# QG self consistency and cut-offs

1. Apr 26, 2006

### arivero

I'd like to expand a bit about a comment I did in the other subforum. I find it amusing that while Planck scale is a good cutoff for Gauge theories, it is not a good one for gravity?

The point is that for a cutoff $$\Delta x$$ in position space, we expect to use a cutoff $$\Delta p= \hbar /\Delta x$$ in momentum space. If the dynamics comes from a force with coupling constant about unity (ie, hc) at the cutoff scale, then both cutoffs are consistent.

This is because the change in momenta is equal to the force times the time interval. As space and time are treated in the same footing, we have that at a distance $$\Delta x$$ we expect momentum exchange of order $$F(\Delta x) {\Delta x \over c }$$

Say otherwise, if we want to regulate with Planck scale, we will ask for space intervals up to Planck Length $$l_P$$ and forces up to Planck Force $$F_P \equiv {\hbar c \over l_P^2}$$

Now, consider two particles at Planck Distance, this is, in the limit of the spatial cut-off. Its gravitational force is

$$F_G= \hbar c {m_1 m_2 \over M_P^2} {1 \over l_P^2}$$

and it does not saturate the Force cut-off because elementary particles are far small than Planck mass. Thus consistency of the Planck length as regulator implies the existence of other forces beyond gravity: the Gauge forces. For instance if 1 and 2 have electromagnetism, then an extra force

$$F_E= \alpha(\l_p) {1 \over l_P^2}$$

saturates the force cut-off.

Now, string theory automatically provides other forces (but when it provides them via Kaluza Klein it should have the same problem; I think modern strings use other techiques to induce the gauge group). As for LQG, it is sometimes touted as a "gravity only" theory, so I can not see how does it bypass this consistency problem.

Last edited: Apr 26, 2006
2. Apr 26, 2006

### hossi

Not neccessarily, depends on the uncertainty relation and does not go along with GR. Reason: is \delta x of a particle with momentum p smaller than its Schwarzschild horizon, then it makes a black hole and vanishes behind a horizon of size propto p, which will increase with p. Best, B.

3. Apr 27, 2006

### Chronos

I may be missing something [not that would be shocking], but, are we not talking about sub-planckian masses here? If so, I do not understand how the Schwarzschild limit is in play.

Last edited: Apr 27, 2006
4. Apr 27, 2006

### arivero

Yes, gravity provides a different way of buildig cutoffs: For a mass m you can build the quantity Gm/r that has dimensions of speed^2, and then $$r={G m \over c^2}$$. And also obviusly I can build the the quantity $$p= m c$$ so its combination could be used for spatial and momentum cutoff respectively, or
$$\Delta x = {G \over c^3 } \Delta P$$

and when we compare with
$$\Delta x = \hbar {1 \over \Delta P}$$

We see that it "does not go along". I suppose you refer to this. But if we want to have QM and GR, we need them to go along.

Now, this implies that the only possible cut-off scale if we insist on regulators is
$$\hbar {1 \over \Delta P}={G \over c^3 } \Delta P$$

$$\Delta P= \sqrt { c^3 \over G \hbar}$$
and
$$\Delta x = {\hbar \over \Delta p} = \sqrt { G \hbar \over c^3}$$

So the compatibility between Schwartzchild cutoff and Planck cutoff condition is just the condition to have the cutoff at Planck Length. Now the funny thing I was pointing out is that this condition does not seem to work for gravity of elementary paticles, because the bounds of the cutoff are not simultaneusly saturated.

At Planck distance, the force

$$F_G= \hbar c {m_1 m_2 \over M_P^2} {1 \over l_P^2}$$

would let only for momentum exchanges

$$\Delta P= F_G \Delta t = F_G {l_P \over c}= \hbar {m_1 m_2 \over M_P^2} {1 \over l_P}$$

a lot smaller that the maximum $$\Delta P$$.

Or, if I saturate the momentum space cutoff, then from

$${\hbar \over l_p} = F_G {l_P \over c} = \hbar {m_1 m_2 \over M_P^2} {l_P \over (\Delta x)^2}$$

I get

$$\Delta x = \sqrt{m_1 m_2 \over M_P^2} l_P$$

and I am exploring spaces way down Planck length.

Last edited: Apr 27, 2006
5. Apr 27, 2006

### arivero

Yes, at least such was my idea, than $$m_1 m_2$$ are typical elementary particle masses. Perhaps the argument is flawed because I am sketching it in the old no-covariant no-Minkowskian formalism, and it could be (I do not expect it, but I have not calculated it) that when going covariant all these masses are ultrarelativistic ones and then with energies hitting planck mass values. Hmm.

6. Apr 27, 2006

### hossi

Okay, maybe I miss the point as well, but I am still wondering what happend to the uncertainty principle? Or, what do you mean with 'saturating the momentum space cut off' - how can you do this without leaving the sub-planckian regime Best,

B.

7. Apr 27, 2006

### arivero

Hmm do you mean that if I integrate in momentum space up to $$\hbar / l_P$$ then the masses in F=G mm/r^2 at such momentum scale are not sub-planckian anymore but relativistic masses? That could be.

8. Apr 27, 2006

### hossi

Yeah, essentially that would be my worry. \delta x \delta p should always be larger than \hbar -- with the argument I gave above is should even increase again for large p close by the Planck scale (p not \delta p). Best,

S.

9. Apr 28, 2006

### arivero

I see. On my side, the argument was that while two particles interacting electromagnetically at a distance of Planck suffer a force of about the Planck force, the same particles interacting via gravity at the same distance seem to suffer a force a lot smaller than Planck force. You seem to imply that it is not so because in such case the particles should be on the plankian regime and thus gravity force and electromagnetic force should be of about the same magnitude.

I will try to rethink the argument using the covariant formalism, if not full GR.