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Short answer:

Force is not a Lorentz invariant and neither is acceleration. The protons repel each other with a force that combines the electric and magnetic components of the Lorentz force and depends on the frame of reference of the observer, but which is maximised in their rest frame.

Details:

Exactly your question is dealt with in Purcell & Morin "Electricity & Magnetism" 3rd ed. p.264.

The problem you may be having is in thinking that force is a relativistic invariant - it is not.

Electric and magnetic fields are transformed when looking at them from a different frame of reference.

In the stationary frame of the protons then there is just the Coulomb repulsion between them given by $$F_{\rm rest} = e\vec{E}_{\rm rest} = \frac{e^2}{4\pi \epsilon_0 r^2}\hat{r} ,$$ where $\vec{E}_{\rm rest}$ 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 $\vec{E}_{\rm lab} =\gamma \vec{E}_{\rm rest}$, where $\gamma = (1-v^2/c^2)^{-1/2}$ and where $\gamma \geq 1$. At the same time, there is a magnetic field caused by the motion of the protons and this contributes a force $e \vec{v} \times \vec{B}_{\rm lab}$, where $\vec{B}_{\rm lab}$ is the B-field measured in the lab frame.

The lab B-field is found using the appropriate transform as $$ \vec{B}_{\rm lab} = -\frac{\gamma}{c^2} \vec{v} \times \vec{E}_{\rm rest}$$

Thus the force between the protons on the lab frame is $$F_{\rm lab} = e(\vec{E}_{\rm lab} + \vec{v}\times \vec{B}_{\rm lab}) = e (\gamma \vec{E}_{\rm rest} - \frac{\gamma v^2}{c^2} \vec{E}_{\rm rest}) = \frac{e \vec{E}_{\rm rest}}{\gamma} = \frac{F_{\rm rest}}{\gamma}. $$ 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.

If you had arranged it so your proton beams travelled in parallel lines then you must also have arranged for some force to act in the direction opposing the proton's mutual repulsion. This force would transform in exactly the same way, so that if there was no net acceleration in the lab frame then there would be no net acceleration in the proton rest frame either.

Short answer:

Force is not a Lorentz invariant and neither is acceleration. The protons always repel each other, with a force that combines the electric and magnetic components of the Lorentz force and depends on the frame of reference of the observer, but which is maximised in their rest frame and which approaches zero as the protons become ultra-relativistic.

Details:

Exactly your question is dealt with in Purcell & Morin "Electricity & Magnetism" 3rd ed. p.264.

The problem you may be having is in thinking that force is a relativistic invariant - it is not.

Electric and magnetic fields are transformed when looking at them from a different frame of reference.

In the stationary frame of the protons then there is just the Coulomb repulsion between them given by $$F_{\rm rest} = e\vec{E}_{\rm rest} = \frac{e^2}{4\pi \epsilon_0 r^2}\hat{r} ,$$ where $\vec{E}_{\rm rest}$ 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 $\vec{E}_{\rm lab} =\gamma \vec{E}_{\rm rest}$, where $\gamma = (1-v^2/c^2)^{-1/2}$ and where $\gamma \geq 1$. At the same time, there is a magnetic field caused by the motion of the protons and this contributes a force $e \vec{v} \times \vec{B}_{\rm lab}$, where $\vec{B}_{\rm lab}$ is the B-field measured in the lab frame.

The lab B-field is found using the appropriate transform as $$ \vec{B}_{\rm lab} = -\frac{\gamma}{c^2} \vec{v} \times \vec{E}_{\rm rest}$$

Thus the force between the protons on the lab frame is $$F_{\rm lab} = e(\vec{E}_{\rm lab} + \vec{v}\times \vec{B}_{\rm lab}) = e (\gamma \vec{E}_{\rm rest} - \frac{\gamma v^2}{c^2} \vec{E}_{\rm rest}) = \frac{e \vec{E}_{\rm rest}}{\gamma} = \frac{F_{\rm rest}}{\gamma}. $$ 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.

