# Homework Help: Transform of a Line Element

1. Oct 23, 2012

### soothsayer

1. The problem statement, all variables and given/known data
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
$t =\left (\frac{c}{g} + \frac{x'}{c} \right )sinh\left (\frac{gt'}{c} \right)$
$x =\left (\frac{c}{g} + \frac{x'}{c} \right )cosh\left (\frac{gt'}{c} \right) - \frac{gt'}{g}$
y = y', z = z'

for constant g with dimensions of acceleration

2. Relevant equations
$ds^2 = -(cdt)^2 + dx^2 + dy^2 + dz^2$

3. The attempt at a solution
After taking the typical steps to transform a line element, I came up with:

$(cdt)^2 = g^2 \left (\frac{c}{g} +\frac{x'}{c}^2 \right) cosh^2 \left(\frac{gt'}{c} \right)(dt')^2 + sinh^2 \left (\frac{gt'}{c}\right) (dx')^2 + 2g \left (\frac{c}{g} +\frac{x'}{c} \right) sinh \left (\frac{gt'}{c} \right) cosh \left(\frac{gt'}{c} \right) dt'dx'$

$(dx)^2 = g^2 \left (\frac{c}{g} +\frac{x'}{c}^2 \right) sinh^2 \left(\frac{gt'}{c} \right)(dt')^2 + cosh^2 \left (\frac{gt'}{c}\right) (dx')^2 - 2g \left (\frac{c}{g} +\frac{x'}{c} \right) sinh \left (\frac{gt'}{c} \right) cosh \left(\frac{gt'}{c} \right) dt'dx'$

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,

$(ds)^2 =-g^2\left(\frac{c}{g} +\frac{x'}{c}\right)^2(dt')^2 -4g\left(\frac{c}{g} +\frac{x'}{c}\right)sinh \left(\frac{gt'}{c} \right ) cosh\left(\frac{gt'}{c} \right )dt'dx' + (dx')^2 +(dy')^2 + (dz')^2$

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?

2. Oct 23, 2012

### clamtrox

There's clearly something wrong with your units. In the coordinate transformations you have same units for t and x, but then in the metric you have an extra conversion factor c.

What is the extra t' doing in the coordinate transform for x? Is it just a typo since it's not in the later stages of your calculation?

3. Oct 23, 2012

### Mentz114

$$x =\left (\frac{c}{g} + \frac{x'}{c} \right )cosh\left (\frac{gt'}{c} \right) - \frac{gt'}{g}$$
Should the last term in this be gt'/c ? The c factors are not right, as clamtrox suggests.

I can see your mistake ( I think). The minus on the cross term in dx2 is wrong.

If you still have a problem, I have a Maxima script for this now.

Last edited: Oct 23, 2012
4. Oct 23, 2012

### soothsayer

Whoa, yeah, sorry. That last term at the end of x should be c2/g, NOT gt'/g. I got carried away, or mixed up in the LaTeX or something. Anyway, it's a constant, and disappears in the differentiation.

Mentz, you're totally right. I made the rookie mistake of assuming that the derivative of coshx was -sinhx. I really should have looked that up while I was doing the problem yesterday XP

Ok, so the cross term in dx2 is now positive, so the cross terms in ds2 vanish. Thanks, guys.

so my new line element becomes:

$ds^2 = g^2 \left(\frac{c}{g} +\frac{x'}{c} \right)^2 \left[sinh^2 \left (\frac{gt'}{c} \right) -cosh^2 \left(\frac{gt'}{c} \right )\right]dt'^2 + \left[cosh^2 \left (\frac{gt'}{c} \right) -sinh^2 \left(\frac{gt'}{c} \right )\right]dx'^2 + dy'^2 + dz'^2$

Which simplifies to:

$ds^2 = -g^2 \left(\frac{c}{g} +\frac{x'}{c} \right)^2 dt'^2 + dx'^2 + dy'^2 + dz'^2$

So, when the problem asks me to take the gt'/c << 1 limit, what does that even mean to the above metric? This is supposed to result in a metric detailing a constant acceleration in the Newtonian limit.

Last edited: Oct 23, 2012
5. Oct 24, 2012

### Staff: Mentor

Are you sure it's not asking for the limit when gx'/c2<<1?

6. Oct 25, 2012

### soothsayer

Yeah, I also thought it might have asked for gx'/c << 1 when I was typing this and didn't have the book on hand, but no, I checked it again and it definitely asked for gt'/c << 1. So I went back to the coordinate transform equations and took the gt'/c << 1 limit of those, instead of the line element, but try as I did, I couldn't seem to find any physically significant results.

7. Oct 25, 2012

### Staff: Mentor

The units on the first term on the rhs of the equation for x in your original post don't seem right. There seems to be a factor of c missing out front. If you make this change and use

sinh(gt'/c) → gt'/c at gt'/c <<1
and
cosh(gt'/c) → 1 + (gt'/c)2/2 at gt'/c <<1

see what you get.