# I Covariant derivative of the contracted energy-momentum tensor of a particle

#### sergiokapone

The energy-momentum tensor of a free particle with mass $m$ moving along its worldline $x^\mu (\tau )$ is

T^{\mu\nu}(y^\sigma)=m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}.

Let contract $T^{\mu\nu}$:

T(y) = g_{\mu\nu}(y)T^{\mu\nu}(y^\sigma)=m\int d \tau g_{\mu\nu}(y)\frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau} = m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}.

Lets get covariant derivative

\nabla_{\nu} T = \nabla_{\nu} \left(m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\right) = \partial_{\nu} \left(m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\right) = m\int d \tau \partial_{\nu} \left(\frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\right) = 0.

Does it wright $\nabla_{\nu} T = 0$? Or I'm wrong?

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#### haushofer

Well, these kind of calculations tend to get messy, as you want to differentiate trajectories and hence involve delta "functions". But you should get

$$\nabla_{\mu}T^{\mu\nu} = 0$$

and that is something different than your result

$$\nabla_{\mu}(T^{\rho\lambda}g_ {\rho\lambda})= 0$$

So I'd say you're wrong. To be honest, I'd have to check a textbook how you should take this divergence properly of this wordline-integral, so maybe someone else knows.

#### PeterDonis

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The energy-momentum tensor of a free particle with mass mm moving along its worldline $x^\mu (\tau )$ is
This doesn't look correct. First, I'm not sure why you would be taking an integral, since the SET is supposed to be a local quantity at a single spacetime point, not an integral quantity. Second, your integral doesn't make sense because there are two factors of $d\tau$ in the denominator and only one in the numerator, and $1 / d\tau$ is not a well-defined integration measure.

Let contract
I'm not sure why you want to contract the tensor. The covariant divergence, as @haushofer has pointed out, is $\nabla_\mu T^{\mu \nu}$.

#### Orodruin

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First, I'm not sure why you would be taking an integral, since the SET is supposed to be a local quantity at a single spacetime point, not an integral quantity.
The given expression is a local quantity at every spacetime point. The 4D-delta in the integrand is only non-zero whenever you are at the point of integration. The integration along the world-line means that the stress-energy tensor is a non-zero quantity along the world-line and has a behaviour similar to a three-dimensional delta function in four dimensions (e.g., a line charge or similar).

Consider a spatial hypersurface $S$ with a fixed time coordinate $t_0$ and spatial coordinates $y^i$. You would then have
$$\int_S T^{\mu\nu}dS_\mu = \int_S m \int_\tau \delta^{(3)}(y^i - x^i(t)) \delta(t_0-t(\tau)) U^\mu U^0 d\tau d^3y.$$
Changing variables from $\tau$ to $t$, we have $U^0 d\tau = (dt/d\tau) d\tau = dt$ and therefore
$$\int_S T^{\mu\nu}dS_\mu = m U^\mu = P^\mu$$
evaluated at the point where the world-line crosses $S$, which is clearly the correct result.

#### sergiokapone

This doesn't look correct. First, I'm not sure why you would be taking an integral, since the SET is supposed to be a local quantity at a single spacetime point, not an integral quantity. Second, your integral doesn't make sense because there are two factors of dτdτd\tau in the denominator and only one in the numerator, and 1/dτ1/dτ1 / d\tau is not a well-defined integration measure.
This definition I saw in different sources, for example here, also in Einstein Gravity in a Nutshell Zee

I'm not sure why you want to contract the tensor. The covariant divergence, as @haushofer has pointed out, is $\nabla_{\mu}T^{\mu\nu}$ .
I know, how to derive $\nabla_{\mu}T^{\mu\nu}$ (see, here), my purpose is to derive $\nabla_{\mu}T$ (where $T = g_{\rho\nu}T^{\rho\nu}$) for some reason.

#### PeterDonis

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The given expression is a local quantity at every spacetime point.
I don't see how. It's integrating over $\tau$ and the integrand is a function of $\tau$.

