1. Nov 16, 2014

### jbmolineux

Imagine a stream of photons moving from the sun to the earth (A, B, C, D, E, F, G, etc.).

Now imagine 2 different men measuring the speed of the photons, and (somehow) recording which photon is hitting them. They are in the same position, but one is traveling 1/2 the speed of light (relative to the earth) toward the sun, and the other is travelling 1/2 the speed of light away from the sun. Say they are giving each other a high-five as they pass and doing the measurement at the same time that their hands slap.

The question I want to ask, is which photon in the stream are each of the measuring? For the guy who is moving toward the sun, the photon-stream is moving towards him at the speed of light (but 1.5 the speed of light relative to the earth, since he's moving 1/2 the speed of light toward the sun). But for the guy moving away from the sun, the photon stream is moving toward him also at the speed of light, but only 1/2 the speed of light relative to the earth.

Therefore, aren't they going to be measuring two totally different photons to be in the same place at the same time as they pass each other? Won't the first guy be measuring a much earlier-in-the-stream photon, while the guy moving away measuring a later-in-the-stream photon? How is this explained?

2. Nov 16, 2014

### Staff: Mentor

You specified that the two men are right next to each other when they are donig their measuring, so they must be measuring the same burst of photons.

3. Nov 16, 2014

### jbmolineux

Yes, but given that the photons are moving 1.5 SOL relative to the earth for the first man, and .5 SOL relative to the earth for the second man, the photons would be getting there faster for the first man.....right? And so shouldn't it therefore mean that the first man would measure a photon much earlier in the stream?

4. Nov 16, 2014

### Staff: Mentor

No, you always measure the speed of the photons to be C (the speed of light) regardless of the speed of the observer. That's one of the groundbreaking parts of Special Relativity:
http://en.wikipedia.org/wiki/Special_relativity
(note, the speed of the source and speed of the observer are interchangeable.)
Again: if the photons and men are all in the same place, they will detect the particular group photons at the same time. But a bit later, the first man will pull ahead of the second and will measure the next group at different times.

5. Nov 16, 2014

### jbmolineux

Yes, I know. That's what creates the paradox.

Observer 1 - moving toward the sun at 1/2 SOL, for him the velocity of the photon stream is the SOL, and by extension, the velocity RELATIVE TO THE EARTH is 1.5 SOL

Observer 2 - moving away from the sun at 1/2 SOL, for him the velocity of the photon stream is also SOL, and so by extension the velocity for him relative to the earth is .5 SOL

Given the fact that each observer experiences the velocity of the photon beam to be the same velocity (SOL) despite his movement, it must FOR HIM be moving faster relative to the earth according to his own movement in relation to the earth.

And when the stream is moving faster for one observer compared to the other, a different photon in the stream would be measured at the same time.

6. Nov 16, 2014

### Staff: Mentor

If we call the first photon emitted A, the second B, and so forth...

Suppose that the inbound and the outbound observers pass each other just as B comes by, so they'll both detect B at their common location as the same time as they exchange their high-five.

The observer on earth will order the various events as follows:
1) A reaches the outbound observer
2) A reaches the earth
3) A reaches the inbound observer
4) B and the outbound observer reach the earth from one direction while the inbound observer reaches the earth from the opposite direction as the two observers exchange high-fives.
5) C reaches the inbound observer
6) C reaches the earth
7) C reaches the outbound observer.

For simplicity, let's say that the distance between sun and earth is ten light-minutes. At the moment that photon B is emitted according to the earthbound observer, the outbound observer would have to be five light-minutes away from the sun and traveling outwards at .5c while the inbound observer is 15 light-minutes away from the sun and traveling inwards at .5c. Otherwise you can't get them both to arrive at the earth at the same time for the high-five.

7. Nov 16, 2014

### Staff: Mentor

There is no paradox here, just a misunderstanding.
That doesn't make sense. You didn't say the velocity of what, but I assume based on the other scenario you mean the velocity of the observer relative to the earth. The earth and sun are not moving linearly relative to each other, so if a person is moving toward the sun at .5C they are moving away from the earth at .5C, not 1.5C. In either case, the velocity of anything relevant to the earth doesn't have anything to do with the problem.
What is "it"? Are you talking about the speed of light in the Earth's reference frame as calculated by the observer? Sure, you can get odd results if you calculation the speed of light for one frame from another. Not sure what this has to do with anything...
Again, no: you specified that the two observers are next to each other. They can't be receiving different packets of photons if they are at the same place at the same time. There is no paradox here, what you are saying is just wrong.

