# Light clocks' time

1. Mar 2, 2015

### latter

I was looking at this because well i wanted to.
http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/Special_relativity_clocks_rods/index.html
And i worked something that if it was right i would be fantastic.
But i cant really believe its this easy.
[Quote
First, we will take the simple case of a light clock
whose motion is perpendicular to the rod. The
light clock will function as before. But now there is an
added complication. The light signal leaves one end of
the rod and moves toward the other end. But since the
rod is moving rapidly, the light signal must chase after
the other end as it flees. As a result, the light signal
requires more time to reach the other end of the rod
That means that the moving light clock ticks more slowly than one at rest.
rod is moving rapidly, the light signal must chase after]

But if you fire the photon from the opposite end of the
light rod the other end will get there quicker causing time to be faster .how can we have 2 clocks going at same speed with
different results.

2. Mar 2, 2015

### robphy

Actually, you've only observed the half-ticks of that light-clock. (You need to also consider the second-half-of-the-tick to arrive at time dilation.)
What you have observed with these two first-half-ticks is related to the relativity of simultaneity.
(Look at my avatar.... if it's not animated, click on it.)

3. Mar 2, 2015

### latter

the second half of the click .Sorry i have clicked your avatar i dont see any help there sorry.
simultaneity i did think of this but i have had no luck.all though that is a bit beyond me.
i dont get the difficulty either because the end is moving (away) it takes longer for the light to arrive.
so doesnt it stand to reason that if this happened the other way round the end was moving (towards) the light would get there early.

i am imaging the length of the rod to be going in the direction of motion.but instead of 1 rod i have 2 rods. one is firing photon from bottom the other top.Now from my understanding the one fired from bottom takes a little longer to reach top.so if its fired from top it is going to arrive sooner isnt it

i was reading about Relativity of Relativity of Simultaneity,but i still do not see connection here.
2 observers can not agree.
in my thought experiment the photons do not arrive at the same time and we wouldnt think they would in the first place.

and i dont see any need to bring in obsevers into it,i may be wrong but thats what i think.
so how is it Relativity of Simultaneity

Last edited: Mar 2, 2015
4. Mar 3, 2015

### Ibix

You are only considering half-ticks of the clock and, as you rightly point out, the two halves of the tick are different lengths in most frames. Since the other half-tick is the same as the first half-tick in the opposite direction, the total tick time is the same starting from either end.

That one half-tick is a different length doesn't matter. There are no physical consequences to it. You can't have one clock at both ends because you can't beat light to the other end. If you have two clocks, the relativity of simultaneity comes into play and you don't expect the two clocks to be in sync for everyone.

Do you know the Lorentz transforms? If not, look them up. Imagine a rod of length 2L centered at the origin, at rest, and lying along the x-axis. Two pulses of light are emitted, one from one end of the rod (t=0, x=L) and one from the other (t=0, x=-L). The pulses travel along the rod and arrive at the opposite ends at (t=2L/c, x=-L) and (t=2L/c, x=L). Feed those four pairs of numbers through the Lorentz transforms and you'll get the (t',x') coordinates of the four events, which are what is measured by someone who sees the rod moving at -v. You'll see the values fit your description.

Now feed in the return journey. You'll find that, although the two clocks don't always tick at the same time, they do tick at the same rate.

5. Mar 5, 2015

### latter

I will dismiss the first and third paragraph because i was not thinking of the photon going back again.

A photon leaves the middle

!______________!_____________!

and travels to either end (not back again).
My point is that if the rod is moving length ways with the direction of motion the photons will arrive at different times.
The hole rod is not in 2 separate reference frames the hole rod is in the same frame,so does relatively of similarity apply..All measuring instruments will be effected in the same way.
Ok each end can not verify to the other end.
So what have i got wrong here.

Last edited: Mar 5, 2015
6. Mar 6, 2015

### Ibix

The rod is in all reference frames. In the one in which it is at rest, the light pulses arrive simultaneously. In other frames they do not. That's the relativity of simultaneity.

7. Mar 6, 2015

### A.T.

This might help to visualize this:

8. Mar 6, 2015

### Staff: Mentor

There's no such thing as something being "in" one reference frame instead of another. Everything is always in all frames all the time - the frame is just a rule for assigning x and t coordinates to events. Thus:

Yes, relativity of simultaneity applies here. The arrival of the light signals is simultaneous if we use a frame in which the rod is at rest but not simultaneous if we use a frame in which the rod is not at rest.

