既然如此，让我们来发现微观世界的运行规律吧！证法如下：
∵量子集团的自旋为匀变速圆周运动
∴Vt=V0+at（Vt是粒子α在受力t时间后的自旋线速度，V0为α的初始自旋线速度）
∵a=ω2r（a为α受力下自旋的加速度，ω为粒子自旋的角速度，r为粒子的半径）
∴Vt=V0+ω2r
at=a0+ω2r
∵α未受力是为匀速圆周运动
∴a0=0
又∵Mt2=hvt2=ms2（M为力矩，h为普朗克常量，v为频率，m为质量，s为路程长）
（α做匀速圆周运动时，a0=0，但v0≠0）
当α为一个能量子时，E=hv，E=M
以上证明的公式统称为“芥旋定律”，其物理意义是：对于任意一个量子集团，其自旋角速度的平方与半径之积，等于真空中光速的平方与时间和√G（G=hv/m）的乘积之比。
既然我们已经决定否认光是概率波的观点，那么我们可以用另一种方式证明一些理论了：
如图所示，我们假设有一个发光源，每t时间发射一个光子，同时发光源以速度V在光子运动方向所在的直线上运动。当发光源发射出一个光子后，在t时间内运动了一定距离s后才发射出下一个光子。我们设发光源速度为0时相邻t时间内发射的两个光子之间的距离为“初始波长”，用λ0表示；发光源速度为V时相邻两光子之间的距离为“变速波长”，用λv表示。
通过分析，我们可以得到如下的等式：
λv-λ0=Vt
因为我们承认物质波理论的正确性，所以
∴
下面，我们根据h的单位为J.s,t的单位为s，将等式两边同时除以t则有:
E=mvc+mc^2
=mc(v+c)
彐V>0(以光子发射方向为正方向），E>mc^2
彐V=0，E=mc^2
彐V＜0，E＜mc^2
奇迹出现了！我们通过经典物理波动模型，证明出了爱因斯坦的质能方程！并且假如我们深入研究的话，我们就会有如下的发现：
若v1>v2>0，
则λv1>λv2，Vv1>Vv2（v表示频率）
又∵v1>v2>0
则Ev1＞Ev2＞mc^2
整合，得：Vv1>Vv2 Ev1＞Ev2
由上式我们可以得知：光子的能量只与光的频率有关，V越大，E越大，与普朗克给出的关系式E=hv不谋而合！从上式我们已经可以预见：我们所得出的超阶质能关系式极有可能是正确的！
接下来，我们就利用上面证明出的关系式接着探索：
∵E=mvc+mc^2
P=mv
∴E=Pc+mc^2
当α为量子时，
E=hv
∴hv=Pc+mc^2
这里居然出现了一个差式！这显然与我们得出的光子相对论质量表达式m=hv/c^2不一致！这是怎么回事？我们的质能方程错了吗？与此同时，我们想到了最初的双缝干涉实验，联系不确定性关系，我们作出了一个大胆的猜想：任何微粒在空间中运动时，在其周围都会被激发出虚拟质量
mx，微粒的实际质量=总质量-mx，即：
mx=P/c,mT=hv/c^2-P/c(其中mt为微粒的实际质量，我们称之为本源质量）
M=mT+P/c
∴E=（mT+P/c）c^2
∵E=mc（v+c）
∴E=（mT+P/c）c（v+c）
=（mTc+P）（v+c）
整合后，我们总结出了关于能量的三个方程：
1.E=mc（v+c）
2.E=（mT+P/c）c^2
3.E=（mTc+P）（v+c）
现在，我们已经完全深入到物理学的深层次了，我们继续发现质能之间的种种关系吧！
在之前我们已经证明出方程
有没有会的呀~~~
In that case, let's find the running law of world!Proofs are as follows:
∵ spin quantum group for uniform variable circular motion
∴ n = where V0 + ats (Vt is alpha particles after stress t time spin velocity, where V0 initial spin velocity of alpha)
∵ a = 2 r omega (a for alpha force under spin acceleration, omega for spin angular velocity of the particles, r is the radius of the particle)
∴ n = where V0 + 2 r omega
The at = a0 + 2 r omega
∵ alpha force is not uniform circular motion
∴ a0 = 0
And ∵ Mt2 = hvt2 = ms2 (M for the moment, h is Planck's constant, v for frequency, M for quality, s for longer distance)
(alpha do uniform circular motion, a0 = 0, but where v0 indicates zero)
When alpha is an energy, E = hv, E = M
Above prove that the formulas are collectively referred to as the "law of mustard spin", its physical meaning is: for an arbitrary quantum group, the square of the spin angular velocity and the radius of the product, is equal to the square of the speed of light in vacuum with time and square root of G (G = hv/m) ratio of the product.
