tobeint39r

2022-04-30

Does protons moving parallel to each other exert magnetic force?

If two protons are moving parallel to each other in same direction with equal velocities then do they exert magnetic force on each other.? As protons are moving they should produce magnetic field and there should be some magnetic force + electric force on each of them. But if look from their refrence frame then both the protons are at rest and hence there will be no magnetic force (as magnetic field is created by a moving charge), only electric force on each of them. How can this be possible that force(magnetic force) on a particle becomes different on changing the reference frame even though the reference frames are non accelerating with respect to each other?

If two protons are moving parallel to each other in same direction with equal velocities then do they exert magnetic force on each other.? As protons are moving they should produce magnetic field and there should be some magnetic force + electric force on each of them. But if look from their refrence frame then both the protons are at rest and hence there will be no magnetic force (as magnetic field is created by a moving charge), only electric force on each of them. How can this be possible that force(magnetic force) on a particle becomes different on changing the reference frame even though the reference frames are non accelerating with respect to each other?

elseptimopc7

Beginner2022-05-01Added 14 answers

Whether they exert both electric and magnetic forces on each other, or only electric forces, depends on which reference frame you use. Electric and magnetic fields transform into each other under changes of reference frame. This is why we talk about the electromagnetic field as if it were one unified tensor field $\mathbf{F}$ rather than two different vector fields $\overrightarrow{E}$ and $\overrightarrow{B}$

Cristian Rosales

Beginner2022-05-02Added 26 answers

Force, like all vectors, is not a Lorentz invariant. It is therefore not surprising that the force between two protons depends on which frame of reference is considered.

The calculation is simple to do by calculating the electric and magnetic fields observed in the frame in which the protons are seen to be moving. Both fields appear different in the moving frame.

You will find the repulsive force between the protons is still there, but reduced by a factor of $\gamma =(1-{v}^{2}/{c}^{2}{)}^{-1/2}$

Details:

In the stationary frame of the protons then there is just the Coulomb repulsion between them given by

${F}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}=e{\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}=\frac{{e}^{2}}{4\pi {\u03f5}_{0}{r}^{2}}\hat{r},$

${F}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}=e{\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}=\frac{{e}^{2}}{4\pi {\u03f5}_{0}{r}^{2}}\hat{r},$

where ${\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}$ is the E-field of a stationary proton.

In the lab frame, the electric field in the direction between the two protons is increased by the Lorentz factor to ${\overrightarrow{E}}_{\mathrm{l}\mathrm{a}\mathrm{b}}=\gamma {\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}$, where $\gamma =(1-{v}^{2}/{c}^{2}{)}^{-1/2}$ and where $\gamma \ge 1$. At the same time, there is a magnetic field caused by the motion of the protons and this contributes a force $e\overrightarrow{v}\times {\overrightarrow{B}}_{\mathrm{l}\mathrm{a}\mathrm{b}}$, where ${\overrightarrow{B}}_{\mathrm{l}\mathrm{a}\mathrm{b}}$ is the B-field measured in the lab frame.

The lab B-field is found using the appropriate transform as

${\overrightarrow{B}}_{\mathrm{l}\mathrm{a}\mathrm{b}}=-\frac{\gamma}{{c}^{2}}\overrightarrow{v}\times {\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}$

Thus the force between the protons on the lab frame is

${F}_{\mathrm{l}\mathrm{a}\mathrm{b}}=e({\overrightarrow{E}}_{\mathrm{l}\mathrm{a}\mathrm{b}}+\overrightarrow{v}\times {\overrightarrow{B}}_{\mathrm{l}\mathrm{a}\mathrm{b}})=e(\gamma {\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}-\frac{\gamma {v}^{2}}{{c}^{2}}{\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}})=\frac{e{\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}}{\gamma}=\frac{{F}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}}{\gamma}.$

This is exactly as required by the rules for transforming forces under special relativity. The force acting between the two protons is smaller in the lab frame and approaches zero as the protons become more and more relativistic.

The same thing would be true for two uncharged particles connected by a spring.

The calculation is simple to do by calculating the electric and magnetic fields observed in the frame in which the protons are seen to be moving. Both fields appear different in the moving frame.

You will find the repulsive force between the protons is still there, but reduced by a factor of $\gamma =(1-{v}^{2}/{c}^{2}{)}^{-1/2}$

Details:

In the stationary frame of the protons then there is just the Coulomb repulsion between them given by

${F}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}=e{\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}=\frac{{e}^{2}}{4\pi {\u03f5}_{0}{r}^{2}}\hat{r},$

${F}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}=e{\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}=\frac{{e}^{2}}{4\pi {\u03f5}_{0}{r}^{2}}\hat{r},$

where ${\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}$ is the E-field of a stationary proton.

In the lab frame, the electric field in the direction between the two protons is increased by the Lorentz factor to ${\overrightarrow{E}}_{\mathrm{l}\mathrm{a}\mathrm{b}}=\gamma {\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}$, where $\gamma =(1-{v}^{2}/{c}^{2}{)}^{-1/2}$ and where $\gamma \ge 1$. At the same time, there is a magnetic field caused by the motion of the protons and this contributes a force $e\overrightarrow{v}\times {\overrightarrow{B}}_{\mathrm{l}\mathrm{a}\mathrm{b}}$, where ${\overrightarrow{B}}_{\mathrm{l}\mathrm{a}\mathrm{b}}$ is the B-field measured in the lab frame.

The lab B-field is found using the appropriate transform as

${\overrightarrow{B}}_{\mathrm{l}\mathrm{a}\mathrm{b}}=-\frac{\gamma}{{c}^{2}}\overrightarrow{v}\times {\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}$

Thus the force between the protons on the lab frame is

${F}_{\mathrm{l}\mathrm{a}\mathrm{b}}=e({\overrightarrow{E}}_{\mathrm{l}\mathrm{a}\mathrm{b}}+\overrightarrow{v}\times {\overrightarrow{B}}_{\mathrm{l}\mathrm{a}\mathrm{b}})=e(\gamma {\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}-\frac{\gamma {v}^{2}}{{c}^{2}}{\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}})=\frac{e{\overrightarrow{E}}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}}{\gamma}=\frac{{F}_{\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{t}}}{\gamma}.$

This is exactly as required by the rules for transforming forces under special relativity. The force acting between the two protons is smaller in the lab frame and approaches zero as the protons become more and more relativistic.

The same thing would be true for two uncharged particles connected by a spring.

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