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Relative velocity and speed of light 
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#1
Jun1713, 03:12 PM

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1) On a recent exam, one of the questions asked to find the minimum speed at which you would have to fire two protons headon if you wanted them to get them within [itex]10^15m[/itex].
To make things a bit easier, I treated one proton as being stationary and then found the energy needed to bring the other from infinity to the specified distance by integrating the coulomb force over that distance. From that I got the velocity which was larger than c but double the minimum speed of each proton. Is this approach valid? I wasn't too sure because of accelerating frames of references involved... 2) It also got me thinking about what would happen if two objects were moving towards each other at, say, 2/3 the speed of light. Wouldn't each appear to the other as moving faster than light? 


#2
Jun1713, 04:35 PM

P: 419

You are probably making mistakes because this problem requires the use of special relativity. There are different "relativistic" expressions for momentum, kinetic energy, etc. when you are dealing with velocities on the order of c. Another interesting aspect of special relativity is that it resolves the issue you mention: two particles each moving at 2/3 c toward one another do not "see" one another as moving at 4/3 c. Because of special relativity and the phenomena of time dilation and length contraction, from each particle's reference frame, the other is observed moving slower than c.



#3
Jun1813, 02:20 AM

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In order for the blue observer to measure the speed of the red observer relative to himself, he can send out radar signals that hit the red observer and reflect back to himself. He notes the time he sent each signal and the time the reflection got back to him. This allows him to measure a distance applied at a particular time and by doing this twice, he can calculate the speed of the red observer. Let's see how this works: The blue observer sends out his first radar signal at his Proper Time of 50 nsecs and receives its echo at 2 nsecs. He calculates the average time between those two events which is 26 nsec. He assumes that the signal spent half its time getting to the red observer and half its time getting back and since it was gone for 48 nsec, he assumes that the red observer was 24 nsec at the speed of light away or 24 feet away from him at his Proper Time of 26 nsecs. He does a similar calculation for his second radar signal which he sent out at his Proper Time of 25 nsecs and received the echo at 1 nsecs. He calculates the average time between those two events which is 13 nsec. He again assumes that the signal spent half its time getting to the red observer and half its time getting back and since this time it was gone for 24 nsec, he assumes that the red observer was 12 nsec at the speed of light away or 12 feet away from him at his Proper Time of 13 nsecs. Now to calculate the speed, he takes the difference between the distances divided by the difference in the times. The two distances were 24 and 12 feet with a difference of 12 feet. The two times were 26 and 13 nsecs with a difference of 13 nsecs. Thus, he has determined that the speed of the red object is 12/13 feet per nsecs relative to himself. The negatives sign means that he is coming toward the blue observer. The fraction evaluates to 0.923c. Now we can transform the coordinates of this IRF into one moving at 0.667c which will show us the rest frame of the blue observer: Now the calculations that the blue observer made of the red observer's speed are more obvious but no better than in the first IRF. I didn't bother to label any of the Proper Time dots for the blue observer since they always match the Coordinate time. Does this help answer your question? 


#4
Jun1813, 08:38 AM

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Relative velocity and speed of light
Once you have found the energy then you can solve for the velocity in the rest frame of the other proton. Specifically, once you have found the velocity in the rest frame of the other proton you need to use the relativistic velocity addition formula to find the velocity of both protons in the frame where their velocities are equal and opposite. 


#5
Jun1813, 10:01 AM

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#6
Jun1813, 12:38 PM

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#7
Jun1813, 01:10 PM

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Further hint: the one (!) physical constant you'll ever need, to work out atomic physics problems is ħc = 197.5 MeVfm. So  at closest approach the Coulomb energy is V = e^{2}/r = (e^{2}/ħc)/(r/ħc) = (1/137)/(.01 fm/197.5 MeVfm). You're welcome to multiply this out and get the answer immediately in terms of MeV. Set this potential energy V equal to the initial kinetic energy of the two protons, i.e. 2(γ 1) mc^{2} where mc^{2} of course is the proton's rest energy, 940 MeV or whatever it is. Solve this for γ, then use the relativistic expression for γ to solve for v. That's all there is to it. 


#8
Jun1813, 01:26 PM

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To see my point, imagine replacing one proton with a superheavy particle with same charge as a proton. Then, this heavy particle will have a reasonable inertial frame (effectively the same as the COM frame), and all the energy needed for close approach is allocated to the proton. Thus, the COM frame is where you can easily justify splitting the energy for equal mass particles, giving all to the light one for extreme mass ratio, or compute exactly the division if e.g. one is 3x the mass of the other. 


#9
Jun1813, 02:16 PM

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I don't see it.
Assume the protons, mass M, are each acted on by an external force which is exactly equal and opposite to the Coulomb force. Assume one proton is at rest and the other has some small velocity towards the one at rest. The work done by the external force on the at rest one is 0, and the work done on the moving one, W, is given by integrating the Coulomb force over distance as described. So in the limit as the small velocity goes to 0 the total energy of the system at the end is W+2Mc˛. The final state of the system is both protons at rest the required distance apart, i.e. the same as the final state in the COM frame. Since the states are the same we immediately conclude that the total energy of the system in the COM frame is also given by W+2Mc˛. EDIT: actually, my previous descriptions do seem to be wrong. This procedure (integrating the Coulomb force) gives you the energy in the COM frame already. There is no need to transform to the COM frame, it is already there. Perhaps that is what you were objecting to? 


#10
Jun1813, 02:36 PM

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#11
Jun1813, 02:46 PM

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#13
Jun1813, 02:54 PM

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#14
Jun1813, 02:54 PM

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#15
Jun1813, 03:14 PM

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#16
Jun1813, 03:43 PM

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However, the relativistic formula is [tex]\frac{v1+ v2}{1+ v1v2/c^2}[/tex] which, with v1= v2= (2/3)c is [tex]\frac{\frac{2c}{3}+ \frac{2c}{3}}{1+ \frac{2c}{3}\frac{2c}{3}/c^2}= \frac{\frac{4}{3}c}{1+ \frac{4c^2}{9}/c^2}[/tex] [tex]= \frac{4c}{3}{\frac{13}{9}}= \frac{\frac{4c}{3}}{\frac{9}{13}}= \frac{12c}{13}[/tex] slightly less than the speed of light. 


#17
Jun1813, 04:02 PM

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#18
Jun1813, 04:18 PM

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The only frame in which you can compute an electrostatic potential, then just distribute it to the charges to get speed at infinity is the COM frame. The rule for dividing (given unequal masses) it is obviously such that sum of the momenta at infinity is zero. 


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