# Why does spin 0 have to be a true scalar?

1. Oct 20, 2007

### pellman

Of course, a spin 0 field, by definition, has to be a scalar field under rotations. But what about boosts?

Compare to classical charge density $$\rho$$, which must satisfy

$$\frac{\partial\rho}{\partial t}+\nabla\cdot J=0$$

That is, $$\rho$$ is the time-component of a four-vector under boosts. But under rotations it is a scalar and in that sense, $$\rho$$ has "spin 0".

One might say, well the Klein-Gordon equation satisfied by spin-0 fields is a scalar equation. That is, all the operations in the equation transform as a scalars, under rotations and boosts both. True, but that does not mean it has to act on scalars. Scalar operators can act on anything. For instance, each of the four components of the Dirac field independently must satisfy the Klein-Gordon equation.

If your first reaction is to respond, "You don't seem to have a good grasp of what a scalar under boosts means," you're right. Please enlighten me.

Todd

2. Oct 20, 2007

### javierR

In quantum field theory, a "scalar" means it is Lorentz invariant. Lorentz transformations consist of spatial rotations and boosts. As you said, the spin of an object comes from how it transforms under the spatial rotations. But our "scalar" also transforms as a scalar (that is, like a function) under boosts. Boosting the coordinate frame takes our coordinates from $x^{\mu}$ to $x'^{\mu}=\Lambda^{\mu}_{\nu}x^{\nu}$ while the scalar field
$\phi(x)$ goes to the new function of the new coordinates $\phi'(x')=\phi(\Lambda^{-1}x')$. That is, if I ask about the field value at a point x' in the new coordinate frame, you can still use the old function, as long as you unboost to the old coordinates. The Lagrangian density of a given field theory is also constructed to transform like this so that it is a Lorentz scalar.
A Lorentz vector representing a massive field, $v_{\mu}$, is spin 1 because it transforms like a vector under spatial rotations, and under boosts it transforms as $(\Lambda^{-1})^{\nu}_{\mu}A_{\nu}(\Lambda x)$.
In general, the Lorentz tranformations R act on a *rank-N Lorentz tensor field* A as $A'_{\rho\cdots \sigma}=R^{\mu\cdots \nu}_{\rho\cdots \sigma} A_{\mu\cdots\nu}$.
Note: for massless fields of spin not equal to zero, there is a helicity instead of spin, which is dictated by how the field transforms under planar spatial rotations (SO(2) rather than SO(3)).

3. Oct 20, 2007

### robphy

When one talks about "scalars" and "vectors", one should really specify the underlying space. A scalar field in an observer's instantaneous 3-dimensional Euclidean space is different from a scalar field in 4-dimensional Minkowski spacetime.

4. Oct 22, 2007

### pellman

I think I understand what you are saying all right. But is there a theoretical justification for it? As I point out, spin=0 by itself is not enough to give scalars under boosts. And the Klein-Gordon equation does not demand a scalar field either.

A macro fluid has a density field. And it is "spin 0" under rotations. But under a boost it will get a gamma factor at every point due to "foreshortening", right? Why doesn't a quantum scalar field behave this way?

5. Oct 22, 2007

### robousy

How do spinors and the graviton transform, and from the result how can you say one has spin 1/2 and the other spin 2?

6. Oct 22, 2007

### Hans de Vries

It's summarized in this table:

$$\begin{tabular}{|r|cllllc|r|l|} \hline &&&&&&&&\\ covariant Tensor && T'_{\ \mu\nu}&=&\Lambda^{-2} &T_{\mu\nu}&& 16 comp.& spin 2 \\&&&&&&&&\\ covariant Vector && A'_{\ \mu} &=&\Lambda^{-1} &A_{\mu} && 4 comp.& spin 1 \\&&&&&&&&\\ covariant Spinor && S'_{\ } &=&\Lambda^{-1/2}&S_{ } && 2 comp.& spin \frac{1}{2}\\&&&&&&&&\\ Constant && C' &=&\Lambda^0 &C && 1 comp.& spin 0 \\&&&&&&&&\\ contravar.Spinor && S'^{\ } &=&\Lambda^{1/2} &S^{ } && 2 comp.& spin \frac{1}{2}\\&&&&&&&&\\ contravar.Vector && A'^{\ \mu} &=&\Lambda^1 &A^{\mu} && 4 comp.& spin 1 \\&&&&&&&&\\ contravar.Tensor && T'^{\ \mu\nu}&=&\Lambda^2 &T^{\mu\nu}&& 16 comp.& spin 2 \\&&&&&&&&\\ \hline \end{tabular}$$

Now the fact that the higher order structures need a higher order general Lorentz transform is pure math. The connection to spin would be justified by the observation that spin 1/2 fermions can pair up ("form products") that behave like spin 0, spin 1 or other higher order spin constructs.

