# A Case Where Modus Ponens Can't be True.

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If you have for a proposition P: P AND P -> ~P you can't conclude ~P. The premises is a contradiction.

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16 minutes ago, Willem F Esterhuyse said:

If you have for a proposition P: P AND P -> ~P you can't conclude ~P. The premises is a contradiction.

How do you make this out ?

(I agree that the conclusion of the proposition is invalid)

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If each logical step of a derivation is written on a separate line, and preferably numbered, then it is easier to understand and discuss.

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46 minutes ago, studiot said:

How do you make this out ?

Because P -> ~P = ~P (by truth table) and this contradicts P.

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I'm pretty sure (P XOR ~P) is a fundamental axiom of binary logic, and your proposition contradicts it.

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16 minutes ago, Genady said:

If each logical step of a derivation is written on a separate line, and preferably numbered, then it is easier to understand and discuss.

Very much so.  +1

2 minutes ago, Willem F Esterhuyse said:

Because P -> ~P = ~P (by truth table) and this contradicts P.

It's a bit stuffy, but adopting Genady's suggestion.

Premises

1) P is a valid statement.

2) We are forming the conjunction of P and itself.  (This is a valid statement)

Conclusion

The statement

1 hour ago, Willem F Esterhuyse said:

P AND P -> ~P

is invalid

For the reason you stated

7 minutes ago, Willem F Esterhuyse said:

Because P -> ~P = ~P (by truth table) and this contradicts P.

Not because the premises are contradictory, as you originally stated.

10 minutes ago, Lorentz Jr said:

I'm pretty sure (P XOR ~P) is a fundamental axiom of binary logic, and your proposition contradicts it.

You are correct, but haven't followed through explicitly using your method for the benefit of the OP.

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18 hours ago, Willem F Esterhuyse said:

If you have for a proposition P: P AND P -> ~P you can't conclude ~P. The premises is a contradiction.

I suspect this is in reference to the BH thread you had started, and the supposed contradiction due to different observers disagreeing on in-fall times. I remind you once again that the two reference frames in question do not deal with the same premise, because they do not share the same notion of ‘time’. To cast it into the terms used here, the P belonging to the frame of the in-falling particle is not the P belonging to the frame of the distant stationary observer - you are comparing proper and coordinates times, but those are very different things. To make a valid comparison, you need to choose a physical premise that does not depend on your choice of observer - in this case the obvious choice is the length of the in-falling particle’s world line, ie proper in-fall time. For the in-falling observer, this world line is of finite length; for the distant stationary observer, this same world line is also of finite length.

They both agree that the particle reaches the horizon, if you use the correct premise. The only difference is that the distant observer never sees this happening (which should be obvious, pardon the pun), whereas the infalling observer does.

This is why it is important to grasp the very crucial difference between ‘global’ and ‘local’.

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For a god-like global observer it is contradictory.

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16 minutes ago, Willem F Esterhuyse said:

For a god-like global observer it is contradictory.

This begs the question that,

1) This meta-observer must exist

2) Its observations satisfy a Boolean algebra

Why should we believe that? Quantum mechanics is not Boolean. Why should "God" see the world in a Boolean way?

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2 hours ago, Willem F Esterhuyse said:

For a god-like global observer it is contradictory.

No, it is not, because one statement is not a negation of the other.

Plus, being (or not) a contradiction is a logical rather than physical attribute and as such it does not depend on observer.

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4 hours ago, Willem F Esterhuyse said:

For a god-like global observer it is contradictory.

No it is not. As has already been explained multiple times now, the two observers do not use the same notion of time when they are discussing in-fall time in their respective frames - hence you are not comparing the same quantities, and thus there can’t be a contradiction.

Let’s be explicit here, so we can put this baby to bed once and for all. Suppose we have a particle in Schwarzschild spacetime starting out at rest at some initial radial position $$r_0$$, which we can take to be far from the black hole, so that it initially corresponds to a far-away stationary Schwarzschild observer. It is then released and begins a radial free fall towards the horizon at $$r_S$$.

