# An equation invariant under change of variable

• I
It's said that the below equation is invariant under a substitution of ##-\theta## for ##\theta## ,

##\frac{d^{2} u}{d \theta^{2}}+u=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right)##

I can't understand this how this is so. It's supposed to be obvious but I can't see it.

PeroK
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It's said that the below equation is invariant under a substitution of ##-\theta## for ##\theta## ,

##\frac{d^{2} u}{d \theta^{2}}+u=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right)##

I can't understand this how this is so. It's supposed to be obvious but I can't see it.

You have to ask yourself what does "the equation is invariant under a substitution of ##-\theta## for ##\theta##" actually mean? Unless you know what that means, how can you see it?

Kashmir
You have to ask yourself what does "the equation is invariant under a substitution of ##-\theta## for ##\theta##" actually mean? Unless you know what that means, how can you see it?
I don't know. I tried searching on the internet but didn't find it. Can you tell me what that means?

PeroK
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I don't know. I tried searching on the internet but didn't find it. Can you tell me what that means?
Let's look at the general case. We have a function ##u(\theta)## which satisfies a differential equation. We then look at a substitution, which in general is of the form ##\theta = f(\alpha)## or ##\alpha = g(\theta)##. Typical examples might be ##\theta = a\alpha + b## or ##\alpha = \cos(\theta)##. In this case we have the substitution ##\alpha = -\theta##.

Note that if you don't "see" what's happening, then it can be confusing to try to work with variables ##\theta## and ##-\theta##, so it's conceptually clearer to let ##\alpha = -\theta##.

A second point is that physicists especially tend to use the same function for both the original ##u(\theta)## and the new ##u(\alpha)##. This is a notational shortcut, but in fact we really want to look at a new function, which we'll call ##v##, where: $$v(\alpha) = u(\theta) = u(f(\alpha))$$ This shows that ##v## is not the same mathematically defined function as ##u##, but the composition of ##u \circ f##. That said, it's seen by physicists as the same "physical" function, so usually they don't bother with this subtlety.

Then, we have to look for the equation that is satisfied by the function ##v(\alpha)## - and it may or may not take the same form as the original equation. Often these substitutions are used to try to change the equation to one that is easier to solve. In this case, however, it is to show that the system has some sort of symmetry in the variable ##\theta##.

The full, rigorous way to tackle this, therefore, is to use the substitution ##\alpha = -\theta## and see what becomes of the original equation. We can start with the first term and look at the second derivative:
$$\frac{dv}{d\alpha} = \frac{dv(\alpha)}{d\alpha} = \frac{dv(\alpha)}{d\theta}\frac{d\theta}{d\alpha} = \frac{du(\theta)}{d\theta}(-1) = -\frac{du}{d\theta}$$
And, if we take the second derivative, we get another factor of ##-1##, which cancels out and we find that:
$$\frac{d^2v}{d\alpha^2} = \frac{d^2u}{d\theta^2}$$
I'll leave you to work on the terms on the RHS.

As I said, this is the full rigorous way to see what's happening. It's certainly quicker just to say that it's obvious!

Delta2 and Kashmir
Let's look at the general case. We have a function ##u(\theta)## which satisfies a differential equation. We then look at a substitution, which in general is of the form ##\theta = f(\alpha)## or ##\alpha = g(\theta)##. Typical examples might be ##\theta = a\alpha + b## or ##\alpha = \cos(\theta)##. In this case we have the substitution ##\alpha = -\theta##.

Note that if you don't "see" what's happening, then it can be confusing to try to work with variables ##\theta## and ##-\theta##, so it's conceptually clearer to let ##\alpha = -\theta##.

A second point is that physicists especially tend to use the same function for both the original ##u(\theta)## and the new ##u(\alpha)##. This is a notational shortcut, but in fact we really want to look at a new function, which we'll call ##v##, where: $$v(\alpha) = u(\theta) = u(f(\alpha))$$ This shows that ##v## is not the same mathematically defined function as ##u##, but the composition of ##u \circ f##. That said, it's seen by physicists as the same "physical" function, so usually they don't bother with this subtlety.

Then, we have to look for the equation that is satisfied by the function ##v(\alpha)## - and it may or may not take the same form as the original equation. Often these substitutions are used to try to change the equation to one that is easier to solve. In this case, however, it is to show that the system has some sort of symmetry in the variable ##\theta##.

The full, rigorous way to tackle this, therefore, is to use the substitution ##\alpha = -\theta## and see what becomes of the original equation. We can start with the first term and look at the second derivative:
$$\frac{dv}{d\alpha} = \frac{dv(\alpha)}{d\alpha} = \frac{dv(\alpha)}{d\theta}\frac{d\theta}{d\alpha} = \frac{du(\theta)}{d\theta}(-1) = -\frac{du}{d\theta}$$
And, if we take the second derivative, we get another factor of ##-1##, which cancels out and we find that:
$$\frac{d^2v}{d\alpha^2} = \frac{d^2u}{d\theta^2}$$
I'll leave you to work on the terms on the RHS.

As I said, this is the full rigorous way to see what's happening. It's certainly quicker just to say that it's obvious!
Thank you so much. I'll work out all the terms and see what I end up with.
Thank you again. :)

Let's look at the general case. We have a function ##u(\theta)## which satisfies a differential equation. We then look at a substitution, which in general is of the form ##\theta = f(\alpha)## or ##\alpha = g(\theta)##. Typical examples might be ##\theta = a\alpha + b## or ##\alpha = \cos(\theta)##. In this case we have the substitution ##\alpha = -\theta##.

Note that if you don't "see" what's happening, then it can be confusing to try to work with variables ##\theta## and ##-\theta##, so it's conceptually clearer to let ##\alpha = -\theta##.

A second point is that physicists especially tend to use the same function for both the original ##u(\theta)## and the new ##u(\alpha)##. This is a notational shortcut, but in fact we really want to look at a new function, which we'll call ##v##, where: $$v(\alpha) = u(\theta) = u(f(\alpha))$$ This shows that ##v## is not the same mathematically defined function as ##u##, but the composition of ##u \circ f##. That said, it's seen by physicists as the same "physical" function, so usually they don't bother with this subtlety.

Then, we have to look for the equation that is satisfied by the function ##v(\alpha)## - and it may or may not take the same form as the original equation. Often these substitutions are used to try to change the equation to one that is easier to solve. In this case, however, it is to show that the system has some sort of symmetry in the variable ##\theta##.

The full, rigorous way to tackle this, therefore, is to use the substitution ##\alpha = -\theta## and see what becomes of the original equation. We can start with the first term and look at the second derivative:
$$\frac{dv}{d\alpha} = \frac{dv(\alpha)}{d\alpha} = \frac{dv(\alpha)}{d\theta}\frac{d\theta}{d\alpha} = \frac{du(\theta)}{d\theta}(-1) = -\frac{du}{d\theta}$$
And, if we take the second derivative, we get another factor of ##-1##, which cancels out and we find that:
$$\frac{d^2v}{d\alpha^2} = \frac{d^2u}{d\theta^2}$$
I'll leave you to work on the terms on the RHS.

As I said, this is the full rigorous way to see what's happening. It's certainly quicker just to say that it's obvious!
The rhs converts to :
##\frac{m}{l^{2}} \frac{d}{d v(-\theta)} V\left(\frac{1}{v(-\theta)}\right)##

And the total equation can be written as ##\frac{d^{2}v(-\theta) }{d(-\theta)^{2}}+v(-\theta)=-\frac{m}{l^{2}} \frac{d}{d v(-\theta)} V\left(\frac{1}{v(-\theta)}\right)## And noting that ##u(\theta)=v(-\theta)## we Finally have
##\frac{d^{2}}{d(-\theta)^{2}} u(\theta)+u(\theta)=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right)##

Is this correct then?

PeroK
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we Finally have
##\frac{d^{2}}{d(-\theta)^{2}} u(\theta)+u(\theta)=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right)##

Is this correct then?
It's true that if you replace ##\theta## by ##-\theta## in the first term that the equation still holds. But, you were supposed to check that the whole equation was still valid with ##\theta## replaced by ##-\theta##.

Whether you use ##u(-\theta)## or ##v(-\theta)##, you need to end up with ##-\theta## everywhere.

That's why I said to be careful not to confuse ##\theta## with ##-\theta##. If the substitution was ##\alpha = \cos \theta##, you clearly need to end up with ##\alpha## or ##\cos \theta## in the equation. You couldn't just leave it as ##\theta## in that case. Do you see that?

It's true that if you replace ##\theta## by ##-\theta## in the first term that the equation still holds. But, you were supposed to check that the whole equation was still valid with ##\theta## replaced by ##-\theta##.

Whether you use ##u(-\theta)## or ##v(-\theta)##, you need to end up with ##-\theta## everywhere.

That's why I said to be careful not to confuse ##\theta## with ##-\theta##. If the substitution was ##\alpha = \cos \theta##, you clearly need to end up with ##\alpha## or ##\cos \theta## in the equation. You couldn't just leave it as ##\theta## in that case. Do you see that?
I'm sorry for taking up your time but I did not simply replace ##\theta## by ##-\theta##. I used the equations you proved and the one I proved as the RHS then I proceeded on. Should I add all of my steps if that makes it clear?

PeroK
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I'm sorry for taking up your time but I did not simply replace ##\theta## by ##-\theta##. I used the equations you proved and the one I proved as the RHS then I proceeded on. Should I add all of my steps if that makes it clear?
I know, but in the end you quoted an equation with mostly ##\theta## as the variable and not ##-\theta##. If you stick with ##\alpha## and put a note after the equation that says "where ##\alpha = -\theta##" you can't go wrong. Note that ##-\theta \neq \theta##.

