Time Dilation Equations and Varying Interpretations

Question

I have been seeing problems involving time dilation but the interpretation used by some people is not always the same. $$ \tag{1} \Delta t' = \gamma \Delta t $$ If we think about the twin B going into a spaceship in space and $\gamma$ =2, some people say proper time is always shorter. Then if $\Delta$t=8 then $\Delta t'$=16, then the interpretation for eq.(1) is when B travels 16 years in earth frame (A) when it comes back it will have aged by 8 years because in A frame time is dilated when he observes B which is coherent with the fact that moving clocks run slower.

The second way that I have seen people interprating the time dilation equation is by using: $$ \tag{2} \Delta t = \gamma \Delta t' $$ Then the interpretation is that if B travels 8 years then A will age by 16 years. As we can see the two equations are not really coherent with each other and have different intepretations. Then what should be the correct equation to use ?

**EDIT:just to be sure, for eq.(1) it is the same as if we were in a lab and we have a stationnary particle and a moving particle. john in the lab with the stationnary particle will decay after 10 seconds($\Delta t$) and the moving particle after 20 seconds ($\Delta t'$) so it take 20 seconds in John frame for particle B to decay which normally takes 10 seconds.

Maybe I am wrong about how myself interpret $\Delta t$, is it the time that the spaceship mesure in his frame then $\Delta t'$ is the time observed by A. How I view it is that $\Delta t$ is only the time between two time events with a normal tick rate so if 8 years is the normal tick rate time between two events time then the gamma factor the dilated time will be 16y so it takes 16 years in A frame for B to 'achieve' the 8 years if it was ticking at a normal rate.

Answer 1

In SR (special relativity) a stationary observer measures a clock in a moving frame to tick slower. In the twin paradox the twin on the earth measures the twin travelling forth and back to earth younger. However the time dilation equation is not applicable by the twin going forth and back as, because of the turn back, the reference frame is not inertial.

To answer to the question the correct equation is the (1).

However to avoid confusion it should be written as:
$\Delta t = \gamma \Delta \tau$
where:
$t$ time in the stationary, but inertial frame (SR assumptions applicable)
$\tau$ proper time in the moving frame

Answer 2

Often, a labeled spacetime diagram helps clarify the interpretation of equations and their symbols.

Here is a spacetime diagram to explain time-dilation.
Time runs upwards and drawn in the reference frame of Alice (an inertial observer who is at rest in this frame). Bob travels to the right with constant velocity. For simplicity, Alice and Bob meet at event O.

Alice wants to determine the elapsed time on her clock from event O to distant event Q.
Alice uses the elapsed proper time on her clock from event O to local event P (on her worldline) since Alice says "events P and Q are simultaneous" (i.e., spacelike segment $PQ$ is Minkowski-perpendicular to the timelike segment $OP$ on her worldline).

As a Minkowski right triangle, $OQ$ is the hypotenuse, $OP$ is the adjacent side, and $PQ$ is the opposite side. The velocity $\beta=(PQ)/(OP)=\tanh\theta$, where $\theta$ is the rapidity [angle] between worldlines. The time dilation factor is $\gamma=\cosh\theta$.

So, time dilation uses the fact that the adjacent side $OP$ is larger than the hypotenuse $OQ$ in Minkowski spacetime geometry. (Recall $\cosh\theta \geq 1$.) In symbols [with extra notation for clarity], $$ \begin{eqnarray*} (t_{Q}^{Alice}-t_{O}^{Alice})= \Delta t_{OQ}^{Alice} = \Delta t_{OP}^{Alice} &=& \tau_{OP} \\ &=& \cosh\theta\ \tau_{OQ} \\ &=& \gamma\ \tau_{OQ} \\ &=& \gamma\ \Delta t_{OQ}^{Bob} \\ \end{eqnarray*} $$

Similarly, for Bob, with $ON$ as the hypotenuse, $OR$ as the adjacent side, and $RN$ as the opposite side, $$ \begin{eqnarray*} (t_{N}^{Bob} -t_{O}^{Bob} )= \Delta t_{ON}^{Bob} = \Delta t_{OR}^{Bob} &=& \tau_{OR} \\ &=& \cosh\theta\ \tau_{ON} \\ &=& \gamma\ \tau_{ON} \\ &=& \gamma\ \Delta t_{ON}^{Alice} \\ \end{eqnarray*} $$

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Admittedly, it might be difficult to visualize that $OP$ is a longer segment than $OQ$ in special relativity (in Minkowski spacetime geometry), or that $OR$ is Minkowski-perpendicular to $ON$.

So, here is a spacetime diagram on "rotated graph paper" [which helps us visualize the ticks along segments and the Minkowski-orthogonality of segments]. In the diagram, we use $\beta=6/10$ so that $\gamma=10/8$.

Answer 3

Here's another way to look at it. Consider the Lorentz transformation $ct'=\gamma (ct-\beta x),\,x'=\gamma (x-\beta ct)$. We can invert this as $ct=\gamma (ct'+\beta x'),\,x=\gamma (x'+\beta ct')$. Thus $\gamma=(\partial_t t')_x=(\partial_{t'} t)_{x'}$, where the subscripts denote what variable is held constant in the definition of partial derivatives. By contrast, $1/\gamma=(\partial_t t')_{x'}=(\partial_{t'} t)_x$; the choice of subscript matters. (Since $ct=(ct'+\beta x)/\gamma$, it's easy to see $(\partial_{t'} t)_x=1/\gamma$.) In particular $(\partial_t t')_x(\partial_{t'} t)_x=1\ne(\partial_t t')_x(\partial_{t'} t)_{x'}$. You can understand length contraction the same way.

This subscript subtlety is in analogy with the results $\frac{x}{r}=(\partial_r x)_\theta,\,\frac{r}{x}=(\partial_r x)_y$ for $2$-dimensional space. (For a proof use $x=r\cos\theta,\,x^2=r^2-y^2$.) Again, $(\partial_r x)_\theta(\partial_x r)_\theta=1\ne(\partial_r x)_\theta(\partial_x r)_y$.