B Can there be time without mass?

[guptasuneet]

TL;DR Summary

According to Relativity, any object traveling at c will not experience time, because of time dilatation. It will seem to the object that it has travelled instantly while ages may pass outside. Further, an object without rest mass will always travel at c. Therefore, if there were no objects with a rest mass in this universe (only photons), would there be time?

According to Relativity, any object traveling at c will not experience time, because of time dilatation. It will seem to the object that it has traveled instantly while ages may pass outside. Further, an object without rest mass will always travel at c. Therefore, if there were no objects with a rest mass in this universe (only photons), would there be time?

 

[PeroK]

The theory of relativity assumes that spacetime is a 4D manifold. It doesn't directly depend on particles, massive or massless, for the existence of spacetime itself.

 

[Nugatory]

According to Relativity, any object traveling at c will not experience time, because of time dilatation.

This is a common misconception, one of those things that everyone “knows” that is pretty much completely wrong.

One way of seeing the problem with this line of thinking is to consider that if it were correct (it’s not!) then if something were traveling at the speed of light relative to you we could just as well say that you are traveling at the speed of light relative to it - and you’d be the one experiencing no time.

Your best bet will be to find a good textbook (Taylor and Wheeler’s “Spacetime Physics” is my favorite and especially good for your particular question) and start over from a more solid foundation.

 

[guptasuneet]

I understand the difference between traveling in a spaceship and staying on land, and why time contracts for the spaceship traveller and not for the land dweller.

All of relativity (distance compression, time contraction, etc.) is explained at relativistic velocities and never at c. However, according to the Lorentz Transformation, if I am traveling at c with respect to another frame of reference outside, then the distance will tend towards zero and accordingly time taken will also tend towards zero. While in the frame of reference my motion will be measured at c, and will take time, according to my perspective it will be practically instantaneous.

I am trying to visualise a universe where all objects travel at c relative to each other. How will time be measured in such a universe? And what will be a static frame of reference?

 

[PeterDonis]

according to the Lorentz Transformation, if I am traveling at c with respect to another frame of reference

You can't; this is impossible. The Lorentz Transformation does not allow one frame of reference to move at $c$ relative to another. The transformation between inertial frames is only valid for $v < c$.

Please read our forum FAQ on this topic:

https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/

 

[guptasuneet]

You can't; this is impossible. The Lorentz Transformation does not allow one frame of reference to move at $c$ relative to another. The transformation between inertial frames is only valid for $v < c$.

Please read our forum FAQ on this topic:

https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/

I agree. There can be no rest frame of a photon, because a photon has no rest mass and always travels at c. Similarly, there cannot be a rest frame of any other object that also has zero rest mass.

My question is that in a theoretical universe composed only of massless objects (i.e. with zero rest mass), how will we measure time? And in such a scenario will time (and space) have any meaningful and measurable definition?

 

[PeterDonis]

in a theoretical universe composed only of massless objects (i.e. with zero rest mass), how will we measure time?

In such a universe there would be no "we" to measure time. We are made of massive objects, not massless ones.

in such a scenario will time (and space) have any meaningful and measurable definition?

A spacetime geometry can be defined just fine even if the only source of stress-energy is massless objects. (Look up "radiation dominated universe".) "Time" and "space" come from the spacetime geometry.

 

[Nugatory]

However, according to the Lorentz Transformation, if I am traveling at c with respect to another frame of reference outside, then the distance will tend towards zero and accordingly time taken will also tend towards zero.

The Lorentz transformations are derived using assumptions that are equivalent to assuming that $v\lt c$, so are not meaningful when $v=c$. The infinities that appear when we set $v=c$ are how the math tells us that we’ve made a mistake and our results are not valid.

My question is that in a theoretical universe composed only of massless objects (i.e. with zero rest mass), how will we measure time? And in such a scenario will time (and space) have any meaningful and measurable definition?

If you won’t let me have any matter I won’t be able to build any measuring devices. That doesn’t meant that time and space aren’t defined; the Taylor and Wheeler book I mentioned above describes space and time in a way that does not require any physically realizable measuring devices, just that we analyze what would happen if we do build such devices.

 

[phinds]

I understand the difference between traveling in a spaceship and staying on land, and why time contracts for the spaceship traveller and not for the land dweller.

Wrong. Time dilation is seen by the land dweller as happening to the spaceship traveler and is seen by the spaceship traveler as happening to the land dweller.

 

[PeroK]

I understand the difference between traveling in a spaceship and staying on land, and why time contracts for the spaceship traveller and not for the land dweller.

According to the first postulate of SR, the laws of physics are the same in all inertial reference frames. There is no difference, therefore, between traveling in a spaceship at constant velocity relative to the Earth and staying on Earth (which is approximately inertial). Velocity-based time dilation is reciprocal and just as the spaceship time is dilated in the Earth frame, so Earth time is dilated in the spaceship frame.

All inertial motion is relative and there is no such thing as absolute velocity. In other words, there is no experiment that can give you an absolute velocity - only your velocity relative to something else. As an example, if you take a plane flying east, then you speed up relative to an inertial frame where the Earth is at rest and spinning. And, if you take a plane west, then you slow down relative to this frame. But, there is no sense in which you are absolutely speeding up or slowing down. Relative to a frame of reference where the Sun is at rest, whether you are speeding up or slowing down might be different again.

 

[guptasuneet]

Wrong. Time dilation is seen by the land dweller as happening to the spaceship traveler and is seen by the spaceship traveler as happening to the land dweller.

Yes, but in the end the space traveler lands back on land and returns to the reference frame of land (or he lands on some distant planet which is in the same reference plane as the land from which he started). Therefore, the apparent time contraction is seen by the land dweller happening for the space traveler.

If there could be a spaceship million of miles long, and a provision for the land dweller to board and exit the spaceship at two different points in space-time, then the space traveler would see the time contraction happening to the land dweller.

 

[guptasuneet]

A spacetime geometry can be defined just fine even if the only source of stress-energy is massless objects. (Look up "radiation dominated universe".) "Time" and "space" come from the spacetime geometry.

Thanks, I'll lookup on the radiation dominated universe. I know theoretically that space and time evolve from space-time geometry. However, I am not able to give to myself a practical interpretation without mass/energy.

It is like the chicken and egg problem. Which came first, space-time or energy/mass? Does space-time geometry form the stage on which energy/mass acts, or is the space-time geometry itself formed due to the presence of energy/mass? Can there be a space-time geometry without energy/mass?

I guess it is futile to talk about them separately. However, I posed this question to understand which of the two is more fundamental in relativity (or is this another of those unanswerable or undefined questions)?

 

[PeroK]

Yes, but in the end the space traveler lands back on land and returns to the reference frame of land (or he lands on some distant planet which is in the same reference plane as the land from which he started). Therefore, the apparent time contraction is seen by the land dweller happening for the space traveler.

This is not correct. There is no sense in which the space traveller's time is really dilated - although a lot of popular science books give that impression.

Instead, we have differential ageing, not time dilation. This is a function of the complete path through spacetime taken by each person - which must include acceleration and/or deceleration phases or instantaneous changes of velocity.

 

[vanhees71]

I agree. There can be no rest frame of a photon, because a photon has no rest mass and always travels at c. Similarly, there cannot be a rest frame of any other object that also has zero rest mass.

My question is that in a theoretical universe composed only of massless objects (i.e. with zero rest mass), how will we measure time? And in such a scenario will time (and space) have any meaningful and measurable definition?

That's a question hard to answer, because it's a different universe.

We can however discuss another aspect of this question, whether we can have a physics without any scale. In the quantum-field theoretical context one should be aware that scale invariance of massless theories is anomalously broken. One example is massless QCD, which consists of massless gauge bosons (gluons) and massless quarks.

One way to see that we have to introduce an energy-momentum scale is already because if you calculate higher-order Feynman diagrams, containing loops, you have to renormalize the usual UV divergences. To do this systematically you have to renormalize the proper vertex functions, i.e., the connected one-particle irreducible truncated diagrams, which leads to wave-function and coupling-constant renormalization. To this end you have to choose a renormalization scale, i.e., an energy-momentum scale at which you define your finite quantities or at which you subtract the infinities in the counterterm approach to renormalization. You have to do that in a spacelike region of four-momenta. Because (sic!) the theory is massless you cannot subtract at the point where all external four-momenta are zero and thus you have to introduce necessarily an energy-momentum scale. In the usually applied (modified) minimal-subtraction scheme, where you use dimensional regularization as a convenient intermediate step (where the dimensionful quantity has to be introduced for the more formal reason that you want to keep the coupling constant a dimensionless quantity as it is naturally in (1+3) dimensions in all spacetime dimensions), this introduces $\Lambda_{\text{QCD}}$ as a scale, and the renormalized coupling becomes a function of this scale ("running coupling"). This is also called "dimensional transmutation", i.e., first you have a dimensionless coupling constant in a theory without an explicit energy-momentum scale but after renormalization you have a energy-momentum scale as a parameter with the coupling constant being a function of it.

