Dynamic Mass of Photon: E=mc^2 & hv

[DonnieD]

i think E=mc^2
and E=hv so dynamic mass of photon is m=hv/c^2 ??

 

[masudr]

E=m_0 c^2 does not apply to photons.

The general equation is

$$E^2 = m_0^2c^4 + \vec{p}^2c^2.$$

As you can see, in the centre of momentum frame, where \vec{p}=0, the equation reduces to E=m_0 c^2, but there is no inertial frame where the photon is at rest.

Furthermore, m_0 = 0 for the photon, so the equation becomes

$$E=|\vec{p}|c$$.

 

[DonnieD]

but the relativistic mass m=hv/c^2?? (not the rest mass m_0)

 

[cesiumfrog]

Yes. It seems like this helps to understand momentum etc, but it is also equivalent to trivially expressing energy in different units.

 

[rbj]

E=m_0 c^2 does not apply to photons.

but the equation E=m c^2 works perfectly well for photons when

$$m = \frac{m_0}{\sqrt{1 - v^2/c^2}}$$

is the relativistic mass or the inertial mass (m = p/v) as observed by someone in a frame that is moving at velocity v with respect to the mass that is m_0 in its own frame.

The general equation is

$$E^2 = m_0^2 c^4 + |\vec{p}|^2 c^2 .$$

and that is compatible with

$$E=m c^2$$
$$m = \frac{m_0}{\sqrt{1 - |\vec{v}|^2/c^2}}$$
and
$$\vec{p} = m \vec{v}$$

but I'm glad you were clear using the m_0 notation instead of just the "m" for rest mass.

 

[pmb_phy]

i think E=mc^2
and E=hv so dynamic mass of photon is m=hv/c^2 ??

Yes. That is quite correct so long as you understand that the "m" you're using is inertial mass (aka "relativistic mass").

Pete

 

[pmb_phy]

..., but it is also equivalent to trivially expressing energy in different units.

That is quite untrue. E = mc² holds only in special circumstances such as isolated systems. IT wouldn't, say, work in a rod under stress. In such case the "relativistic mass" would have a different value than the energy would. For proof see the web page I constructed to prove this point and to give an example. See

http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

Pete

 

[cesiumfrog]

Such erroneous conclusions may come about due to the lack of application in special relativity to anything which can [?] be treated as a particle.

Your example depends on external forces (and worse, on comparing those in different reference frames, which for me sets off alarm bells: you'd be surprised how many long standing paradoxes are actually coordinate transformation errors), can you provide something more concrete?

 

[nakurusil]

but the equation E=m c^2 works perfectly well for photons when

$$m = \frac{m_0}{\sqrt{1 - v^2/c^2}}$$

is the relativistic mass or the inertial mass (m = p/v) as observed by someone in a frame that is moving at velocity v with respect to the mass that is m_0 in its own frame.



and that is compatible with

$$E=m c^2$$
$$m = \frac{m_0}{\sqrt{1 - |\vec{v}|^2/c^2}}$$
and
$$\vec{p} = m \vec{v}$$

but I'm glad you were clear using the m_0 notation instead of just the "m" for rest mass.

The photon is a massless particle.
Relativistic mass is a frowned upon concept.
Therefore it is a bad idea to speculate about the relativistic mass of the photon

 

[country boy]

The photon is a massless particle.
Relativistic mass is a frowned upon concept.
Therefore it is a bad idea to speculate about the relativistic mass of the photon

Right. For photons, just stick to energy and momentum.

Relativistic mass is, indeed frowned upon by purists. But purists may not be the ones who come up with the new ideas. It's still good to kick the concept around a bit.

 

[nakurusil]

Right. For photons, just stick to energy and momentum.

Correct.

Relativistic mass is, indeed frowned upon by purists. But purists may not be the ones who come up with the new ideas. It's still good to kick the concept around a bit.

Doesn't follow logically, why wouldn't the purists be the ones that come up with new ideas. Rigor and creativity go hand in hand, they aren't opposites.

 

[pmb_phy]

I don't know where you got that comment that you quoted but I made no comment such as that in this thread.

Your example depends on external forces (and worse, ...

And that was the point. I maqde that quite clear. Not all systems are closed you know. In fact it is the same problem (worked a bit differently) that Einstein published in either 1906 or 1907.

