The Road to Rindler

This story takes place in two dimensions, (t) and (x).

We are choosing (-,+,+,+) for the signature of our metric.

The color curves in the picture are hyperbolas.

Our goal here is to show that we can start with several assumptions that we say are basic assumptions and get to two equations that contain hyperbolic cosine and hyperbolic sine. Our reason to do this is that we want those mathematical expressions to “make sense” to you, as opposed to being two more things to memorize.

DEFINITIONS

These are to help answer the questions you will have about what choices we made when defining things and what symbols we used.

$\vec{U} = U^t \tilde{\vec{e_t}} + U^x \tilde{\vec{e_x}}$ $\frac {d}{d \tau} \vec{S} = c \frac {dt}{d \tau} \vec{e_t} + \frac {dx}{d \tau} \vec{e_x}$ $U^t = \frac {d}{d \tau} ct$ $A^t = \frac {d^2}{d \tau^2} ct$

The basis vectors that we use, $\vec{e_t}$ and $\vec{e_x}$ come from the inertial frame. If you are using a story that there is an observer on the ground and an observer on a spaceship (accelerating with constant acceleration) then in your story the $\vec{e_t}$ and $\vec{e_x}$ basis vectors come from the observer on the ground.

See Appendix A for an explanation on why $\vec{A}$ only has two terms instead of the usual four.

RESTRICTIONS AND RESULTS OF RESTRICTIONS

At one point we will want to write in two initial conditions: $U^x(\tau) = 0$ when $\tau$ is zero. If you haven’t used initial conditions before to help solve a math problem we should probably have a simple example for you to examine.

PREVIOUSLY DETERMINED TRUTHS

$\vec{U} \cdot \vec{U} = - c^2$
$\vec{U} \cdot \vec{U} = - (U^t)^2 + (U^x)^2$
$\vec{A} \cdot \vec{A} = \alpha^2$
$\vec{A} \cdot \vec{A} = - (A^t)^2 + (A^x)^2$
$\vec{U} \cdot \vec{A} = 0$
$\frac {d}{d \tau} (U^t) = (A^t)$
$\frac {d}{d \tau} (U^x) = (A^x)$

We start with $\vec{U} \cdot \vec{A} = 0$ . Because of our choice for the metric (-,+,+,+) the dot product gives the following…

$\vec{U} \cdot \vec{A} = - (U^t)(A^t) + (U^x)(A^x)$

Since we have $\vec{U} \cdot \vec{A} = 0$ we can make…

$- (U^t)(A^t) + (U^x)(A^x) = 0$

…and this converts to…

$(U^x)(A^x) = (U^t)(A^t)$

…and we want to square everything…

$(U^x)^2(A^x)^2 = (U^t)^2(A^t)^2$

We have the option to replace $(U^t)$ and $(A^t)$ with expressions containing either ‘x’ or nothing at all.

…we can now take what we have…

$(U^x)(A^x) = (U^t)(A^t)$

…and change it to…

$(U^x)(A^x) = ((U^x)^2 + (c^2))((A^x)^2 - (\alpha^2))$

We can do a binomial expansion on the right…

$(U^x)^2(A^x)^2 = (\alpha^2)(U^x)^2 + (c^2)(A^x)^2 - (\alpha^2)(U^x)^2 - (\alpha^2)(c^2)$

The first term on the right will cancel with the single term on the left:

$0 = (c^2) (A^x)^2 - (\alpha^2) (U^x)^2 - (\alpha^2)(c^2)$

We want the two terms with $c^2$ on the same side and we’ll put the third term on the other side.

$(\alpha^2)(U^x)^2 = (A^x)^2 (c^2) - (\alpha^2)(c^2)$

…and then we do the factoring…

$(\alpha^2)(U^x)^2 = (c^2) [ (A^x)^2 - (\alpha^2) ]$

Now we replace $(A^x)^2 - (\alpha^2)$ with $(A^t)^2$ .

