# How to reconcile time dilation equation with concept

So I must be misunderstanding time dilation, but according to the Lorentz transformation $t_v=\frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}$, where $t_v$ is the time on the object moving relative to the measuring observer, and $t_o$ is the time measured on the object still relative to the measuring observer. But by this, one could deduce that $t_v>t_o$ and thus the time on the moving object is greater, implying it is running faster than that of a still object. This seems to be inconsistent with the theory of time dilation as a concept, which suggests that time would move slower on the moving object.

There is an example question which I cannot resolve which seems to illustrate this:

An astronaut set out in a spaceship from Earth orbit to travel to a distant star in our galaxy. The spaceship travelled at a speed of 0.8c. When the spaceship reached the star the onboard clock said the journey took 10 years. An identical clock remained on Earth. What time in years had elapsed on this clock when seen from the astronaut’s spaceship?

If this is resolved with Earth as $t_v$, since it is moving relative to the measurement of the astronaut, we get an answer of 16.7 years. This implies, however, that more time has passed on Earth during the flight and thus Earth time was running faster. If resolved in the alternate way, we get a value of 6 years, however this also seems inconsistent as Earth was set as the stationary reference frame

• t_v is actually not greater than t_o, when V is zero (both objects relatively stationary), t_v = t_o. But as V increases, t_o increases as well since it's denominator is decreasing. Look into the Twin Paradox – Seth Taddiken Jun 15 '17 at 0:52

The formula you have needs to be interpreted carefully. It depends on who is looking at which clock and what is meant by "how fast time is moving." In your formula, let's have $t_o = 1$ second. When you calculate $t_v$ your are calculating how much time has to pass in the rest frame for the moving clock to tick one second. So, moving clocks tick slower, so it takes longer for a moving clock to indicate one second. That's why the formula results in a longer time.

Another way to write the formula is $$t_v = t_o\sqrt{1-\frac{v^2}{c^2}}.$$ In this formula, $t_o$ is a time measured in the rest frame and $t_v$ is the amount of time indicated on the moving clock.

In short, your formula tells how long an at-rest observer has to wait in order to see a moving clock tick $t_o$ time. The second formula tells how much time a moving clock as ticked after an at-rest clock has ticked $t_o$ time.

• It might be good to mention that the OP has written the formula for $\gamma$, which is likely what he/ she's confusing the right formula with. – Selene Routley Mar 16 '17 at 10:24
• @WetSavannaAnimalakaRodVance After looking at the wikipedia article for time dilation, I realized there's more than one way to define the concept. Answer has been edited. – Mark H Mar 16 '17 at 10:28
• Yes, that's a good answer. Usually one knows which way around the scaling goes, so one doesn't do it the wrong way around, but when you're reading the instructions for the first time and don't know which way the scaling goes, you do indeed need to interpret carefully. – Selene Routley Mar 16 '17 at 10:36

Formulas alone might be hard to implement.

A spacetime diagram on rotated graph paper might be useful here, together with some geometrical intuition.

Bob travels at (4/5)c (given by the Minkowski right-triangle with hypotenuse OB). Event Z is 10 ticks along Bob's worldline.

Time dilation relates two timelike segments from event O with endpoints that are simultaneous to the observer making the measurement.

So, Alice says BsimA is simultaneous with B. She measures the time-dilation factor $\gamma=\frac{OBsimA}{OB}=5/3$.

Alice also says ZsimA is simultaneous with Z. She measures the time-dilation factor $\gamma=\frac{OZsimA}{OZ}=16.666/10=5/3$. (Indeed, these Minkowski-right triangles are similar by scaling.)

Trigonometrically, there is an angle $\theta$ between the worldlines (called the rapidity... the arc-length subtended on the unit hyperbola centered at O). These ratios can be thought of as $ADJACENT/HYPOTENUSE=\cosh\theta=\gamma$ (where $OPPOSITE/ADJACENT=\tanh\theta=v$, the velocity).

Bob says ZsimB is simultaneous with Z. He measures the time-dilation factor $\gamma=\frac{OZ}{OZsimB}=10/6=5/3$. Indeed, this Minkowski-right triangle (with Minkowski-right-angle at Z and hypotenuse $OZsimB$) is similar to Alice's Minkowski-right triangles.

It is often said, that single clock dilates relatively to a set of spatially separated and Einstein – synchronized clocks.

