Understanding Time Dilation: How Passing Photons Affect Time Measurement

[Speady]

TL;DR Summary

We perceive an event because photons emitted from the event pass by and hit us. An event has a start moment and an end moment. An event emitted by a source in a time span of T seconds can be seen as a swarm of photons with a length of T x c. km. which passes us in T seconds with a speed c, because (T x c)/T = c.

Now if two different time durations are measured for one and the same event by two different observers, for example T+1 and T-1 seconds. Is the speed of passage then (T x c)/(T+1) and (T x c)/(T-1) respectively? So not c?

You may be wondering…, and yes, there is an example of it!

 

[BvU]

Hi,

I thought an event has no duration ?

$\$

 

[malawi_glenn]

An event emitted by a source

Events are not emitted by anything.
Event is just a point in space-time.

 

[Nugatory]

TL;DR Summary: We perceive an event because photons emitted from the event pass by and hit us. An event has a start moment and an end moment. An event emitted by a source in a time span of T seconds can be seen as a swarm of photons with a length of T x c. km. which passes us in T seconds with a speed c, because (T x c)/T = c.

A bit of a digression here, but none of this is right, mostly because photons are not what you think they are. Explaining what they are is a bit difficult in a B-level thread because they don't act like anything else that we are familiar with, but you'll find some attempts in older threads here. For now, when you're thinking about relativity, your best bet is to try to forget that you ever heard the word "photon" - you can remember it again when you get to quantum electrodynamics - and until then say "flash of light", "pulse of light", "light signal" or something similar.

Now if two different time durations are measured for one and the same event....

Can't happen because event has no size. It is a single point in spacetime the same way that (x=1,y=1) is a single point in the Cartesian plane. To have a distance we need two points, and then we can talk about the distance between them.

yes, there is an example of it!

Show us and we may be able better help yoo with the question here

 

[Ibix]

Now if two different time durations are measured for one and the same event

I think you are thinking of an event as something like clapping your hands, which has a small but finite duration and an extent in space. That is not what the word means in relativity. An event, as @Nugatory says, is the 4d equivalent of a point - so in the clapping example the point where your hands first meet at the instant they touch is an event. We do often approximate something like a clap or an explosion by an event, but this is only an approximation valid when the size and duration is negligible.

yes, there is an example of it!

Again as Nugatory says, state you example and we can work out what you're confused about.

 

[Speady]

Hi,

I thought an event has no duration ?

$\$

from start to finish is a certain length of time

 

[Speady]

Events are not emitted by anything.
Event is just a point in space-time.

not the event, but the photons are emitted
event can also be a collection of movements with a start and an end

 

[jbriggs444]

not the event, but the photons are emitted
event can also be a collection of movements with a start and an end

In physics and, specifically in the physics of special and general relativity, the term "event" means something different than it means in track and field.

 

[Speady]

A bit of a digression here, but none of this is right, mostly because photons are not what you think they are. Explaining what they are is a bit difficult in a B-level thread because they don't act like anything else that we are familiar with, but you'll find some attempts in older threads here. For now, when you're thinking about relativity, your best bet is to try to forget that you ever heard the word "photon" - you can remember it again when you get to quantum electrodynamics - and until then say "flash of light", "pulse of light", "light signal" or something similar.

OK, I say it differently. My source sends out a pulse of light and T seconds later a second pulse of light. The light pulses are on their way to me with speed c (km/s). The distance in space between the two light pulses is T x c km. I observe the light pulses and measure a time of T seconds between the two light pulses. Agree?

 

[Ibix]

OK, I say it differently. My source sends out a pulse of light and T seconds later a second pulse of light. The light pulses are on their way to me with speed c (km/s). The distance in space between the two light pulses is T x c km. I observe the light pulses and measure a time of T seconds between the two light pulses. Agree?

Assuming you are at rest with respect to the source, yes I agree. If you are moving with respect to the source, no.

 

[jbriggs444]

I was going to quibble that the distance between source and receiver would need to be greater than $Tc$ as measured in an inertial frame where both source and receiver are at rest.

We are starting with the simple scenario where both source and receiver are at rest in some chosen inertial frame, right?

 

[Speady]

Assuming you are at rest with respect to the source, yes I agree. If you are moving with respect to the source, no.

