- #1
soothsayer
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Homework Statement
Transform the line element of special relativity from the usual (t, x, y, z) coordinates rectangular coordinates to new coordinates (t', x', y', z') related by
[itex] t =\left (\frac{c}{g} + \frac{x'}{c} \right )sinh\left (\frac{gt'}{c} \right) [/itex]
[itex] x =\left (\frac{c}{g} + \frac{x'}{c} \right )cosh\left (\frac{gt'}{c} \right) - \frac{gt'}{g}[/itex]
y = y', z = z'
for constant g with dimensions of acceleration
Homework Equations
[itex] ds^2 = -(cdt)^2 + dx^2 + dy^2 + dz^2 [/itex]
The Attempt at a Solution
After taking the typical steps to transform a line element, I came up with:
[itex] (cdt)^2 = g^2 \left (\frac{c}{g} +\frac{x'}{c}^2 \right) cosh^2 \left(\frac{gt'}{c} \right)(dt')^2 + sinh^2 \left (\frac{gt'}{c}\right) (dx')^2 + 2g \left (\frac{c}{g} +\frac{x'}{c} \right) sinh \left (\frac{gt'}{c} \right) cosh \left(\frac{gt'}{c} \right) dt'dx'[/itex]
[itex](dx)^2 = g^2 \left (\frac{c}{g} +\frac{x'}{c}^2 \right) sinh^2 \left(\frac{gt'}{c} \right)(dt')^2 + cosh^2 \left (\frac{gt'}{c}\right) (dx')^2 - 2g \left (\frac{c}{g} +\frac{x'}{c} \right) sinh \left (\frac{gt'}{c} \right) cosh \left(\frac{gt'}{c} \right) dt'dx'[/itex]
As you can see, if you plug these expressions into the line element equation above, (dy = dy' and dz = dz'), the cross terms don't cancel out, but actually add together, as so,
[itex](ds)^2 =-g^2\left(\frac{c}{g} +\frac{x'}{c}\right)^2(dt')^2 -4g\left(\frac{c}{g} +\frac{x'}{c}\right)sinh \left(\frac{gt'}{c} \right ) cosh\left(\frac{gt'}{c} \right )dt'dx' + (dx')^2 +(dy')^2 + (dz')^2 [/itex]
which seems wrong, but at the same time, the next part of the question asks me to find the gt'/c << 1 limit, which seems to imply that not all the hyperbolic terms cancel out, which would clearly be the case if the cross terms vanish. Have I made an error in calculating the line element somewhere?