Relative velocity and speed of light

by chipotleaway
Tags: light, relative, speed, velocity
 P: 170 1) On a recent exam, one of the questions asked to find the minimum speed at which you would have to fire two protons head-on if you wanted them to get them within $10^-15m$. 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?
 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.
PF Gold
P: 4,681
 Quote by chipotleaway 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?
Here's a spacetime diagram showing two objects (or observers) approaching each other at 2/3 the speed of light. I'm using the speed of light as one foot per nanosecond to make things easy. Since each observer is traveling at 0.667c in this Inertial Reference Frame, their Proper Time is dilated and takes longer than the Coordinate Time as shown by the spacing of the dots which represent one-nanosecond intervals of Proper Time:

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.

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Mentor
P: 16,947
Relative velocity and speed of light

 Quote by chipotleaway 1) On a recent exam, one of the questions asked to find the minimum speed at which you would have to fire two protons head-on if you wanted them to get them within $10^-15m$. 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...
Your approach is valid. Has your class covered any relativistic topics? Especially the relativistic formulas for KE and momentum, or (even better) the four-momentum?

Once you have found the energy then you can solve for the velocity in the rest frame of the other proton.

 Quote by chipotleaway 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?
The relativistic "velocity addition" formula ensures that this doesn't happen. Also, you need to use the relativistic velocity addition formula to finish the problem the way you were approaching it.

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.
PF Gold
P: 5,027
 Quote by chipotleaway 1) On a recent exam, one of the questions asked to find the minimum speed at which you would have to fire two protons head-on if you wanted them to get them within $10^-15m$. 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...
 Quote by DaleSpam Your approach is valid.
Is it really valid? Assuming the problem is adequately dealt with by coulomb action at a distance approximation, I would think you still need to solve in the center of mass rest frame. In the starting rest frame of either proton, both protons move; and a frame that stays with one of the protons is non-inertial, thus inertial forces must be treated (even non-relativistically). For a proton and massive nucleus, treating the nucleus as an inertial frame is often accurate enough, but I don't see how it could be valid for two protons.
Mentor
P: 16,947
 Quote by PAllen Is it really valid? Assuming the problem is adequately dealt with by coulomb action at a distance approximation, I would think you still need to solve in the center of mass rest frame. In the starting rest frame of either proton, both protons move; and a frame that stays with one of the protons is non-inertial, thus inertial forces must be treated (even non-relativistically). For a proton and massive nucleus, treating the nucleus as an inertial frame is often accurate enough, but I don't see how it could be valid for two protons.
I do think it is valid (assuming Coulomb is a good approximation, as you said). This is just an energy problem. You want to get the energy of two protons at rest a specified distance apart. That can be calculated by the procedure specified. Then you just equate that energy to the KE and solve for the velocity. Then you transform that to the COM frame. I don't see that acceleration is even involved in the question.
Thanks
P: 4,160
 Quote by chipotleaway 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.
This is valid, but certainly making life difficult for yourself. 1) You don't have to transform to the lab frame and it greatly complicates things! Stay in the center-of-mass frame from beginning to end. 2) Instead of integrating the Coulomb force, use the Coulomb potential. 3) You probably used SI units too.

Further hint: the one (!) physical constant you'll ever need, to work out atomic physics problems is ħc = 197.5 MeV-fm.

So -- at closest approach the Coulomb energy is V = e2/r = (e2/ħc)/(r/ħc) = (1/137)/(.01 fm/197.5 MeV-fm). 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) mc2 where mc2 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.
PF Gold
P: 5,027
 Quote by DaleSpam I do think it is valid (assuming Coulomb is a good approximation, as you said). This is just an energy problem. You want to get the energy of two protons at rest a specified distance apart. That can be calculated by the procedure specified. Then you just equate that energy to the KE and solve for the velocity. Then you transform that to the COM frame. I don't see that acceleration is even involved in the question.
If you integrate coulomb force on one proton for infinity to d (final separation desired), and conclude this is the energy needed for that proton, you have attributed the wrong amount of energy to it. You conclude (ignoring extra consants and conversions factors) q^2/d is the energy of one proton in a frame of the other one at rest (which isn't even an inertial frame). You get the rationale for splitting q^2/d between the two protons only in the COM frame. Then, this energy (q^2/2d) gives the velocity of each in the COM frame.

