Understanding Lorentz Invariants and Momentum for My Exam Tomorrow

In summary, the question involves evaluating the Lorentz invariant P^2 and showing that its derivative with respect to time is equal to zero. The first part yields an answer of E^2 - p^2 c^2 = m^2 c^4, and for part b, the derivative of P^2 must be equal to zero. This is achieved by writing P^2 in terms of energy and three momentum, and using the fact that their derivatives must also be equal to zero.
  • #1
stunner5000pt
1,461
2

Homework Statement


Let P be the 4 momentum
u be the 4 velocity
a) Evaluate the Lorentz invariant [itex]P^2[/itex]
b)Differentiate [itex]P^2=P_{\mu}P_{\mu}[/itex] and show that
[tex]\vec{u}\cdot\frac{d}{dt}\left(\frac{m_{0}\vec{u}}{\sqrt{1-u^2/c^2}}\right)=m_{0}c^2\frac{d}{dt}\left(\frac{1}{\sqrt{1-u^2/c^2}}\right)[/tex]

2. The attempt at a solution

The first part yields an answer of [tex]E^2 -p^2 c^2=m^2 c^4[/tex]

Now for part b. Does the P^2 have anything to do with the equality that needs to be proven? Do i need to differentiate P^2 with respect to time? Do i hav to differentiate
[tex]E^2 -p^2 c^2=m^2 c^4[/tex] with respect to time?

Please help!

Thanks
 
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  • #2
stunner5000pt said:

Homework Statement


Let P be the 4 momentum
u be the 4 velocity
a) Evaluate the Lorentz invariant [itex]P^2[/itex]
b)Differentiate [itex]P^2=P_{\mu}P_{\mu}[/itex] and show that
[tex]\vec{u}\cdot\frac{d}{dt}\left(\frac{m_{0}\vec{u}}{\sqrt{1-u^2/c^2}}\right)=m_{0}c^2\frac{d}{dt}\left(\frac{1}{\sqrt{1-u^2/c^2}}\right)[/tex]

2. The attempt at a solution

The first part yields an answer of [tex]E^2 -p^2 c^2=m^2 c^4[/tex]

Now for part b. Does the P^2 have anything to do with the equality that needs to be proven? Do i need to differentiate P^2 with respect to time? Do i hav to differentiate
[tex]E^2 -p^2 c^2=m^2 c^4[/tex] with respect to time?

Please help!

Thanks

Yes, differentiate with respect to time. The derivative of P^2 obviously gives zero since m^2 c^4 is constant.

Now, write P^2 = E^2 - p dot p c^2

(my small p is the three momentum and dot is the dot product

so d/dt(P^2) = 2 E dE/dt - 2 c^2 p dot dp/dt

This must be zero. Now write E= gamma mc^2 and p = gamma m u where u is the ordinary three velocity
 

FAQ: Understanding Lorentz Invariants and Momentum for My Exam Tomorrow

1. What is momentum?

Momentum is a concept in physics that refers to the quantity of motion possessed by an object. It is calculated by multiplying an object's mass by its velocity.

2. How is momentum conserved?

Momentum is conserved in a closed system, meaning that the total momentum of all objects in the system remains constant. This is due to the law of conservation of momentum, which states that momentum cannot be created or destroyed, only transferred between objects.

3. What is the difference between linear and angular momentum?

Linear momentum refers to the motion of an object in a straight line, while angular momentum refers to the motion of an object around a fixed axis. Linear momentum is calculated by multiplying an object's mass by its linear velocity, while angular momentum is calculated by multiplying an object's moment of inertia by its angular velocity.

4. How does mass affect momentum?

The greater the mass of an object, the greater its momentum will be at a given velocity. This means that a heavier object will be more difficult to stop or change its direction of motion than a lighter object.

5. How does velocity affect momentum?

The greater the velocity of an object, the greater its momentum will be at a given mass. This means that a faster-moving object will have more momentum than a slower-moving object of the same mass.

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