# Unit of Tensors

1. Mar 3, 2004

### Sammywu

Event Vector: [ sec, m, m, m ] or [sec, light-sec, ..]

00th component of metric tensor : m^2/sec^2.
iith components of metric tensor : 1.
0ith or i0th components of metric tensor: m/sec.

4-velocity: [ 1, m/sec, .. ]

4-momentum; [kg, kg*m/sec, .. ]

4-Force : [ kg/sec, kg*m/sec^2, .. ]

kg*m/sec^2 = newton.

E charge-current tensor: [ coulomb/m^3, coulomb/m^2*sec, .. )

2. Mar 3, 2004

### Sammywu

time-time component of EM field strength tensor : newton*m/coulomb*sec.
space-space component of EM field strngth tensor: newton*sec/coulomb*m.
time-space components of EM field strength tensor: newton/colomb.

3. Mar 3, 2004

### Sammywu

Here after, N for newton, C for coulo,b, and J for Joule.

EM vector potential : [ N*m/C, N*sec/C, .. ] or [J/C, J*sec/C*m, .. ]

4. Mar 3, 2004

### Sammywu

This does not look good; it's too complicated. Maybe this will be better, if translating unit of time to unit of space:

Ut: Unit of time and Us : unit of space
c*Ut=Us.
Um : Unit of mass
Uc: Unit of electric charge
Ue: Unit of Energy

Event vector : [ Us, Us , .. ]

4-V : Us/Ut ...

4-M : Um*Us/Ut ...

4-F : Um*Us/Ut^2 ...

metric tensor : 1

Uf=Um*Us/Ut^2

qV vector: Uc*Us/Ut

EM field strength tensor: Uf*Ut/(Uc*Us) ...

EM potential vectot : Uf*Ut/Uc or Ue/Uc ...

Unit of (e-zero) : Uf*Us^2/Uc^2

Unit of (1/mue-zero) : Uf*Us^2/Uc^2

E current density tensor: (Uf*Us^2/Uc^2)*(Uc/Us^3)= Uf/(Uc*Us) ... .

5. Mar 3, 2004

### GRQC

I don't understand the point of this post, except perhaps to generate mass confusion. All components of a vector or tensor must necessarily have the same units. It makes no sense for them not to.

Think about this in three dimensions: would a position vector (x,y,z) have x and y in meters, but z in kg?

The main problem is that you're dropping 'c' from many of the components. For example:

All components of this vector are in meters, since the vector (t,x,y,z) is actually (ct,x,y,z). Same for the rest.

The Maxwell (EM) tensor has units of E and/or B, since these have the same units up to a factor of c. Setting c=1 doesn't "change" the units of different components.

6. Mar 5, 2004

### Sammywu

My attemption was to clear some clouds in my mind.
Since tensors are used for describing Physic Quantity, the more I am reading them, I found I shall make their units clearer.

Like this can be altered, since the denominator of 4-V is written as tau in many places but more likely it's c*tau. If d(tau)^2= g(ij)d(x(i))^2. x(0) is c*t and then tau here has to be c*tau.

4-V : 1 ...

4-M : Um ...

4-F : Um*/Us ...

metric tensor : 1

Uf=Um/Us

qV vector: Uc

EM field strength tensor (F) : Uf/Uc

EM potential vector ( A ) : Uf*Us/Uc or Ue/Uc ...

F = dA

Unit of (1/e-zero) : Uf*Us^2/Uc^2

Unit of (mue-zero) : Uf*Us^2/Uc^2

E current density tensor ( J ) : (Uf*Us^2/Uc^2)*(Uc/Us^3)= Uf/(Uc*Us) ... .

d(*F) = 4.Pie.(*J)

Here, It seems to be interesting to determine a natural unit for E charge. For example, two objects with 1 Uc of E charge exerts 1 Uf of force when they are separated in 1 Us of distance, etc ... .

I have a little trouble to determine the unit of energy-momentum-stress tensor. It seems to be Um/Us^3.

