32
$\begingroup$

It is sometimes said, that if you stand still (in space), you travel through time at the speed of light. On the other side light never stands still, so it always only travels through space (at the speed of light), but not through time. Does that mean, if our universe would be filled with light only, no time would exist? Is the existence of mass therefor necessary for the existence of time?

$\endgroup$
5
  • $\begingroup$ I've asked a question which I think is similar to this one. And as expected, it was closed. physics.stackexchange.com/q271087 $\endgroup$
    – velut luna
    Commented Oct 23, 2016 at 12:39
  • 8
    $\begingroup$ It is sometimes said, that if you stand still (in space), you travel through time at the speed of light. ...what? $\endgroup$
    – valerio
    Commented Oct 23, 2016 at 12:51
  • 13
    $\begingroup$ @valerio92 The OP is referring to the common heuristic explanation of special relativity that everything always "moves through spacetime" at exactly the speed of light. So when you move in space at some speed, you need to "move through time" at a slower speed in order to keep the "total spacetime speed" constant at $c$, resulting in time dilation. It's just a heuristic and ignores subtleties like the indefinite metric, don't try to understand it too deeply. $\endgroup$
    – tparker
    Commented Oct 24, 2016 at 2:58
  • 1
    $\begingroup$ See Penrose’s Conformal Cyclic Cosmology. There are videos on YouTube. $\endgroup$
    – JDługosz
    Commented Oct 25, 2016 at 1:51
  • $\begingroup$ I agree with your conclusions. In a universe filled with only photons, time would not exist. Matter is necessary for the existence of time. $\endgroup$
    – Guill
    Commented Oct 28, 2016 at 21:45

7 Answers 7

36
$\begingroup$

A universe containing only light is simply a radiation dominated FLRW universe. Indeed our universe had approximately this geometry in its radiation dominated era. The FLRW metric is a perfectly good spacetime, so time certainly exists. Moreover the geometry is time dependent so we can use the energy density as a measure of time.

I concede that it's hard to build a device capable of measuring time in a universe containing only light, but to claim that time does not exist in such a universe would be plain wrong.

$\endgroup$
6
  • $\begingroup$ When light travels through the expanding universe it red shifts and loses its energy. Could this way light be aged in a light-only universe? $\endgroup$
    – akaltar
    Commented Oct 24, 2016 at 11:23
  • $\begingroup$ @akaltar: Yes, the energy density of the light will be related to time in the way described by the FLRW metric. An FLRW universe has a natural measure of time called the comoving time, and the light energy density will decrease smoothly with increasing comoving time. $\endgroup$ Commented Oct 24, 2016 at 12:41
  • 5
    $\begingroup$ So what you're saying is that the OP's question is essentially just a complicated reformulation of, "if a tree falls in a forest with no one around, does it make a sound"? $\endgroup$ Commented Oct 25, 2016 at 1:16
  • $\begingroup$ A universe filled only with photons would be homogeneous. There would be no density differences, therefore no measurement of time. $\endgroup$
    – Guill
    Commented Oct 28, 2016 at 21:42
  • $\begingroup$ In en.wikipedia.org/wiki/Photon_epoch it says "the photon epoch ... contained a hot dense plasma of nuclei, electrons and photons." In en.wikipedia.org/wiki/… it says "the constituents of the universe which moved relativistically, principally photons and neutrinos" . That doesn't sound like a radiation dominated universe is equivalent to a universe containing only light. $\endgroup$
    – asmaier
    Commented Jun 30, 2017 at 14:39
22
$\begingroup$

Speed has a definite meaning, i.e. , the absolute value of the velocity vector, which is given as $({\rm d}x/{\rm d}t,~{\rm d}y/{\rm d}t,~{\rm d}z/{\rm d}t).$ There is no mathematical meaning in the phrase "you travel through time at the speed of light".

Does that mean, if our universe would be filled with light only, no time would exist?

No. Entropy would still be defined by its statistical definition, and an arrow of time would exist.

Is the existence of mass therefore necessary for the existence of time?

Not of massive particles. As long as there is energy in its general relativistic meaning

$$m_0^2c^2 = \left(\frac{E}{c}\right)^2 - ||\mathbf p||^2$$ in natural units, where $c =1,$ $$m_0^2 = E^2 - ||\mathbf p||^2\,.$$

there will be an invariant mass of the photons comprising the light. In addition recent studies show that an arrow of time exists just because of gravitational interactions .

