# Stupid question about page 70 in a first course in GR

Terilien
Ok thhe follwoing questio is extremely silly. everything else that I've read in the book makes sense but this part doesn't. i just can't seem to udnerstand it because of a combination of stupidity and notation. Near the end of page 69 an identity is derived to show how the components of the gradients transform. That made perfect sense. He then shows us the new notation for partial derivatives and suddenly jumps to showing us what the basis of the gradient one form is. This didn't really make sense to me. firstly could you expand the sumbol and shoe me exactly what it means? Secondly could you show me why that's the basis for the gradient one form? Please exaplain EVERY step. I just can't seem to understand it. The book assumes that you're moderately smart which I'm not.

This is of course directed towards those who have the book. If you don't have it the part I don't get is on this page.

http://www.mth.uct.ac.za/omei/gr/chap3/frame3.html

I have absolutely no idea how we get from 23 to 24. what does dx^alpha mean?

And please don't start barraging me with questions just because I'm in high school. I really didn't like it. Yes I do know multivariable, i do know some linear algebra and I do know something about differential equations.

THis is a first course in general relativity.

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Homework Helper
He doesn't get "from" 23 to 24. 23 is the basic metric, gij in flat space and 24 is general definition of "dot product": the dot product of (Ai) with (Bj) is $\Sigma g_{ij}A^iB^j$

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Terilien
He doesn't get "from" 23 to 24. 23 is the basic metric, gij[/sub[ in flat space and 24 is general definition of "dot product": the dot product of (Ai) with (Bj) is $\Sigma g_{ij}A^iB^j$

Oh no no! Ut's in the section on gradients.

pmb_phy
Terilien said:
And please don't start barraging me with questions just because I'm in high school. I really didn't like it.
We didn't mean to irritate you with the questions. Its just that in order to explain something to someone one has to have a common language. In this case the common language is math. But if we don't have a subset of math in common the communication may be impossible. The questions were meant to find out where you were with the math so that we would know how to respond to your questions.

He doesn't get "from" 23 to 24. 23 is the basic metric, gij in flat space and 24 is general definition of "dot product": the dot product of (Ai) with (Bj) is $\Sigma g_{ij}A^iB^j$
Huh?? Since when is 23 a metric? That is very far from being a metric as I can see. Please elaborate.

Pete

pmb_phy
Ok the follwoing question is extremely silly.
I don't know about anyone else who posts on the internet but there is no serious question that is silly, never mind "extremely silly." That's how we learn.
everything else that I've read in the book makes sense but this part doesn't.
In cases such as this you really need to post the book you're referring to incase someone who has never read your other posts on this topic can make sense out of it. While I know you're referring to "A first course in general relativity," by Bernard F. Schutz, I doubt several other readers of this thread won't know what book you're referring to.
i just can't seem to udnerstand it because of a combination of stupidity and notation.
Seems to me that you're a sharp kid so stupidity is not a factor here.
Near the end of page 69 an identity is derived to show how the components of the gradients transform.
Please state which equations you are speaking of in the derivation. Do you mean Eq. (3.16) to Eq. (3.17)?
That made perfect sense. He then shows us the new notation for partial derivatives and suddenly jumps to showing us what the basis of the gradient one form is. This didn't really make sense to me.
I don't understand where the difficulty is comming from. All he means is "When you see me write () I really mean this ()." Its simply shorthand notation.

firstly could you expand the sumbol and shoe me exactly what it means?
Eq. (3.19) gives you everything you need to know. Note the tripple parallel bars in that equation. They mean "is defined as". The equal sign is different. It means "is equal to". These have two very different meanings.
Secondly could you show me why that's the basis for the gradient one form? Please exaplain EVERY step. I just can't seem to understand it.
Sure. Eq. (3.16) states the condition that d(phi) (d = gradient operator) must adhere to for d(phi) to be a one-form. This comes from Eq. (3.9) on page 63. If you take a look at Eq. (3.14) on page 67 you'll see that the partial derivatives behave like components of a 1-form. That is the reason for the expression between Eq. (3.16) and (3.17) on page 69. The equation which follows Eq. (3.17), i.e. x^beta = \Gamma ^beta _alpha-bar x^aplha-bar, defines \Gamma ^beta _alpha-bar. Take the partial derivative of each side of that equation with respect to x^alpha-bar and that gives you Eq. (3.18). It might just be fatigue now but I'm not sure how he gets his result Eq. (3.20) just by comparing to Eq. (3.12). I'm sure someone else can fill in the details.
The book assumes that you're moderately smart which I'm not.
I'm sorry to destroy your self image but from our discussions it is clear to me that you're a sharp kid. Someone your age understanding that book to that point and all the math that is required to get there is far beyond what any high schooler knows. I used to tutor college students in the math lab when I was an undergrad. Even the kids who came from high school with straight A's were pretty much lost when it came to college algebra. And here you are doing GR. To me that makes you pretty sharp.

Pete

Terilien
I'm sorry to destroy your self image but from our discussions it is clear to me that you're a sharp kid. Someone your age understanding that book to that point and all the math that is required to get there is far beyond what any high schooler knows. I used to tutor college students in the math lab when I was an undergrad. Even the kids who came from high school with straight A's were pretty much lost when it came to college algebra. And here you are doing GR. To me that makes you pretty sharp.

Pete

He. funny thing is that I don't even get straight A's. They must be pretty dim. Or maybe i just stick to problems way longer. another possiblity is tha they nevr learned math the right way. They had learned it through memeorization.

Exactly. I don't understand how he gets 3.20 by comparing it with 3.12. That's where I'm totally lost. I'm not even sure what the symbol means. Everything else is fairly straightforward.

pmb_phy
He. funny thing is that I don't even get straight A's.
In high school I had a C average. When I got into college it was very very rare for me to get less than an A in my physics and math courses.
They must be pretty dim.
I think that the work they did was easy for them in high school so they weren't challanged. College calculus is very challenging.

Pete

quetzalcoatl9
there are other books out there that would probably make this more clear.

in equation (20) the big alpha is just meant to be some transformation matrix, left to be determined.

in going from (23) to (24), the author is showing that the mapping of a tangent vector in one manifold to another is given by a linear transformation - where the transformation is more precisely the jacobian given by the partial derivatives. this follows as a natural consequence of the chain rule and the definition of a differentiable manifold.

however, since the dx's are not themselves tangent vectors, they are something else and are called one-forms or dual vectors. they have several useful and unique properties, some of which you will be aware of from calculus.

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Klarinettus
Terilien,

I am also a high school student that's beginning to learn some GR. I guess my math background is pretty similar to yours (had multivariate, currently taking diffyq and linear algebra). Might I ask what book you're working from? How well is it working for you? I've been using Spacetime and Geometry: An Introduction to General Relativity, by Sean Carroll, and I think that the explanations are very well done. You might want to check it out some time.

Also, have you been using any online resources? I found a really nice website with lots of lecture notes on tensors, but I accidentally bookmarked the wrong page. If I can find it again, I'll post it here.

I hope everything's going well for you. Maybe we can help each other out with some of this stuff.

Terilien
I'm using a first course in general relativity. Unfortunately though my problem with this thing is really hindering my progress through the book. I've flipped through spacetime and geometry. It seemed good. How old are you? I'm 15.

Terilien
Seriosuly though, I must understand this. This is really getting in my way. first off why is the derivative x alpha with respect to x beta equal to the kornecker delta, or why is it defined as such?

also how does he get to the conclusion that dx^aplha is the basis of the gradient one form, and can someone expand that term and show me EXACTLY what it means. If you know what it means it shouldn't be too hard to expand it. I myself am not sure.

page 70 a first course in general relativity.