$(\alpha^2)(U^x)^2 = (c^2) (A^t)^2$

We are going to square root both sides:

$\sqrt{(\alpha^2)(U^x)^2} = \sqrt{(c^2) (A^t)^2}$

$\alpha (U^x) = \pm c (A^t)$

We will use the positive answer but we feel we should accentuate the “plus minus”:

$A^t= \pm \frac {\alpha}{c} (U^x)$

We also have two equations that define $A^t$ or $A^x$ as being derivatives of 4-Velocity:

$\frac {d}{d \tau} (U^t) = (A^t)$

$\frac {d}{d \tau} (U^x) = (A^x)$

We have…

$A^t = \frac {\alpha}{c} (U^x) \: \to \: \frac {d}{d \tau} U^t = \frac {\alpha}{c} U^x$

$A^x = \frac {\alpha}{c} (U^t) \: \to \: \frac {d}{d \tau} U^x = \frac {\alpha}{c} U^t$

From what we now have, we can build two cascades of math that leads us to two Ordinary Differential Equations that we want…

A^t = A^t — ‘Cascade’ is just a word we use for those times when we think it’s helpful to look at a long list of math changes.

\frac {d}{d \tau} U^t = \frac {\alpha}{c} U^x — ‘Cascade’ is just a word we use for those times when we think it’s helpful to look at a long list of math changes.

We will do some work with that ordinary differential equation that we see on the last line. However, before that we should look at what we have one the second line:

$\frac {d}{d \tau} U^t = \frac {\alpha}{c} U^x$
$\frac {d}{d \tau} U^x = \frac {\alpha}{c} U^t$

The above says “hey! one is the derivative of the other along with a necessary ratio of velocities!”

We know from Calculus that if we need two functions where one is the derivative of the other that we get this technology with hyperbolic sine and hyperbolic cosine. I won’t rule out the existence of something else does the same (caution is a virtue) but we have sufficient reason to declare a hypothesis.

What we now have is two textbook Ordinary Differential Equations that fit the following:

\frac{d^2}{dt^2} f(t) = \frac{a^2}{b^2} f(t)

f(t) = c_1e^{\frac{a}{b}t} + c_2e^{-\frac{a}{b}t}

In work with ordinary differential equations we look for initial conditions that will force the calculation to go from its general form to a specific case. We choose the ones below because (reason forthcoming).

“ $U^x(\tau) = 0$ when $\tau = 0$ ”

$e^0=1$

$U^x(0) = c_1 \: (1) + c_2 \: (1)$

$U^x(0) = c_1 + c_2$

$0 = c_1 + c_2$

$c_2 = - c_1$

The above lets us change…

$U^x(\tau) = c_1e^{\frac{\alpha}{c} \tau} + c_2e^{-\frac{\alpha}{c} \tau}$

…to…

$U^x(\tau) = c_1e^{\frac{\alpha}{c} \tau} - c_1e^{-\frac{\alpha}{c} \tau}$

…and then we can factor out the $c_1$ :

$U^x(\tau) = c_1 [e^{\frac {\alpha}{c} \tau} - e^{- \frac {\alpha}{c}} ]$

Now that we have $U^x(\tau)$ we can calculate $U^t(\tau)$ from $\frac {d}{d \tau} U^x = \frac {\alpha}{c} U^t$ which we change to…

$U^t =\frac {c} {\alpha} \frac {d}{d \tau} U^x$

We calculate…

$\frac {d}{d \tau} (c_1 [e^{\frac {\alpha}{c} \tau} - e^{- \frac {\alpha}{c}} ]) = c_1 [\frac {\alpha}{c} e^{\frac {\alpha}{c} \tau} + \frac {\alpha}{c} e^{- \frac {\alpha}{c} \tau} ]$

And we put it into the previous equation to get…

$U^t =\frac {c} {\alpha} c_1 [\frac {\alpha}{c} e^{\frac {\alpha}{c} \tau} + \frac {\alpha}{c} e^{- \frac {\alpha}{c} \tau} ]$

which we adjust in two steps…

$U^t =\frac {c} {\alpha} c_1 \frac {\alpha}{c} [e^{\frac {\alpha}{c} \tau} + e^{- \frac {\alpha}{c} \tau} ]$

$U^t = c_1 [e^{\frac {\alpha}{c} \tau} + e^{- \frac {\alpha}{c} \tau} ]$

We are going to use our initial conditions again. We said $U^x = 0$ when $\tau = 0$ . You can check for yourself that a zero tau makes $U^x = 0$ .

