# I Shouldn't a moving clock appear to be ticking faster instead of slower?

#### Mentospech

Summary
Moving objects time dilation
Hi.
Im looking into special relativity and everything i found about time dilation on internet seems to say that moving clock appear to tick slower than the stationary one. However what I found about this is following, in § 4. (Physical Meaning of the Equations Obtained in Respect to Moving Rigid Bodies and Moving Clocks) of Einstein's On the electrodynamics of moving bodies (1905):

Now if A is on a train (K') and B is a point on the track with synchronized clock (K), then according to this, the passenger A passing B would see B being ahead of his clock.
Please note it is not necessary for the train to stop for this to occur.
Also there can be any (infinite) number of such points, and they can be at arbitrarily short distances.

Meaning any point in the travel the stationary clock outside would be more and more ahead -> the other clock actually ticks faster, not slower.
So what am I missing?

Edit: from the point of stationary observer, the clock on the train would still be slower.

Last edited:
Related Special and General Relativity News on Phys.org

#### ParticleGinger6

With Special Relativity you can figure that clock A (the clock with velocity) ticks slower just by knowing the equation t(1)=t(0)*(gamma) or t(1) = t(0) / (1-v^2/c^2)^(1/2) . Knowing this the object that is in motion at relativistic speeds would "feel" time moving at a slower rate. This would also show the clock moving at a slower rate then the clock that is at a stand still.

If words are easier to understand then just think of the twin paradox. One twin is shot to out into space to visit a star and come back. The other is to remain on earth. If the twin in space is moving at relativistic speeds then upon their arrival from their journey they will come home to a twin that has aged more then they have. This means that the twin in space experienced time at a slower rate then the twin left on earth. The closer to space the farther in age the twins will be upon the space travelling twins arrival back home.

I hope this was able to help your understanding of why the clock in motion was ticking at a slower rate the clock that was not moving.

#### Mentospech

@ParticleGinger6 So you do agree that from the point of traveller, it is his clock that goes slower than the outside clock? By the way the equation provided by me from the quotations seems to provide different results than yours.

#### ParticleGinger6

@Mentospech The point of view of the person can be considered stationary at any time. This means that if I was on a spacecraft that was travelling at relativistic speeds and I observed you on a space station (imagine the space station has 0 velocity) I would see you as the object in motion from my point of view while I would be the one at velocity = 0. So this means that at anyone's point of view the other person would be holding the clock that is ticking at a slower rate as they would be the one travelling at relativistic speeds to the other person.

I know for me this was a hard concept to first visualize however I think there are youtube videos that will help you visualize the concept better. I can not find the one that helped me understand the concept the best right now. For that I am sorry.

#### Doc Al

Mentor
Meaning any point in the travel the stationary clock outside would be more and more ahead -> the other clock actually ticks faster, not slower.
So what am I missing?
Those "stationary" clocks along the train tracks are synchronized according to observers in the frame of the tracks. However, the moving observers will find that those clocks are not synchronized according to them.

Edit: from the point of stationary observer, the clock on the train would still be slower.
True. And from the point of view of the moving observer, the clocks on the track would be slower. It's completely symmetric: Each observer measures the other's clocks to be running slow.

In addition to time dilation, one must consider the relativity of simultaneity to fully understand how they can both see the other's clocks run slow.

#### Mentospech

@ParticleGinger6 Well isnt there a contradiction? On one hand the passenger on the train must see the outside clocks being more and more ahead, on the other hand you say that each observer sees the other frame's clock ticking slower

#### Mentospech

@Doc Al Correct me if Im wrong here, but time dilation is physical effect with casual consequences for both observers, it is a physical fact. While relativity of simultaneity is just an discrepancy in observation without any casual consequences for anyone.

Doesnt the equation provided in the quotation say that the person on train must observe the other clock being ahead at every point in time, progressively?

#### ParticleGinger6

So the equation at small amount of time basically is saying that for every second (according to him) the "stationary" observer is seeing the observer on the train. The observer on the train is experiencing a fraction of a second (based on their velocity). The dilation would not change unless the train was accelerating. So after a longer time such as an hour, the person on the train might have only experienced say 30 minutes.

#### Doc Al

Mentor
Correct me if Im wrong here, but time dilation is physical effect with casual consequences for both observers, it is a physical fact. While relativity of simultaneity is just an discrepancy in observation without any casual consequences for anyone.
Not sure what you're saying here. If you mean that the clocks on the moving train are the ones that are really running slow, then you are mistaken. Note that motion is relative. (You might want to imagine two rocket ships passing in outer space. Each one views itself as being at rest and the other moving.)
Doesnt the equation provided in the quotation say that the person on train must observe the other clock being ahead at every point in time, progressively?
Yes. From the train observer's viewpoint, the clocks on the tracks are out of synch. So? (Now when the train observers measure how the track clocks are recording time -- taking into account their lack of synchronization -- they will measure that those clocks are running slow.)

#### Mentospech

Yes. From the train observer's viewpoint, the clocks on the tracks are out of synch. So? (Now when the train observers measure how the track clocks are recording time -- taking into account their lack of synchronization -- they will measure that those clocks are running slow.)
The passenger can always look only at one clock at a time - the one hes passing by. So that we can say they are at the same location, so we do not need to talk about their synchronization from frame K'
Also it is apparent that if the passenger were to add up all the time differences from the stationary clocks he would end up with the reading he will find on the last clock in his final destination. So i dont think there are any problems there.

Lastly, the quotation from Einstein's paper on relativity clearly says that the passenger will find their clock lagging behind. So how can you say
taking into account their lack of synchronization -- they will measure that those clocks are running slow

#### ParticleGinger6

What exactly do you mean, based on my equation it comes out to be a ratio that time is multiplied by, there is no second order or above factor in my equation after the ration is found

#### Mentospech

@ParticleGinger6:
With Special Relativity you can figure that clock A (the clock with velocity) ticks slower just by knowing the equation t(1)=t(0)*(gamma)
gamma can go to infinity, no?

#### Nugatory

Mentor
Well isnt there a contradiction? On one hand the passenger on the train must see the outside clocks being more and more ahead, on the other hand you say that each observer sees the other frame's clock ticking slower
There is no contradiction if you allow for relativity of simultaneity. However, you first have to be clear about exactly what it means to say that one clock is “running slower” than another:

Say that you and I are moving relative to one another at speed .8c. As we pass one another (so that we are momentarily at the same place at the same time) we both zero our clocks. One minute later I look at my clock and see that it reads 1:00, and of course you are .8 light minutes away from me. 48 seconds after that (because that’s how long it takes light from your clock to cover the .8 light-minute distance between us) I can see what your clock read at the same time that my clock read 1:00; this turns out gone 0:36 so I correctly conclude that your clock is running slow by a factor of 3\5. That’s what “running slow” means - at the same time that my clock reads something, your once-synchronized clock reads something less. It critically depends on the meaning of “at the same time”.

#### ParticleGinger6

no its restricted from 0 to 1 when using gamma = (1-(v^2/c^2))^(1/2)

#### Mentospech

@ParticleGinger6
no its restricted from 0 to 1 when using gamma = (1-(v^2/c^2))^(1/2)
Still, in the equation i provided the limit seems to be 50% of the time in (K) when the speed approaches c

#### Mentospech

There is no contradiction if you allow for relativity of simultaneity. However, you first have to be clear about exactly what it means to say that one clock is “running slower” than another:

Say that you and I are moving relative to one another at speed .8c. As we pass one another (so that we are momentarily at the same place at the same time) we both zero our clocks. One minute later I look at my clock and see that it reads 1:00, and of course you are .8 light minutes away from me. 48 seconds after that (because that’s how long it takes light from your clock to cover the .8 light-minute distance between us) I can see what your clock read at the same time that my clock read 1:00; this turns out gone 0:36 so I correctly conclude that your clock is running slow by a factor of 3\5. That’s what “running slow” means - at the same time that my clock reads something, your once-synchronized clock reads something less. It critically depends on the meaning of “at the same time”.

I understand how we both can see the other one's clock is running slow, because of the doppler effect, which is exactly what you described. However that is completely different phenomenon that is discussed in the Einstein paper and also here. That is why the thought experiment is described as it is.

#### Nugatory

Mentor
no its restricted from 0 to 1 when using gamma = (1-(v^2/c^2))^(1/2)
$\gamma$ is defined to $\frac{1}{\sqrt{1-\beta^2}}$ where $\beta$ is $v/c$ (and $\beta$ is often just written as $v$ when using units in which $c=1$, such as measuring time in seconds and distance in light-seconds).

Introducing any other definitions of these symbols is asking for confusion, as you will be just about the only person in the world who understands what you mean.

#### ParticleGinger6

@Nugatory how were you able to type those equations into here, I have not been able to find how to do that and i feel like I would have been less open to interpretation if I did an equation equation instead.

