Photons kinetic energy

[Gilded]

Yeah, I've just had my first lessons in kinetic energy (Newtonian mechanics) at high school some time ago. However, quite soon I noticed that for a massless object going at c, the kinetic energy is 0. And so, I went "aha, so that's why you don't need to kick photons around to make them move". However, this made me think: how come photons have energy then? Is it only proportional to the wavelength, and if someone tells you the wavelength of a photon you can tell the energy in it right away? Or should I just go and use ONLY the relativistic mech. way to calculate Ek and get a correct result, as I have no idea how to do it. )

[Martin]

Yeah, I've just had my first lessons in kinetic energy (Newtonian mechanics) at high school some time ago. However, quite soon I noticed that for a massless object going at c, the kinetic energy is 0.

 

the highschool formula for kinetic energy is wrong

it gives the wrong answer

as literally thousands of experiments have shown

it just gives approx. right answers at slow speeds, but

the answer it gives gets wronger and wronger

as speed increases

 

so the moral is

don't believe one half em vee-square

 

there is no problem with photon, all a photon's energy is kinetic energy

 

you just have to use the right formulas (the ones that actually check out with careful measurement)

[Gilded]

Yeah, I knew it was merely an approximation but I wasn't sure how badly it fails in high speeds. Sooo... Perhaps someone could tell me how to calculate Ke in a relativistic way. ) I mean, I know the formula but how do you calculate an object's total energy?

[swansont]

As Martin said, 1/2 mv² is a purely classical equation, and the photon is not a classical particle.

 

A photon's energy is hf, where f is the frequency and h is Planck's constant.

[ed84c]

6.6 * 10 ^ -34

[Martin]

Hi gilded, when I said this

there is no problem with photon' date=' all a photon's energy is kinetic energy

[/quote']

 

it was not careful, and could be misleading. what I should say is that

a photon has only its total energy

E = hf

and there is no way to divide that up between mass energy and kinetic energy----one cannot meaningfully distinguish kinetic energy as separate component

 

there is no formula involving the (rest) mass because a photon has zero (rest) mass

so, for example, when one calculates stuff involving the photon MOMENTUM one says that the momentum is E/c

and this gives the right answers, for example in calculating the pressure of light on a mirror or a solar sail, caused by the momentum of photons bouncing from the reflector.

 

You and I probably agree about the concept of mass (there is some controversy about the word itself but the majority probably think like us): the only meaning of mass is the inertia of a body at rest----one does not need to say "rest" mass because this is already understood---and the idea of "relativistic" mass is not useful and should be discarded. In any case this is a semantic, not a physical, issue.

 

Then I think i can say that for light there simply is no K.E. formula, because that would involve m, and m = 0. But instead there are simply Energy formulas involving the total energy of the photon.

 

But since you are interested in K.E. let us look at the real K.E. formula, for something (not a photon) which can exist at rest and which therefore can have a mass.

 

the damn formula is complicated.

also SFN LaTex is broken now!!!

but anyway I will write some stuff out---I expect you already

are familiar with all or most of it---and I use the abbreviation beta = v/c

 

the total energy (including kinetic)

 

E = mc^2/sqrt(1 - beta^2)

 

the kinetic part, with the rest energy subtracted,

 

K.E. = [mc^2/sqrt(1 - beta^2)] - mc^2

 

= [1/sqrt(1 - beta^2) - 1]mc^2

 

Now it is this thing which, for low beta, is approximately equal to

(1/2)beta^2

and if we throw away the correct thing between the [] brackets and

put in the approximation then we get something that boils down to

the highschool formula---we get

 

[(1/2)beta^2]mc^2

 

and if you put in v/c for beta then the c cancels and you just get

(1/2)m v^2

 

so that is just the approximation, what we really want to look at is

the correct thing:

[1/sqrt(1 - beta^2) - 1]

 

and we want to ask how close this thing comes to it

 

[(1/2)beta^2]

 

that is the same as asking

 

How close does [1 + (1/2)beta^2] come to [1/sqrt(1 - beta^2) ]?

 

(now the square brackets are redundant and can be thrown away)

 

if you explore this with a calculator you find that for small beta, like

beta = 0.001 or 0.01 they are so close the calculator can often not even tell the difference (because of its finite accuracy)

 

but already for beta = 0.1 (which means a speed that is a tenth of the speed of light) you can begin to see a difference.

 

So the mathematical intuition is about plotting two curves which are very nearly identical near zero and finding out

How close, over some range of betas, is 1 + (1/2)beta^2 coming to 1/sqrt(1 - beta^2) ?

 

If you have had Taylor series in school then you can also do some theory exercise to explain why the approximation is so good (near beta = 0)

but it is almost better to just experiment with a calculator and experience how close it is. but also that it goes bad for beta bigger than 0.1.

[Gilded]

OK, I think I got the most of it...

 

So... If a 1g particle is going at 0.5c (so I can use the 1/2 for v/c ) )...

The mass is in kg, right? And c is m/s?

 

[1/sqrt(1 - 1/2^2) - 1]0.001*300 000 000^2

[1/0,8660... - 1]0.001*300 000 000^2

0,15470...*0.001*300 000 000^2

= approx. 13923048454133

 

So the particle has about 13.9230 terajoules of kinetic energy. But is this the amount of energy needed to accelerate the particle to 1/2c? Wasn't there a different formula for that? Damn, these are getting complicated. )

 

With the 1/2mv^2 I got 22500 terajoules. Wtf. I think I can smell a miscalculation in the relativistic calculation...

[timo]

Didn´t bother to check your relativistic calculation but the classical one is flawed, at least:

 

0.5*m*v² = 0.5*(1 g)*(0.5 c)² = 0.5*(1 g)*0.25*c² = (1/8 g)*(3*10^8 m/s)²

= (1/8000 kg)*9*10^16 m²/s² = 9/8*10^13 J = 90/8 TJ

 

= 11.25 TJ < 13 TJ

[Gilded]

Ahh, so the screwup was in the classical one.

 

Apparently, I forgot the press the "=" button, resulting into having an answer equal to 150 000 000^2. ) Remember kids, maths and late hours are a bad combination.

[swansont]

OK' date=' I think I got the most of it...

 

So... If a 1g particle is going at 0.5c (so I can use the 1/2 for v/c ) )...

The mass is in kg, right? And c is m/s?

 

[1/sqrt(1 - 1/2^2) - 1']0.001*300 000 000^2

[1/0,8660... - 1]0.001*300 000 000^2

0,15470...*0.001*300 000 000^2

= approx. 13923048454133

 

[nit]You're carrying way too many significant digits[/nit]

[Gilded]

But digit carrying is so fun. The final approximation is rather silly though, but at least it doesn't have the decimal part. )

[Gilded]

Hey, by the way...

 

"= [1/sqrt(1 - beta^2) - 1]mc^2

 

Now it is this thing which, for low beta, is approximately equal to

(1/2)beta^2"

 

I just realized that I didn't quite get that part... OK I'll have a shot at it.

Mass = 1kg, velocity = 1m/s (for beta^2 I got about 1,111111111^-17)

 

Just to annoy swansont, eventually I got 0,50000000000004000277777777777822.

 

Ek = 0.5*1*1 = 0.5

 

So the approximate is wrong about 0,04 picojoules. Yay. )