# Metric outside a weakly gravitating body (MTW Ex 19.1)

1. Feb 28, 2017

### TerryW

1. The problem statement, all variables and given/known data

This is Exercise 19.1 in MTW - See attachment

2. Relevant equations

See attachment

3. The attempt at a solution

I've worked through 19.3a, 19.3b and 19.3c ( see post by zn5252 back in March 2013 replied to by PeterDonis) and proved them for my my own satisfaction and I've also worked through 19.7a, 19.7b and 19.7c. My problems begin when I try to find h00 and hom.

I have arrived at a solution for h00 but it:

a) only works if T'00 >> T'ii. (I think this is true in geometricised units) and

b) I haven't made any use of 19.7a and 19.7b and I don't see how they would be useful because the expansion of 19.2 for h00 will only include terms with $\bar{T}^{00} (= \frac 1 2 (T'^{00} +T'^{ll})$

The result of my labours for h0m is:

$h_{0m} =-\frac 1{2r}\frac{\partial}{\partial x_m} \int(T'_{00,0})r'^2d^3x' -\frac{\partial}{\partial x_m} \int(T'_{00}+T'_{ll})x^jx'^jd^3x' - \frac{x^m}{r}\frac 1{2r} \int(T'_{00,00})r'^2d^3x'$
$\qquad \quad +\big(\frac {x^k}{r^2} \big)_{,m}\int(T'^{0j}{ }_{,0})x'^{j}x'^kd^3x'$

At this point, I can't see any way of using 19.7c to simplify my equation so I am stuck.

Any help anyone?

TerryW

File size:
822.4 KB
Views:
45
File size:
822.4 KB
Views:
27
2. Mar 1, 2017

### TSny

Hi, Terry.

When working out $h_{00}$ you can use identity 19.7a to write $\int T'_{kk} d^3x'$ in a different way. Thus, let $j = k$ in 19.7a and sum over $k$. Use this to show that $\int T'_{kk} d^3x' = \frac{1}{2} \frac{\partial ^2}{\partial t^2} \int T'^{00}r'^2 d^3x'$. The reason for doing this is that you can relate this expression to the first term in the gauge function $\xi_0$ near the top of page 449. This is how the gauge transformation can cancel out unwanted terms in $h_{00}$.

Similarly, 19.7b can be used to re-express $\int T'_{kk} x^{j \,'}d^3x'$ in a way that will be related to the second term of $\xi_0$.

3. Mar 3, 2017

### TerryW

Hi TSny
I've tried this but without success:

$$T^{kk} = ½(T^{00}x^k x^k)_{,00} + 2(T^{lk}x^k)_{,l}- ½(T^{lm}x^k x^k)_{,lm}$$

$$T^{kk} = ½(T^{00}{ }_{,00}x^k x^k) + 2T^{lk}{ }_{,l}x^k + 2T^{lk}δ^k_l - ½(T^{lm}{ }_{,l}x^k x^k + 2T^{lm}x^kδ^k_l )_{,m}$$

$$T^{kk} = ½(T^{00}{ }_{,00}x^k x^k) + 2T^{lk}{ }_{,l}x^k + 2T^{kk} - ½(T^{lm}{ }_{,l}x^k x^k + 2T^{km}x^k)_{,m}$$

$$T^{kk} = ½(T^{00}{ }_{,00}x^k x^k) + 2T^{lk}{ }_{,l}x^k + 2T^{kk} - ½(T^{lm}{ }_{,lm}x^k x^k + 2T^{lm}{ }_{,l}x^kδ^k_m + 2T^{km}{ }{,m}x^k+2T^{km}δ^k_m)$$

$$T^{kk} = ½(T^{00}{ }_{,00}x^k x^k) + 2T^{lk}{ }_{,l}x^k + 2T^{kk} - ½(T^{lm}{ }_{,lm}x^k x^k + 2T^{lk}{ }_{,l}x^k + 2T^{km}{ }_{,m}x^k+2T^{kk})$$

