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what is the relativistic mass of a single photon

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what is the relativistic mass of a single photon

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Try using mass-energy equivilence.eNathan said:what is the relativistic mass of a single photon

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

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How do I plug in those variables?

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Daniel.

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I said relavistic mass, not energy. Can someone give me an example? Or is that too much to ask?

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Daniel.

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russ_watters

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nothing is too much to ask. but you might want to accept it when people answer.eNathan said:I said relavistic mass, not energy. Can someone give me an example? Or is that too much to ask?

the

[tex] E = m c^2 = \hbar \omega[/tex]

resulting in

[tex] m = \frac{E}{c^2} = \frac{\hbar \omega}{c^2}[/tex]

that's it. the mass of a photon. because the speed of a photon in vacua is [itex] c [/itex]

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

where [itex] m_0 [/itex] is the rest mass and [itex] v [/itex] is the particle's velocity relative to some observer,

we can't spell it out any further for you, as best as i can tell.

r b-j

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As far as I know, units of speed are always in Meters/Second, so why are you using KM?300,000km

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Daniel.

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That's incorrect. relmass is not idetical to energy. This is equality under some specfc casesdextercioby said:Relativistic mass(for a photon,at least)ISenergy.In Heaviside-Lorentz units...

Daniel.

The relativistic mass ofa photo can be expressed in two ways one of which is m = p/c.

pete

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Daniel.

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How is m = p/c for a photon ??? isn't m divergent here ??? How do you renormalize this ???pmb_phy said:The relativistic mass ofa photo can be expressed in two ways one of which is m = p/c.

pete

What is the second expression ???

marlon

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thats a pointless equationmarlon said:How is m = p/c for a photon ??? isn't m divergent here ??? How do you renormalize this ???

What is the second expression ???

marlon

<p> = d/dt(m*x) = mv

m = mv/c, v=c, m=m

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this should explain a lot

the (invariant or rest-) mass of a photon is ZERO, c'est tout, point final

the relativistic mass depends on the frame of reference, but since all frames of reference are equivalent, only the invariant mass matters...

marlon

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No. I don't find it weird at all. If I see Rindler state this in no uncertain terms in his text then who am I to argue with Rindler?dextercioby said:

Daniel.

Besides, this is something that one can demonstrate rather easily by a simple example which I've stated. The only think weird is that nobody chose to actually look at the problem I stated and then find the energy and mass.

You'd agree that a photon has momentum, 'p', would you not? You'd also agree that the speed of a photon is c, do you not? Then the relation m = p/c is well defined.marlon said:How is m = p/c for a photon ??? isn't m divergent here ??? How do you renormalize this ???

The energy of a photon is related to the photons momentum, 'p', by E = pc --> p = E/c. Plug this into "m = p/c" and you'll get m = E/c^2 which is fairly obvious. The relation m = p/c is found in many texts on relativity. E.g. see French's text "Special Relativity" on page 16 where in a foot note French states

*By inertial mass we mean the ratio of linear momentum to velocity.

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http://lanl.arxiv.org/abs/gr-qc/9909014

A general argument is presented that for a closed physical system to which the virial theorem applies, the total "gravitational mass" of the system is going to be the total energy of the system, E - the sum of mc^2 for the constituent particles plus the total energy E

Carlip's "gravitational mass" appears to be identical to pmb's "relativistic mass", it's the mass that appears in the low velocity limit of p=mv or f=ma.

Therfore I think it is quite reasonable to say that relativistic mass is just the energy of a system. The only caveat is that the system has to be closed.

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I created a page a while back as an example. See

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

Pete

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russ_watters

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I use km because its a prettier looking number. Besides, "kilo" is just a prefix, meaning "thousand" - so km are still meters (dropping the "k" and adding 3 zeroes iseNathan said:Russ_Watters, you said

As far as I know, units of speed are always in Meters/Second, so why are you using KM?

And its not polite to be argumentative when asking a question and people are trying to help you.

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It appears to me that Carlip is not speaking of any cases in which the energy defined mass is different than the momentum defined mass so no problem arises in that paper that I can see from a quick read. I read it closely last year and have forgotten the details.pervect said:

Pete

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I'm not so sure your example yields an isotropic relationship between momentum and velocity, however your calculation wasn't very detailed so it's unclear what you think.

But I also think that the correct calculation of inertial/gravitational mass will be the same in all directions - that the relationship between system momentum and system velocity will be a scalar and not a tensor.

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I gave you anpervect said:Well, as I see it, what comes out of Carlips equations is an isotropic gravitational / inertial mass (which are the same because of the principle of equivalence) which is the same in all directions.

"..that simple formulais valid only in certain special cases, e.g. for single particles and for systems of freefree[..]. It isnotgenerally valid for constrained systems.

What part do you not understand. I'll elaborate for you. But the example is quite clear and is not lacking in details.I'm not so sure your example yields an isotropic relationship between momentum and velocity, however your calculation wasn't very detailed so it's unclear what you think.

As for Carlip? ... I'm not 100% sure of the connection but I'll take an educated crack at it: It can be shown that in the week field limit

[tex]\nabla^2 \Phi = \frac{4\pi G}{3}(\rho + 3p^/c^2) = \frac{4\pi G}{3}\rho_{grav }[/tex]

Notice the inertial mass from the result I obtained here, http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm, for

So add one for each dimension, i.e. 3, then you'll end up with the expression above for the week field limit.

But that's just a guess. Seems right though.

Pete

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"It can be shown that the momentum density as measured in S is given by [1].

"It can be shown that" is not a particularly strong derivation of a result, sorry :-).

But it does look like you are both getting the same results. Carlip gets

m_grav = (integral of) T_00 + T_11 + T_22 + T_33 (using geometric units where c=1).

You get m_grav = (integral of) rho + 3P (again, getting rid of the pesky factors of 'c' by using geometric units.). Assuming that the stress energy tensor being used is that of a perfect fulid, then T_00 = rho, and T_11 = T_22 = T_33 = P.

Looks the same to me.

Carlip goes on further to say some more interesting things about how the virial theorem sorts the whole mess out.

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I was addressing that one someone who knew the topic and is familiar with that stress-energy momentum tensor.pervect said:Well, an example of where your exposition could use some clarification is

"It can be shown that the momentum density as measured in S is given by [1].

"It can be shown that" is not a particularly strong derivation of a result, sorry :-).

m_grav = (integral of) T_00 + T_11 + T_22 + T_33 (using geometric units where c=1).

You get m_grav = (integral of) rho + 3P (again, getting rid of the pesky factors of 'c' by using geometric units.). Assuming that the stress energy tensor being used is that of a perfect fulid, then T_00 = rho, and T_11 = T_22 = T_33 = P.

Looks the same to me.

Carlip goes on further to say some more interesting things about how the virial theorem sorts the whole mess out.[/QUOTE]

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