# Does infinite one-way speed of light violate p conservation?

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## Main Question or Discussion Point

Suppose for the sake of argument someone said the outward speed of light is infinite and the return speed is c/2, creating a two-way speed of c.

Wouldn't this violate the conservation of momentum?

p = E/c. That means on the way out, the momentum of light would be zero, but on the way back it would be very much non-zero. Shouldn't the momentum be the same both ways if we're to have conservation of momentum?

Any insight at any level is welcomed. Thanks.

atyy

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mfb
Mentor
You can save it by redefining momentum. Essentially you do a coordinate transformation like (x,t) -> (x,t+xc). It is just a redefinition of the time coordinate without observable consequences.

atyy and Dale
Dale
Mentor
Suppose for the sake of argument someone said the outward speed of light is infinite and the return speed is c/2, creating a two-way speed of c.

Wouldn't this violate the conservation of momentum?

p = E/c.
First, in strange frames like that the usual non-tensor formulas may not hold. You would need to re-derive them.

Second, if you write your laws in terms of tensors then they will hold under all coordinates. So the conservation of four-momentum would still be valid. However, in such coordinates you probably wouldn’t be able to clearly identify one component as the energy and the other components as the momentum.

You can save it by redefining momentum. Essentially you do a coordinate transformation like (x,t) -> (x,t+xc). It is just a redefinition of the time coordinate without observable consequences.
Would that have other theoretical consequences?

First, in strange frames like that the usual non-tensor formulas may not hold. You would need to re-derive them.
Does this mean we'd need a new law in the same way that Einstein redefined momentum by multiplying it by γ(u)?

Second, if you write your laws in terms of tensors then they will hold under all coordinates. So the conservation of four-momentum would still be valid. However, in such coordinates you probably wouldn’t be able to clearly identify one component as the energy and the other components as the momentum.
Is what I'm describing then nothing but a coordinate transformation?

I suppose what I am asking is this: would there be no physical difference in the universe or any other laws if the two-way speed of light had a infinite speed one way and c/2 speed the other way?

PeroK
Homework Helper
Gold Member
Suppose for the sake of argument someone said the outward speed of light is infinite ...
What's the definition of infinite speed?

mfb
Mentor
Would that have other theoretical consequences?
It would make all calculations much more complicated.
I suppose what I am asking is this: would there be no physical difference in the universe or any other laws if the two-way speed of light had a infinite speed one way and c/2 speed the other way?
We live in this universe. If you make a weird choice of a coordinate system (the one I posted). As you can see, it has no observable consequences.

Dale
Mentor
Does this mean we'd need a new law in the same way that Einstein redefined momentum by multiplying it by γ(u)?
Yes, something similar.

Is what I'm describing then nothing but a coordinate transformation?

I suppose what I am asking is this: would there be no physical difference in the universe or any other laws if the two-way speed of light had a infinite speed one way and c/2 speed the other way?
That is correct. There is no physical consequence to the one way speed of light, it is simply a choice of coordinates. In this case, an exceptionally complicated choice of coordinates, but perfectly valid.

What's the definition of infinite speed?
Well,
What's the definition of infinite speed?
Well I guess the speed such that the return trip of light is c/2. One that is only possible for outward going light?

Seems a rather symmetry breaking concept, obviously. As others have informed me, it would require a different definition of momentum and/or nasty change in coordinates, but no detectable differences.

I mean I don’t know how else to define it. I certainly don’t want to define it as a limit, because I suspect that might defeat the purpose of my question.

Yes, something similar.

That is correct. There is no physical consequence to the one way speed of light, it is simply a choice of coordinates. In this case, an exceptionally complicated choice of coordinates, but perfectly valid.
Okay so in this instance, not every observer would agree upon WHICH leg of travel was infinite, but everyone would agree on the two-way speed.

That seems like it would require a lot more than just altering momentum. Wouldn’t that require a complete rework of Maxwell’s equations, too? Granted, I’m shooting in the dark, but I’m aware you can combine them into a wave equation with speed c. It seems there would need to be a great deal of reworking of the framework for the laws of physics even if the end result yields the same observations.

Ibix
What you are doing, exactly, is drawing:
1. Drawing a Minkowski diagram
2. Deleting the horizontal grid lines
3. Drawing 45° lines
4. Shearing the diagram so that the 45° lines are horizontal.
You can always write coordinate transforms to relate vectors and events drawn on one diagram to the other.

