First let me say this: I have seen multiple calculations of the nidavellir feat and none were accurate, they all ignored that the star was an old star that had gone out and had to be reignited. In this calc i will use the actual temperature of the star that is accurate to the visuals of the movie and not the temperature of a young real life neutron star.

Here i scaled the size difference between the star and the beams diameter. The size i assumed (2000 meters) is close to the one you get via pixel scaling but it doesn't matter as the result of this calc solely on the size radio of the beam and the star, the surface area of Thor that is exposed to the beam and the temperature of the star.

Surface area of the star = 4 x 3,1415 x 1000^2 = 12.556.000 m^2.

Going by color temperature of the star it is around 5000 Kelvin.

P = εAσT^{4}

Going by this formula of black body radiation the energy output of the star is: 1 (emissivity of the star) x 12 556 000 (surface area) x σ=5.67 × 10^{-8} W/m^{2}•K^4 (Stefan-Boltzmann constant) x 5000^4 = 4.45307625E+14 watts.

Surface area of the beam = ~396 m^2 going by 3,1415 x r^2.

Heat intensity of the beam = 4.45307625E+14 / 396 = 1.124.514.204.545 watts/m^2.

Thor had to hold it open for a few minutes. Assuming he couldn't do a small part of the process thus no handle, 2 minutes is a safe bet.

60 x 2 (timeframe in seconds) x 1.124.514.204.545 = 134.941.704.545.448 Joules per m^2. So approximately 135 TJ or ~32 Kilotons of TNT since exposed part perpendicular to the beam including the armor should be ~1 m^2.

Correction edit: star itself appears to be fompletely white so 6600 K is more fitting. So 1,35200000E+15 / 396 x 120 = 409.696.969.696.969 joules or ~410 TJ or 98 Kilotons of TNT.

The Updated final version of the calculation is in the Comment #7

@erkan12: Its not a magical place, its a super high tech structure (dyson sphere) and the structure doesn't matter for this calc, unless you think the star is magic as well.

With pixel scaling the star is around 850 pixels and the beam is 8 pixels wide. That means the surface area ratio is 45 156. So the heat is focused that much more compared to the surface of the star.

The surface temperature should be more like 6400-6500K instead of 6600K.

The dyson sphere had small holes in it and some light was definitely escaping. Im assuming 3% of the heat escapes (doesn't matter much since its a tiny portion)

So with all this considered the updated calc is: 95126814 W/m^2 star radiation, multiplied by 45 156 is 4 295 546 412 984 joules per second or 4 295 Gigawatts (roughly 4.3 TW) accounting for the 3% energy escape that is a total of 500 TJ total heat or equal to the total energy output of 119.5 kilotons of TNT.

Additional things to consider:

Thor has decent regeneration so him taking 500 TJ over 2 minutes time span does not necessarily mean he can take a 500 TJ attack that lasts 2 seconds

Blackbody ratiation formulas might not works correctly in some extreme situations like the neutron star being a magnetar or being millions of degrees hot but neither fits the star in the movie so it should not be a problem to use the formula.

So what does this equal to? Some comparisons to irl and MCU things:

Enough to vaporize 14 000 humans each second.

Enough to burn through the energy of 1431 first generation arc reactors each second.

Enough to power over 3 400 000 000 average US houses.

Since its said to be a neutron star it might have some properties like the real ones where it can generate radiation via other mechanics than just blackbody radiation due to the extreme spinning speed and magnetic fields of a neutron star. I personally think the star in Nidavellir is more like a normal star (a basic blackbody) since it did not display rapid rotation or extreme magnetic properties. However these can be due to the Dyson Sphere somehow negating the effects from affecting the outside of the core machine.

Second possible problem is... it being a magic star that doesn't work anything like other stars making it unquantifiable. Main reason for this is it is too small even for a Neutron Star. For all we know it has a "atmosphere" that has extremely low density and that might be what glows making the output far lower.

@erkan12: Its not a magical place, its a super high tech structure (dyson sphere) and the structure doesn't matter for this calc, unless you think the star is magic as well.

It's obvious that's a magical place since Eitri can use dwarven magic and those dwarves created that Star Forge by using the same magic. You can't call it ''Super high tech'' structure when you don't even know how it works, lmao. One thing we know that they have dwarven magic, and a usual neutron star doesn't have a metal core like this;

If it wasn't magical, there is no way this metal core could exist for centuries while the neutron star burns.

