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  • Dheeraj Singh

Time Dilation: The Mesmerising Concepts of Space

Edited by Sky Ye.



Time dilation: a concept that perplexes even the best astronomers and scientists out there. Imagine you’re standing facing perpendicular to a passing train. The train maintains a fixed velocity as it moves past you. Your friend has chosen to be on the train, and they decide to shine a light vertically upward. The last part of this scenario, and perhaps the least realistic part, is to imagine a giant mirror that stays above the train following it as it moves along the tracks. Your friend, in the train, will see their light reach the mirror almost instantly, due to the rapid rate that photons (light particles) travel at. Down comes the light as it reflects off the mirror in the exact same manner that it was emitted in the first place (refer to L in the diagram below)


But what about you? What do you see?


Assuming you’re standing in the exact same position and the train is still moving from left to right, you’ll see a diagonal slope as the light goes up, and thus, will reflect to the normal (L) at the same angle, as the law of reflection states the angle of incidence is equal to the angle of reflection (θr = θi)¹. Although light travels extremely fast, it still takes time to travel. Hence when the flashlight was first turned on, it was at any given point A1, and now it’s at a point that’s slightly further away from you, at point A2 (represented as ½vΔt’ on the diagram below).


NOTE: the blue line with arrows represents the light’s path and the right angle triangle in green is a tool, not a path.



Figure 2: As seen in this diagram, an astronaut would experience a much shorter time as compared to the observer on earth - obtained from https://pressbooks.bccampus.ca/collegephysics/chapter/simultaneity-and-time-dilation/

You may already know that the speed of light, c, is always constant, at 3 . 10⁸ m/s, and that the Pythagorean theorem is a²+b² =c², but have you ever wondered how these two formulae are associated with time dilation? Using the Pythagorean theorem, we can evidently conclude that c² must cover a longer distance than a² or b². Continue to assume you’re in the same position; perpendicular to the train while your friend remains inside. If the speed remains constant but the distance travelled changes, the time it takes for the light to travel is different in your reference frame as compared to your friends.


As seen in the diagram, there is scenario 1 on the left, which is your friend’s reference frame, and scenario 2 on the right, displaying your reference frame. As s = d/t, Δt (a simple formula explained later into the article) has to equal L/c, where L represents the distance travelled, and c denotes the speed of light.


A famous formula comes into play when trying to calculate time dilation, namely the Lorentz factor equation. The formula goes as follows:

Where (using the example scenario above):


Δt’ = time dilation that the observer experiences

Δt = time taken in the reference frame of the person emitting the light in the train

v = speed of the moving person in the train

c = speed of light


Deriving this equation will allow us to really connect the dots. For clarity, I will simplify a few elements of the derivation and give some variables a single letter rather than a series of symbols or letters.


To start, we have the height of your friend in the train, h, and the height of the train itself, t. If we were to subtract these two values (t – h), we would get the distance between the top of the person’s head and the train’s roof. We can call this distance x.


We can then apply this to the simple formula that links speed, distance and time: s = d/t. The speed that is relevant here is the speed of light, c and the distance, as we have previously concluded, is x. By substituting these values into s = d/t, we get c = x/t. We can solve for distance, x, and isolate to get x = ct.


Now we can utilise the Pythagorean theorem to understand where our distance actually comes from. If x²+b² =c² (a² in the typical formula is x²), we need to solve for the hypotenuse because time dilation occurs as a result of the increased distance.

We know that x = ct thus we can rewrite this as:

Now we have to break down b, which is the other side of a right-angled triangle. Again, s = d/t, thus distance is equal to speed multiplied by time (d = st). We can call this other speed v, which is the speed of the train, and we can call this other time t’ (said as “t prime”). The time of t’ will be the time taken on the horizontal plane.


We now obtain a rewritten version of the Pythagorean theorem, in terms of speed, distance and time:

Plug this back into s = d/t:

Square both sides to “remove” the square root on top:

Utilise the power of mathematical manipulation as we inch ever so slightly closer to a familiar formula:

Our penultimate step is to divide the top and bottom by c²:

And for our last step: square rooting both sides of the equation.

