Here’s another great video from the folks at Minute Physics. One of the most famous equations from one of the world’s most famous scientists is a bit more complicated than many people realize. E=mc² only describes objects with mass that aren’t moving. But what about massless particles – like light – that are moving? Check out the video for a quick explanation!

Nancy Atkinson is currently Universe Today's Contributing Editor. Previously she served as UT's Senior Editor and lead writer, and has worked with Astronomy Cast and 365 Days of Astronomy. Nancy is also a NASA/JPL Solar System Ambassador.

E = mc^2 is a well known equation. E = composed of photons or quanta. But how many photons are responsible to create this energy, Actually, we do not know. Because, we don’t know the correct mass of a photon. So, we do not know the internal functions of m during excited state of matter. We can get this answer from Complete Unified Theory (page-424, 1998), written by me.

Real formation of E = mc^2 is:

E = (Avogadro number)^2 x root of mass of populated photons x velocity of light or photon / Curie constant. Where, equation of populated photons = mass of electron x [ (mass of
matter)^2 – (mass of Alfa particle)^2] / Pi^2 x Avo. Number x (mass of Alfa particle)^2
.
In this way matter is converting to energy.

The value of Pi is very important in the case of excited
state. It will differ slightly than normal value of Pi. This will depend on the
mass of matter. Then we will get the result same to E = mc^2.

It’s only incomplete if you consider m to be the rest mass. When writing E = m times c-squared, it is normally assumed that m includes a time-dilation factor, taking into account the increase in inertia as a massive object approaches the speed of light.

lcrowellNovember 1, 2012, 6:54 PM

The equation comes from the invariant interval. Distance in four dimensions is

d^2 = (ct)^2 – x^2 – y^2 – z^2.

This is invariant for all observers in any frame. If you watch a particle in motion and measure off the length of its path in space plus time = spacetime, this appears the same to all possible observers. If we set y = 0 and z = 0 and we consider the motion of a particle along the x direction x = vt we get

d^2 = (ct)^2 – (vt)^2 = (c^2 – v^2)t^2.

If we pull c out of the parentheses we get d^2 = (1 – (v/c)^2)(ct)^2. The distance d is the proper time of a clock on that moving particle d= c? and we see that the proper time is related to our coordinate time as ? = sqrt{1 – (v/c)^2}t, which is the time dilation of relativity.

In physics there are conjugate variables to distance called momentum. The invariant quantity is the rest mass of a particle and this is related to the energy and momentum as

(mc^2)^2 = E^2 – (pc)^2,

where energy is the conjugate variable to time. This leads to the full equation

E = sqrt{(pc)^2 + (mc^2)^2}.

If the momentum part is small pc << mc^2 we may use binomial theorem to write

E = (mc^2) sqrt{1 + (pc)^2/(mc^2)^2} ~= (mc^2)(1 + (1/2)p^2/(mc)^2

= mc^2 + p^2/2m

which is the rest mass part plus the standard classical mechanical energy.

E = (Avogadro number)^2 x root of mass of populated photons
x velocity of light or photon / Curie
constant. Where, equation of populated photons = mass of electron x [ (mass of
matter)^2 – (mass of Alfa particle)^2] / Pi^2 x Avo. Number x (mass of Alfa
particle)^2 .

In this way matter converting to energy.

The value of Pi is very important in the case of excited
state. It will differ slightly than normal value of Pi. This will depend on the
mass of matter. Then we will get the result same to E = mc^2.

Comments on this entry are closed.

That doesn’t make sense. The math is wrong. The equation doesn’t balance.

Which equation? And why, the presented ones do balance and they are known to make sense (describe the physics)?

E = mc^2 is a well known equation. E = composed of photons or quanta. But how many photons are responsible to create this energy, Actually, we do not know. Because, we don’t know the correct mass of a photon. So, we do not know the internal functions of m during excited state of matter. We can get this answer from Complete Unified Theory (page-424, 1998), written by me.

Real formation of E = mc^2 is:

E = (Avogadro number)^2 x root of mass of populated photons x velocity of light or photon / Curie constant. Where, equation of populated photons = mass of electron x [ (mass of

matter)^2 – (mass of Alfa particle)^2] / Pi^2 x Avo. Number x (mass of Alfa particle)^2

.

In this way matter is converting to energy.

The value of Pi is very important in the case of excited

state. It will differ slightly than normal value of Pi. This will depend on the

mass of matter. Then we will get the result same to E = mc^2.

Nirmalendu Das.

Email: nirmalgopa@gmail.com

Dated: 01-11-2012.

It’s only incomplete if you consider m to be the rest mass. When writing E = m times c-squared, it is normally assumed that m includes a time-dilation factor, taking into account the increase in inertia as a massive object approaches the speed of light.

The equation comes from the invariant interval. Distance in four dimensions is

d^2 = (ct)^2 – x^2 – y^2 – z^2.

This is invariant for all observers in any frame. If you watch a particle in motion and measure off the length of its path in space plus time = spacetime, this appears the same to all possible observers. If we set y = 0 and z = 0 and we consider the motion of a particle along the x direction x = vt we get

d^2 = (ct)^2 – (vt)^2 = (c^2 – v^2)t^2.

If we pull c out of the parentheses we get d^2 = (1 – (v/c)^2)(ct)^2. The distance d is the proper time of a clock on that moving particle d= c? and we see that the proper time is related to our coordinate time as ? = sqrt{1 – (v/c)^2}t, which is the time dilation of relativity.

In physics there are conjugate variables to distance called momentum. The invariant quantity is the rest mass of a particle and this is related to the energy and momentum as

(mc^2)^2 = E^2 – (pc)^2,

where energy is the conjugate variable to time. This leads to the full equation

E = sqrt{(pc)^2 + (mc^2)^2}.

If the momentum part is small pc << mc^2 we may use binomial theorem to write

E = (mc^2) sqrt{1 + (pc)^2/(mc^2)^2} ~= (mc^2)(1 + (1/2)p^2/(mc)^2

= mc^2 + p^2/2m

which is the rest mass part plus the standard classical mechanical energy.

LC

Real

formation of E = mc^2 is:

E = (Avogadro number)^2 x root of mass of populated photons

x velocity of light or photon / Curie

constant. Where, equation of populated photons = mass of electron x [ (mass of

matter)^2 – (mass of Alfa particle)^2] / Pi^2 x Avo. Number x (mass of Alfa

particle)^2 .

In this way matter converting to energy.

The value of Pi is very important in the case of excited

state. It will differ slightly than normal value of Pi. This will depend on the

mass of matter. Then we will get the result same to E = mc^2.

Nirmalendu Das

Email: nirmalgopa@gmail.com

Dated: 07-11-2012