1. Jul 20, 2007

### pmb_phy

This came up in a newsgroup but there was only one person responding who would only repeat himself regardless of what was posted. I'm looking for more insight into how people read this regarding scalars.

In Geoetricaal methods of mathematical physics by Bernard F. Schutz the author defines "scalar" as follows. From page 64
Consider the 4-momentum P and the particular basis vector e_0. Do you hold that P*e_0 is a scalar? Yes or no answers prefered.

Pete

2. Jul 20, 2007

### HallsofIvy

Staff Emeritus
I can't give you a "yes or no" answer because the answer depends upon what you mean by "the particular basis vector e_0". If you mean just "a particular vector", then, yes, the "dot product" (tensor product contracted) of two vectors (whether one is a "basis vector" or not) is a scalar. If, on the other hand, you mean "pick a basis and call one of them e_0", then, no, because the result depends upon the choice of basis. Do you see the difference?

3. Jul 20, 2007

### pmb_phy

Of course I see the difference. But I'm trying to keep as precisly to Schutz definition as possible. Recall the definition
Notice the last part where it says "definition does not depend on the choice of any particular basis". For that reason I chose to use the basis vector e_0 but a "particular" one at that. I.e. I said

"Consider the 4-momentum P and the particular basis vectore_0.

I was looking for the correct wording to obtain the response "since P*e_0" depends on the particular basis vector e_0 and changing that basis vector upon a change of coordinates means it will yield a different number then P*e_0 is not a scalar.

How do you see that I worded the question at the bottom incorrectly (which is what you're saying right, i.e. that my question was ambiguous?)

Thanks HallsofIvy

Pete

4. Jul 20, 2007

### HallsofIvy

Staff Emeritus
Well, the use of words "specific basis vector" confused me- might be my fault!

Certainly, the product of two "specific" vectors, whether they are basis vectors or not, is a scalar. In fact, one way of determining whether something IS a vector is seeing if its product with any given vector is a scalar.

5. Jul 20, 2007

### pmb_phy

This is very tricky stuff, yes!

Question: What about the exception that Schutz states regarding whether something is a scalar or not?

Consider this too. A covariant tensor of rank two maps two vectors to a scalar correct? What if the two vectors were e_1 and e_2 in a given basis. Then

T(e_1,e_2) = T_12 = a component of the tensor. Is this a scalar/invariant?

Pete

Last edited: Jul 20, 2007
6. Jul 20, 2007

### robphy

P*e_0 is a scalar [the component of P with respect to e_0].
T(e_1,e_2) = T_12 = the e1-e2-component of the tensor is a scalar.
If you change the vectors (in this case, the basis vectors) as arguments of these non-scalar tensors P and T, then [of course] these resulting scalar values change.

I think the point is:
once there are no more free indices [after contracting with whatever other set of tensors], this tensor is no longer a function of direction... and so is a scalar.

7. Jul 20, 2007

### HallsofIvy

Staff Emeritus
Am I misunderstanding something?

You quote "A covariant tensor of rank two maps two vectors to a scalar" and then ask "What if the two vectors were e_1 and e_2 in a given basis?" Two vectors are two vectors- whether they happen to be in a given basis or not! What is so difficult about that?

8. Jul 20, 2007

### pmb_phy

If you change the basis then the components of the tensor will change. Therefore they can't be scalars, according to definition given by Schutz/

Pete

9. Jul 20, 2007

### robphy

I tried to clarify that they are tensor-components-with-respect-to-a-fixed-set-of-axes (e1 and e2).
Once those are chosen, all other bases used to evaluate that contraction (i.e., when e1 and e2 are expressed in any other basis), the final result will not depend on any basis... but only on T,e1, and e2.

In simpler terms, if e is a unit vector pointing down an incline [fixed once and for all] and N is the normal force. N-dot-e is a scalar (which happens to be zero). It doesn't matter if we try to evaluate N-dot-e using horizontal-and-vertical-axes or use parallel-and-perpendicular-axes... we get the same result.

