Why c2 (speed of light squared)?

[Vorde]

Actually it is simple ... In E=mc2 , c2 doesn't actually represent speed of light squared ! c2 is the energy density for unit mass in kg . instead of taking the dimensions for c2 as m2/s2 , take it as joule/kg . And this energy density (9×1016Joule/kg is constant for any substance . Simply energy = mass × energy per unit mass !

And this comes from...where? I don't see how this works at all, most importantly how you get the idea that the energy density is a constant.

 

[Kaushik96]

You know what this relation means ? Mass and Energy are both the same ... So Mass and Energy are directly proportional . To match the SI units of both mass and energy , mass is multiplied with a constant c2 (energy density).

You ask where this energy comes from ? In the sun , two protons are squeezed to form a helium nucleus . In this process some of the mass of protons gets converted to energy . Striking a match also converts some mass (insignificant amount ) into energy. As I said earlier energy density is 9×10^16Joule/kg . In other words 1kg of a substance is 9×10^16Joules of energy.

Same way energy can also be converted into mass . But that's a lot difficult because trillions of joules of energy is needed to make a mass in the order of 10^-9 kg.

In different processes , different substances are converted to energy . The process in the sun I mentioned involves protons . Striking a match involves electrons . And that's how it works . I hope you understand .

 

[Vorde]

Ah, I see what you are saying. But you have a rather circular argument.

How are you possibly going to show that the constant of proportionality of $E \propto M$ is $9 \times 10^-9$ without invoking $E=mc^2$ first - something which falls out of the equations?

 

[Kaushik96]

Practically this is what the relation means . But i don't know how c^2(speed of light squared) is energy density . Give me some time . Ill check what's with this constant !

 

[Kaushik96]

Check out this link ! http://www.drphysics.com/syllabus/energy/energy.html . You may understand

 

[Dale]

But i don't know how c^2(speed of light squared) is energy density .

Technically the term is "specific energy", not "energy density". Energy density is the energy per unit volume (J/m³). Specific energy is the energy per unit mass (J/kg).

Personally, I think you are on a reasonable track here. If you have some proportionality between energy and mass, E=bm, then just by looking at the units you know that b needs to have units of J/kg=m²/s². The only combination of fundamental constants that has those units is c², so it has to be c².

Of course, that begs the question, why is it 1 c² instead of 5 c²? And why is it E=bm instead of E=bm²? To answer either of those you really need a full derivation.

 

[Kaushik96]

Yeah ... that's right ! :)

 

[Vorde]

Agreed, but I think its a matter of insight to say c^2. To say its the energy density in J/kg seems to me just to be saying the obvious because of the equation it is in, when you say c^2 you are saying something more meaningful about the universe.

 

[Kaushik96]

@Dalespam : the reason why "e" is not directly proportional to "m^2" is because as i said earlier energy is mass and mass is energy . They need to be proportional .So "e" must be proportional to "m" and not "m^2" . If it is "m^2" that will mean mass and energy are different .

 

[Vorde]

@Dalespam : the reason why "e" is not directly proportional to "m^2" is because as i said earlier energy is mass and mass is energy . They need to be proportional .So "e" must be proportional to "m" and not "m^2" . If it is "m^2" that will mean mass and energy are different .

But you are starting from the assumption they are directly proportional. As DaleSpam said, you can't get that assumption without a full derivation.

 

[Kaushik96]

Oh ! I see ...The full derivation of this equation is given in this link ! http://www.drphysics.com/syllabus/energy/energy.html

It will answer your questions !

Also this derivation is made with respect to light .
you asked why this specific energy should be constant for substances right ?

This relation only involves light matter and light energy !
If this relation e=mc^2 is true for light it applies to all matter because light matter is made of light ! so c^2(specific energy) will be the same for all light matter .

 

[Vorde]

This relation only involves light matter and light energy !

That is wrong.

E=mc^2 is valid for all things (on the super-quantum scale, you'll have to ask someone else if it applies equally well to quantum stuff). It is valid for you, for me and for a car. That website has an interesting explanation for the average reader, but as has been said a full derivation can be done with SR that is valid for any mass.

 

[Kaushik96]

Hey Hey ! Thats what I also meant ... Light matter is what you see everyday (ie) any object like a pencil , a bike or a car or even you and me , all are made up of light energy ! there are two types of energies (light energy and dark energy ) .

 

[DiracPool]

Oh ! I see ...The full derivation of this equation is given in this link ! http://www.drphysics.com/syllabus/energy/energy.html

Where did the Doc come up with this little nugget for velocity, v=E/(Mc). It looks as though he is using the definition of E=mc^2 to derive the same equation! Am I reading this wrong, or is this just a circular derivation that tells us nothing?

 

[Nugatory]

Where did the Doc come up with this little nugget for velocity, v=E/(Mc). It looks as though he is using the definition of E=mc^2 to derive the same equation! Am I reading this wrong, or is this just a circular derivation that tells us nothing?

It's not circular, it comes from conservation of momentum and the known (through observation) fact that radiation transfers momentum when it strikes a surface.

 

[Vorde]

there are two types of energies (light energy and dark energy ) .

I don't know where you are getting this from, but I'd be very careful.

There is something called dark energy, but I've never heard anyone call 'the rest of the universe' light energy so I'm thinking you might be referring to something more pseudo-scientific.