...on comparing those in different reference frames, ...

Now I'm curious; What exactly do you think a Lorentz tranasformation does??

... can you provide something more concrete?

I already gave you the most concrete example that can be given, its in the URL I gave you.

Its not like I'm saying something different than you can found in, say, Rindlers 1982 intro to SR text.

Pete

 

[pmb_phy]

Right. For photons, just stick to energy and momentum.

That's a good way to get into trouble when you calculate the mass-density of a gas of photons.

Relativistic mass is, indeed frowned upon by purists. But purists may not be the ones who come up with the new ideas. It's still good to kick the concept around a bit.

What exactly is a "purist"? Is it someone who frowns on rel-mass? Please defined the term "purist". Thanks.

Pete

 

[cesiumfrog]

I don't know where you got that comment that you quoted but I made no comment such as that in this thread.

No, it's from the linked webpage you keep referring to and said you constructed.

And that was the point. I maqde [sic] that quite clear. Not all systems are closed you know. [..] What exactly do you think a Lorentz tranasformation does?? I already gave you the most concrete example that can be given, [..] Its not like I'm saying something different than you can found in, say, Rindlers 1982 intro to SR text.

Sarcasm, eh? I really appreciate the more authorative reference since, as I'm not very clever, I don't trust myself to follow an online derivation as closely as I should. By more concrete, what I was actually thinking of is something like this:

If I gave you a spring (with known spring constant and unstressed rest mass) with an (ideal) mirror at each end, and a pair of lasers (to reflect a continuous stream of photons against the mirrors), could you demonstrate that relativistic mass is not equivalent to energy? If so then in what way would you measure the spring's energy versus relativistic mass?

 

[pmb_phy]

No, it's from the linked webpage you keep referring to and said you constructed.

What's with the attitude. I asked you where it came from. I wrote that web page years ago and I don't remember everything I wrote in all those web pages

Sarcasm, eh?

Not at all. I appologize if my comments came off that way. I never post sarcastic comments intentionaly. I appologize if you thought of that as attitude.

I really appreciate the more authorative reference since, ...

I'm glad that reference was able to help you out. Here is the section of concern here. From page 150 in the section Relativistic Mechanics of Continua

The second relativistic effect to be discussed here concerns mass density \rho. It would seem at first sight that \rho and \rho₀ should be related by

\rho = \gamma²\rho₀,

where one \gamma is due to length contraction affecting what is a unit volume in the rest frame, and the other is due to mass increase according to formula (26.3). But that simple formula is valid only in special cases, e.g. for single particles and for systems of free particles [...]. It is not generally valid for constrained systems.

That comment opened my eyes to the more difficult aspects of mass-energy. Through a long period of study I finally came to understand this. Its not that hard if you're skilled in the physics or if you have someone you trust walk you through it.

...as I'm not very clever, ...

cesiumfrog! Please don't confuse cleverness with knowledge. There's no reason to assume that you're not very clever and simply lacking in the acquired skill required to understand the physics. This has been my experience throughout the last 25 years of my life. I doubt that you'd be posting in a physics forum if you weren't clever in the first place.

If I gave you a spring (with known spring constant and unstressed rest mass) with an (ideal) mirror at each end, and a pair of lasers (to reflect a continuous stream of photons against the mirrors), could you demonstrate that relativistic mass is not equivalent to energy? If so then in what way would you measure the spring's energy versus relativistic mass?

The relativistic mass of what? Of just the spring? This is a subtle question and details are required. Thanks.

Pete

 

[cesiumfrog]

The relativistic mass of what? Of just the spring? This is a subtle question and details are required. Thanks.

What I'm trying to do is express your example from the website (which I find a bit too abstract-mathematical) in terms of a specific physical example (which assists my intuitive understanding). Do you agree your (unspecified) external forces can be replaced with streams of photons (of particular frequency and intensity, from a distant source, symmetric in the box's frame) reflecting off opposite sides of the box? And that the box can be replaced by a spring (of equal dimensions whilst compressed by the external forces), without compromising the example?

 

[pmb_phy]

What I'm trying to do is express your example from the website (which I find a bit too abstract-mathematical) in terms of a specific physical example (which assists my intuitive understanding).

Pushing on the ends with your fingers is a good enough example, don't you agree? My fingers exert pressure as good as if not better than a stream of photons, but if a stream of photons is what you like as an example then who am I to second guess you on your iintuition?