#### Nugatory

Mentor
understand how we both can see the other one's clock is running slow, because of the doppler effect, which is exactly what you described. However that is completely different phenomenon that is discussed in the Einstein paper and also here. That is why the thought experiment is described as it is.
The Doppler effect is completely unrelated to what I described, and also will cause them both to “see” the other’s clocks running faster not slower if they are moving towards one another instead of away. The symmetrical time dilation is what’s left over after you’ve allowed George Doppler and light travel time.

#### Nugatory

Mentor
@Nugatory how were you able to type those equations into here, I have not been able to find how to do that and i feel like I would have been less open to interpretation if I did an equation equation instead.
Check our help/info section for the Latex primer... or if you just click the reply button to a post containing some formatted math, the quoted text will show you how it was done.

#### Mentospech

The Doppler effect is unrelated, and also will cause them both to “see” the other’s clocks running faster not slower if they are moving towards one another instead of away. The symmetrical time dilation is what’s left over after you’ve allowed George Doppler and light travel time.
Your argument still seems full of doppler effect. Why dont you use the thought experiment as described? There is no need to look at clocks at any distance other than 0. Or is your position that you cannot use the synchronized clocks?

#### Mentospech

There is no contradiction if you allow for relativity of simultaneity. However, you first have to be clear about exactly what it means to say that one clock is “running slower” than another:
Note that relativity of simultaneity occurs only if you observe object in a distance, it never once happens in the experiment. And even if it did. The effect of this dilation (which happnes in every point on line AB) is physical effect with casual consequences to all inertial frames. While relativity of simultaneity is just illusion without any casual consequences to anyone. So i do not understand how it can be relevant in explaining these casually important effects

#### Pencilvester

Your argument still seems full of doppler effect. Why dont you use the thought experiment as described? There is no need to look at clocks at any distance other than 0. Or is your position that you cannot use the synchronized clocks?
The effect of time dilation describes what happens when someone looks at a single moving clock. In your scenario, the train observer is looking at snap shots of multiple clocks. And by “look” I of course mean after accounting for travel time of light.

#### Ibix

Why dont you use the thought experiment as described?
Because your experiment as described does not show you time dilation. As @Pencilvester says, time dilation is an effect you get watching a single clock moving with respect to you (after correcting for the changing distance). You are looking at multiple clocks in sequence, thereby combining time dilation with the synchronisation of those clocks. That's fine as long as you are aware that is what you are doing, but it appears to me that your problem is that you think the combined effect is just time dilation alone. It is not.

#### Dale

Mentor
Well isnt there a contradiction? On one hand the passenger on the train must see the outside clocks being more and more ahead, on the other hand you say that each observer sees the other frame's clock ticking slower
There is no contradiction as long as you include the relativity of simultaneity. The way to include all of the relativistic effects (time dilation, length contraction, and relativity of simultaneity) is to use the Lorentz transforms. Using units where c=1 and neglecting y and z we have the forward and inverse transforms:

$t'=\gamma(t-vx)$
$x'=\gamma(-vt+x)$
$t=\gamma(t'+vx')$
$x=\gamma(vt'+x')$

where $\gamma = (1-v^2)^{-1/2} > 1$ and $0<v<c=1$. (Note, this is the Lorentz transform, not the Doppler effect)

Now, for a clock at rest at the origin of the unprimed frame $x=0$. So in the primed frame $t'=\gamma (t-vx) = \gamma t$. Because $\gamma>1$ this implies that the unprimed clock runs slow in the primed frame.

Similarly, for a clock at rest at the origin of the primed frame $x'=0$. So in the unprimed frame $t=\gamma t'$. Again, the primed clock runs slow in the unprimed frame.

What you are looking at as a contradiction is a completely different thing. Going back to considering time in the primed frame, instead of looking at the line $x=0$ (representing a single unprimed clock) you are looking at the line $x'=0$ (representing a multitude of unprimed clocks as they pass by a single primed clock). In other words, instead of considering the rate of a single unprimed clock over time, you are considering the time displayed at a single instant for multiple unprimed clocks. To calculate this we first have to solve $x'=0=\gamma(-vt+x)$ to get $x=vt$. Substituting that in we get

$t'=\gamma(t-vx)=\gamma(t-v^2 t)=\gamma (1-v^2) t = t/\gamma$

Therefore, looking along the line $x'=0$ at the unprimed clocks we see the effect that you are noticing. However, again, this is not the rate of a unprimed clock in the primed frame.

The passenger can always look only at one clock at a time - the one hes passing by.
The passenger can do that, but doing so does not tell the passenger anything about the rate of a clock in the other frame as seen by his frame. See the math above.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving