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poeteye
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I understand that a photon has never been known to decay. Is this because, traveling at the speed they do, time has stopped for them? (relative to what we sub-luminals observe)
Photons are "outside" spacetime ? If so, where are they ?DaveC426913 said:No, photons do not experience time.
It's not that they're outside spacetime, but that the time dilation of a clock approaches infinity as it approaches the speed of light (as seen by any sublight observer). I'm not sure what the connection is between this and the fact that photons don't decay in quantum field theory, though.Rade said:Photons are "outside" spacetime ? If so, where are they ?
DaveC426913 said:No, photons do not experience time.
Rade said:Photons are "outside" spacetime ? If so, where are they ?
A photon does not have "a point of view". Nor does anything that travels at c.haushofer said:From our point of view ofcourse. Because there is no Lorentz transformation taking an observer to the frame of reference of a photon doesn't automatically imply that a photon by itself "doesn't experience time". Maybe the question is just not appropriate.
Thank you, nice explanation. I have a question. Are there any physical entities (or concepts) that have 100 % of 4-velocity in the time-dimension and 0 % in the space dimension ?haushofer said:I would look at it in this way: every system in space-time has the same 4-velocity, namely 1 ( with c=1). This follows from
[tex]
v^{\mu}v_{\mu} = \frac{dx^{\mu}}{d\tau}\frac{dx_{\mu}}{d\tau} = \frac{d\tau ^{2}}{d\tau^{2}} = 1
[/tex]
It depends on the observer, and energy and rest-mass of the system how it divides this 4-velocity over space and over time as observed by that observer. A photon is an object with no rest mass, so it is forced to spent all of it's 4-velocity over the spatial dimensions as observed by all observers, leaving no velocity for the time-dimension. Mathematically this means that we can't transform an observer with a Lorentz transformation to the rest-frame of a photon.
haushofer said:I would look at it in this way: every system in space-time has the same 4-velocity, namely 1 ( with c=1). This follows from
[tex]
v^{\mu}v_{\mu} = \frac{dx^{\mu}}{d\tau}\frac{dx_{\mu}}{d\tau} = \frac{d\tau ^{2}}{d\tau^{2}} = 1
[/tex]
It depends on the observer, and energy and rest-mass of the system how it divides this 4-velocity over space and over time as observed by that observer. A photon is an object with no rest mass, so it is forced to spent all of it's 4-velocity over the spatial dimensions as observed by all observers, leaving no velocity for the time-dimension. Mathematically this means that we can't transform an observer with a Lorentz transformation to the rest-frame of a photon.
does a photon gradually slow downpoeteye said:I understand that a photon has never been known to decay. Is this because, traveling at the speed they do, time has stopped for them? (relative to what we sub-luminals observe)
No.andrewj said:does a photon gradually slow down
andrewj said:does a photon gradually slow down
The amount of 4-velocity in the space dimension vs. the time dimension depends on your choice of reference frame. In any sublight object's own rest frame, 100% of its 4-velocity is in the time dimension in that frame.Rade said:Thank you, nice explanation. I have a question. Are there any physical entities (or concepts) that have 100 % of 4-velocity in the time-dimension and 0 % in the space dimension ?
robphy said:Although a photon [with its zero rest-mass] doesn't have a 4-velocity vector normalized to square-norm 1 (since such a normalization is not Lorentz invariant), its 4-momentum vector is lightlike (with square-norm 0)...
Rade said:Thank you, nice explanation. I have a question. Are there any physical entities (or concepts) that have 100 % of 4-velocity in the time-dimension and 0 % in the space dimension ?
haushofer said:Why shouldn't a normalized 4-velocity for a photon not be Lorent invariant? As far as I know a photon also has a normalized 4-velocity, or am I confusing things? The reason that an observer can't perform a Lorentz transformation to "transform some part of the spatial components of that 4-velocity to the time part" is because the photon has no rest mass.
haushofer said:Ok, here the norm of the velocity-vector is explicitly involved, and in that case I see the problem. How I see it, is that one defines a 4-velocity with respect to the eigentime, and this results automatically in a 4-velocity which is normalized. In
[tex]v^{\mu}v_{\mu} = \frac{dx^{\mu}}{d\tau}\frac{dx_{\mu}}{d\tau} = \frac{d\tau ^{2}}{d\tau^{2}} = 1[/tex]
we have [tex]d\tau^{2}=0[/tex] for light, but is this really a problem here? After all, we divide two the same things which are tending to zero, so shouldn't we be able to state that for a photon
[tex]
\lim_{ d\tau \rightarrow 0} \frac{d\tau^{2}}{d\tau^{2}} = 1
[/tex]
?
Maybe I'm overlooking something very basic here.
As said previously one cannot associate a reference frame with a photon, hence, the question is nonsensical.jlorda said:If an observer was riding a photon what would that obsever see? Red shifted light looking back, blue shifted light looking forward?
Yes.jlorda said:So does this mean it is impossible to determine what an observer would see even hypothetically?
When we start examining the very basic properties of the universe, we can't use our classical common-sense tools anymore. They break down.
I don't know what any of that means.petm1 said:What about thinking about a photon using newer tools like length contracted to a point and time dilated to existing second per second, and even though a photon has no rest mass it is emitted and interacts with matter which does have mass, making it a property of mass.
What you can do is ask what an observer would as they approached arbitrarily close to the speed of light.jlorda said:So does this mean it is impossible to determine what an observer would see even hypothetically?
DaveC426913 said:No, photons do not experience time.
Absorption is a process that evolves over time, which a photon doesn't experience. Without time as the dimension over which the photon begins and ends, it must be space over which it begins and ends.rewebster said:what happens to the photon when its absorbed?
That's a question better answered by someone more in the know than I.jlorda said:OK I am up for that. Then what would they see?
DaveC426913 said:That's a question better answered by someone more in the know than I.
Essentially, all the light in front of you will be blue-shifted way up the scale to nearly infinite energy, whereas all the light behind you will be red-shifted way down the scale. Think of what an ambulance in front of you and another behind you would sound like if you were traveling at mach .99.
There are other details, such as the fact that your destination would get up all in your face as you virtually immediately arrive there.
Well... Don't forget that the universe is shot-through with very low frequency radiation as much as anything else; that would be correspondingly blue-shifted up into the visible spectrum.jlorda said:So we would not see anything because our eyes can only see in the visible spectrum of light?
High energy gamma rays would however excite your retina together with all the other components of your eyes and your vision area in your brain, so I imagine that you would perceive an almost infinitely bright light as if it were coming from all over, at least for an infinitesimal time before your death...jlorda said:So we would not see anything because our eyes can only see in the visible spectrum of light?