For the Schwarzschild observer (stationary and far away), the time it takes for the particle to reach the horizon, as measured on his own clock is (in natural units):

$\Delta t=\int _{r_{0}}^{r_{S}}\left(\frac{dt}{dr}\right) dr=-\int _{r_{0}}^{r_{S}}\frac{1}{\sqrt{2\left( 1+\frac{M}{r}\right)\left( 1-\frac{r_{s}}{r}\right)}} dr\ \rightarrow \infty$

This is coordinate in-fall time, since we’re integrating the ratio dt/dr, the expression for which follows from the equation of radial motion. On the other hand, the in-fall time as measured on the falling particle’s own clock is

$\Delta \tau =\int _{r_{0}}^{r_{S}}\left(\frac{d\tau }{dr}\right) dr=\frac{1}{3}\sqrt{\frac{2}{M}}\left(\sqrt[3]{r_{0}} -\sqrt[3]{r_{S}}\right)$

which is the proper in-fall time, and it is very much finite, as you can hopefully see. This expression comes straight from the metric.

In order for there to be a contradiction, it would have to be the case that the length of the world line of the in-falling particle takes on different values for different observers; however, this is not the case, since $$\Delta \tau$$ is a rank-0 tensor, so it is generally covariant, and thus all observers agree on it. To sum up:

$$\Delta \tau$$ is the length of the in-falling particle’s world line in spacetime, and the same for all observers

$$\Delta t$$ is not in general the length of the in-falling particle’s world line in spacetime, and explicitly depends on which observer is chosen

These just simply aren’t the same physical quantities. Thus $$\Delta \tau \neq \Delta t$$ isn’t a contradiction, but a necessary consequence of being in a curved spacetime, since time here is a local quantity. You can’t compare apples and oranges, and then claim there’s a contradiction because they don’t look the same. You have to compare apples to apples instead:

In the frame of the falling particle, the length of its in-fall world line in spacetime is $$\Delta \tau$$.

In the frame of the distant stationary observer, the length of the particle’s in-fall world line in spacetime is also $$\Delta \tau$$.

It’s precisely that world line and its geometric length that your God-observer would see. No contradictions to be found anywhere - quite the contrary actually, since both observers are in perfect agreement about the length of that world line!

PS. Why is it then that the far-away stationary observer never visually sees the particle reach the horizon? That’s because in-fall geodesics aren’t the same as outgoing geodesics. A photon originating close to the horizon can only escape to infinity via a trajectory that’s “flattened” tangentially to the horizon; the relative angles as seen by Schwarzschild and shell observers are related via

$\tan \theta _{shell} =\sqrt{1-\frac{2M}{r}}\tan \theta _{Schw}$

The closer to the horizon one gets, the more tangential the escape trajectories become, and at the photon sphere the angle is so small that nearly all photons become trapped into unstable orbits. Below this point, from the horizon down, the photon must fall in. This is why the Schwarzschild observer never visually sees anything reaching the horizon - because the closer one gets to the horizon, the fewer photons manage to escape back out to reach that observer, resulting a “dimming” of the object as seen by the outside observer (they are also red-shifted, resulting in further dimming). Once the photon sphere is reached, nearly all emitted photons are either captured into orbits, or must fall in, so the object becomes essentially invisible to outside observers. Only under very special circumstances can individual photons escape from here. Once the horizon itself is reached, no escape is possible at all, not even for photons, because null geodesics are everywhere perpendicular to the radial direction (the horizon is a null surface). Due to the symmetries of Schwarzschild spacetime, the coordinate time of a distant observer is defined such that it reflects precisely this difference between ingoing and outgoing null geodesics - it diverges to infinity precisely because no photon can ever reach this observer from the horizon, so nothing is ever visually seen to be reaching that point. This is in contrast to ingoing geodesics, which can be perfectly radial everywhere. Thus there is no contradiction, because ingoing and outgoing null-geodesics simply aren’t the same, not even in this highly symmetric spacetime.

Edited by Markus Hanke
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17 hours ago, Willem F Esterhuyse said:

For a god-like global observer it is contradictory.

!

Moderator Note

No such thing.

If you want to discuss logic, stick to logic. If you want to discuss something else, don’t try and sneak it in under the guise of something else. We expect discussions to be in good faith, not secretly advancing an agenda

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