# I Clock synchronization

1. Mar 19, 2017

### Dimani4

Hi,
Here me again. I have a question about clock synchronization. Suppose we have train with Jill with two clocks and Jack on the ground. As seen from the ground by Jack, the two clocks on the train (which is moving at v in the x-direction) are running slowly, registering only seconds for each second that passes. Equally important, the clocks—which are synchronized by an observer on the train—appear unsynchronized when viewed from the ground, the one at the back of the train reading seconds ahead of the clock at the front of the train, where L is the rest length of the train (the length as measured by an observer on the train). The question is what sees observer on the train (Jill) and observer on the ground (Jack) when they observe the same clocks? Suppose Jack makes a photo of these clocks from the ground and Jill makes photo the same clocks at the train at the same time. As I understand Jack will see on the photo of the train's clock time dilation between the front and back of the clocks by factor of [PLAIN]http://galileo.phys.virginia.edu/classes/252/synchronizing_files/image029.gif. [Broken] But what see Jill? Jill doesn't know she's moving and she doesn't have to see any dilation because she's moving in the inertial frame of reference.
So what Jill will see?
Thank you

Last edited by a moderator: May 8, 2017
2. Mar 19, 2017

### Staff: Mentor

There is no such thing as a photograph that shows two different objects at the same time (unless they are both exactly the same distance from the camera, which isn't the case here). The problem is that the photograph captures the light that hits the film at whatever moment the camera shutter is open - but that light had to travel from the objects being photographed, and if that travel time is different then the photograph has captured the objects at different times. Similar considerations apply to what you "see", because that depends on light reaching the retina of your eye.
If you properly calculate everything - allowing for time dilation, length contraction, relativity of simultaneity, and light travel time - you will find that it all comes out consistently.

3. Mar 19, 2017

### Staff: Mentor

She sees that the clocks are synchronized. That does not mean that the light reflected from each clock at noon reaches her eyes at the same time or that she can take a photograph in which all the clock faces show noon; it means that for every clock, if the clock is a distance $d$ from her the light reflected from it when it reads noon reaches her eyes at $d/c$ seconds after noon, where $d$ is distance to that clock in light-seconds.

Jack on the ground will do something similar, but he will find that that the light reflected from the clocks when they read noon does not reach him at $d/c$ seconds after noon. It arrives early for some clocks and late for others, so after allowing for light travel time he concludes that they are not synchronized.

If you haven't tried googling for "Einstein train simultaneity" yet, now would be a really good time....

4. Mar 19, 2017

### Janus

Staff Emeritus
Assume that Jill is standing exactly halfway between the clocks on her train, and both Jack and Jill take their photos as they pass each other.
Jill's photo will show both clocks reading the same time. The same light coming from each clock also reaches Jack at that moment, so his photo will also show the same time on both clocks (and the same time as Jill's photo shows.). However, While Jill will say that the light she records left both clocks at the same time and while they were an equal distance from her, Jack will not agree. For him, the light from one clock left before the light from the other left before the other, and one clock was closer to him than the other when the light left it. (consider that according to Jack the two clocks are an equal distance from him when he takes the photo. The light took some non-zero time to get to him from each clock, which means he was not an equal distance from the two clocks when the light left either of them. In order the the light from each clock to reach him at the same time traveling at c from clocks that were not the same distance from him when the light left each, the light from each clock had to leave at different times by his own clock. This, in turn, will be consistent with the fact that, according to him, the clocks ticked slower than his own and were not in sync with each other.

5. Mar 20, 2017

### Bartolomeo

As far as I understand, in this case we consider a picture taken by pinhole camera. Will pictures taken by "moving" camera or camera "at rest" be the same? According to Special Relativity, in both cases pictures must be the same. Is it correct?
We must admit that the camera is at some distance from the couple of clocks at the ends of the train.

Last edited: Mar 20, 2017
6. Mar 20, 2017

### Janus

Staff Emeritus
As long as the cameras are co-located when the picture are taken.

Here's the space-time diagram off this according to Jill's frame. The orange line is Jill's world line, the Blue is Jack's world line and the world lines for the clocks are the green and red lines. The yellow diagonals represent the light seen by the camera if the picture is taken when Jack and Jill pass each other. the picture will show both clocks reading 1. Jill will also conclude that, at the moment the picture is taken, both clocks read 2.

