# Does Minkowski diagram show that time goes faster in “moving” frame of reference?

I have a question re the diagram that appears at:

http://www1.phys.vt.edu/~takeuchi/relativity/notes/section12.html

The diagram purports to show that the observer in each frame of reference will observe that the clock in the other frame of reference is running slow (because he/she is comparing the reading of his/her clock with a reading of the other clock in the past).

However, the first observation diagrammed (i.e., the one by the "stationary" character on the left who states, at time T, "Your clock is running slow compared to mine.") appears to show the opposite, since more time has elapsed in the "moving" frame of reference (i.e, between 0 and T', which forms the hypotenuse of the right triangle) than has in the "stationary" frame of reference (i.e, between 0 and T, which forms a side of the same triangle). Thus, since MORE, not less, time has passed in the "moving" frame of reference, it would appear that time is running FASTER, not slower, in the same.

Am I missing something? If so, what?

• Where you have the "Z" shaped lines between the two, the bottom line involves an observation from right-to-left, the middle line involves an observation from left-to-right, and the top line involves an observation from right-to-left again. It can be seen that the traveller's clock is always behind, but the observation that the traveller makes of the stationary clock (the middle, diagonal part of the "Z") is apparently behind, but only because he's seeing at an out-of-date reading. Having said that, I don't see how these claims about time stay consistent if there is another traveller involved. – Steve Jan 7 '18 at 1:34
• Steve, thanks. However, I read the diagram differently. It is my understanding that, in the "Z" shaped lines, the bottom line involves an observation from left-to-right (i.e., an observation of the moving frame of reference from the stationary frame of reference), the middle line involves an observation from right-to-left (i.e., an observation of the stationary frame of reference from the moving frame of reference), and the top line involves an observation from left-to-right (i.e., a second observation of the moving frame of reference from the stationary frame of reference). -Burch – Burch Jan 7 '18 at 23:00
• perhaps I should recast my vocabulary - in the bottom line, the observation is being made by the person on the left, of the clock on the right. The radiation, however, is moving right-to-left, from clock to observer. If you understand the distinction between the time the visual information is radiated, and when it is received (at a later time), that's how the diagram ties things together. A round trip back and forth along any two legs of the Z always leaves each insisting that the other is behind, because implicitly their local clocks advance before news of the other clock's state arrives. – Steve Jan 8 '18 at 9:20

As others have noted, the geometry of the spacetime diagram is not Euclidean but Minkowskian. (The PHY 101 position-vs-time diagram is also not Euclidean... the analogous hypotenuse has "length" equal to that of the timelike-leg.)

Here are some different ways to visualize this. The Takeuchi example uses v=(1/2)c. Below I will use v=(3/5)c.

• Using two-observer graph paper, where the "moving"-observer's axes are obtained by a Lorentz Transformation of the home-observer's axes. (I'm not sure how well Takeuchi's preceding slide motivates the slide you linked.)
• Using hyperbolic graph paper, analogous to polar-coordinate paper.
At the the intersection points between the moving observer's worldline and the family of hyperbolas, the tangents determine the moving observer's axes in the two-observer graph paper.
• Using Bondi's k-calculus method--the Doppler Effect with the Principle of Relativity. The elapsed time between light-signal emissions is proportional to the elapsed time between receptions.
When the home observer is the emitter, the proportionality constant is $k_{home}$.
When the moving observer is the emitter, the proportionality constant is $k_{mov}$.
But the principle of relativity requires these factors to be equal... so just call them both $k$.
From my example, if the emission period is $T=2$, the received period is $kT$ (where $k$ is not yet known or measured).
When the reflected signal is received by Home, that period must be $k(kT)$ which home measures [for the case of v=(3/5)c] as 8.
Thus $k^2=4$ or $k=2$, thus the elapsed time along the moving observer's segment is $kT=4$.
(In general, $k=\sqrt{\frac{1+(v/c)}{1-(v/c)}}$. With $v=(1/2)c$, we have $k=\sqrt{3}$.)
• Using my new visualization involving rotated graph paper, which makes it easier to visualize the ticks of a light-clock, as traced out by the light-signals in the light clock (where each tick occupies the same area on the diagram, as needed by the Lorentz Transformation). This approach can be motivated using Bondi's method.

The x-t plane is non-Euclidean, it exists in hyperbolic space.

The hypotenuse of the right triangle is actually less time than the vertical axis. If it was 45 degrees it'd be going the speed of light and experience zero proper time.

• Connor, thanks. I am now studying the concept of a calibration hyperbola. – Burch Jan 12 '18 at 20:51

Velocity-related time dilation means that time is passing slower in a moving reference frame.

This is easily visible in the twin paradox where the traveling twin is aging slower than the twin who stayed at home.

The twin paradox requires a change in movement of the traveling twin in order to make meet both twins again. But it can also be shown by the means of the proper time equation:

$$\tau = \frac{1}{\gamma}t = \sqrt{(1-\frac{v^2}{c^2})}\ t$$

In this example you are the observer measuring the coordinate time t of several moving objects. Then the proper time $\tau$ depends on the velocity-dependent Lorentz factor $\gamma$, a higher relative velocity yields a lower proper time, that means: time is going slower for moving objects. The extreme case are lightlike phenomena (v=c), where proper time is reduced to zero.