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Books say that special relativity is indistinguishable from Newtonian mechanics when the speed of the primed frame ($v$) is small compared to the speed of light ($c$). This is what I mean by the "Newtonian limit" property of special relativity. (I don't know the correct name for it). But the transformation of the time coordinate

$$t' = \gamma \left(t + \frac{vx}{c^2}\right)$$

involves the spatial quantity $x$ which, if large enough, could balance out the smallness of $v/c^2$. So why doesn't this observation mess up the "Newtonian limit" property?

Qmechanic
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sean mcilroy
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  • Good point! Note that for a noticeable deviation, x has to be very large if v is small. Please see https://physics.stackexchange.com/q/63045/123208 and https://physics.stackexchange.com/q/267996/123208 – PM 2Ring Sep 21 '19 at 23:06

1 Answers1

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Nice question.

Suppose that you're sitting in your lab on Earth, and you want to plan out an experiment in which you will see a large effect from the $vx/c^2$ term in the Lorentz transformation. Your lab is small, so you'll need to travel some large distance $x$ in order to make this term large. If you travel this distance at velocity $v$, then the travel time required for your experiment will be $t=x/v$. If we want the effect to be large, we will need $vx/c^2$ to be on the same order of magnitude as the time scale $t$ of our experiment. But if $vx/c^2\sim t$ and $t=x/v$, then $v\sim c$, which means that your experiment is not actually being carried out under nonrelativistic conditions.