The 4D-delta in the integrand is only non-zero whenever you are at the point of integration.
But the 4D-delta is a function of $\tau$, because the $x^\sigma$ in its argument is a function of $\tau$. So there is no single point at which the delta function is nonzero; which point that is depends on $\tau$. To single out one particular spacetime point, you would have to pick one particular value of $\tau$, not integrate over $\tau$.

#### Orodruin

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I don't see how. It's integrating over ττ\tau and the integrand is a function of ττ\tau.
This is irrelevant. It is just telling you that the stress energy tensor is non-zero on the world line and zero everywhere else. It is a local function of the coordinates $y^\mu$ all the same.

I don't see how. It's integrating over $\tau$ and the integrand is a function of $\tau$.

But the 4D-delta is a function of $\tau$, because the $x^\sigma$ in its argument is a function of $\tau$. So there is no single point at which the delta function is nonzero; which point that is depends on $\tau$. To single out one particular spacetime point, you would have to pick one particular value of $\tau$, not integrate over $\tau$.
The proper time is just an integration variable here. What really matters is how the stress energy tensor varies with the coordinates $y^\mu$ and the integral over the world line with the delta function makes sure that it is zero everywhere but on the world line. As demonstrated in my post, it has the correct behaviour when considering it as the 4-momentum current, which really is the relevant property.

To single out a particular spacetime point, you need to pick the coordinates $y^\mu$, not $\tau$.

#### PeterDonis

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the integral over the world line with the delta function makes sure that it is zero everywhere but on the world line
So basically, this is just due to the general fact that whenever a delta function appears, it needs to be inside an integral to be meaningful?

#### sergiokapone

So basically, this is just due to the general fact that whenever a delta function appears, it needs to be inside an integral to be meaningful?
I thik the mathematically correct transformations with $\delta$ is inside integral. But if you pull it outside integral

\int d \tau \delta^{(4) }(y^\sigma-x^\sigma(\tau )) = \frac{d\tau}{dt}\delta^{(3)}(\vec{y}-\vec{x}(t)) .

then need to be careful.

#### Orodruin

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So basically, this is just due to the general fact that whenever a delta function appears, it needs to be inside an integral to be meaningful?
I prefer to consider it a distribution, but yes. In the case of a point particle, you expect the stress-energy tensor to be non-zero along the world-line only so it should include some delta distribution. Since the world-line is one-dimensional, you can construct the corresponding 3D delta distribution by integrating the 4D delta along the world-line with the appropriate weight (which the expression has, as shown above).

#### PeterDonis

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Since the world-line is one-dimensional, you can construct the corresponding 3D delta distribution by integrating the 4D delta along the world-line with the appropriate weight (which the expression has, as shown above).
I'm still having trouble wrapping my mind around the idea that we can construct a 3D distribution by integrating a 4D distribution, and the idea that we can obtain a local quantity at a single spacetime point by integrating over a worldline. But I'll keep looking at your post #4 to see if it helps.

#### Orodruin

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Let me take another example unrelated to relativity: the electric current density in an infinite idealised line conductor carrying the current $I$. This is given by
$$\vec J = I \delta(x)\delta(y)\vec e_3.$$
In analogy to the above, this can be written as
$$\vec J = I \int \delta(z-s) \delta(x)\delta(y)\vec e_3 ds,$$
where $s$ is the curve length along the wire.

#### PeterDonis

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Let me take another example unrelated to relativity
Ok, so let me describe what this appears to be telling me.

The first expression tells me that the current density is zero except where $x = y = 0$. It's not clear what, if any, other variable this is supposed to be a function of, but if I view it as a function of $z$, it is telling me that the current density is the same at every value of $z$, i.e., all along the wire.

The second expression is telling me the same thing, but it's making explicit the functional dependence on $z$; it's basically saying that there is a continuous series of x-y delta functions along a line that extends infinitely in the $z$ direction. The "continuous series" part is what the third delta function and the integral are for.

