Special Relativity and momentum

[SDG]

It is easy to show that two inertial observers in relative motion do not agree on the conservation of linear momentum when they observe emission of one photon.
The thought experiment:



An LED flashlight in an intergalactic space emits one photon (wave packet) with the momentum P.
The recoil generated velocity v_recoil represents the conservation of the linear momentum.
The figure represents the LED flashlight rest frame and the recoil direction is through the flashlight center of mass G.
The end result is linear motion of the LED flashlight in the v_recoil direction.

Any comments so far?
Do we have an agreement this is the case?

[Meadmaker]

Originally Posted by SDG

It is easy to show that two inertial observers in relative motion do not agree on the conservation of linear momentum when they observe emission of one photon.
The thought experiment:

https://i.imgur.com/L8MFbHo.png

An LED flashlight in an intergalactic space emits one photon (wave packet) with the momentum P.
The recoil generated velocity v_recoil represents the conservation of the linear momentum.
The figure represents the LED flashlight rest frame and the recoil direction is through the flashlight center of mass G.
The end result is linear motion of the LED flashlight in the v_recoil direction.

Any comments so far?
Do we have an agreement this is the case?

Why is V_recoil orthogonal to P?

[SDG]

Originally Posted by Meadmaker

Why is V_recoil orthogonal to P?

It is the black arrow pointing down.
The thin arrow is vector sign above v, the same as above P.

[smartcooky]

Originally Posted by SDG

It is easy to show that two inertial observers in relative motion do not agree on the conservation of linear momentum when they observe emission of one photon.
The thought experiment:

https://i.imgur.com/L8MFbHo.png

An LED flashlight in an intergalactic space emits one photon (wave packet) with the momentum P.
The recoil generated velocity v_recoil represents the conservation of the linear momentum.
The figure represents the LED flashlight rest frame and the recoil direction is through the flashlight center of mass G.
The end result is linear motion of the LED flashlight in the v_recoil direction.

Any comments so far?
Do we have an agreement this is the case?

If the photon is emitted in the direction of the small arrow above P, why would v_recoil be in the same direction. Shouldn't it it be in the other direction - to the left as we look at the diagram?

What does the double-ended thick back arrow represent?

[Shepherd]

I believe that the thick black arrows are supposed to be the opposite direction momentum vectors and that the other arrows are only the vector notation for the names of the vectors. If the arrow above Vrecoil had been only one character wide and above only the V, maybe it would have looked less confusing.

[Darat]

OK - so SR is wrong and you are the only person in a hundred years to have spotted this, even though you have never ever conducted one single piece of actual research.

And what is your next step?

[SDG]

Originally Posted by smartcooky

If the photon is emitted in the direction of the small arrow above P, why would v_recoil be in the same direction. Shouldn't it it be in the other direction - to the left as we look at the diagram?

What does the double-ended thick back arrow represent?

The position where the recoil happens is very very important.
It will become more obvious in the following analysis.

Now what is important is to agree the recoil happens at the point where the emission happens and the fact the recoil goes through the center of mass and the recoil generates linear momentum on the LED flashlight.

Do we have an agreement?

[SDG]

Originally Posted by Shepherd

I believe that the thick black arrows are supposed to be the opposite direction momentum vectors and that the other arrows are only the vector notation for the names of the vectors. If the arrow above Vrecoil had been only one character wide and above only the V, maybe it would have looked less confusing.

Correct.

[SDG]

Originally Posted by Darat

OK - so SR is wrong and you are the only person in a hundred years to have spotted this, even though you have never ever conducted one single piece of actual research.

And what is your next step?

I did my own research. I've been at it for a long time.
We will make more analysis here, and you will make your own conclusion and research if anybody else came with the same type of thought experiment.

[SDG]

Originally Posted by smartcooky

...
What does the double-ended thick back arrow represent?

The momentum is mass m of the flashlight multiplied by v... mv.

The v_recoil is the velocity vector generated by the recoil.

[Darat]

Originally Posted by SDG

I did my own research. I've been at it for a long time.
We will make more analysis here, and you will make your own conclusion and research if anybody else came with the same type of thought experiment.

As I said never done any actual research. SR isn't accepted because of thought experiments it is accepted because of real world experiments and observations.

And again - so you show us that only you have spotted the error in SR that has eluded millions of people over decades, and then?

