Does Light Have a Measurable Density?

[Revolucien]

Khashishi, "Light traveling in free space has no mass. Light that is trapped in a cavity has invariant mass, and therefore a density."
That's pretty cool, got to go look it up though because I can partially understand it, but I'm curious as to what could be considered a confining cavity.

 

[DaTario]

Photons have 0 rest mass, 0 inertia, 0 gravitational impact on matter. Perhaps the effect of the photonic fields leads to an increase in the mass of the matter it affects, but the fields themselves do not warp space in a gravitational sense.

Sorry If I seem to be stubborn, but according to my (humble) understanding of general relativity the entity which can produce modifications in space-time is the energy density. Matter and light have energy density patterns associated to themselves. Therefore, it would be natural to me the statement that both matter and light can produce effects on space-time. A remarkable fact, however, is that matter are more likely to produce non negligible effects, due to the typical magnitude of its energy density.

 

[DaTario]

It is possible to assign a mass to the whole system (cavity+radiation). This mass increases if you increase the amount of light in it, as the rest energy of the system increases.

Is this statement based on the same principles used in that gedanken experiment with a photon inside a box (it is a demonstration of the relation $E = mc^2$), typically presented in relativity's classes?

 

[Khashishi]

What jerromyjon said was incorrect. Photons do have gravitational impact on matter. It isn't necessary to have mass to do so. Momentum is another component of the stress-energy tensor and changes the gravity.

We don't have to talk about photons. The effects are purely classical. You get a partial standing wave pattern if you shoot a beam of light into a reflective cavity. Now, the question is, what do you call "light"? Do you consider the electromagnetic standing wave pattern light? Because it has mass. I don't think it is correct to say the mass is distributed in the cavity walls, because mass is local in classical physics.

 

[my2cts]

Is this statement based on the same principles used in that gedanken experiment with a photon inside a box (it is a demonstration of the relation $E = mc^2$), typically presented in relativity's classes?

I reached this conclusion based on the definition of mass as energy in the rest frame.
The rest frame is the frame in which the total momentum vanishes.
This definition of rest frame avoids the complication of a massless particle or wave.
I have no reference for these statements.except a letter by myself to the publication my local physics society
and I did not learn this explicitly in a physics class.

 

[mfb]

But photons can't accelerate or decelerate. Their velocity remains the same regardless of "energy" they contain...

Their speed (in vacuum) stays the same, the velocity can change.

So if measured density at this focal point is based on the energy and momentum of the photons, and the frequency determines the energy and momentum of the photon...my question from earlier (#17)- Isn't photon density in direct relation to frequency? A higher frequency will take more energy and momentum creating a higher density at a measured point. Sorry, if I'm missing something basic just trying to understand.

That's like saying "more red cars around means a higher car density, so car density is directly related to red car color". You can have a high and a low photon density with any frequency.

If I took a cylinder 10cm diameter x 1 meter long as my volume measurement, then through one tube had a red beam of light, and through another had a blue beam of light, the blue beam would have a higher energy per volume than the red.

Depends on the unspecified energy density in the cylinders. Take a dim blue LED and a bright red laser...

 

[DaTario]

I reached this conclusion based on the definition of mass as energy in the rest frame.
The rest frame is the frame in which the total momentum vanishes.
This definition of rest frame avoids the complication of a massless particle or wave.
I have no reference for these statements.except a letter by myself to the publication my local physics society
and I did not learn this explicitly in a physics class.

I was mentioning this derivation.
http://galileo.phys.virginia.edu/classes/252/mass_and_energy.html

 

[Revolucien]

Their speed (in vacuum) stays the same, the velocity can change.
That's like saying "more red cars around means a higher car density, so car density is directly related to red car color". You can have a high and a low photon density with any frequency.

The reason I had said red was just to say the difference in wavelength/frequency has different energy--> E=ħƒ where ħ is Planck's Constant or Energy=(6.62607004 × 10-34 m2 kg / s)frequency.

Depends on the unspecified energy density in the cylinders. Take a dim blue LED and a bright red laser...

If you use a red and blue LED of the same value of luminosity, based on the E=ħƒ would the frequency/wavelength determine a different amount of energy for each?

 

[mfb]

The reason I had said red was just to say the difference in wavelength/frequency has different energy--> E=ħƒ where ħ is Planck's Constant or Energy=(6.62607004 × 10-34 m2 kg / s)frequency.

I know, but the following question didn't make sense.