Another way, independent from perturbative renormalization theory, is to realize that scale invariance is anomalously broken, which can be shown with the path-integral formalism, where the transformation of the fields under scale variations leads to an extra factor in the corresponding functional determinant (in a similar way as in the more familiar case of the axial anomaly of the chiral symmetry).

 

[Office_Shredder]

Yes, but in the end the space traveler lands back on land and returns to the reference frame of land (or he lands on some distant planet which is in the same reference plane as the land from which he started). Therefore, the apparent time contraction is seen by the land dweller happening for the space traveler.

If there could be a spaceship million of miles long, and a provision for the land dweller to board and exit the spaceship at two different points in space-time, then the space traveler would see the time contraction happening to the land dweller.

There are a couple other threads floating around about this, including one started by me. So I'm far from an expert, but let me try to distill the thread into the part that made this most intuitively clear for me.

Let's say someone is traveling at near c to a planet 100 light years away. About 50 years into their trip (from Earth's perspective), a person on Earth and a person on the planet both release a flash of light. From the Earth's perspective, these two flashes both occur at the midpoint of the trip, at about the same time as the ship reaches the planet.

What happens from the rockets perspective? We know a couple things happen - Earth and the other planet release flashes of light, and those flashes happen at the midpoint of the Earth and the planet. But when those flashes occur is up for grabs. Since light travels at the speed of light from the rocket's perspective, and the Earth and the other planet are traveling away from the planet with the Earth in front, the light shot from the planet will appear to sloooowly catch up to earth, and the light shot from Earth will appear to be caught by the planet very quickly.

So for these beams of light to meet in the middle, the planet must have fired its beam first. In fact, if the speeds are high enough, from the rocket's frame it will appear as its the planet must have fired its light before the rocket even left!

Now if we consider regular flashes like this to be a clock on each of Earth and the planet, we are basically saying that if the clocks are in sync between the planet and the earth, the rocket will think the planet's clock is way ahead of earth. Then when it turns around, it will think Earth's clock is way ahead of the planet.

Another way of thinking about this (I think) is that time dilation is just an observation that if a line of synchronized clocks is traveling past you, you will think the clocks that haven't reached you yet are running ahead of the ones behind you. As a clock moves from in front of you to behind you, it must change from being ahead of the clock at your location to being behind the clock in your location. So the time on that specific clock appears to run slowly. But if you keep looking out the window and checking the time on the clock that you are passing by right now, you will see that specific sequence of clock times advancing at the right rate.

And hence when the rocket reaches the planet, when they look out the window they see the clock in that planet read 100 years after they left earth. The rocket observer thinks the clock on Earth has fallen way behind since it's behind the rocket now. If the rocket stops, they will then stop seeing the Earth clock as being behind the planet clock (and they would stop seeing hypothetical clocks further ahead in the motion of travel as being ahead of the clock on the planet).

I'll second the recommendation of the book spacetime physics. It is not math heavy, and I think focuses on the right set of things for a first time introduction to the material.

 

[PeroK]

There are a couple other threads floating around about this, including one started by me. So I'm far from an expert, but let me try to distill the thread into the part that made this most intuitively clear for me ...

Too many words, not enough calculation!

 

[PeroK]

... when I was learning SR I became very suspicious of all this "thinks" and "perspective" and "sees". So, I cleared all of that out of my thinking and focused on events, times and locations in various reference frames. Everything else I took to be extraenous material that clouded the issue.

 

[PeroK]

Too many words, not enough calculation!

Let's do some calculations.

Assume we have two planets $15$ light years apart. They have synchronised clocks between them (over the course of a few decades). A rocket passes the first planet, A, and synchronises its clock with planet A. The rocket travels to planet B at a speed of $\frac 3 5 c$ and compares its clock with the clock at planet B when it gets there.

We know is that the $\gamma$ factor between the rocket frame and the planets' frame is $\frac 5 4$, hence the planets are $12$ light years apart in the rocket frame. And the journey takes $20$ years according to the rocket clock.

The clocks on planets A and B are dilated by a factor of $\frac 5 4$, as measured in the rocket frame, so the clock on planet B advances only $16$ years during the journey, as measured in the rocket frame.

