# How do I know which observer is running the time faster or slower?

Ok, I'm not a physicist, so I don't know if my question is silly (probably yes), but there is something in special relativity that I can't understand and I would really like it to be clarified.

If two observers are experiencing different velocities, then they will experience time differently. Ok. But, since there is no absolute referential, ever, how can I tell which one is moving "away" from the other in order to know which one is experiencing time faster than the other? I'll try to be more specific. Let's get the classic experiment of the light clock where a beam of light is moving up and down between two mirrors, and counting time at each passage. Now let's get an exact copy of that mirror and put it side by side with the first one. Now let's start moving the second clock horizontally to the right, away from the first one. In order for the speed of light to be the same to all observers, the moving light clock should experience time slower than an observer attached to the first clock. Ok, got it. Very logical. But here's my doubt: if the entire universe is composed of just these two watches, how can I tell that the second mirror is the one that is moving away from the first to the right, and not the first mirror that is moving away from the second to the left? It's impossible to tell. So, how can I know which one of the clocks is the one "moving away" in order to define which one is experiencing time dilation?

I know the answer must be silly because I simply can't find it. Any help?

• You go wrong exactly here: "In order for the speed of light to be the same to all observers, the moving light clock should experience time slower than an observer attached to the first clock." The correct statement is: "In order for the speed of light to be the same to all observers, the stationary observer should measure the time interval between two ticks of the moving clock as greater than the time interval between two ticks of his own clock." ( I wanted to put "stationary" and "moving" in single quotes, but apparently single quotes don't work very well in comments.) – WillO Feb 4 '16 at 0:18
• A difference is always a difference between two things, it's not a difference between individual systems to some imaginary third. If you want to insert a third observer system, then you can still retrieve the relativistic addition of velocities, if you like. – CuriousOne Feb 4 '16 at 0:24
• Yes, but how can I tell which one is the stationary? If I can say the right mirror is moving to the right, why can't I just say the left mirror is moving to the left instead? How to really tell which one is stationary if my "universe" have just these two mirrors and nothing else? – Richard Gold Feb 4 '16 at 2:15
• There don't exist "stationary" or "moving" systems in an absolute sense. Two systems A and B are stationary or moving with respect to each other. If you would build your house in system A you have all rights to name it stationary and the other system B moving but its motion is with respect to your house in the stationary system A . – Frobenius Aug 4 '17 at 7:59
• @HolgerFiedler that is completely wrong. Where is the absolute reference frame in something like de Sitter space? Or, really, Minkowski space is a perfectly valid solution of Einstein's equation. Only in some solutions do you have anything like a global stationary frame or a gravitational potential. – Jerry Schirmer Aug 4 '17 at 20:44

If two observers are experiencing different velocities, then they will experience time differently.

I'm not sure what you mean by this, but I'm confident you mean something wrong. Everyone "experiences time" at exactly the rate of 1 second per second.

how can I tell which one is moving "away" from the other ... ?

Alice says she is standing still and Bob is moving rightward. Bob says he is standing still and Alice is moving leftward. These are two equally valid descriptions of the same reality, just as when Alice (facing west) says that the North Pole is to the right, while Bob (facing east) says that the North Pole is to the left.

how can I tell that the second mirror is the one that is moving away from the first to the right, and not the first mirror that is moving away from the second to the left?

Alice describes it one way, Bob describes it another way, and that's okay. There's no need for a referee here, and as you seem to understand, there's no such thing as the One True Description.

So, how can I know which one of the clocks is the one "moving away" in order to define which one is experiencing time dilation?

I don't know what you mean by "experiencing time dilation", but the answer to what you seem to be asking is that:

1) according to Alice's description of the Universe, Bob's clock is ticking slower than Alice's

2) according to Bob's description of the Universe, Alice's clock is ticking slower than Bob's

3) Both descriptions are correct (just as Bob and Alice are both correct when one says the North Pole lies to the left and one says the North Pole lies to the right). Once again, there is no One True Description.

