So time is relative, speed will altere the “perception” of time… Right?. If you travel at almost C you “age slower”. And photons travel for around 8 minutes before hitting earth.
So… How this combine together? Is 8 minutes for us a lifetime for that photon or the other way around? Is this even applicable to something without mass?
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Photons, and anything else moving at c, don’t really have a perspective.
An object moving at *near* c would experience almost no time at all during the journey. The distance between Earth and the Sun would, from its perspective, contract such that its near light speed movement allowed it to cross the distance much faster.
Photons, and anything else moving at c, don’t really have a perspective.
An object moving at *near* c would experience almost no time at all during the journey. The distance between Earth and the Sun would, from its perspective, contract such that its near light speed movement allowed it to cross the distance much faster.
For the photon, no time would have passed at all. What that means in practice is that photons can’t change as they travel. The photons we receive are exactly the same as the ones that were sent.
Here’s another example. Back in the 80’s, it was thought that neutrinos were massless, and traveled at the speed of light. Neutrinos are extremely light particles that are emitted from some nuclear reactions – and they come in three “favours”.
Also, back in the 80’s, careful measurements had been done of the number of neutrinos coming from the sun, and the figure was only about 1/3 of what it should have been.
There were a few ideas proposed to explain that. One was that the sun had switched off and we would all freeze to death within 10000 years or so, but another was that some of the neutrinos from the sun were transmuting into the other forms, and we were only detecting the 1/3 that stayed in their original form. However, if neutrinos were traveling at c, that couldn’t happen – **if, for the neutrino, no time had passed, then it couldn’t transmute**, since change needs a passage of time, and for objects at c, no time passes.
Since then, we have confirmed that, in fact, neutrinos do have mass, and don’t travel at c, and so time does pass for them on their journey to us from the sun, so we aren’t doomed to an icy future. This was very exciting for the physics world, and probably won (or will win) someone a Nobel Prize.
However, for light itself, we know it travels at c, and therefore no time passes for the photons as they cross the 8 light-minutes between the sun and us. (For them, it also seems like no distance has been traversed).
For the photon, no time would have passed at all. What that means in practice is that photons can’t change as they travel. The photons we receive are exactly the same as the ones that were sent.
Here’s another example. Back in the 80’s, it was thought that neutrinos were massless, and traveled at the speed of light. Neutrinos are extremely light particles that are emitted from some nuclear reactions – and they come in three “favours”.
Also, back in the 80’s, careful measurements had been done of the number of neutrinos coming from the sun, and the figure was only about 1/3 of what it should have been.
There were a few ideas proposed to explain that. One was that the sun had switched off and we would all freeze to death within 10000 years or so, but another was that some of the neutrinos from the sun were transmuting into the other forms, and we were only detecting the 1/3 that stayed in their original form. However, if neutrinos were traveling at c, that couldn’t happen – **if, for the neutrino, no time had passed, then it couldn’t transmute**, since change needs a passage of time, and for objects at c, no time passes.
Since then, we have confirmed that, in fact, neutrinos do have mass, and don’t travel at c, and so time does pass for them on their journey to us from the sun, so we aren’t doomed to an icy future. This was very exciting for the physics world, and probably won (or will win) someone a Nobel Prize.
However, for light itself, we know it travels at c, and therefore no time passes for the photons as they cross the 8 light-minutes between the sun and us. (For them, it also seems like no distance has been traversed).
NASA: How long does it take light to get out from the inside of the Sun?
According to the famous ‘drunkard’s walk’ problem, the distance a drunk, making random left and right turns, gets from the lamp post is his typical step size times the square root of the number of steps he takes. For the sun, we know how far we want to go to get out….696,000 kilometers, we just need to know how far a photon travels between emission and absorption, and how long this step takes. This requires a bit of physics!
The interior of the sun is a seathing plasma with a central density of over 100 grams/cc. The atoms, mostly hydrogen, are fully stripped of electrons so that the particle density is 10^26 protons per cubic centimeter. That means that the typical distance between protons or electrons is about (10^26)^1/3 = 2 x 10^-9 centimeters. The actual ‘mean free path’ for radiation is closer to 1 centimeter after electromagnetic effects are included. Light travels this distance in about 3 x 10^-11 seconds. Very approximately, this means that to travel the radius of the Sun, a photon will have to take (696,000 kilometers/1 centimeter)^2 = 5 x 10^21 steps. This will take, 5×10^21 x 3 x10^-11 = 1.5 x 10^11 seconds or since there are 3.1 x 10^7 seconds in a year, you get about 4,000 years. Some textbooks refer to ‘hundreds of thousands of years’ or even ‘several million years’ depending on what is assumed for the mean free patch. Also, the interior of the sun is not at constant density so that the steps taken in the outer half of the sun are much larger than in the deep interior where the densities are highest. Note that if you estimate a value for the mean free path that is a factor of three smaller than 1 centimeter, the time increases a factor of 10!