But, in the planets' frame, the journey takes $25$ years, so this is what B's clock reads when the rocket arrives.

The immediate conclusion is that B's clock must have read $9$ years ahead in the rocket frame at the start of the journey. That tells us that we must have a loss of simultaneity of $9$ years in this case.

Can we confirm this conclusion by another method? Let's assume that when the rocket reaches the midpoint between the planets it fires a light signal in each direction. The signals should reach the planets simultaneously in their frame, as they were fired from the midpoint. The planets may use this to synchronise their clocks.

In the rocket frame, planet B is moving towards the light signal, so the signal reaches planet B after only $3.75$ years ($6$ light years divided by $(1 + \frac 3 5)c$.

Planet A is moving away from the light signal, so the signal reaches planet A after $15$ years ($6$ light years divided by $(1 - \frac 3 5)c$.

That gives a difference of $11.25$ years in the rocket frame. But, planet B's clock is dilated in the rocket frame, so advances only $\frac 4 5 \times 11.25 = 9$ years.

Hence, we obtain the same result that the clock on planet B (if synchronised with the clock on planet A in their rest frame) is $9$ years ahead as measured in the rocket frame.

This is how a careful analysis of simultaneity resolves the twin paradox and related problems.

 

[PeterDonis]

I posed this question to understand which of the two is more fundamental in relativity

Since there can be vacuum solutions of the Einstein Field Equation in GR, there can be spacetime without matter/energy. But in GR, you cannot have matter/energy without spacetime. So if one had to choose, I would say spacetime is more fundamental in relativity.

 

[Sagittarius A-Star]

But if you keep looking out the window and checking the time on the clock that you are passing by right now, you will see that specific sequence of clock times advancing at the right rate.

This clock has the same tick-rate as the other mentioned clocks. It can be calculated from the inverted Lorenz-transformation for time (the moving clock rests in the primed frame):
$\Delta t = \gamma (\Delta t' + v \Delta x') = \gamma (\Delta t' + v *0)$
$\Delta t' / \Delta t = 1/\gamma$.

 

[Office_Shredder]

This clock has the same tick-rate as the other mentioned clocks. It can be calculated from the inverted Lorenz-transformation for time (the moving clock rests in the primed frame):
$\Delta t = \gamma (\Delta t' + v \Delta x') = \gamma (\Delta t' + v *0)$
$\Delta t' / \Delta t = 1/\gamma$.

My point here is you never look at the same clock twice. Every time you're looking out the window you see a clock in a different location (different location in the frame where the infinite string of clocks is at rest).

I think it has to be true that if there is a clock halfway between the Earth and the target planet (so 50 light years from each of them), and the rocket travels at almost c, at the time the rocket observer passes that midpoint it should read 50 years after time of departure, and when the rocket observer reaches the other other planet the clock on that planet should say 100 years after time of departure. Even though from the rocket's perspective, the midpoint clock read 50-epsilon when it departed, and reads 50+epsilon when the rocket arrives on the other planet.

 

[PeroK]

I think it has to be true that if there is a clock halfway between the Earth and the target planet (so 50 light years from each of them), and the rocket travels at almost c, at the time the rocket observer passes that midpoint it should read 50 years after time of departure, and when the rocket observer reaches the other other planet the clock on that planet should say 100 years after time of departure. Even though from the rocket's perspective, the midpoint clock read 50-epsilon when it departed, and reads 50+epsilon when the rocket arrives on the other planet.

That is all true. That is the relativity of simultaneity.

 

[PeterDonis]

If there could be a spaceship million of miles long, and a provision for the land dweller to board and exit the spaceship at two different points in space-time, then the space traveler would see the time contraction happening to the land dweller.

Actually, the "land dweller" would have to exit the spaceship first, then board it again some time later.

This is actually an instructive way to construct a "twin paradox" scenario. Suppose we pick a frame which we'll call the "land frame"; in this frame, a very, very long spaceship is moving to the right at some speed $v$ which is close to the speed of light. The clocks on the spaceship are all synchronized with each other as viewed in the spaceship's rest frame.

