de-TEXted but otherwise hopefully straight:- Just interested (flummoxed) - how do you get 4 nice looking equations AND a (fairly) nice solution? I can't see how to start to do that.
Possible approach: The first equation can be substituted into the other three to make them linear equations in x, y, and z. Solving this system will give you x, y, z as functions of t. These can be substituted into the first equation to give you an equation in t, which may or may not be easy to solve.
From the original context we're looking at an expanding 'sphere' of light in different frames. Having chosen the point of origin the 4 equations should (?) be easy (?) - apart from the trick of choosing the origin to get nice integers in all four equations. Edit... knowing the point of origin in the past should give the point of coincidence in the future. Er... maybe.
(0,0,0,5), (4,3,8,7), (−4,7,−8,3), (7,12,−4,4) Determinant is 3600 - a nice number to start with. (Thanks to http://matrix.reshish.com/detCalculation.php ) Applying Cramers rule ( http://www.purplemath.com/modules/cramers.htm ) should (?) solve for something... Determinant of ... (0,0,0,5), (0,3,8,7), (0,7,−8,3), (0,12,−4,4) ... is zero. Not a great start. 0/3600=0.
This looks more context sensitive than I thought - here is the original post A four dimensional universe I'll try to read it a few times before proceeding further.
Easiest substitution for t, assuming t > 5 (the future) x^2+y^2+z^2−(t−5)^2=0 ⇒ t = 5 + √(x^2+y^2+z^2) This makes each quadratic equation reduce to a linear equation in x, y, z, and t (x−4)^2+(y−3)^2+(z−8)^2−(t−7)^2=0 ⇒ (x−4)^2+(y−3)^2+(z−8)^2−(–2 + √(x^2+y^2+z^2))^2 = 0 ⇒ x^2 – 8 x +16 + y^2 – 6 y + 9 + z^2 – 16 z + 64 – 4 + 4 √(x^2+y^2+z^2) – x^2 – y^2 – z^2 = 0 ⇒ –8 x – 6 y – 16 z + 4 t = –65 This makes each quadratic equation reduce to a linear equation in x, y, z, and t (x+4)^2+(y−7)^2+(z+8)^2−(t−3)^2=0 ⇒ (x+4)^2+(y−7)^2+(z+8)^2−(2 + √(x^2+y^2+z^2))^2 = 0 ⇒ x^2 + 8 x + 16 + y^2 – 14 y + 49 + z^2 + 16 z + 64 – 4 – 4 √(x^2+y^2+z^2) – x^2 – y^2 – z^2 = 0 ⇒ 8 x – 14 y + 16 z – 4 t = -145 Combining these two, we have an explicit formula for y. –8 x – 6 y – 16 z + 4 t + 8 x – 14 y + 16 z – 4 t = –65 + -145 ⇒ y = 210/20 = 189/18 = 21/2 Substitute in y = 21/2 t = 5 + √(x^2+y^2+z^2) ⇒ t = 5 + √(x^2+z^2+441/4) Substitute in y = 21/2 –8 x – 6 y – 16 z + 4 t = –65 ⇒ –8 x – 16 z + 4 t = –2 Substitute in y = 21/2 8 x – 14 y + 16 z – 4 t = -145 ⇒ 8 x – 14 y + 16 z – 4 t = 2 (no surprise here) Substitute in y = 21/2 AND t = 5 + √(x^2+z^2+441/4) (x−7)^2+(y−12)^2+(z+4)^2−(t−4)^2=0 ⇒ (x−7)^2+9/4+(z+4)^2−(1 + √(x^2+z^2+441/4))^2 = 0 ⇒ x^2 –14 x + 49 + 9/4 + z^2 + 8 z + 16 – 1 – 2 √(x^2+z^2+441/4) + 10 – 10 – x^2 – z^2 – 441/4 = 0 ⇒ – 14 x + 8 z – 2 t = 34 ⇒ – 28 x + 16 z – 4 t = 68 Combining two, we have an explicit formula for x. –8 x – 16 z + 4 t – 28 x + 16 z – 4 t = –2 + 68 ⇒ x = –66/36 = –33/18 = –11/6 So now our four equations are: x^2+y^2+z^2−(t−5)^2=0 ⇒ t = 5 + √(z^2 + 2045/18), x = –11/6, y = 21/2 And any one of: (x−4)^2+(y−3)^2+(z−8)^2−(t−7)^2=0 ⇒ – 16 z + 4 t = –50/3 (x+4)^2+(y−7)^2+(z+8)^2−(t−3)^2=0 ⇒ 16 z – 4 t = 50/3 (x−7)^2+(y−12)^2+(z+4)^2−(t−4)^2=0 ⇒ 16 z – 4 t = 50/3 By substituting in our expression for t into one of these, squaring both sides, we get a quadratic equation in z. 16 z – 4 t = 50/3 ⇒ 4 z – t = 25/6 ⇒ 4 z – 25/6 = 5 + √(z^2 + 2045/18) ⇒ 4 z – 55/6 = √(z^2 + 2045/18) ⇒ 16 z^2 – (440/6) z + 3025/36 = z^2 + 4090/36 ⇒ 15 z^2 – (440/6) z – 1065/36 = 0 ⇒ z = (44 ± 5 √103)/18 By substituting in our (indefinite) expression for z into one of the linear equations, we get an indefinite expression for t, only one value of which satisfies t>5 4 z – t = 25/6 ⇒ t = 4 z – 25/6 ⇒ t = (101 ± 20 sqrt(103))/18 ⇒ t = (101 + 20 sqrt(103))/18 ⇒ z = (44 + 5 √103)/18 Done.
Just amazing. Thank you. More questions may follow when I can do it without having to peek at rpenner's exemplary solution.