A relativity example: if v is a future unit timelike vector, then g(P,v) is time-component-of-P-observed-by-the-observer-with-4velocity-v... it is a scalar. Everyone else will agree on that (i.e. agree on what that observer measures). What they will disagree on is what the time-component-of-P is.

10. Jul 20, 2007

### pmb_phy

I think I can say something which will clear this up and bring on new views on this.

Suppose I form the scalar product P*e_0. If one changes coordinate systems then there are two ways of doing this. One can express both P and e_0 in terms of the new basis vectors and then form the scalar product again and this operation will leave the value of P*e_0 invariant. Or one could say that P never changes but when we change coordinates we have a new basis and as such P*e_0 becomes P*e'_0 and thus the value changes thus keeping in step with the definition I gave of e_0 being a basis vector in that changing coordinate systems changes basis vectors. Oy! Semantics abound in this one huh?

Now that this is even more confused.... LOL!!

Pete

Last edited: Jul 21, 2007
11. Jul 21, 2007

### robphy

I suspect that part of this issue involves notation... and it might take some time to sort it all out. In sorting things out, it might be helpful to explicitly write out the operations... then, if desired, invent a notation to summarize various expressions.

For me, the components [with respect to an orthonormal basis (say, $$\{\hat x,\hat y\}$$) of vectors] of a vector $$\vec P$$ are obtained from this expansion:
$$\vec P = \hat x (\hat x \cdot \vec P) + \hat y (\hat y \cdot \vec P)$$
which expresses the vector P in terms of the basis vectors and the scalar-products of P with those basis vectors.

So, how do we write this in index-notation?
It'll probably depend on the choice of convention.
(Apologies if my latin-vs-greek choices are opposite of yours.)

Personally, I try to think in abstract-index form... so that, in $$P^a$$, the $${}^a$$ does not label a component, but instead is an abstract-index indicating that it is an element of the space of vectors (i.e., of type (1,0)-tensors). [It is used to label a copy of the vector space on which the tensor algbera is based.]

I would write [using the metric explicitly and keeping the hats to remind me of their unit length]:
$$P^a = \hat x^a (\hat x^b g_{bc} P^c) + \hat y^a (\hat y^b g_{bc} P^c)$$ (note that there is no summation convention necessarily being employed if one regards these as abstract indices).

If I call my basis vectors $$\widehat {e_0}$$ and $$\widehat {e_1}$$,
then I would write:
$$P^a = \widehat {e_0}^a (\widehat {e_0}^b g_{bc} P^c) + \widehat {e_1}^a (\widehat {e_1}^b g_{bc} P^c)$$
where ${}_0$ and ${}_1$ label "which basis vector".

In summation form,
$$P^a = \sum_\alpha \widehat {e_\alpha}^a (\widehat {e_\alpha}^b g_{bc} P^c)$$

[I'll try to look at Schutz' book when I get a chance... but I don't have a lot time right now.]

12. Jul 21, 2007

### pmb_phy

I disagree Rob. I believe that in this instance you're trying to define the components of a vector in a way which is inconsistent with the what I would define as the correct definition as that given in Schutz's GR text. The reason I say this is that it works out just right when you do the calculations.
If you carried out this expansion using Euclidean geometry then you'll get the right answer. If you calculate it using Riemann geometry with metric = $\eta_{\alpha \beta}$ = diag(1, -1, -1, -1) then you'd be wrong, i.e. your definition of the components be as you stated them above, i.e.

$$P^{\alpha} = P*e_{\alpha}$$

gives the wrong results upon calculation. The definition of a vector as I posted if (from Schutz's GR text) is

$$P^{\alpha} = P(\omega^{\alpha})$$

where $\omega^{\alpha}$ is a basis 1-form

Thus this expression will give the correct value for $\alpha$ = 1 gives

$$P^{1} = P(\omega^{1})$$

However your definition gives

$$\sigma(1) = -P^{1}$$

I suggest that you work it out with the ole pen and paper. This is just a suggestion though.

Best regards

Pete

Last edited: Jul 21, 2007
13. Jul 21, 2007

### robphy

With a different signature, you can work in the correct signs... with slightly more notation. If the metric is degenerate, you have a little more work to do. However, the general structure of tensors and components doesn't heavily rely on the signature [although signs and various techniques may]. Since your main issue has to do with what a scalar is, I felt it's better to focus on the simple case and not worry about generalities which only complicate the simple.

