# B Is a "meter" observer dependent?

1. Apr 20, 2017

### ajay.05

SI defines one meter as the distance travelled by light in one upon 299792458(=k) seconds, from which we get that,
D=c*k, where k is marked above.
So, let's consider two frames on at rest wrt the earth(f1) and another moving at, say 0.99*c(f2). SR says that the time dilates for f2 wrt f1. And also say that someone is trying to mark a meter distance, from their own frames. As time is different for them, doesn't it mean that the one meter mark they have made is different? Is this really happens?(I mean, am I correct? Or just I'm confusing things?) Does this occur along with Lorentz contraction?
Thanks in advance.

2. Apr 21, 2017

### andrewkirk

To successfully analyse this question, it's necessary to be absolutely clear about what 'mark a meter distance' means. Different interpretations of that will give different answers. For instance, is an observer trying to use a pen to draw two marks, one metre apart, on a straight rod that is co-moving with her?

3. Apr 21, 2017

### FactChecker

Yes, the one meter marks would be different. Time dilation and length contraction are both consequences of the Lorentz transformation.

4. Apr 21, 2017

### ajay.05

I tried to mean like this....Think of a giant wall(stationary wrt earth) beside both of them, and think that they have the ability to mark exactly as per the SI definition on that wall...Now, when a beam of light passes they do their measurement. Now, as time dilates for the person in motion, shouldn't the meter marked by him would be small?

5. Apr 21, 2017

### ajay.05

I would like to know whether this occurs along with Lorentz contraction due to OUR definition of a meter.

6. Apr 21, 2017

### Ibix

You will get the same result whether the person making the marks defines the distance between the chalk marks using the flight time of light or a copy of some standard metal rod, if that's what you're asking.
Not according to him - the marks will be 1m apart because that's what you have declared that he did. According to an observer at rest on the wall, the marks will be more than a meter apart, I think. Note that you'll have to be careful about the procedure for marking the wall since it's moving with respect to the person doing the marking.

You should use the full Lorentz transforms to see why this all follows.

7. Apr 21, 2017

### ajay.05

Thank you...Helped me a lot

8. Apr 21, 2017

### andrewkirk

You could imagine the following: Alice is moving at 0.8c parallel to the wall and very close to it, while Bob is floating right next to the wall. Both of them have a special marker gun that is 1m long and, when they press the trigger it shoots out thin, powerful laser beams capable of burning a mark on the wall from two places that are exactly a metre apart on the gun. The two laser beams are perpendicular to the length of the gun and parallel to each other.

Bob is holding his gun parallel to the wall with the lasers pointing at the wall.
Alice is heading parallel to the wall towards Bob, just far enough away from the wall that her ship won't hit Bob.
Bob's and Alice's laser markers are both oriented parallel to each other and to Alice's direction of travel.

Alice fires her marker when the forward laser of her marker is level with that of Bob, and Bob fires his marker at the same time (which means something in this case because Alice and Bob are in almost the same location).

Alice's and Bob's lasers will both make a pair of marks on the wall, with Alice's forward mark being level with Bob's. If Bob measures the distance between Alice's marks it will be less that 1m while he will measure exactly 1m between the marks he made. Conversely, to Alice it will look like Bob's marks on the wall are less than 1m apart while hers look like they're exactly 1m apart.

9. Apr 21, 2017

### ajay.05

Wow...Thank you

10. Apr 21, 2017

### Ibix

But it only means something for one end of their meter-long dual-laser contraption. Which means that you have the following bit wrong (Edit: assuming that you intend both ends of each rod to fire simultaneously in their own rest frame):
If $\Delta t'=0$, then $\Delta x=\gamma\Delta x'$.

I agree the rest.

Last edited: Apr 21, 2017
11. Apr 21, 2017

### FactChecker

This may help: If you are stationary and another man is moving, you two can not agree on whether two events that are separated in the direction of motion are "simultaneous". To measure the length of a moving rod, the location of the two rod ends must be marked simultaneously. Suppose the moving man has a rod that he thinks is one meter long and marks both ends of the rod on a stationary wall at what he thinks is the same time. You will not agree that he marked both ends at the same time. You will think that he delayed marking the front end and consequently stretched the marks. So you would say that he cheated and the length of the moving rod is shorter than the marks.

12. Apr 22, 2017

### nitsuj

Light clocks are great for seeing length contraction....visualize it in motion compared to you. Assume a 1 meter (at rest) separation between mirrors (i.e. the meter stick). Seeing that zig-zag motion of the photon while the clock moves (photon motion perpendicular to clock motion.....now turn the clock so the photon travels horizontal to the direction of motion of the clock. From your perspective the length between the mirrors has contracted.

13. Apr 22, 2017

### pervect

Staff Emeritus
There is some linguistic confusion here. A proper meter is observer independent, and I've read at least one paper that mentions that the SI meter (the distance light travels in 1/c of a second) should be understood as talking about proper distance. (I don't recall exactly where I read this, alas).

A proper meter can be thought of as having the proper length in the observers rest frame, though mathematical techniques are well-known as to how to calculate it in any frame or coordinates one wishes to use.

But it's also not uncommon to use non-proper meters, which ARE observer dependent - hence the confusion.

So for this specific question, I would suggest thinking of the SI definition of the meter as being that of an observer-independent proper meter. Operationally, one measures the round-trip time of light in some particular observers frame at a single point, and divides by two to get the one-way travel time. Multiplying half the round trip travel time by the defined constant c gives the proper length.

The round-trip travel time, measured by one physical clock without recourse to any clock synchronization, is a measurement of proper time, and the distance the light travels in this proper time is a proper length.

The term "proper time" is important here, and I've given some hints, but let me try to be more explicit about what makes a time interval "proper". A time interval is proper when it can be measured by a single clock. A time interval measured by synchronizing two clocks is not proper, because clock synchronization is inherently observer dependent in special relativity. See any of the discussions of "Einstein's train" or "the relativity of simultaneity" on this last point.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted