On the Neil Degrasse Tyson Q&A on reddit, someone asked: “Since time slows relative to the speed of light, does this mean that photons are essentially not moving through time at all?”
Tyson responded “yes. Precisely. Which means ----- are you seated?Photons have no ticking time at all, which means, as far as they are concerned, they are absorbed the instant they are emitted, even if the distance traveled is across the universe itself.”
Is this true? I find it confusing. Does this mean that a photon emitted at location A at t0 is absorbed at location B at t0, such that it’s at two places at once? In what sense does the photon ‘travel’ then? Or is the thought that the distance traveled, as well as the time, goes to zero?
I know, I was joking. And it was a good opportunity to link to this (genuinely interesting) paper.
… well, mostly joking. There’s a kernel of truth there. “There are no photons” says more than just banning a word. “Wavepackets of light” don’t exist either. There’s just the electromagnetic field, its intensity changes with time, and the change propagates in space. Looking at it like this may help understand the other responses to the question (which are all correct).
When you think of a photon as a particle flying in space, it’s hard to shake off the feeling that you somehow ought to be able to attach yourself to it and come along for the ride, or to imagine how the particle itself “feels” about its existence, how its inner time passes. And then the answer that for a photon, time doesn’t pass at all, feels weird and counter-intuitive. If you tell yourself there’s no particle, just a bunch of numbers everywhere in space (expressing the EM field) and a slight change in those numbers travels down the line, it may be easier to process. A change is not an object to strap yourself to. It doesn’t have “inner time”.
By this argument, ocean waves don’t exist either. There’s only the sea, its height changes with time, and the change propagates in space.
You say that as a reductio ad absurdum, but it is good for some purposes. Anatoly didn’t claim that one should deny photons for all purposes, but only for the purpose of unasking the original question.
Anatoly didn’t claim that one should deny photons for all purposes, but only for the purpose of unasking the original question.
In this case, unasking the original question is basically an evasion, though, isn’t it?
Denying photons may enable you to unask hen’s literal question, or the unnamed Reddit poster’s literal question, but it doesn’t address the underlying physical question they’re driving at: “if observer P travels a distance x at constant speed v in observer Q’s rest frame, does the elapsed time in P’s rest frame during that journey vanish in the limit where v tends to c?”
I reject the claim that your rephrasing is the “real” question being asked. By rephrasing the question, you are rejecting it just as much as Anatoly. I think it is more accurate to say that you evade the question, while he is up front about rejecting it.
In fact, I think your answer is better and probably it is generally better to rephrase problematic questions to answerable questions before explaining that they are problematic, but the latter is part of a complete answer and I think Anatoly is correct in how he addresses it.
I reject the claim that your rephrasing is the “real” question being asked.
That multipledifferentpeople automatically treated hen’s question like it were my rephrasing backs me up on this one, I reckon.
By rephrasing the question, you are rejecting it just as much as Anatoly. I think it is more accurate to say that you evade the question, while he is up front about rejecting it.
Rephrasing a question can be the first step to confronting it head-on rather than rejecting it. If a tourist, looking for the nearest train station, wandered up to me and asked, “where station is the?”, and I rearranged their question to the parseable “where is the station?” and answered that, I wouldn’t say I rejected or evaded their query.
Other people have explained this pretty well already, but here’s a non-rigorous heuristic that might help. What follows is not technically precise, but I think it captures an important and helpful intuition.
In relativity, space and time are replaced by a single four-dimensional space-time. Instead of thinking of things moving through space and moving through time separately, think of them as moving through space-time. And it turns out that every single (non-accelerated) object travels through space-time at the exact same rate, call it c.
Now, when you construct a frame of reference, you’re essentially separating out space and time artificially. Consequently, you’re also separating an object’s motion through space-time into motion through space and motion through time. Since every object moves through space-time at the same rate, when we separate out spatial and temporal motion, the faster the object travels through space the slower it will be traveling through time. The total speed, adding up speed through space and speed through time, has to equal the constant c.