If you use a red and blue LED of the same value of luminosity, based on the E=ħƒ would the frequency/wavelength determine a different amount of energy for each?

What exactly does "amount of energy" mean here? If the energy density of blue light and red light (energy per volume) is the same, then the "photon density" (problematic concept) is lower for blue light, because the energy per photon is higher and we fixed the overall energy density.

 

[jerromyjon]

based on the E=ħƒ would the frequency/wavelength determine a different amount of energy for each?

It's a simple linear relationship, is it not?

 

[Revolucien]

What exactly does "amount of energy" mean here? If the energy density of blue light and red light (energy per volume) is the same, then the "photon density" (problematic concept) is lower for blue light, because the energy per photon is higher and we fixed the overall energy density.

I did not say the energy density was the same, that is where I am thinking there is a difference. By increasing the dim blue light or decreasing the red so they are equal in luminosity is it possible that they could have different energy per volume based on the wavelength? Does it take more energy to create the same luminosity in different wavelengths?

It's a simple linear relationship, is it not?

That's what I'm thinking.

I appreciate all the input you guys have offered on this, some of the responses have made me look up some really interesting stuff.

 

[jerromyjon]

luminosity

is luminosity a "result" of "total energy", or is it a "result" of "number of photons"? I'm not sure if that is asked clearly or specifically enough...

 

[jerromyjon]

This from wikipedia:
In astronomy, luminosity is the total amount of energy emitted by a star, galaxy, or other astronomical object per unit time.[1] It is related to the brightness, which is the luminosity of an object in a given spectral region.[1]

In SI units luminosity is measured in joules per second or watts. Values for luminosity are often given in the terms of the luminosity of theSun, which has a total power output of 3.846×1026 W.[2] The symbol for solar luminosity is L⊙. Luminosity can also be given in terms of magnitude.

that all makes sense, but this confuses me:
The absolute bolometric magnitude (Mbol) of an object is a logarithmic measure of its total energy emission.

Is that just a loose analogy like "moles of atoms"?

 

[mfb]

Luminosity is directly related to energy density. If you have the same luminosity, you have the same energy density.

that all makes sense, but this confuses me:
The absolute bolometric magnitude (Mbol) of an object is a logarithmic measure of its total energy emission.

Is that just a loose analogy like "moles of atoms"?

A logarithmic scale is just a different way to represent the same thing. "+5", "-3" etc. are more handy than large exponents. The magnitude of stars is also a historic thing.

 

[Revolucien]

I don't want to go too far with the luminosity thing because I was just using that because it seemed easier for explanation and this should be considered for all wavelengths visible or not. I looked up lumens and radiant flux and Light has a measurement for luminosity (lumens or lux) a measurement of luminous flux and also radiant flux. Lumens only cover the visible spectrum, but Spectral flux/spectral power is the radiant flux per unit frequency/wavelength. (Wiki) https://en.wikipedia.org/wiki/Radiant_flux

 

[Bkat2d11]

Welcome to PF!

If you are talking about mass density, no, light does not have mass so it doesn't have density.

If it doesn't have any mass how does light move things?

 

[jbriggs444]

If it doesn't have any mass how does light move things?

It has momentum. Also, to the extent that light is omni-directional, it can have energy density and, hence, mass density. [An ideal laser beam remains massless. It has non-zero energy density but has a matching momentum density according to $E=pc$ so that the 4-vector sum has zero magnitude].

Mass is not additive. Two entities that individually have no mass can form a system with non-zero mass.

A quick trip to Google says that:

"The cosmic (2.7 K) background radiation, for instance, has an energy density of 0.24 eV/cm³"

Edit: Note that a Crookes radiometer (aka "light mill") does not demonstrate light pressure. It operates by a different and more subtle principle. You want a Nichols radiometer to measure light pressure.

 

[vanhees71]

Light is an electromagnetic wave field and as any dynamical "object" it contains and transports energy, momentum, and angular momentum. Any relativistic field can be characterized by two parameters, its mass (and by mass we exclusively mean invariant mass, which is a scalar under Poincare transformations) and its spin. The electromagnetic field is massless and has spin 1. As any massless field it has two polarization degrees of freedom, which can be charactrized by the two possible values of the helicity $h=\pm 1$, corresponding to the left- and right-circular polarized wave modes for any wave vector $\vec{k}$. The masslessness of the em. field implies the dispersion relation $\omega=c k=c|\vec{k}|$ for each field mode.

Energy and momentum of the electromagnetic field are described by the energy-momentum-stress tensor. For details see

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf