# Differentials in a derivation

1. Oct 22, 2008

### yuiop

I am trying to understand a derivation posted by Pervect a long time ago (I suspect Pevect is no longer active) that involves differentiation and I was hoping someone here could fill in some of the steps to make it clearer.

The original posts by Pervect are here:

https://www.physicsforums.com/showpost.php?p=1020560&postcount=5

I have taken the liberty of posting a cut down version of Pervects posts here. I have added equation numbers and comment. The comments in blue are just notes and the comments in magenta are the places where I am stuck and need some more explanation.

Last edited by a moderator: May 3, 2017
2. Oct 23, 2008

### yuiop

Part 2

I had to split my original post into two because the latex was not displaying properly.

This is the second post where Pervect attacks the problem from a slightly different angle, but I am still missing the trick where he obtains the differentials.

Obviously I am missing a trick somewhere. Can anyone enlighten me?

Maybe this should have been posted in the Relativity forum, but I am only interested in the mathematical aspects that seem to require the knowledge of a expert in differential equations so I posted it here.

3. Oct 23, 2008

### HallsofIvy

He is not using differential equations here but just the definition of "differential"

From
$$X+ T= (\frac{r}{2m}-1)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}$$
take the differential of both sides. On the right, we need to apply the product rule to a product of three terms: Since the first and second are functions of r and the third term a function of t,we basically get d(f(r)g(r)h(t))= f'(r)g(r)h(t)dr+ f(r)g'(r)h(t)dr+ f(r)g(r)h'(r)dt.

More specifically,
$$dX+ dT= \frac{1}{2}\left(\frac{r}{2m}-1\right)^{-1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}dr$$
$$+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{r}{4m}}dr$$
$$+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{r}{4m}}dt$$

Do the same with the "X- T" equation and solve the two equations for dr and dt.

Last edited by a moderator: May 3, 2017
4. Oct 23, 2008

### cristo

Staff Emeritus
For the next part: eqns 2.5 and 2.6 are correct, and just arise from the definition of a differential: $$dr=\frac{\partial r}{\partial X}dX+\frac{\partial r}{\partial T} dT$$

Your last question: I'm not sure what you're asking. One can evaluate the above, since one has an expression of r in terms of X and T, so can take the derivatives and plug into the above.

5. Oct 24, 2008

### yuiop

Hi,

thanks for the help and sorry for the delay responding. I have spent the afternoon trying to learn differential calculus, product rules and multivariate calculus from the internet. I had no idea there was so much to it I can not honestly say I understand 10 percent of it but I hope I have picked up enough to address this problem. I found the following resources very helpful:

Multivariate calculus
Rules of calculus
Calculus - Wikibook
Product rule - Wikipedia

Back to the problem in hand.

Applying the product rule to (X+T) with respect to r:

$$X+ T= (\frac{r}{2m}-1)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}$$

I get:

\begin{align*} dX+ dT &= \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{-1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}(dr)&\\ &+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}(dr)&\\ &+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}(dt)& (Eq 4.1) \end{align*}

O.K.

$$X- T= (\frac{r}{2m}-1)^{1/2}e^{\frac{r}{4m}}e^{\frac{-t}{4m}}$$

\begin{align*} dX- dT &= \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{-1/2}e^{\frac{r}{4m}}e^{\frac{-t}{4m}}(dr)&\\ &+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{-t}{4m}}(dr)&\\ &- \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{-t}{4m}}(dt)& (Eq 4.2) \end{align*}

Solve (Eq 4.1) for $$e^{\frac{t}{4m}}$$ :

$$e^{\frac{t}{4m}} = \frac{4m(dX+dT)}{e^{\frac{r}{4m}}\left[ dr\left(\left(\frac{r}{2m}-1\right)^{1/2}+\left(\frac{r}{2m}-1\right)^{-1/2}\right) + dt\left(\frac{r}{2m}-1\right)^{1/2}\right] }}$$

and do the same for (Eq 4.2):

$$e^{\frac{t}{4m}} = \frac{e^{\frac{r}{4m}}\left[ dr\left(\left(\frac{r}{2m}-1\right)^{1/2}+\left(\frac{r}{2m}-1\right)^{-1/2}\right) - dt\left(\frac{r}{2m}-1\right)^{1/2}\right] }{4m(dX-dT)}$$

Since both the above equations are equal to $$e^{\frac{t}{4m}}$$ they can be equated to each other to eliminate the pesky t term and by applying the difference of squares rule the following is obtained:

$$16m^2(dX^2-dT^2) = e^{\frac{r}{2m}}\left[dr^2\left(\left(\frac{r}{2m}-1\right)^{1/2}+\left(\frac{r}{2m}-1\right)^{-1/2}\right)^2 -dt^2 \left(\frac{r}{2m}-1\right) \right]$$

By inspection it is obvious that the last equation is close to a solution and this suggests a shortcut that avoids having to find dr and dt individually.

Multiply both sides by $$\frac{2m }{ r} e^{\frac{-r}{2m}}$$ :

$$\frac{32m^3 e^{\frac{-r}{2m}} }{r} (dX^2-dT^2) = dr^2\left(1-\frac{2m}{r}\right)^{-1} -dt^2 \left(1-\frac{2m}{r}\right)$$

which is the Kruskal-Szekeres metric on the left hand side and the Schwazschild metric on the right hand side.

6. Oct 24, 2008

### yuiop

Hi cristo,

could you flesh this out a bit and explain why this form is used here in this particular example or point to some online reference material? I'm afraid I'm still not "getting it".

Is it because the cooordinate equations for R and T are being treated as parametric equations with respect to r and t for the purposes of differentiation?

I think we have one solution thanks to help from HallsofIvy but I never feel comfortable with a solution until I can get it from at least two different angles. Sorry if I being obtuse about all this but differentiation (at this level of complexity) is all fairly new to me.

<EDIT>
This is how it seems to me but I readily admit I am probably out of my depth here.

Starting with:

$$(X +T ) = (X+T)$$

Differentiating both sides with respect to r:

$$\frac{d}{dr} (X+T) = \frac{\partial X}{\partial r} + \frac{\partial T}{\partial r}$$

$$(dX + dT) = \left( \frac{\partial X}{\partial r} + \frac{\partial T}{\partial r} \right) dr$$

$$\frac{ (dX + dT) }{\left( \frac{\partial X}{\partial r} + \frac{\partial T}{\partial r} \right)} = dr$$

$$\partial r \frac{ ( dX + dT) }{\left( \partial X + \partial T \right)} = dr$$

$$dr = \frac{ \partial r }{\left( \partial X + \partial T \right)}dX + \frac{ \partial r }{\left( \partial X + \partial T \right)}dT$$

...then again..maybe you can not throw differential symbols around like that..

Last edited: Oct 24, 2008
7. Oct 26, 2008

### yuiop

OK, I have finally figured out what you were trying to tell me here. Sorry for being so slow I have posted a few corrections and a complete solution below which I hope will clear up any confusion for anyone hoping to learn anything from this thread. Thanks again Christo (and HallsofIvy) for the pointers in the right direction

I now realise I was on completely the wrong track in my last post and that post should be disregarded... I had the wrong end of the stick

This is the correct way to do it:

Starting with the definition of the Kruskal-Szekeres coordinates:

$$X=e^{(r/4M)}(r/2M-1)^{1/2}cosh(t/4M)$$
$$T=e^{(r/4M)}(r/2M-1)^{1/2}sinh(t/4M)$$

(there was a typo in post #1)

Solving these two equations for r gives:

$$r = 2M LambertW((X^2-T^2)/e) + 2M$$

Note: The lambertW function is sometimes know as the ProductLog or Omega function. Solving for r with complicated expressions like this is best done with mathematical software such as this free online one here: http://www.quickmath.com/

Now dr is found by finding the sum of the partial differential of (2M*LambertW((X^2-T^2)/e) + 2M) with respect to X plus the partial differential of (2M*LambertW((X^2-T^2)/e) + 2M) with respect to T using the product rule

$$dr=\frac{\partial r}{\partial X}dX+\frac{\partial r}{\partial T} dT$$

(Exactly as cristo said)

$$dr=\frac{4M LambertW((X^2-T^2)/e)(XdX-TdT)}{(LambertW((X^2-T^2)/e)+1)(X^2-T^2) }$$

The first two coordinate equations can also be solved for t in terms of X and T to give:

$$t = 2M ln[(X+T)/(X-T)]$$

and as before

$$dt=\frac{\partial t}{\partial X}dX+\frac{\partial t}{\partial T} dT$$

$$dt = \frac{4m(XdT-TdX)}{(X^2-T^2) }$$

Now that expressions for dr and dt in terms of X and T have been obtained they can be plugged in to the Schwarzschild metric

$$dS^2 = \left(1-\frac{2M}{r} \right)^{-1} (dr)^2 - \left(1-\frac{2M}{r} \right)(dt)^2$$

to obtain:

$$dS^2 = \frac{16M^2 (dX^2-dT^2) LambertW((X^2-T^2)/e)}{(LambertW((X^2-T^2)/e)+1)(X^2-T^2)}$$

which is the expression originally obtained by Pervect. (Anyone know what happened to Pervect? I miss his knowledgeable contributions to PF )

By substituting (r/2m-1)exp(r/2m) for (x^2-T^2) and (r/2m-1) for LambertW(X^2-T^2)/e) into the last equation as suggested by Pervect, the Kruskal-Szekeres metric is obtained:

$$dS^2 = \frac{32 M^3 e^{-r/2M}}{r}(-dT^2 + dX^2)$$