This set of synchronized clocks is the REST FRAME of the observer. Observer is not a real physical person, but the whole reference frame, or team of observers. Relativistic observer conducts measurements in his own rest frame.

To measure dilation of moving clock an Observer (or team of observers) must have at least two clock in his frame, let's say clock C1 and C2. When moving clock passes by clock C1, he compares readings in immediate vicinity. When clock passes by clock C2, he compares readings again. If moving clock and C1 showed 12 PM at meeting, moving clock will show 3 PM and C2 shows 6 PM when they meet. This is how time dilation works. SINGLE moving clock dilates relatively to a set of synchronized and spatially separated clocks, not vice versa. Set of clock runs faster from the point of view of single clock.

If "moving" observer wants to measure dilation of another clock, he has to turn himself into one "at rest" by means of introducing his own rest frame. He simply puts another additional clock (at least) and synchronizes it by Einstein signalling method (Einstein clock synchronization convention).

https://en.wikipedia.org/wiki/Einstein_synchronisation

Also:

"Two spatially separated clocks, A and B, record a greater time interval between two events than the proper time recorded by a single clock that moves from A to B and is present at both events."

It is absolutely clear even visually (animation from article Time Dilation in Wikipedia).

[

Notion of Observer in SR https://en.wikipedia.org/wiki/Observer_(special_relativity)

In your case, observer in the reference frame of Earth has to place two Einstein – synchronized clocks. One E1 is at Earth, another E2 is far, far away at the point of arrival of Astronaut. These clocks are Einstein - synchronized. When Astronaut arrives to E2 his clock shows gamma times less time. Obviously, clock E2 shows gamma times more time.

We can imagine a row of synchronized clocks of reference system K. each denoted by letter - A, B, C, D ….. Z. Then a person with single clock on his wrist (A’ for example) moves in this reference system K and compares readings of his clock with these clock A-Z successively. When he comes to the clock Z, his clock A’ shows gamma times less time, than clock Z. Thus, clock Z shows gamma times more time, than his own. At this point Z this clock A’ immediately turns back and starts travelling in reverse direction, passing by clock Z, Y, X ….. C, B and finally arrives into point A of reference frame K. Clock A’ compares readings with clocks Z-A successively again and sees, that it dilates itself gamma times, i.e. every clock on the way shows gamma times more time. When clock A’ arrives into point A, clock A’ shows gamma times less time than clock A, and clock A shows gamma times more time than clock A’.

This discrepancy of clock readings when they meet again is often called clock paradox or Twin paradox.

All that goes straight from the Lorentz transformations.

$$T = \frac {t'_{x'}+ \frac {v'} {c^2} x'} {\sqrt {1-( \frac {v} c)^2}} (1)$$

$T$ is clock readings that belongs to reference frame $K$, taken in point $x'$ at moment of time $t'_{x'}$ of reference frame $K'$, and $t'_{x'}$ reading of clocks that belongs to reference frame $K'$ in the point $x'$ of reference frame $K'$

How to interpret Lorentz transform for time?

Transformation demonstrates, that time $T$ of reference frame $K$ (in which it does not depend of $x$ coordinate or any other coordinate) is universal in reference frame $K$ and each point of this frame.

Now let's fix point $x'$, for example $x'=0$. In this case this transformation will look like that:

$$T= \frac {t'_{0'}} {\sqrt {1-( \frac v c)^2}} (2)$$

$T$ is clock reading of reference frame $K$ taken in point $x'=0$ (in the origin $O'$ of reference frame $K'$), and $t'_{o'}$ is time in the reference frame $K'$, in particular in the origin $O'$.

We can take $dT/dt'$ when $x'$ is fixed and will get ${dT}/{dt'} = 1/{\sqrt {1- \frac {v^2} {c^2}}}$

According to (2) it is not time $t'_{o'}$ which is showed by single clock in the point $O'$ runs slower, but time $T$ , which is "distributed" through all reference frame $K$ and taken in the origin $O'$ of reference frame $K'$ runs faster (relatively to time $t'_{o'}$ that is in the origin $O'$ of frame $K'$). Time dilation comes by means of transformation of (2) into:

$$t'_{o'}=T \sqrt {1- {\frac {V^2} {c^2}}}$$

It is correct that $T>t'$ and $t'<T$. It is also true that $T'>t$ and $t<T'$. But that $t<t'$ and $t'<t$ from different points of view is nonsense.