Talk:Hyperbolic motion (relativity)

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I'm not sure I understand the diagram in the article. It makes it appear as if the world lines of the accelerating particles are outside of the light cone?? (unless I'm missing something fundamental, which may be likely)

If the diagram is correct, then can someone explain what a particle at rest would look like within the diagram? Dmitry Brant (talk) 13:25, 9 April 2008 (UTC) Never mind; I misunderstood the nature of the diagram. Sorry! Dmitry Brant (talk) 13:43, 9 April 2008 (UTC)[reply]

Sure, instead of plotting x^2 - y^2 = 1, just plot (x+1)^2 - y^2 = 1. This will make sure that y = 0 for x = 0. --91.214.89.190 (talk) 00:36, 13 January 2016 (UTC)[reply]

It is not clear whether the author intended to define u as coordinate velocity or proper velocity, although I suspect the former. — Preceding unsigned comment added by 128.230.52.225 (talk) 21:24, 8 May 2013 (UTC)[reply]

See the article on proper velocity. The default meaning, unless otherwise stated, is coordinate velocity. DonQuixote (talk) 00:20, 9 May 2013 (UTC)[reply]

The last two latex formulas show parsing errors like this,

Failed to parse (unknown function "\begin"): \begin{array}{c|c} \begin{aligned}u(t) & =\frac{\alpha t}{\sqrt{1+\left(\frac{\alpha t}{c}\right)^{2}}}\\ & =c\tanh\left(\operatorname{arsinh}\frac{\alpha t}{c}\right)\\ x(t) & =\frac{c^{2}}{\alpha}\left(\sqrt{1+\left(\frac{\alpha t}{c}\right)^{2}}-1\right)\\ & =\frac{c^{2}}{\alpha}\left(\cosh\left(\operatorname{arsinh}\frac{\alpha t}{c}\right)-1\right)\\ c\tau(t) & =\frac{c^{2}}{\alpha}\ln\left(\sqrt{1+\left(\frac{\alpha t}{c}\right)^{2}}+\alpha t\right)\\ & =\frac{c^{2}}{\alpha}\operatorname{arsinh}\frac{\alpha t}{c} \end{aligned} & \begin{aligned}u(\tau) & =c\tanh\frac{\alpha\tau}{c}\\ \\ x(\tau) & =\frac{c^{2}}{\alpha}\left(\cosh\frac{\alpha\tau}{c}-1\right)\\ \\ ct(\tau) & =\frac{c^{2}}{\alpha}\sinh\frac{\alpha\tau}{c}\\ \\ \end{aligned} \end{array}

Could any one check and fix?--

(Also, "better" choices than 0 for the initial values of X(T) and X(τ) would allow elimination of the "-1" in the X(T) and X(τ) equations, but in this case there's no mismatch between the equations as written and the statement on initial values above.) — Preceding unsigned comment added by 98.163.18.220 (talk) 15:53, 21 January 2018 (UTC)[reply]

To be specific (by comparison with the lower set of equations): the current initial value of cτ(T) appears to be (c^2 / α)artanh(u_0 / c), and the current initial value of cT(τ) appears to be (c^2 / α)(u_0 γ_0 / c).

And it seems that the "better" initial value of X(T) and X(τ) would be (c^2 / α)γ_0. — Preceding unsigned comment added by 98.163.18.220 (talk) 16:05, 21 January 2018 (UTC)[reply]

The general expression has the form

${\displaystyle c\tau (T)=c\tau _{0}+{\frac {c^{2}}{\alpha }}\left\{\operatorname {arsinh} \left({\frac {u_{0}\gamma _{0}+\alpha T}{c}}\right)-\operatorname {artanh} \left({\frac {u_{0}}{c}}\right)\right\}}$

setting ${\displaystyle \tau _{0}=u_{0}=0}$ gives

${\displaystyle c\tau (T)={\frac {c^{2}}{\alpha }}\operatorname {arsinh} {\frac {\alpha T}{c}}}$

so I'm not sure how you arrive at your initial conditions. Regarding the "-1" in X(T), this can indeed be removed by setting the initial value ${\displaystyle c^{2}/\alpha }$, as explained in the rapidity section. --D.H (talk) 16:48, 21 January 2018 (UTC)[reply]

Yes, sorry, you're right: my mistake was neglecting that u_0 has been set to 0 when I was looking at the cT(τ) and cτ(T) equations. In general, one would use cT(τ)_0 = (c^2 / α)(u_0 γ_0 / c) and cτ(T)_0 = (c^2 / α)artanh(u_0 / c), but if one has already set u_0 = 0 then this is equivalent to using cT(τ)_0 = 0 and cτ(T)_0 = 0. My bad! — Preceding unsigned comment added by 98.163.18.220 (talk) 19:02, 21 January 2018 (UTC)[reply]