8. Nov 17, 2014

### ghwellsjr

Before we get into the details of your scenario, we have to be clear on how each man is going to measure the speed of a photon. If you're going to measure the speed of anything, you have to detect when that thing passes by a clock and then detect when it passes by a second clock some measured distance away. Unfortunately, we can't detect when a photon passes by unless we stop it and that messes up the measurement. Instead, let's consider, as russ watters did in post #2, a burst of photons so that all we have to do is detect a small percentage of them and let the rest pass by on their way to the second clock. This means a steady stream of photons coming from the sun is not going to do it. We have to block that stream and then briefly let some photons through and then block it again. Since you had the men giving high-fives to each other we can let their hands do this blocking process.

Well that's just one problem. A second problem is that they still can't do the measurement the way I described because it requires two clocks that have to be set to the same time and that is not a trivial task so for the time being, let's change the way they do the measurement from a one-way measurement to a two-way measurement. Now, instead of a clock at some measured distance away, let's put a mirror there and reflect the photons back to the first clock where the man is.

Here is a spacetime diagram showing the rest frame of a man located along the blue line with a clock ticking off 1 nsec intervals of time depicted by the dots and a mirror located 6 feet away depicted by the thick black line. Photons depicted by the thin red line coming from the sun off to the left pass by the man and his hands and his clock when it happens to read 0. Then it hits the mirror 6 nsecs later and reflects back to him when his clock reads 12 nsec:

Since the round-trip distance is 12 feet, he concludes that the speed of the photons is 1 foot per nsec.

Now we have to transform all the coordinates of the events in the above diagram to one in which one man is moving at 0.5c towards the sun:

We see that because of length contraction (the mirror is closer to him) and time dilation (his tick marks are farther apart) and relativity of simultaneity (events that were simultaneous in his rest frame are not in this frame), he still measures 12 nsecs as the time for the round-trip so he comes to the same conclusion that the photons traveled at 1 foot per nsec.

We can do another transformation for the second man moving at 0.5c away from the sun but we'll show him in red with a green mirror:

If you look at the individual diagrams for each man, you can see that the time it takes for the photon to go in each direction has the 3x ratio (1.5/0.5 = 3) that you are considering but it is in the opposite order for the two men since they both are making round-trip measurements. Here is a spacetime diagram showing both of their measurements together:

Actually, they are measuring the same burst of photons, it's just that the photons reflect off the first man's mirror earlier than the second man's mirror.

Now I'm going to show you how the men can make one-way measurements with two clocks that have been synchronized. They have to synchronize their two clocks before they make their measurements and they do it using the same technique that they used for making the round-trip measurements. Look at this diagram:

At some time, the blue man sends a burst of photons to his second clock and then when he sees the reflection off the clock, he also sees what time is on the second clock. He averages the time on his first clock when he sent the burst with the time he sees the reflection and he adjusts the second clock so that he will see that time when he repeats the measurement. Eventually, he will get the two clocks synchronized as shown in the example above. The other man does the same thing.

Now we can see what happens when the two men pass as shown in this diagram:

And once again, they both measure the speed of the photons to be 1 foot per nsec even though the light takes three times longer for one man than the other.

Last edited: Nov 17, 2014
9. Nov 17, 2014

### pervect

Staff Emeritus
For starters, lets replace "photon" with "very short pulse of light", as SR is a classical theory.

As a classical object, hopefully it's clear that if two people are measuring the same pulse of light, say pulse A in your stream, and the two people are in the same place, they are measuring the same pulse, not different ones?

The weirdness of quantum mechanics is a separate weirdness, you can think of the short pulse of light as the ensemble of all the photons in it, but the pulse behaves itself, the collection of photons, behaves in a purely classical manner.

In order to measure the speed of the two pulses as the same, there is one familiar concepts which has to be adjusted, but you've singled out the wrong one to focus on. If you google any of the threads on the topic, or find a textbook to read, or google for the catchphrase I'm about to mention, you'll find that the concept that needs to be adjusted is called "the relativity of simultaneity".

There's even a paper about the topic that I mentioned in another thread, "The challenge of changing deeply held student beliefs about the relativity of simultaneity". I'm not sure if it's the right level for you or not, but at least it points out where the real problem lies.

10. Nov 17, 2014

### jbmolineux

Wow thanks everyone! I don't fully understand these answer, but I do get a clear sense that the answer is basically, "they nonetheless end up measuring the same photon in a way that is well-understood by SL," so there is no paradox. Thanks for taking the time!