Understanding the Derivation of Relativistic Mass in Inelastic Collisions

AI Thread Summary
The discussion focuses on deriving the concept of relativistic mass in the context of inelastic collisions as perceived from different reference frames. The key point is that the relativistic mass is equal to the rest mass multiplied by the gamma factor, which is expressed as sqrt(1-u^2/c^2). Participants explore the transformation of velocities, noting that vertical components are affected by the gamma factor while horizontal components require the Lorentz transformation. There is confusion regarding how to apply these transformations correctly, particularly for components of velocity not aligned with the motion of the reference frame. The conversation references historical papers by Planck and Lewis and Tolman to provide context for the derivation challenges.
mainguy
Messages
15
Reaction score
0
Hi guys, thanks for helping with this! I'm a little stuck with this question about the derivation for relativistic mass.

1. Homework Statement

By considering the inelastic collision of two balls as perceived in different reference frames show that the relativistic mass is equal to the rest mass multiplied by the gamma factor (sqrt(1-u^2/c^2).

Homework Equations


So I know the factor is 1/(1-u2/c2) but proving it is tough.
I've considered a reference frame moving at u in a direction perpendicular to the collision, so basically this:

img_5516-e1504665272755.png

The Attempt at a Solution



The vertical velocity u0 is transformed by a gamma factor, u = u'*gamma
So it slows down slightly as expected in the moving frame

It seems to me that the ball moving sidelong, say B in the first image, will have it's velocity altered in two parts. The vertical component will be multiplied by a gamma factor, and the horizontal component will transform as a lorentz:
u'= (u -v)/(1-uv/c2)

It seems clear to me that the vertical velocities of A and B are identical, and that they transform in an identical manner.

From class I know this isn't true, apparently they are identical velocities but they transform in a different manner. But I don't see how B could be transformed via the Lorentz formula if only a compnent of it's velocity is along the line parallel to the motion of the moving frame. Help would be much appreciated![/B]
 

Attachments

  • img_5516-e1504665272755.png
    img_5516-e1504665272755.png
    11.8 KB · Views: 1,060
Physics news on Phys.org
mainguy said:
Hi guys, thanks for helping with this! I'm a little stuck with this question about the derivation for relativistic mass.

1. Homework Statement

By considering the inelastic collision of two balls as perceived in different reference frames show that the relativistic mass is equal to the rest mass multiplied by the gamma factor (sqrt(1-u^2/c^2).

Homework Equations


So I know the factor is 1/(1-u2/c2) but proving it is tough.
I've considered a reference frame moving at u in a direction perpendicular to the collision, so basically this:

View attachment 221980

The Attempt at a Solution



The vertical velocity u0 is transformed by a gamma factor, u = u'*gamma
So it slows down slightly as expected in the moving frame

It seems to me that the ball moving sidelong, say B in the first image, will have it's velocity altered in two parts. The vertical component will be multiplied by a gamma factor, and the horizontal component will transform as a lorentz:
u'= (u -v)/(1-uv/c2)

It seems clear to me that the vertical velocities of A and B are identical, and that they transform in an identical manner.

From class I know this isn't true, apparently they are identical velocities but they transform in a different manner. But I don't see how B could be transformed via the Lorentz formula if only a compnent of it's velocity is along the line parallel to the motion of the moving frame. Help would be much appreciated![/B]

This is a problem that was treated by Planck in a 1906 paper, but the arguments there are not particularly enlightening. A much nicer and more convincing argument was advanced in a 1909 paper by Lewis and Tolman. You can find this argument nicely laid out more-or-less completely on pages 48--50 of "A History of the Theories of Aether and Electricity, Volume II", by Sir Edmund Whittaker, Harper (1953).
 
According tot he diagram, frame ##S## is moving to the right at speed ##V##, but according to what you wrote, you have it moving vertically at speed ##u##. You also use ##u## for the speed of one of the particles in the collision.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top