- #1

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$$-E'(v)a(1+\frac{cE(v)}{\sqrt{E(v)^2-1}})=vE(v)+c\sqrt{E(v)^2-1}$$

It's not separable in E on one side and v expression on the other.

So I'm looking for methods to solve this maybe changes of coordinates ?

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There is a circular reasoning going on here. The equation is wrong, so changing the variables doesn't help.f

- #1

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$$-E'(v)a(1+\frac{cE(v)}{\sqrt{E(v)^2-1}})=vE(v)+c\sqrt{E(v)^2-1}$$

It's not separable in E on one side and v expression on the other.

So I'm looking for methods to solve this maybe changes of coordinates ?

- #2

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Assuming E(v) is positive we can transform it to

[tex]-FdE=dv[/tex]

where

[tex]F=\frac{a(1+\frac{c}{\sqrt{1-E^{-2}}})}{E(v+c\sqrt{1-E^{-2}})}[/tex]

[tex]-FdE=dv[/tex]

where

[tex]F=\frac{a(1+\frac{c}{\sqrt{1-E^{-2}}})}{E(v+c\sqrt{1-E^{-2}})}[/tex]

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- #3

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This does not help since F still depends on v.

- #4

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Ah, you are right.

Then I try

[tex]E=\coth w[/tex] and get

[tex]a(\sinh w +c\cosh w)\frac{dw}{dv}+v\cosh w+c\sinh w=0[/tex]

It is not separable yet.

E should be dimensionless. Are a,c and v are also dimensionless parameters? If not I am afraid there is something wrong in your equation.

Then I try

[tex]E=\coth w[/tex] and get

[tex]a(\sinh w +c\cosh w)\frac{dw}{dv}+v\cosh w+c\sinh w=0[/tex]

It is not separable yet.

E should be dimensionless. Are a,c and v are also dimensionless parameters? If not I am afraid there is something wrong in your equation.

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- #5

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You're right this equation is wrong. I'll look at the calculation again.

- #6

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$$-E'(v)*t*a(v+\frac{cE(v)}{\sqrt{E(v)^2-1}})=vE(v)+c\sqrt{E(v)^2-1}$$

But it anyhow is wrong since there shall be derivatives towards v and t.

Thanks for your help.

- #7

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- #8

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But the derivation of the equation is wrong. I wrote a coordinate transformation ##x'=A(v,t)x+B(v,t)t##, ##t'=D(v,t)x+E(v,t)t##

The equations are ##c^2t'^2-x'^2=c^2t^2-x^2## and ##\frac{dx'}{dt'}|_{x=0}=-v##

A mistake I made came while computing ##dx'## only with a partial derivative towards ##v##.

- #9

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[tex]x'=Ax+Bct[/tex]

[tex]ct'=Dx+Ect[/tex]

[tex]c^2t'^2-x'^2=(E^2-B^2)c^2t^2-(A^2-D^2)x^2+2(DE-AB)xct[/tex]

So

[tex]E=\cosh \phi[/tex]

[tex]B=\sinh \phi[/tex]

[tex]A=\cosh \psi[/tex]

[tex]D= \sinh \psi[/tex]

[tex]\frac{x'}{ct'}|_{x=0}=\frac{B}{E}=-\frac{v}{c}[/tex]

[tex]\frac{x}{ct}|_{x'=0}=-\frac{D}{A}=\frac{v}{c}[/tex]

So

[tex]\phi = \psi[/tex]

[tex]\tanh\phi=-\frac{v}{c}[/tex]

[tex]E=A= \cosh \phi=\frac{1}{\sqrt{1-v^2/c^2}}[/tex]

[tex]B=D= \sinh \phi=\frac{-v/c}{\sqrt{1-v^2/c^2}}[/tex]

[tex]ct'=Dx+Ect[/tex]

[tex]c^2t'^2-x'^2=(E^2-B^2)c^2t^2-(A^2-D^2)x^2+2(DE-AB)xct[/tex]

So

[tex]E=\cosh \phi[/tex]

[tex]B=\sinh \phi[/tex]

[tex]A=\cosh \psi[/tex]

[tex]D= \sinh \psi[/tex]

[tex]\frac{x'}{ct'}|_{x=0}=\frac{B}{E}=-\frac{v}{c}[/tex]

[tex]\frac{x}{ct}|_{x'=0}=-\frac{D}{A}=\frac{v}{c}[/tex]

So

[tex]\phi = \psi[/tex]

[tex]\tanh\phi=-\frac{v}{c}[/tex]

[tex]E=A= \cosh \phi=\frac{1}{\sqrt{1-v^2/c^2}}[/tex]

[tex]B=D= \sinh \phi=\frac{-v/c}{\sqrt{1-v^2/c^2}}[/tex]

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- #10

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- #11

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- #12

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- #13

Mentor

- 4,736

- 3,732

Could you please tell us the exact source for this problem?

- #14

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If B(v), then allowing to change v (for any kind of ground) induces a dv/dt, then the hypothesis was wrong and in fact B(a,v), aso, such that the ODE is of infinite degree since ##B(\{\frac{d^n v}{dt^n}\}_{n=0}^\infty)##

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