So an object at rest in a particular frame of reference has all its motion along the temporal axis, and no motion at all along the spatial axes. It’s traveling through time at speed c and it isn’t traveling through space at all. If this object starts moving, then some of the temporal motion is converted to spatial motion. It’s speed through space increases, and its speed through time decreases correspondingly, so that the motion through space-time as a whole remains constant at c. This is the source of time dilation in relativity (as seen in the twin paradox) - moving objects move through time more slowly than stationary objects, or to put it another way, time flows slower for moving objects.
Of course, the limit of this is when the object’s entire motion through space-time is directed along the spatial axes, and none of it is directed along the temporal axes. In this case, the object will move through space at c, which turns out to be the speed of light, and it won’t move through time at all. Time will stand still for the object. This is what’s going on with photons.
From this point of view, there’s nothing all that weird about a photon’s motion. From the space-time perspective, which after all is the fundamental perspective in relativity, it is moving pretty much exactly like any other object. It’s only our weird habit of treating space and time as extremely different that makes the entirely spatial motion of a photon seem so bizarre.
That is helpful, and interesting, though I think I remain a bit confused about the idea of ‘moving through time’ and especially ‘moving through time quickly/slowly’. Does this imply some sort of meta-time, in which we can measure the speed at which one travels through time?
And I think I still have my original question: if a photon travels through space at c, and therefore doesn’t travel through time at all, is the photon at its starting and its final position at the same moment? If so, in what sense did it travel through space at all?
[Is] the photon at its starting and its final position at the same moment?
At the same moment with respect to whom? That is the question one must always ask in relativity.
The answer is: no, emission and arrival do not occur at the same moment with respect to any actual reference frame. However, as we consider an abstract sequence of reference frames that move faster and faster approaching speed c in the same direction as the photon, we find that the time between the emission and the reception is shorter and shorter.
Does this imply some sort of meta-time, in which we can measure the speed at which one travels through time?
No it doesn’t. Remember, in relativity, time is relative to a frame of reference. So when I talk about a moving object traveling slowly through time, I’m not relativizing its time to some meta-time, I’m relativizing time as measured by that object (say by a clock carried by the object) to time as measured by me (someone who is stationary in the relevant frame of reference). So an object moving slowly through time (relative to my frame of reference) is simply an object whose clock ticks appear to me to be more widely spaced than my clock ticks. In the limit, if a photon could carry a clock, there would appear to me to be an infinite amount of time between its ticks.
I will admit that I was using a bit of expository license when I talked about all objects “moving through space-time” at the constant rate c. While one can make sense of moving through space and moving through time, moving through space-time doesn’t exactly make sense. You can replace it with this slightly less attractive paraphrase, if you like: “If you add up a non-accelerating object’s velocity through space and its (appropriately defined) rate of motion through time, for any inertial frame of reference, you will get a constant.”
And I think I still have my original question: if a photon travels through space at c, and therefore doesn’t travel through time at all, is the photon at its starting and its final position at the same moment? If so, in what sense did it travel through space at all?
Again, it’s important to realize there are many different “time” parameters in relativity, one for each differently moving object. Also, whether two events are simultaneous is relative to a frame of reference.
Relative to my time parameter (the parameter for the frame in which I am at rest), the photon is moving through space, and it takes some amount of (my) time to get from point A to point B. Relative to its own time parameter, though, the photon is at point A and point B (and every other point on its path) simultaneously. Since I’ll never travel as fast as a photon, it’s kind of pointless for me to use its frame of reference. I should use a frame adapted to my state of motion, according to which the photon does indeed travel in non-zero time from place to place.
Again, this is all pretty non-technical and not entirely precise, but I think it’s good enough to get an intuitive sense of what’s going on. If you’re interested in developing a more technical understanding without having to trudge through a mathy textbook, I recommend John Norton’s Einstein for Everyone, especially chapters 10-12. One significant simplification I have been employing is talking about a photon’s frame of reference. There is actually no such thing. One can’t construct an ordinary frame of reference adapted to a photon’s motion (partly because there is no meaningful distinction between space and time for a photon).
Does this mean that a photon emitted at location A at t0 is absorbed at location B at t0, such that it’s at two places at once?
