Problem with understanding time dilation (moving clocks can run faster?)

Question

I ran into a serious problem with the Lorentz transformation and time dilation. In the standard configuration you have one observer S and another one S' with their x-axis aligned. I assume S to be at rest and S' to be moving in direction of the x-axis at a speed v = 0.5*c (the relativistic gamma is then g = 1.15). Each observer has a clock and they meet when both clocks show 0 s.

Now suppose a firecracker goes off at the origin of S. This happens at a time when the clock of S reads 5 s. So he assigns the event the coordinates x = 0 m and t = 5 s. I want to know what coordinates observer S' assigns the same event. According to the Lorentz transformation, these are x' = -862,500,000 m and t' = 5.75 s.

(Just to check, I inserted x' = -862,500,000 m and t' = 5.75 s into the inverse Lorentz transformation and got x = 0 m and t = 5 s as expected)

So far, so good. But I'm having trouble interpreting this. So observer S says 5 s passed between their meeting and the explosion of the firecracker, observer S' says 5.75 s passed between their meeting and the explosion of the firecracker. That's also fine. But this means that S says the clock of S' is ticking faster while S' says the clock of S is running slower. Shouldn't both say that the clock of the other is ticking at a slower rate? Where is the problem in my logic, what do I misunderstand?

I initially expected that when insert x = 0 m and t = 5 s into the Lorentz transformation, I get t' < t (clock runs slower), but this doesn't happen!

Would be fantastic if somebody could help me here. It's a very exciting topic, but I feel like I have hit a dead end. No matter how I try to resolve it, I always get the same problem. One observer sees time dilation, the other time "acceleration", but I know that it should always be time dilation (moving clocks run slower, the mantra of SR).

Answer 1

"So far, so good. But I'm having trouble interpreting this. So observer S says 5 s passed between their meeting and the explosion of the firecracker, observer S' says 5.75 s passed between their meeting and the explosion of the firecracker. That's also fine. But this means that S says the clock of S' is ticking faster while S' says the clock of S is running slower. Shouldn't both say that the clock of the other is ticking at a slower rate? Where is the problem in my logic, what do I misunderstand?"

You are forgetting to take into account the relativity of simultaneity, which is relevant because each observer is supposed to assign time to events using readings on a network of synchronized clocks at rest at different locations along their own rulers (the clocks synchronized using the Einstein synchronization convention), so that each event's time can be judged by a clock that was right next to the event when it happened. That way you don't have to worry about light delays--for example, if I see light from an explosion 5 light-years away in 2005, and light from an explosion 10 light-years away in 2010, I can look at the clock in my system that was next to each event when it happened, and see that each explosion happened next to a clock that read t=2000 at the moment it happened, so I judge that these events happened simultaneously in my frame despite the fact that I saw them at different times. This page has an illustration of such a lattice of rulers and clocks, which give physical meaning to the position and time coordinates assigned by a given inertial frame:

In your example, there are two events that the observer in S' is assigning coordinates to--the first is the event of the clock belonging to frame S reading t=0, the second is the event of the firecracker going off. S' has a clock #1 at position x'=0 that was next to the first event when it happened, and that clock read t'=0 at that moment; and S' has a separate clock #2 at position x'=-865,426,282 m that was next to the second event when it happened, and that clock read t'=5.77 s (my numbers are slightly different than yours, it looks like you had some roundoff error). The key to understand why this does not contradict the fact that S should see all of the clocks in S' running slow is that from the perspective of frame S, these two clocks in S' are not synchronized, due to the relativity of simultaneity. If the two clocks are a distance L apart in S' and synchronized in that frame, then from the perspective of frame S which judges S' to have velocity v, the back clock's time will be ahead of the front clock's time by an amount Lv/c^2 at any given moment. So in this case, with L=865,426,282 m and v = 0.5c = 149,896,229 m/s, frame S will say that at any given moment, clock #2's time is ahead of clock #1's time by the amount (865,426,282)*(149,896,229)/(299,792,458)^2 = 1.4385 seconds. So at t=0 in frame S, when clock #1 reads t'= 0 s, clock #2 reads t'= 1.4434 s. Then 5 seconds later in frame S, each of the clocks in S' has only advanced forward by 4.3301 s, because time dilation says that the clocks in S' must be running slow in frame S. But since clock #2 already had a head start of 1.4434 s, at t = 5 in frame S, clock #2 will read 1.4434 + 4.3301 = 5.7735 s, and that's the moment at which the firecracker goes off next to it.

Answer 2

In the frame $S$ the two ticks of the clock, at $t = 0$ and $t = 5$, happen at the same place. But in frame $S'$ they happen in different places. So you cannot just take the two points in $S'$, treat them as ticks of a clock and transform back to $S$ to calculate the time dilation. If you want to calculate the time dilation going from $S'$ to $S$ you need to take a clock that is stationary in $S'$.