This is the whole reason why we call it 'relativity'! There is no global unambiguous way to define an absolute velocity, so only the relative velocities matter. Everything else is just from a point of view.

When one says 'Bob moves at 10km/h and Alice is stationary' they are implicitly defining a reference frame. Usuallay in day to day life, we define our motion with respect to the earth, so Bob is moving at 10 km/h with respect to the earth. However, Bob'a point of view is equally valid. From Bob's point of view, he is stationary and the earth is moving under him at -10 km/h.

In physics we need to define what our motion is with respect to in order to avoid ambigious statements. Usually, it's obvious what our reference frame is, but in a universe with just a couple of watches, there's no obvious preferred frame so we have to be careful to stipulate 'from Alice's point of view, she is stationary and Bob moves at 10 km/h'

"if the entire universe is composed of just these two watches, how can I tell that the second mirror is the one that is moving away from the first to the right, and not the first mirror that is moving away from the second to the left? It's impossible to tell"

A point in geometry is a location. It has no size i.e. no width, no length and no depth. It makes no sense to say that two points (or two pointlike clocks) move relatively to each other. Clock moves in a reference frame of observer and dilates relatively to the reference frame (or set of physical reference points) of observer. Hence, observer needs at least two spatially separated clocks, let’s say C1 and C2. Observer synchronizes these clocks by light. If one way velocity of light is considered to be c, this is Einstein synchronization. Moving clock passes by clock C1, and these clocks compare readings in immediate vicinity. Then moving clock passes by C2 and they compare readings again.

Single clock measures shorter time interval, than two spatially separated clocks. Time in reference frame from the point of view of the single clock is running gamma times faster instead.

http://www.pstcc.edu/departments/natural_behavioral_sciences/Web%20Physics/Chapter039.htm "Two spatially separated clocks, A and B, record a greater time interval between two events than the proper time recorded by a single clock that moves from A to B and is present at both events."

All that goes straight from the Lorentz transformations.

$T = \frac {t'_{x'}+ \frac {v'} {c^2} x'} {\sqrt {1-( \frac {v} c)^2}}$ (1)

$T$ is clock readings that belongs to reference frame $K$, taken in point $x'$ at moment of time $t'_{x'}$ of reference frame $K'$, and $t'_{x'}$ reading of clocks that belongs to reference frame $K'$ in the point $x'$ of reference frame $K'$

How to interpret Lorentz transform for time?

Transformation demonstrates, that time $T$ of reference frame $K$ (in which it does not depend of $x$ coordinate or any other coordinate) is universal in reference frame $K$ and each point of this frame.

Now let's fix point $x'$, for example $x'=0$. In this case this transformation will look like that:

$T= \frac {t'_{0'}} {\sqrt {1-( \frac v c)^2}}$ (2)

$T$ is clock reading of reference frame $K$ taken in point $x'=0$ (in the origin $O'$ of reference frame $K'$), and $t'_{o'}$ is time in the reference frame $K'$, in particular in the origin $O'$.

We can take $\frac {dT} {dt'}$ when $x'$ is fixed and will get ${dT}/{dt'} = \frac 1 {\sqrt {1- \frac {v^2} {c^2}}}$

According to (2) it is not time $t'_{o'}$ which is showed by single clock in the point $O'$ runs slower, but time $T$ , which is "distributed" through all reference frame $K$ and taken in the origin $O'$ of reference frame $K'$ runs faster (relatively to time $t'_{o'}$ that is in the origin $O'$ of frame $K'$). Time dilation comes by means of transfromation of (2) into:

$t'_{o'}=T \sqrt {1- {\frac {V^2} {c^2}}}$

It is correct that $T>t'$ and $t'<T$. It is also true that $T'>t$ and $t<T'$. But that $t<t'$ and $t'<t$ from different points of view is nonsense.