Staff Emeritus
Seriosuly though, I must understand this. This is really getting in my way. first off why is the derivative x alpha with respect to x beta equal to the kornecker delta, or why is it defined as such?

Hey, sorry, I didn't realize you were still having troubles with this. Ok, to answer the first question, let's recall what $x^{\alpha}_{,\beta}$ actually is. He uses this to denote $$\frac{\partial x^{\alpha}}{\partial x^{\beta}}[/itex]. Now, what possible values of this can we have? Well, if both indices are the same, (say a=0=b), then this is equal to 1. However, if both indices are different (say a=1, b=0), then we have [tex]\frac{\partial x}{\partial t}=\frac{\partial}{\partial t} x=0$$. And so, the value is 1 if the indices are the same, 0 if they are different-- this is the kronecker delta.

Terilien
Hey, sorry, I didn't realize you were still having troubles with this. Ok, to answer the first question, let's recall what $x^{\alpha}_{,\beta}$ actually is. He uses this to denote $$\frac{\partial x^{\alpha}}{\partial x^{\beta}}[/itex]. Now, what possible values of this can we have? Well, if both indices are the same, (say a=0=b), then this is equal to 1. However, if both indices are different (say a=1, b=0), then we have [tex]\frac{\partial x}{\partial t}=\frac{\partial}{\partial t} x=0$$. And so, the value is 1 if the indices are the same, 0 if they are different-- this is the kronecker delta.

but why are they equal zero? the kornecker delta gives me no more trouble. why does it equal zero when the indices are different? After all, aren't they related by the equation right before 3.18?

The kornecker delat is very simple now and everything else in the chapter makes perfect sense. I have no idea why I can't get this particular thing.

Staff Emeritus
Gold Member
why does it equal zero when the indices are different?

Are you asking why, for example,

$$\frac{\partial x}{\partial t} = 0?$$

What is the definition of a partial derivative?

After all, aren't they related by the equation right before 3.18?

No, 3.18 relates different sets of cooodinates, say {t, x, y, z} as the unbarred coordinates and {T, X, Y, Z} as the barred coodinates. These sets of coordinates aren't independent, they're functions of each other. For exaple,

t = t(T, X, Y, Z).

Consequently, $\partial t / \partial X$ is, in general, not equal to zero.

The kornecker delat is very simple now and everything else in the chapter makes perfect sense. I have no idea why I can't get this particular thing.[/QUOTE]

Staff Emeritus
but why are they equal zero?
This just follows from the normal properties of partial derivatives.
the kornecker delta gives me no more trouble. why does it equal zero when the indices are different? After all, aren't they related by the equation right before 3.18?
Look closely at this equation; the alphas have bars over them. This is because they are coordinates in a second intertial frame, and coordinates in the barred frame are functions of the coordinates in the unbarred frame. Thus, $$\frac{\partial x^{\bar{\alpha}}}{\partial x^{\beta}}$$ is not, in general, zero.

Terilien
so these partial derivatives are the partial derivative of two independent coordinates? Then how do we derive the basis from that?
sorry man. Everything else in the book was fine when i understood the notation and all ,but this seems somehow different.

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Staff Emeritus
Gold Member
so these partial derivatives are the partial derivative of two independent coordinates?

Yes

Then how do we derive the basis from that?

A basis for what?

In any case, the elements of a basis are always independent.

Terilien
Yes

A basis for what?

In any case, the elements of a basis are always independent.

the basis of the gradient how do we derive it from that. and yes I know that the basis has to be independent. How do we derive the baiss by comparing that to 3.12?

Please explain it. I can't wait to continue with this.