Now that we have decided to deny alone probability wave point of view, and then we can use another way to prove some theory:
As shown, we assume that there is a light source, each t time to launch a photon, at the same time light source with speed V on the photon motion direction of linear movement.After the hair light sources emit a photon, movement within the time t after a certain distance from s to emit a photon.When we set light source speed of 0 adjacent t time to launch the distance between the two photons to "initial wavelength", expressed in lambda 0;Light source speed V when the distance between the adjacent two photons for the variable wavelength, with lambda V said.
Through the analysis, we can get the following equation:
Lambda v - lambda 0 = Vt
Because we admit the soundness of matter wave theory
∴
Below, we, according to a unit for j. h t units for s, will be on both sides of the equation is divided by t at the same time:
E = MVC + MC ^ 2
= MC (v + c)
彐 V > 0 (photon emission direction as positive direction), E > MC ^ 2
彐 V = 0, E = MC ^ 2
彐 V < 0, E < MC ^ 2
A miracle!By fluctuations in classical physics model, we prove the Einstein's equation.And if we study, we will have the following findings:
If the v1 > v2 > 0,
The lambda v1 > lambda v2, Vv1 > Vv2 frequency (v)
And ∵ v1 > v2 > 0
The Ev1 > Ev2 > MC ^ 2
Integration, too: Vv1 > > Vv2 Ev1 Ev2
We can learn by the type: photon energy associated with the frequency of light, the greater the V, E, and given the Planck equation E = hv the same view!From the type we can already foresee: we order of super mass energy equation was probably right!
Next, we use the above proof and then explore the relation between the:
∵ E = MVC + MC ^ 2
P = mv
∴ E = Pc + MC ^ 2
When alpha for quantum,
E = hv
∴ hv = Pc + MC ^ 2
Here the emergence of a differential.The quality obviously and we concluded that the photon relativity expression inconsistent m = hv/c ^ 2!This is how to return a responsibility?Our mass energy equation is wrong?At the same time, we came up with the original double-slit experiment, contact the uncertainty relation, we made a bold guess: any particle motion in the space, around the virtual quality will be inspired
Mx, actual quality = total mass of the particles - mx, namely:
Mx = P/c, mT = hv/c ^ 2 - P/c (including the actual quality of mT for particle, we call the source quality)
M = mT + P/c
∴ E = (mT) + P/c c ^ 2
∵ E = MC (v + c)
∴ E = (mT + P/c) c (v + c)
+ c + P = (mTc) (v)
After consolidation, we summarized the three equations about energy:
1. E = MC (v + c)
2. E = (mT) + P/c c ^ 2
3. E = (v + c) + P (mTc)
Now, we have fully into a deeper level of physics, we continue to find the relationship between mass energy!!!!