Do you agree your (unspecified) external forces can be replaced with streams of photons (of particular frequency and intensity, from a distant source, symmetric in the box's frame) reflecting off opposite sides of the box?

Only under certain circumstances. If the force on a rod is such as to act towards pulling the rod apart then there is negative pressure present (called "tension") and a stream of photons can't do that. However in the case you stated the fact that the spring is compressed indicates that you're not asking about this situation but the situation in which the photons are acting to compress the spring.

And that the box can be replaced by a spring (of equal dimensions whilst compressed by the external forces), without compromising the example?

Sure. In any case the sitations leads to the same conclusion and yes I can demonstraight it. However you told me that you are unable to follow this kind of derivation. How do you want me to respond beyond this?

Thanks

Pete

 

[country boy]

Doesn't follow logically, why wouldn't the purists be the ones that come up with new ideas. Rigor and creativity go hand in hand, they aren't opposites.

Read again: "... purists may not be the ones who come up with the new ideas." They can, of course, come up with new ideas. The point is that non-purists mulling over seemingly superfluous notions like photon mass can lead (and has) to new insights like gravitational deflection of photons.

But if you want logic, see the philosophy pages.

 

[cesiumfrog]

Sure. In any case the sitations leads to the same conclusion and yes I can demonstraight it. However you told me that you are unable to follow this kind of derivation. How do you want me to respond beyond this?

OK, sorry you're finding this tedious. Now in this situation, it naively seems to me that the relativistic mass of the compressed spring, in any moving frame, will be simply the rest mass of the compressed spring (ie. the rest mass of the relaxed spring, plus the mechanical potential energy) multiplied by the usual lorentz factor (for the relative velocity of the frame). Is this incorrect?

 

[rbj]

The photon is a massless particle.

not in every sense of the word. they do not have rest mass, but they do have inertial mass of (h \nu)/c^2. Peter Mohr and Barry Taylor (at least at one time they were at NIST) have (along with others) proposed that the definition of kilogram be changed from the one regarding the standard prototype in Paris to

The kilogram is the mass of a body at rest whose equivalent energy corresponds to a frequency of exactly (299792458²/66260693 ×10⁴³) Hz.

another wording they have used is

The kilogram is the mass of a body at rest whose equivalent energy is equivalent to a collection of photons of frequencies that sum to exactly (299792458²/66260693 ×10⁴³) Hz.

now, if you put that collection of photons into a hypothetical massless and perfectly mirrored box and put the box on a scale and put the standard kg prototype on the other platter of the scale, which way would it tip?

Relativistic mass is a frowned upon concept.

depends on whose face. not everybody is frowning.

Therefore it is a bad idea to speculate about the relativistic mass of the photon

i'm not speculating, i am making reference to an oft and recent out-of-favor convention that defines momentum as inertial mass times velocity. then there needs to be a differentiation in concept between inertial mass and invariant mass.

when i look up "Gravitational Red Shift" in my old 3rd semester physics textbook, it says:

"Although a photon has no rest mass, it nevertheless behaves as though it possesses the inertial mass
$$m = \frac{h \nu}{c^2}$$ ."

and then goes on to show how "inertial" energy (m c^2) is traded for potential energy (of a particle of said mass) and with that sum conserved, how inertial energy is less as the photon departs a star resulting in lower frequency and that this red shifting is not to be confused with doppler red shifting if the star is moving away.

it's not such a bad idea. I'm still at a loss to understand why so many insist that it is. there are at least a couple ways that photons behave as if they have mass of some sort.

 

[country boy]

That's a good way to get into trouble when you calculate the mass-density of a gas of photons.
What exactly is a "purist"? Is it someone who frowns on rel-mass? Please defined the term "purist". Thanks.

Pete

Sorry I haven't replied before. You are certainly correct about the gas of photons. But a system of photons can have a rest mass without the individual photons having rest masses.

As for the definition, you might say that "purist" and "non-purist" are like thinking "inside" or "outside" the box.