[Meadmaker]

Originally Posted by SDG

The position where the recoil happens is very very important.
It will become more obvious in the following analysis.

Now what is important is to agree the recoil happens at the point where the emission happens and the fact the recoil goes through the center of mass and the recoil generates linear momentum on the LED flashlight.

Do we have an agreement?

Ok. I think I understand your notation now.


Great. So far so good, except that in the above there's a little bit of unusal phrasing that makes me suspicious of what comes next.

"The recoil goes through the center of mass" as something that is very important is somethig that seems like it might be about to mix up something that happens at very different scales, but whatever. So far so good, but reserving the right to advise and extend my remarks at a later time.

Lead on.

[theprestige]

Light-based propulsion is already known to work. IKAROS has done it. No need to waste a week trying to get agreement on the theory, when the practice has already been done.

[Meridian]

What is m here? Rest mass (invariant mass)? Or `relativistic mass', which seems not to be used much (at least not without a specific label) since it leads to confusion, such as forgetting that it's frame dependent. In SR the formula for momentum uses the latter, or, as normally written, invariant mass times gamma times velocity. See e.g., wikipedia .

What you appear to mean (m_0 v) is not a conserved quantity in SR.

[smartcooky]

Originally Posted by theprestige

Light-based propulsion is already known to work. IKAROS has done it. No need to waste a week trying to get agreement on the theory, when the practice has already been done.

[stanfr]

boy this forum moves slow...after the first post I was anxiously awaiting some amazing new insight and here I am, still left hanging!

[Ziggurat]

Originally Posted by SDG

It is easy to show that two inertial observers in relative motion do not agree on the conservation of linear momentum when they observe emission of one photon.

This is absolutely wrong. But it’s not a terribly common wrong, so I can’t predict what mistakes you have made before you reveal them.

Quote:

The figure represents the LED flashlight rest frame and the recoil direction is through the flashlight center of mass G.

Is your point here solely that you want a situation where the recoil applies no torque to the flashlight? If so, this is a slightly awkward way of phrasing it, but it’s a reasonable constraint to apply for the sake of simplicity.

Assuming that is what you mean, press on so I can find your mistakes.

[Ziggurat]

Originally Posted by Meridian

What is m here? Rest mass (invariant mass)? Or `relativistic mass', which seems not to be used much (at least not without a specific label) since it leads to confusion, such as forgetting that it's frame dependent. In SR the formula for momentum uses the latter, or, as normally written, invariant mass times gamma times velocity. See e.g., wikipedia .

What you appear to mean (m_0 v) is not a conserved quantity in SR.

Egads, NO! Do not use relativistic mass. It has overstayed its welcome. There is a reason modern relativity texts tend not to use it anymore. Make the nonlinearities explicit, don’t hide them. There is never a situation where you need relativistic mass rather than rest mass, and it is more likely to confuse than to illuminate.

[Ziggurat]

Originally Posted by Darat

SR isn't accepted because of thought experiments it is accepted because of real world experiments and observations.

That’s true, BUT…

Thought experiments alone were enough to get every physicist who was paying attention to sit up and take note. Even without doing a single experiment, SR made Maxwell’s equations invariant. And that alone is enough to indicate that it was a theory worth taking a very long look at. So we shouldn’t undersell the usefulness of thought experiments.

That being said, you’ve still got to do them right, and SR skeptics never seem to be able to.

[Reformed Offlian]

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.

[Dave Rogers]

Originally Posted by Reformed Offlian

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.

Don't give the game away, I was betting on that!

Dave

[Crossbow]

Originally Posted by SDG

I did my own research. I've been at it for a long time.
We will make more analysis here, and you will make your own conclusion and research if anybody else came with the same type of thought experiment.

Well then, I do hope that you will be able to accept at how wrong you are.

[SDG]

When the same emission is observed from a moving frame then the photon direction is different due to aberration of light.
The conservation of linear momentum implies the photon recoil does not go through the center of mass G.
The expected effect is a change of angular momentum of the LED flashlight.




The rest frame does not predict the change of angular momentum.
The moving frame predicts a change of angular momentum.

[theprestige]

Originally Posted by SDG

I did my own research. I've been at it for a long time.
We will make more analysis here, and you will make your own conclusion and research if anybody else came with the same type of thought experiment.

