I Continuity equation in GR

dsaun777

Taking the time derivative of the energy density of the energy momentum tensor should equal to the spatial derivatives or divergence of the momentum density components. How do the units work out though? shouldn't a time derivative of the energy density be in kg/xt^3 and the spatial derivative of the momentum density be kg/(x^2 t^2). I thought all components of the energy momentum tensor have the same units or am I missing a physical interpretation of what is going on? In the simplest form continuity of some flux ∇j=∂tρ the units make sense what am i missing?

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Orodruin

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The components of a tensor (in holonomic basis) have the same units only if the coordinates all have the same dimensions. This is only true for the energy momentum tensor in Minkowski coordinates if you use units where c=1 and length and time therefore have the same dimensions. It becomes even worse in GR (or just general curvilinear coordinates where coordinates often are angles etc).

dsaun777

The components of a tensor (in holonomic basis) have the same units only if the coordinates all have the same dimensions. This is only true for the energy momentum tensor in Minkowski coordinates if you use units where c=1 and length and time therefore have the same dimensions. It becomes even worse in GR (or just general curvilinear coordinates where coordinates often are angles etc).
So by equating c=1 taking spatial derivatives are the same as taking the time derivatives?

pervect

Staff Emeritus
Using Wiki's units and assuming a flat Minkowskii space-time, $T^{00} = (1/c^2)$* energy density, and $T_{00} = c^2$* energy density.

There's a certain amount of flexibility as to exactly where one does the accounting for the unit differences, but if you set c=1, you don't have to worry about all the details.

My interpretation of what Wiki does, for instance is that the consider the components of the stress-energy tensor $T^{ij}$ to all have the same units, which are kg.

Similarly, for 4-vectors, Wiki use units of meters, so the time component of a 4-vector becomes $c \tau$ rather than $\tau$.

If you use a different text, different conventions may be used. MTW uses geometric units throughout, which I find is the simplest approach. Then the units of energy, momentum, and pressure all have the same units, geometrodynamic centimeters.

Some common conversion factors for geometrodynamic units from the text,. Note that we assume both G, the gravitational constant and c, the speed of light, are unity in geometric units.

c = $3 \, 10^{10}$ cm/sec
$G/c^2$ = $.75 \, 10^{-28}$ cm/gram
$G/c^4$ = $.8 \, 10^{-49}$ cm/erg

So if you have 1 erg, you multiply by $G/c^4$ to get the value in geometerodynamic centimeters, and if you have 1 cm, you multiply it by $c^4/G$ to get ergs.

dsaun777

Using Wiki's units and assuming a flat Minkowskii space-time, $T^{00} = (1/c^2)$* energy density, and $T_{00} = c^2$* energy density.

There's a certain amount of flexibility as to exactly where one does the accounting for the unit differences, but if you set c=1, you don't have to worry about all the details.

My interpretation of what Wiki does, for instance is that the consider the components of the stress-energy tensor $T^{ij}$ to all have the same units, which are kg.

Similarly, for 4-vectors, Wiki use units of meters, so the time component of a 4-vector becomes $c \tau$ rather than $\tau$.

If you use a different text, different conventions may be used. MTW uses geometric units throughout, which I find is the simplest approach. Then the units of energy, momentum, and pressure all have the same units, geometrodynamic centimeters.

Some common conversion factors for geometrodynamic units from the text,. Note that we assume both G, the gravitational constant and c, the speed of light, are unity in geometric units.

c = $3 \, 10^{10}$ cm/sec
$G/c^2$ = $.75 \, 10^{-28}$ cm/gram
$G/c^4$ = $.8 \, 10^{-49}$ cm/erg

So if you have 1 erg, you multiply by $G/c^4$ to get the value in geometerodynamic centimeters, and if you have 1 cm, you multiply it by $c^4/G$ to get ergs.
Thats great but what does it mean to take spatial and time derivatives when all there is are units of mass and cm?

pervect

Staff Emeritus
Thats great but what does it mean to take spatial and time derivatives when all there is are units of mass and cm?
Space and time are unified in special relativity.

You might want to consider - what does the Lorentz transform mean, when it writes

$$t' = \gamma t - \gamma v x / c^2$$

You see the same "mixing together" (to use the non-technical term) of time and space in the Loretnz transform. t' depends not only on t, it also depends on x. So t' = f(t,x). The basic conclusion that we arrive at with the geometrical approach to relativity is that space and time are unified.