It doesn't strictly require a re-write of Maxwell's equations. It requires you to use the tensor form of those equations, wherein "which coordinate system am I using?" gets abstracted out of the maths. You are, of course, free to unabstract (if that's a word) it again in as many different ways as you like and you will get different forms of the equations. But that's a bit like translating a story into a different language - the letters change but the story doesn't.

What you are doing, exactly, is drawing:
1. Drawing a Minkowski diagram
2. Deleting the horizontal grid lines
3. Drawing 45° lines
4. Shearing the diagram so that the 45° lines are horizontal.
You can always write coordinate transforms to relate vectors and events drawn on one diagram to the other.

It doesn't strictly require a re-write of Maxwell's equations. It requires you to use the tensor form of those equations, wherein "which coordinate system am I using?" gets abstracted out of the maths. You are, of course, free to unabstract (if that's a word) it again in as many different ways as you like and you will get different forms of the equations. But that's a bit like translating a story into a different language - the letters change but the story doesn't.
Would this still apply if all inertial reference frames agree that one leg of a light pulse’s travel is infinite and the other is c/2? (although not necessarily agree which leg is which)

Dale
Mentor
Wouldn’t that require a complete rework of Maxwell’s equations, too? Granted, I’m shooting in the dark, but I’m aware you can combine them into a wave equation with speed c. It seems there would need to be a great deal of reworking of the framework for the laws of physics even if the end result yields the same observations.
If you wrote Maxwell’s equations in the usual way, yes it would require a lot of reworking. However, if you use tensors then the same equations hold in any coordinate system.

Ibix
Would this still apply if all inertial reference frames agree that one leg of a light pulse’s travel is infinite and the other is c/2? (although not necessarily agree which leg is which)
As long as your coordinates are smooth and invertible (there are no discontinuities and no point has more than one coordinate description - I think that's all that's needed) you can simply work with the tensor form of physical laws. You can also grind out the form of the laws in your chosen system explicitly if you wish.

As long as your coordinates are smooth and invertible (there are no discontinuities and no point has more than one coordinate description - I think that's all that's needed) you can simply work with the tensor form of physical laws. You can also grind out the form of the laws in your chosen system explicitly if you wish.

This and all the other posts definitely paint a clear picture.

So in essence, and correct me if I’m wrong, the answer is no: Momentum conservation is NOT violated. It is still conserved in this hypothetical scenario, which means other than simplicity and some assumptions about symmetry in the universe, choosing an equal one way speed and two way speed is just something we choose.

Ibix
Yes. Ultimately this is why all attempts to measure the one-way speed of light fail - the answer is a matter of choice. All of the maths is needlessly more complicated if you make any other choice than the symmetric one, and Occam's Razor favours the symmetric choice, but you are free to make a different choice if you wish.

Yes. Ultimately this is why all attempts to measure the one-way speed of light fail - the answer is a matter of choice. All of the maths is needlessly more complicated if you make any other choice than the symmetric one, and Occam's Razor favours the symmetric choice, but you are free to make a different choice if you wish.
Doesn’t that pose a problem for age of the universe estimates that use distant stars as the minimum earliest time? If that light got here instantaneously, then how can we say it is from an object 13 bullion light years away?

Or in other words, wouldn’t other means of determining the age of the universe refute the notion that light can move with an infinite one way speed?

Dale
Mentor
Doesn’t that pose a problem for age of the universe estimates that use distant stars as the minimum earliest time?
No. The age of the universe means something specific. It is the spacetime interval of a worldline of a hypothetical inertial particle which has been around since the beginning of the universe and for which the CMBR is isotropic. This quantity is an invariant quantity which is defined using tensors, so its value is independent of the coordinate system.

Sorcerer
Ibix
Doesn’t that pose a problem for age of the universe estimates that use distant stars as the minimum earliest time? If that light got here instantaneously, then how can we say it is from an object 13 bullion light years away?

Or in other words, wouldn’t other means of determining the age of the universe refute the notion that light can move with an infinite one way speed?
You're moving beyond SR here, so some care is needed. We aren't using reference frames any more, and speeds over long distances aren't really well defined. However, your point is generally correct - the age of the universe depends on your choice of coordinates because the definition of "the universe" depends on your choice of coordinates. The age usually quoted is the proper time shown by a clock that always saw the universe as isotropic (CMB is the same colour in all directions). That is the maximum age you can get and is invariant, as Dale says.

