Given the energy momentum tensor for the perfect fluid,
(i)
Contracting the above expresion with the metric, we get
And by using the fact the 4-velocity by definition has the
norm
And also in four dimensions
So, we get
(ii)
In the local inertial frame, in which the fluid is at rest, the
energy density
is simply given by
And also, in the local frame, the 4-velocity is given by
And locally, we have
And so, we get in the local inertial frame,
And so, we get
So, in this frame, the trace of the stress-tensor is given by
This is not exactly same as the answer (i). But, it is almost same
in a sense only the energy density is given by the local energy
density
.
(iii)
The stress-energy tensor of a system of particles as given in
equation (7.19) in the question
So, the trace in this is case means
Now, for photons, the norm of the momentum is zero (i.e., photons
are massless), i.e.,
And so, we get
Now, we use the result obtained in part (ii) combined with the
above traceless condition (T=0) to get for photons as a perfect
fluid
This is the equation of state, i.e., the energy density as a
function of the pressure.
Question 5 [12 10 22 marks] (a) In a given inertial reference frame, S', consider a region of space where there is a uniform and constant electric field, E', and zero magnetic field, i.e. B' = 0. The frame S' moves with respect to an observer, in another frame S, with velocity v. Write an expression for the electric field, E, observed in S? Clearly explain any notation (i) and new quantities introduced Write an expression for the magnetic field,...
Α'2 = Σ Λ Α' (4.4) V=1...2 The instructions under the summation symbol tell us to assign the values t, x, y, z to the index v and sum the four terms that result. The value of u is left unspecified: if = 1, then this equation corresponds to the first row of equation 4.1; if u = x, it corresponds to the second line of 4.1, and so on. Equations 4.4 and 4.1 are equivalent. Equation 4.4 can be...
please solve 2 QUESTION 2 a) (5 p) Interpret the rocket equation dv(t)M(1)=-udMO [EQ.1) within the framework of the law of momentan conservation, written in a closed system here M(t) is the rocket mass, at time t, whereas dMt) is by definition, dM(t)-M(t+dt)-M(t): -SM(t)-M(!), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is still by definition, dy(t){t+dt)-v(t), i.e. the increase in the velocity of the...
a) (5 p) Interpret the rocker equation dv(t)M(t)=-udM(t) (EQ.1) within the framework of the law of momentum conservation, written in a closed system, here M(t) is the rocker mass, at time t, whereas M(t) is by definition, dM(t)-M(t+dt)-M(t): - dM(t)-dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is, still by definition, dv(t)-v(t+dt)-v(t), i.e. the increase in the velocity of the rocker through the period...
QUESTION 2 25 a) (5 p) Interpret the rocker equation dv(t)M(t)=-udMO (EQ.1) within the framework of the law of momentum conservation, written in a closed system here Mt) is the rocket mass, at time t, whereas dM(t) is by definition, dMtM(t+dt)-M(t): -dM(t)=dM(1), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is still by definition, dv(t)=v(t+dt)-v(t), i.e. the increase in the velocity of the rocker through...
QUESTION 2 a) (5 p) Interpret the rocket equation dv(OM(t)=-udMO (EQ.1) within framework of the law of momentum conservation, written in a closed system, here M(t) is the rocket mass, at time t, whereas M(t) is by definition, dM(t)=M(t+dt)-M(t): -dM(t)=dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is, still by definition, dv(t)=v(t+dt)-vít).i.e. the increase in the velocity of the rocket through the period of...
QUESTION 2 a) (5 p) Interpret the rocket equation dv(OM(t)=-udMO [EQ.1) within the framework of the law of momentum conservation, written in a closed system, here Mt) is the rocket mass, time t, whereas M(t) is by definition, dM(t)=M(t+dt)-M(t): -dM(t)-dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt: on the other hand, dv(t) is, still by definition, dv(t)v(t+dt)-vít), i.e. the increase in the velocity of the rocket through the period...
Question #9 all parts thanks 9. The wavefunction, p(x,t), of a particle moving along the x-axis, whose potential energy V(x) is independent of time, is described by the one-dimensional non-relativistic Schrödinger equation (where m is its mass, h is the reduced Planck constant, i is the imaginary number): 2m (a) Verify that it is a parabolic equation (page E-1-2). [It has wave-like solutions, however.] (b) Use the substitution Px,t)-Xx)Tt) to separate the equation into two ODEs. (c) Solve for T,...
fr the falling fm . Lerive anl vcloci Pey o 42) assumin 5 usinte equatienmtion (6.5-3), niam ity, average velocity, or force on solid surfaces. tion appear, and In the integrations mentioned above, several constants of integration a the velocit stress at the boundaries of the system. The most commonly used boundae are as follows: using "boundary conditions"-that is, statements about a. At solid-fluid interfaces the fluid velocity equals the velocity with which surface is moving: this statement is applied...
3. Consider the vector space V = R2[x] with its standard ordered basisE = 1,x,x2 and the linear map T :R2[x]−→R2[x], T(p)=p(x−1)−p(0)x2 (a) (1 point) What is [T]E? (b) (1 point) Is T invertible? (c) (6 points) Compute the eigenvalues of T and their algebraic multiplicity. (d) (2 points) Is T diagonalisable? If so, find a matrix Q such that Q−1[T]EQ is diagonal. If not, findQ, so that the above matrix is upper triangular.