How fast does photon spin?

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Offline jartza

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How fast does photon spin?
« on: 15/12/2010 20:31:50 »

Let's radiate a frictionless black wheel with circularly polarized photons.

What is the wheel's spinning rate after very long time of circularly polarized photon radiation?



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Offline JP

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« Reply #1 on: 16/12/2010 01:22:41 »
Each photon if it's aligned with the wheel and perfectly absorbed would impart an angular momentum of +/- hbar, where hbar is the reduced Planck's constant.  If you know how many photons hit the wheel, you know it's angular momentum. 

To get spinning rate from angular momentum, you would need to tell us the moment of inertia of the wheel.

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Offline JP

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« Reply #2 on: 16/12/2010 01:31:59 »
By the way, the thread title is a bit confusing.  Photons aren't like physical tops, spinning around.  We call it spin because large objects carry something called angular momentum when they are physically spinning.  Spin is one way for particles to carry angular momentum as well, but careful tests have shown that they aren't physically spinning like tiny tops.

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Offline jartza

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How fast does photon spin?
« Reply #3 on: 16/12/2010 04:56:04 »
What's the spinning energy of a photon?
How's that question?









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Offline jartza

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How fast does photon spin?
« Reply #4 on: 16/12/2010 05:23:49 »

Let's radiate a frictionless black wheel with circularly polarized photons.

What is the wheel's spinning rate after very long time of circularly polarized photon radiation?


Well, the wheel actually has a terminal spinning rate, where the torque exerted by the spins of the photons is equal to torque that is the result of photons arriving at some angle.


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Offline JP

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« Reply #5 on: 16/12/2010 07:10:01 »

Let's radiate a frictionless black wheel with circularly polarized photons.

What is the wheel's spinning rate after very long time of circularly polarized photon radiation?


Well, the wheel actually has a terminal spinning rate, where the torque exerted by the spins of the photons is equal to torque that is the result of photons arriving at some angle.



I have no idea what you're getting at here.  Each photon has angular momentum.  If this photon is absorbed by the disk, then the disk has to gain angular momentum, which it does by spinning faster.

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Offline Geezer

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How fast does photon spin?
« Reply #6 on: 16/12/2010 07:48:59 »

Let's radiate a frictionless black wheel with circularly polarized photons.

What is the wheel's spinning rate after very long time of circularly polarized photon radiation?


Well, the wheel actually has a terminal spinning rate, where the torque exerted by the spins of the photons is equal to torque that is the result of photons arriving at some angle.



I have no idea what you're getting at here.  Each photon has angular momentum.  If this photon is absorbed by the disk, then the disk has to gain angular momentum, which it does by spinning faster.

I think the photons would have to land smack dab on the axis of the disk.

This might be what jartza was referring to when he mentioned torques, but I think you already addressed that point with "perfect alignment".
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline jartza

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« Reply #7 on: 16/12/2010 12:27:51 »

What if the wheel spins at high RPM and the photon is a low energy photon.

Now there's not enough energy in the photon to exert big enough torque long enough time to
cause the right size of angular momentum increase of the wheel!

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Offline jartza

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How fast does photon spin?
« Reply #8 on: 16/12/2010 12:37:52 »

Let's radiate a frictionless black wheel with circularly polarized photons.

What is the wheel's spinning rate after very long time of circularly polarized photon radiation?


Well, the wheel actually has a terminal spinning rate, where the torque exerted by the spins of the photons is equal to torque that is the result of photons arriving at some angle.



I have no idea what you're getting at here.  Each photon has angular momentum.  If this photon is absorbed by the disk, then the disk has to gain angular momentum, which it does by spinning faster.

I mean: first the wheel's spinning accelerates, then it stops accelerating, unless direction of light is adjusted.

I guess you mean "Where does the angular momentum go!?"

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Offline QuantumClue

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« Reply #9 on: 16/12/2010 12:44:59 »
never mind, should have read the OP.

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Offline yor_on

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« Reply #10 on: 16/12/2010 15:56:09 »
Spin is a intrinsic 'property' of a photon, also called polarization. This one is nice for an idea of how they come to be defined Spin and polarization

When you bombard that wheel you only can 'attack' it by the photons momentum, 'pushing' on it as I see it? and to make it 'turn/spin' I guess you will have to angle the direction of you 'attack' so that the 'bouncing' (sort of:) import a 'force' in the wheels direction of 'turning around'. Which should mean that the best approach would be to 'shoot' at its side. (Think of it as a water-wheel, where will the water 'work' best for that wheel to turn, to see my idea:)

The 'spin' in a electron for example if 'classically seen' would be faster than the speed of light in a vacuum. That's why we call it a 'property'.
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Offline CPT ArkAngel

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How fast does photon spin?
« Reply #11 on: 17/12/2010 02:28:49 »

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Offline JP

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How fast does photon spin?
« Reply #12 on: 17/12/2010 03:43:32 »

Let's radiate a frictionless black wheel with circularly polarized photons.

