神马作文网大学物理相对论论文急需一篇大学物理的相对论论文,可以是copy的但是不要是大家都能找到的资源,原创最好800字左右
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大学物理相对论论文
急需一篇大学物理的相对论论文,可以是copy的但是不要是大家都能找到的资源,原创最好800字左右
急需一篇大学物理的相对论论文,可以是copy的但是不要是大家都能找到的资源,原创最好800字左右
Here we present the derivation of the new set of equations termed, Lorentz transformations, and all the subsequent relations.
LORENTZ TRANSFORMATIONS
We consider two coordinate systems (frames of reference) one stationary S and one moving at some velocity v relative to S, then according to the two 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 accomplished via the formation of two 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, according 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 accordingly. 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 kinetic 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 kinetic 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
LORENTZ TRANSFORMATIONS
We consider two coordinate systems (frames of reference) one stationary S and one moving at some velocity v relative to S, then according to the two 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 accomplished via the formation of two 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, according 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 accordingly. 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 kinetic 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 kinetic 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