At time $t' = 0$ by the spaceship clocks, the "land dweller" jumps out of the spaceship at $x = 0$ in the "land frame" and immediately decelerates to a stop, so he is at rest in the "land frame". (We will say that the land frame clocks read $t = 0$ at this instant, and that the point on the spaceship where the land dweller jumped out is at $x' = 0$.) At this instant the land dweller's clock also reads zero, since it was sychronized with the spaceship clocks while he was on board the spaceship. Then the land dweller floats there and waits for a while, while the spaceship flies past him; then, after some time $T$ has passed on his clock, he immediately accelerates to speed $v$ and boards the spaceship. The time that he sees on the spaceship clock when he boards (this will of course be a different clock than the one he saw when he exited the spaceship, but all of the spaceship clocks are synchronized in the spaceship frame) will be greater than $T$.

 

[Sagittarius A-Star]

My point here is you never look at the same clock twice. Every time you're looking out the window you see a clock in a different location (different location in the frame where the infinite string of clocks is at rest).

O.K. In this case

clock times advancing at the right rate.

... means: by the factor $\gamma$ faster than the traveller's watch rate.

 

[MikeeMiracle]

If all particles present in the universe are massless then surely there are no "clocks" to measure "time" with to begin with? That being the case the question becomes irrelevant does it not?

 

[vanhees71]

... when I was learning SR I became very suspicious of all this "thinks" and "perspective" and "sees". So, I cleared all of that out of my thinking and focused on events, times and locations in various reference frames. Everything else I took to be extraenous material that clouded the issue.

That's an interesting statement. I also have the feeling that we dwell too much on Lorentz transformations and kinematic "paradoxes". Perhaps one should rather concentrate on the derivation of the covariant laws. On the other hand, without the Lorentz transformation it's difficult to understand these kinematic properties, which can be best analyzed using the Lorentz transformation (particularly boosts), and here I think the most straight-forward way still is to introduce it as the invariance group of the Lorentzian fundamental form of Minkowski space and then discuss all the issues with clock synchronization and the kinematical issues with light clocks, relativity of simultaneity, time dilation, and length contraction, but that can be done independently from the twin paradox, which should exclusively defined as the issue of "aging" between two twins starting from one common event and meeting again at another common event, and this can be addressed without any tricky Lorentz transformations by simply calculating the proper times of the twins. There's no need to introduce non-inertial reference frames and complicated issues of clock synchronization (particularly in the rather simple case of a twin, or a muon in a storage ring as an example from a real-world experiment, traveling on a circle.

 

[robphy]

... when I was learning SR I became very suspicious of all this "thinks" and "perspective" and "sees". So, I cleared all of that out of my thinking and focused on events, times and locations in various reference frames. Everything else I took to be extraenous material that clouded the issue.

That's an interesting statement. I also have the feeling that we dwell too much on Lorentz transformations and kinematic "paradoxes". Perhaps one should rather concentrate on the derivation of the covariant laws. On the other hand, without the Lorentz transformation it's difficult to understand these kinematic properties, which can be best analyzed using the Lorentz transformation (particularly boosts), and here I think the most straight-forward way still is to introduce it as the invariance group of the Lorentzian fundamental form of Minkowski space and then discuss all the issues with clock synchronization and the kinematical issues with light clocks, relativity of simultaneity, time dilation, and length contraction, but that can be done independently from the twin paradox, which should exclusively defined as the issue of "aging" between two twins starting from one common event and meeting again at another common event, and this can be addressed without any tricky Lorentz transformations by simply calculating the proper times of the twins. There's no need to introduce non-inertial reference frames and complicated issues of clock synchronization (particularly in the rather simple case of a twin, or a muon in a storage ring as an example from a real-world experiment, traveling on a circle.

Along the lines of "too much on Lorentz transformations",
I am reminded of this quote by Mermin ( "Lapses in Relativistic Pedagogy"
http://aapt.scitation.org/doi/10.1119/1.17728 )

...Lorentz transformation doesn't belong in a first exposure to special relativity. Indispensable as it is later on, its very conciseness and power serve to obscure the subtle interconnnectedness of spatial and temporal measurements that makes the whole business work. Only a loonie would start with real orthogonal matrices to explain rotations to somebody who had never heard of them before, but that's how we often teach relativity. You learn from the beginning how to operate machinery that gives you the right answer but you acquire little insight into what you're doing with it.

However, I do think it is important to define quantities operationally
and to geometrically model the physical measurement processes...
what does it mean "to see (visually)"?
"to determine the distance to a distant event from a worldline"? etc...
And I think these are best understood
by drawing spacetime diagrams [instead of boxcars moving in space]
and trying to reason with the Minkowskian spacetime geometry
visually (light cones, intersections, tangents, etc...) and
algebraically (4-vector addition and their dot products, etc...)
and interpret physically.