Since we also have latex]U^t[/latex] we can use it to force latex]c_1[/latex] to identify itself.

$U^t = c_1 [e^0 + e^0]$

$U^t = c_1 [1 + 1]$

$U^t = c_1 [2]$

$c_1 = \frac {U^t}{2}$

From work above we have…

Since we know that $\vec{U} \cdot \vec{U} = -c^2 = - (U^t)^2 + (U^x)^2$

We take out…

$- (U^t)^2 + (U^x)^2 = -c^2$

and modify it with the initial condition that $(U^x)^2 = 0$ …

$- (U^t)^2 + 0 = -c^2$

$- (U^t)^2 = -c^2$

$(U^t)^2 = c^2$

$(U^t) = c$

We now have what we want…

U^x(\tau) = \frac {1}{2} c [e^{\frac {\alpha}{c} \tau} - e^{- \frac {\alpha}{c}} ]

U^t(\tau) = \frac {1}{2} c [e^{\frac {\alpha}{c} \tau} + e^{- \frac {\alpha}{c} \tau} ]

sinh(\frac {\alpha}{c} \tau) = \frac {1}{2} (e^{\frac{\alpha}{c} \tau} - e^{- \frac{\alpha}{c} \tau})

cosh(\frac {\alpha}{c} \tau) = \frac {1}{2} (e^{\frac{\alpha}{c} \tau} + e^{- \frac{\alpha}{c} \tau})

$U^x(\tau) = c \: sinh(\frac {\alpha}{c} \tau)$

$U^t(\tau) = c \: cosh(\frac {\alpha}{c} \tau)$

It probably feels like getting here was a journey of 250 math steps.

The idea of Rindler coordinates gives us a chance to move away from inertial systems.

We might argue that there are five to seven ideas of basic “high math” that present themselves here and their presence gives us a chance to consider math that we will need farther along.

Appendix A

Normally we would have…

$\vec{A} = \frac {dU^t}{d \tau} \vec{e_t} + U^t \frac {d e^t}{d \tau} + \frac {dU^x}{d \tau} \vec{e_x} + U^x \frac {d e^x}{d \tau}$

…but the fact that we are in an inertial frame means that the basis vectors are constant everywhere and the derivative of a constant is zero, so we have…

$\vec{A} = \frac {dU^t}{d \tau} \vec{e_t} + 0 + \frac {dU^x}{d \tau} \vec{e_x} + 0$

…which simplifies to …

$\vec{A} = \frac {dU^t}{d \tau} \vec{e_t} + \frac {dU^x}{d \tau} \vec{e_x}$

Appendix B

“We are taking the positive square root solution so that our four velocity vectors will point from the past to the future.” -Eigenchris

Eigenchris did his work with +,-,-,- and we went with -,+,+,+ so we had to deal with the differences. That said, the video “105a: Acceleration – Hyperbolic Moton and Rindler” was extremely helpful for providing direction needed for this math journey.

TRAILERS

Trailer 1

We are planning a blog that talks about dot products. Given our assumptions if there was a “B” we would have…

$\vec{B} \cdot \vec{B} = - (B^t)(B^t) + (B^x)(B^x)$

and if we had “A” and “B” we would have…

$\vec{A} \cdot \vec{B} = - (A^t)(B^t) + (A^x)(B^x)$

Trailer 2

You are aware that if we do “squaring” to either -7 or +7 we get a positive 49.

When you see something like $- (x^2)$ , putting either x=-7 or x=+7 into it gives us negative 49.

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