From your question, I gather you consider time and space to be different. But if they are different why do we add the term $\gamma v x / c^2$ in the above transform? I.e. how do we explain that t' depends not only on t, but also on x?

To make a point a bit more relevant to your initial question, suppose we have some stress energy tensor in some frame S. And we write the continuity equations in frame S.

If we assume we have only one space and one time dimension, i.e. only x and t, we have only have 4 terms in the stress-energy tensor in this case, so we can write out the continuity equations longhand

$$\partial_0 T^{00} + \partial_1 T^{10}=0 \quad \partial_0 T^{01} + \partial_1 T^{11}=0$$

here $\partial_0$ means $\frac{\partial}{\partial x^0}$, or, since $x^0=t$, it means $\frac{\partial}{\partial t}$. Similarly, $\partial_1$ means $\frac{\partial}{\partial x}$.

It might be helpful to replacing the numeric symbols 0,1 with the alphabetic symbols t and x, in which case we'd write:

$$\partial_t T^{tt} + \partial_x T^{xt}=0 \quad \partial_t T^{tx} + \partial_x T^{xx}=0$$

Now, suppose we choose some different frame S', moving with respect to S. The important thing we need to have happen is that the equations in S' are satisfied if the equations in S are satisfied. The tensor transformation laws for T (i'm not sure if you've been exposed to them yet, I won't go over the details) combined with the chain rule from multivariable calculus allows us to insure that this is true, that when the equations are true in S, they are also true in S'.

I won't write out the chain rule in details, but I will point out that the chain rule says that $\partial_{t'}$ is a function of BOTH $\partial_t$ AND $\partial_x$.

The reason the stress-energy tensor is set up the way it is, is to make this is manifestly true, to make sure a system that satisfied the continuity laws in one frame satisifies it in all frames.

dsaun777

Space and time are unified in special relativity.

You might want to consider - what does the Lorentz transform mean, when it writes

$$t' = \gamma t - \gamma v x / c^2$$

You see the same "mixing together" (to use the non-technical term) of time and space in the Loretnz transform. t' depends not only on t, it also depends on x. So t' = f(t,x). The basic conclusion that we arrive at with the geometrical approach to relativity is that space and time are unified.

From your question, I gather you consider time and space to be different. But if they are different why do we add the term $\gamma v x / c^2$ in the above transform? I.e. how do we explain that t' depends not only on t, but also on x?

To make a point a bit more relevant to your initial question, suppose we have some stress energy tensor in some frame S. And we write the continuity equations in frame S.

If we assume we have only one space and one time dimension, i.e. only x and t, we have only have 4 terms in the stress-energy tensor in this case, so we can write out the continuity equations longhand

$$\partial_0 T^{00} + \partial_1 T^{10}=0 \quad \partial_0 T^{01} + \partial_1 T^{11}=0$$

here $\partial_0$ means $\frac{\partial}{\partial x^0}$, or, since $x^0=t$, it means $\frac{\partial}{\partial t}$. Similarly, $\partial_1$ means $\frac{\partial}{\partial x}$.

It might be helpful to replacing the numeric symbols 0,1 with the alphabetic symbols t and x, in which case we'd write:

$$\partial_t T^{tt} + \partial_x T^{xt}=0 \quad \partial_t T^{tx} + \partial_x T^{xx}=0$$

Now, suppose we choose some different frame S', moving with respect to S. The important thing we need to have happen is that the equations in S' are satisfied if the equations in S are satisfied. The tensor transformation laws for T (i'm not sure if you've been exposed to them yet, I won't go over the details) combined with the chain rule from multivariable calculus allows us to insure that this is true, that when the equations are true in S, they are also true in S'.

I won't write out the chain rule in details, but I will point out that the chain rule says that $\partial_{t'}$ is a function of BOTH $\partial_t$ AND $\partial_x$.

The reason the stress-energy tensor is set up the way it is, is to make this is manifestly true, to make sure a system that satisfied the continuity laws in one frame satisifies it in all frames.
Yes I understand that time and space are unified i.e. space-time. But you can still calculate local time derivatives and do transformations later so that they are tensorial. So the continuity equation is saying the rate of change of energy density is equal to partial i T^{0i} What is this quantity partial of the time component of T^{00} in terms of dimensional analysis? what is the physical interpretation of the rate of change of energy density of the stress energy tensor?

