Originally posted by JBL
Show those chains hauling stars. You accepted that by word of mouth from a liar, something gladiators son in not.
The **** are you arguing you retard? You're just pulling bullshit up because you don't have what you claimed. Even if I did argue this it's utterly irrelevant to my point. Concessions accepted, liar. Just like your lies about that planet being small.
So I skimmed through this, even ignoring what the writer clearly, implied, everyone is falling for Failberto's trap. The L taker is actually winning for once here because he's distracting from the real point of relevancy. The "quantifiable " feat he argued for Hulk wasn't quantifiable, actually failed, and wasn't even a Hulk feat.
But L-Berto desperately needs this distraction.
Originally posted by Delta1938Not everyone 😖hifty:
So I skimmed through this, even ignoring what the writer clearly, implied, everyone is falling for Failberto's trap. The L taker is actually winning for once here because he's distracting from the real point of relevancy. The "quantifiable " feat he argued for Hulk wasn't quantifiable, actually failed, and wasn't even a Hulk feat.But L-Berto desperately needs this distraction.
Originally posted by qwertyuiop1998
Interesting, I thought its Hulk who was asked to provide a quantifiable strength feat. And the only feat you provided, Its also unquantifiableSuperman wins this by lifting a pair of glasses
https://readcomiconline.to/Comic/Action-Comics-2016/Issue-1000?id=133267#14
Originally posted by Argon123
Hulk should win this one. Superman rarely blitz in-character, and he isn't untouchable to people with zero super speed for some strange reason.
Originally posted by Argon123
Hulk should win this one. Superman rarely blitz in-character, and he isn't untouchable to people with zero super speed for some strange reason.
Originally posted by Bodybuildingbob
Here's the thing , Superman has most impressive feats in all of JLA yet he somehow loses to all of them for some strange reason
Originally posted by Argon123
CIS is allowed though. Superman dies
Ok people, here are the finalised rules as regards PIS, CIS, and everything related to it.PIS is, as always, off unless the thread starter says it's ON.
CIS, as was said before, is now a more diverse term, but is not as vague as before.
While CIS still exists in the form of characters like Rhino (who are just too stupid to know better), it also exists in one other form.
This is known as Character Inhibited Power. This applies to characters that have intelligence, like the Silver Surfer, Superman, and so on and so forth.
As Bada said:
"It's a self imposed limitation in certain circumstances which there is concern for civilians and buildings for the most part. It's not stupidity, it's a limitation set until the threat exceeds a certain threshold."
What this means is that people like the Surfer and Superman and so on will not use the full extent of their powers if it will endanger civilians. It doesn't, though, mean they will fight like idiots. The character's personality is an integral part of the match and dictates how they will perform. This is the crux of the rules we've come up with. It doesn't come down to powers, it comes down to the man or woman that weilds them.
In accordance with this, several factors come in to play in debates:
The Opponent, Basic Information, the Arena and the Character's Personality and Experience
Those four are key.
Example:
If Martian Manhunter fights say, Juggernaut.
MM doesn't personally know Juggernaut (Opponent). So he has Basic Information. This is categorised as being what the general public would know about the Juggernaut. It goes by averages. If that average man or woman on the street knows that Juggernaut is super strong, then MM knows. The average man or woman doesn't know, however, that the Juggernaut is weak against psionics. J'onn would approach with caution, not knowing whether Juggernaut was in his weight class, and not knowing the full extent of the man's powers.
However. If Martian Manhunter went up against Amazo, he would know to go for broke right at the start, because he KNOWS Amazo (Personality and Experience). He will use his speed, his strength, his shapeshifting. This is because if he knows what it takes to bring down Amazo, or he believes his standard attacks won't work. If J'onn was fighting Juggernaut, there would come a point when he would realise that normal attacks won't work, and would up his game. Any character that doesn't suffer from Rhino-esque stupidity is capable of this. Even with this, though, the Arena comes in to play. If civilians are in danger, J'onn won't shapeshift in to a fire breathing dragon. Juggernaut on the other hand doesn't care, so wouldn't hesitate to toss cars and trucks full of civilians at the Martian.
Examples:
Thor knows he can't out-brawl Hulk, so uses exotic powers sooner than he would against the likes of Superman, as Superman is an unknown to him.
Superman would go all out against Doomsday or Despero because he knows how powerful they are. Against the Hulk, he's going to take a few punches before realising he'll have to use something rather than slugging it out. He won't bathe the street in heat vision either, because there are civilians nearby.