Here is the same scenario according to Jack's rest frame:

Notice that the same light carrying the same image of each clock also meets up with Jack;s camera at the moment Jack and Jill pass each other and Jack takes his picture. Thus his picture will also record both clocks reading 1. However, the light from each clock that forms the picture left each clock at different moments by his clock and when the clocks were at different distances from him. Also, according to him, at the moment the picture is taken, the two clocks do not read the same as each other or Jill's clock. Jill's clock reads 2, while one of the clocks only reads 1.5 and the other already reads 2.5.

7. Mar 20, 2017

### Bartolomeo

Yes, I agree, that readings of the clocks will be the same, though cameraman and observer in trains reference frame will make different conclusions about simultaneity when these readings "departed".

Just for clarification:

We consider that pinhole camera (obscura) takes picture of two clocks, that are attached to the ends of relativistic train. The train moves in dark space. At certain moment we highlight these clocks by short flashes. EDIT: These two flashes are in immediate vicinity to the clocks. We launch these flashes simultaneously in train's frame.
There are two cameras: one is at rest in the reference frame of the train, at equal distance from the clocks on the ends, and at distance y from the center.
There is another relatively moving camera. Its coordinate at Y axis is y also. Both cameras take picture when their spatial positions coincide, when center of the train and apertures are at points of closest approach.
We agree, that beams of light simultaneously come to apertures of these cameras and readings of the clocks will be the same on the both pictures. We agree, that the clocks will show the same time on the pictures.

What about distance between clocks at the pictures? Will it be the same in both cases, at picture taken by that camera "at rest" and that one "in motion"?

Last edited: Mar 20, 2017
8. Mar 20, 2017

### Mister T

You can get around the problems with light travel time by placing the cameras immediately in front of the clocks. Or simply have the clocks stamp their readings on a card that gets printed out. Either way, anyone can at a later time examine these photographs or cards and see what those clocks read at the time.

Let's say that Jill has synchronized the clocks according some scheme that Jack agrees is a valid way to synchronize clocks. Suppose they agree beforehand that the cards will get printed when the clocks read 12:00 noon. Both Jack and Jill can go back afterwards and examine the cards that were printed, or the photos that were taken. They will indeed agree that both clocks read 12:00 noon. But Jack will not agree that the cards were printed at the same time. Suppose $\frac{Lv}{c^2}$ equals one hour. Jack will claim that the clock in the rear was ahead of the clock in the front by one hour. He will therefore say that if it was indeed 12:00 noon when the front clock read 12:00 noon, then the rear clock should have read 11:00 am when it printed the card stating that the time was 12:00 noon. He will attribute this discrepancy to the error Jill made when she synchronized the clocks.

By the way $\frac{Lv}{c^2}$ is not called time dilation.

Last edited by a moderator: May 8, 2017
9. Mar 21, 2017

### Bartolomeo

@Janus
NOTE: The question "" Will it be the same in both cases .... " is senseless. Obviously it will be not.
Just: What about distance between clocks at the pictures?

10. Mar 21, 2017

### pervect

Staff Emeritus
I'm not sure I agree, possibly I'm not interpreting the question the same way. I'm afraid I haven't been following the thread in deatil, so possibly my interpretation of the question is wrong :(. But if we imagine we have two space-ships, one moving (relative to the CMB) and one stationary (relative to the CMB), and they both take a picture of the stars, I wouldn't expect the pictures to be the same. Relativistic aberration would change the angles, and there would be frequency shifts (color changes) and relativistic beaming effects (intensity changes) as well.

11. Mar 22, 2017

### Bartolomeo

@Janus means, that readings of the clocks will be the same at the both pictures, if cameras are co-located. Two clocks will show the same time. No doubt.
If certain camera is at rest in the reference frame of the train, image will not be distorted, it is obvious,
But, picture taken by moving camera will be distorted because of Lorentz contraction at least. The problem is, how exactly these distortions will appear on the picture.

https://en.wikipedia.org/wiki/Length_contraction

"Length contraction is the phenomenon of a decrease in length of an object as measured by an observer who is traveling at any non-zero velocity to the object“

Image will be Lorentz –contracted, won‘t it?

12. Mar 22, 2017

### PAllen

If the thing being imaged subtends a reasonably small solid angle, the two images will differ in placement on the sensor, and in color, but will have appear to have essentially the same shape due aberration being a conformal transformation. The placement of image details would be consistent with rotation of the uncontracted body so as to produce a contracted cross section (in the cases where this happens, e.g. a ruler). This is for an idealized pin hole camera, with coincidence being of the pin holes and the sensor being big enough to always capture a complete image. This is all worked out in the various papers on Penrose Terrell rotation.

[edit: corrected a little since first version]

Last edited: Mar 22, 2017
13. Mar 22, 2017

### Bartolomeo

Terrell Penrose rotation is the optical illusion. Object must have depth and shine with reflected light all the time. Different parts of the sphere are at different distances from the aperture. In case of Terrell rotation aperture briefly opens and catches reflected beams, so the sphere appears rotated at $\alpha = \arcsin v/c$

There is nothing to rotate in this case. Flashes are at the same distance from aperture. Train moves in dark space. Beams from flashes on the ends of the train travel by equal hypotenuses and simultaneously come to aperture.

We can also consider two twins at certain distance from each other. Each has a clock and a flash. They launch flashes simultaneously. We want to know what will be distance between twins at picture taken by camera at rest and one in motion, when the center between twins and apertures at points of closest approach.

The beams will simultaneously go through aperture and simultaneously hit the film.

Patches of rays from flashes to aperture and from aperture to film form two similar triangles, in moving camera and in that one at rest.

Let’s we take a picture of a flat square, at each corner of the square we attach a flash. We launch flashes simultaneously in square’s frame. Centroid is in the origin. Aperture comes to point y at the same moment, when beams simultaneously come to point y. Then the beams go through aperture simultaneously hit the film.

After visualization we will see four bright dots on the film. Image will be shifted a bit from the center. But square will turn into rectangle on the picture.

By aspect ratio (vertical vs horizontal) we can measure magnitude of contraction

Last edited: Mar 22, 2017
14. Mar 22, 2017

### Bartolomeo

By the way. In Terrell's paper sphere moves in camera's frame, not vice versa.
The problem is (as I see it) - who contracts? Train? Distance between twins? Or film in the camera?

15. Mar 22, 2017

### pervect

Staff Emeritus
I basically agree with that, and wouldn't want to confuse people by implying that I didn't. But I also wanted to say that thinking the two pictures are exactly the same is misleading, I don't think they will be. Perhaps the simplest appraoch is this. Suppose we have a pair of digital clock, with LED's that show the time numerically, and we take photos of the two digital clocks from a moving and stationary observer.

If the digital clock is small enough, the propagation time from each LED to the camera won't vary much. Then the image will be recognizable, and show the same time in both photos, but it will be distorted differently in the two photographs by relativistic effects. The relativistic effects that come to mind are aberration (which changes the size), doppler shift (which changes the color), and relativistic beaming (which changes the brightness).

I believe that for the moving observer (moving relative to the clock frame, where both clocks are at rest), the front image will wind up smaller, bluer, and brighter, and the rear image to be larger, redder, and dimmer. This is still assuming the approximation I mentioned above. See below.

1
2..............................C
3

If we imagine that 1,2, and 3 are pixels that light up "at the same time" (in the frame of the digital clock), and C is the camera, then the light from pixel 1 will take a bit longer to arrive than the light from pixel 2, because pixel 1 is I've ignored this issue and the resulting image distortions in my remarks. Treating it properly requires us not to think of the clock as one entity, but made of several parts, making the analysis rather muddled.

It's harder to analyze what happens to the image of the clock than to work out the time synchrhonization issues that started the thread. The OP seems to want to use the the "photograph" concepet, but that is making more work for themselves, not making it easier. A better approach, I think, is to imagine point-like clocks emitting radio or radar pulses, and a point-like radio receiver. For instance, one could imagine two radio stations transmitting time signals (like WWV does, if one is familiar with WWV). The human-audible part goes something like this. "At the tone, the time will be <#>, coordinated universal time. BEEEEP".

So we imagine two WWV radio stations in front and behind, and two radio receivers, one stationary (in the tower frame) between them at the center, the other moving.

In this WWV example, which is easier to deal with, the two signals will be doppler shifted differently. For the moving receiver, one signal will be blue shifted, making the carrier frequency higher, and the voice speak faster and higher pitched. The other signal will be red shifted, making the carrier frequency lower and the voice speak lower and slower. The non-moving observer won't see such effects, the non-moving observer will hear both voices normally. But if the stationary receiver is currently receving a "beep" from both radio towers, and the moving observer is at the same place as the stationary observer, it will also be receiving a beep from both towers - it can't be otherwise. If we take the limit for a very short "beep", we get the desired result.