PeterDonis

Mentor
How do the units work out though?
If you use "natural" units in which $c = 1$, which are the preferred units in GR, then the units of a time derivative are the same as the units of a spatial derivative. You are encountering problems because you are trying to use clumsy units.

dsaun777

If you use "natural" units in which $c = 1$, which are the preferred units in GR, then the units of a time derivative are the same as the units of a spatial derivative. You are encountering problems because you are trying to use clumsy units.
So with c=1 and length is time and mass is energy then the energy density is mass density which is also m/t^3?

PeterDonis

Mentor
So with c=1 and length is time and mass is energy then the energy density is mass density which is also m/t^3?
If you want to look at it that way, yes.

dsaun777

If you want to look at it that way, yes.
I get the elegance of putting c=1 but doesn't that completely obscure the practical meaning of what those components are supposed to represent physically. Momentum flux is a thing, energy density is a thing despite the fact you can interchange units they have different describable states. Lets say I'm in a vacuum and want to describe the density of a box of perfect non interacting fluid like mass at rest relative to me. I also want to measure the rate of change of the density of the box if a fluid is being poured into it. wouldn't my best description involve measuring the rate of change of the energy density of the box ie time derivative of the density component? How does letting c=1 help me in reality.

PeterDonis

Mentor
doesn't that completely obscure the practical meaning of what those components are supposed to represent physically
No. It just eliminates confusion (from which you appear to be suffering) due to using units which obfuscate the physics for historical reasons. Physically spacetime is a single 4-d geometry, and the natural way to measure lengths in a geometry is to use the same units in all directions. Using "seconds" for some directions and "meters" for others is no different from using "meters" for some directions and, say, "furlongs" for others. It just introduces superfluous unit conversion factors into the math; it doesn't change any of the physics.

Similarly, physically there is no difference between "mass", "energy", and "momentum" considered simply as quantities. The only reason we ordinarily use different units for them is historical accident--we developed common units for these things before we fully understood the underlying physics.

Similar remarks apply if you take derivatives of any of these quantities; physically a "time derivative" and a "space derivative" are just derivatives with respect to the spacetime geometry. There is no more reason, as far as physics is concerned, to use different units for them than there is to take space derivatives with respect to meters in one direction and furlongs in another.

How does letting c=1 help me in reality.
By eliminating the confusion you are suffering from due to using units which obfuscate the physics. See above.

Orodruin

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I get the elegance of putting c=1 but doesn't that completely obscure the practical meaning of what those components are supposed to represent physically.
Quite the opposite really in my opinion.

Also, in high-energy physics it is also common to use units where $\hbar = 1$, which makes length have units of inverse energy. Energy density then has units of E^4 (or just 4, since there is only one really relevant physical base dimension).

dsaun777

No. It just eliminates confusion (from which you appear to be suffering) due to using units which obfuscate the physics for historical reasons. Physically spacetime is a single 4-d geometry, and the natural way to measure lengths in a geometry is to use the same units in all directions. Using "seconds" for some directions and "meters" for others is no different from using "meters" for some directions and, say, "furlongs" for others. It just introduces superfluous unit conversion factors into the math; it doesn't change any of the physics.

Similarly, physically there is no difference between "mass", "energy", and "momentum" considered simply as quantities. The only reason we ordinarily use different units for them is historical accident--we developed common units for these things before we fully understood the underlying physics.

Similar remarks apply if you take derivatives of any of these quantities; physically a "time derivative" and a "space derivative" are just derivatives with respect to the spacetime geometry. There is no more reason, as far as physics is concerned, to use different units for them than there is to take space derivatives with respect to meters in one direction and furlongs in another.

By eliminating the confusion you are suffering from due to using units which obfuscate the physics. See above.
Yes, mathematically and higher levels of physics there is just spacetime geometry.
Quite the opposite really in my opinion.

Also, in high-energy physics it is also common to use units where $\hbar = 1$, which makes length have units of inverse energy. Energy density then has units of E^4 (or just 4, since there is only one really relevant physical base dimension).
If I'm trying to simply measure the rate of change of energy density of a box being filled with a near perfect fluid in a vacuum at rest to me how why would I do so capitalizing the stress energy tensor?

PeterDonis

Mentor
If I'm trying to simply measure the rate of change of energy density of a box being filled with a near perfect fluid in a vacuum at rest to me how why would I do so capitalizing the stress energy tensor?
What do you mean by "capitalizing the stress energy tensor"?

If you want to measure the rate of change of energy density, just measure it. What's the problem?

dsaun777

What do you mean by "capitalizing the stress energy tensor"?