7. Mar 5, 2004

### pmb_phy

Re: Re: Unit of Tensors

That is incorrect. The metric tensor is a tensor and all of it's components do not always have the same units. E.g. the metric when expressed in polar coordinates have some components which have units of distance^2 whereas other are unitless.

8. Mar 5, 2004

### GRQC

Re: Re: Re: Unit of Tensors

Fundamentallly the metric tensor has the same units, as do the one-forms dxi. You shouldn't be mixing and matching them according to coordinate system. From the definition of the line element, we have:

$$ds^2 = g_{ab} dx^a dx^b$$

which clearly demonstrates the separation of forms (units of distance) from metric.

I don't think coordinate transformation shouldn't be seen as changing the "units" of a tensor, although I suppose ultimately it is a matter of interpretation.

Last edited: Mar 5, 2004
9. Mar 6, 2004

### Sammywu

My position is each component could have its own unit.

For instance, the flat metric tensor could be rewrite as diag(-1/c^2,+1,+1,+1) in this coordinate.

While at the same time, we can try to translate them into the same unit, not only easier to calculate, also could see some fundamental relationship between units of different quantities through this translation.

For now, I try to correct some things here.

4-V : 1 ...

4-M : Um ...

4-F : Um*/Us ...

metric tensor : 1

Uf=Um/Us

qV vector: Uc

EM field strength tensor (F) : Uf/Uc

EM potential vector ( A ) : Uf*Us/Uc or Ue/Uc ...

F = dA

Now , if we set a unit of Uc as two objects with 1 Uc of E charge exerts 1 Uf of force when they are separated in 1 Us of distance, k will be 1.

Unit of k : Uf*Us^2/Uc^2

(1/e-zero) = 4.Pie.k.

(mue-zero) = 4.Pie.k/c^2 = 1.Pie.k since c=1 Us/Ut and Us equals to Ut in this case.

E current density tensor ( J ) : (Uc/Us^3).

d(*F) = 4.Pie.k.(*J)

I tried to make sense of stress tensor or mass tensor. In the way, I realized that I can define a 4-M density tensors which has a unit of Um/Us^4. The integration of it over dtdxdydz will be the 4-M tensor.

Ecah components of 4-M density tensor or 4-M tensor are already the flux of mass to the direction. For example, in its own frame, a mass flows through only the direction of time. For a relative moving observer, the mass flows thru time and space. In a Minkoskwi diagram, that's like a world-shadowed-area.

A question here is this seems to indicate Um is just a projected image of Um*Us on the dxdydz spatial hypersurface. A more natural quantity might be Um*Us ( or Um*Ut more likely ) and maybe Um*Us is the invariant.

10. Mar 6, 2004

### GRQC

I'm very confused by your notation. Are you using * as the Hodge star, or as multiplication (or both)? You seem to be flipping between "classical" notation and differential geometric representation.

I'm interested in what you have to say, but it's a bit awkward to read in that form. If you could use the TeX package, it would be a bit easier for others to read.

11. Mar 6, 2004

### Peterdevis

$$ds^2 = g_{ab} dx^a dx^b$$

the infinete line element and the metric are the same object

$$g_{ab}$$ = a component of the metric
$$dx^a$$ and $$dx^b$$ = base vectors of the (0,2)tensor space

So components are unitless and it are the base vectors who have the units.

12. Mar 6, 2004

### Sammywu

How do I get this Tex package?

I know I messed up the symbol due to the tool issue, unfortunately, but I assume you can follow me very easily.

Any way, if we introduce a unit as Um*Us. This does seem to resolve another issue. Notice Uc.Uc=Uf.Us^2=Um.Us because Uf=Um/Us. Now you can see Uc as the sqrt of Um*Us. This might be able to link a natural definition of Uc.

Another side issue we can think of is what is the natural unit for force , energy, time and space.

The first thing came to my mind is of course the plank space and time interval. I am not sure how they fit, just a hunch.

13. Mar 6, 2004

### GRQC

Check the first sticky in the General Physics forum. It tells you how to use the package.