But $T^{lk}{ }_{,l}x^k = T^{kl}{ }_{,l}x^k = T^{km}{ }_{,m}x^k$, then

$$T^{kk} = ½(T^{00}{ }_{,00}x^k x^k) - ½(T^{lm}{ }_{,lm}x^k x^k) + T^{kk}$$

From $T^{αβ}{ }_{,β} = 0$ we get $T^{lm}{ }_{,lm} = - T^{0m}{ }_{,0m}$

$$T^{kk} = ½(T^{00}{ }_{,00}x^k x^k) + ½(T^{0m}{ }_{,0m}x^k x^k) + T^{kk}$$

Then $T^{00}{ }_{,00} + T^{0m}{ }_{,0m} = T^{0α}{ }_{,0α} = T^{0α}{ }_{α0} = 0$

Leaving $T^{kk} = T^{kk}$

This is my original proof of the identity 19.7a (with a similar process for proving 19.b). At this stage I can't see a way of manipulating 19.7a to allow me to write $\int T'_{kk} d^3x'$ in another way.

What am I not seeing?

TerryW

4. Mar 3, 2017

### TSny

OK. Now consider $\int T'^{kk } d^3x'$. Use the relation above and note that the second and third terms on the right are divergences. So, when integrating you can use the divergence theorem.

5. Mar 5, 2017

### TerryW

Thanks for that, I have now sorted h00. $\quad$ h0m seems to have a large number of bits and pieces to work my way through to get to the required answer. I will crack on and hope I don't have to bother you again.

Regards

TerryW

6. Mar 5, 2017

### TSny

OK. It's no bother.

I got hung up for about 2 days on hom, mainly concerning the term $\frac{x^m}{r} \xi_0$ in the expression for $\xi_m$. It took me quite a while to see why that term must be included. I was overlooking the fact that the various $T \, '$ components are evaluated at the retarded time. So, you have to be careful when evaluating $\xi_{0,m}$ in the gauge transformation.

7. Mar 9, 2017

### TerryW

Hi TSny,

I've been batting this around for a few days now and have reached a bit of a road (mind) block. To try make a bit of progress, I'm going to set out my working in stages, starting with my thoughts on what h0m, ξ0 and ξm are:

$h_{0m} = 4\int(T'_{0m})\big(\frac {1}{r}+\frac{x^jx'^j}{r^3} \big)d^3x'$

(This is as suggested, the n = 0 term of (19.2) - I can't see how this can be simplified! I also observe that $4\int(T'_{0m})\big(\frac{x^jx'^j}{r^3} \big)d^3x'$ can provide the term in $h_{0m}$ that we are seeking!)

I've adopted a slightly different convention in labelling, using $x'^j$ instead of $x^{j'}$ which, to me, makes clearer which point the coordinate refers to.

$ξ_{0} =\frac 1{2r}\frac{\partial}{\partial t} \int(T'^{00})r'^2d^3x'$

$\quad+\frac{x^j}{r^3}\int(T'^{0k}x'^j x'^k - \frac{1}{2}T'^{0j}r'^2)d^3x'$

$\quad+\int(T'_{00}+T'_{ll})\frac{x^j x'^j}{r}d^3x'$

$\quad+\frac{1}{2}\int(T'_{00}+T'_{kk})_{,0}\big(\frac{x^j x'^j}{r^2}\big)^2\big(\frac{1}{r}\big)d^3x'$

$ξ_{m} =4\int(T'_{0m})\big(\frac{x^jx'^j}{r} \big)\big(\frac{1}{r}\big)d^3x''$ (n = 1)

$\quad +4\frac{1}{2}\frac{\partial}{\partial t}\int(T'_{0m})\big(\frac{x^jx'^j}{r} \big)^2\big(\frac{1}{r}\big)d^3x'$

$\quad +\frac {x^m}{r}\frac 1{2r}\frac{\partial}{\partial t} \int(T'^{00})r'^2d^3x'$

$\quad +\frac {x^m}{r}\frac {x^j}{r^3}\frac{\partial}{\partial t} \int(T'^{0k}x'^j x'^k - \frac{1}{2}T'^{0j}r'^2)r'^2d^3x'$

$\quad +\frac {x^m}{r}\int(T'_{00}+T'_{ll})\frac{x^j x'^j}{r^2}d^3x'$

$\quad +\frac {x^m}{r}\frac{1}{2}\int(T'_{00}+T'_{kk})_{,0}\big(\frac{x^j x'^j}{r^2}\big)^2\big(\frac{1}{r}\big)d^3x'$

$\quad -\frac {1}{2}\big(\frac 1{r}\big)_{,m}\int(T'_{00})r'^2d^3x'$

$\quad -\big(\frac{x^k}{r^2}\big)_{,m} \int(T'^{0j}x'^j x'^k - \frac{1}{2}T'^{0k}r'^2)d^3x'$

$\quad -\frac{1}{2}\int(T'^{00}+T'^{kk})\big[\big(\frac{x^j x'^j}{r^2}\big)^2\big(\frac{1}{r}\big)\big]_{,m}d^3x'$

Do these look OK to you?