 

[nakurusil]

not in every sense of the word. they do not have rest mass, but they do have inertial mass of (h \nu)/c^2. Peter Mohr and Barry Taylor (at least at one time they were at NIST) have (along with others) proposed that the definition of kilogram be changed from the one regarding the standard prototype in Paris to

The kilogram is the mass of a body at rest whose equivalent energy corresponds to a frequency of exactly (2997924582/66260693×10^43[) Hz.

another wording they have used is

The kilogram is the mass of a body at rest whose equivalent energy is equivalent to a collection of photons of frequencies that sum to exactly (2997924582/66260693×10^43[) Hz.

now, if you put that collection of photons into a hypothetical massless and perfectly mirrored box and put the box on a scale and put the standard kg prototype on the other platter of the scale, which way would it tip?

This is an old problem, the reason the scale will tip is the vertical component of force exerted by the photons colliding with the walls of the box. This should not be misconstrued as the photons having any type of mass.

depends on whose face. not everybody is frowning.

Yes, I know, there are a few members of this forum that cling to the notion of relativistic mass.

i'm not speculating, i am making reference to an oft and recent out-of-favor convention that defines momentum as inertial mass times velocity. then there needs to be a differentiation in concept between inertial mass and invariant mass.

You certainly not talking about relativistic momentum. I corrected this same exact misconception in another thread.

when i look up "Gravitational Red Shift" in my old 3rd semester physics textbook, it says:

"Although a photon has no rest mass, it nevertheless behaves as though it possesses the inertial mass
$$m = \frac{h \nu}{c^2}$$ ."

Still a bad idea, time to let go of this old concept.

it's not such a bad idea. I'm still at a loss to understand why so many insist that it is. there are at least a couple ways that photons behave as if they have mass of some sort.

Not really, the "mirror box" is a misleading example. I can show you the calculations, the balance tips due to the force (see above), not due to the fact that the photons have mass.

 

[pmb_phy]

Sorry I haven't replied before. You are certainly correct about the gas of photons.

Glad to see that you concur!

But a system of photons can have a rest mass without the individual photons having rest masses.

As the saying goes "That is intuitively obvious even to the most casual observer" I love that saying. I finally got a chance to use it.

As for the definition, you might say that "purist" and "non-purist" are like thinking "inside" or "outside" the box.

Nobody would consider themselves as only thinking "inside the box". I most certainly don't.

Where did you get this 'definition' from? I was actually asking the person who posted the term to define what he/she meant when he/she used it.

Thanks for the response

Kind regards

Pete

 

[country boy]

Not really, the "mirror box" is a misleading example. I can show you the calculations, the balance tips due to the force (see above), not due to the fact that the photons have mass.

So what happens to the scale when the photons are all absorbed by the mirrors? The box will no longer be massless. This is the old energy-mass equivalence question.

 

[nakurusil]

So what happens to the scale when the photons are all absorbed by the mirrors? The box will no longer be massless. This is the old energy-mass equivalence question.

What sort of question is this? The whole gimmick (because this is what this problem is, a gimmick) is that the walls are "perfect" mirrors. Have you forgotten?
As to "photons turning into mass", this will NOT happen in your perfectly mirrored box. The photon absorbtion by already existent matter increases the mass of said matter but this should not be misconstrued as the photons having mass. Because they don't.

 

[rbj]

You certainly not talking about relativistic momentum. I corrected this same exact misconception in another thread.

sure i am. i am certainly talking about relativistic momentum:

$$\vec{p} = \frac{m_0 \vec{v}}{\sqrt{1 - \frac{|\vec{v}|^2}{c^2}}}$$

but we are constructing the concept of this momentum as the product of the same velocity \vec{v} and some other mass m:

$$\vec{p} = m \vec{v}$$

you can deny that it exists, but it has dimension of mass and comes out to be:

$$m = \frac{m_0}{\sqrt{1 - \frac{|\vec{v}|^2}{c^2}}}$$

now the energy a inertial particle has in its own reference frame is

$$E_0 = m_0 c^2$$

but if it is moving relative to me, the energy it has in my reference frame is

$$E = m c^2$$

which is also consistent with

$$E^2 = m_0^2 c^4 + |\vec{p}|^2 c^2$$

it's all consistent. the photons have non-zero E and m, but since they move at speed c, they have no m_0.

the "mirror box" is a misleading example. I can show you the calculations, the balance tips due to the force (see above), not due to the fact that the photons have mass.

the balance does not tip. there is a kilogram prototype on the other platter.

one person's "misleading" is anothers "illustrative".