What is the wheel's spinning rate after very long time of circularly polarized photon radiation?


Well, the wheel actually has a terminal spinning rate, where the torque exerted by the spins of the photons is equal to torque that is the result of photons arriving at some angle.



I have no idea what you're getting at here.  Each photon has angular momentum.  If this photon is absorbed by the disk, then the disk has to gain angular momentum, which it does by spinning faster.

I mean: first the wheel's spinning accelerates, then it stops accelerating, unless direction of light is adjusted.

I guess you mean "Where does the angular momentum go!?"


Sometimes these problems can be tricky--absorbing angular momentum from a photon is a quantum effect--but as long as the photon direction is aligned with the axis of the rotating disk, every one that gets absorbed should impart a tiny bit of angular momentum to the disk.  Neglecting friction, air resistance, etc., the disk will continue to rotate faster and faster without need to adjust the direction of the light beam so long as it keeps absorbing photons. 

The reason for this is simply angular momentum conservation.  The photon has angular momentum, which, when absorbed, has to transfer to the disk.  The only way for the disk to gain angular momentum is to spin faster.

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Offline Geezer

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How fast does photon spin?
« Reply #13 on: 17/12/2010 05:46:08 »
Sometimes these problems can be tricky--absorbing angular momentum from a photon is a quantum effect--but as long as the photon direction is aligned with the axis of the rotating disk, every one that gets absorbed should impart a tiny bit of angular momentum to the disk.  Neglecting friction, air resistance, etc., the disk will continue to rotate faster and faster without need to adjust the direction of the light beam so long as it keeps absorbing photons. 

The reason for this is simply angular momentum conservation.  The photon has angular momentum, which, when absorbed, has to transfer to the disk.  The only way for the disk to gain angular momentum is to spin faster.

JP,

It's likely very dangerous to to superimpose the macroscopic view on the quantum view, but does the theoretical disk gain more angular momentum from a photon that hits the disk "dead center"? I'm thinking there must be some sort of couple that will cancel out almost all of the photon's angular momentum when it encounters the disk anywhere other than dead center.
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline JP

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« Reply #14 on: 17/12/2010 06:15:30 »
Actually, it's probably easier to view it classically.

If you consider just tossing classical particles at the disk, you could do a calculation using conservation of angular momentum for each particle to determine how it effects the disk's rotation.  The net effect of many particles would be to sum of the effects of each particle.

EM radiation is a wave distributed over space, but it still carries momentum.  Classically speaking, tather than being a bunch of discrete particles, you can treat an EM wave as having a smooth distribution of momentum over space, so that each point in the beam carries a tiny bit of momentum.  There's a quantity called the Poynting vector, which basically tells you the momentum of each tiny bit of the beam.  If you do the math, you can show that there is a contribution to the Poynting vector from polarization which accounts for the angular momentum of circularly polarized light--this is usually called the spin component of the Poynting vector.  (The beam's overall structure, independent of polarization, also contributes to it's momentum distribution, so it can carry an angular momentum that is independent of spin.) 

You can always get the right answer for how the disk moves by doing a conservation of angular momentum calculation for each tiny bit of the beam that hits the surface and adding up all the contributions.  If the beam hits dead-center along the disk's axis, then you should be able to make a symmetry argument so that you only need to compute a total angular momentum for the beam about its axis, and consider that transferring entirely to the disk.

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Offline Geezer

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« Reply #15 on: 17/12/2010 06:25:13 »
Thanks!

So, if you bombard the entire surface of the disk, there would be virtually no acceleration, but if you bias the bombardment to one side, there will be?
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline JP

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« Reply #16 on: 17/12/2010 06:38:26 »
It depends on what the Poynting vector looks like.  The circular polarization component of the beam's momentum should look like a bunch of vectors directed in a circle around the beam's axis.  This would cause the disk to rotate if the beam hit dead on. 

Physically, the effect would be similar to immersing the disk in water and sending a vortex of water at the disk.  The cause of this effect is entirely different for the beam, since it's due to polarization which isn't the case in a vortex of water.  (You can introduce a separate effect to the beam which is caused by making a vortex-shaped beam.  This is another way to introduce angular momentum to the beam and it's independent of spin.)