The Socratic method is a toy method that only works in carefully contrived fictional scenarios with a straw student who exists purely as a foil for the author's rhetoric. Real people, who understand and respect that their audience is also real people, don't do it like this. They just lay out their evidence, reasoning, and conclusions all at once. Then they and their audience go from there.

[Dave Rogers]

Originally Posted by SDG

When the same emission is observed from a moving frame then the photon direction is different due to aberration of light.

Perhaps you'd like to clarify that statement.

Dave

[Ziggurat]

Originally Posted by SDG

When the same emission is observed from a moving frame then the photon direction is different due to aberration of light.
The conservation of linear momentum implies the photon recoil does not go through the center of mass G.
The expected effect is a change of angular momentum of the LED flashlight.

https://i.imgur.com/5p3Twdt.png


The rest frame does not predict the change of angular momentum.
The moving frame predicts a change of angular momentum.

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.

[Darat]

Originally Posted by Ziggurat

That’s true, BUT…

Thought experiments alone were enough to get every physicist who was paying attention to sit up and take note. Even without doing a single experiment, SR made Maxwell’s equations invariant. And that alone is enough to indicate that it was a theory worth taking a very long look at. So we shouldn’t undersell the usefulness of thought experiments.

That being said, you’ve still got to do them right, and SR skeptics never seem to be able to.

Oh I wasn't trying to say that thought experiments can't or don't play a very important role in science. But this is someone who claimed to have done research, now if they only meant they had done a search through published papers on SR then fair enough.

[Reformed Offlian]

Originally Posted by Ziggurat

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.

If I'm reading this right, this isn't even a "problem" peculiar to SR, or to photons. Even in Galilean relativity, the lampost I'm driving past has an angular momentum in my reference frame that it doesn't have in its own rest frame.

[Reformed Offlian]

Originally Posted by SDG

When the same emission is observed from a moving frame then the photon direction is different due to aberration of light.
The conservation of linear momentum implies the photon recoil does not go through the center of mass G.
The expected effect is a change of angular momentum of the LED flashlight.

https://i.imgur.com/5p3Twdt.png


The rest frame does not predict the change of angular momentum.
The moving frame predicts a change of angular momentum.

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

[Myriad]

Originally Posted by SDG

When the same emission is observed from a moving frame then the photon direction is different due to aberration of light...

And the direction the flashlight is aimed is different by the same angle, for the same underlying reason (Lorentz transformation). The line of the photon's trajectory still intersects G.

[Ziggurat]

Originally Posted by Reformed Offlian

If I'm reading this right, this isn't even a "problem" peculiar to SR, or to photons. Even in Galilean relativity, the lampost I'm driving past has an angular momentum in my reference frame that it doesn't have in its own rest frame.

Yes, that's absolutely true. The reason this gets easier to mess up in special relativity is that it's easier to forget that emitting the photon requires losing mass, and thus it's easier to lose track of how all the different momenta are actually changing. The same basic dynamics happen if, say, you shoot a bullet, but in the case of a bullet it's easier to see that the gun lost mass.

Originally Posted by Reformed Offlian

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

Are you referring to Terrell rotation? Yes, that would make the flashlight appear to rotate, but that's an optical illusion, not an actual rotation of the flashlight. It's a really cool effect, but probably not what SDG has in mind.

[Ziggurat]

Originally Posted by Darat

Oh I wasn't trying to say that thought experiments can't or don't play a very important role in science. But this is someone who claimed to have done research, now if they only meant they had done a search through published papers on SR then fair enough.

There is a way that, in principle, someone could shoot down a physics theory without doing any physical experiments. All you would need to do is demonstrate an internal inconsistency. Internal consistency isn't enough to prove a theory correct, theories can be internally consistent but not consistent with reality, which is why you still need experiments. But if you can show an internal inconsistency, that's disproof of a theory even without any experiments.

The problem faced by SDG is that relativity is completely internally consistent. This is mathematically provable. If you think you've found an inconsistency in relativity, then you've really only found where your own understanding of the theory fails.

[smartcooky]

Originally Posted by Ziggurat

Yes, that's absolutely true. The reason this gets easier to mess up in special relativity is that it's easier to forget that emitting the photon requires losing mass, and thus it's easier to lose track of how all the different momenta are actually changing. The same basic dynamics happen if, say, you shoot a bullet, but in the case of a bullet it's easier to see that the gun lost mass.

