# Proper time

The proper time $$τ$$ is the time measured by an observer $$O$$ (which can just be a particle) who “stands still” in space relative to a coordinate system. For example, if I am standing still on the surface of the Earth one meter away from a tree (where I am standing still and not moving away from the tree) where the origin of a coordinate system $$x^i$$ is located at the tree (and the coordinate system is fixed at that location), I will measure the proper time in my reference frame. Any other observer $$O'$$ who is moving at a relative velocity away from the tree will measure a time $$t'$$ which is longer than $$τ$$.

Suppose an observer $$O'$$ moves away from the origin of the x-coordinate system through space (along the $$x^1$$ dimension) and time (along the $$x^0=ct=t$$ dimension). We can imagine someone at point A who does not move through any of the spatial dimensions $$x^1$$, $$x^2$$, and $$x^3$$. Because he does not move through any of the spatial dimensions in the x-coordinate system, he is traveling through the time dimension $$x^0$$ at the speed of light and he is measuring the proper time $$τ$$. What about the observer $$O'$$ who is moving through the spatial dimension $$x^1$$ in the x-coordinate system at a velocity $$v$$ away from $$O$$? The observer $$O$$(who is stationary at point A) measures $$O'$$ traveling a distance$$x^1$$ in a time $$t$$. Because $$O'$$ is moving through the spatial dimension $$x^1$$ in the x-coordinate system, he measures the time $$t=γt$$ between A and B. (He also measures a different “x coordinate” which will be $$y^1$$.) In other words, $$O'$$ will measure that it took him less time to go from A to B; $$O$$ will measure that it took $$O’$$ a longer time $$t$$ to go from A to B.

# Time Dilation

Figure 1

In one of Einstein’s original though experiments, he imagined an observer $$O'$$ riding in a train which was moving past another observer $$O$$ standing still on the side of the tracks. The train was moving at a velocity $$\vec{v}$$ relative to $$O$$. Two reference frames $$R'$$ and $$R$$ (which should be thought of as coordinate systems moving through space) are attached to $$O'$$ and $$O$$ respectively. The train is moving along only the x-axis in the $$R$$-frame; in the $$R'$$-frame no part of the train is moving through space along the x'-, y'- and -z'axes and is at rest. Right at the moment when $$O'$$ is at $$x=0$$ both observers clocks are synchronized and $$t'=t=0$$. At this moment in time a light pulse is emitted from a light source $$S'$$. This light pulse travels upwards along the vertical, bounces off of the mirror, and then arrives back at $$S'$$. Let the distance between $$S'$$ and the mirror be $$d$$. In the $$R'$$-frame the light pulse travels along the y-axis at a constant speed $$c$$. Let the time interval for the light pulse to travel from $$S'$$ to the mirror and then back to $$S'$$ (in the $$R'$$-frame) be $$Δt'$$. Then the amount of time necessary for the light pulse to travel from $$S'$$ to the mirror (the “half-way distance”) in the $$R'$$-frame is $$\frac{Δt'}{2}$$. Since the speed of the light pulse is constant relative to $$R'$$ then, from kinematics, we know that $$d=c\frac{Δt'}{2}$$ and that

$$Δt'=\frac{2d}{c}.$$

We shall now see that the constancy of the speed of light with respect to both reference frames combined with the fact that the light pulse must travel through a greater distance with respect to the $$R$$-frame leads to $$O$$ measuring a longer time interval $$Δt$$ between event 1 (when the light pulse is emitted) and event 2 (when the light pulse arrives back at $$S'$$). If $$Δt$$ is the time interval measured in the $$R$$-frame for the light pulse to travel from $$S'$$ to the mirror and back to $$S'$$, then the time interval measured in that frame for the light pulse to go from $$S'$$ to the mirror is $$\frac{Δt}{2}$$. By the time the light pulse reaches the mirror the train (and the mirror) will have moved a distance $$v\frac{Δt}{2}$$. Thus the light pulse must have traveled a horizontal distance $$v\frac{Δt}{2}$$ and a vertical distance $$d$$ in this time interval. Using the Pythagorean Theorem the total distance the light pulse traveled is related to the horizontal and vertical distances by $$x^2=(v\frac{Δt}{2})^2+d^2$$. The speed of light is a constant $$c=3×10^8\frac{m}{s}$$ relative to $$R$$ (as it is for any frame) and because this speed is constant it follows (from kinematics) that $$x=c\frac{Δt}{2}$$ and $$(c\frac{Δt}{2})^2=(v\frac{Δt}{2})^2+d^2$$. If we solve for $$Δt$$ we get

$$Δt=\frac{2d}{\sqrt{c^2-v^2}}=\frac{2d}{c\sqrt{1-\frac{v^2}{c^2}}}.$$

Since $$Δt'=\frac{2d}{c}$$ we have

$$Δt=\frac{Δt'}{\sqrt{1-\frac{v^2}{c^2}}}$$

where

$$γ=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.$$

In the $$R'$$-frame events 1 and 2 occur at the same points in space (since there $$(x',\text{ ,}y'\text{, }z')$$ points are identical) as shown in Figure 1. Any observer who measures the time interval between two events in a frame in which those two events occur at the same points in space are said to be measuring the proper time interval $$Δτ$$ between those two events. We see that $$Δt'=Δτ$$ and that

$$Δt=γΔτ.$$

Because $$γ$$ is always greater than one, it follows that $$Δt$$ is always greater than $$Δτ$$. Fundamentally, this means that whenever an observer is measuring the time interval between two events in a frame that are not at the same two spatial points, a time interval $$Δt$$ will be measured that is always greater than $$Δt$$.

It is important to mention that this effect only becomes important when $$γ$$ significantly deviates from equaling one which only happens, roughly speaking, when $$v>0.01c$$. If $$O'$$ started walking “carrying” his coordinate system (which is attached to him) around with him, he would be moving at a speed $$v$$ relative to another frame (corresponding to someone “standing still” on the train) measuring $$Δτ$$ and thus he would measure a slightly longer time. But since his walking speed is much slower than $$0.01c$$ the effect of time dilation is negligible. Oftentimes, for practical purpose, if $$v<0.01c$$, then we can assume that $$Δt=Δτ$$. Only when $$v>0.01c$$ does $$γ$$ start to significantly deviate from being one.

On the most fundamental level, photons are emitted from atoms composing the light source $$S'$$. Those photons then travel through space, bounce off the atoms composing the mirror, and arrive back at the atoms composing $$S'$$. All chemical and biological processes are, at the most fundamental level, due to atoms interacting with other atoms via photons of light and electromagnetic radiation. The fact that it takes a longer time $$Δt$$ for any two atoms to interact with one another via a photon of light/radiation going from one atom to another means that it takes longer for chemical, and therefore biological, interactions to occur. Therefore, $$O$$ in the $$R$$-frame will see all physical processes take a longer time to happen and the march of time will progress more slowly in his reference frame.