In the photon’s own subjective experience? Yes. (Not that that’s possible, so this statement might not make sense). But as another commenter said, certainly the limit of this statement is true: as your speed moving from point A to point B approaches the speed of light, the subjective time you experience between the time when you’re at A and the time when you’re at B approaches 0. And the distance does indeed shrink, due to the Lorentz length contraction.
In what sense does the photon ‘travel’ then?
It travels in the sense that an external observer observes it in different places at different times. For a subjective observer on the photon… I don’t know. No time passes, and the universe shrinks to a flat plane. Maybe the takeaway here is just that observers can’t reach the speed of light.
The first thing to say is that “at t0” means different things to different observers. Observers moving in different ways experience time differently and, e.g., count different sets of spacetime points as simultaneous.
There is a relativistic notion of “interval” which generalizes the conventional notions of distance and time-interval between two points of spacetime. It’s actually more convenient to work with the square of the interval. Let’s call this I.
If you pick two points that are spatially separated but “simultaneous” according to some observer, then I>0 and sqrt(I) is the shortest possible distance between those points for an observer who sees them as simultaneous. The separation between the points is said to be “spacelike”. Nothing that happens at one of these points can influence what happens at the other; they’re “too far away in space and too close in time” for anything to get between them.
If you pick two points that are “in the same place but at different times” for some observer, then I<0 and sqrt(-I) is the minimum time that such an observer can experience between visiting them. The separation between the points is said to be “timelike”. An influence can propagate, slower than the speed of light, from one to the other. They’re “too far away in time and too close in space” for any observer to see them as simultaneous.
And, finally, exactly on the edge between these you have the case where I=0. That means that light can travel from one of the spacetime points to the other. In this case, an observer travelling slower than light can get from one to the other, but can do so arbitrarily quickly (from their point of view) by travelling very fast; and while no observer can see the two points as simultaneous, you can get arbitrarily close to that by (again) travelling very fast.
Light, of course, only ever travels at the speed of light (you might have heard something different about light travelling through a medium such as glass, but ignore that), which means that it travels along paths where I=0 everywhere. To an (impossible) observer sitting on a photon, no time ever passes; every spacetime point the photon passes through is simultaneous.
So: does the distance as well as the time go to 0? Not quite. Neither distance nor time makes sense on its own in a relativistic universe. The thing that does make sense is kinda-sorta a bit like “distance minus time” (and more like sqrt(distance-squared minus time-squared)), and that is 0 for any two points in spacetime that are visited by the same photon.
(Pedantic notes: 1. There are two possible sign conventions for the square of the interval. You can say that I>0 for spacelike separations, or say that I>0 for timelike separations. I arbitrarily chose the first of these. 2. There may be multiple paths that light can take between two spacetime points. They need not actually have the same “length” (i.e., interval). Strictly, “interval” is defined only locally; then, for a particular path, you can integrate it up to get the overall interval. 3. In the case of light propagating through a medium other than vacuum, what actually happens involves electrons as well as photons and it isn’t just a matter of a photon going from A to B. Whenever a photon goes from A to B it does it, by whatever path it does, at the speed of light.)
The Lorentz factor diverges when the speed approaches c. Because of Length contraction and time dilation, both the distance and the time will appear to be 0, from the “point of view of the photon”.
(the photon is “in 2 places at once” only from the point of view of the photon, and it doesn’t think these places are different, after all they are in the same place! This among other things is why the notion of an observer traveling at c, rather than close to c, is problematic)
You can’t build a clock with an electron either. You can build one with a muon though, since it will decay after some interval. It’s not very accurate, but it’s something.
In general, you cannot build a clock moving at light speed. You could build a clock with two photons. Measure the time by how close they are together. But if you look at the center of mass of this clock, it moves slower than light. If it didn’t, the photons would have to move parallel to each other, but then they can’t be moving away from each other, so you can’t measure time.
I’m not sure what the significance of building a clock is...but then, I’m not sure I understand what clocks are. Anyway, isn’t ‘you can’t build a clock on a photon’ just what Tyson meant by ‘Photons have no ticking time at all’?
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.