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

I read somewhere that if photons had mass, it would invalidate quantum electrodynamics theory, and would impact on well established laws such as Coulombs law.

Or are you referring to some sort of mass ~ energy thing here?

[Crossbow]

Originally Posted by smartcooky

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

I read somewhere that if photons had mass, it would invalidate quantum electrodynamics theory, and would impact on well established laws such as Coulombs law.

Or are you referring to some sort of mass ~ energy thing here?

If I may chime in here ...

Although I am sure that 'Ziggurat' can answer your question far better than I can, but I may be to answer your question all the same.

Photons do not have mass.

However, Photons do have Momentum. As such, when Photons are produced by a LED flashlight, a candle, a lighting bug, etc., there is a corresponding loss of mass by the object that produced the Photons in question.

I hope this helps.

[SDG]

Originally Posted by Ziggurat

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.

Thanks, there is a lot said in this post.
1. In order to emit a photon, the mass of the flashlight must decrease.
How does this work? What frame do you have in mind?
Photon, massless particle, leaves the flashlight, what mass change?

[Ziggurat]

Originally Posted by SDG

Thanks, there is a lot said in this post.
1. In order to emit a photon, the mass of the flashlight must decrease.
How does this work? What frame do you have in mind?

The necessity for mass to energy conversion applies in all frames, but the calculations are easiest in the rest frame first and then transformed, rather than trying to calculate in the moving frame. You can do it either way, but the latter is a bit more work.

Quote:

Photon, massless particle, leaves the flashlight, what mass change?

E=mc², remember? Mass gets converted into energy to create the photon.

And before you ask, that’s the equation for rest mass energy in the rest frame of the object, so don’t expect to find an error there. In a frame where the object is moving, it’s E² = p²c² + m²c⁴.

[Reformed Offlian]

Originally Posted by SDG

Thanks, there is a lot said in this post.
1. In order to emit a photon, the mass of the flashlight must decrease.
How does this work? What frame do you have in mind?
Photon, massless particle, leaves the flashlight, what mass change?

So, you "did your own research" and "have been at this for some time", yet in all that time of doing your research, you never learned this? That photons have momentum and that objects emitting them lose mass (and thus momentum) as they do so?

[smartcooky]

Originally Posted by Crossbow

If I may chime in here ...

Although I am sure that 'Ziggurat' can answer your question far better than I can, but I may be to answer your question all the same.

Photons do not have mass.

However, Photons do have Momentum. As such, when Photons are produced by a LED flashlight, a candle, a lighting bug, etc., there is a corresponding loss of mass by the object that produced the Photons in question.

I hope this helps.

Hmmm. Ok, I think so

Originally Posted by Ziggurat

E=mc², remember? Mass gets converted into energy to create the photon.

A penny just dropped... I think?

Correct me here if I have this wrong

1. LED torch, when switched off, has no photons.

2. Switch LED torch on - photon is "created" (it takes energy to do this).

3. LED torch loses this energy in the form of mass (energy ~ mass conversion as a result of E=MC²).

4. The photon cannot remain stationary, it must move at the speed of light in a vacuum, therefore the photon gains momentum.

5. The result is a recoil force in the opposite direction to which the photon is emitted.

Amirite... even partly?

[SDG]

Originally Posted by Ziggurat

The necessity for mass to energy conversion applies in all frames, but the calculations are easiest in the rest frame first and then transformed, rather than trying to calculate in the moving frame. You can do it either way, but the latter is a bit more work.



E=mc², remember? Mass gets converted into energy to create the photon.

And before you ask, that’s the equation for rest mass energy in the rest frame of the object, so don’t expect to find an error there. In a frame where the object is moving, it’s E² = p²c² + m²c⁴.

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

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

As you wrote E² = p²c² + m²c⁴.
Is this equation true?
Let us assume it is.
There is a spontaneous hydrogen atom emission, m is constant, c is constant then we have to conclude p changed.
How p got changed? In what frame, why?

[SDG]

Originally Posted by Reformed Offlian

So, you "did your own research" and "have been at this for some time", yet in all that time of doing your research, you never learned this? That photons have momentum and that objects emitting them lose mass (and thus momentum) as they do so?

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