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Offline JP

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« Reply #17 on: 17/12/2010 06:48:47 »
This has a good explanation without fancy math:
http://people.physics.illinois.edu/Selvin/PRS/498IBR/Twist.pdf

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Offline yor_on

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« Reply #18 on: 17/12/2010 07:27:05 »
I didn't know that it was possible JP. To use light to make something turn at a right angle of the impact, using its momentum and polarization. I think that idea of a vortex was rather illuminating, but weird :) I enjoy light but I doubt I ever will understand it.

And CPT, thanks for that link :)
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Offline yor_on

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How fast does photon spin?
« Reply #19 on: 17/12/2010 07:35:29 »
Have anyone counted on how fast a photons spin would be if 'described classically'? I know it's probably a 'insane' number but still? The 'force' of its polarization is definitely more than what I thought?
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Offline yor_on

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« Reply #20 on: 17/12/2010 09:04:56 »
Had forgotten all about optical tweezers. Ah well.

"Probably only physicists know that circularly polarized light carries with it an angular momentum that results from the spin of individual photons. Few physicists realize, however, that a light beam can also carry orbital angular momentum associated not with photon spin but with helical wavefronts. Beams of this type have been studied only over the last decade. In many instances orbital angular momentum behaves in a similar way to spin. But this is not always so: orbital angular momentum has its own distinctive properties and its own distinctive optical components..."

(Birefringent = Relating to or characterized by double refraction. Splitting a ray into two parallel rays polarized perpendicularly (..getting opposite 'spins'?))

"To rotate objects via the spin angular momentum the light beam needs to be circularly polarized and the sample objects need to be either fully absorptive or birefringent. The latter has interesting physics: if the birefringent particle acts as a quarter-wave plate it can take spin angular momentum from a circularly polarized beam. This experiment was recently reported as an undergraduate laboratory.

We have been able to reproduce the experiments of Ref. with not much difficulty. We used crushed calcite crystals immersed in water and put a quarter-wave plate in the path of the laser beam to change its linear polarization to circular. Inexpensive demonstrations of transfer of orbital angular momentum are difficult to perform because the inexpensive methods to produce beams carrying orbital angular momentum use inefficient binary forked gratings. Efficient gratings in the form of phase holograms used in the early demonstrations are not easy to make. An expensive alternative that is very popular today is to use of spatial light modulators. Spiral phase plates are starting to become available, but in most cases they have to be special ordered."

From Light with a twist in its tail.

And they mention the force too.
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Offline JP

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« Reply #21 on: 17/12/2010 10:37:06 »
Birefringent here means essentially that it's going to change the polarization from circular to linear or from linear to circular.  Doing so creates or removes angular momentum from the beam, so by conservation of angular momentum, that angular momentum has to go somewhere, so the birefringent object rotates...

Going into the details of why this works is a bit complicated.  You can think of a circularly polarized beam as the combination of two linearly polarized beams--one of which is delayed by 1/4 of a wavelength with respect to the other.  If you have a crystal that delays one linear polarization more than the other, it will change the polarization of the light, which amounts to changing its angular momentum.

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Offline QuantumClue

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« Reply #22 on: 17/12/2010 12:57:00 »
JP is right, and I'd like to add something. It was going to be in a previous post here, but I decided to keep it out as it never addressed the OP.

Spin isn't even a true angular physical spin, not the kind of spin we attribute to planets and stars. Such spin cannot exist due to the size of particles. To rotate a particles spin 360 degrees would in normal classical physics, orientate the particle back to where it began, as you would expect changing the orientation of any geometrical object 360 degrees to return it back to the frame it began. However, rotating a pointlike particle 360 degrees does not rotate the object back to its original orientation.

Instead we need to rotate the pointlike object a further 360 degrees before it even settles back the way it began, so if it contained a classical spin, it would need to spin faster than light, and this is strictly forbidden by relativity. Instead of a classical spin, spin now in quantum mechanics is called an instrinsic property which takes the form very similar to an angular momentum. Actually interestingly enough, all the mathematics you need to know about a classical spin is pretty much identical to the mathematics required to understand angular momentum.

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Offline yor_on

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« Reply #23 on: 17/12/2010 14:53:17 »
'K JP You're the Boffin on light..
Why does it do that?

What QuantumClue said?
Isn't that forbidden? It should be, 'screwing around' like that?

Puts a new spin on that old adage.
"How many *** does it take to screw in a lightbulb?"

(Just replace lightbulb with 'photon')

I should really, really, get some sleep here :)
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Offline QuantumClue

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« Reply #24 on: 17/12/2010 15:07:55 »
'K JP You're the Boffin on light..
Why does it do that?