@erkan12: Its not a magical place, its a super high tech structure (dyson sphere) and the structure doesn't matter for this calc, unless you think the star is magic as well.

It's obvious that's a magical place since Eitri can use dwarven magic and those dwarves created that Star Forge by using the same magic (That assumption is baseless, as well as being contradicted by the vfx team saying its a dyson sphere). You can't call it ''Super high tech'' structure when you don't even know how it works, lmao. One thing we know that they have dwarven magic, and a usual neutron star doesn't have a metal core like this (Thats the dyson sphere itself, the things around it are orbital rings);

If it wasn't magical, there is no way this metal core could exist for centuries while the neutron star burns (Look up Dyson Spheres, thats what they basically are, structures covering the entire star to take its energy).

@nwname: I am not talking about the rings, I am talking about the metal core which stands there in the middle of the space even when there is no power. And another fact is that Eitri uses dwarven magic to create weapons.

@erkan12: ... Did you even look up what a dyson sphere is? It IS the sphere core, not the rings around it. Rings are just additional tech. Yes, it normally would collapse on top of the star ut with sufficient tech (like gravity negation which mcu ships have already shown) it can stay there.

Whole theme of Asgard was tech combined with magic not magic alone. The structure = super high tech, The enchantments on the weapons = magic.

First I would strongly disagree with your decision to call that star "white". Ignore the arrows, text (its not 24ft I know), lines, etc. for now, but this is directly from the movie and it clearly has a yellow/orange tint to it.

Second, you calculated the total power output of the star and then you made an obviously incorrect assumption that all of that power was focused through the beam. By dividing the total power output by the beam cross-sectional area you assume the entire star output is going through that beam which is blatantly false if you simply look at the image I have here. Clearly you can see that the star is outputting energy in all directions despite being enclosed. It is very inaccurate to make this false assumption. The more accurate way to calculate this would be to take the equation you used:

P = εAσT4

and divide A out from both sides so that you have P/A = epsilon*sigma*T^4 . With this new equation you would be calculating the watts per square meter of the whole star. Then you would multiple (P/A) by the area of the beam cross section and that would give a much more accurate answer to the power of the beam. I went and did that calculation (and I'm using a beam temperature of 5500 since that has a yellow tint) and I got 51,884,043.75 W/m^2 for the power output of the star. Onto problem 3.

Third, your beam diameter is much bigger than what I calculated it to be. It's not 24 feet like my I used picture suggests but I got a smaller diameter nonetheless. How did you calculate the beam diameter? I used a photo from the movie using Thor's body as a reference. The black bars represent his height, since he is not standing straight (shorter than normal) I estimated him to be 6feet tall instead of his actual height. The beam is the size of the ring that the bars almost reach, so add an additional foot on each side to get ~62 feet in diameter. This is compared to your 2447.8 cm which is equal to 80.3 feet. Nevertheless this step is unnecessary actually since you can just multiply the power per square meter by 1 to get the power hitting Thor.

Fourth, I would argue ardently that the amount of time Thor was in the beam was actually closer to ~40 seconds, not 2 minutes. That two minutes is only based on Etri's statement that it would take a couple minutes, and not the observations we can make from the film. The entire Nidavellir forge scene seemed to be very time consistent between cuts based on character reactions, locations and the melting process. Therefore it is more accurate to measure the time he spent in the beam using the movie run time. So from when we see the beam start to stop on-screen it is about ~40 seconds, Not 2 full minutes.

Now for the revised calculation:

We will use 51,884,043.75 W/m^2 for the power output of the star, as previously calculated in this post.

We then multiply by 1 square meter to account for Thor's backside area, to get 51,884,043.75 W.

Then we multiply by 40 seconds to get 2,075,361,750 Joules, or 2075 MegaJoules

This amount of energy is equal to 0.000496 Kilotons of TNT versus your 119.5 Kilotons of TNT.