Then my initial question: the same source, the same pulses, the same T, but now there are two other observers and they measure an interval of T+1 s and T-1 s. Are the measured speeds of the pulses then (Txc)/T+1) km/s and (Txc)/(T-1) km/s? And not c?

 

[jbriggs444]

Then my initial question: the same source, the same pulses, the same T, but now there are two other observers and they measure an interval of T+1 s and T-1 s. Are the measured speeds of the pulses then (Txc)/T+1) km/s and (Txc)/(T-1) km/s? And not c?

So now you have moving observers. An observer moving toward the source and measuring a reduced time. And an observer moving away from the source and measuring an increased time. (A Doppler shift as noticed by @Sagittarius A-Star)

All three observers calculate the same speed. c.

The other two observers do not agree that the two pulses were emitted T seconds apart. That is the relativity of simultaneity in action. Edit: same emission event, same emission time. They do not agree on elapsed time until reception. That is some combination of time dilation and relativity of simultaneity. Nor do they agree on the distance covered. That is length contraction in action. The effects conspire so that the calculated speed of light is invariant.

Each of the three frames makes identical predictions for every local measurement that is performed. Each of the three frames explains those identical predictions differently. If you lay out coordinates for all of the events of interest in one of the three frames, the Lorentz transformations can give you the corresponding coordinates in the other two frames.

The events of interest are:

Source sends first pulse.
Source sends second pulse
Receiver 1 receives first pulse
Receiver 1 receives second pulse
Receiver 2 receives first pulse
Receiver 2 receives second pulse
Receiver 3 receives first pulse
Receiver 3 receives second pulse

If you line things up nicely, receiver 1, 2 and 3 can all receive the first pulse at the same event.

 

[Sagittarius A-Star]

Then my initial question: the same source, the same pulses, the same T, but now there are two other observers and they measure an interval of T+1 s and T-1 s. Are the measured speeds of the pulses then (Txc)/T+1) km/s and (Txc)/(T-1) km/s? And not c?

No. Light always moved with c, in all inertial frames. The other observers see a relativistic Doppler shift of the frequency $\frac{1}{T}$if they move away from the source or towards the source.

 

[PeroK]

Then my initial question: the same source, the same pulses, the same T, but now there are two other observers and they measure an interval of T+1 s and T-1 s. Are the measured speeds of the pulses then (Txc)/T+1) km/s and (Txc)/(T-1) km/s? And not c?

The invariance of the speed of light, $c$, is manifestly incompatible with classical physics. In particular, classical notions of time and space. Something's got to give in your classical view of physics to accommodate an invariant speed, $c$.

 

[Ibix]

Then my initial question: the same source, the same pulses, the same T, but now there are two other observers and they measure an interval of T+1 s and T-1 s. Are the measured speeds of the pulses then (Txc)/T+1) km/s and (Txc)/(T-1) km/s? And not c?

No. The pulses pass at the same speed, but were not emitted at the same distance according yo this frame. One pulse had less distance to travel than the other so the pulse spacing is different. Time dilation will also affect this.

 

[Speady]

No. The pulses pass at the same speed, but were not emitted at the same distance according yo this frame. One pulse had less distance to travel than the other so the pulse spacing is different. Time dilation will also affect this.

The pulse distance was Txc km when sending, and doesn't just change length, okay? Doesn't this exactly the same length of Txc km now pass the observers with different durations?

 

[Ibix]

The pulse distance was Txc km when sending, and doesn't just change length, okay?

The length measured depends on the speed of the observer with respect to the emitter. So it does change because you changed the speed of the emitter.

 

[Ibix]

Just for information of all responders, this poster has asked these questions before and didn't listen to our answers then, either.

 

[jbriggs444]

The length measured depends on the speed of the observer with respect to the emitter. So it does change because you changed the speed of the emitter.

It changed because you changed the frame used to judge the speed of the emitter (as you know).

The measured length of an extended object is not a direct observable. It is the result of a calculation. It is: "The difference in the coordinates of the right end and the left end at the same time".

Under Newtonian physics, we took "at the same time" for granted. We assumed that all clocks everywhere could be synchronized and could remain synchronized regardless of their state of motion.

Under Special Relativity, time becomes another coordinate. There is room for disagreement about which event over here is simultaneous with which event over there. Simultaneity now corresponds to "has the same value for the $t$ coordinate".

