The passage of time is relative depending on whether one is the stationary observer or the object/particle traveling at the speed of light (or close to it). I get this, kind of. But, when we talk about "it takes light X [units of time] to travel Y [units of distance]", where does time dilation come in? In the case of light's journey from the Sun to Earth, if we imagine the photons doing the actual speed-of-light travelling, is it 8 minutes for "them"? Or is it 8 minutes for us to observe their journey? Apologies in advance if it's a stupid question.
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13$\begingroup$ There is no reference frame in which the photon would be at rest. The speed of the photon is c in every reference frame. So a photon perspective does mot exist $\endgroup$– ThomasCommented Mar 27, 2022 at 16:17
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3 Answers
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If I interpreted this article correctly, then the answer to my question should be: 8 minutes is what we perceive, whereas for the photon the journey is instantaneous, due to the fact that it travels AT the speed of light, not close to it, at which point "the photon itself experiences none of what we know as time: it simply is emitted and then instantaneously is absorbed, experiencing the entirety of its travels through space in literally no time. Given everything that we know, a photon never ages in any way at all."
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20$\begingroup$ That is a good self answer. One thing to be careful of: don't try to imagine yourself moving "at the same speed as a photon". That's not just impossible, its non-sensical. Such a frame of reference is singular and you can't use it to make any predictions. One further point is that not only is there no time passing but space is infinitely contracted so "from the point of view of a a photon" the Sun and Earth are in the same place. This is why we can't really talk about what it experiences... $\endgroup$– James KCommented Mar 26, 2022 at 18:39
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10$\begingroup$ I disagree with this answer in the sense that there is no physical reference frame comoving with a photon. "What a photon sees" is not merely practically unobservable, but undefined even in principle. $\endgroup$– d_bCommented Mar 27, 2022 at 3:30
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4$\begingroup$ @d_b You’re probably right. However, the „photon perspective“ makes the answer so interesting that I would rather not remove that part. $\endgroup$ Commented Mar 27, 2022 at 9:30
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3$\begingroup$ @d_b I don't know enough of the mathematics to bring any rigour to this, but I believe you can sort of take the limit as speed approaches the speed of light, and see what that does to reference frames. My understanding is that would make the travel time and travel distance measured by someone moving in that frame get arbitrarily close to zero. So if you want to be mathematically rigorous yes, there is no reference frame co-moving with a photon. But there are ones that are arbitrarily close, and considering them might be a useful intuition-building exercise. $\endgroup$– BenCommented Mar 28, 2022 at 2:16
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4$\begingroup$ @Ben: There's one big problem with that limit: You're applying a mathematical law to an empirically derived formula, while that formula is not validated in the range where you are applying that limit. That's the sort of math that led to the UV catastrophe. $\endgroup$– MSaltersCommented Mar 28, 2022 at 13:45
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When people talk about the time taken for light to go from the Sun to the Earth, they're generally just considering classical Newtonian mechanics, not relativity. So we simply divide the distance by the speed of light to get the time taken.
There are several other variables that affect this time, since the distance between the Earth and Sun is not constant (the orbit is an ellipse, and it's perturbed by other planets' (mostly Jupiter) gravity), and Earth is moving in its orbit. Unless you describe the time taken with extreme precision, all these effects, including time dilation, are negligible.
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$\begingroup$ Roger that. So, regardless of what the proton might or might not "experience", what we're talking about is how long is the journey observed by a stationary observer. $\endgroup$ Commented Mar 27, 2022 at 13:00
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6$\begingroup$ There’s no absolute frame of reference and therefore no such thing as a “stationary observer”. I guess you mean “observed in the Earth’s reference frame” (which is usually assumed to be inertial, even though it is not really). Also, I assume you are talking about photons, not protons. :) $\endgroup$ Commented Mar 27, 2022 at 13:43
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7$\begingroup$ Even in a relativistic sense, the speed of light is the same in every frame, so the answer is "in every frame where the distance between Sun and Earth is about 93 million miles." Due to relativistic length contraction, that excludes frames in which the Earth / Sun pair is moving sufficiently fast. $\endgroup$ Commented Mar 27, 2022 at 15:07
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$\begingroup$ @BrianDrake Yep, both of those assumptions are correct :) $\endgroup$ Commented Mar 27, 2022 at 15:47
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To measure the time light takes to travel, we run into an obstacle: if the observer is on Earth, they can not actually measure the point of time at which the photon is emitted from the Sun. Conversely, if the observer is the Sun (let's assume they have a magical spacesuit that protects them), they can measure the point of time at which the photon is emitted, but not the point of time at which it arrives on Earth.
To solve this, we might send one observer to the Sun, and one to Earth, give them synchronous clocks and have them both record the time at which the photon was sent and received. The problem with that is that 'synchronous clocks' is itself not well-defined if the clocks are at different locations. The best thing we can do, is give the two observers synchronous clocks while they are still on Earth, and then send one of them to the Sun, at a non-relativistic speed. After the measurement, this observer can return to Earth, and the observers can compare their observations. The recorded times between the emission and receiving of the photons will then be roughly 8 minutes (the relativistic corrections are small at this scale).
An easier way is to put a mirror on the sun and send a photon to it. The photon will return after 16 minutes.
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2$\begingroup$ "An easier way is to put a mirror on the sun". John Cena "Are you sure about that?" meme $\endgroup$ Commented Mar 29, 2022 at 4:02