This new value is only 0.000415% of your calculation. Most of this discrepancy comes for your assumption (whether purposeful or not) that the beam contained all the star's output. Even if we make it the full two minutes the energy is only 3x times greater, which is 0.001245% of 119.5 at 0.001488 Kilotons of TNT. EVEN if we use the 6600 Kelvin Temperature that you claimed (which I disagree that the star is white as apparent by the images), then the total energy is 0.0029 Kilotons of TNT AND that is over the full 2 minutes you claimed! 0.0029 Kilotons is still only 0.00243% of your claimed 119.5 Kilotons of TNT. This is insanely small, and quite frankly makes the feat look pathetic. That's only 3,373 kWh, which is NOT a lot of energy. A typical coal power plant at 600MW will produce 14,400kWh of energy in one day of continuous operation. Another comparison: the W-78 (a common US bomb) Nuclear Warhead designed for three of them to be fitted on a Minuteman III Missle, has a yield of 335-350 Kilotons of TNT per Warhead meaning that one missile would have a combined payload of just over 1 Megaton of TNT, compared to 0.0029Kt. And remember these real world comparisons are compared to 2 minute exposure at 6600degK, versus what I see as more accurate as 40 seconds at 5500degK. All this is in addition to your problems you stated yourself about the emissivity maybe being less than one, which would decrease the energy, and then that the star is impossibly small.

Conclusion: Thor was hit by far less than even 1Kt of energy and NOT anywhere NEAR 119.5 Kt of energy.

TL;DR: Original calc makes bad assumptions that result in HUGE differences in total energy hitting Thor 0.000496 Kt (new calc) vs 119.5 Kt (old calc).

First I would strongly disagree with your decision to call that star "white". Ignore the arrows, text (its not 24ft I know), lines, etc. for now, but this is directly from the movie and it clearly has a yellow/orange tint to it.

This is not the color of the star tho thats the beam that contains the energy of the star (whatever kinda particles its made out of are heated to around 5500K). You can see the color of the star (its surface exposed from the open iris) in the last pic in the OP.

Second, you calculated the total power output of the star and then you made an obviously incorrect assumption that all of that power was focused through the beam. By dividing the total power output by the beam cross-sectional area you assume the entire star output is going through that beam which is blatantly false if you simply look at the image I have here. Clearly you can see that the star is outputting energy in all directions despite being enclosed. It is very inaccurate to make this false assumption. The more accurate way to calculate this would be to take the equation you used:

Its a dyson sphere. Its entire purpose is to collect nearly 100% of the output and the beam should contain the total output of the star as it is the only part of the machine that gives energy to the forge. What do you think the machine does to the collected heat if not used in the beam? It has no connection to any other structure either. The beam is not the visualisation of the radiation coming out of the exposed part alone, the beam did not immediately come out as Thor slowly opened it as at the beginning there was only some light and some negligable amount of particles coming out. And after the thing is fully opened however, the particle beam started and made a shockwave. How does that picture prove the beam only contains the energy coming from the exposed part directly under the opened wall? And the star outputting energy in all directions? Like 90+% of the surface is completely covered by apaque metal how can it come out from the entire surface of the star?

P = εAσT4

and divide A out from both sides so that you have P/A = epsilon*sigma*T^4 . With this new equation you would be calculating the watts per square meter of the whole star. Then you would multiple (P/A) by the area of the beam cross section and that would give a much more accurate answer to the power of the beam. I went and did that calculation (and I'm using a beam temperature of 5500 since that has a yellow tint) and I got 51,884,043.75 W/m^2 for the power output of the star. Onto problem 3.

Third, your beam diameter is much bigger than what I calculated it to be. It's not 24 feet like my I used picture suggests but I got a smaller diameter nonetheless. How did you calculate the beam diameter? I used a photo from the movie using Thor's body as a reference. The black bars represent his height, since he is not standing straight (shorter than normal) I estimated him to be 6feet tall instead of his actual height. The beam is the size of the ring that the bars almost reach, so add an additional foot on each side to get ~62 feet in diameter. This is compared to your 2447.8 cm which is equal to 80.3 feet. Nevertheless this step is unnecessary actually since you can just multiply the power per square meter by 1 to get the power hitting Thor.

I assumed 2km diameter for the star which isi like i said in the OP, inconsequential as the beam intensity comes from the surface area ratio between the star and the beam and not the surafce area.

Fourth, I would argue ardently that the amount of time Thor was in the beam was actually closer to ~40 seconds, not 2 minutes. That two minutes is only based on Etri's statement that it would take a couple minutes, and not the observations we can make from the film. The entire Nidavellir forge scene seemed to be very time consistent between cuts based on character reactions, locations and the melting process. Therefore it is more accurate to measure the time he spent in the beam using the movie run time. So from when we see the beam start to stop on-screen it is about ~40 seconds, Not 2 full minutes.

