Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 38E / b

Table of contents
- Solution b

Solution

Put

B^{2} = \frac{( 1 - A ) ^{2}}{( 1 + A ) ^{2}}, C = (1 + A)^{4} .

The rest-frame metric can be written as

C d \overset{s}{ˉ}_{Minkowski}^{2} + (C - B^{2}) d \overset{ˉ}{t}^{2}

. The Minkowski term is invariant under the Lorentz transformation, while

d \overset{ˉ}{t} = γ (d t - v d x)

. Therefore

f (A) = (1 + A)^{4}, g (A) = (1 + A)^{4} - \frac{( 1 - A ) ^{2}}{( 1 + A ) ^{2}},

and

A = \frac{m}{2 γ ^{2} ( x - v t ) ^{2} + y ^{2} + z ^{2}} .

These substitutions give precisely

d s^{2} = f (A) (- d t^{2} + d x^{2} + d y^{2} + d z^{2}) + γ^{2} g (A) (d t - v d x)^{2} .

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version