Before we have shown that out of the equation
∵ spin quantum group for uniform variable circular motion
∴ n = where V0 + ats (Vt is alpha particles after stress t time spin velocity, where V0 initial spin velocity of alpha)
∵ a = 2 r omega (a for alpha force under spin acceleration, omega for spin angular velocity of the particles, r is the radius of the particle)
∴ n = where V0 + 2 r omega
The at = a0 + 2 r omega
∵ alpha force is not uniform circular motion
∴ a0 = 0
And ∵ Mt2 = hvt2 = ms2 (M for the moment, h is Planck's constant, v for frequency, M for quality, s for longer distance)
(alpha do uniform circular motion, a0 = 0, but where v0 indicates zero)
When alpha is an energy, E = hv, E = M
Above prove that the formulas are collectively referred to as the "law of mustard spin", its physical meaning is: for an arbitrary quantum group, the square of the spin angular velocity and the radius of the product, is equal to the square of the speed of light in vacuum with time and square root of G (G = hv/m) ratio of the product.
Now that we have decided to deny alone probability wave point of view, and then we can use another way to prove some theory:
As shown, we assume that there is a light source, each t time to launch a photon, at the same time light source with speed V on the photon motion direction of linear movement.After the hair light sources emit a photon, movement within the time t after a certain distance from s to emit a photon.When we set light source speed of 0 adjacent t time to launch the distance between the two photons to "initial wavelength", expressed in lambda 0;Light source speed V when the distance between the adjacent two photons for the variable wavelength, with lambda V said.
Through the analysis, we can get the following equation:
Lambda v - lambda 0 = Vt
Because we admit the soundness of matter wave theory
∴
Below, we, according to a unit for j. h t units for s, will be on both sides of the equation is divided by t at the same time:
E = MVC + MC ^ 2
= MC (v + c)
彐 V > 0 (photon emission direction as positive direction), E > MC ^ 2
彐 V = 0, E = MC ^ 2
彐 V < 0, E < MC ^ 2
A miracle!By fluctuations in classical physics model, we prove the Einstein's equation.And if we study, we will have the following findings:
If the v1 > v2 > 0,
The lambda v1 > lambda v2, Vv1 > Vv2 frequency (v)
And ∵ v1 > v2 > 0
The Ev1 > Ev2 > MC ^ 2
Integration, too: Vv1 > > Vv2 Ev1 Ev2
We can learn by the type: photon energy associated with the frequency of light, the greater the V, E, and given the Planck equation E = hv the same view!From the type we can already foresee: we order of super mass energy equation was probably right!
Next, we use the above proof and then explore the relation between the:
∵ E = MVC + MC ^ 2
P = mv
∴ E = Pc + MC ^ 2
When alpha for quantum,
E = hv
∴ hv = Pc + MC ^ 2
Here the emergence of a differential.The quality obviously and we concluded that the photon relativity expression inconsistent m = hv/c ^ 2!This is how to return a responsibility?Our mass energy equation is wrong?At the same time, we came up with the original double-slit experiment, contact the uncertainty relation, we made a bold guess: any particle motion in the space, around the virtual quality will be inspired
Mx, actual quality = total mass of the particles - mx, namely:
Mx = P/c, mT = hv/c ^ 2 - P/c (including the actual quality of mT for particle, we call the source quality)
M = mT + P/c
∴ E = (mT) + P/c c ^ 2
∵ E = MC (v + c)
∴ E = (mT + P/c) c (v + c)
+ c + P = (mTc) (v)
After consolidation, we summarized the three equations about energy:
1. E = MC (v + c)
2. E = (mT) + P/c c ^ 2
3. E = (v + c) + P (mTc)
Now, we have fully into a deeper level of physics, we continue to find the relationship between mass energy!!!!