Does the following help you understand the subject better?

$$\vec P = \displaystyle \frac{\hat x (\hat x \cdot \vec P)}{\hat x \cdot \hat x } + \frac{\hat y (\hat y \cdot \vec P)}{{\hat y \cdot \hat y }}$$

$$P^a =\displaystyle \frac{ \hat x^a (\hat x^b g_{bc} P^c)}{\hat x^d \hat x^e g_{de}} + \frac{\hat y^a (\hat y^b g_{bc} P^c)}{\hat y^d \hat y^e g_{de}}$$

Last edited: Jul 21, 2007
14. Jul 21, 2007

### pmb_phy

Okey Dokey Rob. I just wanted to clarify that. Why? I dunno.

Pete

15. Jul 21, 2007

### robphy

I recently added to my previous post to address your case.
I hope it helps.

16. Jul 22, 2007

### pmb_phy

I believe I quoted Schutz somewhere his math text in which he emphasizes the fact that a scalar may not depend on the coordinate system. By definition of the components of a tensor they must change as one changes coordinate systems. This is true because when you change the coordinate system then the set of basis vectors also change. This is the tricky part. You have to look at the context in which one is plugging in vectors. If one looks at e_1 and e_2 only as vectors then they are geometric objects which do not change under a coordinate transformation. However if you define e_1 and e_2 as basis vectors then they must change upon transformation as well as T_12. This the context is extremely important.

See why the this pretty tricky?

If I didn't post that quote from Schutz then perhaps I should. Would you care to read it?

Pete

17. Jul 22, 2007

### pmb_phy

I recall now why I made that comment regarding components. When I was discussing this in sci.physics.relativity someone jumped down my throat with his usual demeaning tone and "corrected" me regarding the definition of a component. I knew I made a mistake anyway but I always find it best to correct any mistake I make ASAP. That's was what I was thinking. But then again that's not something I'd have to watch out for when I discussing things with you. You and that person are like night and day. He's arrogant and condescending. You are neither my friend.

Best regards

Pete

ps - I will consider the component thing a done with subject since it is irrelevant to the purpose of this thread.

18. Jul 22, 2007

### nrqed

I did not read the whole thread, only the first few posts so I apologize if I am being redundant here. But it seems to me that the key issue is the distinction between vector and basis vector. A basis vector obviously depends on the coordinate system used so the above expression is not a scalar in that case. However, a true vector *is* independent of the basis used and in that case P*v is a scalar. The same thing with feeding two basis vectors e_1 and e_2 to a tensor. The result won't be a scalar because it will depend on the coordinate system used.

Just my two cents...

19. Jul 23, 2007

### pmb_phy

You've hit the nail square on the head Sir. Bravo!

Suppose I am given the two 4-vectors A = (1, 0, 0, 0) and B = (0, 1, 0, 0). Are these basis 4-vectors or simpy 4-vectors. That is the important question here. The only way to know the answer is from the context of the problem at hand. If I declare that both A and B are basis vectors for the current coordinate system then they must change according to the transformation rule for basis vectors when I change the coordinates. Define s as s = T(A,B). Then when the coordinate system changes then so too will the basis vectors A and B. Since s is dependant on the coordinate systems through the dependance on the basis vectors A and B then s is not an invariant quantity and thus it is not a scalar.

However if A and B are simpy 4-vectors defined independantly of the coordinate system then s will be an invariant and so it will also be a scalar.

Thanks for your input. It appears as if we are both on the same wave length.

Best regards

Pete

20. Jul 23, 2007

### ObsessiveMathsFreak

To be pedantic, this definition appears to be wrong. Perhaps I'm misreading the meaning here, but it appears to me that a scalar is dependent on a choice of basis. Mass will have a different value whether you are measuring in kilograms or pounds. In other words, scalar values too have basis vectors, namely the units they are measured in. Or is Schutz talking about scalars in the sense of a mole?