No I meant the general relativistic 4 volumeWWGD said:I don't know the details here but yours seems to be an Euclidean volume element, not the one associated with space time I believe is used in Relativity, QFT, etc. Maybe @PeterDonis can clarify this?
dsaun777 said:I meant the general relativistic 4 volume
Yes I am aware of the d^4x notation but what is this used in generally speaking.PeterDonis said:The general invariant 4-volume element in GR is ##\sqrt{-g} d^4 x##, where ##g## is the determinant of the metric (and will always be negative for a Lorentzian spacetime, hence the minus sign under the square root) and ##d^4 x## is the usual way of writing what you wrote as ##dx dy dz dt##. Note that there is no requirement for the coordinates to be called ##x##, ##y##, ##z##, and ##t##, and the determinant of the metric is what accounts for the curvature of the spacetime.
dsaun777 said:what is this used in generally speaking
Yes I am aware of the lagrangian density integral. I was just wondering what other uses it has specifically qft and GR.pervect said:One use of the four-volume element is that one integrates the Lagrangian density ##\mathcal{L}##, which is a tensor density, over the four volume to get the action S.
see https://en.wikipedia.org/w/index.php?title=Lagrangian_(field_theory)&oldid=892036798#Action
The integral of the Lagrangian density over a spatial 3-veolume at some time t t is the Lagrangian for the 3-volume. Integrated over the 4-volume you get the action, the integral of Ldt.
And of course one can use the principle of stationary action to do a lot of physics.
dsaun777 said:Yes I am aware of the lagrangian density integral. I was just wondering what other uses it has specifically qft and GR.
I want to nominate this as the best post on this forum!vanhees71 said:Usually in relativistic physics you deal with fields as the dynamical objects, like the electromagnetic field, various currents of (conserved) charges, the fluid-four-velocity field, the energy-momentum-stress tensor etc. etc.
All these are "local" quantities, but sometimes you also want more global quantities like the total energy, momentum, and angular momentum of some dynamical system described by fields, and to make sense of such quantities you have to integrate, but when integrating it should make physical sense, and the integrals should be interpretable as physical quantities, which implies that you should get well defined covariant objects (in special relativity everything should be covariant under proper orthochronous Poincare transformations, in GR even generally covariant, i.e., under any diffeomorphism of (local) coordinates).
That's why you restrict your integrals by definition to such integrals that lead to covariant objects. Let's take the most simple example and consider special relativity, using Minkowski coordinates ##x^{\mu}## (inertial frames) with the pseudometric represented by the components ##\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)## (west-coast convention). If you have a current four-vector ##j^{\mu}## that obeys the local conservation law,
$$\partial_{\mu} j^{\mu}=0.$$
What's the meaning of this? First of all you have ##(j^{\mu})=(c \rho,\vec{j})##, where ##\rho## is the charge density and ##\vec{j}## is the current density of this charge (e.g., electric charge and the electric current density).
Now let's first discuss this in terms of the non-covariant usual 3D Euclidean vector calculus. The it's clear that the charge contained in some volume ##V## of an observer's 3D space who is at rest with respect to the chosen inertial reference frame is given by
##Q_V=\int_V \mathrm{d}^3 x \rho(\vec{x})##
and that the amount of charge running through the boundary ##\partial V## of the volume per unit time is given by
$$I=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{j}.$$
The surface-element vectors ##\mathrm{d}^2 \vec{f}## are by definition pointing out of the volume everywhere. Then you can apply Gauss's integral theorem and write
$$I=\int_{V} \mathrm{d}^3 x \vec{\nabla} \cdot \vec{j}.$$
Due to the conservation law (equation of continuity)
$$\partial_{\mu} j^{\mu}=\frac{1}{c} \partial_t (c \rho)+\vec{\nabla} \cdot \vec{j}=\partial_t \rho + \vec{\nabla} \cdot \vec{j}=0.$$
Thus we have
$$I=-\int_V \mathrm{d}^3 x \partial_t \rho.$$
If now the volume and its boundary is at rest in the considered reference fram, we can put the time derivative out of the integral, and we finally get
$$I=-\dot{Q}_V,$$
which says that the change of the amount of charge contained in ##V## is due to the current of charge flowing through the surface, i.e., no charge is created or destroyed but all change of the charge content is due to flow. The minus sign is due to our standard definition of the orientation of the surface by choosing the surface-normal vectors to point out of the volume, i.e., a positive current means that charge flows out of the volume and thus the charge contained in the volume decreases.