 

[DiracPool]

It's not circular, it comes from conservation of momentum and the known (through observation) fact that radiation transfers momentum when it strikes a surface.

Hmmm, OK, can you give me link where I can see this equation, v=E/(Mc), used somewhere by someone with a straight face who derived it independently from Einstein's derivation, and for some other purpose? I mean, all we have to do is set the velocity here to c and we have E=mc^2. I think Einstein had conservation of momentum and radiation transfer in mind when he derived it too. What I'm taking issue with is that it seems the DrPhysics is using E=mc^2 to derive E=mc^2. I don't see what novel insight he is bequeathing upon us here. Have I missed something?

 

[Kaushik96]

Initial momentum or the momentum with which the light hits the cylinder will be equal to TOTAL ENERGY OF THE LIGHT / SPEED OF LIGHT (E/c).

Final momentum or the momentum with which the cylinder moves after colliding with light is just THE MASS OF THE CYLINDER * VELOCITY WITH WHICH IT MOVES (Mv)

According to Momentum conservation principle Initial momentum will be equal to Final momentum .So E/c = Mv (OR) v=E/Mc

 

[Vorde]

That's a nice point. You are missing a factor of gamma on the right side though - so it doesn't quite work. As a novice's derivation it works well though.

 

[Kaushik96]

You are missing a factor of gamma on the right side though

I can't quite understand . What do you mean by this ?

 

[Vorde]

In relativistic physics momentum isn't defined as $p=mv$ but rather as $p= \gamma m v$ where $\gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}$. So your derivation doesn't quite work. It's been a while since I've had to derive E=mc^2 (I had to do it for an extra credit problem on a test) so I'm not sure whether or not the derivation on that site can be simply expanded to be correct or not, but in it's presented form it does not work.

As I said though I approve of it as a teaching method - it's quite simple - as long as readers are aware that a more thorough proof is required.

 

[Kaushik96]

all we have to do is set the velocity here to c and we have E=mc^2. I think Einstein had conservation of momentum and radiation transfer in mind when he derived it too. What I'm taking issue with is that it seems the DrPhysics is using E=mc^2 to derive E=mc^2. I don't see what novel insight he is bequeathing upon us here. Have I missed something?

You can't do something like that because In E=mc^2 relation 'm' stands for the mass of light ! And in this equation v=E/Mc , 'M' stands for the mass of the cylinder ! Even if you put 'c' in the place of 'v' it won't become E=mc^2

 

[kostas230]

Landau's derivation of the equations of relativistic mechanics in his book Classical Field Theory might help you better understand relativity.

 

[Vorde]

In E=mc^2 relation 'm' stands for the mass of light !

You are fundamentally wrong about this, and you won't get anywhere with relativity unless you do research (wikipedia is a great place to start) and realize that the mass can be anything.

 

[Kaushik96]

You are fundamentally wrong about this, and you won't get anywhere with relativity unless you do research (wikipedia is a great place to start) and realize that the mass can be anything.

In the derivation from the external site 'm' refers to the mass of the light . Thats what i meant

 

[Kaushik96]

In relativistic physics momentum isn't defined as $p=mv$ but rather as $p= \gamma m v$ where $\gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}$.

You are right ! Maybe i have to search some more about this derivation .

 

[DiracPool]

You can't do something like that because In E=mc^2 relation 'm' stands for the mass of light !

That's about as wrong as it gets in special relativity. E=mc^2 is actually the abbreviated form of the full equation which is E=mc^2+pc. The abbreviated version is used in cases that deal with energies that specifically EXCLUDE light photons, that is, rest mass. Situations dealing with light quanta will typically exclude the mc^2 term and be in the form E=pc.

And in this equation v=E/Mc , 'M' stands for the mass of the cylinder ! Even if you put 'c' in the place of 'v' it won't become E=mc^2

Of course it will, this is simple algebra.

 

[Kaushik96]

Of course it will, this is simple algebra.

You actually mistook the whole derivation . v=E/Mc not E/mc ! 'M' and 'm' are different ... Go through the derivation once more . This derivation is not circular .

 

[WannabeNewton]

The Lagrangian $L = -mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}}$, $p = \frac{\partial L}{\partial v} = \frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$, $E = p\cdot v - L = \frac{mv^{2}}{{\sqrt{1 - \frac{v^{2}}{c^{2}}}}} + mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}} = \frac{mc^{2}}{{\sqrt{1 - \frac{v^{2}}{c^{2}}}}}$ so in the rest frame we have that $E = mc^{2}$. The speed of light squared quantity is just a unit conversion. If you work in natural units then you won't even see it explicitly.

 

[Kaushik96]

The Lagrangian $L = -mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}}$, $p = \frac{\partial L}{\partial v} = \frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$, $E = p\cdot v - L = \frac{mv^{2}}{{\sqrt{1 - \frac{v^{2}}{c^{2}}}}} + mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}} = \frac{mc^{2}}{{\sqrt{1 - \frac{v^{2}}{c^{2}}}}}$ so in the rest frame we have that $E = mc^{2}$. The speed of light squared quantity is just a unit conversion. If you work in natural units then you won't even see it explicitly.

Agreed . But c^2 is not just an unit conversion . It is also known "the specific energy" .

And I don't know what is "ρ" in the equation . Could you explain?

 

[Vorde]

P is momentum: which is standard notation.

 

[Kaushik96]

P is momentum: which is standard notation.

Sorry bro ! I mistook it as "rho(ρ)"