Regards

TerryW

8. Mar 9, 2017

### TSny

OK. The first term on the right in your expression for $h_{0m}$ can be shown to equal zero via MTW (19.4a). Yes, the second term can be rewritten in terms of the angular momentum $S^k$ and an extra term that will get canceled in the gauge transformation. Also, $h_{0m}$ will include the summation over n for n ≥ 1. But this will get canceled in the gauge transformation.

I have not checked all your terms in $ξ_{0}$ and $ξ_{m}$. I did not modify any of the terms in these expressions as given in MTW. I've attached a figure that shows the role of the various terms of the gauge transformation for $h_{0m}$ as I see it.

#### Attached Files:

• ###### MTW 19.1.png
File size:
70.2 KB
Views:
37
9. Mar 14, 2017

### TerryW

Hi TSny,

I've worked my way through your rationale for getting rid of the unwanted terms in $h^{new}_{0m} = h^{old}_{0m} - (ξ_{0,m}+ξ_{m,0})$ but I think that there are still quite a few bits hanging around!

1. The expressions in the orange boxes cancel.... agree

2. ....and the expressions in the purple boxes cancel - I think that the sum of the purple boxes is -

$$\sum_{n=0}^\infty \frac{1}{n!}\frac{\partial^{n-1}}{\partial t^{n-1}}\int(T'_{00}+T'_kk)_{,m}\frac {(r-|x-x'|)^n}{|x-x'|} d^3 x' \quad (A)$$

3. The contribution from the dark blue box is zero due to choosing the origin at the centre of mass....agreed.

4. The expression of the brown boxes can be shown to cancel using identity (19.7b)....agreed.

5. The expression in the green box will cancel expressions in $h_{0m}$ .... $h_{0m}$ contains only terms with $T'{0m}$ so $h^{new}_{0m}$ will still end up with a term -
$$\frac{2x^j}{r^3}\int T'_{00,0}x^j x^{m'} d^3 x' \quad (B)$$

6. The expressions in the light blue boxes will be of order 1/r3 and can be ignored. I don't think this is correct. In 19.5, the term $\frac{x^l}{r^3}$ isn't ignored. The unprimed co-ordinates refer to the location of the observer and as such, some of the values of xl will be of the same order as r. You could maybe argue that all the terms in $ξ_{0,m}$ disappear by taking the partial derivative under the integral sign and then use the divergence theorem but then you would bring back terms from $ξ_{m,0}$ that were eliminated above!

7. The expression in the red box will cancel terms in $ξ_{0,m}$ due to the dependence on the T's on retarded time. If you had made this your final statement, it would have become invalid because there are no parts of $ξ_{0,m}$ left to cancel. It just appears to add four more terms to my collections of terms not yet eliminated.

To summarise, I still have the following terms hanging around:

From $ξ_{0,m}$ :
$$\frac {\partial}{\partial x^m}'\big( \frac {x^j}{r^3} \int (T'^{0k} x'^k x'^j - \frac {1}{2} T'^{0j} r'^2 )d^3x' \big)\quad(1)$$

$$\frac {\partial}{\partial x^m}\big( \int (T'_{00} + T'_{ll})(\frac {3x'^j x'^k - r'^2 δ_{jk})x^j x^k}{2r^4} d^3x' \big) \quad(2)$$

$$\sum_{n=0}^\infty \frac{1}{n!}\frac{\partial^{n-1}}{\partial t^{n-1}}\int(T'_{00}+T'_kk)_{,m}\frac {(r-|x-x'|)^n}{|x-x'|}d^3 x' \quad (3)$$

From $ξ_{m,0}$ :
$$\frac{2x^j}{r^3}\int T'_{00,0}x^j x^{m'} d^3 x' \quad(4)$$

$$\frac {x^m}{r} ξ_{0,0} \quad (5)$$

If it is indeed OK to take the partial derivative behind the integral sign and then use the divergence theorem, then we can eliminate (2), but that still leaves four terms to argue away, with $\frac {x^m}{r} ξ_{0,0}$ looking particularly challenging.

