- #1

olgerm

Gold Member

- 519

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Hi.

If 2 bodies with charge q are in rest then both have electric force ##F_1=\frac{q*q*k_q}{|\vec{r}|^2}##.

But in another frame of reference, that is moving with velocity v relative to first frame of reference, they feel both magnetic and electric force ##F_2=|\vec{F_{electric}}+\vec{F_{magnetic}}|=|\vec{F_{electric}}|+|\vec{F_{magnetic}}|=|q*(\vec{E}+\vec{v}\times \vec{B})|=|q*(\frac{q*k_q}{r^2}+\vec{v}\times \frac{q*\mu_0*\vec{v}}{4*\pi*|\vec{v}|^2})|=q*(\frac{q*k_q}{|\vec{r}|^2}+\frac{q*\mu_0*|\vec{v}|^2}{4*\pi*|\vec{v}|^2})=\frac{q^2*(k_q+\mu_0*|\vec{v}|^2)}{|\vec{r}|^2}##

So forces are not same in both frames of reference. Is it correct?

##\vec{v}## is crosswise to ##\vec{r}##.

Some source said that I should use relation

##E_2=E_1+v \times B_1##

##B_2=B_1-v/c^2\times E_1##

but that would mean that ##E_2\not=\frac{q*k_q}{r^2}## Is it correct? Does that mean that there is free electromagnetic field(electromagnetic wave(s)) in second frame of reference?

If 2 bodies with charge q are in rest then both have electric force ##F_1=\frac{q*q*k_q}{|\vec{r}|^2}##.

But in another frame of reference, that is moving with velocity v relative to first frame of reference, they feel both magnetic and electric force ##F_2=|\vec{F_{electric}}+\vec{F_{magnetic}}|=|\vec{F_{electric}}|+|\vec{F_{magnetic}}|=|q*(\vec{E}+\vec{v}\times \vec{B})|=|q*(\frac{q*k_q}{r^2}+\vec{v}\times \frac{q*\mu_0*\vec{v}}{4*\pi*|\vec{v}|^2})|=q*(\frac{q*k_q}{|\vec{r}|^2}+\frac{q*\mu_0*|\vec{v}|^2}{4*\pi*|\vec{v}|^2})=\frac{q^2*(k_q+\mu_0*|\vec{v}|^2)}{|\vec{r}|^2}##

So forces are not same in both frames of reference. Is it correct?

##\vec{v}## is crosswise to ##\vec{r}##.

Some source said that I should use relation

##E_2=E_1+v \times B_1##

##B_2=B_1-v/c^2\times E_1##

but that would mean that ##E_2\not=\frac{q*k_q}{r^2}## Is it correct? Does that mean that there is free electromagnetic field(electromagnetic wave(s)) in second frame of reference?

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