If, for example, I'm in a train traveling on earth at $.99c$, and I drop a ball from a $1~\text{m}$ height inside that train how long would it take to hit the ground from the ball's frame of reference? I can see two ways at going about this question, and they produce two different results $$D = \frac {1}{2}at^2 \\ \implies 1 = \frac {1}{2}at^2 \\ \implies t=\sqrt\frac{2}{a}$$

Now $a =\dfrac{f}{m}$ and $f = \dfrac{gmm}{d^2}$ relative to the ball the earth is moving at $.99c$ and so the mass increases by a factor of $\dfrac{1}{\sqrt {1-(.99)^2}}$ which causes acceleration to increase proportionally This means

$$t = \sqrt{\frac{2}{9.8} \cdot\sqrt{1-(.99^2)}}$$ Alternatively we can work it out using the earth's FOR and then convert using time dilation formula; $$1=\frac{1}{2}\cdot 9.8\cdot t^2 \\ \implies t = \sqrt {\frac{2}{9.8}}$$ So from the trains FOR

$$t = \sqrt{\frac{2}{9.8}\cdot\sqrt{1-(.99^2)}}$$ Now this value is different to the other value by a factor of $\sqrt {\sqrt{1-.99^2}}$ or the $\sqrt[4]{1-.99^2}.$ Can someone please tell me which answer is correct and why the other method didn't work (Or that neither is correct and what the correct method is)?

  • $\begingroup$ How long from who's point of view? You, the ball or an observer watching from a platform? Also if the train is Amtrak it'll be late. You need to take that into account too. $\endgroup$ – Robinson Aug 18 '15 at 10:47
  • $\begingroup$ The ball's, sorry I forgot to mention that $\endgroup$ – The Lemon Aug 18 '15 at 10:49
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    $\begingroup$ If people tell you that relativity is as easy as changing mass with speed they are lying to you and it will create a thousand different problems in a thousand different ways. Just say no to relativistic mass. $\endgroup$ – Timaeus Aug 18 '15 at 18:01


To simplify notation let's denote $\gamma = \frac{1}{\sqrt{(1-0.99^2}}$, and let $\tx{m}{ball}, \tx{m}{Earth}$ be the rest (invariant) masses of the ball and the Earth. The way you write the gravitational force on the ball in the two FORs is:

In the train: $$\tx{F}{train} = g \frac{\tx{m}{ball}(\tx{m}{Earth}\gamma)}{d^2} = \tx{m}{ball} \tx{a}{train},$$ hence $$\tx{a}{train} = \gamma\;9.8 \;\&\; \tx{T}{train}=\sqrt{\frac{2}{9.8\gamma}}.$$

On Earth: $$\tx{F}{Earth} = g \frac{(\tx{m}{ball}\gamma)\tx{m}{Earth}}{d^2} = (\tx{m}{ball}\gamma) \tx{a}{Earth},$$ so $$\tx{a}{Earth} = 9.8\;\& \;\tx{T}{Earth} = \sqrt{\frac{2}{9.8}} \neq \gamma \tx{T}{train}.$$

Note that you actually have $\tx{F}{train} = \tx{F}{Earth}$ and the difference in the ball's acceleration in the two frames appears to come from its rest vs. relativistic mass. However, the transformation of forces perpendicular to the direction of motion reads $$ \tx{F}{moving-frame} = \frac{\tx{F}{rest-frame}}{\gamma}\;\; or \;\; \tx{F}{rest-frame} = \gamma \tx{F}{moving-frame} $$ See last sec. in this lecture: 12.3 Transformation of forces (MIT course notes). With this correction we get:

On Earth, as before: $$ \tx{F}{Earth} = g \frac{(\tx{m}{ball}\gamma) \tx{m}{Earth}}{d^2} = (m_{ball}\gamma)\tx{a}{Earth} $$ therefore $$\tx{a}{Earth} = 9.8 \;\&\;\tx{T}{Earth} = \sqrt{\frac{2}{9.8}}.$$ Time dilation implies $$\tx{T}{train} = \tx{T}{Earth}/\gamma.$$

In the train (rest frame; technically not, but good approximation): $$ \tx{F}{train} = \gamma \tx{F}{Earth} = \tx{m}{ball} \tx{a}{train} $$ hence $$\tx{a}{train} = \gamma \tx{F}{Earth}/\tx{m}{ball} = \gamma^2\;9.8 \;\&\;\tx{T}{train}=\sqrt{\frac{2}{9.8\gamma^2}} = T_{Earth}/\gamma;$$ check out.

Catch: Treating the Newtonian gravitational force between two objects in motion at high relative velocity like any other force in Special Relativity and applying the force transformation to it is tricky. The reason is that the force seen by an object is always greater in its rest frame. For the gravitational force above, this means that the force on the ball has to be greater in the train FOR than in the Earth FOR, while the force on the Earth has to be greater in the Earth FOR and smaller in the train FOR. But in both FORs the force is the same on the two objects! The way out is to consider the force on the ball in the Earth FOR as generated by a uniform force field and so to bypass the Newtonian inverse square law entirely. Then the force transformation can be applied safely and things remain consistent within SR.


If the ball has initially a horizontal velocity larger than 11 km/s, its path around Earth's center of mass is a parabola or hyperbola and it will not intersect with the ground.


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