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.
 Mentor P: 16,947 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+2Mc². 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+2Mc². 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?
PF Gold
P: 5,027
 Quote by DaleSpam 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?
Yes. Also, the for two protons shot towards each other, there is no single (inertial) frame one of the protons remains at rest. The frame in which both end up at rest is the COM frame. If you then wanted a frame in which one of the protons was initially at rest and the other approaching from far, you would need to transform from this COM frame.
Mentor
P: 16,947
 Quote by PAllen Yes.
OK, thanks for catching my error!

 Quote by PAllen Also, the for two protons shot towards each other, there is no single (inertial) frame one of the protons remains at rest.
Sure, but I don't think that is relevant to the problem. All you need to do is find the initial velocity from the energy, you don't actually need to do any calculations during the acceleration. And the calculation to determine the energy doesn't require any acceleration either. That is what I was getting hung up on from your responses.
Thanks
P: 4,160
 Quote by DaleSpam 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?
See #7 above.
C. Spirit
Thanks
P: 5,410
 Quote by PAllen If you then wanted a frame in which one of the protons was initially at rest and the other approaching from far, you would need to transform from this COM frame.
What's the problem with starting from a frame in which one proton is at rest and the other one is moving in from infinity, right from the start (meaning principal problems and not because of extra work involved in solving it this way)? Sure the COM frame is an inertial frame itself and sure the frame fixed to one of the two protons is not inertial, but what's the issue there with regards to the problem at hand?
Mentor
P: 16,947
 Quote by Bill_K See #7 above.
Yes, I agree. It is not the easiest approach. I would have used the Coulomb potential to determine the energy but then four-vectors to figure out the speed. Your approach is even simpler than that. I just have a tendency to run to four-vectors with little or no provocation!
PF Gold
P: 5,027
 Quote by WannabeNewton What's the problem with starting from a frame in which one proton is at rest and the other one is moving in from infinity, right from the start (meaning principal problems and not because of extra work involved in solving it this way)? Sure the COM frame is an inertial frame itself and sure the frame fixed to one of the two protons is not inertial, but what's the issue there with regards to the problem at hand?
There is no problem, in principle. It is only somewhat more complicated if you do it in the one inertial frame where one proton is initially at rest (you can't just use electrostatic potential at closest approach, because at closest approach, both of the protons have KE in this frame).
Math
Emeritus
Thanks
PF Gold
P: 39,316
 Quote by chipotleaway 1) On a recent exam, one of the questions asked to find the minimum speed at which you would have to fire two protons head-on if you wanted them to get them within $10^-15m$. 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?
If object 1 is moving directly toward object 2 with speeds relative to some third observer of v2 and v2, then classical physics would say that each is approaching the other at speed v1+ v2. With v1= v2= (2/3)c, that would indeed be (4/3)c, greater than the speed of light. Of course, that does not conflict with classical physics.

However, the relativistic formula is
$$\frac{v1+ v2}{1+ v1v2/c^2}$$
which, with v1= v2= (2/3)c is
$$\frac{\frac{2c}{3}+ \frac{2c}{3}}{1+ \frac{2c}{3}\frac{2c}{3}/c^2}= \frac{\frac{4}{3}c}{1+ \frac{4c^2}{9}/c^2}$$
$$= \frac{4c}{3}{\frac{13}{9}}= \frac{\frac{4c}{3}}{\frac{9}{13}}= \frac{12c}{13}$$
slightly less than the speed of light.
C. Spirit
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P: 5,410
 Quote by PAllen There is no problem, in principle. It is only somewhat more complicated if you do it in the one inertial frame where one proton is initially at rest (you can't just use electrostatic potential at closest approach, because at closest approach, both of the protons have KE in this frame).
Oh was that your initial objection? Then yeah there must be care taken with regards to that, if one wishes to stick to that method. If we stay in the inertial frame which was initially comoving with one of the two protons right when they were fired at each other, then when the two protons come to rest momentarily in the COM frame, they will (at this same moment) certainly have a collective KE in that initially comoving frame so that will have to be taken into account if conservation of energy is used exclusively in that frame. I certainly agree with that.