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Tweefo
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« on: August 01, 2013, 08:19:42 AM »

I've been following this guy on Google+ and can only recommend him.
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Brian KoberleinShared publicly  -  Yesterday 3:31 PM
Dying Light

Photons are massless.  We know they are massless because particles with mass can’t move at the speed of light.  We know that special relativity works, and the speed of light is the same in all frames of reference, and special relativity only works if photons are massless.  Except...

What if photons had a tiny mass.  If it was small enough, then light could travel at almost the speed of light, and special relativity would almost be true.  How confident are we that the photon doesn’t have mass?  After all, we once thought neutrinos didn’t have mass and we now know they do have mass.

Just to be clear, we know a lot more about light than we do about neutrinos.  Neutrinos are notoriously difficult to detect, while photons are quite easy to observe (you’re probably doing it now!)  We’ve conducted tests on light in lots of different ways, and in every experiment the “massless photon” model works.  We have no reason to assume that the mass of a photon is anything other than zero.

Still, it is worth testing that assumption.  We know that if photons have mass it would have to be very small.  There are theoretical models that allow for photon mass.  For very small masses these models look very similar to the ones we use (such as Maxwell’s equations and the constant speed of light), but at larger masses they predict effects we would have seen by now. 

One of these effects would be seen in the cosmic microwave background.  As outlined in a recent paper in Physical Review Letters (http://goo.gl/ImVeyz), the cosmic background wouldn’t match a blackbody curve if the photon had mass. (I’ve talked about the blackbody curve before: http://goo.gl/RZuCT, and you can see the curve in the figure below.) 

The energy of photons depends upon its color.  Photons at the blue end of the spectrum have more energy than photons at the red end of the spectrum.  If photons have mass, then they don’t all move at the constant “speed of light”, but instead the speed of light would depend on the wavelength of light.  Large wavelength (reddish) light has less energy than short wavelength (bluish) light.  This means instead of being a perfect blackbody, the cosmic microwave background would be brighter than expected at long wavelengths and dimmer than expected at short wavelengths.  The larger the photon mass, the bigger this effect would be.

The cosmic background matches a blackbody so perfectly that the photon mass can be no larger than a hundredth the mass of an electron.  That’s pretty tiny, but other optical experiments require that the photon have a mass no larger than a trillionth of a trillionth of the mass of an electron.  That’s a septillionth of an electron mass.  So if the cosmic background puts less of a limit on photon mass than other experiments, what’s the big deal?

If the photon has mass, even a septillionth of an electron mass, it is possible that the lightest neutrino has an even smaller mass.  We know that neutrinos have mass, as I written about before (http://goo.gl/2ZkNs5).  There are actually three types of neutrinos, and we know the sum of their masses can be no larger than 2 millionths of an electron mass.  But we don’t know what their masses actually are.  If the lightest neutrino has a mass even smaller than the photon, then it would be possible for photons to decay into neutrinos.

That would mean photons wouldn’t last forever.  Instead they would have a half life.  This would have huge cosmological consequences, because it would mean that over billions of light years some of the light would decay into neutrinos.  The distant regions of the universe would appear dimmer than they should, and since the brightness of things like supernovae are used to measure distance, the universe would appear larger than it actually is.

But the photons wouldn’t decay equally.  Because bluish photons would have a greater speed than reddish photons, the time dilation effect of special relativity would mean that bluish photons would last longer than reddish photons.  And that means the light from the cosmic microwave background would again be distorted

You can see this in the image below, where the observed blackbody spectrum (the dots) is plotted against different half-lifes for the photon.  (Just as a side note, those error bars are actually 1000 times larger than the actual ones, which would be too small to see on the graph).  The lines are plotted for different half-life values. 

Based on observation of the cosmic microwave background, the half-life of a photon must be at least three years.  That doesn’t seem like a long time, but light travels so fast that time dilation would make them last much longer.  In the visible spectrum their effective half-life would be about a quintillion years, or about a billion times longer than the age of the universe. 

In other words, we can experimentally demonstrate that if photons have mass, then it is so extremely tiny that it is effectively massless, and if they can decay into neutrinos their average lifetime is so incredibly long that it is effectively forever.  So the assumption that the photon is massless and never decays is a good one.

Again, just to be clear, there is absolutely no evidence that photons are anything other than massless, and there are important theoretical reasons to presume photons are massless.  But experiments like these are good because they test the limits of our assumptions, and help keep us honest.