Multiple-representations can help support the various interpretations of what is going on.
(Appealing to certain aspects of one's Euclidean trigonometric intuition
[and demoting other aspects] is a powerful, but overlooked tool if used correctly.)

 

[vanhees71]

Well, but how do you understandably argue for the construction of the Minkowski diagram? You can of course deal without Lorentz transformations by just using Einstein's "two postulates" (validity of the special principle of relativity, independence of the speed of light on the uniform motion of the light source) to get the Minkowski pseudometric. Then you can construct the Minkowski diagram (of course restricting yourself to 1+1 spacetime dimensions) and use the geometry of hyperbolae and light cones to define the spatiotemporal measures. Then you can construct everything geometrically, including clock synchronization of clocks at different places being at rest relative to each other, demonstrating the non-synchronicity of clocks in relative uniform motion to each other, and all the kinematics. Also the Lorentz transformation follows from such constructions.

On the other hand, I always struggle with Minkowski diagrams, because it's hard to forget the Euclidean plane, we are all used to from early schooldays on. I always find an algebraic approach much more clear. I usually calculate everything algebraically and then put the reesults in a Minkowski diagram to visualize it.

 

[.Scott]

... an object without rest mass will always travel at c. Therefore, if there were no objects with a rest mass in this universe (only photons), would there be time?

I think this question needs to be examined in a different light (sorry) than what has been posted so far.
A universe with only photons will still have mass. And because of photon-photon interactions it will, in due course, also have fermions with non-zero rest mass.

But there are a couple of things that are unclear to me:

1) Can a universe of photons become more disordered? If not, then time cannot move forward. If so, would the rate at which entropy increased be so low as to challenge the definition of time?

2) A common (but critically oversimplified) way of modeling a photon is to imagine a particle traveling from a source to a target not much differently than how a baseball travels. But many experiments, including double slit and interferometer experiments, refute this. A "traveling photon", is not spatially-specific - it doesn't have a baseball-like trajectory. In a universe consisting only of photons, each photon might persist quadrillions of years without colliding with another photon - but during this life, its cumulative gravitational affect would be an important contributor to the geometry of the universe. And in a broad statistical sense, that contribution would betray its location.

 

[DaveC426913]

If all particles present in the universe are massless then surely there are no "clocks" to measure "time" with to begin with? That being the case the question becomes irrelevant does it not?

In such a thought experiment, you don't need to consider every practical issue, such as can we make clocks?
Any two particle interactions can serve as a clock.

Let's phrase it another way: in this universe where all particles are massless, is there anything that prevents every interaction that ever did happen to will happen - from happening all at once? Would a universe where everything happened all at once even last a non-zero length of time?

 

[PeterDonis]

Can a universe of photons become more disordered?

Of course. You yourself described one mechanism for this: photon-photon interactions will produce fermion-antifermion pairs.

If not, then time cannot move forward.

Sure it can: the classical GR model of a radiation dominated universe still has a perfectly good definition of "time" that moves forward. (Note that in such a model, which idealizes "photons" as null radiation, there are no photon-photon interactions and no pair production.)

would the rate at which entropy increased be so low as to challenge the definition of time?

Any rate that isn't zero would still give you a second law "arrow" of time. But note that this is just a definition of which direction in time is the "forward" direction; it's not the same as a definition of time.

In a universe consisting only of photons, each photon might persist quadrillions of years without colliding with another photon - but during this life, its cumulative gravitational affect would be an important contributor to the geometry of the universe. And in a broad statistical sense, that contribution would betray its location.

Not necessarily; in the classical GR model of a radiation dominated universe, all you have is the energy density of radiation; there are no individual "photons" with locations.

 

[PeterDonis]

in this universe where all particles are massless, is there anything that prevents every interaction that ever did happen to will happen - from happening all at once?

Yes. The fact that arc length along null worldlines is zero does not mean the universe does not have 4 dimensions. Interactions still happen at distinct events and those distinct events are still distributed through 4 dimensions of spacetime.

 

[.Scott]

Of course. You yourself described one mechanism for this: photon-photon interactions will produce fermion-antifermion pairs.

OK. I didn't immediately recognize that as an increase in entropy. But it certainly would be. In due time, an equilibrium would be reached where the rate of fermion pair creation would equal the rate for pair destruction (not to mention other events). So the early pair productions would represent moves toward that equilibrium and thus an increase in entropy.