It ALL comes down to the CHARACTER, not the POWERSET.
Originally posted by StiltmanFTW
Bricks can always match Clark's speed (or even blitz him, lol), unless they are named Rex or Junior 😛
Originally posted by StiltmanFTW
Hulk eats himSuperman gets eaten
Thor ceases to exist the moment he enters the fight
Originally posted by Astner
Then you're defining the event horizon as the point-of-no-return for particles of the same charge as the black hole. Which makes the event horizon smaller than the Schwarzschild radius.But the Schwarzchild radius is still acting as the point-of-no-return for light, which, according to you wouldn't be part of the black hole's "size," which seems rather arbitrary. And it's completely contradicted when you consider the ergosphere which is completely independent of charge and the definition of the event horizon you're using.
If you don't trust me, I hope you at least trust Misner-Thorne-Wheeler? Let me quote them (from page 879 of MTW: Gravitation):
"The horizon is located at r+ = M + (M^2 - Q^2 - a^2)^½. As with the Schwarzschild horizon of a nonrotating black hole, so also here, particles and photons can fall inward through the horizon; but no particle or photon can emerge outward through it. The horizon is "generated" by outgoing null geodesics (outgoing photon world lines)."
(Emphases mine.)
Originally posted by Magnon
Again, no. The (charge-dependent) r+ which I showed you specifies event horizon in the usual sense. The equation for r+ follows in a very straight-forward way from the space-time metric itself (when expressed in Boyer-Lindquist coordinates), i.e. it's an intrinsic property of the Reissner-Nordström geometry without any reference to a test particle or its charge.
Because if the charge of the black hole is negative, an electron (of sufficiently high velocity, traveling tangential to to the black hole) will be able to enter- and escape the Schwarzschild radius (as long as it's outside the event horizon), because the electromagnetic potential counteracts the gravitational potential.
In fact, let's refer back to the metric tensor expressing the energy in a spherical [r,θ,Φ] geometry that you referred to.
Since photons and neutrons have q = 0, we can, we can rewrite this accordingly.
And you'll recognize this expression as the metric tensor for the expressing the energy of a non-charged black hole. And if you don't, you can crack up any textbook to verify it.
So the Event Horizon you refer to is the point-of-no-return for same charged particles.
Originally posted by Magnon
If you don't trust me, I hope you at least trust Misner-Thorne-Wheeler? Let me quote them (from page 879 of MTW: Gravitation):
"The horizon is located at r+ = M + (M^2 - Q^2 - a^2)^½. As with the Schwarzschild horizon of a nonrotating black hole, so also here, particles and photons can fall inward through the horizon; but no particle or photon can emerge outward through it. The horizon is "generated" by outgoing null geodesics (outgoing photon world lines)."(Emphases mine.)
In fact, let me draw you a picture.
Originally posted by Astner
For the third ****ing time, then you're referring to r+ as particles of same charge as the as the black hole, the ones that will actually be repelled by its charge. If you want to use this as the definition, that's perfectly fine. But that's what you're saying, whether you're grasping or not.Because if the charge of the black hole is negative, an electron (of sufficiently high velocity, traveling tangential to to the black hole) will be able to enter- and escape the Schwarzschild radius (as long as it's outside the event horizon), because the electromagnetic potential counteracts the gravitational potential.
In fact, let's refer back to the metric tensor expressing the energy in a spherical [r,θ,Φ] geometry that you referred to.
Since photons and neutrons have q = 0, we can, we can rewrite this accordingly.
And you'll recognize this expression as the metric tensor for the expressing the energy of a non-charged black hole. And if you don't, you can crack up any textbook to verify it.
If you don't have access to a GR book such as MTW, here's a pdf document detailing this:
https://www.diva-portal.org/smash/get/diva2:912393/FULLTEXT01.pdf
See page 31 of this document for the metric tensor of the Reissner-Nordström geometry. The g_tt and g_rr components depend on Q, which is the electric charge of the black hole.
If we then consider motion of a test particle with charge q (note: now we use small q as opposed to the capital Q above) around the black hole, we can do it e.g. as shown in the above document in Chapter 5.3 (pp. 38). As you can see, in the resulting Euler-Lagrange equations of motion, both Q (the charge of the black hole) and q (the charge of the test particle) now appear.
The event horizon, as a purely geometrical feature of the Reissner-Nordström geometry, depends on Q (but not on q). I have already given the expression r+ for the location of the horizon earlier (which, just like the metric tensor itself, depends on Q but not on q).