# Does time pass for photons?

I understand that a photon has never been known to decay. Is this because, travelling at the speed they do, time has stopped for them? (relative to what we sub-luminals observe)

DaveC426913
Gold Member
No, photons do not experience time.

No, photons do not experience time.
Photons are "outside" spacetime ? If so, where are they ?

JesseM
Photons are "outside" spacetime ? If so, where are they ?
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.

haushofer
No, photons do not experience time.

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.

haushofer
Photons are "outside" spacetime ? If so, where are they ?

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

$$v^{\mu}v_{\mu} = \frac{dx^{\mu}}{d\tau}\frac{dx_{\mu}}{d\tau} = \frac{d\tau ^{2}}{d\tau^{2}} = 1$$

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.

DaveC426913
Gold Member
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.
A photon does not have "a point of view". Nor does anything that travels at c.

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
$$v^{\mu}v_{\mu} = \frac{dx^{\mu}}{d\tau}\frac{dx_{\mu}}{d\tau} = \frac{d\tau ^{2}}{d\tau^{2}} = 1$$
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.
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
Homework Helper
Gold Member
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

$$v^{\mu}v_{\mu} = \frac{dx^{\mu}}{d\tau}\frac{dx_{\mu}}{d\tau} = \frac{d\tau ^{2}}{d\tau^{2}} = 1$$

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.

You probably meant to say that that
every observer has its 4-velocity (a timelike vector) normalized so that its norm has the same value of 1. (Of course, this means that, in any frame, its timelike-part has a larger absolute magnitude than its spacelike-part.) 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)... so that, in any frame, its spacelike-part and its timelike-part have equal absolute magnitudes in that frame. Finally, since Lorentz Transformations preserve the square-norm of a 4-vector, no Lorentz Transformation can boost a timelike vector into a null vector. Thus, there's no way to boost from the reference frame of a timelike observer into that of a photon.

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I understand that a photon has never been known to decay. Is this because, travelling at the speed they do, time has stopped for them? (relative to what we sub-luminals observe)
does a photon gradually slow down

DaveC426913
Gold Member
does a photon gradually slow down
No.

Though they do redshift over very long distances.

Slowing it down

does a photon gradually slow down

I believe they have done experiments where they have slowed the photon.
It does not slow down in a vacuum, however.

JesseM
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 ?
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.

Let us not forget the dual character of photon as particle or wave. I think the best way to imagine a photon is to associate it with the coherence length. withing the coherence length it is a wave, outside the coherence length a particle.

Regards
David

haushofer
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)...

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
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 ?

Like it was said, in your rest frame you always have this situation.

Maybe the analogy between 2 dimensional rotations and Lorentz transformations helps. If I have a vector on a piece of paper, I can always draw a frame in which the vector lies completely along one axis, say the x-axis. The vector has in that particular frame only an x-component; the other component ( say, the y-component ) is zero. Ofcourse, I can rotate this frame to describe the vector with a nonzero x and nonzero y component. This doesn't change the vector itself, but only the description of it. A Lorentz transformation can be regarded as a 4-dimensional rotation in space time.

robphy
Homework Helper
Gold Member
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.

The way one usually normalizes a vector $$v^a$$ is to divide it by the square-root of its [absolute] square-norm:
i.e. the "unit" vector $$\hat v^a = \frac{v^a}{\sqrt{g_{bc}v^bv^c}}$$.
Since a null-vector has zero square-norm, that method of "normalization" can't be applied. There is no natural candidate [that depends only on $$v^a$$ and the metric] to use instead. You'll break Lorentz invariance if you pick (say) a particular non-null vector to use in the denominator.

haushofer
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

$$v^{\mu}v_{\mu} = \frac{dx^{\mu}}{d\tau}\frac{dx_{\mu}}{d\tau} = \frac{d\tau ^{2}}{d\tau^{2}} = 1$$

we have $$d\tau^{2}=0$$ 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

$$\lim_{ d\tau \rightarrow 0} \frac{d\tau^{2}}{d\tau^{2}} = 1$$

?

Maybe I'm overlooking something very basic here.

robphy
Homework Helper
Gold Member
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

$$v^{\mu}v_{\mu} = \frac{dx^{\mu}}{d\tau}\frac{dx_{\mu}}{d\tau} = \frac{d\tau ^{2}}{d\tau^{2}} = 1$$

we have $$d\tau^{2}=0$$ 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

$$\lim_{ d\tau \rightarrow 0} \frac{d\tau^{2}}{d\tau^{2}} = 1$$

?

Maybe I'm overlooking something very basic here.

If $$v^a$$ is a 4-vector tangent to the worldline of a photon,
then it is a null-vector (a vector with zero-square norm using the Minkowski metric $$g_{ab}$$):
$$v^a v_a = v^a (g_{ab}v^b)= v^a g_{ab}v^b =0$$.
If you wish to use another norm, you'll have to use a different metric tensor [which is likely not to have any applicable physical interpretation].

If an observer was riding a photon what would that obsever see? Red shifted light looking back, blue shifted light looking forward?

Hootenanny
Staff Emeritus
Gold Member
If an observer was riding a photon what would that obsever see? Red shifted light looking back, blue shifted light looking forward?
As said previously one cannot associate a reference frame with a photon, hence, the question is nonsensical.

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So does this mean it is impossible to determine what an observer would see even hypothetically? This may be way over my head but im not sure why the question is non-nonsensical? Two negitives = a positive? Im just trying to make sense of this no frame of reference thing in my brain.

DaveC426913
Gold Member
So does this mean it is impossible to determine what an observer would see even hypothetically?
Yes.

When we start examining the very basic properties of the universe, we can't use our classical common-sense tools anymore. They break down.

If you try to answer this question, you will get paradoxes (i.e. non-sensical answers). And if you try to resolve the paradoxes, you will get more. The correct way to resolve the apparent paradoxes is to acknowledge that the initial premise upon which they are constructed is false.

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When we start examining the very basic properties of the universe, we can't use our classical common-sense tools anymore. They break down.

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.

DaveC426913
Gold Member
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.
I don't know what any of that means.