I'll continue to explore this to see if 19.7a,b and c can provide further assistance.

Regards

TerryW

10. Mar 14, 2017

### TSny

Every term in $\xi_0$ produces a "retarded time" term when evaluating $\xi_{0, m}$. All of these retarded time terms are canceled by terms generated by the red box term in $\xi_m$ when evaluating $\xi_{m, 0}$. So, the orange and purple boxes cancel after taking this retarded time cancellation into account. Thus, your extra term that you are getting from the purple boxes is canceled by a particular term generated by the red box when evaluating $\xi_{m, 0}$. [Edited to correct subscripts.] (The summation in your left-over purple term should start at n = 2.)

More explicitly, in your left-over term you have $(T'_{00}+T'_{kk})_{,m}$. Show that due to the retarded time dependence, this is equal to $\left( -\frac{x^m}{r} \right) (T'_{00}+T'_{kk})_{,0}$. Then show that the red box will generate a term that will cancel your left-over term.

Similarly, all the terms in $\xi_{0,m}$ that are due to the dependence on the retarded-time will be canceled by terms in $\xi_{m, 0}$ that come from the red box expression in $\xi_m$. [Edited for clarity.]

Show that this equals zero due to choosing the center-of-momentum frame.

Right, $\frac{x^l}{r^3}$ is of order 1/r2 for a least one value of $l$. However, note that $\left( \frac{x^l}{r^3} \right)_{,m} = \frac{\delta_{ml}}{r^3} - \frac{3x^lx^m}{r^5}$, which is of order 1/r3. [Edited to correct a typo.] The expressions in the light blue boxes are of order 1/r2. But when taking the spatial derivative with respect to $x^m$, you either get terms of order 1/r3, or you get "retarded-time" terms that are canceled by terms generated by the red box.

In taking the spatial derivatives with respect to $x^m$ in (1) and (2), you will either make the terms of order 1/r3 or you will produce "retarded-time" terms that will be canceled by terms in (5). Likewise, (3) gets canceled by (5). (4) is zero in the center-of-momentum frame.

Last edited: Mar 14, 2017
11. Mar 14, 2017

### TSny

I have updated my figure to make the comments more accurate.

#### Attached Files:

• ###### MTW 19.1b.png
File size:
71 KB
Views:
31
12. Mar 17, 2017

### TerryW

Hi TSny,

Your previous post and updated figure has clarified the process of elimination and I can now see that the terms:
$$\frac {\partial}{\partial x^m}'\big( \frac {x^j}{r^3} \int (T'^{0k} x'^k x'^j - \frac {1}{2} T'^{0j} r'^2 )d^3x' \big)\quad(1)$$
$$\frac {\partial}{\partial x^m}\big( \int (T'_{00} + T'_{ll})(\frac {3x'^j x'^k - r'^2 δ_{jk})x^j x^k}{2r^4} d^3x' \big) \quad(2)$$
disappear.

But I still think that there are two terms unaccounted for.

1. In one of my earlier posts, I had incorrectly written the first term of the green box as
$$-\frac {2x^j}{r^3} \int T_{00} x^{j}x^{m'} d^3 x'$$
which you commented on and said that it is zero in the centre of momentum frame. A close look at MTW shows this first term to be
$$-\frac {2x^j}{r^3} \int T_{00} x^{j'}x^{m'} d^3 x'$$
which I don't believe is zero in the centre of momentum frame.

2. I still don't agree that the expressions in the purple boxes cancel. The retarded time term from $ξ_{0,m}$ has already been cancelled by a term from the red box, but it will still produce two terms when the differentiation wrt xm is carried into the integral. One of these terms is cancelled by the purple box on the right hand side, but we are still left with:
$$\sum_{n=2}^\infty \frac{1}{n!}\frac{\partial^{n-1}}{\partial t^{n-1}}\int(T'_{00}+T'_{kk})_{,m}\frac {(r-|x-x'|)^n}{|x-x'|}d^3 x'$$

I've tried to find a way to use 19.7 a,b and c to get these two terms to cancel without success.

If you can help me to resolve this, I promise not to do a post if I can't sort $h_{mn}$!

