I took a stab at deriving the time dilation formula for velocity. This is entirely my own approach though I suspect it is nothing new/special in terms of how it is derived. Just wanted to have some fun and see if i could figure it out for myself. So I knew it was all based on the idea that the speed of light is a constant in any frame of reference. So I drew this diagram showing two frames of reference.

time-dilation diagram

So what I did was I tried to calculate the amount of time it would take for light to travel between the two people in the spaceship (A and B). The distance between them is $D_x$. That's the distance light travels in the rest frame. Then I calculated the distance it would travel as an outside observer would see it (accounting for the velocity of the space ship). For that the distance was longer, but the speed of light the same. The ratio between the difference between the apparent time from the two observers would be the time dilation. Here is my math.

$$ T_0 = \frac{D_x}{C} $$

$$ D_y = V_y \cdot T_0 $$

$$ D_y = V_y \cdot \frac{D_x}{C} $$

$$ D_1 = \sqrt{{D_{x}}^2 + {D_{y}}^2} $$

$$ D_1 = \sqrt{{D_{x}}^2 + {(V_y \cdot \frac{D_x}{C})}^2} $$

$$ T_1 = \frac{D_1}{C} $$

$$ T_1 = \frac{\sqrt{{D_{x}}^2 + {(V_y \cdot \frac{D_x}{C})}^2}}{C} $$

$$ T_1 = \frac{\sqrt{{D_{x}}^2 + {V_y}^2 \cdot \frac{{D_x}^2}{C^2} }}{C} $$

$$ T_1 = \frac{\sqrt{{D_{x}}^2 + {V_y}^2 \cdot \frac{1}{C^2} \cdot {D_x}^2 }}{C} $$

$$ T_1 = \frac{D_x \cdot \sqrt{1 + {V_y}^2 \cdot \frac{1}{C^2} }}{C} $$

$$ T_1 = \frac{D_x \cdot \sqrt{1 + \frac{{V_y}^2}{C^2} }}{C} $$

$$ \Delta T = \frac{T_0}{T_1} $$

$$ \Delta T = \frac{\frac{D_x}{C}}{\frac{D_x \cdot \sqrt{1 + \frac{{V_y}^2}{C^2} }}{C}} $$

$$ \Delta T = \frac{D_x}{C} \cdot \frac{C}{D_x \cdot \sqrt{1 + \frac{{V_y}^2}{C^2} }} $$

$$ \Delta T = \frac{C D_x}{C D_x \cdot \sqrt{1 + \frac{{V_y}^2}{C^2} }} $$

$$ \Delta T = \frac{1}{\sqrt{1 + \frac{{V_y}^2}{C^2} }} $$

So just some notes, subscript 0 here, such as $T_0$ refers to the time in the rest frame, and subscript 1 here, such as $T_1$ represents the frame of reference where the ship is moving.

As you can see the final equation I get is almost exactly right but i seem to have resulted in a plus sign where there should be a minus sign. After going through my work I can't find the source of my error. What did I do wrong?


2 Answers 2


This approach is actually well-thought out and it's pretty much on point. The problem, though, occurs very early.

There are two inertial frames that we're considering. I'll refer to one as observer frame and the other one as the ship frame.

$$T_{0} = \frac{D_{x}}{C}$$

So you define $T_{0}$ to be the time for light to go between the two people with regards to the ship frame.

$$D_{y} = V_{y} \cdot T_{0}$$

If I am not mistaken, I would guess that the reasoning for this was as follows: given the ship frame, in the time it took for light to go between the two people, they flew a distance $D_{y}$.

Unfortunately, $D_{y}$ is defined as the vertical distance with respect to the observer frame. This does not hold for the ship frame.

To explain this a little further, in relativity we let go of certain assumptions. One of the assumptions is that time is universal in all frames. The things is, though, that once you let go of that assumption, there are other consequences.

One of those consequences is that distance is not the same in all frames, just like time. More specifically, if a ship is moving in the $y$-axis, length parallel to the $y$-axis is not the same in all frames (although length parallel to the $x$-axis is okay).

Going back to the derivation, the approach is the right idea; you just have to avoid the mistake above. The distance $D_{x}$ applies to both frames, but $D_{y}$ applies to the observer frame only.

  • 1
    $\begingroup$ That makes perfect sense. The distance the ship travels along Y is the time it takes the light to travel but not in the time in the ship frame but the time in the observer frame. So my assumption there is wrong. Though it does surprise me that despite this mistake the answer I got was still so close to the correct answer. I guess im left wondering how I can actually fix this "proof" to be correct now without rewriting the whole thing (if possible). Regardless this answer looks valid to me. $\endgroup$ Commented May 11, 2017 at 1:15

I think your mistake may be in the first few lines. You set your distance in the Y direction equal to your velocity in the Y direction times the time interval as experienced by the people in the rest frame; shouldn't it be equal to the Y velocity in the moving frame times the time experienced in the moving frame? I may be entirely off-base here though. I did my own similar derivation and ended up with the right answer (the time dilation factor is the same as yours except there's a negative sign instead of a positive sign). My phone's just charging up now so I'll post an image of the derivation shortly (my phone's the only thing I can take a picture with, and I did the derivation on a piece of graphing paper). I'm really sorry if my writing is terrible, or if it's not entirely clear. My writings always been abysmal, in high school I just did everything on the computer since nobody could read my writing. Note; as a previous comment stated, the distance in the X direction is the same in both reference frames since there the velocity is purely in the Y-direction in the second frame. That's an assumption when I replace D squared by C squared times Time in Frame 1 (T1) squared. My quick and dirt derivation

  • $\begingroup$ Oh thanks, I'll look this over tomorrow, but it does indeed look very similar to what I was doing. $\endgroup$ Commented May 11, 2017 at 3:54
  • 1
    $\begingroup$ It is pretty similar. I'll try to make one with neater writing tomorrow and post it as well. $\endgroup$ Commented May 11, 2017 at 3:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.