# Time dilation and Lorentz transformation

I have trouble understanding the Lorentz transformation to proof the dilation of time.

If I use:

$$dt ^{'} = \frac{dt}{\sqrt{(1-(v^2/c^2))}}$$

Understanding that $S^{'}$ is the moving frame of reference.

We can see that the square root in the denominator is smaller when $v \rightarrow c$ and we get that if $v \rightarrow c$ then $dt' \rightarrow \inf$... something different to the expected result that time goes slower in S'.

What am I doing wrong?

Whenever you're working with the Lorentz transformations the best way to proceed is to define the spacetime points of interest in your starting frame then transform to the other frame and see what happens.

In this case we'll assume we are at rest on the Earth watching a rocket zoom past at velocity $v$. Start in the rocket's rest frame and assume that in that frame there is a clock that ticks once at time zero then again at $t=T$. So in the rocket frame the two ticks are are the points $(0,0)$ and $(T,0)$.

In the rocket's frame the Earth is zomming past at velocity $-v$. We'll assume the rocket and Earth origins coincide at the moment of the first tick, so we just hav to work out when the second tick happens in the Earth frame. The Lorentz transformation for time is:

$$t_\text{Earth} = \gamma\left(t_\text{rocket} - \frac{vx}{c^2}\right)$$

The second tick happens at $(t=T,x=0)$ and putting these values into the equation we get:

$$t_\text{Earth} = \gamma T$$

For any speed greater than zero $\gamma \gt 1$ so the time between the two ticks in the Earth frame is greater than the time between the ticks in the rocket frame. Conversely, for every $n$ seconds that pass on the Earth only $n/\gamma$ seconds pass on the rocket.

• So, the observer on Earth sees the clock on the rocket running slower. But an observer on the coasting rocket thinks it is the Earth that has the velocity, and he sees the clock on Earth running slower. How can both clocks run slower than each other? This is known as the Twin Paradox, and is only resolved when the rocket turns around and returns to Earth so the two clocks can be placed at rest next to each other. The rocket clock will then show fewer seconds than the Earth clock. – Gary Godfrey Feb 26 '16 at 9:18
• There are of course a few twin paradox related posts on the site already - just the 271 of them and counting :-) – John Rennie Feb 26 '16 at 9:22
• The OP was puzzled by his expectation that clocks run slower in the particular boosted frame S'. The Twin Paradox illustrates that this idea is not a good one. Instead, each twin (S and S') sees the other twin's clock running slower. This is why I expanded your answer by dredging up the paradox again. – Gary Godfrey Feb 26 '16 at 11:45
• Gary Godfrey - Based on Lorentz transformation equation, if rocket was 300,000 km long and velocity was 260,000 km/s, and clocks were located at opposite ends, clock at rear would be ahead of clock at front by 0.866 sec, and both clocks would tick at half speed of clocks on Earth. Spaceship also contracts to a 150,000 km spatial length. If these clocks and the spaceships length are used as measurement instruments to measure the rate at which an Earth clock ticks, as the rocket passes by this Earth clock, it appears as though the Earth clock is ticking at half speed, even though it is not. – Sean Feb 26 '16 at 23:55
• OK, so that works one way - but not the other. Say, we were starting with the Earth's frame first; then, x would be vt for Earth, since the clock has moved by the amount vt along with the ship. That would mean $t_\text{rocket} =γt_\text{Earth}(1-\frac{v^2}{c^2}). That is clearly different from the standard time dilation formula – Max Jun 7 '17 at 21:35 I think there are two confusions in your question. 1) Time does not go slower in S' or in S. This is a wrong concept. Instead, if the clock is boosted wrt you, the clock appears to be going slower. 2) You have written the Lorentz transformation for boosting the S coordinate system to become S', and the clock remains at rest. This is known as the passive point of view. The S coordinate system records dt seconds between the space-time points of two ticks. As the S' coordinate system rushes away from the clock, the S' coordinates show dt' between the two clock ticks. The clock appears to be going slower (dt'>dt). You are surprised by this result because you expect "time goes slower in S' ". If you boosted a second identical clock so that it was at rest in S', would time be going slower in S or S'? In the active point of view the clock is boosted, and there is only the S coordinate system which remains at rest. Before boosting the clock, the S frame records dt between the clock ticks. After boosting the clock, the same S frame records dt' between the clock ticks. The clock appears to be going slower (dt'>dt). In the active point of view, there is no second frame S' for time to go slower in. Time does go slower in$S^{'}$. As$v$increases$dt^{'}$also increases. Which is time dilation. Time intervals increase in the moving frame as is the expected result. I have trouble understanding the Lorentz transformation to proof the dilation of time. Let's use finite differences instead and, further, the entire expression for$\Delta t'$from the Lorentz transformation $$\Delta t' = \gamma \left(\Delta t - \frac{v}{c^2}\Delta x \right) = \frac{\Delta t - \frac{v}{c^2}\Delta x}{\sqrt{1 - \frac{v^2}{c^2}}}$$ where$v$is the relative speed of the primed and unprimed systems. Now, assume$\Delta x = 0$which means that the two events are co-located in the unprimed system. So, for example, this would be the case for a clock at rest in the unprimed system. It follows that$\Delta t$is, in this case, the elapsed time according to a clock at rest in the unprimed system. Now, this clock at rest in the unprimed system has speed$v$in the primed system thus, •$\Delta t$is the elapsed time according to a clock moving with speed$v$in the primed system. Then, according to the equation above $$\Delta t' = \frac{\Delta t - \frac{v}{c^2}\cdot 0}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}}$$$\Delta t$is smaller than the elapsed time according to clocks at rest in the primed system. Once again,$\Delta t$is the elapsed time according to a clock moving with speed$v\$ in the primed system and, according to clocks at rest in the primed system, this elapsed time is less than the elapsed time in the primed system.

In other words, moving clocks run slower than clocks at rest. This is time dilation (due to uniform relative motion).