Kashmir
I know, but in the end you quoted an equation with mostly ##\theta## as the variable and not ##-\theta##. If you stick with ##\alpha## and put a note after the equation that says "where ##\alpha = -\theta##" you can't go wrong. Note that ##-\theta \neq \theta##.
So have i ended up wrong with the final result?

PeroK
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So have i ended up wrong with the final result?
Yes. For example:
$$\frac{d\sin(-\theta)}{d(-\theta)} \neq \frac{d\sin(\theta)}{d(-\theta)}$$

Yes. For example:
$$\frac{d\sin(-\theta)}{d(-\theta)} \neq \frac{d\sin(\theta)}{d(-\theta)}$$
I'll Redo it.

PeroK
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I'll Redo it.
You don't have to. This was correct:

And the total equation can be written as ##\frac{d^{2}v(-\theta) }{d(-\theta)^{2}}+v(-\theta)=-\frac{m}{l^{2}} \frac{d}{d v(-\theta)} V\left(\frac{1}{v(-\theta)}\right)##
A physicist would simply replace ##v## by ##u## there, but you have to leave the ##-\theta## everywhere, because that was the substitution after all!

Kashmir
You don't have to. This was correct:

A physicist would simply replace ##v## by ##u## there, but you have to leave the ##-\theta## everywhere, because that was the substitution after all!
So My final result is ok?

PeroK
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So My final result is ok?

Kashmir
By intermediate you mean ##\frac{d^{2}v(-\theta) }{d(-\theta)^{2}}+v(-\theta)=-\frac{m}{l^{2}} \frac{d}{d v(-\theta)} V\left(\frac{1}{v(-\theta)}\right)##

PeroK
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By intermediate you mean ##\frac{d^{2}v(-\theta) }{d(-\theta)^{2}}+v(-\theta)=-\frac{m}{l^{2}} \frac{d}{d v(-\theta)} V\left(\frac{1}{v(-\theta)}\right)##
Yes.

You don't have to. This was correct:

A physicist would simply replace ##v## by ##u## there, but you have to leave the ##-\theta## everywhere, because that was the substitution after all!
isn't that what I did in the final equation? Just replaced u by v everywhere and got ##\frac{d^{2}}{d(-\theta)^{2}} u(\theta)+u(\theta)=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right)## ?

PeroK
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isn't that what I did in the final equation? Just replaced u by v everywhere and got ##\frac{d^{2}}{d(-\theta)^{2}} u(\theta)+u(\theta)=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right)## ?
No, because you also replaced ##-\theta## with ##\theta##. A physicist wouldn't use ##v## at all, but just use ##u## for both functions - it saves having to think of a new letter!

And, as long as you are careful it works out. It just doesn't quite stand up to mathematical scutiny. Some physics books mention it and say "technically we should introduce a new function here, but ..." and other books don't say anything.

It's useful to know that there are two functions involved, even if you choose to use the same letter for both.

No, because you also replaced ##-\theta## with ##\theta##. A physicist wouldn't use ##v## at all, but just use ##u## for both functions - it saves having to think of a new letter!

And, as long as you are careful it works out. It just doesn't quite stand up to mathematical scutiny. Some physics books mention it and say "technically we should introduce a new function here, but ..." and other books don't say anything.

It's useful to know that there are two functions involved, even if you choose to use the same letter for both.
I'll study all your replies once again. Thank you so much :)

If this is correct :##\frac{d^{2}v(-\theta) }{d(-\theta)^{2}}+v(-\theta)=-\frac{m}{l^{2}} \frac{d}{d v(-\theta)} V\left(\frac{1}{v(-\theta)}\right)##

And also we started as ##u(\theta)=v(-\theta)##

Then if I just use it in the first equation I will get ##\frac{d^{2}}{d(-\theta)^{2}} u(\theta)+u(\theta)=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right)## but you said its wrong.
I can't see what went wrong?

PeroK
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Then if I just use it in the first equation I will get ##\frac{d^{2}}{d(-\theta)^{2}} u(\theta)+u(\theta)=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right)## but you said its wrong.
I can't see what went wrong?
It's not wrong, but it's not what you were asked to show. You were asked to carry out the substitution ##\theta \rightarrow -\theta##.

I cannot stress enough (and this is the third time I've told you) that you must use a different variable ##\alpha = -\theta## to resolve these sort of conceptual problems.

The question then asks you to carry out the substitition ##\alpha = -\theta##. Your final answer must involve ##\alpha## and only ##\alpha##. It can no longer have ##\theta## in it.

If someone asked you to translate "I don't know" into French, then a correct answer is "je ne sais pas". If you translate it back and give your answer as "I don't know", then of course that means the same thing, but it's no longer in French - it's back into English again - and you were asked to give an answer in French.