1
$\begingroup$

I asked a question on the philosophy SE - here. However, I realized this question is better suited in a Physics SE but I would rephrase it and explain my question very precisely. First, I'll mention three remarks and then I will ask the question.

First of all, in Philosophy we say a proposition can either be correct or incorrect. In that sense, CM is not correct. Of course, it's very accurate, very relevant and very significant but it is not correct. For example, in the case of high-speed particles CM fails to give us the correct answer(to the extent we can measure) and we need to use SR. Similarly, for microscopic particles, we use QM (or some might say that even QM is wrong and we use QFT and even SR is wrong and we use GR).

Secondly, some people might argue that none of the scientific theories are correct. All are approximately correct including QFT or GR. They all have different domains of applicability. I agree with this to some extent that yes, they have different domains of applicability. However, there is a notable difference between CM and SR/GR. GR is reduced to SR in the case of flat space. SR is reduced to CM in the case of low-speed particles. But SR still works even in the case of low-speed particles, you can still use SR formula to calculate the speed of your car. CM is just a subset of SR which is itself a subset of GR. Likewise, though I haven't studied QFT, but as far as I know QFT is a superset of QM. In a non-relativistic case, QFT will be reduced to our known QM.

Thirdly, QFT and GR are two of the most accurately tested theories in Physics. We say these are correct to the extent we can measure. QFT and GR, both have their shortcomings such as there is no quantized version of GR. So, one might say QFT is still wrong because it only works in its own domain and fails to explain things where we need GR.

Now my question is, while we CAN use SR or GR instead of using CM where CM works perfectly, can we say the same thing about QFT and GR? Can we even use QFT instead of GR (doesn't matter whether it gives us accurate results) or vice versa? So for example, can we even use GR in a problem of QFT (however difficult it is)? or GR and QFT are completely designed for different purposes, and trying to replace one with the other would be using a toothbrush to perform an air strike (complete nonsense)?

$\endgroup$
14
  • 1
    $\begingroup$ Comment to the post (v2): Consider to reformulate the title. The analogies do not make much sense. $\endgroup$
    – Qmechanic
    Jun 22 at 9:58
  • $\begingroup$ The title is reformulated. Now it's not a yes/no question. $\endgroup$
    – Earmen
    Jun 22 at 10:46
  • 1
    $\begingroup$ @StephenG-HelpUkraine AdS/CFT is one instance in which we can "recover" GR from QFT. $\endgroup$
    – d_b
    Jun 22 at 14:55
  • 2
    $\begingroup$ QFT, quantum field theory, is based on the postulates of quantum mechanics. CM SR and GR are based on space time coordinates and the particular magnitudes of the phase spaces covered, in a sense from top down, SR emerges from GR and CM from SR. Quantum theories depend on probilities , one event has not meaning, one has to accumulate events to see the fit of the particular QFT with data. GR is not yet definitively quantized, $\endgroup$
    – anna v
    Jun 22 at 15:22
  • 1
    $\begingroup$ in the future one might have a theory where Gravity is quantized, it will include in its mathematics GR and the calculations will depend on the phase space studied, but QFT is a modeling tool that uses GR and SR in its mathematics. $\endgroup$
    – anna v
    Jun 22 at 15:22

0

Browse other questions tagged or ask your own question.