If you had arranged it so your proton beams travelled in parallel lines then you must also have arranged for some force to act in the direction opposing the proton's mutual repulsion. This force would transform in exactly the same way, so that if there was no net acceleration in the lab frame then there would be no net acceleration in the proton rest frame either.

Exactly your question is dealt with in Purcell & Morin "Electricity & Magnetism" 3rd ed. p.264.

The problem you may be having is in thinking that force is a relativistic invariant - it is not.

Electric and magnetic fields are transformed when looking at them from a different frame of reference.

In the stationary frame of the protons then there is just the Coulomb repulsion between them given by $$F_{\rm rest} = e\vec{E}_{\rm rest} = \frac{e^2}{4\pi \epsilon_0 r^2}\hat{r} ,$$ where $\vec{E}_{\rm rest}$ 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 $\vec{E}_{\rm lab} =\gamma \vec{E}_{\rm rest}$, where $\gamma = (1-v^2/c^2)^{-1/2}$. At the same time, there is a magnetic field caused by the motion of the protons and this contributes a force $e \vec{v} \times \vec{B}_{\rm lab}$, where $\vec{B}_{\rm lab}$ is the B-field measured in the lab frame.

The lab B-field is found using the appropriate transform as $$ \vec{B}_{\rm lab} = -\frac{\gamma}{c^2} \vec{v} \times \vec{E}_{\rm rest}$$

Thus the force between the protons on the lab frame is $$F_{\rm lab} = e(\vec{E}_{\rm lab} + \vec{v}\times \vec{B}_{\rm lab}) = e (\gamma \vec{E}_{\rm rest} - \frac{\gamma v^2}{c^2} \vec{E}_{\rm rest}) = \frac{e \vec{E}_{\rm rest}}{\gamma} = \frac{F_{\rm rest}}{\gamma}. $$ 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.

If you had arranged it so your proton beams travelled in parallel lines then you must also have arranged for some force to act in the direction opposing the proton's mutual repulsion. This force would transform in exactly the same way, so that if there was no net acceleration in the lab frame then there would be no net acceleration in the proton rest frame either.

Short answer:

Force is not a Lorentz invariant and neither is acceleration. The protons repel each other with a force that combines the electric and magnetic components of the Lorentz force and depends on the frame of reference of the observer, but which is maximised in their rest frame.

Details:

Exactly your question is dealt with in Purcell & Morin "Electricity & Magnetism" 3rd ed. p.264.

The problem you may be having is in thinking that force is a relativistic invariant - it is not.

Electric and magnetic fields are transformed when looking at them from a different frame of reference.

In the stationary frame of the protons then there is just the Coulomb repulsion between them given by $$F_{\rm rest} = e\vec{E}_{\rm rest} = \frac{e^2}{4\pi \epsilon_0 r^2}\hat{r} ,$$ where $\vec{E}_{\rm rest}$ 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 $\vec{E}_{\rm lab} =\gamma \vec{E}_{\rm rest}$, where $\gamma = (1-v^2/c^2)^{-1/2}$ and where $\gamma \geq 1$. At the same time, there is a magnetic field caused by the motion of the protons and this contributes a force $e \vec{v} \times \vec{B}_{\rm lab}$, where $\vec{B}_{\rm lab}$ is the B-field measured in the lab frame.

The lab B-field is found using the appropriate transform as $$ \vec{B}_{\rm lab} = -\frac{\gamma}{c^2} \vec{v} \times \vec{E}_{\rm rest}$$

Thus the force between the protons on the lab frame is $$F_{\rm lab} = e(\vec{E}_{\rm lab} + \vec{v}\times \vec{B}_{\rm lab}) = e (\gamma \vec{E}_{\rm rest} - \frac{\gamma v^2}{c^2} \vec{E}_{\rm rest}) = \frac{e \vec{E}_{\rm rest}}{\gamma} = \frac{F_{\rm rest}}{\gamma}. $$ 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.

