Proof Pascal's Law: Math Fluid Mech Homework Help

AI Thread Summary
The discussion centers on understanding and proving Pascal's Law in Fluid Mechanics, particularly through a prism-based proof. The original poster expresses confusion about the proof's reliance on the direction of pressure, questioning its validity. They attempt to create their own proof but acknowledge a mathematical flaw, specifically regarding the necessity of multiple equalities to demonstrate that pressure is independent of direction. The conversation notes that the original poster has not returned for over three years, and participants suggest creating new threads for unresolved questions. The topic highlights the complexity of mathematical proofs in fluid mechanics and the challenges faced by learners.
LFS
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Homework Statement


My son was given a proof of Pascal's Law and told to learn it for his course in Fluid Mechanics.
a. It was done with a prism. I cannot type in a link, but google: pascal law prism and pick the first link and scroll down half a page to pressure and you will see this proof.
b. For the life of me, I cannot understand how this is possibly a proof since the direction of the pressure is NOT arbitrary.
c. So I attempted my own proof... However, I am a mathematician - not an engineer. I fixed the arbitrary direction problem. But I still do not understand something critical.

Homework Equations


I have posted my proof on scribd (remove the spaces):
scribd .com /doc /94146673

However, in my opinion my "proof" still has a mathematical flaw.

The Attempt at a Solution

Let M be a point in a static fluid and let p(n,M) be the pressure at a point M in the direction of an arbitrary vector n to M . Presumably - to prove Pascal's Law - I must show that p(n,M) does not depend on n.

Now, using the techniques of the prism proof, I "show" geometrically that p(n,M)=p(x,M)=p(y,M)=p(z,M) where p(x,M) is the pressure at point M in the direction of "positive x-axis", p(y,M) is the pressure at point M in the direction of positive y-axis, ...

Question: Mathematically, it seems to me that I need only ONE of these 3 equalities, e.g. p(n,M)=p(x,M). Using it, I could conclude that given another vector t to M p(t,M)=p(x,M) and conclude that p(t,M)=p(x,M)=p(n,M) and thus p(M) does not depend on my choice of vector.

I am certain that I need all 3 equalities, but WHY? Maybe something with the limits or even with the definition of p(n,M)?

Thanks for any help!
 
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Hello LFS, were you able to figure this out? I would love to know!
 
The original post is more than 3 years old, and the OP has not checked back in that time.

I think it's safe to say that you probably won't hear back from the OP.

PF also requests that you don't "necropost" to these older threads. It's perfectly fine for you to create your own thread if you have a question, and you can reference any other relevant threads.
 
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