14. Mar 7, 2004

### pmb_phy

Re: Re: Re: Re: Unit of Tensors

That is very incorrect. First off the topic is about the components of tensors. And the components of the metric that I was explaining to you. The components of the metric are equal to the coefficients of the dxu. The units of the coefficients must be such that of a single factor guvdxu dxv in the sum has units of distance-squared. That means that if dx1 is an angle then g11 must have units of distance squared.

Last edited: Mar 7, 2004
15. Mar 12, 2004

### Sammywu

Unit of Plank Constant h: Um.Ut

16. Mar 12, 2004

### Sammywu

We can even try a little bit of this, before I learn how to use Tex.

t=i.(1/c).x, where i is the sqrt(-1).

Unit of h is Um.Us and unit of i.h is Um.Ut, where h is the planck constant divided by 2.Pie; that will make both sides of Schrodinger EQ. match and make sense.

geometric tensor could even be rewritten as diag(-1,1,1,1)=diag(i,1,1,1)xdiag(i,1,1,1).

Do you know where I can find a good and thorough discussion of the Dirac's derivation of necessary electron spins and g-factor from Schrodinger EQ?

17. Mar 16, 2004

### DW

http://electron6.phys.utk.edu/qm2/modules/m9/dirac.htm

18. Jul 12, 2004

### Sammywu

I found the natural unit for tensors of rank (n,m) is actually length^(n-m); only in this way the tensor will obey the standard tensor transformation law.

Tensors that do not use this natural unit such as the stress tensor are just a convinient Physic way in dealing with tensor equations.

DW,

I have finally followed thru the Dirac EQ. posted in your reply.

I realized that I need to assume the quantum statespace in here is a 4-component 3-freedom space, instead of the usual 1-component 3-freedom space.

I found a problem in getting this important eq. ; would you mind help clarifying this:

I translated
$$\alpha \cdot p$$
to
$$\alpha_x p_x + \alpha_y p_y + \alpha_z p_z$$

and tried to check the commuting eq. between
$$\sigma_x ' and H_D$$ in page 4.

$$\sigma_x ' ( c ( \alpha_x p_x + \alpha_y p_y + \alpha_z p_z ) + \beta mc^2 + q \phi )$$
-
$$( c ( \alpha_x p_x + \alpha_y p_y + \alpha_z p_z ) + \beta mc^2 + q \phi ) \sigma_x ' =$$

$$c ( \sigma_x ' \alpha_y p_y - \alpha_y p_y \sigma_x ' ) + c ( \sigma_x ' \alpha_z p_z - \alpha_z p_z \sigma_x ') + mc^2 ( \sigma_x ' \beta - \beta \sigma_x ') =$$

$$c ( 2 \sigma_x ' \alpha_y p_y ) + c ( 2 \sigma_x \alpha_z p_z) + mc^2 ( 2\sigma_x ' \beta ) =$$
$$2 \imath c \alpha_z p_y - 2 \imath c \alpha_y p_z + 2 mc^2 \sigma_x ' \beta$$

See, I have this additional item
$$2 mc^2 \sigma_x ' \beta$$
coming from the fact that $$\sigma_x ' and \beta$$ anticommute as well.

Thanks & Regards

Last edited: Jul 12, 2004
19. Jul 14, 2004

### Haelfix

SammyWu, what you have written above is goobledygook, and completely meaningless.

"I realized that I need to assume the quantum statespace in here is a 4-component 3-freedom space, instead of the usual 1-component 3-freedom space."

Umm, I don't know where you got this idea from, but I think you might want to go all the way back to rereading texts on quantum mechanics, b/c relativistic equations like the Dirac equation start using highly abstract quantities and the notation gets cumbersome.. If you're not 100% comfortable with the original material, you will get lost.

Ultimately GRCQ is right, you want your tensor equations to have the same units.

Yes, i'm aware that flipping units over on the other side of the equality, and/or masking them in other terms gives you the right answer if you're careful about it (since they are basically just sets of equations when it comes down to it).. But that can lead to much confusion, and probably not a good practise unless you're a specialist. Not to mention, you won't get anywhere if youre trying to generalize tensors into something more elegant.. Like differential forms.

Besides, most theoreticians make their units dimensionless as a matter of practise.