How Fast Does Time Elapse for Us From a Photon's POV?

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How fast does time elapse for us from the point of view of a photon watching us?

Thanks,

Jake
 
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Fredrik said:
See https://www.physicsforums.com/showthread.php?p=2650120.

(Also, your question kind of contradicts itself, since "for us" sounds like reference to our point of view. I'm guessing you meant to ask how fast our clocks are from a photon's point of view).

Yes, that's what I meant: "how fast our clocks are from a photon's point of view" and I read some of the thread you referred to. So let's change the question: How do our clocks appear to a particle other than a photon accelerating and getting very close to the speed of light? From that particle's point of view, are our clocks moving faster than they are in our point of view?

Thanks,

Jake
 
From the point of view of a particle moving very close to the speed of light, our clocks are moving extremely slow.

Time dilation, not time contraction. ;)
 
Matterwave said:
From the point of view of a particle moving very close to the speed of light, our clocks are moving extremely slow.

Time dilation, not time contraction. ;)

I didn't say time contraction.
 
I know, but the implication that to the particle our clocks moving faster would imply time contraction.
 
Matterwave said:
I know, but the implication that to the particle our clocks moving faster would imply time contraction.

Is it true that if an astronaut, who has been traveling near the speed of light relative to the Earth for a long time, comes back to earth, time will have elapsed slower for him than for us? Doesn't that necessitate our clocks moving faster in his point of view while he was moving faster? How else would we be older and him younger?
 
If person A were moving very fast relative to person B then, yes, person B would see Person A's clocks moving more slowly than his and see person A aging more slowly.

But it is also true that, from A's reference system, B is moving very fast relative to A. Person A would see Person B's clocks moving more slowly and see person B aging more slowly.

But unless A and B can "get together" in the same reference frame (same speed) there is no paradox. And to get into the same reference frame, one must accelerate. That breaks the symmetry.

(Well, of course, they could accelerate symmetrically- then, when they got into the same reference frame, they would find their clocks and their aging to be the same.)
 
So the astronaut thing you hear so many times is a myth?
 
jaketodd said:
Is it true that if an astronaut, who has been traveling near the speed of light relative to the Earth for a long time, comes back to earth, time will have elapsed slower for him than for us? Doesn't that necessitate our clocks moving faster in his point of view while he was moving faster? How else would we be older and him younger?
If we insist on defining his "point of view" at any event on his world line (the curve in spacetime that represents his motion) as the co-moving inertial frame at that point, then he would "see" the clocks on Earth tick ahead with an enormous rate as he slows down to a stop (relative to Earth) and starts speeding up in the opposite direction. (I don't mean that this this is what he would see through a telescope. I'm talking about how he would describe what happens on Earth if he records what he sees through the telescope and then compensates for light travel time).

This is a spacetime diagram I made for another one of these threads a couple of years ago, which shows the Earth twin's point of view, and explains how the other twin would describe things some of the events on his world line (when we use co-moving inertial frames to define the "point of view").

[PLAIN]http://web.comhem.se/~u87325397/Twins.PNG
Fredrik said:
I'm calling the twin on Earth "A" and the twin in the rocket "B".
Blue lines: Events that are simultaneous in the rocket's frame when it's moving away from Earth.
Red lines: Events that are simultaneous in the rocket's frame when it's moving back towards Earth.
Cyan (light blue) lines: Events that are simultaneous in Earth's frame.
Dotted lines: World lines of light rays.
Vertical line in the upper half: The world line of the position (in Earth's frame) where the rocket turns around.
Green curves in the lower half: Curves of constant -t^2+x^2. Points on the two world lines that touch the same green curve have experienced the same time since the rocket left Earth.
Green curves in the upper half: Curves of constant -(t-20)^2+(x-16)^2. Points on the two world lines that touch the same green curve have experienced the same time since the rocket turned around.
It actually makes more sense to define his "point of view" as the coordinate system constructed using the synchronization procedure I mentioned in the thread I linked to. (This procedure only produces inertial frames when it's applied to an object that never accelerates (and never rotates)). See this article for more about this definitio of "point of view", and how to use it to resolve the twin paradox.

If we don't care about "points of view" and only want to know how SR predicts that the astronaut twin will be younger, the answer is that it follows immediately from an axiom of the theory: A clock measures the proper time of the curve in spacetime that represents its motion.

jaketodd said:
So the astronaut thing you hear so many times is a myth?
That's not what he said at all.
 
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Fredrik said:
If we insist on defining his "point of view" at any event on his world line (the curve in spacetime that represents his motion) as the co-moving inertial frame at that point, then he would "see" the clocks on Earth tick ahead with an enormous rate as he slows down to a stop (relative to Earth) and starts speeding up in the opposite direction. (I don't mean that this this is what he would see through a telescope. I'm talking about how he would describe what happens on Earth if he records what he sees through the telescope and then compensates for light travel time).

First of all, thank you. I have some questions: What do you mean by "light travel time" in the above quote? Do you mean he compensates for the time it took the light from Earth to reach him at his current, far away location? Because he is younger when he returns, could you say that he "saw" the Earth's events go in fast forward?

Thanks again,

Jake
 
jaketodd said:
So the astronaut thing you hear so many times is a myth?

No, it isn't a myth, it was verified experimentally. Google the "Haefele-Keating" experiment.
 
I am confused. If someone can go into space and return younger than their twin who stayed on earth, then from the astronaut's point of view, events on Earth must have progressed faster than from the twin on Earth point of view. Right? How else could the astronaut return and find their twin older unless the passage of time was different between them? And what does Fredrik mean by "compensates for light travel time"? Does he mean the astronaut compensates for the time it took the light from Earth to reach him at his current, far away location? Since the astronaut is younger when he returns, could you say that he "saw" (with the compensation) the Earth's events go in fast forward when he was in space?

Thanks for bearing with me guys, I really want to understand this,

Jake
 
Since the astronaut is younger when he returns, could you say that he "saw" (with the compensation) the Earth's events go in fast forward when he was in space?
He sees (without compensation) Earth events go slow motion on the outbound trip, and fast forward on the inbound. Net effect is fast forward.
With compensation: He sees Earth eventsin slow motion on the outbound trip, and slow motion on the inbound trip. During turnaround, he has to adjust his compensation procedure, such that directly after turnaround (with the new procedure) the Earth is quite a bit older than directly before turnaround.
Note: that's a calculated, adjusted "time warp", not something observed. Nothing jumps into the future, the astronaut simply uses a different coordinate system after turnaround.
 
Ich said:
He sees (without compensation) Earth events go slow motion on the outbound trip, and fast forward on the inbound. Net effect is fast forward.

Why would the net effect be fast forward? You're saying on the inbound trip the magnitude of fast forward is greater than the magnitude of slow motion on the outbound trip? How do you arrive at this?

Thanks!

Jake
 
jaketodd said:
Why would the net effect be fast forward? You're saying on the inbound trip the magnitude of fast forward is greater than the magnitude of slow motion on the outbound trip? How do you arrive at this?
You can use the time dilation formula or calculate the proper time of the spaceship's world line to see that only 24 years passed on the ship.

A different (and less exact) way is to imagine lines parallel to the lower dotted line in the diagram, that intersect the t axis at the beginning of each year. Those lines represent light pulses sent from Earth once a year. It should be obvious from the diagram that the spaceship will (literally) see a lot more of them on the return trip than on the outbound trip. If you count the number of pulses that reach the spaceship during the two parts of the trip, you can determine the "speed up" and "slow down" factors as "pulses received"/12.
 
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How do you arrive at this?
Sorry, I forgot http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html" .
Read all the explanations there, it should help.
 
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