# In E=mc^2, why do we square c?

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#### Thebox

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##### In E=mc^2, why do we square c?
« on: 24/06/2016 11:29:07 »
Today for some reason I understood E=mc², however after considering this I do not ''see'' why we square c.

Should this equation not be

E=$$\frac{hf}{S}$$*c?
« Last Edit: 24/06/2016 16:37:28 by chris »

#### PmbPhy

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##### Re: In E=mc² , why do we square c?
« Reply #1 on: 24/06/2016 12:55:12 »
Quote from: PmbPhy
It appears that you're using the equation for the energy of a photon in terms of its wavelength. In that case the correct relationship is

E = hf

If you want the relativistic mass of the photon the start with the relationship between the photons energy and its momentum which is E = pc. The relation between its momentum and speed is p = mc. Solve for m and you get E = mc2.
I was in a hurry when I wrote that so let me finish it by working out the details.

If E is the energy of a photon and f is the frequency of the same photon then the energy is related to its frequency by E = hf. If p is the momentum of the photon and m is the photons relativistic mass then p = mc. The relationship between a photons energy and its momentum is E = pc. If we substitute p = mc into E = mc we get E = (mc)c = mc2.

If the object is not a photon then the same relation holds true. The derivation is too long to put into this post. However I wrote up a derivation and placed it on my website for just this kind of reason. See:
http://www.newenglandphysics.org/physics_world/sr/mass_energy_equiv.htm

Quote from: Thebox
Yes I know you think E=hf but in reality PE=hf   the ''energy'' is not ''created'' until the interaction with ''matter''.  It remains potential ''energy'' until interaction.
Leave it to you to make such erroneous claims. Its for this exact reason that I stopped trying to help you. Not only do you not know what you're talking about but you've never been willing to learn from others who know these subjects so much better than you do. In this case it's been well-established that E = hf is a valid relationship and E is the kinetic energy of the photon and not its potential energy. E = hf is defined as the total energy of a photon in an inertial frame of reference and as such the gravitational potential energy of the photon is zero leaving only kinetic energy. It's impossible for a photon to have any electric potential energy U because U is proportional to the charge of the particle in question. Since photons have zero charge that leaves U = 0 which leaves E = K = hf. Anybody who knows quantum mechanic will tell you that.

I've deleted the rest of your garbage and adding you to my ignore list with the other members here who don't have the ability to either learn physics or play nice.

How do all of you folks tolerate this person?
« Last Edit: 24/06/2016 14:34:27 by PmbPhy »

#### Thebox

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##### Re: In E=mc² , why do we square c?
« Reply #2 on: 24/06/2016 13:17:01 »
Today for some reason I understood E=mc², however after considering this I do not ''see'' why we square c.

Should this equation not be

E=$$\frac{hf}{S}$$*c?
It appears that you're using the equation for the energy of a photon in terms of its wavelength. In that case the correct relationship is

E = hf

If you want the relativistic mass of the photon the start with the relationship between the photons energy and its momentum which is E = pc. The relation between its momentum and speed is p = mc. Solve for m and you get E = mc2

Yes I know you think E=hf but in reality PE=hf   the ''energy'' is not ''created'' until the interaction with ''matter''.  It remains potential ''energy'' until interaction.

hf is absorbed by ''matter'' to create ''energy'' when two or more ''photons'' occupy the same space. Momentum of a ''Photon'' surely does not matter?

If a Photon was travelling at 1mph , it will still be absorbed and converted into ''energy'' regardless of speed.

So E must equal PE (hf) /  S because the laws of thermodynamics allow ''things'' to share entropy hf gain?

added- yes ''compressed'' light , wavelength is ''compressed'' in matter to create more E.

« Last Edit: 24/06/2016 13:22:48 by Thebox »

#### timey

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##### Re: In E=mc² , why do we square c?
« Reply #3 on: 24/06/2016 14:11:14 »
Why do we square c?

Good question Box!!!

A good question because... squaring c is equal to the proportions of a second squared, and the acceleration of g is given in meters per second squared...
Particles are very helpful, they lend themselves to everything...

#### evan_au

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##### Re: In E=mc^2, why do we square c?
« Reply #4 on: 25/06/2016 12:56:15 »
Quote from: TheBox
If a Photon was travelling at 1mph
But a photon cannot travel at 1 mph in a vacuum - it must always travel at 300,000 km/s when you measure it in a vacuum in your lab.

#### Thebox

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##### Re: In E=mc^2, why do we square c?
« Reply #5 on: 25/06/2016 15:32:32 »
Quote from: TheBox
If a Photon was travelling at 1mph
But a photon cannot travel at 1 mph in a vacuum - it must always travel at 300,000 km/s when you measure it in a vacuum in your lab.

That is why I put the word if, momentum has no part to play in the absorption of ''energy'', momentum through a ''medium'' has a part to play in ''energy'' by the compression of ''light'' and more ''Photons'' occupying per parts of the volume/entropy of the ''medium''.

I objectively ''see'' that c is the rate of arrival of ''photons'' at the ''their'' destination, I do not ''see'' why c is being squared or the relationship.

hf arrives and is absorbed or a percent is absorbed, then the ''energy'' build up inside the ''medium'' builds up by ''stored'' Photons that is greater than the ''mediums'' entropy loss rate.

+E=W=+q=+ve=$$\frac{>hf*c}{S}$$ where W is work ?

-E=W=-q=-ve=$$\frac{<hf*c}{S}$$?

« Last Edit: 25/06/2016 15:36:37 by Thebox »

#### jeffreyH

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##### Re: In E=mc^2, why do we square c?
« Reply #6 on: 25/06/2016 19:04:21 »
The definition of momentum is mass times velocity (mv). If we replace v with c (mc) then we have defined an upper bound on momentum. Acceleration can replace c to ma which signifies force. It is a simple step to multiply by a distance over which the force operates to get the energy (mc^2). I could have broken it down into simpler steps but didn't.

#### PmbPhy

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##### Re: In E=mc^2, why do we square c?
« Reply #7 on: 26/06/2016 03:23:37 »
Thebox
Quote from: jeffreyH
The definition of momentum is mass times velocity (mv). If we replace v with c (mc) then we have defined an upper bound on momentum.
I'm not clear on what you mean here, Jeff. There is no upper bound on the momentum of photons. While the velocity does have an upper bound the mass does not. Therefore the momentum doesn't have an upper bound.

Quote from: jeffreyH
Acceleration can replace c to ma which signifies force.
Special relativity deals only with inertial frames of reference. As such a photon cannot accelerate. Is that what you meant by an upper bound for momentum. i.e. that when the photon is subject to a gravitational force it will accelerate? If so then there are instances when there is no upper bound on the velocity of light. A good example is when the photon is moving higher and higher in a uniformly accelerating frame of reference. The velocity increases without bound.

Quote from: jeffreyH
It is a simple step to multiply by a distance over which the force operates to get the energy (mc^2). I could have broken it down into simpler steps but didn't.
It's not that simple in GR. You can't simply multiply the distance times the force to get E = mc^2. You're trying to use your Newtonian intuition to solve a problem in general relativity.

#### jeffreyH

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##### Re: In E=mc^2, why do we square c?
« Reply #8 on: 26/06/2016 10:39:57 »
I am explaining something to Thebox remember.

#### Thebox

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##### Re: In E=mc^2, why do we square c?
« Reply #9 on: 26/06/2016 11:15:20 »
The definition of momentum is mass times velocity (mv). If we replace v with c (mc) then we have defined an upper bound on momentum. Acceleration can replace c to ma which signifies force. It is a simple step to multiply by a distance over which the force operates to get the energy (mc^2). I could have broken it down into simpler steps but didn't.

I have no idea of what that means Jeff, I have no idea why speed is involved either, I dont see eye to eye on this obviously.

I think maybe I need to understand what energy light has while passing through space, I know there is 0q, so I do not consider it is ''active'' energy but rather potential energy  ''reactant'' that is converted into ''energy'',

For purpose of conversation, could we imagine ''light'' to be like a chemical and then there is a ''chemical'' reaction that creates ''energy'' when the ''chemical'' of light interacts with the ''chemical'' of ''matter''?

So what element in ''matter'' causes the ''chemical'' reaction to create ''energy'' when the ''light'' ''chemical'' interacts?

PE(c)=$$\frac{hf}{S}$$ delta ?

« Last Edit: 26/06/2016 11:22:40 by Thebox »

#### jeffreyH

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##### Re: In E=mc^2, why do we square c?
« Reply #10 on: 26/06/2016 12:53:34 »
OK forget the above. You are over-thinking this. Instead of mass lets call it physical 'stuff'. If you want to move that stuff in a particular direction at a particular speed then you have to apply a force. This is what you understand as F = ma. The acceleration of stuff is moving it from being stationary to some non stationary speed. If the force is maintained then that speed continues to increase so we have a continued acceleration. If the force is constant then the acceleration of the stuff is constant. It gets faster and faster. So in order to apply this force we need to add energy. Just as we expend energy when push starting a car. Now acceleration is stuff times distance divided by time squared. If we have a constant speed then there can be no acceleration of our stuff since the speed isn't changing. Therefore a non accelerating amount of stuff moves via stuff times velocity. Velocity is simply speed in a particular direction. So coming back to mass instead of stuff we have p = mv where p is the momentum. Momentum can be imagined as constant speed with no force applied. We stop pushing the car and it keeps moving on its own.
« Last Edit: 26/06/2016 12:58:56 by jeffreyH »

#### PmbPhy

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##### Re: In E=mc^2, why do we square c?
« Reply #11 on: 26/06/2016 13:45:14 »
I am explaining something to Thebox remember.
I know. However when you multiply force times distance in Newtonian mechanics you get kinetic energy, not total inertial energy. So force times distances to get mc^2 is wrong no matter how you think of it. Let's not make the box worse than necessary.

#### Thebox

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##### Re: In E=mc^2, why do we square c?
« Reply #12 on: 26/06/2016 18:28:50 »
OK forget the above. You are over-thinking this. Instead of mass lets call it physical 'stuff'. If you want to move that stuff in a particular direction at a particular speed then you have to apply a force. This is what you understand as F = ma. The acceleration of stuff is moving it from being stationary to some non stationary speed. If the force is maintained then that speed continues to increase so we have a continued acceleration. If the force is constant then the acceleration of the stuff is constant. It gets faster and faster. So in order to apply this force we need to add energy. Just as we expend energy when push starting a car. Now acceleration is stuff times distance divided by time squared. If we have a constant speed then there can be no acceleration of our stuff since the speed isn't changing. Therefore a non accelerating amount of stuff moves via stuff times velocity. Velocity is simply speed in a particular direction. So coming back to mass instead of stuff we have p = mv where p is the momentum. Momentum can be imagined as constant speed with no force applied. We stop pushing the car and it keeps moving on its own.

Force , acceleration, constant speed, mv, none of that relates to E=? ,  I have no idea why you keep mentioning momentum, a ''stationary'' object can gain ''energy'' and only involves the entropy of the object absorbing ''light'' without any complication of momentum which is seemingly not needed?

« Last Edit: 26/06/2016 18:33:19 by Thebox »

#### evan_au

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##### Re: In E=mc^2, why do we square c?
« Reply #13 on: 26/06/2016 22:56:36 »
Quote from: TheBox
a ''stationary'' object can gain ''energy'' and only involves the entropy of the object absorbing ''light'' without any complication of momentum which is seemingly not needed?
When an object absorbs light and heats up, this energy goes into increasing the velocity of the constituent atoms and molecules. This increases both kinetic energy and momentum of the particles.

Because these velocities are in random directions, the object as a whole does not move (it just expands a bit with temperature). This increase in temperature has an impact on Entropy.

When General Relativity talks about momentum and kinetic energy, it is usually talking about motion of the object as a whole.

As I understand it, GR prefers to talk about momentum, because this can be compared more easily in different frames of reference than kinetic energy.

#### PmbPhy

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##### Re: In E=mc^2, why do we square c?
« Reply #14 on: 26/06/2016 23:19:30 »
There are instances where E != mc2. To explain this I'll have to use math. I'm sorry but I know of no way around it. That it does require math should be obvious since we're talking about mathematical quantities.

E != mc2 in such cases as a charged particle moving in an electromagnetic field or a non-charged particle moving in a static gravitational field.

Mass is defined as the quantity m such that p = mv and as such is proportional to the time component of the particles 4-momentum. The time component is not conserved under such conditions and is not proportional to the inertial energy of the particle. However the time component of the corresponding 1-form is conserved and equals the energy.

I came to realize this when I was studying the concept of mass in relativity. I discussed it in the paper I wrote on the subject. It turned out that almost none of the experts that I know personally and speak to have never heard of this. However I later came to find that it was published in the article On the concept of energy in classical relativistic physics by Ludger Hannibal. Eur. J. Phys., 121(1991) 283-285. This paper can be downloaded from: http://booksc.org/book/45962156

Quote
The relativistic concept of energy is discussed under two important aspects: the behaviour of the energy under coordinate transformations and its general definition in the presence of external gravitational and electromagnetic fields. On the basis of energy conservation and a nonrelativistic approximation it is argued that only the zeroth component of the covariant 'generalized' momentum should be called 'energy'.
« Last Edit: 26/06/2016 23:25:36 by PmbPhy »

#### Thebox

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##### Re: In E=mc^2, why do we square c?
« Reply #15 on: 27/06/2016 01:22:27 »

When an object absorbs light and heats up,

Sorry that sounds incomplete, should it not say when the object absorbs light and the light is converted into +q it creates work that causes vibration and an eventual likewise charge repel and expansion?

If not can you please explain the light to energy conversion process?

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this energy goes into increasing the velocity of the constituent atoms and molecules. This increases both kinetic energy and momentum of the particles.

Kinetic energy? momentum?   In an inertial reference frame on a stationary object?  or do you mean molecule displacement?

Quote
Because these velocities are in random directions, the object as a whole does not move (it just expands a bit with temperature). This increase in temperature has an impact on Entropy.

Yes ok , sorry you answered my above, I did not read this part yet.

$$\frac{>hf*c}{S}$$=>kE=>+q=>W=>+ve=>4/3 pi r³?

Quote
When General Relativity talks about momentum and kinetic energy, it is usually talking about motion of the object as a whole.

Ok