Regards

TerryW

13. Mar 17, 2017

### TSny

OK. But I don't understand the prime on the derivative operator in (1).

Right, this term is not zero. But when forming $\xi_{m,0}$, this term will cancel an unwanted term in $h_{0m}$ when doing the gauge transformation.

I think this is the retarded time term that gets canceled by a term from the red box.

14. Mar 17, 2017

### TerryW

Sorry, that's a typo!

It can't. All terms in $h_{0m}$ contain $T_{0m}$ and the second term in your green box covers all the terms in $h_{0m}$ for n>0.

I don't think it is. The retarded time term is the term in the left hand purple box differentiated wrt x0 (t) and it includes the -$\frac {x^m}{r}$ term, like the terms in the red box. My term arises from the differentiation of the product $(T'_{00}+T’_{kk})\frac {(r-|x-x'|)^n}{|x-x'|}$ wrt xm which produces two parts, only one of which is cancelled your right hand purple box.

15. Mar 17, 2017

### TSny

After some manipulations and the use of (19.7c) I get (to order 1/r^3) the following:

$h_{om} = -2 \epsilon_{mkl}S^k\frac{x^l}{r^3} - 2 \frac{x^j}{r^3}\int \left( T^{00 '}x^{m'}x^{j'} \right)_{,0} d^3x'$ + a summation term over n.

The middle term is canceled by the first term of the green box and the summation term is canceled by the second term of the green box.

All of the terms that I call "retarded time terms" are terms that come from differentiation of the retarded time with respect to $x^m$ when forming $\xi_{0,m}$. Thus,

$\frac{\partial}{\partial x^m} T^{00'} = -\frac{x^m}{r} T^{00'}\, _{,0}$.

Similarly for differentiating any other $T '$ component with respect to $x^m$.

So, when forming $\xi_{0, m}$, the purple box will yield a contribution containing $(T'_{00}+T’_{kk})_{,m} =- \frac{x^m}{r} (T'_{00}+T’_{kk})_{,0}$. This well get canceled by a term generated by the red box when calculating $\xi_{m,0}$.

Last edited: Mar 17, 2017
16. Mar 19, 2017

### TerryW

Hi TSny,

I probably won't be able to get back to you for a few days - partly other commitments and partly trying to work a few things out, particularly the 'retarded time' effect.

regards

TerryW

17. Mar 22, 2017

### TerryW

Hi TSny

I've spent a bit of time working out that:
If ξm = T0k
then $ξ_{0,m} = \frac{\partial}{\partial x^m} T^{0k} - \frac{x^m}{r}\frac{\partial}{\partial t} T^{0k}$

From this, using the fact that $\int \frac{\partial}{\partial x^m} T^{0k} d^3x' = 0,$ I get:

$ξ_{0,m}$ :
$$\big( \frac{1}{2r}\big)_{,m}\frac{\partial}{\partial t}\int T'^{00}r'^2 d^3x'\quad (1)$$
$$- \frac{x^m}{r}\frac{1}{2r}\frac{\partial}{\partial t}\int T'^{00}\ _{,0} r'^2 d^3x'\quad(2)$$
$$+\big(\frac{x^j}{r^3}\big)_{,m}\int (T'^{0k} x'^k x'^j - \frac {1}{2} T'^{0j} r'^2 )d^3x'\quad(3)$$
$$-\frac{x^m}{r}\big(\frac{x^j}{r^3}\big)\int (T'^{0k} x'^k x'^j - \frac {1}{2} T'^{0j} r'^2 )_{,0}d^3x'\quad(4)$$
$$-\frac{x^m}{r} \int (T'_{00} + T'_{ll})_{,0}\big[\frac {x^{j}x'^{j}}{r^2}+\frac {(3x'^j x'^k - r'^2 δ_{jk})x^j x^k}{2r^4}\big] d^3x'\quad(5)$$
$$-\frac{x^m}{r} \sum_{n=2}^\infty \frac{1}{n!}\frac{\partial^{n-1}}{\partial t^{n-1}}\int(T'_{00}+T’_{kk})_{,0}\frac {(r-|x-x'|)^n}{|x-x'|}d^3 x' \quad (6)$$

