[yor_on (OP)]

Imagine that you have a 'perfect football' (sphere), instead of air you will fill it with photons.
How you do that is your problem

Now, photons can be superimposed right?
And their momentum (vectors) will take each other out inside that ball so the net result as I understand it will be no momentum left inside that ball. But they will still have their energy, won't they?

So now we have two parameters instead of one, right?

We have the momentum which is the sum of the photons vectors (direction/magnitude) zeroing out as described inside that ball. And we have the 'energy' of the photons, that then will increase, and as I understand it also can be expressed as their 'relative mass'? Now, you keep on pumping in that photon gas. As they are superimposed I coolly expect you to do so until your natural death, or at least until you get hungry.

How is that possible? Not that you get hungry, I would too, but how could I now suddenly split momentum from the photons 'relative mass'? As you de facto can can express the energy as mass. But as photons intrinsically are defined as being massless, momentum would be a better definition to me, until now that is

And as the momentum is the 'pressure' acting on the balls walls, can I now pump it forever?
And will it then become infinitely heavy due too?

Momentum? 'relative mass'? energy?

[JP]

Your question is a bit confusing to me.  There will be momentum carried by the light still.  (By the way, you don't need photons to describe this, so it's easiest to use classical EM waves.)  Whenever a wave bounces off the wall, it reverses its direction and therefore its momentum.  The conservation of momentum requires that the wall absorb this momentum, so the walls are all getting pushed outwards.  In other words, as you fill this ball with light, the pressure on the walls goes up.  Even if the walls are perfectly reflecting and don't heat up from the absorption of energy, they'd eventually fail just from that pressure, so you couldn't just keep adding light.

[Ron Hughes]

If your experiment were possible then sooner or later you would start creating electrons and protons.

[yor_on (OP)]

Well JP, as I understood this, and I agree, I too find it confusing. You have a sphere, that you pump in photons in. As the photons fill up the sphere they will have a pressure due to their momentum just as you say, but it is also so that their momentum will take out each other vectors, and so you can apparently say that the momentum is 'gone'. So suddenly it seems as if you mathematically have negated the momentum. As I see it, this is a 'trick' involved here. You will still have the pressure from the momentum working on the walls and as you say build up as the photons or waves gets added.

I've had this discussion before and I didn't get it then either. So I invited you clever heads to give it a shot And yes 'perfectly reflecting is a must here'

[lightarrow]

You don't need net momentum to have pressure in this case: you only need energy. Remember also that it's the total momentum which is zero, not the one colliding with the container's wall; with a great number of atoms it would be the same: zero total momentum of the atoms, not zero momentum of the atoms colliding with the wall (for n atoms going right toward a wall there are n going left towards the opposite wall.)

Also, you could consider two equal iron balls fired in opposite directions: the total momentum of the system is zero, but when the balls collide with you...ouch! []

[yor_on (OP)]

Yes, that's my view too.. *Ouch*

But when counting on the mass in this sphere it seems to give you a sudden 'rest mass' as those vectors get canceled out?

So yeah, it confuses me in many ways?
==
You wrote "zero total momentum of the atoms" Lightarrow.

I have to be thick because when I read a statement like that I do not see individual atoms 'bouncing around' each one with its own momentum intact, instead I see a Bose Einstein condensate. And neither can I see how they are thought to take each other out practically? All of them will at some time reach the wall, and each one of them will still have a momentum? So why do we use that definition, it doesn't seem to describe the situation?

[lightarrow]

Yes, that's my view too.. *Ouch*

But when counting on the mass in this sphere it seems to give you a sudden 'rest mass' as those vectors get canceled out?

Of course.

So yeah, it confuses me in many ways?
==
You wrote "zero total momentum of the atoms" Lightarrow.

I have to be thick because when I read a statement like that I do not see individual atoms 'bouncing around' each one with its own momentum intact, instead I see a Bose Einstein condensate. And neither can I see how they are thought to take each other out practically? All of them will at some time reach the wall, and each one of them will still have a momentum? So why do we use that definition, it doesn't seem to describe the situation?

If you don't want to see light as particles in this case, you can simply use a classical description, as JP wrote; EM waves do have momentum, just because they convey energy towards the walls: moving energy --> momentum; the momentum is E/c; this wouldn't happen only in the case energy would not propagate (for example a stationary clump of matter).

[yor_on (OP)]

No disrespect Lightarrow but I'm still unable to see how it suddenly becomes 'restmass'. Seen as photons they won't interfere with each other, right? Seen as waves there is a weak possibility of it happening, as I understands it, not a strong but a weak? So they seem to my eyes to be able to move quite unfettered inside our perfectly reflecting sphere?

When we say a thing like their vectors taking each other out, magically transforming momentum to so called 'rest masss' or 'proper invariant mass' aka 'matter' I still want to understand why. Looking at the math doesn't tell me a thing there Lightarrow.

What would mathematicians do if someone constructed an experiment tomorrow, proving that they don't transform into so called 'rest mass'? Throw away the math, swearing to never more trust in it?

Nope, they would adapt the math to the facts, wouldn't they?
So what I'm asking for is why I should see it as rest mass suddenly?
What are the 'mechanism(s)' behind it, thought to create this remarkable feat?

---Some of my other questions--

Okay, do a photon have a size, if not, do it have a dimensionality?
How do we define a dimensionless particle as carrying an energy?

And what does that mean if so? If you can have something dimensionless interacting with us, behaving as if it has a velocity inside our SpaceTime, following its geodesics, as well as our arrow of time macroscopically then you have a well defined object it seems? But without interactions, like in deep space unable to observe, right?

