4 ms·

> A photon is moving at the speed of light. Things moving at the speed of light do not experience distance or time (and have no mass), for them their creation a

• zaarn 8y ago
  >There is still a traversal of spacetime at a limited speed between the creation and destruction. From your point of view, that is correct. From the hypothetical point of view of a photon, an object moving at the speed of light, time does not pass and the distance gets squished into zero. You can observe a partial effect by looking at cosmic rays. We can detect several decay products of cosmic rays on the ground despite these decay products being far too short lived and too slow to have actually reached the ground. However, due to their high speed, their internal clock runs "slower" compared to ours and the distance between the top of the atmosphere where they are created in a collision or decay and the ground is much shorter than they can travel at their speed. A photon experiences the same effect times infinity; distance and time no longer have a reasonable meaning other than being 0.
  • omarchowdhury 8y ago
    I'm sorry, I still don't understand and if you can provide me with further material to research I would appreciate it. From my point of view, for there to be 0 time and 0 distance is equivalent to saying the speed of light is infinite, ie. the point of emission is equivalent to its annihilation, but that is traversed at the limited quantity of c.
    • zaarn 8y ago
      It's a bit of a hard subject to digest, I admit that. I guess one thing would be to accept that distance and experienced time aren't objective. If you move in a train, the time you spend inside the train moving and the distance you cover from your perspective will differ from the time that is shown on the clock at the station as well as the distance between the stations. By a very tiny amount. This process is exponential, so if you double the speed you travel at (as a number relative to four) the effect on your time and distance covered will quadruple. When you travel at C, then this effect becomes infinite. Time passes infinitely slow to the point that the time passed becomes zero. Same for distance. It also doesn't mean the speed of light is infinite, rather, from the perspective of the photon, ignoring the time part, the universe is exactly 0 in size. In turn that means travelling to some point in the universe, from the perspective of that photon, is instant. From the perspective of the outside observer, the photon is not traveling 0 distance, it's moving at the speed of light to a far away destination. From the photon's perspective, it's speed is not infinite, it's not moving at all, it's created and annihilated at the same point in space and time. From ours it's moving at the speed of light for a long time. Because these are different frames of reference, it's not as easy to compare speeds experienced in one to the other, especially if one is moving at C relative to the other. (Simplified) The entire affair is necessary because only things not having mass can move at C (and in turn, must move a C, a massless particle like a photon must always move at C). The entire rest of the universe has mass. From the photons frame of reference, everything else is moving at C. But that isn't allowed because that stuff has mass. So space itself is compressed into a 0 sized point to stop things from moving at C (from the perspective of the photon). Now, the photon however wouldn't be moving because in 0 sized space, you can't be moving at the speed of light. So in turn, time is experienced instantly and the photon is annihilated instantly so it can actually move at C (because it didn't move at all, it was destroyed instantly). In short; the photon doesn't experience time or distance because if it did it would be slower than C, it must always move exactly at C, therefore the universe arranges a situation in which the photon doesn't have to move at all, from it's own perspective.
      • omarchowdhury 8y ago
        Thank you. What would you recommend for me to study to learn more about this behavior of photons, specifically?
        • db48x 8y ago
          Minute Physics has some good videos about relativity.
      • raattgift 8y ago
        We can parameterize any curve in 4-d spacetime with a smooth and monotonic function that maps points on the curve to real numbers. In fact, you're doing that in the text above without realizing it: the function in question is proper time. It's fine -- commonplace, even, especially in the uncurved spacetime of Special Relativity -- to use proper time to parameterize a timelike geodesic. However, those are far from the only curves in a metric-equipped spacetime. In particular, for a null geodesic -- a freely-falling path taken by a photon -- proper time won't do as a parametrization because proper time is zero everywhere along it. However, there's nothing particularly unusual about a photon's worldline: it's just a curve where at every point along it the tangent vector is a light-like vector. On the other hand, along such curves we can use an affine parameter. Wald does this in his _General Relativity_ textbook this way: a geodesic is a curve whose tangent vector satisfies T^{a}\nabla_{a}T^{b} = 0 \RightArrow T^{a}\nabla_{a}T^{b} = \alpha T^{b}, where \alpha is an arbirary function on the curve. We can always reparameterize the second version into the first -- any parameter satisfying the first version is an affine parameter. We can equivalently say g_{\mu\nu}\frac{dx^\mu}{ds}\frac{dx^\nu}{ds} = 0 for all s, where g is the metric tensor and s is an affine parameter along the geodesic x(s). There's some greater depth starting at https://en.wikipedia.org/wiki/Geodesic#Affine_geodesics https://en.wikipedia.org/wiki/Geodesic#Affine_geodesics Affine parameterization lets one calculate the behaviour of photons moving through curved spacetime. If we define momentum with respect to the affine parameter derivative like this: k^{\mu} = \dot{x}^{\mu} then we can take the affine parameter value at one point on the geodesic and use that value's derivative at a different point on the geodesic, giving us a momentum difference between the points that is physically the gravitational redshift or blueshift. Additionally, if \alpha = f(x) then k^{\mu}(x) encodes the photon's momentum components, and from them we can extract the relation E = pc. It doesn't matter that the photon has no proper time because the proper time isn't really physically meaningful. We can still slice up the photon's spacetime into time-indexed 3d spacelike hypervolumes and (for a Eulerian observer at rest in this time-indexing) see that the photon is more red in the slice at time t and more blue in the slice at time t'. We can moreover, thanks to the affine parameterization, calculate how much redder it is at time t and at time t' for arbitrary observers, just like we can calculate the length contraction of a massive object or the time dilation of a clock using their respective proper times. > the universe arranges a situation in which the photon doesn't have to move at all, from it's own perspective No, you've chosen to use concepts from Special Relativity in a limit in which you get yourself into trouble. The photon has a non-zero-length worldline, and you can slice it arbitrarily; you're not restricted to the one unique slicing in which you are least able to talk about its evolution.