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i think E=mc^2

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

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

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In summary, the conversation discusses the equations E = mc^2 and E = hv, and how they relate to the dynamic mass of a photon. It is noted that while E = m_0 c^2 applies to particles at rest, there is no inertial frame where a photon is at rest. The conversation also explores the concept of relativistic mass and its equivalence to energy, with some disagreement among participants. Overall, it is concluded that for photons, it is best to stick to using energy and momentum rather than discussing relativistic mass.

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i think E=mc^2

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

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

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The general equation is

[tex]E^2 = m_0^2c^4 + \vec{p}^2c^2.[/tex]

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

Furthermore, [itex]m_0 = 0[/itex] for the photon, so the equation becomes

[tex]E=|\vec{p}|c[/tex].

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

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masudr said:[itex]E=m_0 c^2[/itex] does not apply to photons.

but the equation [itex]E=m c^2[/itex] works perfectly well for photons when

[tex]m = \frac{m_0}{\sqrt{1 - v^2/c^2}}[/tex]

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

The general equation is

[tex]E^2 = m_0^2 c^4 + |\vec{p}|^2 c^2 .[/tex]

and that is compatible with

[tex]E=m c^2[/tex]

[tex]m = \frac{m_0}{\sqrt{1 - |\vec{v}|^2/c^2}}[/tex]

and

[tex] \vec{p} = m \vec{v} [/tex]

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

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Yes. That is quite correct so long as you understand that the "m" you're using isDonnieD said:i think E=mc^2

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

Pete

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That is quite untrue. E = mccesiumfrog said:..., but it is also equivalent to trivially expressing energy in different units.

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

Pete

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pmb_phy said: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?

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rbj said:but the equation [itex]E=m c^2[/itex] works perfectly well for photons when

[tex]m = \frac{m_0}{\sqrt{1 - v^2/c^2}}[/tex]

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

and that is compatible with

[tex]E=m c^2[/tex]

[tex]m = \frac{m_0}{\sqrt{1 - |\vec{v}|^2/c^2}}[/tex]

and

[tex] \vec{p} = m \vec{v} [/tex]

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

The photon is a

Therefore it is a bad idea to speculate about the

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nakurusil said:The photon is amasslessparticle.

Relativistic massis a frowned upon concept.

Therefore it is a bad idea to speculate about therelativistic 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.

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country boy said: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.

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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.cesiumfrog said:Your example depends on external forces (and worse, ...

Now I'm curious; What exactly do you think a Lorentz tranasformation does??...on comparing those in different reference frames, ...

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

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

Pete

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That's a good way to get into trouble when you calculate the mass-density of a gas of photons.country boy said:Right. For photons, just stick to energy and momentum.

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

Pete

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No, it's from the linked webpage you keep referring to and said you constructed. :rofl:pmb_phy said:I don't know where you got that comment that you quoted but I made no comment such as that in this thread.

pmb_phy said: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?

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What's with the attitude. Icesiumfrog;1249ite366 said:No, it's from the linked webpage you keep referring to and said you constructed.

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 asSarcasm, eh?

I'm glad that reference was able to help you out. Here is the section of concern here. From page 150 in the sectionI really appreciate the more authorative reference since, ...

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.The second relativistic effect to be discussed here concerns mass density [itex]\rho[/itex]. It would seem at first sight that [itex]\rho[/itex] and [itex]\rho[/itex]_{0}should be related by

[itex]\rho[/itex] = [itex]\gamma[/itex]^{2}[itex]\rho[/itex]_{0},

where one [itex]\gamma[/itex] 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 offreeparticles [...]. It isnotgenerally valid for constrained systems.

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

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

Pete

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pmb_phy said: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?

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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?cesiumfrog said: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).

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.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?

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

Thanks

Pete

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nakurusil said: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.

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OK, sorry you're finding this tedious. Now in this situation, it naively seems to me that the relativistic mass of the compressed spring, inpmb_phy said: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?

