# How to apply the time dilation formula?

I am learning special relativity and still not sure how to correctly apply the time dilation formula. Take for instance the following example:

A spaceship leaves earth and travels to Alpha Centauri 4.3 ly away with the speed $$v = 0.8c$$. Seen from earth, the trip takes $$\Delta{t} \approx5.3$$ years. Now with the time dilation formula we can see that the austronaut experienced the time $$\Delta{t'} = \frac{\Delta t}{\gamma} \approx 3.12 \ \text{years}$$ Now what I'm tempted to do is to say that time dilation is symmetric and so, from the perspective of the austronaut, the time elapsed on earth is $$\Delta{t'} = \frac{3.12 \ \text{years}}{\gamma} \approx 1.82 \ \text{years}$$ which is obviously nonsense, however I can't say why this reasoning is flawed.

• Simultaneity! :) Commented Jan 23 at 0:39
• "I can't say why this reasoning is flawed." What reasoning? All I see is an equation. Commented Feb 2 at 11:35
• Did you also read the sentence above the equation? :) Commented Feb 2 at 21:50
• The sentence above the equation says you're going to apply a formula. Where is the reasoning that justifies that? Commented Feb 4 at 13:39

Let's say a clock on Earth reads $$t = 0$$ when the spaceship takes off. According to an observer on Earth, the clock reads $$t = 5.3\text{ years}$$ when the ship reaches Alpha Centauri.

According to an observer on the ship, the same clock reads $$t = 0$$ when the spaceship takes off, and $$t = \text{ 1.82 years}$$ when the ship reaches Alpha Centauri, as you have calculated.

There is no contradiction here. What's happening is that the observers don't agree on which event (the clock showing $$\text{5.3 years}$$ or $$\text{1.82 years}$$) happens at the same time as the event of the spaceship reaching Alpha Centauri.

If two events happen simultaneously according to observer 1, observer 2 who is in motion with respect to observer 1 will in general conclude that the two events are not simultaneous. There is no concept of absolute simultaneity for spatially separated events in special relativity.

Following the same approach as in your second result, one could then use the equation again to calculate a new ship time, etc. in an infinite number of steps toward 0, which you recognize as not correct. There is one basic fact about the time dilation equation that explains this.

In the usual derivation of time dilation, two events are considered and in one frame of reference the events occur at different locations whereas in the other frame they are at the same location. The time interval between the two events in the latter case is referred to as a proper time interval. In your example, you say that the trip takes 5.3 years in the Earth frame. In the ship's frame, both events take place at the same location, i.e. the ship. The time dilation equation relates these two time intervals, non-proper and proper, and shows that the ship's proper time interval is 3.12 years.

You cannot now use 3.12 years in the time dilation equation to find the proper time interval on Earth because 3.12 is a proper time interval in the ship's frame. Using the inverse Lorentz transformation, where the primed frame is the ship's frame:

t = $$\gamma$$ (t'+ vx'/$$c^2$$)

since in the ship's frame, x$$_2'-x_1'$$=0 we find (t$$_2-t_1$$) = $$\gamma$$(t$$_2'-t_1'$$) so we get back that (t$$_2-t_1$$) = 5.3 years.

To supplement the answer by @Puk,
and to address what you actually calculated and why the reasoning is flawed,
let's consider the Euclidean version of your question.

A line that intersects two vertical parallel lines separated by 4.3 units has a slope of 0.8.
Projecting the sloped-segment between the parallels onto the left parallel,
the BASE of the associated right-triangle has a size $$ADJ=OPP/SLOPE=(4.3/0.8)=5.375$$ units.
Now with the "definition of cosine in a right-triangle'' where $$SLOPE=\tan\phi$$ and $$\cos\phi=\frac{1}{\sqrt{1+({\rm SLOPE})^2}}=\frac{1}{\sqrt{1+(4/5)^2}}=\sqrt{25/41}\approx 0.7809$$ we can see that the length along the sloped-segment (the hypotenuse, opposite the right-angle), is $$HYP=\frac{ADJ}{\cos\phi}=\frac{5.375}{0.7809}\approx 6.883$$ Now what I'm tempted to do is to say that "definition of cosine is a right-triangle" is symmetric and so, from the perspective of the sloped-line, a "vertical segment projected onto the sloped-line" (also a hypotenuse) is $$HYP=\frac{ADJ}{\cos\phi}=\frac{6.883}{0.7809}\approx 8.815$$ which is obviously nonsense, however I can't say why this reasoning is flawed.

This diagram might help visualize the calculation.
I think you can see that the "symmetry of the definition of cosine in a right-triangle" is about similar right-triangles associated with two intersecting lines. The calculations of the hypotenuses above involve two similar but not congruent right-triangles where
the base (ADJ) was 5.375 in the first calculation,
and
the base (ADJ) was 6.883 in the second calculation.

In the above, "perpendicular" is defined as being tangent to the "circle".
So, the blue dotted line is perpendicular to the blue solid line because the blue dotted line is tangent to a circle of radius 6.883 with center at the intersection of the lines. (It's also parallel to the tangent on the unit-circle where the blue solid line meets the unit circle.)

So, here is an analogous "position-vs-time diagram" (a "spacetime diagram") associated with your original calculation, with smaller round-off errors than in your original calculation. In this diagram, the “circle” is a hyperbola.

With $$SLOPE=\tanh\theta$$, $$\gamma=\cosh\theta=\frac{1}{\sqrt{1-({\rm SLOPE})^2}}=\frac{1}{\sqrt{1-(4/5)^2}}=\sqrt{\frac{25}{9}}=\frac{5}{3}$$

$$HYP=\frac{ADJ}{\cosh\theta}=\frac{5.375}{5/3}=3.225$$

$$HYP=\frac{ADJ}{\cosh\theta}=\frac{3.225}{5/3}=1.935$$

To connect with @Puk's explanation,
[using the "circle" in that geometry] each solid line can construct its own perpendicular (dotted line) through that point (vertical=5.375, horiz=4.3). Those perpendiculars intersect a line in generally more than one point.