The influence of GR gravity from a mass is infinite. Does the same apply in the quantized description? My thinking is, in the QG case,   there must be a smallest value for a graviton,  and therefore maximum distance it can influence? 

The 'graviton' would be a massless virtual particle, that is a manifestation of the quantized gravitational field.
Off the top of my head, I would think that the distance these virtual particles can 'affect', is determined by the Heisenberg uncertainty principle; massive virtual bosons like the W and Z bosons of the weak interaction can only travel a shrt distance, as they move subluminally and must relinquish their energy/momentum after a certain period of time.
The weak interaction, as a result, has a limited range.

Gravitons, like photons, would be massless and travel at c , so the 'range'  of their interaction could be up to infinity.

range of force is the particle momentum and mean lifetime of the mediator. So a graviton would need to be stable as per what Migl stated above. The HUP does also apply in the decay rates of particles

I knew I shouldn't have asked that question... I'll take it as a "No" then.  

well a graviton could be a stable spin 2 massless particle in that case the range of the force would be infinite as per classical gravity. Assuming that caveat which is also the common hypothesis for the graviton then the answer is yes it can have the same range. As we can only conjecture on what the likely-hood of the graviton properties are that's about as accurate an answer as can be given

25 minutes ago, Mordred said:

well a graviton could be a stable spin 2 massless particle in that case the range of the force would be infinite as per classical gravity. Assuming that caveat which is also the common hypothesis for the graviton then the answer is yes it can have the same range. As we can only conjecture on what the likely-hood of the graviton properties are that's about as accurate an answer as can be given

Thanks. I know QM is work in progress. My thinking was that if the emitted gravitons, being discrete, follow the inverse square law as in EM waves, then there might be a point where they get out of range of influence with each other and the gravitational field strength falls off sharply. In a sense, gravitational influence being proportional  to the collective graviton density in a given space. Gravity emerges from that interaction between them until they are mutually out of range. 

Edited January 8, 20232 yr by StringJunky
Added stuff

you would still have the same classical effects as per Newtons gravitational laws as well as per the EM field with the photon. Range of force isn't quite the same as strength of force in that regard. Its better thought of as "At what range from the source can the force  potentially be given sufficient energy"  with the weak field its range is limited regardless of the energy density of the source.

Edited January 8, 20232 yr by Mordred

12 minutes ago, Mordred said:

you would still have the same classical effects as per Newtons gravitational laws as well as per the EM field with the photon. Range of force isn't quite the same as strength of force in that regard. Its better thought of as "At what range from the source can the force exerted by that force potentially be given sufficient energy 

Yeah, but GR waves diminish infinitesimally and graviton density diminishes in the same way with a cutoff. That's what I was thinking. Do any of the subatomic forces have a discrete range of influence?

Edited January 8, 20232 yr by StringJunky

That would still be accurate as a quantum theory would have to be divergence free. In essence normalized. QFT for example has the infrared and ultraviolet cutoffs for the \[SU(3)\otimes SU(2)\otimes U(1)\] gauges of the SM model.

Edited January 8, 20232 yr by Mordred

Does normalize mean translate into other terms that we have a handle on? What is divergence?

Edited January 8, 20232 yr by StringJunky

23 minutes ago, StringJunky said:

Does normalize mean translate into other terms that we have a handle on? What is divergence?

In math language normalized units  such as c=h=k=g=1. If you have a field such as the EM field you have the normalization via the quanta given Planck units. That's included in the previous expression. Divergence means diverges to infinity where you typically have two points on a graph that this applies.

 \[0\leftarrow x \rightarrow\infty\]

QFT has a regulator to provide an effective cutoff before those two points apply. The infrared regulator and the ultraviolet regulator. these will vary depending on the theory. The more common is dimensional regularization.

Edited January 8, 20232 yr by Mordred

2 minutes ago, Mordred said:

In math language normalized units  such as c=h=k=g=1. If you have a field such as the EM field you have the normalization via the quanta given Planck units. That's included in the previous expression. Divergence means diverges to infinity where you typically have two points on a graph that this applies.

 

0←x→∞

 

QFT has a regulator to provide an effective cutoff before those two points apply. The infrared regulator and the ultraviolet regulator. these will vary depending on the theory. The more common is dimensional regularization.

That's outside my pay grade but thanks for trying.  

No its actually very easy think of a graph.  Now set the max value at 1 (max value for a normalized unit. Now take the bell curve as it trends to 0 and 1 on that graph. Lets use the infrared end, divide 1 by a 1/2 with each result being further divided by 1/2 until you reach 0. You will never reach 0 that is an example of diverging to infinity. You have an infinite number of 1/2 divisions. So you apply some cutoff. (reqularization

Edited January 8, 20232 yr by Mordred

8 minutes ago, Mordred said:

No its actually very easy think of a graph.  Now set the max value at 1 (max value for a normalized unit. Now take the bell curve as it trends to 0 and 1 on that graph. Lets use the infrared end, divide 1 by a 1/2 with each result being further divided by 1/2 until you reach 0. You will never reach 0 that is an example of diverging to infinity. You have an infinite number of 1/2 divisions. So you apply some cutoff. (reqularization

Right. Got you. Thank you. That's the way theorists are trying to quantize it by applying cutoffs on the curve?

In essence yes though the methods vary. I can't recall the theorem name but any infinite set has a finite portion. All gauge groups are finite

Edited January 8, 20232 yr by Mordred

 In this large-distance, static limit, the quantum behavior of the sources does not contribute to the quantum corrections of the potential. The most sensitive gravity sensors use atom interferometry on atoms cooled close to absolute zero and placed in a free fall. In a free fall.