Velocity addition and conservation of the energy

[sha1000]

Hello everyone,

For some time I'm a little bit confused about (at the first view) a very simple question, which is about the conversation of the energy of moving objects (in terms of special relativity).

As an example let's talk about firearms. If the mass of the gun M₁ is infinitely higher than the mass of the bullet M₂, then all the kinetic energy of the shot will be imparted to the bullet. Now, let's take the case of the moving gun-bullet system (in X-direction). In the frame of the static observer (S) the total velocity of the bullet (or kinetic energy) will be different as a function of the direction of the shot (transverse or longitudinal); since sqrt(Vx² + Vy²) < Vx + Vy.

As far as I understand the total kinetic energy of the system is independent of the direction of the shot, then how one can explain the total energy difference between transverse and longitudinal shot? My first guess was that the answer was hiding in the distribution of the energy between the gun and the bullet; however I estimate that this guess is wrong since M₁ (gun) >>>>> M₂ (bullet) ->> all the energy of the shot is imparted to the bullet.

 

[Ibix]

Your first guess is correct. If $M_1$ is infinite, then in any frame where the gun is moving the kinetic energy is infinite both before and after the shot is fired. You are unlikely to get sensible results.

Try again with a finite mass for the gun.

 

[jartsa]

Hello everyone,

For some time I'm a little bit confused about (at the first view) a very simple question, which is about the conversation of the energy of moving objects (in terms of special relativity).

As an example let's talk about firearms. If the mass of the gun M₁ is infinitely higher than the mass of the bullet M₂, then all the kinetic energy of the shot will be imparted to the bullet. Now, let's take the case of the moving gun-bullet system (in X-direction). In the frame of the static observer (S) the total velocity of the bullet (or kinetic energy) will be different as a function of the direction of the shot (transverse or longitudinal); since sqrt(Vx² + Vy²) < Vx + Vy.

As far as I understand the total kinetic energy of the system is independent of the direction of the shot, then how one can explain the total energy difference between transverse and longitudinal shot? My first guess was that the answer was hiding in the distribution of the energy between the gun and the bullet; however I estimate that this guess is wrong since M₁ (gun) >>>>> M₂ (bullet) ->> all the energy of the shot is imparted to the bullet.

Let's first forget all relativistic complications.

Non-moving gun-bullet system:

If the mass ratio is x, then the ratio of accelerations is x, ratio of final speeds is x, ratio of distances traveled is x², ratio of energies gained or lost is x²Moving gun-bullet system:

Energy of the shot is imparted mostly to the object that moves the most.

Ratio of energies = ratio of distances traveled

 

[jartsa]

If the mass ratio is x, then the ratio of accelerations is x, ratio of final speeds is x, ratio of distances traveled is x², ratio of energies gained or lost is x²

Actually ratio of distances traveled is x, not x²

And also ratio of energies gained or lost is x, not x²

Ratio of energies = ratio of distances traveled

You see, this basic formula holds:
E=F*d

 

[sha1000]

Thank you for the responses. Indeed, problem was coming from "infinite" mass assumption.

 

[sha1000]

Your first guess is correct. If $M_1$ is infinite, then in any frame where the gun is moving the kinetic energy is infinite both before and after the shot is fired. You are unlikely to get sensible results.

Try again with a finite mass for the gun.

Hello again,

I would like to ask a more practical question about the energy conservation.

FIRST SCENARIO: Again let's take as an example a gunshot. After the shot, bullet attains the speed V_y = 0.7c in the frame S (approximation = almost all the energy of the shot goes to the bullet).
The kinetic energy of the bullet can be calculated though the formula: KE = mc² - m₀c²; if we put c = 1 and m₀= 1; then KE = m_Vy - 1.
Here, KE = 1/sqrt(1 - V_y²) - 1 = 0.4003.SECOND SCENARIO: Now the exact same replica of the system is moving with the speed V_r = 0.9c in the frame S. The same shot is realized in the transverse direction. The velocity of the bullet is V_y=0.7c in the S' frame and V_y' in the S frame. The transverse velocity transformation formula is V_y' = sqrt(1 - V_r²)*V_y. That gives us V_y' = 0.3051. After the shot, the total velocity of the bullet in S frame equals to: V_tot = sqrt(V_y'² + V_r²) = 0.9503.
The KE of the bullet gained from the shot equals to: KE = m_Vtot - m_Vr = 0.9183.