Getting this property for electromagnetic waves was one of the main things that led Einstein to develop Special Relativity: he looked at waves and thought, “If we do a Galileian transform so that light is standing still, the resulting field is an invalid electrostatic field”
On the Neil Degrasse Tyson Q&A on reddit, someone asked: “Since time slows relative to the speed of light, does this mean that photons are essentially not moving through time at all?”
Tyson responded “yes. Precisely. Which means ----- are you seated?Photons have no ticking time at all, which means, as far as they are concerned, they are absorbed the instant they are emitted, even if the distance traveled is across the universe itself.”
Is this true? I find it confusing. Does this mean that a photon emitted at location A at t0 is absorbed at location B at t0, such that it’s at two places at once? In what sense does the photon ‘travel’ then? Or is the thought that the distance traveled, as well as the time, goes to zero?
There are no photons. There, you see? Problem solved.
(no, the author of the article is not a crank; he’s a Nobel physicist, and everything he says about the laws of physics is mainstream)
Problem evaded. Banning a word fails to resolve the underlying physical question. Substitute “wavepackets of light” for “photons”; what then?
I know, I was joking. And it was a good opportunity to link to this (genuinely interesting) paper.
… well, mostly joking. There’s a kernel of truth there. “There are no photons” says more than just banning a word. “Wavepackets of light” don’t exist either. There’s just the electromagnetic field, its intensity changes with time, and the change propagates in space. Looking at it like this may help understand the other responses to the question (which are all correct).
When you think of a photon as a particle flying in space, it’s hard to shake off the feeling that you somehow ought to be able to attach yourself to it and come along for the ride, or to imagine how the particle itself “feels” about its existence, how its inner time passes. And then the answer that for a photon, time doesn’t pass at all, feels weird and counter-intuitive. If you tell yourself there’s no particle, just a bunch of numbers everywhere in space (expressing the EM field) and a slight change in those numbers travels down the line, it may be easier to process. A change is not an object to strap yourself to. It doesn’t have “inner time”.
I feel I should let this go, and yet...
But we can make them! On demand, even.
By this argument, ocean waves don’t exist either. There’s only the sea, its height changes with time, and the change propagates in space.
You say that as a reductio ad absurdum, but it is good for some purposes. Anatoly didn’t claim that one should deny photons for all purposes, but only for the purpose of unasking the original question.
In this case, unasking the original question is basically an evasion, though, isn’t it?
Denying photons may enable you to unask hen’s literal question, or the unnamed Reddit poster’s literal question, but it doesn’t address the underlying physical question they’re driving at: “if observer P travels a distance x at constant speed v in observer Q’s rest frame, does the elapsed time in P’s rest frame during that journey vanish in the limit where v tends to c?”
I reject the claim that your rephrasing is the “real” question being asked. By rephrasing the question, you are rejecting it just as much as Anatoly. I think it is more accurate to say that you evade the question, while he is up front about rejecting it.
In fact, I think your answer is better and probably it is generally better to rephrase problematic questions to answerable questions before explaining that they are problematic, but the latter is part of a complete answer and I think Anatoly is correct in how he addresses it.
That multiple different people automatically treated hen’s question like it were my rephrasing backs me up on this one, I reckon.
Rephrasing a question can be the first step to confronting it head-on rather than rejecting it. If a tourist, looking for the nearest train station, wandered up to me and asked, “where station is the?”, and I rearranged their question to the parseable “where is the station?” and answered that, I wouldn’t say I rejected or evaded their query.
Oh. Great!
Other people have explained this pretty well already, but here’s a non-rigorous heuristic that might help. What follows is not technically precise, but I think it captures an important and helpful intuition.
In relativity, space and time are replaced by a single four-dimensional space-time. Instead of thinking of things moving through space and moving through time separately, think of them as moving through space-time. And it turns out that every single (non-accelerated) object travels through space-time at the exact same rate, call it c.
Now, when you construct a frame of reference, you’re essentially separating out space and time artificially. Consequently, you’re also separating an object’s motion through space-time into motion through space and motion through time. Since every object moves through space-time at the same rate, when we separate out spatial and temporal motion, the faster the object travels through space the slower it will be traveling through time. The total speed, adding up speed through space and speed through time, has to equal the constant c.