If you want to measure the rate of change of energy density, just measure it. What's the problem?
How do I theoretically calculate that using the stress energy tensor?

PeterDonis

Mentor
How do I theoretically calculate that using the stress energy tensor?
You take the appropriate derivative. What's the problem?

dsaun777

You take the appropriate derivative. What's the problem?
Since c=1 what would that be?

PeterDonis

Mentor
Since c=1 what would that be?
Um, taking the derivative with respect to whatever unit of time corresponds to $c = 1$? So if you are measuring distance in meters, a meter of time is about 3.3 nanoseconds. Or if you measure time in seconds, a second of distance is 299,792,458 meters.

I think you're making this a lot harder than it needs to be.

pervect

Staff Emeritus
The continuity equations in tensor form aren't that much diffrent than they are in non-tensor form.

Look at my simplified 1space+1time example.

The first of the two continuity equations is

$$\partial_t T^{tt} + \partial_x T^{xt} = 0$$

This is basically the same as the classical fluid-flow continuity equation with different notation.

$$\partial_t \, \rho + (\partial_x + \partial_y + \partial_z) (\rho \vec{u}) = 0$$

$\rho$ in the classical equation is the density, kg/volume, and $\vec{u}$ is the fluid velocity.

The classical fluid flow equation is 3d, so you'd suppress the $\partial_y$ and $\partial_z$ terms for comparison purposes to my 1d example.

Basically, the only difference betwen the two is the notation.

We call the energy density in the relativistic version $T^{00}$ or in my notation $T^{tt}$ rather than $\rho$, and and we call the momentum density $T^{xt}$ or $T^{10}$ rather than $\rho \, \vec{u}$.

Additionally, though, you'll note that we had two equations in the relativistic version not just one. The first equation is the energy conservation equation, the second is the momentum conservation equation. It basically shows how pressure differences (denoted by $T^{xx}$ in the relativistic version) cause changes in momentum (T^{xt} = T^{tx}).

I'm not quite sure what the correct classical analogous equation is to the second of my two equations, to be honest. It's probably Bernouli's equations or Euler's equations.

The relativistic version is actually simpler to my mind, as reading over the Wiki on Euler's equations, they divide u the energy into "internal" energy terms and "flow" terms, whereas the two are just combined in the relativistic equation. Also, we do not use non-relativistic equations like E=1/2 m v^2 or p=mv, instead we use the relativistic equivalents.

So far, this doesn't have anything much to do with your units question though. It's all just notational differences. I would say that the basic reason for the choice of units with c=1 is to make transforms between different frames via the Lorentz transform easier.

This may not be of interest to you at the moment, but eventually you may appreciate how much easier it is to do the appropriate Lorentz transforms in units when c=1. If you don't like it, you probably still need to learn it if you ever want to read any of the literature that uses the relativistic approach, such as all of General relativity.

I am unsure how familar you are with energy-momentum 4-vectors. These are discussed in SR textbooks like Taylor's "Space-time physics". A key point here is that momentum and energy "mix together" in different frames the same way that space and time do, because the energy-momentum 4-vector also transforms via the Lorentz transform. So in relativity, we tend to view momentum and energy as two aspects of the same thing, just as we view space and time as two aspects of the same thing.

So in my opinion the underlying issue here is the Lorentz transform, though that's not what you're actively looking at at the moment. If you come to terms with the notational differences, though, I think you will see that the relativistic continuity equations aren't any different than the non-relativistic ones,.

dsaun777

The continuity equations in tensor form aren't that much diffrent than they are in non-tensor form.

Look at my simplified 1space+1time example.

The first of the two continuity equations is

$$\partial_t T^{tt} + \partial_x T^{xt} = 0$$

This is basically the same as the classical fluid-flow continuity equation with different notation.

$$\partial_t \, \rho + (\partial_x + \partial_y + \partial_z) (\rho \vec{u}) = 0$$

$\rho$ in the classical equation is the density, kg/volume, and $\vec{u}$ is the fluid velocity.

The classical fluid flow equation is 3d, so you'd suppress the $\partial_y$ and $\partial_z$ terms for comparison purposes to my 1d example.

Basically, the only difference betwen the two is the notation.

We call the energy density in the relativistic version $T^{00}$ or in my notation $T^{tt}$ rather than $\rho$, and and we call the momentum density $T^{xt}$ or $T^{10}$ rather than $\rho \, \vec{u}$.