No. The age of the universe means something specific. It is the spacetime interval of a worldline of a hypothetical inertial particle which has been around since the beginning of the universe and for which the CMBR is isotropic. This quantity is an invariant quantity which is defined using tensors, so its value is independent of the coordinate system.
You're moving beyond SR here, so some care is needed. We aren't using reference frames any more, and speeds over long distances aren't really well defined. However, your point is generally correct - the age of the universe depends on your choice of coordinates because the definition of "the universe" depends on your choice of coordinates. The age usually quoted is the proper time shown by a clock that always saw the universe as isotropic (CMB is the same colour in all directions). That is the maximum age you can get and is invariant, as Dale says.
Darn. I was hoping for a silver bullet lol. I suppose it hasn't been done yet and it probably never will.

stevendaryl
Staff Emeritus
You have to connect speeds to something concretely measurable. Otherwise, you can achieve your alternative physics by just redefining your coordinates.

Start with coordinates $x, t$. Define a new time coordinate $T$ via:

$T = t - \frac{x}{c}$

In the new coordinate system, $x,T$, the speed of light will be $\infty$ in one direction and $\frac{c}{2}$ in the other direction.

pervect
Staff Emeritus
Suppose for the sake of argument someone said the outward speed of light is infinite and the return speed is c/2, creating a two-way speed of c.

Wouldn't this violate the conservation of momentum?
I assume you're just trying to think about how physics would work with a clock synchornization convention that's not Einsteinian. But perhaps I'm wrong about that.

Assuming that that is your motivation, though, as a matter of semantics I would say that momentum still exists as a conserved physical quantity, it's just no longer given by the relationship p=mv, but by some other relationship.

For any massive object, for instance, it's possible to define momentum in a relativistically correct manner as mass * proper velocity, where proper velocity is the distance an object moves (measured in a particular frame of reference) divided by the proper time takes for the object to move said distance. Note that this proper time interval does not depend on any frame of reference, as it's independent of the observer.

If the concept of proper velocity is unfamiliar, wiki has a short discussion of it <<link>>.

This won't work for calculating the momentum of light, but it will give the correct answer for momentum for any object moving slower than light, in a manner that's independent of any clock synchronization convention used. You didn't mention clock synchronization explicity, but of course it's the Einstein clock synchronziation convention that makes the speed of light the same in both directions.

It's routine to define the energy-momentum 4-vector as mass * proper velocity. And it's fairly routine to note that the energy momentum 4-vector contains the energy and the momentum 3-vector. So a lot of it is semantics and what one is used to.

I assume you're just trying to think about how physics would work with a clock synchornization convention that's not Einsteinian. But perhaps I'm wrong about that.

Assuming that that is your motivation, though, as a matter of semantics I would say that momentum still exists as a conserved physical quantity, it's just no longer given by the relationship p=mv, but by some other relationship.

For any massive object, for instance, it's possible to define momentum in a relativistically correct manner as mass * proper velocity, where proper velocity is the distance an object moves (measured in a particular frame of reference) divided by the proper time takes for the object to move said distance. Note that this proper time interval does not depend on any frame of reference, as it's independent of the observer.

If the concept of proper velocity is unfamiliar, wiki has a short discussion of it <<link>>.

This won't work for calculating the momentum of light, but it will give the correct answer for momentum for any object moving slower than light, in a manner that's independent of any clock synchronization convention used. You didn't mention clock synchronization explicity, but of course it's the Einstein clock synchronziation convention that makes the speed of light the same in both directions.

It's routine to define the energy-momentum 4-vector as mass * proper velocity. And it's fairly routine to note that the energy momentum 4-vector contains the energy and the momentum 3-vector. So a lot of it is semantics and what one is used to.
Thanks for the reply. My motivation is that the notion of light not moving the same speed both ways violates my sense of symmetry and seems vaguely offensive for reasons I can’t explain. But if I understand what many have pointed out here it’s more or less a choice of coordinates.

Nugatory
Mentor
the notion of light not moving the same speed both ways violates my sense of symmetry and seems vaguely offensive for reasons I can’t explain.
And you're in good company too.... I'd expect that most people will make a similar aesthetic judgment.
It just turns out that it's an aesthetic judgment about descriptions of the universe, not about the universe itself.

Ibix
Ibix
The fundamental truth is that light, in vacuum, follows null paths. That's a statement about geometry and completely independent of coordinates or frames or whatever. But how you choose to divide spacetime into space and time is up to you, and the projection of those null paths onto your choice of space depends on it.

If you choose to make your spatial "planes" orthogonal to your time direction then the paths are isotropic because the "planes" are symmetric under any rotation about the time axis. If you choose them to be non-orthogonal the paths are not isotropic because the "planes"
are no linger symmetric under rotation. Naturally, picking orthogonal planes is nicer.

pervect
Staff Emeritus