 

[.Scott]

Sure it can: the classical GR model of a radiation dominated universe still has a perfectly good definition of "time" that moves forward. (Note that in such a model, which idealizes "photons" as null radiation, there are no photon-photon interactions and no pair production.)

In such a case, how would entropy increase? Are Black Holes allowed to form? Does the distribution of matter get clumpy? If nothing statistically irreversible ever happens, you may have a minimal definition of time, but it will lack a direction.

 

[PeterDonis]

Are Black Holes allowed to form? Does the distribution of matter get clumpy?

There is no "matter" in such a universe, only radiation, which doesn't clump gravitationally the way matter (more precisely, non-relativistic matter) does, but that doesn't mean it doesn't clump at all; if it has density variations, there will be some clumping, yes.

If nothing statistically irreversible ever happens

The expansion of the universe itself is statistically irreversible if it is going to continue expanding forever.

 

[guptasuneet]

Actually, the "land dweller" would have to exit the spaceship first, then board it again some time later.

This is actually an instructive way to construct a "twin paradox" scenario. Suppose we pick a frame which we'll call the "land frame"; in this frame, a very, very long spaceship is moving to the right at some speed $v$ which is close to the speed of light. The clocks on the spaceship are all synchronized with each other as viewed in the spaceship's rest frame.

At time $t' = 0$ by the spaceship clocks, the "land dweller" jumps out of the spaceship at $x = 0$ in the "land frame" and immediately decelerates to a stop, so he is at rest in the "land frame". (We will say that the land frame clocks read $t = 0$ at this instant, and that the point on the spaceship where the land dweller jumped out is at $x' = 0$.) At this instant the land dweller's clock also reads zero, since it was sychronized with the spaceship clocks while he was on board the spaceship. Then the land dweller floats there and waits for a while, while the spaceship flies past him; then, after some time $T$ has passed on his clock, he immediately accelerates to speed $v$ and boards the spaceship. The time that he sees on the spaceship clock when he boards (this will of course be a different clock than the one he saw when he exited the spaceship, but all of the spaceship clocks are synchronized in the spaceship frame) will be greater than $T$.

Exactly, if the land dweller exits the ship, come to an immediate stop and jumps back in after some time (immediately going up to the ship's velocity), then the clocks on the spaceship will be ahead of the watch on his hand, because time moved faster for him, relative to the frame of reference of the spaceship.

 

[guptasuneet]

Yes. The fact that arc length along null worldlines is zero does not mean the universe does not have 4 dimensions. Interactions still happen at distinct events and those distinct events are still distributed through 4 dimensions of spacetime.

While the events will be distributed at distinct points in the 4-dimensional spacetime, for the massless photons, all interactions should still apparently happen instantaneously.

 

[Ibix]

While the events will be distributed at distinct points in the 4-dimensional spacetime, for the massless photons, all interactions should still apparently happen instantaneously.

No. Time isn't defined along null worldlines, so it's wrong to use time words like "instantaneously" for things that follow them.

 

[Nugatory]

While the events will be distributed at distinct points in the 4-dimensional spacetime, for the massless photons, all interactions should still apparently happen instantaneously.

That is a common misunderstanding, and one that you will have to give up at some point. The problem is that you are assuming the existence of an inertial frame in which the speed of the photon (properly speaking, "flash of light" - a photon is not what you think it is, but that's a different conversation) is zero, but the speed of light is $c$ in all inertial frames so there can be no such thing.

 

[jbriggs444]

Exactly, if the land dweller exits the ship, come to an immediate stop and jumps back in after some time (immediately going up to the ship's velocity), then the clocks on the spaceship will be ahead of the watch on his hand, because time moved faster for him, relative to the frame of reference of the spaceship.

Time dilation means that his watch ticks slow as reckoned in the rest frame of the [very long] spaceship.

The phrase "time moved faster for him" is inapt. Time moved at the same rate it always does for him. One second proper time per second of proper time. Just as it does for everyone.

 

[PeterDonis]

for the massless photons, all interactions should still apparently happen instantaneously.

For photons, there is no such thing as "apparently happen". The concept of "experienced time" simply does not apply to photons.

 

[Wes Tausend]

Strictly speaking, I would say No.

