Reduced mass for ##m_1## larger than ##m_2##

[Arman777]

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 ?

 

[mfig]

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 $m_1 + m_2$?

 

[Arman777]

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 $m_1 + m_2$?

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]

I don't understand why you say, "But also..." in response to a question. What is your answer to the question I asked?

 

[Arman777]

I don't understand why you say, "But also..." in response to a question. What is your answer to the question I asked?

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 $$μ≈m_2$$

 

[Arman777]

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]

"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]

"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.

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.

 

[PeroK]

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 ?

You could let $m_1 = km_2$, where $k$ is the dimensionless ratio of the two masses.

The case $m_1 >> m_2$ corresponds to the case where $k$ is large.

 

[DrClaude]

And I am saying that you cannot just say $$m_1m_2/m_1=m_2$$ Thats mathematically incorrect.

I don't get this.

You wrote

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 $$μ≈m_2$$

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
$$
\frac{m_1 m_2}{m_1} = m_2
$$

Also, I would say that dropping terms of the order $O(m_2/m_1)$ is not the same as taking the limit $m_2/m_1 \rightarrow 0$.

 

[Arman777]

he second equality is incorrect. It should be

Its not incorret I just put $=$ instead of $≈$.

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

Clearly any of you don't know math.I already explained the case. If you don't get it learn the math and limits.

Lets think we have a limit where
$$lim(x→∞)=xy/(x+y)$$ Then YOU GUYS are saying that since x goes to ∞ we can say that $$lim(x→∞)=xy/x$$ and then $$lim(x→∞)=y =1 $$?
You guys even didnt apply the x goes to infinity case. If you do on the first step it goes to $(∞/∞)$?

This is total rubbish. Math doesn't allow us to do something like that. If you apply the rule that x goes to infinity at that point. YOU HAVE TO ALSO CONSİDER THE NUMERATOR PART. Not just the Denominator. And by doing that equation goes to $$(∞/∞)$$ and that's uncertain.

You cannot just say well since $m_1$ is largen the $m_1$ let's make denominator $m_1$.

This is the proper approach
$$ylim(x→∞)x/(x(1+y/x))=ylim(x→∞)1/(1+y/x)=ylim(x→∞)1/(1+0)=y$$ here $x$ is $m_1$ and $y$ is $m_2$

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.

Its the same approximation.

 

[Arman777]

Math has certain rules. We cannot approximate and cancel stuff for our wishes

 

[Vanadium 50]

Clearly any of you don't know math.

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? (One might also wonder if we are so stupid, why are you asking us for help?)

Dr. Claude gave an excellent answer - I started typing in an answer and quickly realized that I was just writing what he had already written. If there is something you don't understand, "I don't understand" is likely to get you farther than :you don't [sic] know math":

 

[mfig]

I do not understand the aggression towards people's suggestions. The method I and Dr. Claude showed is perfectly acceptable and in fact is quite common in physics. To add rigor, simply fix $m_2$ and consider the reduced mass a function of $m_1$ only.

$\mu = \mu(m_1) = \frac{m_1 m_2}{m_1 + m_2}$

Now for $m_1>>m_2$ simply find $\lim_{m_1\to\infty} \mu(m_1) = \lim_{m_1\to\infty} \frac{m_1 m_2}{m_1 + m_2}$

Since we have $\frac{\infty}{\infty}$ we apply L'Hospital and the final result is immediately apparent.

 

[Arman777]

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
$$
\frac{m_1 m_2}{m_1} = m_2
$$

Is this looks like L'Hosptial ? No

I have no arguments against L'Hospital. Thats another approach but you guys did not do L'Hospital.

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?

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.

But also there's m1m2m1m2m_1m_2 ?

As I said there's m1m2m1m2m_1m_2 part. We can just consider the denominator part.

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.

 

[vela]

Is this looks like L'Hosptial ? No

DrClaude wasn't taking a limit, so why would he use L'Hospital's rule?

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.

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.

 

[Arman777]

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 asked my classical mechanic prof and he said there's no such approximation as you guys said.

 

[Arman777]

He said we should use binomial approximation. For $$m_2/m_1<<1$$

 

[Ibix]

He said we should use binomial approximation. For $$m_2/m_1<<1$$

I'm very confused about this thread. Apart from writing equals instead of approximately equals, you did this in #5.

You still end up making the statement that since $m_2/m_1\ll 1$ then $m_2/m_1$ (plus higher powers thereof) is negligible, so this is exactly the same approximation as others have made in this thread. It's just an unnecessarily cumbersome way to go about it. If this is problem 1 in a problem sheet with more complex problems where a series approximation is necessary then I have some sympathy - starting with a problem where the answer is obvious is fair enough. Otherwise, it's just a case of insisting on using a mechanical digger when a shovel would have done the job quicker.