Animation from Wikipedia

Let two systems $\mathrm{S}$ and $\mathrm{S'}$. Suppose that system $\mathrm{S'}$ is in uniform and translational motion with velocity $\mathbf{v}$ with respect to $\mathrm{S}$.

Now, there exists a false and misunderstanding meaning of time dilation adopted by many users who : (1) read popularized physics books (2) have total ignorance of the Lorentz Transformation (3) build, upon this misunderstanding, thought experiments that lead unavoidably to "paradoxes" and "contradictions".

This false and misunderstanding meaning of time dilation is summarized in the following PROPOSITION F :

PROPOSITION F (FALSE):

If any two events e1 and e2 take place in system $\mathrm{S'}$ apart by a time interval $$\Delta t' = t'_{2}-t'_{1} \tag{01}$$ then in system $\mathrm{S}$ these events take place apart by the dilated time interval $\Delta t = t_{2}-t_{1}$ where
$$\Delta t =\gamma_{v}\Delta t'>\Delta t'\, , \qquad \gamma_{v}=\dfrac{1 \vphantom{\frac12}}{\sqrt{1\!-\!\dfrac{v^{2}}{c^{2}}}}>1 \tag{02}$$

If you adopt PROPOSITION F as true then at once your first result is that all synchronized clocks in $\mathrm{S}$ (those counting the $\mathrm{S}-$time) run faster than the synchronized clocks in $\mathrm{S'}$ (those counting the $\mathrm{S'}-$time) and after that you puzzled with this thought : But the system $\mathrm{S}$ is in uniform and translational motion with velocity $\mathbf{v'}=-\mathbf{v}$ with respect to $\mathrm{S'}$ so according to PROPOSITION F
$$\Delta t' =\gamma_{v'}\Delta t>\Delta t\, ,\quad \text{since} \quad \gamma_{v'}=\gamma_{v}=\dfrac{1 \vphantom{\frac12}}{\sqrt{1\!-\!\dfrac{v^{2}}{c^{2}}}}>1 \tag{03}$$ which contradicts(02). Next you wonder "which clocks run faster: the $\mathrm{S}$es or the $\mathrm{S'}$es ???" a dilemma like "which came first: the chicken or the egg ???".

The true meaning of time dilation, a so so so simple consequence of the Lorentz Transformation is summarized in the following PROPOSITION T :

PROPOSITION T (TRUE):

If two events e1 and e2 take place ON THE SAME SPACE POINT(1) in system $\mathrm{S'}$ apart by a time interval $$\Delta t' = t'_{2}-t'_{1} \tag{04}$$ then in system $\mathrm{S}$ these events take place apart by the dilated time interval $\Delta t = t_{2}-t_{1}$ where
$$\Delta t =\gamma_{v}\Delta t'>\Delta t'\, , \qquad \gamma_{v}=\dfrac{1 \vphantom{\frac12}}{\sqrt{1\!-\!\dfrac{v^{2}}{c^{2}}}}>1 \tag{05}$$

Here there is no symmetry : If to the configuration of systems-events, $$\left\lbrace\mathrm{e1},\mathrm{e2}, \mathrm{S'},\gamma\left(\mathbf{v}\right),\mathrm{S} \right\rbrace \tag{06}$$ on which PROPOSITION T is applicable, you interchange the roles of the two systems $$\mathrm{S'}\Longleftrightarrow \mathrm{S} \tag{07}$$ then to the new configuration of systems-events $$\left\lbrace\mathrm{e1},\mathrm{e2}, \mathrm{S},\gamma\left(-\mathbf{v}\right),\mathrm{S'} \right\rbrace \tag{08}$$ PROPOSITION T is NOT APPLICABLE SINCE THE TWO EVENTS e1 and e2 DON'T TAKE PLACE ON THE SAME SPACE POINT IN SYSTEM $\mathrm{S}$(2). So you could not produce an equation like (03) to contradict equation (05).

(1) or more generally on the same plane normal to $\mathbf{v}$.

(2) or more generally on the same plane normal to $\mathbf{-v}$.