system
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human
[Star Trek] Why were weapons of mass destruction not used during the Dominion War?

I remember 1 instance where a Changeling stole a Roundabout with a tri-colbalt device, and almost blew up the Bajoran sun along with the entire Federation, Romulan, and Klingon fleets. Why were tri-cobalt sun crusher torpedoes not commonly used in the Dominion War. I mean you could sneak 1 or 2 ships under cloak into an strategic enemy systems, and then just blow up the star, killing everyone. Tri-cobalt warheads are not too difficult to manufacture, and the Federation Alliance certainly had cloaking technology, so why didn't they do it. Of course you could argue that the Federation has morals, but what about the Klingons and the Romulans. They could have potentially lighted up the entire Dominion.
gpt
By the 24 th century the explosive yields of weapons are described in terms of isotons as compared with today's convention of using kilotons. However,given that we know the quantity of antimatter and matter used, we should be able to make some assumptions about the comparative yields of the weapons of the Federation.

The current convention is to measure yields in multiples of the explosive yield of 2,4,6-trinitrotoluene CH 3 C 6 H 2 (NO 2 ) 3 . We know that TNT has an explosive yield of 700 calories per gram,(Encyclopaedia Britannica) so we can begin by converting into SI units; specifically the joule. To do this we will multiply the given figure by 4.1868.

1 gram of TNT = 700 calories = 700*4.1868 joules. 1 gram of TNT = 2930.76 joules

Now we need to know how this works in tons. Given that 1 ton is equal to 2240 pounds or 1016.046909 kg we should be able to scale up the explosive yield simply.

Note It is convention to use ton compared to short ton,net ton American or tonne (1000 kilogrammes.)

1 kg of TNT = 2930.76j*1000g = 2930760 joules. 1 ton = 2930760j*1016.046909kg = 2,977,789,639.02084 joules. 1 megatonne = 2,977,789,639,020,840 joules.

According to Einsteins’s theory of special relativity the enrgy E,and momentum,p,of a particle are related by the equation :

E 2 =p 2 c 2 +m 2 c 4

However,given that the antimatter is injected into the matter and held by fields around each packet of antimatter until the detonation signal it would seem that the matter is not in movement and therefore p tends towards zero.

E 2 =0c 2 +m 2 c 4

E=mc 2

Where c is the velocity of light and m is the mass of the particle when at rest.

What we will do is calculate for a detonation of 100% efficiency for simplicity. The amount of matter is therefore 3kg (1.5 antimatter mass combined with 1.5 mass of matter,energy must be conserved in any reaction) and the constant c is the speed of light which we will approximate to 299,792,500 ms -1

Using the mks unit system for SI measurements:

E (joules) = 3 (kg) * 299 792 500 2 (ms -1 )

E (joules) = 3 (kg) * 89,875,543,056,250,000

E (joules) = 269,626,629,168,750,000 joules.

E (joules) = 2.6962662916875*10 17 joules.

Now if 1 megaton = 2,977,789,639,020,840 joules; then 1 photon torpedo yields 90.5458953968987946834375526278729 megatons. (90.55 to 2 D.P.)

This has been upgraded recently to increase yield by 5%. (Zimmerman et al. 1998)

This gives 95.73 (to 2 D.P.) megaton yield which rates at 18.5 isotons. This would mean that 1 isoton is approximately 5.14 megatons to 2D.P.

A typical quantum torpedo has increased the typical yield to 21.8 isotons approximately 112.03 (to 2 D.P.) megatons but by using the zero point motion the yield increases to an equivalent of at least fifty isotons. This would give a megaton yield of 257 megatons. In 1987 the Soviet and American stockpile of nuclear arsenals measured 15 gigatons. This is only 58 times the yield of a single quantum torpedo. In fact the detonation of a single torpedo would release more energy than 13 times the explosives ever used in war from the discovery of gunpowder to the 1990’s.

TLDR: a photon torpedo is a weapon of mass drestruction by todays standards.