I usually tend to avoid on-screen timeframe as they are often not even relatively accurate due to cuts. But i dont remember the scene entirely so it might be time consistent.

Now for the revised calculation:

We will use 51,884,043.75 W/m^2 for the power output of the star, as previously calculated in this post.

We then multiply by 1 square meter to account for Thor's backside area, to get 51,884,043.75 W.

Then we multiply by 40 seconds to get 2,075,361,750 Joules, or 2075 MegaJoules

This amount of energy is equal to 0.000496 Kilotons of TNT versus your 119.5 Kilotons of TNT.

So you just multiplied the surface emission with time? Thats basically what it would take to stay near a 5500K star without any dyson sphere caused effects.

This new value is only 0.000415% of your calculation. Most of this discrepancy comes for your assumption (whether purposeful or not) that the beam contained all the star's output.

It is very much purposeful as the feat involves a machines with one and only purpose of collecting the stars total output as much as possible and only has a single way it uses the collected energy, powering the forges, with the only connection to it being a particle beam(that is clearly not just pure radiation from the exposed area only with visible movement speed and a instantly created shockwave) directed at the forge.

Even if we make it the full two minutes the energy is only 3x times greater, which is 0.001245% of 119.5 at 0.001488 Kilotons of TNT. EVEN if we use the 6600 Kelvin Temperature that you claimed (which I disagree that the star is white as apparent by the images), then the total energy is 0.0029 Kilotons of TNT AND that is over the full 2 minutes you claimed! 0.0029 Kilotons is still only 0.00243% of your claimed 119.5 Kilotons of TNT. This is insanely small, and quite frankly makes the feat look pathetic. That's only 3,373 kWh, which is NOT a lot of energy. A typical coal power plant at 600MW will produce 14,400kWh of energy in one day of continuous operation. Another comparison: the W-78 (a common US bomb) Nuclear Warhead designed for three of them to be fitted on a Minuteman III Missle, has a yield of 335-350 Kilotons of TNT per Warhead meaning that one missile would have a combined payload of just over 1 Megaton of TNT, compared to 0.0029Kt. And remember these real world comparisons are compared to 2 minute exposure at 6600degK, versus what I see as more accurate as 40 seconds at 5500degK. All this is in addition to your problems you stated yourself about the emissivity maybe being less than one, which would decrease the energy, and then that the star is impossibly small.

Conclusion: Thor was hit by far less than even 1Kt of energy and NOT anywhere NEAR 119.5 Kt of energy.

TL;DR: Original calc makes bad assumptions that result in HUGE differences in total energy hitting Thor 0.000496 Kt (new calc) vs 119.5 Kt (old calc).

How is the assumption bad? Why even build a machine that has the purpose of collesting the entire output if the force is only going to use the energy intensity avaible at the surface of the star?

TL,DR: The beam has a color temperature of 5500K not the star's surface, and the beam contains near the total output of the star as the multi km^2 structure (the dyson sphere) would be usesless otherwise. 40 seconds might be better tho so 40 kt might be more accurate than 120 kt.

Also 2 GJ is hilarious. Dwarves should have just used gasoline or other cheap fuels right?

@erkan12: ... Did you even look up what a dyson sphere is? It IS the sphere core, not the rings around it. Rings are just additional tech. Yes, it normally would collapse on top of the star ut with sufficient tech (like gravity negation which mcu ships have already shown) it can stay there.

Whole theme of Asgard was tech combined with magic not magic alone. The structure = super high tech, The enchantments on the weapons = magic.

I did. They surround a natural star in theory. A natural neutron star.

This is not a neutron star;

It's dead. Thor said the star is gone out. There is no power in it. And Eitri said the star needed to be awakened. It only works via magic when they move the rings. It's probably related to the dark energy and works via the same magic with the Bifrost.

They also said that the Dyson Sphere must be as big as a solar system, while Nidavellir was as big as a moon at best. It's completely different.

Eitri also said that Thor was taking the full force of a star, while Thor was taking only a very small part of it on screen, that's not possible if magic is not involved. Based on what Eitri said, it's magical and you can't use scientific calculations on it, it's ridiculous.

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