QUESTION: I expected that the kinetic energy of the bullet gained from the shot would be equal in both scenarios because of the conservation of the energy. Is there any mistake in my calculations or in my reasoning?

Thank you

 

[Mister T]

I expected that the kinetic energy of the bullet gained from the shot would be equal in both scenarios because of the conservation of the energy.

You've given an example of the same scenario analyzed in two different frames. When a quantity is the same in two different frames we say it's invariant. On the other hand, when a quantity stays the same in one frame we say it's conserved.

 

[sha1000]

You've given an example of the same scenario analyzed in two different frames. When a quantity is the same in two different frames we say it's invariant. On the other hand, when a quantity stays the same in one frame we say it's conserved.

I'm a little bit confused... If we analyze this situation only from the point of view of the observer in the frame (S). In one hand we have a gun which doesn't' move and in the over hand an equal gun which is moving with the speed V_r within S.

If the observer wants to measure KE of the bullet gained from the shot he should find the same energy for both stationary and moving systems (because of the conservation of energy)?

 

[jbriggs444]

What about the work done on the gun/planet by the bullet? If you fire a shot in the forward direction, this quantity is non-zero.

 

[sha1000]

What about the work done on the gun/planet by the bullet? If you fire a shot in the forward direction, this quantity is non-zero.

I thank you for your reply. That's why I restrained the problem only to the transverse shot, in order to simplify the calculations.

 

[jartsa]

I thank you for your reply. That's why I restrained the problem only to the transverse shot, in order to simplify the calculations.

The transverse shot is not really a transverse shot, because the longitudinal momentum of the bullet increases, as the gunpowder gives some of its longitudinal momentum to the bullet. The gunpowder has less mass and less longitudinal momentum after the firing.

 

[PeroK]

I'm a little bit confused... If we analyze this situation only from the point of view of the observer in the frame (S). In one hand we have a gun which doesn't' move and in the over hand an equal gun which is moving with the speed V_r within S.

If the observer wants to measure KE of the bullet gained from the shot he should find the same energy for both stationary and moving systems (because of the conservation of energy)?

If you have the bullet fired in the $y$ direction at speed $v_y$, then the gain in KE is

$(\gamma_y -1)mc^2$.

If you analyse this in a frame where the experiment is moving in the $x$ direction at speed $v_x$ then the gain in KE is:

$\gamma_x (\gamma_y -1)mc^2$

The two coincide, therefore, only when $v_x$ in non relativistic.

Note: It might be a useful exercise to derive the above expressions.

 

[PAllen]

If you have the bullet fired in the $y$ direction at speed $v_y$, then the gain in KE is

$(\gamma_y -1)mc^2$.

If you analyse this in a frame where the experiment is moving in the $x$ direction at speed $v_x$ then the gain in KE is:

$\gamma_x (\gamma_y -1)mc^2$

The two coincide, therefore, only when $v_x$ in non relativistic.

Note: It might be a useful exercise to derive the above expressions.

hmm. I get product of gammas -1, rather than with 1 in parentheses as you have:

$(\gamma_x \gamma_y -1)mc^2$

 

[PeroK]

hmm. I get product of gammas -1, rather than with 1 in parentheses as you have:

$(\gamma_x \gamma_y -1)mc^2$

I got $\gamma' = \gamma_x \gamma_y$ for the gamma factor in the new frame.

In that frame the initial energy is $\gamma_x mc^2$.

 

[PAllen]

I got $\gamma' = \gamma_x \gamma_y$ for the gamma factor in the new frame.

In that frame the initial energy is $\gamma_x mc^2$.

Ah, you want increase in KE, not total KE. I was computing total KE.

(fyi, if you don't know it, deriving using 4 vectors the general case is trivial:

$\gamma' = \gamma_u \gamma_v$ (1-u⋅v)

where bolded are 3 vectors.
)

 

[Mister T]

If the observer wants to measure KE of the bullet gained from the shot he should find the same energy for both stationary and moving systems (because of the conservation of energy)?