Because if we stopped testing our ideas, that really would be a dying of the light.
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Mefiante
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« Reply #1 on: August 01, 2013, 08:46:44 AM »

The bottom line is that Einstein’s second special relativity postulate is not just an assumption; there is much empirical evidence for it and a superbly cohesive narrative around the observed facts to support it.  It’s the cohesive narrative (i.e., coherent with observed facts and other bits of established science) that pseudoscientists typically eschew.

Just one minor quibble, however, where Koberlein writes in his opening paragraph that “… the speed of light is the same in all frames of reference…”  That should be “… the speed of light is the same in all inertial frames of reference…”  An inertial frame of reference is an unaccelerated one.  General relativity considers the case of non-inertial frames of reference and is quite happy with measuring differing values of c in such frames.

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BoogieMonster
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« Reply #2 on: August 01, 2013, 10:49:46 AM »

I didn't finish reading the entire post, apologies. However something mefiante said has piqued my interest.

Because the way I understood it c was an absolute maximum regardless of frame of reference.

If you have two accelerated frames of reference, does that mean you could have a relative c that exceeds your own c?

According to my severely limited knowledge of this, relativiy prohibits that?
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Mefiante
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« Reply #3 on: August 01, 2013, 11:53:15 AM »

The value of c an observer measures in their own non-inertial frame of reference will depend on how it is measured.  If you measure it over a sufficiently short time within a small enough region of such a frame, you’ll get arbitrarily close to special relativity’s normal c.  You can read a short overview here and if you’ve got a mind for some arcane symbols ( Wink ), there’s more detail here.

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cr1t
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« Reply #4 on: August 01, 2013, 12:19:33 PM »

This is where my brain stops working. Because we have e = mc2.

So we could get the mass of an object if we have it's energy i.e m = e/c2.

So then a photon should has a mass because it has energy?
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Mefiante
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« Reply #5 on: August 01, 2013, 12:59:01 PM »

So then a photon should has a mass because it has energy?
When physicists talk about a particle having mass, they always mean rest mass — i.e. the mass that would be measured by someone relative to whom the particle is standing still.  A photon has momentum instead of mass (more here).  For ordinary everyday moving objects, momentum is the product the object’s mass and its velocity (roughly speaking only, it’s slightly more complicated).  Because a photon’s speed is constant and non-zero in any given medium including a vacuum, it cannot be held stationary in order to measure a rest mass and so it makes much more sense to speak of a photon’s momentum.  Also, the mathematics of special relativity breaks down for the case where an object’s velocity is equal to c, a problem that is obviated only by assuming that a photon’s rest mass is exactly zero.

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Rigil Kent
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« Reply #6 on: August 01, 2013, 13:02:01 PM »

So we could get the mass of an object if we have it's energy i.e m = e/c2.
I think the formula E=mc^2 refers to the specific case where mass is wholly transformed into energy. It's a bit like a currency converter. Where mass and energy exist simultaneously, their relationship is given by the object's momentum.

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Mefiante
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« Reply #7 on: August 01, 2013, 16:34:06 PM »

E = mc2 means that mass and energy are physically the same property measured in different units.  With reference to the usual distinction we draw between mass and energy, one could think of mass as “frozen” energy and energy as “animated” mass.  At the subatomic scale, complete mass energy interchanges occur regularly.  The formula also gives the total energy (= rest mass energy + energy of motion) of a body in special relativity.  The value of the term m depends on which inertial frame it was measured in, and therefore so does E.

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brianvds
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« Reply #8 on: August 01, 2013, 17:32:29 PM »

With reference to the usual distinction we draw between mass and energy, one could think of mass as “frozen” energy and energy as “animated” mass.  At the subatomic scale, complete mass energy interchanges occur regularly.

Is this where some of those bazillions of different particles come from? Energy of the subatomic collision in an accelerator frozen into mass? If not, where exactly are all those particles in an intact atom? I never could understand any of that stuff.

Another thing that I have wondered about: in a normal exothermic chemical reaction, say, hydrogen burning in oxygen, is there any mass/energy conversion involved? I.e. do the bonds in molecules have any mass, that gets converted into energy when they break/reform (depending on the type of chemical and reaction)?

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Mefiante
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« Reply #9 on: August 01, 2013, 17:55:33 PM »

Is this where some of those bazillions of different particles come from? Energy of the subatomic collision in an accelerator frozen into mass?
Essentially yes, but on the subatomic scale everything is quantised, including energy and mass.  That is why the Standard Model of QM can predict what particles occur at what collision energies.  Those massive and exotic particles that are produced in collisions are also typically very short-lived (half-lives in the pico/femto/atto–second ranges), quickly decaying into more familiar and stable particles plus lots of energetic photons (X– and gamma rays).  In fact, it’s often those characteristic decay events that are looked for because the massive and/or exotic particles cannot be observed directly.