$ξ_{m,0}$ :
$$-\frac{2x^j}{r^3}\int T'_{00,0}x^{j'} x^{m'} d^3 x' \quad(7)$$
$$+4 \sum_{n=1}^\infty \frac{1}{n!}\frac{\partial^{n}}{\partial t^{n}}\int T'_{0m}\frac {(r-|x-x'|)^n}{|x-x'|}d^3 x' \quad (8)$$
$$-\big( \frac{1}{2r}\big)_{,m}\frac{\partial}{\partial t}\int T'^{00}r'^2 d^3x'\quad (9)$$
$$-\big(\frac{x^j}{r^2}\big)_{,m}\int (T'^{0j} x'^j x'^k - \frac {1}{2} T'^{0k} r'^2 )_{,0} d^3x'\quad(10)$$
$$-\sum_{n=2}^\infty \frac{1}{n!}\frac{\partial^{n-1}}{\partial t^{n-1}}\int(T'_{00}+T’_{kk})\big[\frac {(r-|x-x'|)^n}{|x-x'|}\big]_{,m}d^3 x' \quad (11)$$

$\frac {x^m}{r} ξ_{0,0}:$
$$\frac{x^m}{r}\frac{1}{2r}\frac{\partial}{\partial t}\int T'^{00}\ _{,0} r'^2 d^3x'\quad(12)$$
$$+\frac{x^m}{r}\big(\frac{x^j}{r^3}\big)\int (T'^{0k} x'^k x'^j - \frac {1}{2} T'^{0j} r'^2 )_{,0}d^3x'\quad(13)$$
$$+\frac{x^m}{r} \int (T'_{00} + T'_{ll})_{,0}\big[\frac {x^{j}x'^{j}}{r^2}+\frac {(3x'^j x'^k - r'^2 δ_{jk})x^j x^k}{2r^4}\big] d^3x'\quad(14)$$
$$+\frac{x^m}{r} \sum_{n=2}^\infty \frac{1}{n!}\frac{\partial^{n-1}}{\partial t^{n-1}}\int(T'_{00}+T’_{kk})_{,0}\frac {(r-|x-x'|)^n}{|x-x'|}d^3 x' \quad (15)$$

My expression for $h_{0m}$ is:

$h_{0m} = 4\int T'_{0m} \big( \frac{1}{r} +\frac{x^j x'^j}{r^3} +O\big( \frac{1}{r^3} \big) \big) +4 \sum_{n=1}^\infty \frac{1}{n!}\frac{\partial^{n}}{\partial t^{n}}\int T'_{0m}\frac {(r-|x-x'|)^n}{|x-x'|}d^3 x'\quad(16)$

I am now working on these expressions to see if I can get the desired result.
For the moment, do you agree that a) (1) to (16) are correct and that b) there are no other terms?

Regards

TerryW

18. Mar 22, 2017

### TSny

I'm not sure I'm following your way of writing it.

In the expressions $ξ_0$ and $ξ_m$, all of the $T ^ \prime$ components are evaluated at the primed spatial coordinates $x^{j \, \prime}$ and at the retarded time $t - r$. For example, the functional dependence of $T^{0k \, \prime}$ is $T^{0k \, \prime} \left(t-r, x^{1'}, x^{2'}, x^{3'} \right)$. So, when taking the derivative of $T^{0k \, \prime}$ with respect to the unprimed spatial coordinate $x^m$, the only contribution is from the dependence of the retarded time on $x^m$ (via $r$). So, $\frac{\partial}{\partial x^m} T^{0k \prime}= - \frac{x^m}{r}\frac{\partial}{\partial t} T^{0k}$. Maybe that's the same thing as you are saying.

All your expressions (1) through (16) look correct to me. So, hopefully, things will now simplify to what you need.

19. Mar 24, 2017

### TerryW

I don't think there are two sets of co-ordinates here. I think MTW is using the prime to indicate the volume round the origin containing the matter, and the non-primed co-ordinates belong to the stationary observer. So when we are taking derivatives $\frac{\partial x'^i}{ \partial x^m} = δ^i _m \quad$ but $\frac{\partial x^i}{ \partial x^m} = 0$ because the observer is stationary.

