1
$\begingroup$

I am currently taking a course in Modern Physics mainly focused on Special Relativity and came up with the following problem:

Let $O_1$ be an observer at rest. A second observer $O_2$ is moving to the right at speed $v$ with respect to $O_1$, and a third observer $O_3$ is moving also to the right at speed $v'$ but relative to $O_2$. During $\Delta t_3$ an event occurs in $O_3$'s frame that is measured as $\Delta t_2=\gamma'\Delta t_3$ where: $$\gamma'=\frac{1}{\sqrt{1-\frac{v'^2}{c^2}}}$$

Similarly, $\Delta t_2$ is measured by $O_1$ as $\Delta t_1=\gamma\Delta t_2$ where: $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ Hence, $\Delta t_1=\gamma\gamma'\Delta t_3$ : $$\gamma\gamma'=\frac{1}{\sqrt{1-v^2/c^2 - v'^2/c^2 +v^2v'^2/c^4}}$$

Now if we want to directly measure $\Delta t_1$ from $\Delta t_3$ without going through $\Delta t_2$ we would first have to express the speed of $O_3$ in $O_1$'s frame using the Lorentz transformation: $$v''=\frac{v+v'}{1+vv'/c^2}$$

Then applying the formula for time dilation: $$\Delta t_1=\gamma''\Delta t_3$$ where : $$\gamma''=\frac{1}{\sqrt{1-v''^2/c^2}}$$ $$=\frac{1}{\sqrt{1-(\frac{v+v'}{1+vv'/c^2})^2/c^2}} $$

By identification $\gamma''$ is supposed to be equal to $\gamma\gamma'$ which is clearly not the case. I have tried expanding $\gamma''$ and replacing terms with their series expansions to no avail. I can't seem to figure out what the fallacy is.

$\endgroup$
2
$\begingroup$

The basic confusion is in your comment that "During $\Delta t_3$ an event occurs in $O_3$'s frame that is measured as $\Delta t_2 = \gamma^\prime \Delta t_3$". In SR a single "event" occurs at a single point in spacetime--an interval like $\Delta t$ can only describe the time between a pair of events. What's more, the time dilation formula $\Delta t_B = \gamma \Delta t_A$ can only be used if $\Delta t_A$ refers to the time between a pair of events in the frame where both events occurred at the same spatial location in the frame A that measures the time interval $t_A$, like two different readings of a clock that's at rest in frame A (so that $t_A$ would also be the proper time of the clock). If that's not true you have to use the full Lorentz transformation formula rather than the time dilation formula (see my answer here). So it doesn't make sense to combine the formulas $\Delta t_2 = \gamma^\prime \Delta t_3$ and the formula $\Delta t_1 = \gamma \Delta t_2$, as if $\Delta t_2$ referred to the same interval between the same pair of events in both cases, since the events in question can't have been located at the same position from the perspective of both frame $O_3$ (implied by the first equation) and frame $O_2$ (implied by the second equation), given that the frames are moving relative to each other along a single axis.

$\endgroup$
  • $\begingroup$ Great explanation! But is the second way I tried to find $\Delta t_1$ correct? (I understand I could've used the Lorentz transformation for $t$ but is this acceptable?) $\endgroup$ – GeorgSaliba Mar 17 '15 at 19:11
  • $\begingroup$ Yes, that second method does work, assuming the time intervals $\Delta t_1$ and $\Delta t_3$ are describing the time in each frame between a pair of events that occurred at the same position in $O_3$. $\endgroup$ – Hypnosifl Mar 17 '15 at 19:15
1
$\begingroup$

As @Hypnosifl points out, a single event cannot have $\Delta t$. You must define two events such as a light flashing or the beginning and end of a process or a particle moving from point A to point B. You must the define the pair of events (call them $A$ and $B$ with both a time coordinate and a spacial coordinate, e.g. $$ \pmatrix{ct_A \\ x_A} \text{ and } \pmatrix{ct_B \\ x_B}.$$ From this one can define the differences $$\pmatrix{c\Delta t \\ \Delta x}.$$

I'm adding this answer to the list to illustrate how to use matrix notation to do Lorentz transformations. This is only one spatial dimension. Let's take your example now and transforms the events in $\mathcal{O}_3$ to $\mathcal{O}_2$. $$\pmatrix{c\Delta t_2 \\ \Delta x_2}=\pmatrix{\gamma ' & \gamma ' \beta '\\ \gamma ' \beta ' & \gamma '}\pmatrix{c\Delta t_3 \\ \Delta x_3} = \pmatrix{c\gamma '\Delta t_3+ \gamma ' \beta '\Delta x_3\\c \gamma ' \beta '\Delta t_3+ \gamma '\Delta x_3}.$$ (If we were transforming events from one frame to another which was moving in the +x direction, the sign of $\beta'$ would be changed, such as going from $\mathcal{O}_2$ to $\mathcal{O}_3$).

From this you can immediately see that $\Delta t_2=\gamma '\Delta t_3$ only if the events occur at the same location in $\mathcal{O}_3$ ($\Delta x_3=0$). Note that $\Delta x_2 \ne 0$, so you can't use the time dilation form to take these events from $\mathcal{O}_2$ to $\mathcal{O}_1.$

You can also transform velocities this way. $\mathcal{O}_3$ is moving with $\beta ' = v'/c$ with respect to $\mathcal{O}_2$. So, in $\mathcal{O}_2$ the "Lorentz" velocity of $\mathcal{O}_3$ (actually a truncated 4-vector) could be written $$\gamma ' c\pmatrix{1\\ \beta '}$$ where $\gamma '$ is calculated using $\beta '$. When we do the Lorentz transformation on this velocity we are going from $\mathcal{O}_2$ to $\mathcal{O}_1$ so we get $$\gamma '' c\pmatrix{1\\ \beta '' }=\gamma ' c\pmatrix{\gamma & \gamma\beta \\ \gamma \beta & \gamma}\pmatrix{1\\ \beta '}$$ or $$ \gamma '' = \gamma ' \gamma\left(1+\beta\beta' \right)$$ $$ \gamma '' \beta '' = \gamma ' \gamma\left(\beta + \beta ' \right)$$ Take the ratio of these to find $\beta ''$ which agrees with your $v ''$. I love the beauty and symmetry of using these matrix transformations, plus the Lorentz matrix is a lot easier to remember (for me, anyway) than a set of equations.

$\endgroup$
  • $\begingroup$ Shouldn't it be $c\Delta t$ instead of just $\Delta t$? Because otherwise the matrix wouldn't correspond to the transformations $\endgroup$ – GeorgSaliba Mar 17 '15 at 21:20
  • $\begingroup$ Technically yes, but there is a system of units in which c=1 and time is measured in meters. I wasn't consistent with that and mixed the two systems. I'll go edit it. Thanks for pointing that out. $\endgroup$ – Bill N Mar 17 '15 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.