燕大教授谈相对论论文答辩

根据钟慢效应，实验室测得的寿命与固有寿命的比值是洛伦兹因子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)

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; ., 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. ., E = K + Eo = γm0c2 = γE0, (A-29) where E0 = m0c2 and E = mc2. 给分吧

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

第二宇宙速度为千米/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！

相对论是这样一个现实，每个老师都认为自己正确理解了相对论，但有些老师认为相对论是完全错误的；有些老师认为相对论是有问题的，需要修正；有些老师虽然认为相对论正确，但同一道问题，也会给出不同答案。 所以要练习，找你老师要，他给你打分，判断你的对错。

物理学和应用物理学两个专业都要学习，只是不是专门学，而是作为一科的一部分。当然一般大学里面对一些专业将其作为选修课来开的，一些非物理专业的学生也可以通过选修来“粗糙”的学习，其实物理专业学的也很浅。 劝你不要报物理专业，很没有前途的，除非你能考上硕士研究生，或者到更高的层次经行学习。

百度文库的干活

其实相对论非常的容易理解，例如狭义相对论中的光速不变性原理相对速度公式，就是通过迈克尔逊—莫雷实验的几何关系得到的，而相对论的洛仑兹座标变换公式可以通过上式进行微分变换得到。剩下的长度，时间，质量都是可以跳过洛伦兹变换得到，我这里有狭义相对论的课件，如果需要的话就告诉我

三个考点 1、时间关系式 2、长度关系式 3、质速公式、质能公式和相对论动能（当然你把它拆成三个考点也行）