Regards, Hans.

Last edited: Oct 22, 2007
7. Oct 22, 2007

### robousy

Wow, great table! Thats going up in my little grad student office! :)

8. Oct 22, 2007

### robousy

What do the powers of the LT mean?

9. Oct 23, 2007

### pellman

Getting off topic, but how do you get 16 components ==> spin 2? A spin 2 object would have 5 components. If it is analogous to Dirac spinors, describing both particles and anti-particles, then that would still give only 10 components.

10. Oct 23, 2007

### blechman

The "spin" collumn on that table is a little funny. I think a better way to put it is that an integer spin-s particle can be expressed in terms of a tensor of rank s (rank == number of indices). So spin-0 fields have 0 indices, spin-1 has one index, etc. Generalization to spin-1/2 objects with spinor indices follows as well.

As to reducing the number of degrees of freedom: that happens from the actual field equations of motion. So for example: a massless vector field (photon) is described by a rank-1 tensor (4 components). But a photon only has two degrees of freedom, you say! Well, remember that there is also a gauge freedom that allows you to set any one of those components equal to zero, so you really have three components. But now, that component plays the same role as a Lagrange Multiplier in mechanics, enforcing an equation of constraint, which subtracts yet another degree of freedom. So in the end, a massless vector boson has 4-2=2 degrees of freedom, as it should!

In the massive case, things are more complicated, since the mass breaks the gauge symmetry. So for the massive photon (Z-boson, for instance) the gauge constraint actually becomes an equation of motion. Therefore you don't subtract 2, but only 1 and you get 4-1=3 degrees of freedom for a massive spin-1 field, as you should.

For the graviton (spin-2), it's the same story, but the gauge symmetry of Einstein is much more complicated - it's given by a VECTOR rather than a scalar. So the metric is a 4x4 symmetric tensor (rank-2) = 10 components. Now there are FOUR gauge degrees of freedom (each a Lagrange multiplier of 4 constraint equations) = 8. Therefore a massless spin-2 object has 10-8 = 2 degrees of freedom, as it should. For the massive spin-2 case, it's the same story, but you also have to subtract the trace of the tensor, so it's 10-4-1=5 (yay!!)

11. Oct 23, 2007

### pellman

Thanks, blechman. I've wondered about that for a while.

12. Nov 1, 2007

### pellman

I think I figured this out but I am posting it just in case anyone else wants to add (which would be appreciated).

For a Klein-Gordon field, the conserved charge current is proportional to

$$j_\mu=\phi^\dag\partial_\mu\phi-(\partial_\mu\phi^\dag)\phi$$.

Since j must satisfy the conservation law

$$\partial_\mu j^\mu=0$$

then $$j_\mu$$ must be a contravariant vector. But the partial derivative operator in j is by itself a contravariant vector. Therefore, $$\phi$$ must be a scalar.

Later edit

On second thought this is not necessarily the case. The zero on the RHS of

$$\partial_\mu j^\mu=0$$

need not be a zero scalar. If it is a zero tensor of any rank, it is still a good invariant tensor relation.

So the question as to the theoretical requirement that phi be a scalar is still open in my mind.

Last edited: Nov 1, 2007
13. Nov 1, 2007

### blechman

"spin" (by definition) is a description of how a field transforms under the Lorentz Group, the homogeneous isometries of Minkowski spacetime. Look at the Klein-Gordon action, and ask yourself how the KG field must transform under Lorentz transformations in order to keep the action invariant. It is pretty clear that the only way to keep the action invariant is to have the KG field transform trivially:

$$\phi(x)\rightarrow\phi'(x')\equiv\phi(x')$$

This transformation rule for the KG field is a symmetry of the action.

The trivial representation of the Lorentz group, often written (0,0), is the spin-0 representation. Therefore the "scalar field" and a "spin-0 field" are truly one and the same.

The point is: $\phi$ must transform as a scalar under Lorentz transformations. That's the symmetry of the action.

This is how I always thought about it. Is there a flaw in the above argument??