If we change our frame of reference, we systematically change the synchronization of all of our clocks along the axis of relative motion. This opens up a loophole in our definition of length. The measured length depends on "at the same time" which depends on how we synchronize our clocks. And how we synchronize our clocks depends on what we decide to call "at rest".

The reason for the systematic offset mentioned above can be traced to how we physically synchronize clocks. We use Einstein synchronization as the basis for this. But any other method will reach the same result. For Einstein synchronization, we send a light signal on a round trip from A to B and back and measure the round trip time. Then we send a light signal from A to B with a time stamp. B sets his clock to that time stamp plus half of the measured round trip time.

This corresponds to us deciding that the speed of light is the same out and back.

Any other observer using a different standard of rest can complain "but your upstream signal actually took longer than your downstream signal, so your clocks are not actually synchronized". Welcome to the relativity of simultaneity.

 

[Ibix]

It changed because you changed the frame used to judge the speed of the emitter (as you know).

That is a more precise way of saying it, yes.

This is a phenomenon that does not occur in Newtonian physics, where no finite velocity is invariant. In that case the velocity must change and also the frequency (or time between pulses), but the distance between pulses is invariant. However, that is not consistent with observation. Rather, we live in a relativistic universe where $c$ is invariant and the frequency and wavelength both change.

I suppose one could propose a third system of physics where the frequency is the invariant, but I don't know if it could lead to consistent laws. Certainly it can't be consistent with the principle of relativity.

 

[Dale]

from start to finish is a certain length of time

Events don’t have a start or a finish. Events are points in spacetime. Points don’t have a start or a finish.

 

[Speady]

To all posters: If you define a certain speed, for example of light in a vacuum, as invariant, then you can adjust all mathematical formulas for describing what happens to it. Distances and durations will then be molded to fit that invariant velocity. However, this gets a bit complicated to explain if two observers are watching the same movie, one seeing an apparently sped-up shot that is shorter for him, and the other seeing an apparently slowed-down shot that is longer for him, and you have to persevere that the film really turned as fast for one person as for the other. For whom that is no problem, I wish good luck in his wonderful relativistic world.

 

[Ibix]

However, this gets a bit complicated to explain if two observers are watching the same movie, one seeing an apparently sped-up shot that is shorter for him, and the other seeing an apparently slowed-down shot that is longer for him, and you have to persevere that the film really turned as fast for one person as for the other.

I am not sure what you think you are describing here, but understanding how two frames can have different descriptions of a sequence of events is not all that hard. The best tool for visualising it is, IMO, the Minkowski diagram. The transformations between frames are only a little more complex than rotations. You just have to put in the time to learn it, rather than asserting (in the face of over a century of evidence) that it can't possibly make sense.

 

[Speady]

The best tool for visualising it is, IMO, the Minkowski diagram.

This is exactly what I mean, with the math created to mold distances and durations to fit an invariant velocity. It creates its own reality. For over a hundred years. It is a choice to accept reality that way.

 

[Frabjous]

This is exactly what I mean, with the math created to mold distances and durations to fit an invariant velocity. It creates its own reality. For over a hundred years. It is a choice to accept reality that way.

How do you distinguish good math from bad math?

 

[Speady]

How do you distinguish good math from bad math?

There is no "good math" or "bad math". Mathematics serves. You can use it to create your own reality.

 

[Frabjous]

There is no "good math" or "bad math". Mathematics serves. You can use it to create your own reality.

That’s good to know. I just went and rebalanced my checkbook. I am now a millionaire.

 

[Drakkith]

This is exactly what I mean, with the math created to mold distances and durations to fit an invariant velocity. It creates its own reality. For over a hundred years. It is a choice to accept reality that way.

Can you elaborate? It's not clear what you're getting at here. My 'choice' is to accept that the math accurately describes reality as verified by experimental results. Are you saying something different?

 

[Ibix]

This is exactly what I mean, with the math created to mold distances and durations to fit an invariant velocity.

Actually you can derive relativity without assuming an invariant velocity - just the principle of relativity (see Palash Pal's paper "Nothing but relativity", for example). You are led to either Newtonian physics or Einsteinian relativity, and we know Newtonian physics is not correct from experiment.

So the maths and the existence of an invariant speed actually follow from the principle of relativity.

