Einstein's formula e = MC2, the energy of matter is equal to the mass times the square of the speed of light?

Einstein's formula e = MC2, the energy of matter is equal to the mass times the square of the speed of light?

The first step: to discuss the change of energy with mass, we need to know the idea from the dimension
Energy dimension [e] = [M] ([l] ^ 2) ([t] ^ (- 2)), that is, energy dimension is equal to the product of the square of mass dimension and length dimension and the negative quadratic of time dimension
We need to reduce the function of energy to mass to the simplest, then we need to have only one other variable except mass in the energy function
By simplifying ([l] ^ 2) ([t] ^ (- 2)), we can get only one dimension velocity [v]_ ]In the form of:
[V_ ]*[V_ ].
That is [e] = [M] [v]_ ]*[V_]
It can be seen that the simplest way to discuss the mass energy relationship is from the velocity v_ Start
----------------------------------------------------
Step 2: consider the change of energy first
The change of energy is related to the transformation of various energy forms, among which work is directly related to mass
Then consider the effect of workmanship on energy change
When the external force F_ (added after)_ When acting on a particle with a stationary mass of M0, every DS is generated_ (displacement s)_ The energy of the body increases with the displacement of the body
dE=F_ *ds_ (* denotes dot product)
When the direction of the external force and displacement are the same, the above formula becomes
dE=Fds
----------------------------------------
Step 3: how to connect the work done by force with the change of velocity v? In other words, how to get the change of velocity through the effect of force?
We know that the impulse of a force on an object is equal to the increment of its momentum
F_ dt=dP_ =mdv_
----------------------------------------
Step 4: in the above formula, we obviously need to refer to the variable m mass, but we don't want the mass to complicate the relationship between force and velocity. We want to find a way to reduce m, so that we can get the pure relationship between velocity and force
Reference de = FDS and F_ dt=dP_ We know, V_ =ds_ /dt
Then you can get it
dE=v_ *dP_
If we consider the simplest form: when the velocity and momentum change in the same direction:
dE=vdP
---------------------------------
Step 5: change the above formula into the relationship between energy, mass and velocity (because we are going to discuss this form at first)
De = VD (MV) -- because DP = D (MV)
---------------------------------
Step 6: decompose the above formula according to differential multiplication
dE=v^2dm+mvdv
This formula shows that the increment of energy contains two parts: the increment of energy produced by the increase of mass due to the increase of velocity DM and the increment of energy produced by the increase of velocity alone, This is one of the biggest mistakes in classical physics
---------------------------------
Step 7: we don't know how the increment DM of mass increases with speed. Now we need to study how it increases with speed (that is, the direct relationship between the increment DM of mass and the increment DV of speed)
According to Lorentz transformation, the formulas of static mass and moving mass are derived
m=m0[1-(v^2/c^2)]^(-1/2)
Reduced to integer power form:
m^2=(m0^2)[1-(v^2/c^2)]
In this way, we can get the pure functional form of motion mass m with respect to velocity change and static mass
(m^2)(c^2-v^2)=(m0^2)c^2
Use the above formula to derive the velocity V to get DM / DV (the reason for doing this is to find the most direct relationship between the mass increment DM and the velocity increment DV. This is the fundamental purpose of our step)
D [(m ^ 2) (C ^ 2-V ^ 2)] / DV = D [(M0 ^ 2) C ^ 2] / DV
Namely
[d(m^2)/dv](c^2-v^2)+m^2[d(c^2-v^2)/dv]=0
Namely
[m(dm/dv)+m(dm/dv)](c^2-v^2)+(m^2)[0-2v]=0
Namely
2m(dm/dv)(c^2-v^2)-2vm^2=0
It's about 2m away from the common factor (it's definitely not 0, the motion quality is 0? Never heard of it)
The results are as follows
(dm/dv)(c^2-V^2)-mv=0
Namely
(dm/dv)(c^2-V^2)=mv
Because DV is not equal to 0 (we study the case of non-stationary, the increment of the velocity of the moving system to the stationary system is not zero of course)
(c^2-v^2)dm=mvdv
This is the direct relationship between DM and DV
--------------------------------------------
Step 8: with the function of DM, let's go back to the energy increment in step 6
dE=v^2dm+mvdv
=v^2dm+(c^2-v^2)dm
=c^2dm
This is the differential form of the relationship between mass and energy. It shows that the increment of mass is directly proportional to the increment of energy, and the proportional coefficient is constant C ^ 2
------------------------------------------
The last step is to deduce the total energy increment in the process of the object moving from static to velocity v
The result of the previous step is integrated, and the integral interval is from the static mass M0 to the moving mass M
∫dE=∫[m0~m]c^2dm
Namely
E=mc^2-m0c^2
This is the total energy increment in the process of a body moving from rest to velocity v
among
E0 = m0c ^ 2 is called the static energy of an object at rest
EV = MC ^ 2 is called the total kinetic energy (the total energy of motion) of an object
Conclusion: for any object with known mass m, its kinetic energy can be calculated directly by E = MC ^ 2