Before we have shown that out of the equation
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第1个回答 2015-01-02
In that case, let us find microcosm of run rules now! Card method following: ¡ß quantum group of since spin for uniform speed circle movement ∴ Vt=V0+at (Vt is particles Alpha in by force t time Hou of since spin line speed, V0 for alpha of initial since spin line speed) ¡ß a= ω 2R (a, for Alpha by force Xia since spin of acceleration, ω for particles since spin of angle speed, r for particles of RADIUS) ∴ Vt=V0+ ω 2rat=A0+ ω 2R ¡ß Alpha not by force is for uniform circular motion ∴ A0=0 and ¡ß Mt2=hvt2=ms2 (m for torque, H for General lang grams constants, v for frequency, m for quality, s for away long) (Alpha do uniform circular motion Shi, A0=0, but v0 ≠ 0) Dang Alpha for a can quantum Shi, E=hv,E=M above proved of formula collectively for "mustard spin law", its physical meaning is: for arbitrary a quantum group, its since spin angle speed of square and radius of product, is equal to vacuum in the speed of square and time and √ g (G=hv/m) of product of than. Now that we have decided to deny the probability wave view alone, then we can prove some theory in a different way: as shown in the figure, we assume that there is a light source, every t time launching a photon, and light source direction with speed v in the photon motion in a straight line movement. When the light source emits a photon after, during t hours sport a s some distance before firing the next photon. Luminous source adjacent t speed: 0 o'clock our time launching of the distance between the two photons for "initial wavelengths", expressed by λ 0; light source speed v when the distance between adjacent two-photon as "variable-wavelength", use λ vSaid. According to the analysis, we can get the following equation: λ λ 0=Vt v-because we recognize that matter-wave theory was correct, so below ∴ we according to the units to J.s,t units for the s h, divide both sides of this equation by the t include:E=mvc+mc^2=mc (v+c), JI V>0 (with photon emission direction as the positive direction), E>mc^2 JI V=0,E=mc^2 JI V<0,E<mc^2 a miracle happened! We use classical physics model, show Einstein's equation of mass and energy! And if we in-depth research words, we on will has following of found: If V1>V2>0, is λ v1> λ v2,Vv1>Vv2 (v said frequency) and ¡ß V1>V2>0 is Ev1>Ev2>mc^2 integration, have: Vv1>Vv2 Ev1>Ev2 by Shang type we can learned that: photon of energy only and light of frequency related, v more big, e more big, and General lang grams to out of relationship type E=hv coincides with! From the above we can already foresee: we have gotten Super band-mass-energy relation could very well be right! Next, we use top prove out relationship and continued to explore: ¡ß E=mvc+mc^2P=mv ∴ E=Pc+mc^2 when quantum is α, E=hv ∴ hv=Pc+mc^2-an error occurred! Obviously we had the relativistic mass of the photon expressions m=HV/c^2 different! Is this all about? Our mass-energy equation wrong? Meanwhile, we came up with the original two-slit interference experiment, the associated uncertainty relation, we have made a bold conjecture: any particle moving in space, around it were inspired virtual mass MX, the actual quality of the particles = total mass of-MX, namely: mx=P/c,MT=hv/c^2-P/c (where MT is the actual mass of the particles, which we call original quality) M=mT+P/c ∴ E= (mT+P/c) c^2 ¡ß E=mc (v+c) ∴ E= (mT+P/c) c (v+c) = (mTc+P) (v+c) integration, we have summarized on the energy of the three equations: 1. E=mc（v+c）2. E= (mT+P/c) c^23.E= (mTc+P) (v+c), now we are deep into a deeper level of physics, we continue to discover relationships between mass and energy bar! Previously we have shown that equations.
第2个回答 2015-01-18
∵ quantum spin Group uniform circular motion speed
∴Vt = V0 + at (Vt particle α is the linear velocity of the stress time t after the spin, V0 is initial spin α linear velocity)
∵a = ω2r (a force to spin under acceleration α, ω is the angular velocity of the particle spin, r is the radius of the particle)
∴Vt = V0 + ω2r
at = a0 + ω2r
∵α unstressed for uniform circular motion
∴a0 = 0
And ∵Mt2 = hvt2 = ms2 (M is moment, h is Planck's constant, v is the frequency, m is mass, s for long distance)
(Α do uniform circular motion, a0 = 0, but v0 ≠ 0)
When α is an energy neutrons, E = hv, E = M
The formula proved above are collectively referred to as "spin mustard law" and its physical meaning is: For any quantum group, the plot of the square of the radius of its spin angular velocity equal to the speed of light squared vacuum with time and √G (G = hv / m) the product of the ratio.