Typical uncertainties based on ‘order of magnitude’ estimation can lead to travel times 100 times longer or more. Most astronomers are not too interested in this number and forgo trying to pin it down exactly because it does not impact any phenomena we measure with the exception of the properties of the core region right now. These estimates show that the emission of light at the surface can lag the production of light at the core by up to 1 million years.
The point of all this is that it takes a LONG time for light to leave the sun’s interior!!
To a photon, there’s no such thing as time. From its point of view, it would be instant. One way of thinking about this is that in special relativity, distances shrink by a factor of sqrt(1-v^2/c^2). That is zero for a photon where v equals c, so from the photon’s point of view, the distance between the Earth and the Sun is zero.
More technically, “age” is a funny concept in relativity. Time is seen differently by different observers. You need to specify both where and when something happened (not just “Alice met Bob at the corner of Main and Elm, but Alice met Bob at the corner of Main and Elm at 4:30 on Friday). If you take two different events at different times and different places, different observers won’t agree on how far apart they are or when they happened, but they will always agree on the difference between distance squared and time multiplied by the speed of light squared (dx^2-c^2 dt^2). Since a photon moves at the speed of light the distance it moves dx in a time dt is just speed times dt, or dx=cdt, so dx^2-c^2 dt^2=0. As far as the universe is concerned, the distance between a photon leaving the sun and that photon hitting the earth is exactly zero. In our frame, that means dx is 93 million miles and dt is 8 minutes, but to the photon dx=0 and dt=0. There’s no “right” answer for the age of the photon, since every frame is valid, but if you ask the photon, you’ll get zero.
To a photon, there’s no such thing as time. From its point of view, it would be instant. One way of thinking about this is that in special relativity, distances shrink by a factor of sqrt(1-v^2/c^2). That is zero for a photon where v equals c, so from the photon’s point of view, the distance between the Earth and the Sun is zero.
More technically, “age” is a funny concept in relativity. Time is seen differently by different observers. You need to specify both where and when something happened (not just “Alice met Bob at the corner of Main and Elm, but Alice met Bob at the corner of Main and Elm at 4:30 on Friday). If you take two different events at different times and different places, different observers won’t agree on how far apart they are or when they happened, but they will always agree on the difference between distance squared and time multiplied by the speed of light squared (dx^2-c^2 dt^2). Since a photon moves at the speed of light the distance it moves dx in a time dt is just speed times dt, or dx=cdt, so dx^2-c^2 dt^2=0. As far as the universe is concerned, the distance between a photon leaving the sun and that photon hitting the earth is exactly zero. In our frame, that means dx is 93 million miles and dt is 8 minutes, but to the photon dx=0 and dt=0. There’s no “right” answer for the age of the photon, since every frame is valid, but if you ask the photon, you’ll get zero.
Zero seconds.
The closer you get to light speed, the slower you perceive time. Once you hit light speed, time stops for you.
A trip to the opposite side of the universe would be an instant to a photon, but billions of years to an outside observer.
Zero seconds.
The closer you get to light speed, the slower you perceive time. Once you hit light speed, time stops for you.
A trip to the opposite side of the universe would be an instant to a photon, but billions of years to an outside observer.
NASA: How long does it take light to get out from the inside of the Sun?
According to the famous ‘drunkard’s walk’ problem, the distance a drunk, making random left and right turns, gets from the lamp post is his typical step size times the square root of the number of steps he takes. For the sun, we know how far we want to go to get out….696,000 kilometers, we just need to know how far a photon travels between emission and absorption, and how long this step takes. This requires a bit of physics!
The interior of the sun is a seathing plasma with a central density of over 100 grams/cc. The atoms, mostly hydrogen, are fully stripped of electrons so that the particle density is 10^26 protons per cubic centimeter. That means that the typical distance between protons or electrons is about (10^26)^1/3 = 2 x 10^-9 centimeters. The actual ‘mean free path’ for radiation is closer to 1 centimeter after electromagnetic effects are included. Light travels this distance in about 3 x 10^-11 seconds. Very approximately, this means that to travel the radius of the Sun, a photon will have to take (696,000 kilometers/1 centimeter)^2 = 5 x 10^21 steps. This will take, 5×10^21 x 3 x10^-11 = 1.5 x 10^11 seconds or since there are 3.1 x 10^7 seconds in a year, you get about 4,000 years. Some textbooks refer to ‘hundreds of thousands of years’ or even ‘several million years’ depending on what is assumed for the mean free patch. Also, the interior of the sun is not at constant density so that the steps taken in the outer half of the sun are much larger than in the deep interior where the densities are highest. Note that if you estimate a value for the mean free path that is a factor of three smaller than 1 centimeter, the time increases a factor of 10!
Typical uncertainties based on ‘order of magnitude’ estimation can lead to travel times 100 times longer or more. Most astronomers are not too interested in this number and forgo trying to pin it down exactly because it does not impact any phenomena we measure with the exception of the properties of the core region right now. These estimates show that the emission of light at the surface can lag the production of light at the core by up to 1 million years.
The point of all this is that it takes a LONG time for light to leave the sun’s interior!!
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