So an object at rest in a particular frame of reference has all its motion along the temporal axis, and no motion at all along the spatial axes. It’s traveling through time at speed c and it isn’t traveling through space at all. If this object starts moving, then some of the temporal motion is converted to spatial motion. It’s speed through space increases, and its speed through time decreases correspondingly, so that the motion through space-time as a whole remains constant at c. This is the source of time dilation in relativity (as seen in the twin paradox) - moving objects move through time more slowly than stationary objects, or to put it another way, time flows slower for moving objects.
Of course, the limit of this is when the object’s entire motion through space-time is directed along the spatial axes, and none of it is directed along the temporal axes. In this case, the object will move through space at c, which turns out to be the speed of light, and it won’t move through time at all. Time will stand still for the object. This is what’s going on with photons.
From this point of view, there’s nothing all that weird about a photon’s motion. From the space-time perspective, which after all is the fundamental perspective in relativity, it is moving pretty much exactly like any other object. It’s only our weird habit of treating space and time as extremely different that makes the entirely spatial motion of a photon seem so bizarre.
That is helpful, and interesting, though I think I remain a bit confused about the idea of ‘moving through time’ and especially ‘moving through time quickly/slowly’. Does this imply some sort of meta-time, in which we can measure the speed at which one travels through time?
And I think I still have my original question: if a photon travels through space at c, and therefore doesn’t travel through time at all, is the photon at its starting and its final position at the same moment? If so, in what sense did it travel through space at all?
At the same moment with respect to whom? That is the question one must always ask in relativity.
The answer is: no, emission and arrival do not occur at the same moment with respect to any actual reference frame. However, as we consider an abstract sequence of reference frames that move faster and faster approaching speed c in the same direction as the photon, we find that the time between the emission and the reception is shorter and shorter.
No it doesn’t. Remember, in relativity, time is relative to a frame of reference. So when I talk about a moving object traveling slowly through time, I’m not relativizing its time to some meta-time, I’m relativizing time as measured by that object (say by a clock carried by the object) to time as measured by me (someone who is stationary in the relevant frame of reference). So an object moving slowly through time (relative to my frame of reference) is simply an object whose clock ticks appear to me to be more widely spaced than my clock ticks. In the limit, if a photon could carry a clock, there would appear to me to be an infinite amount of time between its ticks.
I will admit that I was using a bit of expository license when I talked about all objects “moving through space-time” at the constant rate c. While one can make sense of moving through space and moving through time, moving through space-time doesn’t exactly make sense. You can replace it with this slightly less attractive paraphrase, if you like: “If you add up a non-accelerating object’s velocity through space and its (appropriately defined) rate of motion through time, for any inertial frame of reference, you will get a constant.”
Again, it’s important to realize there are many different “time” parameters in relativity, one for each differently moving object. Also, whether two events are simultaneous is relative to a frame of reference.
Relative to my time parameter (the parameter for the frame in which I am at rest), the photon is moving through space, and it takes some amount of (my) time to get from point A to point B. Relative to its own time parameter, though, the photon is at point A and point B (and every other point on its path) simultaneously. Since I’ll never travel as fast as a photon, it’s kind of pointless for me to use its frame of reference. I should use a frame adapted to my state of motion, according to which the photon does indeed travel in non-zero time from place to place.
Again, this is all pretty non-technical and not entirely precise, but I think it’s good enough to get an intuitive sense of what’s going on. If you’re interested in developing a more technical understanding without having to trudge through a mathy textbook, I recommend John Norton’s Einstein for Everyone, especially chapters 10-12. One significant simplification I have been employing is talking about a photon’s frame of reference. There is actually no such thing. One can’t construct an ordinary frame of reference adapted to a photon’s motion (partly because there is no meaningful distinction between space and time for a photon).
In the photon’s own subjective experience? Yes. (Not that that’s possible, so this statement might not make sense). But as another commenter said, certainly the limit of this statement is true: as your speed moving from point A to point B approaches the speed of light, the subjective time you experience between the time when you’re at A and the time when you’re at B approaches 0. And the distance does indeed shrink, due to the Lorentz length contraction.
