Why Sun light can reach us if Time is dilating? [closed]

I understand that if something is moving with constant speed in respect to an observer, the time of the moving one runs slower, so the more your speed is, the more your time ticks slower.

Which means if i were on the Moon and went to Earth in slow speed, I will get to their in like 3 days, but if i were to go to earth almost in the speed of light, my time will almost stop, which means 1 second to me will equal to like 50000 years to those on earth, so according to an observer on Earth I will disappear and reappear after 50000 years!

Am I understanding this right? If this is right, how is the light from the Sun can reach us in just 8 minutes? Shouldn't the time of the photon be dilating which means 1 minute in the frame of the photon be like a million years in our frame?

• Related: physics.stackexchange.com/q/29082/2451 and links therein. Jan 3 '14 at 0:25
• Do you agree that $c\simeq3\times10^8$ m/s? Jan 3 '14 at 1:38
• Yes and? how does that solve my confusion? Jan 3 '14 at 1:46
• The equations of time dilation imply that time does not exist for a photon. Feb 5 '21 at 18:59

how is the light from the sun reaching us in just 8 minutes,

Because it travels with speed of light, obviously. By definition, speed of light is the speed with which the photons travel.

• Thanks for the reply but that's not my question. Jan 4 '14 at 0:39

Your question seems to talk about massive objects so I'll discuss massive objects. You have a misconception. If you start out $3 \cdot 10^9$ meters away from me, and head towards me at $.99c\approx 3\cdot 10^8$ meters per second (close enough for our purposes!), I will observe that you get here in 10 seconds. If I look at a clock you're holding (say you're a group of particles that have a specific half life, and when you get to me I'll measure how much of that original particle is left), I will observe that only $t/\gamma$ seconds have elapsed for you. In this case, $t=10\text{s}$ and $\gamma=(1-.99^2)^{-1/2}=10$, so only one second will have elapsed for you.

But this is absolutely symmetric. Nothing is accelerating, so you have full rights to look at the situation as if I'm the one traveling towards you. From your perspective, your time isn't dilated at all, it's my time that is dilated, and it's dilated by the exact same factor as before. So IF the distance between you and I from your frame was still $3\cdot 10^9$ meters, we would have a contradiction, because as discussed before my frame would observe that you had $1$ second of decay occur, while your frame would observe it took me ten seconds to get to you, so $10$ seconds of decay would occur. However, this isn't a contradiction because we haven't factored in length contraction!

From my perspective you're travelling at $3\cdot 10^8$ meters per second towards me, starting from a distance of $3\cdot 10^9$ meters. But from your perspective, I'm travelling towards you at a speed of $3\cdot 10^8$ meters per second, starting out at a distance of $3\cdot 10^9/\gamma=3\cdot 10^8$ meters. That speed at that velocity will get me to you in $1$ second. No contradiction.

All that this means is that you can get to the earth - from your perspective - as fast as you want. As you accelerate to relativistic speeds you'll find the distance between you and the earth shrinking (length contraction) and so the trip will seem to take almost no time at all, provided you accelerate enough. On earth, however, you get closer and closer to $c$ and may have more and more energy, but your speed stays approximately constant as you get very close to $c$ and so it will always take $d/c$ seconds, NEVER 5,000 years!

So, one might realize that to get really meaningful effects one shouldn't consider constant velocity problems. If you introduce acceleration, then you can get meaningful effects. Considering the twin paradox for example, one finds $t_{\text{earth}}=\gamma t_{\text{spaceship}}$. So the distance traveled (in Earth's frame of reference), if $v\approx c$, is $t_{\text{earth}}c=\gamma t_{\text{spaceship}} c$. If the distance to work with (to the moon and back) is a constant $d$, that is, if $\gamma t_{\text{spaceship}}c=d$, then considering only large very relativistic velocities with $v/c\approx 1$, $t_{\text{earth}}=\gamma t_{\text{spaceship}}=d/c$. The time will be constant again! But, of course, we already knew that, and didn't need relativity to answer it. To increase this value one must increase the distance. Increasing the velocity while keeping the distance constant won't help. While this only applies to massive objects, it might be alright to hand wave and say something like, "This is why the infinite time dilation of a photon doesn't matter when asking how fast does the photon get here". (in quotes because "infinite time dilation of a photon" is used here as a VERY hand-wavey concept.)

• Thanks very much for your efforts, i never considered the length contraction to be an important factor, my confusion about this subject is now clear thanks to you : ) Jan 4 '14 at 0:51

The key point that you must remember is that to maintain such a dilated time from the perspective of the earth-bound observer you must be moving relative to that observer. Therefore, if you leave the moon, heading towards the Earth, traveling at nearly the speed of light, you will reach the earth very quickly, so no one will see you travelling for the 50,000 years (or what have you). You'll get there when you get there: very soon.

Edit: I say this in my comment, but I might as well make it clear here too. It doesn't take you 1 second to read Earth from your perspective, it takes much less time because the distance between the moon and the earth gets Lorentz contracted. On the other hand, someone observing you from Earth will see you take 1 second to reach him from the moon, but he will observe your clock as having only progressed the .002 seconds that you yourself observe as the amount of time it takes to cross the distance (because your time has become dilated to him). The trouble you're having is that you are mixing up the relativistic effects of different reference frames.

• I know i will reach the earth very quickly but that's according to my time, but since 1 second in my clock equals a million year on earth, i will get to earth in one second to find all life has vanished! Jan 3 '14 at 0:53
• In your reference frame you aren't moving: the earth is moving. From the perspective of an observer on the earth, you are moving at 299,792 kilometers-per-second so you will reach earth in about 1.2 seconds. But from your perspective, the earth is rushing up to meet you at 299,792 kilometers-per-second but the distance between you and the earth is Lorentz contracted, so from your perspective it only takes .00224 seconds. That's the point: time slows down for you from the Earth's perspective, but space compresses from your perspective. Jan 3 '14 at 6:23
• @Apastrix: you're running into an issue with simultaneous events. There is only one reference frame in which the events where you start your stopwatch on the spaceship and in which you start the stopwatch on Earth are simultaneous. In the boosted reference frame, they are not, and that's enough to account for the gap in time you are describing. Jan 3 '14 at 17:02
• @Jerry Schirmer Indeed i was mixing them Geoffrey, and i never payed the contraction factor any attention but now i see it, Thanks guys, this is just great! Jan 4 '14 at 0:57

The light from the Sun reaches us in 8 minutes, because it travels at the speed of light, and light travels the distance between the Sun and the Earth (approximately 1.496*10^11 m) at speed $$c \approx 3\cdot 10^8 {m \over s}$$ in about 8 minutes (the Sun is therefore 8 light minutes away from the Earth). Time dilation (or length contraction) plays no role at all in this statement.

On the other hand, due to length contraction shrinking the distance in the direction of motion to 0, a photon will take zero photon time to travel any distance. A photon can travel the whole universe in 0 seconds of its own time, because as viewed from the outside its time is completely standing still, and from its perspective, the universe has shrunk to a size of 0.