Now that we have decided to deny the probability wave viewpoint alone, then we can use another way to prove some theories of:
As shown, we assume that there is a light source, emitting a photon every time t, while emitting source movement velocity V in the direction of the photon linear motion is located. When the illumination source emits a photon in the t s time to exercise a certain distance before the next photon is emitted. "Wavelength shift distance between the adjacent two-photon emitting source at a speed of V; we set the distance between adjacent light emitting source speed 时 0 t within the time between the emission of two photons for the" initial wavelength ", represented by λ0 ", represented by λv.
Through analysis, we can get the following equation:
λv-λ0 = Vt
Because we recognize the validity of the theory of matter waves, so
∴
Below, we h according to the unit for Js, t units of s, while the both sides of the equation are divided by t are:
E = mvc + mc ^ 2
= mc (v + c)
Ji V> 0 (in photon emission direction is positive), E> mc ^ 2
Ji V = 0, E = mc ^ 2
Ji V <0, E <mc ^ 2
A miracle happened! We volatility model of classical physics to prove Einstein's mass-energy out of the equation! And if we delve into it, we will have the following findings:
If v1> v2> 0,
The λv1> λv2, Vv1> Vv2 (v indicates frequency)
Also ∵v1> v2> 0
The Ev1> Ev2> mc ^ 2
Integration, too: Vv1> Vv2 Ev1> Ev2
We can learn from the above equation: energy photons only concerned with the frequency of the light, V the larger, E bigger, given the Planck relation E = hv coincide! From the above equation, we can already foresee: the results of our super-order mass-energy relationship is likely to be correct!
Next, we use to prove the above and then explore the relationship:
∵E = mvc + mc ^ 2
P = mv
∴E = Pc + mc ^ 2
When α is quantum,
E = hv
∴hv = Pc + mc ^ 2本回答被提问者采纳
∴Vt = V0 + at (Vt particle α is the linear velocity of the stress time t after the spin, V0 is initial spin α linear velocity)
∵a = ω2r (a force to spin under acceleration α, ω is the angular velocity of the particle spin, r is the radius of the particle)
∴Vt = V0 + ω2r
at = a0 + ω2r
∵α unstressed for uniform circular motion
∴a0 = 0
And ∵Mt2 = hvt2 = ms2 (M is moment, h is Planck's constant, v is the frequency, m is mass, s for long distance)
(Α do uniform circular motion, a0 = 0, but v0 ≠ 0)
When α is an energy neutrons, E = hv, E = M
The formula proved above are collectively referred to as "spin mustard law" and its physical meaning is: For any quantum group, the plot of the square of the radius of its spin angular velocity equal to the speed of light squared vacuum with time and √G (G = hv / m) the product of the ratio.
Now that we have decided to deny the probability wave viewpoint alone, then we can use another way to prove some theories of:
As shown, we assume that there is a light source, emitting a photon every time t, while emitting source movement velocity V in the direction of the photon linear motion is located. When the illumination source emits a photon in the t s time to exercise a certain distance before the next photon is emitted. "Wavelength shift distance between the adjacent two-photon emitting source at a speed of V; we set the distance between adjacent light emitting source speed 时 0 t within the time between the emission of two photons for the" initial wavelength ", represented by λ0 ", represented by λv.
Through analysis, we can get the following equation:
λv-λ0 = Vt
Because we recognize the validity of the theory of matter waves, so
∴
Below, we h according to the unit for Js, t units of s, while the both sides of the equation are divided by t are:
E = mvc + mc ^ 2
= mc (v + c)
Ji V> 0 (in photon emission direction is positive), E> mc ^ 2
Ji V = 0, E = mc ^ 2
Ji V <0, E <mc ^ 2
A miracle happened! We volatility model of classical physics to prove Einstein's mass-energy out of the equation! And if we delve into it, we will have the following findings:
If v1> v2> 0,
The λv1> λv2, Vv1> Vv2 (v indicates frequency)
Also ∵v1> v2> 0
The Ev1> Ev2> mc ^ 2
Integration, too: Vv1> Vv2 Ev1> Ev2
We can learn from the above equation: energy photons only concerned with the frequency of the light, V the larger, E bigger, given the Planck relation E = hv coincide! From the above equation, we can already foresee: the results of our super-order mass-energy relationship is likely to be correct!
