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## Main Question or Discussion Point

i think E=mc^2

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

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

- Thread starter DonnieD
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- #1

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i think E=mc^2

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

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

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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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but the equation [itex]E=m c^2[/itex] works perfectly well for photons when[itex]E=m_0 c^2[/itex] does not apply to photons.

[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 withThe general equation is

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

[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 isi think E=mc^2

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

Pete

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That is quite untrue. E = mc..., 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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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?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.

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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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Right. For photons, just stick to energy and momentum.The photon is amasslessparticle.

Relativistic massis a frowned upon concept.

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

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

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

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: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.

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

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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?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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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.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.

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, inSure. 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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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 toThe photon is amasslessparticle.

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?

depends on whose face. not everybody is frowning.Relativistic massis a frowned upon concept.

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

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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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.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

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

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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.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?

Yes, I know, there are a few members of this forum that cling to the notion of relativistic mass.depends on whose face. not everybody is frowning.

You certainly not talking abouti'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.

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

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.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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Glad to see that you concur!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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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.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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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?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.

As to "photons turning into mass", this will NOT happen in your perfectly mirrored box. The photon absorbtion by

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