$\endgroup$
5
  • $\begingroup$ + vote for addressing an aspect of time I did not cover. The arrow of time and entropy are curiously tied to gravity through Bekenstein-Bousso bounds. This is a measure of the number of quantum states which can define a black hole or a cosmology. There is though the subtle question of whether we would have these sorts of physical configurations if the universe contained only light. If there were no massive particles in the universe would there be black holes? $\endgroup$ Commented Oct 23, 2016 at 15:33
  • $\begingroup$ @LawrenceB.Crowell see the answers here physics.stackexchange.com/questions/288287/… $\endgroup$
    – anna v
    Commented Oct 23, 2016 at 15:48
  • $\begingroup$ If you are using 4-velocity which is the derivation of $(ct, x_1, x_2, x_3)$, then the speed is equal to $c$ in case if $x_1,x_2,x_3$ doesn't change. That's probably the mentioned meaning of "moving at speed of light when staning still". $\endgroup$
    – Džuris
    Commented Oct 23, 2016 at 17:28
  • $\begingroup$ @ Anna, I thought an hour after writing this of the EM geon, which is basically what this Kugelblitz is. It has an unstable fixed point. If there is insufficient EM energy it flies apart, and too much it implodes into a black hole. $\endgroup$ Commented Oct 23, 2016 at 19:57
  • $\begingroup$ @Juris zero mass particles have no proper time to define a four velocity en.wikipedia.org/wiki/… . $\endgroup$
    – anna v
    Commented Oct 24, 2016 at 4:20
14
$\begingroup$

I think you are becoming confused with the notion of proper time. Proper time is a measure of how much passage of time is experienced by an object. When you say that an object sitting still travels through time at the speed of light, you are kind of right. Let me start by explaining this.


"Speed through time"

If we're not experiencing any forces (gravity does not count) due to other objects pushing us or somesuch, then we may consider ourselves as being still and everything else as moving. This is called a choice of reference frame. In picking this reference frame, we have our own measurements of spacial coordinates $x,y,z$ and of how much time has passed-- i.e. by looking at our wristwatch. We call these coordinates $(x,y,z,t)$.

An object's trajectory through spacetime can be represented by a velocity in this choice of coordinates. Without going into too much detail, the most natural way to express this velocity is by how far through space and how far through our time coordinate an object is travelling per second of time that it experiences. Remember, this is called proper time and we denote it $\tau$. This is how we would express an object's velocity through spacetime:

$$\left( \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau},\frac{dt}{d\tau} \right)$$

Again without going into too much detail or technicalities (I can if you want me to, but I suspect that you probably don't have the necessary background), special relativity tells us the amount of time we have to wait for an object to 'experience' one second ($dt/d\tau$) as a formula of its speed as measured by us ($dx/dt, dy/dt, dz/dt$). This factor is often denoted $\gamma$.

$$\gamma = \frac{dt}{d\tau} = \frac{1}{\sqrt{1 - \left(\frac v c \right)^2}} \quad \text{ where } \quad v = \sqrt{\frac{dx}{dt}^2 + \frac{dy}{dt}^2 + \frac{dz}{dt}^2}, \quad c = \text{speed of light}$$

What you call an object's 'speed through time' may be thought of as the amount of proper time this object experiences per second of our own time. In other words, how many minutes pass on a moving stopwatch per minute that passes on our own? This quantity is $d\tau/dt = 1/\gamma$.

$$\frac 1 \gamma = \sqrt{1 - \left(\frac v c \right)^2}$$ This is actually the equation of a circle! $$\left(\frac 1 \gamma \right)^2 + \left(\frac v c \right)^2 = 1$$

So an object's 'speed through time' and speed through space (as measured by us) always lies on the edge of this circle:


                                               enter image description here


Would time stop without mass?

As you can see, an object can either be moving quickly through space (again, according to us), or quickly through time, but not both. A photon has no mass and a consequence of this is that it is always moving at the speed of light. As a result, its arrow tip in the circle above is always pointing straight up, and so it experiences no passage of time; in other words, no proper time. (An interesting note as an aside here is that if the photon were theoretically an unstable particle which decayed, we could never observe the decay because time never passes for the photon!)

So what you are asking is whether time would 'stop' if the universe was all photons. While true that time would never pass for any of these photons, there are still some aspects of the universe's spacetime which we do not understand, and which depend on time. For instance, the universe is expanding. If the universe were all photons, then this could still happen, and would involve something changing about the universe as the time coordinate advanced.

So without any mass in the universe, there would be no passage of proper time (i.e. watches would never tick; biological processes would freeze) but there would still be a notion of time because the universe itself is changing with it.

