Transform of Line Element in Special Relativity

[soothsayer]

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?

 

[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?

 

[Mentz114]

[tex]
x =\left (\frac{c}{g} + \frac{x'}{c} \right )cosh\left (\frac{gt'}{c} \right) - \frac{gt'}{g}
[/tex]
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.

 

[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:

[itex] 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 [/itex]

Which simplifies to:

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

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.

 

[Chestermiller]

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:

[itex] 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 [/itex]

Which simplifies to:

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

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.

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

 

[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.

 

[Chestermiller]

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.