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nakurusil said:The photon is amasslessparticle.

not in every sense of the word. they do not have rest mass, but they do have inertial mass of [itex](h \nu)/c^2[/itex]. 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

another wording they have used is

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 massis a frowned upon concept.

depends on whose face. not everybody is frowning.

Therefore it is a bad idea to speculate about therelativistic 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:

"

[tex] m = \frac{h \nu}{c^2} [/tex] ."

and then goes on to show how "inertial" energy ([itex]m c^2[/itex]) 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.

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pmb_phy said: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.

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rbj said:not in every sense of the word. they do not have rest mass, but they do have inertial mass of [itex](h \nu)/c^2[/itex]. 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

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

[tex] m = \frac{h \nu}{c^2} [/tex] ."

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.

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Glad to see that you concur!country boy said:Sorry I haven't replied before. You are certainly correct about the gas of photons.

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

Nobody would consider themselves as only thinking "inside the box". I most certainly don't.As for the definition, you might say that "purist" and "non-purist" are like thinking "inside" or "outside" the box.

Where did you get this 'definition' from? I was actually asking the person who posted the term to define what

Thanks for the response

Kind regards

Pete

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nakurusil said: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.

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country boy said: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

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nakurusil said:You certainly not talking aboutrelativisticmomentum. I corrected this same exact misconception in another thread.

sure i am. i am certainly talking about

[tex] \vec{p} = \frac{m_0 \vec{v}}{\sqrt{1 - \frac{|\vec{v}|^2}{c^2}}} [/tex]

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

[tex] \vec{p} = m \vec{v} [/tex]

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

[tex] m = \frac{m_0}{\sqrt{1 - \frac{|\vec{v}|^2}{c^2}}} [/tex]

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

[tex] E_0 = m_0 c^2 [/tex]

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

[tex] E = m c^2 [/tex]

which is also consistent with

[tex] E^2 = m_0^2 c^4 + |\vec{p}|^2 c^2 [/tex]

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

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

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 [itex]c^2[/itex] 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 [itex] E = m c^2 = h \nu [/itex], itself is misleading. [itex](h \nu)/c^2 [/itex] 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 "

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nakurusil said:The photon absorbtion byalready existent matterincreases 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 [itex] (h \nu)/c^2 [/itex]?

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

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nakurusil said: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 byalready existent matterincreases 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.

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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 -- arbj said: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" ? it's inclusive and accurate.cin any inertial reference frame).

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

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

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 [itex]\nu_1[/itex] and [itex]\nu_2[/itex] having a head-on collision. Conservation of energy-momentum tells us that the rest mass of the resulting particle is

[tex] m_0 = \frac{2 h}{c^2} \sqrt{\nu_1 \nu_2} [/tex]

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

[tex]

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}

[/tex]

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So you're also unconvinced by Pete's contrary proof?Hurkyl said:We already have a word for energy: energy. We don't need another synonym for it...

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

[tex] m_0 = \frac{2 h}{c^2} \sqrt{\nu_1 \nu_2} [/tex]

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. [itex]\nu_1=\nu_2[/itex]) therefore [itex] m_0 = \frac{2 h \nu}{c^2} [/itex]

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rbj said:sure i am. i am certainly talking aboutrelativisticmomentum:

[tex] \vec{p} = \frac{m_0 \vec{v}}{\sqrt{1 - \frac{|\vec{v}|^2}{c^2}}} [/tex]

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

[tex] \vec{p} = m \vec{v} [/tex]

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

[tex] m = \frac{m_0}{\sqrt{1 - \frac{|\vec{v}|^2}{c^2}}} [/tex]

Good, much better than what you wrote in the other thread when you were trying to explain why massive particles traveling at c would have an infinite momentum.