$\endgroup$
1
  • 2
    $\begingroup$ As a layman this answer gives me the most insight. $\endgroup$
    – MrPaulch
    Commented Oct 24, 2016 at 12:44
5
$\begingroup$

The symmetry of spacetime is the Lorentz group. This is the set of three boosts and three spatial rotations of special relativity. Light rays form a projective Lorentz group. In geometry a projective space is the set of rays that "blow up a point," where all of those rays are equivalent under length change. Light rays emanating from a point, both into the future and those coming in from the past have zero length. The zero length condition is equivalent to saying there is no proper time associated with a light ray or photon. This is the light cone of special relativity. This defines them as the projective spacetime defined at a point of spacetime.

In general relativity spacetime consists of patches of locally flat regions with Lorentz symmetry. The same holds here, where a set of overlapping regions with a projective Lorentz symmetry define general curved spacetime. As a result we can consider spacetime with projective symmetry or according to light cones. Spacetime symmetry itself is then a fibration over the projective spacetime.

We can then see the question is a bit nuanced. If we consider only light rays, we then have only zero proper time for massless particles that might exist on some of these null rays. We might have an operational definition of there not existing time. However, the projective spacetime has this principal bundle construction which defines spacetime as well. This means that while time has no operational definition according to a massive particle along a timelike geodesic or curve, time still has a sort of emergent definition.

Many solutions to the Einstein field equations are vacuum solutions that have no source. This means spacetimes, not all of them necessarily flat, can exist in general relativity that are vacuums with no mass-energy source for the spacetime curvature.

$\endgroup$
4
  • 2
    $\begingroup$ "Light rays form a projective Lorentz group." What does this mean? What is a "projective Lorentz group"? "Spacetime symmetry itself is then a fibration over the projective spacetime. " What does this mean? What exactly is your base here, and what is the fiber? What has any of this to do with the passage of time the question asks about? $\endgroup$
    – ACuriousMind
    Commented Oct 23, 2016 at 12:45
  • $\begingroup$ The projective Lorentz group is $PSL(2,\mathbb C)$. This is well known. Read Ward and Wells book on Twistor geometry. $\endgroup$ Commented Oct 23, 2016 at 13:14
  • $\begingroup$ Not that well-known (relevant Google search). And we have that $\mathrm{PSL}(2,\mathbb{C})\cong \mathrm{SO}^+(3,1)$, i.e. it's not the "projective" Lorentz group, it's just the connected component of the identity (i.e. proper, orthochronous transformations). It acts on the null cone/light rays, but how do the rays "form" the group, as you say? What is this "projective spacetime" you're talking about? $\endgroup$
    – ACuriousMind
    Commented Oct 23, 2016 at 14:08
  • 2
    $\begingroup$ Projective spacetime aka a light cone. My answer here was informal which I think corresponded to the level of the question. $\endgroup$ Commented Oct 23, 2016 at 14:20
1
$\begingroup$

It is sometimes said, that if you stand still (in space), you travel through time at the speed of light.

This is not correct. Space is not the same thing as time, and the symmetry between space and time is limited to Lorentz symmetry. That means that for some specific purposes we may compare the time dimension with the space dimensions. But time is fundamentally different with respect to space. The assertion of a travel at speed of light through time is true if you set c=1 as it is currently practiced for certain purposes.

On the other side light never stands still, so it always only travels through space (at the speed of light), but not through time.

There is some confusion which you can avoid by reading the second postulate of special relativity: light is observed as traveling at c, but it is not said that light is traveling at c. That means that you get two points of view: On one hand, light is observed as traveling through spacetime at c, that means through space and through time. On the other hand, the lightlike spacetime interval of light is zero. That means that the point of emission and the point of absorption are directly adjacent in spacetime.

Does that mean, if our universe would be filled with light only, no time would exist? Is the existence of mass therefor necessary for the existence of time?

The proper time of massless particles and of lightlike movements in general is zero, they are not generating any proper time. This is why massless particles alone cannot build up any time. One mass particle in the universe would change the situation: The mass particle would interact with the massless particles, and the interaction would take place on the basis that light is observed as traveling at c. There would be time.

But you can also think of time in a universe without any particles at all. Space expansion is a process which is working like a clock.

$\endgroup$
-1
$\begingroup$

The time dimension would still exist. There just wouldn’t be any energy in it and nothing to experience the passage of time.

$\endgroup$
-3
$\begingroup$

You seem to presuppose that time slows down for moving objects and stops for light, but this contradicts any theory. For instance, special relativity predicts that time SPEEDS UP for moving objects/observers:

http://www.people.fas.harvard.edu/~djmorin/chap11.pdf David Morin, Introduction to Classical Mechanics With Problems and Solutions, Chapter 11, p. 14: "Twin A stays on the earth, while twin B flies quickly to a distant star and back. [...] For the entire outward and return parts of the trip, B does observe A's clock running slow..."

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.