Staff Emeritus
Ok, I'll give my best attempt at trying to explain (3.20). So, we have $$\frac{\partial x^a}{\partial x^b}=\delta^a_b$$ and we are told to compare this to (3.12) which is $$\tilde{\omega}^a(\vec{e}_b)=\delta^a_b$$ Now, this gives us $$\tilde{\omega}^a(\vec{e}_b)=\frac{\partial x^a}{\partial x^b}$$. Now, (3.15) gives us the general form for the gradient; expressing this in another way $$(\tilde{d}\phi)_i=\frac{\partial \phi}{\partial x^i}$$ where on the LHS I've abused notation a little-- it means simply the ith component of the gradient of phi. Now, $$(\tilde{d}x^a)_b=\tilde{d}x^a\frac{\partial}{\partial x^b}=\frac{\partial x^a}{\partial x^b}=\delta^a_b$$. Since $$\frac{\partial}{\partial x^i}$$ are the basis vectors, this gives us $$\tilde{\omega}^a\left(\frac{\partial}{\partial x^b}\right)=\tilde{d}x^a\frac{\partial}{\partial x^b}$$ from which we obtain $\tilde{d}x^a\equiv \tilde{\omega}^a$

Does this make any more sense?

Edit: If this doesn't make any more sense (and that will probably be due to my bad explanations!) then I should stress that, whilst it is admirable to want to know this, it's not overly important. It would suffice to know that dx^a are basis one-forms, and d/dx^a are basis vectors. If, however, you are interested in the more mathematical foundations of the subject, then Schutz has written another book "Geometrical methods of mathematical physics" which goes into this in far greater detail, which you may be interested in studying.

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Terilien
Ok, I'll give my best attempt at trying to explain (3.20). So, we have $$\frac{\partial x^a}{\partial x^b}=\delta^a_b$$ and we are told to compare this to (3.12) which is $$\tilde{\omega}^a(\vec{e}_b)=\delta^a_b$$ Now, this gives us $$\tilde{\omega}^a(\vec{e}_b)=\frac{\partial x^a}{\partial x^b}$$. Now, (3.15) gives us the general form for the gradient; expressing this in another way $$(\tilde{d}\phi)_i=\frac{\partial \phi}{\partial x^i}$$ where on the LHS I've abused notation a little-- it means simply the ith component of the gradient of phi. Now, $$(\tilde{d}x^a)_b=\tilde{d}x^a\frac{\partial}{\partial x^b}=\frac{\partial x^a}{\partial x^b}=\delta^a_b$$. Since $$\frac{\partial}{\partial x^i}$$ are the basis vectors, this gives us $$\tilde{\omega}^a\left(\frac{\partial}{\partial x^b}\right)=\tilde{d}x^a\frac{\partial}{\partial x^b}$$ from which we obtain $\tilde{d}x^a\equiv \tilde{\omega}^a$

Does this make any more sense?

Edit: If this doesn't make any more sense (and that will probably be due to my bad explanations!) then I should stress that, whilst it is admirable to want to know this, it's not overly important. It would suffice to know that dx^a are basis one-forms, and d/dx^a are basis vectors. If, however, you are interested in the more mathematical foundations of the subject, then Schutz has written another book "Geometrical methods of mathematical physics" which goes into this in far greater detail, which you may be interested in studying.

wait what does dx^a on it's own mean? That's all i need to know. thanks we're very close. also i thought that the gradient acted on a tanget vector to a curve, how do we know that the partials are the basis in that case? Aren't the partial derivatives just a basis for the directional derivative of some function on a manifold with respect to the parameter of a curve?

Please exaplin. sorry but if I have the slightest doubts about something I can't use it.

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Terilien
I think that after i understand this i'll give up for now. I probably don't have all the prerequisite skills for this kind of thing anyway. I'm only 15 afterall and have run into some annoyances for studying on my own.

Staff Emeritus
d/dx^a are the basis vectors for the tangent vector space at a point on the manifold. dx^a are the basis one-forms for the space dual to the tangent vector space.

I'm not sure how to explain this further, and I would re-emphasise that if you want to learn differential geometry strictly, then you should use a book written explicitly on that topic (note that the chapter in "a first course" is entitled "tensors in special relativity").