 

[Dale]

There is no "good math" or "bad math". Mathematics serves. You can use it to create your own reality.

You really cannot use math to create your own reality. No amount of mathematical wrangling will change the outcome of an experiment. We use the math of relativity because it correctly predicts the outcome of such a wide variety of experiments:

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

To the best of our knowledge there is no other math that correctly predicts the outcomes of all of these experiments. I.e. no other math describes reality.

this gets a bit complicated to explain if ...

Yes, sometimes reality is a bit complicated to explain.

 

[Frabjous]

We correct GPS for relativity. If relativity was wrong, we would be making the wrong corrections.
Here’s an excerpt of relativistic effects for GPS https://www.astronomy.ohio-state.edu/pogge.1/Ast162/Unit5/gps.html

To achieve this level of precision, the clock ticks from the GPS satellites must be known to an accuracy of 20-30 nanoseconds. However, because the satellites are constantly moving relative to observers on the Earth, effects predicted by the Special and General theories of Relativity must be taken into account to achieve the desired 20-30 nanosecond accuracy.

Because an observer on the ground sees the satellites in motion relative to them, Special Relativity predicts that we should see their clocks ticking more slowly (see the Special Relativity lecture). Special Relativity predicts that the on-board atomic clocks on the satellites should fall behind clocks on the ground by about 7 microseconds per day because of the slower ticking rate due to the time dilation effect of their relative motion [2].

Further, the satellites are in orbits high above the Earth, where the curvature of spacetime due to the Earth's mass is less than it is at the Earth's surface. A prediction of General Relativity is that clocks closer to a massive object will seem to tick more slowly than those located further away (see the Black Holes lecture). As such, when viewed from the surface of the Earth, the clocks on the satellites appear to be ticking faster than identical clocks on the ground. A calculation using General Relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day.

The combination of these two relativitic effects means that the clocks on-board each satellite should tick faster than identical clocks on the ground by about 38 microseconds per day (45-7=38)! This sounds small, but the high-precision required of the GPS system requires nanosecond accuracy, and 38 microseconds is 38,000 nanoseconds. If these effects were not properly taken into account, a navigational fix based on the GPS constellation would be false after only 2 minutes, and errors in global positions would continue to accumulate at a rate of about 10 kilometers each day! The whole system would be utterly worthless for navigation in a very short time.

 

[Nugatory]

OK, I say it differently. My source sends out a pulse of light and T seconds later a second pulse of light.

Saying it that way is much clearer. Now we have four events:
A) first flash is emitted at the source.
B) first flash is received by you.
C) second flash is emitted at the source.
D) second flash is received by you.

The time between A and C can be directly measured by a clock at the emitter.
The time between B and D can be directly measured by your wristwatch.
These are proper times: invariant, the same for all observers, the same no matter which frame we’re using when we analyze the problem (these are different ways of saying the same thing).
These two times will be equal only if the you and the emitter are at rest relative to one another, meaning that no matter which inertial frame we choose they are both moving in the same direction at the same speed (which may be zero, in which we would say that they are both at rest in that frame.

The time between A and B and the time between C and D cannot be measured. Instead it is something we calculate by making an assumption about what your wristwatch reads at the same time that event A or C happens, and then subtracting that from the time on your wristwatch when flashes are received. “At the same time” means different things in different frames so we’ll be making different assumptions when we use different frames; the result of this calculation will depend on our choice of frame and doesn’t tell us much of anything about anything.

The light pulses are on their way to me with speed c (km/s). The distance in space between the two light pulses is T x c km. I observe the light pulses and measure a time of T seconds between the two light pulses. Agree?

This will be true only if we choose to analyze the problem using one particular frame, the one in which you and the source are both at rest, and then choose to use that frame’s natural definition of “at the same time”.

 

[Nugatory]

For over a hundred years. It is a choice to accept reality that way.

Experiments and observations tell us what reality is so we don’t have a choice to accept it one way or another, we have to accept it the way it is. The choice we do have is to use math that accurately describes reality, or to use math that does not accurately describe reality. This does not seem to be a particularly difficult or controversial choice….

 

[PeterDonis]

It is a choice to accept reality that way.

No, it's not, because reality tells us which way it is through experiments. Experiments have told us to very high accuracy that relativity is correct and Newtonian physics is not.