Now, how does another observer (let's call him Bob) moving against our just considered observer (let's call her Alice) describe this situation? With our 3D calculus we are quite lost here, because what we considered, even when only considering the most simple case of Bob being at rest in another inertial frame moving against Alice's inertial frame with constant velocity. The difficulty is that we argued completely in terms of Bob's coordinates and considered a volume and its boundary at rest relative to Bob. The spatial integrals were thus taken at a fixed coordinate time in Bob's frame. How should Alice describe the very same situation? She has to integrate somehow also such that the condition that the integrals are taken at Bob's fixed time ##t##, but since ##V## and ##\partial V## are extended regions in (Bob's!) space what's at some constant time ##t## for Bob is not at equal times for Alice. Nevertheless Alice should get the same quantities ##Q## and ##I## as Bob since charge inside a volume and current running through this volume's boundary should be the same for Alice as for Bob, but she has to integrate somehow over the same ##V## and ##\partial V## though both are moving relative to Alice, and she cannot simply integrate at a fixed coordinate time ##t'## with respect to her inertial frame.
As usual, to analyse this in the non-covariant (1+3)-formalism is a night-mare. That's why one defines a formalism for covariant integrals in (1+3)D Minkowski space. It's somehow clear that as in usual Euclidean 3D space, where you have volume integrals (3D integrals), surface integrals (2D integrals), and line integrals (1D integrals), where you integrate scalar fields over a volume and vector fields over surfaces or along lines such that you get scalar quantities. Of course in 3D Euclidean space by "scalar" we mean quantities that are invariant under rotations (indeed vectors are invariant under rotations, the volume element ##\mathrm{d}^3 x##, where ##x## are Cartesian coordinates is invariant, ##\mathrm{d}^2 \vec{f}## behaves like a vector under rotations (but not under space reflections, where it is an axial vector!) as well as the line element ##\mathrm{d} \vec{x}## are also vectors, i.e., invariant under rotations, i.e., the integrals all are scalars under rotations. Using the Euclidean coordinates that's all due to the fact that the involved SO(3) matrices representing rotations leave the scalar product ##\mathrm{d} \vec{x} \cdot \vec{V}=\mathrm{d} x_j V_j## (Euclidean Einstein-summation convention implied) invariant. Also since ##\mathrm{det} \hat{O}=1## for ##\hat{O} \in \mathrm{SO}(3)## both ##\mathrm{d}^3 x## is invariant (it's a pseudoscalar under space reflections tough!) and also the orientation of ##\mathrm{d}^2 \vec{f}## doesn't change.
Now the very same arguments hold in (1+3)D Minkowski space. The only thing one has to be careful with are some signs, because now we must distinguish upper and lower indices when dealing with vector and tensor components, and the pseudometric has negative signature, particularly ##\det (\eta_{\mu \nu})=-1##. Note that there are different conventions in the literature concerning the signs, and it's totally arbitrary how you choose the convention but you must stay consistent within your convention. I use the one I'm used to and define the contravariant components of the Levi-Civita tensor such that
$$\epsilon^{\mu \nu \rho \sigma}=\begin{cases} 1 &\text{if} \quad (\mu,\nu,\rho,\sigma) \quad \text{is an even permutation of (0,1,2,3)},\\
-1 &\text{if} \quad (\mu,\nu,\rho,\sigma) \quad \text{is an odd permutation of (0,1,2,3)} \\
0 & \text{otherwise}. \end{cases}$$
In other words ##\epsilon^{0123}=1## and ##\epsilon^{\mu \nu \rho \sigma}## is totally antisymmetric under exchange of the order of the indices.
The great thing is that ##\epsilon^{\mu \nu \rho \sigma}## are invariant tensor components under proper Lorentz transformations, i.e., such Lorentz transformations for which the corresponding matrix ##\hat{\Lambda} \in \mathrm{SO}(1,3)## with ##\mathrm{det} \hat{\Lambda}=+1## since
$$\epsilon^{\prime \alpha \beta \gamma \delta}={\Lambda^{\alpha}}_{\mu} {\Lambda^{\beta}}_{\nu} {\Lambda^{\gamma}}_{\rho} {\Lambda^{\delta}}_{\sigma} \epsilon^{\mu \nu \rho \sigma}=\det \hat{\Lambda} \epsilon^{\alpha \beta \gamma \delta}.$$
However the covariant components are flipped in sign since the pseudometric's determinant is ##-1##:
$$\epsilon_{\mu \nu \rho \sigma}=\det(\hat{\eta}) \epsilon^{\mu \nu \rho \sigma}=-\epsilon^{\mu \nu \rho \sigma}.$$
It's now trivial to see, but now you can define integrals over 4D, 3D, 2D, and 1D submanifolds (or hypersurfaces) of Minkowski space by defining
$$\mathrm{d}^4 x=-\epsilon_{\mu \nu \rho \sigma} \mathrm{d} x_1^{\mu} \mathrm{d} x_2^{\nu} \mathrm{d} x_3^{\rho} \mathrm{d} x_4^{\sigma}=\mathrm{d} x^0 \mathrm{d} x^1 \mathrm{d} x^2 \mathrm{d} x^3,\\
\mathrm{d}^3 \sigma^{\mu} =-\epsilon^{\mu \nu \rho \sigma} \mathrm{d} x_{1\nu} \mathrm{d} x_{2\rho} \mathrm{d} x_{3\sigma},\\
\mathrm{d}^2 \sigma^{\mu \nu} = -\epsilon^{\mu \nu \rho \sigma} \mathrm{d} x_{1\rho} \mathrm{d} x_{2\sigma}, \\
\mathrm{d} \sigma^{\mu \nu \rho}=-\epsilon^{\mu \nu \rho \sigma} \mathrm{d} x_{\sigma}.
