大学物理相对论论文2500字

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大学物理相对论茄吵论文2500字

根据钟慢效应，实验室测得的寿命李州与固有寿命的比颤扰侍值是洛伦兹因子γ=1/√1-v^2/c^2，根据题意γ=n，根据以上两式可以得到粒子的速度v=(√n^2-1)c/n。所以此粒子的相对论动能p=γm0v=n×m0×(√n^2-1)c/n=m0c(√n^2-1)

大学物理相对论论文

Here we present the derivation of the new set of equations termed, Lorentz transformations, and all the subsequent relations.
LORENTZ TRANSFORMATIONS
We consider o coordinate systems (frames of reference) one stationary S and one moving at some velocity v relative to S, then aording to the o postulates of Relativity, stated in the main text, the displacement in both frames is of the same form.
Therefore, we have
(A-1)
(A-2)
We should note here that in the old Galilean transformations these equations would be
(A-3)
which is in direct contradiction to Postulate 2, a firm experimental fact.
Equations (A-1) and (A-2) can be written as
(A-4)

(A-5)
That is,
(A-6)
We are interested in finding and in terms of x and t. That is,
= (x, t) (A-7)
= (x, t) (A-8)
This is acplished via the formation of o linear simultaneous equations as follows:
(A-9)
(A-10)
where a11, a12, a21, and a22 are constants to be evaluated. It is required that the transformations are linear in order for one event in one system to be interpreted as one event in the other system; quadratic transformations imply more than one event in the other system.
Solution of problems involving motion begins with an assumption of their initial conditions; i.e., where does the problem begin?
The classical assumption is to set = 0 at = 0. Therefore, aording to S, the system appears to be moving with a velocity v, so that x = vt. We can obtain this from Equa. (A-9) by writing it in the form = a11(x - vt) so that, when = 0, x = vt. Therefore, we conclude that a12 = -va11. We can write Equations (A-9) and (A-10) as
(A-11)
(A-12)
Substituting and into Equation (A-6) and rearranging, we get
(A-13)
Since this equation is equal to zero, all the coefficients must vanish. That is,
(A-14)
(A-15)
(A-16)
Solving these equations we obtain
(A-17)
(A-18)
where β = v/c and .
Thus, substituting these values in Equas. (A-11) and (A-12) we obtain the famous Lorentz coordinate transformation equations connecting the fixed coordinate system S to the moving coordinate system :
(A-19)
(A-20)
We may also obtain the inverse transformations (from system to S) by replacing v by –v and simply interchanging primed and unprimed coordinates. This gives,
(A-21)
(A-22)
VELOCITY TRANSFORMATIONS
As a direct consequence to these new transformations, all the other mathematical operations and physical variables follow aordingly. For example, the velocity equations (though still the derivatives of the displacement) assume a new form, so the Lorentz form of the velocities is:
From Equas. (A-19) and (A-20) we have:
(A-23)
(A-24)
Therefore:
(A-25)
ENERGY CONSIDERATIONS
Consider a particle of rest mass m0 being acted by a force F through a distance x in time t and that it attains a final velocity v. The kiic energy attained by the particle is defined as the work done by the force F. The applicable equations are,
(A-26)
We note that
and that
Substituting d(γv) in Eq. (45) and integrating, we obtain
(A-27)
That is,
(A-28)
This says that K = (m – m0)c2 and finally one sees that the total energy is equal to the sum of the kiic energy K and the rest energy E0 = m0c2.
i.e., E = K + Eo = γm0c2 = γE0, (A-29)
where E0 = m0c2 and E = mc2.
给分吧

大学物理相对论

E=E0/√(1-1/4)=2E0/√3
W=E-E0=E0(2/√3-1)= 938(2/√3-1) MeV

大学物理相对论习题求解

第二宇宙速度为11.2千米/秒。
相对质量公式为：
M=Mo/√(1-v^2/c^2)
Mo是物体静止时的质量，M是物体运动时的质量，v是物体速度，c是光速。由此可知速度越大，物体质量越大，当物体以光速运动，物体的质量为正无穷。
你把11.2代入公式，得出运动时的质量，减去原来的质量即可。
记得把100t化为千克，1t=1000千克。
敲得真辛苦啊！希望你看得懂！