#### Orodruin

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Ok, so let me describe what this appears to be telling me.

The first expression tells me that the current density is zero except where $x = y = 0$. It's not clear what, if any, other variable this is supposed to be a function of, but if I view it as a function of $z$, it is telling me that the current density is the same at every value of $z$, i.e., all along the wire.

The second expression is telling me the same thing, but it's making explicit the functional dependence on $z$; it's basically saying that there is a continuous series of x-y delta functions along a line that extends infinitely in the $z$ direction. The "continuous series" part is what the third delta function and the integral are for.
Right, this is the correct interpretation. It is the same idea as for the stress-energy tensor of a point particle as presented in the OP - except that one is in curved space, general coordinates, and you have four currents (one for each component of the 4-momentum).

#### PeterDonis

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this is the correct interpretation

#### Orodruin

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Thanks! It is one of my favourite examples.
Note how $\nabla\cdot \vec J = 0$ and that it gives you exactly what you need as the source in Ampere’s law for it to be satisfied by the magnetic field we are so used to from an infinite straight wire.

#### haushofer

I still have some questions, but maybe I'm missing the point:

[*] Why does TS want to show $\nabla_{\mu} T = 0$? What is this "for some reason"? And why would this be true in the first place?
[*] How do you show that $\nabla_{\mu} T^{\mu\nu} = 0$ for this point particle? How do you take the covariant derivatives of the delta distribution, and these 4-velocities which are only defined on the worldline?

Just for closure ;)

#### Orodruin

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[*] Why does TS want to show $\nabla_{\mu} T = 0$? What is this "for some reason"? And why would this be true in the first place?
Beats me.

[*] How do you show that $\nabla_{\mu} T^{\mu\nu} = 0$ for this point particle? How do you take the covariant derivatives of the delta distribution, and these 4-velocities which are only defined on the worldline?
The derivative must (of course) be seen in a distributional sense, just as the tensor itself. Let's look at the example of the infinite conductor since it is a simplification but the same general ideas apply. You would then be interested in computing $\nabla \cdot \vec J$. With the second (i.e, the integral) expression for $\vec J$, you would obtain (in Cartesian coordinates)
$$\nabla\cdot \vec J = I \int (\vec e_1 \partial_x + \vec e_2 \partial_y + \vec e_3 \partial_z) \cdot \vec e_3 [\delta(z-s) \delta(x) \delta(y)] ds = I\int \delta'(z-s) \delta(x) \delta(y) ds = I \delta(x) \delta(y) \int \delta'(z-s) ds = 0,$$
since the integral of $\delta'$ is zero. In essence it boils down to the total current not changing along the wire. If you did everything for a general wire with a current that a priori could be taken as dependent on the position along the wire you would obtain
$$I'(s) = 0.$$

#### sergiokapone

[*] Why does TS want to show $\nabla_{\mu} T = 0$? What is this "for some reason"? And why would this be true in the first place?
Reason: I try to check consequences some type of $\Lambda(R)$ theory (if you are interested, I will show the equation of the therory later), there I obtain such derivative. Please, check my calculations:

\begin{multline}
\nabla_{\mu} T = \partial_{\mu}\int d\tau \frac{m\,\delta^{(4)}(x- X(\tau))}{\sqrt{-g(x)}} = \\
= \int d\tau \left( \frac{m}{\sqrt{-g(x)}} \partial_{\mu} \delta^{(4)}(x- X(\tau)) + \delta^{(4)}(x- X(\tau))\partial_{\mu} \frac{m}{\sqrt{-g(x)}}\right) = \\
= \int d\tau \left( -\delta^{(4)}(x- X(\tau)) \partial_{\mu} \frac{m}{\sqrt{-g(x)}} + \delta^{(4)}(x- X(\tau))\partial_{\mu} \frac{m}{\sqrt{-g(x)}}\right) = 0
\end{multline}

Last edited:

#### haushofer

Reason: I try to check consequences some type of $\Lambda(R)$ theory (if you are interested, I will show the equation of the therory later), there I obtain such derivative. Please, check my calculations:

\begin{multline}
\nabla_{\mu} T = \partial_{\mu}\int d\tau \frac{m\,\delta^{(4)}(x- X(\tau))}{\sqrt{-g(x)}} = \\
= \int d\tau \left( \frac{m}{\sqrt{-g(x)}} \partial_{\mu} \delta^{(4)}(x- X(\tau)) + \delta^{(4)}(x- X(\tau))\partial_{\mu} \frac{m}{\sqrt{-g(x)}}\right) = \\
= \int d\tau \left( -\delta^{(4)}(x- X(\tau)) \partial_{\mu} \frac{m}{\sqrt{-g(x)}} + \delta^{(4)}(x- X(\tau))\partial_{\mu} \frac{m}{\sqrt{-g(x)}}\right) = 0
\end{multline}
You cannot do a partial integration like you do. You differentiate w.r.t. the coordinate x, but integrate w.r.t. the worldine coordinate tau. So you cannot throw away boundary terms like that; it's circular reasoning. The term you throw away is the very first term you start out with, so trivially you obtain your result.

I'm not familiar with higher order extensions of GR. In GR, the divergence of T is zero due to the Bianchi identities. What identities do you use to conclude that $\nabla_{\mu} T=0$? It must follow from some identity.

#### haushofer

Beats me.

The derivative must (of course) be seen in a distributional sense, just as the tensor itself. Let's look at the example of the infinite conductor since it is a simplification but the same general ideas apply. You would then be interested in computing $\nabla \cdot \vec J$. With the second (i.e, the integral) expression for $\vec J$, you would obtain (in Cartesian coordinates)
$$\nabla\cdot \vec J = I \int (\vec e_1 \partial_x + \vec e_2 \partial_y + \vec e_3 \partial_z) \cdot \vec e_3 [\delta(z-s) \delta(x) \delta(y)] ds = I\int \delta'(z-s) \delta(x) \delta(y) ds = I \delta(x) \delta(y) \int \delta'(z-s) ds = 0,$$
since the integral of $\delta'$ is zero. In essence it boils down to the total current not changing along the wire. If you did everything for a general wire with a current that a priori could be taken as dependent on the position along the wire you would obtain
$$I'(s) = 0.$$
Ok, I like that approach. You made me fan of this example too. ;)

But now the follow-up: how would you evaluate an expression like (just to pick one)

$$\frac{dx^{\mu}}{d\tau}\nabla_{\mu} \frac{dx^{\nu}}{d\tau}$$

under the integral sign? My guess is that you use the geodesic equation for the particle, but my discomfort is more with what the covariant derivative of such a wordline-quantity actually means. (or to put it in fancy language: how do you differentiate a field on the worldline (depending only on $\tau$) w.r.t. the targetspace coordinates $x^{\mu}$? Just using the chainrule?)

If you know a reference which covers this, I'd be happy to read about it. I've never really understood this. :)

#### PeterDonis

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In GR, the divergence of T is zero due to the Bianchi identities.
To be clear, in GR the covariant divergence of $T^{\mu \nu}$, i.e., $\nabla_\mu T^{\mu \nu}$, is zero due to the Bianchi identities. As you note, that is not the same as $\nabla_\mu T = \nabla_\mu T^\nu{}_\nu$ being zero.

#### Orodruin

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To be clear, in GR the covariant divergence of $T^{\mu \nu}$, i.e., $\nabla_\mu T^{\mu \nu}$, is zero due to the Bianchi identities. As you note, that is not the same as $\nabla_\mu T = \nabla_\mu T^\nu{}_\nu$ being zero.
To be even clearer, that is the total stress-energy tensor. You can also have interactions between different contributions to the stress-energy tensor that transfer energy and momenta between two fields (for example). The stress-energy tensor of each individual field is then not necessarily divergence free. This is just a generalisation of Newton’s second law.

"Covariant derivative of the contracted energy-momentum tensor of a particle"

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