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Consider a particle moving near the speed of light (let's say 90% of c). It collides with a wall of much higher mass and bounces back. How would relativity affect that impulse value? The two observers would have different values of time for the collision. Does that mean the average force differs between perspective? I understand that the speed at which the particle is travelling is arbitrary, since the effects of special relativity scale all the way to zero velocity. But what would it look like, considering that the effects are so exaggerated at 90% c?

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  • $\begingroup$ I would guess that the collision would be elastic, if that's what you're asking. $\endgroup$ – Corsair64 Aug 10 '17 at 22:04
  • $\begingroup$ I do not understand what you mean by "different values of time for the collision". Do you mean they assign different durations to the collision? That's not true; the collision is instantaneous in either frame. Or do you mean they assign different time coordinates to the collision event? That's true but I can't imagine any reason you'd think it was relevant. In fact, we don't need relativity for this: Two observers, one on eastern standard time and one on greenwich mean time, will also assign different times to the event. So what? $\endgroup$ – WillO Aug 10 '17 at 22:06
  • $\begingroup$ The collision is not instantaneous in any frame. The classic force versus time graph used to illustrate impulse comes to mind. The collision takes place over a very small amount of time, yes, but it is still not instant. $\endgroup$ – Corsair64 Aug 10 '17 at 22:09
  • $\begingroup$ Apologies, I was rushed for time on my earlier comment, and it was inadequate. I will attempt an answer asap. $\endgroup$ – user163104 Aug 10 '17 at 22:14
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Yes, indeed, three-force is frame dependent. Look up the Wikipedia article on four-vectors, for example: it gives you a list of quantities that transform in the same way as vectors joining events in Minkowski spacetime.

The four-momentum of a particle is

$$\mathbf{p}=(E,\,p_x,p_y,p_z)=m_0\,\gamma\,(1,\,v_x,\,v_y,\,v_z)$$

where $E$ is particle's total energy, $v_i$ the Cartesian three-velocity components and $\gamma = 1/\sqrt{1-v^2}$ the Lorentz factor (velocities are all normalized so that $c=1$ here).

If this four-momentum is changed by an impulse through an interaction, we get the four force, which is the rate of change of this momentum with respect to the particle's proper time:

$$\mathbf{F} = \mathrm{d}_\tau \mathbf{p}=\gamma\,(P,\,\mathbf{f})$$

where $\mathbf{f}$ is the "everyday" three force, or the laboratory rate of imparting of impulse to the particle and $P$ the laboratory power, or rate of working by the three force. The components of the vector $\gamma\,(P,\,\mathbf{f})$ transform exactly as event spacetime co-ordinates do, i.e. by the Lorentz transformation, so that the three force is frame dependent and gets mixed with the power of the three force when we observe it from different inertial frames. As you can see from the presence of the $\gamma$ factor, the laboratory impulse of the force varies a great deal under relativistic-speed boosts.

How would it look? Bad. Like: really, really BAD!

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  • $\begingroup$ Thanks. I do understand the physics and the idea on paper, but I can't wrap my mind around the two experiencing differing forces. When we measure impulse in the lab, what frame would you say we are measuring from? The wall? The particle? Something different? I suppose that in lab conditions, relativity hardly applies, but I wonder which frame would be most "appropriate" to measure from. $\endgroup$ – Corsair64 Aug 12 '17 at 15:06
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Quantum Theory (lab observation): It would perform the bounce as a particle if there was no observers, but as there are two like you said, the particle would perform as a wave.

https://www.sciencedaily.com/releases/1998/02/980227055013.htm

Being near of speed of light is not being at the speed of light.

Best regards,

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  • $\begingroup$ Instead of giving a minus, do answer the question. Is my answer outdated? $\endgroup$ – Methark Aug 10 '17 at 22:18
  • $\begingroup$ ...What answer? $\endgroup$ – WillO Aug 10 '17 at 23:48
  • $\begingroup$ haha...you are funny. $\endgroup$ – Methark Aug 10 '17 at 23:49

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