# The Arthur Beiser Logic

1. Jan 8, 2012

### bibhu1984

In his book 'Concepts of MODERN PHYSICS', Chapter 1, Section 1.7, page# 22to 24, Arthur Beiser tries to derive an equation for relativistic momentum, which he finally does. But I found the situation considered by him inappropriate so is with the way he deals with it.
Can anyone please tell me if I am wrong in comprehending his method?

I've attached two snaps of the concerned pages containing the full derivation concerned.

There are some specific assumptions that appear quite normal at the first reading. A careful observation of which is what creates doubt.

1-When the author says both the balls were thrown from rest in their respective frames at the same instant, what instant is he talking about? There's no common instant between both the frames apart from the one when they started moving relative to each other which is not the one he’s pointing at.

2-The orientation of S and S' as shown in the diagram on the right is not correct because if the balls are thrown at such an orientation they will never hit each other because B should be at the left of A before collision as B is travelling along X axis towards right.

3-The balls will never collide at y₌Y/2 in any frame because of a simple logic. Consider frame S. Here A was at rest. After it was thrown with a speed of VA for T seconds it travels for a distance yA ₌ VAT . However B’s speed in S is VB which is <V’B. As previously assumed VA₌V’B. Hence even if B travels for the same time T in S, it will cover a distance yB₌VBT < yA. As seen here if the two balls travel for the same time interval in S they will cover unequal distances. Hence they won’t collide midway.

4- Consider any frame (say S). Let’s focus on only Y axis movement. If we consider relativistic effects on velocities for sure and apply formulas from elastic collison.
Assumption 1: Relativistic effect on mass is not considered.

Two balls of equal mass collide with different velocities as VA>VB.Their final velocities will be simply their initial velocities exchanged. Not as the Author says simply reversed.

Assumption 2: Relativistic effect on mass is considered

Two different balls collide with different initial velocities. Their final velocities cannot be their initial velocities simply reversed.

Where do we see that A will bounce back with simply its initial velocity reversed as does B?

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2. Jan 8, 2012

### MikeLizzi

Yep. That's the example I was thinking of when I posted a reply in your thread in the ScienceForums. Unfortunately, as often happens, that thread got sidetracked by people who are good at broadcasting what they know but not good at listening to the OP. If the same happens here try sending me a Visitor Message by going to my user profile and clicking on "Visitor Messages".

3. Jan 8, 2012

### PAllen

Is there any way you can enlarge or brighten those images? I can't make them out enough to compare your understanding versus what might be the author's intent.

[Edit: Welcome to physicsforums!]

Last edited: Jan 8, 2012
4. Jan 8, 2012

### PAllen

Ok, I saved the images, and then could adjust size and brightness enough to make most of it out (still pretty blurry in parts). I'll use your numbering.

1) I agree that this wording is sloppy. I interpret the author's intent more precisely as follows:

A is thrown in the +y direction with speed V, per S, at some time per S.
B is thrown in the -y direction with speed V, per S', at some time per S'.

These times are such that the balls collide. The collision is an invariant feature of the universe, and must occur in both frames. I am guessing (at this point, since I haven't looked at the rest) that the aim is to show that if conservation laws are required to hold in S and in S', the formula for relativistic momentum follows.

2) See my answer to (1). There is a bit of common convention being used in these diagrams. The picture is meant to show the orientation of axes and relative motion between S and S', not their actual relative positioning at any specific time of the problem setup. This is very common convention, that the author assumes you know. While I think (1) shows definitely sloppy language for introducing relativity, given relativity of time and simultaneity, diagram conventions of this type are something you will need to get used to.

That's all I have time for right now. Hopefully someone will review your other points.

FYI: I've never seen or heard of this book.

5. Jan 8, 2012

### PAllen

3) There is no length difference in the direction transverse to the relative motion of two inertial frames. Thus, despite any differences in time and x coordinate (and differences in ball speed and direction in S versus S'), if S sees the balls start Y apart in the y direction, S' will also see them start at Y apart in the y' direction. It also follows that if they collide at some Y/k in S, they will also collide at y'=Y/k in S'. I read the language at this point in the discussion effectively as: let's consider the hypothesis that k=2, and see if it works out.

In your discussion at (3), I see a problem with your statement:

However B’s speed in S is VB

B's speed in S' is VB. It has not been derived what B's speed in S is, and it is certainly not VB, so I assume the rest of your argument here is flawed since it follows from this starting point.

Last edited: Jan 8, 2012
6. Jan 8, 2012

### PAllen

I do see a lot that is not justified in Beiser's derivation, that could not be taken for granted for a presentation at this level. Besides asking you to assume k (as I described it in my earlier post) is two, you are asked to accept, without justification, that in S (for example), ball A rebounds with the exact speed and reverse trajectory, same for B. This can be justified with a time symmetry argument, but I really think that should be made explicit in a presentation at this level.

So far, I don't see anything wrong in the Beiser's derivation, but it is not my favorite approach.