16. Mar 22, 2017

### PAllen

To talk in any meaningful way about momentarily colocated cameras in relative motion, aperture must open and close effectively instantaneously; else colocation is nonsense. I was speaking about colocated, relatively moving, cameras imaging a clock. Having established that instant aperture opening is required, let's think about the flash. If a flash illuminating the clock is effectively instantaneous, timed such that some light from it hitting the clock reaches the colocated apertures, then the most that can be imaged for a general solid object is is 1 dimensional contour curve. To speak in any way of imaging the clock as a whole, the flash must be long enough so that some light from every part of the clock reaches the colocated apertures (this would be extremely short as far as real world flashes go, but, speaking in principle, this point is needed). Realizing these facts, everything I said about Terrell rotation applies. If you were talking about something other than imaging the clock, I have no idea what it was.

Note, also, that while the first paper on this involved a sphere, the effect applies to objects of all shapes as demonstrated in many later papers. Right here on physics forums, we worked out the case of rulers in detail, demonstrating the effect from first principles.

The upshot is that the the two cameras will image the same time on the clock, but the clock's position on the CCD detector, color, and apparent visual orientation (as captured on the detector) will be different.
This appears nonsensical. Simultaneous per who? If the light goes through two essentially colocated apertures at one event, and the cameras are relative motion, the events of reaching films are separated in space and time. They cannot be simultaneous in both frames.
It is true that for flat object, you can get a contracted image outline in one camera compared to the other. However, if there are features on the flat object, the image will look like the reason for the flattened cross section is simply that the object was turned.

Last edited: Mar 22, 2017
17. Mar 23, 2017

### Bartolomeo

The subject of the discussion were readings of the clocks and distance between them in case if flashes are simultaneous in trains frame.
Appearance of the clocks does not affect these parameters. Image may be fuzzy. In case of square I replaced clocks with lamps.
We work in train‘s frame. Simultaneously in reference frame of the train. Yes, moving cameraman will make a conclusion that flashes were not synchronous and asymmetrical.
If cameraman wants to get „right“ image, he has to launch flashes simultaneously in his frame. That means, that not camera moves in train‘s frame, but vice versa. It is another problem. Also quite entertaining. At least, readings of clocks on the picture will be different then. At least.
I also think so. Will distance between clocks at picture taken by moving camera be shorter than at a picture taken by one at rest?

Last edited: Mar 23, 2017
18. Mar 23, 2017

### Janus

Staff Emeritus
To answer this, let's look at another set of space-time diagrams.

Here will will assume that Jack has his own identical train which is at rest with respect to the tracks. Each world-line in these diagrams are separated by 1 "train car length" as measured in each rest frame.
First we have Jill's frame.

The flashes leave the clocks at simultaneously from a distance of 5 her train cars from Jill. Note when the flash leaves the right clock it is just a bit shy of 3 of Jack's train car lengths from Jack and when the flash leave the left clock, that clock is between the 8th and 9th od Jack's train car lengths. Thus is one direction, she will see ~3 of Jack's train cars stretched out over a distance of 5 of hers and in the other a over 8 of Jack's train cars compressed over a distance of five of hers. This puts a bit over 11 of Jack's train cars that fit between a 10 train car distance for Jill.

Now consider Jack's frame

Here we see that The flashes do not leave simultaneously. However the right flash still leaves while 5 of Jill's train cars from Jill, and a bit less than 3 of Jack's train cars from Jack, while the left flash leaves 5 of Jill train cars from Jill and over 8 of Jack's train cars from Jack. This means that in Jack's photo, the right clock will appear to less than 3 train cars away while the left clock will appear to be over 8 train cars away, and the distance between the clocks will be a bit over 11 train cars apart. To the right he will see 5 of Jill's train cars compressed into a distance of 3 of his, and to the left he will see 5 of Jill's cars stretched out over a distance of better than 8 of his. This of course doesn't directly tell him how far apart the clocks are as measured in his frame, because the light from each clock left at different times. Between the time a flash left the left clock and one left the right clock, the right clock had moved relative to Jack. To get a picture that shows how far apart the clocks are in his frame, the flashes would have to leave the clocks at the same time in his frame. In which case he would see the 10 of Jill's train cars contracted to fit into a distance taken up by 8.66 of his train cars.

19. Mar 25, 2017

### Bartolomeo

That means, that if flashes work simultaneously in train’s frame, on the picture taken by Jack (who is moving with camera) the train appears gamma times stretched and the clocks show the same time.

If flashes work simultaneously in camera’s frame, the train appears gamma times contracted and the clocks show different time. In this case Jack is at rest and the train is in motion.