Next, we use to prove the above and then explore the relationship:
∵E = mvc + mc ^ 2
P = mv
∴E = Pc + mc ^ 2
When α is quantum,
E = hv
∴hv = Pc + mc ^ 2本回答被提问者采纳
第3个回答 2015-01-27
Spin quantum group dreams for uniform variable circular motion
* Vt=V0+at (Vt is the particle alpha in the stress time after t spin line speed, V0 alpha initial spinspeed line)
Dreams of Omega a= 2R (a for the alpha force spin acceleration, Omega is the angular velocity,particle spin r particle radius)
* Vt=V0+ 2R
At=a0+. 2R
Dreams are not under stress is alpha for uniform circular motion
R a0=0
And dreams (M Mt2=hvt2=ms2 for the moment, h is the Planck constant, V frequency, m quality,s as long)
(alpha do uniform circular motion, but a0=0, V0 = 0)
When alpha is an energy neutrons, E=hv, E=M
To prove the above formula are collectively referred to as "Mustard rotation law", its physical meaning is: for an arbitrary quantum group, square and the radius of the spin angular velocity is equal to the product of the velocity of light in a vacuum, and the square of time and the root G(G=hv/m) product ratio.
Now that we have decided to deny that light is the probability wave point of view, so we can prove some theories in another way:
As shown in the figure, we assume that there is a light source of T, each time the emission of a photon, linear and light source with a velocity V in the direction of motion on the motion ofphotons. When the light source emits a photon, in t time movement distance s only after the nextphoton emitted. We design for light speed between two photon emission of adjacent time t 0 when the distance of the "initial wavelength", represented by lambda 0; light source for V speedbetween two adjacent photon distance as "transmission wavelength", represented by lambda v.
Through the analysis, we can get the following equation:
Lambda v- lambda 0=Vt
Because the right for us to recognize the matter wave theory and so
*
Below, we according to the H unit for J.s, t for s units, the both sides of the equation are alsodivided by t:
E=mvc+mc^2
=mc (v+c)
V>0 (been to photon emission in forward direction), E>mc^2
Been V=0, E=mc^2
Been V < 0, E < mc^2
A miracle occurred! We through the classical physical wave model, proved that Einstein's mass energy equation! And if we further research, we will be as follows:
If v1>v2>0,
Then lambda v1> lambda V2, Vv1>Vv2 (V represents the frequency)
And dreams v1>v2>0
Ev1 > Ev2 > mc^2
Integration: Vv1>Vv2, Ev1 > Ev2
Was informed by the type we can: the photon energy is only related to the frequency of light, the greater the greater the V, E, E=hv and Planck introduced the relationship agree without prior without previous consultation! From the type we can already foresee: We obtained super orderrelation is most likely correct!