To illustrate this consider the following:

Take your clock stationary at the origin of $S$ as before, so the two ticks are at the spacetime points $(0, 0)$ and $(5, 0)$. But now have a second firecraker at $t = 5$ but at a distance $4c$ from the origin in $S$ i.e. the spacetime point for the second firecracker is $(5, 4c)$. In $S$ both firecrackers go off when the clock read $5$ seconds. But if we Lorentz transform the two points $(5, 0)$ and $(5, 4c)$ into $S'$ we get:

$$ (5, 0) \rightarrow (\gamma 5, \gamma -5v) $$

$$ (5, 4c) \rightarrow (\gamma (5 - \frac{4vc}{c^2}), \gamma (4c - 5v)) $$

So though the two firecrackers go off simultaneously in $S$ they do not go off simultaneously in $S'$. So should you calculate the time dilation using the firecracker at the origin or the one at $x = 4c$? Obviously you use the one at the origin because you need the clock to be stationary.

The point of this is that when you take the two transformed points in $S'$ and transform them back to $S$ you are effectively trying to calculate the time dilation of a clock that is moving in $S'$. the initial tick and the final tick don't happen at the same place.

Answer 3

Time dilation effect applies when an observer moving with respect to a pair of events (S') measures the time interval with the clock resting in his reference frame, and the events are at the same location in the other rest frame (S). The S' observer then compares his time (5.77 s) with the clock which is at rest with respect to both events (the clock in S). S' will have more time elapsed on the clock with the inference that "the clock in S must be slow," the popularized (but misleading) idea of time dilation.

The clock in S' is moving with respect to the two events, so S will not observe a time dilation effect for the S' clock measuring those events.

Answer 4

The mantra 'moving clocks run slow' is one of the biggest causes of misunderstanding of SR. The time dilation formula in SR applies only when you have two events that occur in one place in one inertial frame and in two places in another- in that case, the time interval between the two events is always shorter in the frame in which the events are in the same place.

So in the example you site, the event t=0 and the event of the firecracker going off are in the same place in S and in two different places in S', so the time interval will be longer in S'.

Note also, that the clocks do not 'run slow'. It is the actual period of time that is shorter in one frame than in the other, and any accurate clock will properly measure the shorter time interval as such.

Answer 5

Time dilation happen to a moving frame in which event takes place. So event in given example took place in $S$, but observer in $S'$ seen $S$ moving in opposite direction of its actual motion. So clock of $S$ appears slow to clock in $S'$. If similar event happened in $S'$, then its clock appear slow to $S$. The frame in which event takes place or measurement has done is called as proper frame. This example also shows that time dilation and length contraction are apparent and not actual or real because clock of stationary frame seems slower and not of moving one.

Answer 6

Geometric representation of an event that happens in M, $x_{M}=0$ and $ct=5s$​​ (explosion of a firecracker) : the wave only arrives at the observer at rest 5 seconds after the meeting of the two frames of reference

• Lorentz vector transformations* (LVT) are $$\begin{cases}t'=\gamma (t- \frac {\vec{r}. v\,\vec {i} }{c^{2}})\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(a)\\\\ \vec{r'} =\vec {r} +(\gamma -1)(\vec {r} \cdot \vec {i} )\vec {i} -\gamma tv \vec {i}\;\;\;\;\;\;(b)\tag{1.1}\end{cases}$$

where $$\begin{cases} \vec{r} =\vec {r} _{\perp }+\vec{r} _{||}=\vec{y}+\vec{x}=\vec{y}+\vec{0}\\ \vec {r'} =\vec{r'} _{\perp }+\vec{r'}_{||}=\vec{y'}+\vec{x'}\\ \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\end{cases}$$ see

I) We suppose that $\vec{r}$ is orthogonal to the axis of direction vector $\vec{i}=\vec{i'}=\frac{\vec{v}}{v}$, equation (a) becomes: $$t'=\gamma[t-v(\vec{x}+\vec{y}).\vec{i}]$$ by supposing the projection of $\vec{r}$ on the axis of direction vector $\vec{i}$ is $\vec{x}=\vec{0}$ et $\vec{y}.\vec{i}=0$, Which give
$$t'=\gamma t\tag{1.2}$$ and equation (b) is written $$\vec{r'}=\vec{x'} + \vec{y'}=\vec{x}+ \vec{y}+(\gamma -1)[(\vec{x}+\vec{y}).\vec{i}] \vec{i}-\gamma vt\vec{i}\;\;\;\;\;\;(b)$$

with:$$\vec{y}\,'= \vec{y}$$

which results with (1.2)
$$\vec{x}'= - vt'\,\vec{i}=-\gamma vt\,\vec{i}=-\gamma vt\,\vec{i}\,' \;\;\;\;\;\;(b)\tag{1.3}$$

we have $$\vec{r'}=\vec{x'}+ \vec{y'}=\vec{x}'+ \vec{y}$$ $$r'^{2}=y^{2}+x'^{2}=y^{2}+v^{2}t'^{2}$$

for light, the previous equation gives

$$c^{2}t'^{2}=c^{2}t^{2}+v^{2}t'^{2}$$
$$t'=\gamma t $$ The two equations of the system (1.1) give the same result.

Pytagorean theorem implies that $ct' > ct$

*https://en.wikipedia.org/wiki/Lorentz_transformation