The conservation law tells you that something you determine using one frame of reference will have the same value both before and after a process. It does not tell you that the value of something will be the same when determined in one frame of reference as it is when determined in another frame of reference.

 

[Dale]

I expected that the kinetic energy of the bullet gained from the shot would be equal in both scenarios because of the conservation of the energy. Is there any mistake in my calculations or in my reasoning?

You are in general correct (although I did not check your numbers). The KE gained will be different in different frames. Energy is conserved, but it is not invariant.

 

[jartsa]

A laser-gun moves slowly (10 km/s) along x-axis, shoots along y-axis. No noticeable relativistic Doppler-effect can be observed. During the firing the gun does just a normal amount work in order to produce the laser beam, like a still standing laser gun.

A rifle moves slowly (10 km/s) along x-axis, shoots along y-axis. No noticeable changes on the work done on the bullet can be observed. During the firing the rifle does just a normal amount work on the bullet, same amount as a still standing rifle does.

I think sha100 is thinking about that kind of thing as in the previous paragraph.

Well the above is not strictly true, because of the tiny relativistic effects. A tiny relativistic Doppler-effect in the laser gun case. And some similar effect in the rifle case.

Hey, how about such rifle where a bulled is propelled by photon gas. The photon gas experiences a relativistic Doppler shift if the rifle moves at relativistic speed.

 

[Ibix]

FIRST SCENARIO:
KE = 1/sqrt(1 - V_y²) - 1 = 0.4003.

SECOND SCENARIO: Now the exact same replica of the system is moving with the speed V_r = 0.9c in the frame S.
The KE of the bullet gained from the shot equals to: KE = m_Vtot - m_Vr = 0.9183.

Note that $0.4003\gamma=0.9183$ for a frame moving at 0.9c.

What are the contributions to the total energy of the system before the shot, in either frame? What about after the shot?

 

[sha1000]

If you have the bullet fired in the $y$ direction at speed $v_y$, then the gain in KE is

$(\gamma_y -1)mc^2$.

If you analyse this in a frame where the experiment is moving in the $x$ direction at speed $v_x$ then the gain in KE is:

$\gamma_x (\gamma_y -1)mc^2$

The two coincide, therefore, only when $v_x$ in non relativistic.

Note: It might be a useful exercise to derive the above expressions.

Indeed the KE equations which I used: KE = m_Vy - 1 and KE = m_Vtot - m_Vr can be derived into the expressions you wrote above $(\gamma_y -1)mc^2$ and $\gamma_x (\gamma_y -1)mc^2$.

In this form it becomes much more relevant. As Ibix noticed when using my numbers I obtain: 0.4003$\gamma_x$ = 0.9183 (for V_x = 0.9c).

My conclusion (correct me pls If I'm wrong): 1) My first thought was that the KE increase gained from the shot must be the same both in stationary gun and moving gun, because I was thinking that the potential energy of the gunpowder was the same in both cases ; 2) Now I see that this idea was wrong since we demonstrated that the KE energy of the bullets is related through the expression 0.4003$\gamma_x$ = 0.9183; 3) Does this mean that the potential energy contained in the gunpowder increase in the same was as the relativistic mass (kinetic energy)? =====> Kinetic and potential energy both contribute to the mass of the system (mass and energy is basically the same thing). As the velocity of the system increases the "relativistic mass" increases through the m= $\gamma_x$m₀. This also applies to the potential energy, that explains the results of my calculations. I think I got it.

Thank you all for your help!

 

[sha1000]

You are in general correct (although I did not check your numbers). The KE gained will be different in different frames. Energy is conserved, but it is not invariant.

I need to mediate on this for a while: "Energy is conserved, but it is not invariant.".

 

[PeroK]

I need to mediate on this for a while: "Energy is conserved, but it is not invariant.".

Another example is momentum. Imagine any system where momentum is conserved. In the centre of momentum frame the total momentum is zero, by definition. But, in any other frame the total momentum will not be zero.

Thus momentum is conserved but not invariant. likewise energy.

Rest mass, for example, is not conserved, but it is invariant (the same in all frames).

Electric charge, for example, is both invariant and conserved.