I have read in the answers here that an electron traveling backwards in time would behave as a positron. I also read in another there that antimatter is matter traveling backwards in time

As far as I know as something travels faster time slows down according to the laws of relativity. For something traveling at the speed of light time stops.

So if antimatter is matter traveling backwards in time wouldn't that mean antimatter is traveling faster than light?

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    $\begingroup$ Nothing is traveling backwards in time and there is no difference between matter and antimatter in this respect. This particular meme is about as false as things can get in physics. It is derived from a symmetry property of Feynman diagrams, but those are little pictures that make life easier for physicists who have to calculate perturbation series in particle physics, they are not actual physical reality. What it really means is that the modified process to the one in a diagram is also possible if we replace particles in the diagram with their anti-particles and reverse the directions. $\endgroup$
    – CuriousOne
    Commented Mar 22, 2016 at 4:32
  • $\begingroup$ There is only one arrow of time-forward.....antimatter or otherwise, nothing can travel 'backward' in time. $\endgroup$
    – GRrocks
    Commented Mar 22, 2016 at 4:40
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    $\begingroup$ In addition to the above, traveling backwards in time isn't the same as traveling faster than light. Particles that travel "backwards in time", which is more of a statement based on models than a statement of fact, still have to obey the laws of relativity and can't travel faster than light. Imagine filming a car driving down the road, now, run the film backwards. In neither viewing is the car traveling faster than light, it's simply reversed direction. $\endgroup$
    – userLTK
    Commented Mar 22, 2016 at 4:49
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    $\begingroup$ Good point @userLTK but I believe the OP is talking about the relativistic transformations which mathematically (makes no physical sense) allow for time to stop at speed of light, and move backward if you exceed it. Keyword being, 'mathematically'. $\endgroup$
    – GRrocks
    Commented Mar 22, 2016 at 5:02
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    $\begingroup$ @GRrocks: There is no such transformation in special relativity, folks are just not looking carefully enough at signs and resulting imaginary constants from square roots when they claim that there is. $\endgroup$
    – CuriousOne
    Commented Mar 22, 2016 at 5:11

1 Answer 1


There is no massive fundamental/elementary "(anti)particle" (or better field) in our Universe that can move at the speed of light. The fastest known massive fundamental (anti)particles are ultra high energy astrophysical neutrinos which can move almost at speed of light but not exactly. The reason for this is that first of all they are the lightest known fundamental (anti)particles in Nature and are given high boost factors at some astrophysical sources as they're produced. So, there is no way for any massive "particle" in our Universe to travel yet faster than speed of light. The only (anti)particle that can travel at the speed of light (but not faster than that) is light itself, namely photon field. Since photon is its own anti particle, we can conclude that the only anti-matter travelling at speed of light is 'light.'

By the construction of Big Bang theory and creation of Universe from a singularity, the arrow of time has been always forwards in an expanding Universe. So, nothing in Nature can move opposite to that arrow (or backward in time as you called it.) The Feynman diagrams--which have been used to work out perturbation theories--make use of the time symmetry where as you mentioned you can replace an incoming electron with an outgoing positron and that has nothing to do with the reality of the processes being involved. They are just mathematical constructs.

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    $\begingroup$ There was an edit suggestion from an anon user, saying: "This is true for the classical view of physics, not the quantum mechanical view." Anon user: please register to the site, and collect 50 rep. So you will be able to comment answers. $\endgroup$
    – peterh
    Commented Sep 24, 2017 at 18:26

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