Time is merely a fundamental quality only assigned to the interaction of two or more masses in motion, or simply an event. Without involving mass nothing would happen, no event could occur and no space between particles would be present to form space-time. It makes no sense for time to exist in a universe devoid of particles and therefore devoid of space 'between' or motion 'of'.

Wes

 

[jbriggs444]

Strictly speaking, I would say No.

Time is merely a fundamental quality only assigned to the interaction of two or more masses in motion, or simply an event. Without involving mass nothing would happen, no event could occur and no space between particles would be present to form space-time. It makes no sense for time to exist in a universe devoid of particles and therefore devoid of space 'between' or motion 'of'.

Wes

An "event" is a four-dimensional coordinate in space-time. Nothing has to occur there for it to exist.

 

[Ibix]

Strictly speaking, I would say No.

Time is merely a fundamental quality only assigned to the interaction of two or more masses in motion, or simply an event. Without involving mass nothing would happen, no event could occur and no space between particles would be present to form space-time. It makes no sense for time to exist in a universe devoid of particles and therefore devoid of space 'between' or motion 'of'.

Wes

As Peter pointed out in #7, we're effectively talking about a radiation dominated FLRW spacetime here. The energy density depends on time, so is a clock in of itself.

 

[Tomas Vencl]

Strictly speaking, I would say No.

Time is merely a fundamental quality only assigned to the interaction of two or more masses in motion, or simply an event. Without involving mass nothing would happen, no event could occur and no space between particles would be present to form space-time. It makes no sense for time to exist in a universe devoid of particles and therefore devoid of space 'between' or motion 'of'.

Wes

As Peter pointed out in #7, we're effectively talking about a radiation dominated FLRW spacetime here. The energy density depends on time, so is a clock in of itself.

Yes, in this framework is effective to work with paradigma that interactions are happening in spacetime.
But I think that original question goes deeper, in fact, to unknown quantum gravity theory. And here, I agree with Wes, that I can imagine, that spacetime itself is gradually generated and formed by interactions. But this is far over current knowledge.

 

[PeterDonis]

I think that original question goes deeper, in fact, to unknown quantum gravity theory.

Discussion of quantum gravity theories is off topic in this forum; it belongs in the Beyond the Standard Model forum.

 

[guptasuneet]

Strictly speaking, I would say No.

Time is merely a fundamental quality only assigned to the interaction of two or more masses in motion, or simply an event. Without involving mass nothing would happen, no event could occur and no space between particles would be present to form space-time. It makes no sense for time to exist in a universe devoid of particles and therefore devoid of space 'between' or motion 'of'.

Wes

Does a photon traveling at c experience time?

A muon generated in the upper atmosphere (from cosmic rays) and traveling at 0.995c experiences time contraction of ~10 times, i.e. it travels the 15 km distance to Earth in 5 microseconds internal time, as compared to 50 microseconds external time. Similarly, the muon experiences the 15 km distance as 1.5 km from its frame of reference.

A photon traveling at c should experience instantaneous traveling in its internal frame of reference, with all external lengths contracted to zero.

Similarly, a massless particle traveling at c will not experience the passage of time. That's what I meant with time not existing without mass.

 

[Ibix]

Does a photon traveling at c experience time?

Time cannot be defined for things traveling at $c$.

A photon traveling at c should experience instantaneous traveling in its internal frame of reference, with all external lengths contracted to zero.

This is not correct, I am afraid. The formulae for length contraction and time dilation are not valid at $c$ because assumptions in their derivation are violated at that speed.

 

[phinds]

Does a photon traveling at c experience time?

No

A photon traveling at c should experience instantaneous traveling in its internal frame of reference, with all external lengths contracted to zero.

Similarly, a massless particle traveling at c will not experience the passage of time. That's what I meant with time not existing without mass.

All of that is wrong and is based on a common misconception which is that a massless object traveling at c can even HAVE a "frame of reference". It cannot.

https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/

 

[Ibix]

That's what I meant with time not existing without mass.

The problem here is one of writing precisely. "Time" is a word with several different meanings in relativity and you need to be careful with it.

If what you meant is that proper time cannot be defined along null worldlines then you are correct, although your reasoning from time dilation formulae isn't really valid as I noted above. But time can also mean coordinate time and it is perfectly possible to define coordinate time in terms of a grid of crossing light rays. In fact that's what the light clock does, as can be seen by extending virtual light rays through the mirrors on a Minkowski diagram. You get a grid of diamonds whose vertical diagonals are regular intervals of time.