7. Jan 8, 2012

### PAllen

A further observation - that k=1/2 can also be justified by a symmetry argument. The situation as viewed by S' is the reflection around the line y=Y/2 compared to the situation for S (see Beiser's illustrations on the second page of your image). If k does not equal 1/2, you would then have collision occurring at different y' than y. This is impossible because the Lorentz transform (which is assumed at this point), does not change y (or y').

I don't know the rest of this book. If the author has been stressing such arguments earlier in the book, it is reasonable to expect the reader to come up with them. Such reasoning is crucial in physics. Still, I would expect some at least some references to symmetry in justifying parts of the derivation, for an introductory book.

8. Jan 9, 2012

### bibhu1984

Thank you very much PAllen for going through the pictures painstakingly and your analytic replies which are being highly appreciated here. I've understood your explanation which clarified most of my doubts in this context.The remaining is being put as follows.

The flaw you found in my statement is not a flaw. It's a corollary of the author's prime assumption VA=V'B. It’s in the text. It’s the author who states so and not me.
To compensate for the blurry pictures I will write again that the author considers B is initially at rest in S’. B’s downward velocity in S’ is denoted by V’B. B’s velocity in S is denoted by VB.
The balls are identical in the frames in which they are at rest. Hence the author assumes that the velocity with which A is thrown in S is the same as the velocity with which B is thrown in S’.
or simply put VA=V’B
However VB=V’B√(1-(V2/C2))
or VB=VA√(1-(V2/C2))
where V is the relative speed between the frames S and S’.
Hence VB < VA
And that’s where my doubt remains. Two bodies(identical or not) colliding at different individual velocities cannot bounce back with their individual velocities simply reversed.
But the author simply assumes that after collision A bounces down with VA and B bounces up with V'B. One infers that if you are in S, to you A will bounce back with VA and B will glance up with a +ve vertical velocity of VB but the magnitudes of both the velocities remain unchanged throughout.

Thanks again.
P.S. If Arthur Beiser is right then to the one in S the collision will appear supernatural.

Last edited: Jan 9, 2012
9. Jan 9, 2012

### PAllen

Ok, I see between blurriness not reading the primes in your argumeent right, I agree what you say about VB versus VB' is correct.

Then your main remaining issue is, as you say:

"And that’s where my doubt remains. Two bodies(identical or not) colliding at different individual velocities cannot bounce back with their individual velocities simply reversed."

Here, I have to agree that Beiser seems to make an unstated assumption (which is not necessary, and not used in similar, slightly longer, derivations I've seen in other books). Beiser is assuming the vertical components of momentum for balls A and B in S are identical in magnitude (though their vertical speeds are not). This would make all the rest of his argument follow, but I don't see any justification he gives for this.

Other derivations I've seen work as follows, instead:

Assume two identical balls in one frame (S) are moving the same speed (in S), with same x component velocity (in S), and opposite y component velocity. Upon collision, the x components of velocity are unchanged, and now you can definitely assume the y components are simply reversed. Assume only that the magnitude of relativistic momentum ([edit:] here we mean 3-momentum - see next post) is some unknown function of rest mass and speed, and is directed in the direction of the particle's motion. Then, require conservation of momentum in both S and S', where S' is an arbitrary alternate reference frame moving in the x direction relative to S. You then derive the unique form for the relativistic momentum satisfying these conditions.

Thus, Beiser seems to want to simplify the derivation via an assumption he does not justify. Even though he gets the right answer, I agree this is not a satisfying derivation.

Last edited: Jan 9, 2012
10. Jan 9, 2012

### PAllen

One final observation is that Beiser's assumption about momentum is correct - he just provides no justification for it.

For the OP, a justification would take the form of arguing that momentum should be a 4-vector (so that any law involving momentum is intrinsically true in any frame if true in one), and, given the transformation law for 4-vectors, the y component of a 4-vector will be unchanged between S and S'.

11. Jan 9, 2012

### bibhu1984

thank you very much. I got the loop now.

12. Jan 10, 2012

### MikeLizzi

Nice job, PAllen.

13. Jan 11, 2012

### PAllen

Not to beat a dead horse, but I thought of a purely physical argument why the two balls must have equal and opposite y/y' component momentum in both frames (despite different speeds in the two frames). Consider replacing the balls with 'clay' balls that stick together (completely elastic rather than inelastic collision). The the merged ball will have the total momentum. Because of the mirror symmetry around Y/2 in the two frames, this would mean (if the total momentum were not zero) that the merged ball would be moving up in one frame and down in the other. This is absurd for two frames with only x direction relative motion. It would imply frame dependence of which objects collide (in one frame the merged ball could collide with something at y=y'=Y, in the other frame at y=y=0).

The upshot is I quite like Beiser's derivation if it were supplemented by two key justifications that I think are required for the careful reader at the intended level of student. (Symmetry about Y/2; argument for equal and opposite y component momentum).