# A beginner's question on time dilation [closed]

I have just started understanding relativity so the question might look absurd.

I know that the faster we move through space , the slower we move through time. But I haven't understood it fully.

I think of this situation: I am travelling in a direction with a speed of c/2 and a beam of light is following me . The initial separation is quite large. Now from classical mechanics the velocity of light with respect to me is c/2.

But it should be c.

      c/2 = s/t

c=s/(t/2)


But this means that time is moving faster for me as compared to a stationary observer and not slower.

Please tell me where I went wrong. Thank youo

## closed as off-topic by WillO, Jon Custer, Kyle Kanos, Phonon, YashasMay 31 at 16:25

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – WillO, Jon Custer, Kyle Kanos, Phonon, Yashas
If this question can be reworded to fit the rules in the help center, please edit the question.

• Firstly, you have not taken into account the fact that you are moving. This brings in ideas of length contraction. Also, the relative speed compared to that of light is always c. You have also halved the time, so time is moving slower for you relative to an observer, not faster. – Physics May 25 at 8:32
• You have not taken into account time dilation of beam of light's internal clock. Let's label your frame S and light beam's frame $\bar {S}$. Then $\bar {t} = \gamma t$, where $t$ is the beam internal clock (proper time) and $\bar {t}$ is your time. Note that t is shorter than $\bar {t}$ by the factor $\gamma$ – JD_PM May 25 at 8:57

The beam of light would easily overtake you,but it would be red-shifted. Time aboard your spaceship would seem to you to be perfectly normal,but if you could contact an external observer who sees himself as being at rest,his time would appear to be slowed. Viewed from the observer's frame of reference,it is your time which which would appear slow. According to Einstein,movement is relative & there is no such thing as absolute motion or absolute rest,only motion or rest relative to some object or objects which form a frame of reference. The motion of the railway station past the train is just as relevant as the motion of the train past the railway station. Physicists talk glibly about spaceships moving at high fractions of c,but I don't see any way this can be achieved & in my opinion it never will be.

I am deriving this using classical mechanics which what you did.

Remember this is not going to be correct (in numerics and problem solving) due to the use of classical mechanics but at least give you a sense of the idea.

Now in your frame of reference: $$s = \frac{c}t$$ let us see the ray of light from two point of references:

(a) ground frame

let the time measurement in the ground frame be $$\tau$$ Therefore: $$s = c\cdot \tau$$ $$\frac sc = \tau$$

let the time measurement in the your frame be $$t$$ Therefore: $$s = \frac c2\cdot t$$ $$\frac {2\cdot s}c = t$$ On taking the ratio of τ and t (to compare the quantities) $$\frac{\tau}t = \frac{\frac sc}{\frac {2\cdot s}c}$$ $$\frac{\tau}t = \frac{1}2$$ $$2\cdot \tau = t$$ This is same as your argument which you gave in the question.
Using actual relativity; $$t = \frac{2\cdot \tau}{\sqrt 3}$$ where t is your time frame and τ is ground time frame.
Which turns out to be $$t \approx 1.1547005384\cdot \tau$$