You can, however, learn special relativity without knowing anything about differential forms, and just treat them as covectors, and treat vectors as "contravarient vectors."

You shouldn't be put off by being a little confused by this; I for one found this topic one of the hardest when I first learned it. Where you go from here basically depends on what you want to learn. For example, are you more interested in Physics or mathematics?

I probably don't have all the prerequisite skills for this kind of thing anyway. I'm only 15 afterall and have run into some annoyances for studying on my own.

I don't know what prerequisite skills you have, but you seem to be coping well. You're bound to run into problems when self-studying (IMHO) since you have no-one to ask for clarification. Did you ever get in contact with someone at your local university (I'd forgotten about your other thread, and just looked at it now).

shoehorn
so these partial derivatives are the partial derivative of two independent coordinates? Then how do we derive the basis from that?
sorry man. Everything else in the book was fine when i understood the notation and all ,but this seems somehow different.

May I suggest that you learn a bit of linear algebra before attempting differential geometry? It doesn't sound like you really understand the topic at all.

Terilien
May I suggest that you learn a bit of linear algebra before attempting differential geometry? It doesn't sound like you really understand the topic at all.

I already know how to solve linear equations, multiply matrices, compute inverses and I have some grasp of vector spaces, though I must admit I didn't study them within a formal context, I just pciked some stuff up myself.

what makes you think that anyway? look I'm just having considreable dificulty with this one topic. The other ones were mostly a matter of notation, and the fact that I encountered many new topic I had never heard of.

Maybe I'm just not used to learning in a step by step manner, or from a book. I'm used to figuring everything out on my own.

anyway everything else regarding basis eems to make sense. Anyway when I do learn, I tend to learn from math books but I couldn't find a book on the subject that was not overly formal!

This is the same thing all over again.

Ps. Come to think of it i never really studied it at all. I just kind of "picked it up". anyway i grow tird of this thred. I'm reading a pdf version of schutz mathematical physics book. I'm afraid that someone is going to question me like last time. anyway please lock or delete this thread.

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Staff Emeritus
wait what does dx^a on it's own mean? That's all i need to know. thanks we're very close. also i thought that the gradient acted on a tanget vector to a curve, how do we know that the partials are the basis in that case? Aren't the partial derivatives just a basis for the directional derivative of some function on a manifold with respect to the parameter of a curve?

Please exaplin. sorry but if I have the slightest doubts about something I can't use it.

I haven't been following this thread, but $dx^a$ on its own is a one-form. One forms are maps from vectors (like for instance $$\frac{\partial}{\partial x^a}$$ - formally, vectors are just partial derivatives) to scalars, and are also known as "duals" of vectors.

It should be obvious, I hope, that $\frac{\partial}{\partial x^a} dx^a$ = 1.

where uses the term "tangent vectors" for vectors and "cotangent vectors" for their duals (i.e. for one-forms, like dx^a)

Terilien
Well it all makes sense right now. i haven't been sleeping properly for awhile but it has been clarified. I guess it looked weird from a computational standpoint.

pmb_phy
look I'm just having considreable dificulty with this one topic. The other ones were mostly a matter of notation, and the fact that I encountered many new topic I had never heard of.
That's why as an undergrad we are forced to take certain courses before others are taken. In the case of a physicist a person would have to have taken a course such as Mathematical Methods of Physics (or something akin to it) and basic mechanics and EM before they would be allowed to take GR. They don't "recommend" it. They demand it. Fortunately for me I had two majors as an undergrad, Physics and Math, so I was well equipped for the math tossed at me in physics.

And I too would appreciate it if people would not cross-examine Terilien. People tend to disappear of this forum under similar conditions. At best post your unsolicited advice and even better make it a general comment. If you feel the need to interigate him then you are free to do so, in PM!

Pete