sure, there are many ways of looking at it. but to say that photons are utterly "massless" without qualification is what is misleading. there are many physicists who still write stuff that contains the qualification "rest mass" just to make sure. indeed that proposed definition of the kilogram from Mohr and Taylor uses the term and clearly implied an equivalence of "mass" in a sense (via an equivalence of energy) between a kilogram and a collection of photons.

the quantity c^2 is a dimensionful conversion factor between mass quantities and energy quantities. no one is saying that photons have no energy, have no momentum, have no active or passive effect of gravity, all of these things that particles having mass have, yet are doggedly so committed to say that while photons have all of these properties in common with particles having mass, that they are nonetheless "massless" without qualification is inconsistent, at best.

while you are free to adopt whatever convention you want or is popular at the moment (that when "mass" is mentioned, it is only "invariant mass"), to say, without qualification, that photons are simply massless despite E = m c^2 = h \nu, itself is misleading. (h \nu)/c^2 is a quantity. and it is a property of the photon (and dimensionful universal constants that really are just manifestations of our system of units). and that quantity is of dimension [M].

is it too inconvenient to qualify sweeping statements? instead of the unqualified "photons are massless particles", is it not as simple to say, " photons have no rest mass (because they travel at speed c in any inertial reference frame)." ? it's inclusive and accurate.

 

[rbj]

The photon absorbtion by already existent matter increases the mass of said matter but this should not be misconstrued as the photons having mass. Because they don't.

and by how much do they increase the mass of said matter? perhaps by (h \nu)/c^2?

still another example of mass-like properties of these "massless" particles.

 

[country boy]

What sort of question is this? The whole gimmick (because this is what this problem is, a gimmick) is that the walls are "perfect" mirrors. Have you forgotten?
As to "photons turning into mass", this will NOT happen in your perfectly mirrored box. The photon absorbtion by already existent matter increases the mass of said matter but this should not be misconstrued as the photons having mass. Because they don't.

Yes, the problem as posed is a sleight of hand. But it does lead us back to the traditional photon-in-a-box insight into the relation between energy and mass.

Sorry. Just thinking outside the box. If I forget anything, please remind me.

 

[Hurkyl]

is it too inconvenient to qualify sweeping statements? instead of the unqualified "photons are massless particles", is it not as simple to say, " photons have no rest mass (because they travel at speed c in any inertial reference frame)." ? it's inclusive and accurate.

I shall be so bold as to say yes. Unless the context explicitly indicates that the notion of relativistic mass is being discussed, one should not need to use any qualification when talking about "mass", just like the physicist can use "metric" without qualification to refer to a symmetric bilinear form -- a metric tensor -- without worrying that someone might think they were talking about the distance function of a metric space.

We already have a word for energy: energy. We don't need another word for it...

and by how much do they increase the mass of said matter? perhaps by (h \nu)/c^2?

still another example of mass-like properties of these "massless" particles.

especially when it induces silly mistakes like this. For a simple case, consider two photons of frequency \nu_1 and \nu_2 having a head-on collision. Conservation of energy-momentum tells us that the rest mass of the resulting particle is

$$m_0 = \frac{2 h}{c^2} \sqrt{\nu_1 \nu_2}$$


Or, if we take a stationary particle of rest mass m_0 absorbing a photon of frequency \nu, its resulting rest mass m_\Sigma is:

$$m_\Sigma = \left( m_0^2 + \frac{2 m_0 h \nu}{c^2} \right)^{1/2} \approx m_0 + m_0 \frac{h \nu}{c^2}$$

 

[cesiumfrog]

We already have a word for energy: energy. We don't need another synonym for it...

So you're also unconvinced by Pete's contrary proof?

especially when it induces silly mistakes like this. For a simple case, consider two photons of frequency \nu_1 and \nu_2 having a head-on collision. Conservation of energy-momentum tells us that the rest mass of the resulting particle is
$$m_0 = \frac{2 h}{c^2} \sqrt{\nu_1 \nu_2}$$

I don't see any mistake by rbj. Technically he gave the amount each photon contributes to the relativistic mass, but he was referring to a situation with zero net total momentum. Or in the language of your simple case, he chose the frame where the resulting particle is stationary (i.e. \nu_1=\nu_2) therefore m_0 = \frac{2 h \nu}{c^2}