Special Relativity and momentum

[SDG]

...
A decrease in mass produces a decrease in momentum when the velocity is constant. So the sideways linear momentum HAS decreased in this scenario, and it hasn't decreased uniformly across the flashlight.
...

Are you sure?

I did not use 'prime' for P in the diagrams but the second figure P is bigger than the first figure P.
This is Doppler effect so when transformation happens from the moving frame to the rest frame the value is correct.
There is P_recoil x component in the moving frame and y components are equal.

 

[SDG]

Wouldn't aberration cause an apparent rotation in the orientation of the flashlight, too?

What apparent rotation?
There is an inertial observer who is part of a bigger grid of inertial observers, no relative motion between them.
It appears to me that if talk about an apparent rotation we are making one preferred inertial observer from his grid of inertial observers.
The angular velocity of apparent rotation is not the same for all observers in the grid.
The grid is broken.
The preferred observer is not even inertial anymore, the observer rotates, to follow the apparent rotation.

 

[SDG]

I hope that the OP takes into account that two relatively-moving observers will not agree on the value of P (the momentum of the photon). Because Doppler shift.

Right, so are we going to ignore the P_recoil x component when the Doppler effect takes care of the y component?

 

[Dave Rogers]

Are you sure?

[qimg]https://i.imgur.com/HpboITi.png[/qimg]

[qimg]https://i.imgur.com/5p3Twdt.png[/qimg]

I did not use 'prime' for P in the diagrams but the second figure P is bigger than the first figure P.
This is Doppler effect so when transformation happens from the moving frame to the rest frame the value is correct.
There is P_recoil x component in the moving frame and y components are equal.

As someone pointed out up thread, this is equally true of classical mechanics; your description of the experiment would be equally applicable to a bullet fired from a rifle at non-relativistic speed.

Dave

 

[Ziggurat]

A hydrogen atom emits a photon going from n=2 to n=1.
Does the rest mass of the hydrogen atom change?

Yes.

Even better, just simple electron, Bremsstrahlung, what is the electron mass change when a photon is emitted?

Just an electron cannot emit Bremsstrahlung radiation. It needs to interact with other matter. And in this case, you are trading off kinetic energy to create the photon, not rest mass energy.

As you wrote E² = p²c² + m²c⁴.
Is this equation true?

Yes. Even more importantly, though, it is part of the theory of special relativity. So if you want to evaluate SR, you need to account for it.

Let us assume it is.
There is a spontaneous hydrogen atom emission, m is constant,

m is not constant. That's the whole point. You have started with a false assumption.

 

[Ziggurat]

Are you sure?

Yes, I am sure.

You made a mistake. You didn't realize that the mass changed. And you didn't account for the effects of this change. That is all that's going on here. You did not find a flaw in special relativity.

 

[Ziggurat]

SDG: do this problem with a gun and a bullet, using Newtonian mechanics. Figure out why the gun doesn't twist when you move to a translating frame. The basics are the same, you need to account for the loss of mass.

 

[Reformed Offlian]

Objects (electrons) emitting photons do not lose mass, they lose momentum.

Yes, objects *do* lose mass when they emit photons. Electrons do not; as Ziggurat points out, they trade off kinetic energy for momentum. However, the atoms to which they are bound *do* change mass, even if their constituent particles do not. Avoid the fallacy of composition here.

 

[Ziggurat]

A hydrogen atom emits a photon going from n=2 to n=1.
Does the rest mass of the hydrogen atom change?

Let me follow up with another example where the numbers are more dramatic and have been directly measured.

Deuterium (one proton, one neutron, one electron - an isotope of hydrogen) has a mass of 2.0141 atomic units. Helium (two protons, two neutrons, two electrons) has a mass of 4.0026 au.

If you squish two deuterium atoms together, they will fuse to form helium. But two deuterium atoms have a total mass of 4.0282 au. That's 0.0256 au more than helium. Squishing them together makes them lose mass. Where does that mass go?

It gets converted into energy. E=mc². 0.0256 atomic units converted to energy might not look like much at first glance, but it's actually a huge amount. And that's basically what powers a hydrogen bomb.

It's very hard to directly measure the mass loss of hydrogen going from an n=2 excited state to an n=1 ground state because the mass loss is so tiny, but it's still happening.

 

[The Man]

A hydrogen atom emits a photon going from n=2 to n=1.
Does the rest mass of the hydrogen atom change?

Yep, as a bound state the hydrogen atom has less rest mass than the sum of the free rest masses of all it components. By emitting a photon and dropping to a lower energy state the electron and subsequently the hydrogen atom as a whole loses mass. As this involves the atom as a whole it reduces the atoms rest mass.


Please see Mass defect in regard to binding energy.

 

[ponderingturtle]

Photons have mass? I thought they were zero-mass particles?

Now we get into various definitions of mass. They do not have a rest mass but some people use relativistic mass which they do have.

 

[The Man]

Photons have mass? I thought they were zero-mass particles?

Now we get into various definitions of mass. They do not have a rest mass but some people use relativistic mass which they do have.

As already noted in this thread photons have both energy and momentum. In emitting a photon a particle losses both energy and momentum. As momentum is mass time velocity, if velocity of the emitting particle remains constant then mass must change. For complex assemblages like an atom where an electron emits a photon by dropping into a lower energy state both the energy and momentum of the electron change (hence the lower state) but the overall binding energy goes up. It now takes even more energy to free that electron, as such the rest mass of the atom drops by the mass equivalent of that photon energy. So while rest mass reduces the photon only carries energy and momentum no rest mass itself as it has no rest or comoving state.

 

[Reformed Offlian]

Right, so are we going to ignore the P_recoil x component when the Doppler effect takes care of the y component?

I don't understand this question. My point was simply that you have to define in which reference frame P is being measured. Even a transversely moving observer will measure it differently (transverse doppler shift is a thing in SR).

 

[SDG]

And... here's the mistake.

In order to emit a photon, the mass of the flashlight must decrease. A decrease in mass produces a decrease in momentum when the velocity is constant. So the sideways linear momentum HAS decreased in this scenario, and it hasn't decreased uniformly across the flashlight. The previous center of mass is no longer the current center of mass. And relative to the stationary point located where the center of mass was when the photon was emitted, the flashlight now has nonzero angular momentum.

And that's the tricky bit, and you don't need special relativity for it. A non-rotating uniformly moving body can have angular momentum, depending on where you measure from.

As I mentioned, there is a lot said in this post.
2. A decrease in mass produces a decrease in momentum when the velocity is constant.

Momentum has to be conserved.
If we consider the photon energy leaving the flashlight system as system mass reduction then the velocity v will be bigger.
Conservation of momentum is the input factor, the velocity is the output effect.


3. So the sideways linear momentum HAS decreased in this scenario, and it hasn't decreased uniformly across the flashlight.
Well, and the force does not propagate instantaneously in SR.
The flashlight has to start rotation at the point of emission regardless where the theoretical center of mass is located.


The force will propagate through the center of mass in the rest frame eventually, not causing any rotation.
A rotation starts right after emission in any other moving frame.

 

[Reformed Offlian]

What apparent rotation?
There is an inertial observer who is part of a bigger grid of inertial observers, no relative motion between them.
It appears to me that if talk about an apparent rotation we are making one preferred inertial observer from his grid of inertial observers.
The angular velocity of apparent rotation is not the same for all observers in the grid.
The grid is broken.
The preferred observer is not even inertial anymore, the observer rotates, to follow the apparent rotation.

My point (and I believe Myriad's) is that the axis of the "recoil" trajectory will still appear to point back through G, even for a laterally-moving observer.

 

[Ziggurat]

As I mentioned, there is a lot said in this post.
2. A decrease in mass produces a decrease in momentum when the velocity is constant.

Momentum has to be conserved.

Total momentum is conserved. But it can be apportioned differently between different parts of the system. The momentum of just the flashlight changes. It is not conserved.

In the moving frame, some of the momentum to the side is carried by the photon after it is emitted. But the photon carried no momentum before it was emitted. Therefore, the flashlight must have less momentum to the side after emitting the photon in order to keep total momentum conserved. But it can have less momentum to the side without changing velocity because its mass decreased.

3. So the sideways linear momentum HAS decreased in this scenario, and it hasn't decreased uniformly across the flashlight.
Well, and the force does not propagate instantaneously in SR.
The flashlight has to start rotation at the point of emission regardless where the theoretical center of mass is located.

It doesn't rotate at all. That's the whole point. The contradiction you imagined doesn't exist. The change in momentum is due to a change in mass, and because that change in mass is not uniform across the body of the flashlight (at least not instantly), the change in momentum isn't either. The "torque" accounts for how ONE SIDE of the flashlight lost mass and therefore momentum, and acquired angular momentum about its FORMER center of mass while not rotating. The former center of mass is not the new center of mass.

You can't apply rules for rigid mass-conserved bodies to a body whose mass is changing and expect everything to be the same, even in Newtonian mechanics. Nor do you seem to understand how angular momentum works when you're looking at something other than the center of mass. That's excusable, since it's tricky business even in Newtonian mechanics, but you're still getting it wrong.

A rotation starts right after emission in any other moving frame.

No. A rotation doesn't start at all, in any frame. You don't understand what's happening yet.

 

[Ziggurat]

My point (and I believe Myriad's) is that the axis of the "recoil" trajectory will still appear to point back through G, even for a laterally-moving observer.

No, there is in fact a torque on the system in this scenario. He's right about that, but wrong about basically everything else.

The flashlight experiences a change in angular momentum without rotating, because its mass changes and so does its center of gravity. When viewed from the initial center of gravity, the flashlight had no angular momentum before emission, but it has angular momentum after emission, because there's more mass on one side than the other after (well, mass times length, but I'm simplifying). When viewed from the final center of gravity, the flashlight has angular momentum before emission, because (again) there's more mass on one side than the other, but no angular momentum after. Either way, the angular momentum changes, due to the torque, without any rotation occurring. What you aren't allowed to do is change where you're measuring your angular momentum from without accounting for that change, which is what SDG is doing without even realizing it. And again, this isn't peculiar to special relativity, the exact same problem comes up in Newtonian physics.

 

[Reformed Offlian]

No, there is in fact a torque on the system in this scenario. He's right about that, but wrong about basically everything else.

The flashlight experiences a change in angular momentum without rotating, because its mass changes and so does its center of gravity. When viewed from the initial center of gravity, the flashlight had no angular momentum before emission, but it has angular momentum after emission, because there's more mass on one side than the other after (well, mass times length, but I'm simplifying). When viewed from the final center of gravity, the flashlight has angular momentum before emission, because (again) there's more mass on one side than the other, but no angular momentum after. Either way, the angular momentum changes, due to the torque, without any rotation occurring. What you aren't allowed to do is change where you're measuring your angular momentum from without accounting for that change, which is what SDG is doing without even realizing it. And again, this isn't peculiar to special relativity, the exact same problem comes up in Newtonian physics.

Ah, OK, thanks.

 

[Ziggurat]

This is one of those "a little knowledge is a dangerous thing" situations. SDG knows just enough to set up a problem with some interesting subtleties, but not enough to actually figure out those subtleties.

It's actually not a bad problem to illustrate some of these subtle issues (sort of like the barn door/ladder problem for showing relativity of simultaneity). I could probably give this problem to upper division physics majors in college on a test and they wouldn't all get it right. Probably only half of them would figure it out without an explanation. And basically nobody who hasn't studied physics is going to figure this out. So getting it wrong is not a sign of stupidity on his part, not by any stretch of the imagination. It really isn't a mark against him.

But when you run into a problem like this where the answer doesn't appear to make sense to you, or you think you've found some sort of contradiction, it's rather arrogant to assume that you not only got it right, but that you discovered something that generations of the brightest minds failed to notice or understand. The far more likely explanation, in every case, is that you're missing something, not that everyone else did. And indeed, SDG did miss something. Which, again, is completely understandable. But that's all it is, nothing more. This isn't the flaw in Special Relativity that SDG thought, it's just an example of his own imperfect understanding of Special Relativity.

 

[slyjoe]

This is one of those "a little knowledge is a dangerous thing" situations. SDG knows just enough to set up a problem with some interesting subtleties, but not enough to actually figure out those subtleties.

It's actually not a bad problem to illustrate some of these subtle issues (sort of like the barn door/ladder problem for showing relativity of simultaneity). I could probably give this problem to upper division physics majors in college on a test and they wouldn't all get it right. Probably only half of them would figure it out without an explanation. And basically nobody who hasn't studied physics is going to figure this out. So getting it wrong is not a sign of stupidity on his part, not by any stretch of the imagination. It really isn't a mark against him.

But when you run into a problem like this where the answer doesn't appear to make sense to you, or you think you've found some sort of contradiction, it's rather arrogant to assume that you not only got it right, but that you discovered something that generations of the brightest minds failed to notice or understand. The far more likely explanation, in every case, is that you're missing something, not that everyone else did. And indeed, SDG did miss something. Which, again, is completely understandable. But that's all it is, nothing more. This isn't the flaw in Special Relativity that SDG thought, it's just an example of his own imperfect understanding of Special Relativity.

Nice polite explanation