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I have recently been thinking about the equations of special relativity and have run across something interesting. I was thinking about the following scenario: one person is standing stationary on a road, another person is driving at 0.125c, and a third person is driving at 0.25c.

For the rest of this post, I will use $\gamma_v$ to denote the relativistic factor for a speed v and $t_v$ to represent the time measured by the person moving at speed v.

So, I thought of two ways of analyzing the time dilation when comparing the stationary person and the person moving at 0.25c. The simple way is:

$t_0= \gamma_{0.25c}t_{0.25c}$ (1)

I then thought about using the time measured by the person moving at 0.125c to determine the time dilation. According to the person moving at 0.125c, the person moving at 0.25c is only moving at 0.125c, so:

$t_{0.125c}=\gamma_{0.125c}t_{0.25c}$ (2)

Additionally, the following should be true:

$t_{0}=\gamma_{0.125c}t_{0.125c}$ (3)

Rearranging equation 3 gives:

$t_{0.125c}=t_{0} / \gamma_{0.125c}$ (4)

Plugging equation 4 into equation 2 gives:

$t_{0} / \gamma_{0.125c} = \gamma_{0.125c}t_{0.25c} $ (5)

Rearranging equation 5 gives:

$t_{0} = {\gamma_{0.125c}}^2 t_{0.25c}$ (6)

Finally, plugging equation 6 into equation 1 gives:

${\gamma_{0.125c}}^2 t_{0.25c} = \gamma_{0.25c}t_{0.25c}$

${\gamma_{0.125c}}^2 = \gamma_{0.25c}$ (7)

However, in carrying out the calculations, equation 7 appears to be false. It appears as if I have made a mistake, but I am not sure where. Any help would be very much appreciated. Thank you!

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    $\begingroup$ You forgot about the relativistic addition of velocities: If somebody is moving at 0.25c relative to the person on the road, it doesn't mean that he moves at 0.125c relative to somebody else who is in turn moving at 0.125c relative to the same person on the road. $\endgroup$ – udrv Apr 13 '17 at 2:06
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There are two errors:

  • Relative-velocity is not subtractive--instead, it's $$v_{BA}=\frac{v_{BO}-v_{AO}}{1-v_{BO}v_{AO}}.$$
  • Time-dilation factors are not multiplicative--there is an extra factor $$\gamma_{BA}=\gamma_{BO}\gamma_{AO}(1-v_{BO}v_{AO}).$$

These are easier to recognize if you work with rapidities (Minkowski angles), where $v_{BO}=\tanh\theta_{BO}$, $\gamma_{BO}=\cosh\theta_{BO}$, etc...
Relative-rapidity is $\theta_{BA}=\theta_{BO}-\theta_{AO}$.
Relative-velocity is $v_{BA}=\tanh(\theta_{BA})=\tanh(\theta_{BO}-\theta_{AO})=\frac{\tanh\theta_{BO}-\tanh\theta_{AO}}{1-\tanh\theta_{BO}\tanh\theta_{AO}}$.
Relative-time-dilation-factor is $\begin{align}\gamma_{BA}=\cosh(\theta_{BA})=\cosh(\theta_{BO}-\theta_{AO})&=\cosh\theta_{BO}\cosh\theta_{AO}-\sinh\theta_{BO}\sinh\theta_{AO}\\&= \cosh\theta_{BO}\cosh\theta_{AO}(1-\tanh\theta_{BO}\tanh\theta_{AO})\end{align}$.


Here's an example with a spacetime diagram on rotated graph paper.
Here $v_{CA}=-3/5$ and $v_{BA}=3/5$--so $\gamma_{CA}=5/4$ and $\gamma_{BA}=5/4$.
So, $v_{BC}=\frac{v_{BA}-v_{CA}}{1-v_{BA}v_{CA}}=15/17$.
Note that $\gamma_{BC}=17/8$ (graphically)
and $\gamma_{BA}\gamma_{CA}(1-v_{BA}v_{CA})=(\frac{5}{4})(\frac{5}{4})(1-(\frac{3}{5})(\frac{-3}{5}))=17/8$.

spacetime diagram on rotated graph paper

postscript: This diagram shows that although Alice says that events N and Q are simultaneous, they are not simultaneous according to Bob or Carol.

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  • $\begingroup$ Thank you so much! I am not an expert in special relativity so I was unaware of these concepts. I'll be sure to research them. Thanks again! $\endgroup$ – dts Apr 13 '17 at 3:12

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