e=mc2 = wrong? Photon mass = 0 yet has energy? Need help Please., page 3

reply to post by lordtyp0

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I think your explanation 2 males sense up to the point you start discussing photons.

Regarding your explanation 1, why not just say E=mc^2 describes how much energy you get out of a mass when you convert mass to energy?
The Sun converts 564 million tons of hydrogen into 560 million tons of helium every second. Where does the missing 4 million tons of mass go? that's released by the sun as energy. that's what e=mc^2 is all about. I think your photon comment has little to do with the equation e=mc^2 because the formula has mass in it and as you correctly point out, the photon is massless.

I would say that for a massless particle, the formula e=mc^2 is not wrong, it's just not applicable, since there's no mass. So you have to look for other formulas to describe the photon. And mass isn't the only form that energy can take, that's another reason why the formula simply doesn't apply to massless particles.

[edit on 4-11-2009 by Arbitrageur]

reply to post by Arbitrageur

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This is going to sound strange, but 0 is still a value in an equation. When applied to zero mass it essentially any amount of energy will accelerate the particle to maximum velocity (c). The inverse of that is obviously: as mass increases it requires more energy to affect it.


That is basically what I was trying to convey, but yes, it is a clunky way of doing it, feel like I am babbling right now as well. Hope it makes sense.

Originally posted by lordtyp0
reply to post by Arbitrageur

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This is going to sound strange, but 0 is still a value in an equation. When applied to zero mass it essentially any amount of energy will accelerate the particle to maximum velocity (c).

You are correct about zero being a valid value in an equation. You are incorrect about the mathematics of how the equation works when zero is inserted.

substitute zero for m in the equation E=mc^2 and you get E=(zero) x c^2

which is correct because it says the amount of energy you can extract from zero mass, is zero.

The equation says nothing about the maximum velocity of a photon, how much energy it takes to accelerate the photon, or anything else you have attributed to the equation as it relates to a photon. The only thing it implies about a photon, is how much mass you could possibly get if you were somehow able to convert the energy of the photon into mass, by re-arranging the equation into the form m=E/(c^2)

Like I said the e=mc^2 is is otherwise not applicable to photons, but the applicable formula for photons is instead E = pc :

Photons

In empty space, the photon moves at c (the speed of light) and its energy and momentum are related by E = pc, where p is the magnitude of the momentum vector p.

The energy and momentum of a photon depend only on its frequency (ν) or equivalently, its wavelength (λ):

And you'll have to see the source for the rest of the equations that apply to the photon because I'm too lazy to copy them here for you.

Lastly to further prove you're barking up the wrong tree trying to relate the E=mc^2 equation to the photon's travel at the speed of light, note that the term in the equation other than energy and mass is NOT the speed of light, it's the speed of light squared! So using your arguments why wouldn't imparting energy to the photon make it travel at the speed of light squared? Obviously it doesn't but that's just further proof it's the wrong equation for what you are trying to do.

reply to post by TheRedneck

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Thanks for the link. I am going thru it now. Does it not make your head hurt?

reply to post by lordtyp0

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From a beginner ATS member, I thought your explanation was thorough. I have also been looking at Nassim Haramein's Grand Unified Theory videos on YouTube.

www.youtube.com...

I think there are about 45 videos on Nassim. I have watched about half of them so far. FASCINATING Stuff. He covers Einstein's theory, quantum physics, sacred geometry...very comprehensive!

Photon energy comes from the frequency.
Frequency of what.
The ether.
The ether has been shown to exhibit pressure waves, see Tesla
"Experiments...", and thus can interact with matter.
We can't talk about such things as the apple cart will be upset.
Ed:
en.wikipedia.org...
E^{2} = p^{2} c^{2} + m^{2} c^{4}.
but see the following lines:

E = hbar omega = h nu = h c \ lambda

p = hbar k,

but frequency is c/lambda so E = h f

So ether frequency gives you the energy for the UFO.
All done by Tesla long ago.
ED+: A little check of the momentum,p, formula you
see how instant velocity is obtained by the UFO.
So with 2 p's, up and sideways you have the UFO propulsion.

[edit on 11/6/2009 by TeslaandLyne]

[edit on 11/6/2009 by TeslaandLyne]

Looks like I was thinking of the relativistic version of F = MA. Probably just associated due to the relativity function. I was thinking since e=0 then anything >0 would propel to max velocity.

The excerpt below is from: math.ucr.edu...
-----------
The concept of relativistic mass is neatly encapsulated in the expression F = d(mv)/dt, where m is relativistic mass. This says that an impulse F dt causes an infinitesimal increase in a body's relativistic momentum mv.

Besides this definition and use of relativistic mass, we wish here to write down the relativistic version of Newton's second law, F = ma. In Newton's mechanics, this equation relates vectors F and a via the mass m of the object being accelerated, which is invariant in Newton's theory. Because m is just a number, in Newton's theory the force on an object is always parallel to the resulting acceleration.

The corresponding equation in special relativity is a little more complicated. It turns out that the force F is not always parallel to the acceleration a. We can express this fact using matrix notation. Let m be the rest mass, and v be the velocity as a column vector, whose entries are expressed as fractions of c and whose magnitude v is the speed as a fraction of c. Let vt be the velocity as a row vector, and let 1 be the 3 × 3 identity matrix. As usual, set γ = (1 – v2) –1/2. The relativistic version of F = ma turns out to be

F = (1 + γ2 v vt) γ m a

and

a = (1 – v vt) F
——————————
γ m

So defining mass via force and acceleration isn't as simple as it was for Newton (although it is simple, in principle, to define the mass as relating impulse and momentum increase, as mentioned a few lines up). Nevertheless, the three components of the two expressions above share a factor of γ m, and the rest mass m only ever appears in both expressions accompanied by γ. The acceleration is not necessarily parallel to the force that produced it, and it's not hard to see from the above equations that it's easier to accelerate a mass sideways to its motion than it is to accelerate it in the direction of its motion. This is how relativity reproduces Lorentz's original concepts of longitudinal and transverse masses; they are actually contained in these equations. The directional dependence that the newtonian meaning of mass has now taken on is neatly contained in the matrices 1 + γ2 v vt and 1 – v vt, and the remaining factor γ m is the relativistic mass. Taking our cue from the equations like this, to isolate quantities that might prove useful, is a powerful tool in mathematical physics.
reply to post by Arbitrageur

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reply to post by Arbitrageur

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Teslaandlyne, the frequency that determines the energy of the photon is the photon's frequency. If it's a photon for red light, it will have less energy than a photon for blue light since blue light has a higher frequency and is therefore more energetic than red light. That is independent of the frequency of the ether if there is such a thing, (which Einstein said there isn't).

Sorry about that.
I got old school, Tesla's 'medium', into an explanation about the
new school of Einstein. Tesla had a short discussion about light
before the photon decree of Einstein perhaps as Tesla's interest
was making light in air, gases and in the medium of a highly
evacuated bulb. Tesla use vacuum equipment to take everything
out of a bulb so there was nothing left. The nothing lit the bulb.
So naturally Tesla says something is left inside to light the bulb.

Tesla continued to push the nothing, or medium, with high pressure
coils (low power oscillations that grew higher and higher pressure)
to transmit electrical pressure and force in some cases.
An electric induction process not magnetic in any way.
So far what can deduce from the Tesla research controversy.
So the photon analogy seems to fit in some way.

reply to post by DaRAGE

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Hi, allow me to give you a conceptual image of a personal theory I have that might explain this.

C is a universal constant for the speed of light.

In the beginning, the universe was a singularity before the big bang, but then it exploded, and began to expand much like a balloon. However, to understand the physical dynamic of this balloon you almost have to think 4th dimensionally, but I will explain it like this. The skin of the balloon is the substance of space in which the universe exists. The skin is the Present. Everything inside the balloon is the past and everything outside the balloon is the future. if I draw a bunch of dots on the balloon and call it matter (Stars etc), you will notice that they move away from each other as the balloon expands. Now think of mass as one of those dots where i'm inside the balloon pinching the skin from the inside so that it lags behind in expansion with the rest of the balloon, forming a dimple. The more it lags the more massive it gets, until it stops expanding all together and becomes stuck in the past. This is what a black hole is. The skin of the balloon itself is light or C. If mass could travel at the rate of expansion of the rest of the balloon, it would no longer be mass, it would be at the border between mass and energy, where they are interchangeable. If I then push past this barrier, so that the dimple of skin is now outward and not inward, the dimple no longer is light but at the point where it breaks out of the skins plain (MC2) it becomes energy. We only experience in skips, which we see as waves or the peaks of waves. (If i took a string and stetched it it breaks into a segments, if i mush it together it becomes more dense/massive. One is energy the other matter). So light is basically that middle point where it is both energy and matter at the same time. It has no mass because it is at the speed of light or C. Hope that helps.

Photons have mass. JAXA is currently sailing a spaceship to Venus using light pressure as its only propulsion. Only massive objects exert pressure.

Photons only have zero mass at rest. But photons are never at rest.* If they stop moving they become some other form of energy. So photons have mass all the time.

The mass of a photon is found by dividing its energy by c^2.

The point here is that things increase in mass as they move faster and faster.

I'm sure this has all been pointed out earlier in the thread; I just thought a summing-up might be useful.

What is the mass of a photon

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*Yeah, I know that physicists have 'stopped light in its tracks', according to the popular media. But they haven't, really; that's just an oversimplification for consumers.

[edit on 22/6/10 by Astyanax]

Originally posted by Astyanax
Photons only have zero mass at rest. But photons are never at rest.*
What is the mass of a photon

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Well, that's what I thought too. But according to your source:

It is almost certainly impossible to do any experiment that would establish the photon rest mass to be exactly zero. The best we can hope to do is place limits on it. A non-zero rest mass would introduce a small damping factor in the inverse square Coulomb law of electrostatic forces. That means the electrostatic force would be weaker over very large distances.

Likewise, the behavior of static magnetic fields would be modified. An upper limit to the photon mass can be inferred through satellite measurements of planetary magnetic fields.

I guess I don't know as much as they do about Coulomb's law like how Coulomb's law can be used to place an upper limit on the rest mass of a photon. I read their explanation and re-read all the Coulombs law math on wiki and I'm not sure how they are getting this "dampening field" from the rest mass of a photon.

It seems easier to just say what you said, that the photon is never at rest so that's why it doesn't have a rest mass, I don't see why they couldn't just use that explanation.

Actually this was something I thought I understood but reading your link has confused me

Originally posted by Arbitrageur
I guess I don't know as much as they do about Coulomb's law like how Coulomb's law can be used to place an upper limit on the rest mass of a photon. I read their explanation and re-read all the Coulombs law math on wiki and I'm not sure how they are getting this "dampening field" from the rest mass of a photon.

Let me ask you this: if a photon has no charge, and zero (rest) mass, how can anyone tell its there?

The evidence of photon existence comes for its effect in ovserbable and measurable physical magnitudes. In the standard-model (the most accurate description of the nature currently known, yet incomplete), the photon is seen as the mediator of the electromagnetic interaction. This is to say that interaction between charged particles happens because they are interchanging photons.

When the electrostatic force is calculated using the standard-model point of view, the Coulomb's law as we know it comes up. But this only happens if the rest mass of the photon, being the mediator of the electromagnetic interaction interaction, is exactly zero. If a non-zero rest mass of the photon would exist, the "strength of the interaction" would be much, much lower, and charged particles wouldn't be aware of other charged particles unless they were very, very close.

Summarizing, Coulomb's law can be regarded as a consecuence of a more general theory, the Standard-Model, in which the value of the photon's rest mass is zero. The verification of Coulomb's law, therefore, supports this value of zero; and if it really were not zero, deviations of the law would help to determine the true value.

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