photon frequency?

[ydoaPs]

do photons have mass? i always assumed that since they have no rest mass, they have no mass. i wanted to find out so i did a little math.

 

i took [math]E=hf[/math] and [math]E^2=m^2c^4+p^2c^4[/math] and put them togather. i got [math]m=\frac{hf}{{c^2}{(1+c^2)^{\frac{1}{2}}}}[/math]. when i solved, i found that the mass was zero. i took a closer look and saw that that wasn't true, for if i put 0 in for m, the photon would have 0 energy.

 

then i decided to put in a stupidly high value for the frequency and i found it DID have mass. it is just so close to 0 at normal frequencys that you don't need to worry about it.

 

i decided to pick a mass and find the frequency. i chose 1 nanogram. i got 4.07x10^46hz. that brings me to my question.

 

the highest frequencys i know of are terahertz rays. is there an upper limit on photon frequency?

[timo]

a) E² = (mc²)² + (pc)², not (mc²)²+(pc²)²

b) also it´s [math] E = \hbar \omega [/math] or E = hf

c) plus: Your math is wrong. The minus sign must be in the numerator and I can´t see where the momentum went. 1-c² doens´t make much sense, either.

 

No, there´s no upper limit on photon frequency. Extremely high energies will lead to some QFT effects like constant creation of particle-antiparticle effects, though.

[fuhrerkeebs]

You solved your equation wrong. Working backwards with your equation you get E²=m²c⁴-c²m², which isn't right. Where did your momentum go?

[ydoaPs]

oops, hit the wrong button,the - was supposed to be a +. i fixed it. i changed the momentum to mv and the v to c

 

b) also it´s [math]E=\hbar\omega[/math'] or E = hf

 

f and [math]\omega[/math] are the same thing; h and [math]\hbar[/math] aren't.

[timo]

If h and hbar aren´t the same thing and related by a factor of 2pi what might that say for f and omega ?

[ydoaPs]

f and omega both represent frequency

[timo]

Indeed, that´s why they are only different in the dimensionless constant 2pi

 

More helpfull: A photon´s momentum is not m*v (at least not if m is supposed to be the rest mass). The momentum of a photon is [math]p = \hbar k [/math] or [math] |p| = E/c = \hbar \omega / c [/math].

[ydoaPs]

ok different momentum formula gives [math]m=\frac{\sqrt{hf(1-c)}}{c^2}[/math]

[timo]

It´s getting warmer. If you now consider point a) in my 1st post and check your equation again (the c³ was a typo, wasn´t it?) you´ll come up with m=0.

 

EDIT: Which is not much of a surprise since |p| = E/c implies m=0.

[ydoaPs]

oops didn't see that. are you sure about that? i have always seen it m^2c^4+p^2c^4. i'll check when i get home.

 

edit:even in the updated way, the photons will still have mass.

[timo]

I am absolutely sure. Check the units if you don´t believe it. That´s also what my statement "1-c² doesn´t make much sense" meant: you can´t substract meters/second (with any nonzero exponent) from a number.

[ydoaPs]

ok i fixed it in post 8. is it right yet?

[timo]

no, let´s just do the math:

E² = (mc²)² + (pc)²

=> (mc²)² = E² - (pc)²

with |p| = E/c

=> (mc²)² = E² - (E/c * c)² = E² - E² = 0

=> m = 0

[ydoaPs]

wow, i guess this thread was pointless.

[5614]

you know there's all different masses; inertial mass, relativistic mass, rest mass... dont photons have one of those masses (ie. it isnt zero for one of them), I know it's not the rest mass, but which one is it and how do you calculate it?

[swansont]

f and omega both represent frequency

 

f represents linear requency, and ^{[math]\omega [/math]}represents angular frequency. They are different by 2^{[math]\pi [/math]}, and the energy equations imply

[timo]

It´s the relativistic mass that´s nonzero (might also have other names). The relativistic mass is simply the energy divided by c² - or in other words: E = mc².

[ydoaPs]

i went back and assumed nothing about mass. it reduced to [math]h^2f^2=2m^2c^4[/math] i solved for m and got [math]m=\frac{hf\sqrt{2}}{2c^2}[/math] once again, REALLY close to 0, but not quite.