Look, I am not interested in yet another debate on relativistic mass. I know that the photon has no rest mass (there is ample experimental proof on that, have a look at Roderik Lakes' paper) and trying to split the hairs in terms of speculating about whether its imparting energy after collision with a massive particle is a sign of it having "relativistic mass" is another thing that I am not interested in.

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

[tex] E_0 = m_0 c^2 [/tex]

but if it is moving relative to me, the energy it has inmyreference frame is

[tex] E = m c^2 [/tex]

Yes, I know all this, I prefer the form [tex]E=\gamma m c^2[/tex]. Together with [tex]p=\gamma mv [/tex] it produces the nice invariant [tex]E^2-(pc)^2=m^2c^4[/tex].

sure, there are many ways of looking at it. but to say that photons are utterly "massless" without qualification is what is misleading.

Don't think so: there is only one type of mass: invariant mass. So, when a particle (like the photon) has zero invariant mass, it means it is "massless".

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 [itex] E = m c^2 = h \nu [/itex], itself is misleading.

I am afraid that you are wrong on this one. This wiki article opens with "the photon is massless". It took a long time (and many fights, including a lot of back and forth about the dreaded "photon in a box") to craft but it attained the status of high quality article.

As a concession to the people supporting the photon having some sort of mass the article includes a paragraph on photons "adding" and "subtracting" [tex]E/c^2[/tex] to the invariant mass of a system but one can just as well (and more naturally) say that it adds/subtracts

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" ? it's inclusive and accurate.cin any inertial reference frame).

Because this is precisely what QED says. Because this is what the wiki article says. Beacuse it does not reference the silly "relativistic mass"

All these are pretty plenty for me. "Relativistic mass " is an anacronism, whether we talk about massive or massless particles. There is only one type of mass, invariant mass . QED predicts that the photon has zero invariant mass, experiment confirms it so the photon is a "massles particle" Good enough for me.

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country boy said: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.

I do not need sleigh of hand problems to clarify my understanding of the relation between mass and energy. There are many realistic issues that provide this clarification a lot better than the "photon in a box"

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Please pause to realize that your understanding of the "photon in a box example" is incorrect. Are you suggesting that a system of particles can generate a net force on itself? Of course not. (Or at least I really hope you aren't suggesting that.)nakurusil said: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.

Also, as mentioned before, the invarient mass of the box will increase when the photon is in it. So claiming the effect is just an illusion due to collision with the walls is missing the point.

How I prefer to view the term relativistic mass is:

It is a defined concept. Most physicists will know what is meant if they see the phrase. To pretend the

But do I feel it is a particularly useful concept? No. And the rarity of use of the term in current publications suggests most physicists agree.

To point out math errors seems fine, but to argue till you are faint to try to remove a concept seems like trying to remove an english word. Some people are going to use the concept, get over it.

Since you brought up QED I'd like to point out that no, there is more than one concept of mass. There are "intrinsic masses" for the particles, and in QED you often integrate over the momentum of particles with invarient mass not equal to their intrinsic mass. So QED doesn't predict that the photon has zero invariant mass, and in fact actuallynakurusil said:There is only one type of mass, invariant mass . QED predicts that the photon has zero invariant mass...

I put intrinsic masses in scare quote because I am not sure if that is the appropriate term. That is the term my prof used, but the book we used never gave it a name. It someone could confirm or tell me the common terminology for this it would be appreciated.

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rbj said:The kilogram is the mass of a body at rest whose equivalent energy corresponds to a frequency of exactly (299792458^{2}/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 (299792458^{2}/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?

We need to be careful here with the switch from mass to weight, that is, from inertial to accelerated frames. [This is similar to the error Bohr found in Einstein's argument about the uncertainty principle.] Special relativity no longer strickly applies and one has to account for what happens to photon frequencies (among other things) and where the measurements are made in the gravitational field. Note that the definition above says "a body at rest." Is that in an inertial frame or gravitational field? Weighing on a scale would imply the latter, but then is it at one-g? How does the definition change at other positions in the field? Is this a good definition?

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