We can also consider these pictures this way:

Imagine that observers $A$ and $B$ with synchronous clocks (in the Einsteinian sense) are located at points $a$ and $b$ on the $X$ -axis of the coordinate system $K$ symmetrically about the point of origin $O$. The coordinates of points $a$ and $b$ are, respectively, equal to $-x$ and $x$. On the $Y$ -axis at point $d$ with the coordinate $y$ there is a camera obscura with its opening directed to the point of origin $O$. Observers $A$ and $B$, who are in range of vision the camera obscura, wish to take their simultaneous photos. To do so, each of them at a point of time $t=0$ activates the flash. The distance from each observer to point $d$ is equal to $\sqrt{x^2+y^2}$. Therefore, after a time equal to $\sqrt{x^2+y^2}/c$., the light pulses reflected from the observers simultaneously fall into the opening of the camera obscura, and this pair of observers $A$ and $B$ is captured on the film. The photo is not distorted. The height of the observers, their cross-sectional dimensions and the distance between them on the photo are proportional to their actual size.

image hosting without registration

Now imagine that the camera obscura is not at rest at point $d$, but moving parallel to the $X$-axis with velocity $v$, getting to point $d$ at the moment of time to $\sqrt{x^2+y^2}/c$. The light reflected from the observers, which at this time has simultaneously come to point $d$, falls into the aperture of the moving camera obscura, and instantly falls on the film, thus producing an image. We believe that the dimensions of the camera obscura in the direction transverse to its motion are small enough and in the direction of its motion are large enough for the film not to go beyond the incident image falling on the film during the light travel period from the aperture to the film.

What kind of distortions appear in the photos and why?

Because of the reduction of the film moving in the coordinate system $K$ the image obtained after visualisation will be stretched in the transverse direction (in the direction of motion) $1/\sqrt{1-v^2/c^2}$ times. The figures of the observers will thicken, and the distance between them will increase. If on the photo one could see the clocks on the observers' hands or close to them, these clocks will show the same time, i.e. that the observers' wish to be photographed at the same moment in time would be fulfilled.

Yet the observers within the reference frame $K'$ where the camera obscura is at rest and where the observers $A$ and $B$ are moving, will think that the photo is "wrong".

They will assume that the flashes were activated asynchronously and asymmetrically to the place of exposure from the points located at different distances from the place of exposure. From their point of view, the light pulses, having simultaneously passed the aperture of the camera obscura at different angles to the plane of the film, did not simultaneously fall on the film and formed the image not at the moment in time $t’$, but during period of time $\Delta t’$.

Let's assume now that in order to obtain a "correct" picture though not satisfying the customers, the observers A and B, the observers within the reference frame $K’$ will activate the flashes simultaneously at the moment in time $t’$, such that the pulses reflected from the observers $A$ and $B$ moving within the reference frame $K'$ will simultaneously arrive at the aperture of the camera obscura. As the distance between the observers and their transverse dimensions are $1/\sqrt{1-v^2/c^2}$ times less than their proper distance and dimensions, the image on the film becomes contracted, so the observers $A$ and $B$ on the photo become "thinner" and the distance between them becomes less. The clocks located close to the observers will show different time, i.e. the observers' wish to be photographed at the same moment in time would not be fulfilled.

The observers within the reference frame $K$, wherein the camera obscura is moving, will assume that the light coming from the flash was emitted asynchronously and asymmetrically in relation to the place of exposure from points located at different distances from the place of exposure. They will assume that the light pulses, having simultaneously passed the aperture of the camera obscura, fell on the film asynchronously and formed the image not at the moment in time $t$, but during period of time $\Delta t$

According to said above, is this statement correct? Length contraction is the phenomenon of a decrease in length of an object as measured by an observer who is traveling at any non-zero velocity relative to the object." https://en.wikipedia.org/wiki/Length_contraction.

Also, which picture is the correct one? Does Jack's picture have physical sense? I think it does, because:

Readings of the clocks on the ends of moving train are always different. The rays went through aperture simultaneously. But on the picture readings are the same. Hence, I have taken a picture of Jill's train which is “at rest”. But, we are in a state of relative motion. Thus, I with my photo camera move myself. Since I and my film in the camera contract ourselves. Since film in photo camera contracts, Jill's train appears stretched after visualization. All that means, that I was in a state of relative motion to Jill and took a picture of her train “at rest”.

Do you agree with that?