[timo]

Where does the hf = 2(mc²)² come from?

 

EDIT: And please, for the future: Add the steps that led you to a conclusion. You cannot expect any better answer than "you are wrong" when you don´t.

[Johnny5]

More helpfull: A photon´s momentum is not m*v (at least not if m is supposed to be the rest mass). The momentum of a photon is [math]p = \hbar k [/math] or [math] |p| = E/c = \hbar \omega / c [/math].

 

Why can't the momentum of a photon be mv?

[5614]

Because it's mass is 0 so it's momentum would therefore also be zero.

 

E^2 = (pc)^2 + (mc^2)^2

 

and it's mass and momentum were both 0 then it'd have 0 energy which is wrong.

 

It's also to do with momentum in relativity and the different momentum in classical mechanics. Relativity's momentum is more general form and it can be carried by massless particles.

[Johnny5]

Because it's mass is 0 so it's momentum would therefore also be zero.

 

E^2 = (pc)^2 + (mc^2)^2

 

and it's mass and momentum were both 0 then it'd have 0 energy which is wrong.

 

It's also to do with momentum in relativity and the different momentum in classical mechanics. Relativity's momentum is more general form and it can be carried by massless particles.

 

I thought so. Well what is the mass of a photon in a frame of reference in which it's center of mass is at rest? Because that is a frame where its speed is zero. And in a frame where it's speed is zero, instead of c, the photon momentum mv=0, while P=E/v is infinite, since v=0. But the photon could still have a nonzero rest frequency in this frame, in which case its energy hf, would be nonzero. So I am still confused by the statement that mv doesn't apply to photons.

 

Let me run through your argument quickly.

 

For any object of rest mass m, and momentum p, the following equation is a true statement which is frame dependent:

 

EE=ppcc + mccmcc

 

Where E is the total energy of the object in a reference frame in which the center of inertia of the object has speed v, where:

 

p = Mv

 

where

 

M = m (1-v^2/c^2)^-1/2

 

Then you hold that for a photon, m=0, from which it follows that for a photon:

 

EE=ppcc

 

From which it follows that for a photon:

 

E=pc

 

(where I have ignored E=-pc) From which it follows that for a photon:

 

E=mv (1-v^2/c^2)^-1/2

 

And in any inertial reference frame the speed v of a photon is c, so for a photon traveling in an IRF the following statement is true:

 

E=mc (1-c^2/c^2)^-1/2

 

From which it follows that:

 

E=mc (0)^-1/2

 

From which it follows that:

 

E=mc/0

 

Which is the division by zero error of algebra unless m=0, whence you must insist that for a photon in an inertial reference frame m=0, which you did do, but then the energy of a photon in an IRF is indeterminate i.e. E=0/0, whence you must resort to quantum physics to escape...

 

E = hf for a photon

[5614]

Well what is the mass of a photon in a frame of reference in which it's center of mass is at rest? Because that is a frame where its speed is zero.

I maintain that a photon's speed is c in all frames so I can't answer your question.

 

Which is the division by zero error of algebra unless m=0, whence you must insist that for a photon in an inertial reference frame m=0, which you did do, but then the energy of a photon in an IRF is indeterminate i.e. E=0/0, whence you must resort to quantum physics to escape...

 

E = hf for a photon

Yeah, I'd agree with that

 

====

 

So now I'm thinking how can you calculate the momentum of a photon?

(It must be possible, after all a photon has energy and for a photon ee=ppcc so p must be more than 0)

[Johnny5]

I maintain that a photon's speed is c in all frames so I can't answer your question.

 

 

Do you literally mean all frames that I can conceive of, or do you mean all inertial reference frames, or do you mean all frames be they inertial or non-inertial frames?

[Johnny5]

 

So now I'm thinking how can you calculate the momentum of a photon?

(It must be possible' date=' after all a photon has energy and for a photon ee=ppcc so p must be more than 0)[/i']

 

Well you go with quantum physics which says:

 

p = h/l

 

But I am still held up by the assertion that p=mv does not hold for photons.