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Imagine looking at a quasar millions of light years away. From your perspective, a photon emitted by the quasar has spent millions of years travelling through space, and its trajectory has been affected by the gravity of all the galaxies between the quasar and you. In non-Euclidean space-time, the path it has travelled along is a straight line.

However, the photon has no mass and travels at the speed of light. According to the theory of relativity, from the photon's perspective, no time passes between the moment it is emitted in the quasar and the moment it is detected in your eye. From the photon's perspective, there is no distance between the quasar and your eye. There is thus no path for it to have followed: for it, the curvature of space-time is not even a straight line; it is a single point.

Any particle which has mass can follow the same "path" but it sees a very different universe.

Is space-time a function of mass?

E = mc^2 expresses energy in terms of mass, time and space. Is there a way to reformulate this so that the observed value of the speed of light can be deduced from the existence of mass?

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  • $\begingroup$ Note that $E=mc^2$ is a simplification, the real formula is $E^2=p^2c^2+m^2c^4$ (cf. this post). $\endgroup$ – Kyle Kanos Nov 12 '14 at 14:46
  • $\begingroup$ Does including the momentum of the photon change the duration or length of its trajectory, when viewed from its own frame of reference? If so, how? $\endgroup$ – James Newton Nov 12 '14 at 14:54
  • $\begingroup$ The mass of a photon is zero, so according to your simplified version, there is no energy. This is demonstrably false. $\endgroup$ – Kyle Kanos Nov 12 '14 at 14:58
  • $\begingroup$ Your description is correct - photons don't use spacetime for their displacement. But your conclusions don't seem to follow from your description. $\endgroup$ – Moonraker Nov 13 '14 at 8:18
  • $\begingroup$ Side note, according to the theory of relativity, the photon has no perspective $\endgroup$ – Jim Jan 12 '15 at 14:12
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From the photon's perspective, there is no distance between the quasar and your eye. There is thus no path for it to have followed: for it, the curvature of space-time is not even a straight line; it is a single point.

There is no photon's perspective. See Does a photon in vacuum have a rest frame? .

Any particle which has mass can follow the same "path" but it sees a very different universe.

A massive particle can't follow the same world-line in four dimensions. The world-line of a massless particle is a lightlike curve; that of a massive particle is timelike.

Is space-time a function of mass?

No, because a photon doesn't have a rest frame, and any two massive particles can have the same rest frame.

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  • $\begingroup$ Yes, this about sums it all up. Very succinct! +1 $\endgroup$ – Jim Jan 12 '15 at 14:14
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You don't need a massless photon to (at least asymptotically, in a limiting sense) make up zero spacetime distances. The faster a massive particle moves, the less spacetime it travels (not space alone, not time alone, but spacetime). Put it mathematically, $$E^2=p^2c^2+m^2c^4$$ and you can see that the value of m gets overshadowd by a large momentum, when traveling fast enough, approaching a photon-like regime. As an example, the time it takes for a massive traveler (say a spacecraft) to travel to a star 100 lightyears away is ~14 years if traveling at 99% of c and ~4 years if traveling at 99.9% of c, as measured in its own frame of reference.

Therefore, to the extent you can achieve the same thought conclusion (near zero spacetime distance) using a massive particle, mass does not appear to be creating spacetime. Mass is responsible for creating spacetime curvature instead, according to general relativity, but that is not what you asked.

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