# 4-volume is a rank (0,0) tensor?

1. Feb 2, 2004

### turin

How can I prove whether or not d4x is a 0th rank tensor? It seems strange that I should be so when it is the product of a 0th, 1st, 2nd and 3rd component, dx0dx1dx2dx3. I heard that the proof involves the Jacobian. I don't get it.

2. Feb 3, 2004

### matt grime

I'm notm sure my interpretation is correct, but in differential forms, d^2 is identically zero. so d^4(x) is zero.

3. Feb 3, 2004

### lethe

d^4x is shorthand for the invariant measure in Minkowski 4 space. it is the volume 4 form, not the exterior derivative applied 4 times.

you can explicitly show that it is invariant, just by applying the change of variables formula (involving a Jacobian, as turin correctly suggests).

4. Feb 3, 2004

thank you

5. Feb 3, 2004

### lethe

in fact, a better notation for the measure would probably be dx^4. but we are stuck with this notation.

6. Feb 4, 2004

### turin

You don't think that would be confused with the 4th component of a tensor (superscript instead of exponent)? I think the notation itself is confusing (the one physicists use). That's why I'm trying to make the transition to using the mathematician's notation, and then instigate a revolution to make all physicists use it under penalty of death by strenuous physical activity.

OK, so d4x' = |&part;(x&mu;')/&part;(x&nu;)|d4x
?

I'm understanding
|&part;(x&mu;')/&part;(x&nu;)|
to be the Jacobian.
Is this the determinant of the transformation matrix? I took multivariable calculus years ago, and I don't remember this stuff.

Last edited: Feb 4, 2004
7. Feb 4, 2004

### lethe

in this case, i actually prefer the physicists notation. i agree with you that dx^4 might be misinterpreted as integration with respect to x^4. perhaps (dx)^4 would be best. anyway, to make a fair comparison, i will tell you the mathematicians notation for the volume form:

vol or else Vol.

so this notation is pretty bad, i think.

Edit: whoops. corrected LaTeX error

Last edited: Feb 4, 2004
8. Feb 4, 2004

### turin

It says the latex source is invalid. I see a \vol here. I'm assuming that is supposed to be some sort of a "v" or a fancy "vol."

9. Feb 4, 2004

### lethe

yeah, its the determinant of the Jacobian. and in the case of a Lorentz transformation, this determinant is 1.

10. Feb 4, 2004

### turin

OK, lethe,
This is probably going to start getting on your nerves, but this termonology (I hope it's just the termonology) is killing me, and I think that you are not quite appreciating how much I am dying from it.

In my first post, I asked how to prove that d4x is a rank (0,0) tensor. (I am of course considering special relativity so the Minkowski metric tensor.) However, I see that you seem to be calling it a 4-form, which I understand to be a rank (0,4) tensor. Maybe I should say a rank (4,0) tensor. Anyway, I understand a 4-form to not be a rank (0,0) tensor.

Do you have any idea what is making me so confused? Am I just being incredibly stubborn about this? Is the distinction trivial or something?

11. Feb 5, 2004

### lethe

you probably just say to yourself: "Self, its invariant under Lorentz transformations, so its a scalar, aka (0,0) tensor"

however, as i have said a few times, any physical/geometric object is invariant, it is only the components which may not be invariant.

what makes the case of the volume form a little tricky is that, unlike the case of other tensors, the volume form only has one component. so that component must therefore be invariant, just like a scalar. in fact, using the physicists definition of a tensor, where a tensor is just a matrix which follows a particular set of transformation rules, then this is indistinguishable from a scalar; it is a 1x1 matrix that does not change under Lorentz transformations.

furthermore, the space of scalars is dual to the space of volume forms.

however, strictly speaking, the volume form is not a scalar, it is indeed a tensor. no harm will come of thinking of it like a scalar, but you do not integrate scalars, you integrate forms.

12. Feb 6, 2004

Hey, consistency in formalism should never go unrewarded. Turin you're right, it's a (0,4) tensor.

If you built it up as $dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$, then it alternates, i.e. $dx^1 \wedge dx^0 \wedge dx^2 \wedge dx^3$, or any other exchange of the 1-forms, multiplies the volume by -1. A true scalar would have no dependence on the order.

Nonetheless, it has only the one component, so it tends to be thought of as a scalar. "Pseudoscalar" is a term sometimes used.

13. Feb 6, 2004

### Peterdevis

The naive volume element d[^n]x is itself a density rather than a n-form.A tensor density transform likes a tensor multiplate with the determinant. you can construct a invariant volume element by multiplying the naive volume element with the root of the determinant of the metric.
For more information see Chapter 2 Manifolds page 20 and 21 http://

Last edited by a moderator: Apr 20, 2017