So we have something that in fact only exist in its interactions, at least it seems so to me. And when it interacts it can do so both as a wave and as a particle, superimposing itself into simultaneous different paths in its mediation with matter, as a wave with several 'real' paths simultaneously as I understands it. As a particle with a probability of those paths, falling out different at each 'real' measurement into only one of those. And depending on definition you can either see it as being 'mediated' by virtual particles or reflected, diffracted, refracted etc.

So, how does it move? in a virtual sea of energy (vacuum energy). That would make sense from a particle view, wouldn't it? But from a wave perspective then? Or could it be that as its frame intrinsically is timeless it doesn't have to care for our definitions of distance at all?

But then we still need to explain the ocean of light making up our reality, acting inside our arrow of time. Where does its wave behavior come in mathematically, starting as a single photon directly? Or is the idea a superposition even from a single photon perspective mathematically? Also, as I understand it, a photon has no magnetic field (it's zero) and without it, and being timeless intrinsically, how is it thought to create any EM fluctuations (waves) mathematically?

-

So yes, I'm still flummoxed here

[Farsight]

Yor_on: start with a cannonball in space coming at you at 10m/s. It's hard for you to decelerate it because it's got considerable momentum. Now imagine the cannonball is at rest, and you're passing it at 10m/s. It's hard for you to accelerate it because it's got considerable inertia. The difference depends on who you say is moving. Apply this to a single photon, considering yourself to be the electron in the Compton Effect:



Photons move at c, delivering a "kick" of momentum. But imagine what this would feel like if you said it was the electron moving instead of the photon. You'd bounce off it, and it would feel like inertia. Now think about those photons in your sphere. They're moving at c bouncing back and forth, but they aren't going anywhere, so they're a bit like the non-moving photon. Whilst they're waves and they're still moving at c, if you want to move the sphere you have to push against them, and the more there are and the more energy they have, the harder it is to move the sphere. Another way to think of is as a cube in space full of richochetting bullets. You have to push against only the bullets hitting the front face only to get it moving. Keep the same number of bullets but make them go faster. The faster the bullets hit that front face, the harder it is to get the cube moving. Hence it feels more massive just because the bullets are moving faster. Once you've got the cube moving moving, it'll keep moving until you give it an opposite push.     

 

[lightarrow]

When we say a thing like their vectors taking each other out, magically transforming momentum to so called 'rest masss' or 'proper invariant mass' aka 'matter' I still want to understand why. Looking at the math doesn't tell me a thing there Lightarrow.

If I have correctly understood you, the problem is to see a body's inertia as a consequence of the fact the specific region of space occupied by that body has energy inside of it?

[yor_on (OP)]

I think i can see how you think Lightarrow, but doesn't the movements inside that sphere take each other out statistically, meaning that the 'bouncing' is on all parts of the inner 'surfaces' which to me seems to indicate that there won't be any noticeable 'new inertia' to the sphere if you tried to move it?

Where do I go wrong?
(And don't you tell me it's due to my misspent youth:)
==

Are you thinking of it in terms of energy=mass?

But if you consider the situation inside and look at the individual 'photons'. They all carry a momentum and it is that you will see be converted into 'force' here, their 'energy' notwithstanding. I mean, even if each one of the photons represent an energy that at some level of concentration may express itself as mass, isn't it the momentum that deliver the 'push' here, as all photons are intrinsically massless?

[Robro]

Wow, this is a good question! I really don't have much to add, but I wonder what would happen if you suddenly opened the mirrored bag, would a flash of light come out, or a bolt of lightning!

[Robro]

Yup, at the right frequency the photons could possibly become phase locked into electrons and positrons.

If your experiment were possible then sooner or later you would start creating electrons and protons.

[lightarrow]

I think i can see how you think Lightarrow, but doesn't the movements inside that sphere take each other out statistically, meaning that the 'bouncing' is on all parts of the inner 'surfaces' which to me seems to indicate that there won't be any noticeable 'new inertia' to the sphere if you tried to move it?

Where do I go wrong?
(And don't you tell me it's due to my misspent youth:)

If you absolutely want to see it in terms of photons' bouncing, you have to consider relativity: the frequency and the intensity of light coming towards a wall moving towards it are  increased (increased intensity = increased number of photons); at the opposite wall is the opposite, but Farsight already talked about this.

Are you thinking of it in terms of energy=mass?

But if you consider the situation inside and look at the individual 'photons'. They all carry a momentum and it is that you will see be converted into 'force' here, their 'energy' notwithstanding. I mean, even if each one of the photons represent an energy that at some level of concentration may express itself as mass, isn't it the momentum that deliver the 'push' here, as all photons are intrinsically massless?

Yes but for massless particles (be them photons or gluons or supposed gravitons or else) the momentum is the flow of the energy: p = E/c. They have momentum because they move and have energy (= because energy is moving).

[yor_on (OP)]

Yes, you are perfectly right there, there will be a resistance as the photons have a shorter distance relative the wall coming towards them and therefore a higher momentum (blue shift as from us moving the 'wall'). As for the other one I'm not sure how you see it? The momentum isn't any rest mass? As for why it's acting 'inertially' that is explained by the very thing you corrected me with, I think?

[lightarrow]

The momentum isn't any rest mass?

No, you don't need rest mass to have momentum, you only need a flux of energy in a specific direction.

p = (K_e/c² + m)*v

p = momentum
K_e = kinetic energy
m = mass
v = velocity.

If m = 0 you still have momentum.

In the last case, the particle have to move at v = c, so:
p = (K_e/c²)*c = K_e/c
but: K_e = (by definition) = E - mc² = (in this case) = E, so:
p = E/c.