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Consider the path $x^\mu(u)$ in Minkowski space; such that:

$$t = \frac{a}{c} \sinh(u) , \quad x = a \cosh(u) ,\quad y = 0 ,\quad z = 0 $$

where $a$ is a positive constant and $u$ is a parameter

Use equation: $$ c \nabla \tau = ds = \sqrt{\eta_{\mu\nu} \dot{x}^\mu \dot{x}^\nu} du $$ to find the proper time elapsed along the path starting from $u = 0$, as a function of $u$.

I'm having quite a bit of trouble with using the notation if someone could help me please?

So far I have established: $$ d_u x^{\mu} = \left(\frac {a}{c}\cosh(u),\, a\sinh(u),\,0,\,0\right)) $$

Then:

$$\eta_{\mu\nu} d_u x^{\mu} d_u x^{\nu} = \sum_{\mu =0}^{3} \sum_{\nu =0}^3 \eta_{\mu\nu} d_u x^{\mu} d_u x^{\nu} $$

Then I get a bit lost...

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  • $\begingroup$ Can you calculate $\dot{x}^\mu(u) = dx^\mu(u)/du$? This is the first step (and almost the last one). $\endgroup$ – Vibert Jan 18 '14 at 12:15
  • $\begingroup$ Yep, Sorry I was updating what I've done, I've got to the step above, then got a bit lost. (I'm trying to follow an example) $\endgroup$ – Sarah Jayne Jan 18 '14 at 12:22
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    $\begingroup$ I really wish people would refrein from voting to close such legitimate technical questions ... $\endgroup$ – Dilaton Jan 18 '14 at 21:12
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Let us assume that you use the metric: $$\left\{ \eta_{\mu \nu} \right\} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} $$ (Note that you could also use: $$\left\{ \eta_{\mu \nu} \right\} = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ which is just a matter of conventions.)

Now, we interpret the notations as follows (as you have mentioned in your comment): \begin{equation} \sqrt{\eta_{\mu \nu} \dot{x}^\mu\dot{x}^\nu} = \sqrt{\sum\limits_{\mu=0}^3\sum\limits_{\nu=0}^3\eta_{\mu \nu} \dot{x}^\mu \dot{x}^\nu} \tag{1} \end{equation} In order to evaluate this, it is important to realize: $$ \eta_{\mu \nu} = 0 \text{ if } \mu \neq \nu$$ Therefore, writing out equation $(1)$ fully: \begin{equation} \begin{aligned} \sqrt{\eta_{\mu \nu} \dot{x}^\mu \dot{x}^\nu} & = \sqrt{ \eta_{00}\dot{x}^0 \dot{x}^0 + \eta_{11}\dot{x}^1 \dot{x}^1 + \eta_{22}\dot{x}^2 \dot{x}^2 +\eta_{33}\dot{x}^3 \dot{x}^3} \\& = \sqrt{ \dot{x}^0 \dot{x}^0 - \dot{x}^1 \dot{x}^1 -\dot{x}^2 \dot{x}^2 -\dot{x}^3 \dot{x}^3} \end{aligned} \end{equation} Furthermore: \begin{equation} \left\{ x^\mu \right\} = \begin{pmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{pmatrix} = \begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix} \tag{2} \end{equation} At this point it is just substituting everything in the formula and evaluating it.

Edit:

One of the most important properties of special relativity is that all inertial reference frames are physically equivalent. This means that if one observer sees $ct,x,y,z$ and the other sees $ct',x',y',z'$, which may in general not be equal, we still have: \begin{equation} c^2 t^2-x^2 - y^2 - z^2 = c^2 t'^2-x'^2 - y'^2 - z'^2 \end{equation} This can be written more compactly (and manifestly Lorentz invariant) by the way equation $(2)$ is defined: $$ \eta_{\mu \nu} x^\mu x^\nu = \eta_{\mu \nu} x'^\mu x'^\nu $$ This is why equation $(2)$ is defined the way it is.

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    $\begingroup$ Can you add a factor of $c^2$ in $\eta_{00}$? This is needed in OP's conventions. $\endgroup$ – Vibert Jan 18 '14 at 17:26
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    $\begingroup$ Thankyou so much for your help! I really appreciate it. Can I ask why you add the factor of c^2 ? $\endgroup$ – Sarah Jayne Jan 19 '14 at 11:32
  • $\begingroup$ @SarahJayne I have edited my original message. Hopefully this answers your question. $\endgroup$ – Hunter Jan 19 '14 at 12:59
  • $\begingroup$ Happy to hear that. $\endgroup$ – Hunter Jan 19 '14 at 13:43
  • $\begingroup$ @SarahJayne: the factor of $c^2$ is always implied in the time component of the metric; this is simply dimenional analysis, as velocity*time = distance. But some authors set $c=1$, as you can always find the correct powers of $c$ in your final answer by dimensional analysis. In your definition of the path $x^\mu(u)$ there's a factor of $c$ included; if you write forgot to compensate in the metric, you end up with a very ugly answer. $\endgroup$ – Vibert Jan 19 '14 at 18:26

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