Action / reaction

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  • #1
DesertFox

Main Question or Discussion Point

Hello, buddies!

I read a piece of text, where I found a puzzling interpretation of Newton's third law (F1=-F2).

The author gives the following representation: F=Rv(m1m2/l^2)

R - reaction;
m1, m2 - masses;
v - velocity of reaction;
l - distance between the two objects.

Please, help me with explanations about the equation, suggestons about its derivation? I will be grateful for every comment.

The formula reminds me of Newton's law of universal gravitation: F=k(m1m2/l^2)

F - force between the masses;
k - gravitational constant;
m1, m2 - masses;
l - distance between the centers of the masses.

But what is the reason for replacing k (gravitational constant) with Rv .... any ideas?

It is puzzling matter, this author keeps me thrown in confusion...
 

Answers and Replies

  • #2
Orodruin
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I read a piece of text
Please give a proper reference to this text. "I read somewhere that" is not an acceptable reference on PF.
 
  • #3
DesertFox
The original text is written in Bulgarian. So, i'm translating it in English.

Proper reference to the text is not the issue; I put the things clearly enough and I asked my question clearly enough.
If you don't agree- I can try to make my question even more clear.

Thank you for being concerned with my thread!
 
  • #4
vanhees71
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Not knowing the text, only from your statement, I'd advice you to forget this text for good and read a real textbook.
 
  • #5
DesertFox
Please, be more specific about the equation (the formula). What is wrong with it?

I'm not advocating the author or the text itself. But I'm trying to grasp his idea. Maybe it is a kind of speculation, but he is using the wrong mathematical expressions?

It's eminent bulgarian intellectual, a philosopher PhD, but he is famous to have solid grounding in physics too. I am not informed whether he has graduadet Physics or not, but that's not so important. If there is flagrant mistakes in his formula, please- state it specifically and detailed. I am not trying to argue, beleive me, just trying to grasp someone's idea.....

Thank you everybody for the replies!
 
  • #6
jbriggs444
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Please, be more specific about the equation (the formula). What is wrong with it?
Without a usable reference, we have no idea what the equation is trying to say. As it stands it is nonsense, not even wrong. [That said, with better context, sense might possibly be made of it].

Among other problems...

1. There is no obvious relationship to Newton's third law.
2. We have no idea what is meant by the "velocity" of a reaction.
3. For that matter, we have no idea what is meant by a "reaction". A chemical reaction? The reaction in Newton's third law?
4. Not all forces are inverse square.
5. There is no reason to expect that force is proportional to mass.
6. No constant of proportionality is given.
 
  • #7
Ibix
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The problem is that without a reference it's like you asking us if "where" is mis-spelled. It might be spelled correctly or it might be a typo for "were". But without being able to see the context of the word - or the formula in this case - it's impossible to determine.
 
  • #8
DesertFox
Actually, I translated and give you the whole passage, the whole context.... the other parts of the text treat completely different matters.
So, in my first post I have translated the whole setting of the author....

I am puzzled by such terms as "reaction" and "velocity of reaction", too... I see that they are not commonly accepted terms and I don't know what he means exactly by them. The big problem is that: the Bulgarian word can be translated in English both as "reaction" and "resistance"... we have one and the same word for them. I think he means REACTION, simply because RESISTANCE doesn't make sense in the context. As far as I know: resistance (R) is part of the electromagnetism and has nothing to do with force, action and reaction.....
 
  • #9
Orodruin
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Proper reference to the text is not the issue
Of course it is. Without knowing the context and what the text says, it can be anything from a misunderstanding on your part to explicit errors int the text. We have no way of knowing. If that is all you have it makes no sense and I suggest you get a proper textbook.
 
  • #10
DesertFox
Of course it is. Without knowing the context and what the text says, it can be anything from a misunderstanding on your part to explicit errors int the text. We have no way of knowing. If that is all you have it makes no sense and I suggest you get a proper textbook.
That's right- this is all we have, the rest of the text (before and after this passage) is devoted to absolutely different matter. I am concerned with that small paragraph (passage), which I tried to translate the best way I can. The rest- it is in the author's head :biggrin:.

I am upset by the the fact I can't go into the heart of the authors idea... even with your help. That's a mysterious idea.... Or probably that's non-sense for real and there's a big error in his physics' setting....
 
  • #11
vanhees71
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If this is really what's in this text, it's just wrong. The equation simply doesn't make any sense.

The Third law, applied to a closed two-particle system, says that the force on particle 1 is opposite to the force on particle 2. This implies, due to the first law
$$\dot{\vec{p}}_1=\vec{F}_1, \quad \dot{\vec{p}}_2 = \vec{F}_2=-\vec{F}_1.$$
Adding both equations leads to
$$\dot{\vec{p}}_1+\dot{\vec{p}}_2=\dot{\vec{P}}=0 \; \Rightarrow \vec{P}=\text{const}.$$
Thus the 3rd Law is equivalent to the conservation of the total momentum of a closed system (it's easily generalized to the case of more than 2 particles).
 
  • #12
DesertFox
If this is really what's in this text, it's just wrong. The equation simply doesn't make any sense.

The Third law, applied to a closed two-particle system, says that the force on particle 1 is opposite to the force on particle 2. This implies, due to the first law
$$\dot{\vec{p}}_1=\vec{F}_1, \quad \dot{\vec{p}}_2 = \vec{F}_2=-\vec{F}_1.$$
Adding both equations leads to
$$\dot{\vec{p}}_1+\dot{\vec{p}}_2=\dot{\vec{P}}=0 \; \Rightarrow \vec{P}=\text{const}.$$
Thus the 3rd Law is equivalent to the conservation of the total momentum of a closed system (it's easily generalized to the case of more than 2 particles).
I'm focused on the following part of his equation: Rv
It represents: RESISTANCE x VELOCITY OF RESISTANCE
It's really hell of mysterious concept.... :(
 
  • #13
vanhees71
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As I said, I have no clue what this formula is about. Without context it doesn't make sense, and it has nothing to do with standard Newtonian mechanics.
 
  • #14
Orodruin
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I'm focused on the following part of his equation: Rv
It represents: RESISTANCE x VELOCITY OF RESISTANCE
It's really hell of mysterious concept.... :(
If what you have quoted is all there is. Stop reading that text and get a proper textbook.
 
  • #15
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the Bulgarian word can be translated in English both as "reaction" and "resistance"...
Mostly out of curiosity, I'm going to ask how the English "inertia" and "momentum" translate into Bulgarian?
 
  • #16
DesertFox
Mostly out of curiosity, I'm going to ask how the English "inertia" and "momentum" translate into Bulgarian?
We have one word for them. In terms of Physics: we use an adjective form of the word, so we DO make difference between "inertia" and "momentum". :biggrin:
 
  • #17
Vanadium 50
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You've started three threads on this book, and every one of them is infused with innumerate nonsense. Here's what you should do with that book:

 
  • #18
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The original text is written in Bulgarian. So, i'm translating it in English.
Can you still provide the original text (in Bulgarian). Maybe we would be able to make sense of it anyway.
There's also a little chance that the other parts of the text do provide the necessary context (even if it might not be obvious).
 
  • #19
Drakkith
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Thread locked for moderation.
 
  • #20
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We will not discuss this book further at PF
 

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