What QuantumClue said?
Isn't that forbidden? It should be, 'screwing around' like that?

Puts a new spin on that old adage.
"How many *** does it take to screw in a lightbulb?"

(Just replace lightbulb with 'photon')

I should really, really, get some sleep here :)


Sorry, a bit lost.

What is forbidden?

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Offline yor_on

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« Reply #25 on: 17/12/2010 16:40:25 »
The double turn, and I was joking of course.
Still it would be interesting to see a explanation of how to see it.
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Offline jartza

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« Reply #26 on: 18/12/2010 16:09:35 »
Sometimes these problems can be tricky--absorbing angular momentum from a photon is a quantum effect--but as long as the photon direction is aligned with the axis of the rotating disk, every one that gets absorbed should impart a tiny bit of angular momentum to the disk.  Neglecting friction, air resistance, etc., the disk will continue to rotate faster and faster without need to adjust the direction of the light beam so long as it keeps absorbing photons. 

The reason for this is simply angular momentum conservation.  The photon has angular momentum, which, when absorbed, has to transfer to the disk.  The only way for the disk to gain angular momentum is to spin faster.


Do you want to know where the angular momentum goes?
It is radiated away with the radiation that is radiated away from the wheel.

Or mass of wheel increases.



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Offline jartza

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« Reply #27 on: 20/12/2010 06:01:09 »
Let's say we are traveling through a galaxy very fast. In our frame of reference most photons are moving approximately along our line of motion. So most photon's spins are aligned approximately along our line of motion. So if these photons give us some net angular momentum, that angular momentum's direction is along the line of our motion.



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Offline Geezer

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« Reply #28 on: 20/12/2010 06:14:46 »
Or mass of wheel increases.


Can you explain how that might happen?
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline jartza

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« Reply #29 on: 20/12/2010 06:21:34 »
Mass increase happens when the wheel absorbs photons but does not radiate photons.


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Offline JP

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« Reply #30 on: 20/12/2010 06:28:42 »
Well basically if the wheel starts rotating, and you assume E=mc2, then the wheel gains mass.  But the energy we're discussing here is rotational energy, which is also explained in terms of angular momentum transfer.

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Offline Geezer

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« Reply #31 on: 20/12/2010 06:41:22 »
Well basically if the wheel starts rotating, and you assume E=mc2, then the wheel gains mass.  But the energy we're discussing here is rotational energy, which is also explained in terms of angular momentum transfer.

So, is it safe to assume that, typically, wheels don't gain mass?
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline jartza

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« Reply #32 on: 20/12/2010 07:54:55 »
JP, When a photon hits a black object, the mass of the object increases.




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Offline JP

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« Reply #33 on: 20/12/2010 07:57:42 »
To Geezer: Generally, no, unless you're in a case where relativistic effects are important.  Even then, defining mass from E=mc2 makes the assumption that you're dealing with relativistic mass rather than invariant mass (you have two kinds of mass in relativity).

To jartza:  The object gains energy.  If you define mass from E=mc2, then the object gains mass.  You've posted a few times about a law of conservation of mass.  This law simply doesn't exist.  It's conservation of energy and you've chosen to define mass as proportional to energy.
« Last Edit: 20/12/2010 08:01:22 by JP »

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Offline jartza

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« Reply #34 on: 20/12/2010 08:14:41 »
JP, I remember conservation of mass discussion. It ended when you said: OK I admit such law exists.


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Offline JP

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« Reply #35 on: 20/12/2010 08:48:27 »
It went pretty much the way this discussion is going.  I said you were talking about conservation of energy, but that if you wanted to call it conservation of mass, using E=mc2, you could, since energy is proportional to mass if you define it that way. 

That's also an answer to Geezer's question.  The mass gained in this definition is proportional to the energy gained divided by the speed of light squared.  The speed of light is a huge number, so the mass gained is negligible in most cases.

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Offline peppercorn

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« Reply #36 on: 20/12/2010 09:43:56 »
JP, I remember conservation of mass discussion. It ended when you said: OK I admit such law exists[, but by scientific convention is called 'conservation of energy']

Then why are you going over the same ground again? Can't you just move on to pastures new?

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Offline jartza

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« Reply #37 on: 21/12/2010 00:39:34 »
Let's call the "law of conservation of mass" "alphagammasuperduperlaw".
Let's call the "mass" that is conserved in alphagammasuperduperlaw "Crazy Jartza's Silly Mass"

When a photon hits a black object the Crazy Jartza's Silly Mass of the object increases.
When the Crazy Jarza's Silly Mass of an object increases the moment of inertia of the object increases.

