- #1

- 190

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Hello there,

please tell me whether [tex]e^{iS} = 0[/tex] when [tex]S = +\infty[/tex]

S is the action of a particle [tex] = \int L dt[/tex]

e.g. when the particle goes to infinity and comes back, kinetic energy blows up,

then S blows up.

- Thread starter kakarukeys
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- #1

- 190

- 0

Hello there,

please tell me whether [tex]e^{iS} = 0[/tex] when [tex]S = +\infty[/tex]

S is the action of a particle [tex] = \int L dt[/tex]

e.g. when the particle goes to infinity and comes back, kinetic energy blows up,

then S blows up.

- #2

LeonhardEuler

Gold Member

- 859

- 1

It does'nt:kakarukeys said::grumpy: :yuck:

Hello there,

please tell me whether [tex]e^{iS} = 0[/tex] when [tex]S = +\infty[/tex]

[tex]e^{ix}=\cos{x}+i\sin{x}[/tex]

[tex]\lim_{x\rightarrow \infty}e^{ix}=\lim_{x\rightarrow \infty}\cos{x}+i\lim_{x\rightarrow \infty}\sin{x}[/tex]

But niether the real nor imaginary parts of the limit exist; as x approaches infinity they oscilate between -1 and 1.

- #3

- 190

- 0

That's what my math teacher told me,

but my physics teacher told me entirely different thing.

but my physics teacher told me entirely different thing.

- #4

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,916

- 19

The map

However...

As is often the case, there is probably some related concept that is practical to use here, and whether he knows it or not, it's what your physics professor meant.

- #5

- 190

- 0

Do you mean, for computation purpose, we can define the limit to be zero

it can't be proven?

it can't be proven?

- #6

- 190

- 0

this is no college level!

Feymann Path Integral!

- #7

- 574

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kakarukeys said:

this is no college level!

Feymann Path Integral!

College=university. Don't get freaked out.

- #8

- 190

- 0

Feynmann Path Integral is at graduate level :rofl:

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