Time dilates!

Time runs more slowly for people traveling at constant velocity relative to your position! This sounds like science fiction. It’s not. What’s more, it’s not too hard to demonstrate.

Outside of normal math, there is just one thing we need to know:

Theorem 1 The speed of light is constant to all inertial observers.

Imagine you are in a car traveling with velocity {v} and you shine a torch at the front windscreen. You measure the speed of light and call it {c}. You have a friend on the pavement, and you are moving directly away from them. They measure the speed of light coming from the torch and they get {c} – exactly the same!

This is counter to normal intuition. If this was done with a ball, and you throw a ball at the windscreen with velocity {s} then your friend would record the ball traveling at {v+s}. But this is not the case at speeds close to the speed of light. Just accepted it. It comes from the rules of electromagnetism. It’s what this implies that is surprising.

The distance {x} something travels in time {t} moving with velocity {v} is {x=vt}.The light from your torch travels a distance of {ct}. If the light particle has coordinates {(x,y,z)} and we use standard Pythagoras to calculate the distance to a particle then we know distance = {\sqrt{x^2 + y^2 + z^2}}.

What we now have are two descriptions of where a light particle is. They must be equal. In the frame of the car we have:

\displaystyle  \begin{array}{rcl}  ct = \sqrt{x^2+y^2+z^2} \implies c^2t^2 = x^2+y^2+z^2\\ \implies c^2t^2 - (x^2+y^2+z^2)=0 \end{array}

Let’s say that your friend measures their time as {t'} and space coordinates as {(x',y',z')}. Imagine you shine the light as you pass your friend. This way both of you agree, at this moment, {t=t'=0} and {x=x'=0}. Your friend would measure the distance to the light particle as {ct'} (note the {c} is the same) and would also calculate its distance from coordinates as {\sqrt{x'^2 + y'^2 + z'^2}}. They would have an expression similar to the one above.

\displaystyle  \begin{array}{rcl}  ct' = \sqrt{x'^2+y'^2+z'^2} \implies c^2t'^2 = x'^2+y'^2+z'^2\\ \implies c^2t'^2 - (x'^2+y'^2+z'^2) = 0 \end{array}

If both these terms are equal to {0} then we set them equal to each other.

\displaystyle  c^2t^2 - (x^2+y^2+z^2) = c^2t'^2 - (x'^2+y'^2+z'^2) \ \ \ \ \ (1)

This gives a relationship between your perspective in the car and your friends’ perspective on the footpath.

Let’s use this relationship more generally. Say you are moving away from your friend in the {x} direction with velocity {v} and your friend is not moving. You look back and measure your friends coordinates, in your space, as {x=vt, y=0, z=0}. From your friends perspective, in their coordinates, they are not moving so they record {x'=y'=z'=0}. Putting these conditions into Equation 1 gives:

\displaystyle  \begin{array}{rcl}  c^2t^2 - (v^2t^2)= c^2t'^2 \implies t^2(1-v^2/c^2) = t'^2\\ \implies t \sqrt{(1-v^2/c^2)} = t' \end{array}

This is an expression for time in your friends reference frame as you see them move away from you. Say you are traveling at 0.50 x speed of light. This means {v/c=0.50}. You see your friend traveling away from you at this speed. If you observe them for 10 seconds in your time then in their time {t'} is the number of seconds that has passed.

\displaystyle  \begin{array}{rcl}  t' = 10 \sqrt{1-0.50^2} = 10 \times 0.866 = 8.66 \; \text{seconds} \end{array}

Your friends time passes more slowly than yours! Yes, they age less as they move away from you. All this from the rule of triangles that you learned at school and a statement that there is a universal speed constant between all inertial frames, the speed of light. It only violates our intuition because we travel at miniscule fractions of the speed of light as animals.

Time is relative!

If you have the appetite for more rigorous proof to the general Special Relativity equations known as Lorentz Transformations, have a look at this article : Lorentz Transformations in Special Relativity

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