Last edited: Nov 1, 2007
14. Nov 1, 2007

### pellman

I don't know about that.

Without giving much thought to it, I notice right away that there is a term in the action proportional to $$\int \phi^\dag \phi d^4 x$$. Under a spatial rotation $$d^4 x' = d^4 x$$ so phi transforming as a scalar under rotations does indeed leave the action invariant under rotations.

But under boosts don't we have $$d^4 x' \ne d^4 x$$ due to "foreshortening" and time-dilation? So wouldn't that necessarily require $$\phi(x)\rightarrow\phi '(x')\ne\phi(x')$$ to maintain invariance of the action? I'm sort of guessing here since I am not all that familiar with the effect of boosts on the volume element.

15. Nov 1, 2007

### Hans de Vries

The power of the Lorentz transform determines the structure of the table.
You see that it runs from -2 to +2 and can have a value of 1/2 as well in
the case of spinors. (A spinor is a structure in the "amplitude" domain and
is squared to become an observable) Any integer or half integer value is
possible beyond the -2..+2 range.

A Lorentz transform transforms one vector into another. Two Lorentz
transforms are needed to transform a tensor into another. A Lorentz
scalar must be frame independent (0th order Lorentz transform).

What's behind this all is, is how these quantities arise, for instance:
grad(scalar) = vector, grad(grad(scalar)) = tensor.
The gradient is taken by differentiating against the base vectors
x, y and z which Lorentz transform. Therefor: something which
is the gradient of a gradient needs a Lorentz transform squared.

The spin gets into this table because of the way the spins of multiple
particles combine. Two spin 1/2 particles can become a spin-0 particle
represented by a scalar field, or a spin-1 particle represented by a
vector field. Four fermions can become a spin-2 particle represented
by a tensor field. They form "products" and the Lorentz transforms
required become products of Lorentz transforms.

Regards, Hans

Last edited: Nov 1, 2007
16. Nov 1, 2007

### George Jones

Staff Emeritus
Under a standard boost in the $x^1$ direction, what is

$$dx'^0 \wedge dx'^1$$?

17. Nov 1, 2007

### blechman

You have to be very careful: you're INTEGRATING over coordinates. Therefore, if you perform a coordinate transformation that changes the measure, but you can then perform a coordinate relabeling that returns the measure back. So no, there's no change in the measure.

To be specific: consider an integral $\int dx f(x)$. Now perform a symmetry transformation: x ->x'; f(x) -> f'(x'). Then the new integral is $\int dx' f'(x')$. But x' is just a dummy variable, since you're integrating over it. So relabel it x and you get $\int dx f'(x)$. If this is supposed to be the same action, you are left with f'(x)=f(x), the condition for a scalar. QED.

18. Nov 1, 2007

### George Jones

Staff Emeritus
The Jacobian of the transformation of is also needed, i.e.,

$$\int f(x) dx = \int f'(x') \frac{dx}{dx'} dx',$$

assuming $f$ is a real-valued function of a single real variable.

For spacetime,

$$dx'^0 dx'^1 dx'^2 dx'^3 = \frac{\partial \left( x'^0, x'^1, x'^2, x'^3 \right)}{\partial \left( x^0, x^1, x^2, x^3 \right)} dx^0 dx^1 dx^2 dx^3.$$

If, for example, the $x$'s are time and inertial spatial coordinates, and the $x'$'s are time and spherical coordiantes, then the Jacobian is not equal to unity, and

$$dx'^0 dx'^1 dx'^2 dx'^3 \ne dx^0 dx^1 dx^2 dx^3.$$

In the the case of (restricted) Lorentz transformations between inertial coordinate systems, the Jacobian is equal to unity. In order yo show this, it is sufficient to demonstrate that boosts and rotations individually have unit Jacobians.

In post #16, the differentials and the antisymmetric wedge product automatically take of the Jacobian:

$$dx'^0 \wedge dx'^1 = \gamma^2 \left(dx^0 - v dx^1 \right) \wedge \left(dx^1 - v dx^0 \right) = dx^0 \wedge dx^1.$$

Last edited: Nov 1, 2007
19. Nov 1, 2007

### blechman

absolutely, George. I was being a little sloppy, but my conclusion is (fortunately) still true.

thanx for cleaning it up! ;-)

20. Nov 2, 2007

### pellman

Thanks, George and blechman. This is pretty much sewed up.

One rather broad question remains in my mind: why do we require the action to be an invariant?

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