If you had arranged it so your proton beams travelled in parallel lines then you must also have arranged for some force to act in the direction opposing the proton's mutual repulsion. This force would transform in exactly the same way, so that if there was no net acceleration in the lab frame then there would be no net acceleration in the proton rest frame either.

Exactly your question is dealt with in Purcell & Morin "Electricity & Magnetism" 3rd ed. p.264.

The problem you may be having is in thinking that force is a relativistic invariant - it is not.

Electric and magnetic fields are transformed when looking at them from a different frame of reference.

In the stationary frame of the protons then there is just the Coulomb repulsion between them given by $$F_{\rm rest} = e\vec{E}_{\rm rest} = \frac{e^2}{4\pi \epsilon_0 r^2}\hat{r} ,$$ where $\vec{E}_{\rm rest}$ 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 $\vec{E}=\gamma \vec{E}_{\rm rest}$. At the same time, there is a magnetic field caused by the motion of the protons and this contributes a force $e \vec{v} \times \vec{B}$, where $\vec{B}$ is the B-field measured in the lab frame.

The lab B-field is found using the appropriate transform as $$ \vec{B} = -\frac{\gamma}{c^2} \vec{v} \times \vec{E}_{\rm rest}$$

Thus the force between the protons on the lab frame is $$F_{\rm lab} = e(\vec{E} + \vec{v}\times \vec{B}) = e (\gamma \vec{E}_{\rm rest} - \frac{\gamma v^2}{c^2} \vec{E}_{\rm rest}) = \frac{e \vec{E}_{\rm rest}}{\gamma} = \frac{F_{\rm rest}}{\gamma}. $$ This is exactly as required by the rules for transforming forces under special relativity.

If you had arranged it so your proton beams travelled in parallel lines then you must also have arranged for some force to act in the direction opposing the proton's mutual repulsion. This force would transform in exactly the same way, so that if there was no net acceleration in the lab frame then there would be no net acceleration in the proton rest frame either.

Exactly your question is dealt with in Purcell & Morin "Electricity & Magnetism" 3rd ed. p.264.

The problem you may be having is in thinking that force is a relativistic invariant - it is not.

Electric and magnetic fields are transformed when looking at them from a different frame of reference.

In the stationary frame of the protons then there is just the Coulomb repulsion between them given by $$F_{\rm rest} = e\vec{E}_{\rm rest} = \frac{e^2}{4\pi \epsilon_0 r^2}\hat{r} ,$$ where $\vec{E}_{\rm rest}$ 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 $\vec{E}_{\rm lab} =\gamma \vec{E}_{\rm rest}$, where $\gamma = (1-v^2/c^2)^{-1/2}$. At the same time, there is a magnetic field caused by the motion of the protons and this contributes a force $e \vec{v} \times \vec{B}_{\rm lab}$, where $\vec{B}_{\rm lab}$ is the B-field measured in the lab frame.

The lab B-field is found using the appropriate transform as $$ \vec{B}_{\rm lab} = -\frac{\gamma}{c^2} \vec{v} \times \vec{E}_{\rm rest}$$

Thus the force between the protons on the lab frame is $$F_{\rm lab} = e(\vec{E}_{\rm lab} + \vec{v}\times \vec{B}_{\rm lab}) = e (\gamma \vec{E}_{\rm rest} - \frac{\gamma v^2}{c^2} \vec{E}_{\rm rest}) = \frac{e \vec{E}_{\rm rest}}{\gamma} = \frac{F_{\rm rest}}{\gamma}. $$ 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.

If you had arranged it so your proton beams travelled in parallel lines then you must also have arranged for some force to act in the direction opposing the proton's mutual repulsion. This force would transform in exactly the same way, so that if there was no net acceleration in the lab frame then there would be no net acceleration in the proton rest frame either.