# Does time pass for photons?

1. Jan 26, 2008

### poeteye

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)

2. Jan 26, 2008

### DaveC426913

No, photons do not experience time.

3. Jan 26, 2008

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

4. Jan 26, 2008

### JesseM

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.

5. Jan 26, 2008

### haushofer

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.

6. Jan 26, 2008

### haushofer

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.

7. Jan 27, 2008

### DaveC426913

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

8. Jan 27, 2008

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 ?

9. Jan 27, 2008

### robphy

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.

Last edited: Jan 27, 2008
10. Jan 27, 2008

### andrewj

does a photon gradually slow down

11. Jan 27, 2008

### DaveC426913

No.

Though they do redshift over very long distances.

12. Jan 27, 2008

### poeteye

Slowing it down

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

13. Jan 27, 2008

### JesseM

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.

14. Jan 28, 2008

### MeChaState

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

15. Jan 28, 2008

### haushofer

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.

16. Jan 28, 2008

### haushofer

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.

17. Jan 28, 2008

### robphy

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.

18. Jan 28, 2008

### 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.

19. Jan 28, 2008

### robphy

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].

20. Jan 28, 2008

### jlorda

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