Special Relativity II: Time Dilation Part 2

Hello, and welcome back to MPC! A few days ago, we discussed time dilation from a conceptual point of view. Today, we will look at a more mathematical approach to time dilation. If mathematics is not your best subject, I encourage you to at least try reading this post as it will provide you with insight into why and when time dilation occurs. Let’s get started!

Imagine a scenario in which a person, person 1, is standing on Earth, and another person, person 0, is flying through the sky on a rocket ship. For simplicity, let’s say that person 0 is moving at a constant speed, v, in his rocket ship (note that this means that person 0 is not accelerating). Let’s also pretend that person 0 has a particle of light that is bouncing up and down inside of his rocket ship. Of course, this particle of light is moving at the speed of light, c. Here is a diagram of the situation:

TD2_1

Figure 1: A diagram of the scenario we will be discussing

Now, let’s think about how long it will take for that particle of light to travel from the bottom of the rocket ship to the top of the rocket ship once. First, let’s look at the particle of light from person 0’s perspective:

TD2_2

Figure 2: The movement of the particle of light from person 0’s perspective

Of course, person 0 just sees the particle move from the bottom of the rocket ship to the top of the rocket ship. The particle of light is simply traveling a distance of L at a speed of c. Therefore, the time it takes for the particle to travel from the bottom of the rocket ship to the top of the rocket ship, from person 0’s perspective, is:

TD2_E1.png

It should be noted that this equation was derived using the definition of velocity (velocity is distance divided by time).

Now, let’s talk about the particle’s motion from the perspective of person 1. It is the same right? Well, not exactly. We have to remember that person 1 is watching the rocket ship move at a speed of v. Because of this, to person 1, the particle’s motion looks like the following:

TD2_3

Figure 3: The movement of the particle of light from person 1’s perspective

It must be emphasized that, now, the vertical component of the particle of light’s motion (how quickly it is moving up if you were to completely ignore horizontal motion) is not c — the diagonal motion is c. Why is this the case? Recall that the speed of light is constant, so if the vertical component of the particle’s motion were c (as it currently is to person 0), the particle’s diagonal motion would be greater than c according to person 1 (by the Pythagorean theorem) — a direct violation of the principles of relativity!

You may notice that this scenario is very similar to our billiards example in the previous post. Indeed, it is essentially the same! Let’s break it down! In the billiards example, one thing that we noticed is that the ball traveling diagonally traveled a greater distance than the ball traveling in a straight line. This is, once again, the case here. Let’s “zoom in” on this situation:

TD2_4.png

Figure 4: A magnified view of the movement of the particle of light from person 1’s perspective

It should be noted that in Figure 4, the letters represent distances, whereas in Figure 3, they represented speeds. In any case, this picture should make sense. The particle of light is traveling a vertical distance of L. Furthermore, it is traveling a distance of vt1 horizontally (by the definition of velocity), where v is the velocity of the rocket ship and t1 is how long it takes for the particle of light to move from the bottom of the rocket ship to the top of the rocket ship according to person 1.

It is very important that we have made a distinction between t0 and t1As discussed in last week’s post, person 0 may claim that it takes 5 seconds for an event (such as the particle traveling to the top of the rocket ship) to happen while person 1 may claim that it only takes 1 second for that same event to happen. Because t0 and t1 may be different, we must represent them with different variables.

So, according to person 1, what distance, d, is the particle of light traveling to reach the top of the rocket ship? By the Pythagorean theorem:

TD2_E2.png

Using the definition of velocity, t1 is:

TD2_E3

And there we go! Already, we can see that person 1’s measurement of the amount of time that it will take for the particle of light to reach the top of the rocket ship will be greater than person 0’s measurement (we know this because the numerator of equation 3 will always be greater than the numerator of equation 1).

Now, let’s make things more interesting: how much longer will person 1 say it takes for the particle of light to reach the top of the rocket ship than person 0 says? We can calculate this. To see how, let’s rearrange equation 3:

TD2_E4.png

Right now, we have two “problems.” First of all, we are trying to relate t0 and t1, so it would certainly be useful if our equation contained both of those variables. Also, the L in the equation is rather bothersome. Could the relationship between person 0 and person 1’s time measurements really depend on the height of the rocket ship? If we want to compare each person’s times, do we actually have to measure the height of the rocket ship? We may just have to. For now, though, it is in our best interest to see if we can get rid of that L. It may be possible, it may not be possible.

By rearranging equation 1 we get:

TD2_E5

If we substitute equation 5 into equation 4, we get:

TD2_E6.png

Look at that! We have killed two birds with one stone: our equation contains both t0 and t1 and no longer contains L. This was by no means guaranteed to happen. Nature has just been too kind to us!

Now, our next goal should be to either write t0 in terms of t1 or t1 in terms of t0. If we can achieve this, then it will be much simpler to calculate one of the times given the other. Let’s try using some algebra:

TD2_E6.png

TD2_E7.png

TD2_E8

TD2_E9.png

TD2_E10.png

TD2_E11.png

TD2_E12.png

TD2_E13.png

TD2_E14.png

Viola! We have done it! We now have t1 in terms of t0. So, if we were given t0, v, and c (which is a well-known constant), we would be able to calculate t1. We can also do this in the opposite direction — if we were given t1, v, and c, it would not be too difficult to calculate t0.

Well, we are almost done. Many physicists think that equation 7 is too ugly, so they have done some extra work to make it neater:

TD2_E14

TD2_E15

TD2_E16.png

TD2_E17

Are we done now? Almost. Physicists also hate that ugly square root in the denominator, so they have defined the following variable:

TD2_E18.png

That letter that looks like a y is actually the Greek letter gamma. When gamma is defined as it is in equation 9, it is known as the relativistic factor. Using the relativistic factor in equation 8, we get the following:

TD2_E19.png

That looks pretty neat to me! With that, we are done. We have successfully (and neatly) written t1 in terms of t0. Now, you may be wondering: Who cares? So what if one person says that it only takes 1 seconds for the particle of light to reach the top of the rocket ship while someone else says that it takes 5 seconds? You’re right, that information is pretty worthless. However, we can use it to uncover other, more interesting phenomena about our world. That will be the focus of next week’s post: the consequences of time dilation. See you then!

P.S.: This post may have been a lot to digest. I recommend that you think it over for the next few days. If you have any questions, feel free to post them in the comments! Also, if you would like this proof in video format, you can find a great version over here.

(featured image: https://cdn.shutterstock.com/shutterstock/videos/589834/thumb/1.jpg)

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5 thoughts on “Special Relativity II: Time Dilation Part 2

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