# I Why can't you use Simultaneity when doing Length Contraction

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1. Jun 30, 2016

### Afterthought

Suppose that frame O' moves at speed v = 0.6c relative to frame O. A rod with two balls is attached to its ends is 10 meters long in its rest frame, O'. Length contraction will tell you that in frame O, the rod is 8 meters long.

But aren't the two balls at the ends "events"? They are clearly located at some location and some time. In which case relativity of simultaneity would tell you that in frame O, you don't see both balls at the same time! Instead this would imply the bizarre scenario that O doesn't see the the entire rod at some instance, but rather sees it slowly materialize.

I feel like I'm missing something here..

2. Jun 30, 2016

### Ibix

The balls aren't events.

An event is a point in space at a point in time, and the balls don't exist at one instant in time - they exist for an extended period. They are described by worldlines, which is the line you get when you join up all the events that each ball passes through.

3. Jun 30, 2016

### Staff: Mentor

Each ball is at some point in space at any given time. So....
No, because at any moment both balls are somewhere in both frames so both observers can always see the entire rod. Relativity of simultaneity just says that when the two balls are at two particular points in space at the same time according to one observer, the other observer will find that they are at those two points, but not at the same time.

4. Jun 30, 2016

### m4r35n357

Yes, but stick with it. Length contraction is quite an abstract concept, and you certainly don't "see" it in the sense that you seem to mean. You can only "reckon" or "calculate" it, but you have to be careful interpreting what it represents (different parts of the rod really do exist at different times according to a moving frame, that is exactly what the equations say). The word "see" is bandied around freely in introductory SR texts and courses, but working out what you really see requires further processing to take light travel time into account.

For now just treat it as a calculation and take the numbers literally without trying to interpret it as seeing. You cannot "see" simultaneity, (which is spacelike), you can only see down null intervals (which are timelike).

Sorry I couldn't be more constructive, but I find your post to be a textbook illustration of my personal issues with how SR is taught. Perhaps someone else can do better and I will see (sic) the light!

5. Jun 30, 2016

### Afterthought

Ah, so the balls aren't events. But if the rod magically appeared and disappeared an infinitesimally small moment later in frame O', wouldn't the balls qualify as events then, leading you to seeing a materializing rod in frame O?

Can you clarify what you mean by "at any moment both balls are somewhere in both frames"?

6. Jun 30, 2016

### Ibix

The balls are still not events, but the ball at the time it was attached to the rod is an event, yes. If the rod appears and disappears in O' its appearance will not be simultaneous along its length in O. In O, you'd see the left hand end (for the sake of argument) then a bit to the right of that, then a bit to the right of that, and so on. It would look rather like a piece of rod moving from one ball to the other, although careful observation would tell you that you're seeing a continuously changing segment of the rod.

7. Jun 30, 2016

### Afterthought

This is what I meant by "materializing". Anyways, thank you, I think I'm slowly starting to get it.

Let's say the rod is a light source whose frequency increases by time but not by distance in the frame of O'. When you see the rod in the frame of O at any instance, would you see a spectrum of frequencies, with the "left" side having the greatest frequency, and the "right" side having the smallest frequency? (left and right meaning in the normal way people would draw this space time diagram).

8. Jun 30, 2016

### Ibix

Anything that's simultaneous in one frame is non simultaneous in another. So if, in one frame, the rod is changing colour smoothly and simultaneously then in any other frame the colours will flow along its length. Spacetime diagrams are an excellent tool for understanding this if you know how to draw them. Draw a set of parallel sloped lines of different colours, then move a horizontal ruler up the page. Colours will flow along the edge of the ruler.

Note: that isn't quite what you would see - the spacetime diagram doesn't include effects of the lightspeed delay. What you would actually see depends on the lightspeed delay (as @m4r35n357 noted in #4), which depends on where you are.

9. Jun 30, 2016

### Afterthought

Thank you again! I'd consider my question to be answered now that I know that.

10. Jul 3, 2016

### m4r35n357

This is aimed at the OP, but follows from your post.

I decided to do a diagram for a pair of 4 unit length rods (blue) moving at $v = 0.6c$ through the observer's frame (black), to illustrate how the light delay is handled, and how it depends on whether the rod is approaching or receding. The rods are shown every four time units in their own frame. There is a minimum of annotation, but hopefully it should be clear, as it is all to "scale". The observation is made (red for light) when the near edge of the approaching rod is equidistant to the trailing edge of the receding rod.

I have highlighted the figure for "length contraction" in magenta (3.2), and the distances of what you see in green (8.0 and 2.0, depending on which rod you are looking at).

There are many readings of time that can be made in both frames, but I shall not cover them here! I just thought it might help to show a bit of basic geometry.

Last edited: Jul 3, 2016
11. Jul 3, 2016

### Mister T

You don't need to think about them magically appearing and disappearing. All you need to do is note their positions at some time. Say that in O' one ball has a position $x'_1=0$ at time $t'_1=0$. And at the same time the second ball is located at $x'_2=10$ at $t'_2=0$.

What a theory of relativity does is provide the mechanism for calculating the values of $x$ and $t$ in frame O for each of these two events.

12. Jul 3, 2016

### Staff: Mentor

Make an image file out of it. For this type of image (line drawings with lots of white space), GIF is best because it makes the smallest files. JPG, PNG and some other formats also work. Then use the UPLOAD button like you did with the PDF file. After uploading, you'll see buttons labeled THUMBNAIL and FULL IMAGE. The FULL IMAGE button inserts the image directly into the post.

13. Jul 3, 2016

### m4r35n357

Thanks for the encouragement ;) I tried it earlier, but it looked rubbish, because of the alpha channel. Now edited!

14. Jul 3, 2016

### FactChecker

"some time" is ambiguous. O and O' can not agree on what "some time" means. Suppose each frame has independently synchronized a string of its clocks, clocks C for O and C' for O', using light signals, along the entire coordinate line. That defines what "simultaneous'" means in each coordinate system. Suppose they have arranged so that their clocks are exactly matched when the lead ball is at a point A=A' ( That is, C(A) = C'(A') when the lead ball is at A and at A' ). Then their clocks will not agree at the trailing ball. Each frame thinks that the other frame clocks are more wrong the farther away they are from the matched pair.