- #1

Alexandre

- 29

- 0

How can show that momentum is the gradient of the action for the free particle? I tried it like this for one dimensional case:

[tex] s=\int Ldt [/tex]

[tex] ds=Ldt [/tex]

[tex] ds=\frac{mv^2}{2}dt\:[/tex]

Velocity is constant right? So I should be able to to this:

[tex] \frac{ds}{dx}=\frac{mv^2}{2}\frac{dt}{dx} [/tex]

I'm not sure about the following step where I use: [tex] \frac{dt}{dx}=\frac{1}{v} [/tex] is this correct?

[tex] \frac{ds}{dx}=\frac{mv^2}{2}\frac{1}{v} [/tex]

And finally I get:

[tex] \frac{ds}{dx}=\frac{mv}{2} [/tex]

It differs from momentum by the factor of 1/2, where did I make the mistake?

[tex] s=\int Ldt [/tex]

[tex] ds=Ldt [/tex]

[tex] ds=\frac{mv^2}{2}dt\:[/tex]

Velocity is constant right? So I should be able to to this:

[tex] \frac{ds}{dx}=\frac{mv^2}{2}\frac{dt}{dx} [/tex]

I'm not sure about the following step where I use: [tex] \frac{dt}{dx}=\frac{1}{v} [/tex] is this correct?

[tex] \frac{ds}{dx}=\frac{mv^2}{2}\frac{1}{v} [/tex]

And finally I get:

[tex] \frac{ds}{dx}=\frac{mv}{2} [/tex]

It differs from momentum by the factor of 1/2, where did I make the mistake?

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