I derived my formula for $ξ_{0,m}$ as follows:

If $ξ_{m}= T'_{mk} x'^k$ (I've changed this to get the indices in balance)
then
$Δξ_{m} = \frac{\partial {(T'_{mk} x'^k)}}{ \partial x^m} dx^m + \frac {\partial {(T'_{mk} x'^k)}}{ \partial x^0} \frac {\partial |x-x'|}{ \partial x^m} dx^m$

i.e. the change in $ξ_{m}$ has a component arising from the change of position and a component arising from the need to account for the retarded time effect at the observer.

So

$Δξ_{m} = \frac{\partial {(T'_{mk} x'^k)}}{ \partial x^m} dx^m + \frac {\partial {(T'_{mk} x'^k)}} { \partial x^0} \frac {\partial [(|x-x'|)^2]^½} { \partial x^m} dx^m$

$\quad \frac {\partial [(|x-x'|)^2]^½}{ \partial x^m} = \frac {1}{2} \frac {1}{(x^i - x'^i)^½} \frac {\partial }{ \partial x^m}[x^i x^i - 2x^i x'^i +x'^i x'^i]$

$\quad\frac {\partial [(|x-x'|)^2]^½}{ \partial x^m} = \frac {1}{2} \frac {1}{|x^i - x'^i|} [- 2x^iδ^i _m+x'^i δ^i _m]$ (Observer is stationary)

$\quad\frac {\partial [(|x-x'|)^2]^½}{ \partial x^m} = \frac {x^m - x'^m}{|x^i -x'^i|}$

$\quad$ and as x >> x'

$\quad\frac {\partial [(|x-x'|)^2]^½}{ \partial x^m} = - \frac {x^m}{r}$

Therefore:

$\frac {\partial {ξ_{m}}}{\partial x^m} = \frac{\partial {(T'_{mk} x'^k)}}{ \partial x'^m} - \frac {\partial {(T'_{mk} x'^k)}}{ \partial x^0} \frac {x^m}{ r}$

I'm pleased that you agree I've now correctly identified the necessary components for the problem, but so far, I haven't had any success boiling them down to the required result.

Regards

TerryW

20. Mar 24, 2017

### TSny

We agree that there is only one coordinate system. The primed coordinates and the unprimed coordinates are coordinates in this one coordinate system. Nevertheless, the primed coordinates are independent of the unprimed coordinates, unless there is some sort of constraint that imposes a functional dependence of the primed and unprimed coordinates. In the expressions in $\xi_m$ and $\xi_0$, there is one such constraint. The primed time in the $T’$ components is constrained to be the retarded time. That is, we have the functional constraint $t’ = t – r$. This is the only constraint. There is no constraint that imposes any functional dependence of the primed spatial coordinates and the unprimed spatial coordinates. So, $\frac{\partial x \, ' ^i}{\partial x^m} = 0$ for all $i, m = 1, 2, 3$.

Also, $\frac{\partial x^i}{\partial x^j} = \delta^i_j$. This is true whether or not the observer is stationary. It is just an expression of the independence of the three spatial coordinates of the observer.

Note that in the equation above, the left-hand side is $Δξ_{m}$, which is not summed over $m$. But each term of the right-hand side has $m$ occurring three times. So, it is not clear if there is supposed to be a summation over $m$.

To avoid confusion, suppose we want the change in $ξ_{m}$ due to a small change in the particular unprimed coordinate $x^3$, say. I would write this as

$Δξ_{m} = \frac{\partial {(T'_{mk} x'^k)}}{ \partial x^3} dx^3$

Then, we note that the only dependence of $T'_{mk} x'^k$ on $x^3$ is in the $r$ in the retarded time $t' = t - r$. So, we can write

$Δξ_{m} = \frac{\partial {(T'_{mk} x'^k)}}{ \partial t'} \frac {\partial t'}{ \partial x^3} dx^3 = \frac{\partial {(T'_{mk} x'^k)}}{ \partial t'} \left( - \frac {\partial r}{ \partial x^3}\right)dx^3 =\frac{\partial {(T'_{mk} x'^k)}}{ \partial t'} \left( - \frac{x^3}{r} \right) dx^3$

Finally, we can replace $\partial t'$ by $\partial t$ since $t' = t - r$. So, we finally arrive at

$Δξ_{m} = - \frac{x^3}{r} ξ_{m,0}dx^3$

This is equivalent to $\frac{\partial{ξ_{m}}}{\partial x^3} = - \frac{x^3}{r} ξ_{m,0}$

Last edited: Mar 24, 2017