Assume there are observers at A and B, sitting at rest relative to each other. The distance between them as seen by them is X. Their watches are synchronized. Alice, sitting at A, emits a particle when her watch says t0; Bob, sitting at B, receives it when his watch says t1. Define T = t1-t0. The speed of the particle is V = X/T.
If the particle is massive, then V is always smaller than c (the speed of light). We can imagine attaching a clock to the particle and starting it when it is emitted. When Bob receives it, the clock’s time would read a time t smaller than T, given by the equation:
t = T (1 - V^2/c^2)^(1/2) (this is the Lorentz factor equation mentioned by Plasmon).
As the speed V of the particle gets closer and closer to c, you can see that the time t that has passed “for the particle” gets closer and closer to 0. One cannot attach a clock to a photon, so the statement that “photons are not moving through time” is somewhat metaphoric and its real meaning is the limiting statement I just mentioned. The photon is not “at two places at once” from the point of view of any physical observer, be it Alice and Bob (for whom the travel took a time T = X/c) or any other moving with a speed smaller than c (for whom the time taken may be different but is never 0).
Thanks, it sounds like Tyson just said something very misleading. I looked up the Lorentz factor equation on Wiki, and I got this:
gamma = 1/[(1 - V^2/c^2)^(1/2)]
Is that right? If that’s right, then the Lorentz transformation (I’m just guessing here) for a photon would return an undefined result. Was Tyson just conflating that result with a result of ‘zero’?
Your equation for the gamma factor is correct. You are also correct in saying that they Lorentz transformation becomes undefined. The significance of this is that it makes no sense to talk about the “frame of reference of photon”. Lorentz transformation equations allow us to switch from some set of time and space coordinates to another one moving at speed V < c relative to the first one. They make no sense for V = c or V > c.
I think that what Tyson meant by his somewhat imprecise answer was what I said in my comment above: if you take the equation t *gamma = T (that relates the time t that passes between two events for an object that moves with V from one to the other, with the time T that passes between between the events on a rest frame) and take the limit V approaching c for finite T, you get t = 0. If you want to keep the meaning of the equation in this limit, you have then to say that “no time passes for a photon”. The issue is that the equation is just a consequence of the Lorentz transformations, which are inapplicable for V = c, and as a consequence the words “no time passes for a photon” do not have any clear, operational meaning attached to them.
Assume there are observers at A and B, sitting at rest relative to each other. The distance between them as seen by them is X. Their watches are synchronized. Alice, sitting at A, emits a particle when her watch says t0; Bob, sitting at B, receives it when his watch says t1. Define T = t1-t0. The speed of the particle is V = X/T.
If the particle is massive, then V is always smaller than c (the speed of light). We can imagine attaching a clock to the particle and starting it when it is emitted. When Bob receives it, the clock’s time would read a time t smaller than T, given by the equation:
t = T (1 - V^2/c^2)^(1/2) (this is the Lorentz factor equation mentioned by Plasmon).
As the speed V of the particle gets closer and closer to c, you can see that the time t that has passed “for the particle” gets closer and closer to 0. One cannot attach a clock to a photon, so the statement that “photons are not moving through time” is somewhat metaphoric and its real meaning is the limiting statement I just mentioned. The photon is not “at two places at once” from the point of view of any physical observer, be it Alice and Bob (for whom the travel took a time T = X/c) or any other moving with a speed smaller than c (for whom the time taken may be different but is never 0).
Thanks, it sounds like Tyson just said something very misleading. I looked up the Lorentz factor equation on Wiki, and I got this:
gamma = 1/[(1 - V^2/c^2)^(1/2)]
Is that right? If that’s right, then the Lorentz transformation (I’m just guessing here) for a photon would return an undefined result. Was Tyson just conflating that result with a result of ‘zero’?
Your equation for the gamma factor is correct. You are also correct in saying that they Lorentz transformation becomes undefined. The significance of this is that it makes no sense to talk about the “frame of reference of photon”. Lorentz transformation equations allow us to switch from some set of time and space coordinates to another one moving at speed V < c relative to the first one. They make no sense for V = c or V > c.
I think that what Tyson meant by his somewhat imprecise answer was what I said in my comment above: if you take the equation t *gamma = T (that relates the time t that passes between two events for an object that moves with V from one to the other, with the time T that passes between between the events on a rest frame) and take the limit V approaching c for finite T, you get t = 0. If you want to keep the meaning of the equation in this limit, you have then to say that “no time passes for a photon”. The issue is that the equation is just a consequence of the Lorentz transformations, which are inapplicable for V = c, and as a consequence the words “no time passes for a photon” do not have any clear, operational meaning attached to them.
I think I understand. Thanks very much for taking the time.