Additionally, though, you'll note that we had two equations in the relativistic version not just one. The first equation is the energy conservation equation, the second is the momentum conservation equation. It basically shows how pressure differences (denoted by $T^{xx}$ in the relativistic version) cause changes in momentum (T^{xt} = T^{tx}).

I'm not quite sure what the correct classical analogous equation is to the second of my two equations, to be honest. It's probably Bernouli's equations or Euler's equations.

The relativistic version is actually simpler to my mind, as reading over the Wiki on Euler's equations, they divide u the energy into "internal" energy terms and "flow" terms, whereas the two are just combined in the relativistic equation. Also, we do not use non-relativistic equations like E=1/2 m v^2 or p=mv, instead we use the relativistic equivalents.

So far, this doesn't have anything much to do with your units question though. It's all just notational differences. I would say that the basic reason for the choice of units with c=1 is to make transforms between different frames via the Lorentz transform easier.

This may not be of interest to you at the moment, but eventually you may appreciate how much easier it is to do the appropriate Lorentz transforms in units when c=1. If you don't like it, you probably still need to learn it if you ever want to read any of the literature that uses the relativistic approach, such as all of General relativity.

I am unsure how familar you are with energy-momentum 4-vectors. These are discussed in SR textbooks like Taylor's "Space-time physics". A key point here is that momentum and energy "mix together" in different frames the same way that space and time do, because the energy-momentum 4-vector also transforms via the Lorentz transform. So in relativity, we tend to view momentum and energy as two aspects of the same thing, just as we view space and time as two aspects of the same thing.

So in my opinion the underlying issue here is the Lorentz transform, though that's not what you're actively looking at at the moment. If you come to terms with the notational differences, though, I think you will see that the relativistic continuity equations aren't any different than the non-relativistic ones,.
Yes, that does clear things up a little I appreciate the response. I am just trying to mess around with tensor derivatives and would like to see what the math on paper represents in the real world. A throwback to Einstein and his thought experiments.

romsofia

I am just trying to mess around with tensor derivatives and would like to see what the math on paper represents in the real world.
Most of the time it's nothing physical in our universe. It is more productive to come up with a physical situation, then try to describe it mathematically (if you're a physicist).

dsaun777

Most of the time it's nothing physical in our universe. It is more productive to come up with a physical situation, then try to describe it mathematically (if you're a physicist).
I have a physical idea in mind and it involves derivatives I think...

pervect

Staff Emeritus
There is sometimes a bit of wiggle room in how the quantities are defined, so one has to look carefully at the defintiions of any particular source, or problem.

So what I say won't necessarily always be exactly in agreement with any particular textbook, but it should give you some basic insight as to what is going on.

I'll also limit myself to the 1-space, 1-time case, for ease of exposition. After you understand that case, you can move on to the more general case.

So in the one-space, one-time case, we have 4 terms in the stress-energy tensor, $T^{00}, T^{01}, T^{10},$ and $T^{11}$. Right off the bat, we can cay that $T^{01}=T^{10}$ because the stress energy tensor is symmetric. So we only have three terms to worry about.

At this point I'll mention that the time coordinate is t, and is related to the 0 superscript, while the space coordinate is x, and is related to the 1 superscript.

Aditionally I'll talk about the significance of the quantities in the rest frame of an object. Some of my remarks won't apply if the object is not in its rest frame.

Given all these cautions (plus any I forgot to mention :) ), $T^{00}$ is basically the energy density. In my mind this is true even if (or even when) we divide it by c^2 to account for mass differences, because $T^{00}$ includes things like "binding energy" that better fit the mold of "energy density" that the idea of mass as a "quantity of material" which is the Newtonian idea. But it may be possible to argue the point.

$T^{01}$ is basically momentum density. (In general, we'd say it's the "1" or x component of the mention, but because there is only one spatial dimension, we can just say that it's the momentum density without further specification).

$T^{11}$ is a bit tricky. If we have a rope under tension, $T^{11}$ is negative. If we have a column under compression, $T^{11}$ is positive. So it's closely related to stress, or pressure.

Understanding the stress-tensor in three dimensions, $T^{ij}$ eventually becomes necessary, but it's beyond the scope of what I want to present, other than to refer the reader to the classical stress tensor. Understanding the physical significance of $T^{11}$ as tension or compression in the 1 d case (for an object in it's rest frame! Not an object in a general frame) is a good start, though, I think.

"Continuity equation in GR"

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