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Nashita
05-18-2018, 01:49 PM
Assalamualaikum

I am a first year student, my course of study is physics honours.

In Laplace's eqn in spherical polar coordinates with azimuthal symmetry , while solving it using the method of separation of variables why do we equate the radial part and the theta part with a constant ?

P.S. I looked for it in Google but no where it's written why they equate it with a constant!
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Zzz_
05-18-2018, 03:08 PM
wa'alaikum as'salaam,

have you tried youtube? It's a good place to go for these questions.

you can even ask questions under the video of interest.

https://www.youtube.com/results?sear...sperical+polar

laplace equation sperical polar - YouTube
Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube....
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Nashita
05-18-2018, 05:32 PM
Originally Posted by Zzz_
wa'alaikum as'salaam,

have you tried youtube? It's a good place to go for these questions.

you can even ask questions under the video of interest.

https://www.youtube.com/results?sear...sperical+polar

laplace equation sperical polar - YouTube
Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube....
Yes I have tried YouTube, I think I need to see some more videos on this topic.

Thank you so much!
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Futuwwa
05-31-2018, 09:38 PM
I'm not entirely sure of what you're talking about, but if it is what it sounds like to me, the reason is the following:

By doing it in those coordinates, you can rearrange the equation so that all the stuff with the variable r is on one side of the equals sign, and all the stuff with the theta is on the other side. Let's call the stuff on both sides as follows (i.e. left-hand side is a function of r only, a right-hand side a function of theta only):

LHS(r) = RHS(theta)

Thing is, the equation is true for any values of r and theta, you can literally plug in any two and it's still true. If you, then, keep r constant and change theta, then LHS(r) remains so too, and because of it being equal to it, RHS(theta) will be constant too, even as you change theta. From that, it follows that RHS(theta) must be equal to a constant and not change value depending on theta. The same argument in reverse for r and LHS(r). We can call this constant C.

Thus, we get LHS(r) = RHS(theta) = C.

Which we can split up into two single-variable differential equations that we can solve separately:
LHS(r) = C
RHS(theta) = C


I hope that makes sense.
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Nashita
06-01-2018, 06:10 AM
Originally Posted by Futuwwa
I'm not entirely sure of what you're talking about, but if it is what it sounds like to me, the reason is the following:

By doing it in those coordinates, you can rearrange the equation so that all the stuff with the variable r is on one side of the equals sign, and all the stuff with the theta is on the other side. Let's call the stuff on both sides as follows (i.e. left-hand side is a function of r only, a right-hand side a function of theta only):

LHS(r) = RHS(theta)

Thing is, the equation is true for any values of r and theta, you can literally plug in any two and it's still true. If you, then, keep r constant and change theta, then LHS(r) remains so too, and because of it being equal to it, RHS(theta) will be constant too, even as you change theta. From that, it follows that RHS(theta) must be equal to a constant and not change value depending on theta. The same argument in reverse for r and LHS(r). We can call this constant C.

Thus, we get LHS(r) = RHS(theta) = C.

Which we can split up into two single-variable differential equations that we can solve separately:
LHS(r) = C
RHS(theta) = C


I hope that makes sense.
Absolutely correct! Thank you so much .
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Grandad
06-01-2018, 06:23 AM
Originally Posted by Nashita
Yes I have tried YouTube, I think I need to see some more videos on this topic.

Thank you so much!
Can I uncross my eyes now? :/::
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Nashita
06-01-2018, 06:27 AM
Originally Posted by Grandad
Can I uncross my eyes now? :/::
-__________-


What do you mean??
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Grandad
06-01-2018, 06:50 AM
Originally Posted by Grandad
Can I uncross my eyes now? :/::
Sorry. My post was meant for Brother Futuwwa's post. In the UK, whenever someone comes up with a complex statement that the reader is not smart enough to understand he (or she) will say: 'I've gone cross-eyed over this'.....meaning, argghhh what on earth was that??? It's a light-hearted joke against oneself (and a compliment to the other).

I hope this help.
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Nashita
06-01-2018, 06:55 AM
Originally Posted by Grandad
Sorry. My post was meant for Brother Futuwwa's post. In the UK, whenever someone comes up with a complex statement that the reader is not smart enough to understand he (or she) will say: 'I've gone cross-eyed over this'.....meaning, argghhh what on earth was that??? It's a light-hearted joke against oneself (and a compliment to the other).

I hope this help.
LOL [emoji23][emoji23][emoji23][emoji23][emoji23][emoji23]

This forum thread is for those who understand math and physics, if you aren't into math or physics it's obvious that you won't be able to understand anything clearly!
And this forum thread is on Laplace's equation and solving it by spherical polar coordinates. If you want to know more about Laplace's equation then Google it!
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Grandad
06-01-2018, 06:58 AM
Originally Posted by Grandad
Sorry. My post was meant for Brother Futuwwa's post. In the UK, whenever someone comes up with a complex statement that the reader is not smart enough to understand he (or she) will say: 'I've gone cross-eyed over this'.....meaning, argghhh what on earth was that??? It's a light-hearted joke against oneself (and a compliment to the other).

I hope this help.
You could so easily confuse a stupid (old) person! :facepalm:;D
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anatolian
06-01-2018, 06:36 PM
I studied engineering but physics is still my passion. It is totally a different story..
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