Where is the inside of the Tardis? Is it a world in itself? Is it part of a different world?

I would say it's a bit of independent space that's "glued" to ordinary space at the entry door. That also explains how time travelling works: That independent space is "unglued" from space, and then "glued" again to space at another time.

The rest of the police box is just a bit of camouflage in ordinary space to hide the glueing (because otherwise it would look quite strange; not to mention that accidentally walking into the glueing position from the side might turn out deadly because your body gets cut in half).

I think I've answered here https://scientific-speculation.codidact.com/posts/292119 . I'll re-post for easy access.

Setup

I realized that it's possible to design a specific box that's bigger on the inside. Wanted to write this down.

Relating Mass to Lorentz Contraction

As velocity approaches the speed of light, and as the force of gravity approaches the Schwazchild radius, the length (as measured by an independent outside observer) of a spaceship decreases. Passengers inside the ship don't notice the warped space time. This is documented by the Lorentz contraction ( $ \beta $ ). Such that new length = $ \beta \times $ original length.

The Lorentz contraction is described as a function of velocity-

$$ \beta = \sqrt{ 1 - \%c^{2}} $$

Where %c is the speed you are going (as a percentage of the speed of light).

The Schwarzchild radius provides for us a distance from a mass at which the Lorentz contraction should be equal to 0 (the Event Horizon, or Schwarzchild Radius). This is connected to mass by this equation.

$$ r = G M \div c^{2} $$

Since r is a radius, we get double the benefit. A box holding a sphere of radius "r" is a cube "2r" long on each side. $ d = 2r $

This equation is a re-ordering of the orbital velocity equation, with v set to c

$$ v_{orbit} = \sqrt{ G M \div r } $$

Designing a 40 meter room (d1) that only takes up 10 meters (d2) of space

Figure out the Lorentz contraction from the new and old dimensions

$$ d_{2} = 0.25 d_{1} \rightarrow d_{2} = (0.25)(2)(r_{1}) $$$$ \therefore \beta = 0.5 $$

Use the Lorentz contraction to figure out the velocity

$$ 1 - \beta^{2} = \%c^{2} \rightarrow \sqrt{ 1 - \beta^2} = \%c $$

Use the velocity and desired new radius to find mass

$$ r = G M \div (\%c \cdot c)^{2} $$$$ 10 = G M \div (0.866 \cdot c)^{2} $$$$ M = 1.01 \times 10^{28} $$

For reference, our 10 meter room contains a gravity generator generating 1% of the mass of the Sun

This isn't actually too bad. Kilonovas (neutron star collisions) are believed to create a lot of the heavier elements (gold for example) and spew them far enough away to get picked up by places like Earth.

It's entirely possible that black holes have similar events that spew out very fine material that is crushed beyond neutron degeneracy pressure (black hole dust, for a simple word). So you might just be able to pick it up somewhere.

Only a few centimeters of this material has enough mass to do the job.

What about Hawking radiation? The calculation for 1 cm thick wall of black hole dust has a temperature of almost 0 Kelvin, and will take longer to evaporate from Hawking radiation than the life of the Universe.

Problems Left to Solve

Being able to generate enough spacebending force to create our bigger on the inside room doesn't mean the field is flat. Special equipment, or thoughtful design of where to put our black hole material would be necessary.

Also, special equipment will be required to get in and out of a gravity well so steep.

And shielding to keep this from dragging all the planets within 1 light year into orbit around the new room.