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Griffiths, Introduction to electrodynamics, in problem 12, shows that if a man starts running at moment $t=0$ from position $x>0$, under the influence of a constant force, will never be reached by a light ray that passes at the origin at time $t>0$.

Shown in the picture is the man as the hyperbolic line, photons as dashed lined -

enter image description here The picture is of course pretty clear and it is obvious the man and the ray will never cross paths because they are separated by the trajectory of another.

But on the other hand, the man of course sees the light and the light will always travel faster the him, so how come the light will never reach him?

Qmechanic
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proton
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    "... the man of course sees the light..." No he doesn't, that would mean the photon caught up to him. An accelerating object develops an event horizon behind it past which events cannot be casually connected. That is what this example is trying to show. – Asher Sep 16 '17 at 20:08
  • @Asher the light speed is certainly greater than the man's speed, so the light must reach at some point. why is it not true? – proton Sep 16 '17 at 20:17
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    as you can see in the diagram, the man's speed asymptotically approaches the speed of light before the photon reaches him. – Asher Sep 16 '17 at 20:24
  • if you're running at a growing speed and I'm running after you at a speed that is and will always be greater than yours, I will eventually reach you. Why doesn't it work at the speed of light? – proton Sep 16 '17 at 20:30
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    Because there is an accelerated frame in play here. Mathematically the time in the future at which the two meet is infinite, which is an indicator that it never actually happens. – Asher Sep 16 '17 at 20:35

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Isn't this diagram wrong? I mean, following the reasoning, you can draw a straight line inside the cone in such way that the inclination is greater than the bisector line. Tracing a vertical line over this diagram, it will intersect our line and the line bisector line in such way that the x point of intersection is greater in our line than in the bisector line, which means we have constructed the line representing a particle moving with speed greater than c.

In another words, shouldn't the hyperbole be symmetric/invariant under reflection wrt the ct axis, and not wrt the x axis, as is show here?

Now, i know this is an older question, but this solution and the book is famous enough to more people come eventually come here to ask that as time passes, so i think it is ok to ask this here.

LSS
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  • The diagram is correct. Just because $x_{\text{hyperbola}} > x_{\text{line}}$ doesn't mean that the particle following the hyperbola moves with speed greater than $c$. You should look at the derivatives, and $\dot{x}{\text{hyperbola}} < \dot{x}{\text{line}} = c.$ – md2perpe Nov 18 '21 at 16:50
  • It shouldn't be symmetric w.r.t. the $ct$ axis. That would show two particles accelerating in opposite directions. Now it's one particle first coming closer and slowing down, then moving further away and speeding up. – md2perpe Nov 18 '21 at 16:53
  • @md2perpe I don't think i got it. How it doesn't mean the particle has moved with speed greater than c? I mean, take the bisector of the plane, which represents the relation between ct and x for light. Now, take another line C inside the cone, at the same region the hyperbola is inside, but make this line have inclination greater than the bissector. So, in this way, if we trace a vertical line it will intersect both bisector and the C line at the same ct point, but x for the C line will be greater than the x corresponding for the bissector/light line. – LSS Nov 19 '21 at 00:32
  • @md2perpe Also, search any minkowski diagram on google, we will see that when the axis ct is vertical, the hyperbole is with its focus on the vertical axis too, that is, the ct axis. Outisde this cone, that is, outise the cone where the focus is at the ct axis, the region is not allowed for any path followed by real particle. But it is precisely this region that the hyperbole is draw in the diagram above. – LSS Nov 19 '21 at 00:35
  • It might be more common to have $ct$ on the vertical axis and $x$ on the horizontal, but that doesn't really matter. In this diagram it's the other way around, and we have to follow that. – md2perpe Nov 19 '21 at 00:39
  • Consider a space ship with position $x(t) = \sqrt{x_0^2 + ct^2}.$ What is the speed at different times? Show that $|\dot{x}(t)| < c$ for all $t$. Will the light signal $x(t)=ct$ ever reach the space ship? – md2perpe Nov 19 '21 at 00:42