Next, we use the relationship between the above demonstrated then explore:
Dreams E=mvc+mc^2
P=mv
R E=Pc+mc^2
When alpha for quantum,
E=hv
R hv=Pc+mc^2
Here there's a difference type! This is clearly and photon we derive the relativistic massexpression m=hv/c^2 inconsistent! This is how to return a responsibility? Our mass energy equation is wrong? At the same time, we think of the original double slit interference experiment,contact the uncertainty relation, we made a bold guess: any particle movement in space, around which are excited by the virtual mass
MX, the actual quality = total mass of -mx particles, i.e.:
Mx=P/c, mT=hv/c^2-P/c (where MT is the actual quality of particles, we call the source quality)
M=mT+P/c
* E= (mT+P/c) c^2
Dreams (v+c E=mc)
* E= (mT+P/c) C (v+c)
= (mTc+P) (v+c)
After the integration, we summarize three equation of energy:
1.E=mc (v+c)
2.E= (mT+P/c) c^2
3.E= (mTc+P) (v+c)
* Vt=V0+at (Vt is the particle alpha in the stress time after t spin line speed, V0 alpha initial spinspeed line)
Dreams of Omega a= 2R (a for the alpha force spin acceleration, Omega is the angular velocity,particle spin r particle radius)
* Vt=V0+ 2R
At=a0+. 2R
Dreams are not under stress is alpha for uniform circular motion
R a0=0
And dreams (M Mt2=hvt2=ms2 for the moment, h is the Planck constant, V frequency, m quality,s as long)
(alpha do uniform circular motion, but a0=0, V0 = 0)
When alpha is an energy neutrons, E=hv, E=M
To prove the above formula are collectively referred to as "Mustard rotation law", its physical meaning is: for an arbitrary quantum group, square and the radius of the spin angular velocity is equal to the product of the velocity of light in a vacuum, and the square of time and the root G(G=hv/m) product ratio.
Now that we have decided to deny that light is the probability wave point of view, so we can prove some theories in another way:
As shown in the figure, we assume that there is a light source of T, each time the emission of a photon, linear and light source with a velocity V in the direction of motion on the motion ofphotons. When the light source emits a photon, in t time movement distance s only after the nextphoton emitted. We design for light speed between two photon emission of adjacent time t 0 when the distance of the "initial wavelength", represented by lambda 0; light source for V speedbetween two adjacent photon distance as "transmission wavelength", represented by lambda v.
Through the analysis, we can get the following equation:
Lambda v- lambda 0=Vt
Because the right for us to recognize the matter wave theory and so
*
Below, we according to the H unit for J.s, t for s units, the both sides of the equation are alsodivided by t:
E=mvc+mc^2
=mc (v+c)
V>0 (been to photon emission in forward direction), E>mc^2
Been V=0, E=mc^2
Been V < 0, E < mc^2
A miracle occurred! We through the classical physical wave model, proved that Einstein's mass energy equation! And if we further research, we will be as follows:
If v1>v2>0,
Then lambda v1> lambda V2, Vv1>Vv2 (V represents the frequency)
And dreams v1>v2>0
Ev1 > Ev2 > mc^2
Integration: Vv1>Vv2, Ev1 > Ev2
Was informed by the type we can: the photon energy is only related to the frequency of light, the greater the greater the V, E, E=hv and Planck introduced the relationship agree without prior without previous consultation! From the type we can already foresee: We obtained super orderrelation is most likely correct!
Next, we use the relationship between the above demonstrated then explore:
Dreams E=mvc+mc^2
P=mv
R E=Pc+mc^2
When alpha for quantum,
E=hv
R hv=Pc+mc^2
Here there's a difference type! This is clearly and photon we derive the relativistic massexpression m=hv/c^2 inconsistent! This is how to return a responsibility? Our mass energy equation is wrong? At the same time, we think of the original double slit interference experiment,contact the uncertainty relation, we made a bold guess: any particle movement in space, around which are excited by the virtual mass
MX, the actual quality = total mass of -mx particles, i.e.:
Mx=P/c, mT=hv/c^2-P/c (where MT is the actual quality of particles, we call the source quality)
M=mT+P/c
* E= (mT+P/c) c^2
Dreams (v+c E=mc)
* E= (mT+P/c) C (v+c)
= (mTc+P) (v+c)
After the integration, we summarize three equation of energy:
1.E=mc (v+c)
2.E= (mT+P/c) c^2
3.E= (mTc+P) (v+c)
第4个回答 2014-12-27
at=a0+ω2r
∵α未受力是为匀速圆周运动？？？？？追问
∵α未受力是为匀速圆周运动？？？？？追问
是α未受力时是匀速直线运动，打错了，不好意思。
