-the-university-of-oxford/2016/1/a/solution

d τ = d t 1 - v^{2} / c^{2} = d t / γ

U = d X / d τ

and

A = d U / d τ

. Differentiating

U = γ (c, v)

gives

A^{μ} = (γ^{4} \frac{v \cdot a}{c}, γ^{2} a + γ^{4} \frac{v \cdot a}{c ^{2}} v) .

U^{2} = c^{2}, P^{2} = m^{2} c^{2}, P = m U .

For the explicitly requested contraction,

U \cdot A = γ^{5} v \cdot a - γ^{3} v \cdot a - γ^{5} \frac{v ^{2}}{c ^{2}} v \cdot a = γ^{3} v \cdot a [γ^{2} (1 - v^{2} / c^{2}) - 1] = 0.

Thus four-acceleration is orthogonal to four-velocity. Equivalently this follows by differentiating the constant Lorentz scalar

U^{2} = c^{2}

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