A photon has a constant amount of angular momentum, called "spin".

When a photon hits a spinning black object it increases the angular momentum and the moment of inertia of the object.

Oh yes, when a photon hits a black object the increase of Crazy Jartsa's Silly Mass is proportional to the frequency of the photon.
 
So, when a very low frequency photon hits a spinning black object, the increase of moment of inertia can be ignored as very small, and the spinning rate of the object increases.

When a very high frequency photon hits a spinning black object, the increase of angular momentum can be ignored as very small, and the spinning rate of the object decreases.[whitespace removed]
« Last Edit: 21/12/2010 01:36:29 by peppercorn »

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Offline peppercorn

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« Reply #38 on: 21/12/2010 01:43:43 »
[-take your pick from any of it-]
Silly new words, same old waffle.

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Offline JP

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« Reply #39 on: 21/12/2010 08:03:06 »
If you put a disk on a frictionless axle through its center and bombard it with photons carrying angular momentum then it will get contributions to its energy from rotation.  It will also get pushed onto its axis, but since this axis is presumably bolted to the earth somehow, this push gets transferred into the earth.  This contribution to energy is lost.  Therefore if you just look at the disk itself, energy isn't conserved.  Angular momentum is. 

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Offline jartza

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« Reply #40 on: 21/12/2010 10:41:49 »
Hey JP that went wrong

But I'll fix it for you.

If you just look at the disk, neither energy nor angular momentum is conserved. (both increase)
If you look at the disk and the photon, both energy and angular momentum are conserved.
If you look at the disk and the photon, some linear momentum disappears, it goes to the continental plate, through the bolts.

OK?




(if you are looking from a frame where disk and continental plate that the disk is bolted on are moving, then the continental plate receives an energy force times distance, where force is radiation pressure and distance is the distance that the continental plate moved while the force was pushing )


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Offline yor_on

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« Reply #41 on: 21/12/2010 20:32:00 »
I don't agree there Jartza. Energy and mass is not the exact same. If we want to keep a 'transforming' universe we need a law of conservation of energy, that as energy is the smallest ah, 'property' we know of :)

But matter is different, exactly how it is we don't really know, but there is none that have made a lasting piece of matter (out of light) that I can hold in my hand yet. And very few particles of spontaneous, or not, pair-production stay without transforming almost instantly, well, as I know. Then there are other arguments too for stating the same.
« Last Edit: 21/12/2010 20:33:45 by yor_on »
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Offline JP

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« Reply #42 on: 22/12/2010 01:20:45 »
If you look at the disk and the photon, both energy and angular momentum are conserved.

That's not true.  Angular momentum is conserved in the special case that your light beam strikes the coin in the exact center.  Energy isn't conserved since some is transferred into the earth.  Momentum conservation still holds, as long as you don't consider the components directed into the earth.  You can do this because momentum has a direction.

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Offline jartza

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« Reply #43 on: 22/12/2010 01:47:34 »
JP, when a photon hits a solar panel, it's not the Earth that gains energy, it's the panel.

 

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Offline JP

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« Reply #44 on: 22/12/2010 03:32:06 »
JP, when a photon hits a solar panel, it's not the Earth that gains energy, it's the panel.

Not if that panel's connected to the earth by an axle, which is what we've been assuming in this problem.

You're making some very basic mistakes

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Offline jartza

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« Reply #45 on: 22/12/2010 08:16:24 »
What's the matter JP, it's an odd claim that a car wheel does not get hot in the sun, if the  bearing of the wheel is very good, isn't it?. 

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Offline JP

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« Reply #46 on: 22/12/2010 15:48:45 »
What's the matter JP, it's an odd claim that a car wheel does not get hot in the sun, if the  bearing of the wheel is very good, isn't it?. 

What does this have to do with my claim that energy is transferred into the car's axle and into the earth?  It still is.

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Offline jartza

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« Reply #47 on: 22/12/2010 20:38:02 »
I think I need some more info.

A continuous one watt laser beam hits a wheel with frictionless bearing, and the axle is bolted to the ground.
How many Joules does the Earth receive in a second? Is it closer to 0 Joules or 1 Joules?





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Offline Geezer

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How fast does photon spin?
« Reply #48 on: 22/12/2010 20:43:04 »
0 J

(Not Simpson)
There ain'ta no sanity clause, and there ain'ta no centrifugal force ćther.

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Offline peppercorn

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« Reply #49 on: 22/12/2010 23:58:57 »
I think I need some more info.
I think you need some help (mathematical obviously) [::)]