Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

At the bottom of pg. 3, Kiritsis states the following

To appreciate the difficulties with the quantization of Einstein gravity, we look at a single-graviton exchange between two particles (Fig. 1.1a)). Then, the amplitude is proportional to $E^2/M_P^2,$ where $E$ is the energy of the process and $M_P$ is the Planck mass, $M_p\sim 10^{19}\, \mathrm{GeV}$. . . . Therefore, the graviational interaction is irrelevant in the IR ($E\ll M_P$) but strongly relevant in the UV. In particular, this implies that two-graviton exchange diagram (Fig. 1.1b)) is proportional to the dimensionsless ratio $$ \frac{E^2}{M_P^4}\int _0^\Lambda d\tilde{E}\tilde{E}\sim \frac{\Lambda ^2E^2}{M_P^4}. $$

My question is: How does he obtain the $E^2/M_P^2$ and $\Lambda ^2E^2/M_P^4$ with so little effort? What is this physicist black-magic?

I would have to at least:

  1. Write down the interaction terms in the Lagrangian.

  2. Write down the kinetic term for the graviton in the Lagrangian.

  3. Deduce the graviton propagator (my attempt at this physicist black-magic is telling me that it is just proportional to $1/k^2$?).

  4. Deduce the interaction Feynman rule.

  5. Write down the amplitude.

  6. Determine the superficial degree of divergence.

  7. Write down the result.

And even upon trying to do this in my head, I still didn't get the right $\Lambda$ dependence. (In the two-graviton case, there is $1$ loop, $2$ bosonic propagators, and $2$ fermionic propagators, so the superficial degree of divergence should be $4\cdot 1-2\cdot 2-2\cdot 1=-2$, no? The only thing I can think of is that the scalar interaction term would contain derivatives, increasing the superficial degree of divergence, but I don't see how this would happen with fermions.)

share|cite|improve this question
Both statements are just dimensional analysis. For instance that the graviton exchange between two particles goes like $E^2/M_P^2$ can be derived as follows. The amplitude is dimensionless. Since the coupling to gravity goes like $1/M_P^2$ there must be something with dimensions of [$Energy^2$] in the numerator to compensate. The only Lorentz invariant quantity with energy dimensions is the energy in the center of mass frame $E$, thus the amplitude goes like $E^2/M_P^2$ – whistles Feb 14 '14 at 2:03
up vote 1 down vote accepted

This is not a rigourous (white-magic) answer, but maybe this will give ideas.

Suppose $E > M_p >>\Lambda$ The part of amplitude due to one particle has $2$ vertex and $1$ propagators, and you may estimate this to :

$$\frac{p(p-\tilde E)}{M_p}\frac{1}{(p-\tilde E)^2}\frac{(p-\tilde E)(p- E)}{M_p}\tag{1}\sim \frac{E^2}{M_p^2}$$

The part of the amplitude due to the gravitons may be estimated as :

$$\int_0^\Lambda d^4 \tilde E \frac{1}{\tilde E^2(E -\tilde E)^2} \tag{2}$$

The total amplitude would be :

$$ A\sim(\frac{E^2}{M_p^2})^2\int_0^\Lambda d^4 \tilde E \frac{1}{\tilde E^2(E -\tilde E)^2} \tag{3}$$

With $E >> \Lambda$, we have $E - \tilde E \sim E$, so with $d^4E' \sim E'^3 dE'$, we have :

$$ A\sim\frac{E^2}{M_p^4}\int_0^\Lambda d \tilde E ~~\tilde E \tag{4}$$

share|cite|improve this answer
Are you assuming that the particle is a scalar, so that the interaction is of the form $\partial ^\mu \phi \partial ^\nu \phi g_{\mu \nu}$? This is the only way I could see you getting momenta coming into what appears to be your vertex factors. But if my dimensional analysis is correct, $[\partial ^\mu \phi \partial ^\nu \phi g_{\mu \nu}]=L^{-4}$, so that the corresponding coupling constant would be dimension-less, and in particular, could not be proportional to $1/M_P$. What's going on? – Jonathan Gleason Sep 2 '13 at 20:12
Yes, I supposed a scalar field. The interacting term in the Lagrangian is $\frac{1}{M_p}h_{\mu \nu}\partial^\mu \phi \partial ^\nu \phi$. Here $h_{\mu \nu}$ is a bosonic field, so its dimension is $[h] =M=L^{-1}$, so you need to divide by a mass coupling constant, which is $M_p$. $h_{\mu \nu}$ is not a metrics, and so is not dimensionless. At first order, the relation with the metrics is : $g_{\mu\nu} = \eta_{\mu\nu} + \large \frac{h_{\mu\nu}}{M_p}$ – Trimok Sep 3 '13 at 4:51
One last question, and then I think I got it. How do we determine the kinetic term for $h_{\mu \nu}$? (I ask because this will allow me to determine the appropriate dimensions of a spin $2$ field, and hence the appropriate dimensions of the coupling constant.) – Jonathan Gleason Sep 3 '13 at 7:02
The general expression is that of a bosonic field : $\partial h \partial h$. For a graviton theory, you may demand that the action is invariant under the gauge symmetry $h_{\mu\nu} \to h_{\mu\nu} + \partial_\mu \epsilon_\nu+\partial_\nu \epsilon_\mu$. This leads to the lagrangian density : $\frac{1}{2}\partial_\lambda h^{\mu\nu}\partial^\lambda h_{\mu\nu} - \frac{1}{2}\partial_\lambda h^\mu_\mu\partial^\lambda h_\nu^\nu -\partial_\lambda h^{\lambda\nu}\partial^\mu h_{\mu\nu}+ \partial^\nu h^\lambda_\lambda\partial^\mu h_{\mu\nu}$ – Trimok Sep 3 '13 at 8:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.