It travels in the sense that an external observer observes it in different places at different times. For a subjective observer on the photon… I don’t know. No time passes, and the universe shrinks to a flat plane. Maybe the takeaway here is just that observers can’t reach the speed of light.
Not quite either of those.
The first thing to say is that “at t0” means different things to different observers. Observers moving in different ways experience time differently and, e.g., count different sets of spacetime points as simultaneous.
There is a relativistic notion of “interval” which generalizes the conventional notions of distance and time-interval between two points of spacetime. It’s actually more convenient to work with the square of the interval. Let’s call this I.
If you pick two points that are spatially separated but “simultaneous” according to some observer, then I>0 and sqrt(I) is the shortest possible distance between those points for an observer who sees them as simultaneous. The separation between the points is said to be “spacelike”. Nothing that happens at one of these points can influence what happens at the other; they’re “too far away in space and too close in time” for anything to get between them.
If you pick two points that are “in the same place but at different times” for some observer, then I<0 and sqrt(-I) is the minimum time that such an observer can experience between visiting them. The separation between the points is said to be “timelike”. An influence can propagate, slower than the speed of light, from one to the other. They’re “too far away in time and too close in space” for any observer to see them as simultaneous.
And, finally, exactly on the edge between these you have the case where I=0. That means that light can travel from one of the spacetime points to the other. In this case, an observer travelling slower than light can get from one to the other, but can do so arbitrarily quickly (from their point of view) by travelling very fast; and while no observer can see the two points as simultaneous, you can get arbitrarily close to that by (again) travelling very fast.
Light, of course, only ever travels at the speed of light (you might have heard something different about light travelling through a medium such as glass, but ignore that), which means that it travels along paths where I=0 everywhere. To an (impossible) observer sitting on a photon, no time ever passes; every spacetime point the photon passes through is simultaneous.
So: does the distance as well as the time go to 0? Not quite. Neither distance nor time makes sense on its own in a relativistic universe. The thing that does make sense is kinda-sorta a bit like “distance minus time” (and more like sqrt(distance-squared minus time-squared)), and that is 0 for any two points in spacetime that are visited by the same photon.
(Pedantic notes: 1. There are two possible sign conventions for the square of the interval. You can say that I>0 for spacelike separations, or say that I>0 for timelike separations. I arbitrarily chose the first of these. 2. There may be multiple paths that light can take between two spacetime points. They need not actually have the same “length” (i.e., interval). Strictly, “interval” is defined only locally; then, for a particular path, you can integrate it up to get the overall interval. 3. In the case of light propagating through a medium other than vacuum, what actually happens involves electrons as well as photons and it isn’t just a matter of a photon going from A to B. Whenever a photon goes from A to B it does it, by whatever path it does, at the speed of light.)
Thanks, that was very helpful, especially the explanation of timelike and spacelike relations.
The Lorentz factor diverges when the speed approaches c. Because of Length contraction and time dilation, both the distance and the time will appear to be 0, from the “point of view of the photon”.
(the photon is “in 2 places at once” only from the point of view of the photon, and it doesn’t think these places are different, after all they are in the same place! This among other things is why the notion of an observer traveling at c, rather than close to c, is problematic)
You can’t build a clock with a photon.
You can’t build a clock with an electron either. You can build one with a muon though, since it will decay after some interval. It’s not very accurate, but it’s something.
In general, you cannot build a clock moving at light speed. You could build a clock with two photons. Measure the time by how close they are together. But if you look at the center of mass of this clock, it moves slower than light. If it didn’t, the photons would have to move parallel to each other, but then they can’t be moving away from each other, so you can’t measure time.
I’m not sure what the significance of building a clock is...but then, I’m not sure I understand what clocks are. Anyway, isn’t ‘you can’t build a clock on a photon’ just what Tyson meant by ‘Photons have no ticking time at all’?
Yes. I meant that he meant that.
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.
Getting this property for electromagnetic waves was one of the main things that led Einstein to develop Special Relativity: he looked at waves and thought, “If we do a Galileian transform so that light is standing still, the resulting field is an invalid electrostatic field”