Another thing that I have wondered about: in a normal exothermic chemical reaction, say, hydrogen burning in oxygen, is there any mass/energy conversion involved?
Yes, a water molecule has slightly less mass than the sum of the masses of two oxygen atoms and a hydrogen atom.  The mass difference is equal to the total binding energy.

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Mefiante
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« Reply #10 on: August 02, 2013, 11:40:30 AM »

Oops, spot the silly mistale in my prior post. Embarrassed

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brianvds
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« Reply #11 on: August 02, 2013, 14:39:49 PM »

Oops, spot the silly mistale in my prior post. Embarrassed

'Luthon64

I wouldn't have spotted it if you didn't make a fuss of it, and now I spot another silly typo in THIS post! :-)

I would assume that the mass of the binding energy in atoms is extremely small?

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Mefiante
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« Reply #12 on: August 02, 2013, 15:53:51 PM »

… now I spot another silly typo in THIS post! :-)
That one was quite deliberate to see who’s paying attention, so well done! Tongue

I would assume that the mass of the binding energy in atoms is extremely small?
Yes, it’s minuscule indeed.  You can work it out per molecule, per mole or per unit mass from the molar masses of the molecule versus the sum of its constituent atoms.  In the case of water, H2O ≡ 18.01528 g/mol (source), while H ≡ 1.007947 g/mol and O ≡ 15.99943 g/mol (source: atomic molar masses).  The molar mass deficit ∆M = 2×1.007947+15.99943–18.01528 = 18.01532–18.01528 = 0.00004 g/mol.  Using E = mc2, this comes to about 3.6 Gigajoules per mole, or almost 200 MJ/g.  (If this energy were released into the water entirely as heat, it would increase the water’s temperature to almost 48,000,000 °C!)  A small amount of mass translates into an enormous amount of energy.  Since the mass difference is so tiny, molar masses for compounds are, for most practical purposes, calculated sufficiently accurately simply by summing up those of the compounds’ atomic constituents.

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« Reply #13 on: August 05, 2013, 06:41:30 AM »

… now I spot another silly typo in THIS post! :-)
That one was quite deliberate to see who’s paying attention, so well done! Tongue

I would assume that the mass of the binding energy in atoms is extremely small?
Yes, it’s minuscule indeed.  You can work it out per molecule, per mole or per unit mass from the molar masses of the molecule versus the sum of its constituent atoms.  In the case of water, H2O ≡ 18.01528 g/mol (source), while H ≡ 1.007947 g/mol and O ≡ 15.99943 g/mol (source: atomic molar masses).  The molar mass deficit ∆M = 2×1.007947+15.99943–18.01528 = 18.01532–18.01528 = 0.00004 g/mol.  Using E = mc2, this comes to about 3.6 Gigajoules per mole, or almost 200 MJ/g.  (If this energy were released into the water entirely as heat, it would increase the water’s temperature to almost 48,000,000 °C!)  A small amount of mass translates into an enormous amount of energy.  Since the mass difference is so tiny, molar masses for compounds are, for most practical purposes, calculated sufficiently accurately simply by summing up those of the compounds’ atomic constituents.

'Luthon64

Holy crap!  Minuscule correctly spelt[1] in a post on an internet forum!  I'm in a happy place.

1. Or spelled, take your pick.
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Mefiante
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« Reply #14 on: August 05, 2013, 08:49:22 AM »

Holy crap!  Minuscule correctly spelt[1] in a post on an internet forum!  I'm in a happy place.

1. Or spelled, take your pick.
We aim to please where we can, when we can. Wink

The spelling may be correct but the calculation is highly misleading.  It illustrates (1) the danger of uncritical use of ostensibly reliable data sources, and (2) that you get dodgy results when you approach a problem from the wrong end.

It’s much more accurate to start with the bond dissociation energy because the relative error in determining this quantity is significantly smaller than that of molar masses.  Bond dissociation energy is the average per-bond energy needed to break up a compound into its atomic constituents.  For water, this quantity is about 459 kJ/mol, and water has two bonds so the dissociation energy is 2×459 = 918 kJ/mol.  In addition, there’s a further ~23 kJ/mol in the form of hydrogen bonds between individual water molecules (without which water would be a gas at normal atmospheric conditions).  In total, then, water has about 940 kJ/mol of binding energy.  Expressed as mass, this binding energy comes to a much more modest ∆M = 0.0000000104 g/mol, or 0.000000058% of the constituent atoms’ mass.  Still, as heat, it would raise the water’s temperature to about 12,500 °C.

'Luthon64
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