# Help me reconcile this

1. Jul 26, 2011

### Jon Bright

I'm at a bit stumped with respect to simple thought experiment.

Imagine I fire a photon at a mirror and observe the photon returned. First, I send off a photon. It traverses space to the mirror. It bumps into an electron in the mirror, gets absorbed, the electron has no stable orbit for the new energy level, so returns to its previous state emitting the excess energy as a photon. This new photon happens to come back at me, and I observe it. Lets simplify it to three stages: travel1, bump&grind, travel2.

Now, when I actually observe the returned photon, is when the mirror emitted it. In other words, travel2 is zero (by definition). There is no distinction in my timeframe. If I measure the total time that passed however, and subtract the time it takes for the mirror to send me a photon in return (i.e. subtract the b&g part), I know this equals travel1+travel2. But travel2 is 0, leaving me with a travel1 too big (by a factor of 2) and hence an overestimate the distance to the mirror.

I'm not sure where I'm failing. Is it simply that I didn't bother to Lorentz transform? If I transformed my and the mirror frames, would I find that the distance is half that which would be suggested by 'travel1'? But then, wouldn't I also find that time has elapsed between the mirror's emission event and my observing it? Wouldn't this contradict defining travel2 as zero?

Note that I'm not so much interested in actual measurements, or making it add up, as I am in figuring out the correct relative timeline for the events - in my time frame.

2. Jul 26, 2011

### Mentz114

I don't follow why you say travel2 is zero. Surely if the mirror is distance d from your eye, the travel time is d/c ?

3. Jul 26, 2011

### Jon Bright

But until I observe it emitting a photon, has it actually done so?

By correlation, say if I observe a supernova 450000 ly away. In my timeframe, did it occur 450000 years ago, or does it occur when I observe it?

4. Jul 26, 2011

### Mentz114

Yes.
It would take the light about 450000 years ( with some model dependency) to reach you, so it happened in your distant past.

5. Jul 26, 2011

### pervect

Staff Emeritus
This seems to be mostly about quantum mechanics, you might want to repost in a QM forum.

One way of dealing with quantum mechanics is to think of particles as having multiple histories. But once you've observed the photon, the only possible history is that it was emitted. This emission however happened in the past in your reference frame.

6. Jul 26, 2011

### ghwellsjr

There's no way to know how the total round trip measured time is partitioned by nature between t1 and t2 and so you are free to partition them any way you like but if you are going to follow Einstein's convention, then you will divide the measured time in half and assign each half to t1 and t2.

I am assuming the bump&grind time to be negligibly small compared to the total measured time.

7. Jul 26, 2011

### Jon Bright

My issue with it having already occurred (i.e. there's a distinction between a information arrival event and a past event the information is about) is that in the mirror case I know in advance, with 100% mathematical certainty, that I'll be getting a photon back - eventually if I repeat sufficiently many times. (I.e. for any given run the probability is non-zero.) But how can I know this, and when it occurs, if the information hasn't arrived yet? I guess the reconciliation is that a prediction even with 100% certainty is not the same as information, or observation. So 'travel2' is really a period of prediction, which then retroactively is confirmed when the information arrives. (So the terminology I'd use to describe the timeline may vary depending on where I'm at. I.e., my observer state. This makes good sense.)

8. Jul 26, 2011

### Staff: Mentor

Why pick on 'travel2'? 'travel1' is just as much a prediction until you confirm that you've received the reflection.