9. Mar 6, 2015

### 1977ub

I always think that if you think of the rod having clocks at both ends or something like this where it is transforming it can help to remind one of the RoS.

In one reference frame, the clocks at the rod's ends are measured to be in sync, and in other frames they are not.

10. Mar 6, 2015

### Staff: Mentor

How can it be a clock then? How would the middle know when to emit the next pair of photons for the next tick?

11. Mar 11, 2015

### latter

Ive worked out what i think you are saying to me.But it is not what i am saying a least i think it isnt. You are saying this

and saying that because i am saying that these 2 photons (in my rod idea) arrive at different times,thats fine but only from my frame of reference. From somebody elses its different.

This is my rod idea put enougher way.imagine 2 gps satalites or atomic clocks flying through space at constant speed.Then checking the time. Now they should be the same time but there not.

i see the rod as 2 clocks i dont think you are seeing the rod as 2 clocks.Your seeing them as objects like lightening moving.I could be completely wrong but this is what i see at the moment

12. Mar 11, 2015

### Ibix

If your two atomic clocks are travelling at the same speed, then all observers (as long as they aren't accelerating, anyway) will agree that the two clocks tick at the same rate as each other. They will not necessarily agree what rate that is (due to time dilation), and they will not necessarily agree how far apart the two clocks are (due to length contraction), and they will not necessarily agree that the clocks show the same time (due to the relativity of simultaneity).

Before Einstein, we knew that "here" was a concept that changed according to who was talking. For me, "here" means my house. For you, it means your house, or whereever you are now. For somebody on a train, "here" is moving towards their destination at 60mph. And just for fun, the passenger could say that his "here" isn't moving and ours is moving away from the destination at 60mph. However, we thought that "now" was different - something everyone agreed on. It's 08:27 GMT as I type this - you would agree, so would the person on the train. Martians could look through their telescopes at Earth, work out when the Sun was directly overhead in Greenwich, correct for the light speed delay, and set the clocks on their war machines to GMT for convenience when invading.

However, one of the things Einstein realised as a consequence of his "light travels at the same speed in all inertial reference frames" idea was that it meant that there is no universal concept of "now", any more than there is a universal concept of "here". This has a number of consequences, one of which is that there is no universal agreement on what "synchronised clocks" are. What's hurting your head is that you are thinking clocks can't be synchronised and unsynchronised at the same time. The problem with that statement is the last four words - in relativity it's quite easy to disagree about what "at the same time" means.

It does all add up to a coherent picture. You probably won't be able to see that without messing around with the Lorentz transforms, however, as I recommended in my first post. You don't need more than high school maths for it.

13. Mar 11, 2015

### Staff: Mentor

I agree. In one frame they arrive at the same time. In other frames they arrive at different times. This is the relativity of simultaneity.

I don't think it is even 1 clock, let alone 2 clocks. How can it be a clock? A clock is a physical mechanism designed to "tick" at a regular interval. You forbid the light pulse from returning, so how does the clock know when to "tick" next?

To make your point you have gotten rid of the mechanism's ability to tick at a regular interval. It is therefore not a clock.

14. Mar 11, 2015

### A.T.

And a new(?) philosophical conundrum is born:

If a clock ticks only once, is it still a clock?

15. Mar 12, 2015

### latter

Just divide the segment or add segments. The photon simply triggers or passes over the marker.the time will be a huge number depending on how long ago you started the clock.i wouldnt of posted this if i didnt think i was talking about a clock i know a little more than to do that.

In the example below a photon would be fired from either end with a observer sitting in the middle to see that 1 photon hit before the other.
|__|__|__|__|__|__|

Reality of simultaneity kicks in.But what happens if the observers in other frames use a clock like the above.A

observer would get a different

result while still standing in

his same frame depending on

what direction the clock is

facing.
If you dont think this is a

clock could you say why.
If this passes as a clock what

am i getting wrong now.

16. Mar 13, 2015

### Ibix

This still isn't a clock. A clock has to have a cyclical process in it (a pendulum, or a vibrating quartz crystal, or something) and you still don't have one. All you've done by adding multiple checkpoints is give yourself multiple examples of the relativity of simultaneity. You can put a clock at each gate if you want, and everyone will agree what the clocks read when the pulses pass, but not whether or not the clocks are in sync.

Reversing the situation with pulses coming inwards doesn't help you either. It would be like running a video of the original experiment backwards. If the pulses arrive at the middle simultaneously for one observer then they must arrive simultaneously for all observers. They must, therefore, disagree about whether or not the pulses were emitted simultanously.

There is only a problem if two things which happen simultaneously and in the same place in one frame do not happen simultaneously and in one place in another frame. When the two "simultaneous" things aren't in the same place it doesn't matter whether they are simultaneous or not because there is no universal way to check whether or not they happened simultaneously.

For the third time - look up the Lorentz transforms. They'll show you what's happening from the perspective of any frame you want.

17. Mar 13, 2015

### Staff: Mentor

I still don't think this is a clock. It can still only tick once at each segment. Another name for a clock is a "frequency standard". What you describe cannot be used, as far as I can tell, to compare against any other frequency standard. It simply isn't a clock if the light pulses don't bounce back to where they started from.

I think it is pretty clear why it is not a clock. It doesn't repeat. It ticks once at a given location, and that is it.

If you have a light clock which is 150 m long then it will tick at a definite frequency of 1 MHz. You can set an oscilloscope at 1 MHz on one end and get a tick every cycle. With your clock, it doesn't matter how long or short it is or where you put your oscilloscope, you only get one tick where you are at and you cannot use it to tune your oscilloscope.

18. Mar 15, 2015

### latter

The pulse will just hit the first segment to tell it and a number (1)to represent time.
A new pulse is sent out the other side.This will take some time to happen i dont see this as a problem yet,(or whether this could be got round, the next segment is hit a number (2) appears each segment would have to be preprogrammed to produce a number.
Why does it have to go back to the one it came from it can clearly go from segment too segment whether it can be the same light pulse i dont know yet perhaps this could be a problem because the light has stopped and restarted.I dont know yet,whether that could bugger it up.

In your second paragraph i think you have gone over to the waveform.Surely we are only dealing with the point of impact.The leading edge.The front edge would have to react with a sensor.
You seem to be saying in your second paragraph that a oscilloscope gets a tick ,but what exactly is it in the wave thats reacting with the
oscilloscope.Surely its still the leading edge of something in the wave.

19. Mar 15, 2015

### Ibix

But what would you do with those numbers? If it doesn't tell anybody that "1 second has passed, 2 seconds have passed" and so on then it's not actually doing anything. And the only way it could tell anybody that "1 second has passed" would be to emit a light pulse, which would take time to get to the user. And then we're back to what DaleSpam and I have been telling you - you need an outbound leg and an inbound leg to make a clock.

You can use an electronic signal or something instead of the light pulse; you get the same problem.

The pulse can go on. The problem is that if something doesn't go back to the origin there is no way for anyone to measure that the pulse has passed each point.

DaleSpam isn't taking a position on the nature of light here. He's simply noting that a light clock 150m long (a pair of mirrors 150m apart with a light pulse bouncing between them) will "tick" one million times per second because light, with a speed of 300 million meters per second, will make the 300m there-and-back trip one million times per second. He triggers his oscilloscope every time the pulse leading edge (or peak, or whatever) returns to one end of the setup.

You are describing sensors set up 150m away, 300m away, 450m away and so on. You can't trigger an oscilloscope off that without putting in really long wires, connecting each of the sensors back to the oscilloscope. And once you've done that you've got a return journey, the length of which varies in moving frames such that everyone agrees that your oscilloscope will receive signals from opposite-pointing setups simultaneously. They won't agree on why they are simultaneous, but will agree that they are simultaneous.

You drew an ASCII art diagram earlier:
I have a feeling that you are imagining watching the light pulses travelling along between these posts and seeing each one raising a flag as the pulse passes, and arguing with a moving observer about whether or not they were simultaneous. But you can't do that unless you are in the diagram somewhere! And that means that you cannot neglect the travel time of light from each marker to the point where you are standing.

20. Mar 15, 2015

### Staff: Mentor

And how could I use those numbers to see if another clock is accurate? Say the other clock is located at rest at the first segment and is supposed to tick at a frequency of 1 Hz. How can you use your apparatus to confirm whether or not this other clock is actually ticking at the claimed rate of 1 Hz.

So that it can repeat and thereby generate a frequency. Remember, the key thing in a clock which makes it a clock is a frequency standard. You can think, clock = frequency standard.

It doesn't matter if it is the leading edge, the trailing edge, a peak, or a zero crossing. The point is that it is something which repeats at a regular frequency, aka, a frequency standard. Your device doesn't do that. It triggers once at any given location and stops. There is no repetition, and therefore no frequency, so it isn't a clock.