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Reduced mass can be defined as
##μ= \frac {m_1m_2} {m_1+m_2}## For ##m_1>>m_2##, ##μ_2≈m_2##
How can we prove this ?
##μ= \frac {m_1m_2} {m_1+m_2}## For ##m_1>>m_2##, ##μ_2≈m_2##
How can we prove this ?
But also there's ##m_1m_2## ? Our teacher used the binomial expansion. I don't think your arpproach is true.How can we do it by using it ?mfig said:If ##m_1>>m_2##, then how would you approximate ##m_1 + m_2##?
For example, say ##m_1=10^7## and ##m_2=1##. What, approximately, is ##m_1 + m_2##?
Okay I did it.mfig said:I don't understand why you say, "But also..." in response to a question. What is your answer to the question I asked?
As I said there's ##m_1m_2## part. We can just consider the denominator part.mfig said:I don't understand why you say, "But also..." in response to a question. What is your answer to the question I asked?
And I am saying that you cannot just say $$m_1m_2/m_1=m_2$$ Thats mathematically incorrect. I can understand you don't worry about that, but it seems you didnt understand me. Even if we try to take it as a limit we get $$ylim(x→∞)=x/(x+y)$$ and then we cannot put just infinity since ##(∞/∞)## is uncertain. So you can do either L' Hospital's Rule or we can write as $$ylim(x→∞)x/(x(1+y/x))=ylim(x→∞)1/(1+y/x)=ylim(x→∞)1/(1+0)=y$$ which that's what I wrote.mfig said:"As I said there's m1m2" role="presentation">m1m2 part. We can just consider the denominator part."
I was going to walk you through it (Step1, then Step 2, etc.). I guess you misunderstood what was happening, which is why you didn't answer my simple question. Of course step 2 involves cancelling...
Anyway, glad you got it.
Arman777 said:Reduced mass can be defined as
##μ= \frac {m_1m_2} {m_1+m_2}## For ##m_1>>m_2##, ##μ_2≈m_2##
How can we prove this ?
I don't get this.Arman777 said:And I am saying that you cannot just say $$m_1m_2/m_1=m_2$$ Thats mathematically incorrect.
The second equality is incorrect. It should beArman777 said:Okay I did it.
$$μ=\frac {m_1m_2} {m_1+m_2}=m_2(\frac {1} {1+m_2/m_1})$$ then
$$m_2(1+m_2/m_1)^{-1}=m_2(1-m_2/m_1)$$ and then we can say $$m_2/m_1≈0$$ since, ##m_1>>m_2 → 1>> m_2/m_1## Hence $$μ≈m_2$$
Its not incorret I just put ##=## instead of ##≈##.DrClaude said:he second equality is incorrect. It should be
Clearly any of you don't know math.I already explained the case. If you don't get it learn the math and limits.DrClaude said:I don't see the difference between what you did and using m1+m2≈m1m1+m2≈m1m_1 + m_2 \approx m_1 to write
DrClaude said:Also, I would say that dropping terms of the order O(m2/m1)O(m2/m1)O(m_2/m_1) is not the same as taking the limit m2/m1→0m2/m1→0m_2/m_1 \rightarrow 0.
Arman777 said:Clearly any of you don't know math.
DrClaude said:I don't get this.
The second equality is incorrect. It should be
$$m_2(1+m_2/m_1)^{-1} \approx m_2(1-m_2/m_1)$$
I don't see the difference between what you did and using ##m_1 + m_2 \approx m_1## to write
$$
\frac{m_1 m_2}{m_1 + m_2} \approx \frac{m_1 m_2}{m_1}
$$
One the approximation is made, normal math takes over and one can write
$$
\frac{m_1 m_2}{m_1} = m_2
$$
Vanadium 50 said:Do you think this is going to help? Likewise, do you think YELLING at the people who are trying to help you is a good strategy?
Arman777 said:But also there's m1m2m1m2m_1m_2 ?
Arman777 said:As I said there's m1m2m1m2m_1m_2 part. We can just consider the denominator part.
Arman777 said:And I am saying that you cannot just say $$m_1m_2/m_1=m_2$$ Thats mathematically incorrect. I can understand you don't worry about that, but it seems you didnt understand me. Even if we try to take it as a limit we get $$ylim(x→∞)=x/(x+y)$$ and then we cannot put just infinity since ##(∞/∞)## is uncertain. So you can do either L' Hospital's Rule or we can write as $$ylim(x→∞)x/(x(1+y/x))=ylim(x→∞)1/(1+y/x)=ylim(x→∞)1/(1+0)=y$$ which that's what I wrote.
You started to convo with what is the value of ##m_1+m_2## its not hard to see what you are going to do.
DrClaude wasn't taking a limit, so why would he use L'Hospital's rule?Arman777 said:Is this looks like L'Hosptial ? No
People read what you wrote and don't agree with you. There's nothing mathematically wrong with the method mfig and DrClaude used. Neither was taking a limit, so your claim that they're overlooking an indeterminate form doesn't apply. As DrClaude noted earlier, neglecting higher order terms isn't the same as taking a limit.I was not yelling. I was trying make a point there. I wrote the same things during the conversation and non of you tried to read it. And still you guys are posting stuff just to say that I was wrong and, I am not wrong.. Its so sad to see such things in people that when someone makes a mistake no one have guts to accept it.
I asked my classical mechanic prof and he said there's no such approximation as you guys said.vela said:DrClaude wasn't taking a limit, so why would he use L'Hospital's rule?People read what you wrote and don't agree with you. There's nothing mathematically wrong with the method mfig and DrClaude used. Neither was taking a limit, so your claim that they're overlooking an indeterminate form doesn't apply. As DrClaude noted earlier, neglecting higher order terms isn't the same as taking a limit.
To put it another way, you seem to be claiming that you can't say
$$\frac x{1000001} = \frac x{1000000+1} \cong \frac x{1000000}$$ without knowing what the value of ##x## is, which is ridiculous. Just as dividing by 1000000 will give you close to the same answer as dividing by 1000001 because ##1000000 \gg 1## regardless of the value of ##x##, dividing by ##m_1## will give you approximately the same answer as dividing by ##m_1+m_2## when ##m_1 \gg m_2## whether or not ##m_1## appears in the numerator.
I'm very confused about this thread. Apart from writing equals instead of approximately equals, you did this in #5.Arman777 said:He said we should use binomial approximation. For $$m_2/m_1<<1$$
Reduced mass is a concept in physics that is used to describe the effective mass of a two-body system. It takes into account the masses of both bodies and their distance from each other.
Reduced mass is important because it allows us to simplify the equations used to describe the motion of a two-body system. It also helps us understand the relationship between the masses of the two bodies and how it affects their motion.
The reduced mass is calculated by taking the product of the two masses and dividing it by their sum. In mathematical terms, it can be represented as μ = (m1m2) / (m1 + m2).
When ##m_1## is larger than ##m_2##, the reduced mass will be closer to the value of ##m_2##. This means that the motion of the two-body system will be more heavily influenced by the smaller mass.
No, reduced mass cannot be negative as it is a measurement of mass and mass cannot be negative. It can, however, be zero if one of the masses in the system is zero.