$$
Here we assume that we have parametrized the hypersurfaces by the appropriate number of parameters, e.g., for the 3D-hypersurface by ##(q^1,q^2,q^3)##. Then we define ##\mathrm{d} x_{j}^{\mu} =\mathrm{d} q_j \partial_{q_j} x(q)^{\mu}## (where here we don't sum over ##j## of course).
Now you can integrate scalar fields, vector fields, antisymmetric 2nd-rank tensor fields, and totally antisymmeric 3rd-rank tensor fields and get scalars. Of course the latter case is just the same as integrating the Hodge dual of the 3rd-rank totally antisymmetric tensor field along a line with the integration element ##\mathrm{d} x^{\mu}##.
Now, how about the situation with the charge and current denstiy, ##j^{\mu}## which is a four-vector field, taking Bob's fixed-time integrals? Bob is characterized by being at rest in his frame. In a covariant way this is described by Bob's four-velocity vector ##u_{\text{B}}^{\mu}=(1,0,0,0)##.
It's very easy to make the volume integral covariant. Bob defines his volume at a fixed time ##t## of his frame, i.e., it's characterized by the implicit equation ##\Phi(x)=u_{\mu} x^{\mu}=x^0=c t=\text{const}##, and as parameters Bob may choose his spatial coordinates ##\vec{x}##. The "Minkowski-orthogonal" hypersurface vector is given by ##\partial_{\mu} \Phi(x)=u_{\mu}## and thus his hyper-surface element is simply
$$\mathrm{d}^3 \sigma^{\mu}=\mathrm{d} x^1 \mathrm{d} x^2 \mathrm{d} x^3 u^{\mu}.$$
Now the charge contained in his volume ##V## is indeed manifestly covariantly calculated by
$$Q=\frac{1}{c} \int_V \mathrm{d}^3 \sigma^{\mu} j_{\mu}=\int_{V} \mathrm{d}^3 \vec{x} \rho(t,\vec{x}).$$
For the evaluation of the current ##I## the argument is very similar:
In the 4D covariant integration formalism the integration element over a 2D hypersurface is a 2nd-rank. Bob only integrates over spatial-plane hypersurface elements, i.e., ##\mathrm{d}^2 \sigma^{0\nu}=u_{\mu} \mathrm{d}^2 \sigma^{\mu \nu}##. It's pretty easy to see that indeed
$$I=\int_{\partial V} \mathrm{d}^2 \sigma^{\mu \nu} u_{\mu} j_{\nu}.$$
Now it's very easy to calculate the same expressing the various vectors and fields in terms of Alice's components. You just use the Lorentz transformation for the various vectors. Of course Alice can use the same parameters as Bob to parametrize the 3D- and 2D-hypersurfaces. These parameters can be considered as scalars.
Of course there are also the analogues of Gauss's integral theorem for the various integrals. E.g., the 4D integral over the divergence of a vector field is
$$\int_{V^{(4)}} \mathrm{d}^4 x \partial_{\mu} A^{\mu} = \int_{\partial V^{(4)}} \mathrm{d}^3 \sigma^{\mu} A_{\mu}.$$
This can be used to easily show from ##\partial_{\mu} j^{\mu}=0## that the total charge, i.e.,
$$Q_{\text{tot}}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \rho(t,\vec{x})=\text{const}$$
is a Lorentz scalar.
Yes. The energy momentum tensor is composed of the currents corresponding to energy and momentum.dsaun777 said:Would the same reasoning follow for calculating the total energy momentum flow across the surface boundary in the case of the energy momentum tensor?
A 4-dimensional element volume refers to the amount of space occupied by a four-dimensional object. In other words, it is the measure of the three-dimensional space within a four-dimensional object.
A 4-dimensional element volume is different from a 3-dimensional volume because it takes into account an additional dimension. While a 3-dimensional volume measures the space occupied by a three-dimensional object, a 4-dimensional element volume measures the space within a four-dimensional object.
It is difficult for humans to visualize a 4-dimensional element volume as we are only able to perceive three dimensions. However, there are mathematical models and computer simulations that can help us understand and visualize four-dimensional objects and their volumes.
The 4-dimensional element volume is calculated using mathematical formulas and equations specific to four-dimensional space. These calculations involve considering the length, width, height, and an additional dimension of the object.
Understanding 4-dimensional element volume is important in fields such as physics, engineering, and mathematics. It can also be useful in understanding complex systems and patterns in nature, such as weather patterns and the behavior of particles in quantum mechanics.