[Note that (velocity) v ≤ 3 . 10⁸ m/s; you cannot have a value that exceeds the speed of light]


Just like that, we have arrived at our initial equation for time dilation.


Now that we’ve derived the formula and understood why it’s true, what does this actually mean?


To put this into perspective, let’s take a simple example. Following my previous scenario, say the time measured in your friend's reference frame was 5 seconds, and the speed of the train was 60km/h, which is roughly equal to 16.66 m/s.

Although the result contradicts a lot of the information given prior to this calculation, it gives us an explanation for why time dilation is an extremely rare occurrence on Earth, which is why it is more apparent in space. 


Technically, time dilation in the given scenario would still occur, but the decimal after 5 would be an extremely small value, meaning that the value approaches so close to 5 that we can just ignore it. From this, we can deduce that the train will have to be moving at an extremely fast rate before we can observe noticeable time dilation. As distance vt’ increases, it would take a longer time for the observer to see the light travel. If the train was moving at 30,000,000 m/s, it would now take 5.0251 seconds in my reference frame according to the calculation below:

Let’s take a look at another example. Say, theoretically, a spaceship can travel at 70% of the speed of light, at 0.7c. We want to travel to Sirius, the brightest star in the night sky, in the constellation of Canis Major. The distance from Earth to Sirius is 8.7 light years. To make sure we have an accurate calculation, we need to ensure that 8.7 light years is the length contracted distance, using the formula below:

To clarify this calculation, length contraction is the phenomenon where a moving object’s length is shortened as compared to when the object is at rest.


Figure 3: A diagram showing how an object’s length at rest differs from it’s length when it is in motion, especially when it is moving fast - obtained from https://commons.wikimedia.org/wiki/File:Length-contraction.svg

Using s = d/t, we can conclude that t = d/s, thus it would take 279960344.6 seconds, which is roughly equal to 8.877 years. After reaching Sirius, the spaceship’s crew decides to return back to Earth, thus the distance is doubled, totalling to 17.754 years. This is the time that the crew experiences, but what about you?

This means that, on Earth, 24.86 years will pass before they return, but they only experience 17.754 years in space.


Interestingly enough, this same concept applies to gravity, and this process is called “gravitational time dilation.” As a matter of fact, gravity possesses the ability to slow down time, especially when an object moves nearer to a very strong gravitational force.


Figure 4: A picture showing how gravitational time dilation can occur - obtained from https://www.quora.com/Science-What-is-the-fabric-of-the-space-time-and-how-it-is-related-to-the-dark-matter

To visualise this, imagine a piece of paper lying horizontally while being secured between two tables with most of the paper filling in the gap between them. If we decide to put a metal sphere in the centre of the paper, as a result of gravity acting on the sphere’s mass, the paper will curve downwards. Similar to this, spacetime does the same thing, which usually means that if an object is denser and heavier, it will have a stronger gravitational force. Thus, it is able to bend spacetime even more, which means that time would gradually slow down as the observer sees an object approaching the gravitational force.


For instance, if you were falling into a black hole and at a safe distance away, I was observing it, I would see it happening in the stretched out time it takes for it to happen. Comparatively, you would experience the “actual time” when you have reached past the event horizon and started to fall into the black hole. In this example we assume that the human body is able to tolerate the force, meaning your body won’t disintegrate within a matter of seconds.


Perhaps when we say that our week went by so slowly; that it “felt like forever”², a small part of that statement is actually true. Our perception of time and gravity is constantly changing as we discover new areas of astrophysics, which is what makes this field of study so mesmerising and absorbing.


 

Reference:

  1. 1.2 The Law of Reflection - University Physics Volume 3. (2016, September 29). OpenStax. Retrieved October 15, 2022, from https://openstax.org/books/university-physics-volume-3/pages/1-2-the-law-of-reflection

  2. (2013, June 24). Does time go faster at the top of a building compared to the bottom? Retrieved October 15, 2022, from https://www.wtamu.edu/~cbaird/sq/2013/06/24/does-time-go-faster-at-the-top-of-a-building-compared-to-the-bottom/

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